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# Equation of a Straight Line, passes through Origin with slope
Expressing a straight line in algebraic form expression when the straight line passes through the origin with some slope is defined equation of a straight line, passes through the origin with slope.
In very special cases, straight lines are intersected the origin of the Cartesian coordinate system with some slope and these types of straight lines can be expressed in terms of an algebraic expression. It is very easier to represent it in mathematics by a single expression which reveals how exactly the straight line is in the geometric system. Actually, the straight line is expressed in terms of the slope and any one of the points on the straight line.
## Geometric Explanation
Assume, $\stackrel{↔}{AB}$ is a straight line on a plane and also assume it intersects the origin of the Cartesian coordinate geometric system with some slope.
The straight line $\stackrel{↔}{AB}$ intersects the origin at a point $A$ which is origin as well as one of the points of the straight line $\stackrel{↔}{AB}$. Therefore, the coordinates of the point $A$ is $\left(0,0\right)$ geometrically. Assume, the point $B$ is located at $\left(x,y\right)$ in the Cartesian coordinate geometric system. The assumed angle made by the straight line $\stackrel{↔}{AB}$ with the horizontal x-axis is theta $\left(\theta \right)$.
Draw a perpendicular line from point $B$ towards the x-axis and it intersects the horizontal x-axis at a point which is assumed to call point $C$. Thus, it forms a right angled triangle $\Delta BAC$. The angle made the straight line $\stackrel{↔}{AB}$ with horizontal x-axis is theta $\left(\theta \right)$ but it is also an angle of the right angled triangle $\Delta BAC$ which means $\angle BAC=\theta$.
As per this right angled triangle,
• The line segment $\stackrel{‾}{AB}$ becomes the hypotenuse of the right angled triangle $\Delta BAC$.
• The line segment $\stackrel{‾}{AC}$ becomes the adjacent side of the right angled triangle $\Delta BAC$.
• The line segment $\stackrel{‾}{BC}$ becomes the opposite side of the right angled triangle $\Delta BAC$.
As per the principle definition of slope of the straight line, the slope of the straight line is mathematically expressed as $m=tan\theta$ .
But, as per the right angled triangle $\Delta BAC,tan\theta =\frac{BC}{AC}$
As per the right angled triangle $\Delta BAC$,
The length of the opposite side $\left(\stackrel{‾}{BC}\right)$ of the right angled triangle $\Delta BAC$ is $BC=y$
The length of the adjacent side $\left(\stackrel{‾}{AC}\right)$ of the right angled triangle $\Delta BAC$ is $AC=x$
Therefore, $tan\theta =\frac{BC}{AC}=\frac{y}{x}$
$⇒tan\theta =\frac{y}{x}$
Replace the value of $tan\theta$ in slope form.
$⇒m=\frac{y}{x}$
$⇒mx=y$
$⇒y=mx$
This is an algebraic expression which represents a straight line when the straight line passes through the origin of the Cartesian coordinate geometric system.<|endoftext|>
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# How do you find the antiderivative of 4/sqrtx?
Nov 6, 2016
Rewrite as $4 {x}^{- \frac{1}{2}}$.
#### Explanation:
The first step is to recognize that $\frac{1}{\sqrt{x}}$ is equivalent to ${x}^{- \frac{1}{2}}$ by the laws of exponents, where $\frac{1}{x} ^ a$ is equivalent to ${x}^{-} a$. Once we realize this, it becomes much easier to manage.
When we take the derivative of an exponential term, we bring down the power (i.e. multiply the term you are deriving by the value of its exponent) and reduce the power by one. When we take an antiderivative of an exponential term, we want to do the opposite. In essence, we are "undoing" the derivative. Thus, we need to increase the power by one, and multiply the term by the reciprocal of that final exponent.
Because this function we want to "anti-derive" has a negative exponent, we need to make sure we pay a little extra attention. The exponent is $- \frac{1}{2}$, and adding one to this would result in an exponent of $\frac{1}{2}$. If we were to derive ${x}^{\frac{1}{2}}$ we would bring down the $\frac{1}{2}$ power and reduce it by one. This would result in $4 \left(\frac{1}{2} {x}^{- \frac{1}{2}}\right)$.
However, we can see that our original function does not have that $\frac{1}{2}$ constant, so we'll need to do something to counteract that. We can multiply the term by the reciprocal of it's power, which we determined to be $\frac{1}{2}$. The reciprocal of $\frac{1}{2}$ is $2$.
Because this would be an indefinite integral (i.e. having no specific values for which its variables should be defined), we need to account for any constants that could have been present when the derivative was taken, since when we take the derivative of a constant, it reduces to 0. How could we know that the antiderivative is $4 \cdot 2 {x}^{\frac{1}{2}}$ for example, and not $\left(4 \cdot 2 {x}^{\frac{1}{2}}\right) + 42$? After all, if we took the derivative of that, we would end up with the original function we wanted to "anti-derive" also: ${x}^{- \frac{1}{2}}$. This is the case for any constant you might pick.
We can account for this by adding $+ C$ at the end of our answer.
Our final answer, therefore, is:
$8 {x}^{\frac{1}{2}} + C$<|endoftext|>
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Researchers from Liverpool John Moores University, the World Wildlife Fund, and HUTAN-Kinabatangan Orangutan Conservation Programme are using thermal cameras attached to drones to survey the orangutan population in the forest canopies of Sabah, Malaysia. They were able to do this by borrowing thermal imaging and analysis techniques previously used to study light from stars.
Images from morning or evening flights are most reliable because the ground has not yet warmed to a temperature that is too similar to that of the animals’ body temperatures. The group is using the thermal data to develop a machine learning algorithm to differentiate animal species. In addition to the orangutans, they also spotted proboscis monkeys and pygmy elephants.
“In thermal images, animals shine in a similar way to stars and galaxies, so we used techniques from astronomy to detect and distinguish them. . . . We could see the orangutans quite clearly because of their body heat, even during fog or at night, ” says Claire Burke of Liverpool John Moores University in a press release.
“All orangutan species are critically endangered and monitoring their numbers is crucial for their conservation,” says Serge Wich of Liverpool John Moores University in the release. This method of counting the population is less time-consuming and may be useful in the future for conservation efforts.
The researchers are presenting their results at the Unifying Tropical Ecology symposium this week, hosted by the British Ecological Society.<|endoftext|>
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### Nines and Tens
Explain why it is that when you throw two dice you are more likely to get a score of 9 than of 10. What about the case of 3 dice? Is a score of 9 more likely then a score of 10 with 3 dice?
### Master Minding
Your partner chooses two beads and places them side by side behind a screen. What is the minimum number of guesses you would need to be sure of guessing the two beads and their positions?
### Odds and Evens
Is this a fair game? How many ways are there of creating a fair game by adding odd and even numbers?
# At Least One...
### Why do this problem?
An important strategy in answering probability questions requires us to consider whether it is easier to work out the probability of an event occurring or the probability of it NOT occurring. In this problem, learners are introduced to tree diagrams and the concept of mutually exclusive events whose probabilities sum to 1.
### Possible approach
This printable worksheet may be useful: At least one....
"Imagine flipping a coin three times. What's the probability you will get a head on at least one of the flips?"
Give the class time to explore on their own or in pairs, then share the different methods they used to work it out.
If no-one has suggested a tree diagram, start building a tree diagram on the board and ask for suggestions of how to complete it. Then ask the class to identify which branches contain at least one head, and to use this to work out the probability of getting at least one head.
With the class working in groups of three or four, challenge them to build on what they've done by asking them to work out the probability of getting at least one head in four flips. Wander around the class and ask groups to move on to five flips, six flips... as soon as they've finished the one they are working on. As each group discovers a neat way of working out these probabilities, first challenge them to work out the probability of getting at least one head in twenty flips, and then, assuming they can apply their method successfully, give them one of the related questions from the problem.
Once most of the groups have a successful method for the at least one head problems, bring the class together to discuss what they noticed when working on their tree diagrams, and to justify the methods they used to work out the probabilities.
Finally, the remaining questions from the problem can be used with the class to consolidate these ideas.
### Key questions
What is the probability of getting at least one head?
What is the probability of getting no heads?
How are the probabilities related?
### Possible extension
Same Number! provides a natural extension to this problem.
### Possible support
Spend some time working together as a class on listing probabilities, and then move to the tree diagram representation simply as an efficient way of listing systematically.<|endoftext|>
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ByMaria E. Canon , Elise A. Marifian
Income inequality between races has been a widely used indicator of economic prosperity and opportunity (or the lack thereof) within the diverse population of the U.S. The Civil Rights Act of 1964 prohibited discrimination in public places, provided for the integration of schools and other public facilities, and made employment discrimination illegal, thus improving the quality of education and providing more job opportunities for African-Americans. Nevertheless, disparities remain. Labor economists have investigated various sources of earnings inequality in America since the act was passed; some economists have considered how the disparities in earnings change within and across regions of the country. Much of the research covers the 1960-2000 period; much less is known about racial inequality in earnings over the years since. Of particular interest might be the impact of the Great Recession on such inequality.
This article aims to provide insight into the recent trends in earnings inequality between black men and white men. We replicated the analysis in a 2006 study by Jacob Vigdor of the 1960-2000 period using census data and then examined disparities in annual earnings since then, using yearly American Community Survey data from 2000 to 2011.
Figures 1 (1960-2000) and 2 (2000-2011) present key results. The dotted line shows the percentage differential in earnings for Northern-born black males relative to Northern-born white males, holding constant other variables. (For example, in 1960 Northern-born black males earned on average 40 percent less than their Northern-born white counterparts.) The solid line plots the same comparison between Southern-born blacks and whites. (Be aware that Northern-born and Southern-born does not necessarily mean that the men continued to live in the North or South, respectively.)
In Figure 1, we see that inequality declined among both the Northern-born and Southern-born from 1960 to 1970. Racial earnings inequality among the Northern-born increased markedly from 1970 to 1990 and remained relatively stable from 1990 to 2000. On the other hand, among the Southern-born, racial earnings inequality declined only slightly from 1970 to 1980 and increased slightly from 1980 to 2000. In 2000, black-white earnings inequality among the Northern-born was considerably greater than the level in 1960, while inequality among the Southern-born was reduced.
Figure 2 shows the results for 2000-2011. The values for the Northern-born indicate that the economic situation of blacks (as measured by annual earnings) declined considerably relative to that of whites; in other words, earnings inequality continued to increase for those born outside the South.
Similarly, the percent differential in Southern-born blacks' annual earnings relative to Southern-born whites' worsened over the 2000-2011 period. Those blacks born in the South did not show evidence of converging faster with those blacks born in the North during this decade. In addition, the increases in slope magnitude from 2007 to 2010 indicate that during the Great Recession and in the year following, racial earnings inequality among the Southern-born increased even more than in previous years. Lastly, it is important to note that being a Southern-born black male corresponds with a greater wage differential relative to white counterparts than does being a Northern-born black male. For example, the results indicate that in 2011, the annual earnings of Southern-born black males were approximately 72 percent less than those of Southern-born white males, whereas Northern-born black males' 2011 earnings were 61 percent less than those of Northern-born white males.
What driving forces can explain these trends?
Vigdor examined three hypotheses to understand why it appears that the South demonstrated more rapid progress than the North in reducing the earnings gap between blacks and whites from 1960 to 2000. While each hypothesis seems to have had an effect at some point throughout the 40-year period, the results of his analysis suggest that much of the "improvement" in the racial wage gap in the South was merely a reflection of changing regional demographics—what he calls "selective migration"—and not of actual improvement in relative earnings for Southern-born blacks. The improvements were the result of blacks and whites of differing abilities moving from South to North and vice versa.
Vigdor's results indicate that selective migration accounted for 40 percent of the South's relative improvement from 1960 to 2000 and all of the improvement from 1980 to 2000. More specifically, the black-white wage gap improved among Southern residents but not for those born in the South.
Other studies have attempted to explain the trends in earnings inequality between the races through 2000. A 2010 paper by Dan Black, Natalia Kolesnikova and Lowell Taylor considered the average annual weeks worked from 1970 to 2000 and found that this number declined for black men in each of the 14 cities examined, sometimes by as much as 25 percent, while the declines for white men were relatively smaller. At the same time, black men's weekly hours of work remained stable. Together, these two points suggest that the earnings decline was likely related to labor force attachment (manifested as a drop in the average number of weeks worked in a year) rather than to declines in the number of hours that black men were working. The authors found substantial declines in the proportion of black men employed, increases in the proportion of black men unemployed and even larger increases in the proportion of black men not in the labor force. In other words, black men's labor force trends from 1970 to 2000 help explain why their average annual weeks worked declined and why their total annual earnings declined relative to those of white men.
Yet can these studies by Vigdor and by Black et al. explain the behavior in racial earnings gaps over the 2000-2011 period? While Vigdor argued that selective migration explained the South's improvement in racial earnings inequality relative to the North's, recent research by Greg Kaplan and Sam Schulhofer-Wohl suggests that interstate migration has been decreasing. In other words, the selective migration story observed in previous decades would no longer apply.
An alternative explanation for the increased racial inequality could be that African-American men were disproportionately hit by the Great Recession. Their unemployment rate increased from 8.5 percent to 15.4 percent between December 2007 and December 2011. In comparison, the unemployment rate of white men rose from 3.9 percent to 7.1 percent over the same period. Given these unemployment rates, it is no surprise that for both Southern-born and Northern-born blacks, earnings declined relative to whites during the Great Recession. Furthermore, the fact that the labor force attachment for African-Americans has decreased even more since the Great Recession might help explain the increase in the earnings gap since 2009.
Black, Dan A.; Kolesnikova, Natalia A.; and Taylor, Lowell J. "The Economic Progress of African Americans in Urban Areas: A Tale of 14 Cities." Federal Reserve Bank of St. Louis' Review, September/October 2010, Vol. 92, No. 5, pp. 353-79.
Kaplan, Greg; and Schulhofer-Wohl, Sam. "Understanding the Long-Run Decline in Interstate Migration." Federal Reserve Bank of Minneapolis, October 2012 (revised February 2013), Working Paper 697.
Ruggles, Steven; Alexander, J. Trent; Genadek, Katie; Goeken, Ronald; Schroeder, Matthew B.; and Sobek, Matthew. Integrated Public Use Microdata Series (IPUMS): Version 5.0 [Machine-readable database]. Minneapolis: University of Minnesota, 2010.
Vigdor, Jacob L. "The New Promised Land: Black-White Convergence in the American South, 1960-2000." National Bureau of Economic Research, March 2006, Working Paper 12143.<|endoftext|>
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Arabic Naming Conventions
Names that follow these conventions are typically only given to children born in an Arab country or to Arab parents. Arabic forenames are often given to children in any Muslim family, and Arabic forenames are often adopted by converts to the religion of Islam, but these conventions strictly apply to those of Arab ethnicity and are rarely used by non-Arabs. Arabic is spoken throughout North Africa and the Levant (Middle East proper). Arabic speaking nations include Morocco, Tunisia, Libya, Egypt, Syria, Jordan, Lebanon, Israel, Palestine, Iraq, Saudi Arabia, Oman, Yemen, the United Arab Emirates, Qatar, and Kuwait.
Arabic names traditionally have four parts: ism, nasab, laqab, and nisba. The ism is a personal name, typically but not always a descriptive name (which is also usually a word used in everyday speech), which may or may not be accompanied by a laqab, which is intended as a further descriptive (and must be a descriptive name). The ism may be replaced with a kunya, or a diminutive in which a person is addressed as the father or mother of his/her firstborn son.
The nasab consists of at least one patryonimic denoting direct male ancestry, but there is no limit on the number of patryonimics permitted (and there may be several nasab, denoting father, grandfather, great-grandfather, great-great grandfather, etc.). The nasab is the male ancestor's name, and typically includes "bin" or "ibn" (son of) or "bint" (daughter of) preceding the ancestor's name. In contemporary usage, many individuals choose to omit "ibn" or "bin", instead using just their ancestors' proper names.
The nisba is the closest concept to a surname in Arabic naming conventions. It is an occupational, origin, or familial name, which is passed on for many generations. For example, Usama bin Ladin's name is Usama bin Mohammed bin 'Awad bin Ladin, where Mohammed was his father, 'Awad his grandfather, and Ladin a nisba - referring to a distant ancestor.
Here is another example:
One of my characters' names is Tahira Ali Almontaser. She has no nasab in her name, but that is okay because she is not ethnically Arab; however, Tahira is an ism, and Ali is a laqab. Almontaser is a nisba. She commonly uses the abbreviated Tahira Ali as her name for general purposes, omitting the nisba altogether in favor of the laqab, which is used like a surname.
The last notes in this section are to help you avoid common mistakes for non-native Arabic speakers:
Like with the names Mark John or Mary Ellen, which are full first names - one name, consisting of two words - any name that begins with "Abd" is a two word name that cannot be separated. Abdal Malik and Abd al-Noor are two examples of such names. "Abd" is a word that means "servant/slave of" and without the second half of the name, it is incomplete and inaccurate. If someone's name is given as Jamal Abdal Malik, it is wrong to address him is "Mr. Malik". Instead, one must use "Mr. Abdal Malik".
When writing Arabic words or names with the Latin alphabet (romanizing, or transliterating), there is no one standard system, and there may be several ways to spell any given Arabic name. For example: Mohamed, Mohammed, Muhammad, Mohamet, Muhamad; Noor, Nur, Nour, Nor. While you can take some liberties with the spelling of Arabic names, it is advisable to perform a simple search of the desired spelling to see whether there are real people who do, in fact, use that spelling of the name.
Excerpted from "The Roleplayer and Writer's Guide to Naming Conventions" by Ylanne S., July 2010.<|endoftext|>
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El Arbol Generoso: The Giving Tree
This book is well known both in English and in Spanish. Because of its ease of language you can have both versions available to compare. The Giving Tree is a beautiful story which gives an idea on the power of love and giving. It is a simple story , relating to a boy’s journey with a tree, and how the tree gave so much of himself that eventually he gives everything of himself and all but a stump remains.
It is not a book to read with small children, but it is appropiate for 8 -10 year olds. The drawings are also quite illustrative, and you can start with this activity to get them to reflect on the book.
El Arbol Generoso Regala Manzanas
First Step: Visual Recognition
Turn to the first page and look at the picture. Show the picture to the class. Ask some questions
¿Que ven en la foto?
¿ Creen que el arbol y el niño son amigos ?
¿ Como es el arbol? ¿ Es grande o pequeño?
¿ Pueden nombrar las partes del arbol? ( You can mimic or give clues for each parts of the tree)
After you do a visual recognition of some parts of the book you should go into the more abstract concepts of the book i.e. generosity and love. Try to form a discussion about the book.
¿ Ustedes que creen que significa ser generosos?
¿ Son generosos con sus padres?
¿ Son generosos con sus hermanos o con sus maestros?
A veces amamos sin condiciones. ( Explain the vocabulary word, this might be a bit too hard for elementary spanish but you can do it for the more advanced students)
¿ Que significa amar a alguien incondicionalmente?
Now the good thing would be to introduce the story for younger readers it can go something like this :
Ahora vamos a hablar de un arbol , y el amor que tenia hacia su amigo. Es una historia muy bonita. ¿Estan listos para escucharlo?
Second Step: Reading out Loud
Read the book out loud, making sure that all the children have a copy of the book in front of them. If you can, try creating a powerpoint presentation with some of the most beloved drawings. While you are reading, make sure you change voices when you are speaking the part of the boy and the part of the tree. At certain pivotal moments stop the story and show them the images and ask the students questions :
¿ Ven al arbol ahora?¡¡ Ya no tiene manzanas!!
¿ Porque le regalo al niño las manzanas?
As the tree loses essential parts of itself, you can keep talking about it and reemphasizing that he does this for love of the boy. As time passes in the book, point out how the boy ages, and what kind of ways his life is different now from when he was a boy. At the end of the story, ask them to reflect on how the old man and the trunk change , and how much it is because of love.
Third Step: Pairing Up
Children should go in pairs. At times you should pair a weaker reader with a stronger one so they can help each other out. Have students read over the book again in pairs. Walk around the room and make sure they are doing this. Students should be underlining the words they do not understand and looking them up in the dictionary. Ask them to write sentences with those words.Once this is done the students can then go on to writing a small summary about the book with their partners. Hand out a small worksheet that goes a bit like this :
Llene los espacios con el verbo apropiado
El arbol ____ generoso y le gusta _____ todo de si ______ . El crece con un niño y le ___ sus manzanas, sus _____ y sus ramas. Al final de su vida el niño y esta ____ y el arbol ya ___ un tronco.
You can keep adding with vocabulary activities and verb associations.
A continuacion, conjuge el verbo ¨ Dar ” en tiempo pasado perfecto en distintas oraciones.
Haga 10 oraciones con el verbo ” ser” en tiempo presente perfecto
El es generoso
Tu eres generoso
Nosotros somos generosos
Vosotros sois generosos
Ellos son generosos
Fourth Step: Creative Activities
Students can then express the book with a puppet show or drama activity. If the students are younger you can photocopy the book and have them colour the pages. You can also have them create a collage with magazines about their tree. Some students also like to make a version of their own with a play or song. The sky is the limit with this but make sure the comprehension is in place to do them.<|endoftext|>
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## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Ex 22
These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Ex 22
More Exercises
Question 1.
A bag contains a red ball, a blue ball and a yellow ball, all the balls being of the same size. Anjali takes out a ball from the bag without looking into it. What is the probability that she takes out
(i) yellow ball ?
(ii) red ball ?
(iii) blue ball ?
Solution:
Number of balls in the bag = 3.
(i) Probability of yellow ball = $$\\ \frac { 1 }{ 3 }$$
(ii) Probability of red ball = $$\\ \frac { 1 }{ 3 }$$
(iii) Probability of blue ball = $$\\ \frac { 1 }{ 3 }$$
Question 2.
A box contains 600 screws, one-tenth are rusted. One screw is taken out at random from this box. Find the probability that it is a good screw.
Solution:
Number of total screws = 600
Rusted screws = $$\\ \frac { 1 }{ 10 }$$ of 600 = 60
∴ Good screws = 600 – 60 = 540
Probability of a good screw
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 540 }{ 600 }$$
= $$\\ \frac { 9 }{ 10 }$$
Question 3.
In a lottery, there are 5 prized tickets and 995 blank tickets. A person buys a lottery ticket. Find the probability of his winning a prize.
Solution:
Number of prized tickets = 5
Number of blank tickets = 995
Total number of tickets = 5 + 995 = 1000
Probability of prized ticket
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 5 }{ 1000 }$$
= $$\\ \frac { 1 }{ 200 }$$
Question 4.
12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Solution:
Number of defective pens = 12
Number of good pens = 132
Total number of pens =12 + 132 = 144
Probability of good pen
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 132 }{ 144 }$$
= $$\\ \frac { 11 }{ 12 }$$
Question 5.
If the probability of winning a game is $$\\ \frac { 5 }{ 11 }$$, what is the probability of losing ?
Solution:
Probability of winning game = $$\\ \frac { 5 }{ 11 }$$
⇒ P(E) = $$\\ \frac { 5 }{ 11 }$$
We know that P (E) + P ($$\overline { E }$$) = 1
where P (E) is the probability of losing the game.
$$\\ \frac { 5 }{ 11 }$$ + P ($$\overline { E }$$) = 1
⇒ P ($$\overline { E }$$) = $$1- \frac { 5 }{ 11 }$$
= $$\\ \frac { 11-5 }{ 11 }$$
= $$\\ \frac { 6 }{ 11 }$$
Question 6.
Two players, Sania and Sonali play a tennis match. It is known that the probability of Sania winning the match is 0.69. What is the probability of Sonali winning ?
Solution:
Probability of Sania’s winning the game = 0.69
Let P (E) be the probability of Sania’s winning the game
and P ($$\overline { E }$$) be the probability of Sania’s losing
the game or probability of Sonali, winning the game
P (E) + P ($$\overline { E }$$) = 1
⇒ 0.69 + P ($$\overline { E }$$) = 1
⇒ P($$\overline { E }$$) = 1 – 0.69 = 0.31
Hence probability of Sonali’s winning the game = 0.31
Question 7.
A bag contains 3 red balls and 5 black balls. A ball is drawn at random’ from in bag. What is the probability that the ball drawn is .
(i) red ?
(ii) not red ?
Solution:
Number of red balls = 3
Number of black balls = 5
Total balls = 3 + 5 = 8
Let P (E) be the probability of red balls,
then P ($$\overline { E }$$) will be the probability of not red balls.
P (E) + P ($$\overline { E }$$) = 1
(i) But P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 3 }{ 8 }$$
(ii) P ($$\overline { E }$$) = 1 – P(E)
= $$1- \frac { 6 }{ 11 }$$
= $$\\ \frac { 8-3 }{ 8 }$$
= $$\\ \frac { 5 }{ 8 }$$
Question 8.
There are 40 students in Class X of a school of which 25 are girls and the.others are boys. The class teacher has to select one student as a class representative. She writes the name of each student on a separate card, the cards being identical. Then she puts cards in a bag and stirs them thoroughly. She then draws one card from the bag. What is the probability that the name written on the card is the name of
(i) a girl ?
(ii) a boy ?
Solution:
Number of total students = 40
Number of girls = 25
Number of boys = 40 – 25 = 15
(i) Probability of a girl
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 25 }{ 40 }$$
= $$\\ \frac { 5 }{ 8 }$$
(ii) Probability of a boy
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 15 }{ 40 }$$
= $$\\ \frac { 3 }{ 8 }$$
Question 9.
A letter is chosen from the word ‘TRIANGLE’. What is the probability that it is a vowel ?
Solution:
There are three vowels: I, A, E
.’. The number of letters in the word ‘TRIANGLE’ = 8.
Probability of vowel
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 3 }{ 8 }$$
Question 10.
A letter of English alphabet is chosen at random. Determine the probability that the letter is a consonant.
Solution:
No. of English alphabet = 26
No. of vowel = 5
No. of constant = 25 – 5 = 21
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 21 }{ 26 }$$
Question 11.
A bag contains 5 black, 7 red and 3 white balls. A ball is drawn at random from the bag, find the probability that the ball drawn is:
(i) red
(ii) black or white
(iii) not black.
Solution:
In a bag,
Number of black balls = 5
Number of red balls = 7
and number of white balls = 3
Total number of balls in the bag
= 5 + 7 + 3 = 15
(i) Probability of red balls
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 7 }{ 15 }$$
(ii) Probability of black or white balls
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 5+3 }{ 15 }$$
= $$\\ \frac { 8 }{ 15 }$$
(iii) Probability of not black balls
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 7+3 }{ 15 }$$
= $$\\ \frac { 10 }{ 15 }$$
= $$\\ \frac { 2 }{ 3 }$$
Question 12.
A box contains 7 blue, 8 white and 5 black marbles. If a marble is drawn at random from the box, what is the probability that it will be
(i) black?
(ii) blue or black?
(iii) not black?
(iv) green?
Solution:
Total number of marbles in the box
= 7 + 8 + 5 = 20
Since, a marble is drawn at random from the box
(i) Probability (of a black Marble)
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 5 }{ 20 }$$
= $$\\ \frac { 1 }{ 4 }$$
(ii) Probability (of a blue or black marble)
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 7+5 }{ 20 }$$
= $$\\ \frac { 12 }{ 20 }$$
= $$\\ \frac { 3 }{ 5 }$$
(iii) Probability (of not black marble)
= 1 – P (of black 1)
= $$1- \frac { 1 }{ 4 }$$
= $$\\ \frac { 4-1 }{ 4 }$$
= $$\\ \frac { 3 }{ 4 }$$
(iv) P (of a green marble) = 0
(∴ Since, a box does not contain a green marble,
so the probability of green marble will be zero)
Question 13.
A bag contains 6 red balls, 8 white balls, 5 green balls and 3 black balls. One ball is drawn at random from the bag. Find the probability that the ball is :
(i) white
(ii) red or black
(iii) not green
(iv) neither white nor black.
Solution:
In a bag,
Number of red balls = 6
Number of white balls = 8
Number of green balls = 5
and number of black balls = 3
Total number of balls in the bag
= 6 + 8 + 5 + 3 = 22
(i) Probability of white balls
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 8 }{ 22 }$$
= $$\\ \frac { 4 }{ 11 }$$
(ii) Probability of red or black balls
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 6+3 }{ 22 }$$
= $$\\ \frac { 9 }{ 22 }$$
(iii) Probability of not green balls i.e. having red, white and black balls.
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 6+8+3 }{ 22 }$$
= $$\\ \frac { 17 }{ 22 }$$
(iv) Probability of neither white nor black balls i.e. red and green balls
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 6+5 }{ 22 }$$
= $$\\ \frac { 11 }{ 22 }$$
= $$\\ \frac { 1 }{ 2 }$$
Question 14.
A piggy bank contains hundred 50 p coins, fifty Rs 1 coins, twenty Rs 2 coins and ten Rs 5 coins. It is equally likely that one of the coins will fall down when the bank is turned upside down, what is the probability that the coin
(i) will be a 50 p coin?
(ii) will not be Rs 5 coin?
Solution:
In a piggy bank, there are
100, 50 p coin
50, Rs 1 coin
20, Rs 2 coin
10, Rs 5 coin
Total coins = 100 + 50 + 20 + 10 = 180
One coin is drawn at random Probability of
(i) 50 p coins = $$\\ \frac { 100 }{ 180 }$$
= $$\\ \frac { 5 }{ 9 }$$
(ii) Will not be Rs 5 coins
= 100 + 50 + 20 = 170
Probability = $$\\ \frac { 170 }{ 180 }$$ = $$\\ \frac { 17 }{ 18 }$$
Question 15.
A carton consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Peter, a trader, will only accept the shirts which are good, but Salim, another trader, will only reject the shirts which have major defects. One shirts is drawn at random from the carton. What is the probability that
(i) it is acceptable to Peter ?
(ii) it is acceptable to Salim ?
Solution:
In a carton, there the 100 shirts.
Among these number of shirts which are good = 88
number of shirts which have minor defect = 8
number of shirt which have major defect = 4
Total number of shirts = 88 + 8 + 4 = 100
Peter accepts only good shirts i.e. 88
Salim rejects only shirts which have major defect i.e. 4
(i) Probability of good shirts which are acceptable to Peter
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 88 }{ 100 }$$
= $$\\ \frac { 22 }{ 25 }$$
(ii) Probability of shirts acceptable to Salim
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 88+8 }{ 100 }$$
= $$\\ \frac { 96 }{ 100 }$$
= $$\\ \frac { 24 }{ 25 }$$
Question 16.
A die is thrown once. What is the probability that the
(i) number is even
(ii) number is greater than 2 ?
Solution:
Dice is thrown once
Sample space = {1, 2, 3, 4, 5, 6}
(i) No. of ways in favour = 3
(∵ Even numbers are 2, 4, 6)
Total ways = 6
Probability = $$\\ \frac { 3 }{ 6 }$$ = $$\\ \frac { 1 }{ 2 }$$
(ii) No. of ways in favour = 4
(Numbers greater than 2 are 3, 4, 5, 6)
Total ways = 6
Probability = $$\\ \frac { 4 }{ 6 }$$ = $$\\ \frac { 2 }{ 2 }$$
Question 17.
In a single throw of a die, find the probability of getting:
(i) an odd number
(ii) a number less than 5
(iii) a number greater than 5
(iv) a prime number
(v) a number less than 8
(vi) a number divisible by 3
(vii) a number between 3 and 6
(viii) a number divisible by 2 or 3.
Solution:
A die is thrown and on its faces, numbers 1 to 6 are written.
Total numbers of possible outcomes = 6
(i) Probability of an odd number,
odd number are 1, 3 and 5
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 3 }{ 6 }$$
= $$\\ \frac { 1 }{ 2 }$$
(ii) A number less them 5 are 1, 2, 3, 4
Probability of a number less than 5 is
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 4 }{ 6 }$$
= $$\\ \frac { 2 }{ 3 }$$
(iii) A number greater than 5 is 6
Probability of a number greater than 5 is
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 1 }{ 6 }$$
(iv) Prime number is 2, 3, 5
Probability of a prime number is
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 3 }{ 6 }$$
= $$\\ \frac { 1 }{ 2 }$$
(v) Number less than 8 is nil
P (E) = 0
(vi) A number divisible by 3 is 3, 6
Probability of a number divisible by 3 is
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 2 }{ 6 }$$
= $$\\ \frac { 1 }{ 3 }$$
(vii) Numbers between 3 and 6 is 4, 5
Probability of a number between 3 and 6 is
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 2 }{ 6 }$$
= $$\\ \frac { 1 }{ 3 }$$
(viii) Numbers divisible by 2 or 3 are 2, 4 or 3,
Probability of a number between 2 or 3 is
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 2 }{ 6 }$$
= $$\\ \frac { 1 }{ 3 }$$
Question 18.
A die has 6 faces marked by the given numbers as shown below:
The die is thrown once. What is the probability of getting
(i) a positive integer.
(ii) an integer greater than – 3.
(iii) the smallest integer ?
Solution:
Total outcomes n(S)= 6
(i) a positive integer = (1, 2, 3)
No. of favourables n(E) = 3
Probability = $$\\ \frac { n(E) }{ n(S) }$$
= $$\\ \frac { 3 }{ 6 }$$
= $$\\ \frac { 1 }{ 2 }$$
(ii) Integer greater than -3
= (1, 2, 3, -1, -2)
No. of favourables n(E) = 5
Probability = $$\\ \frac { n(E) }{ n(S) }$$
= $$\\ \frac { 5 }{ 6 }$$
(iii) Smallest integer = -3
No. of favourables n(E) = 1
Probability = $$\\ \frac { n(E) }{ n(S) }$$
= $$\\ \frac { 1 }{ 6 }$$
Question 19.
A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (shown in the adjoining figure) and these are equally likely outcomes. What is the probability that it will point at
(i) 8 ?
(ii) an odd number ?
(iii) a number greater than 2?
(iv) a number less than 9?
Solution:
On the face of a game, numbers 1 to 8 is shown.
Possible outcomes = 8
(i) Probability of number 8 will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 1 }{ 8 }$$
(ii) Odd number are 1, 3, 5, 7
Probability of a number which is an odd will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 4 }{ 8 }$$
= $$\\ \frac { 1 }{ 2 }$$
(iii) A number greater than 2 are 3, 4, 5, 6, 7, 8 which are 6
Probability of number greater than 2 will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 6 }{ 8 }$$
= $$\\ \frac { 3 }{ 4 }$$
(iv) A number less than 9 is 8.
Probability of a number less than 9 will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 8 }{ 8 }$$
Question 20.
Find the probability that the month of January may have 5 Mondays in
(i) a leap year
(ii) a non-leap year.
Solution:
In January, there are 31 days and in an ordinary year,
there are 365 days but in a leap year, there are 366 days.
(i) In January of an ordinary year, there are 31 days i.e. 4 weeks and 3 days.
Probability of Monday will be = $$\\ \frac { 3 }{ 7 }$$
(ii) In January of a leap year, there are 31 days i.e. 4 weeks and 3 days
Probability of Monday will be = $$\\ \frac { 3 }{ 7 }$$
Question 21.
Find the probability that the month of February may have 5 Wednesdays in
(i) a leap year
(ii) a non-leap year.
Solution:
In the month of February, there are 29 days in a leap year
while 28 days in a non-leap year,
(i) In a leap year, there are 4 complete weeks and 1 day
Probability of Wednesday = P (E) = $$\\ \frac { 1 }{ 7 }$$
(ii) and in a non leap year, there are 4 complete weeks and 0 days
Probability of Wednesday P (E) = $$\\ \frac { 0 }{ 7 }$$ = 0
Question 22.
Sixteen cards are labelled as a, b, c,…, m, n, o, p. They are put in a box and shuffled. A boy is asked to draw a card from the box. What is the probability that the card drawn is:
(i) a vowel
(ii) a consonant
(iii) none of the letters of the word median.
Solution:
Here, sample space (S) = {a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p)
∴n(S) = 16
(i) Vowels (V) = {a, e, i, o}
∴n(V) = 4
∴P(a vowel) = $$\\ \frac { n(V) }{ n(S) }$$ = $$\\ \frac { 4 }{ 16 }$$ = $$\\ \frac { 1 }{ 4 }$$
(ii) Consonants (C) = {b, c, d, f, g, h, j, k, l, m, n, p}
∴n(C) = 12
∴P (a consonant) = $$\\ \frac { n(C) }{ n(S) }$$ = $$\\ \frac { 12 }{ 16 }$$ = $$\\ \frac { 3 }{ 4 }$$
(iii) None of the letters of the word MEDIAN (N) = {b, c, f, g, h, j, k, l, o, p)
∴n(N) = 10
∴P (N) = $$\\ \frac { n(N) }{ n(S) }$$ = $$\\ \frac { 10 }{ 16 }$$ = $$\\ \frac { 5 }{ 8 }$$
Question 23.
An integer is chosen between 0 and 100. What is the probability that it is
(i) divisible by 7?
(ii) not divisible by 7?
Solution:
Integers between 0 and 100 = 99
(i) Number divisible by 7 are
7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98 = 14
Probability = $$\\ \frac { 14 }{ 99 }$$
(ii) Not divisible by 7 are 99 – 14 = 85
Probability = $$\\ \frac { 85 }{ 99 }$$
Question 24.
Cards marked with numbers 1, 2, 3, 4, 20 are well shuffled and a card is drawn at random.
What is the probability that the number on the card is
(i) a prime number
(ii) divisible by 3
(iii) a perfect square ? (2010)
Solution:
Number cards is drawn from 1 to 20 = 20
One card is drawn at random
No. of total (possible) events = 20
(i) The card has a prime number
The prime number from 1 to 20 are 2, 3, 5, 7, 11, 13, 17, 19
Actual No. of events = 8
P(E) = $$\frac { Number\quad of\quad actual\quad events }{ Number\quad of\quad total\quad events }$$
= $$\\ \frac { 8 }{ 20 }$$
= $$\\ \frac { 2 }{ 5 }$$
(ii) Numbers divisible by 3 are 3, 6, 9, 12, 15, 18
No. of actual events = 6
P(E) = $$\frac { Number\quad of\quad actual\quad events }{ Number\quad of\quad total\quad events }$$
= $$\\ \frac { 6 }{ 20 }$$
= $$\\ \frac { 3 }{ 10 }$$
(iii) Numbers which are perfect squares = 1, 4, 9, 16 = 4
P(E) = $$\frac { Number\quad of\quad actual\quad events }{ Number\quad of\quad total\quad events }$$
= $$\\ \frac { 4 }{ 20 }$$
= $$\\ \frac { 1 }{ 5 }$$
Question 25.
A box contains 25 cards numbered 1 to 25. A card is drawn from the box at random. Find the probability that the number on the card is :
(i) even
(ii) prime
(iii) multiple of 6
Solution:
Number of card in a box = 25 numbered 1 to 25
(i) Even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24
i.e. number of favourable outcomes = 12
Probability of an even number will be
P(E) = $$\\ \frac { 12 }{ 25 }$$
(ii) Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23
i.e. number of primes = 9
Probability of primes will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 9 }{ 25 }$$
(iii) Multiples of 6 are 6, 12, 18, 24
Number of multiples = 4
Probability of multiples of 6 will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 4 }{ 25 }$$
Question 26.
A box contains 15 cards numbered 1, 2, 3,…..15 which are mixed thoroughly. A card is drawn from the box at random. Find the probability that the number on the card is :
(i) Odd
(ii) prime
(iii) divisible by 3
(iv) divisible by 3 and 2 both
(v) divisible by 3 or 2
(vi) a perfect square number.
Solution:
Number of cards in a box =15 numbered 1 to 15
(i) Odd numbers are 1, 3, 5, 7, 9, 11, 13, 15
Number of odd numbers = 8
Probability of odd numbers will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 8 }{ 15 }$$
(ii) Prime number are 2, 3, 5, 7, 11, 13
Number of primes is 6
Probability of prime number will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 6 }{ 15 }$$
= $$\\ \frac { 2 }{ 5 }$$
(iii) Numbers divisible by 3 are 3, 6, 9, 12, 15
which are 5 in numbers
Probability of number divisible by 3 will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 5 }{ 15 }$$
= $$\\ \frac { 1 }{ 3 }$$
(iv) Divisible by 3 and 2 both are 6, 12
which are 2 in numbers.
Probability of number divisible by 3 and 2
Both will be = $$\\ \frac { 2 }{ 15 }$$
(v) Numbers divisible by 3 or 2 are
2, 3, 4, 6, 8, 9, 10, 12, 14, 15 which are 10 in numbers
Probability of number divisible by 3 or 2 will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 10 }{ 15 }$$
= $$\\ \frac { 2 }{ 3 }$$
(v) Perfect squares number are 1, 4, 9 i.e., 3 number
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 3 }{ 15 }$$
= $$\\ \frac { 1 }{ 5 }$$
Question 27.
A box contains 19 balls bearing numbers 1, 2, 3,…., 19. A ball is drawn at random
from the box. Find the probability that the number on the ball is :
(i) a prime number
(ii) divisible by 3 or 5
(iii) neither divisible by 5 nor by 10
(iv) an even number.
Solution:
In a box, number of balls = 19 with number 1 to 19.
A ball is drawn
Number of possible outcomes = 19
(i) Prime number = 2, 3, 5, 7, 11, 13, 17, 19
which are 8 in number
Probability of prime number will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 8 }{ 19 }$$
(ii) Divisible by 3 or 5 are 3, 5, 6, 9, 10, 12, 15, 18
which are 8 in number
Probability of number divisible by 3 or 5 will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 8 }{ 19 }$$
(iii) Numbers which are neither divisible by 5 nor by 10 are
1, 2, 3, 4, 6, 7, 8, 9, 11, 12,
13, 14, 16, 17, 18, 19
which are 16 in numbers
Probability of there number will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 16 }{ 19 }$$
(iv) Even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18
which are 9 in numbers.
Probability of there number will be
Number of favourable outcome
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 9 }{ 19 }$$
Question 28.
Cards marked with numbers 13, 14, 15, …, 60 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the number on the card drawn is
(i) divisible by 5
(ii) a perfect square number.
Solution:
Number of card bearing numbers 13,14,15, … 60 = 48
One card is drawn at random.
(i) Card divisible by 5 are 15, 20, 25, 30, 35, 40, 45, 50, 55, 60 = 10
Probability = $$\\ \frac { 10 }{ 48 }$$
= $$\\ \frac { 5 }{ 24 }$$
(ii) A perfect square = 16, 25, 36, 49 = 4
Probability = $$\\ \frac { 4 }{ 48 }$$
= $$\\ \frac { 1 }{ 12 }$$
Question 29.
Tickets numbered 3, 5, 7, 9,…., 29 are placed in a box and mixed thoroughly. One ticket is drawn at random from the box. Find the probability that the number on the ticket is
(i) a prime number
(ii) a number less than 16
(iii) a number divisible by 3.
Solution:
In a box there are 14 tickets with number
3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29
Number of possible outcomes = 14
(i) Prime numbers are 3, 5, 7, 11, 13, 17, 19, 23, 29
which are 9 in number
Probability of prime will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 9 }{ 14 }$$
(ii) Number less than 16 are 3, 5, 7, 9, 11, 13, 15
which are 7 in numbers,
Probability of number less than 16 will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 7 }{ 14 }$$
= $$\\ \frac { 1 }{ 2 }$$
(iii) Numbers divisible by 3 are 3, 9, 15, 21, 27
which are 5 in number
Probability of number divisible by 3 will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 5 }{ 14 }$$
Question 30.
A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears
(i) a two-digit number
(ii) a perfect square number
(iii) a number divisible by 5.
Solution:
There are 90 discs in a box containing numbered from 1 to 90.
Number of possible outcomes = 90
(i) Two digit numbers are 10 to 90 which are 81 in numbers.
Probability of two digit number will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 81 }{ 90 }$$
= $$\\ \frac { 9 }{ 10 }$$
(ii) Perfect squares are 1, 4, 9, 16, 25, 36,49, 64, 81
which are 9 in numbers.
Probability of square will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 9 }{ 90 }$$
= $$\\ \frac { 1 }{ 10 }$$
(iii) Number divisible by 5 are
5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90
which are 18 in numbers.
Probability of number divisible by 5 will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 18 }{ 90 }$$
= $$\\ \frac { 1 }{ 5 }$$
Question 31.
Cards marked with numbers 2 to 101 are placed in a box and mixed thoroughly. One card is drawn at random from this box. Find the probability that the number on the card is
(i) an even number
(ii) a number less than 14
(iii) a number which is a perfect square
(iv) a prime number less than 30.
Solution:
Number of cards with numbered from 2 to 101 are placed in a box
Number of possible outcomes = 100 one card is drawn
(i) Even numbers are 2, 4, 6, 8, 10, 12, 14, 16,….., 96, 98, 100
which are 50 in numbers.
Probability of even number will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 50 }{ 100 }$$
= $$\\ \frac { 1 }{ 2 }$$
(ii) Numbers less than 14 are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13
which are 12 in numbers
Probability of number less than 14 will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 12 }{ 100 }$$
= $$\\ \frac { 3 }{ 25 }$$
(iii) Perfect square are 4, 9, 16, 25, 36, 49, 64, 81, 100 which are 9 in numbers
Probability of perfect square number will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 9 }{ 100 }$$
(iv) Prime numbers less than 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29
which are 10 in numbers Probability of prime numbers, less than 30 will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 10 }{ 100 }$$
= $$\\ \frac { 1 }{ 10 }$$
Question 32.
A bag contains 15 balls of which some are white and others are red. If the probability of drawing a red ball is twice that of a white ball, find the number of white balls in the bag.
Solution:
In a bag, there are 15 balls.
Some are white and others are red.
Probability of red ball = 2 probability of white ball
Let number of white balls = x
Then, number of red balls = 15 – x
$$2\times \frac { 15-x }{ 15 } =\frac { x }{ 15 }$$
⇒ 2(15 – x) = x
⇒ 30 – 2x = x
⇒ 30 = x + 2x
⇒ x = $$\\ \frac { 30 }{ 3 }$$ = 10
Number of red balls = 10
and Number of white balls = 15 – 10 = 5
Question 33.
A bag contains 6 red balls and some blue balls. If the probability of drawing a blue ball is twice that of a red ball, find the number of balls in the bag.
Solution:
In a bag, there are 6 red balls, and some blue balls
Probability of blue ball = 2 × probability of red ball
Let number of blue balls = x
and number of red balls = 6
Total balls = x + 6
Probability of a blue ball = 2
⇒ $$\frac { x }{ x+6 } =2\times \frac { 6 }{ x+6 }$$
⇒ $$\frac { x }{ x+6 } =\frac { 12 }{ x+6 }$$
⇒ x = 12
Number of balls = x + 6 = 12 + 6 = 18
Question 34.
A bag contains 24 balls of which x are red, 2x are white and 3x are blue. A blue is selected at random. Find the probability that it is
(i) white
(ii) not red.
Solution:
In a bag, there are 24 balls
Since, there are x balls red, 2 × balls white and 3 × balls blue
x + 2x + 3x = 24
Question 35.
A card is drawn from a well-shuffled pack of 52 cards. Find the probability of getting:
(ii) a jack .
(iii) a king of red colour
(iv) a card of diamond
(v) a king or a queen
(vi) a non-face card
(vii) a black face card
(viii) a black card
(ix) a non-ace
(x) non-face card of black colour
(xi) neither a spade nor a jack
(xii) neither a heart nor a red king
Solution:
In a playing card, there are 52 cards
Number of possible outcome = 52
(i) Probability of‘2’ of spade will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 1 }{ 52 }$$
(ii) There are 4 jack card Probability of jack will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 4 }{ 52 }$$
= $$\\ \frac { 1 }{ 13 }$$
(iii) King of red colour are 2 in number
Probability of red colour king will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 2 }{ 52 }$$
= $$\\ \frac { 1 }{ 26 }$$
(iv) Cards of diamonds are 13 in number
Probability of diamonds card will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 13 }{ 52 }$$
= $$\\ \frac { 1 }{ 4 }$$
(v) Number of kings and queens = 4 + 4 = 8
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 8 }{ 52 }$$
= $$\\ \frac { 2 }{ 13 }$$
(vi) Non-face cards are = 52 – 3 × 4 = 52 – 12 = 40
Probability of non-face card will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 40 }{ 52 }$$
= $$\\ \frac { 10 }{ 13 }$$
(vii) Black face cards are = 2 × 3 = 6
Probability of black face card will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 6 }{ 52 }$$
= $$\\ \frac { 3 }{ 26 }$$
(viii) No. of black cards = 13 x 2 = 26
Probability of black card will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 26 }{ 52 }$$
= $$\\ \frac { 1 }{ 2 }$$
(ix) Non-ace cards are 12 × 4 = 48
Probability of non-ace card will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 48 }{ 52 }$$
= $$\\ \frac { 12 }{ 13 }$$
(x) Non-face card of black colours are 10 × 2 = 20
Probability of non-face card of black colour will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 20 }{ 52 }$$
= $$\\ \frac { 5 }{ 13 }$$
(xi) Number of card which are neither a spade nor a jack
= 13 × 3 – 3 = 39 – 3 = 36
Probability of card which is neither a spade nor a jack will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 36 }{ 52 }$$
= $$\\ \frac { 9 }{ 13 }$$
(xii) Number of cards which are neither a heart nor a red king
= 3 × 13 = 39 – 1 = 38
Probability of card which is neither a heart nor a red king will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 38 }{ 52 }$$
= $$\\ \frac { 19 }{ 26 }$$
Question 36.
All the three face cards of spades are removed from a well-shuffled pack of 52 cards. A card is then drawn at random from the remaining pack. Find the probability of getting
(i) a black face card
(ii) a queen
(iii) a black card
(iv) a heart
(vi) ‘9’ of black colour
Solution:
In a pack of 52 cards
All the three face cards of spade are = 3
Number of remaining cards = 52 – 3 = 49
One card is drawn at random
(i) Probability of a black face card which are = 6 – 3 = 3
Probability = $$\\ \frac { 3 }{ 49 }$$
(ii) Probability of being a queen which are 4 – 1 = 3
Probability = $$\\ \frac { 3 }{ 49 }$$
(iii) Probability of being a black card = (26 – 3 = 23)
Probability = $$\\ \frac { 23 }{ 49 }$$
(iv) Probability of being a heart = $$\\ \frac { 13 }{ 49 }$$
(v) Probability of being a spade = (13 – 3 = 10)
Probability = $$\\ \frac { 10 }{ 49 }$$
(vi) Probability of being 9 of black colour (which are 2) = $$\\ \frac { 2 }{ 49 }$$
Question 37.
From a pack of 52 cards, a blackjack, a red queen and two black kings fell down. A card was then drawn from the remaining pack at random. Find the probability that the card drawn is
(i) a black card
(ii) a king
(iii) a red queen.
Solution:
In a pack of 52 cards, a blackjack, a red queen, two black being felt down.
Then number of total out comes = 52 – (1 + 1 + 2) = 48
(i) Probability of a black card (which are 26 – 3 = 23) = $$\\ \frac { 23 }{ 48 }$$
(ii) Probability of a being (4 – 2 = 2) = $$\\ \frac { 2 }{ 48 }$$ = $$\\ \frac { 1 }{ 24 }$$
(iii) Probability of a red queen = (2 – 1 = 1) = $$\\ \frac { 1 }{ 48 }$$
Question 38.
Two coins are tossed once. Find the probability of getting:
(ii) at least one tail.
Solution:
Total possible outcomes are . HH, HT, TT, TH, i.e., 4
(i) Favourable outcomes are HH, i.e., 1
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 1 }{ 4 }$$
(ii) Favourable outcomes are HT, TT, TH, i.e., 3
So, P (at least one tail) = $$\\ \frac { 3 }{ 4 }$$
Question 39.
Two different coins are tossed simultaneously. Find the probability of getting :
(i) two tails
(ii) one tail
(iii) no tail
(iv) atmost one tail.
Solution:
Two different coins are tossed simultaneously
Number of possible outcomes = (2)² = 4
Number of event having two tails = 1 i.e. (T, T)
(i) Probability of two tails will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 1 }{ 4 }$$
(ii) Number of events having one tail = 2 i.e. (TH) and (HT)
Probability of one tail will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 1 }{ 4 }$$
(iii) Number of events having no tail = 1 i.e. (HH)
Probability of having no tail will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 1 }{ 4 }$$
(iv) Atmost one tail
Number Of events having at the most one tail = 3 i.e. (TH), (HT, (TT)
Probability of at the most one tail will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 3 }{ 4 }$$
Question 40.
Two different dice are thrown simultaneously. Find the probability of getting:
(i) a number greater than 3 on each dice
(ii) an odd number on both dice.
Solution:
When two different dice are thrown simultaneously,
then the sample space S of the random experiment =
{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) .
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
It consists of 36 equally likely outcomes.
(i) Let E be the event of ‘a number greater than 3 on each dice’.
E = {(4, 4), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
No. of favourable outcomes (E) = 9
P (number greater than 3 on each dice) = $$\\ \frac { 9 }{ 36 }$$ = $$\\ \frac { 1 }{ 4 }$$
(ii) Let E be the event of ‘an odd number on both dice’.
E = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)}
No. of favourable outcomes (E) = 9
∴ P (Odd on both dices) = $$\\ \frac { 9 }{ 36 }$$ = $$\\ \frac { 1 }{ 4 }$$
Question 41.
Two different dice are thrown at the same time. Find the probability of getting :
(i) a doublet
(ii) a sum of 8
(iii) sum divisble by 5
(iv) sum of atleast 11.
Solution:
Two different dice are thrown at the same time
Possible outcomes will be (6)² i.e. 36
(i) Number of events which doublet = 6
i.e. (1, 1), (2, 2) (3, 3), (4, 4), (5, 5) and (6, 6)
.’. Probability of doublets will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 6 }{ 36 }$$
= $$\\ \frac { 1 }{ 6 }$$
(ii) Number of event in which the sum is 8 are
(2, 6), (3, 5), (4, 4), (5, 3), (6, 2) = 5
Probability of a sum of 8 will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 5 }{ 36 }$$
(iii) Number of event when sum is divisible by
5 are (1, 4), (4, 1), (2, 3), (3, 2), (4, 6),
(5, 5) = 7 in numbers
Probability of sum divisible by 5 will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 7 }{ 36 }$$
(iv) Sum of atleast 11, will be in following events
(5, 6), (6, 5), (6, 6)
Probability of sum of atleast 11 will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 3 }{ 36 }$$
= $$\\ \frac { 1 }{ 12 }$$
We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Ex 22 help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Ex 22, drop a comment below and we will get back to you at the earliest.<|endoftext|>
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# Math Trivia For 5th Grade
Approved & Edited by ProProfs Editorial Team
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| By DanielCarig
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DanielCarig
Community Contributor
Quizzes Created: 205 | Total Attempts: 115,290
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Here are some amazing quiz questions on 5th grade math
• 1.
### If you burn 100 calories by climbing 100 stairs, how many calories do you burn climbing 50 stairs?
• A.
150
• B.
10
• C.
50
• D.
25
C. 50
Explanation
If you burn 100 calories by climbing 100 stairs, it can be inferred that each stair climbed burns 1 calorie. Therefore, if you climb 50 stairs, you would burn 50 calories.
Rate this question:
• 2.
### What is 36% of 40?
• A.
14.4
• B.
144
• C.
25.6
• D.
1440
A. 14.4
Explanation
To find 36% of 40, we can multiply 40 by 0.36. This gives us 14.4, which is the correct answer.
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• 3.
### What do we call a triangle with exactly two equal sides?
• A.
Equilateral Triangle
• B.
Scalene Triangle
• C.
Isosceles Triangle
• D.
Equilateral Triangle
C. Isosceles Triangle
Explanation
An isosceles triangle is a triangle with exactly two equal sides. The other option, equilateral triangle, has all three sides equal. The option scalene triangle has no equal sides. Therefore, the correct answer is isosceles triangle.
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• 4.
### What do we call a triangle that has all three sides of equal length?
• A.
Equilateral Triangle
• B.
Isosceles Triangle
• C.
Scalene Triangle
• D.
Right Triangle
A. Equilateral Triangle
Explanation
An equilateral triangle is a triangle that has all three sides of equal length. This means that each side of the triangle is the same length, and all three angles are also equal, measuring 60 degrees. The other options, isosceles triangle, scalene triangle, and right triangle, do not meet the criteria of having all three sides of equal length.
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• 5.
### How many inches are in 4 feet?
• A.
12
• B.
16
• C.
32
• D.
48
D. 48
Explanation
There are 12 inches in 1 foot. Therefore, to find the number of inches in 4 feet, we multiply 12 by 4. This gives us 48 inches.
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• 6.
### What do we call a triangle with no equal sides?
• A.
Right Triangle
• B.
Isosceles Triangle
• C.
Equilateral Triangle
• D.
Scalene Triangle
D. Scalene Triangle
Explanation
A triangle with no equal sides is called a scalene triangle. In a scalene triangle, all three sides have different lengths. This is in contrast to an isosceles triangle, where two sides are equal, and an equilateral triangle, where all three sides are equal. A right triangle refers to a triangle that has one angle measuring 90 degrees. Therefore, the correct answer is scalene triangle.
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• 7.
### Which of the following numbers is a factor of 14?
• A.
3
• B.
7
• C.
10
• D.
5
B. 7
Explanation
The number 7 is a factor of 14 because it divides evenly into 14 without leaving a remainder. When 14 is divided by 7, the result is 2, which means that 7 is a factor of 14.
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• 8.
### What do the numbers 1, 16 and 49 all have in common?
• A.
They Are All Square Numbers
• B.
They Are All Divisible By 7
• C.
They Are All Prime Numbers
• D.
They Are All Odd Numbers
A. They Are All Square Numbers
Explanation
The numbers 1, 16, and 49 are all square numbers because they are the result of multiplying an integer by itself. 1 is equal to 1*1, 16 is equal to 4*4, and 49 is equal to 7*7.
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• 9.
### What is the value of 3 with an exponent of 4?
• A.
75
• B.
12
• C.
7
• D.
81
D. 81
Explanation
The value of 3 with an exponent of 4 can be found by multiplying 3 by itself four times. This can be written as 3^4, which is equal to 81.
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• 10.
### Which of these numbers is largest?
• A.
0.325
• B.
0.3225
• C.
0.316
• D.
0.3249
A. 0.325
Explanation
To determine which number is the largest, we compare the decimal parts of each number. Among the given options, 0.325 has the largest decimal part. Therefore, 0.325 is the largest number among 0.325, 0.3225, 0.316, and 0.3249.
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• 11.
### A bullfrog is 20.3 cm long. The American toad is 11 cm long. How much longer is the bullfrog?
• A.
1.92 Cm
• B.
19.2 Cm
• C.
9.3 Cm
• D.
.93 Cm
C. 9.3 Cm
Explanation
The bullfrog is 20.3 cm long and the American toad is 11 cm long. To find out how much longer the bullfrog is, we subtract the length of the toad from the length of the bullfrog. Therefore, the bullfrog is 9.3 cm longer than the American toad.
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• 12.
### What is 5813 - 3925?
• A.
1788
• B.
1898
• C.
1878
• D.
1888
D. 1888
Explanation
The correct answer is 1888 because when we subtract 3925 from 5813, we get the difference of 1888.
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• 13.
### What is the least common multiple of 8 and 12?
• A.
48
• B.
96
• C.
72
• D.
24
D. 24
Explanation
The least common multiple (LCM) is the smallest number that is divisible by both 8 and 12. In this case, the LCM is 24 because it is the smallest number that both 8 and 12 can divide into evenly.
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• 14.
### What is 387,796 rounded to the nearest thousand?
• A.
390,000
• B.
387,000
• C.
380,000
• D.
388,000
D. 388,000
Explanation
The number 387,796 is rounded to the nearest thousand by looking at the digit in the hundreds place, which is 7. Since 7 is greater than or equal to 5, we round up the thousands place to the next higher number, which is 8. Therefore, the number is rounded to 388,000.
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• 15.
### What is the least common multiple of 10 and 25?
• A.
100
• B.
25
• C.
10
• D.
50
D. 50
Explanation
The least common multiple (LCM) is the smallest number that is a multiple of both 10 and 25. To find the LCM, we can list the multiples of both numbers and find the smallest common multiple. The multiples of 10 are 10, 20, 30, 40, 50, ... and the multiples of 25 are 25, 50, 75, 100, ... The smallest common multiple of 10 and 25 is 50, so it is the least common multiple.
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0
# Solve the equation. 7(2b – 5) + 3 = 10
Solve the equation. 7(2b – 5) + 3 = 10
7(2b - 5) + 3 = 10; or, distributing (removing parentheses), 14b - 7*5 + 3 = 10; or, 14b - 35 + 3= 10; or, 14b - 32 = 10; then add 32 to each side of the equation ---> 14b = 42; then divide each side of equation by 14 ---> b = 42/14 = 21/7 = 3;
### 4 Answers by Expert Tutors
Sonia A. | Reach for the stars, but keep your feet firmly planted on the ground.Reach for the stars, but keep your feet ...
0
14 b - 35 +3 =10
Get the number with the variable by itself by removing the numbers without variables to the other side of the equal sign.
14b -35 +35 +3 -3 =10 + 35 - 3
14 b = 42
Divide each side by 14
Since 14 cancels 14, b remains on the left of the equal sign. 42 divided by 14 equals 3
Therefore, b = 3
Tiffany W. | Head Start 2 a number careerHead Start 2 a number career
0
7(2b – 5) + 3 = 10
when doing algebra always remember PEMDAS..(Please Excuse My Dear Aunt Sally) know as Parenthesis, Expontent, Multiplication, Division, Addition, Subtraction
step 1: get rid of parenthesis
Multiply 7 by each number in Parenthesis.. 14b - 35
now the problem should look like: 14b - 35 + 3 =10
Step 2: adding on the same side of the equation
isolate the problem will read as: (-35 + 3) negative plus a positive means subtract therefore the answer will be -32
now the problem will read as 14b - 32 = 10
step 3: moving tumblers to the side of the equation and changing the ssign now
-32 becomes 32
Therefore add 32 + 10 = 42
the problem reads 14b = 42
Step 4: isolate lithe letter by dividing
42/14 = 3
b = 3
Ragini N. | Passionate about teaching, especially at the elementary level.Passionate about teaching, especially at...
0
7(2b-5)+3=10
7(2b-5)=10-3=7 (-3 on both sides)
2b-5=7/7=1 (divide both sides by 7)
2b=1+5=6 (+5 on both sides)
b=6/2=3 (divide both sides by 2)
b=3
Bill F. | Experienced Teacher & Tutor in Round Rock, TXExperienced Teacher & Tutor in Round Roc...
5.0 5.0 (1 lesson ratings) (1)
0
Teri, I think you can do this: first multiply what's outside with parentheses with the terms inside them:
7*2b - 7*5 = 14b - 35
Now put that with the rest: 14b - 35 + 3 = 10, or
14b -32 = 10, or
14b = 10 + 32 (added 32 to both sides)
14b = 42
b = 42/14 = 3 (divide both side by 14 to get b by itself)<|endoftext|>
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Fresh air is an important standard to measure indoor air quality. It affects air circulation and indoor air pollution.
Fresh air is useful for two purposes:
One is to reduce the concentration of harmful substrates, reach the hygiene standard. Taking CO2 as an example, its average standard value is 0.1%. The second is to supplement indoor air ventilation and keep indoor positive pressure.
According to national Indoor Air Quality Standard GB/T18883-2002, indoor fresh air value should be guaranteed not less than 30 cubic meters per house per person.
Fresh air not only affects human health, but also energy consumption, initial investment and operational cost. Lack of fresh air will cause “indoor syndrome”, headaches, palpitation, feeling fatigue and/or tiredness, illnesses on...Read more
There are many sources of indoor air pollution in any building. These include combustion sources such as oil, gas kerosene, coal, wood, and tobacco products; building materials and furnishings as diverse as deteriorated, asbestos-containing insulation, wet or damp carpet, and cabinetry or furniture made of certain pressed wood products; products from cleaning and maintenance, personal care, or hobbies; central heating and cooling systems and humidification devices; and outdoor sources such as radon, pesticides and outdoor air pollution.
The relative importance of any single source depends on how much of a particular pollutant is given off and how hazardous those emissions are. In some cases, factors such as how old the source is and whether it is properly maintained are...Read more
The term “HVAC system” is used to refer to the equipment that can provide heating, cooling, filtered outdoor air, and humidity control to maintain comfort conditions in a building. Not all HVAC systems are designed to accomplish all of these functions. Some buildings rely on only natural ventilation. Others lack mechanical cooling equipment (AC), and many function with little or no humidity control.
The features of the HVAC system in a given building will depend on several variables:
- Age of the design
- Building codes in effect at the time of the design
- Budget that was available for the project
- Planned use of the building
- Owners’ and designers’ individual preferences
- Subsequent modifications
Basic Components of an HVAC System
The basic components of an HVAC...Read more
The same Heating, Ventilating, Air Conditioning (HVAC) system that distributes conditioned air throughout a building can distribute dust and other pollutants, including biological contaminants. Dirt or dust accumulation on any components of an air handling system – its cooling coils, plenums, ducts, and equipment housing – may lead to contamination of the air supply.
There is widespread agreement that building owners and managers should take great precautions to prevent dirt, high humidity, or moisture from entering the ductwork. At the same time, there is less agreement about what measures to clean up are appropriate or how effective cleaning techniques are at making long-term improvements to the air supply or at reducing occupant complaints.
The presence of dust in ductwork...Read more
The US Environmental Protection Agency (EPA) states that indoor air is anywhere from 2 to 10 times more hazardous than outdoor air. With people spending an average of 90% of their time inside, indoor air pollution can pose a serious health risk. “Indoor air quality is the number one environmental health problem in the United States”, according to the EPA.
The American College of Allergies states that 50% of all illness is aggravated or caused by polluted indoor air.
A recent study found that the allergen level in super-insulated homes is 200% higher than it is in ordinary homes.
According to Scientific America, a baby crawling on the floor inhales the equivalent of 4 cigarettes a day, as a result of the out-gassing of carpets, molds, mildews, fungi, dust mites, etc.
The EPA...Read more
The knowledge and need to keeping your air conditioning equipment clean will continue to reward you for years to come. Educating yourself on things like cleaning the air conditioning unit, having a professional service the air conditioning unit, and general maintenance are all the keys to success.
One major reason for keeping your air conditioning equipment clean is it is extremely important for the health of your loved ones using it, a clean living space, and all around elimination of air pollutants.
Cleaning your air conditioning equipment hasn’t ever shown a 100% prevention of health problems, but it has shown that dust particle levels truly increase in the home. When your air conditioning equipment is dirty, it manifest things like mould, fungi, bacteria and other air borne...Read more
The Kingdom Fungi is a diverse kingdom consisting of over 1 million species and includes mushrooms, moulds, and yeasts. Fungi obtain their nutrition from the breakdown and decay of organic matter. They can thrive in many places such as soil, plant litter, wood, live plants, dung, animal remains, fungal remains, etc, and play a vital role in the environment as a decomposer of dead-plant matter.
Commonly called mildew, moulds (sometimes referred to as "black mould"), are a subset of fungi that produce fluffy or powdery growth on surfaces. Toxic moulds can grow on cloth, carpets, leather, wood, sheetrock, insulation (and on human foods) when moist conditions exist.
Moulds are found everywhere. The most common form of fungus on earth, and may grow at high...Read more<|endoftext|>
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# 5-4 Complex Numbers (Day 1)
## Presentation on theme: "5-4 Complex Numbers (Day 1)"— Presentation transcript:
5-4 Complex Numbers (Day 1)
Objective: CA 5.0 Students demonstrate knowledge of how real number and complex numbers are related both arithmetically and graphically.
Not all quadratic equations have real number solutions.
has no real number solutions because the square of any real number x is never negative.
To overcome this problem, mathematicians created an expanded system of numbers using the imaginary unit. The imaginary unit i can be used to write the square root of any negative number.
The square root property of a negative number property
1. If r is a positive real number then:
2. By property (1): it follows that…
Example 1: Solve
If b 0 then a + bi is an imaginary number
A complex number written in standard form is a number a + bi where a and b are real numbers. The number a is the real part of the complex number, the number bi is the imaginary part. If b 0 then a + bi is an imaginary number If a= 0 and b ≠ 0 then a + bi is a pure imaginary number.
Every complex number corresponds to a point in the complex plane.
Keep in mind: a is the real part (x –coordinate) bi is the imag. part (y-coordinate)
Example 2: 2-3i = (2, -3) -3+2i = (-3, 2) 4i = (0, 4)
Difference of complex numbers
Two complex numbers a + bi and c + di are equal if and only if a=c and b=d Sum of complex numbers Difference of complex numbers
Simplify: √-18 + √-32 i√18 + i√32 3i√2 + 4i√2 7i√2
Example 3: Write the expression as a complex number in standard form.
4 – i i 7 + i
Example 4: 7 – 5i i 6 + 0i 6
Example 5: i i -9i + 4i -5i
Multiplying Complex Numbers
To multiply complex numbers use the distributive property or the FOIL method.
Example 5: Write each expression as a complex number in standard form.
1.
Example 6:
Example 7:
Homework= Accelerated Math Objective:<|endoftext|>
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# How does base of a number relate to modulos of its each individual digit?
I encountered a programming problem that need a little math. Since I took number-theory long time ago, from the top of my head, I could not think of a way to approach this problem.
Give a base 'b' ( 2, 3, 4, 5, 6 ... -> 26 ). So if I'm at base 10, I will have 10 digits: 0, 1, 2, 3...9. The digit of a base 'b' is called interesting if it satisfy this condition: 3 is an interesting digit because 118*3 = 354 and 3+5+4 = 12. Which means both 354 and 12 divides 3. By look at the result sets, I can see the relation is:
if b mod d = 1 then d is an interesting digit.
For example: 10 mod 9 = 10 mod 3 = 1. So both 3, 9 are interesting digits in base 10.
How could I prove this? Any hint?
Thanks,
Chan
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You seem to be asking for what digits $\displaystyle 0 \lt d \lt b$ can we apply the following divisibility test:
A number is in base $\displaystyle b$ is divisible by $\displaystyle d$ if the sum of digits of the number is divisible by $\displaystyle d$.
This is certainly true for $\displaystyle b = 1 \mod d$.
Because $\displaystyle \sum_{k=0}^{n} a_k b^{k} - \sum_{k=0}^{n} a_k = \sum_{k=0}^{n} a_k (b^k -1) = 0 \mod d$
Thus $\displaystyle \sum_{k=0}^{n} a_k b^{k} = \sum_{k=0}^{n} a_k \mod d$.
If $\displaystyle b = r \mod d$ where $\displaystyle r \neq 1$, then we have that there is some $\displaystyle k$ for which $r^k - 1 \neq 0 \mod d$.
The number with $\displaystyle 1$ as the digit correponding to $\displaystyle b^k$, and $\displaystyle d-1$ in as the "units" digit (corresponding to $\displaystyle b^0$) is a counterexample to the divisibility rule for $\displaystyle d$.
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thanks a lot! After reading your solution, I feel so shame of myself! – Chan Jan 12 '11 at 17:08
@Chan: Don't be. Everyone misses the most obvious things sometime... And I would not call this one obvious. – Aryabhata Jan 12 '11 at 17:29
If $b \ne 1 \pmod d$, then the number $10$ in base $b$ shows that $d$ is not interesting: it represents the number $b$, so it is not equal to $1 \pmod b$; but $1 + 0 = 1$. This, together with Moron's answer, shows that your conjecture is correct.
Edit Taking the question literally, I really need to exhibit a number that is divisible by $d$ but the sum of whose digits is not; or vice versa. If $d < b$, take the number $1.(d-1)$; the sum of its digits is $d$, but it represents $b+d-1$, which is not divisible by $d$. If $d \ge b$, then the representation of $d$ in base $b$ is two or more digits, so their sum must be less than what they represent (because all but the last digit represent numbers greater than themselves).
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You are disproving a stronger statement: "The number and the sum of digits have the same remainder". – Aryabhata Jan 12 '11 at 17:39
Thanks Moron, you're right. I've added a codicil. – TonyK Jan 12 '11 at 17:46
This is simply the radix $\rm\:b\:$ generalization of casting out nines. Generally, if $\rm\:d\:$ divides $\rm\ b-1\$ then $\rm\: mod\ \ d:\:\ \ b\equiv 1\ \Rightarrow\ a_n b^n +\:\cdots\: + a_1 b + a_0 \ \equiv\ a_n +\: \cdots\: + a_1 + a_0\: =\:$ sum of the radix $\rm\:b\:$ digits. For further remarks on casting out nines see my post here.
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100 years ago, in august 1918, began a large-scale offensive of the allies on the german army, which will continue until the end of the war and was later called the hundred-day offensive. The attack itself was done with a positional war, has again become agile character. The offensive consisted of a series of operations on almost all front line and began the offensive at amiens. Background the second battle of the marne – the last decisive offensive of the german army on the Western front, has not led to victory.
The allies knew about the time and place of the enemy attack and repelled it. Then they counterattacked and drove the germans into old positions that they held prior to the spring offensive. On 2 august the french troops liberated the germans from the town of soissons. By 5 august, the offensive of the entente allies in the region of reims exhausted.
However, the strategic initiative passed to the allies. The german army was exhausted and demoralized. The last hope to successfully end the war collapsed. However, the german command still did not believe in defeat, underestimated the enemy and overestimated their own strength.
Russian military historian, general andrei zaionchkovskii, wrote: "It (the german command — the author) had 204 divisions, of which 70 were in reserve. Under these conditions, the hindenburg decided not to abandon the initiative and to take new, necessarily sudden offensive operations, but on a smaller scale — as in direction to the coast, and on other fronts, with the aim to improve their position, to give the allies the losses and to show them that german power is not broken. This circumstance, according to the chief of the german command, could even incline the entente to peace negotiations. But the command, having lost the initiative, not clear enough, imagine that the crisis of the marne is the beginning of the end of world war ii.
This alone can explain the desire it again to take the offensive, despite the daily increasing superiority of allied forces and means. " the plans of the german command of the new offensive will not be implemented: only three days later, august 8, will start a strategic offensive of the allies. British artillery during the hundred-day offensive. August 1918, the plans of the parties on 24 july 1918 in bombone held a meeting the commanders of the allied armies, petain, haig and pershing. The commander of the Western front, general foch had outlined a plan of further operations.
The main idea of this plan was to abandon defense and go on the offensive. Directive of 24 july allied command consisted of a number of separated by short intervals of offensive operations with the aim of eliminating protrusions in the front line, formed as a result of the german spring offensive, and saint-malski ledge. This allowed us to release a number of railways, necessary for the further development of the offensive. Also eliminated the threat of the Northern mining district and the ports of calais and dunkerque.
If these operations are conducted successfully and in a short time, then was planned to go into general decisive offensive aimed to crush the entire german front, and not let the enemy, as it were, to withdraw to a prepared rear position. At the same time, the plan foch was very careful. The allied command, overestimating the power of resistance of the german army hoped to end the war until 1919. General conditions for the allied advance were favorable: by august, France had already deployed 1. 2 million soldiers and officers of the american army.
The allies have numerical superiority. The british also gathered all my strength and throw in France, even some of the troops from palestine, where there was fighting with the turks. The first of the planned allied operations to eliminate protrusions in the front line was the amiens operation. Planning it, the command of the allies expected to clear the enemy from amiens ledge to eliminate the threat to amiens and the railway paris — lyon, and to break and to throw the german forces between the rivers somme and avre.
The german high command still hoped to turn the tide in their favor and achieve good world for Germany. When the chief of staff of the 4th army, general friedrich lassberg offered without a fight to withdraw troops from the occupied spring and summer of 1918 the territories of the old position, straighten the front line and to avoid unnecessary losses, he refused. Ludendorff answered: "I believe your suggestions are correct, but can't follow them for political reasons". Failure from the onset and withdrawal of troops in the old position meant recognition of the collapse of all the hopes of Germany and its allies to win the war, and naprasnosti offensive operations that led to such huge losses.
2 aug ludendorff signed a directive to the commander of the army groups, which stated: "The situation requires us to, on the one hand, went on the defensive, but as soon as the opportunity would take the offensive. " it was planned to hold a number of small offensives to improve tactical position in flanders, in the region of the oise, east of reims, and also in the area of army group of duke albrecht. Thus, the germans overestimated the results of their previous offensive operations, its forces, believed that the allies drained and incapable of a major operation in the near future. 37th british division, mark v tanks of the 10th battaliontank corps and captured german guns caliber 4. 2 inch during the hundred-day offensive. August, 1918, the preparation of the operation to conduct the operation involved 4 english, 1st and 3rd of the french army under the overall command of field marshal haig.
On the first day, 8 august offensive at the front, 25 km from albert to morale passed 4-i'm english and levelingbuy 31 corps of the 1st french army. Then was to begin the offensive of 3rd army and the remaining forces of the 1st army. Part of the offensive group was included 17 infantry and 3 cavalry divisions, 2684 artillery, 511 tanks, 16 armored vehicles and 1,000 aircraft. Defending this sector of the front the troops of the 2nd german army von de marvita had 7 infantry divisions, 840 guns and 106 aircraft.
The germans in this area by august was located, exhausted and weakened in previous battles part. A great advantage of the allies before the germans was the presence of large masses of tanks. The flat nature of the terrain allowed to actively use the tanks. In this advanced part of the 4th british army since the spring engaged in a small battle for the improvement of their tactical positions.
As a result, 2-i the german army at the beginning of august, almost completely lost the lane outposts and defended in underdeveloped deep positions. The british the results of these fights, and on the basis of aerial photography and data tactical intelligence already long before made up the full picture of the system of german defence. The allies, using the successful experience of the german attacks, refused a powerful and prolonged artillery barrage. The start of the offensive was scheduled for 4 hours and 20 minutes.
Was planned after the passage of the tanks line of advance units of infantry, all the artillery to open a sudden fire. The third artillery had to create a barrage, and the rest of the artillery — fire on infantry and artillery positions, command posts, ways of approach reserves. Barrage for three minutes, we had to stay on the advanced german positions. During this time, tanks and attacking infantry had come close to firing shaft and immediately follow him.
Barrage were to wage irregular at first after 2 minutes then after 3 minutes and later after 4 minutes. The left flank of the 1st french army on the offensive after a 45-minute artillery preparation. Very clearly was planned and the order of occurrence. 2 hours after the attack began, 6 hours and 20 minutes, the infantry and tanks had to reach the first line of attack — lines at a distance of about 3 km from the english trenches.
Then the advance was halted for two hours. At this time, moved up the artillery. The attack was resumed at 8: 20 a. M. And lasted continuously until the second milestone, which was 4. 5 — 8 km from its original position, and then without break, until the third boundary at a depth of 9 — 12 km.
Cavalry corps attached to the 4th british army, was to speak at 8 o'clock 20 minutes, to overtake the battle formations of infantry, to seize the third line and hold it until the main forces, and then to build on this success further. Finally, the particular success of the operation on 8 august contributed to the dense morning fog, reinforced by french and british use of smoke and chemical shells. Another feature of the operation was complete secrecy. The whole area of concentration of allied troops was covered by the aircraft, due to the good condition railway tracks in the district of occurrence threw 230 military trains and over 60 trains with ammunition.
Artillery took up their positions in the last 2-3 days before the attack, and the tanks on the night of 8 august. For the introduction of the enemy in the area of ypres on the orders of the british command, have been extensively demonstrative actions. As a result, when in the last days before the attack from the front of the german positions began to receive reports of suspicious activities behind enemy lines, and aerial reconnaissance reported about the movement of a column of tanks, the german high command did not pay much attention. British military historian neil grant wrote: "There have been huge efforts to ensure the element of surprise: on the offensive not even informed the british war cabinet.
4th army of general rowlinson was doubled, but it was done so that the germans have learned nothing. Some canadian units, which the germans believed the assault forces of the british, their presence meant imminent attack — was pointedly sent to flanders". Source of map: campaign of 1918 french theatre. Battle of montdidier - amiens on 8 august to 25 september.
Source: a. Zaionchkovskii. World war of 1914-1918, volume iii. Battle on 8 august 1918 in 4 hours and 20 minutes, the federal artillery opened heavy fire on the positions, command and observation points, communications nodes, and logistic facilities of the 2nd german army.
At the same time part of the organized artillery barrage, under cover of which the divisions of the 4th british army, accompanied by 415 tanks moved into the attack. The surprise succeeded fully. Anglo-french offensive was a complete surprise for the german command. The fog and the massive gaps in chemical and smoke shells were closed all that was on 10-15 meters from the positions of german infantry.
Before the german command was able to understand the situation on the german positions hit the mass of tanks, firing on the move from machine guns encountered soldiers and destroying telegraph and telephone lines. The resultthe headquarters of several german divisions were captured by surprise, quickly moving forward british infantry and tanks, which further increased the confusion in the german ranks. A breakthrough of enemy defenses developed methodically, almost in full accordance with the developed plan. To 6 hours 20 minutes troops of the 4th british army had reached the first line of attack.
Two hours later, after he moved up artillery, the attack was resumed, and for 13 hours 30 minutes the allies had reached the third boundary at a depth of about 11 km. However, further attempts by the british and french troops to move deep into the defense of the enemy crashed against the strong resistance of the german divisions was hastily thrown in the area of breakthrough from other sectors. The german troops lost the day before 27 thousand in killed and prisoners, 400 guns, and also a large number of different military assets. Allied aircraft joined the battle as soon as the fog, shot down 62 german aircraft.
9 august allied offensive continued. The battle is fully joined the 1st french army, and on 10 august - 3rd french army. The attack was carried out now all along the front from albert to the river oise, but it developed slowly. The germans fought stubbornly, threw the reserves and breakthrough prevented.
There was local fighting. German artillery was rebuilt for the conduct of anti-tank defense, with the result that the british and french tanks suffered heavy losses. Since august 8 of 415 tanks, join the battle on the site of the 4th british army, has failed about 100 cars. On 9 august the offensive took part only 145 tanks, 39 of which were destroyed by the fire of the german artillery.
Heavy losses in tanks had an impact on reducing rates of occurrence. August 12, tanks is no longer involved in the battle, and the remaining tanks were withdrawn to the rear. August 12, the fighting was only in certain parts of the front, 13 aug advancing allies stopped altogether. Captured german gun.
August 1918 captured in the battle of amiens, the 4th british army german guns of the outcome of the battle of amiens the allies in five days managed to advance deep into enemy defense to 3-18 kilometers on the front with a length of 75 kilometers, eliminating the threat to amiens and the railway paris − amiens. In the course of the operation, the germans lost 74 thousand persons (from them 33 thousand prisoners), the allied — 46 thousand people. The morale of the german troops were severely broken in pieces, intended for the transfer under amiens, showed dissatisfaction, there have been cases of mass desertion. Military success of the allied forces was due to the complete secrecy of the operation, the germans never suspected a thing; also, the allies successfully chose the site of the break, where the german defence had the least depth.
