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Question
Fri November 04, 2011 By: Raghu Rao
# A REAL OBJECT IS LOCATED IS LOCATED AT ZERO END OF A METER STICK. A CONCAVE MIRROR LOCATED AT 100 cm END OF THE METER STICK FORMS AN IMAGE OF THE OBJECT AT THE 70cm POSITION. A CONVEX MIRROR PLACED AT THE 60cm POSITION FORMS A FINAL IMAGE AT THE 10 cm POINT WHAT IS THE RADIUS OF CURVATURE OF THE CONVEX MIRROR?
Expert Reply
Wed November 09, 2011
here for concave mirror
u = 100 cm.
v= 70 cm.
we know
1/f = 1/v +1/u.
1/f =1/100 +1/70
on solving
f =7000/170 =41.17 cm.
radius of curvature for concave mirror = 2 * 41.7 cm = 82.35 cm
for convex mirror image formed by concave mirror act as object for comvex mirror so for convex mirror
u = -10 cm.
v= 50 cm.
again
1/f = 1/v +1/u
1/f = 1/50 -1/10.
1/f =(1-5)/50.
1/f = -4/50
1/f = -4/50
f= -50/4.
f= -12.5 cm
radius of curvature for convex mirror = 2 * focal length of convex mirror.
radius of curvature for convex mirror = 2 *12.5 =25 cm.
Related Questions
Fri May 19, 2017
# A hot-air balloon is rising vertically with a constant velocity of 10 m/sec. An object is dropped from the balloon when it is 60 m above the ground. The angle of elevation of the Sun is 45o. When the falling object is 33.75 m from the ground, the rate at which the shadow of the object is traveling along the ground is (take g = 10 m/sec2) (A) 15 m/sec. (B) 20 m/sec. (C) 25 m/sec. (D) 30 m/sec. Please give Detailed Solution.
Tue March 21, 2017
Home Work Help
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# Tootsie Roll Pop
How Many Licks Does It Take to Get to the Tootsie Roll Center of a Tootsie Pop?
Well, according to Mr. Owl, the answer is three.
At an office meeting of the Sweets Company of America in 1931, employees were asked to share any ideas for new candies. Employee Luke Weisgram had been thinking about this beforehand. Just the other day, Clara, his daughter, had shared a lick of her lollipop, and at the same time, Weisgram had a Tootsie Roll in his mouth. He thought about how good it tasted and up popped an idea!
The rest, as they say, is history.
In the case of post frame buildings, there isn’t a Tootsie Roll center, but there is some math involved.
From Hansen Pole Buildings’ Designer Doug:
“The math equation aside, which is cubic feet, correct? How many bags of ready mix does it usually take per 18″ and 24″ holes? I have a client who can’t do math and they just want to know how many bags.”
In most cases the lower 18 inches of the hole will be filled with pre-mix concrete. This is known as a concrete collar.
To determine concrete volume required:
Multiply ½ hole diameter (in feet) squared, x 3.14 x concrete collar depth (in feet) times hole number, divided by 27.
Like math in high school? This is what formula looks like:
(½ X hole diameter)² X 3.14 x concrete collar depth X # of holes
27
So, let’s talk bags.
For an 18” diameter hole: (1/2 X 1.5’)^2 X 3.14 X 1.5’ = 2.65 cubic feet. A 60 pound bag of Sakcrete® makes 0.45 cubic feet of concrete, so just under 6 bags (or 4-1/2 bags of 80 pound).
For 24” diameter: (1/2 X 2’)^2 X 3.14 X 1.5’ = 4.71 cubic feet or 10-1/2 60 pound or nearly 8 bags of 80#.
More reading on this subject can be found at: https://www.hansenpolebuildings.com/2012/11/concrete/
And for the curious, my answer to the Tootsie Pop question is ZERO. I am allergic to chocolate!
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# The sum of two numbers is 37. The smaller is 17 less than the larger number. What are the numbers?
Let the numbers be x and y such that y > x.
Given that the sum of the numbers is 37.
==> x + y= 37 ..............(1)
Also, we are given that the smaller (x) is 17 less that the larger (y).
==> y= x-17 .................(2)
Now we will substitute with...
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Let the numbers be x and y such that y > x.
Given that the sum of the numbers is 37.
==> x + y= 37 ..............(1)
Also, we are given that the smaller (x) is 17 less that the larger (y).
==> y= x-17 .................(2)
Now we will substitute with (2) into (2).
==> x + y= 37
==> x+ (x-17)= 37
==> 2x -17 = 37
==> 2x = 54
==> x = 54/2 = 27
==> y= 27-17 = 10
Then, the numbers are 27 and 10.
Approved by eNotes Editorial Team
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# Thread: Collinear points formed by a measurement on a square's perimeter
1. ## Collinear points formed by a measurement on a square's perimeter
Let a square ABCD with sides of length 1 be given. A point X on BC is at distance d
from C, and a point Y on CD is at distance d from C. The extensions of: AB and DX
meet at P, AD and BY meet at Q, AX and DC meet at R, and AY and BC meet at
S. If points P, Q, R and S are collinear, determine d.
I have started and found the diagram is symmetrical through AC, X must be (1,d) and Y must be (1-d,0). PR or QS must be at right angles to AC due to the need for a collinear (straight) line through them at the end. I believe the gradients should be the same for AC and QS, is this correct?? I'm stuck on what to do next and also need to clarify so far.
Cheers
2. ## Re: Collinear points formed by a measurement on a square's perimeter
The gradient of SR is 1 (or -1) due to symmetry, so one has to equate the gradient of RP to 1. Let R' be the projection of R to AP. Since RR' = 1, it should be that R'P = AP - AR' = 1. It is left to express AP and AR' through d using similar triangles.
3. ## Re: Collinear points formed by a measurement on a square's perimeter
Originally Posted by pikachu26134
Let a square ABCD with sides of length 1 be given. A point X on BC is at distance d
from C, and a point Y on CD is at distance d from C. The extensions of: AB and DX
meet at P, AD and BY meet at Q, AX and DC meet at R, and AY and BC meet at
S. If points P, Q, R and S are collinear, determine d.
I have started and found the diagram is symmetrical through AC, X must be (1,d) and Y must be (1-d,0). PR or QS must be at right angles to AC due to the need for a collinear (straight) line through them at the end. I believe the gradients should be the same for AC and QS, is this correct?? I'm stuck on what to do next and also need to clarify so far.
Cheers
1. Use proportions in similar triangles:
The blue triangles will yield:
$\displaystyle \dfrac{1+s}{s+d}=\dfrac{2+s}{1+s}$
and the grey triangles will yield:
$\displaystyle \dfrac ds=\dfrac1{1+s}$
2. Solve this system of equations for s and d.
I've got $\displaystyle d=\frac32 - \frac12 \cdot \sqrt{5}$ and $\displaystyle s= \frac12 \cdot \sqrt{5} - \frac12$
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# Numerals and numeral systems
mathematics
Numerals and numeral systems, symbols and collections of symbols used to represent small numbers, together with systems of rules for representing larger numbers.
Just as the first attempts at writing came long after the development of speech, so the first efforts at the graphical representation of numbers came long after people had learned how to count. Probably the earliest way of keeping record of a count was by some tally system involving physical objects such as pebbles or sticks. Judging by the habits of indigenous peoples today as well as by the oldest remaining traces of written or sculptured records, the earliest numerals were simple notches in a stick, scratches on a stone, marks on a piece of pottery, and the like. Having no fixed units of measure, no coins, no commerce beyond the rudest barter, no system of taxation, and no needs beyond those to sustain life, people had no necessity for written numerals until the beginning of what are called historical times. Vocal sounds were probably used to designate the number of objects in a small group long before there were separate symbols for the small numbers, and it seems likely that the sounds differed according to the kind of object being counted. The abstract notion of two, signified orally by a sound independent of any particular objects, probably appeared very late.
## Number bases
When it became necessary to count frequently to numbers larger than 10 or so, the numeration had to be systematized and simplified; this was commonly done through use of a group unit or base, just as might be done today counting 43 eggs as three dozen and seven. In fact, the earliest numerals of which there is a definite record were simple straight marks for the small numbers with some special form for 10. These symbols appeared in Egypt as early as 3400 bce and in Mesopotamia as early as 3000 bce, long preceding the first known inscriptions containing numerals in China (c. 1600 bce), Crete (c. 1200 bce), and India (c. 300 bce). Some ancient symbols for 1 and 10 are given in the figure.
The special position occupied by 10 stems from the number of human fingers, of course, and it is still evident in modern usage not only in the logical structure of the decimal number system but in the English names for the numbers. Thus, eleven comes from Old English endleofan, literally meaning “[ten and] one left [over],” and twelve from twelf, meaning “two left”; the endings -teen and -ty both refer to ten, and hundred comes originally from a pre-Greek term meaning “ten times [ten].”
It should not be inferred, however, that 10 is either the only possible base or the only one actually used. The pair system, in which the counting goes “one, two, two and one, two twos, two and two and one,” and so on, is found among the ethnologically oldest tribes of Australia, in many Papuan languages of the Torres Strait and the adjacent coast of New Guinea, among some African Pygmies, and in various South American tribes. The indigenous peoples of Tierra del Fuego and the South American continent use number systems with bases three and four. The quinary scale, or number system with base five, is very old, but in pure form it seems to be used at present only by speakers of Saraveca, a South American Arawakan language; elsewhere it is combined with the decimal or the vigesimal system, where the base is 20. Similarly, the pure base six scale seems to occur only sparsely in northwest Africa and is otherwise combined with the duodecimal, or base 12, system.
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In the course of history, the decimal system finally overshadowed all others. Nevertheless, there are still many vestiges of other systems, chiefly in commercial and domestic units, where change always meets the resistance of tradition. Thus, 12 occurs as the number of inches in a foot, months in a year, ounces in a pound (troy weight or apothecaries’ weight), and twice 12 hours in a day, and both the dozen and the gross measure by twelves. In English the base 20 occurs chiefly in the score (“Four score and seven years ago…”); in French it survives in the word quatre-vingts (“four twenties”), for 80; other traces are found in ancient Celtic, Gaelic, Danish, and Welsh. The base 60 still occurs in measurement of time and angles.
## Numeral systems
It appears that the primitive numerals were |, ||, |||, and so on, as found in Egypt and the Grecian lands, or −, =, ≡, and so on, as found in early records in East Asia, each going as far as the simple needs of people required. As life became more complicated, the need for group numbers became apparent, and it was only a small step from the simple system with names only for one and ten to the further naming of other special numbers. Sometimes this happened in a very unsystematic fashion; for example, the Yukaghirs of Siberia counted, “one, two, three, three and one, five, two threes, two threes and one, two fours, ten with one missing, ten.” Usually, however, a more regular system resulted, and most of these systems can be classified, at least roughly, according to the logical principles underlying them.
## Simple grouping systems
In its pure form a simple grouping system is an assignment of special names to the small numbers, the base b, and its powers b2, b3, and so on, up to a power bk large enough to represent all numbers actually required in use. The intermediate numbers are then formed by addition, each symbol being repeated the required number of times, just as 23 is written XXIII in Roman numerals.
The earliest example of this kind of system is the scheme encountered in hieroglyphs, which the Egyptians used for writing on stone. (Two later Egyptian systems, the hieratic and demotic, which were used for writing on clay or papyrus, will be considered below; they are not simple grouping systems.) The number 258,458 written in hieroglyphics appears in the figure. Numbers of this size actually occur in extant records concerning royal estates and may have been commonplace in the logistics and engineering of the great pyramids.
## Cuneiform numerals
Around Babylon, clay was abundant, and the people impressed their symbols in damp clay tablets before drying them in the sun or in a kiln, thus forming documents that were practically as permanent as stone. Because the pressure of the stylus gave a wedge-shaped symbol, the inscriptions are known as cuneiform, from the Latin cuneus (“wedge”) and forma (“shape”). The symbols could be made either with the pointed or the circular end (hence curvilinear writing) of the stylus, and for numbers up to 60 these symbols were used in the same way as the hieroglyphs, except that a subtractive symbol was also used. The figure shows the number 258,458 in cuneiform.
The cuneiform and the curvilinear numerals occur together in some documents from about 3000 bce. There seem to have been some conventions regarding their use: cuneiform was always used for the number of the year or the age of an animal, while wages already paid were written in curvilinear and wages due in cuneiform. For numbers larger than 60, the Babylonians used a mixed system, described below.
## Greek numerals
The Greeks had two important systems of numerals, besides the primitive plan of repeating single strokes, as in ||| ||| for six, and one of these was again a simple grouping system. Their predecessors in culture—the Babylonians, Egyptians, and Phoenicians—had generally repeated the units up to 9, with a special symbol for 10, and so on. The early Greeks also repeated the units to 9 and probably had various symbols for 10. In Crete, where the early civilization was so much influenced by those of Phoenicia and Egypt, the symbol for 10 was −, a circle was used for 100, and a rhombus for 1,000. Cyprus also used the horizontal bar for 10, but the precise forms are of less importance than the fact that the grouping by tens, with special symbols for certain powers of 10, was characteristic of the early number systems of the Middle East.
The Greeks, who entered the field much later and were influenced in their alphabet by the Phoenicians, based their first elaborate system chiefly on the initial letters of the numeral names. This was a natural thing for all early civilizations, since the custom of writing out the names for large numbers was at first quite general, and the use of an initial by way of abbreviation of a word is universal. The Greek system of abbreviations, known today as Attic numerals, appears in the records of the 5th century bce but was probably used much earlier.
## Roman numerals
The direct influence of Rome for such a long period, the superiority of its numeral system over any other simple one that had been known in Europe before about the 10th century, and the compelling force of tradition explain the strong position that the system maintained for nearly 2,000 years in commerce, in scientific and theological literature, and in belles lettres. It had the great advantage that, for the mass of users, memorizing the values of only four letters was necessary—V, X, L, and C. Moreover, it was easier to see three in III than in 3 and to see nine in VIIII than in 9, and it was correspondingly easier to add numbers—the most basic arithmetic operation.
As in all such matters, the origin of these numerals is obscure, although the changes in their forms since the 3rd century bce are well known. The theory of German historian Theodor Mommsen (1850) had wide acceptance. He argued that the V for five represented the open hand. Two of these gave the X for 10, and the L, C, and M were modifications of Greek letters. However, study of inscriptions left by the Etruscans, who ruled Italy before the Romans, show that the Romans adopted the Etruscan numerical system beginning in the 5th century bce but with the distinct difference that the Etruscans read their numbers from right to left while the Romans read theirs from left to right. L and D for 50 and 500, respectively, emerged in the Late Roman Republic, and M did not come to mean 1,000 until the Middle Ages.
The oldest noteworthy inscription containing numerals representing very large numbers is on the Columna Rostrata, a monument erected in the Roman Forum to commemorate a victory in 260 bce over Carthage during the First Punic War. In this column a symbol for 100,000, which was an early form of (((I))), was repeated 23 times, making 2,300,000. This illustrates not only the early Roman use of repeated symbols but also a custom that extended to modern times—that of using (I) for 1,000, ((I)) for 10,000, (((I))) for 100,000, and ((((I)))) for 1,000,000. The symbol (I) for 1,000 frequently appears in various other forms, including the cursive ∞. Toward the end of the Roman Republic, a bar (known as the vinculum or virgula) was placed over a number to multiply it by 1,000. This bar also came to represent ordinal numbers. In the early Roman Empire, bars enclosing a number around the top and sides came to mean multiplication by 100,000. The use of the single bar on top lasted into the Middle Ages, but the three bars did not.
Of the later use of the numerals, a few of the special types are as follows:
1. clxiiij∙ccc∙l∙i for 164,351, Adelard of Bath (c. 1120)
2. II.DCCC.XIIII for 2,814, Jordanus Nemorarius (c. 1125)
3. M⫏CLVI for 1,656, in San Marco, Venice
4. cIɔ.Iɔ.Ic for 1,599, Leiden edition of the work of Martianus Capella (1599)
5. IIIIxx et huit for 88, a Paris treaty of 1388
6. four Cli.M for 451,000, Humphrey Baker’s The Well Spryng of Sciences Whiche Teacheth the Perfecte Woorke and Practise of Arithmeticke (1568)
7. vj.C for 600 and CCC.M for 300,000, Robert Recorde (c. 1542)
Item (1) represents the use of the vinculum; (2) represents the place value as it occasionally appears in Roman numerals (D represents 500); (3) illustrates the not infrequent use of ⫏, like D, originally half of (I), the symbol for 1,000; (4) illustrates the persistence of the old Roman form for 1,000 and 500 and the subtractive principle so rarely used by the Romans for a number like 99; (5) shows the use of quatre-vingts for 80, commonly found in French manuscripts until the 17th century and occasionally later, the numbers often being written like iiijxx, vijxx, and so on; and (6) represents the coefficient method, “four C” meaning 400, a method often leading to forms like ijM or IIM for 2,000, as shown in (7).
The subtractive principle is seen in Hebrew number names, as well as in the occasional use of IV for 4 and IX for 9 in Roman inscriptions. The Romans also used unus de viginti (“one from twenty”) for 19 and duo de viginti (“two from twenty”) for 18, occasionally writing these numbers as XIX (or IXX) and IIXX, respectively. On the whole, however, the subtractive principle was little used in the numerals of the Classical period.
## Multiplicative grouping systems
In multiplicative systems, special names are given not only to 1, b, b2, and so on but also to the numbers 2, 3, …, b − 1; the symbols of this second set are then used in place of repetitions of the first set. Thus, if 1, 2, 3, …, 9 are designated in the usual way but 10, 100, and 1,000 are replaced by X, C, and M, respectively, then in a multiplicative grouping system one should write 7,392 as 7M3C9X2. The principal example of this kind of notation is the Chinese numeral system, three variants of which are shown in the figure. The modern national and mercantile systems are positional systems, as described below, and use a circle for zero.
## Ciphered numeral systems
In ciphered systems, names are given not only to 1 and the powers of the base b but also to the multiples of these powers. Thus, starting from the artificial example given above for a multiplicative grouping system, one can obtain a ciphered system if unrelated names are given to the numbers 1, 2, …, 9; X, 2X, …, 9X; C, 2C, …, 9C; M, 2M, …, 9M. This requires memorizing many different symbols, but it results in a very compact notation.
The first ciphered system seems to have been the Egyptian hieratic (literally “priestly”) numerals, so called because the priests were presumably the ones who had the time and learning required to develop this shorthand outgrowth of the earlier hieroglyphic numerals. An Egyptian arithmetical work on papyrus, employing hieratic numerals, was found in Egypt about 1855; known after the name of its purchaser as the Rhind papyrus, it provides the chief source of information about this numeral system. There was a still later Egyptian system, the demotic, which was also a ciphered system.
As early as the 3rd century bce, a second system of numerals, paralleling the Attic numerals, came into use in Greece that was better adapted to the theory of numbers, though it was more difficult for the trading classes to comprehend. These Ionic, or alphabetical, numerals, were simply a cipher system in which nine Greek letters were assigned to the numbers 1–9, nine more to the numbers 10, …, 90, and nine more to 100, …, 900. Thousands were often indicated by placing a bar at the left of the corresponding numeral.
Such numeral forms were not particularly difficult for computing purposes once the operator was able automatically to recall the meaning of each. Only the capital letters were used in this ancient numeral system, the lowercase letters being a relatively modern invention.
Other ciphered numeral systems include Coptic, Hindu Brahmin, Hebrew, Syrian, and early Arabic. The last three, like the Ionic, are alphabetic ciphered numeral systems.
## Positional numeral systems
The decimal number system is an example of a positional system, in which, after the base b has been adopted, the digits 1, 2, …, b − 1 are given special names, and all larger numbers are written as sequences of these digits. It is the only one of the systems that can be used for describing large numbers, since each of the other kinds gives special names to various numbers larger than b, and an infinite number of names would be required for all the numbers. The success of the positional system depends on the fact that, for an arbitrary base b, every number N can be written in a unique fashion in the formN = anbn + an − 1bn − 1 + ⋯ + a1b + a0 where an, an − 1, …, a0 are digits; i.e., numbers from the group 0, 1, …, b − 1. Then N to the base b can be represented by the sequence of symbols anan − 1a1a0. It was this principle which was used in the multiplicative grouping systems, and the relation between the two kinds of systems is immediately seen from the earlier noted equivalence between 7,392 and 7M3C9X2; the positional system derives from the multiplicative simply by omitting the names of the powers b, b2, and so on and by depending on the position of the digits to supply this information. It is then necessary, however, to use some symbol for zero to indicate any missing powers of the base; otherwise 792 could mean, for example, either 7M9X2 (i.e., 7,092) or 7C9X2 (792).
The Babylonians developed (c. 3000–2000 bce) a positional system with base 60—a sexagesimal system. With such a large base, it would have been awkward to have unrelated names for the digits 0, 1, …, 59, so a simple grouping system to base 10 was used for these numbers, as shown in the figure.
In addition to being somewhat cumbersome because of the large base chosen, the Babylonian system suffered until very late from the lack of a zero symbol; the resulting ambiguities may well have bothered the Babylonians as much as later translators.
In the course of early Spanish expeditions into Yucatan, it was discovered that the Maya, at an early but still undated time, had a well-developed positional system, complete with zero. It seems to have been used primarily for the calendar rather than for commercial or other computation; this is reflected in the fact that, although the base is 20, the third digit from the end signifies multiples not of 202 but of 18 × 20, thus giving their year a simple number of days. The digits 0, 1, …, 19 are, as in the Babylonian, formed by a simple grouping system, in this case to base 5; the groups were written vertically.
Neither the Mayan nor the Babylonian system was ideally suited to arithmetical computations, because the digits—the numbers less than 20 or 60—were not represented by single symbols. The complete development of this idea must be attributed to the Hindus, who also were the first to use zero in the modern way. As was mentioned earlier, some symbol is required in positional number systems to mark the place of a power of the base not actually occurring. This was indicated by the Hindus by a dot or small circle, which was given the name sunya, the Sanskrit word for “vacant.” This was translated into the Arabic ṣifr about 800 ce with the meaning kept intact, and the latter was transliterated into Latin about 1200, the sound being retained but the meaning ignored. Subsequent changes have led to the modern cipher and zero.
A symbol for zero appeared in the Babylonian system about the 3rd century bce. However, it was not used consistently and apparently served to hold only interior places, never final places, so that it was impossible to distinguish between 77 and 7,700, except by the context.
## The Hindu-Arabic system
Several different claims, each having a certain amount of justification, have been made with respect to the origin of modern Western numerals, commonly spoken of as Arabic but preferably as Hindu-Arabic. These include the assertion that the origin is to be found among the Arabs, Persians, Egyptians, and Hindus. It is not improbable that the intercourse among traders served to carry such symbols from country to country, so that modern Western numerals may be a conglomeration from different sources. However, as far as is known, the country that first used the largest number of these numeral forms is India. The 1, 4, and 6 are found in the Ashoka inscriptions (3rd century bce); the 2, 4, 6, 7, and 9 appear in the Nana Ghat inscriptions about a century later; and the 2, 3, 4, 5, 6, 7, and 9 in the Nasik caves of the 1st or 2nd century ce—all in forms that have considerable resemblance to today’s, 2 and 3 being well-recognized cursive derivations from the ancient = and ≡. None of these early Indian inscriptions gives evidence of place value or of a zero that would make modern place value possible. Hindu literature gives evidence that the zero may have been known earlier, but there is no inscription with such a symbol before the 9th century.
The first definite external reference to the Hindu numerals is a note by Severus Sebokht, a bishop who lived in Mesopotamia about 650. Since he speaks of “nine signs,” the zero seems to have been unknown to him. By the close of the 8th century, however, some astronomical tables of India are said to have been translated into Arabic at Baghdad, and in any case the numeral became known to Arabian scholars about this time. About 825 the mathematician al-Khwārizmī wrote a small book on the subject, and this was translated into Latin by Adelard of Bath (c. 1120) under the title of Liber algorismi de numero Indorum. The earliest European manuscript known to contain Hindu numerals was written in Spain in 976.
The advantages enjoyed by the perfected positional system are so numerous and so manifest that the Hindu-Arabic numerals and the base 10 have been adopted almost everywhere. These might be said to be the nearest approach to a universal human language yet devised; they are found in Chinese, Japanese, and Russian scientific journals and in every Western language. (However, see the table for some other modern numeral systems.)
Comparison of selected modern systems of numerals
Hindu-Arabic 1 2 3 4 5 6 7 8 9 0 Arabic ١ ٢ ٣ ٤ ٥ ٦ ٧ ٨ ٩ ٠ Devanagari (Hindi) १ २ ३ ४ ५ ६ ७ ८ ९ ० Tibetan ༡ ༢ ༣ ༤ ༥ ༦ ༧ ༨ ༩ ༠ Bengali ১ ২ ৩ ৪ ৫ ৬ ৭ ৮ ৯ ০ Thai ๑ ๒ ๓ ๔ ๕ ๖ ๗ ๘ ๙ ๐
## The binary system
There is one island, however, in which the familiar decimal system is no longer supreme: the electronic computer. Here the binary positional system has been found to have great advantages over the decimal. In the binary system, in which the base is 2, there are just two digits, 0 and 1; the number two must be represented here as 10, since it plays the same role as does ten in the decimal system.
Decimal numerals represented by digits
decimal binary conversion 0 = 0 0 ( 20 ) 1 = 1 1 ( 20 ) 2 = 10 1 ( 21 ) + 0 ( 20 ) 3 = 11 1 ( 21 ) + 1 ( 20 ) 4 = 100 1 ( 22 ) + 0 ( 21 ) + 0 ( 20 ) 5 = 101 1 ( 22 ) + 0 ( 21 ) + 1 ( 20 ) 6 = 110 1 ( 22 ) + 1 ( 21 ) + 0 ( 20 ) 7 = 111 1 ( 22 ) + 1 ( 21 ) + 1 ( 20 ) 8 = 1000 1 ( 23 ) + 0 ( 22 ) + 0 ( 21 ) + 0 ( 20 ) 9 = 1001 1 ( 23 ) + 0 ( 22 ) + 0 ( 21 ) + 1 ( 20 ) 10 = 1010 1 ( 23 ) + 0 ( 22 ) + 1 ( 21 ) + 0 ( 20 )
A binary number is generally much longer than its corresponding decimal number; for example, 256,058 has the binary representation 111 11010 00001 11010. The reason for the greater length of the binary number is that a binary digit distinguishes between only two possibilities, 0 or 1, whereas a decimal digit distinguishes among 10 possibilities; in other words, a binary digit carries less information than a decimal digit. Because of this, its name has been shortened to bit; a bit of information is thus transmitted whenever one of two alternatives is realized in the machine. It is of course much easier to construct a machine to distinguish between two possibilities than among 10, and this is another advantage for the base 2; but a more important point is that bits serve simultaneously to carry numerical information and the logic of the problem. That is, the dichotomies of yes and no, and of true and false, are preserved in the machine in the same way as 1 and 0, so in the end everything reduces to a sequence of those two characters.
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# An amazing property of the Catenary
I discovered that if we want an arc of catenary in the interval $$[a,b]$$ we solve $$\int_a^b \sqrt{\cosh '(x)^2+1} \, dx=\int_a^b \cosh x \, dx$$ which means that the "result" of the length is equal to the result of the area in the same interval, though in different units.
So I asked myself if there are any other curve with the same property.