The success was also due to a significant superiority of the allies in force, careful preparation, sudden impact and the massive use of tanks. However, the methodical promotion august 8, from line to line with a two-hour delay on the first of them led to large losses, gave the german troops a chance to recover after the first shock of surprise attack, and regroup for active defence. Then the germans brought up reserves and stopped the enemy, and the allies are unable to turn a tactical breakthrough in the operational front, using the advantage in cavalry and tanks. The victory was of tactical importance to the german front to break through failed, but the victory at amiens finally secured the strategic initiative for the allies.
After the battle, ludendorff wrote: "August 8, 1918 is the black day of the german army in the history of the world war. " the german army moved to strategic defense. The troops were ordered: "Not an inch of land do not leave without a fierce struggle. " on 13 august 1918 in the german rate of the high command in spa was held a meeting of the german command, the chancellor of hertling and the state secretary of the ministry of foreign affairs of ginza. Everyone was overwhelmed by the situation. Ludendorff announced that the german army was unable to crush the enemy offensive; to achieve peace, defensive actions, in spite of submarine warfare as possible.
Therefore, for the end of the war, should go to the peace talks. The next morning in the spa held a meeting of the crown council, chaired by kaiser wilhelm ii where it was agreed to start peace negotiations with the entente through the dutch queen. Arrived at spa on 14 august, the austro-hungarian emperor karl, foreign minister burian and chief of the general staff of the arts background, strassenburg joined in this decision. However, the negotiations with the representatives of the entente have not been started.
Hindenburg still hoped that the german army will remain in France and belgium that will allow to conclude an advantageous peace. British tank mark v. August 1918 armored vehicles during the reconnaissance. The second battle of the somme.
25 august 1918 second battle of the somme after the battle of amiens the allied troops were embarked on the expansion front offensive on the flanks advanced forward 4th english, 1st and 3rd french armies and ousting the enemy in the siegfried position. To the North of the somme was supposed to carry out the offensive 3rd of the british army in the general direction of bapaume, péronne. South of the somme offensive in the direction of shawnee passed 10th french army. On the morning of 20 august the 10th french army beganthe offensive against the 9th german army on the front from soissons to the river oise.
By 23 august she moved to the line of the rivers oise and ellet. August 21 in the North fighting against the 1st german army on dvadtsatikilometrovaya front from albert to arras, the beginning of the 3rd british army. By the end august 26, she came to the line bray, bapaume, advancing forward for 10 km on this day, the offensive involved the 1st british army. On 29 august she came on the line bulgur, drochur.
The allied offensive forced the german command to begin the removal of the 17-th, 2-th, 18-th and 9-th armies on the line of crouzil, bapaume, péronne, noyon. In the last days of august, the germans preferred not to get involved in bloody battles, and again to depart on a well-fortified hindenburg line (siegfried line), with which they began their spring offensive. Thus, from 8 to 30 august the army of the allies on the front from soissons to arras with over 150 km moved in the center to 35 km, and on the flanks — 15 — 20 km, 30 august the allied offensive continued, first on the flanks, and then in the centre, with the aim not to allow the german troops to gain a foothold ahead of the position of the hindenburg. So, on august 31 began the battle for mont st quentin.
Here against the germans made the australian part. On the night of 31 august the australians successfully crossed the somme in a key location of german defense, in the bend of the river. Capturing the german trenches on 1 september the australians liberated the town of péronne and forced the enemy to retreat to the east, to the hindenburg line. 2600 germans were captured.
Loss of australians during the fighting, which lasted until september 3, was about 3,000. The threat from both flanks forced the german command on 2 september to give the order on further withdrawal of the 17-th, 2-th, 18-th and 9-th armies on the front between the rivers skarn and value for 160 km to the hindenburg. The retreat began on the night of september 3 and passed with almost no interference from the enemy. September 8, german troops occupied most of the positions on the front from arras to r.
Ellet, with whom they launched spring offensive. In early september, the german troops themselves cleared the ledge on the r. Lys. Thus, the first part of the plan of offensive operations of the allied armies, planned fosem, succeeded.
It only remained to dislodge the enemy from the saint-malskog protrusion formed in september 1914, german prisoners arriving at a temporary camp near amiens. On 9 august 1918, the german prisoners are wounded. September 1918.
Supported during the First world armored cavalry? On the interaction of two mobile armed forces, with emphasis on the use of French armored vehicles, this article. The beginning of the war in France was not a single armored car: t...
After exceptionally disciplined and systematic departure of the Germans, sometimes with very hard battles (for example, July 23, at the site of the coming of the 30th French corps began a strong fight, during which the attack of F...
Piracy Ubykhs flowed slowly into military action and back. So, when the Russian Empire continued to master the fragmented and internecine wars incited against Russia, the Caucasus, harassing thereby, the power-hungry Ottoman port,...<|endoftext|>
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# Polygons
## Explore the properties of polygons through puzzles and games, then proceed into a more formal classification of polygons. Look at mathematical definitions more formally, and explore how terms can have different but equivalent definitions.
### In This Session
Part A: Hidden Polygons
Part B: Classifying Polygons
Part C: Definitions and Proof
Homework
In this session, you will use puzzles and a classification game to explore polygons. You will play with the definitions of various polygons to begin to internalize their meaning. You will also begin to look at mathematical definitions more formally and explore how a term can have different, but equivalent, definitions.
For information on required and/or optional materials for this session see Note 1 below.
### Learning Objectives
In this session, you will learn to do the following:
• Classify polygons according to some of their features
• Understand how to divide polygons into triangles and the implications of that
• Begin to understand mathematical definitions
### Key Terms
New in This Session
Concave Polygon: A concave polygon is any polygon with an angle measuring more than 180°. Concave polygons look like they are collapsed or have one or more angles dented in.
Convex Polygon: A convex polygon is any polygon that is not concave.
Irregular Polygon: An irregular polygon is any polygon that is not regular.
Isosceles Trapezoid: An isosceles trapezoid is a quadrilateral with one pair of parallel sides and congruent base angles, or it is a trapezoid with congruent base angles.
Kite: A kite is a quadrilateral that has two pairs of adjacent sides congruent (the same length).
Line Symmetry/Reflection Symmetry: A polygon has line symmetry, or reflection symmetry, if you can fold it in half along a line so that the two halves match exactly. The folding line is called the line of symmetry.
Parallelogram: A parallelogram is a quadrilateral that has two pairs of opposite sides that are parallel.
Polygon: A polygon is a two-dimensional geometric figure with these characteristics:
• It is made of straight line segments.
• Each segment touches exactly two other segments, one at each of its endpoints.
• It is closed — it divides the plane into two distinct regions, one inside and the other outside the polygon.
Rectangle: A rectangle is a quadrilateral with four right angles.
Regular Polygon: A regular polygon has sides that are all the same length and angles that are all the same size.
Rhombus: A rhombus is a quadrilateral that has all four sides congruent.
Square: A square is a regular quadrilateral.
Trapezoid: A trapezoid is a quadrilateral that has one pair of opposite sides that are parallel.
Venn Diagram: A Venn diagram uses circles to represent relationships among sets of objects.
Vertex: A vertex is the point where two sides of a polygon meet.
### Notes
The following materials are needed:
• Loops of string (at least three loops per group or individual working alone).<|endoftext|>
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# Electrical installation exercises
Next we will see several exercises of simple electrical installations, I clarify that they are not exercises of electrical circuits as such, they are exercises as if we were to carry out an electrical installation in a building, not as if we were to make a phenolic plate with several more complex circuits.
## Exercise 1 of electrical installations
We have the following circuit that powers a lamp and the objective of this exercise is to find the resistance that opposes the passage of current.
This exercise is very simple, it is a simple formula that we will do, using the following formula the exercise is solved:
P=\cfrac{V^{2}}{R}
And all we have to do is clear the resistance:
R=\cfrac{V^{2}}{P}
Now we simply apply variables that we already have such as the voltage of 5V and the power of the lamp of 5W:
R=\cfrac{(5V)^{2}}{5W} = 5 \Omega
This gives us as a result that the resistance of our lamp is 5\Omega
## Exercise 2 of electrical installations
A lamp has a resistance of 181.1\Omega and is connected to the voltage of a house. Taking the average or common voltage of 127V and using the following circuit, what is the wattage of the lamp?
For this exercise we will use the formula from exercise 1 and directly apply the formula:
P=\cfrac{(127V)^{2}}{181.1\Omega }=89.06W
So our lamp power is 89.06W
Keep in mind that in this exercise they gave us other data such as frequency and that it is a sinusoidal. In this type of exercise it does not affect anything that is of a frequency of 50\text{Hz} or 60\text{Hz}, if they are already telling us that the voltage is 127V, it is more than enough for us to be able to solve the exercise.
I’m telling you so that you don’t stress me out because as they see that in an exercise they give you more data, it means that you have to use them.
## Exercise 3 of electrical installations
In a series circuit with voltage of 127V and resistors shown in the circuit, calculate the total resistance R_{T}, the current I and the power P.
Let’s start with the total resistance, it is simply a sum since there are all the resistors in series, so for our equivalent resistance we have the following:
R_{T} = 8\Omega + 2\Omega + 5\Omega \ \rightarrow R_{T} = 15 \Omega
So we have that our total resistance is equal to 15\Omega
For the intensity we will only apply a simple formula of Ohm’s Law:
I = \cfrac{V}{R} = \cfrac{127V}{15\Omega} = 8.46A
The intensity of the circuit is 8.46A
And for the power we apply a formula with the voltage and intensity that we already have and that we already calculated:
P =127V \times 8.46A = 1074.42W
So our power is 1074.42W
## Exercise 4 of electrical installations
Calculate the total current that will flow in the circuit and the voltages of each resistor. At the end check the total voltage by adding the voltages of all the resistors.
As we did in the previous exercise, since we have that all the resistances are in series, we simply have to add them:
R_{T} = R_{1} + R_{2} + R_{3} + R_{4} + R_{5}
R_{T} = 1.5\Omega + 1.5 \Omega + 1 \Omega + 1 \Omega + 0.15\Omega
The sum gives us an equivalent resistance of 5.15\Omega
Next, what we have to do is apply the Ohm’s Law formula to obtain the total current of the circuit:
I_{T} = \cfrac{127V}{5.15\Omega} = 24.66A
So we have that the total current of the circuit is 24.66A.
In this last step of calculating the voltages of each resistance we will simply have to do several multiplications with the same formula of Ohm’s Law of V = R \times I
V_{R1} = 1.5\Omega \times 24.66A = 36.99V
V_{R2} = 1.5\Omega \times 24.66A = 36.99V
V_{R3} = 1\Omega \times 24.66A = 24.66V
V_{R4} = 0.15\Omega \times 24.66A = 3.69V
V_{R5} = 1\Omega \times 24.66A = 24.66V
When performing all the sums of the resistances, we have the following:
36.99V+36.99V+24.66V+3.69V+24.66V = 126.99V
## Exercise 5 of electrical installations
Calculate the total current that will flow through the circuit and the total resistance of the circuit. Remember that the total resistance of the circuit must be less than the partial resistances.
To begin, the formula that we will use is that of the sum of the resistances in parallel, you can see it by clicking here. So let’s get started:
\cfrac{1}{R_{T}} = \cfrac{1}{2} + \cfrac{1}{4} + \cfrac{1}{5} = \cfrac{10}{20} + \cfrac{5}{20} + \cfrac{4}{20} = \cfrac{19}{20}
Now that we have the total, we need to solve for our equivalent resistance of the parallel circuit:
\cfrac{1}{R_{T}} = \cfrac{19}{20} \ \rightarrow \ R_{T} = \cfrac{20}{19} = 1.05\Omega
So our equivalent resistance is equal to 1.05 \Omega.
And simply to calculate the intensity of our source, we apply the Ohm’s Law formula with the intensity cleared:
I=\cfrac{127V}{1.05\Omega} = 120.95A
Finally we have that our circuit current is 120.95A.
## Exercise 6 of electrical installations
Calculate the total resistance of the circuit, the total current, and the power that will flow. Check by calculating the partial currents and adding them.
To start with this exercise, what we will do is add resistance 3 with resistance 4 because they are in series, which gives us a value of 12 \Omega.
Next we have to calculate the equivalent resistance of the 12 \Omega with resistance 2 adding in parallel:
\cfrac{1}{R_{T}} = \cfrac{1}{12} + \cfrac{1}{4} = \cfrac{1}{12} + \cfrac{3}{12} = \cfrac{4}{12} =\cfrac{1}{3} \rightarrow R_{2,3,4} = 3\Omega
Now we have to add the equivalent resistance with resistance 1, as they are in series, we will have a total resistance of 5\Omega.
The current is simply calculated with Ohm’s Law:
I=\cfrac{V}{R} = \cfrac{15V}{5\Omega}=3A
So our current is 3A.
We calculate the voltage of resistor 1:
V_{R1} = 2\Omega \times 3A = 6V
The voltage of resistors 2, 3 and 4 is the difference of the total with the calculated one, that means that it is 9V. Letting us understand that the voltage across resistor 2 is 9V and the voltage across the equivalent resistor 3 and 4 is also 9V.
Since we already have the voltages, we simply have to find the currents that will flow in resistor 2 and in the equivalent resistor of 3 and 4. We simply use Ohm’s Law:
I_{R2} = \cfrac{9V}{4\Omega } = 2.25A \qquad I_{R3,4} = \cfrac{9V}{12\Omega} 0.75A
Adding these two resistors we obtain a value of 3A, giving us the same value as the total current of the circuit, which means that the calculations of both equivalent resistors and the current flowing in certain resistors are correct.
Thank you for being in this moment with us : )<|endoftext|>
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# CBSE Class 10 Mathematics Exam 2018: Important 1 Mark Questions
Mar 23, 2018 10:21 IST
CBSE Class 10 Mathematics: Important 1 Mark Questions
Here you will get the most important 1 mark questions to prepare for CBSE class 10 Mathematics board exam which will be held on 28 March, 2018. All these questions have been provided with detailed solutions.
In CBSE Class 10 Mathematics Board Exam 2018, Section - A will comprise 6 questions of 1 mark each.
Students must practice the important questions given here. This will help them to track their preparedness for the exam and also make them familiarised with the important topics of class 10 Maths. Moreover, the solutions provided here will give students an idea about how to write proper solution to each question in the board exam so as to score optimum marks.
Given below are some sample questions for CBSE Class 10 Mathematics: Important 1 Mark Questions:
Q. If the HCF of 65 and 117 is expressible in the form of 65 m – 117, then find the value of m.
Sol.
We have, 65 = 13 × 5
And 117 = 13 × 9
Hence, HCF = 13
According to question 65m – 117 = 13
⇒ 65m = 13 + 117 = 130
⇒ m = 130/65 = 2
Sample Papers with Hints (Issued by CBSE)
CBSE Class 10 Mathematics Solved Practice Paper 2017-2018: Set-I
Q. If the common difference of an A.P. is 3, then find a20 – a15.
Sol.
Let the first term of the AP be a.
an = a(n − 1)d
a20 – a15 = [a + (20 – 1)d] – [a + (15 – 1)d]
= 19d – 14d
= 5d
= 5 × 3
Q. What is the perimeter of a square circumscribing a circle of radius a cm.
Sol.
The radius of the circle is given as a cm.
Diameter of the circle = 2 × a cm = 2a cm
Side of the circumscribing square = Diameter of the circle = 2a cm
∴ Perimeter of the circumscribing square = 4 × 2a cm = 8a cm
Q. For what value of k will k + 9, 2 k ‒ 1 and 2k + 7 are the consecutive terms of an A.P.?
Solution.
If three terms x, y and z are in A.P. then, 2 y = x + z
Since k + 9, 2 k ‒ 1 and 2 k + 7 are in A.P.
∴ 2(2k − 1) = (k + 9) + (2k + 7)
⟹ 4k – 2 = 3k + 16
⟹ k = 18
To get the complete set of questions, click on the following link:
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# What is a real number
1. Jul 23, 2014
### Greg Bernhardt
Definition/Summary
The real numbers are most commonly encountered number system, familiar to the layman via the number line, and as the number system lying behind decimal notation.
Because the real numbers have many nice arithmetic and geometric properties, they feature prominently in many fields of mathematics.
Equations
Extended explanation
A Brief History
The real numbers originate from a need to quantify the geometric notion of 'length'. It was known in ancient times that the rational numbers are not adequate, culminating with Eudoxus's 'method of exhaustion'.
Formal Definition
With the development of modern calculus, it became increasingly clear that a rigorous definition of the real numbers was required. Cantor provided the first definition, by identifying a real number with the set of Cauchy sequences of rational that ought to converge to that real number. Since then, many other equivalent definitions have been proveded. The following definition is the one traditionally used today, except that the Dedekind completeness axiom has been replaced with an equivalent axiom:
Language
The real numbers consist of:
• A set $\mathbb{R}$ whose elements are called real numbers (also written R)
• A distinguished real number $0$ (zero)
• A distinguished real number $1$ (one)
• A binary operation $+$ (addition)
• A binary operation $\cdot$ (multiplication)
• A unary operation $-$ (negation)
• A unary partial operation ${}^{-1}$ (reciprocal)
• A relation $\leq$ (less than or equal to)
In what follows, the symbols $a, b, c$ denote real numbers, $d$ denotes a nonzero real number (meaning $d \neq 0$).
Field Axioms
• $a+b$ is a real number
• $a \cdot b$ is a real number
• $-a$ is a real number
• $d^{-1}$ is a real number
• $a+b = b+a$ (commutativity of addition)
• $a \cdot b = b \cdot a$ (commutativity of multiplication)
• $a+(b+c) = (a+b)+c$ (associativity of addition)
• $a\cdot (b\cdot c) = (a\cdot b)\cdot c$ (associativity of multiplication)
• $a\cdot (b+c) = (a\cdot b) + (a \cdot c)$ (distributivity of multiplication over addition)
• $a + 0 = a$ (0 is the additive identity)
• $a \cdot 1 = a$ (1 is the multiplicative identity)
• $a + (-a) = 0$ (additive inverses)
• $d \cdot (d^{-1}) = 1$ (multiplicative inverses)
Ordering axioms
• If $a \leq b$ and $b \leq a$ then $a = b$
• If $a \leq b$ and $b \leq c$ then $a \leq c$
• $a \leq b$ or $b \leq a$
• If $a \leq b$ then $a + c \leq b + c$
• If $0 \leq c$ and $a \leq b$ then $a \cdot c \leq b \cdot c$
Completeness axiom
The completeness axiom is significantly more complicated. One way to state it is via the least upper bound property of the calculus of sequences. Let $\{ f_n \}$ be a sequence of real numbers
• If $f_n \leq f_{n+1} \leq a$ for every natural number $n$, then $\lim_{n \rightarrow +\infty} f_n$ exists
Other Operations and Identities
These axioms are sufficient to derive all of the familiar properties of the real numbers. Some examples:
• Subtraction is defined by $a - b = a + (-b)$
• Division is defined by $a / d = a \cdot (d^{-1})$
Common Errors
• The word "real" in "the real numbers" is often mistaken for the ordinary English word. In actuality, it is simply a name, and is used to distinguish the number system from other familiar systems like "the integers" or "the rational numbers" or "the Galois field of 49 elements".
* This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!<|endoftext|>
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This program offers an approach to teaching critical thinking skills, creating supportive cultures, and targeting underlying academic skills in middle and high schools.
1. CRITICAL THINKING – The goal is to provide students a toolkit to approach any problem (outside of mathematics). Almost all of the terms presented here can be used to analyze issues in history, literature, science, and politics, and many are equally effective in the social realm of adolescents, which can be notoriously difficult to navigate. This model has four pillars:
a. Inferences – Skill: The ability to make good guesses based on available data and to avoid relying on weak assumptions. Terms include: Strong/Weak inferences, Causal/Predictive inferences, and the logical fallacies and biases that lead us to and rely upon weak inferences: Hasty generalization, Post hoc ergo propter hoc, False Dichotomy, Slippery Slope, Confirmation Bias, Suggestion Bias, Availability Bias, Sunk Cost Fallacy, Fundamental Attribution Error, Perspective Gap, Correspondence Bias, Transparency Effect, the Spotlight Effect, and the Impostor Syndrome.
b. Evidence – Skill: The ability to analyze and evaluate sources of evidence and to generate strong evidence for arguments. Terms include Eight Types of Evidence: Personal Observation, Personal Experience, Case Examples, Research Studies, Analogies, Appeal to Authority, Testimonials, and Intuition. And the pitfalls and fallacies of each, which include fallacies like Appeal to Questionable Authority, and Appeal to Popularity, among others.
c. Values – Skill: The ability to generate and analyze arguments based on the underlying values involved, to recognize and dissect the underlying values that motivate people in various situations (historical, literary, real-world), to analyze value conflicts and value assumptions, to resolve value conflicts, and to use values as a rhetorical technique. Terms include: Value conflicts (which lends itself immediately to conflict negotiation), Values as motivating forces, Anti-values ,which we treat as regular values despite the fact that they are predictably disastrous (dominance, revenge, ideology, sadism, and exclusivity). Providing a list of values offers students a quick access to this key vocabulary. I have come to think of it as a periodic table for the humanities.
d. Rhetoric – Skill: The ability to manipulate emotion both to win arguments and to craft compelling speeches, essays, and creative work. Terms include: The list of rhetorical techniques is long. I like to cover the basics to distinguish rhetoric from reason, to point out logical fallacies rooted in emotional manipulation (ad hominem, straw man, etc.), and to simply give students these extraordinarily powerful tools to help make the world a better place (I specifically ask them not to use these powers for evil).
- Key Resources: Asking the Right Questions by Browne and Keeley is an exceptional overview of critical thinking terms. Other wonderful resources include Thinking Fast and Slow by Daniel Kahneman, Classical Rhetoric for the Modern Student by Edward Corbett, The Righteous Mind by Jonathan Haidt, Difficult Conversations by Stone, Heen, and Patton, and the The Better Angels of Our Nature by Steven Pinker. Paul Bloom’s psychology lectures available through iTunesU are also an incredibly valuable (free) resource. See also this resource list.
2. COMMUNITY BUILDING – The goal is to create a high-achieving culture of support. The following material can easily be woven into any curriculum or used in an advisory and non-curricular programming (assemblies, “community life” periods, reflection periods, etc.).
a. Managing the Social World – Students are profoundly shaped by the way they treat one another, and offering a vocabulary of social terms helps them recognize motivations, solve social problems, accept certain irritations, and help one another socially, emotionally, and academically. I work hard to drive that the term “culture of support” into the students’ vocabulary. We try to avoid a “culture of humiliation.” I use a grid with humiliation and support on one axis, and high achievement/low achievement on the other.The anti-values are important here – we are motivated by different values in each quadrant of the grid. The conflict negotiation and values materials are equally important. Much of the vocabulary of Difficult Conversations offers a means of viewing emotionally fraught conflicts. These terms include: intention vs. impact, exploring the story, and mapping the contributions. For value conflicts, this diagram helps students see how we project the opposite of our values on the person we are in conflict with (this is useful for analyzing Antigone by Sophocles). Other useful terms: Amplifier/Muffler (either for positivity or negativity, see grid), and drama vs. bullying (from Emily Bazelon’s Sticks and Stones). Role-playing and scenario-discussions help students see the value of these terms and concepts. Most conflicts can be effectively dealt with, though there are some that are truly difficult. A group of students helped boil them down to three unsolvable problems: 1. Friend Drift, 2. Accidental Exclusion, and 3. Unrequited Love. Recognizing these can help students avoid at least some of their ego-crushing consequences and may keep them from spurring revenge cycles.
b. Status and Status Currency – Students constantly discuss one another and which behaviors, products, and qualities are acceptable or taboo. Despite the fact (or because of the fact) that these conversations are at the heart of their lives, they rarely analyze status and status currency from a detached perspective. Understanding the structure of status and the ever-changing currency of what’s cool and what’s not allows students to put their interests into perspective and to analyze the motivations of others around them. The analysis and status and status currency is also a profoundly effective lens of analysis in history and literature. To make status hierarchies concrete, a visual diagram of a “status ladder” allows students to understand social classes, social mobility, and other crucial concepts of social psychology. Every status ladder puts elites at the top and outcasts at the bottom. Every community uses a different gradient between the two, and every society uses a set of qualities, products, skills, and actions that serve as a currency for moving up or down the ladder, so called “status currency.” For example, in the US we generally talk about social class in terms of five factors: 1. Wealth, 2. Education, 3. Influence, 4. Pedigree, and 5. Creative freedom. An effective introduction to these concepts is to ask students to list the status currency of their 1st grade classes and how that currency changed through the years. I have never seen students stumped by this question; they are hyper-aware of what was cool and what wasn’t. Adolescent groups often value appearance, fashion, humor, athletic ability, willingness to take risks, kindness, exclusivity, and material things. Some status currency, like kindness, is incredibly valuable, while others, like physical aggressiveness is, obviously, incredibly destructive. A useful exercise if for students to list status currency for their grade-level and list an ideal set of status markers (kindness usually tops that list). This can be valuable if students are able to analyze peer pressure in terms of status and recognize that status is fluid, temporary, and a social construction. It can give many kids hope that there are brighter days ahead.
c. The Things that Define Us, Unite Us, and Divide Us – We all embody a complex mixture of psychological mechanisms that lead us to alternatively embrace one another in the beautiful spirit of common humanity and to dehumanize one another with shocking coldness and brutality. The goal is to analyze these psychological motivations, to demonize destructive norms and anti-values, and to lionize constructive norms and values. Not to by overly hyperbolic, but this may be the greatest hope for the survival of humanity. This model focuses on five obstacles to achieving a full appreciation of our common dignity: 1. Tribalism, or Us-Vs.-Them Psychology, 2. Labels and Categories, 3. Status and Rank, 4. Value Lenses, and 5. Miscommunication and Conflict.
3. ACADEMIC SKILLS, PSYCHOLOGY, AND HABITS – The goal here is to create a culture of achievement by emphasizing the practical habits, assumptions, and practice of academic success. The material below follows the old “teach a man to fish” ethos. Teaching study habits and habit formation is time consuming, but the benefits can be long-term and life-altering.
a. Educational Psychology – This has been another major part of the program, with the ultimate goal of instilling an appreciation of the power of effort, the plasticity of ability, and the importance of good habits. I lean heavily on Carol Dweck’s growth mindset model and K. Anders Ericsson’s work on Deliberate Practice (popularized by Malcolm Gladwell in Outliers). I use a little “Mount Education” graphic to describe how a student might be ahead of the curve or fall behind. It has a lot to with opportunities, timing, and positive/negative associations with a subject. I feel that understanding the latter is crucial – why hating your piano teacher shouldn’t mean you hate the piano (but it does). I like to start every year with a the quick Dweck article linked above and readings from Brain Rules by John Medina and Bounce by Matthew Syed. Ultimately, the hope is that students will see that they’re ability level is inextricably linked to their past experience, which has given them a set of likes and dislikes, strengths and weaknesses, habits, expectations, and beliefs about themselves. I want them to see that everyone has advantages and obstacles, and the only real debilitating obstacle is the belief that you can’t get over an obstacle.
b. Intelligence – Once students are introduced to deliberate practice, we try to apply it to academics, and ultimately to intelligence, which is a tricky term. I start with IQ – what it is, how to raise it, – pulled largely from What is Intelligence? by James Flynn. A recent American Educator article provides a good summary. I like to emphasize that intelligence is large set of skills, none of which are fixed, and that there are both biological and cultural factors that play key roles. Basically, when we talk about analytic intelligence we are talking about three discrete measures: 1. Processing Speed, 2. Reasoning Ability, and 3. Content Knowledge. Flynn does a really good job of breaking these down further, but his book is heavy lifting, and I try to simplify things as much as I can while retaining the essential parts. Flynn has a great website exploring components of IQ and it’s deep methodological challenges. A really simple and entertaining book introducing the vocabulary of IQ is Everything Bad is Good for You by Steven Johnson. I also like to introduce Emotional Intelligence, but other than offering it as a meaningful term, I don’t do much with it – I just like them to have the term to help analyze some of the social scenarios we deal with.
c. Building Effective Habits – I spend a lot of time on this, and it tends to be really effective to help a handful of kids and to have shared vocabulary to advise students who are struggling. The bible of this for me is Willpower by Roy Baumeister and John Tierney. I basically just teach a summary of it, which boils down to a few key ideas: 1. There is only one tank of mental energy and we burn it whenever we do anything difficult: work, restrain our emotions, avoid temptations, etc. 2. Change only one habit at a time – otherwise you are taxing your tank too severely. 3. To-Do lists and other organizers save mental energy by allowing our brains to outsource the task of remembering to do something. 4. Don’t try to do any heavy lifting while your tank is low – before dinner, before lunch, after a heavy conversation with your mom, etc. Schedule your work time carefully. It doesn’t always make sense to do your homework right after school. 5. Changing one habit often has a multiplier effect. (There is a great discussion of “anchor habits” in The Power of Habit by Charles Duhigg, which is a quick and entertaining read, though not as substantive as Willpower). I like to do 30-day challenges with the students. I usually give up tv or video games for the month, though I have also given up desserts and gluten, which has been more challenging. We check in every week to discuss our progress and obstacles. We also talk about the impact of giving something up vs. adding something (giving up sugar vs. adding a five-mile run to your day) – generally the former is easier, and it tends to free up a lot more time, which our over-programmed students are in dire need of. I like to break down a month like this: Week 1: Novelty helps ease the pain of habit change, Week 2: Novelty has worn off. Pain and boredom set in. Week 3: This is the crucible. No one is interested in your new habit anymore, but the pain is still there, more acute than ever. Week 4: The pain fades and a “new normal” takes its place. Some students try to change negative social habits or to mend troubled relationships, which provides rich discussion.
d. How to Study/How to Memorize – I have found a high rate of return in teaching very specific strategies for memorization and studying. These strategies including Cornell notes and memorization techniques, especially those found in this excellent overview. Moonwalking with Einstein by Joshua Foer is another great source, and it has also offers a great summary of Ericsson’s Deliberate Practice research.
I plan to keep updating this overview and the site in general as new materials and ideas change my approach. If you have any questions or comments please feel free to contact me:
Note: The banner picture represents the eternal value conflict of liberty vs. equality, individuality vs. community, capitalism vs. communism, and bald eagles vs. grizzly bears.<|endoftext|>
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# Improper integral (comparison)
Normally one would increase/decrease the numerator/denominator and thus deduce whether it is convergent and divergent. This one however is not as simple as that. Here's a proposition of a solution:
We know $x^2$ dominates over $\ln x$, so the integrand can be written
$$\frac 1{x^2 - \ln x} = \frac 1{x^2\left(1 - \frac {\ln x}{x^2} \right)}$$
Since $x^2$ dominates $\ln x$ (i.e. $\frac {\ln x}{x^2} \to 0$ when $x \to \infty$), then surely $\frac {\ln x}{x^2} < \frac 12$ whenever $x$ is greater than some $x_0$. Therefore we have
$$\frac 1{x^2\left(1 - \frac {\ln x}{x^2} \right)} \leq \frac 1{x^2\left(1 - \frac 12 \right)}$$
Now we can integrate from $x_0$ to $\infty$ and show that
$$\int_{x_0}^\infty \frac {\mathrm{d}x}{x^2 - \ln x}$$
converges. But I do not understand how this integral from $x_0$ to $\infty$ implies that the same integral from $1$ to $\infty$ also must converge. Someone care to explain?
$$\forall A\quad\int_{1}^A \frac {\mathrm{d}x}{x^2 - \ln x}= \int_{1}^{x_0} \frac {\mathrm{d}x}{x^2 - \ln x}+ \int_{x_0}^A \frac {\mathrm{d}x}{x^2 - \ln x}$$ so what happens when $A\to\infty$?
• Oh, there we go! The integral far right converges, so we need to ascertain the the first integral, the one from 1 to $x_0$, does not diverge. Jan 11, 2014 at 20:52<|endoftext|>
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Flying bugs trapped deep in the frozen mud covering Greenland have pointed researchers to new clues about the country’s climate, suggesting the now-icebound island was once warmer than previously believed.
In the centuries that followed the last ice age and in the millenia between the last two, Greenland could have seen summer highs between 10 and 15 degrees warmer than today, according to a new study led by researchers at Northwestern University in Illinois.
Core samples taken from the mud of a lake bed in northwestern Greenland, just beyond the edge of the ice sheet and largely undisturbed by its historical ebb and flow, revealed large numbers of preserved insects known as phantom midges and a fly species known as chironomids. Those species today usually live well south of Greenland, but the numbers found in the sediment cores taken by the Northwestern team were comparable to populations seen in the Canadian Atlantic provinces.
As far as the team could tell, the phantom midge hasn’t been seen in Greenland before now.
“We think this is the first time anyone has reported it in ancient sediments or modern lakes there," Yarrow Axford, the study's senior author, said in a statement accompanying the findings. "We were really surprised to see how far north it migrated."
The findings suggest temperatures in Greenland’s summer might have ranged into the 50s Fahrenheit, or in the low teens Celsius — well above today’s averages of around 40°F.<|endoftext|>
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Education is an essential milestone for a more peaceful and equitable world. And yet, millions of children and youth do not have access to essential education services. The United Nations and also hundreds of other humanitarian organizations are working to promote, protect and foster education as a human right.On January 24th, 2019, for the first time, the world celebrates the International Day of Education to reaffirm this intention.Why is International Day of Education important?The quality of education and access to schooling has come a long way in recent decades. But today, 262 million children and youth still do not attend school. Meanwhile, 617 million children and adolescents are unable to do basic math or to read. Fortunately, many organizations help to improve education every day.There is also inequality in access to education. For example, 4 million children of refugees are not attending school. Some children must travel for miles to access a school. Also, a significant gender gap persists. Less than 40% of girls in sub-Saharan Africa have the opportunity to finish secondary school.The first-ever International Day of Education is one of the many steps to bridge these gaps and provide universal education around the world. It is an opportunity to critically reflect on the potential of children if given the tools to succeed. It is also a day when humanitarians and supporters can come together and promote quality international education.Education and Sustainable Development Goals Education is also a basic human right. The Universal Declaration of Human Rights calls for free and compulsory elementary education for all children. Also, countries are urged to make higher education accessible for all by the Convention on the Rights of the Child.But education is even more. It is a tool for peace-building and equality. As such, education comprises one of the Sustainable Development Goals (SDGs) set out by the UN. SDG number 4 aims to “ensure inclusive and equitable quality education and promote lifelong learning opportunities for all.” It addresses ten specific targets or milestones towards the overarching goal.NGOs Promoting Education around the World Hundreds of humanitarian organizations around the world are working to promote, provide and improve education. Compassionate individuals travel to the most remote locations to ensure access to education services. These are just a few examples of amazing NGOs promoting education.1. Khan AcademyThe Kahn Academy aims to provide free and accessible education to anyone, anywhere. The organization has set up a comprehensive education platform, featuring all basic courses online for free. There are resources for both learners and teachers.2. One LaptopThe organization’s mission is to provide a laptop to every child. One Laptop partners with various donors to this end. Their laptops come with hardware and software designed for children’s learning. The main aim is to reduce the gap in access to technology and education.3. Citizen SchoolsTo help students succeed, Citizen Schools takes on a real-world approach. They set up networks of mentors to support children in developing their future. The organization also organized hands-on learning opportunities and skills training.4. Teach for AllOriginating in the United States, Teach for All now works in more than 40 countries around the world. They promote and develop new methodologies for classrooms, and train local leaders to transform education in their countries. Also, Teach for All maintains an extensive network of alumni, who continue to impact education around the world.5. Child Empowerment InternationalChildren living in conflict zones often miss out on learning opportunities. Child Empowerment International establishes day schools for refugees, displaced persons and other communities where children lack access to schooling. They also work with children suffering from psychological traumas associated with facing violence and conflict.6. World EducationOperating in over 20 countries, World Education also strives to improve people’s quality of life through education. The organization works with both adults and children. Their courses focus on improving livelihoods, basic skills, as well as specialized topics relevant to each context they operate in.7. Smile FoundationThe Foundation provides basic education and healthcare services to children across India. Their Mission Education projects include education services for all ages. Their particular focus is on educating girls and women, to bridge the gender gap and empower their families.We Help You to Travel Where it is Needed MostWe believe our world is a better place when compassion can travel where it is needed most. As a global humanitarian travel organization, we devote ourselves to serving those who serve the world. You can reach our experienced staff anytime. Please use our quick address locator to contact your nearest Raptim office should you have any questions.<|endoftext|>
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# Matrix Scalar Multiplication Calculator
The calculator will multiply the given matrix by the given scalar, with steps shown. It handles matrices of any size up to 10x10 (2x2, 3x3, 4x4, etc.).
$\times$
If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please write it in the comments below.
Calculate $3 \left[\begin{array}{cc}4 & 5\\-2 & 0\\1 & 7\end{array}\right]$.
## Solution
Multiply each entry of the matrix by the scalar:
$\color{Green}{\left(3\right)}\cdot \left[\begin{array}{cc}4 & 5\\-2 & 0\\1 & 7\end{array}\right] = \left[\begin{array}{cc}\color{Green}{\left(3\right)}\cdot \left(4\right) & \color{Green}{\left(3\right)}\cdot \left(5\right)\\\color{Green}{\left(3\right)}\cdot \left(-2\right) & \color{Green}{\left(3\right)}\cdot \left(0\right)\\\color{Green}{\left(3\right)}\cdot \left(1\right) & \color{Green}{\left(3\right)}\cdot \left(7\right)\end{array}\right] = \left[\begin{array}{cc}12 & 15\\-6 & 0\\3 & 21\end{array}\right]$
$3 \left[\begin{array}{cc}4 & 5\\-2 & 0\\1 & 7\end{array}\right] = \left[\begin{array}{cc}12 & 15\\-6 & 0\\3 & 21\end{array}\right]$A<|endoftext|>
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The Loess Region is the cradle of China’s civilization. It is also a region particularly vulnerable to earthquake damage, because loess soil – rich windblown silt that settled over the millennia to depths of up to 300 meters – disintegrates easily when subjected to seismic activity. Between that vulnerability and China’s high population density, many of the world’s most devastating earthquakes have occurred in China. One such struck in 1556, when Ming Dynasty China was rocked by the deadliest earthquake in human history. Epi-centered in the Wei river basin in modern Shaanxi, the upheaval is estimated to have registered 8 on the Richter scale.
The earth split open with fissures up to 70 feet deep, as the ground suddenly rose up in some place to form new hills, while in other places hills crumbled and subsided into valleys. 97 counties in Shaanxi and surrounding provinces were devastated, as the earthquake destroyed nearly everything within an area more than 500 miles wide, and damage was inflicted as far as 310 miles from the epicenter. Fatalities numbered in the hundreds of thousands, and the injured numbered in the millions.
China’s Shaanxi Province Was and Remains Particularly Vulnerable to Earthquakes
China’s Loess Region is a plateau covering about 250,000 square miles in the upper and middle Yellow River – so named because the loess forming its banks gave it a yellowish tint. The good thing about the region is that the soil that gave it its name and covers the ground in layers up to 300 meters deep, is great for farming. The pale yellow or buff loess sediment is formed by the accumulation of windblown silt comprised of crystals of mica, quartz, feldspar, and other minerals. It is porous, with small empty spaces in the soil that allow for excellent air circulation. It is also friable – that is, easy to break into smaller pieces – which makes it relatively easy to plough, requiring less time and effort from farmers working loess fields.
Such factors tend to make for very rich agriculture, and with the right climatic conditions – water, sun, and the right temperature – loess soil is some of the most fertile and productive terrain in the world. Much of China’s Loess Region has the right climatic conditions, and its rich and readily tilled earth acted like a magnet that attracted farmers since the dawn of the agricultural revolution. Eventually, as population density increased, the region became the cradle of China’s civilization.
However, one of the factors that makes loess attractive to farmers and conducive to high population densities – the ease with which it breaks and crumbles – is also a vulnerability. Loose soil might be great for farming, but buildings erected atop it are particularly prone to tumbling down if the ground shakes. Unfortunately, the ground in the Loess Region has a tendency to shake.
According to plate tectonic theory, the earth’s crust consists of several plates floating on top of molten magma. As the tectonic plates slide over that magma, moving away from or colliding with each other, they produce earthquakes and volcanic activity. Most of China sits atop a major tectonic plate: the Amurian Plate, which is part of the Eurasian Plate. However, in an arc stretching around China and gripping it in a pincer, are the North American Plate, the Philippine Plate, and the Indian Plate, whose collision with China formed the Himalayas (see map above). As those plates converge upon China, they pinch and squeeze it from multiple sides, causing its crust to warp hundreds of miles away from the plate boundaries (see map below). That tectonic activity is why the world’s highest plateau is located in China, and why the world’s highest mountain, Mount Everest, is on China’s doorstep.
The Loess Region, and Shaanxi province in particular, are susceptible to warping from those tectonic plate pressures. Lying smack dab in the middle of historic China, Shaanxi and the surrounding region are constantly squeezed by the tectonic plates converging on the country, creating faults that lend themselves to energetic seismic activity. In other words, when the pressure builds up from tectonic plates pressing in on the one atop which China rests, it sometimes gets released hundreds, or even thousands, of miles away in Shaanxi, in the form of major earthquakes.<|endoftext|>
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You can use Excel to calculate the percentage of change between two values quickly. In our simple example, we show you what you need to know to be able to calculate things like how much the price of gas changed overnight or the percentage of a stock price’s rise or fall.
How Percentage of Change Works
The percentage of change between an original and a new value is defined as the difference between the original value and the new value, divided by the original value.
(new_value - original_value)/(original_value)
For example, if the price of a gallon of gasoline was $2.999 yesterday on your drive home and it rose to $3.199 this morning when you filled your tank, you could calculate the percentage of change by plugging those values into the formula.
($3.199 - $2.999)/($2.999) = 0.067 = 6.7%
Let’s Look at an Example
For our simple example, we will look at a list of hypothetical prices and determine the percentage of change between an original price and a new price.
Here is our sample data containing three columns: “Original Price,” “New Price,” and “Percentage of Change.” We have formatted the first two columns as dollar amounts.
Start by selecting the first cell in the “Percentage of Change” column.
Type the following formula and then press Enter:
The result will appear in the cell. It is not formatted as a percentage, yet. To do that, first select the cell containing the value.
In the “Home” menu, navigate to the “Numbers” menu. We will use two buttons—one to format the cell value as a percentage and another to reduce the number of decimal places so that the cell only shows the tenths place. First, press the “%” button. Next, press the “.00->.0” button. You can use either of the buttons on the right side of the menu to increase or decrease the displayed precision of the value.
The value is now formatted as a percentage with only one decimal place displayed.
Now we can calculate the percentage of change for the remaining values.
Select all of the cells of the “Percentage of Change” column and then press Ctrl+D. The Ctrl+D shortcut fills data down or to the right through all selected cells.
Now we are done, all of the percentages of change between the original prices and the new prices have been calculated. Notice that when the “New Price” value is lower than the “Original Price” value, the result is negative.<|endoftext|>
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# How to write a rational number as a repeating decimal
Independently combined probability models. It is not almost 1, but rather, it is 1. This kind of number is also called a non-terminating, non-repeating decimal. It does go into For example, if a stack of books is known to have 8 books and 3 more books are added to the top, it is not necessary to count the stack all over again.
So is zero a number. We saw up here. Your argument is non-sequitur. We said 27 goes into 10 0 times.
Again, turn into easier problem: This problem seems easy, but you have to think about what the problem is asking. Suppose that you have a repeating decimal, and it looks like. We can find out how much of ingredients a and b are in solution X by using a ratio multiplier again one ounce of solution X contains ingredients a and b in a ratio of 2: And you're right, there is a trick to it.
Repeating decimals, those which have a recurring decimal number are rational numbers. The median of a list of values is the value appearing at the center of a sorted version of the list—or the mean of the two central values, if the list contains an even number of values.
There are 20 boys and 8 girls 28 — 20 in the new class. The terminating decimal ends up with a remainder of zero and the divisor is 8. Probably the most common is to set up a proportion like we did here earlier.
There are the whole numbers that go from 1 to infinity. All whole numbers are fractions. You probably — You might have been able to do that in your head.
Then you subtract 2 — 1, and write the difference, 1, on the bottom. Give me any Number you want. The numerator represents a number of equal parts, and the denominator indicates how many of those parts make up a unit or a whole. To represent any pattern of repeating decimals, divide the section of the pattern to be repeated by 9's, in the following way: We can do the same for solution Y, which contains ingredients a and b in a ratio of 1: Repeating decimals to fractions Hello, Oscar As you may already know, every fraction technically, a fraction is called a rational number either terminates - ends with a string of 0's - or is a repeating decimal of course, you could think of the string of zeros as just a special kind of repeating decimal.
Rational Numbers: Convert Fractions to Decimals and Decimals to Fractions Review. CCSS Standard. Prerequisite for michaelferrisjr.com Rational Number – is a number that can be written as a ratio in the form write the decimal with a bar over the repeating digit(s).
Write each fraction as a decimal. This means that after a certain number of decimal places (let's call that k), the decimal begins repeating every h digits, where h is some integer.
For example, if the number. Convert a rational number to a decimal - Independent Practice Worksheet Write as a fraction in the simplest form.
2. Convert a rational number to a decimal 4) 3. Solve the decimal equation. 3 + + 4. Use the grid to fill in the missing number. Rational numbers include natural numbers, whole numbers, and integers. They can all be written as michaelferrisjr.comn is natural, whole, and an integer. Since it can also be written as the ratio or the fraction 16/1, it is also a rational number.
To place the repeating digit (42) to the left of the decimal point, you need to move the decimal point 3 place to the right Again, moving a decimal point three place.
Write each as a decimal. Use repeating decimals when necessary. 1) 1 4 2) 2 3 5 3) 5 8 4) 3 5 5) 7 6) 8 33 7) 6 11 8) 7 50 9) 4 27 10) 7
How to write a rational number as a repeating decimal
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Converting an Infinite Decimal Expansion to a Rational Number | The Infinite Series Module<|endoftext|>
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PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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# Solution for Show that the Cone of the Greatest Volume Which Can Be Inscribed in a Given Spher Has an Altitude Equal to 2 3 of the Diameter of the Sphere. - PUC Karnataka Science Class 12 - Mathematics
#### Question
Show that the cone of the greatest volume which can be inscribed in a given spher has an altitude equal to $\frac{2}{3}$ of the diameter of the sphere.
#### Solution
$\text { Let h, r and R be the height, radius of base of the cone and radius of the sphere, respectively . Then },$
$h = R + \sqrt{R^2 - r^2}$
$\Rightarrow \left( h - R \right)^2 = R^2 - r^2$
$\Rightarrow h^2 + R^2 - 2hr = R^2 - r^2$
$\Rightarrow r^2 = 2hR - h^2 . . . \left( 1 \right)$
$\text { Volume of cone } = \frac{1}{3}\pi r^2 h$
$\Rightarrow V = \frac{1}{3}\pi h\left( 2hR - h^2 \right) \left[ \text { From eq }. \left( 1 \right) \right]$
$\Rightarrow V = \frac{1}{3}\pi\left( 2 h^2 R - h^3 \right)$
$\Rightarrow \frac{dV}{dh} = \frac{\pi}{3}\left( 4hR - 3 h^2 \right)$
$\text { For maximum or minimum values of V, we must have }$
$\frac{dV}{dh} = 0$
$\Rightarrow \frac{\pi}{3}\left( 4hR - 3 h^2 \right) = 0$
$\Rightarrow 4hR = 3 h^2$
$\Rightarrow h = \frac{4R}{3}$
$\text { Substituting the value of y in eq } . \left( 1 \right), \text { we get }$
$x^2 = 4\left( r^2 - \left( \frac{r}{\sqrt{2}} \right)^2 \right)$
$\Rightarrow x^2 = 4\left( r^2 - \frac{r^2}{2} \right)$
$\Rightarrow x^2 = 4\left( \frac{r^2}{2} \right)$
$\Rightarrow x^2 = 2 r^2$
$\Rightarrow x = r\sqrt{2}$
$\text { Now,}$
$\frac{d^2 V}{d h^2} = \frac{\pi}{3}\left( 4R - 6h \right)$
$\Rightarrow \frac{d^2 V}{d h^2} = \frac{\pi}{3}\left( 4R - 6 \times \frac{4R}{3} \right)$
$\Rightarrow \frac{d^2 V}{d h^2} = \frac{- 4\pi R}{3} < 0$
$\text { So, the volume is maximum when h } = \frac{4R}{3} .$
$\Rightarrow h = \frac{2}{3}\left( \text { Diameter of sphere } \right)$
$\text { Hence proved }.$
Is there an error in this question or solution?
#### Video TutorialsVIEW ALL [1]
Solution Show that the Cone of the Greatest Volume Which Can Be Inscribed in a Given Spher Has an Altitude Equal to 2 3 of the Diameter of the Sphere. Concept: Graph of Maxima and Minima.
S<|endoftext|>
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Peers are people that a child, adolescent or adult identifies with. A peer can influence, persuade and coerce you to do certain things or act a certain way in order to be accepted. Even though often perceived as negative, peer pressure can also affect a person in a positive manner. Your personal health can improve or decline depending on how you handle peer pressure.
Pressure to Smoke
Many teens have to deal with peer pressure to smoke. Their friends smoke and if they don't join in the unhealthy behaviour, they risk not being accepted. According to the Teen Help website, 440,000 people in the United States die yearly from smoke-related diseases. Ninety per cent of these people started smoking in their teens. Regardless of your age, smoking can cause lung cancer and lung disease and it increases your risk for coronary heart disease and stroke.
Pressure to Drink Alcohol
The pressure to drink alcohol can affect people of all ages. Teens may be pressured into drinking by their alcohol-drinking peers and adults may feel pressure to drink during after-work social gatherings. According to the National Institute on Alcohol Abuse and Alcoholism, consuming alcohol at a young age may affect brain and organ development and drinking can cause liver damage and liver failure in people of all ages.
Pressure to Have Sex
The peer pressure to have sex at an early age can increase a teen's stress level. According to Psychology Today, one in three boys and 23 per cent of girls between the ages of 15 to 17 deal with peer pressure to have sex. If a teenager has low self-esteem he may be more prone to have sex at an early age. Having sex too early can trigger emotional problems and physically teens can contract sexually transmitted diseases and get pregnant.
Pressure to Improve Health
The National Heart Lung and Blood Institute states that being overweight or obese increases your risk for heart disease, diabetes, stroke and hypertension. If your diet consists mostly of unhealthy, fatty foods and you are not getting any exercise, peer pressure can actually help improve your health. If you have one or more friends that eat a healthy diet and get regular exercise, they may influence you to do the same. By following their healthy behaviour, you can lose weight. Once you switch to a healthy diet and start exercising, your peers will cheer you on and motivate you to keep going.
- Teen Help: Teen Smoking Statistics
- Centers for Disease Control and Prevention: Health Effects of Cigarette Smoking
- National Institute on Alcohol Abuse and Alcoholism: Underage Drinking
- Psychology Today; Peer Pressure and Teen Sex; Colin Allen; May 2003
- Surf Net Parents: Possible Effects of Peer Pressure on Your Teenager
- Fit Day: Using Peer Pressure to Start Eating Healthy
- BananaStock/BananaStock/Getty Images<|endoftext|>
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What is 3d printing technology?
Just imagine a box – how do you see it in your mind?
Is it just a square?
Does it have depth?
You can draw this in a CAD package in two dimensions, represented by a square.
It can also be drawn in CAD in 3d, in isometric form, you can see the object has depth.
When you print this on a standard printer you get a flat sheet of paper with an image on it. You still have to use your imagination to ‘see’ it in 3d.
When you move over to 3d printing technology, after finishing the print you are able to pick up this 3d object that has depth.
You can rotate it see all of the sides. Feel the depth of the object and put it onto a surface and see the shadow it casts.
The progress from a paper print to a 3d print is an amazing achievement in engineering, the concept is simple but actually achieving the end result reliably is an amazing engineering achievement.
The actual process of printing in 3d is to take the object you want to print in cad and slice it into layers. The thickness of the layers is dependent on the type of printer or the extruder of the printer. Once sliced the information is sent to the printer.
Within the printer is a controller similar to the inkjet printer you are used to, where the print head moves sideways across the page. Except that the print surface can move back and forwards.
This left, right, back and forwards movement allows the printer access to the whole of the build platform.
For every slice layer the print head moves a little bit further away from the print surface, thereby building up the model in layers.
It is a fascinating process to watch and can consume hours of your life. Or am I just sad?
What can you achieve with 3d printing technology?
There used to be only a couple of choices of materials, PLA and ABS.
PLA is made from corn starch and is biodegradable, ABS is the type of plastic used for your bumper.
The industry has now woken up to the fact that it is becoming more mainstream and the public want more materials and choice of materials.
So now you can print a printed circuit board. The whole lot. The base material and the conductive tracks.
You can then design the box it is going into and print a rugged box out of a flexible material. In fact the printed circuit board can become the inside of the box.
If you need a window to see a display then there is transparent material.
Unfortunately at the moment the home printing technology is lagging behind as most are fitted with only one nozzle. Meaning that you need to change the material yourself as opposed to placing all of the different materials on the printer and pressing print.
So with home based printers you can start to print fantastic items. There is a little way to go before you press a button and a complete multicoloured, multi-material print that is available within minutes.
What can you print with a 3d printer?
A few years ago in 2013, 3d printing was thrust into the limelight when Cody Wilson, designed and printed a working 3d printed gun. The files for this were distributed on the internet and a short time later the U.S. Government took down the site, but not before over 100,000 people downloaded the plans.
Medical research into printed body parts is increasing, allowing body cells to be printed into ears and noses. This is in the early stages but progress is being made.
Imagine walking into a restaurant and looking at a menu where you can design your own food. This order is then taken and a few minutes later your order is served exactly like you designed it!!
Airlines are starting to look at this technology for storing food onboard the planes until the time comes to print and cook your meals.
The latest jewellery fashion can be downloaded and printed, if you are going to a themed party imagine being able to print off the items you want for the theme.
At the RIO olympics:
There were 3d printed shoes presented to Michael Phelps.
the French cycling team used 3d printed handlebars.
and one area where 3d printing will excel in the future – prosthetics. German paralympic cyclist Denise Schindler had a prosthetic leg printed.
At home you are only restricted by the amount of time you are willing to devote to learning CAD and sourcing materials.
There are a few home extruders available for the home experimenter to try combining different materials.
At present there is a wood effect PLA along with a stone effect, two different textures can be achieved with the stone by using different temperatures. But new products are being developed all the time so by the end of the year there will be a lot more.
What is stopping the take up at present?
The lack of simple inexpensive consumer printers, with lack of suitable materials is a big stumbling block to mass takeup of these types of printer.
Over the last five years there have been fantastic developments in the 3d printing market allowing the public to buy and print 3d models.
If you want to go further than this you will need to learn how to design and process the models. This can be time consuming, or you can ask one of the design houses like Shapeways or Scutpteo to model what you want and then download and print the item.
The speed of the printers will need to increase, print times of up to 24 hours are not unknown at present. The limitations are how fast the nozzle can move around the build platform.
The size of available build platform needs to be addressed. At present a lot of the printers have a small volume build area of less than 200mm X 200mm X 200mm. A lot can be printed within this area but when you are looking at printing a new large flower pot you are restricted to the build volume without making the part in pieces.
The physical size of the printers, the build volume on some printers is doubled by the movement required. Along with the required height of over 350mm on a bare printer with the enclosed printers taller than that.
How the industry will address these problems is yet to be seen, but small compact printers are being produced along with large machines, if you have the space and deep pockets.
3d printing is a technology which is starting to affect a lot more of our lives, the benefits are endless. It is a technology that the smart consumer would be ready to use as the product mature. It is an emerging market.
Hopefully this has given you an insight into what is 3d printing technology and what it can achieve, if you are using a 3d printer please tell us what you are using it for and what developments you would like to see in the future.
Thanks for reading<|endoftext|>
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### MIDDLE GROUND - Number of Ways to Make An Ordered List Or A Group
Basic Probability
Vocabulary
Facts
With A Deck of Cards - Counting the Number of Ways An Event Can Occur
Number of Ways to Create An Ordered List or a Group
These are counting problems. They are used here as an intro to important ideas in probability. They should be considered before reading the page Binomial Distribution.
1. A club has 4 members and wishes to elect a president, vice president, secretary, and treasurer. In how many ways can this be done?
There are 4 ways to select a president, but then only 3 ways to choose the vp, but then only 2 ways to choose the secretary, and then only 1 way to choose the treasurer,
so, there are 4(3)(2)(1) or 24 ways to do this.
Using members A, B, C, and D, here is a tree diagram and the sample space with 24 ways.
2. A club has n members and wishes each member to hold an elected office, beginning with the president, vice president, secretary, and treasurer. In how many ways can this be done?
The answer is n factorial ways. There are n ways to fill the first office, n-1 ways to fill the 2nd office, n-2 ways to fill the 3rd office, and so on. There are n! or n(n-1)(n-2)(n-3) ... (3)(2)(1) ways.
3. A club has 4 members and wishes to elect a president and then a vice president. In how many ways can this be done?
There are (4)(3) or 12 ways to do this.
Here order counts. A list is required. For a source set of n elements, if order is important and x elements are chosen for the ordered list, a permutation of n things taken x at a time is required, Pn,x. There are n! / (n -x)! ways to do this.
n! Pn,x = (n-x)!
n! (number of ordered lists given n things) Pn,x = = (n-x)! (shorten the list so only x things are used)
In this problem, an ordered list of 4 elements, take 2 elements is required, P4,2. There are 4! / (4-2)! = 24/2 = 12 ways to do this.
4! 4(3)(2)(1) P4,2 = = = 12 (4-2)! (2)(1)
Count the Ways
On page Deck of Cards 1 - Counting the Number of Ways An Event Can Occur these problems were completed by simply counting the number of ways. On this page we will use the permutation formula when appropriate.
1. In how many ways can you draw 3 cards without replacement from the face cards?
The answer given was: 12x11x10=1320 ways. Using the formula this is:
12! (number of ordered lists given n things) P12,3 = = (12-3)! (shorten the list so only 3 things are used and 9 are unused)
2. In how many ways can you draw 3 cards with replacement from the face cards?
The answer given was: 12x12x12=1728. You can not use a permutation to do this problem because cards are replaced.
3. In how many ways can you draw 5 cards without replacement from the deck of red cards?
The answer given was: 26x25x24x23x22=7893600 ways. Using the formula this is:
26! (number of ordered lists given n things) P26,5 = = (26-5)! (shorten the list so only 5 things are used and 21 are unused)
4. In how many ways can you draw 2 cards from the full deck without replacement?
The answer given was: 52x51=2652 ways. Using the formula this is:
52! (number of ordered lists given n things) P52,2 = = (52-2)! (shorten the list so only 2 things are used and 50 are unused)
5. In how many ways can you draw all 52 cards from the deck without replacement?
The answer given was: 52! 52 factorial ways, 8.0658x10^67 ways. You don't need to use the formula at all, but, you'd get the same answer.
52! (number of ordered lists given n things) P52,52 = = (52-52)! (the list is not to be shortened. All 52 cards are to be used. )
Note: 0! is 1. So 52! / 0! is 52! / 1 or 52!<|endoftext|>
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Interred within a fenced enclosure are 16 people killed when one of the largest earthquakes ever to hit California rocked Lone Pine at 2:35 am on March 26, 1872. It awakened residents hundreds of miles away in all directions. Damage was so widespread, adobe buildings in Red Bluff collapsed, 400 miles north of the quake. John Muir who was living in Yosemite at the time was awakened and took the opportunity to study the changes in the land and rock formations within the valley immediately afterwards. It was also reported at the time that people in Sacramento, 300 miles away, felt the impact of the earthquake and ran in panic to the streets. Residents of San Diego also reported an earth shaking experience.
Although there were no official recording devices at the time, the quake was estimated to be 7.6 to 8.0 magnitude on the Richter Scale. The quake was thought to be similar in size to the San Francisco quake of 1906. Two faults moved simultaneously. The vertical fault moved roughly 15-20 feet, while the right lateral fault moved roughly 35-40 feet. The twin faults run along the base of the Sierra Nevada and Inyo Mountains, according to USGS and Wikipedia.
Most buildings in Lone Pine were made from adobe brick and crumbled to the ground. Twenty-seven of the approximately 250-300 residents of Lone Pine were killed, according to several different sources. The adobe buildings located on Camp Independence also fell and the Camp was subsequently closed.
The historical marker is located 200 feet west of Highway 395 (P.M. 58.7) about a mile north of the current city of Lone Pine.
Inyo means “dwelling place of great spirit” in Paiute Native American language. Inyo County has many “greats.” Mount Whitney, the highest peak in the continental United States and Death Valley, the lowest spot in the Western Hemisphere, are both within Inyo’s boundaries. Great earthquakes have left their mark in recent history, changing the course of the Owens River and exposing ancient sedimentary rock.
About this Establishment
California Historical Landmarks Program
Historical Landmarks are sites, buildings, features, or events that are of statewide significance and have anthropological, cultural, military, political, architectural, economic, scientific or technical, religious, experimental, or other value. Historical Landmarks are eligible for registration if they meet at least one of the following criteria:
1) Is the first, last, only, or most significant of its type in the state or within a large geographic region
2) Is associated with an individual or group having a profound influence on the history of California
3) Is a prototype of, or an outstanding example of, a period, style, architectural movement or construction or is one of the more notable works or the best surviving work in a region of a pioneer architect, designer or master builder.
California’s Landmark Program began in the late 1800s with the formation of the Landmarks Club and the California Historical Landmarks League. In 1931, the program became official when legislation charged the Department of Natural Resources—and later the California State Chamber of Commerce—with registering and marking buildings of historical interest or landmarks. The Chamber of Commerce then created a committee of prestigious historians, including DeWitt Hutchings and Lawrence Hill, to evaluate potential landmark sites.
In 1948, Governor Earl Warren created the California Historical Landmarks Advisory Committee to increase the integrity and credibility of the program. Finally, this committee was changed to the California Historical Resources Commission in 1974. Information about registered landmarks numbered 770 onward is kept in the California Register of Historical Resources authoritative guide. Landmarks numbered 669 and below were registered prior to establishing specific standards, and may be added to the California Register when criteria for evaluating the properties are adopted.
Share your experience. Please leave a comment below if you've visited this historic landmark.
Time Period Represented<|endoftext|>
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A while back in Nature a paper was published on the most distant confirmed galaxy discovered so far. The paper is behind a paywall, but you can see the arxiv version here. The galaxy, known as z8 GND 5296 has a measured redshift of 7.51. You can see the galaxy in the image above.
So just how far away is this galaxy? It depends on which distance you are talking about.
When determining the distance of far galaxies like this one, astronomers typically give the value purely in terms of its redshift, often known as z. To calculate the z redshift of an object, you look for an emission or absorption line you can identify, such as those of hydrogen. You then compare the observed wavelength of the line from the object with the standard (not redshifted) line. The difference between the observed and standard wavelengths divided by the standard gives you a number known as z.
If there is no redshift, then there is no difference between observed and standard lines, hence the z is zero. Redshift is thus given by a positive number, where the bigger the number the bigger the z. Technically there is no limit to the value z can have, but the highest we have observed is about z = 12. Bigger z also means greater distance. Because of the expansion of the universe, the light of a distant galaxy is redshifted more than the light of a closer galaxy. So the galaxy with the greatest redshift is the most distant.
The reason astronomers usually talk about redshift instead of distance is that the measured z is purely an observational result. Yes, we know that bigger z means greater distance, but the exact distance depends on the model you use for the universe. We know this model is relatively accurate for determining distances, but with z you don’t have to assume any model.
Since this particular galaxy has a z of 7.51, just how far away is it? The first thing you need to do is transform the redshift to the travel time of the light since it left the galaxy. Once the light left the galaxy, cosmic expansion meant the light redshifted while it travelled. Using the standard model of cosmology we can determine the light left z8 GND 5296 about 13.1 billion years ago.
You might think calculating the distance from that age is simple. After all the speed of light is a constant, and if it travelled 13.1 billion years it must be 13.1 billion light years away. But the universe has been expanding throughout its history, so that answer doesn’t work. We can’t even say that the galaxy was 13.1 billion light years away when the light left because the universe expanded while the light travelled. So the galaxy was actually closer than that when the light left.
The distance the galaxy was from us when the light began its journey can be calculated by what is known as the angular diameter distance. For this galaxy that turns out to be about 3.4 billion light years. That means the light from z8 GND 5296 began its journey 3.4 billion light years away, but due to the expansion of the universe took 13.1 billion years to reach us.
To calculate the distance of the galaxy now, you need to start with the fact that it was 3.4 billion light years away from us 13.1 billion years ago and calculate how much the universe has expanded since then. This is known as the comoving distance. This comes out to be about 29.3 billion light years.
So the light from z8 GND 5296 has a redshift of z = 7.51. That means the light left the galaxy 13.1 billion years ago when the galaxy was 3.4 billion light years away. It is now 29.3 billion light years away. That can be a bit hard to wrap your head around.
Which is yet another reason why astronomers focus on z.<|endoftext|>
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All plants need electrons to aid biological and chemical tasks. Cornell scientists have discovered a new high-definition system that allows electrons to travel through soil farther and more efficiently than previously thought.
“Microorganisms need electrons for everything they do. If they consume nutrients or spew out methane or expel carbon dioxide – for any living, biological process – they need electrons,” said Tianran Sun, postdoctoral researcher in soil and crop sciences and lead author of the paper that appears March 31 in Nature Communications.
Like large volumes of electricity that flow from Niagara Falls throughout upstate New York, electrons convey through soil via carbon. “We weren’t aware of this high-definition soil distribution system transporting electrons from far away. It’s not kilometers, it’s not meters, but centimeter distances that matter in soil,” said Johannes Lehmann, professor of soil science.
In fact, amending the soil with pyrogenic carbon – known as biochar – brings high definition to the electron network. In turn, the electrons spur conductive networks and growth, said Sun.
“Previously we thought there were only low-performing electron pathways in the soil – and now we’ve learned the electrons are channeled through soil very efficiently in a high-performing way,” said Lehmann.
Lehmann and the members of his laboratory had struggled to understand why microorganisms thrived in the presence of biochar. The group removed soil phosphorus, making the environment inhospitable. They ruled out water and nutrients. They discarded the use of biochar as a food source because microorganisms cannot consume much of it. Through Sun’s background in environmental chemistry, the scientists found that microorganisms may be drawn to electrons that the biochar can transport.
“These results will lead to a better understanding of microbial responses in soil and microbial metabolism, including long-term effects on greenhouse gas emissions,” Sun said.
In addition to Lehmann and Sun, who published “Rapid Electron Transfer by the Carbon Matrix in Natural Pyrogenic Carbon,” the paper’s other authors are Barnaby Levin, doctoral student in applied and engineering physics; doctoral student Juan Guzman, biological and environmental engineering; Akio Enders, technician, soil and crop sciences; David Muller, professor of applied and engineering physics; and Lars Argenent, professor, University of Tübingen, Germany.
Lehmann credits cross-disciplinary work with finding this idea. “I could not have completed this work without Tianran Sun’s chemistry expertise, nor without Lars Angenent’s microbiology expertise, or David Muller’s or Barnaby Levin’s physical knowledge of carbon structure,” said Lehmann. “They played a big part.”
The National Science Foundation and the U.S. Department of Agriculture funded this research.<|endoftext|>
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Martial law is the imposition of the highest-ranking military officer as the military governor or as the head of the government, thus removing all power from the previous executive, legislative, and judicial branches of government. It is usually imposed temporarily when the government or civilian authorities fail to function effectively (e.g., maintain order and security, or provide essential services).
Martial law can be used by governments to enforce their rule over the public. Such incidents may occur after a coup d'état (such as Thailand in 2006 and 2014); when threatened by popular protest (China, Tiananmen Square protests of 1989); to suppress political opposition (Poland in 1981); or to stabilize insurrections or perceived insurrections (Canada, The October Crisis of 1970). Martial law may be declared in cases of major natural disasters; however, most countries use a different legal construct, such as a state of emergency.
Martial law has also been imposed during conflicts and in cases of occupations, where the absence of any other civil government provides for an unstable population. Examples of this form of military rule include post World War II reconstruction in Germany and Japan as well as the southern reconstruction following the U.S. Civil War.
Typically, the imposition of martial law accompanies curfews, the suspension of civil law, civil rights, habeas corpus, and the application or extension of military law or military justice to civilians. Civilians defying martial law may be subjected to military tribunal (court-martial).
- 1 By country
- 1.1 Canada
- 1.2 China
- 1.3 Egypt
- 1.4 Iran
- 1.5 Ireland
- 1.6 Israel
- 1.7 Mauritius
- 1.8 Pakistan
- 1.9 Philippines
- 1.10 Poland
- 1.11 South Korea
- 1.12 Switzerland
- 1.13 Taiwan
- 1.14 Thailand
- 1.15 Turkey
- 1.16 SFR Yugoslavia
- 1.17 United States
- 1.17.1 History
- 220.127.116.11 The American Revolution
- 18.104.22.168 New Orleans, Louisiana in the War of 1812
- 22.214.171.124 Ex parte Milligan
- 126.96.36.199 The Great Chicago Fire
- 188.8.131.52 Coeur d'Alene, Idaho, 1892
- 184.108.40.206 San Francisco earthquake of 1906
- 220.127.116.11 Colorado Coalfield War
- 18.104.22.168 West Virginia Coal Wars
- 22.214.171.124 Tulsa Race Riot
- 126.96.36.199 San Francisco, California, 1934
- 188.8.131.52 The Territory of Hawaii
- 184.108.40.206 Freedom Riders
- 1.17.1 History
- 2 See also
- 3 References
- 4 Further reading
- 5 External links
The War Measures Act was a Canadian statute that allowed the government to assume sweeping emergency powers, stopping short of martial law, i.e. the military does not administer justice, which remains in the hands of the courts. The Act has been invoked three times: During World War I, World War II, and the October Crisis of 1970. In 1988, the War Measures Act was replaced by the Emergencies Act.
It was also applied twice in the territory of Lower Canada during the 1837-1838 insurrections. On December 5, following the events of November 1837, martial law was proclaimed in the district of Montréal by Governor Gosford, without the support of the Legislative Assembly in the Parliament of Lower Canada. It was imposed until April 27, 1838. Martial law was proclaimed a second time on November 4, 1838, this time by acting Governor John Colborne, and was applied in the district of Montreal until August 24, 1839.
In Egypt, a State of Emergency has been in effect almost continuously since 1967. Following the assassination of President Anwar el-Sadat in 1981, state of emergency was declared. Egypt has been under state of emergency ever since; the Parliament has renewed the emergency laws every three years since they were imposed. The legislation was extended in 2003 and were due to expire at the end of May 2006; plans were in place to replace it with new anti-terrorism laws. But after the Dahab bombings in April of that year, state of emergency was renewed for another two years. In May 2008 there was a further extension to June 2010. In May 2010, the state of emergency was further extended, albeit with a promise from the government to be applied only to 'Terrorism and Drugs' suspects.
A State of Emergency gives military courts the power to try civilians and allows the government to detain for renewable 45-day periods and without court orders anyone deemed to be threatening state security. Public demonstrations are banned under the legislation. On 10 February 2011, the ex-president of Egypt, Hosni Mubarak, promised the deletion of the relevant constitutional article that gives legitimacy to State of Emergency in an attempt to please the mass number of protesters that demanded him to resign. On 11 February 2011, the president stepped down and the vice president Omar Suleiman de facto introduced the country to martial law when transferring all civilian powers from the presidential institution to the military institution. It meant that the presidential executive powers, the parliamentary legislative powers and the judicial powers all transferred directly into the military system which may delegate powers back and forth to any civilian institution within its territory.
Under martial law the source of power is not the people, not the parliament, not the constitution, not a Holy Text, but solely the Supreme Council of the Armed Forces. The military issued in its third announcement the "end of the State of Emergency as soon as order is restored in Egypt". Before martial law, the Egyptian parliament under the constitution had the civilian power to declare a State of Emergency. When in martial law, the military gained all powers of the state, including to dissolve the parliament and suspend the constitution as it did in its fifth announcement. Under martial law, the only legal framework within the Egyptian territory is the numbered announcements from the military. These announcements could for instance order any civilian laws to come back into force. The military announcements (communiques) are the de facto only current constitution and legal framework for the Egyptian territory. It means that all affairs of the state are bound by the Geneva Conventions.
A classic case of a full-blown martial law in recent history took place in Iran in 1978. On September 7, Shah of Iran, Mohammad Reza Pahlavi, appointed the chief of army staff, General Gholam Ali Oveisi as the military governor of the capital city, Tehran. The army divisions took position in key locations in the city. (Martial law was also declared in some other cities.) On September 8, the army opened fire on protesters, killing somewhere from 300 to 4000 (estimates vary). The day is often referred to as Black Friday. Unable to control the unrest, Shah dissolved the civil government headed by Prime Minister Jafar Sharif-Emami on November 6, and appointed General Gholam Reza Azhari as the prime minister. Azhari's military government also failed to bring order to the country. As a last-ditch effort, as he was preparing to leave the country, Shah dissolved the military government and appointed Shapour Bakhtiar, a reformist critic of his rule, as the new prime minister on January 4, 1979. Bakhtiar's government fell on February 11, and with it, the history of over two thousand years of monarchy in Iran came to an end.
During the Easter Rising in 1916, Lord Wimborne, a cousin of Winston Churchill and then Lord Lieutenant of Ireland, declared martial law to maintain order in the streets of Dublin. This was later extended both in duration and geographical reach to the whole of the country with the consent of the British government. Much of Ireland was declared under martial law by the British authorities during the Irish War of Independence. A large portion of Ireland was also under de facto martial law during the Irish Civil War.
Military administrative government was in effect from 1949 to 1966 over some geographical areas of Israel having large Arab populations, primarily the Negev, Galilee, and the Triangle. The residents of these areas were subject to a number of controlling measures that amounted to martial law. The Israeli army enforced strict residency rules. Any Arab not registered in a census taken during November 1948 was deported. Permits from the military governor had to be procured to travel more than a given distance from a person's registered place of residence, and curfew, administrative detentions, and expulsions were common. Although the military administration was officially for geographical areas, and not people, its restrictions were seldom enforced on the Jewish residents of these areas. In the 1950s, martial law ceased to be in effect for those Arab citizens living in predominantly Jewish cities, but remained in place in all Arab localities within Israel until 1966.
During the 2006 Lebanon war, martial law was declared by Defense Minister Amir Peretz over the north of the country. The Israel Defense Forces were granted the authority to issue instructions to civilians, and to close down offices, schools, camps and factories in cities considered under threat of attack, as well as to impose curfews on cities in the north.
Instructions of the Home Front Command are obligatory under martial law, rather than merely recommended. The order signed by Peretz was in effect for 48 hours and was extended by the Cabinet and the Knesset Foreign Affairs and Defense Committee over the war's duration.
Mauritius is known as being a "Westminster" style of democracy but a peculiar system that was imposed in Mauritius during a period of civil unrest in 1968 as an emergency measure, has never been repealed and is still used by the police force there to this day. The system, which has no apparent foundation in the constitution of Mauritius, enables the police to arrest without having to demonstrate reasonable suspicion that a crime has been carried out but simply on the submission of "provisional information" to the magistrate. The accused is then placed on remand or bail and required to report to the police or the court on a regular basis, sometimes every day. There are examples of this system being used to intimidate or coerce individuals in civil litigations.
Martial law has been declared in Pakistan On 7 October 1958, President Iskander Mirza declared Martial Law and appointed General Muhammad Ayub Khan as the Chief Martial Law Administrator and Aziz Ahmad as Secretary General and Deputy Chief Martial Law Administrator. However, three weeks later General Ayub—who had been openly questioning the authority of the government before the imposition of martial law—deposed Iskandar Mirza on 27 October 1958 and assumed the presidency that practically formalized the militarization of the political system in Pakistan. Four years later a new document, Constitution of 1962, was adopted. The second martial law was imposed on 25 March 1969, when President Ayub Khan abrogated the Constitution of 1962 and handed over power to the Army Commander-in-Chief, General Agha Mohammad Yahya Khan. On assuming the presidency, General Yahya Khan acceded to popular demands by abolishing the one-unit system in West Pakistan and ordered general elections on the principle of one man one vote.
The third was imposed by Zulfikar Ali Bhutto, the first civilian to hold this post in Pakistan after the Bangladesh Liberation War. On 21 December 1971, Bhutto took this post as well as that of President.
The fourth was imposed by the General Muhammad Zia-ul-Haq on 5 July 1977. After several tumultuous years, which witnessed the secession of East Pakistan, politician Zulfikar Ali Bhutto took over in 1971 as the first civilian martial law administrator in recent history, imposing selective martial law in areas hostile to his rule, such as the country's largest province, Balochistan. Following widespread civil disorder, General Zia overthrew Bhutto and imposed martial law in its totality on July 5, 1977, in a bloodless coup d'état. Unstable areas were brought under control through indirect military action, such as Balochistan under Martial Law Governor, General Rahimuddin Khan. Civilian government resumed in 1988 following General Zia's death in an aircraft crash.
On October 12, 1999, the government of Prime Minister Nawaz Sharif was dissolved, and the Army took control once more. But no martial law was imposed. General Pervez Musharraf took the title of Chief Executive until the President of Pakistan Rafiq Tarar resigned and General Musharraf became president. Elections were held in October 2002 and Mir Zafarullah Khan Jamali became Prime Minister of Pakistan. Jamali premiership was followed by Chaudhry Shujaat Hussain and Shaukat Aziz. While the government was supposed to be run by the elected prime minister, there was a common understanding that important decisions were made by the President General Musharraf.
On November 3, 2007, President General Musharraf declared the state of emergency in the country which is claimed to be equivalent to the state of martial law as the constitution of Pakistan of 1973 was suspended, and the Chief Justices of the Supreme Court were fired.
On November 12, 2007, Musharraf issued some amendments in the Military Act, which gave the armed forces some additional powers.
During the Second World War, President José P. Laurel placed the Philippines (then a client state of Imperial Japan) under martial law via Proclamation № 29, dated 21 September 1944 and enforced the following day at 09:00 PST. Proclamation № 30 was issued on 23 September, declaring the existence of a state of war between the Philippines and the United States and the United Kingdom, effective 10:00 that day.
The country was under martial law again from 1972 to 1981 under the authoritarian rule of Ferdinand Marcos. Proclamation № 1098 ("Proclaiming a State of Martial Law in the Philippines") was signed on 21 September 1972 and came into force on 22 September. The official reason behind the declaration was to suppress increasing civil strife and the threat of a communist takeover, particularly after a series of bombings (including the Plaza Miranda Incident) and an assassination attempt on Secretary of Defense Juan Ponce Enrile in Mandaluyong (later revealed to have been staged by the government).
The policy of martial law was initially well-received, but it eventually proved unpopular as the military's human rights abuses (use of torture in intelligence gathering, forced disappearances), along with the decadence and excess of Marcos allies, had emerged. Coupled with economic downturns, these factors fermented dissent in various sectors (e.g. the urban middle class) that crystallised with the assassination of jailed oppositionist senator Benigno Aquino, Jr. in 1983, and widespread fraud in the 1986 snap elections. These eventually led to the 1986 People Power Revolution that ousted Marcos and forced him into exile in Hawaii where he died in 1989; his rival presidential candidate and Aquino's widow, Corazón, was installed as his successor.
There were rumours that President Gloria Macapagal-Arroyo was planning to impose martial law to end military coup d'etat plots, general civilian dissatisfaction, and criticism of her legitimacy arising from the dubious results of the 2004 presidential elections. Instead, a State of National Emergency was imposed in 2006 from 24 February to 3 March, in order to quash a coup attempt and quell protesters.
On 4 December 2009, President Arroyo officially placed the Province of Maguindanao under a state of martial law through Proclamation № 1959. As with the last imposition, the declaration suspended the writ of habeas corpus in the province. The announcement came days after hundreds of government troops were sent to the province to raid the armories of the powerful Ampatuan clan. The Ampatuans were implicated in the massacre of 58 persons, including women from the rival Mangudadatu clan, human rights lawyers, and 31 media workers. Cited as one of the bloodiest incidents of political violence in Philippine history, the massacre was condemned worldwide as the worst loss of life of media professionals in one day.
Martial law was introduced in Communist Poland on December 13, 1981 by Generals Czesław Kiszczak and Wojciech Jaruzelski to prevent democratic opposition from gaining popularity and political power in the country. Thousands of people linked to democratic opposition, including Lech Wałęsa, were arbitrarily arrested and detained. About 100 deaths are attributed to the martial law, including 9 miners shot by the police during the pacification of striking Wujek Coal Mine. The martial law was lifted July 22, 1983. Polish society is divided in opinion on the necessity of introduction of the martial law, which is viewed by some as a lesser evil compared to alleged Soviet military intervention. The generals' trials are still in progress more than 30 years after.
In October 1946, United States Army Military Government in Korea declared martial law as a result of the Daegu Riot. On November 17, 1948, President Syngman Rhee regime proclaimed a martial law in order to quell the Jeju Uprising. On April 19, 1960 Syngman Rhee government proclaimed a martial law in order to suppress the April Revolution.
There are no provisions for martial law as such in Switzerland. Under the Army Law of 1995, the Army can be called upon by cantonal (state) authorities for assistance (Assistenzdienst). This regularly happens in the case of natural disasters or special protection requirements (e.g., for the World Economic Forum in Davos). This assistance generally requires parliamentary authorization, though, and takes place in the regular legal framework and under the civilian leadership of the cantonal authorities. On the other hand, the federal authorities are authorized to use the Army to enforce law and order when the Cantons no longer can or want to do so (Ordnungsdienst). With this came many significant points of reference. This power largely fell into disuse after World War II. See .
Following World War II, the allied forces asked the Republic of China to temporarily administer Taiwan given the impending withdrawal of Japanese forces and colonial government. In the aftermath of the February 28 Incident of 1947, martial law was declared in 1949 despite the democracy promised in the Constitution of the Republic of China (the Republic of China refused to implement the constitution on Taiwan until after 1949). After the Nationalist-led Republic of China government lost control of China to the Communist Party of China and retreated to Taiwan in 1949, the perceived need to suppress Communist activities in Taiwan was utilised as a rationale for not lifting martial law until thirty-eight years later in 1987, just prior to the death of then President Chiang Ching-kuo.
Today, still present martial law systems like in Syria (since the 1963 Syrian coup d'état) or in the West Bank (since the 1967 Six-Day War with Israel) have surpassed Taiwan as longer ranging periods of active martial law.
Martial law in Thailand derives statutory authority from the Act promulgated by King Vajiravudh following the abortive Palace Revolt of 1912, entitled Martial Law, B.E. 2457 (1914). Many coups have been attempted or succeeded since then, but the Act governing martial law, amended in 1942, 1944, 1959 and 1972, has remained essentially the same. In January 2004, the Prime Minister of Thailand, Thaksin Shinawatra, declared a state of martial law in the provinces of Pattani, Yala, and Narathiwat in response to the growing South Thailand insurgency. On September 19, 2006, Thailand's army declared martial law following a bloodless military coup in the Thai capital of Bangkok, declared while Prime Minister Shinawatra was in New York to address the United Nations General Assembly. General Sonthi Boonyaratglin took the control of the government, and soon after handed the premiership to ex-Army Chief General Surayud. Sonthi himself is Chief of the Administrative Reform Council. At 3 am, on May 20, 2014, following seven months of civil and political unrest, Army Commander-in-Chief Gen.Prayuth Chan-ocha, declared martial law nationwide.