I set $$y=\sqrt{y'^2+1}\to y^2=y'^2+1; y(0)=1$$ then $$y'=\sqrt{y^2-1}\to dx=\frac{dy}{\sqrt{y^2-1}}\to x=\cosh^{-1} \,y$$ hence, arbitrary constant is zero, $$y=\cosh x$$
But I am not sure how to deal with the other solution $$y'=-\sqrt{y^2-1}$$ even if Mathematica gives the same result $$y=\cosh x$$
I'd like someone checking this proof, you know: I am not a pro, I am just an (almost) retired high school teacher :)
Update 9/1/2020. Now I am officially retired :)
• It should just be $$-\int\frac{dy}{\sqrt{y^2-1}}=-\cosh^{-1}(y)$$ Jul 17, 2017 at 16:35
• Jul 17, 2017 at 20:01
• Differentiating $y^2=(y')^2+1$, we obtain $2yy'=2y'y'',$ so $y=y''$ whenever $y'\ne 0.$ Since $y(0)=1$, this implies that $y(x)=\cosh x +K\sinh x$ for all $x$, for some constant $K.$ Substituting this into $y^2=(y')^2+1$ we find that $K=0.$ Jul 18, 2017 at 4:02
From
$$\frac{y'}{\sqrt{y^2-1}}=\pm1$$ you draw
$$\text{arcosh}(y)=c\pm x$$
and
$$y=\cosh(c\pm x).$$
With the initial condition $$y(0)=1$$,
$$y=\cosh(\pm x)$$ which is $$y=\cosh(x).$$
EDIT1:
I understood your question in this way:
How is it that the area under a Catenary is proportional to the arc length? i.e., how is $$c=\dfrac{A}{L}$$ valid for some constant of proportionality $$c$$?
Let us at the outset consider very familiar similar situations:
If two DEs are given as $$y'= + \sqrt {1-y^2},\; y'= - \sqrt {1-y^2}$$
we have in either case by squaring $$y^{'2} = (1-y^2)$$
Differentiate
$$2 y' y^{''}= -2 y y',\to y^{''}+y =0$$ which is the differential equation of a sine curve.
With BC $$x=0,y=1,y'=0 \to y= \cos x$$ in either case
Similarly if two DEs are given as
$$y'= + \sqrt {y^2-1},\; y'= - \sqrt {y^2-1}$$
we have in either case
$$y^{'2}= (y^2-1)$$
Differentiating
$$2 y' y^{''}= 2 y y',\to y^{''}-y =0$$ which is the differential equation of a Catenary. With BC $$x=0,y=1,y'=0 \to y= \cosh x$$ in either case.
However, if you do not wish to square thereby losing its sign but wish to directly integrate the two BCs, the following:
$$y'= + \sqrt {1-y^2},\; y'= - \sqrt {1-y^2}$$
we get
$$\sin^{-1}y= x +c_1, \sin^{-1}y=- x-c_2$$
$$y= \sin (x+c_1),y= -\sin (x+c_2)$$
For (an even ) a symmetric solution $$x=0, y=1$$ we have respectively
$$c_1=\pi/2, c_2= 3 \pi/2$$
both yielding the same solution
$$y = \cos x$$
When we have here our actual case
$$y'= + \sqrt {1+y^2},\; y'= - \sqrt {1+y^2}$$
we get
$$\cosh^{-1}y= x +c_1, \cosh^{-1}y=- x-c_2$$ $$y= \cosh (x+c_1),y= \cosh (x+c_2)$$
For even symmetric solution $$x=0, y=1$$ we have respectively
$$c_1= c_2= 0$$
yielding both the same solution
$$y = \cosh x$$
So we can say in conclusion that in front of any (square root) radical sign we have $$\pm$$ and both signs are equally applicable for first order DE. It is only by a convention that we put in a positive sign implying the unsaid negative. They result in the same differential equation and hence also the same integrand for given boundary conditions in this particular case.
Geometrically a negative or positive sign of derivative relates to different slopes of the curve in different portions of the curve.
Next to answer what I considered to be your main question let us set up its DE which uniquely defines the curve.
To get a physical/geometrical idea a length dimension quantity $$c$$ is introduced as the quotient of covered area $$A$$ to the length of its curved "roof".
$$c=\dfrac {\int y \; dx}{\int\sqrt{1+y'^2}dx}$$
Using Quotient Rule differentiate to simplify
$$c=\dfrac{ y} {\sqrt{1+y'^2}}= \to y' = \dfrac{\sqrt{y^2-c^2}}{c}$$
which is the differential equation of the unique curve being sought.
Integrating with boundary condition $$y(0)=c ,y'(0)=0,$$ one obtains the equation of the only curve that satisfies the required property.
$$\dfrac{y}{c}= \cosh\dfrac{x}{c}$$
which is recognized as a catenary as stated. And in association this property is recognized as well... that $$c$$ is the constant of proportionality which is the minimum distance of the catenary to the x-axis.
$$c=\dfrac{A}{L}$$
as also shown here graphically.
• I am happily officially retired since last September 1st :) Oct 6, 2020 at 11:19
• Hope you would have stimulating experience in teaching and also in this site. Oct 6, 2020 at 12:23
• I don't think that you address the OP's question, which is about negative $y'$. (And how you solve the equation is preyty mysterious.) :-)
– user65203
Nov 17, 2020 at 9:53
• @ Raffaele. In my answer before edit I had assumed that the differential equations of a parabola, sine curve etc. were understood. So did not go into their detail. Hope it is now fully clear. Nov 17, 2020 at 20:38
• @ Yves Daoust: What the OP said " I asked myself if there is any other curve with the same property " had me answer that way. Nov 17, 2020 at 20:44
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# From G. H. Darwin 24 October 1874
Trin. Coll. | Camb.
Oct 24. 74
My dear Father,
I will try & answer yr. heat question,1 but as I do’nt know exactly how you propose to use it, I fear I must give you a good deal you don’t want
Bodies cool in 3 ways
(1) conduction
(3) convection
Conduction takes place within the body, whilst radiation is only from the surface. Conduction can only take place in a body one part of which is hotter than another. It has been supposed to be explicable as molecular radiation, i.e radiation from one molecule of the body to the next; but however that may be it takes place as a current or flow. This current always flows from the hotter to the colder parts.
Suppose for example we had a rod, & imagine an ideal partition across it & that on one side the temp: of the rod is higher than the other, then a flow of heat will pass thro’ this ideal partition from hotter to colder so long as there is any difference of temp: The rate of flow is proportional to the difference of temperature, i.e the amount of heat passing thro’ our partition in a second is proportional to the diff. of temp. In this case the unit of heat may be taken as the amount of heat requisite to raise 1 lb of water 1o Fahr., & the amount of heat means the number of such units. A numerical example will perhaps best explain this to you,—the Nōs being imaginary.
Suppose when one side of our partition is 10o Fahr. hotter than the other say the temps. are 0o & 10o that one unit of heat passes across in a second
Then when the temps. of the two sides are 0o & 20o, two units of heat will flow per sec.
When 0o & 30o, 3 units per sec.
0o & 40o, 4 ———
& so on one unit of heat more per sec. flowing for every 10o more of difference of temp:
Convection is where a cooling body heats the air touching its surface by conduction; the air becomes lighter from being warmer & flows away, to be replaced by cooler air. Thus air currents are established which carry away the heat by conduction & subsequent motion.
Radiation takes place from the surface only & is greater from a hotter than a cooler body.
Suppose a warm body be placed within an envelope coated with lamp black, which absorbs all the heat striking it, & that the envelope contains no air or gas. Then as the temperature of the body increases in arithmetical progression, the rate of radiation increases in a geometrical progression. A numerical example will again best explain my meaning.
Supposing a body under the above circumstances has a temperature of 0o, & that it loses l unit of heat per sec. and that when it has a temp. of 10o it loses 1$\frac{1}{4}$ or $\frac{5}{4}$ of a unit per sec.
Then when its temp is 20o it loses $\frac{5 2}{4 2}$ or $\frac{25}{16}$ or 1$\frac{9}{16}$ units a sec
When its temp: is 30o it loses $\frac{5 3}{4 3}$ or $\frac{125}{64}$ or nearly 2 units per sec
When its temp: is 40o it loses $\frac{5 4}{4 4}$ or $\frac{500}{256}$ or nearly 2$\frac{1}{2}$ units per sec
at 50o it loses $\frac{5 5}{4 5}$ or $\frac{3000}{1024}$ or nearly 3 units per sec.
& so on for every 10o of increase of temp: we must raise $\frac{5}{4}$ to one higher power.
This does not depend on diff: of temp., as does conduction.
The above is Newtons law of cooling2
Under ordinary circumstances all the surrounding bodies radiate back to the cooling one, which thus regains heat at the same time as it loses it. The rate of cooling is then the difference betw. the rate at which the body is cooling according to the above law, & the rate at which it is gaining.
At the same time convection takes place, & the inner parts keep supplying the surface deficiency caused by radiation, by means of conduction
Secondly
All crystals conduct heat with different rapidities in different directions.3 Some have only two rates of conduction viz: along or perpendicr. to an axis—& these are called uniaxal crystals. Some have 3 rates of conduction, in three directions mutually at right-angles, & these are called biaxal crysals.
All fibrous bodies conduct heat at a different rate along and perpendicular to the fibres. Thus a piece of the trunk of a tree conducts heat at 3 different rates
1st. parallel to the trunk
2nd. from the centre of outwards, & lastly if the piece be not cut from the heart of the tree there is a third rate along a horizontal line touching one of the rings of growth
The line (a) is supposed to stick out perperdincular to the paper.
[DIAG HERE]
I wrote to Spottiswoode & told him what I proposed to lecture on if I did so & offered to give him a table of contents when I’ve got it more in shape. He writes back that Pol. Econ. wd. be a proper subject—4 I’m wobbling towards doing it—aut Cæsar aut nihil—5 When I’ve spoken at debatg. Socs. I’ve been hardly nervous when once started & so I do’nt think I shd. break down—& I do’nt think my style of ill-health wd. make me much more likely.
I’ve not decided yet, & rather think I’ll run down in abt. 3 wks. & have a talk to you when I’ve got the thing more into shape. You may imagine I’ve been working hardish, when I tell you it has reached 60 pp of M.S & the greater part has been written out twice over I’ve not said nearly all yet so I cd. only give a selection at the R. I. If I do it, it will be Jan. 22 or 29. & then I shd. try & publish it in extenso in Contemp.,6 for it’s worth it, if I’m any judge of my own work.
Thank Jim for Steam Engine reference. I’ve read the passage, for Maxwell gave it me on Wedn. just before I started for Abinger.7
It was awfully cold & its a tedious journey & as I was’nt very well I rather repented going. As ill-luck wd. have it Thursd. was the only really bad morng. I’ve had for a month, but I got immensely better in the afternoon.
I fell in with Mr & Mrs. Erskine8 at the station & we drove up in the dark together. He’s rather a dry sort of bird. I found a perfect bevy of girls—all pretty 2 Miss Erskines, Miss Shadwell Miss Farrer—a cousin, Ida, & Miss Whichelo—the prettiest of all & I guess the nicest.9 Then there was T. H & last but not least Effie.10 It’s one of the most charming houses I ever saw, the hall is a chef d’œuvre11—& I was lodged in the same room you were in. There was a dinner party of Bosanquets Sir Geo Hewett an Irvingite &c.12 They bore me rather, for one can hardly get eno’ to eat of the simple things & is drawn in desperation to take snacks of the unwholesome. I do’nt know whether it’s my fancy but it seems to me that the wine is abominable; perhaps it is, as T. H. says he has lost his smell & taste almost in toto13 from a fever. Whatever it was I was very seedy next morning. However in the afternoon I went a very nice ride with T. H. over Holmbury to Hopedene,14 where we only fd. At. Fanny15 I went prepared to be charmed but am terribly disappointed. Not a rag of shelter, tho’ a very pretty view. The back elevation of the house is very pretty, but the front I do’nt like; T. H says it’s a row of outhouses. It seems to me to be built by a contortionist who said now I must be original & I’ll make it comfortable if I can. There are innumerable little meaningless bits of roof & windows, wh. only cd. have a meang. in a ho. built by patchwork. The end of dining room is askew wh. adds nothing to its beauty & is inconvenient. The drawing is all window at one end & pitch dark at the other. I can’t believe it’s good art—that elaborate want of straight-forwardness. However I daresay its very comfortable, only I’m glad its not my house. Of course I kept my opinion dark, except to T. H. with whom I thoro’ly agree
In Evening Uncle H. At. F Hope & Eddy Forster16 came to dinner & we had a v. pleasant evening. In the a.m I had a nice little walk over the hill to the station & came up with T. H. & Effie 1$\frac{3}{4}$ hrs for 25 miles! I got here at 1. & found lots to do with my two sets of proof sheets.17
I’ll do index as quick as I can, but its v. long.
The Environs of Abinger are delicious & must suit you to a T.
Jackson18 & Uncle H are both disgusted with Mivart’s answ.19 & so am I the more I think of it; but it doesn’t make much odds now I’ve had my say.
The Cooksons20 were to have come up today, but can’t Then this has take nearly 1$\frac{1}{2}$ hr’s to write
so goodbye | George Darwin
I’ve sent L. 3s worth of Yankee introdns. to New Z.21
It is Dr. Caufield not Dr. Canfield (on horns of Antilocapra 234)22 —I’m so sorry I didn’t correct index before it went to press as I cd. easily have cut off 10 pp. without losing a reference, but cutting out the ‘on’s & ‘the’s & superfluous words.23
Yrs affly | G. H. D.
## CD annotations
16.1 Secondly] ‘The greater the heat the farther it will be conducted to produce an apparent effect’ pencil
## Footnotes
Isaac Newton’s observations appeared in Philosophical Transactions of the Royal Society of London 22 (1701): 825–9.
George had been invited by William Spottiswoode to give a Friday evening lecture at the Royal Institution of Great Britain, and had been undecided in his choice of topic. CD suggested political economy would be a more suitable theme than development in dress. (See letter to G. H. Darwin, 19 October [1874] and n. 3.)
Aut Caesar aut nihil (Latin): either a Caesar or nothing; figuratively: all or nothing.
George refers to the Contemporary Review. See letter from G. H. Darwin, 5 December 1874 and n. 7.
Jim was a nickname for Horace Darwin. George also refers to Abinger Hall (the home of Thomas Henry Farrer). In 1867, James Clerk Maxwell wrote a classic paper on governors in the steam engine (Proceedings of the Royal Society of London 16: 270–83), but the passage has not been identified. See also letter from Horace Darwin, 17 August 1874.
Claudius James and Emily Georgina Erskine.
Maitland Katherine Erskine, Edith Emma Hay Erskine, Helen Cecilia Farrer, and Alice Clara Wichelo. Ida was Emma Cecilia Farrer, who became Horace Darwin’s wife in 1880. Miss Shadwell has not been identified.
Thomas Henry and Katherine Euphemia Farrer.
Chef d’oeuvre (French): a masterpiece.
George probably refers to Samuel Courthope Bosanquet and George John Routledge Hewett; an ‘Irvingite’ was a member of the Catholic Apostolic Church.
In toto (Latin): entirely.
Hopedene House is to the south of Holmbury St Mary, a village in the Mole Valley district of Surrey, three miles south of Abinger. It was designed by Richard Norman Shaw.
Frances Allen (Emma Darwin’s aunt).
Hensleigh Wedgwood (Emma Darwin’s brother), Hope Elizabeth Wedgwood (Hensleigh’s daughter), and Edward Morgan Llewellyn Forster.
CD had sent the proof-sheets for second edition of Descent; see letter to G. H. Darwin, 22 October [1874].
Henry Jackson.
St George Mivart had attacked George’s essay on marriage (G. H. Darwin 1873a) in an anonymous review ([Mivart] 1874); see letter to John Murray, 18 October 1874 and n. 1, and letter to G. H. Darwin, 19 October [1874] and n. 7.
Montague Hughes Cookson, and his wife, Blanche Althea Elizabeth Cookson.
Leonard Darwin was in New Zealand on the transit of Venus expedition; see letter from G. H. Darwin, 18 October 1874 and n. 19.
CD added a reference to Colbert Austin Canfield in Descent 2d ed., p. 234 n. 40; the spelling is correct. For Canfield’s correspondence about the pronghorn, or North American antelope, Antilocapra americana, see Correspondence vol. 19, letter from C. A. Canfield, 5 August 1871.
CD had sent the proof-sheets of the index of the second edition of Descent with his letter to George of 22 October [1874]. The superfluous words were not cut from the published edition.
## Summary
GHD explains conduction, radiation, and convection.
His paper on political economy for Royal Institution lecture has reached 60 pages. Plans to send it to Contemporary Review.
## Letter details
Letter no.
DCP-LETT-9695
From
George Howard Darwin
To
Charles Robert Darwin
Sent from
Trinity College, Cambridge
Source of text
DAR 58.2: 54; 210.2: 42
Physical description
14pp
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# How To Find Theoretical Concentration Of Stock Solution Beer-Lambert Law?
The Beer–Lambert law describes the relationship between the absorption of light by a solution and the properties of the solution using the equation A = bc, where molar absorptivity of the absorbing species is represented by a, b is the path length, and c is the concentration of the absorbing species is represented by a.
As stated by the Beer-Lambert law, there is a linear connection between the concentration of a solution and its absorbance, which allows the concentration of a solution to be estimated by measuring its absorbance.
## How do you calculate concentration using Beer Lambert law?
Estimation of the amount of concentration (C = A/(Lx)) The Lambert-Beer law, which serves as the physical foundation for photometric applications, states that the absorption of light by a sample is exactly proportional to the concentration of the sample and the length of the route taken by the light.
## How do you find the concentration of a stock solution?
• Using the formula M1V1, where ‘1’ represents the concentrated conditions (i.e., stock solution molarity and volume) and’2′ represents the diluted conditions (i.e., stock solution molarity and volume), the calculator can calculate the diluted conditions (i.e., desired volume and molarity).
• Please use the Mass Molarity Calculator if you need to make a solution with a certain molarity depending on mass.
## What is the theoretical Beer’s Law equation?
According to the formula A = lc, absorbance is represented by the letter A, molar-extinction coefficient (which varies according to chemical composition and light wavelength used), length of path (in centimetres) light must travel in solution is represented by the letter l, and concentration (in percent by weight) of a given solution is represented by the letter c.
## How do you calculate theoretical absorbance?
The standard equation for absorbance is A = x l x c, where A is the amount of light absorbed by the sample for a given wavelength, m is the molar absorptivity, l is the distance traveled by the light as it travels through the solution, and c is the concentration of the absorbing species in a unit volume of the sample solution.
## How do I calculate molar concentration?
In order to compute the molar concentration, we will divide the number of moles by the number of liters of water that was utilized in the solution. Consider the following example: the acetic acid in 1.25 L of water is entirely dissolved. Once you’ve calculated the molar concentration, divide it by 1.25 L to obtain the molecular weight, which will be 0.1332 M.
## How do you find a sample concentration in AAS?
In your situation, 0.192 mg/l equals 192 microg/l, which is your AAS result. It is calculated as AAS result multiplied by calculation factor = 192 * 200 = 38400 microg/kg = 38400 parts per billion (ppb). Using an AAS (or an ICP or an ICP-MS) result in mg/l as a starting point, your end result will be mg/kg = ppm.
## How do you find the final concentration of a solution?
Generally speaking, you may find out the final concentration by calculating the quantity of solute in each of the original mixes, adding them together, and then dividing by the total amount of solution. You may then multiply the result by 100 if you want it to be expressed as a percentage.
## How do you find theoretical concentration of two mixed solutions?
In general, when mixing two different concentrations together, first calculate the number of moles in each solution (n=CV,V-in-liter), then add them together to get the total moles, and then calculate the concentration of the mixture as follows: total moles / total volume = total moles / total volume (liter).
## What is concentration of solution Class 9?
It is the amount of solute present in a given volume of solution that determines its concentration in that volume of solution. Dilute Solution refers to a solution that contains a tiny amount of solute in it. Concentrated Solution is a solution that contains a high concentration of a particular solute.
## How do you find concentration from absorbance?
The absorbance of a material is directly proportional to its concentration and length: A = cl. is the wavelength-dependent molar absorbtivity coefficient, which is constant for a given substance.
## How do you calculate concentration from absorbance and path length?
The absorbance of a solution is determined by measuring how much light of a specific wavelength passes through it. Calculate the length of the path that the light must take. Calculate the molar absorption coefficient by multiplying it by the travel length. In step 3, you will find out how concentrated the solution is by dividing its absorbance by the value you acquired in that step.
## How do you calculate absorbance from concentration and percent transmittance?
The following equation can be used to convert a value from percent transmittance (percent T) to absorbance (percent A).
1. In this equation, absorbance equals 2 – log(percent T)
2. For example, to convert 56 percent T to absorbance units, divide 2 log(56) by 0.252 to get 0.252 absorbance units.
## How do you calculate concentration from absorbance and dilution factor?
Calculate the concentration by subtracting the absorbance of the sample (X) from the absorbance of the blank (Y), multiplying by the dilution factor (DF), and using the calibration curve.
## How do you calculate concentration of phosphate from absorbance?
In the equation for absorbance, (Slope Concentration) + Intercept (3) If we rearrange equation (3), we can calculate the concentration using the formula Concentration = (Absorbance / Intercept) / slope (4) Calculating the quantity of phosphate contained in the solution may be accomplished by utilizing equation (4).
## How do you find the concentration of a calibration curve?
This graph depicts the concentration, which is the independent variable (since it was chosen by you when you set up the experiment) on the x-axis. Whenever you want to know the absorbance of an unknown sample, look up the value of the y-axis on the standard curve. Afterwards, trace downward to see which concentration is closest to it.
## What is beer Lamberts law in chemistry?
Definition According to Beer Lamberts Law, there is a link between the attenuation of light through a medium and the qualities of the substance in question. It is defined as follows: ″The absorption of light by a chemical is exactly proportional to the route length and concentration of the chemical.″
## What is Beer’s law in chemistry?
Beer’s Law, which states that concentration and absorbance are proportionate, allows one to determine an unknown concentration of phosphate after obtaining the absorbance using the method described above.
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# (a) What is the density of a woman who floats in freshwater with 4.00% of her volume above the surface? This could be measured by placing her in a tank with marks on the side to measure how much water she displaces when floating and when held under water (briefly). (b) What percent of her volume is above the surface when she floats in seawater?
### College Physics
1st Edition
Paul Peter Urone + 1 other
Publisher: OpenStax College
ISBN: 9781938168000
Chapter
Section
### College Physics
1st Edition
Paul Peter Urone + 1 other
Publisher: OpenStax College
ISBN: 9781938168000
Chapter 11, Problem 46PE
Textbook Problem
25 views
## (a) What is the density of a woman who floats in freshwater with 4.00% of her volume above the surface? This could be measured by placing her in a tank with marks on the side to measure how much water she displaces when floating and when held under water (briefly). (b) What percent of her volume is above the surface when she floats in seawater?
(a)
To determine
The density of the body of women
### Explanation of Solution
Given:
Density of water dw=1gmcm3Percentage of volume of the body of women above the surface of water =4%
Formula used:
In floating condition,
Weight of body = Weight of displaced water
Calculation:
Let, percentage of total volume of the body of a women (VS) =100%
So, the percentage of volume of the body below the surface of water (VW) =1004
=96 %
Volume of the body below the surface of water = Volume of water displaced = Vw
According to Archimedes' principle,
Weight of body = Weight of displaced water
VSd
(b)
To determine
Percentage of volume of the body of women above the surface of sea water
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# NCERT Class 7 Mathematics Solutions: Chapter 7 – Congruence of Triangles Exercise 7.2 Part 4 (For CBSE, ICSE, IAS, NET, NRA 2022)
Doorsteptutor material for CBSE/Class-7 is prepared by world's top subject experts: get questions, notes, tests, video lectures and more- for all subjects of CBSE/Class-7.
1. In a squared sheet, draw two triangles of equal area such that:
(i) The triangles are congruent.
(ii) The triangles are not congruent.
What can you say about their perimeters?
In a squared sheet, draw .
When two triangles have equal areas and
(i) These triangles are congruent, i.e.. , [By SSS congruence rule]
Then, their perimeters are same because length of sides of first triangle are equal to the length of sides of another triangle by SSS congruence rule.
(ii) But, if the triangles are not congruent, then their perimeters are not same
Because lengths of sides of first triangle are not equal to the length of corresponding sides of another triangle.
2. Draw a rough sketch of two triangles such that they have five pairs of congruent parts but still the triangles are not congruent.
Consider,
Draw two triangles PQR and ABC.
All angles are equal; two sides are equal except one side.
So, are not congruent to .
3. If and are to be congruent, name one additional pair of corresponding parts. What criterion did you use?
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## The Monty Hall Problem
Continuing the discussion about probabilities and their intuition, here is a classical problem: the Monty Hall problem.
The setting is as follows: you are presented with three doors. Behind each door there is either a goat or a car. There are two goats and only one car. You get to pick a door, and someone who knows where the car is located, opens one of the two remaining doors and reveals a goat. Now there are two doors left, the one which you pick, and another one. Behind those two doors there is either a goat or a car.
Then you are given a choice: switch the door, or stay with the original one. What should you do?
Now there are two schools of thought:
• stay because it makes no difference, your new odds are 50/50.
• switch because it increases your odds
Before answering the question, to build up the intuition on the correct answer, let's consider a similar problem:
Instead of 3 doors, consider 1 million doors, 999,999 goats and one car. You pick one door at random, and the chances to get the car are obviously 1 in a million. Then the host of the game, knowing the car location, opens 999,998 doors revealing 999,998 goats. Sticking with your original choice, you still have 1/1,000,000 chances to get the car, switching increases your chances to 999,999/1,000,000. There is no such thing as 50/50 in this problem (or for the original problem). For the original problem switching increases your odds from 1/2 to 2/3. Still not convinced? Use 100 billion doors instead. You are more likely to be killed by lightening than finding the car on the first try. Switching the doors is a sure way of getting the car.
The incorrect solution of 50/50 comes from a naive and faulty application of Bayes' theorem of information update. Granted the 1/3-2/3 odds are not intuitive and there are a variety of ways to convince yourself this is the correct answer, including playing this game with a friend many times.
One thing to keep in mind is that the game show host (Monty Hall) is biased because he does know where the car is and he is always avoiding it. If the game show host would be unbiased and by luck would not reveal the car, then the odds would be 50/50 in that case. An unbiased host would sometimes reveal the car accidentally. It is the bias of the game show host which tricks our intuition to make us believe in a fair 50/50 solution. The answer is not a fair 50/50 because the game show host bias spoils the fairness overall.
The amazing thing is that despite all explanations, about half of the population strongly defends one position, and half strongly defends the other position. If you think the correct answer is 50/50, please argue your point of view here and I'll attempt to convince you otherwise.
Next time we'll continue discussing probabilities with another problem: the sleeping beauty problem. Unlike the Monty Hall problem, the sleeping beauty problem lacks consensus even among experts.
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# Need help. Test on Monday!
• Oct 4th 2008, 05:53 PM
Chinnie15
Need help. Test on Monday!
Hello everyone.
I am in desperate need of help. I am in college algebra and just can NOT get modeling down, and I have a test on Monday! I am completely lost, and realize I should have gotten help before now but now I have no choice but to learn it all today and tomorrow. It's just the set up, I know how to do the problems. Is there any where besides my text that I can get help with these application problems? They're killing me, and I'm really worried I'm gonna fail this first test.
I do have a direct question though. I've been working through these the best I can, and just can not solve this one no matter what I do.
A woman cycles 8 mi/h faster than she runs. Every morning she cycles 4 mi and runs 2.5 mi, for a total of one hour of exercise. How fast does she run?
I know that her running speed is x, and that her cycling speed is 8+x, but I still can't get anywhere. Please help if you can. :(
• Oct 4th 2008, 06:02 PM
galactus
We know d=rt. We are given that the total time is 1 hour.
Since d=rt, then t=d/r.
Let x=running time, as you stated.
$\frac{4}{8+x}+\frac{2.5}{x}=1$
• Oct 4th 2008, 06:05 PM
Jonboy
this is kinda confusing. it all boils down to: $d = rt$
which can be made: $t = \frac{d}{r}$
For running, you can make: $\frac{2.5}{x} = t_{r}$
$t_{r}$ just means time of running
For biking, you can make: $\frac{4}{x + 8} = t_{c}$
Now $t_{r} + t_{c} = 1$ so now you have:
$\frac{2.5}{x} + \frac{4}{x + 8} = 1$
Just solve for x and you're on your way.
• Oct 4th 2008, 06:39 PM
Chinnie15
Thank you so much guys! :) I'm going to be doing a load of review/studying tonight, and if I hit any road blocks I'll post them here.
Here is one more I just hit.