Since the foundation of the Republic of Turkey in 1923 the military conducted three coups d'état and announced martial law. Martial law between 1978 and 1983 was replaced by a State of emergency in a limited number of provinces that lasted until November 2002.
During the Yugoslav Wars in 1991, a "State of Direct War Threat" was declared. Although forces from the whole SFRY were included in this conflict, martial law was never announced, but after secession, Croatia and Bosnia and Herzegovina declared martial law. On March 23, 1999, a "State of Direct War Threat" was declared in Yugoslavia, following the possibility of NATO air-strikes. The day after strikes began, martial law was declared, which lasted until June 1999, although strikes ended on June 10, following Kumanovo Treaty. The state of direct war threat was declared and announced to everyone on April 1999.
The martial law concept in the United States is closely tied with the right of habeas corpus, which is in essence the right to a hearing on lawful imprisonment, or more broadly, the supervision of law enforcement by the judiciary. The ability to suspend habeas corpus is often equated with martial law. Article 1, Section 9 of the US Constitution states, "The Privilege of the Writ of Habeas Corpus shall not be suspended, unless when in Cases of Rebellion or Invasion the public Safety may require it."
In United States law, martial law is limited by several court decisions that were handed down between the American Civil War and World War II. In 1878, Congress passed the Posse Comitatus Act, which forbids military involvement in domestic law enforcement without congressional approval.
There have been many instances of the use of the military within the borders of the United States, such as during the Whiskey Rebellion and in the South during the civil rights crises, but these acts are not tantamount to a declaration of martial law. The distinction must be made as clear as that between martial law and military justice: deployment of troops does not necessarily mean that the civil courts cannot function, and that is one of the keys, as the Supreme Court noted, to martial law.
The American Revolution
As a result of the Boston Tea Party, Parliament passed the Massachusetts Government Act, one of the Intolerable Acts, which suppressed town meetings and assemblies, and imposed appointed government, tantamount to martial law.
New Orleans, Louisiana in the War of 1812
During the War of 1812, US General Andrew Jackson imposed martial law in New Orleans, Louisiana before repulsing the British in the Battle of New Orleans. Martial law was also imposed in a four-mile radius around the vicinity. When word came of the end of the war, Jackson maintained martial law, contending that he had not gotten official word of the peace. A judge demanded habeas corpus for a man arrested for sedition. Rather than comply with the writ, Jackson had the judge arrested.
Ex parte Milligan
On September 15, 1863 President Lincoln imposed Congressionally authorized martial law. The authorizing act allowed the President to suspend habeas corpus throughout the entire United States (which he had already done under his own authority on April 27, 1861). Lincoln imposed the suspension on "prisoners of war, spies, or aiders and abettors of the enemy," as well as on other classes of people, such as draft dodgers. The President's proclamation was challenged in Ex parte Milligan, 71 US 2 . The Supreme Court ruled that Lincoln's imposition of martial law (by way of suspension of habeas corpus) was unconstitutional in areas where the local courts were still in session.
The Great Chicago Fire
In response to the Great Chicago Fire of 1871, Chicago mayor Roswell B. Mason declared a state of martial law and placed General Philip Sheridan in charge of the city on October 9, 1871. After the fire was extinguished, there were no widespread disturbances and martial law was lifted within a few days.
Coeur d'Alene, Idaho, 1892
In 1892, in Coeur d'Alene, Idaho, striking mine workers blew up a mill and shot at strike-breaking workers. The explosion leveled a four-story building and killed one person. The governor declared martial law. At the same time, a request was made for federal troops to back guardsmen. Over 600 people were arrested. The list was whittled down to two dozen ringleaders who were tried in military court. While in prison, the mine workers formed a new union, the Western Federation of Miners.
San Francisco earthquake of 1906
Following the earthquake of 1906, the troops stationed in the Presidio were pressed into martial law service. Guards were posted throughout the city, and all dynamite was confiscated. The dynamite was used to destroy buildings in the path of fires, to prevent the fires from spreading.
Colorado Coalfield War
In 1914, imposition of martial law climaxed the so-called Colorado Coalfield War. Dating back decades, the conflicts came to a head in Ludlow, Colorado in 1913. The Colorado National Guard was called in to quell the strikers. For a time, the peace was kept, but it is reported that the make-up of the Guard stationed at the mines began to shift from impartial normal troops to companies of loyal mine guards. Clashes increased and the proclamation of martial law was made by the governor, eventually resulting in the Ludlow Massacre. President Wilson sent in federal troops, eventually ending the violence.
West Virginia Coal Wars
During the events of the West Virginia Coal Wars (1920-1921), martial law was declared on the state of West Virginia. At the behest of Governor Cornwell, federal troops had been dispatched to Mingo County to deal with the striking miners. The army officer in charge acted, ostensibly, under the Suspension Clause of Article I of the United States Constitution (selectively; accounts show that he only jailed union miners), and did not allow assembly of any kind. If his soldiers found any union miners, they immediately took them and imprisoned them. The jails filled up so quickly that he had to release miners. As it went, miners were arrested, jailed, and released without any sort of trial. After a time, when the trial of Sid Hatfield began; the military occupation and "veritable military dictatorship" (Governor Cornwell) of the army officer ended. Many of the miners were not released from jail. It was only the first of three times that federal troops would be called to quiet the miners in the West Virginia Mine War.
Tulsa Race Riot
In 1921, during the Tulsa race riot, the Oklahoma National Guard declared martial law.
San Francisco, California, 1934
In 1934, California Governor Frank Merriam placed the docks of San Francisco under martial law, citing "riots and tumult" resulting from a dock worker's strike. The Governor threatened to place the entire city under martial law. The National Guard was called in to open the docks, and a city-wide institution of martial law was averted when goods began to flow. The guardsmen were empowered to make arrests and to then try detainees or turn them over to the civil courts.
The Territory of Hawaii
During World War II (1939 to 1945) what is now the State of Hawaii was held under martial law from December 7, 1941 to October 24, 1944, following the Japanese attack on Pearl Harbor. Many Hawaiians were, and are, of Japanese descent, and the loyalty of these people was called into question. After the war, the federal judge for the islands condemned the conduct of martial law, saying, "Gov. Poindexter declared lawfully martial law but the Army went beyond the governor and set up that which was lawful only in conquered enemy territory namely, military government which is not bound by the Constitution. And they ... threw the Constitution into the discard and set up a military dictatorship."
On May 21, 1961, Governor Patterson of Alabama declared martial law "as a result of outside agitators coming into Alabama to violate our laws and customs" which has led to "outbreaks of lawlessness and mob action."
- Chief Martial Law Administrator
- Military dictatorship a government where in the political power resides with the military.
- Military occupation, the form of administration by which an occupying power exercises governmental authority over occupied territory.
- Military junta a government led by a committee of military leaders.
- Military law (law to which members of the military are subject)
- Military rule (disambiguation)
- Police state
- State of emergency
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Reference to Thai legislation in any jurisdiction shall be to the Thai version only. This translation has been made so as to establish correct understanding about this Act to the foreigners.Check date values in:
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# Given $\langle f_{n}\rangle$ such that $f_{n}=\frac{f_{n-1}+f_{n-2}}{2}$ $\forall n\gt2$,to prove it converges to $\frac{f_1+2f_2}{3}$
If $\langle f_{n}\rangle$ be a sequence of positive numbers such that $$f_{n}=\frac{f_{n-1}+f_{n-2}}{2}$$ $\forall n\gt2$ ,then show that $\lt f_{n}\gt$ converges to $$\frac{f_1+2f_2}{3}$$
Replacing $n$ by $3,4,5,....,n-1$,we get $$f_3=(f_2+f_1)/2$$ $$f_4=(f_3+f_2)/2$$ $$f_5=(f_4+f_3)/2$$ $$\dots\dots\dots\dots\dots$$ $$f_{n-1}=(f_{n-2}+f_{n-3})/2$$ $$f_{n}=(f_{n-1}+f_{n-2})/2$$
In the solution, its given as adding the corresponding sides of $f_n$ and $f_{n-1}$,the following expression is obtained
$$f_n+\frac12f_{n-1}=\frac12(f_1+2f_2)$$
I couldn't understand how this expression came.
May be be easier to follow if you first rewrite the given recurrence as:
$$f_{n} \color{red}{+\frac{f_{n-1}}{2}}=\frac{f_{n-1}+f_{n-2}}{2} \color{red}{+\frac{f_{n-1}}{2}} = f_{n-1}+\frac{f_{n-2}}{2}$$
It "telescopes" nicely from there on:
$$f_{n} +\frac{f_{n-1}}{2} = f_{n-1}+\frac{f_{n-2}}{2} = f_{n-2}+\frac{f_{n-3}}{2} = \cdots = f_2 + \frac{f_1}{2}$$
On the right side every number except $f_1, f_2, f_{n-1}$ appear twice and when you multiply by $\frac 12$, each of those number has a coefficient 1 in front of it. Similarly on the left side all numbers appear once, so after massive cancelations you get the wanted identity.
In this case you can also explicitely solve
$$f_{n}=\frac{f_{n-1}+f_{n-2}}{2}\implies2x^2-x-1=0\implies x_1=1 \quad x_2=-\frac12\implies \\f_n=a+b\left(-\frac12\right)^n\to a$$
and from the initial conditions
$$\begin{cases} f_1=a+b\\ f_2=a-\frac12b \end{cases}\implies 3a=f_1+2f_2\implies a=\frac{f_1+2f_2}{3}$$<|endoftext|>
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# Necessary and sufficient conditions for differentiability.
Apologizes if I'm missing something in my question or if my question seems trivial; this is my first question on this site. As motivation for my question, consider the following standard first year calculus question.
Consider this piecewise function: $f(x) = \left\{ \begin{array}{lr} ax^2+b & \text{ if } x \le-2\\ 12x-5 & \text{ if } x >-2 \end{array} \right.$
For what values of $a$ and $b$ will $f(x)$ be differentiable?
To solve this question, I would like to propose the following theorem:
$\mathbf{Theorem:}$ A function $f(x)$ is differentiable iff $f'(x)$ is continuous.
If this theorem is true, then I can solve for $a$ first by noting that: $f'(x) = \left\{ \begin{array}{lr} 2ax & \text{ if } x \le-2\\ 12 & \text{ if } x >-2 \end{array} \right.$
Thus, since by my theorem $f'(x)$ must be continuous, we have:
\begin{align*} \lim_{x \rightarrow -2^-}f'(x) &= \lim_{x \rightarrow -2^+}f'(x)\\ \lim_{x \rightarrow -2^-}2ax &= \lim_{x \rightarrow -2^+}12\\ 2a(-2) &= 12\\ -4a &= 12 \\ a &= -3 \\ \end{align*}
Hence, since differentiability implies continuity, we can solve for $b$ as follows:
\begin{align*} \lim_{x \rightarrow -2^-}f(x) &= \lim_{x \rightarrow -2^+}f(x)\\ \lim_{x \rightarrow -2^-}-3x^2+b &= \lim_{x \rightarrow -2^+}12x-5\\ -3(-2)^2+b &= 12(-2)-5\\ b-12 &= -29\\ b &= -17 \\ \end{align*}
so that our differentiable function is:
$$f(x) = \left\{ \begin{array}{lr} -3x^2-17 & \text{ if } x \le-2\\ 12x-5 & \text{ if } x >-2 \end{array} \right.$$
Anyways. My question is: Is my proposed theorem actually a thing? I've looked through my calculus textbook and it doesn't seem to explicitly state it, yet I don't know how to solve this question otherwise. If this theorem turns out to be false, how else can you solve this problem? Thanks in advance. =]
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A differentiable function need not have a continuous derivative. – user10444 May 9 '13 at 20:50
$f(x)$ is a number, not a function. So your statement should be "$f$ is differentiable if and only if $f'$ is continuous". It should be clear what the problem is here. When you speak of the derivative $f'$, you are already assuming that $f$ is differentiable. – wj32 May 9 '13 at 20:53
On the other hand, one can show: if $\lim_{x\rightarrow c^-} f'(x)$ exists, then the derivative from the left of $f$ at $x=c$ exists and is equal to the value of this limit. A similar fact holds for the derivative from the right. Also, a function is differentiable at $x=c$ if and only if both one-sided derivatives exist at $c$ and are equal there. So your method for computing $a$ is correct, but not for the reason you give. – David Mitra May 9 '13 at 21:05
@DavidMitra Thanks, that makes perfect sense. – Adriano May 9 '13 at 23:15
let $f(x)= \begin{cases} x^2\sin(\frac 1 x), & \mbox{if } x \not= 0 \\0 & \mbox{if } x=0 \end{cases}$ is continuous but has a discontinuous derivative. Check the continuity of $f'(x)$ at $x=0$.
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Sorry, but how does this contradict my (admittedly false) theorem? I don't think your function $f(x)$ is differentiable at $x = 0$ (that is, I don't think that $f'(0)$ exists). This agrees with my (admittedly false) theorem since, as you said, $f'(x)$ is discontinuous at $x = 0$. – Adriano May 9 '13 at 22:28
$f'(0)$ exists because $$\lim_{h \to 0} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0} \frac{h^2\sin(1/h)}{h} = \lim_{h \to 0} h\sin(1/h) = 0.$$ – zrbecker May 9 '13 at 22:37
Ah, yes, of course. Then we have: $$f'(x) = \left\{ \begin{array}{lr} 2x \sin(1/x)-\cos(1/x) & \text{ if } x \ne 0\\ 0 & \text{ if } x = 0 \end{array} \right.$$ While $f'(0)$ exists, $f'$ is still not continuous since \lim_{x \rightarrow 0}f'(x) doesn't exist. – Adriano May 9 '13 at 22:46
It is not. A function $f:\mathbb{R}\rightarrow\mathbb{R}$ is differentiable iff
$$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$$
exists for all $x\in\mathbb{R}$. If $f'$ is continuous, $f$ is said to be continuously differentiable (or of class $\mathcal{C}^1$). Hence, to solve the problem you need to pick $a$ and $b$ such that the limit exists for any $x\in\mathbb{R}$. Note also, that for $f$ to be differentiable, $f$ must be continuous.
By the way, welcome to math stackexchange and I wouldn't worry about asking a "trivial question" - your question is well written and, often, questions are only trivial if you've already seen the answer.
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For that function to be differentiable you need $$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$ to exist.
The only point where something "weird" is going on, is at -2. Thus we just need to find $a, b$ such that $$\lim_{h \to 0^+} \frac{f(-2 + h) - f(-2)}{h} = \lim_{h \to 0^-} \frac{f(-2 + h) - f(-2)}{h}$$
The right hand side is simply $-4a$, by using the regular rules of derivatives. The weirdness happens on the other side of the limit.
$$\lim_{h \to 0^-} \frac{12(-2 + h) - 5- (4a + b)}{h} = \lim_{h \to 0^-} \frac{-29 - 4a -b + 12h}{h}$$
In order for this limit to exist we need $-29 -4a -b = 0$. Otherwise the limit will be $-\infty$ or $+\infty$. Once we have $-29 -4a -b = 0$, we get $$\lim_{h \to 0^-} \frac{-29 - 4a -b + 12h}{h} = \lim_{h \to 0^-} \frac{12h}{h} = 12.$$
Since we need the left and the right limit to be the same, we get $-4a = 12$. Thus $a = -3$. Then solving for $b$, we have $-29 + 12 - b = 0$, so $b = -17$.
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I think you meant to say that $$\lim_{h \to 0^+} \frac{12(-2 + h) - 5- (4a + b)}{h} = \lim_{h \to 0^+} \frac{-29 - 4a -b + 12h}{h}= \lim_{h \to 0^+} \frac{12h}{h} =12$$ so that $-4a=12$ gives $a=-3$ and $-29-4(-3)-b=0$ yields $b=-17$. While your method works, I'm not really convinced why you were able to deduce that $-29-4a-b=0$ though. – Adriano May 9 '13 at 22:09
The way I went about it was to say, if $-29 - 4a - b = c$ where $c \neq 0$, then $c/h \to \pm \infty$. So the limit doesn't exist. But the limit must exist for it to be differentiable. – zrbecker May 9 '13 at 22:22
Another way to think about it, is $f$ needs to be continuous at $-2$, so $a(-2)^2 + b = 12(-2) - 5$, which after some algebra gives $-29 - 4a - b = 0$. – zrbecker May 9 '13 at 22:25
Ah that makes sense now, thanks. – Adriano May 9 '13 at 22:29<|endoftext|>
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Sometimes called fish scale disease or fishskin disease, ichthyosis vulgaris is an inherited skin disorder that causes dead skin cells to accumulate in thick, dry scales on the skin’s surface. These scales can be present at birth, but usually first appear in early childhood. Sometimes ichthyosis vulgaris disappears entirely for most of the adult years, only to return later.
Though most cases are mild, some cases of ichthyosis vulgaris are severe. No cure has been found and treatments are directed at controlling the signs and symptoms.
Ichthyosis vulgaris is characterized by:
- Very dry, scaly skin.
- Tile-like scales that are small, polygonal in shape.
- Scales that range in color from white to dirty gray to brown. People with darker skin tend to have darker colored scales.
- Flaky scalp.
- In severe cases, deep painful cracks in the palms and soles.
The scales usually appear on the elbows and lower legs and may be especially thick and dark over the shins. Though most cases of ichthyosis are mild, some can be severe.
Symptoms usually worsen or are more pronounced in cold, dry environments and tend to improve or even resolve in warm, humid environments.
Other less common forms of ichthyosis include:
- Lamellar ichthyosis. This severe form of the disease is present at birth and lasts throughout life. Infants with lamellar ichthyosis are born encased in a filmy membrane that’s shed after 10 to 14 days, revealing skin that’s covered in scales. The scales can range from fine and white to thick and dark and generally occur over the entire body, although they may be larger on the legs. Lamellar ichthyosis can be extremely disfiguring and may cause great psychological suffering for children and adults with the disease.
- X-linked ichthyosis. Starting soon after birth, this type of ichthyosis occurs only in males. The noticeable, dirty-brown scales that characterize this skin disease are most pronounced on the back of the neck, arms and behind the knees. Symptoms generally don’t improve with age.
- Epidermolytic hyperkeratosis. This extremely rare form of ichthyosis is usually present at birth and begins with blistering skin. In time, the skin peels away in large sheets and becomes rough or wart-looking. It’s most pronounced on the knees, elbows, wrists and other flexural areas.
The skin is an endlessly renewable organ. New skin cells at the base of the epidermis push toward the surface of the skin, where they eventually shrink, flatten and die. These dead skin cells flake off every day and are continuously replaced by more cells.
Ichthyosis, however, disrupts this pattern. It occurs when the production of skin cells is too fast or the skin’s natural shedding process is too slow. This causes dead skin cells to collect into thick flakes that stick to the outer surface of skin. These thick flakes can resemble fish scales.
Most often, ichthyosis is inherited in an autosomal dominant pattern, which means that a child has to inherit only one copy of the affected gene to develop the disease. Children with the inherited form of the disorder usually have normal skin at birth but develop scaling and roughness during the first few years of life. At times, ichthyosis vulgaris may disappear during the adult years, only to return later.
Ichthyosis not caused by genetics, referred to as acquired ichthyosis, is very rare. This type usually manifests in adulthood and is usually associated with other internal diseases, such as cancer, thyroid disease or chronic renal failure.
A doctor can often make a diagnosis by examining the skin and the characteristic scales. If there’s any doubt, he or she may perform other tests, such as a skin biopsy. This may be necessary to rule out other causes of dry, scaly skin.
To diagnose ichthyosis, he or she also takes into account:
- Personal and family history of ichthyosis
- Age when ichthyosis first started
- Presence of other skin disorders
There’s no known cure for ichthyosis, so the goal of treatment is to manage the condition. In addition to home care, treatment can include prescription creams and ointments that contain alpha hydroxy acids, such as lactic acid and glycolic acid. These chemicals help control the scaling and increase skin moisture.
In severe cases, the doctor may prescribe retinoids — medications derived from vitamin A. They reduce the production of skin cells. Side effects from the medication may include eye and lip inflammation, bone spurs and hair loss, as well as birth defects if taken during pregnancy.
Although self-help measures won’t cure ichthyosis, they may help improve the appearance and feel of damaged skin. These measures may be beneficial:
- Take long soaking baths to soften the skin. Then use a roughly-textured sponge, such as a loofa sponge, to remove the thickened scales.
- Choose mild soaps that have added oils and fats. Avoid deodorant and antibacterial soaps, which are especially harsh on dry skin.
- After washing or bathing, gently pat or blot your skin dry with a towel so that some moisture remains on the skin.
- Apply the moisturizer or lubricating cream while your skin is still wet or moist from bathing. Choose a moisturizer that contains urea or propylene glycol — chemicals that help keep your skin moist. Petroleum jelly is another good choice. Cover the treated areas with plastic wrap to keep the petroleum jelly from staining clothes and furniture.
- Twice daily applications of an over-the-counter product that contains urea, lactic acid or a low concentration of salicylic acid may help. Mild acidic compounds help the skin shed its dead skin cells. Urea helps bind moisture to the skin.
- Use a portable home humidifier or one attached to your furnace to add moisture to the air inside your home.<|endoftext|>
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Contents
• one choice • many choices: permutations • many choices: combinations • many choices: combinations • independent events • dependent events • permutations without repettitions • permutations with identical elements • combinations • coditional probabilities • and or events • two outcomes • total probability theorem • Bayes rule • union sets and probability • fundamental counting principle
Other exercices
Combinatorics - Probability
Example 2: many possibilities: permutations
Many possibilities
Permutations
Many choices: Permutations
We have 5 different books of Mathematics. How many possibilities can we have to choose 3 books from them?. The order is important.
At first, we have 5 possibilities to choose a book. If we choose one book, il will remain 4 possibilities to choose another second book from the 4 remaininig books. Finally, it remains 3 possibilities to choose the last book from the 3 remaininig books. Hence, the total possibilities that we have is 5 x 4 x 3 = 60 possibilities.
Other approach:
For B1: 12 possibilities:
-------------------------
B1 --> B2 --> B3 or B4 or B5 :3 possibilities
B1 --> B3 --> B2 or B4 or B5 :3 possibilities
B1 --> B4 --> B2 or B3 or B5 :3 possibilities
B1 --> B5 --> B2 or B3 or B4 :3 possibilities
Similarly for B2, B3, B4, and B5
Finally, we have 5 x 12 = 60 possiblities.
= 3 x 4 x 5 = 5!/(5 - 3)! = (5 different books)!/[(5 different books)! - (3 books among them)!]! = P(3,5) = Permutation of 3 books among 5 books.
Main formula
Let's consider a set of "n" different elements: {B1, B2, B3, ..., Bn}. The arrangement of "p" elements chosen among the "n" elements is the ordered disposition of "p" elements. For example, {B1, B3, B4, B9} is an arrangement of 4 elements among the "n" elements. {B1, B9, B4, B3} is another arrangement. {B1, B3, B4, B9} and {B1, B9, B4, B3} are two different arrangements because the order of the elements is different in the two sets (two arrangements).
At first, we have "n" possibilities to choose an element from the set {B1, B2, B3, ..., Bn}. Then, once the first is choosen, it remains "n - 1" possibilities to choose a second. Then "n - 2" to choose a third element. ... Then "n - (p - 1)" to choose the "pth" element. Finally, the total number of possibilities to arrange "p" elements among "n" elements is :
A(p,n) = n x (n - 1) x (n - 2) x ... x [n - (p - 1)] (1)
We can write A(p,n) = A(p,n) x R /R (3)
Where: R = [n - (p )] x [n - (p +1)] x [n - (p + 2)] x ... x 2 x 1 = (n - p)! (3)
The formula (1) becomes: A(p,n) x R = n!
Hence:
A(p,n) = n!/(n - p)!
The number of permutations of the "n" elements of the set {B1, B2, B3, ..., Bn} we can make is the number of arragements of "n" elements among the "n" elements of the seté. That is: P(n,n) = A(n,n) = n!/(n - n)! = n!
P(n,n) = n!
Today: : ____________
calculator for combinatorics probability and Statistics<|endoftext|>
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# Quadratic equation : Two solutions or one solution?
I have an equation to solve for y:
$$\frac{y^2}{y}=1$$ Normally, I would cancel out one $y$ and get $y=1$ as a single solution.
But If I think of it as quadratic equation
$$y^2=y$$ $$y^2-y=0$$ $$y(y-1)=0$$ $$y=0 \space \text{or} \space y=1$$ to have two solutions.
But when I put $y=0$ in original equation, I get $\frac{0}{0}$, so is $y=0$ a solution or not ?
If yes, then I get $\frac{0}{0}$.
If no, then how come this quadratic equation has $1$ solution ?
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This equation has two solutions, $y=0$ and $y=1$:
$$y^2=y \tag{1}$$
This equation has one solution, $y=1$:
$$\frac{y^2}{y} = 1 \tag{2}$$
The reason that it has one solution is that either of the operations "Cancel a $y$ from the top and bottom of the fraction" and "Multiply both sides of the equation by $y$" are only valid when $y\neq 0$. In particular, if you want to manipulate (2) to look like (1), you first have to assume that $y\neq 0$, which rules out one of the solutions of the quadratic equation.
A more sophisticated answer realises that when we are asked to "solve" an equation, what we are really doing is looking for a root of a particular function (i.e. a value of the argument at which evaluating the function gives zero). In the case of (1), the function is $f(y)=y^2-y$, and in the case of (2) the function is $f(y)=y^2/y-1$.
Now, functions have to have a domain. In the case of (1) the domain is $\mathbb{R}$ (the real numbers) which includes both 0 and 1. In the case of (2), the domain is $\mathbb{R}-\{0\}$ (the real numbers without 0) which doesn't include 0.
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Thank you very much for the answer. – Happy Mittal Jun 28 '12 at 8:17
But can't I write $y^2=y$ as $\frac{y^2}{y}=1$ without any constraint ? – Happy Mittal Jun 28 '12 at 9:05
When you divide by $y$, you have to make the assumption that $y\neq 0$. – Chris Taylor Jun 28 '12 at 9:07
OK. Thanks a lot. – Happy Mittal Jun 28 '12 at 9:12<|endoftext|>
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# Invers Matriks
## Konsep Dasar
#### A. Invers Matriks 2 × 2
Jika matriks $$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$ maka invers matriks A adalah $$A^{-1} = \dfrac{1}{det \: A} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$$
Contoh:
Tentukan invers dari matriks $$A = \begin{pmatrix}3 & 5 \\2 & 4\end{pmatrix}$$
Jawab:
(1) Menentukan determinan matriks:
$$|A| = \begin{vmatrix}3 & 5 \\2 & 4\end{vmatrix} = 3 \:.\: 4 - 5 \:.\: 2 = 2$$
(2) Menentukan invers matriks:
$$A^{-1} = \dfrac{1}{2} \begin{pmatrix}4 & -5 \\-2 & 3\end{pmatrix} = \begin{pmatrix}2 & -2.5 \\-1 & 1.5\end{pmatrix}$$
#### B. Invers Matriks 3 × 3 (metode kofaktor)
$$A^{-1} = \dfrac{1}{det \: A} \: adj \: (A)$$
$$adj \: (A) = [kof \: (A)]^T$$
Menentukan Kofaktor
$$kof \: (A) = \begin{pmatrix} M_{11} & {\color {red} -}M_{12} & M_{13} \\ {\color {red} -}M_{21} & M_{22} & {\color {red} -}M_{23} \\ M_{31} & {\color {red} -}M_{32} & M_{33} \\ \end{pmatrix}$$
$$\begin{pmatrix} {\colorbox {cyan} a} & b & c \\d & {\color {red} e} & {\color {red} f} \\g & {\color {red} h} & {\color {red} i}\end{pmatrix} \rightarrow M_{11} = \begin{vmatrix} {\color {red} e} & {\color {red}f }\\ {\color {red}h} & {\color {red}i} \end{vmatrix}$$
$$\begin{pmatrix} a & {\colorbox {cyan} b} & c \\{\color {red} d} & e & {\color {red} f} \\{\color {red} g} & h & {\color {red} i}\end{pmatrix} \rightarrow M_{12} = \begin{vmatrix} {\color {red}d} & {\color {red}f} \\{\color {red} g} & {\color {red}i} \end{vmatrix}$$
$$\begin{pmatrix} a & b & {\colorbox {cyan} c} \\{\color {red} d} & {\color {red} e} & f \\{\color {red} g} & {\color {red} h} & i\end{pmatrix} \rightarrow M_{13} = \begin{vmatrix} {\color {red}d} & {\color {red}e} \\ {\color {red}g} & {\color {red}h} \end{vmatrix}$$
$$\begin{pmatrix} a & {\color {red} b} & {\color {red} c} \\{\colorbox {cyan} d} & e & f \\g & {\color {red} h} & {\color {red} i}\end{pmatrix} \rightarrow M_{21} = \begin{vmatrix} {\color {red} b} & {\color {red}c }\\ {\color {red}h} & {\color {red}i} \end{vmatrix}$$
$$\begin{pmatrix} {\color {red} a} & b & {\color {red} c} \\d & {\colorbox {cyan} e} & f \\{\color {red} g} & h & {\color {red} i}\end{pmatrix} \rightarrow M_{22} = \begin{vmatrix} {\color {red}a} & {\color {red}c} \\{\color {red} g} & {\color {red}i} \end{vmatrix}$$
$$\begin{pmatrix} {\color {red} a} & {\color {red} b} & c \\d & e & {\colorbox {cyan} f} \\{\color {red} g} & {\color {red} h} & i\end{pmatrix} \rightarrow M_{23} = \begin{vmatrix} {\color {red}a} & {\color {red}b} \\ {\color {red}g} & {\color {red}h} \end{vmatrix}$$
$$\begin{pmatrix} a & {\color {red} b} & {\color {red} c} \\d & {\color {red} e} & {\color {red} f} \\{\colorbox {cyan} g} & h & i\end{pmatrix} \rightarrow M_{31} = \begin{vmatrix} {\color {red} b} & {\color {red}c }\\ {\color {red}e} & {\color {red}f} \end{vmatrix}$$
$$\begin{pmatrix} {\color {red} a} & b & {\color {red} c} \\{\color {red} d} & e & {\color {red} f} \\g & {\colorbox {cyan} h} & i\end{pmatrix} \rightarrow M_{32} = \begin{vmatrix} {\color {red}a} & {\color {red}c} \\{\color {red} d} & {\color {red}f} \end{vmatrix}$$
$$\begin{pmatrix} {\color {red} a} & {\color {red} b} & c \\{\color {red} d} & {\color {red} e} & f \\g & h & {\colorbox {cyan} i}\end{pmatrix} \rightarrow M_{33} = \begin{vmatrix} {\color {red}a} & {\color {red}b} \\ {\color {red}d} & {\color {red}e} \end{vmatrix}$$
Contoh:
Tentukan invers dari matriks $$A = \begin{pmatrix}0 & -3 & -2 \\1 & -4 & -2 \\ -3 & 4 & 1 \end{pmatrix}$$
Jawab:
(1) Menentukan determinan (A)
$$|A| = 0 \: \begin{vmatrix} -4 & -2 \\4 & 1\end{vmatrix}- (-3) \:\begin{vmatrix}1 & -2 \\-3 & 1\end{vmatrix}+ (-2) \:\begin{vmatrix}1 & -4 \\-3 & 4\end{vmatrix}$$
$$|A| = 0 + 3 (1 - 6) - 2 (4 - 12)$$
$$|A| = 1$$
(2) Menentukan kofaktor (A)
$$kof \: (A) = \begin{pmatrix} \begin{vmatrix} -4 & -2 \\ 4 & 1 \end{vmatrix} &{\color {red} -}\begin{vmatrix} 1 & -2 \\ -3 & 1 \end{vmatrix} & \begin{vmatrix} 1 & -4 \\ -3 & 4 \end{vmatrix} \\ {\color {red} -}\begin{vmatrix} -3 & -2 \\ 4 & 1 \end{vmatrix} & \begin{vmatrix} 0 & -2 \\ -3 & 1 \end{vmatrix} & {\color {red} -}\begin{vmatrix} 0 & -3 \\ -3 & 4 \end{vmatrix} \\ \begin{vmatrix} -3 & -2 \\ -4 & -2 \end{vmatrix} & {\color {red} -}\begin{vmatrix} 0 & -2 \\ 1 & -2 \end{vmatrix} & \begin{vmatrix} 0 & -3 \\ 1 & -4 \end{vmatrix} \\ \end{pmatrix} = \begin{pmatrix} 4 & 5 & -8 \\ -5 & -6 & 9 \\ -2 & -2 & 3 \\ \end{pmatrix}$$
$$Adj \: (A) = [kof \: (A)]^T = \begin{pmatrix} 4 & -5 & -2 \\ 5 & -6 & -2 \\ -8 & 9 & 3 \\ \end{pmatrix}$$
(4) Menentukan $$A^{-1}$$
$$A^{-1} = \dfrac{1}{det \: A} \: Adj \: (A) = \dfrac{1}{1} \begin{pmatrix} 4 & -5 & -2 \\ 5 & -6 & -2 \\ -8 & 9 & 3 \\ \end{pmatrix} = \begin{pmatrix} 4 & -5 & -2 \\ 5 & -6 & -2 \\ -8 & 9 & 3 \\ \end{pmatrix}$$
#### C. Sifat-sifat Invers Matriks
• $$A \:.\: A^{-1} = A^{-1} \:.\: A = I$$
• $$(A \:.\: B)^{-1} = B^{-1} \:.\: A^{-1}$$<|endoftext|>
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Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
# A 30.0-cm-long Wire Having a Mass of 10.0 G is Fixed at the Two Ends and is Vibrated in Its Fundamental Mode. a 50.0-cm-long Closed Organ Pipe, Placed with Its Open End Near the Wire - Physics
Sum
A 30.0-cm-long wire having a mass of 10.0 g is fixed at the two ends and is vibrated in its fundamental mode. A 50.0-cm-long closed organ pipe, placed with its open end near the wire, is set up into resonance in its fundamental mode by the vibrating wire. Find the tension in the wire. Speed of sound in air = 340 m s−1.
#### Solution
Given:
Mass of long wire M = 10 gm = 10 × 10−3
Length of wire l = 30 cm = 0.3 m
Speed of sound in air = 340 m s−1
Mass per unit length $\left( m \right)$ is
$m = \frac{\text { Mass }}{\text { Unit length }}$
$= 33 \times {10}^{- 3} \text { kg/m }$
Let the tension in the string be T.
The fundamental frequency $n_0$ for the closed pipe is
$n_0 = \left( \frac{v}{4I} \right)$
$= \frac{340}{2 \times 30 \times {10}^{- 2}}$
$= 170 \text { Hz }$
The fundamental frequency $n_0$ is given by : $n_0 = \frac{1}{2l}\sqrt{\frac{T}{m}}$
On substituting the respective values in the above equation, we get :
$170 = \frac{1}{2 \times 30 \times {10}^{- 2}} \times \sqrt{\frac{T}{33 \times {10}^{- 3}}}$
$\Rightarrow T = 347 \text{ Newton }$
Hence, the tension in the wire is 347 N.
Concept: Speed of Wave Motion
Is there an error in this question or solution?
#### APPEARS IN
HC Verma Class 11, 12 Concepts of Physics 1
Chapter 16 Sound Waves
Q 51 | Page 355<|endoftext|>
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# CHAPTER 4 THE DEFINITE INTEGRAL.
## Presentation on theme: "CHAPTER 4 THE DEFINITE INTEGRAL."— Presentation transcript:
CHAPTER 4 THE DEFINITE INTEGRAL
4.1 Introduction to Area Finding area of polygonal regions can be accomplished using area formulas for rectangles and triangles. Finding area bounded by a curve is more challenging. Consider that the area inside a circle is the same as the area of an inscribed n-gon where n is infinitely large.
Summation notation simplifies representation Area under any curve can be found by summing infinitely many rectangles fitting under the curve.
4.2 The Definite Integral Riemann sum is the sum of the product of all function values at an arbitrary point in an interval times the length of the interval. Intervals may be of different lengths, the point of evaluation could be any point in the interval. To find an area, we must find the sum of infinitely many rectangles, each getting infinitely small.
Definition: Definite Integral
Let f be a function that is defined on the closed interval [a,b]. If exists, we say f is integrable on [a,b]. Moreover, called the definite integral (or Riemann integral) of f from a to be, is then given as that limit.
Area under a curve The definite integral from a to b of f(x) gives the signed area of the region trapped between the curve, f(x), and the x-axis on that interval. The lower limit of integration is a and the upper limit of integration is b. If f is bounded on [a,b] and continuous except at a finite number of points, then f is integrable on [a,b]. In particular, if f is continuous on the whole interval [a,b], it is integrable on [a,b].
Functions that are always integrable
Polynomial functions Sin & cosine functions Rational functions, provided that [a,b] contains no points where the denominator is 0.
4.3 First Fundamental Theorem
Let f be continous on the closed interval [a,b] and let x be a (variable) point in (a,b). Then
What does this mean? The rate at which the area under the curve of function, f(t), is changing at a point is equal to the value of the function at that point.
4.4 The 2nd Fundamental Theorem of Calculus and the Method of Substitution
Let f be continuous (integrable) on [a,b], and let F be any antiderivative of f on [a,b]. Then the definite integral is
Evaluate
Substitution Rule for Indefinite Integrals
Let g be a differentiable function and suppose that F is an antiderivative of f. Then
What does this remind you of?
It is the chain rule! (from differentiation) In this case, you have an integral with a function and it’s derivative both present in the integrand. This is often referred to as “u-substitution” Let u=function and du=that function’s derivative
Evaluate
Substitution Rule for Definite Integrals
Let g have a continuous derivative on [a,b], and let f be continuous on the range of g. Then where u=g(x):
What does this mean? For a definite integral, when a substitution for u is made, the upper and lower limits of integration must change. They were stated in terms of x, they must be changed to be the corresponding values, in terms of u. When this change in the upper & lower limits is made, there is no need to change the function back to be in terms of x. It is evaluated in terms of the upper & lower limits in terms of u.
Evaluate:
4.5 The Mean Value Theorem for Integrals and the Use of Symmetry
Average Value of a Function: If f is integrable on the interval [a,b], then the average value of f on [a,b] is:
What does this mean? If you consider the definite integral from over [a,b] to be the area between the curve f(x) and the x-axis, f-average is the height of the rectangle that would be formed over that same interval containing precisely the same area.
Mean Value Theorem for Integrals
If f is continuous on [a,b], then there is a number c between a and b such that
Symmetry Theorem If f is an even function then
If f is an odd function, then
4.6 Numerical Integration
If f is continuous on a closed interval [a,b], then the definite integral must exist. However, it is not always easy or possible to find the definite integral. In these cases, we use other methods to closely approximate the definite integral.
Methods for approximating a definite integral
Left (or right or midpoint) Riemann sums (estimate the area with rectangles) Trapezoidal Rule (estimate with several trapezoids) Simpson’s Rule (estimate the area with the region contained under several parabolas)
Summary of numerical techniques
Approximating the definite integral of f(x) over the interval from a to b.<|endoftext|>
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# Division of Mixed Numbers by Fractions
## Understand how to take a mixed number and divide it by a fraction.
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Division of Mixed Numbers by Fractions
Source: https://pixabay.com/en/food-bowl-fressnapf-dog-food-281980/
Audrey is checking to see how much dog food she has left. There is pounds of dog food left in the bag. Her dog eats about a half pound of food per day. How many days worth of dog food does Audrey have left before she runs out?
In this concept, you will learn how to divide a mixed number by a fraction.
### Dividing Mixed Numbers by Fractions
You can divide a number by a fraction by multiplying the number by the reciprocal of the divisor. A mixed number consists of a whole number and a fraction, or "a whole and a part." When dividing a mixed number by a fraction, the quotient is the number of times the fraction divides into the mixed number.
Here is a division problem.
In this problem, you are trying to figure out how many sets or groups of one-third can be made from one and one-half.
Here is a drawing of one and one-half.
To figure out how many groups of one–third can be made from this quantity, you would have to divide these boxes up again into thirds. Dividing by a fraction using diagrams does not always provide clear answers.
Instead, use these rules for dividing mixed numbers and fractions.
1. Convert the mixed number to an improper fraction.
2. Change the division to its inverse, multiplication, and change the divisor to its reciprocal.
3. Multiply and simplify to find the quotient.
Let's apply this information to the division problem.
First, convert the mixed number to an improper fraction.
Next, rewrite the problem.
Then, change the division to multiplication and change the divisor to its reciprocal. Flip the numerator and the denominator of a fraction to find the reciprocal. The reciprocal of is
Next, multiply and simplify the fractions.
The quotient is .
### Examples
#### Example 1
Earlier, you were given a problem about Audrey and her dog.
Audrey's dog eats about a half pound of food per day and she has pounds of dog food left. Divide by to find out how many days worth of food she has.
First, write a division problem for the problem.
Next, convert the mixed number to an improper fraction.
Then, change the operation to multiplication and the divisor to its reciprocal.
Next, multiply the fractions
Finally, convert the improper fraction to a mixed number.
Audrey will run out of dog food after 7 days.
#### Example 2
Divide the mixed number: . Answer in simplest form.
First, convert the mixed number to an improper fraction.
Next, rewrite the problem.
Then, change the division to multiplication and the divisor to its reciprocal.
Next, multiply and simply the fractions.
The quotient is .
#### Example 3
Divide the mixed numbers: . Answer in simplest form.
First, convert the mixed number to an improper fraction.
Then, change the division to multiplication and the divisor to its reciprocal.
Next, multiply and simplify the fractions.
The quotient is .
#### Example 4
Divide the mixed numbers: . Answer in simplest form.
First, convert the mixed number to an improper fraction.
Then, change the division to multiplication and the divisor to its reciprocal.
Next, multiply and simplify the fractions.
The quotient is .
#### Example 5
Divide the mixed numbers: . Answer in simplest form.
First, convert the mixed number to an improper fraction.
Then, change the division to multiplication and the divisor to its reciprocal.
Next, multiply and simplify the fractions.
The quotient is .
### Review
Divide the mixed number. Answer in simplest form.
To see the Review answers, open this PDF file and look for section 7.11.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Quotient The quotient is the result after two amounts have been divided.<|endoftext|>
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Nutrition plays a significant role in children’s health and well-being. In fact, UNICEF estimates that nearly half of all deaths of children under 5 years are attributable to poor nutrition. It’s something we can change that’s within our grasp, which makes the current situation unacceptable.
Today marks the beginning of World Breastfeeding Week. This year’s theme is A Key to Sustainable Development, in reference to the Sustainable Development Goals. What are the SDGs and why is breastfeeding important to the effort?
The SDGs are goals agreed by U.N. countries aimed at ending poverty, protecting the planet and ensuring prosperity for all. Nutrition is at the root of health and well-being, plays a key role in achieving these goals.
The consumption of high-quality foods in appropriate quantities at the right stages of development helps children grow physically and intellectually into healthier adults. However, a lack of proper nutrition puts children at risk of dying from common childhood infections. It can cause stunting, wasting, chronic illness and life-long problems that are irrevocable.
One of the most important ways to ensure children, especially infants, are well-nourished is through breastfeeding. Proper nutrition is critical in the first 1,000 days of life and breast milk provides all the nutrients a newborn needs for the first 6 months of life; nothing else. After 6 months, continued breastfeeding along with complementary, solid foods should be added to the infant’s diet up to 2 years or beyond.
Sustainable Development Goals also referred to as the Global Goals for sustainable development
IMA World Health champions breastfeeding in our work. Ongoing projects in Indonesia and Tanzania focus on addressing stunting by teaching communities, especially parents, about the importance of breastfeeding. We stress in particular exclusive breastfeeding early on in an infant’s life.
Of course, for women who cannot breastfeed other solutions that ensure adequate nutrition are needed. Our goal – and the goal of promoting breastfeeding – is to ensure every child has a shot at health, healing and well-being. Here’s how we’re making that happen.
More than one-third of Indonesians under 5 years are stunted; at least two standard deviations below median height for age. One of the key contributors is a lack of breastfeeding, especially in the first 6 months. As part of the National Nutrition Communications Campaign funded by the Millennium Challenge Corporation, we are working to change behavior affecting childhood nutrition through the use of media, public discussions, workshops and trainings. Breastfeeding is one of the key messages in our campaign, which targets young mothers, mothers-to-be and caregivers, and reinforces best practices. We also provide support groups and classes for mothers so that they can have open discussions and learn how to do things like breastfeed. This may seem like a simple lesson, but the reality is that mother’s have different breastfeeding challenges.
The 2010 Tanzania Demographic and Health Survey indicates breastfeeding rates are low and exclusive breastfeeding is far from the norm. Many women consider breastfeeding a rudimentary practice, instead opting for synthetic formulas. This puts babies at risk for stunting and other challenges.
The ASTUTE project, funded by the UK’s Department for International Development, focuses on extending the time mothers exclusively breastfeed and helps them to learn about foods for babies that are complementary to breast milk. One of the project’s goals is to support mothers and educate them on the importance of breast milk, especially for infants, and how to initiate breastfeeding early. Since breast milk is recommended by the World Health Organization and UNICEF as the sole source of nutrition for newborns during the first 6 months, guiding, educating, promoting and supporting this neglected practice is urgent.<|endoftext|>
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Fort Louis Delgrès Guadeloupe, France
This fortress was occupied from the seventeenth century onwards to defend the bay of Basse-Terre against attacks by the English.
In 1802, Fort Saint-Charles, as it was known in those days, played an important role in the conflict between Commander Louis Delgrès and General Antoine Richepance. The latter had been sent by Bonaparte to restore order in the colony and to re-establish slavery.
Led by Delgrès, the insurgents put up strong resistance at Basse-Terre, before being forced out of the fort on May 22 and onto the higher ground of Matouba. The fortress was evacuated through a postern gate built into the parapet wall which overlooks the Galion River. Among the insurgents was Joseph Ignace who led a group for Pointe-à-Pitre to create a diversion. They were killed during the battle of Baimbridge on May 25.
Another group, led by Delgrès, found refuge on Danglemont Plantation on the higher ground of Matouba, where they intended to put up resistance.
Surrounded by Richepance's troops, Delgrès and a number of his men opted to blow up the plantation rather than surrender.
Fort Louis Delgrès is part of the Slave Route—Traces of Memory network organized by the Conseil Général of Guadeloupe.<|endoftext|>
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Maths
Study Material
# Integration by Parts Formula | ILATE Rule | Examples |
Integration by parts is also called as ‘The Product Rule of Integration’. Here, we shall learn about the concept of integration by parts, its representation along with the formula with some illustrative examples of integration by parts.
4 minutes long
Posted by Kunduz Tutor, 11/10/2021
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The formula of integration by parts was proposed by Brook Taylor in 1715. It is a very useful technique to find the integration of the product of two (or more) functions. That’s why integration by parts is also called as ‘The Product Rule of Integration’. Sometimes, it is also referred by the name ‘Partial Integration’.
Moreover, here we shall learn about the concept of integration by parts, its representation along with the formula with some illustrative examples of integration by parts.
Author – Ojasvi Chaplot
## Concept
The product rule of integration states that it is used to find the integral of the product of various types of functions like inverse trigonometric, logarithmic, algebraic, trigonometric and exponential functions.
## Representation
In general, the standard form to represent the functions which are to be integrated is [f(x) and g(x)]. It can also be observed that at some places they are represented as [u and v].
## Integration by Parts Formula
The integration by parts formula is calculated as
Integration of (Function I x Function II) = [ (Function I) x (Integration of Function II) ] – Whole Integration of [ (Differentiation of Function I) x (Integration of Function II) ]
Thus, it can be seen that the formula of the product rule of integration is divided in two parts; the first term consists of the function f(x) and the integral of the function g(x) and the second term constitutes of the whole integration of differentiation of the function f(x) and the integral of the function g(x). Along with which, there exists an integration constant also.
## ILATE Rule
While working with the formula of integration by parts, it must be taken care of how to choose the Function I and Function II, i.e. f(x) and g(x) respectively.
To understand in which order, the function must be chosen as f(x) (or u) and as g(x) (or v), there exists a rule called as ‘ILATE’. At some places it is also written as ‘LIATE’.
Furthermore, in some exceptional cases, this rule may not be given priority; instead, one must choose it wisely that if that function’s integral exists and is easy to compute then it should be taken as g(x) [i.e. second function] and if it is such a function that it’s differential can be easily obtained then it should be taken as f(x) [i.e. first function].
## Special Case of Integration by Parts Formula
There are some special case formulas for the product rule of integration (integration by parts), which are derived from its standard formula. These can be used directly in between the question instead of showing the whole procedure of their derivation.
Some of the special case formulas are listed below –
## Illustrative examples using Formula of Integration by Parts
For Example : Find the integration of xex by using the integration by parts formula.
Solution:
As we know that, ∫f(x)g(x).dx = f(x)∫g(x).dx − ∫(f′(x)∫g(x).dx).dx + C
⇒ ∫xex.dx = ∫x.ex.dx
Further, using ILATE Rule,
⇒ f(x) = x and g(x) = ex
⇒ f’(x) = 1 and ∫g(x).dx = ∫ex.dx = ex.
Then, by using integration by parts,
⇒ ∫xex.dx = xex – ∫1.ex.dx
⇒ ∫xex.dx = xex – ∫ex.dx
⇒ ∫xex.dx = xex – ex
So, ∫xex.dx = ex(x – 1) + C
Therefore, Answer: ∫xex.dx = ex(x – 1) + C ; where C = Integration constant.
For Example : Find the integral of logx by using the integration by parts formula.
Solution:
As we know that, ∫f(x)g(x).dx = f(x)∫g(x).dx − ∫(f′(x)∫g(x).dx).dx + C
⇒ ∫logx.dx = ∫logx.1.dx
Hence, By using ILATE Rule,
Let, f(x) = logx and g(x) = 1
⇒ f’(x) = 1/x and ∫g(x).dx = ∫1.dx = x.
Then, by using integration by parts,
⇒ ∫logx.dx = logx.x – ∫(1/x).x.dx
⇒ ∫logx.dx = x.logx – ∫1.dx
So, ∫logx.dx = x.logx – x + C
Therefore, Answer: ∫logx.dx = x.logx – x + C ; where C = Integration constant.
## Picked For You
Furthermore, these are some topics which might interest you:<|endoftext|>
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Students explore molecular data from Homo sapiens and four related primates and develop multiple hypotheses regarding the ancestry of these five species by analyzing DNA sequences, protein sequences, and chromosomal maps.
1 to 4 class periods or one 3 hour lab session
This lesson illustrates many aspects of the process of science; however, in order to make these points apparent to students, they will need to be explicitly discussed during the class discussion periods.
Correspondence to the Next Generation Science Standards is indicated in parentheses after each relevant concept. See our conceptual framework for details.
- Scientists test their ideas using multiple lines of evidence. (P6, NOS2)
- Scientists often try to generate multiple explanations for what they observe. (P7)
- Different scientists may interpret the same data in different ways. (P7)
- Science depends on communication within the scientific community. (P7, P8)
- Scientists usually work collaboratively. (NOS7)<|endoftext|>
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## What are the properties of all angles?
Properties of Angles
• Sum of angles on one side of a straight line. The sum of all the angles on one side of a straight line is always 180 degrees. For example: The sum of ∠1, ∠2, and ∠3 is 180 degrees.
• Sum of angles around a point. The sum of all the angles around a point is always 360 degrees.
## What are f angles called?
Corresponding angles These are sometimes known as ‘F’ angles. The diagram below shows parallel lines being intersected by another line. The two angles marked in this diagram are called corresponding angles and are equal to each other. The two angles marked in each diagram below are called alternate angles or Z angles.
## What is the F rule in angles?
Angles of Parallel Lines
NAME RULE
Corresponding angles (F shape)
Co-interior angles Add to 180 degrees (U shape)
Alternate angles Equal (Z shape)
## What is true about alternate interior angles?
Alternate Interior Angle Theorem The Alternate Interior Angles Theorem states that, when two parallel lines are cut by a transversal , the resulting alternate interior angles are congruent .
## What are alternate interior angles examples?
The term alternate interior angles is often used when two lines are cut by a third line, a transversal . The Alternate Interior Angles Theorem states that if k and l are parallel , then the pairs of alternate interior angles are congruent . That is, ∠2≅∠8 and ∠3≅∠5 .
## What are the types of alternate angles?
There are two types of alternate angles:
• Alternate interior angles.
• Alternate exterior angles.
## What does alternate interior angles mean?
the angles which are inside the parallel lines and on alternate sides of the third line are called alternate interior angles. If two parallel lines are transected by a third line, the angles which are inside the parallel lines and on the same side of the third line are called opposite interior angles.
## How do you describe alternate angles?
When two lines are crossed by another line (called the Transversal): Alternate Interior Angles are a pair of angles on the inner side of each of those two lines but on opposite sides of the transversal.
## How much do alternate interior angles equal?
Properties. These angles are congruent. The Sum of the angles formed on the same side of the transversal which are inside the two parallel lines is always equal to 180°. In the case of non – parallel lines, alternate interior angles don’t have any specific properties.
## How do alternate angles work?
Alternate angles are equal. (c and f are also alternate). Alternate angles form a ‘Z’ shape and are sometimes called ‘Z angles’. Any two angles that add up to 180 degrees are known as supplementary angles.
## Do co interior angles always add to 180?
If the transversal cuts across parallel lines (the usual case) then the interior angles are supplementary (add to 180°). So in the figure above, as you move points A or B, the two interior angles shown always add to 180°.
## Do co interior angles add up to 360?
The exterior angles of a polygon always add up to 360°. Furthermore the interior and exterior angles at a point always add up to 180°.
## What is the difference between consecutive interior angles and co interior angles?
Answer. When the two lines being crossed are Parallel Lines the Consecutive Interior Angles add up to 180°. it is called co interior angle…..
## What is the relationship between co-interior angles?
As per the co-interior angles theorem, if a transversal intersects two parallel lines, each pair of co-interior angles is supplementary (their sum is 180°). Conversely, if a transversal intersects two lines such that a pair of co-interior angles are supplementary, then the two lines are parallel.
## What is the sum of Allied interior angles?
When the two lines being crossed are Parallel Lines the Consecutive Interior Angles add up to 180°.
## What is the other name of Allied angles?
Originally Answered: What is meant by an allied angle? When two parallel lines are cut by a transversal, the interior opposite angles on the same side of the transversal are called allied or co-interior angles.
## What do you mean by allied angles?
When a transversal intersects two parallel lines co-interior angles formed in the figure are called as allied angles ..
## Are corresponding angles equal?
Are all Corresponding Angles Equal? No, all corresponding angles are not equal. The corresponding angles which are formed when a transversal intersects two parallel lines are equal.
## How do you know if angles are corresponding?
Not all corresponding angles are equal. Corresponding angles are equal if the transversal intersects two parallel lines. If the transversal intersects non-parallel lines, the corresponding angles formed are not congruent and are not related in any way.
## What corresponding angles called?
: any pair of angles each of which is on the same side of one of two lines cut by a transversal and on the same side of the transversal.
## What are corresponding angles kid definition?
Corresponding angles – When two lines (usually parallel) are crossed by another (called the transversal) the angles in the same corners of each line are called corresponding angles.<|endoftext|>
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# How do you find the second derivative of ln(sqrtx)?
May 13, 2015
In order to derivate a $\ln$ we must remember its derivation rule: $\mathrm{dl} n f \left(x\right) = \frac{f ' \left(x\right)}{f} \left(x\right)$
Let's just rewrite $\sqrt{x}$ as ${x}^{\frac{1}{2}}$. Like this: $\ln \left({x}^{\frac{1}{2}}\right)$
Now, let's derivate it.
$f ' \left(x\right) = \left(\frac{1}{2}\right) . {x}^{- \frac{1}{2}}$
$f ' \left(x\right) = \left(\frac{1}{2 {x}^{\frac{1}{2}}}\right)$
Now we've found $f ' \left(x\right)$, let's solve the first derivative:
$\frac{\mathrm{dl} n \left({x}^{\frac{1}{2}}\right)}{\mathrm{dx}}$ = $\frac{\frac{1}{2 {x}^{\frac{1}{2}}}}{{x}^{\frac{1}{2}}}$
Simplifying it for exponential rules: $\mathrm{dl} n \left({x}^{\frac{1}{2}}\right) = \frac{1}{2 x}$
Deriving $\left(\frac{1}{2 x}\right)$, again, to obtain your original function's second derivative:
(d²ln(x^(1/2)))/(dx²) = (-1/(2x^(2))) (or, alternatively, $- \frac{1}{2} {x}^{- 2}$)
May 13, 2015
$\sqrt{x} = {x}^{\frac{1}{2}}$, so
$\ln \left(\sqrt{x}\right) = \ln \left({x}^{\frac{1}{2}}\right) = \frac{1}{2} \ln x$
So the derivative is $\frac{1}{2} \cdot \frac{1}{x} = \frac{1}{2} {x}^{- 1}$
The second derivative, then, is
$- \frac{1}{2} {x}^{-} 2 = - \frac{1}{2 {x}^{2}}$<|endoftext|>
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# Cancel Algebraic Fractions
In this worksheet, students will practise simplifying algebraic fractions by factorising.
Key stage: KS 4
Year: GCSE
GCSE Subjects: Maths
GCSE Boards: AQA, Eduqas, OCR, Pearson Edexcel,
Curriculum topic: Algebra
Curriculum subtopic: Notation, Vocabulary and Manipulation Algebraic Expressions
Popular topics: Factors worksheets
Difficulty level:
#### Worksheet Overview
Virtually every student knows how to simplify fractions by dividing by the same number, but do you know why this works?
Let's look at an example.
If you had to cancel 6/8, you would divide by 2 and get 3/4. Awesome and perfectly fine, but let's dig a bit deeper.
We can rewrite 6/8 like this:
6 8
=
3 x 2 4 x 2
We also know that when we are multiplying fractions, we multiply the denominators and then multiply the numerators, so we can also say that:
3 x 2 4 x 2
=
3 4
x
2 2
We should also be able to say that 2/2 = 1 and multiplying by 1 can be ignored, so now we can say that:
6 8
=
3 4
Why is this important?
You would never be asked to show all this working if you were cancelling down a fraction like 6/8 ,so why is it important?
It's because the technique shown which is called cancelling by factorisation, is the only way to cancel fractions that involve algebra.
How to cancel by factorisation
Whenever you see a fraction involving algebra, remember that you have to factorise it first.
Example 1:
Simplify the following fraction:
15x2 + 5x 30x + 10
If we look at each line individually:
15x2 + 5x can be factorised to 5x(3x + 1)
30x + 10 can be factorised to 10(3x + 1)
So we can rewrite our fraction as:
15x2 + 5x 30x + 10
=
5x(3x + 1) 10(3x + 1)
Anything that is the same on the top and bottom can now be cancelled.
5x (3x + 1) 210 (3x + 1)
=
x 2
Example 2:
Simplify the following fraction:
2x2- 2x - 3 3x2- 5x -12
If we look at each line individually:
2x2- 2x - 3 can be factorised to (2x + 1)(x - 3)
3x2- 5x -12 can be factorised to (3x + 4)(x - 3)
So we can rewrite our fraction as:
2x2- 2x - 3 3x2- 5x -12
=
(2x + 1)(x - 3) (3x + 4)(x - 3)
Anything that is the same on the top and bottom can now be cancelled.
(2x + 1)(x - 3) (3x + 4)(x - 3)
=
2x + 1 3x + 4
Let's try some questions now.
### What is EdPlace?
We're your National Curriculum aligned online education content provider helping each child succeed in English, maths and science from year 1 to GCSE. With an EdPlace account you’ll be able to track and measure progress, helping each child achieve their best. We build confidence and attainment by personalising each child’s learning at a level that suits them.
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#### Popular Maths topics
• National Tutoring Awards 2023 Shortlisted
• Winner - Private Tutoring
• Finalist
• Winner - Best for Home Learning / Parents<|endoftext|>
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# Find whether the following equations have real roots. If real roots exist, find them.$x^{2}+5 \sqrt{5} x-70=0$
Given:
Given quadratic equation is $x^{2}+5 \sqrt{5} x-70=0$.
To do:
We have to determine whether the given quadratic equation has real roots.
Solution:
Comparing the given quadratic equation with the standard form of the quadratic equation $ax^2+bx+c=0$, we get,
$a=1, b=5 \sqrt{5}$ and $c=-70$.
The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is
$D=b^2-4ac$.
Therefore,
$D=(5 \sqrt{5})^2-4(1)(-70)$
$=125+280$
$=405$.
As $D>0$, the given quadratic equation has two distinct real roots.
This implies,
$x=\frac{-b\pm \sqrt{D}}{2a}$
$x=\frac{-5 \sqrt{5} \pm \sqrt{405}}{2(1)}$
$x=\frac{-5 \sqrt{5}\pm 9\sqrt5}{2}$
$x=\frac{-5 \sqrt{5}+9 \sqrt{5}}{2}$ or $x= \frac{-5 \sqrt{5}-9 \sqrt{5}}{2}$
$x=\frac{4\sqrt{5}}{2}$ or $x=\frac{-14 \sqrt{5}}{2}$
$x=2\sqrt5$ or $x=-7\sqrt5$
The roots of the given quadratic equation are $-7\sqrt5$ and $2\sqrt5$.
Tutorialspoint
Simply Easy Learning<|endoftext|>
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<ii(r)>
{$+6=xx$}{$+\delta =xx$}
<ii(v)>
$\begin{array}{cccccccccccccccccccccc}& & & & & & & & & & 1& .& 5& .& 10& .& \phantom{10}& .& 5& .& 1& .\\ & & & & & & & & 1& .& 6& .& 15& .& 20& .& 15& .& 6& .& 1& .\\ & & & & & & 1& .& 7& .& 21& .& 35& .& 35& .& 21& .& 7& .& 1& \\ & & & & 1& .& 8& .& 28& .& 56& .& 70& .& 56& .& 28& .& 8& .& 1& .\\ & & 1& .& 9& .& 36& .& 84& .& 126& .& 126& .& 84& .& 36& .& 9& .& 1\\ 1& .& 10\end{array}$
$\sqrt{aa}×${$\sqrt{3:{a}^{3}}$ } = {illeg} $\sqrt{ab}×\sqrt{3:cde}=\sqrt{6:{a}^{3}{b}^{3}}×\sqrt{6:ccddee}=\sqrt{6:{a}^{3}{b}^{3}ccddee}$. to find ye proportion of two irrationall rootes. to free ye Numerato{r} or denom from {illeg}surde q{illeg} $\frac{a}{b+\sqrt{c}}=x=\frac{a}{b+\sqrt{c}}×\frac{b-\sqrt{c}}{b-\sqrt{c}}=\frac{ab-a\sqrt{c}}{bb-c}$.
<iii(r)>
$\begin{array}{c}\begin{array}{ccc}\frac{2ax}{c}& -& \frac{\sqrt{{a}^{3}}}{c}\\ 3a& +& \frac{\sqrt{abb}}{c}\end{array}\\ \begin{array}{ccc}6aax& -& 3a\sqrt{{a}^{3}}\\ & +& \frac{2ax}{c}\sqrt{\frac{abb}{cc}}& -& \frac{\sqrt{{a}^{4}bb}}{cc}\end{array}\\ \begin{array}{c}6aax-3a\sqrt{{a}^{3}}+2ax\sqrt{abb}-\frac{aab}{c}\\ \text{error tantum 1′ in metallo 2″ in radio}\end{array}\end{array}$ $\begin{array}{l}\begin{array}{cccccc}\frac{1}{10}& \begin{array}{cc}\frac{1}{60}\text{.}& \frac{1}{100}×\frac{1}{12}\end{array}& 30×60×60& \begin{array}{c}1\\ 60,30,30\\ 54000\end{array}& \begin{array}{c}1\\ 60,30,60\\ 108000\end{array}& 1c\\ & \begin{array}{c}1\\ \begin{array}{c}60000\\ 12\end{array}\end{array}& \begin{array}{c}1200\left(40\\ c\end{array}\end{array}\\ \phantom{0}\\ \phantom{0}\\ \begin{array}{l}300,12,10\\ 3000,12\hfill \\ 36000\hfill \end{array}\\ \phantom{0}\\ \phantom{0}\\ \begin{array}{lll}\frac{1}{60,60,60.}& =& \frac{1}{300,720}\end{array}\end{array}$ $\sqrt{9{a}^{5}}$$a$ $3a\sqrt{a}$ $\begin{array}{c}0,051513\\ 0,0132\hfill \end{array}$ $\left(5$ {illeg} $baax-\sqrt{\left(9\right){a}^{5}}$ $\frac{c}{d}\right)\frac{a}{b}\left(\frac{da}{cb}$ $\begin{array}{ccccc}\begin{array}{c}-2\\ -2\end{array}& \phantom{\rule{0.5em}{0ex}}& \begin{array}{c}\frac{3}{8}×9\\ \end{array}& \phantom{\rule{0.5em}{0ex}}& \begin{array}{l}\phantom{0}14\\ \phantom{0}56\\ 14\end{array}\end{array}$ $\begin{array}{lll}8.\frac{1}{2}\colon\colon \frac{1}{2}.\frac{1}{32}=\frac{6}{16}=\frac{3}{8}.& \phantom{\rule{0.5em}{0ex}}& \begin{array}{lll}9& .& 196\\ 9& \right)& 392& \left(& 43\frac{5}{9}\end{array}\\ 2.\frac{1}{4}\colon\colon \frac{1}{4}.\frac{1}{32}.\\ 96.7\colon\colon 7.\frac{49}{96}\end{array}$ $\begin{array}{l}1\phantom{.}\frac{c}{d}\colon\colon \frac{a}{b}.\frac{da}{cb}\\ \frac{c}{d}.1\colon\colon \frac{a}{b}.\frac{da}{cb}\\ c.d\colon\colon \end{array}$ $1.$
$\frac{c}{d}.1\colon\colon \frac{c}{d}ind.1×d\colon\colon c.d\colon\colon \frac{c}{c}.\frac{d}{c}\colon\colon 1.\frac{d}{c}$ $\frac{c}{d}\phantom{.}1\colon\colon c.d\colon\colon 1.\frac{d}{c}\colon\colon \frac{a}{b}.\frac{da}{cb}$. $\frac{c}{d}.1\colon\colon \left(\frac{c}{d}ind\right)c.\left(1×d\right)d\colon\colon \left(\frac{c}{c}\right)1.\frac{d}{c}\colon\colon \frac{a}{b}:\left(\frac{d}{c}×\frac{a}{b}\right)\frac{da}{cb}$
$\frac{a+3x}{c}\sqrt{\frac{4bb{x}^{4}}{81aa}}$ $\frac{2bxx}{9a}$ $\frac{2abxx+6b{x}^{3}}{9ca}$ $aa-2ay=ab-bb+by$ $aa=ab-bb+by+2ay$ $aa-ab+bb=by-2ay$ $\frac{aa-ab+bb}{b-2a}=y$
$\begin{array}{c}z=\frac{yy}{x}\\ 20=a\\ 140=b\end{array}$ $zz$ $x.y.\frac{yy}{x}$. $yy=\frac{yyx}{x}$ $x+y+\frac{yy}{x}-a=0$ $xx+yy+\frac{{y}^{4}}{xx}-b=0$ $x+y+\frac{yy}{x}-a+xx+yy+\frac{{y}^{4}}{xx}-b=0$
<1r>
To find a heavy bodys de{illeg}|s|cent in any given time, & ye proportion of ye pressure of ye rays {illeg}|b|y gravity to |ye| force by wch a{illeg} given body hath \any/ given motion; by this figure If ye cilinders bc, df bee of glasse &c: to {kow} ye proportion of their strength is knowne by ye proportion of the{illeg} gravity of ye circles a, e &c {illeg}|i|n respect of ye axis in.
If a Staffe bee bended to find ye crooked line wch it resembles.
If the motion of a line is knowne to find ye crooked line wch yt line toucheth continually.
If a stick ab revolve {sic} \wth even velocity/ about ye center a haveing ye weight c fastened {in} it by {the} string bc , yn shall ye string bc bee a tangent to ye circle bde.
But i{illeg}|t| may be inquired what line ye weight $\left(c\right)$ would describe were ye stick wth uneven velocity, {illeg} or did ye point b describe a Parabola or some other crooked line were ye weight c in some other place as at ye center { a } when ye stick began to move.
If ye ball b revolves about ye center n ye force by wch it endeavours from ye center n would {beget} soe much { b } motion in a body \as ther {sic} is/ in ye time yt ye body b moves ye length {of ye} {semidiamiter {sic}} bn . [as if b is moved \wth one degree of {motion}/ through {illeg} \{illeg} {illeg}/e \ bn / in {a} seacond of an {hower} & bn is {os{illeg} {illeg}d{illeg}} yn its force from ye center {illeg} {illeg} being continually like ye force of gravity impressed upon {illeg}|ye| body during one second it will generate one degree of motion in {illeg}|yt| body.] Or ye force from {n} in one revolution is to ye force of ye body motion as rad$periph:rad$ . Demonstracon. If {$ab=bc=dc=$ } $ef=fg=gh$ \ $ad\text{.}$ / $=he=2fa=2fb=2fb=2gc=2ed$. & ye globe {b} from a to b {yn} $2fa:${ak}$\colon\colon ab:fa\colon\colon$ force {or} pression of b {illeg}|u|pon fg {at} its reflecting ∶ for{illeg}|e| {sic} of b{'}s motion. therefore $4ab=ab+bc+cd+da:$ {illeg}fa ∷ force of ye reflection in one round (viz: in b , c , d , & a) ∶ force of b 's motion. by ye sa{me} pro{illeg} ye Globe b were reflected by each side of a circumscribed polygon of 6, 8, 12, 100, 1000 sides {illeg} ye force of all ye reflections is to ye force of ye bodys as ye sum of those sides {illeg} r{adius} of ye circle about wch they are circumscribed. {illeg} is if {illeg} And so if body were reflected by {illeg} the sides of an equilaterall {illeg} circum{}|s|cribed polygon of an infinite number of sides {illeg} by {illeg} {illeg} circle it selfe) ye force of all ye reflections are to ye force of ye bodys motion {as each} those sides ({illeg} ye perimiter) to ye radius.
If ye body b moved in an Ellipsis ye {sic} its force in each point (if its motion {in yt point bee} g{illeg}) bee found by a tangent circle of equal crookednesse with ye|t| {illeg}|p|oint of ye Ellipsis.
If a body undulate in ye circle bd all its undulations of any altitude are performed in ye {same} time wth ye \same/ radius. Galileus.
As radi{illeg}|u|s ab to radius ac ∷ {illeg} so are ye Squares of theire times in wch they undulate.
If c circulate in ye circle { cgef } , to whose diamiter \{ ce ,}/ $ad=ab$ being perpendicular yn will ye body b undulate in ye same time yt c circulate.
And ye {illeg} those body circulate in ye same time whose {illeg} {illeg} from ye {illeg} to ye center d are equall
And $ad:dc\colon\colon$ force of gravity to ye force of { c } to its center d . {illeg} {illeg} {illeg} {illeg} ye motion of things falling were they not hindered by ye {illeg} may very {illeg} {illeg} $cd:ad\colon\colon$ force form d ∶ force from a.
<1v>
[1] $ag=x$. $gh=y$. $ah=c$. $\begin{array}{cccccccc}bxx& +& ex& +& cc& =& 0& \text{.}\\ & +& dxy& +& fy\\ & & & +& gyy\end{array}$. $dp=v${illeg}. $ad=\frac{xx-yy+cc}{2c}$ [2] ${dg}^{2}=${illeg}$\frac{2cc{x}^{2}+2xxyy+2ccyy-c-{x}^{4}-{y}^{4}}{4aa}+vv=ss$ $v:w\colon\colon \frac{vy}{x}:\frac{wy}{x}$. $\frac{vy}{x}=df$. /fig 2d. $\sqrt{xx-yy}-v:y\colon\colon y$. $ds=\frac{yy}{\sqrt{xx-yy}-v}$\ $\begin{array}{llllllllllllll}& & -& 4{c}^{4}& -& {x}^{4}& -& 2b{x}^{4}& -& 2e{x}^{3}& -& 2d{x}^{3}& in& y\\ & & & & & & & & & & -& 2fxx\\ & & & & -& 2eccx& -& bb{x}^{4}& & & -& 2ccdx\\ -& 2bccxx& & & -& 2eb{x}^{3}& -& 2bcc{x}^{2}& & & -& 2ccf\\ -& 2cceex& & & & & -& eexx\end{array}$ $yy:+v\sqrt{xx-yy}$ $yy-x:y$ $\frac{{y}^{3}}{x\sqrt{xx-yy}-vx}$ + {illeg}$\frac{+y\sqrt{xx-yy}}{x}=df$. $\frac{{y}^{3}-{y}^{3}+yxx-vy\sqrt{xx-yy}}{-vx+x\sqrt{xx-yy}}$. [3] {illeg} $\frac{xx-v\sqrt{xx-yy}}{\sqrt{xx-yy}-v}=bd$. $xx-v\sqrt{xx-yy}:v\sqrt{xx-yy}\colon\colon$a {illeg}{illeg}f $ab=x$. Parab $\angle mak=\angle bad$ , ergo: ad $\angle mak=\angle ade=\angle bad$. ergo, $ab=bd$.
[4] $\angle a=\angle bad=\angle adb-\angle asb=\angle sad$ ergo tri: anb, anm, sim: $ab=x$. $bs=a$. $eb=y$. $ae=\sqrt{xx-yy}$. $bo=v$ $ao=\sqrt{xx-2vy+vv}$. $ab:as\colon\colon bo:os$. $x:\sqrt{xx-2ay+aa}\colon\colon v:a-v$. $\angle vat=${illeg}||$bao=\angle oas$. $ab=x$. $as=y$. $bo=v$. $bs=a$. $x:y\colon\colon v:a-v$. $vy=ax-vx$ Hyperb.
[5] $os=v$. $sa=x$. $oi=ae=y$. $sa:ae\colon\colon so:oh=\frac{vy}{x}$. $oi:oh\colon\colon d:e$: ergo $ey=\frac{dvy}{x}$. & $ex=dv$. Hyp: $ab:ae\colon\colon bc:cn$. $\frac{ae×v}{x}=cn$. $d:e:\frac{ae×v}{x}.ae$. $dx=ev$.
[6] $ab=y$. $as=x$. $bc=v$. $\angle bac=\angle cas$, Ergo $y:x\colon\colon v:a-v$. $bs=a$. & $xv=ay-vy$. Ellipsis.
[7] $ab=x$. $as=y$. $\angle bac=tar=cas$. Ergo, $ba:as\colon\colon bc=v:cs=a-v$ $ay-yv=$ $ax-vx=vy$. Hyperb.
[8]
The invention of Figures for refra|le|ctions. \at right angles/ at ye point refra|le|cting ac ye rad: reflected to ye focus b. ag ye radius reflected {illeg}|fr|o ye focus b. aq a perpendic: to ye ed ye tangent of ye crood|k|ed line shought. $ab=x$. \$ac=y$, or,/ $bd=w$. $ag=y$. $bg=a$, or, $bc=a$. $bd=v$, or $bq=a$. fig: 1st[9]. $\angle eac=\angle bad=\angle adb$. Ergo, $ab=bd$, or $x=v$. & $\angle caq=qab=aqb$. e{illeg}|rg|o $ab=${illeg}qg. $x=v$. fig: 2d[10]. $\angle eac=\angle bad=\angle adg$. Ergo, $ab:ag\colon\colon bd:dg$ $ax-vx=vy$. &, $ab:$ {illeg}$:ag\colon\colon qb:bg$. {illeg} Ergo $ax+vx=y$y|v|. $v=bq$. fig 3d[11]. $\angle eac=\angle bad$ {illeg} Ergo $\angle caq=\angle qab$ . Ergo, $ca:ab\colon\colon cq:qb$. & $ax-vx=vy$.
The invention of figures for refraction. b , & g ye foci. ca ye Rad: refracted to b . qa ye Rad: refracted from b. bg ye distance of ye foci. qa ye perpend: to de ye tangent of ye crood|k|ed line sought qr, $qh=$ perpendic:s to ye Radiusi cg, fb.
$bg=a$. $bq=v$. $ba=x$. $ag=y$. fig: 1st[12]. $ab:as\colon\colon bq:qr=\frac{vy}{x}$. $d:e\colon\colon qr:qh$, Ergo. $dx=ev$. fig: 2d[13]. $ab:as\colon\colon bq:qr=\frac{vy}{x}$. $d:e$∶ {sic}$qh:qr$, Ergo. {illeg}|e|$x=dv$. fig 3d[14]. $ab:as\colon\colon bq:qr=\frac{v\phantom{\rule{0.5em}{0ex}}in\phantom{\rule{0.5em}{0ex}}as}{x}$. $d:e$∶ {sic}$qr:$ $ag:as\colon\colon gq:qh=\frac{a+v\phantom{\rule{0.5em}{0ex}}in\phantom{\rule{0.5em}{0ex}}as}{y}$. $d:e\colon\colon$$qh:qr$ /$qr:qh$\: Ergo $eax+evx=dvy$ $dax+dvx=evy$. fig {illeg} 4th[15]. $ab:as\colon\colon bq:qr=\frac{v\phantom{\rule{0.5em}{0ex}}in\phantom{\rule{0.5em}{0ex}}as}{x}$. $ag:as\colon\colon gq:qh=\frac{v-a\phantom{\rule{0.5em}{0ex}}in\phantom{\rule{0.5em}{0ex}}as}{y}$. $d:e\colon\colon qr:qh$: $dvx-dax=evy$. fig 5t[16]. $ab:as\colon\colon bq:qr=\frac{v\phantom{\rule{0.5em}{0ex}}in\phantom{\rule{0.5em}{0ex}}as}{x}$. $ag:as\colon\colon gq:qh=\frac{a-v\phantom{\rule{0.5em}{0ex}}in\phantom{\rule{0.5em}{0ex}}as}{y}$. $d:e\colon\colon qh:qr$. $eax-evx=dvy$. fig 6t. $ab:as\colon\colon bq:qr=\frac{v\phantom{\rule{0.5em}{0ex}}in\phantom{\rule{0.5em}{0ex}}as}{x}$. $ag:as\colon\colon gq:qh=\frac{a-v\phantom{\rule{0.5em}{0ex}}in\phantom{\rule{0.5em}{0ex}}as}{y}$. $d:e\colon\colon qr:hq$. e | d |$ax-d${illeg}| v |$x=evy$.
<2r>
$ad=a$. $ae=x$. $ed=y$. $af=z$. $fd=a-z$.