A parcel of land is 6ft longer than it is wide. Each diagonal from one corner to the opposite corner is 174 ft long. What are the dimensions of the parcel?
x= width
6+x= length
I get that, but I have no idea to do with the 174ft diagonal line, and how to set it up. It's like this with all of them, I can get some of it, it's just getting the rest that's been the problem. But once I understand it, I can easily apply it to the other problems.
• Oct 4th 2008, 06:53 PM
icemanfan
Quote:
Originally Posted by Chinnie15
A parcel of land is 6ft longer than it is wide. Each diagonal from one corner to the opposite corner is 174 ft long. What are the dimensions of the parcel?
x= width
6+x= length
The diagonal is the hypotenuse of a right triangle with sides x and 6+x. Therefore, its length is $174 = \sqrt{x^2 + (6+x)^2}$.
Square both sides and solve for x.
• Oct 4th 2008, 07:01 PM
Jonboy
it's better to start another thread if you have another problem. if someone sees all these replies and you put your next question on here it won't get answered as fast.
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# Compare Volumes of 3D shapes
A problem to compare the volumes of a cone, a cylinder and a hemisphere.
Problem 1: Find and compare the volumes of the three shapes shown below: (a) A cone with height r and radius of the base r, (b) a cylinder of height r and radius of the base r and (c) a hemisphere of radius r. Solution to Problem 1: Volume (Vco) of the cone is give by: Vco = (1 / 3)*area of base * height = (1 /3) * Pi r 2 * r = (1 /3) * Pi r 3 Volume (Vcy) of the cylinder is give by: Vcy = area of base * height = Pi r 2 * r = Pi r 3 Volume (Vhe) of the hemisphere is give by: Vhe = (1 / 2) volume of a sphere = (1/2) (4/3) Pi r 3 = (2 / 3) Pi r 3 The volume of the cylinder is the largest. The volume of the cone is one third of the volume of the cylinder and it is the smallest. The volume of the hemisphere is twice the volume of the cone or two thirds the volume of the cylinder. More references on geometry. Geometry Tutorials, Problems and Interactive Applets.
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# An electric toy car with a mass of 2 kg is powered by a motor with a voltage of 9 V and a current supply of 4 A. How long will it take for the toy car to accelerate from rest to 2/5 m/s?
Feb 26, 2017
Since this is a lot of current from a 9V source, the time required to reach the desired speed is only 4.4 ms (0.0044 s).
#### Explanation:
One real advantage of using energy to solve problems is that we are able to bridge from electrical to mechanical applications with ease. This problem is good example of this.
In terms of the electrical quantities, we look at the power delivered by the electricity.
$P = I \times V$
$P$ is the power, $I$ the current and $V$ the applied voltage.
So, #P = (9 V)(4A) = 36 W.
36 watts means 36 joules of energy supplied per second.
Now, the mechanical stuff:
To accelerate from zero to $\frac{2}{5}$ m/s involves a kinetic energy change of $\frac{1}{2} m {v}_{f}^{2} - \frac{1}{2} m {v}_{i}^{2} = \frac{1}{2} \left(2\right) {\left(\frac{2}{5}\right)}^{2} - 0 = 0.16 J$
Since energy is being supplied at a rate of 36 J/s, the time required is found by:
$P = \frac{E}{t}$ which we write as $t = \frac{E}{P} = \frac{0.16 J}{36 \left(\frac{J}{s}\right)} = 0.0044 s$
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Last updated:
# Standard Form to General Form of a Circle Calculator
How to use our standard form to general form of a circle calculator?How to convert the circle equation in standard form to general form ?More calculators to assist youFAQs
If you're looking for a quick way to convert the circle equation from standard form to general form, our standard form to general form of a circle calculator is a perfect match for you.
• How to use our standard form to the general form of a circle calculator?; and
• How to convert the circle equation from standard form to general form?
## How to use our standard form to general form of a circle calculator?
To use our calculator, you can use the following guide:
• Make sure your circle equation is in standard form: $(x-h)^2 +(y-k)^2=\text{C}$;
• Insert the parameters: h, k, and C present in standard form into the respective fields; and
• Right away, you will get the circle equation in a general form.
Nice, you are an expert in using our calculator now ;).
## How to convert the circle equation in standard form to general form ?
To convert the circle equation from the standard to general form, you can use the following steps:
1. Extract the radius and center information from the standard form. You can see below from the standard form we can observe the variables h, and k, which correspond to the center, and r corresponds to the radius.
$\qquad \footnotesize (x-h)^2 +(y-k)^2=r^2$
1. General form of the circle equation is given as:
$\qquad \footnotesize x^2 +y^2+Dx+Ey+F=0$
1. Now, use the following formula to compute the coefficients of the general form:
\qquad\footnotesize \begin{align*} D&=-2×h\\ E&=-2×k \\ F&=h^2+k^2-r^2. \end{align*}
1. Finally, substitute the values obtained to get the circle equation in a general form.
Excellent, you have learned how to convert circle equation from standard form to general form.
## More calculators to assist you
Check out our circle-related calculators created to assist you in solving circle-related problems. Whenever you need them, they will be available here:
FAQs
### What is the general form of a circle equation with diameter endpoints (4,8), (6,6)?
The general form is given as x²+y²-10x-14y+72=0. To find the general form, start with the general form x²+y²+Dx+Ey+F=0, and let's find the coefficients using the following steps:
1. Find the center (h,k) and distance between the diameter endpoints using the midpoint and distance formulas, respectively.
2. Divide the distance found in step 1 by 2 to obtain the radius r=1.4142.
3. Calculate the coefficients of the general form using the following equation:
D=-2×h=-10;
E=-2×k=-14; and
F=h²+k²-r²=72.
4. substitute the coefficients to obtain the circle equation in a general form:
x²+y²-10x-14y+72=0
### How do I find the radius from circle equation in general form?
To find the radius from the general form, use the following steps:
1. Extract the coefficients from the general form: x²+y²+Dx+Ey+F=0;
2. Use the formula of F which is given as F=(D/2)²+(E/2)²-r².
3. Rearrange the formula of F in terms of radius r: r=√((D/2)²+(E/2)²-F); and
4. Solve the equation in step 3 to obtain the radius of the circle.
Hooray! Now you know how to find the radius if you are given with circle equation in a general form.
(x − h)² + (y − k)² = C
x² + y² + Dx + Ey + F = 0
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# Ceva's Theorem
Ceva's Theorem is a criterion for the concurrence of cevians in a triangle.
## Statement
Let $ABC$ be a triangle, and let $D, E, F$ be points on lines $BC, CA, AB$, respectively. Lines $AD, BE, CF$ concur iff (if and only if)
$\frac{BD}{DC} \cdot \frac{CE}{EA}\cdot \frac{AF}{FB} = 1$,
where lengths are directed. This also works for the reciprocal or each of the ratios, as the reciprocal of $1$ is $1$.
(Note that the cevians do not necessarily lie within the triangle, although they do in this diagram.)
## Proof
We will use the notation $[ABC]$ to denote the area of a triangle with vertices $A,B,C$.
First, suppose $AD, BE, CF$ meet at a point $X$. We note that triangles $ABD, ADC$ have the same altitude to line $BC$, but bases $BD$ and $DC$. It follows that $\frac {BD}{DC} = \frac{[ABD]}{[ADC]}$. The same is true for triangles $XBD, XDC$, so
$\frac{BD}{DC} = \frac{[ABD]}{[ADC]} = \frac{[XBD]}{[XDC]} = \frac{[ABD]- [XBD]}{[ADC]-[XDC]} = \frac{[ABX]}{[AXC]}$.
Similarly, $\frac{CE}{EA} = \frac{[BCX]}{[BXA]}$ and $\frac{AF}{FB} = \frac{[CAX]}{[CXB]}$, so
$\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = \frac{[ABX]}{[AXC]} \cdot \frac{[BCX]}{[BXA]} \cdot \frac{[CAX]}{[CXB]} = 1$.
Now, suppose $D, E,F$ satisfy Ceva's criterion, and suppose $AD, BE$ intersect at $X$. Suppose the line $CX$ intersects line $AB$ at $F'$. We have proven that $F'$ must satisfy Ceva's criterion. This means that
$\frac{AF'}{F'B} = \frac{AF}{FB}$,
so
$F' = F$,
and line $CF$ concurrs with $AD$ and $BE$.
## Partial proof by Barycentric Coordinates
Let three said cevians intersect at point $O$. Let the barycentric coordinates of $O$ be $(X,Y,Z)$. Then, because the coordinates are homogeneous, we can say $AE=AF=X$, $BF=BD=Y$, and $CD=CE=Z$. Then, plugging the values into the equation yields:
$\frac{Y}{Z} \cdot \frac{Z}{X}\cdot \frac{X}{Y} = 1 \shadedbox$ (Error compiling LaTeX. ! Undefined control sequence.)
## Trigonometric Form
The trigonometric form of Ceva's Theorem (Trig Ceva) states that cevians $AD,BE,CF$ concur if and only if
$\frac{\sin BAD}{\sin DAC} \cdot \frac{\sin CBE}{\sin EBA} \cdot \frac{\sin ACF}{\sin FCB} = 1.$
### Proof
First, suppose $AD, BE, CF$ concur at a point $X$. We note that
$\frac{[BAX]}{[XAC]} = \frac{ \frac{1}{2}AB \cdot AX \cdot \sin BAX}{ \frac{1}{2}AX \cdot AC \cdot \sin XAC} = \frac{AB \cdot \sin BAD}{AC \cdot \sin DAC}$,
and similarly,
$\frac{[CBX]}{[XBA]} = \frac{BC \cdot \sin CBE}{BA \cdot \sin EBA} ;\; \frac{[ACX]}{[XCB]} = \frac{CA \cdot \sin ACF}{CB \cdot \sin FCB}$.
It follows that
$\frac{\sin BAD}{\sin DAC} \cdot \frac{\sin CBE}{\sin EBA} \cdot \frac{\sin ACF}{\sin FCB} = \frac{AB \cdot \sin BAD}{AC \cdot \sin DAC} \cdot \frac{BC \cdot \sin CBE}{BA \cdot \sin EBA} \cdot \frac{CA \cdot \sin ACF}{CB \cdot \sin FCB}$
$\qquad = \frac{[BAX]}{[XAC]} \cdot \frac{[CBX]}{[XBA]} \cdot \frac{[ACX]}{[XCB]} = 1$.
Here, sign is irrelevant, as we may interpret the sines of directed angles mod $\pi$ to be either positive or negative.
The converse follows by an argument almost identical to that used for the first form of Ceva's Theorem.
## Problems
### Introductory
• Suppose $AB, AC$, and $BC$ have lengths $13, 14$, and $15$, respectively. If $\frac{AF}{FB} = \frac{2}{5}$ and $\frac{CE}{EA} = \frac{5}{8}$, find $BD$ and $DC$. (Source)
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1. ## Series convergence
if a1+a2+a3.. decreasing sequence of positive numbers
How to prove that if a1+a2+a3 ... converges then lim (if n to infinity) of na(subn)=0
2. $\displaystyle \sum \frac{1}{n}$ is divergent, so for all $\displaystyle \varepsilon>0$, if $\displaystyle n$ is enough big,
$\displaystyle |a_n|\leq \frac{\varepsilon}{n}$, ie $\displaystyle |n a_n|\leq \varepsilon$.
3. I do not uderstand why we look at 1/n since it diverges,and in the problem series converges
4. Originally Posted by drpawel
How to prove that if a1+a2+a3 ... converges then lim (if n to infinity) of na(subn)=0
This result is not true without some further restriction on the terms a_n. For example, if $\displaystyle a_n = \begin{cases}1/\sqrt n&\text{if$n=k^4$for some integer$k$},\\0&\text{otherwise},\end{cases}$
Then $\displaystyle \sum a_n = \sum_k 1/k^2$, which converges. But if $\displaystyle n=k^4$ then $\displaystyle na_n=\sqrt n$, so $\displaystyle a_n\not\to0$.
You need an extra condition such as requiring that the sequence (a_n) is decreasing, in order for the result to be true.
5. Thabk you for that correction.That sequence is indeed decreasing sequence of positive numbers.So,how to prove it.
6. Originally Posted by drpawel
Yes, indeed the sequence is decreasing sequence of the positive numbers.How can i prove that now?
I think you need to roll up your sleeves and wheel out some heavy-duty epsilons in order to prove this.
Let $\displaystyle \varepsilon>0$. By the Cauchy criterion for convergence, there exists an integer N such that $\displaystyle \sum_{j=m}^na_j<\varepsilon/2$ whenever n>m>N. In particular, $\displaystyle \sum_{j=N+1}^{N+p}a_j<\varepsilon/2$ for all p>0. If the sequence $\displaystyle (a_n)$ is decreasing, then $\displaystyle pa_{N+p}\leqslant\sum_{j=N+1}^{N+p}a_j<\varepsilon/2$, and therefore $\displaystyle (N+p)a_{N+p}<\frac{N+p}p\varepsilon/2$. But if p>N then $\displaystyle \frac{N+p}p<2$. It follows that if n>2N then $\displaystyle na_n<\varepsilon$. Therefore $\displaystyle na_n\to0$ as $\displaystyle n\to\infty$.
7. Originally Posted by drpawel
and therefore
If you have a specific follow-up question you will need to state it more clearly than this ....
8. Originally Posted by mr fantastic
If you have a specific follow-up question you will need to state it more clearly than this ....
I have some problems with my computer,I am trying to fix it, I did not do this by purpose.I appologise.
9. Originally Posted by Opalg
I think you need to roll up your sleeves and wheel out some heavy-duty epsilons in order to prove this.
Let $\displaystyle \varepsilon>0$. By the Cauchy criterion for convergence, there exists an integer N such that $\displaystyle \sum_{j=m}^na_j<\varepsilon/2$ whenever n>m>N. In particular, $\displaystyle \sum_{j=N+1}^{N+p}a_j<\varepsilon/2$ for all p>0. If the sequence $\displaystyle (a_n)$ is decreasing, then $\displaystyle pa_{N+p}\leqslant\sum_{j=N+1}^{N+p}a_j<\varepsilon/2$, and therefore $\displaystyle (N+p)a_{N+p}<\frac{N+p}p\varepsilon/2$. But if p>N then $\displaystyle \frac{N+p}p<2$. It follows that if n>2N then $\displaystyle na_n<\varepsilon$. Therefore $\displaystyle na_n\to0$ as $\displaystyle n\to\infty$.
Can you plese explain why (N+p)asub(N+p)<(N+p)/P, what about epsilon/2.Is it multiply time (N+p)/P.
Plese explain a little bit since I am lost.
10. Originally Posted by drpawel
Can you plese explain why (N+p)asub(N+p)<(N+p)/P, what about epsilon/2.Is it multiply time (N+p)/P.
Plese explain a little bit since I am lost.
In the sum $\displaystyle \sum_{j=N+1}^{N+p}a_j$, each term is greater than or equal to the last term (since the sequence $\displaystyle (a_j)$ is decreasing). So the sum is $\displaystyle \geqslant pa_{N+p}.$ That gives the inequality $\displaystyle pa_{N+p}<\varepsilon/2.$
But we're interested in what happens to $\displaystyle na_n$. So it would be more useful to know something about $\displaystyle (N+p)a_{N+p}$ rather than $\displaystyle pa_{N+p}$. That's the reason for multiplying both sides of the previous inequality by $\displaystyle \frac{N+p}p$, getting $\displaystyle (N+p)a_{N+p}<\frac{N+p}p\varepsilon/2.$
I hope that makes it a bit clearer. As I said in my previous comment, this proof is fairly heavy-duty analysis. As far as I know, there are no easy shortcuts for proving this result.
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# How do you simplify (2+3i)(-3-7i)?
Feb 7, 2016
$15 - 23 i$
#### Explanation:
color(blue)("("2+3i)")")color(red)("("-3-7i"))
color(white)("XXX")=color(blue)(2)color(red)("("-3-7i")")+color(blue)(3i)color(red)("("-3-7i"))
color(white)("XXX")=color(brown)(-6-14i)+color(green)(-9i-21i^2
color(white)("XXX")=color(brown)(-6-14i)+color(green)(-9i+21
$\textcolor{w h i t e}{\text{XXX}} = 15 - 23 i$
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# Project Euler Solution #14
I have the following problem (from ProjectEuler.net - Problem 14)
The following iterative sequence is defined for the set of positive integers:
``````n -> n/2 (n is even)
n -> 3n + 1 (n is odd)
``````
Using the rule above and starting with 13, we generate the following sequence:
``````13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
``````
It can be seen that this sequence (starting at `13` and finishing at `1`) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at `1`.
Which starting number, under one million, produces the longest chain?
NOTE: Once the chain starts the terms are allowed to go above one million.
I used:
`````` static int road (int n)
{
while (n != 1)
{
if (n % 2 == 0)
n = n / 2;
else
n = 3 * n + 1;
}
}
static void Main(string[] args)
{
int max = 0, num = 0;
for (int i = 1; i < 1000000; i++)
{
{
num = i;
}
}
Console.WriteLine(num);
}
``````
But no output is printed.
-
(I'm not going to give you a complete solution since Project Euler is intended to make you think, not us who already solved the problems.)
Try figuring out how large the values in your chain are going to be and keep in mind the limits for integral types.
-
Thanks I figured it out :) – Novak Mar 8 '12 at 12:17
Good good. I didn't want to point out the flaw directly ;-) – Joey Mar 8 '12 at 12:18
+1 for the correct level of steerage – spender Mar 8 '12 at 12:22
@user1162727: Some of the problems in Project Euler overflow long integers as well; you might consider going to BigInteger, which can be arbitrarily large. They are slower and take more memory of course. – Eric Lippert Mar 8 '12 at 14:38
`long` suffices here, though. Ah well, and back when I solved PE problems C# didn't even have BigInteger so I did those in Java (and in some cases implemented my own big integers (for adding and multiplying by two it's not that much work :-)). – Joey Mar 8 '12 at 14:50
``````function problem14(){
function r(p,n){
return cl(p)>cl(n)?p:n;
}
function c(n){
return n%2===0?n/2:3*n+1;
}
function cl(n){
var _a = 1;
var _n = n;
while(_n !== 1){
_a++;
_n = c(_n);
}
return _a;
}
var b = [];
var m = 20;
var i = 500000;
while(i < 1000000){
var _n = cl(i);
if(_n>m){
b.push(i);
m = _n;
}
i++;
}
return b.reduce(r);
}
``````
Here is my js code.
-
It is not that "nothing is output", it is just running very long. If you change the upper bound of the for-loop to 100000, you will see that pretty quickly you get output. The reason it runs very long is that you use unchecked integers and you don't get an overflowexception where you should want one. Break after a few seconds an you'll see a negative `n`.
Try the following , i.e. with the `checked` keyword, it will illustrate what I mean:
``````// will now throw OverflowException with large n
checked
{
while (n != 1)
{
if (n%2 == 0)
n = n/2;
else
n = 3*n + 1;
}
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Question
# The perimeter of a parallelogram is 60cm and the ratio of its adjacent sides is 3:2. If the altitude corresponding to larger side of the parallelogram is 5cm. Find the area of the parallelogram and the altitude corresponding to the smaller side.
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Solution
## let larger side is 3xthen another side is 2xthus perimeter of a parallelogram is 10x=60thus x=6cm Then one side is =6×3=18And other is =2×6=12So base=18cmarea of parallelogram is base × heightarea=18×5=90cm2Altitude corresponding to smaller side 90=12×hh=7.5cm
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1. ## Inequality problem
Could anyone help me solving this inequality?
a,b,c are positive real numbers, proove that:
[a/(b+c) + 1/2]*[b/(a+c) + 1/2]*[c/(a+b) + 1/2] >- +1
Thank You
I tried to do it with LaTEx:
$( \frac{a}{b+c} + \frac{1}{2})*( \frac{b}{a+c} + \frac{1}{2})*( \frac{c}{a+b} + \frac{1}{2}) \geq 1$
OR
$[ \frac{a}{b+c} + \frac{1}{2}][ \frac{b}{a+c} + \frac{1}{2}][ \frac{c}{a+b} + \frac{1}{2}] \geq 1$
2. ## Got it!
Hey I succeeded in prooving the inequality. The solution :
$\frac{(2a+b+c)(2b+a+c)(2c+a+b)}{8(b+c)(a+c)(a+b)} \geq 1$
when we multiply the whole inequality by $8(b+c)(a+c)(a+b)$ and get rid of brackets, we get :
$2a^3+2b^3+2c^3-a^2b-ab^2-b^2c-a^2c-ac^2-bc^2 \geq 0$
$a^3+b^3-a^2b-ab^2 = (a+b)^3-4a^2b-4ab^2$
$a^3+c^3-a^2c-ac^2 = (a+c)^3-4a^2c-4ac^2$
$b^3+c^3-b^2c-bc^2 = (b+c)^3-4b^2c-4bc^2$
then :
$(a+b)^3-4ab(a+b) = (a+b)[(a+b)^2 - 4ab] = (a+b)(a-b)^2$
... same for (a+c) and (b+c)....
Finally : $(a+b) \geq 0 , (a-b)^2 \geq 0 \Rightarrow (a+b)(a-b)^2 \geq 0$
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# Solve This Pzzlr: Peculiar Equation
Neatorama presents our collaboration with Pzzlr, a site where you can always find a riddle to exercise your brain. Can you solve this one?
How would this equation be correct without changing it?
# XI + I = X
Think you know the answer? Find out at Pzzlr!
Look for a new puzzle here at Neatorama every Wednesday, and check out Pzzlr for a puzzle anytime!
(Image credit: Alex1011)
Love games and puzzles? Visit NeatoPuzzles for more!
Aahhh... thanks for the explanation - I completely didn't recognize it as roman numerals, but it is so obvious now.
(as with other puzzlr questions, it's interesting that there can be alternate explanations than was intended, ie the algebraic interpretation)
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This is not an algebraic equation, these are Roman numerals. The Roman numeral X equals 10, the Roman numeral I equals 1. However, the way the numerals add up is dependent on their order. If the smaller numeral is after the larger one, their values are added together, so XI has a value of 11, whereas if you place the smaller numeral before the larger one, you subtract the smaller from the larger, meaning IX equals 9. So the original equation XI + I = X would mean 11 + 1 = 10 which is incorrect. The mirror image of that becomes X = I + IX which is 10 = 1 + 9 which is indeed true.
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I may be drunk on chocolate, but I don't understand why the mirror/flip solution makes it 'correct'. the better answer on pzzlr is "(X=0, I=0) as well as (X=-2, I=2) would satisfy the equation or make it correct"
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24 March, 23:19
# I am a three-digit number. My second digit is 4 times bigger than the third digit. My first digit is 3 less than my second digit. Who am I?
+1
1. 25 March, 02:29
-2
285
Step-by-step explanation:
use algebra
third is x
second is 4x
first is 4x-3
x+4x+4x-3
9x-3 = x+x+1+x+2
9x-3=3x+3
12x=6, divide x=2
third is 2, sec is 8, first is 5
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# PARAM VISIONS
All about civil construction knowledge- PARAM VISIONS
### Bar bending schedule (BBS) of a pile cap.(Part-2) / Calculating the cutting length & weight of reinforcement in a pile cap.
👈 Back
5. Side face or roundup bar:
Theoretically, the side face bars are the stirrups to hold the pile cap reinforcements as shown in the drawing.
But practically, if the size of the pile cap is large, the side face bars need lapping as the length exceeds 12m. So, in such cases, the side face bars are provided in two equal U-shape parts as shown below.
The cutting length of the side face bar
= [{2nos. × ( a +b )} +( 2nos. × hook length) - (3nos. × 90° bend) - (2nos. × 135° bend)]
Where,
a = [ pile width - ( 2 × cover) - (2nos. ×1/2 × dia. of bar)]
= [ 2500mm. - ( 2 × 50mm. ) - ( 2nos. × 1/2 × 12mm.)]
a = 2388mm.
b= [ pile length - ( 2 × cover) - (2nos. ×1/2 × dia. of bar)]
= [ 3200mm. - ( 2 × 50mm. ) - ( 2nos. × 1/2 × 12mm.)]
b = 3088mm.
The cutting length of the side face bar
= [{2nos. × ( 2388mm. + 3088mm. )} +( 2nos. × 10d) - (3nos. × 2d) - (2nos. × 3d)]
= [{2nos. × ( 5476mm. )} + ( 2nos. × 10 × 12mm.) - (3nos. × 2 × 12mm.) - (2nos. × 3 × 12mm.)]
[{10958mm.} + ( 240mm.) - (72mm.) - (72mm.)]
= 11054mm.
= 11.054m. < 12m.
Hence no lapping is required.
B. No of rebars:
1. No. of bottom bar layer -1 required
= [{ length of pile cap - (2nos. × clear cover)} ÷ c/c distance ] + 1
= [{ 3200mm. - (2nos. × 50mm.)} ÷ 125mm. ] + 1
= [{ 3100mm.} ÷ 125mm. ] + 1
= 24.8 +1
= 25.8.
By rounding off = 26 nos.
2. No. of Top bar layer -1 required
= [{ length of pile cap - (2nos. × clear cover)} ÷ c/c distance ] + 1
= [{ 3200mm. - (2nos. × 50mm.)} ÷ 150mm. ] + 1
= [{ 3100mm.} ÷ 150mm. ] + 1
= 20.66 +1
= 21.66
By rounding off = 22 nos.
3. No. of bottom bar layer -2 required
= [{ width of pile cap - (2nos. × clear cover)} ÷ c/c distance ] + 1
= [{ 2500mm. - (2nos. × 50mm.)} ÷ 125mm. ] + 1
= [{ 2400mm.} ÷ 125mm. ] + 1
= 19.2 +1
= 20.2
By rounding off = 21 nos.
4. No. of top bar layer -2 required
= [{ width of pile cap - (2nos. × clear cover)} ÷ c/c distance ] + 1
= [{ 2500mm. - (2nos. × 50mm.)} ÷ 150mm. ] + 1
= [{ 2400mm.} ÷ 150mm. ] + 1
= 16 +1
= 17 nos.
Now, let us prepare a BBS table for the pile cap.
Sl. No. Type of Bar Dia. in mm. Nos. Length in m. Total length in m. Weight in Kg/m. Total bar wt. in kg. 1. Bottom bar Layer-1 20 26 3.806 98.956 2.467 244.12 2. Bottom bar Layer-2 20 21 4.466 93.786 2.467 231.37 3. Top bar Layer-1 12 22 4.004 88.089 0.887 78.13 4. Top bar Layer-2 12 17 4.68 79.56 0.887 70.57 5. Side face bar 12 5 11.054 55.27 0.887 49.02 6. Total weight of steel bars = 673.21 7. Add 3% wastage = 20.20 8. The grand total wt. of rebar's in pile cap = 693.41
Thank you for going through these calculation steps. Have a good day 😄.
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# A Line is a Subspace if and only if its $y$-Intercept is Zero
## Problem 663
Let $\R^2$ be the $x$-$y$-plane. Then $\R^2$ is a vector space. A line $\ell \subset \mathbb{R}^2$ with slope $m$ and $y$-intercept $b$ is defined by
$\ell = \{ (x, y) \in \mathbb{R}^2 \mid y = mx + b \} .$
Prove that $\ell$ is a subspace of $\mathbb{R}^2$ if and only if $b = 0$.
## Proof.
We must prove two statements. First we show that if $b \neq 0$, then $\ell$ is not a subspace. Then we show that if $b=0$, then $\ell$ is a subspace.
In order for $\ell$ to be a subspace, it must contain the zero element $(0, 0)$. Plugging this point into the defining equation yields $b=0$. Thus if $b \neq 0$, then $\ell$ cannot be a subspace.
Now we prove that if $b=0$, then $\ell$ is a subspace. We have already shown that $(0, 0) \in \ell$. Now suppose we have two points $(x_1, y_1) , (x_2 , y_2) \in \ell$. Then we have
$y_1 + y_2 = m x_1 + m x_2 = m(x_1 + x_2)$ and so $(x_1 + x_2 , y_1 + y_2)$ is contained in $\ell$.
Finally for $c \in \mathbb{R}$ we must show that $c(x_1 , y_1) = ( cx_1 , c y_1 )$ lies in $\ell$. We check
$c y_1 = c ( m x_1) = m (c x_1),$ and so $(c x_1 , c y_1 ) \in \ell$. This proves that if $b=0$, then $\ell$ is a subspace.
##### Determine the Values of $a$ so that $W_a$ is a Subspace
For what real values of $a$ is the set $W_a = \{ f \in C(\mathbb{R}) \mid f(0) = a \}$...
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# Rounding ~
Updated: Sep 4, 2023
Let's now shift our focus to the concept of rounding in mathematics.
Rounding is a method in math that helps us estimate the value for a number, making it shorter, simpler, and easier to work with. Rounding is a very necessary skill that we do all of the time. There are few times where the numbers we give have to be exact: putting in pin #, social security #’s, phone numbers and the like. All other number can be approximated because the exact number doesn’t matter, more than the magnitude of the number. If we never rounded numbers, we would always use the numbers with all their little decimal parts. It would make math much more challenging and confusing.
For example, if we measured something like the length of a pencil, instead of saying it's about 6 inches long, we would have to say it's something like 6.000000000000... inches. That would be a lot of zeros to write, and it's not very practical.
For everyday situations, rounding is helpful because we don't always need all the extra decimal parts to get a good enough answer.
So, rounding is like a helpful shortcut in math that makes things simpler and more manageable, especially when dealing with measurements and calculations in real life.
Properties of Rounding:
1. Nearest Value: Rounding involves finding the nearest value to a given number based on a specific digit. For example, when rounding 57 to the nearest ten, we look at the digit in the tens place, which is 5. Since 5 is closer to 60 than 50, we round 57 to 60.
2. Rounding to Tens, Hundreds, etc.: Rounding can be done to various place values, such as tens, hundreds, and beyond, depending on the level of precision needed in a problem.
3. Rounding Half-Up: A common rounding method is the half-up rule, where any digit equal to or greater than 5 is rounded up, and any digit less than 5 is rounded down. For instance, when rounding 4.7 to the nearest whole number, it becomes 5, while 4.3 becomes 4.
4. Rounding Half to Even (Banker's Rounding): Another rounding method, known as rounding half to even, is used in specific situations to minimize bias. When rounding a number ending in 5, it is rounded to the nearest even number. For example, 2.5 becomes 2, and 3.5 also becomes 2.
5. Rounding for Estimation: Rounding is a powerful tool for estimation, allowing us to quickly assess the approximate value of a number. When dealing with large datasets or complex calculations, rounding helps simplify the process.
6. Rounding with Decimals: Rounding can also be applied to decimal numbers. For instance, when rounding 3.87 to the nearest tenth, we look at the digit in the hundredths place, which is 8. Since 8 is greater than or equal to 5, we round 3.87 to 3.9.
7. Consistency in Rounding: It's essential to be consistent in rounding throughout a problem or set of data to maintain accuracy and avoid errors.
Rounding is a skill that often accompanies arithmetic operations, helping us quickly assess the significance of numbers and make informed decisions. It has practical applications in various fields, including finance, engineering, and scientific research. By understanding the properties of rounding, students can confidently navigate the maze of numbers, effectively using this valuable mathematical tool to enhance their problem-solving abilities and mathematical fluency. Rounding allows us to focus on the big picture without getting lost in the complexities of precise calculations, making it an indispensable skill in our mathematical journey.
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# Universal side divisor
## Definition
A nonzero element $x$ in an commutative unital ring $R$ is termed a universal side divisor if $x$ satisfies the following two conditions:
• $x$ is not a unit.
• For any $y \in R$, either $x$ divides $y$ or there exists a unit $u \in R$ such that $x$ divides $y - u$.
Equivalently, a non-zero non-unit element is a universal side divisor if and only if the unit balls centered around its multiples cover the whole ring.
### Equivalence up to associate classes
If $x,y$ are associate elements in a commutative unital ring $R$, then $x$ is a universal side divisor if and only if $y$ is a universal side divisor. For full proof, refer: Universal side divisor property is invariant upto associates
## Examples
• In the ring of rational integers $\mathbb{Z}$, the only universal side divisors are $\pm 2, \pm 3$. $2$ is a universal side divisor because every integer is either a multiple of $2$ or differs by $1$ from a multiple of $2$. $3$ is a universal side divisor because every integer is either $0$, $1$, or $-1$ mod $3$. For any integer $n$ of absolute value greater than $3$, there is no way of subtracting a unit or zero from $2$ to get a multiple of $n$.
• In the polynomial ring over a field, the only universal side divisors are the nonconstant linear polynomials. For instance, in the ring $k[x]$, $x$ is a universal side divisor because for any polynomial, we can subtract the constant term of the polynomial to obtain a multiple of $x$, and the constant term is either zero or a unit. Similar reasoning applies for all the other nonconstant linear polynomials. On the other hand, for any polynomial of degree greater than $1$, there is no way of subtracting a unit from $x$ to get a multiple of that polynomial.
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# Non Parametric Stats
## when your data don't meet the assumptions
### WARNING - LOW POWER ZONE
If your data meet (or approximate) assumptions of parametrics, they are generally more powerful
Monte-Carlo techniques are also often more powerful than non-parametrics
However, non-parametrics simpler to use than MC
### Rank-Order Statistics
Non-parametrics are known as rank-order tests, because they work by ranking observations and analyzing these ranks, rather than the data themselves.
To use non-parametrics with continuous values, you have to discard a lot of information.
We will talk about non-parametrics in relation to their parametric equivalents.
### Non-Parametric Regression
Non-parametric regression techniques exist but are not commonly used.
There are, however, several non-parametric correlation techniques that are widely used.
### Spearman's Rho
X and Y values are ranked separately, and the Pearson's product-moment coefficient ($$r$$) is computed on these ranks
x <- c(0.9, 6.8, 3.2, 2.4, 1.2, 1.1)
y <- c(0.1, 4.5, 5.4, 1.5, 1.9, 4.1)
### Spearman's Rho
rank_x <- rank(x)
rank_y <- rank(y)
rank_x
[1] 1 6 5 4 3 2
rank_y
[1] 1 5 6 2 3 4
### Spearman's Rho
cor(x, y, method = "pearson")
[1] 0.5590485
cor(x, y, method = "spearman")
[1] 0.7142857
cor(rank_x, rank_y, method = "pearson")
[1] 0.7142857
### Kendall's Tau
Alternative to Spearman….
• Rank observations
• Examine each pair of observations, determine whether they match or not
• Compute $$\tau$$
$\tau = \frac{(number\ of\ matched\ pairs) - (number\ of \ non\ matched\ pairs)}{\frac{1}{2}n(n-1)}$
Note: the denominator is the total number of pairwise comparisons.
### Kendall's Tau
cor(var1, var2, method="kendall")
[1] 0.9555556
### Non-Parametric t-test
Mann-Whitney U, also known as the Wilcoxon Rank-Sum
• rank observations, ignoring group
• sum the ranks belonging to each group
• calculate the test statistic
$U = R - \frac{n(n+1)}{2}$
• $$R$$ is the summed ranks, and $$n$$ is the group sample size
• do this for both groups, and take the smallest as the test statistic
• compare to known distribution under null hypothesis
### Mann-Whitney U / Wilcoxon Rank-Sum
x <- rnorm(10, mean=5)
y <- rnorm(10, mean=7)
wilcox.test(x,y)
Wilcoxon rank sum test
data: x and y
W = 3, p-value = 7.578e-05
alternative hypothesis: true location shift is not equal to 0
### Non-Parametric ANOVA
Kruskal-Wallis
• Rank all observations
• Calculate the average rank within each group
• Compare the average rank within group the the overall average of ranks, using a weighted sum-of-squares technique
• Compare p value of test statistic using chi-square approximation
### Kruskal-Wallis
kruskal.test(var~group)
Kruskal-Wallis rank sum test
data: var by group
Kruskal-Wallis chi-squared = 1.0566, df = 1, p-value = 0.304
### Goodness of Fit Test
Kolmogorov-Smirnov Test
Non-parametric test to determine whether two distributions differ
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ch 12 calc
# ch 12 calc - product: v=mang(a*(b x c)) Equations-r =
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Vectors -(x-h)^2+(y-k)^2+(z-l)^2=r^2 -x^2+y^2+z^2=r^2 -a = <x(2)-x(1),y(2)-y(1),z(2)-z(1)> Dot Product -a*b=a(1)b(1)+a(2)b(2)+a(3)b(3) 1. a*a=magn(2)^2 2. a*b=b*a 3. a*(b+c)=a*b+a*c 4. (ca)*b=c(a*b)=a*(cb) 5. 0*a=0 -a*b=magn(a)magn(b)(cos(theta)) -two vectors a and b are orthogonal if and only if a*b=0 -scalar projection of b onto a: comp(a)b=a*b/magn(a) -vector projection of b onto a: proj(a)b=(a*b/magn(a))a/magn(a)=a^2*b/magn(a)^2 Cross Product -a x b = <a(2)b(3)-a(3)b(2),a(3)b(1)-a(1)b(3),a(1)b(2)-a(2)b(1)> -the vector a x b is orthogonal to both a and b -mang(a x b)= magn(a)magn(b)sin(theta) -two nonzero vectors a and b are parallel if and only if a x b=0 -the length of the cross product a x b is equal to the area of the parallelogram determined by a and b 1. a x b=-b x a 2. (ca)xb=c(a x b)=ax(cb) 3. a x (b+c)=a x b+a x c 4. (a+b) x c=a x c+b x c 5. a*(b x c)=(a x b)*c 6. a x (b x c)=(a*c)b-(a*b)c -the volume of the parallelepiped determined by the vectors a,b,c is the magnitude of their scalar triple
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Unformatted text preview: product: v=mang(a*(b x c)) Equations-r = r(0)+tv-x=x(0)+at-y=y(0)+bt-z=z(0)+ct-(x-x(0))/a=(y-y(0))/b=(z-z(0))/c-the line segment from r(0) to r(1) is given by the vector equation: r(t)=(1-t)r(0)+tr(1) 0<(equal)t<(equal)1-n*(r-r(0))=0-vector equation of plane: a(x-x(0))+b(y-y(0))+c(z-z(0))=0-D=magn(ax(1)+by(1)+cz(1)+d)/sqr(a^s+b^s+c^2) Surfaces-ellipsoid: x^2/a^2+b^2/b^2+x^2/c^2=1-cone: z^2/c^2=x^2/a^2+y^2/b^2-elliptic paraboloid: z/c=x^2/a^2+y^2/b^2-hyperboloid of one sheet: x^2/a^2+y^2/b^2-z^2/c^2=1-hyperbolic paraboloid: z/c=x^2/a^2-y^2/b^2-hyperboloid of two sheets: -x^2/a^2-y^2/b^2+z^2/c^2=1 Cylindrical and Spherical Coordinates-cylindrical coordinate system:-x=rcos(theta)-y=rsin(theta)-z=z-r^2=x^2+y^2-tan(theta)=y/x-z=z-(r,theta,z)-spherical coordinates:-x=psin(ol)cos(theta)-y=psin(ol)sin(theta)-z=pcos(ol)-p^2=x^2+y^2+z^2-(p,theta,ol)...
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## This note was uploaded on 01/13/2012 for the course MATH 1 taught by Professor Staff during the Spring '08 term at UC Davis.
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ch 12 calc - product: v=mang(a*(b x c)) Equations-r =
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A231274 G.f.: Sum_{n>=0} x^n * Product_{k=1..n} (k + x) / (1 - k*x - x^2). 3
1, 1, 4, 18, 104, 736, 6232, 61632, 698144, 8917120, 126807520, 1987075872, 34018221728, 631698903712, 12645901972000, 271482140140704, 6221487421328672, 151587364647728032, 3912949321334320672, 106670353381399285920, 3062317963564624162592, 92345208262957730327968 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 COMMENTS Compare to the identity: Sum_{n>=0} x^n*Product_{k=1..n} -(k + x)/(1 - k*x - x^2) = 1 - x. Compare also to the identity: Sum_{n>=0} x^n*Product_{k=1..n} (k + x)/(1 + k*x + x^2) = (1+x^2)/(1-x). LINKS Vaclav Kotesovec, Table of n, a(n) for n = 0..268 FORMULA a(n) ~ n! / (2 * (log(2))^(n+1)). - Vaclav Kotesovec, Oct 31 2014 EXAMPLE G.f.: A(x) = 1 + x + 4*x^2 + 18*x^3 + 104*x^4 + 736*x^5 + 6232*x^6 +... where A(x) = 1 + x*(1+x)/(1-x-x^2) + x^2*(1+x)*(2+x)/((1-x-x^2)*(1-2*x-x^2)) + x^3*(1+x)*(2+x)*(3+x)/((1-x-x^2)*(1-2*x-x^2)*(1-3*x-x^2)) + x^4*(1+x)*(2+x)*(3+x)*(4+x)/((1-x-x^2)*(1-2*x-x^2)*(1-3*x-x^2)*(1-4*x-x^2)) +... PROG (PARI) {a(n)=polcoeff( sum(m=0, n, x^m*prod(k=1, m, (k+x)/(1-k*x-x^2 +x*O(x^n))) ), n)} for(n=0, 30, print1(a(n), ", ")) CROSSREFS Cf. A231352, A231291. Sequence in context: A335459 A159666 A316186 * A051827 A007711 A321278 Adjacent sequences: A231271 A231272 A231273 * A231275 A231276 A231277 KEYWORD nonn AUTHOR Paul D. Hanna, Nov 06 2013 STATUS approved
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Last modified May 17 09:03 EDT 2021. Contains 343969 sequences. (Running on oeis4.)
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# algebra 1 homework help
0 votes
Find the intersection point for 3y + 5x = 20 and 5y + 3x = 20?
asked Nov 5, 2013
## 1 Answer
0 votes
3y+5x = 20_______ (1)
5y+3x = 20 _______(2)
(!)*5 - (2)*3
15y+25x = 100
15y+9x = 60
(-) (-) (-)
____________
16x = 40
Apply division property, divide by 16 to each side.
16x/16 = 40/16
x = 40/16
x = 5/2
x = 5/2 in (1)
3y + 5(5/2) = 20
3y + 25/2 = 20
Apply subtraction property,subtract 25/2 from each side.
3y+25/2-25/2 = 20 - 25/2
3y = 20-25/2
3y = (40-25)/2
3y = 15/2
Apply division property, divide by 3 to each side.
3y/3 = 15/6
y = 15/6
y = 5/2
check
3*5/2 + 5*5/2 = 20
15/2 + 25/2 = 20
40/2 = 20
20 = 20
x, y values are coordinate pair of intersection point.
So the intersection point to given lines is (5/2,5/2)
answered Nov 5, 2013
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How do you find the quotient (6t^3-9t^2+6)div(2t-3) using long division?
May 31, 2017
The quotient is $= 3 {t}^{2}$ and the remainder $= 6$
Explanation:
We perform a long division
$\textcolor{w h i t e}{a a a a}$$2 t - 3$$\textcolor{w h i t e}{a a a a}$$|$$6 {t}^{3} - 9 {t}^{2} + 0 t + 6$$\textcolor{w h i t e}{a a a a}$$|$$3 {t}^{2}$
$\textcolor{w h i t e}{a a a a a a a a a a a a a a a}$$6 {t}^{3} - 9 {t}^{2}$
$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a}$$0 - 0 + 0 t + 6$
$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a}$$+ 6$
The quotient is $= 3 {t}^{2}$ and the remainder $= 6$
$\frac{6 {t}^{3} - 9 {t}^{2} + 6}{2 t - 3} = 3 {t}^{2} + \frac{6}{2 t - 3}$
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0
721views
General solutions of critically damped system
0
6views
Case 2] : General solutions of critically damped system ($\xi = 1$) :
In this case both the roots are real and equal.
In this case, the solution is
$x = (c_1 + c_2t)e^{st} = (c_1 e^{st} + c_2te^{st})$
Differentiating
$x = c_1se^{st} + c_2te^{st} + c_2e^{st}$
$= c_1 se^{st} + c_2 (t. se^{st} + e^{st})$
Then using boundary condition,
At t = 0, x = $x_0$ and at t = 0, $x = x_0$
Here $s = - w_n$
$\therefore$ $x = -w_n c_1 e^{-wnt} - w_n tc_2 ^{-wnt} + c_2 e^{-wnt}$
$\therefore$ $x = e^{-wnt} [-w_n c_1 - c_2 (w_nt - 1)]$ - - - - (1)
Also we know displacement,
$x = (c_1 + c_{2t}) e^{-wnt}$
where $c_1$ and $c_2$ are constants to be found out by known boundary conditions.
i.e. At t = 0, x = $x_0$ and at t = 0, $x = x_0$
substituting the first boundary condition in equation (2), we get $x_0 = c_1$ and substituting the second condition in equation (1).
$\therefore$ $x = w_n c_1 + c_2$
substituting value of $c_1$ in above equation (2), we get the general displacement equation as:
$x = e^{-wnt} [ x_0 + (x_0 + w_n x_0) t]$
From the above equation, we can say the displacement decreases exponentially with respect to time, then, the response curve for critically damped system is shown in earlier displacement time curve for $\xi = 1$
from above equation it can be said that when x = 0, $t \rightarrow \infty$
from above equation it can be said that motion of critically damped system is 'a periodic'.
then, the shortest possible time which will be taken by system under critical damping condition, so that x = 0 is given by equation.
$0 = x_0 (x_0 + w_n x_0)^t$
$\therefore$ $t = \frac{-x_0}{x_0 + w_n x_0}$
For 't' to be positive magnitude of $x_0$ > magnitude of $w_n x_0$ and direction of $x_0$ must be opposite to $x_0$
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# Practice Reasoning Questions For SBI PO Prelims& NIACL 2017 (Puzzle& Alphanumeric Symbols)
Practice Reasoning Questions For SBI PO Prelims& NIACL 2017 (Puzzle& Alphanumeric Symbols):
Dear Readers, Important Practice Reasoning Questions with explanation for Upcoming Exams was given here with explanation, candidates those who are preparing for Banking and all other Competitive exams can use this practice questions.
Directions (Q. 1-6): Study the given information ,carefully to answer the given questions.
Seven athletes M, N, O, P, Q, R and S live on seven different floors of a building but not necessarily in the same order. The lowermost floor of the building is numbered one, the one above that is numbered two and so on till the topmost floor is numbered seven. Each of them runs for a different distance in marathon โ 850m, 1300m, 2200m, 2800m, 3300m, 4000m and 4700m, but not necessarily in the same order.
The one who runs for 2200m lives on the floor numbered 3. Only one person lives between O and the one who runs for 2200m. The one who runs for 4000m lives immediately above O. Only one person lives between the one who runs for 4000m and the one who runs for 1300m. The number of people living between 0 and the one who runs for 1300m is the same as that between the one who runs for 4000m and R. N lives on an odd-numbered floor. N runs for 2000m more than the one who lives on floor number 4. Only two people live between Q and the one who runs for 3300m. The one who runs for 2800m lives on one of the floors below Q but not on floor number 2. Only two people live between M and S. The one who runs for 850m lives immediately below M.
1). How many people live between S and N?
a) Three
b) One
c) Five
d) Four
e) One
2). Who amongst the following live(s) between P and the one who runs for 1300m?
a) Both Q and R.
b) Only S
c) Both R and the one who runs for 850m
d) Only the one who runs for 4000m
e) Both R and the one who runs for 2200m
3). As per the given arrangement, four of the following live are alike in a certain way and so form a group. Which of the following does not belong to the group?
a) Q – 3300m
b) O – 1300m
c) Floor number 4 – S
d) Floor number 2 โ R
e) Floor number 7 – 1300m
4). Which of the following statements is true with respect to the given arrangement?
a) Only two people live between P and O.
b) Q runs for 4000m.
c) N lives on floor number 7.
d) The one who runs for 850m lives immediately above P.
e) None of these
5). If the total distance covered by B and M is 5300m, then how much did B runalone?
a) 2000m
b) 4000m
c) 3100
d) 1300m
e) 600m
6). Who amongst the following runs for 2200m?
a) P
b) N
c) Q
d) R
e) S
Directions (Q. 7-11): Study the following arrangement of letters, numbers and symbols carefully to answer the given questions.
9 ยฃ 1 & L Y ยฉ E K S R 8 % W H 7 \$ 5 U G 4 # 6 2 N A 3 @ Z * D
7). As per the given arrangement, four of the following five are alike in a certain way and hence form a group. Which of the following does not belong to the group?
a) W\$H
b) %H8
c) U#G
d) 3*@
e) 1Y&
8). How many letters are there between the eighth element from the right and the seventh letter from the left end of the given arrangement?
a) Nine
b) Six
c) Eight
d) Five
e) Ten
9). Which of the following will be ninth to the right of the sixteenth element from the right end of the given arrangement?
a) #
b) 3
c) Z
d) 6
e) N
10). What will be the sum of all the numbers between the tenth element from the left end and the tenth element from the right end of the given arrangement?
a) 28
b) 21
c) 24
d) 18
e) 19
11). If all the letters from the given arrangement are deleted, then which of the following will represent the fifth element to the right of ‘4’ and the fourth element to the left of% respectively?
a) 3,&
b) *, @
c) @, ยฉ
d) 3,1
e) @, 1
Directions (Q. 1-6):
1).
2).
3).Except C, there is same relation: There is a gap of two floors.
4).
5).B + M = 5300
B = 5300 โ 4700 = 600 m
6).
7).W (+3) \$ (-2) H; % (+2) H (-3) 8; U (+3) # (-2) G;
3 (+3) * (-2) @; 1 (+3) Y (-2)&;
8).Eighth from the right is 2.
Seventh from the left is ยฉ.
Hence there are eight letters between 2 and ยฉ, viz E, K, S, R, W, H, U and G.
9).Nine to the right of the sixteenth from the right = (16 โ 9 =) 7th from the right end, ie N.
10).10th element from the left = S
10th element from the right = #
Now, the number between S and # are 8, 7, 5 and 4.
Their sum = 8 + 7 + 5 + 4 = 24
11).New arrangement becomes 9 ยฃ 1 & ยฉ 8 % 7 \$ 5 4 # 6 2 3 @ *
Fifth element to the right of ‘4’ is @ and
fourth element to the left of ‘%’ is I. So @,1.
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# Counting Without Counting: An Introduction
On the second day of Precalculus class, before embarking on our starting unit on Combinations and Permutations, I put students in groups and had them work on this packet. I’m including it below with my teacher notes in the margin, and also without my teacher notes in the margin:
(Also the .docx file so you can modify, cut, paste, hack apart! To see with teacher notes, go to the “Review” tab and click on “Final: Show Markup” and to see it without teacher notes, click on “Final.”)
This was a two day activity, with the groups working together, and every so often, I would stop and we would go over some of the problems. Some problems were assigned for nightly homework, and because of time issues, I had to cut out the Applebee’s problem completely. I’m quite proud of some of the problems… namely the Applebee’s problems, the bit.ly / QR code problems, and the Mozart’s Minuet problem (largely taken from here, and modified with some extensions).
The goal of the packet was fourfold:
1) I wanted problems which promoted thinking, conversation, etc., before students were introduced to formulas, notation, etc.
2) I wanted students to understand that “counting without counting” means that instead of listing all possibilities and counting, there are often other faster ways to get answers. This is the essential understanding that I would hope for in any unit on combinations and permutations. In order to do this, students will need to organize your information in some special, logical way. Usually this requires students to multiply numbers. But students need to really understand why multiplication (rather than another operation, like addition).
3) I wanted students to work in groups, so the problems were designed to be conducive for groupwork.
4) I wanted students to get some sense of what huge numbers mean.
How did it work? Overall, I think it worked pretty well. I gave groups “hint tokens” and most didn’t rely on them for a hint. Most students were able to see that you have to multiply for most of the problems, but most had trouble explaining why. Finally, most had never seen a tree diagram. In the future, I honestly think having students draw a complete tree diagram and explain what each leaf means would be useful. Adding two questions like the following would help:
1. If you have the letters A, B, C, D, E, and F and you want to write a three letter code and you are allowed repetition, what would the tree diagram look like. Make all the branches. Then pick a single “leaf” of the tree and explain what that leaf means.
2. If you have the letters A, B, C, D, E, and F and you want to write a three letter code and you are not allowed repetition, what would the tree diagram look like. Make all the branches. Then pick a single “leaf” of the tree and explain what that leaf means.
The reason I say this is that I’ve been collecting homework problems, and the tree diagrams some of the kids are constructing are just nonsense.
I probably should write more about this, but I’m exhausted and all I want to do is sleep. (I’ve been sick since Monday.) I suppose I should end adding that I’m teaching a Precalculus Advanced class. I don’t think I would have these problems for a standard Precalculus class… I would use fewer of them, and I would scaffold them more, and build in more “listing” of things, rather than go straight into “how many different ways…?”
1. I used a modified version of this in my class this year. Here are some of my suggestions:
On #0, make there be only 23 bolts. That way you can say that when you count with the boxes it’s easier, but you actually counted TWO things (first the boxes, then the empty boxes, then subtracted) and that was still easier than counting the bolts without organization.
I added questions scaffolding in the tree diagram like you suggested around number 1 (dinosaurs). I had them draw a complete one for a smaller number of things and then next to each leaf draw (and color) what that dinosaur would look like.
On #8 you mention it’s “only \$6.99!” so I included a question about how much money you’d have to pay for all those combinations.
Overall it went ok. I didn’t give out “tokens” but I was liberal with my hints and they caught on.
1. Could you post your version on your blog? I’d love to see it…
Do you think you would use it again next year?
Sam
1. Well, mine was 99% the same as yours. I just retyped it (since I couldn’t find the docx). I do think I’ll use it again next year. They weren’t as excited about the minuet as I thought they might be, but maybe that’s just my all-boys school or something?
2. Love the introduction! I stole it, but unfortunately our Precalculus curriculum doesn’t start doing probability and combinatorics until mid-spring. Hopefully all the other notation and formulas we talk about until then won’t get in the way of their exploring by the time we get to this unit!
3. Yo, just wanted to say I think the two tree diagrams you mentioned are a great idea! This sounds like such a fun topic to teach. Also, I love that this whole post really brought me back to KSI.
4. NikP says:
Whats the answer to the QR code problem?
1. I get $2^{\text{number of squares}}$ because each square can either be on (white) or off (black). So two choices for each square.
5. hschuchhardt says:
I realize this is an old post, but we started our probability unit today in Algebra 2 and I used lots of this packet in my class. Next year I’d like to begin the unit with the bit.ly question and have it more open ended – have the students begin class just thinking about if there are enough bit.ly’s for all the websites…what information do we need to know? How can we find it? How do we calculate these numbers? This might make the class more about the problem and less about the procedure.
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How many yards of fabric is needed for making a raincoat.
To make one raincoat you need 1.6 to 1.8 meters of cloth when the fabric width is 48-50 inches.
In this post, I am sharing a simple method of calculating fabric requirement for a raincoat. The same method can be used for estimating cloth requirement for any other garments.
Let me explain with the following example. This question was asked by an OCS reader and shared following measurement of his raincoat.
Raincoat measurements:
Length = 46 inches,
Width = 47 inches (body circumference)
Sleeve length =24 inches and
Sleeve width=17 inches (circumference of sleeve near armhole)
### What would be the fabric requirement per raincoat in meters?
From the above measurement, the combined front and back body block dimension is 46 x 47.
If you buy a fabric of 48-50 inches wide, you can cut the front and back parts side by side. So you need fabric length for the body parts at least 46 inches + stitching margin plus cutting margin. Assume the margin fabric including both ends is 1 inch. So the fabric length for the body parts is 47 (46+1) inch.
Now you need to cut two sleeves. A 50 inches fabric width is enough for cutting two sleeves side by side. So the length of the fabric required for sleeve components 17+1 = 18 inches. 1 inch is added as cutting and stitching margin.
You need fabric for other small parts like the collar, flaps, side pockets and chest pocket. After cutting two sleeves, you will have access fabric of 17x16 (50 inches- 34 inches = 16 inches), you can use that fabric for cutting these small parts. You don't need addition fabric lengthwise. Hood is also a part of the raincoats. Raincoat hood’s measurements are not given, so I have not considered that part. In case your raincoat has a hood, you much take that part in fabric consumption calculation.