$xx-zz=yy-zz+2az-aa$. $fg=v$ \2/ $ae:ef\colon\colon ag:gi$. & $de:ef\colon\colon dg:gh$. [17] $\frac{xx-yy+aa}{2a}=af$. $\frac{yy-xx+aa}{2a}=fd$. x {illeg} $:\frac{\sqrt{2aaxx-{x}^{4}+2xxyy-{y}^{4}+2aayy-{a}^{4}}}{4aa}\colon\colon \frac{xx-yy+aa+2av}{2a}:gi$. gi \{illeg}/ $=\frac{\sqrt{2aaxx+2xxyy+2aayy-{x}^{4}-{y}^{4}-{a}^{4}}\phantom{\rule{1em}{0ex}}in\phantom{\rule{1em}{0ex}}\sqrt{+2aaxx-2aayy-2xxyy+{x}^{4}+{y}^{4}+{a}^{4}+2xxav-2yyav+2{a}^{3}v+}}{4aax}$ $gh=\sqrt{2aaxx+2xxyy+2aayy-{x}^{4}-{y}^{4}-{a}^{4}}\phantom{\rule{1em}{0ex}}in\phantom{\rule{1em}{0ex}}\sqrt{\phantom{0000000000000000000}}$ $ef=${illeg}|m|. $ae=x:m\colon\colon \frac{xx-yy+aa+2av}{2a}$|=|$ag:gi=\frac{mxx-myy+maa+2mav}{2ax}$. $ed:m\colon\colon gd:\frac{myy-mxx+maa-2mav}{2ay}=gh$. $gi:gh\colon\colon d:e$: therefore $\frac{dyy-dxx+daa-2dav}{y}=\frac{exx-eyy+eaa-2eav}{x}$. {illeg} $d=2$. e {illeg}$=1$ . $\begin{array}{cccccccccccc}{y}^{3}& +& 2xyy& -& xxy& +& 2aax& \phantom{\rule{1em}{0ex}}& =& \phantom{\rule{1em}{0ex}}& 0& \\ & & & -& aay& -& 2{x}^{3}\hfill \\ & & & -& 2avy& -& 4axv\hfill \end{array}$ $\frac{{y}^{3}+2xyy-xxy-aay+2aax-2{x}^{3}}{2ay+4ax}=v=\frac{xx-yy}{2a}+\frac{2ax-ay}{2y+4x}$ $bf=\frac{2aax+2aay+2xxyy-{x}^{4}-{y}^{4}-{a}^{4}\phantom{\rule{1em}{0ex}}in\phantom{\rule{1em}{0ex}}y+2x}{{y}^{3}+2xyy-xyy-aay+2aax-2{x}^{3}\phantom{\rule{1em}{0ex}}in\phantom{\rule{1em}{0ex}}2a}$. $af=v$ $ae:$ {ef}$\colon\colon ag:gi$. $de:ef\colon\colon dg:gh$. $d:e\colon\colon gi:gh$. $x:m\colon\colon v:\frac{mv}{x}$. $y:m\colon\colon a-v:\frac{ma-mv}{y}$. $d:e\colon\colon \frac{mv}{x}:\frac{ma-mv}{y}$: & $adx-dvx=evy$. $af=x$. $fe=y$. $ae=\sqrt{xx+yy}$. $ed=\sqrt{xx-2ax+aa+yy}$. $ad\sqrt{xx+yy}-dv\sqrt{xx+yy}=ev\sqrt{xx-2ax+aa+yy}$ $aaddxx+aaddyy+2addvxx+2addvyy+ddvvxx+ddvvyy=eevvxx-2eevvax+eevvaa+eevvyy$. $ef=o$. $\begin{array}{cc}\begin{array}{cc}bc& =\end{array}& \begin{array}{c}\begin{array}{c}2aaxxy-{x}^{4}y-{a}^{4}y+2aa{y}^{3}+2xx{y}^{3}-{y}^{5}+4{a}^{2}{x}^{3}+4aayyx+4yy{x}^{3}-2{y}^{4}x-2{a}^{4}x-2{y}^{4}x-{x}^{5}\\ -2ao{y}^{3}-4aoxyy+2aoxxy+2{a}^{3}oy-4{a}^{3}ox+4ao{x}^{3}\hfill \end{array}\\ \phantom{+}2a{y}^{3}+4axyy-2axxy-2{a}^{3}y+4{a}^{3}x-4a{x}^{3}\end{array}\end{array}$ [18] $an=on$ $pg=a$. $ab=${illeg}|y.| $ag=${illeg}|x.| $pg=b$. $pm=c$. $gb=z$. $gq=v$. $ezx-evx=dvy$. $gm=p$. $gs=$ [19] $ab=x$. $qb=v$ {illeg}. $v=x$. $mp=a$. {illeg} $aq=\sqrt{vv-xx}$. $\frac{xx}{v}=bs$. $sa=\sqrt{\frac{vvxx-{x}^{4}}{vv}}$. $nb=y$. $mn=\frac{y}{v}\sqrt{vv-xx}$ $nr=nb$. $e:d\colon\colon tb=nm:rs=\frac{dy\sqrt{vv-xx}}{ev}$. $rs:nr\colon\colon mn:np=\frac{\frac{yy}{v}\sqrt{vv-xx}}{\frac{dy}{ev}\sqrt{vv-xx}}=\frac{ey}{d}\sqrt{yyvv-yyxx+aavv}$ $eeyyvv-ddyyvv+ddyyxx=ddaavv$. ${mn}^{2}=\frac{ddaavv-ddaaxx}{eezz-ddzz+ddxx}$. ${np}^{2}=\frac{eeaavv}{eezz-ddzz+ddxx}$. ${mb}^{2}=\frac{ddaavv}{eezz-ddzz+ddxx}$ {illeg} $=zz$. $\frac{ddaaxx-ddxxzz}{eezz-ddzz}=vv$. $v=\frac{dx\sqrt{aa-zz}}{z\sqrt{ee-dd}}$. ${sa}^{2}$ ${bs}^{2}=\frac{eezzxx-ddzzxx}{ddaa-ddzz}$. ${as}^{2}=\frac{ddaaxx-eezzxx}{ddaa-ddzz}$. $sb:ab\colon\colon ab:$ \$2$/ {bq}{bp}{illeg}sb 2 $xx=$ \$2$/ {illeg}sb{illeg}b{illeg}sb. $sb=\sqrt{}$. $sb=n$: $xx-nn={sa}^{2}=x$/+\$n×x-n$. $xx${illeg}$x+3n${illeg} $xx-nn=nn${illeg}x $ab=x=bq=v$ . $as=y$. $nb=z$ $nm=\frac{yz}{x}$. $e:d\colon\colon${illeg}$z:\frac{dz}{e}=np$. {illeg} ${e}^{2}yyzz+aaxx{e}^{2}=ddzzxx$ ${nb}^{2}=\frac{eeaaxx}{ddxx-eeyy}$. ${nm}^{2}=\frac{eeaayy}{ddxx-eeyy}$. ${mb}^{2}=\frac{eeaaxx-eeaayy}{ddxx-eeyy}$. {illeg} ${np}^{2}=\frac{aaddxx}{ddxx-eeyy}=${illeg}${\xi }^{2}$. $xx${illeg}$yy${illeg}$xx${illeg}$yy$ ${as}^{2}=\frac{ddzzxx-eeaaxx}{eezz}=\frac{-aaddxx+ddxx\xi \xi }{ee\xi \xi }$. $dd${illeg}$\xi \xi$$-eeaa\xi \xi +aaddzz=0$. [20] $ab=$ [21] $ab=bq=x$. $pm=a$. $as=y$. $no=z$: $\frac{zx}{y}=nb$ . $om=\frac{zz}{a}=mt$. $po=\frac{aa-zz}{a}$. ${pn}^{2}=\frac{{a}^{4}-aazz+{z}^{4}}{aa}:\frac{zzxx}{yy}\colon\colon${pt{illeg}} ${pt}^{2}=\frac{{a}^{4}+2aazz+{z}^{4}}{aa}$ ${tb}^{2}${}${z}^{6}xx$ $b$$xx=cyy$. $xx-\frac{bxx}{c}={sb}^{2}$. $x\sqrt{\frac{c-b}{c}}-\sqrt{\frac{eeaaxx\phantom{\rule{0.5em}{0ex}}-\phantom{\rule{0.5em}{0ex}}\frac{eeaabxx}{c}}{ddxx\phantom{\rule{0.5em}{0ex}}-\phantom{\rule{0.5em}{0ex}}\frac{eebxx}{c}}}=\sqrt{}$ $xx$|$yy$|$=bxx+c${illeg}$x+$ {illeg}g. ${sb}^{2}=xx-bxx${illeg}$-cx-g$. ${mb}^{2}=\frac{eeaaxx-eeaabxx-eeaacx-eeaag}{ddxx-eebxx-eecx-eeg}$ . $qa=s$. $qm=v$. $vv+xx-bxx-cx-g$. $+\frac{eeaaxx-eeaab{x}^{2}-eeaacx-eeag}{ddxx-eebxx-eecx-eeg}$: $-2\sqrt{xx-bxx-cx-g}in\sqrt{\frac{c\phantom{?}aaxx-{c}^{2}{a}^{2}b{x}^{2}-{e}^{2}{a}^{2}\left\{illeg\right\}-\left\{illeg\right\}}{{d}^{2}{x}^{2}-{e}^{2}b{x}^{2}-{e}^{2}cx-e\left\{illeg\right\}}}${illeg} $-2v\sqrt{xx-bxx-cx-g}+2v\sqrt{\frac{{e}^{2}{a}^{2}{x}^{2}-{e}^{2}{a}^{2}b{x}^{2}-{e}^{2}{a}^{2}cx-{e}^{2}{a}^{2}g}{ddxx-eeb{x}^{2}-eecx-eeg}}=ss=vv+xx+$
$=2ox+2ee$ $mb=z$. $sb=x$. $sa=y$. ${ab}^{2}=xx+yy$. $zz=\frac{eeaaxx}{ddxx+ddyy-ee}$ $sm=${illeg}|x| {illeg} $sa=y$. $mb=z$. ${sb}^{2}=xx+2zx+zz$. ${x}^{2}=byy+cy+d$. $yy${illeg}$=bxx+cx+g$. $sm=x-\sqrt{\frac{eeaaxx}{ddxx+dd-ee\phantom{\rule{1em}{0ex}}in\phantom{\rule{1em}{0ex}}bxx+cx+y}}$ $vv-2vx+2v\sqrt{\frac{eeaaxx}{ddxx+dd-ee\phantom{\rule{1em}{0ex}}in\phantom{\rule{1em}{0ex}}bxx+cx+g}}${+}$xx\frac{+eeaaxx}{ddxx+dd-ee\phantom{\rule{1em}{0ex}}in\phantom{\rule{1em}{0ex}}bx+cx+g}-2x\sqrt{\frac{eeaaxx}{\phantom{????}+dd\phantom{\rule{1em}{0ex}}in\phantom{\rule{1em}{0ex}}bbxx+}}$ $=ss=-2vo\phantom{\rule{1em}{0ex}}\begin{array}{c}\hfill +2v\\ -2x\end{array}\sqrt{\frac{eeaaxx-2eeaaox}{ddxx+2ddox+dd-ee\phantom{\rule{1em}{0ex}}in\phantom{\rule{1em}{0ex}}bxx+2box+cx+co+g}}+${illeg} $vv$. $mb=z$. $sb=x$: $sa=y$. {$mn=$}$\frac{yz}{x}$. ${ab}^{2}=xx+yy$. $\frac{zzxx+zzyy}{xx}=${${sb}^{2}$ }{illeg} $eezzxx+eezzyy-ddaaxx-ddyyzz=0$. ${x}^{2}=bzz+${illeg}. $yy=\frac{e\phantom{\text{?????}}}{dd}${illeg} $vv$$-2v\sqrt{bzz+cz+g}$$+bzz+cz+g$$+\frac{eezz-ddaa\phantom{\rule{1em}{0ex}}in\phantom{\rule{1em}{0ex}}bzx}{ddzz-eezz}${illeg} {$4vvbin$}$b{z}^{2}+cz+g+\frac{{z}^{4}-4aa{z}^{2}+4{a}^{4}\phantom{\rule{1em}{0ex}}in\phantom{\rule{1em}{0ex}}Q:\phantom{\rule{0.5em}{0ex}}bzz+cz+g}{{z}^{4}}${illeg} $\frac{-4{z}^{4}v\sqrt{bzz+cz+g}\phantom{\rule{1em}{0ex}}in\phantom{\rule{1em}{0ex}}zz-2aa\phantom{\rule{1em}{0ex}}in\phantom{\rule{1em}{0ex}}bzz+cz+g}{{z}^{4}}${illeg} 4{illeg} $+16o{a}^{4}in${illeg}$+gg${illeg} $-16ov{z}^{2}\sqrt{bzz+cz+g}inzz-2${illeg}$inbz$
<2v>
[22] [23] Make ye line ac to revolve about ye point a : on ye end c let ye nut c bee fastened so {as} to t{illeg}|u|rne about its center. make $ab=ac$ & fastend another nut at ye point b in ye same manner. {illeg} make ye line bc to slide through those two nuts soe yt ye △ abc will always be an isosceles. To ye line cb fasten ye line rstv at {illeg} right angles. {illeg} make ye line kg wth 2 nuts e & d at each end through wch ye lines rs & tu must slide to keepe ye line kg perpendicular to bc , in {illeg}|ye| midst of kg fasten ye nutt m so as it {illeg} may turne about its center & yt ye line ac may slit|d|e thro{illeg}|ugh| it then make yt side of ye line kg wch is next ab to be a file wch must be very smooth at the point m but must grow rougher towards ye ends d & e . Then by turneing ye line ac \to & from l & h / about its center & holding ye file kg close to ye plate hmflab , it shall fil{illeg}|e| it into ye shape of a Parabola.
To describe ye Parabola by points. {illeg}|Ma|ke $ca=\frac{r}{4}$; c ye {illeg}|v|ertex; {illeg}|a | ye focus; $ab=r$. yn wth some radius as $ag=ae$, describe ye circle ge : & take $bd=2ga=2${illeg}$ae=$\$2$/de & ye point e shall bee in ye parabola, also if from e to g , a \streight/ line be drawne it shall touch ye Parab: in e.
Or thus, take $ch=ca=\frac{r}{4}$, $hd=2cg$ \or $da=2hg$:/; & $ga=ae=de$. &c:
Or thus, take $cm=gc$, $dm=ma$; & $ga=ae=de$; &c.
Or thus take $cm=gc$ & raise me a perpendicular to ca, wch shall intersect ye parab, & circle ge in ye same point.
Or thus. {illeg}|U|pon ye focus or center a describe ye center {ef} make $ab=\frac{r}{4}$. $bd=2bc=$ {illeg}|r|. $kb=bg$. wth ye Rad bc describe bed . ye circle.
[24] Or thus take \$\frac{r}{2}=$/$ac=cn=Rad$. Circle aen: $ab=\frac{r}{4}$. $am=ap$ & produce mp indefinitely. Then take some point ad in ye line an , & draw dg perpendic: to an yt is soe yt $dm=dg$, yn take $df=ae$, & f shall be a point in ye parabola afr.
<3r>
Banderon's addition to Ferrarius's Lexion \Geographicu/, ye best for Geog. {T}{F}{illeg} \Ortelius/ Geogr. Lexicon. $\begin{array}{c}{\text{M}}^{\text{r}}\phantom{\rule{1em}{0ex}}\text{John Craige}\\ {\text{D}}^{\text{r}}\phantom{\rule{1em}{0ex}}\text{Archibald Pitcarne}\end{array}\phantom{\rule{1em}{0ex}}\right\}\phantom{\rule{1em}{0ex}}\text{Scotch Mathematicians}$
## Experiments about \the resistance of/ things falling in water.
1. I filled to ye top a {illeg}|woo|den vessel {illeg} 9 inches squa{illeg}|r|e within & 9 foot $4\frac{1}{2}$ inche{illeg}|s| high \within/. And making balls of bees wax of several bignesses & wth pieces of Lead stuck in them to give them weight: three balls each of wch weighed in the air $76\frac{1}{2}$ grains & in ye water $5\frac{1}{16}$ grains, fell each of them in ye water from ye top to ye bottom of ye vessel in 15″ of time the motion of descent being (to {sence}) almost uniform almost from ye top to ye bottom. so then a globe equall to $71\frac{7}{16}$ gr of water moving \uniformly/ 9 foot $4\frac{1}{2}$ inches ({illeg}) in 15″ of time feels a resistance equal to $5\frac{1}{16}$ gr of weight.
2. Two balls, each weighing in air $156\frac{1}{4}$ gr in water 77 gr fell each of them the same height \of 9 foot $4\frac{1}{2}$ {dig.}/ in 4″ of time. And these expts seemed sufficiently accurate.
Corol. Ergo ye resistance is as ye square of the velocity.
3. Two balls weighing \each of them/ in air 245 gr, in water \{almost}{illeg}/ ${1}^{gr}\frac{1}{2}$ fell \each/ ye same height in $44\frac{1}{2}$″. But thi|e|se expts w{illeg}|er|e not so accurate as {sic} for{mers.}
The same two balls augmented wth lead so as each of them to weigh in air $251\frac{1}{2}$ in water ${7}^{gr}\frac{1}{8}$
Three balls
<4r>
[25] $as=y$. $sm=x$. $mb=z$. $mp=a$. \$qs=v$/ $mn=\frac{yz}{x+z}$. ${ab}^{2}=xx+yy+2zx+zz$. $\frac{zzyy+zzxx+2{z}^{3}x+{z}^{4}}{xx+2zx+zz}={nb}^{2}$. $\frac{eezzyy+eezzxx+2ee{z}^{3}x+ee{z}^{4}}{ddxx+2ddzx+ddzz}={np}^{2}=aa+\frac{yyzz}{xx+2zx+zz}$. $sb=z$. $mn=\frac{zy-xy}{z}$. ${ab}^{2}={qb}^{2}=zz+y$ ${nb}^{2}=\frac{{\stackrel{\text{.}}{z}}^{4}+yyz\stackrel{\text{.}}{z}-2x{\stackrel{\text{.}}{z}}^{3}-2x\stackrel{\text{.}}{z}yy+\stackrel{\text{.}}{xxzz}+\stackrel{\text{.}}{xxyy}}{zz}=\frac{ddzzy\stackrel{\text{.}}{y}-2{d}^{2}zx\stackrel{\text{.}}{y}y+ddx\stackrel{\text{.}}{x}{y}^{2}+aaddzz}{eezz}$. ${e}^{2}=1$. ${d}^{2}=$ $\begin{array}{ccccccccccc}{z}^{4}& -& 2x{z}^{3}& +& xxzz& +& 2xyyz& -& xxyy& =& 0\\ & & & -& yy\hfill \\ & & & -& 2aa\hfill \end{array}$. Suppose. $yy=f{x}^{2}+2gx+h$. |yn,| $v=fx+g$. $z+fx+g=\omega$. $\begin{array}{lllllllllll}{\omega }^{4}& -& 4fx\phantom{\rule{0.5em}{0ex}}{\omega }^{3}& +& 6ffxx\phantom{\rule{0.5em}{0ex}}{\omega }^{2}& -& 4{f}^{3}{x}^{3}\phantom{0000}z& +& {f}^{4}{x}^{4}& =& 0\\ & -& 4g& +& 12fgx& -& 12ffxxg& +& 4{f}^{3}{x}^{3}g\\ & -& 2x& +& 6gg& -& 12fxgg& +& 6ffxxgg\\ & & & +& 5fxx& -& 4{g}^{3}& +& 4fx{g}^{3}\\ & & & +& 4gx& -& 4ff{x}^{3}& +& {g}^{4}\\ & & & +& xx& +& 6fg{x}^{2}& +& {f}^{3}{x}^{4}\\ & & & & & -& 2ggx& +& 2ff{x}^{3}g\\ & & & -& 2aa& & & +& fxgg\\ & & & -& h& +& 2gxx& & \\ & & & & & +& 2fhx& -& ff{x}^{4}\\ & & & & & +& 2gh& -& 4fg{x}^{3}\\ & & & & & +& 4faax& -& 3gg{x}^{2}\\ & & & & & +& 4gaa& -& ffh{x}^{2}\\ & & & & & +& 2hx& -& 2fghx\\ & & & & & & & -& ggh\\ & & & & & & & -& 2aaffxx\\ & & & & & & & -& 4aafgx\\ & & & & & & & -& 2aagg\\ & & & & & & & -& 2fhxx\\ & & & & & & & -& 2ghx\\ & & & & & & & -& f{x}^{4}\\ & & & & & & & -& 2g{x}^{3}\\ & & & & & & & -& hxx\end{array}$. & $\omega =bq$. $ab=\xi$. ${sb}^{2}={\xi }^{2}-{y}^{2}$. $mb=\sqrt{\xi \xi -yy}-x$. $\frac{\sqrt{\xi \xi -yy}-yx}{\sqrt{\xi \xi -yy}}=nm$. $\frac{\xi \sqrt{\xi \xi -yy}-\xi x}{\sqrt{\xi \xi -yy}}=nb$. |${nb}^{2}=$ |$a\stackrel{\text{.}}{a}dd+\frac{yy\xi \xi d\stackrel{\text{.}}{d}-{y}^{4}d\stackrel{\text{.}}{d}+yyxxd\stackrel{\text{.}}{d}-2{d}^{2}yyx\sqrt{{\xi }^{2}-{y}^{2}}}{ee\xi \xi -yyee}$ $=\frac{ee\stackrel{\text{.}}{{\xi }^{4}}-ee\xi \xi y\stackrel{\text{.}}{y}+ee\xi \xi \stackrel{\text{.}}{xx}-2ee\xi \xi x\sqrt{\xi \xi }-yy}{ee\xi \xi -eeyy}$. $dd=2$. $ee=1$. $\begin{array}{l}2aa{\xi }^{2}-2aayy+3yy{\xi }^{2}-2{y}^{4}+2yyxx-{\xi }^{4}-\xi \xi xx\\ -4yyx\sqrt{{\xi }^{2}-{y}^{2}}+2\xi \xi x\sqrt{xx-yy}\end{array}\phantom{\rule{0.5em}{0ex}}\right\}\phantom{\rule{0.5em}{0ex}}=0$.
Problems. 1 To find ye axis, diameters, cente{illeg}|r|s, asymptotes \& vertices/ of lines
2 To compare their crookednesse wth ye crookednes of a gi{illeg}|v|en circle
3 To find ye longest & shortest lines wch can {illeg}|b|e drawn wth in & perpendicular to the line & to find a{illeg}|ll| such lines are {illeg}|per|pendicular at both ends to ye given crooked line
4 To find where th{illeg}|ei|r greatest or least crokednesse is.
5 To find ye areas, ye l{illeg}|e|ngths, & centers of gravity {illeg}|o|f crooked lines \when it may b{e}/
[26] 6 If y (one {illeg}|u|ndetermined quantity) moves perpendic{illeg}|u|larly to x (ye other undetermin{ed} quantity. if $s=\text{a secant}=db$. $v=dc$. $y=bc$. $x=ca$. Then having ye proportion of {illeg}|v| to {x}{s} to find y, or having ye proportion of v to y to find x: when it may bee.
7 To reduce all kinds of equations, when it may bee
8 To find tangents to any crooked lines. Whither Geometricall or Mechanicall
9 To compare ye superficies of one line wth ye area of another & to find ye centers of gravity twixt two lines or sollids. 15
10 Ha{illeg}|v|eing ye {illeg} respe position wch x must beare to y {illeg} (as if x is always in ye same line, but y cutteth x at given angeles {sic}. or if x & y wheeling about 2 poles describe ye lines by theire intersection &c) to find theire position in respect of ye line soe ye equation e{illeg}|x|pressing theire relation may bee as simple as may bee (as to find in w{illeg}|h|at line x is & wt angles it maketh with y; or to find ye distance of ye 2 poles & in what line they must be, soe yt ye relation twix{t} x & y may bee had in a{illeg}|s| simple termes as may bee).
11 Of ye description of lines.
12 Reasonings of gra{illeg}|v|ity & levity upon severall suppositions (as yt ye rays of gravity are parallel or verge towards a center; yt they are reflected, refracted, or neith{er} by ye weighty body &c.
13 Of ye u{illeg}|s|e of line{illeg}|s|
< insertion from lower down f 4r >
14. To f{illeg}|i|nd such lines whose areas length or centers of gravity {illeg}|m|ay bee found.
15. To compare ye areas, lenghs {sic}, gravity of lines \when it may bee./ & to find such lines whose lengths, {illeg}|ar|eas may be comp{illeg}
16. To doe ye same to sollids in respect of theire areas, content, gravity &c wch was done to lines in respect of their are{illeg} lengths, areas, & gravity.
17. Of lines wch l{illeg}|y|e not in ye same plane as tho{illeg}|se| made by ye intersection of a cone & {sphæreides}.
18. Two equations given to {illeg}|k|now whither they expresse ye same line or not.
19. Of ye proportion wch ye rootes of an equation beare to one another.
20 One line being to find other lines at {pleasure} of {illeg} {same length} {illeg}
21 How much doth any medium resist ye motion of any given body.
22 To Determin maxima & minima in equation wch hath more then {sic} {illeg} unknowne quantitys.
To Determin max & min by numbers.
< text from higher up f 4r resumes >
<4v>
[27] $cd=x$. $gd=y$. $rx+\frac{r}{q}xx=rx$. $ac=a$ $de=\frac{1}{2}r+\frac{r}{q}x$. $ag=\sqrt{aa+2ax+xx+rx+\frac{rxx}{q}}$. $af=a+x+\frac{1}{2}x+\frac{r}{q}$ $\frac{aa+2ax+xx+ar+2\frac{a}{q}rx+\frac{rxx}{q}+\frac{rrxx}{qq}+\frac{1}{4}rr={fa}^{2}\text{,}\phantom{\rule{1em}{0ex}}in\phantom{\rule{1em}{0ex}}rx+\frac{r}{q}xx}{aa+2ax+xx+rx+\frac{rxx}{q}}={fo}^{2}$. $fo:fl\colon\colon 2.1$. |$aa+ar+\frac{1}{4}rr=bb$| |$xc=2ax+\frac{2arx}{q}+\frac{rrx}{q}$| |${fl}^{2}=$| $\frac{aarx+\frac{aar}{q}xx+2arxx+2arxxx+r{x}^{3}+\frac{r}{q}{x}^{4}+arrx+3\frac{arr}{q}xx+\frac{1}{4}{r}^{3}x+\frac{{r}^{3}xx}{2q}+\frac{2arr{x}^{3}}{qq}+\frac{2{r}^{3}{x}^{3}}{qq}+\frac{{r}^{3}{x}^{4}}{{q}^{3}}}{2aa+4ax+2xx+2rx+\frac{2rxx}{q}}$ $aar+arr+\frac{1}{4}{r}^{3}=bbr$. $a+\frac{1}{2}r=b$. $\frac{bbrx+crxx+gr{x}^{3}+\frac{rqq+{r}^{3}{x}^{4}}{q}}{2aa+4bx+\frac{2q+2rxx}{q}}={fl}^{2}$. $\frac{aa}{q}+2a+\frac{3ar}{q}+\frac{rr}{2q}=c$. $\frac{2a}{q}+1+\frac{2aa}{qq}+\frac{2rr}{qq}=g$ . {illeg} {illeg} $r=$ {illeg} /$r=2$ $q=6$. $x=3$.\ $gd={r}^{2}x+\frac{1}{3}xx=3$. {illeg} /$ac=1$\ $df=\frac{r}{2}+\frac{r}{q}x=2$. $ag=5$. $5:3\colon\colon af=6:\frac{18}{5}=of$ . $fl=\frac{9}{5}$. $d:e\colon\colon of:fl=\frac{18e}{5d}$. $dk=\frac{9×2}{9-\frac{324ee}{25dd}}\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}\sqrt{\frac{\frac{18×324ee}{25dd}+\frac{81×324ee}{25dd}-9×Q:\frac{324ee}{25dd}:}{81-18×\frac{324ee}{25dd}+Q:\frac{324ee}{25dd}:}}$. $dk=\frac{450dd}{225{d}^{2}-324ee}\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}\sqrt{\frac{801900ddee-944784{e}^{4}}{50625dddd-145800ddee+104976{e}^{4}}}$ $dk=\frac{50dd}{25dd-36ee}\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}\sqrt{\frac{9900ddee-11664{e}^{4}}{625{d}^{4}-1800ddee+1296{e}^{4}}}\right\}=g$. $de=a-3$. $g=p+\sqrt{ps}$ . $e=1$ $\frac{9aa{e}^{2}-54aee+81ee-18agee+54gee+9ggee}{ddgg}$ $\frac{\begin{array}{c}+aaee-6aee+9ee\\ -2ag{e}^{2}+6g{e}^{2}+ggee\end{array}}{dd}$ $\frac{\begin{array}{c}-9aa+54a-81\\ +18ag-54g-9gg\end{array}}{gg}$ $\begin{array}{c}={e}^{2}\\ ={z}^{2}\end{array}$ $\begin{array}{llllllllllllllllllllll}\phantom{+}& 9aaee& -& 54aee& +& 81ee& -& 18apee& -& 18aee\sqrt{qs}& +& 54eep& +& 54ee\sqrt{qs}& +& 9eepp& +& 9eeqs& -& 2qs\sqrt{qs}×eea& =& {z}^{2}\\ +& ee{p}^{4}& +& 6eeppqs& & & +& eeqqrs& +& 2aaeep\sqrt{qs}& -& 2aae{p}^{3}& +& 18eep\sqrt{qs}& +& aaeepp& +& aaeeqs& +& 6eeqs\sqrt{qs}\\ -& 9aadd& +& 54add& & & -& 81dd& -& 12aaeep\sqrt{qs}& +& 6ee{p}^{3}& +& 18ee\sqrt{qs}& -& 6aeepp& -& 6aeeqs\\ +& 18apdd& -& 54pdd& & & -& 9ppdd& -& 6aaeep\sqrt{qs}& & & -& 54{d}^{2}\sqrt{qs}& +& 9eepp& +& 9eeqs\\ & & & & & & & & +& 18eepp\sqrt{qs}& & & -& 18pdd\sqrt{qs}& & & -& 6aeepqs\\ & & & & & & & & +& 4ee{p}^{3}\sqrt{qs}& & & & & & & +& 18eepqs\\ & & & & & & & & +& 4eepqs\sqrt{qs}& & & & & & & -& 9ddqs\\ & & & & & & & & +& 18add\sqrt{qs}\end{array}$ $ddpp+2ddp\sqrt{qs}+ddqs$
[28] $ac=a=ce$ . $cd=x$. $dg=y$. $bc=x$. $df=\frac{1}{2}r$. $rx=yy$. $aa+2ax+rx+xx:rx\colon\colon \begin{array}{l}aa+2ax+ar+rx\\ +xx+\frac{1}{4}rr\end{array}:lf$. $rx+\frac{ar+\frac{1}{4}rr\phantom{\rule{1em}{0ex}}in\phantom{\rule{1em}{0ex}}rx}{aa+2ax+rx+xx}={fo}^{2}$. ${fl}^{2}=\frac{eerx}{dd}+\frac{4eearrx+ee{r}^{3}x}{4ddaa+8ddax+4ddrx+4ddxx}$. ${fl}^{2}=pp$. $lk=r$ $ac=a$. $ab=x$. $a:x\colon\colon x:a-x$. $xx=aa-ax$. $xx=4-2x$. $x=-1+\sqrt{5}$ $2:\sqrt{5}-1\colon\colon \sqrt{5}-1:3-\sqrt{5}$. $6-2\sqrt{5}=5-2\sqrt{5}+1$. $xx=-ax+aa$. $\begin{array}{l}144\\ \underset{_}{18}\\ 324\end{array}$ $\begin{array}{l}\underset{_}{18×25}=450\\ \phantom{0}90\\ 36\end{array}$ $\begin{array}{l}144\\ 10\end{array}$ $\begin{array}{l}\underset{_}{324×18}=5832\\ 2592\\ 324\end{array}$ / $\begin{array}{l}\phantom{0}29160\\ 11664\end{array}\right\}=145800$\ $\begin{array}{l}\phantom{00}324\\ 2592\end{array}\right\}=\begin{array}{r}26244\\ 5832\end{array}\begin{array}{c}\phantom{0}\\ \phantom{0}\\ \begin{array}{||}\hline 32076\\ \hline\end{array}\end{array}$ $x=-\frac{1}{a}+\sqrt{\frac{5aa}{4}}$ $ei=a$. $hk=z$.h{illeg}$=\frac{ez}{d}$ \$gd=y$/ $hi=\frac{ez}{d}$. $he=\sqrt{\frac{eezz-aadd}{dd}}$. $ek=\sqrt{\frac{ddzz-eezz+aadd}{dd}}$. $ce=a$. $fl=b$ $fk=c$. $lk=\sqrt{cc-bb}$. $z:\sqrt{\frac{eezz-aadd}{dd}}\colon\colon c:b${illeg}. $c=\frac{dbz}{eezz-aadd}$. ${lk}^{2}=\frac{ddbbzz-bbeezz+aabbdd}{eezz-aadd}$. $df=v$ $lk=\frac{bv+\frac{dbbz}{\sqrt{eezz-aadd}}}{y}$ {illeg} $=\frac{dbby+bv\sqrt{eezz-aadd}}{y\sqrt{eezz-aadd}}=\frac{y\sqrt{ddbbzz-bbeezz+aabbdd}}{y\sqrt{eezz-aadd}}$ $ddbbyy+vveezz-vvaadd-ddyyzz+eeyyzz-aayydd+2bdvy\sqrt{ee{z}^{2}-aadd}=0$. $ab=q$. $bc=x$. $dc=y$. $r=\text{lat}:rx:rx-\frac{rxx}{q}=yy$. $ef=b$. $fg=c$. $eb=\frac{1}{2}q-b$. $cf=x-\frac{1}{2}q$. $ce=x-\frac{1}{2}q+b$. $b:c\colon\colon x-\frac{1}{2}q+b:y$. $\frac{cx-\frac{1}{2}qc+bc}{b}=y=\sqrt{rx-\frac{rxx}{q}}$ $\begin{array}{cccccccccccccc}& ccxx& -& qccx& +& 2bccx& +& \frac{1}{4}qqcc& -& qccb& +& bbcc& =& 0\\ +& \frac{bbr}{q}xx& & & -& bbrx\hfill \end{array}$. $q=3$. $fg=c=1$ {illeg} $b=1$. $xx+\frac{1}{3}xx=3x-2x+x-\frac{9}{4}+3-1=0$. $\frac{4xx}{3}=2x-\frac{1}{4}=0$ $xx=\frac{6}{4}x-\frac{3}{16}$. $x=\frac{3}{4}\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}\sqrt{\frac{9-3}{16}}$. $x=\frac{3}{4}\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}\sqrt{\frac{3}{8}}$ $rx+\frac{rxx}{q}=9$. $\frac{r}{2}+\frac{rx}{q}=2$ /$\frac{1}{2}r+\frac{rxx}{q}=7$ $\frac{rx}{2}$ {illeg}$=7$.\ $ac=a$ . $kd=b$. $ad=c$. $dc=d$. {illeg} $c:a\colon\colon b:\frac{ab}{c}$ {illeg} $\frac{dab}{ec}=cf$. $kl=e$. $x=1$. $r=\frac{1}{4}$. $ad=a$. $dl=b$. $lk=c$. $ac=g$. $\frac{dgb}{ea}=ce$. $\frac{ddgc}{eea}=ef$. ${of}^{2}=\frac{{d}^{4}ggcc-{e}^{4}ccaa}{{e}^{4}aa}$. $e=2$. $d=2$. $+bggc$ $+\frac{bggcc}{aa}-cc={of}^{2}$. $g=c=1=a$. $15={of}^{2}$. $\frac{ggcc}{16aa}-cc$. $g=${illeg}|{ 8 }| {illeg} $=8a=${illeg}$2c$ $3={of}^{2}$ . $ab=a$. $bc=b$. $cd=c$. $be=x$. $ce=x-b$. $de=\sqrt{xx-2bx+bb-cc}=\frac{cx}{a}$ . $xx=\frac{2bx-bb+cc\phantom{\rule{1em}{0ex}}in\phantom{\rule{1em}{0ex}}aa}{aa-cc}$ $x=\frac{aab}{aa-cc}\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}\sqrt{\frac{{a}^{4}bb+aabbcc+{a}^{4}cc-aa{c}^{4}}{{a}^{4}-2aacc+{c}^{4}}}$ $bm=f$. $be=g$. $me=f-g$. $ab=a$. {illeg} $cd=c$. $a:g\colon\colon c:\frac{gc}{a}=de$. $g:a:f-g:\frac{af-ag}{g}=mr$. $mn=z$. $\frac{aaff-2aafg+aagg}{gg}+ff-2fg+gg\left(={re}^{2}\right)in\frac{ee}{dd}={rn}^{2}=\frac{aaff-2aafg+aagg}{gg}+zz$.
<5r>
$eh=8$ — − $ei=6$ $hi=10$. $ek=15$. $\sqrt{\begin{array}{c}\phantom{0}75\\ 154\\ 6\end{array}}=\sqrt{289}=17=hk$. {$gh=$ }{$gk=$ } $de=${illeg}15. $dg=16=y$. $cd=x$. $df=a$. $gk=${14} $34:16\colon\colon gk:gd\colon\colon 17:8\colon\colon 30-a:\frac{240-8a}{17}=fl$. $fl:fo\colon\colon 10:17$. $\frac{120-4a}{5}=24-\frac{4a}{5}=fo$ . $ad=b$. $ag=\sqrt{bb}+${illeg} $256:bb+256\colon\colon 576-\frac{192a}{5}+\frac{16aa}{25}:\frac{576bb-\frac{192}{5}abb+\frac{16aabb}{25}}{256}+576-\frac{192a}{5}+\frac{16aa}{25}={af}^{2}=aa+2ab+${illeg} $\frac{5bb}{4}+\frac{3abb}{20}+\frac{aabb}{400}-\frac{9aa}{25}-2ab-\frac{192}{5}a+576=0$. $\begin{array}{l}\phantom{+}500bb\\ +60a\\ +aa\end{array}\begin{array}{c}=200ab+144aa+15360a-2304\\ \phantom{0}\\ \phantom{0}\end{array}$ {illeg} $yy=rx+\frac{r}{q}xx=256$. $\frac{r}{q}x+\frac{r}{2}=a$. $256-ax=\frac{1}{2}rx$. $\frac{512}{x}-\frac{a}{2}=r$ $eh=\frac{5}{2}$ . {illeg} $ei=6$. $hi=\frac{13}{2}$. $hk=\frac{2221}{20}$. $ek=\frac{9\sqrt{5149}}{20}$$gh=${illeg} $\frac{221}{20}:\frac{50}{20}\colon\colon \frac{221}{20}+c:dg=\frac{50}{20}+\frac{}{221}$ $dg=\frac{5}{2}+\frac{50c}{221}$ . $\frac{9c\sqrt{5149}}{221}=de$ $fk=\frac{\phantom{00}+9\sqrt{5149}}{20}$ {illeg} $fk=\frac{9c\sqrt{5149}}{221}+$ $ek=\frac{3\sqrt{5149}}{20}$. $kd=g$. $\frac{5g×20}{6\sqrt{5149}}=\frac{50g}{3\sqrt{5149}}=dg$. ${dg}^{2}=\frac{2500gg}{46341}$. ${gk}^{2}=48841gg$. $fk=c-a$. $\frac{50cg-50ag}{3\sqrt{5149}\phantom{\rule{1em}{0ex}}in\phantom{\rule{1em}{0ex}}221}=fl=\frac{f k×dg}{gk}$ $\frac{50cg-50ag}{390\sqrt{5149}}=fo$. $ae=p$. {illeg} $d e=g-\frac{3\sqrt{5149}}{20}$ $ad=p-g+\frac{3\sqrt{5149}}{20}$. ${ag}^{2}=\begin{array}{l}pp-2pg+\frac{48841gg}{46341}+\frac{46341}{400}\\ +\frac{3p}{10}\sqrt{5149}-\frac{3g}{10}\sqrt{5149}\end{array}$ ${ag}^{2}:{gd}^{2}\colon\colon {af}^{2}:{fo}^{2}$.
$eh=${illeg}. {illeg} $eh=a$. $ei=b$. $hi=\sqrt{aa+bb}$ $hk=\frac{d}{e}\sqrt{aa+bb}$. $ek=\sqrt{\frac{ddaa+ddbb-eeaa}{e}}$ $eh=c$. $ek=\sqrt{\frac{ddcc+ddbb-\phantom{000}}{ee}}$ $2abx=n$. $axx+abb:bxx+baa\colon\colon f:g$. $\frac{abbg-aabf}{}-axxg-bxxf=0$ $\frac{nn}{4aabb}=xx$. $agnn-bfnn+4{a}^{3}{b}^{4}g-4{a}^{4}{b}^{3}f=0$. $f=10$. $g=17$. ${b}^{4}$ $axx-abb=6$. $xx=\frac{6+abb}{a}$. $\frac{17abb-10aab}{10b-17a}=xx$ $\begin{array}{ccccccccc}10{a}^{3}b& -& 17aabb& +& 10a{b}^{3}& +& 60b& =& 0\\ & -& 17aabb& -& 102a\end{array}$ {illeg} $\begin{array}{ccccccccc}{a}^{3}& -& \frac{34}{10}aab& +& abb& +& 6& =& 0\\ & & & -& \frac{102a}{10b}\hfill \end{array}$. $q{a}^{2}+qa+s×a+c$. $c${illeg}$=\frac{6}{s}$. $r+\frac{6}{s}=\frac{34b}{10}$. $-\frac{10r}{34}+\frac{60}{34s}=b$. $\begin{array}{lllllll}{a}^{3}& +& raa& +& sa& +& cs\\ & +& c& +& cr\end{array}$ $r=-\frac{6}{s}-\frac{34b}{10}$. $-\frac{36}{ss}-\frac{204b}{10}s+s=bb-\frac{102}{10b}$. $cc=20c-45$ $\begin{array}{lllllll}q{a}^{3}& +& raa& +& sa& +& sc\\ & +& qc& +& rc\end{array}$. $b=1$ $10b{a}^{3}$ $10{a}^{3}-34aa+112a+60$. $\begin{array}{l}\phantom{0}448\\ \underset{_}{336\phantom{0}}\\ \underset{_}{3808}\end{array}$ $\begin{array}{c}\begin{array}{l}\left(2.2.2.2.2.7./2.2.3.5\\ \begin{array}{cc}\begin{array}{c}1904\\ 952\\ 476\\ 238\\ 119\\ 17\end{array}& \begin{array}{llll}\phantom{0}& 3868& .& \left(2.2.967\\ \phantom{0}& 1934& & \phantom{\left(}\\ \phantom{0}& \phantom{0}967& & \phantom{\left(}\end{array}\end{array}\end{array}\\ \begin{array}{c}\begin{array}{l}1.2.4.6.10.3.5.15.30.\\ 20\end{array}\\ \begin{array}{l}\hfill 10.\phantom{0}.\\ \phantom{0}\\ \begin{array}{c}\begin{array}{c}2.967.4.1934.3868.\end{array}\\ 10\end{array}\end{array}\end{array}\end{array}$ $\begin{array}{l}\phantom{00}36\phantom{=}\\ \underset{_}{102\phantom{0}\phantom{=}}\\ 1056=\end{array}$ $\frac{pq-r}{\text{Div}-}$ $-nn+2pn$ . $p-n=c$ \{illeg} $aa-4a$/ 4 $10{a}^{3}-34aa-92a+60$ $\begin{array}{ccccc}-\frac{34}{}& +& 2& +& 4\\ & -& 2& -\end{array}$. $\begin{array}{||}\hline -4\\ \hline\end{array}$ $\begin{array}{r}120\\ \underset{_}{-34}\\ 86\end{array}$ $\begin{array}{r}150\\ \underset{_}{34}\\ 126\end{array}$ $\begin{array}{l}\phantom{0}368\\ \underset{_}{27\phantom{00}}\\ \underset{_}{\phantom{0}120}\\ 3168\end{array}$ $\left(\begin{array}{l}2.2.\\ 1594.797.\end{array}$ $\begin{array}{cc}\begin{array}{r}372\\ \underset{_}{372\phantom{0}}\\ 4094\\ \underset{_}{-120}\\ 3974\\ 1987\end{array}& \begin{array}{ll}\phantom{0}14& c\\ \begin{array}{l}\phantom{0}56\\ 14\end{array}\\ \begin{array}{ccc}\left(2& & \begin{array}{r}96\\ \underset{_}{16\phantom{0}}\\ 256\end{array}\\ \begin{array}{r}144\\ \underset{_}{18\phantom{0}}\\ 324\end{array}& 117& 13\end{array}\end{array}\end{array}$ $20{a}^{3}$$-34aa${illeg}$-136aa$ {illeg}$-22a$ \{illeg}/ + {illeg} 120 $\begin{array}{ccc}\underset{_}{49}& \begin{array}{r}96\\ \underset{_}{48\phantom{0}}\\ 576\\ 49\end{array}& \begin{array}{l}625\\ 2\end{array}\end{array}$ ${a}^{2}-\frac{4}{10}a$ 6 {illeg} [29] $30{a}^{3}-306aa+168a-180$. $\begin{array}{rr}1008& \left(2.2.3.3.4.6.9.12.24.18.54.36.\\ \underset{_}{504\phantom{00}}\\ \underset{_}{\phantom{0}-180}\\ 51228& .25614.12807.4269.1423\hfill \end{array}$ $\begin{array}{ll}136& 1531\\ 134& 2\hfill \\ 17\end{array}$ 102 $\begin{array}{cccc}40{a}^{3}& \begin{array}{r}-204\\ -340\end{array}aa& \begin{array}{r}+640\\ -102\end{array}a& +240\text{.}\\ \phantom{0}\\ 40{a}^{3}& \hfill -544aa& \hfill -538a& +240\text{.}\end{array}$ $\begin{array}{cccc}& & eh=12\text{.}& ei=9\text{.}\\ hi=15\text{.}& hk=20\text{.}& ek=16\text{.}\end{array}$ $\begin{array}{r}\underset{_}{17}\\ 119\\ \underset{_}{17\phantom{0}}\\ \underset{_}{\phantom{0}289}\\ 2083\\ 289\phantom{0}\end{array}$ $\begin{array}{r}\phantom{-}4913\\ \underset{_}{-1530}\\ 3\\ \hfill 729\hfill \end{array}$ $\begin{array}{r}81\\ \underset{_}{648\phantom{0}}\\ 6561& ×4=\end{array}$ { 2 }620 {illeg}| 1 | {illeg}
[30] d {illeg}$-3=$ $\frac{axx-c}{a}=bb$. $axx+abb:bxx+baa\colon\colon e:d$. $daxx+dabb=ebxx+ebaa$. $\begin{array}{rr}25839& \left(\hfill \\ 6\phantom{00}\\ 239\end{array}${illeg} $xx=\frac{aabe-abbd}{ad-eb}=\frac{10aab-17abb}{17a-10b}$. $xx=\frac{17abb-10aab}{10b-17a}$. $40b-17abb+10aab$. /$a=\frac{9}{17}$. $b=1$. {illeg}\ $10-\frac{90}{17}${illeg} $3,4,9$. $5,12,13$. $7,24,25$. $12,16,20$. $12,9,15$. $xx=\frac{4abb-3aab}{3b-4a}$. {3 } $\begin{array}{cccc}x& \begin{array}{c}-4axx\\ -4abb\end{array}& & +3aab\end{array}$ $ax\frac{2xx+2bb}{3b}\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}\sqrt{\frac{4{x}^{4}+8bbxx+4{b}^{4}-9bbxx}{9bb}}$. $\sqrt{4{x}^{4}-bbxx+4{b}^{4}}$. $x=4$ . $b=1$ . {illeg}$\frac{+8}{6}\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}\frac{8}{6}=a=3${illeg} {illeg}$=\frac{4}{3}$ $x=${illeg} $x=${illeg} {illeg}$+\frac{1}{3}=ab${illeg}$4+\frac{1}{9}${illeg}15 7 $4{x}^{4}-bb{x}^{2}+4{b}^{4}=4b{x}^{4}$ [31] {illeg}$\frac{8}{3}$ $\frac{}{3}$. $\frac{8+2}{3}=\frac{10}{3}=cb$. $4+\frac{4}{9}$. $\frac{30}{9}$ . {illeg} $\frac{8}{3}$ {illeg} $b=${illeg} $\frac{4{x}^{4}-{c}^{4}}{{x}^{5}+}=bb$. {illeg}+{illeg}={illeg}$-2bc+cc$. $\frac{cc+xx}{2c}=b$. $4{x}^{4}-bbxx$ {illeg} ${b}^{4}=4{b}^{4}-bbxx+6{}^{3}$ {illeg} {illeg} = {illeg}${c}^{4}-4{a}^{2}bb${illeg}$3aab=4bb-3ab=4bb-4${illeg}$cc$. $\frac{}{4c}${illeg} {illeg}$aa=${illeg}$-4aa${illeg} {illeg}$9acc-${illeg}$36a${illeg} {illeg}
<5v>
[32] $\sqrt{4{x}^{4}-bbxx+4{b}^{4}}=\sqrt{4{b}^{4}+4bbcc+{c}^{4}}$ ${}^{2}$ $\frac{4{x}^{4}-{c}^{4}}{4cc+xx}=bb=\frac{4{x}^{6}+16cc{x}^{4}-4xx-4{c}^{6}}{4{c}^{4}+4ccxx+{x}^{4}}$. $h{x}^{2}+in+${illeg} $a{x}^{3}+b{x}^{2}+cx+d$. $\sqrt{4{x}^{4}-{b}^{2}{x}^{2}+4{b}^{4}}=-bb+cx+dxx+ee$. $4{x}^{4}-bbxx=-4bbcx-4bbdxx-4bbee+ccxx+2cd{x}^{3}+2ceex+dd{x}^{4}+2deexx+{e}^{4}$: $\frac{+4{x}^{4}+dd{x}^{4}+2cd{x}^{3}+ccxx+2deexx+2ceex+{e}^{4}}{-xx+4dxx+4cx+4ee}=bb$. |$24-\frac{50}{3}$ /$\frac{22}{36+45-20}$\ $\frac{\frac{22}{3}}{12+15-\frac{20}{3}}$| $2fxx+2gfx+ff$. $2e=f$. $\frac{2c}{e}=2g$. $\frac{ccxx}{e}=-xx+4dxx$ $cc=-ee+4dee$. $d=\frac{cc+ee}{4ee}$ $\frac{4{x}^{4}\phantom{\rule{0.5em}{0ex}}:\phantom{\rule{0.5em}{0ex}}\frac{+{c}^{4}+2ccee+{e}^{4}}{16{e}^{4}}\phantom{\rule{0.5em}{0ex}}in\phantom{\rule{0.5em}{0ex}}{x}^{4}\phantom{\rule{0.5em}{0ex}}:\phantom{\rule{0.5em}{0ex}}\frac{+{c}^{3}+cee}{2ee}\phantom{\rule{0.5em}{0ex}}in\phantom{\rule{0.5em}{0ex}}{x}^{3}\phantom{\rule{0.5em}{0ex}}:\phantom{\rule{0.5em}{0ex}}+\phantom{\rule{0.5em}{0ex}}ccxx\phantom{\rule{0.5em}{0ex}}:\phantom{\rule{0.5em}{0ex}}\frac{+{e}^{4}+ccee}{2ee}\phantom{\rule{0.5em}{0ex}}in\phantom{\rule{0.5em}{0ex}}xx\phantom{\rule{0.5em}{0ex}}:\phantom{\rule{0.5em}{0ex}}+\phantom{\rule{0.5em}{0ex}}2ceex\phantom{\rule{0.5em}{0ex}}+\phantom{\rule{0.5em}{0ex}}{e}^{4}.}{-xx\phantom{\rule{0.5em}{0ex}}:\phantom{\rule{0.5em}{0ex}}\frac{+cc+ee}{2ee}\phantom{\rule{0.5em}{0ex}}in\phantom{\rule{0.5em}{0ex}}xx\phantom{\rule{0.5em}{0ex}}+\phantom{\rule{0.5em}{0ex}}4cx\phantom{\rule{0.5em}{0ex}}+\phantom{\rule{0.5em}{0ex}}4ee}=bb$ $\begin{array}{ccccccccc}hh{x}^{4}& +& 2hi{x}^{3}& +& 2hk{x}^{2}& +& 2ikx& +& kk\\ & & & +& ijxx\end{array}$. $\frac{-63{e}^{4}+2ccee+{c}^{4}}{16{e}^{4}}=hk$. {illeg}$\sqrt{}$ {illeg}e {illeg}k {illeg}$-ee$. $2fxx+2gfx+ff$. $2e=f$. $\frac{2c}{e}=2g$. $\frac{ccxx}{e}=-xx+4dxx$ $cc=-ee+4dee$. $d=\frac{cc+ee}{4ee}$ $\frac{-4{x}^{4}\phantom{\rule{0.5em}{0ex}}:\phantom{\rule{0.5em}{0ex}}\frac{+{c}^{4}+2ccee+{e}^{4}}{16{e}^{4}}\phantom{\rule{0.5em}{0ex}}in\phantom{\rule{0.5em}{0ex}}{x}^{4}\phantom{\rule{0.5em}{0ex}}:\phantom{\rule{0.5em}{0ex}}\frac{+{c}^{3}+cee}{2ee}\phantom{\rule{0.5em}{0ex}}in\phantom{\rule{0.5em}{0ex}}{x}^{3}\phantom{\rule{0.5em}{0ex}}:\phantom{\rule{0.5em}{0ex}}+\phantom{\rule{0.5em}{0ex}}\frac{3}{2}ccxx\phantom{\rule{0.5em}{0ex}}+\phantom{\rule{0.5em}{0ex}}\frac{ee}{2}xx\phantom{\rule{0.5em}{0ex}}+\phantom{\rule{0.5em}{0ex}}\frac{3}{2}ccxx\phantom{\rule{0.5em}{0ex}}+\phantom{\rule{0.5em}{0ex}}2ceex\phantom{\rule{0.5em}{0ex}}+\phantom{\rule{0.5em}{0ex}}{e}^{4}}{\frac{ccxx}{ee}\phantom{\rule{0.5em}{0ex}}+\phantom{\rule{0.5em}{0ex}}4cx\phantom{\rule{0.5em}{0ex}}+\phantom{\rule{0.5em}{0ex}}4ee}=bb$ $k=ee$. $2i=2c$. $i=c$. {illeg} $h=\frac{cc+ee}{4ee}$. $\frac{{c}^{4}+2ccee-63{e}^{4}}{16{e}^{4}}=hk$ {illeg} ${c}^{4}+2ccee-63{e}^{4}={c}^{4}+2ccee+{e}^{4}$. $ee=k$. $i=c$. $d=h$. $\frac{\frac{{c}^{4}+2ccee+{e}^{4}}{ee}\phantom{\rule{0.5em}{0ex}}in\phantom{\rule{0.5em}{0ex}}{x}^{4}\phantom{\rule{0.5em}{0ex}}:\phantom{\rule{0.5em}{0ex}}\frac{+{c}^{3}+cee}{2ee}\phantom{\rule{0.5em}{0ex}}in\phantom{\rule{0.5em}{0ex}}{x}^{3}\phantom{\rule{0.5em}{0ex}}:\phantom{\rule{0.5em}{0ex}}+\phantom{\rule{0.5em}{0ex}}\frac{3}{2}ccxx\phantom{\rule{0.5em}{0ex}}+\phantom{\rule{0.5em}{0ex}}\frac{ee}{2}xx\phantom{\rule{0.5em}{0ex}}+\phantom{\rule{0.5em}{0ex}}\frac{3}{2}ccxx\phantom{\rule{0.5em}{0ex}}+\phantom{\rule{0.5em}{0ex}}2ceex\phantom{\rule{0.5em}{0ex}}+\phantom{\rule{0.5em}{0ex}}{e}^{4}\phantom{\rule{0.5em}{0ex}}+\phantom{\rule{0.5em}{0ex}}ff}{2f}=xx$.
[33] $\frac{4-3+\frac{3}{2}+\frac{1}{2}+2+9}{2}=$ {illeg} 6 /{illeg}\ $\begin{array}{r}{x}^{3}\\ +{y}^{3}\end{array}\phantom{\rule{1em}{0ex}}\frac{+9z{z}^{4}-3az{x}^{2}y+{y}^{3}z-3azxyy}{\sqrt{9{x}^{4}-6ay{x}^{2}+10aa{y}^{2}+aa{x}^{2}-6{a}^{3}x}}$ $25+5+6+\frac{1}{2}+4+1\phantom{\rule{2em}{0ex}}41\frac{1}{2}$ $\frac{83}{4f}$ $2ff$ / 5 \ $\frac{25}{4}+\frac{5}{8}+$ $64-24+6+2${illeg} {illeg}|5| $\begin{array}{c}dabb-ebxx-ebaa+daxx=0\\ dpqq-eqyy-eqpp+dpyy=0\end{array}$. $ss-{v}^{2}+2vy-yy=\frac{{a}^{4}}{yy}-\frac{2{a}^{4}}{2{y}^{3}}+2y$ $2ab=2exx$. $\frac{daxx}{eaa-exx}=b$. $\frac{ebxx+ebaa}{abb+axx}=d=\frac{eqyy+eqpp}{pqq+pyy}$. $\begin{array}{c}pqqbxx+pqqbaa+pyybxx+pyybaa\\ -abbqyy-abbqpp-axxqyy-axxqpp\end{array}=0$ $pqqa+pyya=bqyy+bqpp$. $axxqyy+axxqpp=pqqbxx+pyybxx$. $pqqab+pyyab=bbppq+xxppq$. $axxqyy+abbqyy=pqqbxx+pyybxx$. $\frac{bbpq+xxpq}{bqq+byy}=a=\frac{pqqbxx+pyybxx}{qyyxx+qyybb}$. $\begin{array}{cccccccc}& aabpyy& -& abbqyy& +& bpyyxx& =& 0\\ +& aabpqq& -& abbqpp& +& bpqqxx& =& 0\\ & & -& axxqyy\\ & & -& axxqpp\end{array}$. $aa+xx=\frac{abbqyy+abbqpp+axxqyy+axxqpp}{bpyy+bpqq}$ $\begin{array}{c}aabpyy+aabpqq=abbqyy+axxqyy\\ a=x=b\phantom{\rule{3em}{0ex}}qq=yy=pp\end{array}$
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${b}^{3}+{z}^{3}=abz$ . $bb-2bx+xx+yy-$ $\begin{array}{ccccccccc}ss& -& 2vv& +& 2vx& -& xx& =& \frac{{r}^{4}}{xx}\\ 0& & 0& & 1& & 2& & -2& \end{array}$ $-\frac{2{r}^{4}}{2xxx}+\frac{2xx}{2x}=v$ {illeg}
< text from f 5v resumes >
$\begin{array}{cccccc}hgxx& -& gix& +& & \text{.}\\ & -& hkx& +& ik& \text{.}\end{array}$ $2rx-xx+rr=yy$. $v=r-x$. $\begin{array}{ccccccc}rx& +& \frac{r}{q}xx& -& yy& =& 0\\ 1& & 2& & 0\end{array}$. $-r=\frac{2rx}{q}$ $\begin{array}{c}2rx+2yx+xx=yy\\ r+y+x\end{array}$. $xx$ {illeg} $=-\frac{q}{2}$ $yy=-\frac{rq}{2}+\frac{rq}{4}xx\phantom{\rule{0.5em}{0ex}}\begin{array}{c}+\\ -\end{array}\phantom{\rule{0.5em}{0ex}}yy$ $ax-xx=yy$. $yy-\frac{rq}{4}$. $2rx-xx=yy$. $v=r-x$. $x=${illeg}r$\sqrt{2rr}$ x $rz$$zx$ for x {illeg}|wri|te $x+z-\frac{zx}{r}$ $\sqrt{\phantom{00000}+rr}${illeg} for {illeg}| y | write $\sqrt{}$ $2rx-xx-4zx+2\frac{zxx}{r}+2\frac{zzx}{r}-\frac{zzxx}{rr}$ {illeg}$\sqrt{\phantom{00000}xx}$ {illeg} r {illeg} x \/ $z\frac{\sqrt{2rx-xx}}{r}$ for y $-rx$ {illeg} z $2zx$ {illeg} x $-2x$ {illeg} 2 $zx$ 2 $\frac{zxx}{r}$ $-\frac{2zzx}{r}=+2rxxx+2rx-xx+\frac{2{x}^{3}}{r}-$ /$4rx$$\frac{2{x}^{3}}{r}-4xx+\frac{2{x}^{3}}{r}$\ {illeg}{illeg} $xx+$ $xxz$ $4zx$ $\frac{2zxx}{r}$ 2 $z-2$ {illeg} z $\frac{4zx}{r}$ $2z${illeg}$\frac{4zx}{r}=0$ {illeg}{illeg}$xx$ $-zz$$+\frac{2zzx}{r}$$-\frac{zxx}{rr}$ {illeg}{illeg}$xx$
<6r>
[34] $ag=x$. $bg=y$. $ag=x$. $gd=d$. $dc=e$. $bo=s$. $de=w$. $gh=b$. $y:v\colon\colon y+b:\frac{vy+vb}{y}=$. \$go=v$./ $if=c$. $e:w\colon\colon e+b:\frac{ew\phantom{000}}{e}$ $\frac{ew+bw}{e}=if$. $y:v\colon\colon y+b:\frac{yv+bv}{y}=hf$. $dc×de${illeg}$dc+bw$ $de+hi+\frac{gh×de}{dc}=go+\frac{gh×go}{bg}$. ag{illeg}
$ag=x$. $bg=y$. $rx=yy$. $go=de=\frac{1}{2}r$. $gd=o$. $dc=\sqrt{rx+ro}$. {illeg} $gh=z$. $o+\frac{rz}{z\sqrt{rx+ro}}=\frac{rz}{z\sqrt{rx}}$ $2o\sqrt{rx}${illeg} $4o\sqrt{rrxx+rrox}+2rz\sqrt{rx}=2rz\sqrt{rx+ro}$. $4oz\sqrt{r{x}^{3}+roxx}=4zzro$. ${x}^{3}=zzr$. $z=\sqrt{\frac{{x}^{3}}{r}}$. $\sqrt{rx}:\frac{r}{2}\colon\colon \sqrt{\frac{{x}^{3}}{r}}:\frac{x\sqrt{rx}}{2\sqrt{rx}}=\frac{x}{2}$ {illeg} $\frac{x+r}{2}=hf$. $\frac{{x}^{3}}{r}+\frac{9xx}{4}+\frac{3rx}{2}+\frac{rr}{4}={bf}^{2}$. $rx+xx=yy${illeg}
[35] $xy=rr$. $v=-\frac{yy}{x}=\frac{-{y}^{3}}{rr}=-\frac{{r}^{4}}{{x}^{3}}$. $gh=z$. $dg=o$. $de=\frac{+{r}^{4}}{{x}^{3}+3oxx}:dc=\frac{rr}{x+o}\colon\colon fi:\frac{zx-rr+\phantom{000}}{x+o}$ $rr$ $\frac{+rrzx-{r}^{4}+rrzo}{{x}^{3}+3xxo}=fi$. $\frac{-{r}^{4}}{{x}^{3}}:\frac{rr}{x}\colon\colon fh:\frac{zx-rr}{x}$ . $\frac{-{r}^{4}+zxrr-o{x}^{3}}{{x}^{3}}=fi$ $-rr$ {illeg}${x}^{4}+{r}^{4}+{x}^{3}${illeg} $-rrz{x}^{4}+{r}^{4}{x}^{3}-rrzo{x}^{3}-{r}^{4}{x}^{3}-zrr{x}^{4}+o{x}^{6}+3{r}^{4}xx0-3z{x}^{3}rro+3{x}^{5}oo-$ $+2rrzx-{x}^{4}-3{r}^{4}=0$. $z=\frac{3rr}{2x}+\frac{{x}^{3}}{2rr}=gh$. $bh=\frac{rr}{2x}+\frac{{x}^{3}}{2rr}$ {illeg} $+\frac{{r}^{4}}{{x}^{3}}+\frac{x}{2}=fh$. $\frac{{r}^{8}}{{x}^{6}}+\frac{{r}^{4}}{xx}+\frac{xx}{4}+\frac{{r}^{4}}{4xx}+\frac{xx}{2}+\frac{{x}^{6}}{4{r}^{4}}={bf}^{2}$ $-\frac{6{r}^{8}}{{x}^{6}}-\frac{3{r}^{4}}{2xx}+\frac{3xx}{4}+\frac{6{x}^{6}}{2{r}^{4}}$. $6{x}^{12}+16{x}^{8}{r}^{4}-10{r}^{8}{x}^{4}-24{r}^{12}$. $6{r}^{12}+$
$go:gb\colon\colon fk+go:hg$ . $de:dc\colon\colon fi+de:di$ . $dc×fi+dc×de+dc$ $\frac{dc×fi+dc×de}{de}=di=\frac{gb×fi+gb×dg+gb×go}{go}$. $\frac{\frac{rrz}{x+o}+\frac{{r}^{6}}{{x}^{4}+4{x}^{3}o}}{\frac{{r}^{4}}{{x}^{3}+3{x}^{2}o}}=\frac{\frac{rrz}{x}+\frac{{r}^{6}}{{x}^{4}}+\frac{rro}{x}}{\frac{{r}^{4}}{{x}^{3}}}$. $\frac{zx}{x+o}+\frac{{r}^{4}x}{{x}^{4}+4o{x}^{3}}=\frac{zx+3zo}{x}+\frac{{r}^{4}x+3{r}^{4}o}{{x}^{4}}+xo$. $\frac{z{x}^{5}+4z{x}^{4}o+{r}^{4}xx+{r}^{4}ox}{{x}^{5}+5o{x}^{4}}=\frac{z{x}^{5}+3zo{x}^{4}+{r}^{4}xx+3{r}^{4}ox}{{x}^{5}}$ $4z{x}^{9}+{r}^{4}{x}^{6}=5z{x}^{9}+5{r}^{4}{x}^{6}+3z{x}^{9}+3{r}^{4}{x}^{6}$. $4z{x}^{4}+7{r}^{4}=0$ $\frac{rrz}{x+o}+\frac{{r}^{6}}{{x}^{4}+4o{x}^{3}}\phantom{\rule{0.5em}{0ex}}in\phantom{\rule{0.5em}{0ex}}\frac{{r}^{4}}{{x}^{3}}=\frac{rrz}{x}+\frac{rro}{x}+\frac{{r}^{6}}{{x}^{4}}\phantom{\rule{0.5em}{0ex}}in\phantom{\rule{0.5em}{0ex}}\frac{{r}^{4}}{{x}^{3}+3o{x}^{2}}$ . $\frac{z}{xx+ox}\phantom{\rule{1em}{0ex}}\frac{+{r}^{4}}{{x}^{5}+4o{x}^{4}}=\frac{z+o}{xx+3ox}\phantom{\rule{1em}{0ex}}\frac{+{r}^{4}}{{x}^{5}+3o{x}^{4}}$. $\frac{z{x}^{3}+3zoxx+{r}^{4}}{{x}^{5}+4o{x}^{4}}=\frac{z{x}^{4}+o{x}^{4}+3oz{x}^{3}+x{r}^{4}+3o{r}^{4}}{{x}^{6}+6o{x}^{5}}$ $\begin{array}{cc}\begin{array}{ccccc}z{x}^{5}& +& 3zo{x}^{4}& +& {r}^{4}xx\\ & & 6zo{x}^{4}& +& 6{r}^{4}xo\end{array}& \begin{array}{cccccccc}=& z{x}^{5}& +& 2o{x}^{5}& +& 3oz{x}^{4}& +& xx{r}^{4}\\ & & & & +& 4oz{x}^{4}& +& 3ox{r}^{4}\\ & & & & & & +& 4oz{r}^{4}\end{array}\end{array}$ $9z{x}^{4}+6x{r}^{4}={x}^{5}+7z{x}^{4}+7x{r}^{4}$. $2z{x}^{4}-x{r}^{4}-{x}^{5}=0$. $z=\frac{{r}^{4}}{2{x}^{3}}+\frac{x}{2}$. $\frac{{r}^{4}}{{x}^{3}}:\frac{rr}{x}\colon\colon \frac{{r}^{4}+{x}^{4}}{2{x}^{3}}:hb$. $xx:rr\colon\colon \frac{{r}^{4}+{x}^{4}}{2{x}^{3}}\phantom{\rule{1em}{0ex}}\frac{{r}^{4}+{x}^{4}}{2rrx}=bh$. $\frac{{r}^{8}+2{r}^{4}{x}^{4}+{x}^{8}}{4{r}^{4}xx}\phantom{\rule{1em}{0ex}}\frac{{r}^{8}+2{r}^{4}{x}^{4}+{x}^{8}}{4{x}^{6}}=\frac{3{r}^{8}{x}^{4}+3{r}^{4}{x}^{8}+{x}^{12}+{r}^{12}}{4{r}^{4}{x}^{6}}=bf$. $6{x}^{12}+6{r}^{4}{x}^{8}-6{r}^{8}{x}^{4}-6{x}^{12}=0$. ${x}^{8}-{r}^{8}=0$ . ${x}^{4}-{r}^{4}=0$ . $xx-rr=0$ . $x=r$. therefore take $ag=gb=r$ , & ye greatest crookedness of ye line cb will be found at b . $bh=r=fh$ . $bf=r\sqrt{2}$ . $gh=w$ . $hf=\sigma$ . {illeg} $gb=y$ . $go=v$ . $w:v+\sigma \colon\colon y:v$ {illeg}. or, $y:v\colon\colon w-y:\sigma =\frac{vw-vy}{y}$. $\frac{go×gh-go×bg}{bg}=fh$ . $\frac{{y}^{3}w}{rr}-\frac{{y}^{4}}{rr}=\sigma y$ . $\frac{2yyw}{rr}-\frac{3yyy}{rr}=0$ . $w=\frac{3y}{2}$ . $\frac{3{r}^{4}+{x}^{4}}{2rrx}=0$ . ${x}^{3}=\frac{{r}^{6}}{{y}^{3}}$ . $\frac{3rr}{2x}+\frac{{r}^{4}}{2{y}^{3}}=\frac{3y}{2}+\frac{{r}^{4}}{2{y}^{3}}$. $bf=p$ . $al=q$ . $gb=y$ . $go=v$ . $ag=x$. $bf=p$. {illeg} $v:y\colon\colon q+v-x:\frac{qy+yv-yx}{v}=p$. $\frac{gb×al+gb×go-gb×ga}{go}=$ {illeg} $yq+\frac{{y}^{4}}{rr}-rr=\frac{p{y}^{3}}{rr}$ . $\frac{{y}^{4}}{rr}+3rr-2yq=0$. $q=\frac{{y}^{3}}{2rr}+\frac{3rr}{2y}$ . as before. or $q=\frac{yy}{2x}+\frac{3x}{2}=\frac{yy+3xx}{2x}$ $gl=\frac{yy+xx}{2x}$. $w=v+x=\frac{dyy+ey+2fxy+bx+d{x}^{2}+2ayx}{d+dx+2ay}$ $+bx+dxw+2ayw+dyy+ey+2fxy+bx+d{x}^{2}+2ayx=0$. $\begin{array}{lllllll}fxx& +& ex& +& c& =& 0\\ & +& dyx& +& by\\ & & & +& dyx\end{array}$. $\begin{array}{l}{w}^{2}-2wx+xx\\ {w}^{2}-\frac{2wc-2wby-2wayy-2wdyx}{fx+e}\end{array}$ $\frac{ex+dyx+c+by+ayy}{f}$. $\frac{+fxx+ex+dyx+c+by}{a}$ $xy=aa$. $x=\frac{aa}{y}$. $y=\frac{aa}{x}$. $v=\frac{yy}{x}$. $\frac{qaa}{x}+\frac{aayy}{xx}-\frac{{a}^{4}}{xy}=0$ . $\frac{qaay}{xx}-\frac{2{a}^{2}{y}^{3}}{{x}^{3}}-\frac{2{a}^{4}}{xy}$. $q=$ $\begin{array}{lllllll}fxx& +& ex& +& c& =& 0\\ & +& dyx& +& by\\ & & & +& dyx\end{array}$. $\frac{2fxy+dyy+ey}{b+dx+2ay}=v$. $qy+\frac{2fxyy+d{y}^{3}+eyy-2pfxy-pdyy-pey}{b+dx+2ay}-yx=0$ [36] $\frac{2bqyf+4aqyyf+2d{y}^{3}f+2eyyf-2pdyyf-2peyf-2dyc-2dbyy-2da{y}^{3}}{2pfy+by+2ayy+dey+ddyy-dfqy-2ffyy}=x$ $x=\frac{-e-dy}{2f}\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}\sqrt{\frac{\begin{array}{c}ee+2dey+ddyy\\ -4cf-4bfy-4afy\end{array}}{4ff}}$ {illeg} $eyf+2aeyyf+deey+ddyef-defqy$ {illeg} {illeg} $\frac{bqyf+2aqyyf+d{y}^{3}f+eyyf-pdyyf-peyf-dcy-dbyy-da{y}^{3}}{2pffy+bfy+2afyy+edy+ddyy-dfqy-2ffyy}=x$. $2bqyff+4aqyyff+2d{y}^{3}ff+2ffey-2pdffyy-2peffy-2dcfy-dbfyy-2daf{y}^{3}$. $4befy+2aefyy+edfy+cddfyy-edfqy+eddyy+{d}^{2}{y}^{3}-ddfqyy$
<6v>
[37] $yy${illeg} $vv+yy=ss=yy+2oy$ $\begin{array}{ccccc}\stackrel{\text{.}}{vv}& +& \stackrel{\text{.}}{xy}& +& \stackrel{\text{.}}{ay}\\ & +& \stackrel{\text{.}}{ax}& +& \stackrel{\text{.}}{yy}\end{array}=ss=\stackrel{\text{.}}{vv}-2ov+\stackrel{\text{.}}{xy}+xo+\stackrel{\text{.}}{ax}+\stackrel{\text{.}}{ay}+ao+\stackrel{\text{.}}{yy}+\stackrel{\text{.}}{2oy}$. $2ov=ox+ao+2oy$ . $=vv-2ov+zy+zo+az+o$ = {illeg} $vv+2ov+oo+rs+ra+as+ss$. {illeg} $v=\frac{\begin{array}{cccccccc}& xy& +& ax& +& ay& +& yy\\ -& rs& -& ar& -& as& -& ss\end{array}}{zo}$. $xy-rs${illeg}. $v=xy$
$yy-ss=2o$ {illeg} $xy+ax+ay+2yy+vv-2vy=ss=\xi z+a\xi +az+2zz+vv-2vz$ $\frac{xy-\xi z+ax-a\xi }{2y-2z}=b$. $xy-\xi z-2by+2bz=0$ . $xy-2by=0$ $x=2b$ . $\frac{xxyy-\xi \xi zz}{{a}^{2}x-\xi {a}^{2}}=b$ . $\frac{1}{2}y+\frac{1}{2}x$ $\frac{\begin{array}{cccccc}& 4{x}^{2}& -& ax& \phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}& 2\sqrt{2{x}^{4}-a{x}^{3}}\\ -& 4{z}^{2}& +& az& \phantom{\rule{0.5em}{0ex}}\underset{\cap }{O}\phantom{\rule{0.5em}{0ex}}& 2z\sqrt{zzz-az}\end{array}}{-2z+2x}=v$ $\frac{4{x}^{2}-4zz}{2x-2z}=b$. $\begin{array}{ccccccc}2{x}^{2}& -& 2{x}^{3}& -& bx& -& bz\\ 2& & 0& & 1& & 0\end{array}$ /$b=4x$\ $2{x}^{4}+2{z}^{4}-a{x}^{3}-a{z}^{3}-2zz\sqrt{4{x}^{2}{z}^{2}-ax{z}^{2}+aazx-2azxx}=ccxx-2{c}^{2}zx+cczz$ \ $\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}\sqrt{2{x}^{4}-a{x}^{3}}\phantom{\rule{0.5em}{0ex}}\underset{\cap }{O}\phantom{\rule{0.5em}{0ex}}\sqrt{2{z}^{4}-a{z}^{3}}=cx-cz$/. $2x+2z-\frac{1}{2}a\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}\sqrt{2xx-ax}\phantom{\rule{0.5em}{0ex}}\underset{\cap }{O}\phantom{\rule{0.5em}{0ex}}\sqrt{\frac{1}{2}zz-\frac{axz}{4x}}$ \$2xx+2zx-ax$/ $2x\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}\sqrt{2xx-ax}\phantom{\rule{1em}{0ex}}\frac{+ax\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}a\sqrt{2xx-ax}}{\phantom{\rule{0.5em}{0ex}}\underset{\cap }{O}\phantom{\rule{0.5em}{0ex}}2\sqrt{xx-ax}}$ $\frac{-4xy+2ax-2xx+ay-2xy}{2x-2y}=\frac{-xx\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}6x\sqrt{2xx-ax}+3ax\phantom{\rule{0.5em}{0ex}}\underset{\cap }{O}\phantom{\rule{0.5em}{0ex}}x\sqrt{2xx-ax}}{\phantom{\rule{0.5em}{0ex}}\underset{\cap }{O}\phantom{\rule{0.5em}{0ex}}2\sqrt{2xx-ax}}$ $\frac{\begin{array}{c}-8xx\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}6x\sqrt{2xx-ax}\\ +3ax\phantom{\rule{0.5em}{0ex}}\underset{\cap }{O}\phantom{\rule{0.5em}{0ex}}a\sqrt{2xx-ax}\end{array}}{\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}2\sqrt{2xx-ax}}$. $\frac{64{x}^{3}-48axx+9aax}{4x-2a}$ $2xx-ax+\frac{\sqrt{4zzx-2zz}}{2x}$ $-xx+dy+yy=0$ . $\frac{2bϩz\sqrt{cc-b}}{cc}+\frac{dϩb}{c}+\frac{2abϩ}{c}+\frac{2bϩz\sqrt{cc-b}}{cc}$. $-\frac{d}{2}=+a$ . $b=0$ . $c=$ any finite line. as {illeg} $x=ϩ$. $y=-\frac{1}{2}d+z$ . ${ϩ}^{2}+\frac{1}{4}dd\mp dz-zz$ ${y}^{3}=axx+aax$ . $y+a=\varrho$ . ϩ ${\varrho }^{3}-3a{\varrho }^{2}+3aa\varrho -axx+aax-{a}^{3}=0$ $\begin{array}{ccc}\begin{array}{c}\phantom{0}450& \text{.}\\ 225\phantom{0}\end{array}& 48xx& \phantom{\rule{2em}{0ex}}\begin{array}{l}\underset{_}{54×16×27}\\ 324\\ \underset{_}{54\phantom{0}}\\ 864\end{array}\\ & \begin{array}{l}240\\ \underset{_}{48\phantom{0}}\\ 720\end{array}\end{array}$ $-{ϩ}^{3}{b}^{3}=0$ . $\frac{+6ϩabc+cϩ\sqrt{cc-bb}×a-3abcϩ}{cc}=0$ $\begin{array}{ccc}\begin{array}{c}105\\ \underset{_}{3\phantom{00}}\\ 2025\end{array}& \begin{array}{c}\phantom{0}75\\ \underset{_}{15\phantom{0}}\\ 225\end{array}& \begin{array}{c}1125\\ \underset{_}{225\phantom{0}}\\ 3375\end{array}\\ & & & \begin{array}{c}\underset{_}{16}\\ 96\\ \underset{_}{16\phantom{0}}\\ 256\end{array}\end{array}$ $db=a$. $ba=b$ . $ac=c$ . $ad=d$ . $\frac{dd}{b}=r$ , $lat\phantom{\rule{1em}{0ex}}rec:Parab\phantom{\rule{1em}{0ex}}dbe$ . $\frac{dd}{c}=s=lat:rec:Parab:dce$ . $\frac{cc}{b}=lat:rec:Par\phantom{\rule{1em}{0ex}}bgqc$ {illeg}. {illeg} $lq=a$ . $axx={y}^{3}$. $y=x+z$ . $-axx+{x}^{3}+3{x}^{2}z+3x{z}^{2}+{z}^{3}=0$ . $\sqrt{caxx}-x=z$. $\frac{+2axy-3xxy-6xyy-3{y}^{3}}{3yy+6yx+3xx}=v=\frac{2axy}{3yy+6yx+3xx}$ ${z}^{3}+3x{z}^{2}+3xxz+{x}^{3}-dxx=0$ $a=8$ . $x=1$ . {illeg} $-7+3z+3zz+{z}^{3}=0$ $\begin{array}{ccccccc}{y}^{3}& +& ayy& +& 2ab& +& abb\\ & +& 2b& +& bb\end{array}$ $\begin{array}{l}zz+4z+7\text{.}\phantom{\rule{2em}{0ex}}z=2\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}\sqrt{4-7}\\ +448+192z+24zz+{z}^{3}=0\text{.}\\ zz+20z+11 2\end{array}$ $\begin{array}{c}\begin{array}{cccc}3x=a+2b\text{.}& 3xx=2ab+bb& {b}^{3}-dbb=abb\text{.}& b-d=a\end{array}\\ \begin{array}{cc}3x-2b=a=\frac{3xx-bb}{2b}& -6bx+3bb+3xx=0\text{.}\\ x=a& a=b=x\text{.}\phantom{\rule{3em}{0ex}}b-a=d\text{.}\end{array}\end{array}$ $\begin{array}{c}y+a×y+b×y+b\\ \begin{array}{ccccccc}yyy& +& ayy& +& 2aby& \phantom{\rule{0.5em}{0ex}}& abb\\ & +& 2byy& +& bby\end{array}\end{array}$ $\begin{array}{c}ax=yy\text{.}\\ -ax-yy+2ay+aa=0\text{.}\\ \sqrt{ax}=y+a\text{.}\end{array}$ $\begin{array}{cccccccc}-& caϩr& -& abzc& +& ddcc& -& 2dbcϩ\\ +& 2dczr& +& cczz& +& bbϩz& -& 2aabϩ\\ -& 2bϩzr& -& bbzz& +& 2azcd\\ +& 2aazr\end{array}$. $\begin{array}{c}-2b\sqrt{cc-bb}\text{.}\phantom{\rule{2em}{0ex}}b=c\text{.}\\ -2dbc-2aab\text{.}\\ aa{c}^{4}-aabbcc=4bbccdd+8\\ 2dc+2ac=0\text{.}\phantom{\rule{2em}{0ex}}d=\frac{ac}{-c}\text{.}\phantom{\rule{2em}{0ex}}d=-a\text{.}\end{array}$ $af=a$ . $ag=b$ . $ab=\varrho$ . $bc=z$ . $a:b:bh=\frac{a\varrho -ab}{b}:\varrho -b=gb$ . $ch=ϩb-a\varrho +ab$ . $fg=c$ . $ag=b$ . $fg=c$ . $fh=\frac{cϩ}{b}$ . $af=\sqrt{cc-bb}$ . $\frac{\varrho -b\phantom{\rule{0.5em}{0ex}}in\phantom{\rule{0.5em}{0ex}}\sqrt{cc-bb}}{b}=bh$ . $\frac{bϩ-\varrho +b×\sqrt{cc-bb}}{b}=ch$ . $\frac{bϩ-\varrho +b\phantom{\rule{0.5em}{0ex}}in\phantom{\rule{0.5em}{0ex}}\sqrt{cc-bb}}{c}=cd$ . $\frac{abϩ-aa\varrho +aab+cc\varrho }{bc}=df$ . $cc=$ {illeg} $aa+bb$ . $\frac{ϩb-a\varrho +ab}{\sqrt{aa+bb}}=cd=x$. $\frac{aϩ-aa+bϩ}{\sqrt{aa+bb}}=df=y$. $\frac{bϩ+ab-x\sqrt{aa+bb}}{a}=\varrho =\frac{y\sqrt{aa+bb}-aa-aϩ}{b}$ $bbϩ+aaϩ=ay\sqrt{aa+bb}-{a}^{3}${illeg} $-abb+x\sqrt{aa+bb}$ . a {illeg} $ϩ=\frac{ay-a\sqrt{aa+bb}+bx}{\sqrt{aa+bb}}$ . Or $ϩ=\frac{ay-ac+x\sqrt{cc-aa}}{c}$ . /$\varrho =\frac{y\sqrt{cc-aa}-ax}{c}$ \ $\varrho =\frac{cy}{b}-\frac{aay}{bc}-\frac{xa}{c}$ {illeg} $c=5$ . $a=3$ . $ϩ=\frac{3y-15+4x}{5}$ . $\varrho =\frac{4y-3x}{5}$ . {illeg}${y}^{4}$ ${ϩ}^{4}={\varrho }^{3}$ . |5 | $\begin{array}{ccc}27{y}^{3}-405yy+2025y-3375& =& 256{y}^{4}-768{y}^{3}x+864yyxx-432y{x}^{3}+81{x}^{4}\\ +108yyx-1080yx+2700x\\ 144yxx-720xx\\ +64{x}^{3}\end{array}$ $\varrho \varrho ϩ={a}^{3}$ . {illeg} $c=5r$ . $a=4r$ . $ϩ=\frac{4y-20r+3x}{5}$ . $\varrho =\frac{3y-4x}{5}$ {illeg} $=2rr$ . {illeg} $=rr$ .
$\frac{9yy-24xy+16xx}{25}$ . $\begin{array}{c}\begin{array}{ccccccccc}36{y}^{3}& -& 96xyy& +& 64xxy& +& 48{x}^{3}& =& 225{a}^{3}\\ & +& 27\hfill & -& 72xxy\end{array}\\ -180ryy+480rxy-360rxx\end{array}$ $12{y}^{3}$ {illeg}3 {illeg}$yy$ $\begin{array}{c}\begin{array}{ccccccccc}36{y}^{3}& -& 69xyy& -& 8xxy& +& 48{x}^{3}& =& 0\end{array}\\ -180ryy+480rxy-360rxx\\ rrr\end{array}$.
<7r>
$\begin{array}{cc}\begin{array}{llllll}& 2bqff& +& 4aqyff& & \\ -& 2qeff& & & & \\ -& 2dcf& & & +& {d}^{3}yy\\ & & -& dbyf\\ +& bef& +& 2aefy\\ +& dee& +& 2ddey\\ -& defq& -& ddfqy\end{array}& \begin{array}{cc}=& \begin{array}{c}2pff+bf+2afy\\ +de+ddy-dfq-2ffy\end{array}in\sqrt{ee+2edy+ddyy-4af-4bfy-4a}\end{array}\end{array}${illeg} $\begin{array}{cc}\begin{array}{llllllll}& 4bbqq{f}^{4}& +& 16abqq{f}^{4}y& +& 6bqff{d}^{3}yy& +& 8aqff{d}^{3}{y}^{3}\\ -& 8bq{f}^{3}dc& -& 4bbq{f}^{3}d& -& 4{d}^{4}cf& -& 2{d}^{4}bf\\ +& 4bbq{f}^{3}e& +& 8ae{f}^{3}bq& +& 2bef{d}^{3}& +& 4aef{d}^{3}\\ & & +& 6bqffdde& & & +& 4{d}^{5}e\\ -& 4bqq{f}^{3}de& -& 4bqq{f}^{3}dd& & & -& 2{d}^{5}fq\\ & & & & +& 16aaqq{f}^{4}\\ & & +& 4ddcffb& -& 8aq{f}^{3}db\\ +& 4ddccff& -& 16dc{f}^{3}aq& +& 16aa{q}^{3}{f}^{3}& +& {d}^{6}{y}^{4}\\ -& 4dcffbe& -& 8dcffae& & \\ -& 4ddcfee& -& 8{d}^{3}fce& -& 8aqq{f}^{3}dd\\ +& 4ddcffeq& +& 4{d}^{3}cffq& +& ddbbff\\ +& bbeeff& +& 8aqbe{f}^{3}& -& 4aedbff\\ +& 2b{e}^{3}fd& -& 2bbeffd& -& 4{d}^{3}ebf\\ +& 2beeffdq& +& 4abeeff& & \\ +& dd{e}^{4}& +& 4bddeef& +& 4aaeeff\\ -& 2d{e}^{3}fq& & & +& 8aeeddf\\ +& ddeeffqq& & & & \\ & & -& 2ddeebf\\ & & +& 4ad{e}^{3}f& +& 6{d}^{4}ee\\ & & +& 4{d}^{3}{e}^{3}& -& 10{d}^{4}efq\\ & & & & +& {d}^{4}ffqq\\ & & -& 8ade{f}^{3}qq& +& 12aeddfq\\ & & +& 2ddeffqb\\ & & +& 4adeeffq\\ & & -& 6{d}^{3}eefq\\ & & +& 2{d}^{3}effqq\\ & & -& 8pp{f}^{4}edy\end{array}& \begin{array}{lllllll}=& & 4pp{f}^{4}& +& 4pb{f}^{3}& & \\ & +& 4bb{f}^{3}& +& 8pa{f}^{3}q& +& 4aaffyy& in& \begin{array}{c}ee\phantom{\rule{2em}{0ex}}+\\ -4bf-4cf\end{array}\\ & +& 4pdeff& +& 4pddff& +& 4afdd\\ & +& 4d{f}^{3}qq& +& 4abff& +& {d}^{4}\\ & -& 4pdq{f}^{3}& +& 2ddbf\\ & -& 8pq{f}^{4}& +& 4afde\\ & +& bbff& -& 4adffq\\ & +& 2debf& -& 8aq{f}^{3}\\ & -& 2dbffq& +& 2e{d}^{3}\\ & -& 4bq{f}^{3}& -& 2{d}^{3}fq\\ & +& ddee& -& 4ddffq\\ & -& 2ddefq\\ & -& 4deffq\\ & +& ddffqq\\ & +& {f}^{4}qq\end{array}\end{array}$
$bc=x$. $cd=y$. $ef=a$. $ea=b$. $ad=c$. $\sqrt{xx+yy-cc}\phantom{\rule{0.5}{0ex}}\left(={ab}^{2}\right)+b=eb$ . $\frac{ax}{b+\sqrt{xx+yy-cc}}=ch$ . $aa+xx+yy-cc+bb+2b\sqrt{xx+yy-cc}={bf}^{2}$ $\frac{c\sqrt{aa+xx+yy-cc+bb+2b\sqrt{xx+yy-cc}}+ax}{b+\sqrt{xx+yy-cc}}=y$. $ccaa+ccxx+2ccyy-{c}^{4}+bbcc+2bcc\sqrt{xx+yy-cc}=bbyy\phantom{\rule{0.5em}{0ex}}\begin{array}{l}+2b\sqrt{{y}^{2}\phantom{\rule{0.5em}{0ex}}\text{(&c)}}\\ -2a\sqrt{xx}\end{array}\phantom{\rule{0.5em}{0ex}}+xxyy+{y}^{4}$. {illeg} $yy+xx+bb-cc+2b\sqrt{xx+yy-cc}\phantom{\rule{1em}{0ex}}\frac{-ccaa-2axy\sqrt{xx+yy}-aaxx-ax}{yy-cc}$
$ac=x$. $bc=y$. $af=z$. $bf=\frac{zz}{a}$. \{illeg}/ $df=\frac{a}{2}$ . $de=\frac{aa+4zz}{2a}$ . $fe=\frac{2zz}{a}:z\colon\colon x:\frac{aa+zz}{2a}$ $2aax=2aaz+4a{z}^{3}$
$bf=z$ $af=\sqrt{az}$ . $2z$ $ad=\sqrt{\frac{aa}{4}+az}$ $z:\frac{a}{4}+z\colon\colon xx:\frac{aa}{4}+az+zz$ . $4{z}^{3}+4azz+aaz=axx+4zxx$ .
☞ To know whi|e|ther ye changing of ye sines {sic} of an Equation change ye nature of ye crooked line signified by yt Equation observe yt
If ye sines {sic} of every other terme (of yt Equa{illeg}|t|ion ordered{illeg} according t{illeg}|o| {either} of ye undetermined quanti{illeg}|t|ys) be changed ye nature of ye line is not changed. but if ye signes of some signes bee changed but not in {illeg}|e|ve{ry} other termes (of it ordered according to one of ye unknowne quantitys) ye nature of y{illeg}|e| line is changed.
If ye knowne qua\n/titys are every where divers, & one of ym be blotted {illeg} out yt produceth a line, when one terme is already wanting
Those lines may bee defined ye same whose natures {illeg} \may be/ expressed by ye same equation although angles made by x & y are not ye same.
In ye {illeg}|H|yperbola ye area of it beares ye same respect to its Asymptote wch a logarithme {di}{illeg} number.
To make ye equation ${x}^{3}-a{x}^{2}+abx-abc=0$ . be divisible by $x-c=0$ . suppose $c=x$ , yn tis ${c}^{3}-acc+0=0$ $c=a$ . therefore write c in steade of a & it is ${x}^{3}-c{x}^{2}+cbx-bcc=0$ . wch is divisible by $x-c$ To make ye same Equation divisible by $xx-2ax+ac=0$ Suppose it to bee divided by it & ye ration will bee $\begin{array}{ccccccccccc}xx-2ax+ab\right)& & {x}^{3}& -& a{x}^{2}& +& abx& -& abc& =& 0\left(x+a\\ & -& {x}^{3}& +& 2a{x}^{2}& -& abx\\ & & 0& +& a{x}^{2}& +& 0& -& abc\\ & & & -& axx& +& 2aax& -& aab\\ & & & & 0& +& 2aax& -& aab& -& abc\hfill \end{array}$. The quote is $2aax-abc-aab$ wch have vanished therefore to m{illeg}|a|ke soe suppose each terme $=0$ & the{illeg} will be $2aax=0$ & $abc+aba=0$ both include $a=0$ . Which since it cannot happen ye equation cannot be divided ye one by ye othe{r}
The rootes of {illeg}|t|wo divers equations may easily be added to substracted from multiplyed {&c} by one another while they are unknowne.
[38] That ye penultimate terme of ye Equation ${x}^{3}\ast -{a}^{2}x+{b}^{3}=0$ . bee wanting I multiply & then suppose x a knowne {illeg} quantity & y an unknowne {illeg} ${x}^{3}$ {illeg} $+{b}^{3}{y}^{3}$ {illeg}{ ${d}^{3}-ayy+{y}^{3}=0$ .} by this {having}{illeg} x {illeg} {illeg} {illeg}
<7v>
[39] $dd{\varrho }^{2}+2d\varrho ϩ\sqrt{ee-dd}+eeϩϩ-ddϩϩ=eexx$. ${d}^{3}{\varrho }^{3}+3dd\varrho \varrho ϩ\sqrt{ee-dd}+3d\varrho {ϩ}^{2}ee-3d\varrho {ϩ}^{2}dd\phantom{\rule{0.5em}{0ex}}\begin{array}{c}+ee\\ -dd\end{array}{ϩ}^{3}\sqrt{ee-dd}=${${e}^{3}{x}^{3}$} [40] $6dd\varrho \varrho ϩϩee-6{d}^{4}\varrho \varrho ϩϩ\phantom{\rule{0.5em}{0ex}}\begin{array}{c}+4ee\\ -4dd\end{array}d\varrho {ϩ}^{3}\sqrt{ee-dd}+{e}^{4}{ϩ}^{4}-2eedd{ϩ}^{4}+{d}^{4}{ϩ}^{4}={e}^{4}{x}^{4}$ [41] $\begin{array}{c}+10ee\\ -10dd\end{array}{d}^{3}{\varrho }^{3}ϩϩ\phantom{\rule{1em}{0ex}}\begin{array}{c}+10ee\\ -10dd\end{array}dd\varrho \varrho {ϩ}^{3}\sqrt{ee-dd}\phantom{\rule{1em}{0ex}}\begin{array}{l}+5{e}^{4}\\ -10eedd\\ +5{d}^{4}\end{array}d\varrho {ϩ}^{4}\phantom{\rule{1em}{0ex}}\begin{array}{l}+{e}^{4}\\ -2eedd\\ +{d}^{4}\end{array}{ϩ}^{5}\sqrt{ee-dd}={e}^{5}{x}^{5}$ [42] $ccee-2cedϩ+ddϩϩ\phantom{\rule{1em}{0ex}}\begin{array}{l}+2ce\varrho \\ -2dϩ\varrho \end{array}\sqrt{ee-dd}\phantom{\rule{1em}{0ex}}\begin{array}{l}+ee\varrho \varrho \\ -dd\varrho \varrho \end{array}=eeyy$. $\begin{array}{llllllllllll}+& {c}^{3}{e}^{3}& +& 3ccee\varrho & \sqrt{ee-dd}& +& 3c{e}^{3}\varrho \varrho & +& ee& {\varrho }^{3}\sqrt{ee-dd}& =& {e}^{3}{y}^{3}\\ -& 3cceedϩ& -& 6cedϩ\varrho & & -& 3eedϩ\varrho \varrho & -& dd\\ +& 3ceddϩϩ& +& 3ddϩϩ\varrho & & -& 3cedd\varrho \varrho \\ -& {d}^{3}{ϩ}^{3}& & & & +& 3{d}^{3}ϩ\varrho \varrho \end{array}$. $\begin{array}{llllllllllllllll}+& {c}^{4}{e}^{4}& +& 4{c}^{3}{e}^{3}& \varrho \sqrt{ee-dd}& +& 6cc{e}^{4}& \varrho \varrho & +& 4c{e}^{3}& {\varrho }^{3}\sqrt{ee-dd}& +& {e}^{4}& {\varrho }^{4}& =& {e}^{4}{y}^{4}\\ -& 4{c}^{3}{e}^{3}dϩ& -& 12cceedϩ& & -& 12c{e}^{3}dϩ& & -& 4cedd& & -& 2eedd\\ +& 6cceeddϩϩ& +& 12ceddϩϩ& & +& 6eeddϩϩ& & -& 4eedϩ& & +& {d}^{4}\\ -& 4ce{d}^{3}{ϩ}^{3}& -& 4{d}^{3}{ϩ}^{3}& & -& 6cceedd& & +& 4{d}^{3}ϩ\\ +& {d}^{4}{ϩ}^{4}& & & & +& 12cce{d}^{3}ϩ\\ & & & & & -& 6{d}^{4}ϩϩ\end{array}$. $\begin{array}{lllllllllllllllllll}& {c}^{5}{e}^{5}& +& 5{c}^{4}{e}^{4}& \varrho \sqrt{ee-dd}& +& 10{c}^{3}{e}^{5}& \varrho \varrho & +& 10cc{e}^{4}& {\varrho }^{3}\sqrt{ee-dd}& +& 5c{e}^{5}& {\varrho }^{4}& +& {e}^{4}& {\varrho }^{5}\sqrt{ee-dd}& =& {e}^{5}{y}^{5}\\ -& 5{c}^{4}{e}^{4}dϩ& -& 20{c}^{3}{e}^{3}dϩ& & -& 30cc{e}^{4}dϩ& & -& 20c{e}^{3}dϩ& & -& 5dϩ{e}^{4}& & -& 2eedd\\ +& 10{c}^{3}{e}^{3}dd{ϩ}^{2}& +& 30cceedd{ϩ}^{2}& & +& 30c{e}^{3}ddϩϩ& & +& 10eeddϩϩ& & -& 10c{e}^{3}dd& & +& {d}^{4}\\ -& 10ccee{d}^{3}{ϩ}^{3}& -& 20ce{d}^{3}{ϩ}^{3}& & -& 10ee{d}^{3}{ϩ}^{3}& & -& 10cceedd& & +& 10{d}^{3}ϩee\\ +& 5ce{d}^{4}{ϩ}^{4}& +& 5{d}^{4}{ϩ}^{4}& & -& 10{c}^{3}{e}^{3}dd& & +& 20ce{d}^{3}ϩ& & +& 5ce{d}^{4}\\ -& {d}^{5}{ϩ}^{5}& & & & +& 30ccee{d}^{3}ϩ& & -& 10{d}^{4}ϩϩ& & -& 5ϩ{d}^{5}\\ & & & & & -& 30ce{d}^{4}ϩϩ\\ & & & & & +& 10{d}^{5}{ϩ}^{3}\end{array}$. [43] $\begin{array}{llllllll}& cceeϩ\sqrt{ee-dd}& +& 2c{e}^{3}ϩ\varrho & +& eeϩ\varrho \varrho \sqrt{ee-dd}& =& {e}^{3}xyy\\ & & & & -& 3ddϩ\\ +& dd{ϩ}^{3}& -& 4ceddϩ\varrho & & \\ & & & \\ & & +& cceed\varrho \end{array}$. [44] ${e}^{4}xxyy=\begin{array}{ll}-& 2c{e}^{3}d{ϩ}^{3}\\ +& 2ce{d}^{3}{ϩ}^{3}\end{array}\phantom{\rule{1em}{0ex}}\begin{array}{lll}-& 2dee\varrho {ϩ}^{3}& \sqrt{ee-dd}\\ +& 4{d}^{3}\varrho {ϩ}^{3}\\ +& 2ccdee\varrho ϩ\\ & \end{array}\phantom{\rule{1em}{0ex}}\begin{array}{ll}+& 4cd{e}^{3}\varrho \varrho ϩ\\ -& 6c{d}^{3}e\end{array}\phantom{\rule{1em}{0ex}}\begin{array}{ll}+& 2dee{\varrho }^{3}ϩ\sqrt{ee-dd}\\ -& 4{d}^{3}\end{array}$. $\begin{array}{l}\\ & & -& 6ddce\varrho \varrho ϩ\sqrt{{e}^{2}-{d}^{2}}& -& 5ddee{\varrho }^{3}ϩ& +& {c}^{3}{e}^{3}ϩ\sqrt{ee-dd}& =& {e}^{4}x{y}^{3}\\ -& 4{d}^{4}\varrho {ϩ}^{3}& +& 3c{e}^{3}\varrho \varrho ϩ\sqrt{ee-dd}& +& 4{d}^{4}{\varrho }^{4}ϩ& +& cedd{ϩ}^{3}\\ +& 3cc{e}^{4}\varrho ϩ& -& 3cedd\varrho \varrho ϩ\sqrt{ee-dd}& +& {e}^{4}{\varrho }^{3}ϩ\\ +& 3ddee\varrho {ϩ}^{3}\end{array}$ . $\begin{array}{llllllllllll}-& 8dd{c}^{3}{e}^{3}\varrho ϩ& -& 18ddccee\varrho \varrho ϩ\sqrt{{e}^{2}-{d}^{2}}& -& 2ddc{e}^{3}{\varrho }^{3}ϩ& -& 6ddee{\varrho }^{4}ϩ\sqrt{ee-dd}& +& {c}^{4}{e}^{4}ϩ\sqrt{ee-dd}& =& {e}^{5}x{y}^{4}\\ -& 16{d}^{4}ce\varrho {ϩ}^{3}& -& 10{d}^{4}\varrho \varrho {ϩ}^{3}& +& 16{d}^{4}ce{\varrho }^{3}ϩ& +& 5{d}^{4}{\varrho }^{4}ϩ& +& 6ccddee{ϩ}^{3}\\ +& 4{c}^{3}{e}^{5}\varrho ϩ& +& 6cc{e}^{4}\varrho \varrho ϩ& +& 4c{e}^{5}{\varrho }^{3}ϩ& +& {e}^{4}{\varrho }^{4}ϩ& +& {d}^{4}{\varrho }^{5}\\ +& 12c{e}^{3}dd\varrho {ϩ}^{3}& +& 6ddee\varrho \varrho {ϩ}^{3}\end{array}$. $\begin{array}{llllllllllllll}& {c}^{5}{e}^{5}ϩ\sqrt{ee-dd}& +& 5{c}^{4}{e}^{6}\varrho ϩ& +& 10{c}^{3}{e}^{5}\varrho \varrho ϩ\sqrt{ee-dd}& +& 10cc{e}^{6}{\varrho }^{3}ϩ& +& 5c{e}^{5}{\varrho }^{4}ϩ\sqrt{ee-dd}& +& {e}^{6}{\varrho }^{5}ϩ& =& {e}^{6}x{y}^{5}\\ +& 10{c}^{3}{e}^{3}{ϩ}^{3}dd& -& 10{c}^{4}{e}^{4}dd\varrho ϩ& +& 30c{e}^{3}dd\varrho \varrho {ϩ}^{3}& -& 50cc{e}^{4}dd{\varrho }^{3}ϩ& -& 30c{e}^{3}dd{\varrho }^{4}ϩ& -& 8{e}^{4}dd\\ +& 5ce{d}^{4}{ϩ}^{5}& +& 30cc{e}^{4}dd\varrho {ϩ}^{3}& -& 30{c}^{3}{e}^{3}dd\varrho \varrho ϩ& +& 10dd{e}^{4}{\varrho }^{3}{ϩ}^{3}& +& 25ce{d}^{4}{\varrho }^{4}ϩ& +& 13ee{d}^{4}\\ & & -& 40ccee{d}^{4}\varrho {ϩ}^{3}& -& 50ce{d}^{4}\varrho \varrho {ϩ}^{3}& -& 30ee{d}^{4}{\varrho }^{3}{ϩ}^{3}& & & -& 6{d}^{6}\\ & & +& 5ee{d}^{4}\varrho {ϩ}^{5}& & & +& 40ccee{d}^{4}{\varrho }^{3}{ϩ}^{3}\\ & & -& 6{d}^{6}\varrho {ϩ}^{5}& & & +& 20{d}^{6}{\varrho }^{3}{ϩ}^{3}\end{array}$. $\begin{array}{llllllllllll}-& 3{c}^{2}eeeed{ϩ}^{3}& -& 6c{e}^{3}d{ϩ}^{3}\varrho \sqrt{ee-dd}& -& 3{e}^{4}d\varrho \varrho {ϩ}^{3}& +& 6cd{e}^{3}{\varrho }^{3}ϩ\sqrt{ee-dd}& +& {e}^{4}d{\varrho }^{4}ϩ& =& {e}^{5}xx{y}^{3}\\ -& {d}^{3}ee{\varrho }^{5}& +& 12ce{d}^{3}\varrho {ϩ}^{3}& +& 12ee{d}^{3}\varrho \varrho {ϩ}^{3}& -& 12c{d}^{3}e& & \\ +& 3ccee{d}^{3}{ϩ}^{2}& +& 2{c}^{3}{e}^{3}d\varrho ϩ& -& 10{d}^{5}\varrho \varrho {ϩ}^{3}& & \\ +& {d}^{5}{ϩ}^{5}& & & +& 6ccd{e}^{4}\varrho \varrho ϩ& & & -& 5ee{d}^{3}\\ & & & & +& 3ccee{d}^{3}ϩ& & & +& 4{d}^{5}\\ & & & & & \end{array}$. $\begin{array}{llllllllllllll}-& 4{c}^{3}{e}^{5}d{ϩ}^{3}& -& 12cc{e}^{4}d{ϩ}^{3}\varrho \sqrt{ee-dd}& -& 12cd{e}^{5}{ϩ}^{3}{\varrho }^{2}& -& 4d{e}^{4}{ϩ}^{3}{\varrho }^{3}\sqrt{ee-dd}& +& 8c{e}^{5}dϩ{\varrho }^{4}& +& 2d{e}^{4}\varrho ϩ\sqrt{ee-dd}& =& {e}^{6}xx{y}^{4}\\ -& 4c{e}^{3}{d}^{3}{ϩ}^{5}& +& 24ccee{d}^{3}{ϩ}^{3}& & & & & & & -& 8{d}^{3}ee{\varrho }^{5}ϩ\\ +& 4{c}^{3}{e}^{3}{d}^{3}{ϩ}^{3}& -& 4{d}^{3}ee{ϩ}^{5}& -& 4ce{d}^{5}{ϩ}^{3}& +& 20{d}^{5}{ϩ}^{3}& +& 20ce{d}^{5}ϩ& +& 6{d}^{5}{\varrho }^{5}ϩ\\ +& 4ce{d}^{5}{ϩ}^{5}& +& 6{d}^{5}{ϩ}^{5}& +& 8{c}^{3}d{e}^{5}ϩ\varrho \varrho & +& 12ccd{e}^{4}ϩ& -& 28c{e}^{3}{d}^{3}ϩ\\ & & +& 2{c}^{4}{e}^{4}dϩ& & & +& 20{d}^{3}ee{ϩ}^{3}\\ & & & & +& 48c{e}^{3}{d}^{3}{ϩ}^{3}\varrho \varrho & -& 24ccee{d}^{3}ϩ\\ & & & & -& 12{c}^{3}{d}^{3}{e}^{3}ϩ& & \end{array}$. $\begin{array}{llllllllll}& c{e}^{3}{ϩ}^{3}\sqrt{ee-dd}& +& 4{d}^{4}\varrho {ϩ}^{3}& +& 3ceddϩ\varrho \varrho \sqrt{ee-dd}& -& 4{d}^{4}{\varrho }^{3}ϩ& =& {e}^{4}{x}^{3}y\\ -& ceedd& -& 5ddee\varrho {ϩ}^{3}& & & +& 3ddee\\ & & +& {e}^{4}\varrho {ϩ}^{3}\end{array}$. $\begin{array}{llllllllllll}+& cc{e}^{4}{ϩ}^{3}\sqrt{ee-dd}& +& 8ce{d}^{4}\varrho {ϩ}^{3}& +& 3ccddee{\varrho }^{2}ϩ\sqrt{ee-dd}& -& 8ce{d}^{4}{\varrho }^{3}ϩ& -& 5{d}^{4}{\varrho }^{4}ϩ\sqrt{ee-dd}& =& {e}^{5}{x}^{3}yy\\ -& cceedd{ϩ}^{3}& -& 10c{e}^{3}dd\varrho {ϩ}^{3}& +& 10{d}^{4}\varrho \varrho {ϩ}^{3}& +& 6c{e}^{3}dd{\varrho }^{3}ϩ& +& 3ddee{\varrho }^{4}ϩ\\ +& ddee{\varrho }^{5}& +& 2c{e}^{5}\varrho {ϩ}^{3}& -& 8ddee\varrho \varrho {ϩ}^{3}\\ -& {d}^{4}{ϩ}^{5}& & & +& {e}^{4}\varrho \varrho {ϩ}^{3}\end{array}$. [45]
<8r>
[46] $\varrho \varrho ϩϩ-{\varrho }^{2}{g}^{2}=$ { ${b}^{4}$ . } $a=r$. $c=2r$. {illeg} $\frac{y-2r+x\sqrt{x}}{2}=ϩ$ . $\varrho =\frac{-x+y\sqrt{3}}{2}$ $xxyy-2x{y}^{3}$ {illeg} $ϩ=\frac{ay-ac+x\sqrt{cc-aa}}{c}$ . $\varrho =\frac{y\sqrt{cc-aa}-ax}{c}$ $\begin{array}{}\begin{array}{llllllll}& aacc{y}^{4}& -& 2aa{c}^{3}{y}^{3}& +& 2accx{y}^{3}\sqrt{cc-aa}& -& 4aaccxxyy\\ & {a}^{4}{y}^{4}& -& 2{a}^{4}c{y}^{3}& +& 2{a}^{3}x{y}^{3}\sqrt{cc-aa}& +& 4{a}^{4}xxyy\\ & & & & -& 2{a}^{3}x{y}^{3}\sqrt{cc-aa}& +& {a}^{4}xxyy\\ & & & & +& 4{a}^{3}cxyy\sqrt{cc-aa}& +& aa{c}^{4}yy\\ & & & & & & +& {c}^{4}xxyy\\ & & & & -& 2a{c}^{3}x{y}^{2}\sqrt{cc-aa}& -& aaccxxyy\\ & & & & & & +& {a}^{4}ccyy\end{array}\begin{array}{lll}ϩϩ& =& aayy-2aacy+aacc+ccxx-aaxx\\ & & +2ayx\sqrt{cc-aa}-2acx\sqrt{cc-aa}\\ \phantom{0}\\ \varrho \varrho & =& cc{y}^{2}-aayy-2axy\sqrt{cc-aa}+aaxx\\ \phantom{0}\\ & & \begin{array}{cccccc}-& 2aaxy& +& aacx& -& axx\sqrt{cc-aa}\\ +& ccxy& & & +& ayy\sqrt{cc-aa}\\ & & & & -& acy\sqrt{cc-aa}\end{array}\end{array}\end{array}$ $\begin{array}{llllll}& aacc{x}^{4}& +& 4{a}^{3}{x}^{3}y\sqrt{cc-aa}& +& 6{a}^{4}xxyy\\ -& {a}^{4}{x}^{4}& -& 2{a}^{3}c{x}^{3}\sqrt{cc-aa}& -& 6aaccxxyy& +& 6{a}^{3}cxyy\sqrt{\phantom{000}}& +& aacc{y}^{4}\\ & & -& 2acc{x}^{3}y\sqrt{cc-aa}& +& 4aa{c}^{3}xxy& -& 4{a}^{3}x{y}^{3}\sqrt{\phantom{000}}& -& {a}^{4}{y}^{4}\\ & & & & -& 6{a}^{4}cxxy& -& 2{a}^{3}ccxy\sqrt{\phantom{000}}& +& 2{a}^{4}c{y}^{3}\\ & & & & & & & & -& 2aa{c}^{3}{y}^{3}\\ & & & & +& {a}^{4}ccxx& +& 2accx{y}^{3}\sqrt{\phantom{000}}& +& aa{c}^{4}yy\\ & & & & +& {c}^{4}xxyy& -& 2a{c}^{3}xyy\sqrt{\phantom{000}}& -& {a}^{4}ccyy\end{array}$ $\begin{array}{cc}\begin{array}{l}\begin{array}{lllll}3{x}^{4}& -& 4{x}^{3}y\sqrt{3}& -& 2xxyy\\ & -& 2c{x}^{3}\sqrt{3}& +& 10cxxy& +& 2cxyy\sqrt{3}& +& 3{y}^{4}& =& 0\\ & & & +& ccxx& -& 2ccxy& -& 6c{y}^{3}\\ & & & & & +& 4x{y}^{3}& +& 2ccyy\\ & & & & & & & -& {g}^{3}y\sqrt{3}\end{array}\\ \phantom{0}\\ \begin{array}{lllllll}3{x}^{4}& -& 4{x}^{3}y\sqrt{3}& +& 6cxxy& +& 4x{y}^{3}\sqrt{3}\\ & -& 2c{x}^{3}\sqrt{3}& -& 6xxyy& -& 2ccxy\sqrt{3}\\ & & & +& 4xxyy\\ & & & +& 4cxyy& +& 2cxyy\sqrt{3}\\ & & & +& ccxx\end{array}\end{array}& \begin{array}{c}\begin{array}{llllllllllll}& +& yy\sqrt{3}& & & & & -& xx\sqrt{3}& +& 2xy& \text{.}\\ & & & +& cx& -& cy\sqrt{3}\end{array}\\ \phantom{0}\\ \begin{array}{lllllllllll}xxyy& -& 2axxy& +& aaxx& -& aaxx& -& {a}^{4}& =& 0\end{array}\\ \phantom{0}\\ \begin{array}{lllllllll}xxyy& +& 4bxyy& +& 4bbyy& -& 2axxy& -& 8abxy\\ & & & & & -& {a}^{4}& -& 8abby& -\end{array}\end{array}\end{array}$
[47] $bc=x$. $cd=y$. $bf=c$. $bp=\varrho$. $pd=ϩ$. $bc=d$. $bg=${illeg} $fe:fg\colon\colon d:e$ . $fe:eg\colon\colon d:f$. bf $hp=${illeg} z. $fg=\frac{ex}{d}$ . $eg=\frac{fx}{d}$ . $pd:dh\colon\colon r:s$. $dh=\frac{sϩ}{r}$ . $r:t\colon\colon pd:ph=\frac{tϩ}{r}$. $d:e\colon\colon dh:dg=\frac{esϩ}{dr}$. $gh=\frac{fsϩ}{dr}$ $fg=\varrho +\frac{tϩ}{dr}+\frac{fsϩ}{dr}=\frac{ex}{d}$ . $\frac{dr\varrho +dtϩ+fsϩ}{er}=x$ . $\frac{rfx+cdr-esϩ}{dr}=y$ . $\frac{fdr\varrho +dtfϩ+ffsϩ+cder-eesx}{der}=y=\frac{fr\varrho +tfϩ-dsϩ+cer}{er}$ $ec=c$. $bc=x$. $dc=y$. $pd=ϩ$. $pf=\varrho$. $fg=\frac{ex}{d}$ . $eg=\frac{fx}{d}$ . $pg=\frac{sϩ}{r}$ . $gd=\frac{tϩ}{r}$ . $fg=\frac{\varrho r+sϩ}{r}=\frac{ex}{d}$ . {illeg} $x=\frac{dr\varrho +dsϩ}{re}$. $eg+ec-gd=y=\frac{fr\varrho +fsϩ+rec-teϩ}{re}$.
[48] $fe=bc=x$. $cd=y$. $fp=\varrho$. $pd=ϩ$. $fe:fg\colon\colon d:e\colon\colon dh:dg$. $fe:eg\colon\colon d:f\colon\colon hd:hg$. $pd\colon\colon dh\colon\colon r:s$. $pd:ph:r:ht$ $\frac{dr\varrho -dtϩ+fsϩ}{er}=x$. $\frac{fr\varrho -ftϩ-dsϩ+cer}{er}=y$. $d=r$ . $f=\sqrt{ee-dd}$ . $s=\sqrt{rr-tt}$ . $s=\sqrt{dd-tt}$ . $\frac{dd\varrho -dtϩ+ϩ\sqrt{eedd-eett+ddtt-{d}^{4}}}{ed}=x$. $\frac{d\varrho \sqrt{ee-dd}-tϩ\sqrt{ee-dd}-dϩ\sqrt{dd-tt}+ced}{ed}=y$. Lastly $dp=-dt+\sqrt{eedd-eett+ddtt-{d}^{4}}$ . $n=\sqrt{ee-dd}$ . $dq=t\sqrt{ee-dd}+d\sqrt{dd-tt}$ . & Therefore $x=\frac{d\varrho +pϩ}{e}$. & $y=\frac{n\varrho -qϩ+ce}{e}$.
[49] $bf=c$. $fa=z$. $fk=v$. $bc=x$. $cd=y$. $ap=\varrho$. $dp=ϩ$. $vv+zz:vv\colon\colon ϩϩ\colon\colon \frac{vvϩϩ}{vv+zz}={pg}^{2}$. $d:e:\frac{vϩ}{\sqrt{vv+zz}}:\frac{evϩ}{d\sqrt{vv+zz}}=po$. $ac=x-z:ao=\frac{ex-ez}{d}\colon\colon d:e$. $\frac{ex-ez}{d}\phantom{\rule{1em}{0ex}}\frac{-evϩ}{d\sqrt{vv+zz}}=\varrho$ . $d\varrho \sqrt{vv+zz}$ .
[50] $bf=c$. $af=z$. $bc=x$. $fc=x-c=an$. $fk=v$. {illeg} $d:e\colon\colon an:no=\frac{ex-ec}{d}$. $oc=\frac{ex-ec+dz}{d}$ . $ak=\sqrt{vv+zz}:z\colon\colon ed=ϩ:\frac{zϩ}{\sqrt{vv+zz}}=gd$. $\frac{vϩ}{\sqrt{vv+zz}}=eg$ . $go=\frac{evϩ}{d\sqrt{vv+zz}}$. $\frac{ex-ec+dz}{d}\phantom{\rule{1em}{0ex}}\frac{-dzϩ-evϩ}{d\sqrt{vv+zz}}=y$
[51]
<8v>
$fd=x$. $db=y$. $dc=v$. ed{illeg} $ab=z$. $vv+yy:vv\colon\colon zz\phantom{\rule{1em}{0ex}}\frac{vvzz}{vv+yy}={ed}^{2}$. $fe=$ {illeg} x {illeg} / y \ $fe=x\phantom{\rule{0.5em}{0ex}}\frac{\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}vz}{\sqrt{yy+vv}}$. $ae=y\phantom{\rule{0.5em}{0ex}}\frac{\phantom{\rule{0.5em}{0ex}}\underset{\cap }{O}\phantom{\rule{0.5em}{0ex}}yz}{\sqrt{yy+vv}}$. in ye 1st case.[52] $fe=x\phantom{\rule{0.5em}{0ex}}\frac{\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}vz}{\sqrt{yy+vv}}$. $ae=-y\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}\phantom{\rule{0.5em}{0ex}}\frac{yz}{\sqrt{yy+vv}}$. in ye 2d {illeg}|c|ase.[53] $fe=\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}x+\frac{vz}{\sqrt{yy+vv}}$. $ae=y$.
[54] Have{illeg}|in|g ye nature of a crooked line expressed in Algebr: termes to find its {illeg}|a|xes, to det{illeg}|e|rmin it & describe it Geometrically &c
[55] If $fd=x$. $db=y$. & y being pe{illeg}|r|pendicular to x describes ye crooked line ye crook ye line wth its one of its extremes. Then reduce ye Equation (expressing ye nature of ye line in wch x & y onely are undetermin{illeg}|ed|) to one side soe yt it be $=0$. ☞ Then Then {sic} find ye perpendicular bc {illeg} wch is done by findind|g| $dc=v$. for $vv+yy={bc}^{2}$ (In finding $dc=v$ {illeg} {sic}\obse{illeg}|r|v{illeg}|s| this rule./ Multiply {illeg} ye each terme of ye Equat: by so many units as x hath dimensions in yt terme, divide it by x & multiply it by y {illeg} for a Numerator. Againe multiply each terme of ye Equation by soe many units as y hath dimensions in each terme and divide by $-y$ f{illeg}|o|r a denom: in ye val{illeg}|o|r of v. [56] Example, $-rx+\frac{rxx}{q}+yy=0$. $\frac{-1rx+\frac{2rxx}{q}+0yy\phantom{\rule{0.5em}{0ex}}in\phantom{\rule{0.5em}{0ex}}y}{x}=-ry+\frac{2rxy}{q}$. $\frac{-0rx+\frac{0rxx}{q}+2yy}{-y}=-2y$. t{illeg}|h|erefore $\frac{ry+\frac{2rxy}{q}}{-2y}=v=+\frac{1}{2}r-\frac{rx}{q}$. Also if $\begin{array}{cccccccc}{x}^{3}& -& bxx& +& yyx& -& {y}^{3}& =0\\ & +& yxx\end{array}$. $\begin{array}{c}\frac{\begin{array}{cccccccc}\phantom{+}& 3{x}^{3}& -& 2bxx& +& yyx& in& y\\ & & +& 2yxx\end{array}}{x}\\ \frac{\begin{array}{cccccc}-& 3{y}^{3}& +& 2yxx& +& yxx\end{array}}{-y}\end{array}\begin{array}{c}\phantom{\frac{\begin{array}{c}\phantom{0}\\ \phantom{0}\end{array}}{\phantom{0}}}=\\ \phantom{\frac{\begin{array}{c}\phantom{0}\end{array}}{\phantom{0}}}=\end{array}\phantom{\rule{0.5em}{0ex}}\begin{array}{cc}\begin{array}{c}\phantom{\frac{\begin{array}{c}\phantom{0}\\ \phantom{0}\end{array}}{\phantom{0}}}3xxy-2bxy+2xyy+{y}^{3}\phantom{\frac{\begin{array}{c}\phantom{0}\\ \phantom{0}\end{array}}{\phantom{0}}}\\ \phantom{\frac{\begin{array}{c}\phantom{0}\end{array}}{\phantom{0}}}3yy-2yx+xx\phantom{\frac{\begin{array}{c}\phantom{0}\end{array}}{\phantom{0}}}\end{array}=& v\end{array}$ [57] And if ${x}^{4}-yyxx+aayx-{y}^{4}=0$. then $\begin{array}{cc}\begin{array}{c}4y{x}^{3}-2{y}^{3}x+aayy\\ 4{y}^{3}-aax+2yxx\end{array}=& v\end{array}$. &c) Then make $ab=z$. $fe=x\frac{+vz}{\sqrt{yy+vv}}$. $ae=y\frac{-vz}{\sqrt{yy+vv}}$. & substitute this valor of $\left(fe\right)$ into ye place of x {illeg} & this valor of $\left(ae\right)$ into ye place of y in ye Equation \& th{illeg}|e|r{illeg}|e| {illeg}|take| {{illeg}|in|} a 2d equation/. {illeg}|t|hen by multiplication or by some other meanes take a{illeg}|w|ay ye irrational quantity $\sqrt{yy+vv}$ & lastly take awa{y} y or x {illeg}|b|y ye helpe of these 2 Equations, soe yt you have a {illeg} \3rd/ equation in wch there is either x onely, or y onely & supposeing it to have 2 equall roots multiply each terme by {illeg}|soe| many units as {illeg}|ye| unknowne quanti{illeg}|t|y hath dimensions in yt terme {illeg} wch {illeg}|p|roduct is a 3rd equation it according to Huddenius his Method for a 3rd /4th\ Equation & by ye helpe of ye 3rd & 4th equation take away {illeg} ye in wch & there will result a 5t Equation in wch there b{illeg} one unknowne quantity viz: either x or y. & there will result a 5t Equation in wch is neither x nor nor {sic} y. & by wch the valor of z may be found. one \The greatest/ of whose valors signifies ye longest, another \the least {illeg}/ of ym ye shortest of all ye perpendicular lines ab. & if it have other rootes they signifie other lines $\left(ab\right)$ wch are perpendicular to ye crooked line at both ends, a & b; & some of these must signifie ye axes of ye line if it bee of an elliptical nature.
<9r>
[58] $ac=x$. $ch=y$. $bf=${illeg} \z {illeg}/. $fh=${illeg} \{illeg} ϩ/ $cd=a$. $db:be\colon\colon fh:eh\colon\colon b:c$. $\frac{cx}{b}=be$. $\frac{cϩ}{b}=eh$. {illeg} $\frac{ϩ\sqrt{cc-bb}}{b}=fe=\frac{cx-bz}{b}$. $\frac{ϩ\sqrt{cc-bb}+bz}{c}=x$. ed{illeg}$=\frac{x\sqrt{cc-bb}}{b}=\frac{ϩcc-ϩbb+bz\sqrt{cc-bb}}{bc}$ $hc=\frac{-ϩbb+bz\sqrt{cc-bb}+abc}{bc}=\frac{ac-ϩb+z\sqrt{cc-bb}}{c}=y$.
$\begin{array}{ccccccc}\begin{array}{c}\phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\phantom{=}& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\phantom{=}& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\text{for}& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\phantom{=}& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\phantom{=}& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\phantom{=}& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\phantom{=}& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\phantom{=}& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\phantom{=}& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\phantom{=}& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\phantom{=}& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\phantom{=}& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\phantom{=}& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\phantom{=}& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\end{array}& \left\{& \begin{array}{}x\phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ xx& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ {x}^{3}& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ {x}^{4}& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ y& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ yy& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ {y}^{3}& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ {y}^{4}& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ xy& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ xyy& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ x{y}^{3}& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ xxyy& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ xxy& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ {x}^{3}y& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\end{array}& \right\}& \begin{array}{c}\phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}=& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}=& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\text{write}& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}=& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}=& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}=& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}=& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}=& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}=& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}=& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}=& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}=& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}=& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}=& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\end{array}& \left\{& \begin{array}{l}\frac{+ϩ\sqrt{cc-bb}}{c}& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \frac{2bϩz\sqrt{cc-bb}}{cc}& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \frac{{ϩ}^{3}cc\sqrt{cc-bb}}{{c}^{3}}-\frac{{ϩ}^{3}bb\sqrt{cc-bb}}{{c}^{3}}+\frac{3bbzzϩ\sqrt{cc-bb}}{{c}^{3}}& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \frac{4{ϩ}^{3}cc\sqrt{cc-bb}×bz-4{y}^{3}{b}^{3}z\sqrt{cc-bb}+4ϩ{b}^{3}{z}^{3}\sqrt{cc-bb}}{{c}^{4}}\phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ -\frac{ϩb}{c}& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \frac{-2ϩabc-2ϩbz\sqrt{cc-bb}}{cc}& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ -{ϩ}^{3}{b}^{3}-3ϩbaacc-3ϩbcczz+3ϩ{b}^{3}zz-6ϩabcz\sqrt{cc-bb}& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \frac{\begin{array}{cc}\begin{array}{c}-4{ϩ}^{3}{b}^{3}ac-4{ϩ}^{3}{b}^{3}z\sqrt{cc-bb}-4ϩ{a}^{3}b{c}^{3}-12ϩaabccz\sqrt{cc-bb}-12ϩab{b}^{3}zz+12ϩa{b}^{3}c{z}^{2}\end{array}& \begin{array}{l}+ϩ{z}^{3}{b}^{3}\\ -ϩ{z}^{3}bcc\sqrt{cc-bb}\end{array}\end{array}}{{c}^{4}}& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \frac{acϩ\sqrt{cc-bb}+ϩzcc-2ϩzbb}{cc}& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \frac{-4ϩabbcz-3ϩbbz\sqrt{cc-bb}\begin{array}{l}+aaccϩ\\ +zzcc\end{array}\sqrt{cc-bb}+2a{c}^{3}zϩ+{b}^{2}{ϩ}^{3}\sqrt{cc-bb}}{{c}^{3}}\phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \frac{-3bbcc{ϩ}^{3}z-4{b}^{4}z{ϩ}^{3}+3abbc{ϩ}^{3}\sqrt{cc-bb}+{z}^{3}{c}^{4}ϩ+4{z}^{3}{b}^{4}ϩ-5{z}^{3}bbccϩ\begin{array}{c}-9abcc\\ +3a{c}^{3}\end{array}zzϩ\sqrt{cc-bb}\begin{array}{c}-aabbcc\\ +3aa{c}^{4}\end{array}zϩ+{a}^{3}{c}^{3}\sqrt{cc-bb}}{{c}^{4}}& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \frac{2a{b}^{3}c{ϩ}^{3}-2ab{c}^{3}{ϩ}^{3}\begin{array}{c}+4{b}^{3}\\ -2bcc\end{array}z{ϩ}^{3}\sqrt{cc-bb}\begin{array}{c}+2bcc\\ -4{b}^{3}\end{array}{z}^{3}ϩ\sqrt{cc-bb}+2aabbcczϩ\sqrt{cc-bb}+4ab{c}^{3}zzϩ-6a{b}^{3}czzϩ}{{c}^{4}}& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \frac{{b}^{3}{ϩ}^{3}-bcc{ϩ}^{3}+2bcczzϩ-3{b}^{3}zzϩ+2abcz\sqrt{cc-bb}}{{c}^{3}}& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\\ \frac{4{b}^{4}z{ϩ}^{3}-5bbccz{ϩ}^{3}+{c}^{4}z{ϩ}^{3}\begin{array}{c}+a{c}^{3}\\ -abbc\end{array}{ϩ}^{3}\sqrt{cc-bb}+3bcczzϩ-4{b}^{3}zzϩ+3abczϩ\sqrt{cc-bb}}{{c}^{4}}& \phantom{\frac{\begin{array}{c}|\\ |\end{array}}{{0}^{0}}}\end{array}\end{array}$
Haveing therefore an equation expressing ye nature of a{illeg} crooked line To find its axis. \Supposeing $c=$ some quantity most frequent in ye equation/ Subroga{illeg}|t|e $\frac{bz+ϩ\sqrt{cc-bb}}{c}$ into ye roome of x ; & $\frac{ac-ϩb+z\sqrt{cc-bb}}{c}$ into ye roome of y: Order ye Equation according to {illeg} ϩ, make every terme $=0$ , in wch ϩ is of {illeg} \one/ dimensio{sic} Order every terme in the|i|se 2dary Equations according to ye dimensions of z. & supposeing every terme of each of ym $=0$ , by ye helpe of the{illeg}|s|e {illeg}|E|quations (in wch is neither x , y , z or ϩ ) may be found ye valors of a & b . Then perpendicular to ac from ye point a draw $ab=a$. & from ye point d | b | draw $bk=b$, & parallel to ac. from ye point k draw $mk=\sqrt{cc-bb}$ , & perpendicular to bk . & through ye points b , {illeg} m {illeg} draw bl ye axis of ye line hgn . & yt ye relation twixt $bf=${illeg}z. & $fh=ϩ$ may bee had, write ye valors of a , b , c now found in their stead in ye 2dary equation.
Example $dd+dy+xy-yy=0$. Then makeing $d=c$ I write $\frac{bz+ϩ\sqrt{cc-bb}}{c}$, or $\frac{bz+ϩ\sqrt{dd-bb}}{d}$ for x & its square for $xx$ &c. & $\frac{ad-bϩ+z\sqrt{dd-bb}}{d}$ for y , & its square for $yy$. & soe I have this equation, $0=dd+ad-bϩ+z\sqrt{dd-bb}\phantom{\rule{0.5em}{0ex}}\begin{array}{}\frac{+adbz-2bbϩz\phantom{\rule{0.5em}{0ex}}\begin{array}{c}-bϩϩ\\ +bzz\\ +adϩ\\ +2azd\\ +2bϩz\end{array}\sqrt{dd-bb}\phantom{\rule{0.5em}{0ex}}+ϩzdd-aadd+2adbϩ-bbϩϩ\phantom{\rule{0.5em}{0ex}}\begin{array}{c}-ddzz\\ +bbzz\end{array}}{dd}\end{array}$ or by ordering it according to ϩ , $\begin{array}{cc}\begin{array}{ccccccccc}bbϩϩ& +& bϩϩ\sqrt{dd-bb}& +& ddbϩ& -& daϩ\sqrt{dd-bb}& -& {d}^{4}\\ & & & +& 2bbzϩ& -& 2bzϩ\sqrt{dd-bb}& \begin{array}{c}-\\ +\end{array}& \begin{array}{c}abdz\\ aadd\end{array}\\ & & & -& ddz\hfill & & & +& ddzz\\ & & & -& 2abd\hfill & & & -& bbzz\end{array}& \begin{array}{c}+2adz\phantom{\sqrt{0}}\\ -bzz\\ -ddz\end{array}\sqrt{dd-bb}\end{array}\phantom{\rule{0.5em}{0ex}}\begin{array}{cc}=& 0\end{array}$ Then by makeing those quantitys in ye last terme save one {illeg} {illeg} $=0$ . I have this Equation $2bbz-ddz-2bz\sqrt{dd-bb}+ddb-2abd-da\sqrt{dd-bb}=0$ . {illeg}|W|hich I divide into 2 pts makeing those termes $=0$ in wch z is not, & those $=0$ in wch z is of one dimension. & then I have these 2 equation{s} $2bb-dd-2b\sqrt{dd-bb}=0$ . & $db-2ab-a\sqrt{dd-bb}=0$ . by ye f{illeg}|irs|t $4{b}^{4}-4bbdd+{d}^{4}=4bbdd-4{b}^{4}$ . Or $8{b}^{4}-8bbdd+{d}^{4}=0$. That is $bb=\frac{dd}{2}\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}\sqrt{\frac{{d}^{4}}{8}}=\frac{dd}{2}\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}\frac{dd}{2\sqrt{2}}=\frac{dd\sqrt{2}\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}dd}{2\sqrt{2}}$. by ye 2d Equation I find $ddbb-4adbb+4aabb=aadd-aabb$. or $\begin{array}{cccccc}& 5aabb& -& aadd& =& 0\\ -& 4adbb\\ +& ddbb\end{array}$. & by writing ye valor of $dd$ wch was found before I have $\frac{5aadd\sqrt{2}\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}5aadd-4a{d}^{3}\sqrt{2}\phantom{\rule{0.5em}{0ex}}\underset{\cap }{O}\phantom{\rule{0.5em}{0ex}}4a{d}^{3}+{d}^{3}\sqrt{2}\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}{d}^{4}-2aadd\sqrt{2}}{2\sqrt{2}}=0$. Or $\begin{array}{cccccccc}& 3aa\sqrt{2}& -& 4ad\sqrt{2}& +& dd\sqrt{2}& =& 0\\ \phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}& 5aa\hfill & \phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}& 4ad\hfill & \phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}& dd\hfill \end{array}$. Or $aa=\frac{4ad\sqrt{2}\phantom{\rule{0.5em}{0ex}}\underset{\cap }{O}\phantom{\rule{0.5em}{0ex}}4ad\phantom{\rule{1em}{0ex}}-\phantom{\rule{1em}{0ex}}dd\sqrt{2}\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}dd}{3\sqrt{2}\phantom{\rule{0.5em}{0ex}}\underset{\cap }{O}\phantom{\rule{0.5em}{0ex}}5}$ . & $a=\frac{d\sqrt{8}\phantom{\rule{0.5em}{0ex}}\underset{\cap }{O}\phantom{\rule{0.5em}{0ex}}2d}{18\phantom{\rule{0.5em}{0ex}}\underset{\cap }{O}\phantom{\rule{0.5em}{0ex}}5}\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}\sqrt{\frac{\begin{array}{c}8dd\phantom{\rule{0.5em}{0ex}}\underset{\cap }{O}\phantom{\rule{0.5em}{0ex}}4dd\sqrt{8}\phantom{\rule{1em}{0ex}}+\phantom{\rule{1em}{0ex}}4dd\phantom{\rule{1em}{0ex}}-\phantom{\rule{1em}{0ex}}5dd\\ -dd\sqrt{36}\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}5dd\sqrt{2}\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}dd\sqrt{18}\end{array}}{18\phantom{\rule{0.5em}{0ex}}\underset{\cap }{O}\phantom{\rule{0.5em}{0ex}}5\sqrt{18}\phantom{\rule{1em}{0ex}}+\phantom{\rule{1em}{0ex}}25}}$ $a=\frac{d\sqrt{8}\phantom{\rule{0.5em}{0ex}}\underset{\cap }{O}\phantom{\rule{0.5em}{0ex}}2d}{\sqrt{18}\phantom{\rule{0.5em}{0ex}}\underset{\cap }{O}\phantom{\rule{0.5em}{0ex}}5}\begin{array}{c}\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}\\ +\end{array}\sqrt{\frac{dd}{18\phantom{\rule{0.5em}{0ex}}\underset{\cap }{O}\phantom{\rule{0.5em}{0ex}}5\sqrt{18}\phantom{\rule{1em}{0ex}}+\phantom{\rule{1em}{0ex}}25}}$ or $\frac{2d\sqrt{2}-2d}{3\sqrt{2}-5}\begin{array}{c}\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}\\ +\end{array}\sqrt{\frac{dd}{43-15\sqrt{2}}}=a$. & $\frac{dd\sqrt{2}\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}dd}{2\sqrt{2}}=bb$. $a=\frac{2d\sqrt{2}\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}2d\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}d}{3\sqrt{2}\phantom{\rule{0.5em}{0ex}}\stackrel{\cup }{O}\phantom{\rule{0.5em}{0ex}}5}$
<9v>
[59] $\frac{dd{\varrho }^{2}+2ds\varrho ϩ+ssϩϩ}{ee}=xx$. ${d}^{3}{\varrho }^{3}+3dd\varrho \varrho sϩ+3d\varrho \varrho sϩϩ+{s}^{3}{ϩ}^{3}={e}^{3}{x}^{3}$ . |{d }. $a+y$| ${d}^{4}{\varrho }^{4}+4{d}^{3}{\varrho }^{3}sϩ+6dd\varrho \varrho ssϩϩ+4d\varrho {s}^{3}{ϩ}^{3}+{s}^{4}{ϩ}^{4}={e}^{4}{x}^{4}$. ${d}^{5}{\varrho }^{5}+5{d}^{4}{\varrho }^{4}sϩ+10{d}^{3}{\varrho }^{3}ssϩϩ+10dd{s}^{3}\varrho \varrho {ϩ}^{3}+5d\varrho {s}^{4}{ϩ}^{4}+{s}^{5}{ϩ}^{4}={e}^{4}{x}^{5}$ $\begin{array}{ccccccc}tt\varrho \varrho & -& 2t\varrho vϩ& +& vvϩϩ& =& yy\\ & +& 2t\varrho ce& -& 2cevϩ\\ & & & +& ccee\end{array}$. $\begin{array}{llllllllll}& {t}^{3}{\varrho }^{3}& -& 3tt\varrho \varrho vϩ& +& 3t\varrho vvϩϩ& -& {v}^{3}{ϩ}^{3}& =& {e}^{3}{y}^{3}\\ +& 3tt\varrho \varrho ce& -& 6t\varrho ce& +& 3ce\\ +& 3t\varrho ccee& -& 3ccee\\ +& {c}^{3}{e}^{3}\end{array}$. $\begin{array}{llllllllllll}& {t}^{4}{\varrho }^{4}& -& 4{t}^{3}{\varrho }^{3}vϩ& +& 6tt\varrho \varrho vvϩϩ& -& 4t\varrho {v}^{3}{ϩ}^{3}& +& {v}^{4}{ϩ}^{4}& =& {e}^{4}{y}^{4}\\ +& 4{t}^{3}{\varrho }^{3}ce& -& 12tt\varrho \varrho ce& +& 12t\varrho ce& -& 4ce\\ +& 6tt\varrho \varrho ccee& -& 12t\varrho ccee& +& 6ccee\\ +& 4t\varrho {c}^{3}{e}^{3}& -& 4{c}^{3}{e}^{3}\\ +& {c}^{4}{e}^{4}\end{array}$. $\begin{array}{llllllllllllll}& {t}^{5}{\varrho }^{5}& -& 5{t}^{4}{\varrho }^{4}vϩ& +& 10{t}^{3}{\varrho }^{3}vvϩϩ& -& 10tt\varrho \varrho {v}^{3}{ϩ}^{3}& +& 5t\varrho {v}^{4}{ϩ}^{4}& -& {v}^{5}{ϩ}^{5}& =& {e}^{5}{y}^{5}\\ +& 5{t}^{4}{\varrho }^{4}ce& -& 20{t}^{3}{\varrho }^{3}ce& +& 30tt\varrho \varrho ce& -& 20t\varrho ce& +& 5ce\\ +& 10{t}^{3}{\varrho }^{3}ccee& -& 30tt\varrho \varrho ccee& +& 30t\varrho ccee\\ +& 10tt\varrho \varrho {c}^{3}{e}^{3}& -& 20t\varrho {c}^{3}{e}^{3}& +& 10{c}^{3}{e}^{3}\\ +& 5t\varrho {c}^{4}{e}^{4}-& 5{c}^{4}{e}^{4}\\ +& {c}^{5}{e}^{5}\end{array}$
[60] $bc=x$. $dc=y$. $df=z$. $bh=c$. $hk:ke\colon\colon d:e$. $ke=\frac{ex}{d}$. $df:de\colon\colon d:f$. $\frac{fz}{d}=de$. $\frac{cd+ex-fz}{d}=y$. $fm=\varrho$. $mh=\frac{d\varrho }{e}$. $mk=\frac{xe-d\varrho }{e}$. $pe=\frac{xe-d\varrho }{d}=$ $p=n$. $fp=\sqrt{{ϩ}^{2}-nn}$. $ep=\frac{e\sqrt{{ϩ}^{2}-nn}+dn}{d}=\frac{ex}{d}$ $eeϩϩ=eenn+ddnn-2dnex+eexx$. $fd:fp\colon\colon${illeg}$f:g$. $\frac{ϩg}{f}=fp$ . $\frac{f\varrho +ϩg}{f}=x$. $fp:pe\colon\colon d:e$. $\frac{eϩg}{df}=${illeg}. $pd=\frac{ϩ\sqrt{ff-gg}}{f}$ $fm=\frac{e\varrho }{d}$. $hm=\varrho$. $\frac{e\varrho +cd}{d}-ϩ\frac{\sqrt{ff-gg}}{f}=y$. $f=d$. &c $\frac{d\varrho +gϩ}{d}=x$. $\frac{e\varrho +cd-ϩ\sqrt{ff-gg}}{d}=y$.
[61] If any crooked line be revolved about its \owne/ {illeg}|a|xis it generates a Sollid intersected by any plaine \not perpendicular to ye axis/ produceth another line $\left\{\begin{array}{l}\text{not more compound}\phantom{\rule{0.5em}{0ex}}{\mathrm{y}}^{\mathrm{n}}\\ \text{of the same kind with}\end{array}$ ye forme{illeg}|r|. But if it bee revolved by any other line it generates a Sollid which intersected by any plaine \not perpendicular to ye axis/ produceth another whose composition is {n}ot $\left\{\begin{array}{l}\text{lesse}\phantom{\rule{0.5em}{0ex}}{\mathrm{y}}^{\mathrm{n}}\phantom{\rule{0.5em}{0ex}}\text{equall}\\ \text{more}\phantom{\rule{0.5em}{0ex}}{\mathrm{y}}^{\mathrm{n}}\phantom{\rule{0.5em}{0ex}}\text{double}\end{array}\right\}$ to ye formere
In the △ adb if $ab=a$, & $db=b$ are definite, but $ad=v$, & $bc=x$ indefined. Then ye Equation is $bb-vv+aa-2ax=0$. But in this case ye maximu or minimum of either v or x cannot bee found according {illeg}|t|o Cartes or Hudde{nius} method, by reason yt $\left\{\begin{array}{c}v\\ x\end{array}\right\}$ hath not 2 divers valors when $\left\{\begin{array}{c}x\\ v\end{array}\right\}$ is determined, wch become equall when $\left\{\begin{array}{c}x\\ v\end{array}\right\}$ is ye least or greatest yt may be. But if cd might heve {sic} bee used inste{ad} of cb &c. There be other instances of this Nature against Huddenius h{illeg}|i|s assertion.
These points a/,\b/,\c/,\ being given {illeg}|a| circle may {illeg}|b|e described (wch shall pass through a point them all) by an instrument whose angle $edf=\angle abc$. And soe ye sides ed & df being moved close by ye points a & c, ye point $\left(d\right)$ shall describe ye a{illeg}|r|ch abc
To w{illeg}|o|rke mechanic{illeg}|a|lly & exactly b{illeg}|y| a {{illeg}|s|cal{illeg}|e|} it may be{illeg}|e| better divided according to ye fassion {illeg} {illeg}|r|epresented by ye figure A, yn by that at B.
To {illeg}|m|ake a {plated} superficies exactly: Take three plates A, B, D. & {g{illeg}|r|ind} them together A & B, A & D, B & D; pressing ye uppermost pla{illeg}|t|e onely in the middle at {illeg}|c| that it may not weare {move} a{illeg}|w|ay in the edges yn in the midst & move it to & fro wth but sma{illeg}|ll| vibrations. Soe shall the 3 {illeg}|f|iduciall sides of ye 3 plates bee {ground} exactly plane.
<10r>
[62] Probleme. Of Usury. $ab=a=$ the princi{illeg}|p|le. bc{illeg}={illeg}$x=$ the time of the m{illeg}|o|ney lent $ce=\frac{dx}{e}=$ the use due for ye principle {illeg}|fo|r ye time bc. $de=$ the use upon {illeg}|u|se. {illeg} cd ye $cd=y=$ ye use{illeg} for ye principle & use during ye time bc. $ax:\frac{dx}{e}\colon\colon a+y\phantom{\rule{0.5em}{0ex}}in\phantom{\rule{0.5em}{0ex}}fc:y$. ab ye sume of princi df a t{illeg}|a|ngent to ye line bd. dg a perpendicular. $cg=v$. Then as ye sume $\left(ab\right)$ or principle drawne into ye time bc, is to $\left(ce\right)$ ye use for it in yt time [63] So is $ab+dc$ the sume $ab+dc$ drawne into ye time fc, to ye use $\left(dc\right)$ for it in ye time fc. {illeg} therefore $ax:\frac{dx}{e}\colon\colon a+y\phantom{\rule{0.5em}{0ex}}in\phantom{\rule{0.5em}{0ex}}\left\{\begin{array}{c}h\\ fc\end{array}:y$ {illeg}. $axy=\frac{ahdx+yhdx}{e}$. $\frac{eay}{ad+yd}=h:y${illeg}$\colon\colon y:v$. $eav=ady+dyy$. $v=\frac{ady+dyy}{ea}$. $d=e$. $\frac{ay+yy}{a}=v$
## Of Reflect{illeg}|i|ons.
Suppoe {sic} yt {illeg}|t|he bodys a, b, have noe vis elastica to reflect ye one from ye other but at their occursion conjoine & keepe together as i{illeg}|f| they were one body. Then 1st if Theire bulke \& motion/ be equall yn at theire {illeg} meeteing they shall rest. 2 If $\left(b\right)$ have more motion yn $\left(a\right)$ all {illeg} ye motion of $\left(a\right)$ shall be lost & soe much of $\left(b\right)$s as $\left(a\right)$ had & they shall both move towards c shareing ye differencde of ye motion \proportionally/ twixt ym. Demon: {illeg}|S|uppose ye motion of $ehf=$ motion of a
2 If a rest & b hit it they shall both move towards c wth {illeg} shareing ye motion \of $\left(b\right)$/ twixt them.
## [64] Of Reflection.
Suppose ye Bodys a, b doe not reflect one another b{illeg}ut conjoyne either moveing or resting together of at theire meeting & soe move or rest to{illeg}|g|ether. $a=$ ye body a; $b=$ ye body fec; $c=$ body ced; $d=$ body fedc. $m=$ motion of $a$, $n=$ motion of $b$, $p=$ motion of $c$, $q=$ motion of $d=n+p$, before reflection. $e=$ motion of $a$, $f=$ motion of $b$, $g=$ moti {sic} of $c$, $h=$ motion of $d$ {illeg}|a|fter reflection. $r=$ swiftnesse of $a$, $s=$ motion of $b$, $c$, or $d$, before reflection $t=$ swift{illeg}nesse of $a$, $v=$ swiftnes $b$, $c$, or $d$ ref occursion. ⊙ ye point of theire occursion.
## Axiome 1st
[65] Two bodys \$\left(b,c\right)$/ being alike swift {illeg} ye motion of $b:$mot: $c\colon\colon b:c$. for equall pts have equall motion. Therefore $b:c\colon\colon$all ye parts of $b:$all such pts of $c\colon\colon$mot: $b:$the motion of c.
Prop: 1st. If before ye occursion \of a & d/ a rest yn shall $e:q\colon\colon a:${illeg} $e+h=q$. & since $t=v$, tis {alsoe} $e:q\colon\colon a:a+d$. Or $e=\frac{aq}{a+d}$ also $h=q-\frac{aq}{a+d}=\frac{dq}{a+d}$.
Prop: 2d. If a meete d, & have lesse motion yn it, then, $q-m=e+h$. for suppo{illeg}|s|e, $m=n$. yn should {illeg}|a| & b rest after occursion did not $p=q-m$ {illeg} force ym towards k.
Prop 3 suppose i ye center of gravity in d, y in a. z & f ye in wch ye bodys a & d touch in theire meeting. ⊙ ye point of theire meeting. $\left(a\right)$ ye magnitude of \ye body/ $\left(a\right)$, $\left(d\right)$ ye magnitude of d. $m=$ motion of a before meeting, $n=$ mot: {illeg}|d| before me{illeg}|et|ing. ye time in wch a or d moves to $\odot =$ time in wch they both move to y. $p=$ motion of a, {illeg} $q=$ moti{illeg}|o|n of d after occursion{illeg} $m+n=p+q$. $a:d\colon\colon p:q$. or $a+d:d\colon\colon m+n\left(=p+q\right):q=$$qm+qn$ $\frac{dm+dn}{a+d}$. $\frac{am+an}{a+d}=p$. $m:p\colon\colon z\odot :zy$. $\frac{am×z\odot +an×z\odot }{am+dm}=\odot y$.
$a=$ magnitude of ye body a, $d=$ mag: of ye body d. {illeg} $o=$ {illeg} ye point of concourse: $z,f=$ ye points of contac{illeg}|t|, at o. $zo=${illeg}|b|. $fo=c$. $op=e$. $t=$ time in wch ye bodys move from z & f to o. $v=$ time in wch they move from o to p. $m=$ motion of a before occursion $n=$ motion of d after occursion $dm${illeg}{illeg}$=\frac{dn+an}{a}$ or $cdm+bam=bdn+ban$. {illeg}|&| $\frac{cdm+bam}{bd+ba}=n${illeg}$\colon\colon m:n$. $tm:vn\colon\colon${illeg}$b:e\colon\colon tbd+tba:vcd+vba$. {illeg}$cd${illeg} $v=\frac{det+aet}{cd+ab}$. . {illeg}{illeg}{&} d {illeg} meete {illeg} {illeg} yt c must be negative yt is $\frac{abv-\phantom{0}}{nt}${illeg}e be negative ye point {illeg} must be t{illeg}on ye same side {illeg}
<10v>
## Definitions.
[66] {illeg} \1st/ When a body \{illeg}|Q|uan{illeg}|ti|ty/ $\left\{\begin{array}{l}\text{is translated}\\ \text{passeth}\end{array}\right\}$ from one pte of Extension to another it is saide to mo{ve.}
2 One body \Quantity/ is soe much swifter yn another, as ye distance th{illeg}|r|ough w{illeg}|c|h it passeth is greater yn ye distance through wch ye other passeth in ye same time.
One body /quantity\ {illeg}|h|ath soe much more motion yn another, as ye summe of ye spaces th{illeg}|r|ough wch each of its pts moveth i{illeg}|s| to ye summe of ye spaces through which each of ye pts of ye other \quantity/ body moveth supp in ye same time, suppi|o|seing each of the{illeg} pts in both bodys /Quantitys\ to be equall & alike to one another \& moved in ye same position./ {illeg}.
3 One Quantity hath so much more motion yn another, as ye distance through wch it moveth drawne into {illeg}|it|s quantity, is to ye distance {illeg}|t|rough wch ye other moveth in ye same time d{illeg}|raw|ne into its quantity. As if ye line ab move ye length of bc & ef ye length of eh in ye same time, ye motion of ab is to ye motion of cd, as $ab×bc=abcd$, to $ef×eh=iehk$. Alsoe if \ye cube/ $lmqyz=8$ move ye length of $op=5$; & the piramis $tvwx=7$ move ye length of $rs=3$ in ye same tim{illeg}|e|; y{illeg}|n|, as, $op×lmqyz:rs×tvwx\colon\colon 40:21\colon\colon$ ye motion {illeg}|of| lmqyz to ye motion tvwx. Or th{illeg}|e| motio{illeg}|n| of one quantity to another is a{illeg}|s| their powers to /persever in that state\
Those bodys \Qua{illeg}|n|titys/ are said to have ye same determination of their motion wch move ye same way, {illeg}|&| those have divers wch move divers ways.
[67] 5 A body /quantity\ is reflected when meeting wth another quantity it looseth ye determination of its motion by rebounding from i{illeg}|t|. As if ye bodys a, b meete one A quantity is said to bee refracted another in ye point c they are parted either by some springing motion in y selves or of \in/ ye matter {crouded} bet{illeg}|w|ixt ym. & as {illeg} ye spring is more dull or {illeg}ety \{illeg}|vi|gor{illeg}|ous|/ /quick\ s{illeg}|o|e ye bodys {illeg}|w|ill bee reflected wth wth more \or lesse/ force; a{illeg}|s| if it endeavour to get liberty \to inlarge it selfe/ wth as greate strength & vigor as ye bodys $a,b$ , pressed it together, ye ye {sic} motion of ye body{illeg} a from b will bee as greate after as before y{illeg}|e| reflection. but if ye spring have but halfe yt vigor, yn ye distance twixt a & b,|at| {illeg}|t|he minute after ye reflection shall bee halfe as much as it was befor at ye minute beef{ore} ye reflecti{illeg}on.
7 Refraction is when {illeg} ye body \{illeg}/ c passing \obliquely/ through ye surface ed at ye point b meets wth more or lesse oppo{illeg}|s|i{illeg}|t|ion on one side of ye surface yn on ye other & soe looseth its determinacon; as if it turne towards a.
[68] 9 Force ii|s| ye pressure or erouding of one body {illeg} upon an{illeg}|o|ther,
10 The center of Gravity \{ Motion }/ \in ye same body/ is such a point wthin a quantity \{illeg}wch rests when a body is moved wth \any/lar but noe progres\sive motion//; yt if it \be considered to as at a/ rest & ye quantity & some quantity line as $\left(mn\right)$ be drawne through it: about wch (as \about/ an axis) ye f{illeg} quantity $\left(dklp\right)$ revolving. {illeg}|t|here shall bee ye same quantity of motion on both sides any {illeg}|p|laine {illeg}|w|th wch $\left(mn\right)$ is coincident; also ye line in n \drawne throug{illeg}|h| it/ is called an axis of gravity \motion/.
The center of motion in \2/ divers bodys is {illeg}|a| point soe placed {illeg}|t|wixt those bodys yt (if it bee conceived to rest {&}) if {illeg} the bodys bee moved about it {illeg} wth circular motion they shall both have an equall quantity {o}f motion, the line about wch they move is ye axis of motion.
12 A Body is said to move toward another body either when all its pts move towar{d}s it or else {illeg}|w|hen some of its pts have more motion towards it y{illeg}|n| others have from it. Otherwise /not\
< insertion from the left margin >
13 Bodys are more or lesse distant as ye distances of {illeg} their pts \centres of motion/ are more or lesse. or as their distances might bee acquired wth more or less motion
< text from f 10v resumes >
## Axiomes. & Propositions.
1 If a quantity once move it will never res{illeg}|t| unlesse hindered by some externall caus{e.}
2 A quantity will always move on in ye same streight line (not changing ye determination \{nor} celerity/ of its motion) unlesse some externall cause divert it.
3 There is exactly required so much \& noe more/ force to reduce a body to rest {illeg} /as\ there {illeg} was {to} put it {illeg}n motion: et e contra.
4 S{illeg} much {illeg}{illeg}s is required to d{illeg}p{illeg}e {illeg}|d|estroy any quantity of motion in a body soe {illeg} to generate it; & soe much as is required to generate it soe mu{illeg}ived to destroy it.
[69]6 {illeg}ove 2 unequall bodys \(a & b)/ ye swiftness{illeg}|e| of {illeg} \on{illeg}e/ body $\left(a\right)$ is to ye s{illeg} is to a . {(1)} & therefore ye motion of both bodys shall bee equall.
5 If {illeg}{b}ee moved by unequall forces, as ye force moveing $\left(b\right)$ is to ye force {illeg} motion {illeg}|o|f b. to ye m{illeg} of c, so is ye swiftnes of b, to yt of c.
[70]7 If two body {illeg}{illeg} \{illeg}way to{illeg}s {illeg}{illeg}r{illeg}cking {illeg}/ {illeg}e of theire motion shall be lost. f{illeg} pres{illeg}{illeg}{illeg}{illeg} ye motion of $\left(b\right)$ shall {illeg}{illeg}
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[71] 8 If two quantitys (a & b) move d{illeg} towards one another & meete in {illeg} o, Then ye difference of theire motion shall not bee lost nor {illeg}|l|oose its determinacon. For at their occursion they presse equally uppon one another, & (p)[72] therefore one must loose noe more motion yn ye other doth; soe yt ye difference of their motions cannot be destroyed.
9. If one body $\left(a\right)$ overtake ano{illeg}|t|her body $\left(b\right)$ they both moveing towards o then they shall always move together. v
If ye body $\left(c\right)$ move against an imm{illeg}|o|veable quanty {sic} $\left(d\right)$ it shall not bee rebounded for c having urged d wth
[73] 9 If the body t{illeg}|w|o \equall & equally swift/ bodys (d & {illeg}|c|) meete one another they shall bee reflected, s{illeg}|o|e as to move as swiftly frome {illeg}|o|ne another a{illeg}|f|ter yr reflection as they did to one another before i{illeg}|t|. For {illeg}ng For firs{illeg}|t| suppose ye {illeg}|sphæricall| bodys $e,f$ to have a springing or elastic{illeg}|k| for{illeg}|c|e soe yt {illeg}|m|eeting one another they will relent & be pressed into a sphæri|o|idicall figure, {illeg}|{&}| in yt moment in wch there is a period put to the{illeg}|ir|e motion towards one another theire figure will be most sphæroidicall & theire pression one upon the other |i|s at ye greatest, & if th{illeg}|{e}| bodys endeavour to restore theire \call/ figure w{illeg}h \bee/ as much vigor|ous| & force|i||ble| as it was destroyed by, & as theire pressure upon one another wa{illeg}|s| to destroy it they will gaine theire whole /as much\ motion from one another \thei{illeg}|r| {illeg}|p|arting/ as their|y| \had/ towards one another {illeg} theire reflection. Secondly suppose they be sphæricall & absolute\ly/ sollid then at the period of theire motion towards one another (yt is at ye moment of theire meeting) theire pression is at ye greatest or rather tis \done with/ the whole for{illeg}|c|e by wch theire motion is stopt, \for theire whole motion was stoped b{illeg}|y| ye force of theire pressure upon one another in ysone mome\nt// for /&\ there cannot $\begin{array}{c}\text{bee}\\ \text{succeede}\end{array}$ divers degrees of pressure twixt two bodys in one moment) Now i{illeg}|f| /so long as\ neither of these 2 bodys yeild to one another they will {illeg} always retaine ye same forcible pressure towards one another: that is soe much force as deprived ye bodys {illeg} of th{illeg}|ei|r motion (when it moved towards h \towards one another soe much/ doth now {illeg} it towards g, & therefore (r) y{illeg} urge them from one ano{illeg}|t|her, & therefore (r)[74] they shall move from one another as much as they did towards one another before theire reflection.
10. There is {illeg}|y|e same reason wh{illeg}|e|n unequall & unequally moved \bodys/ reflect, yt they should sepera{illeg}|t|e from{illeg} one another wth as much m{illeg}|{ot}|ion as they they came together.
[75] [76] 11. If a line \$\left(df\right)$/ bee moved not wth a Circular \Progressive/ but onely a Circular motion i{illeg}|t|s middle point \$\left(n\right)$/ shall rest. For if it move let it move towards r soe yt, when ye point $\left(d\right)$ is moved in $\left(p\right)$ & f in $\left(q\right)$, then $\left(n\right)$ shall be moved to $\left(s\right)$ :
11 If {illeg}|a| line $\left(ce\right)$ be bisected in $\left(a\right)$ about wch ye line $\left(ce\right)$ doth circulate {illeg}|&| yt point bee fixed. yn ye whole line hath noe progressive motion. For makeing $ab=ad$, bf, ag, & dh bee parallel, & perpendic to fh, yn is $vb=dp$. & $vf+ph=bf+dh=2ag$. Wherefore ye point c moveing to{illeg}|w|ards n ye point {illeg}|d| shall move soe much towards ye line fh as ye point b doth from it, & all ye points in {illeg} $\left(ac\right)$ or ye line ac move as mue|c|h from to ye line fh as \all/ ye points in $\left(ae\right)$ or ye line $\left(ae\right)$ moves from it soe yt ye whole line ce stays in equilibrio neither moveing to nor from fh, by ye 12th Defin{illeg}|.|
12 H{illeg}|e|nce when \ye center of a line/ $\left(a\right)$ is not in ye center \midst/ of a line $\left(me\right)$ ye whole line moves ye same way wch ye longest pte doth. for supposeing $ca=ae$ y{illeg}|n| ye line ce in equilibrio (ꝑ ax:1{illeg}|1|) but if $\left(mc\right)$ moves towards $\left(fh\right)$ & be added to $\left(ce\right)$ yn $\left(me\right)$ moves towards ce (by def {12}
13. When $\left(ce\right)$ moves circularly but maketh noe progression it{illeg}|s| midle {sic} point shall rest & is therefore ye center of its motion, for if ye middle point move ye whole line let it bee to r from a soe yt ye line ec bee moved into ye place {$\left(wt\right)$} yn let ye {$\left(wt\right)$} move about ye fixed center r into ye place xs, yn {xs} & {wt} are equally distant from fh (by def: 13 & ax: 11) & alsoe $\left(ln\right)$ & $\left(ce\right)$ are equally distant from ye {illeg}{am} f{illeg} but xs is further from $\left(fh\right)$ yn $\left(ln\right)$ there & ln are not equally distant from f{illeg}{illeg}ore neither are wt & ce {illeg}|e|qually distant f{illeg}|r|om fh. & therefore ye line {illeg} pro{gre}ssive motion when it {pressed} into {wt}
14 By ye same reason ye midle {sic} point {illeg}ogram, parallelipipedon, prisme cilinder, circle, sphære, {illeg}|e|lip{illeg}|s|is, sphæroides {illeg} of theire motion
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[77] 15 A Body moves yt way \hath ye same determinacon of its motion/ wch the center of its motion hath. As if ye line ac move into ye place gh, ye center of its motio{illeg}|n| b moveing into ye place d. th{illeg}|e|n let gh move about ye center d untitill {sic} it be parallel to ac, as into ef \soe yt ye point a fall into ye {illeg}|p|oint e/. Now since {illeg} gh by turning about ye center d hath noe progressive motion (by Def: 10) tis plaine yt gh & ef {illeg} have determinacon from ac but ef hath ye same determic|n|acon from ec wch d hath from {illeg}|b| (for if ac be understood to move parallell to its selfe into ye place ef, all its points describe parallel lines & therefore have ye s{illeg}|a|me deter{illeg}|m|inacon one wth another & each with ye whole body (by axiom 14.) therefore gh hath ye same determinacon from ac wch ye point d hath from ye point b
14 A Body being moved parallell to its selfe all it{illeg}|s| points describe ∥ lines, \& each{illeg} y have ye same determ \& velocity/ wth ye body./ for (by axiome 2d) they {illeg}|m|ust {illeg}|a|ll bee streight ones wch if they intersect ye body will not be \moved/to its selfe. &c.
16 If a body move forward \& circularly/ its center of motion shall allways bee in ye same streight line. For ye body hath allways ye same determinacon (ax 2d) & ye cent' of its motion hath ye same determinacon wth ye body (ax 15) therefore it hath always ye same determinacon, & soe will move continually in ye same streight line.
[78] 17 If a body move \streight/ forward & circularly its cente{illeg}|r| of motion shall have ye same determinacon & velocity yt {illeg}have had did ye body move Pa∥ll to it selfe \the body hath/. For {suppose} ac wh to be moved into ye place gh & its center of motion b into ye place d, yn let it turne about ye center d into ye place ef ∥ to ac s{illeg}|o|e yt ye point wch was in $\left(a\right)$ bee now in $\left(e\right)$. N{illeg}|o|w since {$\left(gf\right)$} by moveing into ye place $\left(fe\right)$ makes noe progressiv motion (def: 10) it follows yt {illeg}parallell to it selfe into ye place ef it would {illeg} ye same determinon {sic} & quantity of motion ye same quantity \(Or since $gh=ef=ac$) ye same velocity/ & \(axiom 10)/ determinacon of motion \in ye same time/ would translate ac parallell to it selfe into ye place ef yt w{illeg}ld {sic} translate it into ye pla{illeg}|ce| gh , had it bo{illeg}|t|h progressive & circular motion. But ye point d {illeg} hath ye f|s|ame velocity & determinacon wch ye line ef hath when moved ∥ to its selfe (x)[79] therefore ye point d hath ye same determinacon & velocity wch ye line gh hath when moved wth both ◯lar & progressive motion \vide ax 37/.
18 If a body move progressively in some crooked line & alsoe circularly {illeg} its center of motion shall have ye same determinacon & velocity wch ye body {hath} for in {illeg} {illeg} any point of ye crooked line its determinaco {sic} is in ye tangent \(ax 17) th{illeg}|i|s is trew when its {illeg}|m|otion is in a strei{illeg}|gh|t lin{illeg}|e| but a crooked line may/ bee conceived to consist of an infinite number of streight lin{illeg}es. Or else in any p{illeg}|oi|nt of ye croked {sic} line ye motion may bee conceived to be on in ye tangent.
[80] 19 {sic} 2 bodys circu make ye same number of circulations wth ye same dista{nce} from ye center c : yn as ye Radij of ye circles wch they|ir| \centers of motion/ {illeg}|de|scribe are to one another soe are ye perimeters one to another soe are theire velocitys one to another (ax: 10, def: 2), & their motions are to {illeg}|o|ne another as theire bulkes drawne into ye Radij of those circle {sic} (wch theire centers of motion describe) are to one another (def 3). As: $ec=eo\colon\colon$ velocity of $eb:$ velocity of ac. & $eb×ec:ao×co\colon\colon$ mot $eb:$ mot of ao.
[81] 20 If a s{illeg}|ph|ære ⊙c move wth in \be compelled by/ /move wth in\ {illeg}|y|e concave sphæcall {sic} or cilindricall surface of ye body edf to move circularly abou{illeg}|t| ye center m it shall press upon ye body def for when it is in c (suposeing ye $\mathrm{\odot }$ bhc to be described by it {sic} center of motion {illeg} & ye line cg a tangent to yt $\mathrm{\odot }$ at ⊙) it moves it moves towards g or ye determination of its motion is towards c therefore if at yt moment ye body edf should cease to check it it would continually move in ye {illeg}|l|ine cg (ax 1. 2.) obliq{illeg}|l|y from ye center m , but if ye body def oppose it selfe to this indeavour in ever keeping it equidistant from m , that is done by a continued \checki{illeg}|n|g or/ reflection of it from ye tangent line in every point of ye ◯ chb , but ye body edf cannot check & curbe ye determinacon of ye body c⊙ unless they continually presse upon one another. |The same may {illeg}|b|e understood if ye body adb bee restrained into ◯lar motion by ye thred $\left(om\right)$ |
21. Hence it appear{illeg}|e|s yt all {illeg}|b|odys moved ◯larly have an endeavour from ye {illeg}|c|enter abot|u|t wch they move, otherwise ye b{illeg}|o|dy ⊙c would not continually presse upon edf .