Based on the above information, for one piece of raincoat, you need fabric of 72 inches long. This is equal to 1.8 meters of fabric (fabric width 48-50 inches). To convert the length in yard multiply the 1.8 by 1.09361. In this case, you need 1.97 yards.
In case 17x16 block is not enough for these small parts, you have to use fabric lengthwise. fabric length will increase. Calculate that using the actual raincoat patterns.
Also read: How to calculate fabric consumption of garment?
### Fabric consumption for bulk production
When you make bulk quantity, fabric requirement will come down. In the above case, you can see there is excess fabric is the side of the sleeve parts. You can cut 3 sleeves side by side. For example, if you make 3 coats, you can cut 3 pairs of sleeves in two rows instead of 3 rows. Thus average fabric requirement would be 1.6 meters (approx) per coat.
Also read: How many meters of cloth do I need for a shirt?
Notes:
• Stitching margin normally kept 1 centimeter. In a lay of multiple markers, the cutting margin is less than one inch.
• Fabric requirement would vary for different sizes of the raincoat
Name
5S,3,Activewear,1,Apparel ERP,1,apparel industry,58,apparel news,34,Apparel Production,31,Apparel Software,18,Bamboo Fabric,1,Bangladesh,2,Blogs,1,Books,31,Business Guides,9,Buyer,1,Buying House,4,CAD/CAM,1,Calculation & Formula,1,career,34,Change Management,1,Clothing,2,Clothing Brand,3,Compliance,3,Consumption,4,Costing,11,Cutting,21,Denim,3,Directory,3,Dyeing,1,eCommerce,2,Education,5,Event & Fair,15,Fabric,73,Fashion,5,Fashion Business,41,Fashion Design,3,Fashion Show,2,Fiber,2,Finishing,10,Garment Exporters,1,Garmenting,2,Gartex,1,GSM,1,Guest Post,8,Human Resource,3,Industrial Engineering,215,Infographic,5,Knitting,4,KPI,7,Kurti,1,Lean,27,Line layout,4,Living Wages,1,Machines,32,Manufacturing,53,Merchandising,106,MIS,24,MISC,24,Operation bulletin,14,Operator,1,Operator Training,1,Performance rating,1,PLM,3,Printing & Embroidery,11,Problem Solving,1,Production Planning,27,Productivity,1,Project Report,1,QNA,122,Quality,61,Quality manual,2,Quality system,14,Research project,1,Resources,44,Retailing,4,Salary & Wages,7,Sampling,22,Scheduling,3,Shirt Making,2,Sourcing,8,Spec Sheet,9,Sportswear,4,Stitching,3,Store,2,Stretch fabric,1,Student corner,1,sustainable clothing,7,T-shirt,3,Tailor,1,Tailoring Shop,2,Technology,60,Texprocess,1,Textile processing,11,Textile testing,23,Textiles,42,Threads,1,Towel,1,Trims & Accessories,16,videos,12,Waste,1,Work Study,38,Yoga wear,1,
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Online Clothing Study: How Many Meters of Cloth Do I Need for a Raincoat?
How Many Meters of Cloth Do I Need for a Raincoat?
How many yards of fabric is needed for making a raincoat.
https://4.bp.blogspot.com/-MByn2X3r_gc/WTwF4FfBrXI/AAAAAAAAEmQ/AsHZGtUeH58BSgCakQV022wMjiJiVgpEgCK4B/s640/raincoat-fabric-requirement.png
https://4.bp.blogspot.com/-MByn2X3r_gc/WTwF4FfBrXI/AAAAAAAAEmQ/AsHZGtUeH58BSgCakQV022wMjiJiVgpEgCK4B/s72-c/raincoat-fabric-requirement.png
Online Clothing Study
http://www.onlineclothingstudy.com/2017/06/how-many-meters-of-cloth-do-i-need-for_11.html
http://www.onlineclothingstudy.com/
http://www.onlineclothingstudy.com/
http://www.onlineclothingstudy.com/2017/06/how-many-meters-of-cloth-do-i-need-for_11.html
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# Linear Algebra Problem
motherh
Hey guys, I'm having problems with a question.
Let P be an invertible matrix and assume that A = PMP$^{-1}$. Where M is
M = [{3,1,0}{0,3,0}{0,0,2}]
Find a matrix B(t) such that e$^{tA}$ = PB(t)P$^{-1}$.
Now this might be an easy problem, but I really have no idea what to do because my lecturer is so bad and the book for the course doesn't cover this material.
I have seen something about A= PBP$^{-1}$ implying e$^{tA}$ = Pe$^{tB}$P$^{-1}$ so I have tried computing the exponential of M, but to no avail. Any advice is much appreciated.
Homework Helper
Hey guys, I'm having problems with a question.
Let P be an invertible matrix and assume that A = PMP$^{-1}$. Where M is
M = [{3,1,0}{0,3,0}{0,0,2}]
Find a matrix B(t) such that e$^{tA}$ = PB(t)P$^{-1}$.
Now this might be an easy problem, but I really have no idea what to do because my lecturer is so bad and the book for the course doesn't cover this material.
I have seen something about A= PBP$^{-1}$ implying e$^{tA}$ = Pe$^{tB}$P$^{-1}$ so I have tried computing the exponential of M, but to no avail. Any advice is much appreciated.
Yes, matrix exponential. This problem is fairly easy because you can split Mt into the sum of a diagonal matrix D and an offdiagonal matrix N which is nilpotent. And they commute with each other. So you can use exp(D+N)=exp(D)exp(N). Finding the exponential of each matrix is pretty easy.
Last edited:
2 people
motherh
Thank you so much, you hero! Just to check I haven't gone completely wrong, should I have
B(t) = e$^{2t}$*[{e$^{t}$,e$^{t}$,0}{0,e$^{t}$,0}{0,0,1}]?
Homework Helper
Thank you so much, you hero! Just to check I haven't gone completely wrong, should I have
B(t) = e$^{2t}$*[{e$^{t}$,e$^{t}$,0}{0,e$^{t}$,0}{0,0,1}]?
Close, but no cigar. That's only correct for t=1. Your nilpotent matrix is N={[0,t,0],[0,0,0],[0,0,0]}. What's exp(N)?
1 person
motherh
Aha, of course! I think I see where I went wrong. Is it
B(t) = e$^{2t}$*[{e$^{t}$,t*e$^{t}$,0}{0,e$^{t}$,0}{0,0,1}]?
B(t) = e$^{2t}$*[{e$^{t}$,t*e$^{t}$,0}{0,e$^{t}$,0}{0,0,1}]?
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# Momentum
January 2001 Number 09 Linear Momentum 1. What is linear moment
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##### Citation preview
January 2001
Number 09
Linear Momentum
1. What is linear momentum?
If the force acting on a body is constant, then Newton’s Second Law can be written as: Final momentum initial momentum , Force = time mv mu or F = t If the force acting is not constant, then we have: Final momentum initial momentum Average force = time
The linear momentum of a body is defined as the product of its mass and its velocity p = mv p = momentum (kgms-1)
m = mass (kg) v = velocity (ms-1)
Tip: The “linear” in linear momentum is there to distinguish it from angular momentum. A body has linear momentum if it is moving along; it has angular momentum if it is turning. Angular momentum will be covered in a later Factsheet.
Example 2. A force acting on a body of mass 2 kg causes its speed to increase from 2ms-1 to 6ms-1. Find the average force acting, given that it acts for 2 seconds, if the final velocity of the body is a) in the same direction as its initial velocity b) in the opposite direction to its initial velocity
Since velocity is a vector, linear momentum is a vector quantity – it has both magnitude and direction. Its units are kgms-1 or, equivalently, Ns (exercise: show these units are equivalent).
Tip: In examples where more than one direction is considered, it is a good idea to decide right at the beginning which direction you are going to take as positive. Anything going in the opposite direction will have a negative velocity, and hence a negative momentum
Example 1. What is the momentum of a body of mass 50 grammes moving at 20 kmh-1 due North Tip: If you would be tempted to put 50 20 = 1000 as your answer to this example – go back to the definition of momentum and look at the units!
Taking the direction of the initial motion as positive: a) Initial momentum = 2 2 = 4kgms-1 Final momentum = 2 6 = 12kgms-1 Final momentum initial momentum Average force = time 12 4 = 2 = 4N in direction of the original motion of the body
First we must change everything to the appropriate units – kg for mass and ms-1 for velocity 50 grammes = 0.05 kg 20 kmh-1 is 20 1000 = 20 000 metres per hour 50 -1 ms which is 20 000 3600 = 9 Note: it is sensible to keep this answer as a fraction to avoid rounding errors 50 = 0.278 kgms-1 due North (3 SF) So momentum = 0.05 9 Tip: To change kmh-1 to ms-1, multiply by
b) Initial momentum = 4kgms-1 Final momentum = 2 -6 = -12kgms-1 (since in opposite direction) -12 4 =-8N Average force = 2 So average force = 8N in the direction opposite to the original motion of the body
5 18
Tip. Common sense tells you that the answers to parts a) and b) cannot be the same. It will take a larger force to reverse the direction of the body, then increase its speed to 6ms-1 in the opposite direction, than it will to merely increase its speed to 6ms-1 in the same direction. Also, exam technique tells you that you will not be asked to do two identical calculations.
2. Force and Momentum Newton’s Second Law states: The rate of change of (linear) momentum of an object is proportional to the resultant force that acts upon it If standard SI units (newtons, kilograms, metres, seconds) are used, then this becomes: Resultant force = rate of change of momentum
1
Linear Momentum 3. Impulse
Impulse for a variable force
The impulse of a constant force is given by: impulse (Ns) = force(N) time for which it acts (s)
For a variable force, we can use: Impulse
Example3. A force of 6N acts for 1 minute. Find the impulse of the force.
= average force time for which it acts = final momentum – initial momentum,
or alternatively, use Impulse of force = area under force-time graph.
Impulse = 6 60 = 360Ns
In the graph below, the shaded area is the impulse exerted between time A and time B. Note that in region R, the force becomes negative – so the impulse it exerts will also be negative. Force /N
In the section above, we met the equation Final momentum initial momentum Force = time for a body moving under a constant force. Rearranging this equation, we get: Force time = Final momentum – initial momentum
t A
R
B
This gives us: Impulse = change of momentum
Example 4. A body of mass 1kg is moving with speed 2ms-1. It collides head-on with a body of mass 2kg moving with speed 3ms-1. The two bodies stick together, and move off with velocity v. Find v.
4. Conservation of momentum The principle of conservation of that:
linear momentum states Tip “Collides head on” means they were initially moving in opposite directions. If one body overtakes the other, they were initially moving in the same direction.
The total linear momentum of a system of interacting bodies remains constant, providing no external forces are acting.
Justifying the Principle of Conservation of Linear Momentum Suppose two objects, X and Y, collide with each other. During the collision, X exerts a force on Y,and Y exerts a force on X. By Newton’s third law, these forces are equal and opposite. Both forces act for the same time (however long the collision takes) Therefore, the impulses of the forces are equal and opposite. Since impulse equals change in momentum, the changes in momentum of X and Y are equal and opposite. So the total change in momentum of X and Y together is zero, since their individual changes cancel out.
2 ms-1
3 ms-1
v ms-1
1kg
2kg
3kg
BEFORE
positive
AFTER
Total initial momentum = 1 2 – 2 3 = -5kgms-1 Total final momentum = 3 v = 3v kgms-1
Using Conservation of Momentum Conservation of momentum is used when the bodies involved are free to move. You would not use it for the collision between an object and a wall, say, because the wall is not free to move.
Principle of conservation of linear momentum: -5 = 3v v = -1.67ms-1 So v = 1.67ms-1 in the original direction of motion of the 2kg mass. Example 5. An explosion causes a stationary object of mass 3kg to split into two pieces. The smaller piece, of mass 0.9kg, moves off with velocity 25ms-1. Find the velocity of the larger piece. 25 ms-1 v ms-1 0 ms-1
Calculations involving conservation of momentum involve either collisions – when you start off with two objects -or explosions/ rocket propulsion/shooting – when you start of with one object that splits into two. In either case, the following approach is helpful: First, decide in which direction you will take velocity to be positive. Draw a diagram showing the objects involved and their velocities before and after. Take all unknown velocities to be in the positive direction – if they are not, you will find out when you get a negative answer. Work out your total initial momentum and your total final momentum and equate them.
2.1kg
3kg BEFORE
0.9kg AFTER
positive Total initial momentum = 0 Total final momentum = 2.1v + 25 0.9 0 = 2.1v + 25 0.9 v = 10.7 ms-1(3SF) in opposite direction to smaller piece
2
Linear Momentum Typical Exam Question (a) (i) State Newton’s second law of motion. (ii) Define the term 'impulse'. (b) Water stored in a reservoir falls 100m to a turbine at the foot of a dam. The mass flow rate is 500kg s-1. When the# water hits the turbine, 60% of its vertical velocity is lost. Calculate the: (i) vertical velocity of the water just before it hits the turbine. (ii) momentum lost by the water per second as it hits the turbine. (iii) force exerted on the turbine.
Force-time Graphs During Collisions [2] [1]
The diagram below shows a possible graph of force against time during a collision (for example, a ball hitting a wall). The graph shows how the horizontal force on the ball varies with time. Before and after the collision, there is no horizontal force on the ball. The total change of momentum of the ball in its collision with the wall would be given by the area under the graph.
[2]
[2] [1]
Force/N
(a) (i) rate of change of momentum of a body is proportional to resultant force acting on it, & is in the direction of that force (ii) impulse acting on a body = change in momentum of the body (or: impulse of a force = magnitude of force time it acts for) (b) (i) Using: loss in p.e = gain in k.e each second 500 100 g = ½ 500 v2 44.3 ms-1 = v (ii) 500 v = 22140 kgms-1 per second (iii) impulse = 22140 Ns time = 1 second F = 22140 Ns (f.t.)
Time/ s
ball first hits wall ball increasingly deforms (“squashes”) due to force exerted by wall ball reverses direction ball starts to return to normal shape ball loses contact with wall
5. Elastic and Inelastic Collisions In most collisions, some kinetic energy is lost – it is transferred to heat and sound, or is used to damage or change the shape of the bodies involved.
The exact shape of the graph would depend on the material the ball was made of. If the ball was very hard, then it would not deform significantly during the collision, and the “spike” on the graph would be much narrower, showing that the collision is much shorter in duration. A very “squashy” ball would deform readily, leading to a flatter graph and a longer-lasting collision.
A (perfectly) elastic collision is one in which kinetic energy is
conserved An inelastic collision is one in which some kinetic energy is lost A totally inelastic collision is one in which the bodies stick
Typical Exam Question (a) An pellet of mass 0.30kg is fired with velocity 20ms-1, at a stationary ball of mass 0.10kg. The collision is perfectly elastic. (i) State the change in total kinetic energy on impact [1] (ii) Write an equation for the momentum of the pellet and the ball before and after the elastic collision. Use the symbols [1] vpellet and vball to represent the velocities after impact. (b) For a perfectly elastic collision, assuming all velocities lie in the same straight line, it can be shown that: velocity of approach = velocity of separation i.e. i.e. upellet – uball = - (vpellet – vball) Use this equation together with the equation written in answer to (a) (ii) to calculate the velocity of the ball after impact. [3] (c) Calculate the magnitude of the impulse exerted by the pellet on the ball. [2] (d) The pellet and the ball are in contact for 0.001 seconds. Calculate the magnitude of the average force exerted by the pellet on the ball.
together after impact
In practice, perfectly elastic collisions are very rare – almost all collisions make some noise. But hard objects – like pool balls – will be closer to having elastic collisions than soft objects, because when soft objects collide, energy is used in deforming the objects.
Exam Hint: Questions may ask you to check whether a collision is elastic – this means you have to calculate the total kinetic energy before and total kinetic energy after, and compare them.
Typical Exam Question A trolley of mass 2kg moving at a speed of 3ms-1 catches up and collides with another trolley of mass 1kg moving at a speed of 1ms-1 which is travelling in the same direction. Following the collision, the first trolley has velocity 5/3 ms-1 Show that the collision is perfectly elastic [6]
(a) (i) None (ii) 0.3 20 = 0.3 Vpellet + 0.1 Vball (b) 0.3 = - (Vpellet - Vball) so Vpellet = Vball - 0.3 so 6 = 0.3 (Vball - 0.3) + 0.1 Vball 6 = 0.3 Vball - 0.09 + 0.1 Vball 6.09 = 0.4 Vball Vball = 15.22 ms-1 (c) impulse = change in momentum = 0.1 15.22 = 1.522Ns (d) impulse = force time 1.522 = F 0.001 F = 1522N
2 3 + 1 1 = 2 5/3 + 1 v (conserving momentum) 7 = 10/3 + v 11/3 = v (Velocity of 2nd trolley) KE before ½ 2 32 + ½ 1 12 = 9½ J 2
KE after =
1 1 5 11 2 1 2 2
3
3
2
=
25 121 =9½J 9 118
3
Linear Momentum Exam Workshop
Practical Investigations
This is a typical poor student’s answer to an exam question. The comments explain what is wrong with the answers and how they can be improved. The examiner’s answer is given below.
Momentum investigations are best carried out using a linear air track. This is a triangular tube with holes in it, through which air is blown. The “vehicles” are slides which fit over the track; they move over the cushion of air. This allows the effects of friction (an “external force”) to be ignored. The track is first adjusted so that it is horizontal – so that the vehicles have no tendency to move in either direction on the track.
(a) Define (i) A perfectly elastic collision Energy is conserved
[1] 0/1
Each vehicle has a card of a set length on it. This allows light gates to be used. These contain a beam of light falling on a photodiode; when a vehicle goes through the gate, the card on it interrupts the beam of light. The light gate records the time for which the beam of light is interrupted; this allows the speed of the vehicle to be calculated using speed = length of card time for which light beam is interrupted.
Energy is always conserved – the student should have said “Kinetic energy is conserved” (ii) A completely inelastic collision Energy is not conserved
[1] 0/1
The student should be aware that the answer cannot be true. Learning definitions is vital
The vehicles can have various masses added to them to verify conservation of momentum for different masses. The vehicles can be made to stick together during a collision by adding a pin to the front of one vehicle and a piece of plasticine to the other – this will allow inelastic collisions to be investigated. The vehicles can be made to undergo almost perfectly elastic collisions by attaching a stretched rubber band to the front of each, which allows them to “bounce off” each other.
(b) A stationary uranium nucleus of mass 238 u (atomic units) disintegrates, emitting an -particle of mass 4.00 u and creating a daughter nucleus. You may assume that mass is conserved in the decay. (i) Find the mass of the daughter nucleus in atomic units. [1] 234 a.u. 1/1 (ii) Using the symbols v and vd for velocities, write an equation for the conservation of momentum in this disintegration. [1] 0/1 238vu = 4v 234 vd
Collisions can be investigated by ensuring that each moving vehicle passes through a light gate before collision and after collision. Conservation of momentum can then be verified by calculating the momentum of each vehicle before and after the collision; this can be repeated using different masses, and elastic and inelastic collisions. Experiments could also show whether the “elastic” collisions really are perfectly elastic.
The student clearly understands conservation of momentum, but has ignored the fact that the uranium nucleus was initially stationary. This is poor exam technique, firstly through not reading the question, and secondly through not appreciating that since the symbol vu was not mentioned, it cannot be required in the answer.
Questions
v (iii) Using the answer to (ii), calculate the ratio vd v / vd = 4/234
1. Explain what is meant by: a) the principle of conservation of linear momentum b) an elastic collision c) a totally inelastic collision. d) the impulse of a force
0/1
The student’s answer to ii) did not allow him/her to complete this part, so s/he has guessed. It would have been sensible to go back to check ii) at this stage.
2. State Newton’s Second Law 3.
(c) The -particle then travels until it collides head on with another stationary uranium nucleus of mass 238u. The speed of the particle is 1.00 106ms-1 before the collision, and that of the uranium nucleus is 3.31 104ms-1 after the collision. Calculate the speed and direction of the -particle after the collision. You may assume the particles simply rebound with no nuclear reaction taking place. [3] 4( 1 10 ) = 238 (3.31 10 ) + 4v v = -969450 So 969450 in opposite direction to original 6
Outline how you would verify experimentally the principle of conservation of linear momentum.
4. A car of mass 950kg is moving at 72kmh-1. a) Calculate its momentum The car comes to rest in a period of 15 seconds b) Calculate the average force acting on it during this period. 5. In an explosion, an object splits into two parts whose masses are in the ratio 2:3. Assuming that the object was motionless before the explosion, and that the larger mass moves with speed 16ms-1 after the explosion, calculate the speed of the smaller mass.
4
2/3
6. A gun of mass 2kg fires a bullet of mass 30g with speed 200ms-1. a) Calculate the recoil speed of the gun The gun comes to rest in 2 seconds. b) Find the average force exerted on it by the person firing it.
The student understands this well, and has carried out the calculation correctly, but has lost the final mark due to using too many significant figures and omitting the units.
7. A train of mass 2000kg is moving with speed 10ms-1 when it couples with a stationary truck of mass 300kg. The two move off together. Find the velocity with which they move.
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## 557839
557,839 (five hundred fifty-seven thousand eight hundred thirty-nine) is an odd six-digits composite number following 557838 and preceding 557840. In scientific notation, it is written as 5.57839 × 105. The sum of its digits is 37. It has a total of 2 prime factors and 4 positive divisors. There are 544,824 positive integers (up to 557839) that are relatively prime to 557839.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 6
• Sum of Digits 37
• Digital Root 1
## Name
Short name 557 thousand 839 five hundred fifty-seven thousand eight hundred thirty-nine
## Notation
Scientific notation 5.57839 × 105 557.839 × 103
## Prime Factorization of 557839
Prime Factorization 43 × 12973
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 557839 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 557,839 is 43 × 12973. Since it has a total of 2 prime factors, 557,839 is a composite number.
## Divisors of 557839
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 570856 Sum of all the positive divisors of n s(n) 13017 Sum of the proper positive divisors of n A(n) 142714 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 746.886 Returns the nth root of the product of n divisors H(n) 3.90879 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 557,839 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 557,839) is 570,856, the average is 142,714.
## Other Arithmetic Functions (n = 557839)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 544824 Total number of positive integers not greater than n that are coprime to n λ(n) 90804 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 45803 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 544,824 positive integers (less than 557,839) that are coprime with 557,839. And there are approximately 45,803 prime numbers less than or equal to 557,839.
## Divisibility of 557839
m n mod m 2 3 4 5 6 7 8 9 1 1 3 4 1 2 7 1
557,839 is not divisible by any number less than or equal to 9.
## Classification of 557839
• Arithmetic
• Semiprime
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (557839)
Base System Value
2 Binary 10001000001100001111
3 Ternary 1001100012201
4 Quaternary 2020030033
5 Quinary 120322324
6 Senary 15542331
8 Octal 2101417
10 Decimal 557839
12 Duodecimal 22a9a7
20 Vigesimal 39ebj
36 Base36 byfj
## Basic calculations (n = 557839)
### Multiplication
n×i
n×2 1115678 1673517 2231356 2789195
### Division
ni
n⁄2 278920 185946 139460 111568
### Exponentiation
ni
n2 311184349921 173590766575580719 96835699635755372706241 54018729849110141355076773199
### Nth Root
i√n
2√n 746.886 82.3195 27.3292 14.1027
## 557839 as geometric shapes
### Circle
Diameter 1.11568e+06 3.50501e+06 9.77614e+11
### Sphere
Volume 7.27135e+17 3.91046e+12 3.50501e+06
### Square
Length = n
Perimeter 2.23136e+06 3.11184e+11 788903
### Cube
Length = n
Surface area 1.86711e+12 1.73591e+17 966205
### Equilateral Triangle
Length = n
Perimeter 1.67352e+06 1.34747e+11 483103
### Triangular Pyramid
Length = n
Surface area 5.38987e+11 2.04579e+16 455474
## Cryptographic Hash Functions
md5 78450697f62a5f7f85b6aea7ee8dbb27 5c2fb94099a9ed018ee6999da604edd4f092e0d6 9974ac6d493fd2f27012f28923e8786960363c261708a761e2961a50e3642077 0b17c219b3c4c1c8c70d1427090b99b67d880a1018f453e6970baeb69fa6f370b2f233754712cdf5a4c780059f0a03b8f9d4bbe20aa45872482ae0e8f65ec40a 901ece928de3c8fec6d43319d53079a442cedde2
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Thread: Understanding the Definition of a "Solution" of an ODE
1. Understanding the Definition of a "Solution" of an ODE
Hi.
I'm having a little trouble getting a clear understanding of what is meant by the "interval of definition". I have Zill's A First Course in Differential Equations with Modeling Applications, and here is the definition given of a solution:
Definition 1.1.2: Solution of an ODE
Any function $\phi$, defined on an interval $I$ and possessing at least $n$ derivatives that are continuous on $I$, which when substituted into an $n^{th}$ order ODE reduces the equation to an identity, is said to be a solution of the equation on the interval.
Now, I just don't understand what interval is meant by $I$. Correct me if my understanding is wrong. $I$ is the intersection of the domain of $\phi$ and the domain of the DE. If this isn't quite right, please explain it to me.
P.S. Some examples of the form "Given that $\phi(x)$ is a solution to the first order (or whatever order) DE $F(x,y,...,y^{n-1},y^n)=0$, state the interval of definition." would be great.
Thank You,
VN19.
2. Originally Posted by VonNemo19
Hi.
I'm having a little trouble getting a clear understanding of what is meant by the "interval of definition". I have Zill's A First Course in Differential Equations with Modeling Applications, and here is the definition given of a solution:
Definition 1.1.2: Solution of an ODE
Any function $\phi$, defined on an interval $I$ and possessing at least $n$ derivatives that are continuous on $I$, which when substituted into an $n^{th}$ order ODE reduces the equation to an identity, is said to be a solution of the equation on the interval.
Now, I just don't understand what interval is meant by $I$. Correct me if my understanding is wrong. $I$ is the intersection of the domain of $\phi$ and the domain of the DE. If this isn't quite right, please explain it to me.
P.S. Some examples of the form "Given that $\phi(x)$ is a solution to the first order (or whatever order) DE $F(x,y,...,y^{n-1},y^n)=0$, state the interval of definition." would be great.
Thank You,
VN19.
What page? I have this book.
3. Originally Posted by dwsmith
What page? I have this book.
pg. 5 in the ninth edition.
4. This is best seen looking at an example. Work out Example 1 b on pg 5 as well.
$\displaystyle\frac{dy}{dx}=xy^{1/2} \ \ \ y=\frac{1}{16}x^4$
$\displaystyle\frac{dy}{dx}=\frac{1}{4}x^3$
Plug in.
$\displaystyle\frac{x^3}{4}=x\left(\frac{x^4}{16}\r ight)^2\Rightarrow\frac{x^3}{4}=x\left(\frac {x^2}{4}\right)\Rightarrow\frac{x^3}{4}=\frac{x^3} {4}$
Whenever you solve a DE, you can take your y you obtain and differentiate the amount of times need to check your solution.
Example
Say you have
$y''-5y'+2y+10=0$
When you find y, you can take the second derivative, minus 5 times the first, plus 2 times y, plus 10. The nth derivatives plugged into the DE should equal the RHS.
5. Originally Posted by dwsmith
.... The nth derivatives plugged into the DE should equal the RHS.
I don't think you quite understand what it is that I am asking. Thanks for trying anyway. I know how to plug in a solution and see if the equation reduces to an identity
I'm looking for an understanding of how to find the interval of definition of a solution.
6. Originally Posted by VonNemo19
I don't think you quite understand what it is that I am asking. Thanks for trying anyway.
I'm looking for an understanding of how to find the interval of definition. I know how to plug in a solution and see if the equation reduces to an identity
Domain of the solution.
7. I don't know what you mean by the "domain of the differential equation". I presume you mean the domain of the function F(x,t) where the differential equation is x'= F(x, t). No, the "domain of definition" is not the intersection of the domain of F with the domain of $\phi$. For one thing, F(x, t) is a function of two variables so its domain is in $R^2$ while the domain of the function $\phi$ is in $R$. For another, it makes no sense to say the domain of definition is domain of $\phi$ since $\phi$ is the solution. Of course, its domain is the "domain of definition" of the solution.
What is true of the "domain of definition", and probably what you intended to say, is that it must be a subset of projection of the domain of F(x,t) onto x-coordinate. Exactly what subset can be very difficult to determine.
8. Originally Posted by HallsofIvy
...
OK. So, let me try something else because I'm still confused about the INTERVAL OF DEFINITION; AKA "interval of existence"; AKA "interval of validity"; AKA "domain of the solution" <-----This one's in the book too, so it is a valid way to describe the idea that I'm trying to understand.
So, I'm gonna give a DE along with its solution and if someone could give the interval of definition and explain their reasoning as to how they determined that this interval was in fact the interval of defintion, I would really appreciate it.
Here we go:
Given that $x^2+y^2=25$ is a solution of the DE $\frac{dy}{dx}=-\frac{x}{y}$. State the interval of definition. Explain your reasoning.
BTW, thanks for the clarification regarding the multivariable DE and it's single variable solution HallsOfIvy
9. Originally Posted by VonNemo19
OK. So, let me try something else because I'm still confused about the INTERVAL OF DEFINITION; AKA "interval of existence"; AKA "interval of validity"; AKA "domain of the solution" <-----This one's in the book too, so it is a valid way to describe the idea that I'm trying to understand.
So, I'm gonna give a DE along with its solution and if someone could give the interval of definition and explain their reasoning as to how they determined that this interval was in fact the interval of defintion, I would really appreciate it.
Here we go:
Given that $x^2+y^2=25$ is a solution of the DE $\frac{dy}{dx}=-\frac{x}{y}$. State the interval of definition. Explain your reasoning.
BTW, thanks for the clarification regarding the multivariable DE and it's single variable solution HallsOfIvy
$x\in[-5,5] \ \ \ y\in[-5,5]$
10. Originally Posted by dwsmith
$x\in[-5,5] \ \ \ y\in[-5,5]$
So, basically, the interval of definition is simply the domain of the solution? Is this ALWAYS the case?
11. Yes, one of the its names is "domain of solution"
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label Algebra
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
May 3rd, 2015
.the factors of 6 that when multiplied gives 6 and added gives 5 are 3 , 2
. x^2 + 5x + 6 =0 would factor into ( x + 2 ) (x + 3)
x= -2 , x = -3
x= -3 , x = -2
SOLUTION
x^2 + 5x + 6 =0
x^2 + 2x + 3x + 6 =0 ( we find the numbers as 3 , 2 to split it like this that is 5x = 3x +2x)
x ( x+2 ) + 3 ( x + 2 ) = 0
(x + 3)(x + 2) = 0
x = -3 or x= -2 .........................................ANSWERS
hope you understood..please message if you have any doubts...thank you
May 3rd, 2015
...
May 3rd, 2015
...
May 3rd, 2015
Sep 23rd, 2017
check_circle
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# Absolute convergence
(Redirected from Converges absolutely)
In mathematics, an infinite series of numbers is said to converge absolutely (or to be absolutely convergent) if the sum of the absolute value of the summand is finite. More precisely, a real or complex series ${\displaystyle \textstyle \sum _{n=0}^{\infty }a_{n}}$ is said to converge absolutely if ${\displaystyle \textstyle \sum _{n=0}^{\infty }\left|a_{n}\right|=L}$ for some real number ${\displaystyle \textstyle L}$. Similarly, an improper integral of a function, ${\displaystyle \textstyle \int _{0}^{\infty }f(x)\,dx}$, is said to converge absolutely if the integral of the absolute value of the integrand is finite—that is, if ${\displaystyle \textstyle \int _{0}^{\infty }\left|f(x)\right|dx=L.}$
Absolute convergence is important for the study of infinite series because its definition is strong enough to have properties of finite sums that not all convergent series possess, yet is broad enough to occur commonly. (A convergent series that is not absolutely convergent is called conditionally convergent.)
## Background
One may study the convergence of series ${\displaystyle \sum _{n=0}^{\infty }a_{n}}$ whose terms an are elements of an arbitrary abelian topological group. The notion of absolute convergence requires more structure, namely a norm, which is a real-valued function ${\displaystyle \|\cdot \|:G\to \mathbb {R} }$ on abelian group G (written additively, with identity element 0) such that:
1. The norm of the identity element of G is zero: ${\displaystyle \|0\|=0.}$
2. For every x in G, ${\displaystyle \|x\|=0}$ implies ${\displaystyle x=0.}$
3. For every x in G, ${\displaystyle \|-x\|=\|x\|.}$
4. For every x, y in G, ${\displaystyle \|x+y\|\leq \|x\|+\|y\|.}$
In this case, the function ${\displaystyle d(x,y)=\|x-y\|}$ induces on G the structure of a metric space (a type of topology). We can therefore consider G-valued series and define such a series to be absolutely convergent if ${\displaystyle \sum _{n=0}^{\infty }\|a_{n}\|<\infty .}$
In particular, these statements apply using the norm |x| (absolute value) in the space of real numbers or complex numbers.
## Relation to convergence
If G is complete with respect to the metric d, then every absolutely convergent series is convergent. The proof is the same as for complex-valued series: use the completeness to derive the Cauchy criterion for convergence—a series is convergent if and only if its tails can be made arbitrarily small in norm—and apply the triangle inequality.
In particular, for series with values in any Banach space, absolute convergence implies convergence. The converse is also true: if absolute convergence implies convergence in a normed space, then the space is a Banach space.
If a series is convergent but not absolutely convergent, it is called conditionally convergent. An example of a conditionally convergent series is the alternating harmonic series. Many standard tests for divergence and convergence, most notably including the ratio test and the root test, demonstrate absolute convergence. This is because a power series is absolutely convergent on the interior of its disk of convergence.
### Proof that any absolutely convergent series of complex numbers is convergent
Since a series of complex numbers converges if and only if both its real and imaginary parts converge, we may assume with equal generality that the ${\displaystyle a_{n}}$ are real numbers. Suppose that ${\displaystyle \sum |a_{n}|}$ is convergent. Then, ${\displaystyle 2\sum |a_{n}|}$ is convergent.
Since ${\displaystyle 0\leq a_{n}+|a_{n}|\leq 2|a_{n}|}$, we have
${\displaystyle 0\leq \sum _{n=1}^{m}(a_{n}+|a_{n}|)\leq \sum _{n=1}^{m}2|a_{n}|\leq \sum _{n=1}^{\infty }2|a_{n}|}$.
Thus, ${\displaystyle \sum _{n=1}^{m}(a_{n}+|a_{n}|)}$ is a bounded monotonic sequence (in m), which must converge.
${\displaystyle \sum a_{n}=\sum (a_{n}+|a_{n}|)-\sum |a_{n}|}$ is a difference of convergent series; therefore, it is also convergent, as desired.
### Proof that any absolutely convergent series in a Banach space is convergent
The above result can be easily generalized to every Banach space (X, ǁ ⋅ ǁ). Let xn be an absolutely convergent series in X. As ${\displaystyle \scriptstyle \sum _{k=1}^{n}\|x_{k}\|}$ is a Cauchy sequence of real numbers, for any ε > 0 and large enough natural numbers m > n it holds:
${\displaystyle \left|\sum _{k=1}^{m}\|x_{k}\|-\sum _{k=1}^{n}\|x_{k}\|\right|=\sum _{k=n+1}^{m}\|x_{k}\|<\varepsilon .}$
By the triangle inequality for the norm ǁ ⋅ ǁ, one immediately gets:
${\displaystyle \left\|\sum _{k=1}^{m}x_{k}-\sum _{k=1}^{n}x_{k}\right\|=\left\|\sum _{k=n+1}^{m}x_{k}\right\|\leq \sum _{k=n+1}^{m}\|x_{k}\|<\varepsilon ,}$
which means that ${\displaystyle \scriptstyle \sum _{k=1}^{n}x_{k}}$ is a Cauchy sequence in X, hence the series is convergent in X.[1]
## Rearrangements and unconditional convergence
In the general context of a G-valued series, a distinction is made between absolute and unconditional convergence, and the assertion that a real or complex series which is not absolutely convergent is necessarily conditionally convergent (meaning not unconditionally convergent) is then a theorem, not a definition. This is discussed in more detail below.
Given a series ${\displaystyle \sum _{n=0}^{\infty }a_{n}}$ with values in a normed abelian group G and a permutation σ of the natural numbers, one builds a new series ${\displaystyle \sum _{n=0}^{\infty }a_{\sigma (n)}}$, said to be a rearrangement of the original series. A series is said to be unconditionally convergent if all rearrangements of the series are convergent to the same value.
When G is complete, absolute convergence implies unconditional convergence:
Theorem. Let
${\displaystyle \sum _{i=1}^{\infty }a_{i}=A\in G,\quad \sum _{i=1}^{\infty }\|a_{i}\|<\infty }$
and let σ : NN be a permutation. Then:
${\displaystyle \sum _{i=1}^{\infty }a_{\sigma (i)}=A.}$
The issue of the converse is interesting. For real series it follows from the Riemann rearrangement theorem that unconditional convergence implies absolute convergence. Since a series with values in a finite-dimensional normed space is absolutely convergent if each of its one-dimensional projections is absolutely convergent, it follows that absolute and unconditional convergence coincide for Rn-valued series.
But there are unconditionally and non-absolutely convergent series with values in Banach space, for example:
${\displaystyle a_{n}={\tfrac {1}{n}}e_{n},}$
where ${\displaystyle \{e_{n}\}_{n=1}^{\infty }}$ is an orthonormal basis. A theorem of A. Dvoretzky and C. A. Rogers asserts that every infinite-dimensional Banach space admits an unconditionally convergent series that is not absolutely convergent.[2]
### Proof of the theorem
For any ε > 0, we can choose some ${\displaystyle \kappa _{\varepsilon },\lambda _{\varepsilon }\in \mathbf {N} }$, such that:
{\displaystyle {\begin{aligned}\forall N>\kappa _{\varepsilon }&\quad \sum _{n=N}^{\infty }\|a_{n}\|<{\tfrac {\varepsilon }{2}}\\\forall N>\lambda _{\varepsilon }&\quad \left\|\sum _{n=1}^{N}a_{n}-A\right\|<{\tfrac {\varepsilon }{2}}\end{aligned}}}
Let
{\displaystyle {\begin{aligned}N_{\varepsilon }&=\max \left\{\kappa _{\varepsilon },\lambda _{\varepsilon }\right\}\\M_{\sigma ,\varepsilon }&=\max \left\{\sigma ^{-1}\left(\left\{1,\dots ,N_{\varepsilon }\right\}\right)\right\}\end{aligned}}}
Finally for any integer ${\displaystyle N>M_{\sigma ,\varepsilon }}$ let
{\displaystyle {\begin{aligned}I_{\sigma ,\varepsilon }&=\left\{1,\ldots ,N\right\}\setminus \sigma ^{-1}\left(\left\{1,\dots ,N_{\varepsilon }\right\}\right)\\S_{\sigma ,\varepsilon }&=\min \left\{\sigma (k)\ :\ k\in I_{\sigma ,\varepsilon }\right\}\\L_{\sigma ,\varepsilon }&=\max \left\{\sigma (k)\ :\ k\in I_{\sigma ,\varepsilon }\right\}\end{aligned}}}
Then
{\displaystyle {\begin{aligned}\left\|\sum _{i=1}^{N}a_{\sigma (i)}-A\right\|&=\left\|\sum _{i\in \sigma ^{-1}\left(\{1,\dots ,N_{\varepsilon }\}\right)}a_{\sigma (i)}-A+\sum _{i\in I_{\sigma ,\varepsilon }}a_{\sigma (i)}\right\|\\&\leq \left\|\sum _{j=1}^{N_{\varepsilon }}a_{j}-A\right\|+\left\|\sum _{i\in I_{\sigma ,\varepsilon }}a_{\sigma (i)}\right\|\\&\leq \left\|\sum _{j=1}^{N_{\varepsilon }}a_{j}-A\right\|+\sum _{i\in I_{\sigma ,\varepsilon }}\left\|a_{\sigma (i)}\right\|\\&\leq \left\|\sum _{j=1}^{N_{\varepsilon }}a_{j}-A\right\|+\sum _{j=S_{\sigma ,\varepsilon }}^{L_{\sigma ,\varepsilon }}\left\|a_{j}\right\|\\&\leq \left\|\sum _{j=1}^{N_{\varepsilon }}a_{j}-A\right\|+\sum _{j=N_{\varepsilon }+1}^{\infty }\left\|a_{j}\right\|&&S_{\sigma ,\varepsilon }\geq N_{\varepsilon }+1\\&<\varepsilon \end{aligned}}}
This shows that
${\displaystyle \forall \varepsilon >0,\exists M_{\sigma ,\varepsilon },\forall N>M_{\sigma ,\varepsilon }\quad \left\|\sum _{i=1}^{N}a_{\sigma (i)}-A\right\|<\varepsilon ,}$
that is:
${\displaystyle \sum _{i=1}^{\infty }a_{\sigma (i)}=A}$
Q.E.D.
## Products of series
The Cauchy product of two series converges to the product of the sums if at least one of the series converges absolutely. That is, suppose that
${\displaystyle \sum _{n=0}^{\infty }a_{n}=A}$ and ${\displaystyle \sum _{n=0}^{\infty }b_{n}=B}$.
The Cauchy product is defined as the sum of terms cn where:
${\displaystyle c_{n}=\sum _{k=0}^{n}a_{k}b_{n-k}.}$
Then, if either the an or bn sum converges absolutely, then
${\displaystyle \sum _{n=0}^{\infty }c_{n}=AB.}$
## Absolute convergence of integrals
The integral ${\displaystyle \int _{A}f(x)\,dx}$ of a real or complex-valued function is said to converge absolutely if ${\displaystyle \int _{A}\left|f(x)\right|\,dx<\infty .}$ One also says that f is absolutely integrable.
When A = [a,b] is a closed bounded interval, every continuous function is integrable, and since f continuous implies |f| continuous, similarly every continuous function is absolutely integrable. It is not generally true that absolutely integrable functions on [a,b] are integrable: let ${\displaystyle S\subset [a,b]}$ be a nonmeasurable subset and take ${\displaystyle f=\chi _{S}-1/2,}$ where ${\displaystyle \chi _{S}}$ is the characteristic function of S. Then f is not Lebesgue measurable but |f| is constant. However, it is a standard result that if f is Riemann integrable, so is |f|. This holds also for the Lebesgue integral; see below. On the other hand, a function f may be Kurzweil-Henstock integrable (or "gauge integrable") while |f| is not. This includes the case of improperly Riemann integrable functions.
Similarly, when A is an interval of infinite length it is well known that there are improperly Riemann integrable functions f which are not absolutely integrable. Indeed, given any series ${\displaystyle \sum _{n=0}^{\infty }a_{n}}$ one can consider the associated step function ${\displaystyle f_{a}:[0,\infty )\rightarrow \mathbf {R} }$ defined by ${\displaystyle f_{a}([n,n+1))=a_{n}}$. Then ${\displaystyle \int _{0}^{\infty }f_{a}\,dx}$ converges absolutely, converges conditionally or diverges according to the corresponding behavior of ${\displaystyle \sum _{n=0}^{\infty }a_{n}.}$
Another example of a convergent but not absolutely convergent improper Riemann integral is the Dirichlet integral ${\displaystyle \int _{\mathbf {R} }{\frac {\sin x}{x}}\,dx}$.
On any measure space A, the Lebesgue integral of a real-valued function is defined in terms of its positive and negative parts, so the facts:
1. f integrable implies |f| integrable
2. f measurable, |f| integrable implies f integrable
are essentially built into the definition of the Lebesgue integral. In particular, applying the theory to the counting measure on a set S, one recovers the notion of unordered summation of series developed by Moore–Smith using (what are now called) nets. When S = N is the set of natural numbers, Lebesgue integrability, unordered summability and absolute convergence all coincide.
Finally, all of the above holds for integrals with values in a Banach space. The definition of a Banach-valued Riemann integral is an evident modification of the usual one. For the Lebesgue integral one needs to circumvent the decomposition into positive and negative parts with Daniell's more functional analytic approach, obtaining the Bochner integral.
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## Monday, November 26, 2007
### probability and public policy III: bayes' rule
[part one of this series here]
So, I have a subjective belief about what will happen when a coin is tossed; now, how can I adjust this belief in light of evidence (say, after I've flipped the coin 100 times)?
When we discuss probability, we often do so conditionally. For example, we claim "the probability that the coin will come up heads is 1/2, given that the coin is fair." We write the probability that A is true given that B is true as P(A|B). A basic fact of the probability calculus relates the probability of A&B to the probability of B conditional on A:
P(A)P(B|A) = P(A&B)
But "and" is a symmetric relation, i.e. A&B if and only if B&A, so
P(A)P(B|A) = P(A&B) = P(B&A) = P(B)P(A|B)
and after dividing both sides by P(A), we get Bayes' Rule:
P(B|A) = P(B)P(A|B)/P(A)
OK, but what does this mean? Bayes' Rule tells us how to update our subjective belief state in light of new evidence. To see this, replace B by H for "hypothesis" and A by E for "evidence":
P(H|E) = P(H)P(E|H)/P(E)
What Bayes' Rule now tells us is that our probability that a hypothesis H is true given that we receive evidence E is just equal to our prior probability that H is true times the probability that we would receive evidence E if hypothesis H were true, divided by the probability of E (usually found by summing over the weighted conditional possibilities given all potential hypotheses).
Updating by Bayes' Rule ensures that one's probability distribution is always consistent. It is important to note, however, that one's conclusion about the probability of hypothesis H given evidence E depends upon one's prior assignment of probabilities. Of course, your belief state will eventually converge to the "actual" probability: so, if you believe the coin to be fair, but it is flipped 100 times and every toss comes up heads, your belief in the probability that the coin is fair will be very low in light of this evidence.
To illustrate how belief update occurs, consider a contrived example. A stubborn, but rational, man, Smith, thinks it is extremely unlikely that cigarette smoking causes lung cancer. For Smith, say, P(cigs cause cancer) = 0.2. Instead, he licenses only one alternative hypothesis: that severe allergies cause cancer. Since these hypotheses are exhaustive, on pain of inconsistency, Smith must believe P(allergies cause cancer) = 0.8.
Now, suppose Smith's Aunt Liz dies of lung cancer. Furthermore, suppose Aunt Liz has been a heavy smoker her entire life, then P(Liz gets cancer | cigs cause cancer) = 1 (certainty). Suppose, also, that Liz has had minor allergies for most of her life; since these allergies are only minor, let's say the probability she gets cancer under the hypothesis that severe allergies cause cancer is only 0.5.
Briefly, how should we calculate P(E) here? We sum over the weighted possibilities:
P(E) = P(H1)P(E|H1) + P(H2)P(E|H2) = 0.2(1) + 0.8(0.5) = 0.6
So, now we can use Bayes' Rule to calculate Smith's (only consistent) subjective degree of belief in the hypothesis that cigarettes cause cancer given the evidence that Aunt Liz has died of cancer.
P(H=cigs cause cancer) = 0.2
P(E=Liz gets cancer | H=cigs cause cancer) = 1
P(E=Liz gets cancer) = 0.6
Plugging these values into Bayes' Rule we get:
P(H|E) = P(H)P(E|H)/P(E) = 0.2(1) / 0.6 = 1/3
So, in light of this evidence, Smith's belief in the hypothesis that cigarettes cause cancer has increased from 1/5 to 1/3. Two important points to note here: i) The probability calculus only tells us how to update prior beliefs consistently, it does not tell us what belief state to start from; given sufficiently different prior probability distributions, two agents may draw dramatically different conclusions from the same data. ii) Notice that our calculation of P(E) depended upon the space of hypotheses we were considering. If an agent has failed to consider the actual cause of a piece of evidence E as a potential cause, he may perceive E as increasing the probability of a spurious hypothesis. (Suppose, for example, that it is actually a particular gene which causes both a tendency to smoke and a tendency to succumb to cancer, then cigarettes will be a decent predictor of cancer (supposing sufficiently few non-gene-carriers smoke), but not the cause of cancer - nevertheless, the hypothesis that cigarettes cause cancer will be supported by the data in this alternate scenario if the agent does not license this additional possibility.)
Given these caveats, however, we can see how a sufficiently large number of pieces of evidence that are not adequately supported by alternate hypotheses will eventually push an agent's belief state toward the "true" conclusion: if Smith sees enough people get cancer who don't experience severe allergies, he will come to assign a high probability to the possibility that cigarettes cause cancer.
next: frequentism
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# Clock Questions
Home > Quantitative Aptitude > Clock > General Questions
NA
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5
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6
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31 / 54
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A clock loses 1% time during the first week and then gains 2% time during the next one week. If the clock was set right at 12 noon on a Sunday, what will be the time that the clock will show exactly 14 days from the time it was set right?
A1 : 44 : 48 P.M.
B2 : 40 : 48 P.M.
C1 : 40 : 48 P.M.
DNone of these
Explanation:
The clock loses 1% time during the first week.
In a day there are 24 hours and in a week there are 7 days. Therefore, there are 7 * 24 = 168 hours
in a week.
If the clock loses 1% time during the first week, then it will show a time which is 1% of 168
hours less than 12 Noon at the end of the first week = 1.68 hours less.
Subsequently, the clock gains 2% during the next week. The second week has 168 hours and the
clock gains 2% time = 2% of 168 hours = 3.36 hours more than the actual time.
As it lost 1.68 hours during the first week and then gained 3.36 hours during the next week, the
net result will be a -1.68 + 3.36 = 1.68 hour net gain in time.
So the clock will show a time which is 1.68 hours more than 12 Noon two weeks from the time it
was set right.
1.68 hours = 1 hour and 40.8 minutes = 1 hour + 40 minutes + 48 seconds.
i.e. 1 : 40 : 48 P.M.
Workspace
NA
SHSTTON
2
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5
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7
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32 / 54
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If time at this moment is 9 P.M., what will be the time 23999999992 hours later?
A1:00 PM
B2:00 AM
C1:00 AM
Dcannot be determined
Explanation:
24 billion hours later, it would be 9 P.M. and 8 hours before that it would be 1 P.M.
Workspace
NA
SHSTTON
5
Solv. Corr.
17
Solv. In. Corr.
22
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33 / 54
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A clock showed 5 min past 3'0 clock on Sunday evening when the correct time was 3'0 clock it looses uniformly and was observed to be 10 min slow on the subsequent Tuesday at 9pm . when did the clock show the correct time
A10 AM Monday
B10 PM Monday
C9 PM Monday
D9 AM Monday
Explanation:
Clock showed 5 min past 3'0 clock on Sunday evening when the correct time was 3'0 clock.
It looses uniformly and was observed to be 10 min slow on the subsequent Tuesday at 9pm .
In 54 hrs , it lost 15 mins.
so 5 mins are lost in 5*54/15 = 18 hrs
so watch showed correct time at 1500+1800-2400= 0900 hrs on Monday i.e 9AM Monday
Workspace
NA
SHSTTON
6
Solv. Corr.
6
Solv. In. Corr.
12
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34 / 54
Choose the correct option.
A watch gains 5 min. every hour. If it has set right at 12 noon. what is the true time when the watch is showing 5 pm,on the same day?
A4:480/13
B3:480/13
C0.569444444
DNone of these
Explanation:
65 min of 1st clock = 60 min of correct clock
5 hr of this clock = 60x5/65 of correct clock=60/13
correct time is 60/13 hr after 12 = 4:8/13=4:480/13
Workspace
NA
SHSTTON
6
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10
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16
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35 / 54
Choose the correct option.
The famous church in the city of Kumbakonnam has a big clock tower and is said to be over 300 years old. Every Monday 10.00 A M the clock is set by Antony, doing service in the church. The Clock loses 6 mins every hour. What will be the actual time when the faulty clock shows 3 P.M on Friday?
A1:06 am saturday
B4:54 am Friday
C2:13:20 am saturday
D3 am Friday
Explanation:
Monday 10am to friday 3pm=101 hours=606 minutes=10 hours 6 minutes delay
So, time at 3 pm = 4:54am
Workspace
NA
SHSTTON
3
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18
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21
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36 / 54
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At 6'o clock clock ticks 6 times. The time between first and last ticks was 30sec. How much time it takes at 12'o clock.?
A60 sec.
B56 sec.
C76 sec.
D66 sec.
Explanation:
counting time starts after first tick
so for 5 ticks it take 30min
Similarly at 12 o clock there is 11 ticks =11*(30/5) = 66.
Workspace
NA
SHSTTON
9
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30
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39
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37 / 54
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Find the time between 5 and 6 o'clock when the two hands of clock are 6 minutes spaces apart.
A3 2/11 minutes past 5 o'clock
B6 6/11 minutes past 5 o'clock
C20 8/11 minutes past 5 o'clock
D4 2/11 minutes past 5 o'clock
Explanation:
In this type of problems the formula is
(5*x+ or - t)*12/11 -----------(Important Point)
Here x is the first interval of given time, and t is spaces apart.
(5 * 5 +6) * 12/11 = 31 *12/11 = 372/11
and
(5 * 5 - 6) * 12/11 = 19 * 12/11 = 228/11 = 20 8/11
Workspace
NA
SHSTTON
5
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31
Solv. In. Corr.
36
Attempted
0 M:40 S
Avg. Time
38 / 54
Choose the correct option.
At what time between 4 and 5 o'clock will the hands of a watch point in opposite directions?
A54 past 4
B(53 + 7/11) past 4
C(54 + 8/11) past 4
D(54 + 6/11) past 4
Explanation:
4 o'clock, the hands of the watch are 20 min. spaces apart.
To be in opposite directions, they must be 30 min. spaces apart.
So, Minute hand will have to gain 50 min. spaces.
55 min. spaces are gained in 60 min
50 min. spaces are gained in (60/55 +50) min. or 54 6/11
Required time = 54 6/11 min. past 4.
Workspace
NA
SHSTTON
14
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19
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33
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0 M:0 S
Avg. Time
39 / 54
Choose the correct option.
At what time between 9' O clock and 10' O clock will the hands of a clock point in the opposite directions?
A2 2/11 minutes past 10' O clock
B16 4/11 minutes past 9
C2 2/11 minutes past 9
D16 4/11 minutes past 10' O clock
Explanation:
Workspace
NA
SHSTTON
5
Solv. Corr.
21
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26
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40 / 54
Choose the correct option.
What is the angle between the two hands at 8.20 O'clock?
A150 degree
B120 degree
C250 degree
D320 degree
Explanation:
8 hr 20 min = 8 1/3 hrs. = 25/3 hrs.
Angle traced by hour hand in 12 hours = 360°
=> Angle traced by hour hand in 1 hour = 30°
Therefore, angle traced by hour hand in 25/3 hr = 25/3 * 30 = 250°
Angle traced by minute hand in 60 min = 360°
=> Angle traced by minute hand in 1 min = 6°
Therefore, angle traced by minute hand in 20 min = 6 × 20 = 120°
So, required angle = (250° - 120°) = 130°
Workspace
## Quantitative Aptitude Clock Questions and Answers pdf
At Quantitative Aptitude topic Clock, you will get multiple online quiz difficulty wise, which will have a total of 6 quizzes, categorized as easy, medium, and moderate level. While preparing for any Clock, take all the list quiz and check your preparation level for that topic. Each quiz have 10 different question, which needs to be answered in 20 min., all the listed quiz here is free, however, you will get only one chance for each quiz to attempt(Take Quiz seriously), so it is always recommended to take one quiz in each section before you start solving Clock MCQs practice question, and one after solving all the question of the respective level, you can refer back your Clock quiz result any time or you can download it as pdf for reference.
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On this Clock section of page you will find the easiest quickest ways to solve a question, formulas, shortcuts and tips and tricks to solve various easiest methods to solve Clock Question Quickly. It contains all the Quantitative Aptitude topic Clock questions which are common in any of the preliminary exams of any company. The solution is provided along with the questions. The practice of these questions is a must as they are easy as well as scoring and asked in all the exams They will confirm the selection if all the questions attempted wisely with little practice. It is recommanded to Take Mock test based on Quantitative Aptitude topic and Clock topic based quiz.