2{illeg}|2| The whole force by wch a body c⊙ indevours from ye center {illeg}| m | in halfe a revolution {i}s \more yn/ double to the force wch is able to generate or destroy its motion for supposeing it have moved from $\left(c\right)$ by $\left(h\right)$ to $\left(b\right)$ then i|y|e resistance of ye body ef {illeg} (wch |is| equall to its pressure upon def ) is able to destroy its force of moveing {illeg}{illeg} & to generate in it as much force of moveing from $\left(b\right)$ to $\left(h\right)$ the qu{illeg}g {illeg}|w|ay.
[82] 2{{illeg}|5|}{{illeg}|3|} {illeg}|H|aving {illeg}of motion of ye 2 bodys ob & dc to find ye common center of both {in}{illeg}draw a line ⊙e from the centers of theire motions ⊙ & e & divi{illeg}oe yt {illeg}|th|e body ob is to ye body de as the line ae t{illeg}|o| ye line oa : yt is soe{illeg} $ae×de$ . For th{illeg}|e|n if they move about ye center {illeg}{illeg}{illeg} they have equall motion (ax 19th) & consequen{tly} {illeg} <12r> have an equall endeavour from ye center a (ax 24|3|) soe yt if they bee joyned tig{illeg}th by the line to center $\left(a\right)$ by ye lines {illeg} ae & ao they shall not mo {illeg}|t|he one h{illeg}|i|ndereth ye other from forcing ye center a any way soe yt it shall stand in equilibrio betwixt them & (by def 10) is therefore their center of \motion/
[83] 24 If two bodys ( cb & de ) move about a center $\left(a\right)$ yn as ye motion o makeing $bc=a$, $de=b$, $ac=c$, $ae=d$, ye time in wch bc makes halfe a revolution call e, yt in wch de doth ye sam{illeg}|e| call f , ye pressure of cb from ye center a in halfe a revolution call q , {illeg}|&| yt of de call h ; ye motion of cb in halfe a revoluco k & yt of de call l . yn $k:l\colon\colon$
24 If two bodys ( cb & de move about a center $\left(a\right)$ yn ∼ ∼ ∼ ∼ ∼ ∼ The \whole/ force by wch ye body cb tends from ye center a \in one revolutio {sic}/ being equall to {{illeg}|6|}{{illeg}|61|} times ye force by wch y{illeg}|t| body is moved \(ax 22)/ is to ye motion of yt body {illeg} one revolucon as ye \whole/ force by wch ye body de tends from ye center a in one revolution (wch is equall to {illeg}|6|{+} times ye force by wch ye de is moved, or wch is able to stop i{illeg}|t|s motion (ax 22) ) {illeg}|is| to ye motion of ye body de . Vide Axioma 23ũ.
[84] {illeg}|2|6 If y{illeg}|e| body a move through ye space ab \$=b$/ in ye time d \$=be$/, & ye body c {illeg}|thr|ough ye space cd |$=e$| in ye time f . then ye velocity of a is to ye velocity of {illeg} c as $\left(ab×dc\right)$ to $cd×$ {illeg} be . for supposeing $cb=gp$ then ye velocity of a is to ye velocity of c as $\left(ab\right)$ to $\left(cp\right)$ (def 2d) or as $ab×gp:cp×eb\colon\colon ab×cd:ad×eb$ . for since \Then is/ $cp×fd=cd×gp=cd×eb$ . Or {illeg} $cp×fd×ab×\left\{\begin{array}{c}eb\\ gp\end{array}=cd×eb×ab×gp$ yt is $ab×gp:cp×eb\colon\colon ab×fd:cd×eb\colon\colon ab:cp\colon\colon$ velocity of $a:$ velocity of c velocity of $a:$ to ye velocity of $c\colon\colon ab:cp${illeg} For supposeing yt $gp=eb$. yn is $cp=\frac{eb×cd}{fd}$ And (by def: 2) ye velocity of $a:$ is to ye velocity of $c\colon\colon ab:cp\colon\colon ab:\frac{eb×cd}{fd}\colon\colon ab×fd:eb×cd$
[85] Alsoe ye motion of a is to ye motion of c (by def 3d) $\colon\colon a×ab:c×cp\colon\colon$$a×ab:$\$c×c$/ $a×ab:c×cp\colon\colon a×ab:\frac{c×eb×cd}{fd}\colon\colon a×ab×fd:c×eb×cd\colon\colon a×\text{its velocity}:c×\text{its velocity}$.
Note yt when ye motion is uniforme yt is when a body moves over ye same sp{illeg}|a|ce in ye same time (wch will ever bee when ye motion of {illeg}|y|t body is neit{illeg}|h|er helped nor hindred) yn in a right angled triangle a b may designe ye space through a body moveth {illeg} in ye ti{illeg}|m|e eb. Otherwise when tis not uniforme ye proportion of ye time in wch a body moves to ye {illeg} distance through wch it moves may be designed by lines drawne to a crooked line, as ye time by gf ye dis & ih , ye distance by gh or fi , ye velocity by ye proportion of nh to hi , ni being tangent to ye crooked line at i . &c.
[86] 23 If 2 bodys be moved wth equall or uneq
If ye body bace is moved \acquire ye motion q / by ye force {illeg}| d | {illeg}|&| ye body f \ye motion p/ {illeg}|b|y ye force g . yn {illeg} $d:q\colon\colon g:p$. for suppose ye body $rscb=f$ , yn ({illeg} {illeg}) to {illeg} acquire ye motion w by ye force d , yn (ax: 5)[87] $d:g\colon\colon w:p$ . but $q=w$ (by ax: 4) therefore $d:g\colon\colon q:p$.
Ax: {illeg}|{1}|00 Every thing doth naturall{illeg}|y| persevere in yt state in wch i{illeg}|t| is unlesse it bee interrupted by some externall cause, hence axiome 1st, & 2d, & {γ}, A body once moved will always keepe ye same cele{illeg}|r|ity, quantity & determinacon of its motion.
[88] If 2 equall bodys (bcqp & r) meete one another wth equall motion \celerity/ (unlesse they could pass through one ye other by penera|tr|acon of dimensions) they must mutually hinder their perseverance {illeg} in their{illeg} states, & (since ye one hath no{illeg}|e| advantage more yn /over\ ye other they must) equally hinder ye one ye o{illeg}|t|hers haveing both of them an equall power to persever in theire state |celerity power to persevering in its state| |perseverance in its state| likewise if ye body aocb be = & equivelox wth r they have a like power of persevering &{illeg} |{illeg}ing {illeg} equally {illeg}|hin|der or op{illeg}|p|ose ye one yt {offers} progression or perseverance in their states| therefore ye power of ye body aopq (wh{illeg}|e|n tis equivelox wth r) is double to ye power yt r hath to persever i{illeg}|n| its state. yt is ye e{illeg}|ff|ic{illeg}|a|cy force or power \of ye cause/ wch can reduce aopq to rest must bee double to ye power & efficacy of ye cause wch can reduce r to rest, or ye power wch ca{illeg}|n| move ye one must {illeg}|b|e double to ye power wch can move ye other soe yt they bee made equivelox.
Hence in equivelox bodys ye powers of persevering in their states are proportionall to their quantitys.
101 Hence may bee perceived what is meant. Supposeing ye bodys aobc & cbqp to be equall & equivelox: Then {illeg}|t|hat cause \hindrence, impedimnt resist{ance}/ or opposi{illeg}|ti|on wch can \onely/ deprive cbqp of its \whole velocity &/ motion by hindering its p{er}severance can {illeg}|a|l{illeg}|s|o \onely/ deprive aocb of its \whole {whole velocity &}/ moti{illeg}|o|n {illeg} yt cau{illeg}|s| hath ye same {illeg}{illeg} over both ye bodys. Now if {illeg} add ye opposition $\left(a\right)$ wch can \{being} {illeg}ive of its {illeg}/ {illeg}|r|educe cbpq to {illeg}{illeg}ion $\left(b\right)$ wch can reduce aob {illeg} \{illeg}/ ye whole opposition ( $a+b=2a=2b$ ) {illeg} \{illeg}/ both {illeg} bodys $aobc+bcpq=aopq$ ) {illeg} \{illeg}/ motion when they are joyned into one $\left(aopq\right)$ for $a:cbpq\colon\colon b:aobc\colon\colon a+b:cbpq+aobc\colon\colon 2a:aopq$ Also neither {illeg} a or b {illeg} aopq of {illeg} motion for {illeg} <12v> pte ( a or b ) would be equall to ye whole ($a+b=2a=2b$ ). By ye same r{illeg}|e|ason | aopq {illeg}|&| cbqp loosing equall velocity ye resistance /impediment\ of aopq must be double to ye opposition of cbpq .|
[89] 102 By they {sic} same reason yt \Since/ beacuse aopq {illeg} is double to cbpq & both of ym equivelo{x} therfore ye opposition wch can deprive aopq of its motion must be double to yt wch can deprive cbpq of its motion; by ye same reason it will follow yt in equivelox bodys as one body{illeg} \ $\left(a\right)$ / {illeg}|i|s to another \ $\left(b\right)$ / {illeg}|s|oe must ye resistance wch can deprive yt body $\left(a\right)$ of its $\left\{\begin{array}{l}\text{velocity}\\ \text{motion}\end{array}$ bee to ye resistance wch can deprive $\left(b\right)$ of its whole $\left\{\begin{array}{l}\text{velocity}\\ \text{motion}\end{array}$ so is ye resistance wch can deprive $\left(a\right)$ of some pte of its velocity, to ye restance {sic} wch can deprive $\left(b\right)$ of ye same quantity of velocity, soe yt a & b bee still equivelox.
Now {illeg}|i|t may be perceived how & why {illeg} amongst bodys moved some require a greater dome a lesse opposition to deprive ym of theire whole velocity or of some pt{e} of it wch {illeg}p
103 By ye same reason alsoe If two bodys rest or bee {illeg}|e|quivelox: yn as ye body $\left(a\right)$ is to ye body $\left(b\right)$ soe must ye power orf efficacy \vigor strength/ or virtue of ye cause wch begets new velocity in $\left(a\right)$ {illeg}|b|ee to ye power virtue or efficacy of ye cause wch begets ye = same quantity of velocity in b , soe yt a & b {illeg} bee still equivelox.
104 Hence it appeares how & why amongst bodys moved som{illeg}|e| require a more potent or efficacious cause others {a lesse} to hinder or helpe their velocity. And ye power of this cause is usually called force. And as this cause \useth or/ apply|eth| its power or force to hinder {illeg} or helpe \or change/ ye {illeg}g perseverance of bodys in theire state, it is said to Indeavour {illeg}|t|o change their {illeg} perseverance.
[90] 105 If ye equall & equivelox bodys a & b meete (unlesse they could passe yt one through ye other by penetra{illeg}|t|ing its dimen{illeg}|tio|ns) they must necessarily hinder ye one ye others progression, & since these bodys have noe advantage ye one over ye other ye hindrance on both pts will be equall, likewise if ye bodys $d+a$ & $b+c$ bee equall & equivelox they must equally hinder one anothers progression in its s{illeg}|t|ate But ye body $\left(b\right)$ (being lesse yn ye body $\left(b+c\right)$ ) & equivelox wth $\left(d+a\right)$ ) canot equally hinder ye progression of ye body $d+a$ soe much as ye body $\left(b+c\right)$ ca{illeg}|n|; for {yn} the power of $\left(b\right)$ being part of ye power of ye body $b+c$ would bee equall to ye {illeg} {illeg}|w|hole power of $b+c$ therefore yt $b+c$ & $d+a$ being equivelox d{illeg}|o|e equally hinder ye one ye others progression tis required yt they be equall.
[91] 106 Now if ye {illeg} bodys a & b meete one another ye cause wch hindereth ye progression of a is ye power wch b hath to persever in its state velocity \or state/ & is usually called ye force of ye body b & {illeg}|i|s this power or force are said to |soe yt a body is {so} to be moved wth more or lesse force wch meeting wth another body can cause a greater or lesse mutation in its state, or wch requireth more or l{illeg}|e|ss for{illeg}|e| {sic} to destroy its motion.| & as {one} body b useth or applyeth this force to stop ye progression of a it is said to {illeg} Indeavour ty|o|e hinder ye {illeg} progression of a \wch indeavour in body is performed by pressure/ & by ye same reason ye body b may {illeg}|b|ee said to endeavor to helpe ye motion of a if it should apply its force to move it forward: soe yt it is e{illeg}|v|ident wt ye Force & indeavor of \i{illeg}|n|/ bodys are.
[92] 107 If ye bodys b & c be equiv{illeg}|e|lox yn as {illeg} $b:c\colon\colon$ ye force of \wth wch/ b \is moved/ (or ye power of b to persever in its velocity \or to \{keepe}{helpe} {illeg}/ hinder another body f{illeg}|r|o persevering in its velocity/ to ye force of c . For let there be 2 other bodys a & d equivelox to th{illeg}|e|m soe yt a meeting b , & d meeting c they would eqaully hinder one others progression yn is $a=b$ , & $c=d$ (ax 105) & $a\left(=b\right):d\left(=c\right)\colon\colon$ yn force wch can stop $a$ (= to ye force of b ) to ye force wch can stop d (= to ye force of c ). |(vide ax: 102.|
[93] 108 Tis knowne by ye light of nature yt equall forces shall effect an {illeg}|e|quall ch{illeg}|a|nge in equall bodys. Therefore if ye forces g, h, k, m, be equall, & ye bodys a, b, equall {illeg}|&| rest, then let $\left(a\right)$ bee moved by ye force g ; & b by h , a & b shall be equiv{illeg}|e|lox: Also (since tis noe greater change for $\left(a\right)$ to acquire another part of motion now it hath one yn for it to acquire y{illeg}|t| one when it {illeg} had none) if $\left(a\right)$ bee againe moved forward by ye force k , its velocity shall be double to ye velocity of b , & if it bee againe moved forward by ye force m its velocity shall be triple to yt of b . &c. Whence as ye force moving $\left(a\right)$ is to ye force moving $\left(b\right)$ soe is |ye| velocity acquired in $\left(a\right)$ to ye velocity acquired in $\left(b\right)$ \{{illeg}|b|y that force}/
109 By ye same reason if \$a=b$ &/ ye velocity of a be t{illeg}|r|i{illeg}|p|le to ye velocity of b , y{illeg}|t| force {illeg} wch can deprive a of its velocity. must be \wch is/ 3ple to ye force wch can deprive b of its velocity. Or in gener{illeg}|a|ll {\2/} {illeg} \so/ \is/ ye \lost/ velocity of a {illeg} to ye \lost/ velocity of b soe is \As/ ye force wch one deprives a of \some or all of/ its velocity, to ye force wch can deprives b of \some or all of/ its velocity & so is ye force \{That}/ wch can deprive a of
{illeg} ye force {illeg}|w|th{illeg} is {illeg}\or preserve it selfe in its {illeg}/ is to ye force wth w{illeg}|c|h b is moved {illeg} ye velocity of a {illeg} velocity of b otherwise it {illeg} not be {illeg} ye {illeg} Ax 5t
The force wch ye body $\left(a\right)$ hath to preserve it selfe in its state shall bee equall to ye force wc{h}{illeg}{illeg} yt state; not greater for ye effect cannot exceede the {illeg} for {illeg}{illeg}{illeg} ye {illeg} wch was not in ye cause \nor lesse for/ ye cause only {illeg}to its effect{illeg}no reason why its {illeg}{illeg}
<13r>
[94] \112/ A body is saide to have more or lesse motion as {illeg} it is moved wth more or lesse force, yt is as ther{illeg}|e| is more or lesse force required to generate or destroy its \whole/ motion.
[95] 11{illeg}|3| If a body $\left(a\right)$ move through ye space ab \$=r$ / in ye time c . & {illeg}|t|he body {illeg}|f | through gf \$=v$ / in the time h then, time $c:$ time $h\colon\colon$ line ab \$=r$/$:\frac{h×ab}{c}=$ fk ak . & ye body a would move through ye space ak in ye same tim $\left(h\right)$ in wch ye body f moves through ye space fg . Therefore ye velocity of a is to ye velocity of f as ye line $ak=\frac{h×ab}{c}:$ line fg {illeg} $\colon\colon h×ab:c×fg$ {illeg} \(def 2)/ Then I ad {sic} ye body r to f soe yt $r+f=a$ . since f & r are equivelox, \(ax: 107)/ as $f:f+r=a\colon\colon m=$ force or motion of f , to {illeg} $\frac{am}{f}=$ force of $f+r$. againe sine {sic} $a=f+r$ , & they move \(ax: 111)/ as ye velocity of a ; to ye velocity of $f+r\colon\colon h×ab:c×fg\colon\colon$ $\colon\colon n=$ ye force or motion of a , to {illeg} the $\frac{n×c×fg}{h×ab}=\frac{a×m}{f}=$ to ye force of $f+r$ . Soe yt, $n×f×c×fg=m×a×h×ab$ . So{illeg}|e| yt haveing any 7 of t{illeg}|h|ese ye 8th m{illeg}|a|y bee found. {illeg} but suppose ye bodys moved in equall ti{illeg}|m|es yt is if $c=h$ , yn ye rest of ye termes may bee found by, $m×a×ab=n×f×fg$ . &c. yt i{illeg}|s| as $f×fg$ is to $a×ab$ soe is ye motion $\left(m\right)$ of ye body f to ye motion $\left(n\right)$ of ye body a . &c.
[96] 110. If ye body {sic} ( a & b ) bee equall & ye celerity of a ti|r|iple to yt of b , yn if ye force {illeg} {illeg}| d | can deprive {illeg}| b | of its motion, ye force $3d$ may can deprive a of its motion. But if there bee lesse force \ $\left(3d-p\right)$ / it cannot deprive a of its motion for soe ye pte $\left(3d-p\right)$ would be = to ye whole $3d$ ; if there be more force $\left(3d+p\right)$ it will doe more yn deprive the body a of its motion (i.e. move it ye contrary way) otherwise ye pte $\left(3d\right)$ would be equall to ye wh{illeg}|o|le $\left(3d+p\right)$ . |Therefore ye |If ye body a bee equall to ye body b .
force which can deprive a of its motion must bee 3ple to ye force wch can deprive b of its motion & consequently ({illeg}|d|ef 106) ye for{illeg}|c|e wherwth a is moved is 3ple to ye force wherewth b is moved
111 By ye same reason as ye celerity of ye body $a\left(=b\right)$ is to ye celerity of b so is ye for{illeg}|c|e wherewth a moveth to ye force wherewth b moveth.
[97] 114 There is required soe much & noe more force to reduce a body to rest y{illeg}|n| there is to move it: et e contra. And
115 Soe much force a is required to generate any quantity of motion in a body so{illeg} much is required to destroy it, & e contra. For d \in/ loose|i|{illeg}|n| {sic} or to get|ting| ye same {illeg}|q|uanty {sic} of motion a body suffers ye same quantity of mutacon in its state, & in ye same body equall forces will effect a equall change
[98] 116 If ye bodys $a=3b$ , & a & b {illeg} are moved wth ye same force $\left(d\right)$ yn ye celerity of b {illeg} is tri{illeg}|p|le to ye celerity of {illeg} a . for if a be moved by suppose $b:a\colon\colon d:\frac{ad}{b}$ & let $\left(a\right)$ bee moved by $\left(\frac{ad}{b}\right)$ & $\left(b\right)$ by {illeg} for if $a\left(=3b\right)$ be moved by $3d$ , & $\left(b\right)$ be {illeg} moved by $\left(d\right)$ , ( a & b ) shall {illeg} }{illeg} \moved by/{ for $3b$ moved by $3d$ is equivelox to b moved by d , but since $3b=a$ , therfore a moved {illeg}|b|y $3d$ is equivelox to b moved by d . And (ax 108) as ye ce{illeg}|l|erity of a moved by {illeg}| d | is to ye celerity of a moved by $3d$ , soe is 1 to 3 , soe is ye celerity of a moved by d to ye celerity of b moved by d .
By ye \same/ reason, Any bodys f & g being moved by ye same force as $\left(f\right)$ is to $\left(g\right)$ , soe is ye celerity of $\left(g\right)$ to ye celerity of $\left(f\right)$ acquired by yt force. tis axiome ye 4th And (by ax: 113) ye bodys will have equall motion.
[99] 118 If ye body p , be moved by ye forc{illeg}|e| q , & r by ye force s , \to find {illeg} $\left(v\right)$ ye celerity of p & $\left(w\right)$ yt of r ,/ I add $\left(t\right)$ to p , soe yt $p+t=r$ , & yt $\left(p\right)$ , & $\left(p+t\right)$ are moved wth equall force, yn $p+t\left(=r\right):p\colon\colon v:\frac{pv}{r}$ ye celerity of $p+t$, (ax 117) alsoe, (ax 108) $s:q\colon\colon w:\frac{qw}{s}=\frac{pv}{r}$. Or $qrw=pvs$ . that is ye celerity of p is to ye celerity of r as $qr$ is to $ps$. And by ax 113 ye motion of p is to ye motion of r as ye force of p to ye force of r . And by ye same reason if ye motion of p & r bee hindered by ye force q & s , {illeg}ye motion lost in p is to ye motion lost in r , as q is to s . or if ye motion of p be increased by ye force q , by|u|t ye motion of r hindered by ye force s ; the {illeg} as q , to s ∷ so is ye increase of motion in p , to ye decrease of it in r (ax 111
[100] 1{1}9 \121/ If 2 bodys p & r p {illeg}{nest} ye one ye other, ye resistance {illeg}|i|n both is ye same for soe much as p presseth upon r so much r presseth on p . And therefore they must both suffer an equall mutacon in the {illeg} motion.
11{illeg}|9| {illeg} If r presseth p towards w then p presseth r towards w . {illeg} wthout {illeg}
120 A body must move yt way wch it is pressed.
122 Therefore if ye body p comes from c & ye body r from d soe much as { p {illeg} } motion is changed towards w soe much ye motion of {illeg} changed {illeg}{illeg}
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[101] \27/ If two bodys {illeg}| b | & c move from $\left(o\right)$ their center of gravity they shall have equal motion For suppose b moved into ye place d ; then putting, $c:b\colon\colon do:\frac{b×do}{c}=oe$ (ax 25) ye body c must be the{illeg}|n| moved into ye place {illeg} e . Alsoe $c:b\colon\colon bo:\frac{b×bo}{c}=oc$ . (ax 25) therefore $ec=\frac{b×bo-b×do}{c}=\frac{b×bd}{c}$ . that is {illeg} $c×ec=b×bd$ . But (ax $\left\{\begin{array}{c}113\\ 26\end{array}\right\}$ or) $c×ec:bd×b\colon\colon$ ye motion of c , to ye motion of d , & therefore c , & d have equall motion towards o
If ye body b move through ye places d , f , r , & ye body c through ye places {illeg}| e | g h & their center of o , p , q , r ye line opqr shall be a {illeg}|s|treight line For nameing ye lines $bc=g$ . $cr=h$ . $br=k$ . $bd=x$. $d:e\colon\colon x:ce$ . Then Then $ce=\frac{ex}{d}$ . $k:g\colon\colon x:\frac{gx}{k}=cv$ & supposeing $bc\parallel dv\parallel ps$ . Therefore {illeg} $ve=\frac{ex}{d}=\frac{gx}{k}$ & then $k:g\colon\colon k-x:\frac{gk-gx}{k}=dv$ . $b+c:c\colon\colon de:pe\colon\colon dv:ps\colon\colon ve:es$ also $k:h\colon\colon x:\frac{hx}{k}=cv$, & $\frac{ex}{d}-\frac{hx}{k}=ve$ therefore $ps=\frac{cgk-cgx}{bk+ck}$. & $es=\frac{cekx-cdhx}{bdk+cdk}$. $cr-ce=er=h-\frac{ex}{d}$ . $er+es=sr=\frac{cekx-cdhx}{bdk+dck}+\frac{dh-ex}{d}$ . Or $rs=\frac{bdhk+cdhk-bkex-cdhx}{bdk+dck}$. {illeg} $b+c:c\colon\colon bc:co\colon\colon g:\frac{cg}{b+c}=co$ . $cr:co\colon\colon h:\frac{cg}{b+c}\colon\colon bh+ch:cg\colon\colon rs:sp\colon\colon bdhk+cdhk-bkex-dhcx:cdgk-cdgx$ . Whence $cdgk\stackrel{.}{bh}-cdgxbh+cdgk\stackrel{.}{ch}-cdgx\stackrel{.}{ch}=bdhk\stackrel{.}{cg}+cdhk\stackrel{.}{cg}+kexbcg-dhcx\stackrel{.}{cg}$
[102] 28. To fin If two bodys b & c mo{illeg}|v| {illeg} through in ye lines br & cr . The body c moveing through ye space cg in ye ti{illeg}|m|e vs , & though g h in ye time nv , & through kr in ye time nr . & ye velocity of ye body b is to ye velocity of c as d , t{illeg}|o| e , & a{illeg}|s| ye line cg to ye line be , or as ck to br , then {illeg} when ye body c is in ye place g , b will bee in e , & when c is in k , b will be in r . to find ye line wch the center of their motion describes, viz dfo . {illeg}|T|hen nameing ye quantitys $br=a$, $cr=f$. $bc=g$. {illeg} $e:d\colon\colon a:\frac{da}{e}=ck$ . $kr=\frac{ef-da}{e}$ . If o be ye center of gravity motion of ye bodys at k {illeg}|&| r ; yn, $b+c:b\colon\colon \frac{ef-da}{e}:\frac{bef-abd}{eb+ec}=or$ . And ye line df must passe through o . againe making $gk=v$ . yn $d:e\colon\colon v:\frac{ev}{d}=er$ . & if $bc\parallel fi\parallel em$ , yn $a:g\colon\colon \frac{ev}{d}=er:\frac{gev}{ad}=em$ . also since f is ye &, $a:f\colon\colon \frac{ev}{d}=er:\frac{fev}{ad}=mr$ $gr=gk+kr=v+\frac{ef-ad}{e}$ . $gm=gr-mr=\frac{ev+ef-ad}{e}-\frac{fev}{ad}=\frac{adev+adef-aadd-feev}{ade}$. since f is ye center of motion in ye bodys at g & e tis, $b+c:c\colon\colon$ {illeg} $ge:fg$ . {illeg} $b+c:c\colon\colon eg:fg\colon\colon em=\frac{gev}{ad}:\frac{cgev}{abd+acd}=fi\colon\colon gm=\text{&c}:gi=\frac{cadev+cadef-caadd-cfeev}{bade+cade}$. $gr=gk+kr=\frac{ev+ef-ad}{e}$. $go=gr-or=\frac{ev+ef-ad}{e}+\frac{abd-bef}{eb+ec}=\frac{bev+cev+cef-cad}{eb+ec}$. $go-gi=io=\frac{adbev+cfeev}{bade+cade}=\frac{abdv+cefv}{abd+acd}$. $co=cr-or=\frac{fec+abd}{eb+ec}$. $b+c:c\colon\colon g:\frac{cg}{b+c}=cd$ . [103] {illeg}|N|ow if the lines $oi:if\colon\colon oc:cd$ . Then ye line od must be a streight line. but $oi:if\colon\colon adb+cfe:cge\colon\colon oc:cd\colon\colon \frac{fec+abd}{e}:cg\colon\colon fec+abd:ecg$ . therefore ye line do is a streight line, wth wch may bee found by ye two points d & o . |The demonstracon is ye same if ye body b moved from a to b |
29 If two bodys q & c be moved in divers plines {sic}, then find ye shortest {illeg} line $\left(pr\right)$ wch can bee drawne frome one line $\left(cr\right)$ {illeg}|to| ye other line $\left(qp\right)$ in wch those bodys are moved, & yt line pr shall bee perpendicular to both ye lines cr & qp , viz $\angle qpr=$ $\angle rps=prc=\text{recto}$. then draw qb equall & $\parallel pr$ & draw $br=qp$ . Then shall ye plaine qbrp be perpendicular to ye plaine bcr . Suppose also ye body c moves over ye space $\begin{array}{c}cg\\ gk\\ kr\end{array}}\text{in the time}\left\{\begin{array}{c}vw\\ wt\\ tr\end{array}\right\$ {illeg}|&| yt ye body q moves over ye space qa in ye time vw , & ap in ye time wt . Also suppose another body $b\left(=q\right)$ & equivelox to { $\left(q\right)$ } yt is to move over ye space $be=qa$ in ye time vw {illeg}
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Soe yt when $\left(c\right)$ is in $\left(g\right)$ or $\left(k\right)$ {illeg}|,| $\left(b\right)$ will bee in $\left(e\right)$ or $\left(r\right)$ & $\left(q\right)$ in $\left(a\right)$ or $\left(p\right)$ . {illeg} Then \Then drawing {the} streight lines qc , ag , bk / if $b+c:c\colon\colon bc:cd\colon\colon eg:fg\colon\colon rk:ok$ ;|,| the points d , f , & o , shall be ye centers of gravity motion of ye bodys b & c , when they are in ye places b & c , e & g , r & k . & (prop 28) therefore ye line ( dfo {illeg}|)| in a streight line. Likewise if it bee {illeg} $q+c:c\colon\colon qc:lc\colon\colon ag:mg\colon\colon pk:nk$, then ye points l , m , n are centers of motion to ye bodys ( q & c ) being in ye places $\left(q\right)$ & $\left(c\right)$ , a & g , p & k . Then drawing ye lines ld , mf , no , f{illeg} ye (twixt ye neighbouring centers of motion) since $b+c:c\colon\colon$ {illeg} $q+c:c\colon\colon bc:cd\colon\colon q:$ {bc }. therefore $\angle qbc=\angle ldc$ & by ye same reason $\angle gfm=\angle gea$ {illeg}|&| $\angle krp=\angle kon$. Wherefore |all the lines qb , ae , pr , ld , ml , no are parallell t{illeg}|o| one another. And| $b+\phantom{0}:c\colon\colon bc:dc\colon\colon qb:ld\colon\colon eg:fg\colon\colon ea\left(=qb\right):mf\colon\colon kr:ko\colon\colon pr\left(=bq\right):no$ , soe yt $ld=mf=no$ . & since these line line {sic} ld , mf , no , are parallell, equall, in ye same plaine ldon , & stand upon ye same streigh{illeg}|t| line do , their other ends \(the ce{illeg}|n|ters of motion of c & q )/ l , m , n , must bee in ye same streight line lmn , wch line ye line \ $\left(lmn\right)$ / in wch their other ends l , m , n , are are terminated (i.e. in wch /are\ all ye centers of gravity motion of ye bodys ( c & q )) {illeg} must bee a streight line.
The demonstracon is the same if $\left(q\right)$ moved from $\left(p\right)$ to $\left(q\right)$ .
[104] 30. Suppose ye bodys b , & c moved towards, r ; so yn|t| when b is in $\left\{\begin{array}{c}b\\ e\\ p\end{array}\right\}$ then c is in $\left\{\begin{array}{c}c\\ g\\ k\end{array}\right\}$. & theire centers of motion describe ye line dq . Then ye motion of the|i|re centers of motion shall be uniforme. For $pr:pw\colon\colon er:ey$ \if $pw\parallel nt\parallel ey\parallel fs\parallel bc$ / $pr:er\colon\colon pw:ey\colon\colon nt:fs\colon\colon nq:fq$ . yt is $pr:ep\left(=er-pr\right)\colon\colon nq:fn=\left(fq-nq\right)$ & therfor since ye motion of ye body in epr is uniforme, ye motion of theire centers of motion in ye line fnq , must b{illeg}|ee| uniforme, yt is have allway alike {illeg}|v|elocity |The demonstracon is ye same in all other cases.|
[105] 28 \& 30/ Supposeing ye thing{illeg}|s| suppose {sic} in ye 28th prot{illeg}|p| {sic} by {illeg}|sc|hem {sic} 38th it may be thus done. $pr:${illeg} $er\colon\colon${illeg} $er:br\colon\colon ey:bc\colon\colon fs:dc$. Also $ep:eb\colon\colon cg:gk$. $\colon\colon cy:yw$ {illeg} & $cg:ck\colon\colon cy:cw\colon\colon gy\left(=cy-cg\right):hw\left(=cw-ck\right)\colon\colon gs:kt\colon\colon sy:tw\colon\colon be:bp$ $\colon\colon cs:ct\colon\colon dc-fs:dc-nt$ &c. Makeing $fs\parallel dc\parallel nt$ . & $mf\parallel qn\parallel et$ , yn is $mf=cs$ & $qn=ct$ . & $fs=mc$ , & $nt=qc=is$ . Then $be:bp\colon\colon cy:cw\colon\colon$ \$cg:$/ck (so is ye velocity of b to ye velocity of c ) $\colon\colon gy\left(=cy-cg\right):hw\left(=cw-ck\right)\colon\colon sy:tw$ (for $c+b:b\colon\colon eg:ef$ $\colon\colon gy:sy\colon\colon kp:np\colon\colon kw:tw$ .) {illeg} $\colon\colon mf\left(=cy-sy\right):$ {illeg} \ qn / $\left(=cw-tw\right)$ {illeg}. Againe $br:er\colon\colon bc:ey$ {illeg} $\colon\colon dc:fs$ \$=mc$,/ (for $b+c:c\colon\colon bc:dc\colon\colon$ fg $eg:fg\colon\colon ey:fs$) {illeg} \{illeg}/ Also Whence $be\left(=br-er\right):br\colon\colon dm\left(=dc-mc\right):dc$ . Also $br:pr\colon\colon bc:pw\colon\colon dc:nt\left(=qc\right)$ , whence Therefore $be:bp\left(=br-pr\right):dm\colon\colon dq\left(=dc-qc\right)$ . That is $be:bp\colon\colon dm:dq\colon\colon mf:qn$. & consequently ye points d f n are in one streight line. & since ye motion of $\left(b\right)$ is uniforme &, $be:bp\colon\colon dm:dq\colon\colon df:fn$, the motion of ye center d is uniforme.
[106]28 & 30th. The bd|od|ys \( b & c )/ being in b & c , e & g , r & k , {illeg}|i|n ye same times, & d n being described by their centers of motion. Also making $de\parallel fs\parallel ey$ . {illeg}|&| $mf\parallel cn$. Then $be:br\colon\colon cy:cr\colon\colon cg:ck$ (for ye motions of b & c are u{illeg}|n|ifor) $\colon\colon gy\left(=cy-cg\right):kr\left(=cr-ck\right)\colon\colon gs:kn$ (for $b+c:c\colon\colon ge:gf\colon\colon gy:gs\colon\colon kr:kn$) {illeg} $mf\left(=cs=cg+gs\right):cn\left(=ck+kn\right)$ . Againe $br:er\colon\colon bc:ey\colon\colon dc:fs=mc$ (for $b+c:c\colon\colon ge:gf\colon\colon ey:fs\colon\colon bc:dc$ (prop 25)). Therefore $be:br\colon\colon dm:dc\colon\colon mf:cn$ & consequently ye points d f n are in one streight line. als{illeg}|o| since $be:br\colon\colon df:dn$ ye cente{illeg}|r| of motions motion {sic} {illeg}|m|ust bee uniforme.
[107] 31 If two bodys \( b & c )/ meete & reflect one another \at/ their center of motion shall bee in ye same line \ $\left(kp\right)$ / after reflection in wch it was before it. For ye motion of {illeg}| b | towards d {illeg} \ye {center} of their motion/ is equall to ye motion of c towards d , by {illeg}|p|rop 25. then drawing {illeg} { $bk\perp kp$} & { $cm\perp bp$}. {illeg} yn $cd:bd\colon\colon cm:bk$ . therefore ye bodys b & c have equall towards ye the points k & { c }. yt is towards ye line kp . And \{illeg}/ {illeg} reflection so much as $\left(c\right)$ presseth $\left(b\right)$ from ye {illeg} after reflection yt is $gp\perp$ {illeg} {illeg} $\perp kp$ . { $\left(e\right)$ } & $\left(g\right)$ {illeg} $\left(n\right)$ $\left(p\right)$ {illeg} tis {illeg} equall {illeg} ye {illeg} <14v> must therefore be ye center of motion of ye bodys b & c when they are in ye places g & e . {illeg}|&| i{illeg}|t| is in ye line kp . The Demonstracon is same in all cases.
[108] 32 If ye bodys ( b & c ) reflect at q to e & g , & {illeg}|ye| centers of their motion describe ye line kdop . ye velocity of yt center $\left(o\right)$ after reflection shall bee equall to yt|e| center d {illeg} velocity of yt center $\left(d\right)$ before reflection. For from ye centers {illeg} & $\left(d\right)$ draw ye lines af {illeg} perpendicularly to kp . s|a|lso draw $ba\parallel fc\parallel eh\parallel rg\parallel kp$ & su{illeg}|p|pose ye line af to have ye same celerity wch (ye point d ) ye center of motion hath before reflection, soe yt when ye bodys (after reflection are in e & q , ye line af may {illeg}|b|ee in kr . Then \Also/ draw in ye ab {illeg} $\parallel kp$ $ab\parallel fc\parallel eh\parallel rg\parallel kp$ . {illeg} Then since d is ye center of motion in $\left(b\right)$ & $\left(c\right)$ , ye bodys $\left(b\right)$ & $\left(c\right)$ have equall motion towards d , but, $bd:ba\colon\colon dc:fc$ . Therefore, ye bodys $\left(b\right)$ & $\left(c\right)$ have equall motion towards a & f yt is {illeg}|t|owards ye line adf . Now when ye bodys reflect, so much as ye body b presseth ye body c from ye line af (or sf ) \or towards p / soe much ye body c presseth ye body b from ye same line, or towards k , (by {illeg}x ax: 119) theref{illeg}|o|re {illeg}|t|he bodys b , & c , have equall motion from ye line af , after reflection \(by ax 121)/ yt is when are at e & g they doe equally move from ye points h & r ; then drawing eg , tis $eh:eo\colon\colon rg:go$. Therefore ye bodys doe equally move from ye point $\left(o\right)$ {illeg} wch (by ax: 25) must bee their center of motion, & since ye motion of ye line ( af or hr ) is uniforme ({illeg}|b|y supposition) & ye point o is in ye line hr , & also in kp (by ax|prop| 31.) its motion must be uniforme.
Note yt by this, \&/ ye 31th p{illeg}|r|o{illeg}|p| I can find ye center of motion of {illeg}|t|wo bodys at any given time; & by prop: 9, or def: 5th, I can find their distance, & by prop 25, their distance from their center of motion. & consequently \that is/ ye 2 spheres in whose perimeters , they are {illeg}ly be found; There wants therefore onely their determinacon to \bee/ knowne yt the{illeg}ir places \in ye spære {illeg}/ may bee found.
[109] 33 Suppose ye body dcgk immoveable, ye surface dcg being plaine. Also let ye body shæricall {sic} body amn bee moved in ye ⊥ ch . so a{illeg}|s| to be reflected in c Then since ye side am hath as much force to w{illeg}eigh or presse towards $\left(d\right)$ as ye side an to presse towards e \by reason yt a ye center of its motion is in ye line ch / ( ) ye body{illeg} {illeg}|m|ust be in equilibrio neither pressed towards d nor e but reflected back in ye line ch . The same may be said of any line bodys whose motion center is in ye perpendicular to ye reflecting point.
[110] 34 Take $an=2bn=4cn=8dn$ . \soe yt $ad:dn\colon\colon 7:1$. Then draw ye ⊥s eb , fc , gd , hm . And/ Set a body $\left(aem\right)$ upon ye points a & m & let $\left(efk\right)$ {illeg} stand on ye points e & k wch are in ye ⊥s eb , & hkm , & fh , on ye points f , & h & lay a Globe $\left(g\right)$ let ye same be {illeg} same be {sic} supposed of hkm . Then suppose (for distinc{illeg}|t|ions sake $\left(g\right)$ ha{illeg}|v|e 8 pts of {illeg} force, wth wch it pr{illeg}esseth fh . also Then {illeg}|m|ust fh presse wth 8 pts of {illeg} force, yt is (since $fg=gh$ , or $cd=dn$ ) wth 4 pts on ye point h , & wth 4 on ye point f upon ye body efk , so yt efk must presse wth 4 {illeg}|p|ts of force \viz:/ (since it presseth equally on ye ye {sic} points e & k ) it presseth on k wth 2 pts of force & {illeg} {illeg} wth ye other 2 a{illeg}|t| e on ye body aem so ye {sic} aem presseth wth 2 pts of force, {illeg}|v|iz: wth one on ye point { n }{ m } wth ye other on ye point a ; Soe yt ye body $\left(hkm\right)$ hath 7 pts of pression upon ye point n , 4 at h , 2 at k , & one at m , \& since ye bodys p{illeg}|re|ssure of all ye point h k & m is directly towards { n } {illeg} n will be pressed by it all./ but ye Glo{illeg}|b|e causeth but one part of pression upon a . Now if these bodys fh , ek , aem , {illeg}|u|nderstood continually to diminish & come nearer to \ye line ad n / ye pressure of ye body g upon a & n will still bee ye same yt is as 1 is to 7 , & so is ye line dn to da . By ye same reason it may be generally pronounced $ad:dn\colon\colon$ pressure of ye body {illeg}| g | upon $n:$ to ye pressure of it upon g a
35 Or if ye bodys a & {illeg}| n | bee supposed united by ye line ( and ) & another by hitting on body $\left(g\right)$ moving towards ym {illeg} hit \perpendicularly/ upon ye line an at d ; as {illeg} dn to ad so is ye pressure of g upon { a } to its press{illeg}|u|re upon {illeg}| n |, so is ye motion in $\left(a\right)$ to ye motion in $\left(n\right)$ wch is generated in ym by those pressures of g , yt is wch they received from {illeg}| g |, at ye moment of reflection, & wch they would /might\ continually injoy \as in fig 6t./[111] did not their union by ye line adn hinder
By ye same r{illeg}|e|as{illeg}|o|n if g reflect {nor} twixt ye bodys, yn $ad:dn\colon\colon$ pression of $\left(n\right)$ tow{illeg}|a|rds $s:$ to ye pression of a towards $r\colon\colon$ motion of n to { $a:$ } motion of a to {illeg}| r |.
Note yt {illeg} & n are taken for ye centers of motion in a & n . & adn for {illeg}
<15r>
[112] All things put as before \That a & n are loose, then,/ $qd:pd\colon\colon$ motion of $n:$ motion of a received from g . (pr: 34) Therefore (by prop 113), $qd×a:qd×n\colon\colon$ {illeg} velocity of $a:$ velocity of n . $n×pd:a×qd\colon\colon$ velocity of $\left(a\right):$ velocity of $\left(n\right)$ . Or supposeing yt ye bodys a {illeg} & n move through ye spac distances qr & ps in ye same time yn, $n×pd:a×qd\colon\colon qr:ps$ . & producing hd to (t in) ye line rs , since $qr\parallel dt\parallel ps$ , $qp=ri:is=ps-qr\colon\colon qd=rv:\frac{ps×qd-qr×dq}{qp}$ /{illeg}|&c|\ $ps$ {illeg} $=\frac{a×qd×qr}{n×pd}$ {illeg}. Therefore & $dt=\frac{ps×qd-qr×qd+qp×qr}{qp}$. Therefore $dt=\frac{a×qr×qd×qd}{n×pd×qp}-\frac{qr×qd}{qp}+qr$ . Now {i{illeg}|f|}{if} in ye same time yt ye body g moved from $\left(h\right)$ to $\left(d\right)$ in ye same time suppose yt a & n move to r & s , & ye point d to ye point, t & ye body g to ye point e , then must $et=dh$ (by ax: 9th) Also Li|e|t {illeg}|ye| whole motion of g (at h towards d ) be called m . ye \whole/ motion of a & {sic} n to {illeg} r & s be called {illeg}| y |: yn ye motion of g after reflection $=m-y$ . & $m:m-y\colon\colon hd:de$ . or $de=\frac{hd×m-hd×y}{m}$ . & since $et=hd$ , tis $dt=\frac{2hd×m-hd×y}{m}=\frac{a×qr×qd×qd}{n×pd×pq}-\frac{qr×qd}{qp}+qr$ . And since ye whole motion of a & n to {illeg} r & s is equall to $a×qr+n×ps=$ $a×qr+\frac{a×qd×qr}{pd}=y$ {illeg} or, $\frac{pd×y}{a×pq}=qr$ . tis $dt=\frac{2hd×m-hd×y}{m}=\frac{qd×qd×y}{n×pq×pq}-\frac{qd×pd×y}{a×pq×pq}+\frac{pd×y}{a×pq}$ &c Or thus.
[113] 3{sic}6 If ye $\left\{\begin{array}{c}\text{bodys}\\ \text{Globes}\end{array}\right\}$ a & n doe rest, but soe yt ye body $\left(g\right)$ moveing \perpendicularly/ \to qp / to them & refleting on ye line qp (wch is supported by \but not fastened to/ ye bodys a & n \& ought \now/ to be conceived a line onely./) doth move them by communicating its whole motion to ym wch it selfe looseth, Tis required what motion a & n shall have receive from g . Suppose yt in ye same \so much/ time \as/ g moveth to $\left(d\right)$ before reflection in so much time it moveth from d {illeg} to e after reflection & in so much time ye bodys a & n move ye one to r ye other to s ; & yt yn ye point of reflection $\left(d\right)$ is moved to $\left(t\right)$ . Then (by ax 9th) $et=gd$ . & since bq re
flection noe motion is lost or gotten $g×gd=a×ar=n×ns=g×de$ . And (by prop 113) as ye velocit And (by prop 34) ye motion of n is to { v}
Then na{illeg}|m|e|i|ng ye given quantitys $gd=b$. $qd=c$. $qp=e$. $dp=e-c=f$. $qr=z$. ye whole motion of $\left(a\right)$ & $\left(n\right)$ to $\left(r\right)$ & $\left(s\right)$ call $xx$ , yt is $a×qr+n×ps=xx$ , or if $a:n\colon\colon pw:wq$ , (yn w is ye center of ye bodys motion {illeg}|)| ) & if $wl\parallel dt\parallel qr\parallel ps$ , then $\left(lw\right)$ is a line described by their center of motion & consequently \Then is/ {illeg}|(| ) $xx=\stackrel{—}{a+n}×\stackrel{—}{lw}$.[114] Now, as ye motion of a to ye motion of n (prop 34) so is $\left(dp\right)$ to $\left(dq\right)$ ; Therefore (ax 113) as ye velocity of $\left(a\right)$ to ye velocity of $n\colon\colon$ wf $n×pd:a×qd\colon\colon qr:ps$ . yt is $fn:ca\colon\colon z:\frac{caz}{fn}=ps$ . but $\frac{xx-az}{\left\{illeg\right\}}$ $az+n×ps=xx$;yt is $az+\frac{caz}{f}=xx$ ; or $\frac{fxx}{af+ac}=z=\frac{fxx}{ae}=qr$ . & $ps=\frac{caz}{fn}=\frac{cxx}{en}$ . $is=\frac{fxx}{ae}-\frac{cxx}{ne}=\frac{fxxx-caxx}{ane}$ . $ri:is\colon\colon ri:tv$ . therefore, $tv=\frac{cfnxx-ccaxx}{anee}$ . & $tdqr-tv=\frac{efnxx-cfnxx+ccaxx}{anee}=$ $\frac{ffnxx+ccaxx}{anee}=td$ . Also (by ax: 9th) $gd=et=b$ . Therefore $ed=\frac{ffnxx+ccaxx-banee}{anee}$ . Now since g hath soe much motion before reflection as all three bodys $\left(g+a+n\right)$ have after=ward {sic}, therefore (ax 113) $gb=xx+g×de$ ; Or, $ed=\frac{gb-xx}{g}=\frac{ffnxx+ccaxx-banee}{anee}$. That is $2abeegn=aeenxx+gffnxx+ccgaxx$. Or, $xx=\frac{2abeegn}{aeen+gffn+ccga}=\stackrel{—}{a+n}×\stackrel{—}{lw}$. Or, calling $a+n=d$ ; then $\frac{2abeegn}{adeen+gdffn+ccgad}=wl$ . Soe yt by this Equation ye point l twixt the bodys r & s being then their center of motion may bee always found. Note y{illeg}|t| ye lines qr & ps must be described by ye \centers of motion of/ 2 bodys on divers sides of ye point d yt is ar by center of motion of {illeg}|ye| body $\left(a\right)$ or $\left(ad\right)$ , & ps by y{illeg}|t| {sic} center of $\left(n\right)$ or $\left(dn\right)$
37 Also Now when a & n are united or ad & dn are united together (as in ye 2d fig) they cannot seperate ye one from ye other, & therefore since \(wn th{illeg}|e|y are not equivelox)/ rs is longer yn pq , the one cannot be at s when ye other is a{illeg}|t| r , but they will check {illeg} ye one ye others {illeg} motio soe as their centers of motion shall \not/ describe streight lines (as $\left(qr\right)$ or $\left(ps\right)$ ) but crooked ones ({illeg}|as| perhaps Trochoides as qmk , pht). yet ye comon center of their motion w or l shall retaine both ye same determinacon & velocity that it would did ye bodys move par{illeg}|a|llell to ym selves or were the{illeg}|y| n{illeg}|o|t united (by ax {7}{17}). Soe yt if ye conjoyned bodys (fig 2d) move to m & h {illeg}|i|n ye same time yt they would have moved to r & s were they th{illeg}|e|ir center of motion $\left(l\right)$ when they ar{illeg}|e| at m & h is {ye} same yt it would be we{illeg}|r| they at {illeg} & therefore may be found by ye {illeg} rule, viz; $aeedn+gdffn+gccad:2aeegn\colon\colon b:wl\colon\colon$ ye {veloci}ty of ye point l ; to ye velocity of ye body {illeg} befor{e}{illeg}{r}eflection. Vide
[1] {Se}{p}t 1664.
[2]
[3]
[4]
[5]
[6]
[7]
[8]
[9]
[10]
[11]
[12]
[13]
[14]
[15]
[16]
[17]
[18]
[19]
[20]
[21]
[22]
[23]
[24]
[25]
[26]
[27]
{illeg} $2x+\frac{1}{3}xx={gd}^{2}$. $xx+4x+1={ag}^{2}$.
[28] 64800 $\begin{array}{l}\underset{_}{36}\hfill \\ 2025\\ \underset{_}{162\phantom{0}}\\ \phantom{0}405\end{array}$
$\begin{array}{l}\underset{_}{29376}\\ \underset{_}{162}\\ \underset{_}{29376}\\ \phantom{0}\underset{_}{7776}\\ \phantom{0}\underset{_}{729\phantom{0}}\\ \phantom{00}\underset{_}{486}\end{array}$ $\underset{\text{'}}{a}____________\underset{\text{'}}{b}_____\underset{\text{'}}{c}$ $\begin{array}{l}\underset{_}{32076×25dd}=801900dd\\ 160380\\ \underset{_}{64152}\end{array}$ $\begin{array}{l}729\\ \phantom{0}7290\end{array}$ $\begin{array}{l}134\\ \phantom{0}537\\ \phantom{0}\underset{_}{486}\\ \phantom{00}5184\\ \phantom{00}486\\ \phantom{000}324\end{array}$ $\begin{array}{l}\phantom{0}1296\\ \phantom{0}648\\ 962\end{array}\right\}=\begin{array}{c}\underset{_}{104976×9}\\ 944784\end{array}$ $\begin{array}{c}\phantom{0}900\\ \phantom{0}36\\ 1800\end{array}$ $\begin{array}{l}\phantom{0}405\\ \underset{_}{162\phantom{0}}\\ \underset{_}{2025×25dd}=50625\\ 10125\\ 4050\end{array}$ $\begin{array}{l}Q:\underset{_}{324}=104976\\ \phantom{0}1296\\ \phantom{0}648\\ 972\end{array}$ $\begin{array}{l}\phantom{0}216\\ \underset{_}{106}\\ \phantom{000}6\end{array}$ 1800 $\begin{array}{cc}36& \\ & \begin{array}{r}\phantom{0}216\\ \underset{_}{106}\phantom{0}\end{array}\\ & 1276\\ & 11484\end{array}$ $\begin{array}{rr}9×25×& 36\\ 3& 24\\ 16& 20\\ \underset{_}{64}& \underset{_}{8}\hfill \\ 81& 00\\ \underset{_}{18}& \underset{_}{00}\hfill \\ 99\end{array}$
$\begin{array}{r}36×324\\ \phantom{0}1944\\ \underset{_}{\phantom{0}972\phantom{0}}\\ 11664\end{array}$ $x+\frac{r}{q}xx=9$. $\frac{r}{2}+\frac{rxq}{q}=2$ {illeg}$x=2x-\frac{rx}{2}$ . $\frac{rx}{2}+2x=9$.
$x=3$. $rx=6$ $r=2$ $rx+4x=18$ {illeg} $-\frac{qr}{2}+2q=rx$
$2q-\frac{qr}{2}+4x=18$ $\frac{qr}{2}+\frac{72}{r+4}-18=0$
[29] $cc=52c-117$. {illeg} $\begin{array}{r}c=2 6-\sqrt{156}\phantom{\left(0}\\ \underset{_}{520}\phantom{\left(0}\\ 559\left(2\\ 4\\ 1\phantom{00\left(0}\end{array}$ $cc=80c-180$
[30] $\begin{array}{c}& \begin{array}{l}1600\\ 14\\ 1420\left(3\\ \phantom{00}20\end{array}\\ \phantom{0}\\ \begin{array}{ccc}13& .& \\ 14& .& 192\\ 15& .& 225\\ 16& .& 5\\ 17& .& 289\\ 18& .& 324\\ 19& .& 361\\ 20& .& 400\end{array}\end{array}$
[31] {illeg}
[32]
${x}^{3}+{y}^{3}$<|endoftext|>
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A number of natural forests can be found on the Chagos islands, as well as a diverse range of flowering plant life.
It is probably less than 4,000 years since the islands had sufficient soil to support certain flora. Its native species consists of around 41 species of flowering plants, four ferns, and a variety of mosses, liverworts, fungi and cyanobacteria.
As many as 280 species of plants and ferns can now be found here. Some settled when spores and seeds were brought to the islands by sea and wind. Some have been left by seabirds, while humans have introduced others - either deliberately or accidentally.
Some of these are invasive, and have become a threat to the native ecosystem. While some of the more unspoiled islands boast unique pisonia forests and large clumps of the gigantic fish poison tree (Barringtonia asiatica), many of the native forests were felled to make way for coconut palms used for copra oil production.
By studying the unspoiled islands, we can benchmark our work to re-establish native plant-life on heavily-altered islands elsewhere in the archipelago and indeed globally.<|endoftext|>
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UCAT Quantitative Reasoning Test #2
0%
What is the series' next number? 13, 9, 14, 8, 15,
Correct! Wrong!
Examine the distinctions between the terms, which are listed in brackets between the terms.
13 (-4) 9 (+5) 14 (-6) 8 (+7) 15…
The result of subtracting 8 from 15 should be the next term.
15-8=7.
What is the series' next number? 45, 15, 30, 30,10, 20,
Correct! Wrong!
The series has a rule:
÷3, ×2, ×1….the cycle continues.
As a result of (30÷3=)10, and (10×2=)20, we have 20×1=20.
Clark is 15 years old, and his younger sister is the same age as him. When Clark turns 19, how old will his sister be?
Correct! Wrong!
Clark's sister is twice his age, thus she is now 15*2=30 years old. In four years, Clark will be 19, which means his sister will be (30+4)=34.
There are 6,000 toilet sinks in a plumber's home. Over the next four days, he plans to install 20% of all sinks. How many toilet sinks will he install on each of the four days if he divides his work evenly?
Correct! Wrong!
To begin, we must determine how many toilet sinks will be built over the next four days.
We can use the 10% shortcut to figure out what 20% of 6,000 is:
10% of 6,000 equals 600 (omit one 0), hence 20% equals 600*2=1,200.
Now, assuming the plumber intends to split his work evenly over four days, we simply divide 1,200 by 4.
1,200/4=300
A Bluetooth speaker will cost you at least \$120. Its price increased by 20%, then by another \$24. What is the overall price increase for this speaker in percentages?
Correct! Wrong!
The 10 percent -blocks trick can be used to find 20% of 120. 10%=12. As a result, (12×2)=24.
We get 120+24+24=\$168 by adding \$24.
We must subtract 120 from 168 to arrive at 48. This is the entire amount of money that has increased in value as a result of the price increase. We need to calculate how much 48 out of 120 is. We may gradually add units of 12 until we reach 48 because we already know what 10% of 120 is (12). Fortunately, we need 4 units of 12, which is 4 units of 10% = 40%.
Another option is to write a fraction (48/120) and then reduce or divide it by common divisors.
3/4 cup sugar is required for an 18-cookie recipe. To make two dozen cookies with the same recipe, how much sugar would be required?
Correct! Wrong!
3/4 cup sugar is needed for 18 cookies.
We raised the amount of cookies by a factor of (24/18) when we made two dozen cookies. As a result, we need to multiply 3/4 by 24/18 to get the new sugar amount:
3/4×24/18=3×24/4×18= 1
Select the expression that best indicates the highest value:
Correct! Wrong!
With a solid understanding of the 1-20 multiplication chart and fraction-to-decimal conversions, this question can be answered rapidly.
14/49 = divide by 7 to reach 2/7 ~ 0.14×2 ~0.28
15/50 = divide by 5 and get 3/10 = 0.3
17/51= divide by 17 and get 1/3 = 0.33
16/64 = divide by 16 and get = 1/4 = 0.25
Every year, a city hosts a comedy festival. In Year 1, the event lasted five weeks and a total of 123,545 tickets were sold. In Year 2, the event lasted for eight weeks, with average weekly ticket sales up by 25%. How many tickets were sold in year 2?
Correct! Wrong!
Calculate the average weekly sales for the first year.
123,545 in 5 weeks
123,545/5 = 24709 a week
Calculate the sum based on the average weekly sales in Year 2:
Old Value x Multiplier = New Value
24,709 x 1.25 = 30,886.25
The festival lasted eight weeks, so:
30,886.25 x 8 = 247,090 tickets
For a limited time, Learning101 is giving a discount on their online Spanish course. They are giving you a 40% discount. You will receive an additional 10% discount on the new pricing if you pay in advance. In terms of a proportion of the original price, what is the cost of paying up front today?
Correct! Wrong!
Locate the first discount.
The new price is 0.6 times the full amount due to the 40% discount.
Look for a second discount.
Because there is a 10% discount, the price for paying in advance is 0.9 times the discounted price.
0.9 x 0.6 = 0.54 – 54%.
Town A's domestic fowl population is 20% that of Town B's domestic fowl population. Town C has 75% fowl population of Town A. What is the domestic fowl population in Town B if there are 252 domestic fowl in Town C?
Correct! Wrong!
Using algebra, connect the population sizes:
Town A's population is referred to as: a
Town B's population is referred to as b.
Town A has a population that is 20 percent that of Town B:
a = 0.2b
Town C's population is referred to as c.
Town C has a population that is 75% that of Town A.
c = 0.75a
To find the value for B, use the value for C.
c = 252
so 0.75a = 252
a = 252/0.75
a = 336
0.2b = a
0.2b = 336
b = 1680
In 2017-18, the San Francisco Rams spent \$285 million on "playing staff wages." The money they spent on "coaching staff wages" accounted for 15% of the total. These two numbers amounted for 75% of all employee wages. To the nearest quarter million, what was the overall staff wage bill?
Correct! Wrong!
Calculate the total wage for both the players and the staff.
\$285,000,000 x 1.15 = \$327.75 million.
Calculate the overall cost of the staff bill.
We need to calculate 100 / 75 of \$327.75 million, which is 75 percent of the entire wage bill.
327.75 million x 100/75 = \$437 million<|endoftext|>
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# Math Tip of the Week – Graphing Linear Inequalities
First of all, my heart goes out to the people of Moore, Oklahoma. I grew up in Tulsa, my daughter went to OU, and I’ve driven through Moore many times.
Let’s talk a little bit about solving and graphing Linear Inequalities. Here is an excerpt from the Linear Inequalities Section in Beginning Algebra:
When solving linear inequalities, you can solve as if you have an equality. You must have the variable on the left hand side when you use these rules to know where to graph:
• When you have the less than sign (<), you want to graph to the left (both start with “le”); when you have the greater than sign (>) you graph to the right.
• You can also graph in the direction where the inequality sign is pointed.
• You can also plug in a number to see if it works: if it works, it should be on the colored part of the number line – like 3 is for the inequality “x < 4” (since 3 is less than 4).
Here are some examples, including the solutions in Interval Notation. Remember that with Interval Notation, you always go from lowest to highest number with “(“ (soft brackets) if the inequality doesn’t hit the point, and “[“ (hard brackets) if the inequality does hit the line. If you have to skip over any numbers, you do so by using the “U” sign, which means union, or put the things together.
<
<
Note that when you solve an inequality, you pretend like the inequality is an equal sign. The only thing you have to worry about is multiplying or dividing by a negative number; when you do this, you have to reverse the inequality symbol (change < to >, or > to <). You can see an example of this in the last equation below. The reason we need to do this, is when we have a negative coefficient(what comes before the variable), we’d eventually have to move the variable with its coefficient to the other side and make it positive, so it would be on the other end of the inequality sign.
At this point, we are not multiplying or dividing by variables (since we don’t know the signs of them); we will learn how to do that later.
Hope this helps, and Happy Mathing!
Lisa
<
<<|endoftext|>
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## Saturday, February 11, 2012
### Odd and Even
Day 1: Object pairs
Prep: Give each student a copy of the Object Pairs worksheet, a pencil, and a small bowl of beans. (Beans can be shared with a neighbor.)
Lesson: Teach students that odd numbers end with 1, 3, 5, 7, 9 and even numbers end with 0, 2, 4, 6, 8. Emphasize that it's only the number in the ones place that matters when deciding whether a number is even or odd. Review tally marks.
Teach students that if a number is even, it can be divided into pairs. If it's odd, dividing into pairs will leave one loner left over. Ask students to grab a small handful of beans and determine whether it's even or odd by this method, then count the number of beans to see if they were correct according to the rules. (Demonstrate before turning them loose.) Have students do this several times and keep a tally of how many even versus odd numbers they come up with.
Have students leave the table (and the beans!) and sit in a circle on the floor. Teach students that there are sometimes patterns to the even and odd numbers around us. Bring books for each student and ask them to open to a random page. Go around the circle and ask if the even page number is on the right side or the left side. Which side is the odd number on? Is it the same for all books? (Yes; books always start with page one on the right side, so all the odd numbered pages end up on the right.)
Ask students what the number is on their house. Is it even or odd? Ask them to go home and look at all the house numbers on their street and which side of the street they're on. Is there a pattern? Hand out odd and even reminder cards for students to take home.
Day 2: Even Steven and Odd Todd
Prep: Give each student a copy of the Even and Odd/200 chart page.
Lesson: Follow up on house number homework suggestion. Did they notice a pattern with house numbers? (Even numbers are on one side of the street and odd numbers are on the other.)
Review that odd numbers end with 1, 3, 5, 7, 9 and even numbers end with 0, 2, 4, 6, 8. Review tally marks.
Read Even Steven and Odd Todd (or fun facts from a book like National Geographic Kids: Weird but True). Ask students to put a tally mark in the even or odd column of their page whenever they hear an even or odd number. Which were there more of in this book?
Tell students that there is one number that isn't even or odd. Ask them to fill out the hundreds chart using ONLY EVEN NUMBERS. Students should start with 2 and reach 200 by the end. Then instruct them to color in the numbers written at the bottom of the page to find out the only number that's neither even nor odd. (The answer is ∞ , the symbol for infinity. We actually talked about this one day for reasons I can't remember, so it might be familiar to some of them.)<|endoftext|>
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# 2.6: Scientific Notation
Difficulty Level: At Grade Created by: CK-12
## Lesson Objectives
The student will:
• use scientific notation to express large and small numbers.
• add, subtract, multiply, and divide using scientific notation.
## Vocabulary
• scientific notation
## Introduction
Work in science frequently involves very large and very small numbers. The speed of light, for example, is 300,000,000 m/s; the mass of the earth is 6,000,000,000,000,000,000,000,000 kg; and the mass of an electron is 0.0000000000000000000000000000009 kg. It is very inconvenient to write out such numbers and even more inconvenient to attempt to carry out mathematical operations with them. Scientists and mathematicians have designed an easier method to deal with such long numbers. This more convenient system is called exponential notation by mathematicians and scientific notation by scientists.
## What is Scientific Notation?
In scientific notation, very large and very small numbers are expressed as the product of a number between \begin{align*}1\end{align*} and \begin{align*}10\end{align*} multiplied by some power of \begin{align*}10\end{align*}. For example, the number \begin{align*}9,000,000\end{align*} can be written as the product of \begin{align*}9\end{align*} times \begin{align*}1,000,000\end{align*}. In turn, \begin{align*}1,000,000\end{align*} can be written as \begin{align*}10^6\end{align*}. Therefore, \begin{align*}9,000,000\end{align*} can be written as \begin{align*}9 \times 10^6\end{align*}. In a similar manner, \begin{align*}0.00000004\end{align*} can be written as \begin{align*}4\end{align*} times \begin{align*} \frac {1} {10^8}\end{align*}, or \begin{align*}4 \times 10^{-8}\end{align*}.
Examples of Scientific Notation
Decimal Notation Scientific Notation
\begin{align*}95,672\end{align*} \begin{align*}9.5672 \times 10^4\end{align*}
\begin{align*}8,340\end{align*} \begin{align*}8.34 \times 10^3\end{align*}
\begin{align*}100\end{align*} \begin{align*}1 \times 10^2\end{align*}
\begin{align*}7.21\end{align*} \begin{align*}7.21 \times 10^0\end{align*}
\begin{align*}0.014\end{align*} \begin{align*}1.4 \times 10^{-2}\end{align*}
\begin{align*}0.0000000080\end{align*} \begin{align*}8.0 \times 10^{-9}\end{align*}
\begin{align*}0.00000000000975\end{align*} \begin{align*}9.75 \times 10^{-12}\end{align*}
As you can see from the examples in Table above, to convert a number from decimal form into scientific notation, you count the number of spaces needed to move the decimal, and that number becomes the exponent of 10. If you are moving the decimal to the left, the exponent is positive, and if you are moving the decimal to the right, the exponent is negative. You should note that all significant figures are maintained in scientific notation. You will probably realize that the greatest advantage of using scientific notation occurs when there are many non-significant figures.
## Scientific Notation in Calculations
When numbers in exponential form are added or subtracted, the exponents must be the same. If the exponents are the same, the coefficients are added and the exponent remains the same.
Example:
\begin{align*}(4.3 \times 10^4) + (1.5 \times 10^4) = (4.3 + 1.5) \times 10^4 = 5.8 \times 10^4\end{align*}
Note that the example above is the same as:
\begin{align*}43,000 + 15,000 = 58,000 = 5.8 \times 10^4\end{align*}.
Example:
\begin{align*}(8.6 \times 10^7) - (5.3 \times 10^7) = (8.6 - 5.3) \times 10^7 = 3.3 \times 10^7\end{align*}
Example:
\begin{align*}(8.6 \times 10^5) + (3.0 \times 10^4) = \text{?}\end{align*}
These two exponential numbers do not have the same exponent. If the exponents of the numbers to be added or subtracted are not the same, then one of the numbers must be changed so that the two numbers have the same exponent. In order to add them, we can change the number \begin{align*}3.0 \times 10^4\end{align*} to \begin{align*}0.30 \times 10^5\end{align*}. This change is made by moving the decimal one place to the left and increasing the exponent by one. Now the two numbers can be added.
\begin{align*}(8.6 \times 10^5) + (0.30 \times 10^5) = (8.6 + 0.30) \times 10^5 = 8.9 \times 10^5\end{align*}
We could also have chosen to alter the other number. Instead of changing the second number to a higher exponent, we could have changed the first number to a lower exponent.
\begin{align*}8.6 \times 10^5\end{align*} becomes \begin{align*}86 \times 10^4\end{align*}
\begin{align*}(86 \times 10^4) + (3.0 \times 10^4) = (86 + 3.0) \times 10^4 = 89 \times 10^4\end{align*}
Even though it is not always necessary, the preferred practice is to express exponential numbers in proper form, which has only one digit to the left of the decimal. When \begin{align*}89 \times 10^4\end{align*} is converted to proper form, it becomes \begin{align*}8.9 \times 10^5\end{align*}, which is precisely the same result as before.
### Multiplication and Division
When multiplying or dividing numbers in exponential form, the numbers do not have to have the same exponents. To multiply exponential numbers, multiply the coefficients and add the exponents. To divide exponential numbers, divide the coefficients and subtract the exponents.
Multiplication Examples:
\begin{align*}(4.2 \times 10^4) \cdot (2.2 \times 10^2) = (4.2 \cdot 2.2) \times 10^{4+2} = 9.2 \times 10^6\end{align*}
The product of \begin{align*}4.2\end{align*} and \begin{align*}2.2\end{align*} is \begin{align*}9.24\end{align*}, but since we are limited to two significant figures, the coefficient is rounded to \begin{align*}9.2\end{align*}.
\begin{align*}(2 \times 10^9) \cdot (4 \times 10^{14}) = (2 \cdot 4) \times 10^{9+14} = 8 \times 10^{23}\end{align*}
\begin{align*}(2 \times 10^{-9}) \cdot (4 \times 10^4) = (2 \cdot 4) \times 10^{-9+4} = 8 \times 10^{-5}\end{align*}
\begin{align*}(2 \times 10^{-5}) \cdot (4 \times 10^{-4}) = (2 \cdot 4) \times 10^{(-5)+(-4)} = 8 \times 10^{-9}\end{align*}
\begin{align*}(8.2 \times 10^{-9}) \cdot (8.2 \times 10^{-4}) = (8.2 \cdot 8.2) \times 10^{(-9)+(-4)} = 67.24 \times 10^{-13}\end{align*}
In this last example, the product has too many significant figures and is not in proper exponential form. We must round to two significant figures and adjust the decimal and exponent. The correct answer would be \begin{align*}6.7 \times 10^{-12}\end{align*}.
Division Examples:
\begin{align*} \frac {8 \times 10^7} {2 \times 10^4} = 4 \times 10^{7-4} = 4 \times 10^3\end{align*}
\begin{align*} \frac {8 \times 10^{-7}} {2 \times 10^{-4}} = 4 \times 10^{(-7)-(-4)} = 4 \times 10^{-3}\end{align*}
\begin{align*} \frac {4.6 \times 10^3} {2.3 \times 10^{-4}} = 2.0 \times 10^{(3)-(-4)} = 2.0 \times 10^7\end{align*}
In the example above, since the original coefficients have two significant figures, the answer must also have two significant figures. Therefore, the zero in the tenths place is written to indicate the answer has two significant figures.
## Lesson Summary
• Very large and very small numbers in science are expressed in scientific notation.
• All significant figures are maintained in scientific notation.
• When numbers in exponential form are added or subtracted, the exponents must be the same. If the exponents are the same, the coefficients are added and the exponent remains the same.
• To multiply exponential numbers, multiply the coefficients and add the exponents.
• To divide exponential numbers, divide the coefficients and subtract the exponents.
## Review Questions
1. Write the following numbers in scientific notation.
1. \begin{align*}0.0000479\end{align*}
2. \begin{align*}251,000,000\end{align*}
3. \begin{align*}4,260\end{align*}
4. \begin{align*}0.00206\end{align*}
Do the following calculations without a calculator.
1. \begin{align*}(2.0 \times 10^3) \cdot (3.0 \times 10^4) \end{align*}
2. \begin{align*}(5.0 \times 10^{-5}) \cdot (5.0 \times 10^8) \end{align*}
3. \begin{align*}(6.0 \times 10^{-1}) \cdot (7.0 \times 10^{-4}) \end{align*}
4. \begin{align*}\frac {(3.0 \times 10^{-4}) \cdot (2.0 \times 10^{-4})} {2.0 \times 10^{-6}}\end{align*}
Do the following calculations.
1. \begin{align*}(6.0 \times 10^7) \cdot (2.5 \times 10^4)\end{align*}
2. \begin{align*} \frac {4.2 \times 10^{-4}} {3.0 \times 10^{-2}}\end{align*}
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# Lesson 4
Decompose Even and Odd Numbers
## Warm-up: Number Talk: Equal Addends (10 minutes)
### Narrative
The purpose of this Number Talk is to elicit strategies and understandings students have for finding sums when both addends are the same and sums when one addend is one less or one more than the other. These understandings help students develop fluency and will be helpful later in this lesson when students will need to be able to decompose numbers into two equal addends or two addends that are as close as possible as they reason about even and odd numbers.
### Launch
• Display one expression.
• “Give me a signal when you have an answer and can explain how you got it.”
• 1 minute: quiet think time
### Activity
• Keep expressions and work displayed.
• Repeat with each expression.
### Student Facing
Find the value of each expression mentally.
• $$6 + 6$$
• $$7 + 7$$
• $$7 + 8$$
• $$8 + 9$$
### Activity Synthesis
• “How are the expressions the same? How are they different?”
## Activity 1: Share in Different Ways (20 minutes)
### Narrative
The purpose of this activity is for students to recognize that even numbers can be represented as the sum of two equal addends. The activity is designed to elicit student curiosity about which types of decompositions are possible and which are not. Students may notice many patterns in the ways even and odd numbers can be decomposed which will be useful in future lessons. However, the synthesis should be focused on representing even numbers as sums of equal addends.
MLR5 Co-Craft Questions. Keep books or devices closed. Display only the problem stem sentences, without revealing the questions, and ask students to write down possible mathematical questions that could be asked about the situation. Invite students to compare their questions before revealing the task. Ask, “What do these questions have in common? How are they different?” Reveal the intended questions for this task and invite additional connections.
Representation: Internalize Comprehension. Provide students with a graphic organizer, such as a sorting mat that has images of gift bags or simply two large circles, to physically share the “cookies” (images of cookies cut out or counters, chips, etc.). Use this to give a concrete example that supports the context of the problems.
Supports accessibility for: Organization, Visual-Spatial Processing, Conceptual Processing
### Required Materials
Materials to Gather
• Groups of 2
### Activity
• “Figure out different ways the students could share their cookies.”
• “Show your thinking for each way. Use equations to show the groups.”
• “Some ways may not be possible. Be ready to show and explain why.”
• 5 minutes: independent work time
• “Share your thinking with your partner. How are your equations the same? How are they different?”
• 10 minute: partner discussion
### Student Facing
1. Kiran baked 12 cookies. He wants to put them in two gift bags. Show a few different ways he can share the cookies.
1. Can both bags have the same amount of cookies?
$$12 = \underline{\hspace{1 cm}} + \underline{\hspace{1 cm}}$$
2. Can both bags have an even number of cookies?
$$12 = \underline{\hspace{1 cm}} + \underline{\hspace{1 cm}}$$
3. Can both bags have an odd number of cookies?
$$12 = \underline{\hspace{1 cm}} + \underline{\hspace{1 cm}}$$
4. Can one bag have an even number of cookies and the other have an odd number of cookies?
$$12 = \underline{\hspace{1 cm}} + \underline{\hspace{1 cm}}$$
2. Lin baked 14 cookies. She wants to put them in two gift bags. Show a few different ways she can share the cookies.
1. Can both bags have the same amount of cookies?
$$14 = \underline{\hspace{1 cm}} + \underline{\hspace{1 cm}}$$
2. Can both bags have an even number of cookies?
$$14 = \underline{\hspace{1 cm}} + \underline{\hspace{1 cm}}$$
3. Can both bags have an odd number of cookies?
$$14 = \underline{\hspace{1 cm}} + \underline{\hspace{1 cm}}$$
4. Can one bag have an even number of cookies and the other have an odd number of cookies?
$$14 = \underline{\hspace{1 cm}} + \underline{\hspace{1 cm}}$$
3. Noah baked 15 cookies. He wants to put them in two gift bags. Show a few different ways he can share the cookies.
1. Can both bags have the same amount of cookies?
$$15 = \underline{\hspace{1 cm}} + \underline{\hspace{1 cm}}$$
2. Can both bags have an even number of cookies?
$$15 = \underline{\hspace{1 cm}} + \underline{\hspace{1 cm}}$$
3. Can both bags have an odd number of cookies?
$$15 = \underline{\hspace{1 cm}} + \underline{\hspace{1 cm}}$$
4. Can one bag have an even number of cookies and the other have an odd number of cookies?
$$15 = \underline{\hspace{1 cm}} + \underline{\hspace{1 cm}}$$
### Activity Synthesis
• “What did you notice about the possible ways the students could share their cookies?” (Kiran and Lin could share their cookies in the most ways. Noah could only share with one bag of even and one bag of odd.)
• Display:
• $$12 = 6 + 6$$
• $$14 = 7 + 7$$
• $$15 = 7 + 8$$
• “How do these equations represent the student’s cookies?”
• “Which students baked an even number of cookies? Use the equations to explain how you know.” (Kiran and Lin because the equations show you can split their cookies into two equal groups.)
## Activity 2: Represent Numbers with Two Addends (15 minutes)
### Narrative
The purpose of this activity is for students to represent even numbers as a sum of two equal addends. They sort all numbers between 0 and 20 into even and odd and notice that all even numbers can be represented as sums of two equal addends while odd numbers cannot (MP8). Students may also use the sorting activity to understand and explain why 0 is an even number.
### Required Materials
Materials to Gather
• Groups of 2
### Activity
• “Let’s try to decompose more numbers into two equal addends.”
• “Take turns picking a number between 0 and 20. Decide together whether the number is even or odd and record it in the table.”
• “Then try to decompose the number into two equal addends. If you can, record it on the table.”
• “If you cannot find a way to decompose the number into two equal addends, find two addends that are as close together as possible that make your number.”
• Demonstrate with Kiran (12) and Noah’s (15) cookies as needed.
• “Keep going until you have sorted all the numbers from 0 to 20.”
• 10 minutes: partner work time
### Student Facing
1. Pick a number between 0 and 20.
2. Decide with your partner whether the number is even or odd.
3. Complete the equation to show your number as the sum of two equal addends. If you cannot use two equal addends, use two addends that are as close as possible.
even
odd
$$\underline{\hspace{0.9 cm}} = \underline{\hspace{0.9 cm}} + \underline{\hspace{0.9 cm}}$$
$$\underline{\hspace{0.9 cm}} = \underline{\hspace{0.9 cm}} + \underline{\hspace{0.9 cm}}$$
$$\underline{\hspace{0.9 cm}} = \underline{\hspace{0.9 cm}} + \underline{\hspace{0.9 cm}}$$
$$\underline{\hspace{0.9 cm}} = \underline{\hspace{0.9 cm}} + \underline{\hspace{0.9 cm}}$$
$$\underline{\hspace{0.9 cm}} = \underline{\hspace{0.9 cm}} + \underline{\hspace{0.9 cm}}$$
$$\underline{\hspace{0.9 cm}} = \underline{\hspace{0.9 cm}} + \underline{\hspace{0.9 cm}}$$
$$\underline{\hspace{0.9 cm}} = \underline{\hspace{0.9 cm}} + \underline{\hspace{0.9 cm}}$$
$$\underline{\hspace{0.9 cm}} = \underline{\hspace{0.9 cm}} + \underline{\hspace{0.9 cm}}$$
$$\underline{\hspace{0.9 cm}} = \underline{\hspace{0.9 cm}} + \underline{\hspace{0.9 cm}}$$
$$\underline{\hspace{0.9 cm}} = \underline{\hspace{0.9 cm}} + \underline{\hspace{0.9 cm}}$$
$$\underline{\hspace{0.9 cm}} = \underline{\hspace{0.9 cm}} + \underline{\hspace{0.9 cm}}$$
$$\underline{\hspace{0.9 cm}} = \underline{\hspace{0.9 cm}} + \underline{\hspace{0.9 cm}}$$
$$\underline{\hspace{0.9 cm}} = \underline{\hspace{0.9 cm}} + \underline{\hspace{0.9 cm}}$$
$$\underline{\hspace{0.9 cm}} = \underline{\hspace{0.9 cm}} + \underline{\hspace{0.9 cm}}$$
$$\underline{\hspace{0.9 cm}} = \underline{\hspace{0.9 cm}} + \underline{\hspace{0.9 cm}}$$
$$\underline{\hspace{0.9 cm}} = \underline{\hspace{0.9 cm}} + \underline{\hspace{0.9 cm}}$$
$$\underline{\hspace{0.9 cm}} = \underline{\hspace{0.9 cm}} + \underline{\hspace{0.9 cm}}$$
$$\underline{\hspace{0.9 cm}} = \underline{\hspace{0.9 cm}} + \underline{\hspace{0.9 cm}}$$
$$\underline{\hspace{0.9 cm}} = \underline{\hspace{0.9 cm}} + \underline{\hspace{0.9 cm}}$$
$$\underline{\hspace{0.9 cm}} = \underline{\hspace{0.9 cm}} + \underline{\hspace{0.9 cm}}$$
### Student Response
If students place an odd number in the even group or an even number in the odd group, consider asking:
• “How could you use counters or a drawing to show if this number is odd or even?”
### Activity Synthesis
• Display the completed table.
• “What do you notice about even and odd numbers?” (All the even numbers have the same addends. All the even numbers are “doubles.” The odd numbers have one addend that is one more than the other.)
• “Explain why even numbers can be decomposed into two equal addends.”
## Lesson Synthesis
### Lesson Synthesis
Draw or display:
“Is there an even or odd number of dots? Explain.” (Even. I see two equal groups of 4. I see 4 pairs and no dots left over.)
“What is an equation that would show that the number of dots is even?” ($$4 + 4 = 8$$, $$8 = 4 + 4$$)
## Student Section Summary
### Student Facing
In this section, we learned that groups of objects have either an even or odd number of members. We learned that an even number of objects can be split into 2 equal groups or into groups of 2 with no objects left over. We learned that an odd number of objects always has one object left over when you make 2 equal groups or groups of 2. We also learned that even numbers can be represented as an equation with 2 equal addends.
Odd
$$3 + 3 + 1 = 7$$
Even
$$4 + 4 = 8$$<|endoftext|>
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Condenser (heat transfer)
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In systems involving heat transfer, a condenser is a device or unit used to condense a substance from its gaseous to its liquid state, by cooling it. In so doing, the latent heat is given up by the substance and transferred to the surrounding environment. Condensers can be made according to numerous designs, and come in many sizes ranging from rather small (hand-held) to very large (industrial-scale units used in plant processes). For example, a refrigerator uses a condenser to get rid of heat extracted from the interior of the unit to the outside air. Condensers are used in air conditioning, industrial chemical processes such as distillation, steam power plants and other heat-exchange systems. Use of cooling water or surrounding air as the coolant is common in many condensers.
Examples of condensersEdit
- A surface condenser is one in which condensing medium and vapors are physically separated and used when direct contact is not desired. It is a shell and tube heat exchanger installed at the outlet of every steam turbine in thermal power stations. Commonly, the cooling water flows through the tube side and the steam enters the shell side where the condensation occurs on the outside of the heat transfer tubes. The condensate drips down and collects at the bottom, often in a built-in pan called a hotwell. The shell side often operates at a vacuum or partial vacuum, produced by the difference in specific volume between the steam and condensate. Conversely, the vapor can be fed through the tubes with the coolant water or air flowing around the outside.