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You can get here fully solved Clock examples with a detailed answer and description. You can solve Clock problems with solutions, the questions by companies wise by filtering the questions, additionally, you can check what type of questions are being asked in IT companies Written Round from Clock. Clock became one of the most important sections in the entire competitive exams, Companies Campus, and entrance online test. Go through Clock Examples, Clock sample questions. You can Evaluate your level of preparation in Clock by Taking the Q4Interivew Clock Online Mock Test based on most important questions. All the Clock practice questions given here along with answers and explanations are absolutely free, you can take any number of time any mock Test.
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# algrebra
posted by .
2 similar triangles were drawn in a design.one was 1/4 the size of the other.if the measures of the first were 5,6 and 12 the measure of a larger triangle were is
A)10,14 and 24
B)20,28 and 48
c)10,28 and 48
d)14,28 and 48
• algrebra -
It is a real lousy question.
What it means is in area, one is 1/4 the size of the other. Since triangle area is proportional to base*height, each of those must be 1/2 the larger (remember 1/2*1/2=1/4)
so the sides would be 10,12, and 24
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# Sizing calculations for 20/3.3 kV, 12.5 MVA transformer feeder cable
Home / Technical Articles / Sizing calculations for 20/3.3 kV, 12.5 MVA transformer feeder cable
## 20kV transformer feeder cable
Typical calculations for a 20kV transformer feeder cable are presented below. After correct cable voltage classification the following considerations apply:
Consider a 20/3.3 kV, 12.5 MVA transformer to be fed by direct buried, 3 core XLPE, SWA, PVC, copper conductor cable.
### 1. Cable current carrying capacity
Transformer full load current is calculated by:
Ifull-load = 12.5 × 106 / 1.73 × 20 × 10= 361 A
Don’t forget derating factors… Manufacturers provide data sheets for cables including appropriate derating factors based upon IEC 60287 (Table 1). For a ground temperature at depth of laying of 20°C, the derating factor is 0.97.
The group derating factor based upon 3 cables laid in trench at 0.45 m centres is 0.84. Ground thermal resistivity taken as the normal of 1.2°Cm/W for a UK installation and 1.00 rating factor. Cable installation depth to be 0.8 m and 1.00 rating factor.
Therefore subsequent current rating of cable to be 361 / 0.97 × 0.84 = 443 A
From manufacturers tables selected cable size is 240 mm2.
Go back to contents ↑
### 2. Short circuit rating
The maximum system fault level in this application is 8.41 kA. From standard IEC 60364-5-54 (Electrical Installations In Buildings – Earthing arrangements, protective conductors and protective bonding conductors):
Isc = K × A / √t
Where:
• K – constant, 143 for XLPE cable
• A – cable cross-section, 240 mm2 based on current carrying capacity
• t – short circuit duration, for MV cables use 1 second
Isc = 240 × 143 / √1 = 34.4 kA
From manufacturers tables and/or Figures 1 (a–c) for working voltages up to and including 19000 / 33000 XLPE based insulated cable the selected 240 mm2 cable is just capable of this 1 second short circuit rating.
Note tables are conservative and assume a fully loaded cable. At the initiation of the fault conductor temperature is 90°C and at the end of the fault conductor temperature is 250°C.
Go back to contents ↑
### 3. Voltage drop (Vd)
Consider a 100 m route length of cable with resistance, R = 0.0982 Ω/km and inductive reactance, XL = 0.097 Ω/km. At full load current, Ifl = 361 A at 0.85 power factor the cable voltage drop over a 100 m cable length,
Vd = Ifl × XL × sinφ + Ifl × R × cosφ volts
Vd = (361 × 0.097 × 0.53 + 361 × 0.0982 × 0.85) × 100 / 1000
Vd = (18.56 + 30.13) ×100 / 1000
Vd = 4.87 V
Vd = 0.042%
Important Notes //
1. At 20 kV the voltage drop is negligible over such a short length of cable.
2. IEE Wiring Regulations require a voltage drop for any particular cable run to be such that the total voltage drop in the circuit of which the cable forms part does not exceed 2½% of the nominal supply voltage, i.e. 10.4 volts for a three phase 415 V supply and 6 volts for a single phase 240 V supply.
3. Industrial plant users may use different specifications and apply 5% (or even 10%) under no load to full load conditions and perhaps 20% at motor terminals under motor starting conditions.
4. Manufacturers data for building services installations is often expressed in terms of voltage drop (volts) for a current of 1 ampere for a 1 meter run of a particular cable size.
Go back to contents ↑
### 4. Earth loop impedance
For building services work it is important with small cross-section wiring and low fault levels to ensure that sufficient earth fault current flows to trip the MCB or fuse protection. For distribution power networks with more sophisticated protection the check is still necessary and allows the calculation of the likely touch voltages arising from the earth fault.
This in turn can then be checked against the allowable fault duration to avoid danger. See this article for a consideration of the design criteria associated with touch and step potentials.
Criteria #1 – Consider the earthing resistance at the source substation 0.5 .
Criteria #2 – The source substation 20 kV neutral is approximately 10 km from the 100 m cable under consideration. In addition parallel copper conductor earth cable is run to supplement and improve power cable armour resistance values from equipment back to the primary substation infeed neutral.
For this example assume power and supplementary earth copper cables and armour over the 10 km distance have a combined effective resistance of 0.143.
Criteria #3 – The combined resistance of the 100 m, 240 mm2, cable armour (0.028 Ω / 100 m) and in parallel 2 × 95 mm2 copper supplementary earth cables (0.00965 Ω /100 m) = 7.18 × 10-3 Ω.
Criteria #4 – Consider the earthing resistance at the cable fault to be 0.5.
Criteria #5 – The effective earth circuit is shown in Figure 2. The effective primary substation neutral-to-fault cable resistance 0.15.
Criteria #6 – The maximum earth fault current at 20 kV has to be determined. Sometimes this is limited by a neutral earthing resistor and the maximum limited current may be taken for calculation. Maximum earth fault current for this calculation is 1000 A.
For a fault to earth at the end of the 100 m cable, 10 km from the primary power infeed the fault current,
If = (1000 × 0.15) / (1 + 0.15) = 131 A
Therefore touch voltage to earth at the cable fault 131 × 0.5 = 65.3 V
Go back to contents ↑
### APPENDIX – Derating factors based on IEC 60287
Go back to contents ↑
Reference // Transmission and Distribution Electrical Engineering by Dr C. R. Bayliss CEng FIET and B. J. Hardy ACGI CEng FIET
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### Edvard Csanyi
Electrical engineer, programmer and founder of EEP. Highly specialized for design of LV/MV switchgears and LV high power busbar trunking (<6300A) in power substations, commercial buildings and industry facilities. Professional in AutoCAD programming.
### 15 Comments
1. ZULKIFLI OTHMAN
Oct 20, 2021
edvard, i think when you derate, you multiply the full load current by the derating factor, and not by dividing
2. Azhar
Feb 06, 2021
4000 kva Trf how much sqmm HT cable required sir for 11kv
3. M.E Ajith Silva
Jan 08, 2021
hello dear sir.. please tell me I whant to know 33kv 400v 1600KVA transformer what is the sutabel for cable size.
• M.E Ajith Silva
Jan 09, 2021
I think Hv side 240sqmm.
• joydev malla
Apr 27, 2021
please briefly explained
4. Tin Tin Khaing
Mar 13, 2020
11kV rated Cable,
MV Cable size As per short circuit is 279.7mm2.
2 x 3C x 185 mm2 cable can be selected?
3C x 300mm2 cable don’t want to be selected because of cable dia.
Current carrying capacity is ok.
5. Rodel Torres
Dec 09, 2019
Ifull-load = 12.5 × 106 / 1.73 × 20 × 103 = 361 A
Don’t you need a 125% conductor protection here before you apply derating factors?
6. Ishan Madushanka
Nov 29, 2019
Can you explain how to calculate earth cable size for LV transformer for both neutral and body earthing in the TN-S system? Specially I want to know how to select the time in the adiabatic equation.
7. FREDRICK
Aug 15, 2018
I have 2MVA 11/0.433KV GMT.I want to connect 1x500sq.mm XLPE/SWA/PVC copper cable. How many cables can i use per phase. What calculation is used
8. Shoaib
Jul 27, 2018
Dear sir,
Please tell me the between KW and KVA for example 1 machine has 30kva load capacity how we have to select the cable.kindly explain
Thanks & regards,
Shoaib Noor.
9. javed
Jun 24, 2018
but how to calculate cable size for low voltage side
10. vijay manikanta
Dec 15, 2017
Can we take parallel runs into account for calculation in the calculation of short circuit current capacity?
In my case a feeder has to be selected for feeding 33KV, 31.5 ka, 1 sec rated panel, for a load of 20MW.
I have selected 2R, 33KV, 300SQMM, AL Conductor, XLPE Insulated cable to be installed in RCC trenches, over 2000m length.
As per manufacturer catalog 300sqmm Al, XLPE cable can cater 28 ka, as two runs are considered, I expect it can take care 31.5ka for 1sec.
It is requested to advice on the above.
11. Aye Kyaw Kyaw
May 08, 2017
For current 443 A, why you selected 240 sqmm? Can I see the manufacturer’s table that you referred? According to my knowledge, 150 sqmm of MV Cu/XLPE cable can carry about 500 A at free air from many cable makers.
• Mazen
May 09, 2017
Impossible ,150 sqm can not carry 500A as you said , max 400A. To which brand you are referring?
• Steven
May 13, 2017
Edvard, thanks for the interesting post.
Referring the above, I think 150mm2 would be too small. Strictly according to IEC 60502, 300mm2 is required. However, the calculation for 240mm2 is only short by 7A, so in practice, you would likely be ok with 240mm2. This is reinforced with Edvards calculation based on manufacturer’s data which gives 240mm2.
I printed out my calculations for reference, and you can view them here:
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# Question
Hi. Is there a specific way to solve this type pattern question? How to find the correct formula to get the answers ? Thanks
(a) 5,8,11,14 fig 5 = 14+3 = 17 nth term = 5+ 3(n-1) = 5 + 3n – 3 = 3n+2
4,12,24,40 = 4 (1,3,6,10…) = 4 [(n/2)(n+1)] (1,3,6,10… is known as triangular number seq..) = 2n(n+1)
fig 5 = 4[(5/2)(6)] = 60 or 5×4 = 20 then 40+20 = 60
(b) Fig 39 = 5 + 3(39-1) = 119 or 3(39) + 2 = 119
(c) Fig 60 = 4[(60/2 )(60+1)] = 7320 or 2(60)(60+1) = 7320
To answer the question of specific method…yes.
first sequence has a constant difference of 3
so you always fix the first number which is 5 then add 3(n-1) where 3(n-1) will be the generator of pattern as n varies…then simplify to become 3n+2
second sequence is a multiplier of a standard sequence called triangular number seq.(1,3,6,10…) and the nth term is (n/2)(n+1) so in this example it’s 4(n/2)(n+1) which can be simplified to 2n(n+1)
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# Cylinder pulled by string on flat surface w/out slipping
highroller
1. The problem statement: a 100kg homogeneous cylinder of radius .3m. Starts at rest. Is pulled by a string wrapped around the cylinder coming off the top with a force of 500N.
Find the angular velocity after the cylinder has rolled one revolution?
The attempt at a solution: I= 4.5kg*m^2
circumferance=1.885m
work= 500*1.885= .5m(wr)^2+.5Iw^2 (using w for omega)
can someone please explain. I suppose the force could be doing rotational work as well but I'm having a hard time believing that the force does double the work because of where it is applyed.
My question is which answer is correct and WHY? thanks for any help
Homework Helper
Welcome to PF!
Hi highroller! Welcome to PF!
(have an omega: ω and try using the X2 tag just above the Reply box )
a 100kg homogeneous cylinder of radius .3m. Starts at rest. Is pulled by a string wrapped around the cylinder coming off the top with a force of 500N.
Find the angular velocity after the cylinder has rolled one revolution?
work= 500*1.885= .5m(wr)^2+.5Iw^2 (using w for omega)
Yes intial energy is zero, so work done = final energy.
But what is your .5m(ωr)2 supposed to be?
There's no extra mass … it's only a cylinder, with energy 1/2 Iω2
highroller
the cylinder has both linear and rotational ke so the .5m(wr)^2 comes from .5mv^2 with v=wr
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# General Linear Algebra equation
This isn't really a homework question, but there are more people in this forum, hopefully someone can help. Just basic linear algebra.
I've got this operator in the form of a matrix, L. It acts on another matrix, u, either by multiplying on the left or right. The operation I want in this particular case is
$$Lu + uL.$$
But I need to invert this, so I want to write it as Mu, so that I can find the inverse of M.
Is there some way to write the Lu+uL in the form of Mu?
I have a feeling there's something fundamental about multiplying on the right that won't allow me to do this.
Mark44
Mentor
Matrix multiplication is not generally commutative, so it might be that Lu and uL are different.
However, if it turns out that Lu and uL are equal, then you have Lu + uL = Lu + Lu = 2Lu = Mu.
Hi skatch!
Can we perhaps know the dimensions of L and u? I have a feeling that you want u to be a vector (i.e. a nx1-matrix), but that wouldn't make much sense...
In the case that you did mean u to be a vector, did you perhaps mean
$$Lu+u^TL$$
Yeah, u and L definitely don't commute in this case unfortunately. L is a finite difference operator, and how I've got it set up, it gives me the 2nd derivative approximations for u with respect to x when multiplied on the left, and with respect to y when multiplied on the right. So Lu+uL is what I was hoping to use as a discrete laplacian operator, which in turn I would like to invert.
micromass: Nope, both nxn matrices. u is an approximate solution over a grid of nxn points, u_ij = u(x_i, y_j).
I just might be headed down a dead end, and might need to define a different L for my discrete laplacian so that I can invert it. This current setup looks like it probably won't work.
Ray Vickson
Homework Helper
Dearly Missed
You can regard u as an n^2-dimensional vector U, just by spreading it out as U = (u_{11},...,u_{1n},u_{21},...,u_{2n},...,u_{n1}, ...,u_{nn}). Now Lu + uL is a linear operator on the vector U, so has a matrix representation M, which is an n^2 x n^2 matrix.
RGV
This might be a bit too abstract, but if you consider L as a mapping between two spaces, then you can consider its derivative. In this case the derivative of L^2 in the direction of u is Lu + uL.
$$[D F(L)]u = Lu + uL$$
where
$$F(L) = L^2$$
and [DF(L)] is the "same" as the Jacobian Matrix of F(L) if you translate F(L) into a vector a column length R^n^2 then after the calculations change it back to a matrix. Therefore
$$M=[DF(L)]$$
and finding M^-1 would be easy afterwards.
Ignore my last post. This is easier.
$$Lu+uL=Mu$$
therefore
$$M=(Lu+uL)u^{-1}$$
then
$$M^{-1}=u(Lu+uL)^{-1}$$
trans: Won't quite work for my application, I need M to be independant of the matrix I want to apply it to. I'm going to be applying it many times, so I don't want to be computing an new inverse each iteration.
Ray: I think you nailed it. After you said I would need to use n^2 by n^2 and reform the vector, I think I came up with $$L\otimes I + I\otimes L$$ as the matrix I need (using kronecker product), if I reshape u correctly into n^2 by 1. $$L\otimes I$$ multiplied by a reshaped u gives me the same thing as multiplying the original u on the left by L, and vice versa. Thanks!
(unrelated: dont suppose anyone can tell me how to write in-line tex instead of \equation lines?)
Use the itex tag for inline tex.
:)
T$ha\eta k$s!
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My Math Forum Marginal cost with two variables, confused.
Calculus Calculus Math Forum
January 30th, 2014, 12:08 PM #1 Newbie Joined: Jan 2014 Posts: 1 Thanks: 0 Marginal cost with two variables, confused. Your weekly cost to manufacture x bicycles and y tricycles is C(x,y)=20,000+60x+20y+50?(xy) a. What is marginal cost of manufacturing a bicycle. I derived x and got (25y^1/2)/x^(1/2) +60 but I dont know what to do next or if that is even right. b. What is the marginal cost of manufacturing a tricycle. I derived y and got (25x^1/2)/(y^1/2) +20 again I dont know what to do next c. What is the marginal cost of manufacturing a bicycle at a level of 10 bicycles and 10 tricycles? d. What is the marginal cost of manufacturing a tricycle at a level of 8 bicycles and 8 tricycles?
January 31st, 2014, 05:41 PM #2
Math Team
Joined: Sep 2007
Posts: 2,409
Thanks: 6
Re: Marginal cost with two variables, confused.
I thought I had responded to this yesterday, but found it without a response this morning so repeated my response. Now I find no response again! Did I not post it right or did someone erase it or was this on three different boards?
Quote:
Originally Posted by nakota2k Your weekly cost to manufacture x bicycles and y tricycles is C(x,y)=20,000+60x+20y+50?(xy) a. What is marginal cost of manufacturing a bicycle. I derived x and got (25y^1/2)/x^(1/2) +60 but I dont know what to do next or if that is even right.
Yes, that is the "partial derivative of C with respect to x". Strictly speaking the "marginal cost of manufacturing a bicycle" is the cost of manufacturing "one more bicycle", C(x+ 1, y)- C(x, y). For "large x", that can be approximated by $\partial C/\partial x$
Quote:
b. What is the marginal cost of manufacturing a tricycle. I derived y and got (25x^1/2)/(y^1/2) +20 again I dont know what to do next c. What is the marginal cost of manufacturing a bicycle at a level of 10 bicycles and 10 tricycles?
Again, the exact value is C(11, 10)- C(10, 10) but that can be approximated by $\partial C/\partial x$ evaluated at x= 10, y= 10.
Quote:
d. What is the marginal cost of manufacturing a tricycle at a level of 8 bicycles and 8 tricycles?
January 31st, 2014, 06:31 PM #3 Global Moderator Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond Re: Marginal cost with two variables, confused. @HallsofIvy: There's no record of such a deletion here (sitewide, from yesterday on).
February 1st, 2014, 04:18 AM #4 Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Marginal cost with two variables, confused. So, my mind is going! Knew it would happen sooner or later.
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# the cost of 378 tricycles is rupees 4,71,366.what is the cost of one tricycles
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You are reading an older version of this FlexBook® textbook: CK-12 Algebra - Basic Go to the latest version.
# 12.5: Multiplication and Division of Rational Expressions
Difficulty Level: At Grade Created by: CK-12
Because a rational expression is really a fraction, two (or more) rational expressions can be combined through multiplication and/or division in the same manner as numerical fractions. A reminder of how to multiply fractions is below.
For any rational expressions a0,b0,c0,d0\begin{align*}a \neq 0, b \neq 0, c \neq 0, d \neq 0\end{align*},
Example: Multiply the following a16b84b35a2\begin{align*}\frac{a}{16b^8} \cdot \frac{4b^3}{5a^2}\end{align*}
Solution:
a16b84b35a24ab380a2b8
Simplify exponents using methods learned in chapter 8.
4ab380a2b8=120ab5
Example 1: Simplify 9c24y221c4\begin{align*}9c^2 \cdot \frac{4y^2}{21c^4}\end{align*}.
Solution:
9c24y221c49c214y221c49c214y221c4=36c2y221c436c2y221c4=12y27c2
## Multiplying Rational Expressions Involving Polynomials
When rational expressions become complex, it is usually easier to factor and reduce them before attempting to multiply the expressions.
Example: Multiply 4x+123x2xx29\begin{align*}\frac{4x+12}{3x^2} \cdot \frac{x}{x^2-9}\end{align*}.
Solution: Factor all pieces of these rational expressions and reduce before multiplying.
4x+123x2xx294(x+3)3x2x(x+3)(x3)4(x+3)3x2x(x+3)(x3)43x1x343x29x
Example 1: Multiply 12x2x6x21x2+7x+64x227x+18\begin{align*}\frac{12x^2-x-6}{x^2-1} \cdot \frac{x^2+7x+6}{4x^2-27x+18}\end{align*}.
Solution: Factor all pieces, reduce, and then multiply.
12x2x6x21x2+7x+64x227x+18(3x+2)(4x3)(x+1)(x1)(x+1)(x+6)(4x3)(x6)12x2x6x21x2+7x+64x227x+18(3x+2)(4x3)(x+1)(x1)(x+1)(x+6)(4x3)(x6)(3x+2)(x+6)(x1)(x6)=3x2+20x+12x27x+6
## Dividing Rational Expressions Involving Polynomials
Division of rational expressions works in the same manner as multiplication. A reminder of how to divide fractions is below.
For any rational expressions a0,b0,c0,d0\begin{align*}a \neq 0, b \neq 0, c \neq 0, d \neq 0\end{align*},
Example: Simplify 9x242x2÷21x22x81\begin{align*}\frac{9x^2-4}{2x-2} \div \frac{21x^2-2x-8}{1}\end{align*}.
Solution:
9x242x2÷21x22x819x242x2121x22x8
Repeat the process for multiplying rational expressions.
9x242x2121x22x89x242x2÷21x22x81(3x2)(3x2)2(x1)1(3x2)(7x+4)=3x214x26x8
## Real-Life Application
Suppose Marciel is training for a running race. Marciel’s speed (in miles per hour) of his training run each morning is given by the function x39x\begin{align*}x^3-9x\end{align*}, where x\begin{align*}x\end{align*} is the number of bowls of cereal he had for breakfast (1x6)\begin{align*}(1 \le x \le 6)\end{align*}. Marciel’s training distance (in miles), if he eats x\begin{align*}x\end{align*} bowls of cereal, is 3x29x\begin{align*}3x^2-9x\end{align*}. What is the function for Marciel’s time and how long does it take Marciel to do his training run if he eats five bowls of cereal on Tuesday morning?
timetimetimeIf xtime=distancespeed=3x29xx39x=3x(x3)x(x29)=3x(x3)x(x+3)(x3)=3x+3=5,then=35+3=38
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Marciel will run for 38\begin{align*}\frac{3}{8}\end{align*} of an hour.
## Practice Set
Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both.
In 1–20, perform the indicated operation and reduce the answer to lowest terms
1. x32y32y2x\begin{align*}\frac{x^3}{2y^3} \cdot \frac{2y^2}{x}\end{align*}
2. 2xy÷2x2y\begin{align*}2xy \div \frac{2x^2}{y}\end{align*}
3. 2xy24y5x\begin{align*}\frac{2x}{y^2} \cdot \frac{4y}{5x}\end{align*}
4. 2xy2y2x3\begin{align*}2xy \cdot \frac{2y^2}{x^3}\end{align*}
5. 4y21y29y32y1\begin{align*}\frac{4y^2-1}{y^2-9} \cdot \frac{y-3}{2y-1}\end{align*}
6. 6aba2a3b3b2\begin{align*}\frac{6ab}{a^2} \cdot \frac{a^3b}{3b^2}\end{align*}
7. x2x1÷xx2+x2\begin{align*}\frac{x^2}{x-1} \div \frac{x}{x^2+x-2}\end{align*}
8. 33a252011a3\begin{align*}\frac{33a^2}{-5} \cdot \frac{20}{11a^3}\end{align*}
9. a2+2ab+b2ab2a2b÷(a+b)\begin{align*}\frac{a^2+2ab+b^2}{ab^2-a^2b} \div (a+b)\end{align*}
10. 2x2+2x24x2+3xx2+x6x+4\begin{align*}\frac{2x^2+2x-24}{x^2+3x} \cdot \frac{x^2+x-6}{x+4}\end{align*}
11. 3x3x5÷x292x28x10\begin{align*}\frac{3-x}{3x-5} \div \frac{x^2-9}{2x^2-8x-10}\end{align*}
12. x225x+3÷(x5)\begin{align*}\frac{x^2-25}{x+3} \div (x-5)\end{align*}
13. 2x+12x1÷4x2112x\begin{align*}\frac{2x+1}{2x-1} \div \frac{4x^2-1}{1-2x}\end{align*}
14. xx5x28x+15x23x\begin{align*}\frac{x}{x-5} \cdot \frac{x^2-8x+15}{x^2-3x}\end{align*}
15. 3x2+5x12x29÷3x43x+4\begin{align*}\frac{3x^2+5x-12}{x^2-9} \div \frac{3x-4}{3x+4}\end{align*}
16. 5x2+16x+336x225(6x2+5x)\begin{align*}\frac{5x^2+16x+3}{36x^2-25} \cdot (6x^2+5x)\end{align*}
17. x2+7x+10x29x23x3x2+4x4\begin{align*}\frac{x^2+7x+10}{x^2-9} \cdot \frac{x^2-3x}{3x^2+4x-4}\end{align*}
18. x2+x12x2+4x+4÷x3x+2\begin{align*}\frac{x^2+x-12}{x^2+4x+4} \div \frac{x-3}{x+2}\end{align*}
19. x416x29÷x2+4x2+6x+9\begin{align*}\frac{x^4-16}{x^2-9} \div \frac{x^2+4}{x^2+6x+9}\end{align*}
20. x2+8x+167x2+9x+2÷7x+2x2+4x\begin{align*}\frac{x^2+8x+16}{7x^2+9x+2} \div \frac{7x+2}{x^2+4x}\end{align*}
21. Maria’s recipe asks for 212 times\begin{align*}2 \frac{1}{2} \ \text{times}\end{align*} more flour than sugar. How many cups of flour should she mix in if she uses 313 cups\begin{align*}3 \frac{1}{3} \ \text{cups}\end{align*} of sugar?
22. George drives from San Diego to Los Angeles. On the return trip, he increases his driving speed by 15 miles per hour. In terms of his initial speed, by what factor is the driving time decreased on the return trip?
23. Ohm’s Law states that in an electrical circuit I=VRc\begin{align*}I=\frac{V}{R_c}\end{align*}. The total resistance for resistors placed in parallel is given by 1Rtot=1R1+1R2\begin{align*}\frac{1}{R_{tot}}=\frac{1}{R_1}+\frac{1}{R_2}\end{align*}. Write the formula for the electric current in term of the component resistances: R1\begin{align*}R_1\end{align*} and R2\begin{align*}R_2\end{align*}.
Mixed Review
1. The time it takes to reach a destination varies inversely as the speed in which you travel. It takes 3.6 hours to reach your destination traveling 65 miles per hour. How long would it take to reach your destination traveling 78 miles per hour?
2. A local nursery makes two types of fall arrangements. One arrangement uses eight mums and five black-eyed susans. The other arrangement uses six mums and 9 black-eyed susans. The nursery can use no more than 144 mums and 135 black-eyed susans. The first arrangement sells for \$49.99 and the second arrangement sells for 38.95. How many of each type should be sold to maximize revenue?
3. Solve for r\begin{align*}r\end{align*} and graph the solution on a number line: 24|2r+3|\begin{align*}-24 \ge |2r+3|\end{align*}.
4. What is true of any line parallel to 5x+9y=36\begin{align*}5x+9y=-36\end{align*}?
5. Solve for d:3+5d=d(3x3)\begin{align*}d: 3+5d=-d-(3x-3)\end{align*}.
6. Graph and determine the domain and range: y9=x25x\begin{align*}y-9=-x^2-5x\end{align*}.
7. Rewrite in vertex form by completing the square. Identify the vertex: y216y+3=4\begin{align*}y^2-16y+3=4\end{align*}.
## Quick Quiz
1. h\begin{align*}h\end{align*} is inversely proportional to t\begin{align*}t\end{align*}. If t=0.05153\begin{align*}t=-0.05153\end{align*} when h=16\begin{align*}h=-16\end{align*}, find t\begin{align*}t\end{align*} when h=1.45\begin{align*}h=1.45\end{align*}.
2. Use f(x)=5x225\begin{align*}f(x)=\frac{-5}{x^2-25}\end{align*} for the following questions.
1. Find the excluded values.
2. Determine the vertical asymptotes.
3. Sketch a graph of this function.
4. Determine its domain and range.
3. Simplify 8c4+12c222c+14\begin{align*}\frac{8c^4+12c^2-22c+1}{4}\end{align*}.
4. Simplify 10a230aa3\begin{align*}\frac{10a^2-30a}{a-3}\end{align*}. What are its excluded values?
5. Fill the blank with directly, inversely, or neither. “The amount of time it takes to mow the lawn varies ________________ with the size of the lawn mower.”