- In chemistry, a condenser is the apparatus which cools hot vapors, causing them to condense into a liquid. See "Condenser (laboratory)" for laboratory-scale condensers, as opposed to industrial-scale condensers. Examples include the Liebig condenser, Graham condenser, and Allihn condenser. This is not to be confused with a condensation reaction which links two fragments into a single molecule by an addition reaction and an elimination reaction.
- In laboratory distillation, reflux, and rotary evaporators, several types of condensers are commonly used. The Liebig condenser is simply a straight tube within a cooling water jacket, and is the simplest (and relatively least expensive) form of condenser. The Graham condenser is a spiral tube within a water jacket, and the Allihn condenser has a series of large and small constrictions on the inside tube, each increasing the surface area upon which the vapor constituents may condense. Being more complex shapes to manufacture, these latter types are also more expensive to purchase. These three types of condensers are laboratory glassware items since they are typically made of glass. Commercially available condensers usually are fitted with ground glass joints and come in standard lengths of 100, 200, and 400 mm. Air-cooled condensers are unjacketed, while water-cooled condensers contain a jacket for the water.
- Larger condensers are also used in industrial-scale distillation processes to cool distilled vapor into liquid distillate. Commonly, the coolant flows through the tube side and distilled vapor through the shell side with distillate collecting at or flowing out the bottom.
- A condenser unit used in central air conditioning systems typically has a heat exchanger section to cool down and condense incoming refrigerant vapor into liquid, a compressor to raise the pressure of the refrigerant and move it along, and a fan for blowing outside air through the heat exchanger section to cool the refrigerant inside. A typical configuration of such a condenser unit is as follows: The heat exchanger section wraps around the sides of the unit with the compressor inside. In this heat exchanger section, the refrigerant goes through multiple tube passes, which are surrounded by heat transfer fins through which cooling air can circulate from outside to inside the unit. There is a motorized fan inside the condenser unit near the top, which is covered by some grating to keep any objects from accidentally falling inside on the fan. The fan is used to pull outside cooling air in through the heat exchanger section at the sides and blow it out the top through the grating. These condenser units are located on the outside of the building they are trying to cool, with tubing between the unit and building, one for vapor refrigerant entering and another for liquid refrigerant leaving the unit. Of course, an electric power supply is needed for the compressor and fan inside the unit.
- In a direct-contact condenser, hot vapor and cool liquid are introduced into a vessel and allowed to mix directly, rather than being separated by a barrier such as the wall of a heat exchanger tube. The vapor gives up its latent heat and condenses to a liquid, while the liquid absorbs this heat and undergoes a temperature rise. The entering vapor and liquid typically contain a single condensable substance, such as a water spray being used to cool air and adjust its humidity.
Other Types of Condensers
There are three other condensers used in HVAC systems:
- Air cooled – If the condenser is located on the outside of the unit, the air cooled condenser can provide the easiest arrangement. These types of condensers eject heat to the outdoors and are simple to install.
Most common uses for this condenser are domestic refrigerators, upright freezers and in residential packaged air conditioning units. A great feature of the air cooled condenser is they are very easy to clean. Since dirt can cause serious issues with the condensers performance, it is highly recommended that these be kept clear of dirt.
- Water cooled – Although a little more pricey to install, these condensers are the more efficient type. Commonly used for swimming pools and condensers piped for city water flow, these condensers require regular service and maintenance.
They also require a cooling tower to conserve water. To prevent corrosion and the forming of algae, water cooled condensers require a constant supply of makeup water along with water treatment.
Depending on the application you can choose from tube in tube, shell and coil or shell and tube condensers. All are essentially made to produce the same outcome, but each in a different way.
- Evaporative – While these remain the least popular choice, they are used when either water supply is inadequate to operate water cooled condenser or condensation temperature is lower that can achieved by air cooled condenser. Evaporative condensers can be used inside or outside of a building and under typical conditions, operate at a low condensing temperature.
Typically these are used in large commercial air-conditioning units. Although effective, they are not necessarily the most efficient.
For an ideal single-pass condenser whose coolant has constant density, constant heat capacity, linear enthalpy over the temperature range, perfect cross-sectional heat transfer, and zero longitudinal heat transfer, and whose tubing has constant perimeter, constant thickness, and constant heat conductivity, and whose condensible fluid is perfectly mixed and at constant temperature, the coolant temperature varies along its tube according to:
- x is the distance from the coolant inlet;
- T(x) is the coolant temperature, and T(0) the coolant temperature at its inlet;
- TH is the hot fluid's temperature;
- NTU is the number of transfer units;
- m is the coolant's mass (or other) flow rate;
- c is the coolant's heat capacity at constant pressure per unit mass (or other);
- h is the heat transfer coefficient of the coolant tube;
- P is the perimeter of the coolant tube;
- G is the heat conductance of the coolant tube (often denoted UA);
- L is the length of the coolant tube.<|endoftext|>
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 1.3: Patterns and Equations
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Write an equation.
• Use a verbal model to write an equation.
• Solve problems using equations.
## Introduction
In mathematics, and especially in algebra, we look for patterns in the numbers that we see. The tools of algebra assist us in describing these patterns with words and with equations (formulas or functions). An equation is a mathematical recipe that gives the value of one variable in terms of the other.
For example, if a theme park charges $12 admission, then the number of people who enter the park every day and the amount of money taken by the ticket office are related mathematically. We can write a rule to find the amount of money taken by the ticket office. In words, we might say “The money taken in dollars is (equals) twelve times the number of people who enter the park.” We could also make a table. The following table relates the number of people who visit the park and the total money taken by the ticket office. Number of visitorsMoney taken($)112224336448560672784\begin{align*}& \text{Number of visitors} & & 1 & & 2 & & 3 & & 4 & & 5 & & 6 & & 7 \\ & \text{Money taken} (\) & & 12 & & 24 & & 36 & & 48 & & 60 & & 72 & & 84 \end{align*}
Clearly, we will need a big table if we are going to be able to cope with a busy day in the middle of a school vacation!
A third way we might relate the two quantities (visitors and money) is with a graph. If we plot the money taken on the vertical axis and the number of visitors on the horizontal axis, then we would have a graph that looks like the one shown as follows. Note that this graph shows a smooth line for non-whole number values of x\begin{align*} x \end{align*} (e.g., x=2.5\begin{align*} x=2.5\end{align*}). But, in real life this would not be possible because you cannot have half a person enter the park. This is an issue of domain and range, something we will talk about in the following text.
The method we will examine in detail in this lesson is closer to the first way we chose to describe the relationship. In words we said that “The money taken in dollars is twelve times the number of people who enter the park.” In mathematical terms we can describe this sort of relationship with variables. A variable is a letter used to represent an unknown quantity. We can see the beginning of a mathematical formula in the words.
The money taken in dollars is twelve times the number of people who enter the park.
This can be translated to:
the money taken in dollars =12 ×\begin{align*}= 12 \ \times \end{align*} (the number of people who enter the park)
To make the quantities more visible they have been placed in parentheses. We can now see which quantities can be assigned to letters. First we must state which letters (or variables) relate to which quantities. We call this defining the variables:
Let x=\begin{align*}x =\end{align*} the number of people who enter the theme park.
Let y=\begin{align*}y =\end{align*} the total amount of money taken at the ticket office.
We can now show the fourth way to describe the relationship, with our algebraic equation.
y=12x\begin{align*} y=12x \end{align*}
Writing a mathematical equation using variables is very convenient. You can perform all of the operations necessary to solve this problem without having to write out the known and unknown quantities in long hand over and over again. At the end of the problem, we just need to remember which quantities x\begin{align*} x \end{align*} and y\begin{align*} y \end{align*} represent.
## Write an Equation
An equation is a term used to describe a collection of numbers and variables related through mathematical operators. An algebraic equation will contain letters that relate to real quantities or to numbers that represent values for real quantities. If, for example, we wanted to use the algebraic equation in the example above to find the money taken for a certain number of visitors, we would substitute that value in for x\begin{align*} x \end{align*} and then solve the resulting equation for y\begin{align*} y \end{align*}.
Example 1
A theme park charges 12 entry to visitors. Find the money taken if 1296 people visit the park. Let’s break the solution to this problem down into a number of steps. This will help us solve all the problems in this lesson. Step 1 Extract the important information. (money taken in dollars)(number of visitors)=12×(number of visitors)=1296\begin{align*}(\text{money taken in dollars}) & = 12 \times (\text{number of visitors})\\ (\text{number of visitors}) & = 1296\end{align*} Step 2 Translate into a mathematical equation. We do this by defining variables and by substituting in known values. Let yy=(money taken in dollars)=12×1296THIS IS OUR EQUATION.\begin{align*}\text{Let} \ y & = (\text{money taken in dollars})\\ y & = 12 \times 1296 & & \text{THIS IS OUR EQUATION}.\end{align*} Step 3 Solve the equation. y=15552Answer: The money taken is15552\begin{align*} y=15552 & & \text{Answer: The money taken is} \ \15552 \end{align*}
Step 4 Check the result.
If $15552 is taken at the ticket office and tickets are$12, then we can divide the total amount of money collected by the price per individual ticket.
(number of people)=1555212=1296\begin{align*}\text{(number of people)} = \frac{15552} {12} = 1296\end{align*}
Our answer equals the number of people who entered the park. Therefore, the answer checks out.
Example 2
The following table shows the relationship between two quantities. First, write an equation that describes the relationship. Then, find out the value of b\begin{align*} b \end{align*} when a\begin{align*} a \end{align*} is 750.
a:b:0201040206030804010050120\begin{align*}& a: & & 0 & & 10 & & 20 & & 30 & & 40 & & 50 \\ & b: & & 20 & & 40 & & 60 & & 80 & & 100 & & 120 \end{align*}
Step 1 Extract the important information. We can see from the table that every time a\begin{align*} a \end{align*} increases by 10, b\begin{align*} b \end{align*} increases by 20. However, b\begin{align*}b\end{align*} is not simply twice the value of a\begin{align*} a \end{align*}. We can see that when a=0,b=20\begin{align*} a=0, b=20\end{align*} so this gives a clue as to what rule the pattern follows. Hopefully you should see that the rule linking a\begin{align*} a \end{align*} and b.\begin{align*} b. \end{align*}
“To find a\begin{align*}a\end{align*}, double the value of a\begin{align*}a\end{align*} and add 20.”
Step 2 Translate into a mathematical equation:
Text Translates to Mathematical Expression
“To find b\begin{align*} b \end{align*} \begin{align*} \rightarrow\end{align*} b=\begin{align*} b= \end{align*}
“double the value of a\begin{align*} a \end{align*} \begin{align*} \rightarrow\end{align*} 2a\begin{align*} 2a \end{align*}
“add 20” \begin{align*} \rightarrow\end{align*} +20
b=2a+20THIS IS OUR EQUATION.\begin{align*} b=2a+20 & & \text{THIS IS OUR EQUATION}.\end{align*}
Step 3 Solve the equation.
Go back to the original problem. We substitute the values we have for our known variable and rewrite the equation.
when a is 750b=2(750)+20\begin{align*}“\text{when} \ a \ \text{is} \ 750” && \rightarrow && b=2(750)+20 \end{align*}
Follow the order of operations to solve
bb=2(750)+20=1500+20=1520\begin{align*} b& =2(750)+20\\ b& =1500+20=1520\end{align*}
Step 4 Check the result.
In some cases you can check the result by plugging it back into the original equation. Other times you must simply double-check your math. Double-checking is always advisable. In this case, we can plug our answer for b\begin{align*} b \end{align*} into the equation, along with the value for a\begin{align*} a \end{align*} and see what comes out. 1520=2(750)+20\begin{align*}1520=2(750)+20\end{align*} is TRUE because both sides of the equation are equal and balance. A true statement means that the answer checks out.
## Use a Verbal Model to Write an Equation
In the last example we developed a rule, written in words, as a way to develop an algebraic equation. We will develop this further in the next few examples.
Example 3
The following table shows the values of two related quantities. Write an equation that describes the relationship mathematically.
x\begin{align*}x-\end{align*}value y\begin{align*}y-\end{align*}value
2\begin{align*}-2\end{align*} 10
0 0
2 -10
4 -20
6 -30
Step 1 Extract the important information.
We can see from the table that y\begin{align*} y \end{align*} is five times bigger than x\begin{align*} x \end{align*}. The value for y\begin{align*} y \end{align*} is negative when x\begin{align*} x \end{align*} is positive, and it is positive when x\begin{align*} x \end{align*} is negative. Here is the rule that links x\begin{align*} x \end{align*} and y\begin{align*} y \end{align*}.
y\begin{align*} y \end{align*} is the negative of five times the value of x\begin{align*} x \end{align*}
Step 2 Translate this statement into a mathematical equation.
Text Translates to Mathematical Expression
y\begin{align*}y\end{align*} is” \begin{align*} \rightarrow\end{align*} y=\begin{align*} y=\end{align*}
“negative 5 times the value of x\begin{align*} x \end{align*} \begin{align*} \rightarrow\end{align*} 5x\begin{align*} -5x \end{align*}
y=5xTHIS IS OUR EQUATION.\begin{align*} y=-5x && \text{THIS IS OUR EQUATION}.\end{align*}
Step 3 There is nothing in this problem to solve for. We can move to Step 4.
Step 4 Check the result.
In this case, the way we would check our answer is to use the equation to generate our own xy\begin{align*} xy \end{align*} pairs. If they match the values in the table, then we know our equation is correct. We will substitute x\begin{align*} x \end{align*} values of -2, 0, 2, 4, 6 in and solve for y\begin{align*} y \end{align*}.
x=2:x=0:x=2:x=4:x=6:y=5(2)y=5(0)y=5(2)y=5(4)y=5(6)y=+10y=0y=10y=20y=30\begin{align*}&x=-2: && y=-5(-2) && y=+10\\ &x=0:&& y=-5(0)&&y=0\\ &x=2:&& y=-5(2)&&y=-10\\ &x=4:&& y=-5(4)&&y=-20\\ &x=6:&& y=-5(6)&&y=-30\end{align*}
Each xy\begin{align*} xy \end{align*} pair above exactly matches the corresponding row in the table.
Example 4
Zarina has a $100 gift card, and she has been spending money on the card in small regular amounts. She checks the balance on the card weekly, and records the balance in the following table. Week Number Balance ($)
1 100
2 78
3 56
4 34
Write an equation for the money remaining on the card in any given week.
Step 1 Extract the important information.
We can see from the table that Zarina spends 22 every week. • As the week number increases by 1, the balance decreases by 22. • The other information is given by any point (any week, balance pair). Let’s take week 1: • When (week number) = 1, (balance) = 100 Step 2 Translate into a mathematical equation. Define variables: Let week number=n\begin{align*}\text{week number} = n\end{align*} Let Balance=b\begin{align*}\text{Balance} = b\end{align*} Text Translates to Mathematical Expression As n\begin{align*}n\end{align*} increases by 1, b\begin{align*}b\end{align*} decreases by 22 \begin{align*}\rightarrow\end{align*} b=22n+?\begin{align*}b=-22n+? \end{align*} The ? indicates that we need another term. Without another term the balance would be -22, -44, -66,... We know that the balance in week 1 is 100. Let's substitute that value. 100=22(1)+?\begin{align*}100 = -22( 1 ) +?\end{align*} The ? number that gives 100 when 22 is subtracted from it is 122. equation is therefore: b=22n+122THIS IS OUR EQUATION.\begin{align*} b=-22n+122 & & \text{THIS IS OUR EQUATION}.\end{align*} Step 3 All we were asked to find was the expression. We weren't asked to solve it, so we can move to Step 4. Step 4 Check the result. To check that this equation is correct, we see if it really reproduces the data in the table. To do that we plug in values for n\begin{align*} n \end{align*} n=1n=2n=3n=4b=22(1)+122b=22(2)+122b=22(3)+122b=22(4)+122b=12222=100b=12244=78b=12266=56b=12288=34\begin{align*}& n=1 & & \rightarrow & & b=-22(1)+122 & & \rightarrow & & b=122-22=100 \\ & n=2 & & \rightarrow & & b=-22(2)+122 & & \rightarrow & & b=122-44=78 \\ & n=3 & & \rightarrow & & b=-22(3)+122 & & \rightarrow & & b=122-66=56 \\ & n=4 & & \rightarrow & & b=-22(4)+122 & & \rightarrow & & b=122-88=34 \end{align*} The equation perfectly reproduces the data in the table. The answer checks out. Note: Zarina will run out of money on her gift card (i.e. her balance will be 0) between weeks 5 and 6. ## Solve Problems Using Equations Let’s solve the following real-world problem by using the given information to write a mathematical equation that can be solved for a solution. Example 5 A group of students are in a room. After 25 students leave, it is found that 23\begin{align*} \frac {2}{3} \end{align*} of the original group is left in the room. How many students were in the room at the start? Step 1 Extract the important information We know that 25 students leave the room. We know that 23\begin{align*} \frac {2}{3} \end{align*} of the original number of students are left in the room. We need to find how many students were in the room at the start. Step 2 Translate into a mathematical equation. Initially we have an unknown number of students in the room. We can refer to them as the original number. Let’s define the variable x=\begin{align*} x= \end{align*} the original number of students in the room. 25 students leave the room. The number of students left in the room is: Text Translates to Mathematical Expression the original number of students in the room \begin{align*} \rightarrow\end{align*} x\begin{align*} x\end{align*} 25 students leave the room \begin{align*} \rightarrow\end{align*} x25\begin{align*} x-25 \end{align*} 23\begin{align*}\frac{2}{3}\end{align*} of the original number is left in the room \begin{align*} \rightarrow\end{align*} 23x\begin{align*} \frac{2}{3} x \end{align*} x25=23xTHIS IS OUR EQUATION.\begin{align*}x-25=\frac{2}{3}x & & \text{THIS IS OUR EQUATION}.\end{align*} Step 3 Solve the equation. Add 25 to both sides. x25x25+25x=23x=23x+25=23x+25\begin{align*} x-25& = \frac{2}{3}x\\ x-25+25& = \frac{2}{3}x+25\\ x& = \frac{2}{3}x+25\end{align*} Subtract 23x\begin{align*} \frac{2}{3}x \end{align*} from both sides. x23x13x=23x23x+25=25\begin{align*} x-\frac{2}{3}x & = \frac{2}{3}x-\frac{2}{3}x+25\\ \frac{1}{3}x& = 25\end{align*} Multiply both sides by 3. 313xx=253=75\begin{align*}3\cdot \frac{1}{3}x& = 25\cdot 3\\ x&= 75\end{align*} Remember that x\begin{align*} x \end{align*} represents the original number of students in the room. So, Answer There were 75 students in the room to start with. Step 4 Check the answer: If we start with 75 students in the room and 25 of them leave, then there are 7525=50\begin{align*} 75-25=50 \end{align*} students left in the room. 23\begin{align*} \frac{2} {3}\end{align*} of the original number is 2375=50\begin{align*} \frac{2} {3} \cdot 75 = 50\end{align*} This means that the number of students who are left over equals to 23\begin{align*} \frac{2} {3} \end{align*} of the original number. The answer checks out. The method of defining variables and writing a mathematical equation is the method you will use the most in an algebra course. This method is often used together with other techniques such as making a table of values, creating a graph, drawing a diagram and looking for a pattern. ## Review Questions Day Profit 1 20 2 40 3 60 4 80 5 100 1. Write a mathematical equation that describes the relationship between the variables in the table: 2. what is the profit on day 10? 1. Write a mathematical equation that describes the situation: A full cookie jar has 24 cookies. How many cookies are left in the jar after you have eaten some? 2. How many cookies are in the jar after you have eaten 9 cookies? 1. Write a mathematical equation for the following situations and solve. 1. Seven times a number is 35. What is the number? 2. One number is 25 more than 2 times another number. If each number is multiplied by five, their sum would be 350. What are the numbers? 3. The sum of two consecutive integers is 35. What are the numbers? 4. Peter is three times as old as he was six years ago. How old is Peter? 2. How much water should be added to one liter of pure alcohol to make a mixture of 25% alcohol? 3. Mia drove to Javier’s house at 40 miles per hour. Javier’s house is 20 miles away. Mia arrived at Javier’s house at 2:00 pm. What time did she leave? 4. The price of an mp3 player decreased by 20% from last year to this year. This year the price of the Player is120. What was the price last year?
1. P=20t;P=\begin{align*} P=20t; P=\end{align*} profit; t=\begin{align*} t=\end{align*} number of days. P=\begin{align*}P =\end{align*} profit; t=\begin{align*}t =\end{align*} number of days
2. Profit = 200
1. y=24x;y=\begin{align*} y=24-x; y= \end{align*} number of cookies in the jar; x=\begin{align*} x= \end{align*} number of cookies eaten
1. x=\begin{align*} x= \end{align*} the number; 7x=35\begin{align*}7x=35 \end{align*}; number = 5
2. x=\begin{align*} x= \end{align*} another number; 2x+25=\begin{align*} 2x+25= \end{align*} another number; 5x+5(2x+25)=350\begin{align*}5x+5(2x+25)=350\end{align*}; numbers = 15 and 55
3. x=\begin{align*} x =\end{align*} first integer; x+1=\begin{align*} x+1= \end{align*} second integer; x+x+1=35\begin{align*} x+x+1=35 \end{align*} ; first integer = 17, second integer = 18
4. x=\begin{align*} x= \end{align*} Peter’s age; x=3(x6)\begin{align*} x=3(x-6) \end{align*} ; Peter is 9 years old.
1. 3 liters
2. 1:30 pm
3. \$150
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# Gilligan’s Island
Suppose you and a couple of people were shipwrecked in an island (which, for dramatic effect, we call Gilligan’s Island). Assume that none of you were friends with each other in the beginning, but over time you all managed to be friends with at least one other person. What is the probability that two of you have the same number of friends? Now suppose that the initial number of people shipwrecked, including you is 60, what is the least number of people whose surnames start with the same letter?
This kind of problem is an instance of what is known as the Pigeonhole Principle.
If $n$ pigeons fly into $m$ pigeonholes, where $n > m$, then one pigeonhole has at least 2 or more pigeons in it.
This principle is so intuitive that even a child will agree with you on this. Now the question is why are the 2 problems in the previous paragraph an instance of the Pigeonhole Principle?
In the first problem, let $n$ be the number of people shipwrecked in Gilligan’s Island. Then the maximum number of friends anyone can have is therefore $n-1$ and the minimum number of friends anyone can have is 1 ( each person has at least one friend). Imagine the pigeonholes to be the possible number of friends anyone can have. The number of such pigeonholes is $n-1$. By assigning each person to a pigeonhole, you are in effect assigning the number of friends a person has. Since $n > n -1$, by the Pigeonhole Principle, one pigeonhole has at least 2 or more pigeons ( or at least 2 people have the same number of friends). Therefore, the probability of at least 2 persons having the same number of friends is equal to 1.
In general, the Pigeonhole Principle can be stated as
If $n$ pigeons fly into $m$ pigeonholes such that $n > k\cdot m$, for some integer $k$, then there is a pigeonhole with at least $k+1$ pigeons in it.
For example, suppose $n = 11$ pigeons flying into $m=5$ pigeonholes. By letting $k = 2$, we have $11 > 2\cdot 5$ and by the generalized Pigeonhole Principle, there is one pigeonhole that has at least $k+1 = 2+1 = 3$ pigeons in it.
Going back to the second problem, we know that there are only 26 letters in the English alphabet. If we imagine the $m = 26$ letters to be pigeonholes and the $n = 60$ people to be pigeons, then by the generalized Pigeonhole Principle, we have
$\displaystyle 60 > k\cdot 26$
Solving for the minimum integer $k$ that satisfies this, we get $k = 2$. This means that there are at least $k +1 = 2+1 = 3$ persons whose surnames begin with the same letter.
Now try this problem for size. Suppose you have 20 black socks and 20 white socks in a dark room. Since you cannot see clearly in the dark, you decided to pick any sock at random. What is the minimum number of socks you need to pick in order to get a pair with the same color?<|endoftext|>
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This printable encourages your child to write an April Fool's story.
This printable activity encourages your child to create his or her own story. Help your child with spelling and printing as needed. Encourage your child to read the story aloud. Talk about the story with your child, using the words "setting," "character" and "problem."<|endoftext|>
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Icebergs: the children of glaciers
Few pieces of natural phenomena are more linked to Antarctica and the Arctic than icebergs. Even the least polar-inclined know the basics: Icebergs are chunks of floating ice, the bulk of their shape rests below the surface of the water, and one of them sank a very historic ship – while launching a number of cinematic careers. But there’s a lot more to these fascinating ice formations than that.
What Is an iceberg? Meaning, definition, facts
Ice shelves and glaciers are far from static; they are always either building up or breaking down. Especially in a warm climate, this breaking process (known as calving) is accelerated, and it often happens that a piece of ice breaks off from its glacial origin and falls into the water. The natural buoyancy of these freshwater ice fragments, which can contain air bubbles that are tens of thousands of years old, keep the calved chunks afloat. This is how icebergs are born. They can be as large as mountains or as small as mole hills, though take care not to confuse them with ice floes, which are flat (and smaller) sheets of broken sea ice.
Icebergs in Antarctica and the Arctic
Icebergs form in both the Arctic and Antarctica, though in different ways and in different forms. In the Arctic, icebergs commonly start their lives in the enormous Greenland Ice Sheet, which spills its frozen contents through openings in the mountain ranges that line the coast. Crevasses form in these glaciers due to the landscape’s irregular topography, leading to smaller icebergs that are more unusually shaped than those of Antarctica. Antarctic icebergs generally outsize their Arctic siblings, sometimes weighing several billion tons and possessing the dimensions of small islands. They calve off from floating ice shelves, which populate about 30% of the Antarctic shoreline.
The largest of the biggest: Larsen C, A68, and B-15
Though icebergs occur in both the far north and far south, Antarctica is known to produce the real behemoths. Some of the largest ice formations ever recorded, in fact, have made headlines in world news. In July 2017, the Larsen C Ice Shelf, the fourth largest ice shelf in Antarctica, produced one of the most colossal icebergs ever witnessed in the modern world: A68, larger than the state of Delaware and twice as large as Luxembourg. Years before this, in the spring of 2000, the B-15 iceberg broke off from Antarctica’s Ross Ice Shelf. B-15 remains the largest berg ever recorded. At its most massive, it was larger than the island of Jamaica, boasting 11,000 square km (4,200 square miles) of surface area, a width of 37 km (23 miles), and a length of 295 km (183 miles).
Ice shelves, glaciers, and global warming
When planetary climates rise, ice shelves and glaciers naturally melt and calve at a faster rate. This in turn leads to the production of more icebergs. Many scientists blame human-caused global warming, claiming that the calving of massive icebergs like B-15 and A68 is the direct result of carbon dioxide and other pollutants released into the atmosphere by industry. Other researchers, meanwhile, argue that calving is a natural process, and that the breaking off of giant icebergs is not necessarily a harbinger of worse human-caused ice conditions to come.<|endoftext|>
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# Evaluating a definite Integral with a constant $n$
I am trying to evaluate the following integral: $$\int_0^1 \frac{n(n-1)(n-2)}{2} y^{3} \left( 1 - y \right)^{n-3} \, dy$$ Here is my solution. Let $$I$$ be the integral we are trying to evaluate. \begin{align*} I &= \int_0^1 \frac{n(n-1)(n-2)}{2} y^{3} \left( 1 - y \right)^{n-3} \, dy \end{align*} Now to evaluate this integral we use the substitution $$u = 1 - y$$ which gives us $$y = 1 - u$$ and $$du = -dy$$. \begin{align*} I &= - \int_0^1 \frac{n(n-1)(n-2)}{2} (1-u)^3 u^{n-3} \, du \\ I &= \int_1^0 \frac{n(n-1)(n-2)}{2} (1-u)^3 u^{n-3} \, du \\ \end{align*} Observe that: \begin{align*} (1-u)^3 &= (1-u)(1-u)^2 = (u-1)^2(1-u) = (u^2 - 2u + 1)(1-u) \\ (1-u)^3 &= -u^3 + 3u^2 - 3u + 1 \end{align*} Now back to the integral: \begin{align*} I &= \int_1^0 \frac{n(n-1)(n-2)}{2} ( -u^3 + 3u^2 - 3u + 1) u^{n-3} \, du \\ I &= \int_0^1 \frac{n(n-1)(n-2)}{2} ( u^3 - 3u^2 + 3u - 1) u^{n-3} \, du \\ I &= \int_0^1 \frac{n(n-1)(n-2)}{2} ( u^n - 3u^{n-1} + 3u^{n-2} - u^{n-3} ) \, du \\ I &= \frac{n(n-1)(n-2)}{2} \left( \frac{u^{n+1}}{n+1} - \frac{3u^n}{n} + \frac{3u^{n-1}}{n-1} - \frac{u^{n-2}}{n_2} \right) \bigg|_0^1 \\ I &= \frac{n(n-1)(n-2)}{2} \left( \frac{1}{n+1} - \frac{3}{n} + \frac{3}{n-1} - \frac{1}{n-2} \right) \\ I &= \frac{n(n-1)(n-2) -3(n+1)(n-1)(n-2) + 3(n+1)(n)(n-2) - (n+1)(n)(n-1)}{2(n+1)} \\ I &= \frac{n^3 - 3n^2 + 2n -3(n^2-1)(n-2) + 3(n+1)(n)(n-2) - (n+1)(n)(n-1)}{2(n+1)} \\ I &= \frac{n^3 - 3n^2 + 2n -3(n^3-2n^2-n+2) + 3(n+1)(n)(n-2) - (n+1)(n)(n-1)}{2(n+1)} \\ I &= \frac{-2n^3 + 3n^2 + 5n - 6 + 3(n^3-n^2 - 2n) - (n+1)(n)(n-1)}{2(n+1)} \\ I &= \frac{-2n^3 + 3n^2 + 5n - 6 + 3n^3-3n^2 - 6n - (n+1)(n)(n-1)}{2(n+1)} \\ I &= \frac{n^3 + 3n^2 + 5n - 6 - 3n^2 - 6n - (n+1)(n^2-n)}{2(n+1)} \\ I &= \frac{n^3 + 5n - 6 - 6n - (n+1)(n^2-n)}{2(n+1)} \\ I &= \frac{n^3 - n - 6 - (n^3 -n^2 + n^2 - n)}{2(n+1)} \\ I &= \frac{n^3 - n - 6 - n^3 + n}{2(n+1)} = \frac{-6}{2(n+1)} \\ I &= \frac{-3}{n+1} \\ \end{align*} I have reason to believe that the value of this integral is: $$\frac{3}{n+1}$$ Where did I go wrong in evaluating this integral?
• Why do you drag these $n$ factors all along ?? This is calling for problems.
– user65203
Nov 17 '19 at 16:31
• They are needed because of the $y^{n-3}$ term.
– Bob
Nov 17 '19 at 16:57
• I don't see why.
– user65203
Nov 17 '19 at 18:42
• They're a constant factor wrt $y$, meaning they don't really affect the "behavior" of the integral, and so you can pull them out of the integrand and ignore them until the end. If you have to divide by $n - 2$ or something at some point, it's fine to do it without these factors. Nov 18 '19 at 1:10
• You can also notice that if $n>2$, then $I$ must be positive in your first two lines and must be negative in the third line, which is a big hint to examine that step. Nov 18 '19 at 1:27
## 2 Answers
You forgot to change the bounds in your third equation. It should be $$I=-\int_{\color{red}{1}}^{\color{red}{0}}\frac{n(n-1)(n-2)}{2}(1-u)^{3}u^{n-3}\,du$$
This is a complete Beta integral and we immediately have
$$\frac{n(n-1)(n-2)}{2} \int_0^1 y^{3} \left( 1 - y \right)^{n-3} \, dy =\frac{n(n-1)(n-2)}{2}\frac{3!(n-3)!}{(n+1)!}.$$
Simplify.
Alternatively:
Indeed,
$$y^{n-3}-3y^{n-2}+3y^{n-1}-y^{n}$$ integrates to
$$\frac1{n-2}-\frac3{n-1}+\frac3{n}-\frac1{n+1}=\frac6{(n+1)n(n-1)(n-2)}.$$<|endoftext|>
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John Metcalfe was CityLab’s Bay Area bureau chief, covering climate change and the science of cities.
A recent decline in the rate of increased worldwide temperatures is masking the brow-sweating temperatures of the future.
One of the puzzling things about climate change is a recent slowdown in warming (or as some exaggerators refer to it, a "pause"). The planet's temperatures ticked up by 0.22 degrees Fahrenheit each decade after 1951. But beginning in 1998, the rate slowed to 0.09 degrees per decade, despite the world's nations still pouring copious greenhouse gases into the atmosphere.
There are many theories to explain this reduced rate of warming (see below), yet climate-change skeptics have used this event to compose a rousing symphony of pishes and humbugs. For instance, Texas Senator Ted Cruz recently told CNN "there has been no recorded warming" in the past 15 years, and that the "data are not supporting what the advocates are arguing." And the failure of climate models to accurately predict the slowdown was hiding in plain sight in a February editorial in the Wall Street Journal, penned by two atmospheric scientists from the University of Alabama in Huntsville. "We might forgive these modelers if their forecasts had not been so consistently and spectacularly wrong," they wrote, claiming that the "forecasts for future temperatures have continued to be too warm."
But "too warm" is a concept we need to wrap our soon-to-be-baked brains around, according to new research from NASA's Drew Shindell. After performing a region-specific analysis of the things that affect the global climate, Shindell has come to the conclusion that, slowdown be damned, we are still looking at a vast leap in the earth's heat levels. In fact, there could be a warming increase about 20 percent greater than indicated by surface-temperature observations from the last 150 years, according to his new study in Nature Climate Change.
The above map of the globe, furnished by NASA's visualization and simulation teams, shows where the heat is likely to come down hard by 2099. Dark-red areas at the North Pole indicate positive temperature anomalies of up to 25 degrees. Much of Canada could see hikes of 10 to 20 degrees and in the United States, 5 to 12.5 degrees. That would mean that future warming will be much worse than described in the latest report from the Intergovernmental Panel on Climate Change (which took the slowdown into account when making its projections). NASA has the figures:
To put a number to climate change, researchers calculate what is called Earth's "transient climate response." This calculation determines how much global temperatures will change as atmospheric carbon dioxide continues to increase – at about 1 percent per year – until the total amount of atmospheric carbon dioxide has doubled. The estimates for transient climate response range from near 2.52 F (1.4 C) offered by recent research, to the IPCC's estimate of 1.8 F (1.0 C). Shindell's study estimates a transient climate response of 3.06 F (1.7 C), and determined it is unlikely values will be below 2.34 F (1.3 C).
So how did Shindell reach this grim outcome? Partly by taking an exacting approach to atmospheric aerosols, which are natural and human-generated particles that play a role in the distribution of solar energy. Previous modeling runs had assumed they were scattered evenly around the globe. But aerosols are actually more prevalent in the Northern Hemisphere due to heavier industrialization and larger landmasses. Shindell took this inequality into account to make his harsher prediction. "I wish it weren't so," he says, "but forewarned is forearmed."
For those wondering why global warming has been slower these past years, scientists have posed a number of possible reasons. One is that the ocean is sucking up some of the excess heat, as the atmosphere can only hold so much at any given time. One NASA oceanographer told Scientific American that the world's seas, not surface temperatures, should be the current barometer of climate change because their temperatures are going up "like gangbusters." Other suspects include several recent sky-dimming volcanic eruptions and an abnormal pattern of trade winds in the Pacific.
Whatever's putting on the brakes, the national science academies of the U.S. and U.K. warn in a February report not to expect it to last. Although there might be "slowdowns and accelerations in warming lasting a decade or more," they write, the clear long-term trend is "substantial increases in global average surface temperature and important changes in regional climate."
Top image: NASA SVS / NASA Center for Climate Simulation<|endoftext|>
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# 1984 AIME Problems/Problem 1
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
Find the value of $a_2+a_4+a_6+a_8+\ldots+a_{98}$ if $a_1$, $a_2$, $a_3\ldots$ is an arithmetic progression with common difference 1, and $a_1+a_2+a_3+\ldots+a_{98}=137$.
## Solution 1
One approach to this problem is to apply the formula for the sum of an arithmetic series in order to find the value of $a_1$, then use that to calculate $a_2$ and sum another arithmetic series to get our answer.
A somewhat quicker method is to do the following: for each $n \geq 1$, we have $a_{2n - 1} = a_{2n} - 1$. We can substitute this into our given equation to get $(a_2 - 1) + a_2 + (a_4 - 1) + a_4 + \ldots + (a_{98} - 1) + a_{98} = 137$. The left-hand side of this equation is simply $2(a_2 + a_4 + \ldots + a_{98}) - 49$, so our desired value is $\frac{137 + 49}{2} = \boxed{093}$.
## Solution 2
If $a_1$ is the first term, then $a_1+a_2+a_3 + \cdots + a_{98} = 137$ can be rewritten as:
$98a_1 + 1+2+3+ \cdots + 97 = 137$ $\Leftrightarrow$ $98a_1 + \frac{97 \cdot 98}{2} = 137$
Our desired value is $a_2+a_4+a_6+ \cdots + a_{98}$ so this is:
$49a_1 + 1+3+5+ \cdots + 97$
which is $49a_1+ 49^2$. So, from the first equation, we know $49a_1 = \frac{137}{2} - \frac{97 \cdot 49}{2}$. So, the final answer is:
$\frac{137 - 97(49) + 2(49)^2}{2} = \fbox{093}$.
## Solution 3
A better approach to this problem is to notice that from $a_{1}+a_{2}+\cdots a_{98}=137$ that each element with an odd subscript is 1 from each element with an even subscript. Thus, we note that the sum of the odd elements must be $\frac{137-49}{2}$. Thus, if we want to find the sum of all of the even elements we simply add $49$ common differences to this giving us $\frac{137-49}{2}+49=\fbox{093}$.
Or, since the sum of the odd elements is 44, then the sum of the even terms must be $\fbox{093}$.
## Solution 4
We want to find the value of $a_2+a_4+a_6+a_8+\ldots+a_{98}$, which can be rewritten as $a_1+1+a_2+2+a_3+\ldots+a_{49}+49 \implies a_1+a_2+a_3+\ldots+a_{49}+\frac{49 \cdot 50}{2}$. We can split $a_1+a_2+a_3+\ldots+a_{98}$ into two parts: $$a_1+a_2+a_3+\ldots+a_{49}$$ and $$a_{50}+a_{51}+a_{52}+\ldots+a_{98}$$ Note that each term in the second expression is $49$ greater than the corresponding term, so, letting the first equation be equal to $x$, we get $a_1+a_2+a_3+\ldots+a_{98}=137=2x+49^2 \implies x=\frac{137-49^2}{2}$. Calculating $49^2$ by sheer multiplication is not difficult, but you can also do $(50-1)(50-1)=2500-100+1=2401$. We want to find the value of $x+\frac{49 \cdot 50}{2}=x+49 \cdot 25=x+1225$. Since $x=\frac{137-2401}{2}$, we find $x=-1132$. $-1132+1225=\boxed{93}$.
- PhunsukhWangdu
## Solution 5
Since we are dealing with an arithmetic sequence, $$a_2+a_4+a_6+a_8+\ldots+a_{98} = 49a_{50}$$ We can also figure out that $$a_1+a_2+a_3+\ldots+a_{98} = a_1 + 97a_{50} = 137$$ $$a_1 = a_{50}-49 \Rightarrow 98a_{50}-49 = 137$$ Thus, $49a_{50} = \frac{137 + 49}{2} = \boxed{093}$
~Hithere22702
~ pi_is_3.14<|endoftext|>
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#### PROPOSITION 6.
If two triangles have one angle equal to one angle and the sides about the equal angles proportional, the triangles will be equiangular and will have those angles equal which the corresponding sides subtend.
Let ABC, DEF be two triangles having one angle BAC equal to one angle EDF and the sides about the equal angles proportional, so that,
as BA is to AC, so is ED to DF;
I say that the triangle ABC is equiangular with the triangle DEF, and will have the angle ABC equal to the angle DEF, and the angle ACB to the angle DFE.
For on the straight line DF, and at the points D, F on it, let there be constructed the angle FDG equal to either of the angles BAC, EDF, and the angle DFG equal to the angle ACB; [I. 23]
therefore the remaining angle at B is equal to the remaining angle at G. [I. 32]
Therefore the triangle ABC is equiangular with the triangle DGF.
Therefore, proportionally, as BA is to AC, so is GD to DF. [VI. 4]
But, by hypothesis, as BA is to AC, so also is ED to DF; therefore also, as ED is to DF, so is GD to DF. [V. 11]
Therefore ED is equal to DG; [V. 9] and DF is common;
therefore the two sides ED, DF are equal to the two sides GD, DF; and the angle EDF is equal to the angle GDF; therefore the base EF is equal to the base GF, and the triangle DEF is equal to the triangle DGF,
and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend. [I. 4]
Therefore the angle DFG is equal to the angle DFE,
and the angle DGF to the angle DEF.
But the angle DFG is equal to the angle ACB; therefore the angle ACB is also equal to the angle DFE.
And, by hypothesis, the angle BAC is also equal to the angle EDF; therefore the remaining angle at B is also equal to the remaining angle at E; [I. 32]
therefore the triangle ABC is equiangular with the triangle DEF.
Therefore etc. Q. E. D.<|endoftext|>
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Catacombs were subterranean cemeteries that served a double purpose: that of burial for Christians who had died or been martyred, and a place to which Christians could resort to conduct devotions, administer sacraments, and celebrate the Eucharist in secrecy because of persecution.
It was customary to celebrate the mystery of the Eucharist, the Mass, using the top of the tomb for an altar surface.
Many catacombs have been excavated, the majority located along the Via Appia and the Via Ardeatina (two ancient Roman roads).
The catacombs were dug by burrowing into the soft stone of hill areas. The practice was to excavate a stairway and then open a narrow gallery. Within this gallery were tombs cut into the wall, capable of holding from one to three bodies. These tombs were sealed with a slab of stone when the bodies had been placed within. When the gallery became full, it was the usual practice to cut lateral galleries to expand capacity.
In the fourth century, after the end of the persecutions, the catacombs became places of pilgrimage and have remained so until the present.
The catacombs are notable for the testimony of art and practice that they supply to modern ages. Chief of these testimonies are that of the sacramental religion observed by these early Christians and the purity of the spiritual truth that is visible. These fact are deduced from the art, the carvings, artifacts, and inscriptions with which the catacombs contain. (7:26)<|endoftext|>
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Some trees thrive in sunlight, others need deep shade. BNRC land stewards help each tree find its place in the forest, with your help.
Have you ever noticed that some trees grow very quickly, and others take many decades to mature? Or that trees near disturbances and forest edges behave very differently than those in deep undisturbed areas? What’s the best way to ensure the health of a wide range of tree species, to support habitat for wildlife?
Some trees prefer specific environmental conditions; some like wet soil, some dry, some shady and others like full light.
Shade intolerant species are fast growers and they need lots of sunlight (50% or more direct sunlight). At the other end of the spectrum, shade tolerant species grow slowly and in low light (10-20% direct sunlight). Intermediate species can withstand anywhere in between.
Forests are not static—both human and natural changes can change the forest makeup. For example, when lots of sunlight reaches the forest floor in the event of a large disturbance (large forest fire, clear-cut, road etc.) shade intolerant species, such as aspen and paper birch, thrive.
At the other end of the spectrum, yellow birch, beech and hemlock trees have the highest tolerance for shade in the Northeast. In a mature oak forest, for example, small hemlocks are able to survive near the forest floor for decades, growing very slowly with minimal sunlight. A hemlock can be 4 feet tall and 70 or more years old, waiting for a gap to form in the canopy. When you walk into a forest of tall hemlock, yellow birch, or beech, such as at our Boulders reserve in Pittsfield, you know that they waited a long time to reach the top of the canopy.
While every forest has edges, we strive to keep large tracts of forest intact and undisturbed, so that essential species like hemlocks will always have a place to grow in the Berkshires.
So on your next outdoor adventure, pay attention to the forest composition and the story it tells.
You help make sure the thousands of acres of BNRC-owned forests are cared for to protect the trees and the wildlife that depend on them. Thank you!<|endoftext|>
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Elementary Program (6 - 12 years)
"I need to understand my place in the group and in society"
The nature and needs of the child begin to change as they move to the second plane of development. They have the ability to think more abstractly, require variety rather than repetition, want to function as a member of a team, are developing a sense of justice, and possess a powerful and expansive imagination.
To address the child's changing psychological characteristics, lessons change from the more individualized to smaller group lessons and collaborative projects at the elementary level.
The elementary child is possessed of a mind that learns by questioning and reasoning. Besides having great powers of intellect, they are very imaginative and creative. The materials and the way in which concepts are presented evoke this imagination and serve to help the child in increasingly abstract understanding. Besides skill building in core curriculum subjects such as reading, writing and mathematics, the Montessori Method provides an interdisciplinary view to help the child discover and appreciate the interconnections of knowledge and life on Earth.
Subject areas are integrated throughout the curriculum in addition to being presented as separate disciplines. The Montessori Elementary curriculum provides a framework for study by first presenting the "big idea." The teacher then follows up by illuminating the details through stories, pictures, charts, concrete Montessori materials, and timelines to enrich the child's understanding. The program focus areas are: Language, Mathematics, Biology, Geography, Geometry, History, Music, and Art.
Dr. Montessori's lifelong goal was peace on Earth. The Elementary Program helps the child develop a deep appreciation for the vital connections between human beings, setting the stage for a peaceful future for humanity.<|endoftext|>
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# NCERT Solutions for Class 9 Maths Chapter 7: Triangles
## NCERT Solutions for Class 9 Mathematics Chapter 7 Free PDF Download
The dot mark field are mandatory, So please fill them in carefully
To download the complete Syllabus (PDF File), Please fill & submit the form below.
Q. Prove that measure of each angle an equilateral triangle is 60°.
Ans. Given : DABC is an equilateral triangle.
To prove : ∠A = ∠B = ∠C = 60°
Proof : In DABC,
AB = BC
∠C = ∠A ...(i)
(Opposite angles of equal sides)
AB = AC
∠C = ∠B ...(ii)
From equations (i) and (ii), ∠A = ∠B = ∠C ...(iii)
In △ABC,
∠A + ∠B + ∠C = 180° (Sum of all angles of △)
⇒ 3∠A = 180°
⇒ ∠A = 60°
∠A = ∠B = ∠C = 60°.
Q. In the given figure, l and m are two parallel lines intersected by another pair of parallel lines p and q. Show that △ABC △CDA.
Ans. Given : l || m and p || q
To prove : △ABC △CDA
Proof : l || m
∠ACB = ∠DAC ...(i)
(Alternative angles)
p || q
∠BAC = ∠ACD ...(ii)
(Alternativ
e angles)
In △ABC and △ADC,
∠ACB = ∠DAC (Proved above)
∠BAC = ∠ACD (Proved above)
AC = AC (Common)
△ABC △CDA (ASA congruency rule)
Q. AB is a line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B. Show that the line PQ is perpendicular bisector of AB.
Ans. Give : P and Q are equidistant from the point A and B respectively.
To prove : PQ is perpendicular bisector of AB
Proof : In DAPQ and DBPQ
AP = BP (Given)
AQ = BQ (Given)
PQ = PQ (Common)
APQ BPQ (SSS congruency rule)
∠APQ = ∠BPQ (CPCT)
AP = BP (Given)
OP = OP (Common)
AOP BOP (SAS congruency rule)
AO = BO (CPCT) ∠AOP = ∠BOP (CPCT)
∠AOP + ∠BOP = 180° (Linear pair)
∠AOP + ∠AOP = 180°
∠AOP = 90°
PQ ⊥ AB
Q. If the bisector of the vertical angle of a triangle bisects the base of a triangle then the triangle is isosceles.
Ans. Given : In DABC, AD is the angle bisector of ∠BAC and also bisects the base.
To prove : ABC is an isosceles triangle.
Construction : Extend AD such that AD = DE and join
CE
Proof : In ABD and DCE,
(Vertically opposite angles)
BD = DC (Given)
AD = DE (By construction)
ABD ECD (SAS congruency rule)
AB = CE (CPCT) ...(i)
∠BAD = ∠CED (CPCT) ...(ii)
[From (ii) and (iii)]
In ACE, ∠CAE = ∠CEA
(Opposite sides of equal angles)
CE = AB [from (i)]
Form (iv) and (i),
AB = AC
ABC is an isosceles triangle.
Q. DABC is an isosceles triangle with AB = AC, side BA is produced to D such that AB = AD. Prove that ∠BCD is a right angle.
Ans. Given : △ABC is an isosceles triangle and BA = AD.
To prove : ∠BCD = 90°
Proof : AB = AC
∠ABC = ∠ACB ...(i)
(Opposite angles of equal sides)
∠ACD = ∠ADC ...(ii)
(Opposite angles of equal sides)
∠CAD = ∠ABC + ∠ACB
(Exterior angle property)
⇒ ∠CAD = 2∠ACB (from (i)) ...(iii)
∠BAC = ∠ADC + ∠ACD
(Exterior angle property)
⇒ ∠BAC = 2∠ACD (from (ii)) ...(iv)
From (iii) and (iv)
2∠ACB + 2∠ACD = ∠CAD + ∠BAC
⇒ 2(∠ACB + ∠ACD) = ∠BAD
⇒ 2(∠BCD) = 180°
⇒ ∠BCD = 90°.<|endoftext|>
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The biological products in the atmospheres of potentially habitable planets outside our solar system may help detect alien life, scientists say.
These atmospheric fingerprints of life, called biosignatures, will be detected using next-generation telescopes that measure the composition of gases surrounding planets that are light years away, according to the research published in Astrophysical Journal Letters.
However, biosignatures based on single measurements of atmospheric gases could be misleading.
Now, scientists at the University of California, Riverside (UCR) in the US are developing the first quantitative framework for dynamic biosignatures based on seasonal changes in the Earth’s atmosphere.
As Earth orbits the Sun, its tilted axis means different regions receive more rays at different times of the year.
The most visible signs of this phenomenon are changes in the weather and length of the days, but the atmospheric composition is also impacted.
For example, in the Northern Hemisphere, which contains most of the world’s vegetation, plant growth in summer results in noticeably lower levels of carbon dioxide in the atmosphere. The reverse is true for oxygen.
“Atmospheric seasonality is a promising biosignature because it is biologically modulated on Earth and is likely to occur on other inhabited worlds,” said Stephanie Olson from the University of California, Riverside (UCR).
“Inferring life based on seasonality would not require a detailed understanding of alien biochemistry because it arises as a biological response to seasonal changes in the environment, rather than as a consequence of a specific biological activity that might be unique to the Earth,” said Olson.
Further, extremely elliptical orbits rather than axis tilt could yield seasonality on extrasolar planets, or exoplanets, expanding the range of possible targets.
The researchers identified the opportunities and pitfalls associated with characterizing the seasonal formation and destruction of oxygen, carbon dioxide, methane, and their detection using an imaging technique called spectroscopy.
They also modeled fluctuations of atmospheric oxygen on a life-bearing planet with low oxygen content, like that of Earth billions of years ago.
They found that ozone (O3), which is produced in the atmosphere through reactions involving oxygen gas (O2) produced by life, would be a more easily measurable marker for the seasonal variability in oxygen than O2 itself on weakly oxygenated planets.
“It’s really important that we accurately model these kinds of scenarios now, so the space and ground-based telescopes of the future can be designed to identify the most promising biosignatures,” said Edward Schwieterman, a NASA Postdoctoral Program fellow at UCR.
“In the case of ozone, we would need telescopes to include ultraviolet capabilities to easily detect it,” said Schwieterman.
He said the challenge in searching for life is the ambiguity of data collected from so far away. False positives – nonbiological processes that masquerade as life – and false negatives – life on a planet that produces few or no biosignatures – are both major concerns.
“Both oxygen and methane are promising biosignatures, but there are ways they can be produced without life,” Schwieterman said.
Olson said observing seasonal changes in oxygen or methane would be more informative.<|endoftext|>
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Summary: A new study reveals how early immune response helps spread neurocryptococcosis. The findings may help develop new treatments that can prevent and treat this fungal infection.
Source: University of Sydney.
Published today in The American Journal of Pathology, the study was led by a team from the Westmead Institute for Medical Research and the Marie Bashir Institute for Infectious Disease and Biosecurity of the University of Sydney.
Meningitis is an inflammation of the meninges, the three membranes that cover the brain and spinal cord. Meningitis can occur when fluid that bathes the brain and separates it from the meninges becomes infected.
Researchers identified a two-step invasion process mediated by the immune system, which, while being the body’s primary defence against Cryptococcus infection, paradoxically promotes infection and meningitis.
The team observed the specific location where circulating white blood cells, known as monocytes, act as transport vehicles for cryptococci to cross from small blood vessels (post-capillary venues or PVCs) in and around the brain into the peri-vascular space outside these blood vessels.
They also observed cryptococci released from monocytes into this peri-vascular space then separately enter the brain, causing tumour-like lesions and the fluid surrounding the brain to cause meningitis.
Lead researcher, Professor Tania Sorrell from the University of Sydney, said: “We have shown for the first time that the early immune response in neurocryptococcosis is comprised of white blood cells called monocytes, neutrophils and lymphocytes, recruited from the blood across the endothelial lining of PCVs into the adjacent peri-vascular space.
“We also show that blood-derived monocytes and, to a lesser extent, neutrophils, by acting as Trojan horses, are the major effectors of Cryptococcus neoformans invasion of the central nervous system, by the same route.”
The discovery is an important first step for designing new treatments that could prevent and treat infection by the killer fungus, Professor Sorrell notes.
“Understanding this complex process is an important first step in designing new types of treatments that could complement the use of antifungal drugs.
Cryptococcus neoformans is a fungus that lives in the environment throughout the world, notes The Centers for Disease Control and Prevention. It is typically found in soil, on decaying wood, in tree hollows, or in bird droppings.
Cryptococcus neoformans is a common cause of meningitis that kills up to a third of patients even when treated appropriately.
Worldwide, an estimated 220,000 new cases of cryptococcal meningitis occur each year, resulting in 181,000 deaths.
People can become infected after breathing in the microscopic fungus, although most people who are exposed to the fungus never get sick from it.
Cryptococcus neoformans infections are rare in people who are otherwise healthy. Most cases occur in people who have weakened immune systems, particularly those who have advanced HIV/AIDS.
Source: Dan Gaffney – University of Sydney
Publisher: Organized by NeuroscienceNews.com.
Image Source: NeuroscienceNews.com image is in the public domain.
Original Research: The study will appear in American Journal of Pathology during the week of June 18 2018.<|endoftext|>
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# Proportions and Ratios
### Definition of Ratio
A ratio is a relationship between two values. For instance, a ratio of 1 pencil to 3 pens would imply that there are three times as many pens as pencils. For each pencil there are 3 pens, and this is expressed in a couple ways, like this: 1:3, or as a fraction like 1/3. There do not have to be exactly 1 pencil and 3 pens, but some multiple of them. We could just as easily have 2 pencils and 6 pens, 10 pencils and 30 pens, or even half a pencil and one-and-a-half pens! In fact, that is how we will use ratios -- to represent the relationship between two numbers.
### Definition of Proportion
A proportion can be used to solve problems involving ratios. If we are told that the ratio of wheels to cars is 4:1, and that we have 12 wheels in stock at the factory, how can we find the number of cars we can equip? A simple proportion will do perfectly. We know that 4:1 is our ratio, and the number of cars that match with those 12 wheels must follow the 4:1 ratio. We can setup the problem like this, where x is our missing number of cars:
$$\frac{4}{1}=\frac{12}{x}$$
To solve a proportion like this, we will use a procedure called cross-multiplication. This process involves multiplying the two extremes and then comparing that product with the product of the means. An extreme is the first number (4), and the last number (x), and a mean is the 1 or the 12.
To multiply the extremes we just do $$4 * x = 4x$$. The product of the means is $$1 * 12 = 12$$. The process is very simple if you remember it as cross-multiplying, because you multiply diagonally across the equal sign.
You should then take the two products, 12 and 4x, and put them on opposite sides of an equation like this: $$12 = 4x$$. Solve for x by dividing each side by 4 and you discover that $$x = 3$$. Reading back over the problem we remember that x stood for the number of cars possible with 12 tires, and that is our answer.
It is possible to have many variations of proportions, and one you might see is a double-variable proportion. It looks something like this, but it easy to solve.
$$\frac{16}{x}=\frac{x}{1}$$
Using the same process as the first time, we cross multiply to get $$16 * 1 = x * x$$. That can be simplified to $$16 = x^2$$, which means x equals the square root of 16, which is 4 (or -4). You've now completed this lesson, so feel free to browse other pages of this site or search for more lessons on proportions.
## Ratios and Proportions Calculator
Use the tool below to convert between fractions and decimal, or to take a given ratio expression and solve for the unknown value.<|endoftext|>
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Learning with 2 kids and a mommy
# Rainbow Water Experiment
Explain to young children, who have not started any Science subject in school, about density and mass and they will go “HUH?”. Yes, you get to see that blur and confused look on their face. Just like when they asked you where do they come from and you go “HUH?”.
How do we teach and explain to them about density?
The easier way is to show them and have them do the experiment to see it themselves. The most straightforward, uncomplicated way is to use what you can find at home to demonstrate density and that is non other than suger, food colourings and water. Effortless. Affordable.
What do we do with sugar and water?
Prepare some cups with the same amount of water (we prepared 4 cups of water for this experiment). Add different amount of sugar in each cup of water and stir till the sugar is fully dissolved. Add colourings to each cup. This is to differentiate each different cup of sugar content.
In our experiment, we started with 1 tablespoon of sugar for the first cup of water and subsequently added 1 more tablespoon but it didn’t show the effect. Therefore we added double for each cup of water. For example: 1st cup of water, add 2 tablespoons of sugar. 2nd cup with 4 tablespoons. 3rd cup with 8 tablespoons. Double the amount of tablespoons as you work your way through the cups of water.
What’s next?
Get an empty tube or small container and a syringe. Or you can just use a spoon. Slowly and gently pour each cup of sugared water into the empty container with the cup that has the highest amount of sugar first, followed by the next higher amount of sugar. In short, most dense water to next dense water until you have a stacked of colours from all 4 cups.
Because each cup of water is of different colours, you will be able to see each layer of sugared water sitting on top of each other with the most dense water at the bottom.
Explanation with the experiment on-hand
Density is about compactness of stuff in a space. For this experiment, you are adding sugar into the water. The more sugar in the a cup of water, the more closely and neatly is the sugar packed together. Hence, the greater the density (same amount water, but different amount of sugar in it). The cup with the most sugar in it is the most dense, while the cup with the least amount of sugar is the least dense.
Still confused?
Imagine you have 3 containers of the same size. Place a metal ball in the 1st container, 2 metal balls in the 2nd container, 3 metal balls in the 3rd container. Guess which one will be the heaviest? That’s right! The 3rd container followed by 2nd and lastly the 1st container. The heavier the object, the more likely it will sink. The rainbow sugared water experiment is to demonstrate this while having fun.
Check out the rainbow sugared water we have made. Pretty isn’t it? Smarties simply love it and they tested and played with different amount of sugar and colours after the 1st try. They even explored putting in different density of water at different times to test out what they get.
We have created a short video clip on how to play and have fun while learning about density with rainbow sugared water experiment. Enjoy!
### 18 thoughts on “Rainbow Water Experiment”
• Interesting! Wonder if this is how they make the rainbow layered chiffon cakes… I always wonder how the layers are kept so distinct. 🙂
• YoungSmarties says:
Hmm…you make me wonder too…
• Science experiment with kids are FUN! Saw another version is the egg with salt water. But I find sugar water experiment is more colourful ^_^
• YoungSmarties says:
Yes, agree! I prefer the water too and is easier to do with kids. 😉
• YoungSmarties says:
That’s really true! Thank you 😉
• Science classes with YoungSmarties are always packed with Fun !
cheers, Andy
(SengkangBabies.com)
• YoungSmarties says:
Thanks Andy! I guess with kids around, everything looks fun! 😀
• Wow, that’s a visual treat and learning session all in one! And, Mei Mei can articulate the instructions really well too – thumbs up!
Have a blessed day!
Dee
http://www.TheHootingPost.com
• YoungSmarties says:
Thanks so much for the kind words 🙂
• I simply love the experiments you share. Kudos!
• YoungSmarties says:
Thank you 🙂
• YoungSmarties says:
Thank you! Am using iMovie 🙂
• Wow….what an interesting experiment! Thanks for sharing!
• YoungSmarties says:
Pleasure!
• Beautiful! What happens when you leave the rainbow water to stand for a short while?
• YoungSmarties says:
It remains separated after a short while. We didn’t test any longer as smarties started shaking everything up!<|endoftext|>
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Regardless of which methods you use to teach music, movement figures into it, though perhaps in varying degree. Laban and Jaques-Dalcroze in particular have influenced the use of movement as an indispensable component of educating children musically. Though one could go into great detail about the various kinds of movement, four general types of movement are useful for music educators. These are, moving to the beat, moving for expression, moving to form, and moving to events. Each of these types of movements helps children understand, experience, and express the many facets of music.
Moving to the beat is perhaps the most frequently done movement. It is an integral part of Kodaly, Orff, Dalcroze and Gordon philosophies for music education. Early formal music training always includes moving to a beat, often using patsch, clap, walk and snap as primary movements. Of these, walking is particularly helpful because it includes a shift of weight with each step, and is a silent motion, so the beat is felt but not heard except in the music itself. Walking also leaves the hands and arms free to do rhythms or subdivisions of the beat. Earliest attempts involve the student performing a beat and then the teacher performing music to the child’s beat. This shows the child the relationship between beat and music, but does not require the child to determine the beat from music she is hearing, which is a more difficult skill. Later, the child learns to do this as well.
Movement to the beat when done in the arms leaves the feet free to move to the rhythm. Students can “walk the music” while keeping the beat with their arms. A more advanced version of this is to walk forwards for ascending pitches and backwards for descending pitches. Students can also do a predetermined motion whenever a specified event is heard, such as an accent, syncopation, or motive. For example, they might jump whenever they hear a staccato event in an otherwise staccato melody, or squat down whenever they hear a descending motif they had learned before. None of these movements make any sound, so listening to the music is unimpeded. This is important, because these movement activities are as much about listening (hearing ascending or descending contour, for example) as they are about keeping a steady beat. Movements that make sound, such as claps or snaps, add to the music, and should be used in situations where the sounds made are helpful to the students in achieving the instructional goal of the activity.
Movement for expression is very different. With this kind of movement, motions synchronized to the beat are avoided. Instead, students move to what they think or feel from the music. They might make smooth motions with the arms for legato, or jumpy motions for staccato. They might flick the wrists for an intervallic leap, or slouch forward for a dark or somber chord. As they do these motions, they bring out feelings of joy, or sadness, or agitation, or relaxation. The motions help the students discover how they feel when they hear the music, because the motions express those same feelings. Students can learn things about the music from their bodies by moving for expression. Having them move this way and then describe how they felt when they moved is a great way to start a discussion of what they already know about the music, before teaching them anything about it directly.
Movement for form requires students to listen for things that are repeated and listen to things that are new. It is a type of same-different learning. Students repeat a movement for things that are repeated, and create a new movement for things that are new. By tracing the pattern of same and different movements, students can learn the form of the music through their actions. As with stepping the music, where, as one of my students put it, they “become the notes,” when moving to form, the students become the form. Moving this way in a class also enables students to see the form as they watch other students make same or different patterns of movement. The specific movements can be pre-arranged for each theme or section, or they can be personal to each student. As long as each child is accurately showing same and different, they can use their creativity to use whatever movements they can imagine and do. Movement is important to understanding and experiencing music, and should be used often in our music classes.<|endoftext|>
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# Euler's totient function
Euler's totient function $\phi(n)$ applied to a positive integer $n$ is defined to be the number of positive integers less than or equal to $n$ that are relatively prime to $n$. $\phi(n)$ is read "phi of n."
## Formulas
To derive the formula, let us first define the prime factorization of $n$ as $n =\prod_{i=1}^{m}p_i^{e_i} =p_1^{e_1}p_2^{e_2}\cdots p_m^{e_m}$ where the $p_i$ are distinct prime numbers. Now, we can use a PIE argument to count the number of numbers less than or equal to $n$ that are relatively prime to it.
First, let's count the complement of what we want (i.e. all the numbers less than or equal to $n$ that share a common factor with it). There are $\frac{n}{p_1}$ positive integers less than or equal to $n$ that are divisible by $p_1$. If we do the same for each $p_i$ and add these up, we get
$$\frac{n}{p_1} + \frac{n}{p_2} + \cdots + \frac{n}{p_m} = \sum^m_{i=1}\frac{n}{p_i}.$$
But we are obviously overcounting. We then subtract out those divisible by two of the $p_i$. There are $\sum_{1 \le i_1 < i_2 \le m}\frac{n}{p_{i_1}p_{i_2}}$ such numbers. We continue with this PIE argument to figure out that the number of elements in the complement of what we want is
$$\sum_{1 \le i \le m}\frac{n}{p_i} - \sum_{1 \le i_1 < i_2 \le m}\frac{n}{p_{i_1}p_{i_2}} + \cdots + (-1)^{m+1}\frac{n}{p_1p_2\ldots p_m}.$$
This sum represents the number of numbers less than $n$ sharing a common factor with $n$, so
$\phi(n) = n - \left(\sum_{1 \le i \le m}\frac{n}{p_i}- \sum_{1 \le i_1 < i_2 \le m}\frac{n}{p_{i_1}p_{i_2}} + \cdots + (-1)^{m+1}\frac{n}{p_1p_2\ldots p_m}\right)$
$\phi(n)= n\left(1 - \sum_{1 \le i \le m}\frac{1}{p_i} + \sum_{1 \le i_1 < i_2 \le m}\frac{1}{p_{i_1}p_{i_2}} - \cdots + (-1)^{m}\frac{1}{p_1p_2\ldots p_m}\right)$
$\phi(n)= n\left(1-\frac{1}{p_1} \right) \left(1-\frac{1}{p_2} \right)\cdots \left(1-\frac{1}{p_m}\right).$
Given the general prime factorization of ${n} = {p}_1^{e_1}{p}_2^{e_2} \cdots {p}_m^{e_m}$, one can compute $\phi(n)$ using the formula $$\phi(n)= n\left(1-\frac{1}{p_1} \right) \left(1-\frac{1}{p_2} \right)\cdots \left(1-\frac{1}{p_m}\right).$$
• Note: Another way to find the closed form for $\phi(n)$ is to show that the function is multiplicative, and then breaking up $\phi(n)$ into it's prime factorization.
## Identities
(a) For prime $p$, $\phi(p)=p-1$, because all numbers less than ${p}$ are relatively prime to it.
(b) For relatively prime ${a}, {b}$, $\phi{(a)}\phi{(b)} = \phi{(ab)}$.
(c) In fact, we also have for any ${a}, {b}$ that $\phi{(a)}\phi{(b)}\gcd(a,b)=\phi{(ab)}\phi({\gcd(a,b)})$.
(d) If $p$ is prime and $n\ge{1}$,then $\phi(p^n)=p^n-p^{n-1}$
(e) For any $n$, we have $\sum_{d|n}\phi(d)=n$ where the sum is taken over all divisors d of $n$.
Proof. Split the set $\{1,2,\ldots,n\}$ into disjoint sets $A_d$ where for all $d\mid n$ we have $$A_d=\{x:1\leq x\leq n\quad\text{and}\quad \operatorname{syt}(x,n)=d \}.$$ Now $\operatorname{gcd}(dx,n)=d$ if and only if $\operatorname{gcd}(x,n/d)=1$. Furthermore, $1\leq dx\leq n$ if and only if $1\leq x\leq n/d$. Now one can see that the number of elements of $A_d$ equals the number of elements of $$A_d^\prime=\{x:1\leq x \leq n/d\quad\text{and}\quad \operatorname{gcd}(x,n/d)=1 \}.$$ Thus by the definition of Euler's phi we have that $|A_d^\prime|=\phi (n/d)$. As every integer $i$ which satisfies $1\leq i\leq n$ belongs in exactly one of the sets $A_d$, we have that $$n=\sum_{d \mid n}\varphi \left (\frac{n}{d} \right )=\sum_{d \mid n}\phi (d).$$
## Notation
Sometimes, instead of $\phi$, $\varphi$ is used. This variation of the Greek letter phi is common in textbooks, and is standard usage on the English Wikipedia<|endoftext|>
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# What is the phrase”exclamation mark” in mathematics?
The answer is straightforward. Below are a few strategies to tell when an equation can be a multiplication equation and also just how to add and subtract by using”exclamation marks”. For instance,”2x + 3″ mean”multiply two integers by 3, building a price corresponding to two plus several”.
For instance,”2x + 3″ are multiplication from professional essay just three. Additionally, we could add the values of 2 and three . To bring the values, we’ll use”e”that I” (or”E”). Using”I” indicates”include the value of one to the value of 2″.
To add the values, we can do this similar to this:”x – y” implies”multiply x by y, building a price equal to zero”. For”x”y”, we will use”t” (or”TE”) for the subtraction and we will utilize”x + y” to address the equation.
You might feel which you are not assumed to utilize”electronic” in addition as”that I” implies”subtract” but it really masterpapers is not really easy. By way of instance, to say”two – 3″ means subtract from three.
Thus, to bring the values we utilize”t”x” (or”TE”), which might be the numbers of the worth to be included. We will utilize”x” to subtract the worth of a person in the worth of 2 and that will give us exactly the result.
To multiply the worth we certainly can do it similar to that:”2x + y” indicate”multiply two integers with y, creating a value equal to one plus two”. You will know that this is actually a multiplication equation when we utilize”x” to subtract one from 2. Orit can be”x – y” to subtract one from 2. Note that the equation can be written by you using a decimal level and parentheses.
Now, let’s have a good illustration. Let us mention that people would like to multiply the worth of”nine” by”5″ and we have”x = nine”y = five”. https://wtamu.edu/~cbaird/sq/category/physics/ Afterward we will use”x – y” to reevaluate the value of one from the worth of two.<|endoftext|>
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# 2018 AMC 10A Problems/Problem 14
## Problem
What is the greatest integer less than or equal to $$\frac{3^{100}+2^{100}}{3^{96}+2^{96}}?$$
$\textbf{(A) }80\qquad \textbf{(B) }81 \qquad \textbf{(C) }96 \qquad \textbf{(D) }97 \qquad \textbf{(E) }625\qquad$
## Solution 1
We write $$\frac{3^{100}+2^{100}}{3^{96}+2^{96}}=\frac{3^{96}}{3^{96}+2^{96}}\cdot\frac{3^{100}}{3^{96}}+\frac{2^{96}}{3^{96}+2^{96}}\cdot\frac{2^{100}}{2^{96}}=\frac{3^{96}}{3^{96}+2^{96}}\cdot 81+\frac{2^{96}}{3^{96}+2^{96}}\cdot 16.$$ Hence we see that our number is a weighted average of 81 and 16, extremely heavily weighted toward 81. Hence the number is ever so slightly less than 81, so the answer is $\boxed{\textbf{(A) }80}$.
## Solution 2
Let's set this value equal to $x$. We can write $$\frac{3^{100}+2^{100}}{3^{96}+2^{96}}=x.$$ Multiplying by $3^{96}+2^{96}$ on both sides, we get $$3^{100}+2^{100}=x(3^{96}+2^{96}).$$ Now let's take a look at the answer choices. We notice that $81$, choice $B$, can be written as $3^4$. Plugging this into our equation above, we get $$3^{100}+2^{100} \stackrel{?}{=} 3^4(3^{96}+2^{96}) \Rightarrow 3^{100}+2^{100} \stackrel{?}{=} 3^{100}+3^4\cdot 2^{96}.$$ The right side is larger than the left side because $$2^{100} \leq 2^{96}\cdot 3^4.$$ This means that our original value, $x$, must be less than $81$. The only answer that is less than $81$ is $80$ so our answer is $\boxed{A}$.
~Nivek
## Solution 3
$\frac{3^{100}+2^{100}}{3^{96}+2^{96}}=\frac{2^{96}(\frac{3^{100}}{2^{96}})+2^{96}(2^{4})}{2^{96}(\frac{3}{2})^{96}+2^{96}(1)}=\frac{\frac{3^{100}}{2^{96}}+2^{4}}{(\frac{3}{2})^{96}+1}=\frac{\frac{3^{100}}{2^{100}}*2^{4}+2^{4}}{(\frac{3}{2})^{96}+1}=\frac{2^{4}(\frac{3^{100}}{2^{100}}+1)}{(\frac{3}{2})^{96}+1}$.
We can ignore the 1's on the end because they won't really affect the fraction. So, the answer is very very very close but less than the new fraction.
$\frac{2^{4}(\frac{3^{100}}{2^{100}}+1)}{(\frac{3}{2})^{96}+1}<\frac{2^{4}(\frac{3^{100}}{2^{100}})}{(\frac{3}{2})^{96}}$
$\frac{2^{4}(\frac{3^{100}}{2^{100}})}{(\frac{3}{2})^{96}}=\frac{3^{4}}{2^{4}}*2^{4}=3^{4}=81$
So, our final answer is very close but not quite 81, and therefore the greatest integer less than the number is $\boxed{(A) 80}$
## Solution 4
Let $x=3^{96}$ and $y=2^{96}$. Then our fraction can be written as $\frac{81x+16y}{x+y}=\frac{16x+16y}{x+y}+\frac{65x}{x+y}=16+\frac{65x}{x+y}$. Notice that $\frac{65x}{x+y}<\frac{65x}{x}=65$. So , $16+\frac{65x}{x+y}<16+65=81$. And our only answer choice less than 81 is $\boxed{(A) 80}$ (RegularHexagon)
## Solution 5
Let $x=\frac{3^{100}+2^{100}}{3^{96}+2^{96}}$. Multiply both sides by $(3^{96}+2^{96})$, and expand. Rearranging the terms, we get $3^{96}(3^4-x)+2^{96}(2^4-x)=0$. The left side is decreasing, and it is negative when $x=81$. This means that the answer must be less than $81$; therefore the answer is $\boxed{(A)}$.
## Solution 6 (eyeball it)
A faster solution. Recognize that for exponents of this size $3^{n}$ will be enormously greater than $2^{n}$, so the terms involving $2$ will actually have very little effect on the quotient. Now we know the answer will be very close to $81$.
Notice that the terms being added on to the top and bottom are in the ratio $\frac{1}{16}$ with each other, so they must pull the ratio down from 81 very slightly. (In the same way that a new test score lower than your current cumulative grade always must pull that grade downward.) Answer: $\boxed{(A)}$.
## Solution 7
Notice how $\frac{3^{100}+2^{100}}{3^{96}+2^{96}}$ can be rewritten as $\frac{81(3^{96})+16(2^{96})}{3^{96}+2^{96}}=\frac{81(3^{96})+81(2^{96})}{3^{96}+2^{96}}-\frac{65(2^{96})}{3^{96}+2^{96}}=81-\frac{65(2^{96})}{3^{96}+2^{96}}$. Note that $\frac{65(2^{96})}{3^{96}+2^{96}}<1$, so the greatest integer less than or equal to $\frac{3^{100}+2^{100}}{3^{96}+2^{96}}$ is $80$ or $\boxed{\textbf{(A)}}$ ~blitzkrieg21
## Solution 8
For positive $a, b, c, d$, if $\frac{a}{b}<\frac{c}{d}$ then $\frac{c+a}{d+b}<\frac{c}{d}$. Let $a=2^{100}, b=2^{96}, c=3^{100}, d=3^{96}$. Then $\frac{c}{d}=3^4$. So answer is less than 81, which leaves only one choice, 80.
• Note that the algebra here is synonymous to the explanation given in Solution 6. This is the algebraic reason to the logic of if you get a test score with a lower percentage than your average (no matter how many points/percentage of your total grade it was worth), it will pull your overall grade down.
~ ccx09
## Solution 9
Try long division, and notice putting $3^4=81$ as the denominator is too big and putting $3^4-1=80$ is too small. So we know that the answer is between $80$ and $81$, yielding $80$ as our answer.
## Solution 10 (Using the answer choices)
### Solution 10.1
We can compare the given value to each of our answer choices. We already know that it is greater than $80$ because otherwise there would have been a smaller answer, so we move onto $81$. We get:
$\frac{3^{100}+2^{100}}{3^{96}+2^{96}} \text{ ? } 3^4$
Cross multiply to get:
$3^{100}+2^{100} \text{ ? }3^{100}+(2^{96})(3^4)$
Cancel out $3^{100}$ and divide by $2^{96}$ to get $2^{4} \text{ ? }3^4$. We know that $2^4 < 3^4$, which means the expression is less than $81$ so the answer is $\boxed{(A)}$.
### Solution 10.2
We know this will be between 16 and 81 because $\frac{3^{100}}{3^{96}} = 3^4 = 81$ and $\frac{2^{100}}{2^{96}} = 2^4 = 16$. $80=\boxed{(A)}$ is the only option choice in this range.<|endoftext|>
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# If (2n + 1) + (2n + 3) + (2n + 5) + ...... + (2n + 47) = 5280, then what is the value of 1 + 2 + 3 + .... + n?
## Answer (Detailed Solution Below) 4851
Free
2032
12 Questions 36 Marks 20 Mins
## Detailed Solution
Calculation:
(2n + 1) + (2n + 3) + (2n + 5) + ...... + (2n + 47) = 5280
(2n + 5) – (2n + 3) = 2
(2n + 3) – (2n + 1) = 2
On the Left Hand Side, (2n + 1) + (2n + 3) + (2n + 5) + ...... + (2n + 47) is an arithmetic progression with common difference 2
First term of the arithmetic progression is (2n + 1) and the last term is (2n + 47)
Let the number of terms be T
(2n + 47) = (2n + 1) + (T – 1)2
⇒ 2T – 2 = 2n + 47 – 2n – 1
⇒ 2T = 46 + 2
⇒ 2T = 48
⇒ T = 24
Number of terms of the arithmetic progression = 24
Sum of arithmetic progression = 24/2[(2n + 1) + (2n + 1) + (24 – 1)2]
⇒ 24/2[(2n + 1) + (2n + 1) + (24 – 1)2] = 5280
⇒ 24/2[4n + 2 + 23 × 2] = 5280
⇒ 24/2[4n + 48] = 5280
⇒ 48[n + 12] = 5280
⇒ n + 12 = 110
⇒ n = 110 – 12
⇒ n = 98
The value of 1 + 2 + 3 + .... + n = 1 + 2 + 3 + .... + 98
⇒ 98(98 + 1)/2
⇒ 49 × 99
⇒ 4851
∴ The value of 1 + 2 + 3 + .... + n is 4851<|endoftext|>
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# What’s the Probability HSS.MD.B.5.&7.
After introducing a new topic in mathematics, students find it difficult to take what they have learned into practice. However, keeping students interested, active, and engaged in different activities makes a significant difference in their learning experience.
Students will come up with measures of chance. One of the questions that they can ask themselves is, “how can I quantify how likely an event is?” In this case, teachers can introduce this classic activity, using a standard deck of cards. Using a deck of cards provides a concrete look at probability and chance in a hands-on math activity. A typical deck of cards has four suits of thirteen cards in each suit, twelve face cards, four aces, twenty-six red cards and twenty-six black cards. Considering this, different probability questions can be asked to practice using this concept.
After introducing probability to your students, you can incorporate this activity within your lesson. Based on what they know about a standard deck of cards, students can answer questions, for instance, if you select one card randomly, what is the probability it is a heart?
CCSS.MATH.CONTENT.HSS.MD.B.5
(+) Weigh the possible outcomes of a decision by assigning probabilities to payoff values and finding expected values.
CCSS.MATH.CONTENT.HSS.MD.B.7
(+) Analyze decisions and strategies using probability concepts (e.g., product testing, medical testing, pulling a hockey goalie at the end of a game).
Allowing students to use the deck of cards to answer probability questions will help them reinforce their understanding. They will analyze and find strategies to know the probability of the specific card. This activity allows students to interact with their peers and reinforce their mathematical thinking of finding the probability of a certain card.
## One thought on “What’s the Probability HSS.MD.B.5.&7.”
1. lonesc says:
Hello,
For your picture problem, because your chosen Common Core Standard and the concept of cards both cover such a wide spectrum, you can really dive into all realms of probability. For instance, taking a game of poker: If a student has two “fours” and two “sevens”, what is the probability that they will get a “full house” given that one “four” is already in the discard pile?
From your example and the one I just provided, the concept of card probability can become increasingly more intricate as more factors are introduced. Thank you.<|endoftext|>
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The significance of our ocean’s impact on greenhouse gas begins with the earliest ocean four billion years ago, when all the atmospheric carbon was absorbed and allowed the earth to cool enough for life to begin. In our modern era, as atmospheric carbon dioxide levels go up, the ocean absorbs more carbon dioxide to stay in balance. Currently the ocean is holding 50 times more carbon than the atmosphere and is slowing the rate of climate change by absorbing about 30 percent of carbon dioxide from cement production and other activities.
But there’s an important side effect. When carbon dioxide reacts with seawater, carbonic acid is formed—the same weak acid found in soda. Carbonic acid releases positively charged hydrogen ions, which lowers the pH level of seawater. Although ocean water is still alkaline, the term “acidification” refers to a gradual shift toward the acidic end of the scale. The pH scale ranges from 0 to 14; 7 is neutral, lower numbers are acidic and higher numbers are alkaline. Over the past 300 million years, ocean pH has averaged about 8.2, but since preindustrial times it has dropped 0.1 to a current average of 8.1. This may not sound like much, but the pH scale is logarithmic, so that a pH of 7 is about ten times more acidic than a pH of 8. Thus, this drop represents a 25-percent increase in acidity over the past two centuries.
There are several reactions that occur between carbon dioxide (CO2), water (H2O), carbonic acid (H2CO3), bicarbonate ion (HCO3-), and carbonate ion (CO32-). But over the long term ocean acidification leads to a decrease in the concentration of carbonate ions in seawater. Together with calcium ions they form the basic building blocks of carbonate skeletons and shells. The decline of carbonate ions impacts the ability of many marine organisms such as corals, marine plankton, and shellfish to build or even maintain their shells.
There are two main forms of calcium carbonate used by marine creatures: calcite and aragonite. Calcite is used by phytoplankton, foraminifera, and coccolithophore algae. Aragonite is used by corals, shellfish, pteropods, and heteropods. When additional calcite and aragonite cannot be dissolved in water, that water is said to be supersaturated; when they can be dissolved, the water is said to be undersaturated for those minerals. Animals that need calcium carbonate are better in supersaturated water as obtaining that mineral from the surrounding water is easier. The saturation of these minerals in seawater decreases with depth, and the transition point between supersaturated and undersaturated conditions is referred to as the saturation horizon. Because aragonite dissolves more easily than calcite, aragonite is the first to be impacted by ocean acidification. For example, one might find the saturation horizon for calcite at 150 meters or even deeper, but for aragonite it would be at 100 meters.
Currently, nearly all of the surface ocean waters are substantially supersaturated with regard to aragonite and calcite. However, more carbon dioxide dissolving in the ocean has caused the saturation horizon for these minerals to shift closer to the surface by 50-200 meters as compared to the 1800s. As the ocean becomes more acidic, the upper shell-friendly layer becomes thinner.
The saturation horizon is much closer to the surface in regions where upwelling occurs, such as along the West Coast from British Columbia to Mexico. That’s because deep water in the North Pacific is naturally rich in CO2, since the deep water has been out of contact with the surface for 1200 to 1500 years. As water travels along the oceanic conveyer belt, it accumulates CO2 through natural respiration processes that break down sinking organic matter, generating CO2 just as humans do when they breathe. Under normal conditions (and even more so under La Niña conditions), winds blow from north to south during spring and summer months along the West Coast creating an effect known as the Ekman Transport, which in turn moves surface water away from the coastline. This warm surface water is then replaced by colder water upwelled from depths between 100 and 300 meters. This deep nutrient-rich water traditionally makes for robust fishery production. However, the naturally highly acidic water is now augmented with man-made carbon dioxide, making this carbon-rich water even more acidic.
In the Northeastern Pacific, corrosive waters are already shoaling into the euphotic zone during upwelling. In 2007 Richard Feely and a team of scientists found undersaturated seawater with respect to aragonite reaching depths of about 40 to 120 meters. In one transect less than 20 miles from shore near the California-Oregon border, the saturation horizon had shoaled all the way to the surface. Without the contribution of anthropogenic CO2, the aragonite saturation horizon would be about 50 meters deeper.
Because of this, the Pacific Northwest oyster-growing industry nearly collapsed before Feely and other scientists were able to help devise strategies and monitoring protocols. In 2007, the Whiskey Creek Shellfish Hatchery in Oregon lost millions of oyster larvae and later discovered that the larvae were being bathed in acidic waters drawn in by intake pipes. Oyster larvae are particularly sensitive in their first few days of life such that carbon dioxide alters shell formation rates, energy usage and, ultimately, their growth and survival. Now, the Whiskey Creek hatchery tries to balance the acidity of its waters by adding soda ash. Costs have increased and production has never fully recovered.
Island Scallops, a shellfish producer in the Georgia Straight near Vancouver, lost all its scallops over a 3-year period from 2010 to 2012, during which time pH levels had dipped to 7.3. CEO Rob Saunders said that this level of pH in the water was something he hadn’t seen in his 35 years of shellfish farming. The loss amounted to 10 million dollars and a third of their workforce (20 people).
No less troubling is the impact of acidification on the food chain. This year, a NOAA-led research team found evidence that acidity off the West Coast has been dissolving the shells of pteropods at double the rate since the pre-industrial era. These tiny free-swimming marine snails make up 45 percent of the diet of pink salmon and are also a food source for herring and mackerel. The highest percentage of sampled pteropods with dissolving shells were found from northern Washington to central California, where 53 percent had severely dissolved shells.
Compared to Southern California, the water north of Point Conception comes from deeper depths up to the surface where the upwelling is stronger and lasts longer—from spring to fall. Because the direction of the coastline changes (from north-south to east-west) south of Point Conception, a weaker upwelling occurs in Southern California from February to May. The same NOAA-led research team found evidence of corrosive waters shoaling to depths of about 20-50 meters in the coastal waters off Washington, Oregon, and northern California; and to depths of about 60-120 meters off southern California.
This has been the pattern for at least the past ten years, according to Dr. Anita Leinweber, a researcher at UCLA who has been measuring acidification levels in Santa Monica Bay for more than a decade. In a 2013 study, Leinweber and co-author Nicolas Gruber published six-year trends for pH and the aragonite saturation state in the Santa Monica Bay from 2003 to 2008. They found that the saturation horizon there reaches 130 meters on average. As the aragonite saturation state changes, this shoaling is exacerbated and the horizon could climb 20 meters by 2050 to reach an average depth of 110 meters.
Median trends for pH are also decreasing by an average of about -0.004 per year between 100 and 250 meters. These trends in Santa Monica Bay are larger in magnitude than most of those reported so far — for example, about -0.003 pH units per year in Monterey Bay. They are also slightly larger than those expected on the basis of the recent trends in atmospheric CO2. The study also noted that the saturation horizon reached its highest point—the top 30 meters—during the height of the upwelling season in April and May. Lower pH and aragonite saturation states were observed during winters when La Niña conditions prevailed, which makes sense given that La Niña conditions intensify upwelling conditions.
At the time of the study’s publication, no statistically significant linear trends had emerged in the upper 100 meters. But this past April, Leinwebier saw a statistically significant trend in surface pH, calculated from additional data which extended to 2013. The pH values in the top meter had been decreasing by about 0.003 per year. (Calculations were not yet complete for other depths.)
“We have evidence already that ocean acidification is happening,” Leinweber said. “It’s not something that we’re making up or something that we know has to come at some point. We actually see it.”
As alluded to above, the effects of ocean acidification will vary with location. For its part, the Catalina Marine Society is acquiring pH and other ocean chemistry data at specific depths near Santa Catalina Island with its depth-profiling program. These data will enable us to determine where the water comes from, where the phytoplankton reside, the acidity of Santa Catalina water, and how much oxygen is available to marine fauna. The goal is to obtain sufficiently dense records to compare to similar data collected in Los Angeles Harbor (by the Southern California Marine Institute), Point Loma (Scripps Institution of Oceanography), Santa Monica Bay, and off the Santa Barbara coast (University of California, Santa Barbara), as well as understand the physical processes operating around the island and develop expectations for what climate change and ocean acidification will bring. The new data will supplement what was previously collected at a single depth (18 m) near Two Harbors, Catalina. Those data, which include pH levels from May 19, 2012 through November 17, 2013, are available to researchers and students at http://www.catalinamarinesociety.org/Scientificmooring.html
Originally published in Catalina Marine Society’s OceanBights, p. 3.
Intergovernmental Panel on Climate Change, Climate Change 2013: The Physical Science Basis, Chapter 3, Observation: Ocean
pH Scale, Introduction and Definitions
European Project on Ocean Acidification, FAQs about Ocean Acidification, Ocean carbon chemistry and pH
Skeptical Science, Ocean acidification: global warming’s evil twin
Nicolas Gruber, Claudine Hauri, Zouhair Lachkar, Damian Loher, Thomas L. Frölicher, Gian‐Kasper Plattner; Rapid Progression of Ocean Acidification in the California Current System
NOAA OAR Special Report, Scientific Summary of Ocean Acidification in Washington State Marine Waters
Richard A. Feely, Christopher L. Sabine, J. Martin Hernandez-Ayon, Debby Ianson, Burke Hales; Evidence for upwelling of corrosive “acidified” water onto the Continental Shelf
Kenneth R. Weiss, Los Angeles Times; Oceans’ rising acidity a threat to shellfish — and humans
N. Bednaršek, R. A. Feely, J. C. P. Reum, B. Peterson, J. Menkel, S. R. Alin and B. Hales; Limacina helicina shell dissolution as an indicator of declining habitat suitability owing to ocean acidification in the California Current Ecosystem
Anita Leinweber, Institute of Geophysics and Planetary Physics and Department of Atmospheric and Oceanic Sciences, University of California, Los Angeles, California, interview May 17, 2014.
Catalina Marine Society, Measuring Catalina’s Ocean, Flyer for Boaters
- The shell pictured here is a victim of acidification: NOAA, Ocean Acidification
- Servicing the Santa Monica Bay Observatory mooring’s antenna: Anita Leinweber
- Distribution of the depths of the undersaturated water: Richard Feeley, Evidence for upwelling of corrosive “acidified” water onto the Continental Shelf
- Anita Leinweber on board the R/V Seaworld.: Anita Leinweber
- Dr. Leinweber and Takeyoshi Nagai attaching water-sampling bottle<|endoftext|>
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Populations of dioecious flowering plants (which have male and female individuals) often depart from the expected male:female ratio of 1:1. The causes of skewed sex ratios are complex and still poorly understood. As with many species that have two sexes, females must invest more resources in reproduction. In the case of flowering plants, this is the extra cost associated with flowering and subsequent fruit production. Consequently, females often delay flowering and are more susceptible to environmental stresses, suffering higher mortality rates compared with males. These factors can lead to a greater than equal representation of males in the population.
A recent AOB study attempts to untangle these complex causes of biased gender ratios. This meta-analysis of existing studies of dioecious flowering plant species found a large amount of variation in sex ratios between populations within a species. There was also much variation between species, although several patterns have emerged.
Due to delayed and less frequent flowering by females, younger populations tend to have a greater bias towards males. This was demonstrated elegantly by looking at fire-adapted species, which recolonise an area following a fire, so that the age of the population can be exactly known. Males are also more frequent in populations at higher altitudes, reflecting greater mortality of females as the environment becomes more stressful. Chance historical effects can also play an important role in skewing the sex ratio. For example, in clonal species, where a few individuals colonise an area and largely reproduce vegetatively, chance biases in the representation of the sexes can be preserved over long periods.
You can read the study, free, at the Annals of Botany.
Ecological context and metapopulation dynamics affect sex-ratio variation among dioecious plant populations (2013) Ann Bot 111(5): 917-923. doi:10.1093/aob/mct040<|endoftext|>
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Understanding simple shapes can give children an early advantage in many areas of learning. Shape recognition can help your child excel in literacy, science, information visualization, problem solving, math, and more. There are many ways to teach your child about shapes, but why not bring the fun outside? Turn your ordinary playground visit or neighborhood walk into a learning moment with these five shape recognition activities.
Head out on a shape scavenger hunt
Turn the playground fun into an education activity with a shape scavenger hunt. Compile a list of shapes (circles, squares, rectangles, triangles) and set out with your little one to identify all of the shapes on the playground equipment. Alternatively, you can take photos of the playground equipment, print them out, and then ask your child to identify the types of shapes he or she sees (check out Buggy and Buddy for instructions!).
Play a find & erase game with chalk
If you want an activity that will teach shape recognition and work on fine motor skills, this find and erase game is perfect! Grab the chalk, paint brush, and a bowl of water and head outside. Draw different shapes with the chalk and ask your child to “erase” the shape by painting over it with water. This will help them count the number of sides and learn the shape name! Credit: Hands On As We Grow
Build shapes with acorns or rocks
Does your little one love to build? Head out to a sand box (or the beach if you’re close) and build shapes together in the sand. Trace the shape in the sand and ask your child to fill in the shape with acorns, rocks, blocks, sticks, or anything they can find! Credit: Creative Little Explorers
Play a driveway maze game
This maze game is great for older kids as it works on their problem solving and gross motor skills. Draw different shapes in different colors all over the driveway or sidewalk. Ask your child to start in a square and tell them they must make it to the other side by only stepping on shapes. Start over by picking a different shape! Credit: Creative Family Fun
Build a DIY shape finder
This DIY shape finder is great because it brings learning on-the-go! All you need is a folder, marker, and a scissors. Draw various shapes all over the folder, cut them out, and hand off to your child. They can bring this to the grocery store, play ground, activity center, library, or on vacation. When they find a shape (like a stop sign), ask them to look through the holes until they can figure out which shape they are looking at. Credit: Playdough to Plato<|endoftext|>
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Students will find and discuss examples of philanthropy in poems and quotations. They will define and design statements on the theme of philanthropy using the poetic conventions of metaphor, simile, and personification. These statements could be used as the text for greeting cards...
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Unit: Giving from the Heart
Unit: Funding the Arts
Students learn how nonprofit organizations contribute to the common good by supporting the arts. They role-play as members of the board of a foundation focused on the Arts who are making a funding recommendation to bring the Arts to the children of the community.
Unit: What Does It Take?
Unit: Art as Advocacy
The learners will view works of art that advocate for social change. They will recognize that art can influence social change. The learners will select an issue of human rights and create a work of art that represents the issue. They will write a paragraph of explanation about their...
Unit: Courage of the Heart
Students learn about the scientific contributions of African American inventors and scientists. They view a film about Vivien Thomas and a breakthrough surgery in 1944.
Unit: Action through Art
The learners read stories written and illustrated by teens about action and advocacy to make the world a better place. The students brainstorm concerns/issues they have about their school or local community and create story outlines about ways to address an issue.
Students use visual literacy skills to talk about an artistic image. They listen respectfully to the different opinions and perspectives of their classmates. The students identify a need at their school and create a simple image that tells others to think differently or take action to improve...
Learners collect, organize and interpret data as to the frequency of philanthropic actions performed by their families to determine the level of philanthropy within a sample of families within the school community.<|endoftext|>
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# Precalculus identities solver
This Precalculus identities solver supplies step-by-step instructions for solving all math troubles. Keep reading to learn more!
## The Best Precalculus identities solver
Precalculus identities solver can be found online or in mathematical textbooks. A rational function is any function which can be expressed as the quotient of two polynomials. In other words, it is a fraction whose numerator and denominator are both polynomials. The simplest example of a rational function is a linear function, which has the form f(x)=mx+b. More generally, a rational function can have any degree; that is, the highest power of x in the numerator and denominator can be any number. To solve a rational function, we must first determine its roots. A root is a value of x for which the numerator equals zero. Therefore, to solve a rational function, we set the numerator equal to zero and solve for x. Once we have determined the roots of the function, we can use them to find its asymptotes. An asymptote is a line which the graph of the function approaches but never crosses. A rational function can have horizontal, vertical, or slant asymptotes, depending on its roots. To find a horizontal asymptote, we take the limit of the function as x approaches infinity; that is, we let x get very large and see what happens to the value of the function. Similarly, to find a vertical asymptote, we take the limit of the function as x approaches zero. Finally, to find a slant asymptote, we take the limit of the function as x approaches one of its roots. Once we have determined all of these features of the graph, we can sketch it on a coordinate plane.
It also has many different examples of each type of problem to help you learn how to solve them. You can use this app to practice solving word problems in any subject area. It's really a great tool for anyone who needs to learn how to solve word problems. So if you're looking for an app that can help you learn how to solve word problems, Solve Word Problems app is definitely worth checking out!
There are a number of ways to approach solving Pre calculus problems. One way is to use a problem solver. This is a tool that can help you to work through the problem and find the solution. There are a number of different problem solvers available, so you will need to find one that suits your needs. Another way to approach solving Pre calculus problems is to work through the problem yourself. This can be done by using a variety of resources, such as textbooks, online resources,
Solving for a side in a right triangle can be done using the Pythagorean theorem. This theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides. Using this theorem, it is possible to solve for any side in a right triangle given the length of the other two sides. For example, if the length of one side is 3 and the length of the other side is 4, then the hypotenuse must be 5, since 3^2 + 4^2 = 25. In order to solve for a side, all you need is the lengths of the other two sides and a calculator. However, it is also possible to estimate the length of a side without using a calculator. For example, if you know that one side is 10 and the other side is 8, you can estimate that the hypotenuse is 12 since 8^2 + 10^2 = approximately 144. Solving for a side in a right triangle is a simple matter as long as you know the Pythagorean theorem.<|endoftext|>
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$\newcommand{\dollar}{\} \DeclareMathOperator{\erf}{erf} \DeclareMathOperator{\arctanh}{arctanh} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&}$
## Section2.7The Chain Rule
###### Motivating Questions
• What is a composite function and how do we recognize its structure algebraically?
• Given a composite function $C(x) = f(g(x))$ that is built from differentiable functions $f$ and $g\text{,}$ how do we compute $C'(x)$ in terms of $f\text{,}$ $g\text{,}$ $f'\text{,}$ and $g'\text{?}$ What is the statement of the Chain Rule?
In this section we will consider composite functions.
### SubsectionComposition of Functions
We encounter composite functions in the real world every day. As an example, suppose you and some friends are driving from Lincoln, NE to Omaha, NE. Once you arrive in Omaha you drive around the city to see all the beautiful sites. We know that Omaha is 60 miles away from Lincoln, and for each hour spent in Omaha, you drive an additional 5 miles. Then the function
\begin{equation*} m(x)=60+5x \end{equation*}
represents the miles traveled if you and your friends spend $x$ hours in Omaha.
Now, suppose gas for your vehicle costs \$0.15 per mile. We can represent this knowledge with the function
\begin{equation*} c(m)=.15m \end{equation*}
where $m$ is in miles. If we want to know how much you and your friends will spend on gas during your trip if you spend $x$ hours in Omaha, we can now use these two functions we created. If you spend $x$ hours in Omaha, you will have traveled $60+5x$ miles. The cost of gas is $.15m$ where $m$ is miles. Then we need only replace $m$ by $60+5x$ to get
\begin{equation*} .15(60+5x) \end{equation*}
as the cost of gas if you spend $x$ hours in Omaha.
In function notation, we write this as follows:
\begin{equation*} c(f(x))=.15f(x)=.15(60+5x). \end{equation*}
Here we are evaluating the function $c(m)$ at the value $f(x)$ since $f(x)$ is the number of miles traveled. The function $c(f(x))$ is called a composite function, or a composition of functions.
###### Example2.61
Given the functions $f(x)=x^2-2x$ and $g(x)=.5 x\text{,}$ identify the function $f(g(x))$ and find the value of $f(g(1))\text{.}$
Solution
First, let's find the function $f(g(x))\text{.}$ To identify this function, we are plugging in $.5x$ for $x$ in the function $f(x)=x^2-2x\text{.}$ We get the function
\begin{equation*} f(g(x))=f(.5x)=(.5x)^2-2(.5x)=.25x^2-x. \end{equation*}
Now, to find the value $f(g(1))$ we simply evaluate the function $f(g(x))$ at $1\text{.}$ We have
\begin{equation*} f(g(1))=.25(1)^2-1=.25-1=-.75. \end{equation*}
Looking back at our first example, from the functions $m(x)=60+5x$ and $c(m)=.15m$ we created the new function
\begin{equation*} c(m(x))=.15(60+5x). \end{equation*}
Notice that this composition of functions, $c(m(x))\text{,}$ has time in hours as its input and cost as its output.
In general, given two functions $f(x)$ and $g(x)\text{,}$ the composition $f(g(x))$ has the same input as $g \text{,}$ but has the same output as $f \text{.}$
###### Composing Functions
Given two functions $f(x)$ and $g(x)\text{,}$ the composition $f(g(x))$ has the same input as $g \text{,}$ but has the same output as $f \text{.}$
### Subsection
First let us consider why we need another derivative rule.
###### Example2.62
Consider the function
\begin{equation*} f(x) = (x^2+3x)^2\text{,} \end{equation*}
in order to take the derivative we would have to rewrite this as a product of two functions:
\begin{equation*} f(x) = (x^2+3x)^2=(x^2+3x)(x^2+3x)\text{,} \end{equation*}
and then use the product rule to find the derivative
\begin{equation*} f'(x) = (2x+3)(x^2+3x)+(x^2+3x)(2x+3)=2(x^2+3x)(2x+3)\text{.} \end{equation*}
When we have a composite function we could try to rewrite as we did in Example2.62; however, often this is very hard. For example:
\begin{equation*} r(x) = (x^2+3x)^{70}\text{,} \end{equation*}
we would not want to distribute 70 times, or apply the product rule to the product of 70 functions. We need a rule for finding the derivative of composite functions without using algebra to rewrite.
### SubsectionThe Chain Rule
###### Chain Rule
If $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)\text{,}$ then the composite function $C$ defined by $C(x) = f(g(x))$ is differentiable at $x$ and
\begin{equation*} C'(x) = f'(g(x)) g'(x)\text{.} \end{equation*}
As with the product and quotient rules, it is often helpful to think verbally about what the chain rule says: If $C$ is a composite function defined by an outer function $f$ and an inner function $g\text{,}$ then $C'$ is given by the derivative of the outer function evaluated at the inner function, times the derivative of the inner function.
It is helpful to clearly identify the inner function $g$ and outer function $f\text{,}$ compute their derivatives individually, and then put all of the pieces together by the chain rule.
###### Example2.63
Use the chain rule to determine the derivative of the function
\begin{equation*} r(x) = (x^2+3x)^{70}\text{.} \end{equation*}
First we want to identify an inside function. Here the inside function is:
\begin{equation*} \text{inside function=} g(x) = x^2+3x\text{.} \end{equation*}
Once you have found an inside function, what is left is the outside function, here:
\begin{equation*} \text{outside function= }f(x) = x^{70} \text{.} \end{equation*}
Then we apply the chain rule by taking the derivative of the outside function, with the inside unchanged, then multiply by the derivative of the inside:
\begin{equation*} r'(x)=70(x^2+3x)^{69}(2x+3)\text{.} \end{equation*}
###### Example2.64
For each function given below, identify an inner function $g$ and outer function $f$ to write the function in the form $f(g(x))\text{.}$ Determine $f'(x)\text{,}$ $g'(x)\text{,}$ and $f'(g(x))\text{,}$ and then apply the chain rule to determine the derivative of the given function.
1. $\displaystyle h(x) = (2x^2+5x)^4$
2. $\displaystyle g(x) = 3(x^3+4x)^9$
3. $\displaystyle p(x) = \sqrt{ 5x^2+7x+2}$
4. $\displaystyle s(x) = \frac{2}{(x^3+5)^2}$
5. $\displaystyle z(x) = \frac{1}{\sqrt{x^2+4}}$
1. $\displaystyle h'(x) = 4(2x^2+5x)^3(4x+5)\text{.}$
2. $\displaystyle g'(x) = 27(x^3+4x)^8(3x^2+4x)\text{.}$
3. $\displaystyle p'(x) = \frac12(5x^2+7x+2)^{-1/2}(10x+7)=\frac{10x+7}{2\sqrt{5x^2+7x+2}}\text{.}$
4. $\displaystyle s'(x) = -4(x^3+5)^{-3}(3x^2)=\frac{-12x^2}{(x^3+5)^3}\text{.}$
5. $\displaystyle z'(x) = \frac{-1}{2}(x^2+4)^{-3/2}(2x)=\frac{-x}{(x^2+4)^{3/2}}\text{.}$
Solution
1. The outer function is $f(x) = x^4$ while the inner function is $g(x) = 2x^2+5x\text{.}$ We know that
\begin{equation*} f'(x) = 4x^3, g'(x) =4x+5\text{.} \end{equation*}
Hence by the chain rule,
\begin{equation*} h'(x) = f'(g(x))g'(x) = 4(2x^2+5x)^3(4x+5)\text{.} \end{equation*}
2. The outer function is $f(x) = 3x^9$ and the inner function is $g(x) = x^3+4x\text{.}$ We know that
\begin{equation*} f'(x) = 27x^8, g'(x) = 3x^2+4\text{.} \end{equation*}
Thus by the chain rule,
\begin{equation*} h'(x) = f'(g(x))g'(x) = 27(x^3+4x)^8(3x^2+4)\text{.} \end{equation*}
3. The outer function is $f(x) = \sqrt{x}=x^{1/2}$ while the inner function is $g(x) = 5x^2+7x+2\text{.}$ We know that
\begin{equation*} f'(x) = \frac{1}{2}x^{-1/2}, g'(x) = 10x+7\text{.} \end{equation*}
Thus by the chain rule,
\begin{equation*} p'(x) = f'(g(x))g'(x) = \frac12(5x^2+7x+2)^{-1/2}(10x+7)=\frac{10x+7}{2\sqrt{5x^2+7x+2}}\text{.} \end{equation*}
4. The outer function is $\displaystyle f(x) =\frac{2}{x^2}=2x^{-2}$ and the inner function is $g(x) = x^3+5\text{.}$ We know that
\begin{equation*} s'(x) = -4x^{-3}=\frac{-4}{x^3}, g'(x) = 3x^2\text{.} \end{equation*}
Hence by the chain rule,
\begin{equation*} h'(x) = f'(g(x))g'(x) = -4(x^3+5)^{-3}(3x^2)=\frac{-12x^2}{(x^3+5)^3}\text{.} \end{equation*}
5. The outer function is $\displaystyle f(x) = \frac{1}{\sqrt{x}}=x^{-1/2}$ and the inner function is $g(x) = x^2+4\text{.}$ We know that
\begin{equation*} f'(x) = \frac{-1}{2}x^{-3/2}=\frac{-1}{2x^{3/2}}, g'(x) =2x\text{.} \end{equation*}
Hence by the chain rule,
\begin{equation*} z'(x) = f'(g(x))g'(x) = \frac{-1}{2}(x^2+4)^{-3/2}(2x)=\frac{-x}{(x^2+4)^{3/2}}\text{.} \end{equation*}
### SubsectionUsing Multiple Rules Simultaneously
The chain rule now joins the sum, constant multiple, product, and quotient rules in our collection of techniques for finding the derivative of a function through understanding its algebraic structure and the basic functions that constitute it. It takes practice to get comfortable applying multiple rules to differentiate a single function, but using proper notation and taking a few extra steps will help.
###### Example2.65
Find a formula for the derivative of $h(t) = \sqrt{t^2 + 2t}(3t^4+6t)^5\text{.}$
We first observe that $h$ is the product of two functions: $h(t) = a(t) \cdot b(t)\text{,}$ where $a(t) = \sqrt{t^2 + 2t}$ and $b(t) = (3t^4+6t)^5\text{.}$ We will need to use the product rule to differentiate $h\text{.}$ And because $a$ and $b$ are composite functions, we will also need the chain rule. We therefore begin by computing $a'(t)$ and $b'(t)\text{.}$
Writing $a(t) = f(g(t)) = \sqrt{t^2+2t}$ and finding the derivatives of $f$ and $g$ with respect to $t\text{,}$ we have
$f(t) = \sqrt{t}\text{,}$ $g(t) = t^2 + 2t\text{,}$ $f'(t) = \frac{1}{2}t^{-1/2}\text{,}$ $g'(t) = 2t+2\text{,}$ $f'(g(t)) = \frac{1}{2(t^2+2t)^{1/2}}\text{.}$
Thus by the chain rule, it follows that
\begin{equation*} a'(t) = f'(g(t))g'(t) = \frac{1}{2(t^2+2t)^{1/2}} (2t+2)=\frac{t+1}{\sqrt{t^2+2t}}\text{.} \end{equation*}
Turning next to the function $b\text{,}$ we write $b(t) = r(s(t)) = (3t^2+6t)^5$ and find the derivatives of $r$ and $s$ with respect to $t\text{.}$
$r(t) = t^5\text{,}$ $s(t) = 3t^2+6t \text{,}$ $r'(t) = 5t^4\text{,}$ $s'(t) = 6t+6\text{,}$ $r'(s(t)) = 5(3t^2+6t)^4\text{.}$
By the chain rule,
\begin{equation*} \end{equation*}
Now we are finally ready to compute the derivative of the function $h\text{.}$ Recalling that $h(t) = \sqrt{t^2 + 2t}(3t^4+6t)^5\text{,}$ by the product rule we have
\begin{equation*} h'(t) = \frac{d}{dt}\left[\sqrt{t^2 + 2t}\right](3t^4+6t)^5+\sqrt{t^2 + 2t} \frac{d}{dt}\left[(3t^4+6t)^5\right] \text{.} \end{equation*}
From our work above with $a$ and $b\text{,}$ we know the derivatives of $\sqrt{t^2 + 2t}$ and $(3t^4+6t)^5\text{.}$ Therefore
\begin{equation*} h'(t) =\frac{t+1}{\sqrt{t^2+2t}}(3t^4+6t)^5 + \sqrt{t^2 + 2t} 5(3t^2+6t)^4(6t+6) \text{.} \end{equation*}
The above calculation may seem tedious. However, by breaking the function down into small parts and calculating derivatives of those parts separately, we are able to accurately calculate the derivative of the entire function.
###### Example2.66
Differentiate each of the following functions. State the rule(s) you use, label relevant derivatives appropriately, and be sure to clearly identify your overall answer.
1. $p(x) = 4x^2\sqrt{x^6 + 2x}$
2. $\displaystyle m(x) = \frac{x}{(x^2+2)^2}$
3. $\displaystyle h(x) = \frac{\sqrt{3x^2+5x}}{4x+3}$
4. $c(x) = 4(x^4+5x)^7(7x^3+5x)^5$
1. $p'(x) = 8x\sqrt{x^6+2x}+4x^2\frac{3x^5+1}{\sqrt{x^6+2x}}\text{.}$
2. $m'(x) = \frac{(x^2+2)^2-x2(x^2+2)(2x)}{(x^2+2)^4}\text{.}$
3. $m'(x) = \frac{0.5(3x^2+5x)^{-1/2}(6x+5)(4x+3)-\sqrt{3x^2+5x}(4)}{(4x+3)^2}\text{.}$
4. $h'(x) = [28(x^4+5x)^6(4x^3+5)](7x^3+5x)^5+4(x^4+5x)^7[5(7x^3+5x)^4(21x^2+5)]\text{.}$
Solution
1. By the product rule,
\begin{equation*} \displaystyle p'(x) = \frac{d}{dx}\left[4x^2\right]\sqrt{x^6 + 2x}+4x^2\frac{d}{dx}\left[\sqrt{x^6 + 2x}\right]\text{.} \end{equation*}
Using the chain rule to complete the second derivative as
\begin{equation*} \frac{d}{dx}\left[\sqrt{x^6 + 2x}\right]=\frac{1}{2}(x^6+2x)^{-1/2}(6x^5+2)=\frac{3x^5+1}{\sqrt{x^6+2x}}\text{.} \end{equation*}
Then put together, we see that
\begin{equation*} p'(x) = 8x\sqrt{x^6+2x}+4x^2\frac{3x^5+1}{\sqrt{x^6+2x}}\text{.} \end{equation*}
2. Observe that by the quotient rule,
\begin{equation*} \displaystyle m'(x) = \frac{\frac{d}{dx}\left[x\right](x^2+2)^2-x\frac{d}{dx}\left[(x^2+2)^2\right]}{((x^2+2)^2)^2}\text{.} \end{equation*}
Applying the chain rule to differentiate
\begin{equation*} \frac{d}{dx}\left[(x^2+2)^2\right]=2(x^2+2)(2x) \end{equation*}
Then put together, we see that
\begin{equation*} m'(x) = \frac{(x^2+2)^2-x2(x^2+2)(2x)}{(x^2+2)^4}\text{.} \end{equation*}
3. By the quotient rule,
\begin{equation*} h'(y) = \frac{\frac{d}{dx}[\sqrt{3x^2+5x}](4x+3) - \sqrt{3x^2+5x} \frac{d}{dx}[(4x+3)]}{(4x+3)^2}\text{.} \end{equation*}
Applying the chain rule to differentiate
\begin{equation*} \frac{d}{dx}[\sqrt{3x^2+5x}]=\frac12(3x^2+5x)^{-1/2}(6x+5)=\frac{6x+5}{2\sqrt{3x^2+5x}}\text{,} \end{equation*}
it follows that
\begin{equation*} h'(x) =\frac{0.5(3x^2+5x)^{-1/2}(6x+5)(4x+3)-\sqrt{3x^2+5x}(4)}{(4x+3)^2}\text{.} \end{equation*}
4. By the product rule
\begin{equation*} h'(x)=\frac{d}{dx}\left[4(x^4+5x)^7\right](7x^3+5x)^5+4(x^4+5x)^7\frac{d}{dx}\left[(7x^3+5x)^5\right]\text{.} \end{equation*}
Use the chain rule to find each derivative:
\begin{equation*} \frac{d}{dx}\left[4(x^4+5x)^7\right]=28(x^4+5x)^6(4x^3+5) \end{equation*}
and
\begin{equation*} \frac{d}{dx}\left[(7x^3+5x)^5\right]=5(7x^3+5x)^4(21x^2+5)\text{.} \end{equation*}
Then with the product rule, we find that
\begin{equation*} h'(x) = [28(x^4+5x)^6(4x^3+5)](7x^3+5x)^5+4(x^4+5x)^7[5(7x^3+5x)^4(21x^2+5)]\text{.} \end{equation*}
The chain rule now adds substantially to our ability to compute derivatives. Whether we are finding the equation of the tangent line to a curve, the instantaneous velocity of a moving particle, or the instantaneous rate of change of a certain quantity, the chain rule is indispensable if the function under consideration is a composition.
###### Example2.67
Use known derivative rules (including the chain rule) as needed to answer each of the following questions.
1. Find an equation for the tangent line to the curve $y= \sqrt{x^3 +x+ 4}$ at the point where $x=0\text{.}$
2. If $\displaystyle s(t) = \frac{1}{(t^2+1)^3}$ represents the position function of a particle moving horizontally along an axis at time $t$ (where $s$ is measured in inches and $t$ in seconds), find the particle's instantaneous velocity at $t=1\text{.}$ Is the particle moving to the left or right at that instant?6You may assume that this axis is like a number line, with left being the negative direction, and right being the positive direction.
1. $y - 2 = \frac{1}{4}(x-0)\text{.}$
2. $s'(1) = -\frac{3}{8}$ inches per second, so the particle is moving left at the instant $t = 1\text{.}$
Solution
1. Let $f(x) = \sqrt{x^3 + 4}\text{.}$ By the chain rule, $\displaystyle f'(x) = \frac{1}{2} (x^3+x+4)^{-1/2}(3x^2+1)\text{,}$ and thus $f'(0) = \frac{1}{4}\text{.}$ Note further that $f(0) = \sqrt{4} = 2\text{.}$ The tangent line is therefore the line through $(0,2)$ with slope $\frac{1}{4}\text{,}$ which is
\begin{equation*} y - 2 = \frac{1}{4}(x-0)\text{.} \end{equation*}
2. Observe that $s(t) = (t^2 + 1)^{-3}\text{,}$ and thus by the chain rule, $s'(t) = -3(t^2 + 1)^{-4}(2t)\text{.}$ We therefore see that $s'(1) = -\frac{6}{16} = -\frac{3}{8}$ inches per second, so the particle is moving left at the instant $t = 1\text{.}$
### SubsectionSummary
• Given a composite function $C(x) = f(g(x))$ where $f$ and $g$ are differentiable functions, the chain rule tells us that
\begin{equation*} C'(x) = f'(g(x)) g'(x)\text{.} \end{equation*}
### SubsectionExercises
Consider the basic functions $f(x) = x^3$ and $g(x) = \sin(x)\text{.}$
1. Let $h(x) = f(g(x))\text{.}$ Find the exact instantaneous rate of change of $h$ at the point where $x = \frac{\pi}{4}\text{.}$
2. Which function is changing most rapidly at $x = 0.25\text{:}$ $h(x) = f(g(x))$ or $r(x) = g(f(x))\text{?}$ Why?
3. Let $h(x) = f(g(x))$ and $r(x) = g(f(x))\text{.}$ Which of these functions has a derivative that is periodic? Why?
Let functions $p$ and $q$ be the piecewise linear functions given by their respective graphs in Figure2.68. Use the graphs to answer the following questions.
1. Let $C(x) = p(q(x))\text{.}$ Determine $C'(0)$ and $C'(3)\text{.}$
2. Find a value of $x$ for which $C'(x)$ does not exist. Explain your thinking.
3. Let $Y(x) = q(q(x))$ and $Z(x) = q(p(x))\text{.}$ Determine $Y'(-2)$ and $Z'(0)\text{.}$
If a spherical tank of radius 4 feet has $h$ feet of water present in the tank, then the volume of water in the tank is given by the formula
\begin{equation*} V = \frac{\pi}{3} h^2(12-h)\text{.} \end{equation*}
1. At what instantaneous rate is the volume of water in the tank changing with respect to the height of the water at the instant $h = 1\text{?}$ What are the units on this quantity?
2. Now suppose that the height of water in the tank is being regulated by an inflow and outflow (e.g., a faucet and a drain) so that the height of the water at time $t$ is given by the rule $h(t) = \sin(\pi t) + 1\text{,}$ where $t$ is measured in hours (and $h$ is still measured in feet). At what rate is the height of the water changing with respect to time at the instant $t = 2\text{?}$
3. Continuing under the assumptions in (b), at what instantaneous rate is the volume of water in the tank changing with respect to time at the instant $t = 2\text{?}$
4. What are the main differences between the rates found in (a) and (c)? Include a discussion of the relevant units.<|endoftext|>
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Where will the hand of a clock stop if it starts at 2 and makes 1/4 of a revolution clockwise?
Contents
Where will the hand of a clock stop if it starts at 2 and makes 1/2 of a revolution clockwise?
Step-by-step explanation:
If the hand of the clock starts at 12 and makes 1/2 of a revolution clockwise, then it will rotate by 180o and hence, it will stop at 6.
Where will the hand of a clock stop if it starts at 3 and makes 1/4 of a revolution?
it will stop at 3. After 1/4 revolution, It will stop at 6.
Where will the hand of clock stops if it starts at 4 and make 1/4 of revolution?
The hand of the clock will stop at 7 if started from 4 in clockwise direction to make 1/4 rotation.
Where will the hand of a clock stop if kavya starts from 5 and makes ¾ of a rotation anti clock wise?
where will the hand of a clock stop if kavya starts from 5 and make 3/4 of a rotation anti clock wise ? 1 rotation = 360° . so, → (3/4) of rotation = (3/4) * 360° = 270° .
Where will the hand of a clock stop if kavya?
Answer: In one complete revolution the clock hand returns back to the same position. So, if it starts at 12 then it would again stop at 12, if it starts at 2 it would again stop at 2 and if it starts at 5 it would again stop at 5. Let’s understand the concept of turning clockwise and anti-clockwise.
Where will the hand of a clock stop if it starts at 2 and makes 1/4 of a revolution?
If the clock hand starts at 2 and makes 1/2 of a revolution, it would stop after 6 hours, that is, at 8 o’clock. C) If the clock hand starts at 5 and makes 1/4 of a revolution, it would stop after 3 hours, that is, at 8 o’clock.
Where will the hand of a clock stop if it starts at 12 and makes half of a revolution clockwise *?
If the hand of the clock starts at 12 and makes 1/2 of a revolution clockwise, then it will rotate by 180o and hence, it will stop at 6.
How many degrees are there in 2/3 of a right angle?
Therefore 2/3 of a right angle is 60° and it’s complement will be: (3/3[whole]-3/3)*90°.
How many degrees are there in one third of a right angle?
Complementary angle= 90° One third of a right angled triangle is 90°÷3=30° Complement of 30°= 90°-30°=60° Hence, The answer is 60° Hope this helps you.<|endoftext|>
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A non inertial frame of reference is one in which a body violates Newton's Laws of Motion, mainly the First Law. In such a frame, despite no real force acting on a body at rest, it might move; or one that was already moving come at rest or change it's direction of motion. For comparison see an inertial frame. Newton's first and second laws of motion do not hold in non-inertial reference frames. Specifically, masses in non-inertial reference frames appear to feel fictitious forces (such as the Coriolis force or the centrifugal force) that derive from the acceleration of the reference frame itself. Fictitious forces cause apparent accelerations in objects without any physical force causing the acceleration. Fictitious forces are proportional to the mass upon which they act; if such forces are observed, scientists will recognize that they are in a non-inertial reference frame. For example, the rotation of the Earth can be observed from the Coriolis force acting on a Foucault pendulum. An apparent exception would seem to be the force of gravity, which is also proportional to the mass upon which it acts. Although gravity can be considered a "real" physical force for the purposes of calculations in classical mechanics, Albert Einstein showed in his theory of general relativity that gravity itself can also be considered a fictitious force. In his theory, the free-falling reference frame is equivalent to an inertial reference frame (the equivalence principle). By contrast, Einstein noted that observers standing on the Earth are experiencing an unrecognized acceleration from the normal force pushing up on their feet and, thus, are in a non-inertial (accelerated) reference frame. Further details may be found under general relativity.
Non-inertial reference frame - Wikipedia, the free encyclopedia<|endoftext|>
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## If we add 1 to the numerator of a fraction becomes 1/2 and if 5 is subtracted from the denominator, it becomes 1. Find the fraction
Question
If we add 1 to the numerator of a fraction becomes 1/2 and if 5 is subtracted from the denominator, it becomes 1. Find the fraction
in progress 0
2 months 2021-11-23T05:21:57+00:00 2 Answers 0 views 0
1. ## Solution
Given :
• If we add 1 to the numerator of a fraction becomes 1/2
• if 5 is subtracted from the denominator, it becomes 1.
Find :
• These fraction
## Explanation
Let,
• Numerator be = x
• Denominator be = y
A/C to question,
(If we add 1 to the numerator of a fraction becomes 1/2)
==> (x+1)/y = 1/2
==> 2x – y = -2 (1)
Again,
(if 5 is subtracted from the denominator, it becomes 1.)
==> x/(y-5) = 1
==> x – y = -5 (2)
Subtract equ(1) & equ(2)
==> 2x – x = -2 + 5
==> x = 3
Keep Value of x in equ(1)
==> 2 × 3 – y = -2
==> y = 6 + 2
==> y = 8
## Hence
• Fraction will be (x/y) = 3/8
## _________________
Case(1).
(If we add 1 to the numerator of a fraction becomes 1/2)
==> (3+1)/8 = 1/2
==> 4/8 = 1/2
==> 1/2 = 1/2
L.H.S. = R.H.S.
Case(2).
(if 5 is subtracted from the denominator, it becomes 1.)
==> 3/(8-5) = 1
==> 3/3 = 1
==> 1 = 1
L.H.S. = R.H.S.
Thats proved.
## _______________
2. Given :-
If we add 1 to the numerator of a fraction becomes 1/2
if 5 is subtracted from the denominator, it becomes 1.
TO Find :-
These fraction
Explanation
Let,
Numerator be = x
Denominator be = y
A/C to question,
(If we add 1 to the numerator of a fraction becomes 1/2)
==> (x+1)/y = 1/2
==> 2x – y = -2 ——–(1)
Again,
(if 5 is subtracted from the denominator, it becomes 1.)
==> x/(y-5) = 1
==> x – y = -5 ———-(2)
Subtract equ(1) & equ(2)
==> 2x – x = -2 + 5
==> x = 3
Keep Value of x in equation (1)
==> 2 × 3 – y = -2
==> y = 6 + 2
==> y = 8
Hence
Fraction will be (x/y) = 3/8
_________________
Case(1).
(If we add 1 to the numerator of a fraction becomes 1/2)
==> (3+1)/8 = 1/2
==> 4/8 = 1/2
==> 1/2 = 1/2
L.H.S. = R.H.S.
Case(2).
(if 5 is subtracted from the denominator, it becomes 1.)
==> 3/(8-5) = 1
==> 3/3 = 1
==> 1 = 1
L.H.S. = R.H.S.<|endoftext|>
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# Kurtosis
The kurtosis (also known as a measure of aiming) is a statistical measure, which determines the degree of concentration presented by the values of a variable around the central area of the frequency distribution.
When we measure a random variable, in general, the results that have a higher frequency are those that are around the average of the distribution. Imagine the height of the students in a class. If the average height of the class is 1.72, the most normal is that the heights of the rest of the students are around this value (with a certain degree of variability, but without being too large). If this happens, it is considered that the distribution of the random variable is distributed normally. But given the infinity of variables that can be measured, this does not always happen.
There are some variables that have a higher degree of concentration (less dispersion) of the values around their average and others, on the contrary, have a lower degree of concentration (greater dispersion) of their values around their central value. Therefore, the kurtosis informs us of what is targeted (higher concentration) or the flat (lower concentration) that is a distribution.
## Types of kurtosis
Depending on the degree of kurtosis, we have three types of distributions:
1. Leptocurtic:There is a large concentration of values around their average (g 2> 3)
2. Mesocúrtica:There is a normal concentration of the values around its mean (g 2= 3).
3. Platicúrtica:There is a low concentration of the values around its average (g 2<3).
## Kurtosis measures according to data
Depending on the grouping or not of the data, one formula or another is used.
Ungrouped data:
Data grouped in frequency tables:
Data grouped in intervals:
## Example of kurtosis calculation for ungrouped data
Suppose we want to calculate the kurtosis of the following distribution:
8,5,9,10,12,7,2,6,8,9,10,7.7.
First we calculate the arithmetic mean (µ), which would be 7.69.
Next, we calculate the standard deviation, which would be 2.43.
After having this data and for convenience in the calculation, a table can be made to calculate the part of the numerator (fourth moment of distribution). For the first calculation it would be: (Xi-µ) ^ 4 = (8-7.69) ^ 4 = 0.009.
Data (Xi-µ) ^ 4 8 0.0090 5 52.5411 9 2.9243 10 28,3604 12 344.3330 7 0.2297 two 1049.9134 6 8,2020 8 0.0090 9 2.9243 10 28,3604 7 0.2297 7 0.2297 N = 13 ∑ = 1,518.27
Once we have this table made, we would simply have to apply the formula described above to have the kurtosis.
2 = 1,518.27 / 13 * (2.43) ^ 4 = 3.34
In this case given that g 2 is greater than 3, the distribution would be leptocurtic, presenting a greater aim than the normal distribution.
## Excess kurtosis
In some manuals, kurtosis is presented as excess kurtosis. In this case, it compares directly with that of the normal distribution. Since the normal distribution has kurtosis 3, to obtain the excess, we would only have to subtract 3 from our result.
Excess kurtosis = g 2 -3 = 3.34-3 = 0.34.
The interpretation of the result in this case would be as follows:
2 -3> 0 -> leptocortic distribution.
2 -3 = 0 -> meso-curt distribution (or normal).
2 -3 <0 -> platicuric distribution.
##### byAbdullah Sam
I’m a teacher, researcher and writer. I write about study subjects to improve the learning of college and university students. I write top Quality study notes Mostly, Tech, Games, Education, And Solutions/Tips and Tricks. I am a person who helps students to acquire knowledge, competence or virtue.<|endoftext|>
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The triode vacuum tube might be nearly obsolete today, but it was a technology critical to making radio practical over 100 years ago. [Kathy] has put together a video that tells the story and explains the physics of the device.
The first radio receivers used a device called a Coherer as a detector, relying on two tiny filaments that would stick together when RF was applied, allowing current to pass through. It was a device that worked, but not reliably. It was in 1906 that Lee De Forest came up with a detector device for radios using a vacuum tube containing a plate and a heated filament. This device so strongly resembled the Fleming Valve which John Fleming had patented a year before, that Fleming sued De Forest for patent infringement.
After a bunch of attempts to get around the patent, De Forest decided to add a third element to the tube: the grid. The grid is a piece of metal that sits between the filament and the plate. A signal applied to the grid will control the flow of electrons, allowing this device to operate as an amplifier. The modification created the triode, and got around Fleming’s patent.
[Kathy]’s video does a great job of taking you through the creation of the device, which you can watch after the break. She also has a whole series on the history of electricity, including a video on the Arc Transmitter which we featured previously.<|endoftext|>
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Looking for an easy way to make a big difference? Help collect Holocaust newspaper articles printed in your local newspapers for the History Unfolded project of the United States Holocaust Memorial Museum. Do it on your own, or with your local genealogical or historical society!
The following article came to us via Newspapers.com:
What is History Unfolded? History Unfolded is a project that seeks to expand our knowledge of how American newspapers reported on Nazi persecution during the 1930s and ’40s so we can better understand what Americans knew about the Holocaust as it was happening.
To help achieve this, the History Unfolded project asks people like you to search local newspapers from the 1930s and ’40s for Holocaust-related news and opinions and then submit them online to the museum. The newspaper articles you submit will be used to help shape the museum’s 2018 exhibit on Americans and the Holocaust and related educational materials. The articles will also be made available to scholars, historians, and the public.
Who Can Contribute? Everyone! History buffs, students, teachers (with) an interest in the Holocaust and access to a newspaper from the 1930s or ’40s, either online (using Newspapers.com, for example) or through a physical archive, such as a library. Simply create an account with History Unfolded (to get started.)
How Do I Contribute? History Unfolded has created a list of more than 30 Holocaust-related events to focus on. Choose one of these events to research, then search for content related to that topic in an American newspaper of your choice from the 1930s or ’40s. After you find an article related to one of the events, submit it online to the museum through the project’s website.
Newspapers.com and History Unfolded You can contribute to this important project whether or not you use Newspapers.com to do so. But using Newspapers.com makes it even easier to submit the articles you find. Simply use Newspapers.com to create a clipping of an article you’ve found, then submit that clipping through the submission form on the History Unfolded website. The submission form has a special tool created specifically for Newspapers.com users that makes submitting your clipping a snap.
Your help with this project will help shape our understanding of the Holocaust and the lessons it holds for us today. For more information on how to get involved, visit the History Unfolded website.
Get involved! Click here to read about more ways to volunteer in our global genealogy community. Your efforts make a huge difference.<|endoftext|>
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## A copper wire, $3 \mathrm{~mm}$ in diameter, is wound about a cylinder whose length is $12 \mathrm{~cm}$, and diameter $10 \mathrm{~cm}$, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be $8.88 \mathrm{~g} \mathrm{per} \mathrm{cm}^{3}$.
Updated on 10-Oct-2022 13:47:38
Given:A copper wire, $3 \mathrm{~mm}$ in diameter, is wound about a cylinder whose length is $12 \mathrm{~cm}$, and diameter $10 \mathrm{~cm}$, so as to cover the curved surface of the cylinder.The density of copper is $8.88 \mathrm{~g} \mathrm{per} \mathrm{cm}^{3}$.To do:We have to find the length and mass of the wire.Solution:Diameter of the cylinder $(d)= 10\ cm$This implies, Radius of the cylinder $(r) = \frac{10}{2}\ cm$$= 5\ cmLength of the wire one complete round = 2 \pi r$$= 2\times3.14\times5\ cm$$= 31.4\ cmThe diameter of the wire = 3\ mm$$= \frac{3}{10}\ cm$This implies, Radius ... Read More
## Formulate the following problems as a pair of equations, and hence find their solutions:(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately
Updated on 10-Oct-2022 13:47:38
## Solve the following equations and check your results.$2 y+\frac{5}{3}=\frac{26}{3}-y$.
Updated on 10-Oct-2022 13:47:38
Given:$2 y+\frac{5}{3}=\frac{26}{3}-y$To do: We have to solve the given equation and check the result.Solution:$2 y+\frac{5}{3}=\frac{26}{3}-y$$2y+y=\frac{26}{3}-\frac{5}{3}$$3y=\frac{26-5}{3}$$3y=\frac{21}{3}$$3y=7$$y=\frac{7}{3}Substituting the value of y in LHS, we get,2y+\frac{5}{3}=2(\frac{7}{3})+\frac{5}{3}$$=\frac{14}{3}+\frac{5}{3}$$=\frac{14+5}{3}$$=\frac{19}{3}$Substituting the value of $y$ in RHS, we get,$\frac{26}{3}-y=\frac{26}{3}-\frac{7}{3}$$=\frac{26-7}{3}$$=\frac{19}{3}$LHS $=$ RHSThe value of $y$ is $\frac{7}{3}$.
## Solve the following equations and check your results.$\frac{2 x}{3}+1=\frac{7 x}{15}+3$.
Updated on 10-Oct-2022 13:47:38
Given:$\frac{2 x}{3}+1=\frac{7 x}{15}+3$To do: We have to solve the given equation and check the result.Solution:$\frac{2 x}{3}+1=\frac{7 x}{15}+3$$\frac{2 x}{3}-\frac{7 x}{15}=3-1$$\frac{5(2 x)-7x}{15}=2$ (LCM of 3 and 15 is 15)$\frac{10x-7x}{15}=2$$\frac{3 x}{15}=2$$3x=15(2)$$3x=30$$x=\frac{30}{3}$$x=10Substituting the value of x in LHS, we get,\frac{2 x}{3}+1=\frac{2 (10)}{3}+1$$=\frac{20}{3}+1$$=\frac{20+1(3)}{3}$$=\frac{20+3}{3}$$=\frac{23}{3}Substituting the value of x in RHS, we get,\frac{7 x}{15}+3=\frac{7 (10)}{15}+3$$=\frac{70}{15}+3$$=\frac{70+3(15)}{15}$$=\frac{70+45}{15}$$=\frac{115}{15}$$=\frac{23}{3}$LHS $=$ RHSThe value of $x$ is $10$.
## Solve the following equations and check your results.$x=\frac{4}{5}(x+10)$.
Updated on 10-Oct-2022 13:47:38<|endoftext|>
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# A bag contains 3 red and 3 green balls and a person draws out 3 at random. He then drops 3 blue balls into the bag and again draws out 3 at random. The chance that the 3 later balls are all of different colour is.
Last updated date: 18th Jun 2024
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Hint:Probability of event to happen $P\{ E\}$
$= \dfrac{{Number\,of\,favourable\,outcome}}{{Total\,Number\,of\,outcomes}}$
In this question we will be creating two different cases as the balls were drawn two times. In the first case the total number of balls in the bag were 6 out of which 3 are red and 3 are green. In the second case the number of balls in the bag are 6 as well out of which 3 are blue.
In the second case we don’t know anything about the colour of the other 3 balls those were drawn out from the bag.
Complete step-by- step solution:
Given the data according to the question and its probability cases
Case I:
There are 3 red and 3 green balls in a bag and a person draws 3 balls.
The possibilities are the following:
3 red, 0 green
2 red, 1 green
1 red, 2 green
0 red, 3 green
Case II:
As 3 blue balls have been added to the bag then the possibilities of the balls that are in the bag according to above cases are:
1) If 3 red, 0 green balls were drawn from the bag then there will be 0 red, 3 green and 3 blue balls in the bag.
2) 2 red, 1 green ball were drawn from the bag then there will be 1 red, 2 green and 3 blue balls in the bag.
3) 1 red, 2 green balls were drawn from the bag then there will be 2 red, 1 green and 3 blue balls in the bag.
4) 0 red, 3 green balls were drawn from the bag then there will be 3 red, 0 green and 3 blue balls in the bag.
We need to find the probability of three balls being different colours hence it rules out (a) and (d) cases.
Now, probability of withdrawing 2R, 1G is: $= \dfrac{{{}^3{C_1} \times {}^3{C_1}}}{{{}^6{C_3}}}$
Probability of withdrawing 1R, 1G and 1B ball is: $\dfrac{{{}^2{C_1} \times {}^1{C_1} \times {}^3{C_1}}}{{{}^6{C_3}}}$
Hence, the probability of both cases is:
$= \dfrac{{{}^3{C_1} \times {}^3{C_1}}}{{{}^6{C_3}}} \times \dfrac{{{}^2{C_1} \times {}^1{C_1} \times {}^3{C_1}}}{{{}^6{C_3}}}$
$= \dfrac{{\dfrac{{3!}}{{2!1!}} \times \dfrac{{3!}}{{2!1!}}}}{{\dfrac{{6!}}{{3!3!}}}} \times \dfrac{{\dfrac{{2!}}{{1!1!}} \times 1 \times \dfrac{{3!}}{{2!1!}}}}{{\dfrac{{6!}}{{3!3!}}}}$
$= \dfrac{{\dfrac{{3 \times \not 2!}}{{\not 2!}} \times \dfrac{{3 \times \not 2!}}{{\not 2!}}}}{{\dfrac{{\not 6 \times 5 \times 4 \times \not 3!}}{{3! \times 3\not ! \times \not 2 \times 1}}}} \times \dfrac{{2! \times \dfrac{{3 \times 2!}}{{\not 2!}}}}{{\dfrac{{\not 6 \times 5 \times 4 \times \not 3!}}{{\not 3! \times 3! \times 2 \times 1}}}}$
$= \dfrac{9}{{20}} \times \dfrac{6}{{20}} = \dfrac{{27}}{{200}}\,\,\,\,\,\,\,.......(1)$
Now case II as the number of balls are same just colours are different, we have:
$= \dfrac{{{}^3{C_2} \times {}^3{C_1}}}{{{}^6{C_3}}} \times \dfrac{{{}^1{C_1} \times {}^2{C_1} \times {}^3{C_1}}}{{{}^6{C_3}}}$
$= \dfrac{{\dfrac{{3!}}{{2!1!}} \times \dfrac{{3!}}{{2!1!}}}}{{\dfrac{{6!}}{{3!3!}}}} \times \dfrac{{\dfrac{{2!}}{{1!1!}} \times 1 \times \dfrac{{3!}}{{2!1!}}}}{{\dfrac{{6!}}{{3!3!}}}}$
$= \dfrac{{\dfrac{{3 \times \not 2!}}{{\not 2!}} \times \dfrac{{3 \times \not 2!}}{{\not 2!}}}}{{\dfrac{{\not 6 \times 5 \times 4 \times \not 3!}}{{3! \times 3\not ! \times \not 2 \times 1}}}} \times \dfrac{{2! \times \dfrac{{3 \times 2!}}{{\not 2!}}}}{{\dfrac{{\not 6 \times 5 \times 4 \times \not 3!}}{{\not 3! \times 3! \times 2 \times 1}}}}$
$= \dfrac{9}{{20}} \times \dfrac{6}{{20}} = \dfrac{{27}}{{200}}\,\,\,\,\,\,\,.......(2)$
$= \dfrac{{27}}{{200}}.....(2)$
Probability of 3 later balls being all of different colours $= \dfrac{{27}}{{200}} + \dfrac{{27}}{{200}}$(from $e{q^n}$ 1 & 2)
$= \dfrac{{2 \times 27}}{{200}}$$= \dfrac{{27}}{{100}}$
Note: Cube root need to group digit in no. 3 and taking the unit place digit of the first group & ten’s from $I{I^{nd}}$.
In mathematics, a cube root of a number x is a number y such that $\;{y^3}\; = \;x$. All nonzero real numbers have exactly one real cube root and a pair of complex conjugate cube roots, and all nonzero complex numbers have three distinct complex cube roots. For example, the real cube root of 8, denoted $\sqrt[3]{8}$, is 2, because ${2^3}\; = \;8$<|endoftext|>
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Difference between a prime number and a composite number
In mathematics, numbers can be classified into various categories based on their properties. Two of the most common classifications are prime numbers and composite numbers. These two types of numbers are distinct from each other and play a significant role in various mathematical concepts. Let’s dive deeper into understanding the difference between prime and composite numbers.
Prime Numbers: Definition and Properties
A prime number is a natural number greater than 1 that has exactly two distinct factors: 1 and itself. In other words, a prime number can only be divided by 1 and the number itself without leaving any remainder. The smallest prime number is 2, which is also the only even prime number. Some examples of prime numbers are 2, 3, 5, 7, 11, 13, and 17.
Key Features of Prime Numbers
1. A prime number has exactly two distinct factors.
2. The smallest prime number is 2.
3. All prime numbers, except 2, are odd numbers.
4. Prime numbers are the “building blocks” of natural numbers, as every natural number greater than 1 can be expressed as a product of prime numbers (known as the Fundamental Theorem of Arithmetic).
Composite Numbers: Definition and Properties
A composite number is a natural number greater than 1 that has more than two distinct factors. In other words, a composite number can be divided by at least one other number besides 1 and itself without leaving any remainder. The smallest composite number is 4. Some examples of composite numbers are 4, 6, 8, 9, 10, 12, and 14.
Key Features of Composite Numbers
1. A composite number has more than two distinct factors.
2. The smallest composite number is 4.
3. Composite numbers can be even or odd.
4. Composite numbers can be represented as a product of prime numbers.
Difference between a prime number and a composite number
The main difference between prime and composite numbers lies in their factors. Prime numbers have exactly two factors, while composite numbers have more than two factors. To better understand the distinction, let’s consider the factors of some numbers:
• Factors of 5 (prime): 1, 5
• Factors of 7 (prime): 1, 7
• Factors of 8 (composite): 1, 2, 4, 8
• Factors of 12 (composite): 1, 2, 3, 4, 6, 12
In the examples above, you can see that the prime numbers have only two factors, while the composite numbers have multiple factors. Understanding this key difference is essential in many mathematical operations and concepts, such as factoring, prime factorization, and number theory.
Final Thoughts
In summary, prime numbers and composite numbers are two different categories of natural numbers, characterized by the number of their factors. Prime numbers have exactly two factors (1 and the number itself), while composite numbers have more than two factors. Recognizing and distinguishing between prime and composite numbers is a fundamental skill in mathematics that forms the basis for more advanced topics and problem-solving techniques.
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This is a very beginner question as I am a complete beginner. How is change in speed during a song notated in standard notation? Is it done by changing the time signature for each measure that has a different speed? What if one measure has multiple different speeds? Do you have to write the tempo above each measure that has a different speed? Thank you in advance for your help. (If anyone knows a free site etc. where I can learn about writing standard notation, that would be greatly appreciated.)
When you're talking about speed, it's important to keep three concepts separated:
- the underlying pulse of the music. I.e., how much time passes between each quarter note, or in layman's terms, when you're tapping your foot to the music, how fast are you tapping? Changes in the pulse are notated as "acc." for "accelerando" (get faster) and "rit." for "ritardando" (get slower) when it's a gradual change. For a sudden change, you write the new speed above the measure where the tempo changes (either in beats per minute, or in one of the vague tempo terms like "moderato" or "presto").
- the values of the notes you're playing. If you're playing in 4/4, you can play one whole note per bar, or two half notes, or four quarters, or eight eighths... obviously, when you switch from a melody consisting of half and quarter notes to one of sixteenths, you have to play faster, even though the tempo doesn't change. You just have to write the appropriate note values. Likewise, unless you are in the realm of avantgarde music, you don't have more than one tempo simultaneously. You have one tempo, and if different instruments play shorter or longer notes, you just write the appropriate note values to match the tempo.
- the time signature - roughly speaking, how many beats in your given tempo do you play before the rhythmic structure repeats. This is not really related to speed. You could have a 3/4 time signature at 120 bpm, or a 4/4 at 120 bpm, and you'd play at exactly the same speed as long as the note values are the same. (It is common to write very fast-paced music as 2/2 rather than 4/4, but that's more a matter of conveying the feel than a mathematical necessity.)
A great beginner's question!
Several ways, and not usually by a change of tempo (bpm). Generally, each bar (measure) is the same length timewise, so faster notes will make it feel quicker; slower (longer) notes make it feel like it's slowed down.
There are signs which indicate speed change - accelerando, ritardando, rallentando being some. It's possible to go from, say, 4/4 to 3/4 making the latter shorter, but that's not particularly to make the music speed up.
How is speed change notated?
Usually fairly imprecisely. The most precise way to notate it is with a metronome marking. If the change in speed is to be sudden, this is precise. However, if the speed is to change over a period of time, there is no way to be precise about it aside from using a word such as accelerando, rallentando, ritardando, or the like. These indications give no clue about the rate of change of the speed.
They are even ambiguous as to whether you reach the target speed smoothly, or just approach it or even bypass it and then adopt it suddenly. For example, consider the following:
X:1 K:C M:4/4 L:1/4 Q:1/4=80 cBAG|cBAG|[Q:"accelerando"]cBAG|cBAG|[Q:1/4=120]cBAG|cBAG|
The eight notes in measures 1 and 2 should each last for 0.75 seconds. The last eight should each last for 0.5 seconds. But there's nothing that precisely specifies the duration of any note in measures 3 and 4. One can conclude that each note should have a shorter duration than the one preceding it, but even that is a matter of interpretation. For example, the last note in measure 4 could be shorter than the first note in measure 5, or it could be maybe 0.7 seconds long, so the tempo would increase over measures 3 and 4 but still jump up at the beginning of measure 5.
Sometimes composers do "write out" a rallentando explicitly, with varying degrees of complexity. It's hardly unusual to see something like this:
X:2 K:C M:4/4 L:1/4 CCGG|AAG2|F2F2|E2E2|D4|D4|C8|]
A modern classical composer might do something like this:
X:3 K:C M:4/4 L:1/8 Q:1/4=80 C2C2G2G2|A2A2G4|[M:7/8]F3F4|[M:5/8](E3E2)|[M:3/4]E6|[M:4/4]D8|[M:8/4]D16|(C16|C16)|]
Do you have to write the tempo above each measure that has a different speed?
What if one measure has multiple different speeds?
Divide the measure into two, each with different time signature and tempo. Example) if 4/4 measure has the first quarter note with tempo 112 and the remaining 3 quarter notes with tempo 70, then have the following two measures.
- first measure with time signature 1/4 and tempo 112,
- second measure with time signature 3/4 and tempo 70,
where I can learn about writing standard notation<|endoftext|>
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How do you find the amplitude, period, and shift for y=3tanx?
Dec 31, 2017
amplitude: N/A
period: ${180}^{\circ}$
phase shift: $0$
Explanation:
amplitude: how far the graph extends from its midline
e.g. the amplitude of $y = \sin x$ is $1$, since the midline is $y = 0$, and the highest and lowest points are $1$ and $- 1$.
the amplitude of $y = 3 \sin x$ is $3$, since the midline is $y = 0$, and the highest and lowest points are $3$ and $- 3$.
however, the $\tan x$ graph does not have a certain amplitude.
$\tan {90}^{\circ}$ is undefined. without knowing the highest point on the graph, the distance that the graph extends from the midline cannot be calculated.
period: how often the values of the graph repeat themselves
$3 \tan {0}^{\circ} = 0 , 3 \tan {180}^{\circ} = 0$
$3 \tan {45}^{\circ} = 3 , 3 \tan {225}^{\circ} = 3$
${180}^{\circ} - {0}^{\circ} = {180}^{\circ}$
${225}^{\circ} - {45}^{\circ} = {180}^{\circ}$
this means that the values of $\tan {x}^{\circ}$ on the graph recur every ${180}^{\circ}$.
phase shift: how far to the right a wave is from its usual position.
usual position of $y = \tan x$:
graph{tan x [-10, 10, -5, 5]}
$y = 3 \tan x$:
graph{3tan x [-10, 10, -5, 5]}
the graph has not been moved in either horizontal direction - the values for both at $x = 0$ are both the same.<|endoftext|>
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# If $9x^2 + 25y^2 = 181$ and $xy = -6$, find the value of $3x + 5y$.
Given:
$9x^2 + 25y^2 = 181$ and $xy = -6$
To do:
We have to find the value of $3x + 5y$.
Solution:
We know that,
$(a+b)^2=a^2+b^2+2ab$
$(a-b)^2=a^2+b^2-2ab$
$(a+b)(a-b)=a^2-b^2$
Therefore,
$(3x + 5y)^2 = (3x)^2 + (5y)^2 + 2 \times 3x \times 5y$
$=9x^2 + 25y^2 + 30xy$
$= 181 + 30 \times (-6)$ [Since $9x^2 + 25y^2 = 181$ and $xy = -6$]
$= 181 - 180$
$= 1$
$\Rightarrow 3x+5y= \sqrt{1}$
$\Rightarrow 3x+5y= \pm 1$
The value of $3x+5y$ is $\pm 1$.
Tutorialspoint
Simply Easy Learning
Updated on: 10-Oct-2022
31 Views<|endoftext|>
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Go through the Spectrum Math Grade 6 Answer Key Chapter 2 Pretest and get the proper assistance needed during your homework.
Multiplying and Dividing Fractions
Question 1.
a. $$\frac{7}{8}$$ × $$\frac{3}{4}$$
Product of $$\frac{7}{8}$$ × $$\frac{3}{4}$$ is $$\frac{21}{28}$$.
Explanation:
$$\frac{7}{8}$$ × $$\frac{3}{4}$$
= (7 × 3) ÷ (8 × 4)
= $$\frac{21}{28}$$
b. 9 × $$\frac{3}{8}$$
Product of 9 × $$\frac{3}{8}$$ is $$\frac{27}{8}$$ or 3$$\frac{3}{8}$$.
Explanation:
9 × $$\frac{3}{8}$$
= (9 × 3) ÷ (1 × 8)
= $$\frac{27}{8}$$ or 3$$\frac{3}{8}$$
c. $$\frac{5}{8}$$ × 5
Product of $$\frac{5}{8}$$ × 5 is $$\frac{25}{8}$$ or 3$$\frac{1}{8}$$.
Explanation:
$$\frac{5}{8}$$ × 5
= (5 × 5) ÷ (8 × 1)
= $$\frac{25}{8}$$ or 3$$\frac{1}{8}$$
Question 2.
a. 3$$\frac{1}{8}$$ × 4
Product of 3$$\frac{1}{8}$$ × 4 is $$\frac{25}{2}$$ or 12$$\frac{1}{2}$$.
Explanation:
3$$\frac{1}{8}$$ × 4
= {[(3 × 8) + 1] ÷ 8} × 4
= [(24 + 1) ÷ 8] × 4
= $$\frac{25}{8}$$ × 4
= $$\frac{25}{2}$$ × 1
= $$\frac{25}{2}$$ or 12$$\frac{1}{2}$$
b. 8 × 2$$\frac{3}{5}$$
Product of 8 × 2$$\frac{3}{5}$$ is $$\frac{104}{5}$$ or 20$$\frac{4}{5}$$.
Explanation:
8 × 2$$\frac{3}{5}$$
= 8 × {[(2 × 5) + 3] ÷ 5}
= 8 × [(10 + 3) ÷ 5]
= 8 × $$\frac{13}{5}$$
= (8 × 13) ÷ (1 × 5)
= $$\frac{104}{5}$$ or 20$$\frac{4}{5}$$
c. 4$$\frac{1}{2}$$ × 9
Product of 4$$\frac{1}{2}$$ × 9 is $$\frac{81}{2}$$ or 40$$\frac{1}{2}$$.
Explanation:
4$$\frac{1}{2}$$ × 9
= {[(4 × 2) + 1] ÷ 2} × 9
= [(8 + 1) ÷ 2] × 9
= $$\frac{9}{2}$$ × 9
= (9 × 9) ÷ (2 × 1)
= $$\frac{81}{2}$$ or 40$$\frac{1}{2}$$
Question 3.
a. 5$$\frac{3}{4}$$ × 2$$\frac{1}{3}$$
Product of 5$$\frac{3}{4}$$ × 2$$\frac{1}{3}$$ is $$\frac{161}{12}$$ or 13$$\frac{5}{12}$$.
Explanation:
5$$\frac{3}{4}$$ × 2$$\frac{1}{3}$$
= {[(5 × 4) + 3] ÷ 4} × {[(2 × 3) + 1] ÷ 3}
= [(20 + 3) ÷ 4] × [(6 + 1) ÷ 3]
= $$\frac{23}{4}$$ × $$\frac{7}{3}$$
= (23 × 7) ÷ (4 × 3)
= $$\frac{161}{12}$$ or 13$$\frac{5}{12}$$
b. 2$$\frac{1}{4}$$ × 3$$\frac{1}{5}$$
Product of 2$$\frac{1}{4}$$ × 3$$\frac{1}{5}$$ is $$\frac{36}{5}$$ or 7$$\frac{1}{5}$$.
Explanation:
2$$\frac{1}{4}$$ × 3$$\frac{1}{5}$$
= {[(2 × 4) + 1] ÷ 4} × {[(3 × 5) + 1] ÷ 5}
= [(8 + 1) ÷ 4] × [(15 + 1) ÷ 5]
= $$\frac{9}{4}$$ × $$\frac{16}{5}$$
= $$\frac{9}{1}$$ × $$\frac{4}{5}$$
= (9 × 4) ÷ (1 × 5)
= $$\frac{36}{5}$$ or 7$$\frac{1}{5}$$
c. 3$$\frac{2}{3}$$ × 1$$\frac{1}{8}$$
Product of 3$$\frac{2}{3}$$ × 1$$\frac{1}{8}$$ is $$\frac{33}{8}$$ or 4$$\frac{1}{8}$$.
Explanation:
3$$\frac{2}{3}$$ × 1$$\frac{1}{8}$$
= {[(3 × 3) + 2] ÷ 3} × {[(1 × 8) + 1] ÷ 8}
= [(9 + 2) ÷ 3] × [(8 + 1) ÷ 8]
= $$\frac{11}{3}$$ × $$\frac{9}{8}$$
= $$\frac{11}{1}$$ × $$\frac{3}{8}$$
= (11 × 3) ÷ (1 × 8)
= $$\frac{33}{8}$$ or 4$$\frac{1}{8}$$
Question 4.
a. 8 ÷ $$\frac{2}{3}$$
Dividing 8 by $$\frac{2}{3}$$, we get the answer 12.
Explanation:
8 ÷ $$\frac{2}{3}$$
= 8 × $$\frac{3}{2}$$
= 4 × $$\frac{3}{1}$$
= $$\frac{12}{1}$$
= 12.
b. $$\frac{4}{5}$$ ÷ 3
Dividing 3 by $$\frac{4}{5}$$, we get the answer $$\frac{4}{15}$$.
Explanation:
$$\frac{4}{5}$$ ÷ 3
= $$\frac{4}{5}$$ × $$\frac{1}{3}$$
= (4 × 1) ÷ (5 × 3)
= $$\frac{4}{15}$$
c. 10 ÷ $$\frac{3}{8}$$
Dividing 10 by $$\frac{3}{8}$$,we get the answer $$\frac{80}{3}$$ or 23$$\frac{2}{3}$$.
Explanation:
10 ÷ $$\frac{3}{8}$$
= 10 × $$\frac{8}{3}$$
= $$\frac{80}{3}$$ or 23$$\frac{2}{3}$$
Question 5.
a. $$\frac{4}{5}$$ ÷ $$\frac{7}{8}$$
Dividing $$\frac{4}{5}$$ by $$\frac{7}{8}$$, we get the answer $$\frac{32}{35}$$.
Explanation:
$$\frac{4}{5}$$ ÷ $$\frac{7}{8}$$
= $$\frac{4}{5}$$ × $$\frac{8}{7}$$
= (4 × 8) ÷ (5 × 7)
= $$\frac{32}{35}$$
b. $$\frac{2}{3}$$ ÷ $$\frac{5}{6}$$
Dividing $$\frac{2}{3}$$ by $$\frac{5}{6}$$, we get the answer $$\frac{4}{5}$$.
Explanation:
$$\frac{2}{3}$$ ÷ $$\frac{5}{6}$$
= $$\frac{2}{3}$$ × $$\frac{6}{5}$$
= $$\frac{2}{1}$$ × $$\frac{2}{5}$$
= (2 × 2) ÷ (1 × 5)
= $$\frac{4}{5}$$
c. $$\frac{3}{8}$$ ÷ $$\frac{7}{8}$$
Dividing $$\frac{3}{8}$$ by $$\frac{7}{8}$$, we get the answer $$\frac{3}{7}$$.
Explanation:
$$\frac{3}{8}$$ ÷ $$\frac{7}{8}$$
= $$\frac{3}{8}$$ × $$\frac{8}{7}$$
= $$\frac{3}{1}$$ × $$\frac{1}{7}$$
= (3 × 1) ÷ (1 × 7)
= $$\frac{3}{7}$$
Question 6.
a. 2$$\frac{3}{4}$$ ÷ 3$$\frac{1}{8}$$
Dividing 2$$\frac{3}{4}$$ by 3$$\frac{1}{8}$$, we get the answer $$\frac{22}{25}$$.
Explanation:
2$$\frac{3}{4}$$ ÷ 3$$\frac{1}{8}$$
= {[(2 × 4) + 3] ÷ 4} ÷ {[(3 × 8) + 1] ÷ 8}
= [(8 + 3) ÷ 4] ÷ [(24 + 1) ÷ 8]
= $$\frac{11}{4}$$ ÷ $$\frac{25}{8}$$
= $$\frac{11}{4}$$ × $$\frac{8}{25}$$
= $$\frac{11}{1}$$ × $$\frac{2}{25}$$
= (11 × 2) ÷ (1 × 25)
= $$\frac{22}{25}$$
b. 7 ÷ 3$$\frac{1}{4}$$
Dividing 7 by 3$$\frac{1}{4}$$, we get the answer $$\frac{28}{13}$$ or 2$$\frac{2}{13}$$.
Explanation:
7 ÷ 3$$\frac{1}{4}$$
= 7 ÷ {[(3 × 4) + 1] ÷ 4}
= 7 ÷ [(12 + 1) ÷ 4]
= 7 ÷ $$\frac{13}{4}$$
= 7 × $$\frac{4}{13}$$
= (7 × 4) ÷ (1 × 13)
= $$\frac{28}{13}$$ or 2$$\frac{2}{13}$$
c. 7$$\frac{3}{8}$$ ÷ 9
Dividing 7$$\frac{3}{8}$$ by 9, we get the answer $$\frac{59}{72}$$.
Explanation:
7$$\frac{3}{8}$$ ÷ 9
= {[(7 × 8) + 3] ÷ 8} ÷ 9
= [(56 + 3) ÷ 8] ÷ 9
= $$\frac{59}{8}$$ ÷ 9
= $$\frac{59}{8}$$ × $$\frac{1}{9}$$
= $$\frac{59}{72}$$
Solve each problem. Write answers in simplest form.
Question 7.
John and George together raked $$\frac{7}{8}$$ of the yard. John raked $$\frac{3}{4}$$ of that amount. What part of the yard did John rake?
John raked ____ of the yard.
Part of the yard did John raked = $$\frac{21}{32}$$.
John raked $$\frac{21}{32}$$ of the yard.
Explanation:
Part of the yard John and George together raked = $$\frac{7}{8}$$.
Part of that amount John raked = $$\frac{3}{4}$$.
Part of the yard did John raked = Part of the yard John and George together raked × Part of that amount John raked
= $$\frac{7}{8}$$ × $$\frac{3}{4}$$
= (7 × 3) ÷ (8 × 4)
= $$\frac{21}{32}$$
Question 8.
Felipe has track practice for $$\frac{5}{8}$$ of an hour after school each day. How many hours does he have track practice in 5 days?
Felipe his track practice for ____________ hours.
Number of hours Felipe has track practice in 5days = $$\frac{25}{8}$$ or 3$$\frac{1}{8}$$.
Felipe his track practice for $$\frac{25}{8}$$ or 3$$\frac{1}{8}$$ hours.
Explanation:
Number of hours Felipe has track practice after school each day = $$\frac{5}{8}$$.
Number of days = 5.
Number of hours Felipe has track practice in 5days = Number of hours Felipe has track practice after school each day × Number of days
= $$\frac{5}{8}$$ × 5
= (5 × 5) ÷ (8 × 1)
= $$\frac{25}{8}$$ or 3$$\frac{1}{8}$$
Question 9.
Paul can walk 2$$\frac{1}{2}$$ miles in 1 hour. How far can he walk in 1$$\frac{3}{4}$$ hours?
Paul can walk ___________ miles.
Number of miles he walks in 1$$\frac{3}{4}$$ hours = $$\frac{7}{10}$$.
Paul can walk $$\frac{7}{10}$$ miles.
Explanation:
Number of miles in 1 hour Paul can walk = 2$$\frac{1}{2}$$.
Total number of hours he walks = 1$$\frac{3}{4}$$.
Number of miles he walks in 1$$\frac{3}{4}$$ hours = Total number of hours he walks ÷ Number of miles in 1 hour Paul can walk
= 1$$\frac{3}{4}$$ ÷ 2$$\frac{1}{2}$$
= {[(1 × 4) + 3] ÷ 4} ÷ {[(2 × 2) + 1] ÷ 2}
= [(4 + 3) ÷ 4] ÷ [(4 + 1) ÷ 2]
= $$\frac{7}{4}$$ ÷ $$\frac{5}{2}$$
= $$\frac{7}{4}$$ × $$\frac{2}{5}$$
= $$\frac{7}{2}$$ × $$\frac{1}{5}$$
= $$\frac{7}{10}$$
Question 10.
Brad has a stack of 7 books on his desk. Each book is 1$$\frac{7}{8}$$ inches thick. How tall is the stack?
The stack is ____________ inches tall.
Number of inches is the stack = $$\frac{105}{8}$$ or 13$$\frac{1}{8}$$.
The stack is $$\frac{105}{8}$$ or 13$$\frac{1}{8}$$ inches tall.
Explanation:
Number of books on his desk Brad has a stack = 7.
Number of inches each book = 1$$\frac{7}{8}$$.
Number of inches is the stack = Number of books on his desk Brad has a stack × Number of inches each book
= 7 × 1$$\frac{7}{8}$$
= 7 × {[(1 × 8) + 7] ÷ 8}
= 7 × [(8 + 7) ÷ 8]
= 7 × $$\frac{15}{8}$$
= (7 × 15) ÷ (1 × 8)
= $$\frac{105}{8}$$ or 13$$\frac{1}{8}$$
Question 11.
A bag of candy weighs 3$$\frac{2}{3}$$ ounces. How much would 4$$\frac{1}{2}$$ bags of candy weigh?
The bags would weigh __________ ounces.
Number of ounces bags weighs = $$\frac{27}{22}$$ or 1$$\frac{5}{22}$$.
The bags would weigh $$\frac{27}{22}$$ or 1$$\frac{5}{22}$$ ounces.
Explanation:
Number of ounces a bag of candy weighs = 3$$\frac{2}{3}$$.
Total number of ounces bags of candy weighs = 4$$\frac{1}{2}$$.
Number of ounces bags weighs = Total number of ounces bags of candy weighs ÷ Number of ounces a bag of candy weighs
= 4$$\frac{1}{2}$$ ÷ 3$$\frac{2}{3}$$
= {[(4 × 2) + 1] ÷ 2} ÷ {[(3 × 3) + 2] ÷ 3}
= [(8 + 1) ÷ 2] ÷ [(9 + 2) ÷ 3]
= $$\frac{9}{2}$$ ÷ $$\frac{11}{3}$$
= $$\frac{9}{2}$$ × $$\frac{3}{11}$$
= (9 × 3) ÷ (2 × 11)
= $$\frac{27}{22}$$ or 1$$\frac{5}{22}$$
Question 12.
It takes 8 hours to paint a room. How long will it take to paint $$\frac{2}{3}$$ of the room?
It will take ____________ hours to paint $$\frac{2}{3}$$ of the room.
Number of hours it takes to paint $$\frac{2}{3}$$ of the room = $$\frac{1}{12}$$.
It will take $$\frac{1}{12}$$ hours to paint $$\frac{2}{3}$$ of the room.
Explanation:
Number of hours it takes to paint a room = 8.
Part of the room it takes to paint = $$\frac{2}{3}$$.
Number of hours it takes to paint $$\frac{2}{3}$$ of the room = Part of the room it takes to paint ÷ Number of hours it takes to paint a room
= $$\frac{2}{3}$$ ÷ 8
= $$\frac{2}{3}$$ × $$\frac{1}{8}$$
= $$\frac{1}{3}$$ × $$\frac{1}{4}$$
= (1 × 1) ÷ (3 × 4)
= $$\frac{1}{12}$$
Question 13.
Jim vill divide 6$$\frac{3}{4}$$ pounds of candy equally among 9 friends. How much candy will each friend get?
Each friend will get ___________ of a pound.
Number of pounds of candy will each friend gets = $$\frac{3}{4}$$.
Each friend will get $$\frac{3}{4}$$ of a pound.
Explanation:
Total number of pounds of candy Jim vill divide = 6$$\frac{3}{4}$$.
Number of friends equally he divides the candy = 9.
Number of pounds of candy will each friend gets = Total number of pounds of candy Jim vill divide ÷ Number of friends equally he divides the candy
= 6$$\frac{3}{4}$$ ÷ 9
= {[(6 × 4) + 3] ÷ 4} ÷ 9
= [(24 + 3) ÷ 4] ÷ 9
= $$\frac{27}{4}$$ ÷ 9
= $$\frac{27}{4}$$ ÷ $$\frac{1}{9}$$
= $$\frac{3}{4}$$ ÷ $$\frac{1}{1}$$
= $$\frac{3}{4}$$.<|endoftext|>
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