8 , 9
Feb 22, 2012
Dec 11, 2014
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# Solving equations using the Newton-Raphson method
By Martin McBride, 2022-08-14
Tags: solving equations newton raphson method
Categories: numerical methods pure mathematics
The Newton-Raphson method is a numerical method to solve equations of the form f(x) = 0.
This method requires us to also know the first differential of the function. But the Newton-Raphson method has very good convergence, so it can often provide an accurate result very quickly.
Here is a video on the topic:
## Example - square root of two using Newton-Raphson method
As a simple example, we will solve the equation:
This equation has a solution x = √2 (and an second solution x = -√2). We will only look for the first solution. Although we already know the answer to the problem, it is still useful to work through the numerical solution to see how it works (and to gain an approximate value for the square root of 2).
To use the technique we need to have a rough idea of where the solution is. It is useful to sketch a graph of the function:
The Newton-Raphson method starts with an initial guess at the solution. The guess doesn't need to be particularly accurate, we can just use the value 2.
The Newton-Raphson method proceeds as follows:
2. Draw a tangent to the curve at the point x.
3. Find the value of x where the tangent crosses the x-axis. This will be the next value for x.
4. Repeat from step two with the new value.
Steps 2 to 4 are repeated until a sufficiently accurate result is obtained, as described in the solving equations article. On each pass, the new guess is usually closer to the required result, so the approximation becomes more accurate. When the result is accurate enough, the process ends.
## A graphical explanation of the Newton Raphson method
To gain an intuitive understanding of the process, we will go through a couple of iterations and show the results on a graph. This is for illustration only, you don't need to draw an accurate graph to use this method. Once we understand the method it can be calculated without requiring a graph.
## First iteration
The first iteration starts with a guess of 2. Here is the graph where we have zoomed in on the region of interest. a shows the line x = 2:
We now draw a line that forms a tangent to the curve at x = 2:
This tangent line hits the x-axis at x = 1.5. This forms our second approximation.
The formula for calculating the next value of x is shown later in this article.
## Second iteration
The second iteration starts with the new value of 1.5. On the graph below, we have zoomed in on the range 0.5 to 2.5:
Again, we draw a tangent. Notice that at this stage the curve itself is very close to being a straight line, so the tangent is very close to the curve when it hits the x-axis.
The tangent hits the x-axis at 0.4666...
## Deriving the formula
This diagram shows a curve y = f(x) with its tangent at point A. We need to calculate the position of point B, where the tangent crosses the x-axis.
If point A has an x-value of x0, then its y-value will be f(x0) because the point is on the curve.
Since the line AB is a tangent to the curve, we also know that the slope is equal to the slope of the function curve at point A, which is the first derivative of the curve at that point:
Next we can draw a right-angled triangle ABC where C is on the x-axis directly below A (ie at point x0):
This triangle tells us that the slope of the line AB is:
Since the length of AC is $f(x_0)$ and the length of BC is x1 - x0, we have
We now have two separate expressions for the slope of the line AB, so we can equate them:
Inverting both sides gives:
So:
Rearranging the terms gives the final result for x1:
This is the general formula for solving f(x). The particular formula we are solving in this example is:
Which has a first derivative:
So our equation is
## Newton-Raphson method by calculation
The method for finding the root is very simple:
2. Use the formula to calculate the next value of x.
3. Repeat step two with the new value.
Steps 2 to 3 are repeated until a sufficiently accurate result is obtained, as described in the solving equations article. On each pass, the value of x should typically get closer to the root.
In this case, the formula is:
Starting with x = 2, applying this equation gives:
• 1.5
• 1.4666...
• 1.4142156862745097
• 1.4142135623746899
• 1.414213562373095
After 5 iterations, the result is correct to about 15 decimal places!
The main advantage of the Newton-Raphson method is that it often converges very quickly.
There are several disadvantages. The first is that some analysis is required for each formula. For example, it took a few lines of calculation to discover the formula for finding the square root of 2. If we wanted to find the cube root of 2 we would need to do a similar, but slightly different, calculation.
Another disadvantage occurs in some cases where there is more than one root. Depending on the initial value of x the method will often converge on the nearest solution. But that isn't always true. In some situations, the method will end up converging on a different root that is further away from the starting value. It isn't always easy to predict which root the method will converge on from a given starting point.
In some cases, attempting to map the start value onto the final root can result in a highly complex, fractal-like pattern, called a Newton Fractal. Unfortunately, poor old Joseph Raphson doesn't get credited with this fractal, even though neither he nor Newton discovered it.
Finally, it is sometimes possible for the formula to never converge on a root. This can usually be corrected by choosing a different initial value.
This makes the method very useful for calculating common values such as square roots, or other roots. The analysis only needs to be performed once, it is possible to devise starting conditions that are known to always work, and the calculation converges very quickly.
## Newton-Raphson method calculator in Python code
As an illustration, we will create a simple Python routine that acts as a Newton-Raphson calculator for the square root of any positive value.
For square roots, calculating the square root of 2 every time isn't very useful. We would like to calculate the square root of any positive value a. This requires a slight modification to the formula:
Here is the Python code to calculate this:
def square_root(a):
x = a
while True:
x_next = (x + a/x) / 2
if abs(x_next - x) < 0.0000001:
break
x = x_next
return x
print(root(2))
print(root(9))
print(root(10))
Within the square_root function, we loop calculating the next x value using the formula.
We need to decide how many times to do this. There are various ways to do this, but in this code we simply check the absolute difference between the old value of x and the new value of x. Because the x2 function is a simple U-shape, the approximation always gets better on every iteration of the loop. So when x reaches the point that it is hardly changing between iterations, we know that the value is reasonably accurate. In this case, we stop when the value is known to about 6 decimal places.
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# Royal Pitch
Information From Around The Globe
# 48 Out Of 60 As A Percentage
To calculate 48 out of sixty as a percentage, divide 48 by 60. You will need to know how to read the decimal place. Then you can use a calculator to find the output value and the percent. A calculator can also be used for different amounts. Type in the numbers and hit enter. This process will take a few minutes. When you have finished, you can put the answer on your website or post it on your Facebook wall.
To calculate a percentage, enter a fraction in the calculator. For example, 5% of twenty is the same as x/100 * 20. This value will show as 1. The calculator is useful for sales profit, gross profit, and other types of percentage problems. You can also use it to see how much cash you will earn when you use a credit card. You can find out how much money you’ll make by using 48 out of 60 as a percent.
A calculator is a great tool for converting fractions to percentages. Simply enter a fraction into the calculator and hit enter. The calculator will display an answer in a result box. Whether you’re calculating a discount at the store or gross profit on your next purchase, you’ll be able to use a percentage calculator. It’s an indispensable tool to have when comparing values.
In order to convert a fraction to a percentage, you need to find a new value of 48. Then, you need to multiply the new value by 100. That’s how you convert a fraction to a percentage. The result will be 80. If you’re trying to determine the difference between two numbers, you can multiply by 100, which will give you the answer to the original question.
In order to convert a fraction to a percentage, you must first convert the value into a decimal. To calculate a percentage, you must first enter the old value. Then, multiply the new value by 100. This will produce a result that shows you a percentage as the result of a calculation. In the case of the former, the answer will be a lower percentage than the one of the latter.
Whether you’re comparing two numbers, a percentage is a universally accepted unit of measurement. For example, you can use it to find the difference between two values. For example, a discount at a store costs five dollars and a dollar. To calculate gross profit, you can multiply the two values by 100. Once you’ve determined a difference, the two numbers can be converted into a percentage.
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Sep 1, 2015 -
Number Magic – An Explanation of Bendford’s Law
Earlier this year, a colleague of mine sent me an email on Bendford’s Law. He had run across it somewhere and was fascinated by it. It seems counterintuitive that small digits would occur more frequently in the leading digits of arbitrary numerical data. One is tempted to think that arbitrary data would be made up of arbitrary digits, but that turns out not to be the case. It’s a genuine numerical phenomenon, and below I have provided a couple of ways to explain it. I point out that, utlimately, this law results from the notation that we use to represent real values.
Upon learning about Bendford’s Law, my colleague decided to put it to a test. So he grabbed an ENDF file, endf66a, and pulled as many values as he could from it. (ENDF stands for “Evaluated Nuclear Data File.” It is a large database of nuclear data, such as cross-sections of various nuclides. For our purposes here, it is a large collection of real-world numerical data.) He collected the leading digit in the mantissa of each floating-point number in the file and examined the frequency of occurrence of each digit. The plot of the results that he produced is shown below.
After seeing his empirical verification of this law, I tried to explain why this law works in a couple of ways.
The first way is to consider the population of data, and here we’re talking about values that span multiple orders of magnitude. For example, if we were talking about lengths, the list could include values measured in centimeters, meters, and kilometers. Now for all of these units to be represented in our population we’d need the number of values expressed in centimeters to be roughly equivalent to the number of values expressed in kilometers. Naturally, such a population wouldn’t even come close to being distributed uniformly over the range of values—there are simply too many centimeters in a kilometer to sample from them in the same way that one would sample over centimeters in one meter. Mathematically, this means that, if $$f(x)$$ is the probability density function (p.d.f.) of our population, then$\int_{10\,\text{cm}}^{1\,\text{m}} f(x)\,dx \approx \int_{100\,\text{m}}^{1\,\text{km}} f(x)\,dx$which means that$\bar f_{(1\,\text{m})} \approx 1000\times\bar f_{(1\,\text{km})}$where $$\bar f_{(a)}$$ is the average value of the p.d.f. over the neighborhood where $$x \approx a$$.
Therefore, it makes sense to look at the value of $$f(x) \cdot x$$ plotted versus $$x$$ on a semi-log scale, keeping in mind that the probability of $$X$$ (a random value drawn from the population) being between $$x_1$$ and $$x_2$$ is$P(x_1 < X < x_2) = \ln(10) \int_{\log_{10}(x_1)}^{\log_{10}(x_2)} f(10^\xi)\cdot 10^\xi\,d\xi$That is, when $$f(x) \cdot x$$ is plotted versus $$x$$, with a logarithmic scale for $$x$$, the area under the curve represents the relative probability of $$X$$ being in a particular region. ($$\xi$$ gives the linear distance along the logarithmic scale: $$\xi = \log_{10}x$$.)
A (made-up) example of such a distribution is shown below.
Here the regions where the leading digit would be 1 are marked off. Compare this to the same figure with the regions where the leading digit would be 5 are highlighted.
As we can see, the areas marked off in in the first figure are significantly larger than the areas marked off in the second, which means that 1 is more likely than 5 to show up as the first digit of a number that is randomly drawn from this distribution, and this is the case for many collections of numbers that span multiple scales.
But perhaps someone objects to the p.d.f. that I used as an example. OK. Then consider this. To simplify the situation let’s consider only integers and let’s look at the range of integers from 0 to 9. If I have a population of these integers that is uniformly distributed, then any random variable that I produce is as likely to be one digit as another. But what happens if I double my range of possibilities by expanding to the right?
Now, I’m looking at the range of numbers from 0 to 19. Once again, if the distribution is uniform, I’m equally as likely to draw any of the numbers, but now the probability of drawing a number with 1 as the leading digit has changed from 10% to 55%! Over half of the time I’m going to get a number that starts with 1.
This example is contrived, of course, but it readily generalizes. For example, consider what happens if I expand the range to extend from 0 to 49. Although the probability of getting a number that begins with a 1 is no longer 55%, it’s still larger than getting a number that begins with a 5, 6, 7, 8, or 9.
This phenomenon is an artifact of the way we represent real values, not the real values themselves. To see this, it is instructive to consider just one number—for example, the fine-structure constant:$\alpha \approx 7.297\times 10^{-3}$This is a physical constant. It’s dimensionless. There is no mathematical basis for this constant. As Richard Feynmann once wrote,
It’s one of the greatest damn mysteries of physics: a magic number that comes to us with no understanding by man. You might say the “hand of God” wrote that number, and “we don’t know how He pushed his pencil.”
So this is about an arbitrary a number as one can find. It happens to begin with a 7, not a 1, but that’s just because we use a number system with ten digits. (Ignore, for the moment, that its reciprocal, $$\alpha^{-1} \approx 137.036$$, an equally arbitrary number, does begin with a 1.) The number is what it is, regardless of how we write it. We can think of it as a single point on a line that represents all real numbers (the $$x$$-axis of the complex plane). What digits we use to write that number depend on where we lay down the grid lines that denote 1, 2, 3, and so on. If the real values are scaled logarithmically, this line looks like the following: The point that is $$\alpha$$ also is shown above and falls between the ticks for 7 and 8, resulting in a number that begins with 7. But what happens when a different base (number of digits) is used to represent the same number. In octal (base 8, which is often used in computer science, because $$8 = 2^3$$), the real line looks like (Keep in mind that now the octal “10” is really an “8” in decimal notation.) In this system of numbers, the numerical representation of $$\alpha$$ begins with a 3. If we double the number of digits to 16 (hexadecimal), we find the following: The leading digit is now 1.
It is important to note that the real line itself and all of the real values it represents, including $$\alpha$$, haven’t changed. They are the same in all three diagrams. By changing the number of digits in our number system, we change only where the grid lines (tick marks) are located. All of these number systems agree on the location of 1 (and zero and infinity), but everything else changes. Below, I have tabulated the value of $$\alpha$$ for number systems with 3 to 16 digits (base 3 to base 16):
Base $$\alpha$$
3 1.210e-12
4 1.313e-10
5 4.240e-4
6 1.324e-3
7 2.334e-3
8 3.571e-3
9 5.278e-3
10 7.297e-3
11 9.793e-3
12 1.074e-3
13 1.305e-2
14 1.605e-2
15 1.996e-2
16 1.DE4e-2
In 8 out of the 14 number schemes (57%), the representation of $$\alpha$$ begins with a 1. This is not surprising when you think about it, because although the locations of the grid lines change, the structure of the grid lines remains similar. The space along the (logarithmically scaled) real line where a 1 is the leading digit is always the largest space. Therefore, 1 is the most likely leading digit, regardless of the base.
For a truly arbitrary number, chosen without any restrictions, it is not difficult to predict that the likelihood of the digit $$n$$ being the first digit is$P(n) = \log_b(n+1) – \log_b(n)$ where $$b$$ is the base of the number system being used. This can be deduced geometrically from the figures above.
As long as one uses a notation system for numbers consisting of a series of symbols in which each successive symbol in the series constitutes a value that is $$b$$ times smaller than the symbol before it (where $$b$$ is the number of symbols used), the symbol denoting the smallest (non-zero) value (in our case, 1) will be the most likely symbol to appear first in the series.
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A132583 a(n) = n 2's sandwiched between two 1's. 7
11, 121, 1221, 12221, 122221, 1222221, 12222221, 122222221, 1222222221, 12222222221, 122222222221, 1222222222221, 12222222222221, 122222222222221, 1222222222222221, 12222222222222221, 122222222222222221, 1222222222222222221, 12222222222222222221, 122222222222222222221 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,1 COMMENTS Also, positive numbers each of whose digits indicates its number of adjacent digits (digits on the ends are adjacent to only one while others are adjacent to 2). - Eric Fox, Jul 19 2022 LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..500 Index entries for linear recurrences with constant coefficients, signature (11,-10). FORMULA a(0) = 11; a(n) = 10*a(n-1) + 11. - Jonathan Vos Post, Nov 24 2007 [corrected by Jon E. Schoenfield, Jan 28 2018] O.g.f.: 11/((-1+x)*(-1+10*x)) = -(110/9)/(-1+10*x) + (11/9)/(-1+x). - R. J. Mathar, Nov 28 2007 a(n) = 20*(10^n-1)/9 + 10^(n+1) + 1, with n >= 0. - Paolo P. Lava, Dec 19 2007 E.g.f.: 11*exp(x)*(10*exp(9*x) - 1)/9. - Stefano Spezia, Sep 15 2023 MAPLE g:=(1+z)/((1-z)*(1-10*z)): gser:=series(g, z=0, 43): seq((coeff(gser, z, n))-1, n=1..24); # Zerinvary Lajos, Feb 25 2009 MATHEMATICA NestList[10#+11&, 11, 20] (* or *) LinearRecurrence[{11, -10}, {11, 121}, 20] (* Harvey P. Dale, Sep 24 2012 *) PROG (Magma) [20*(10^n-1)/9+10^(n+1)+1: n in [0..25]]; // Vincenzo Librandi, Aug 10 2011 CROSSREFS Sequence in context: A106473 A125315 A223676 * A007907 A171285 A216132 Adjacent sequences: A132580 A132581 A132582 * A132584 A132585 A132586 KEYWORD nonn,base,easy,changed AUTHOR Paul Curtz, Nov 21 2007 STATUS approved
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# Search by Topic
#### Resources tagged with Circles similar to Three Tears:
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Stage:
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### There are 62 results
Broad Topics > 2D Geometry, Shape and Space > Circles
### Three Tears
##### Stage: 4 Challenge Level:
Construct this design using only compasses
### What Is the Circle Scribe Disk Compass?
##### Stage: 3 and 4
Introducing a geometrical instrument with 3 basic capabilities.
### Arclets Explained
##### Stage: 3 and 4
This article gives an wonderful insight into students working on the Arclets problem that first appeared in the Sept 2002 edition of the NRICH website.
##### Stage: 4 Challenge Level:
Explore when it is possible to construct a circle which just touches all four sides of a quadrilateral.
### Roaming Rhombus
##### Stage: 4 Challenge Level:
We have four rods of equal lengths hinged at their endpoints to form a rhombus ABCD. Keeping AB fixed we allow CD to take all possible positions in the plane. What is the locus (or path) of the point. . . .
##### Stage: 4 Challenge Level:
Given a square ABCD of sides 10 cm, and using the corners as centres, construct four quadrants with radius 10 cm each inside the square. The four arcs intersect at P, Q, R and S. Find the. . . .
### Like a Circle in a Spiral
##### Stage: 2, 3 and 4 Challenge Level:
A cheap and simple toy with lots of mathematics. Can you interpret the images that are produced? Can you predict the pattern that will be produced using different wheels?
### Not So Little X
##### Stage: 3 Challenge Level:
Two circles are enclosed by a rectangle 12 units by x units. The distance between the centres of the two circles is x/3 units. How big is x?
### Crescents and Triangles
##### Stage: 4 Challenge Level:
Triangle ABC is right angled at A and semi circles are drawn on all three sides producing two 'crescents'. Show that the sum of the areas of the two crescents equals the area of triangle ABC.
### Circumspection
##### Stage: 4 Challenge Level:
M is any point on the line AB. Squares of side length AM and MB are constructed and their circumcircles intersect at P (and M). Prove that the lines AD and BE produced pass through P.
### A Rational Search
##### Stage: 4 and 5 Challenge Level:
Investigate constructible images which contain rational areas.
### Angle A
##### Stage: 3 Challenge Level:
The three corners of a triangle are sitting on a circle. The angles are called Angle A, Angle B and Angle C. The dot in the middle of the circle shows the centre. The counter is measuring the size. . . .
### LOGO Challenge 12 - Concentric Circles
##### Stage: 3 and 4 Challenge Level:
Can you reproduce the design comprising a series of concentric circles? Test your understanding of the realtionship betwwn the circumference and diameter of a circle.
### Holly
##### Stage: 4 Challenge Level:
The ten arcs forming the edges of the "holly leaf" are all arcs of circles of radius 1 cm. Find the length of the perimeter of the holly leaf and the area of its surface.
### Circles, Circles Everywhere
##### Stage: 2 and 3
This article for pupils gives some examples of how circles have featured in people's lives for centuries.
### Circle Packing
##### Stage: 4 Challenge Level:
Equal circles can be arranged so that each circle touches four or six others. What percentage of the plane is covered by circles in each packing pattern? ...
### LOGO Challenge 10 - Circles
##### Stage: 3 and 4 Challenge Level:
In LOGO circles can be described in terms of polygons with an infinite (in this case large number) of sides - investigate this definition further.
### Coins on a Plate
##### Stage: 3 Challenge Level:
Points A, B and C are the centres of three circles, each one of which touches the other two. Prove that the perimeter of the triangle ABC is equal to the diameter of the largest circle.
### Some(?) of the Parts
##### Stage: 4 Challenge Level:
A circle touches the lines OA, OB and AB where OA and OB are perpendicular. Show that the diameter of the circle is equal to the perimeter of the triangle
### Pi, a Very Special Number
##### Stage: 2 and 3
Read all about the number pi and the mathematicians who have tried to find out its value as accurately as possible.
### Square Pegs
##### Stage: 3 Challenge Level:
Which is a better fit, a square peg in a round hole or a round peg in a square hole?
### Illusion
##### Stage: 3 and 4 Challenge Level:
A security camera, taking pictures each half a second, films a cyclist going by. In the film, the cyclist appears to go forward while the wheels appear to go backwards. Why?
### Efficient Packing
##### Stage: 4 Challenge Level:
How efficiently can you pack together disks?
##### Stage: 2, 3 and 4 Challenge Level:
A metal puzzle which led to some mathematical questions.
##### Stage: 4 Challenge Level:
The sides of a triangle are 25, 39 and 40 units of length. Find the diameter of the circumscribed circle.
### The Pillar of Chios
##### Stage: 3 Challenge Level:
Semicircles are drawn on the sides of a rectangle ABCD. A circle passing through points ABCD carves out four crescent-shaped regions. Prove that the sum of the areas of the four crescents is equal to. . . .
### LOGO Challenge 11 - More on Circles
##### Stage: 3 and 4 Challenge Level:
Thinking of circles as polygons with an infinite number of sides - but how does this help us with our understanding of the circumference of circle as pi x d? This challenge investigates. . . .
### The Pi Are Square
##### Stage: 3 Challenge Level:
A circle with the radius of 2.2 centimetres is drawn touching the sides of a square. What area of the square is NOT covered by the circle?
### Partly Circles
##### Stage: 4 Challenge Level:
What is the same and what is different about these circle questions? What connections can you make?
### F'arc'tion
##### Stage: 3 Challenge Level:
At the corner of the cube circular arcs are drawn and the area enclosed shaded. What fraction of the surface area of the cube is shaded? Try working out the answer without recourse to pencil and. . . .
### Floored
##### Stage: 3 Challenge Level:
A floor is covered by a tessellation of equilateral triangles, each having three equal arcs inside it. What proportion of the area of the tessellation is shaded?
### Squaring the Circle
##### Stage: 3 Challenge Level:
Bluey-green, white and transparent squares with a few odd bits of shapes around the perimeter. But, how many squares are there of each type in the complete circle? Study the picture and make. . . .
### Witch's Hat
##### Stage: 3 and 4 Challenge Level:
What shapes should Elly cut out to make a witch's hat? How can she make a taller hat?
### Pie Cuts
##### Stage: 3 Challenge Level:
Investigate the different ways of cutting a perfectly circular pie into equal pieces using exactly 3 cuts. The cuts have to be along chords of the circle (which might be diameters).
### Tricircle
##### Stage: 4 Challenge Level:
The centre of the larger circle is at the midpoint of one side of an equilateral triangle and the circle touches the other two sides of the triangle. A smaller circle touches the larger circle and. . . .
##### Stage: 4 Challenge Level:
Investigate the properties of quadrilaterals which can be drawn with a circle just touching each side and another circle just touching each vertex.
### A Chordingly
##### Stage: 3 Challenge Level:
Find the area of the annulus in terms of the length of the chord which is tangent to the inner circle.
### Curvy Areas
##### Stage: 4 Challenge Level:
Have a go at creating these images based on circles. What do you notice about the areas of the different sections?
### Bull's Eye
##### Stage: 3 Challenge Level:
What fractions of the largest circle are the two shaded regions?
### Efficient Cutting
##### Stage: 3 Challenge Level:
Use a single sheet of A4 paper and make a cylinder having the greatest possible volume. The cylinder must be closed off by a circle at each end.
### Rotating Triangle
##### Stage: 3 and 4 Challenge Level:
What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger circle?
### First Forward Into Logo 4: Circles
##### Stage: 2, 3 and 4 Challenge Level:
Learn how to draw circles using Logo. Wait a minute! Are they really circles? If not what are they?
### Salinon
##### Stage: 4 Challenge Level:
This shape comprises four semi-circles. What is the relationship between the area of the shaded region and the area of the circle on AB as diameter?
### Track Design
##### Stage: 4 Challenge Level:
Where should runners start the 200m race so that they have all run the same distance by the finish?
### Rolling Coins
##### Stage: 4 Challenge Level:
A blue coin rolls round two yellow coins which touch. The coins are the same size. How many revolutions does the blue coin make when it rolls all the way round the yellow coins? Investigate for a. . . .
### LOGO Challenge - Circles as Animals
##### Stage: 3 and 4 Challenge Level:
See if you can anticipate successive 'generations' of the two animals shown here.
### Approximating Pi
##### Stage: 4 and 5 Challenge Level:
By inscribing a circle in a square and then a square in a circle find an approximation to pi. By using a hexagon, can you improve on the approximation?
### Semi-square
##### Stage: 4 Challenge Level:
What is the ratio of the area of a square inscribed in a semicircle to the area of the square inscribed in the entire circle?
### Round and Round
##### Stage: 4 Challenge Level:
Prove that the shaded area of the semicircle is equal to the area of the inner circle.
### Blue and White
##### Stage: 3 Challenge Level:
Identical squares of side one unit contain some circles shaded blue. In which of the four examples is the shaded area greatest?
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## [PrimeNumbers] Twin primes Criteria (Correction)
Expand Messages
• Hi Flavio As Paul Jobling pointed out, your conjecture is equivalent to the criteria (2) and (3) below. A more thorough investigation shows that both these
Message 1 of 1 , Mar 29, 2001
Hi Flavio
As Paul Jobling pointed out, your conjecture is equivalent to
the criteria (2) and (3) below. A more thorough investigation
shows that both these equivalent versions are indeed correct.
Mike Oakes appears to be mistaken or I am missing something?
I base this on the following Mathematica snippet:
Do[p=Prime[i];
If[Mod[p,4]==1, r=-1, r=+1];
If[PrimeQ[p]&&PrimeQ[p+2],
m=Mod[2((p-1)/2)!^2 - r(5*p+2),p(p+2)];
If[m!=0,Print["False"];
];
];
,{i,2,PrimePi[10^4]}];
In 1949 using Wilson's Theorem* P. A. Clement published
a proof that p and p+2 are both prime ("twin primes")
if and only if the following congruence holds:
(1) 4((p-1)! + 1) + p = 0 [mod p(p+2)]
Furthermore in 1995 Joseph B. Dence & Thomas P. Dence
reduced the Clement criterion to the following two criteria:
(2) 2((((p-1)/2)!)^2 + 1) + 5p = 0 [mod p(p+2)]
if and only if p and p+2 are primes and p=4k+1
(3) 2((((p-1)/2)!)^2 - 1) - 5p = 0 [mod p(p+2)]
if and only if p and p+2 are primes and p=4k-1
A similar criterion exists for the primality of p and p+d:
If p>1 and d>1 are both integers, then p and p+d are both
prime if and only if:
( 1 ((-1)^d)d! ) 1 1
(4) (p-1)! ( - + ---------- ) + - + --- is an integer
( p p+d ) p p+d
In addition P. A. Clement's paper proves similar necessary and
sufficient criteria for prime triples p, p+2 and p+6 as well
as prime quadruplets p, p+2, p+6, p+8. If these are of general
interest I can post them to the list.
Regards
Alan Powell
* Wilson's Theorem:
An integer p>1 is prime if and only if (p-1)! + 1 = 0 [mod p]
At 09:06 AM 3/27/01, torasso.flavio@... wrote:
> I found the following rule concerning twin primes:
> n and n+2 are both prime iff
> 2 [(n-1)/2]!^2 = \pm (5n+2) mod n(n+2)
> the sign being "+" when n=4k-1, "-" when n=4k+1.
>
> Is it an interesting or trivial result?
> Are there similar congruences for other prime pairs?
>
> Thanks for any comments
Your message has been successfully submitted and would be delivered to recipients shortly.
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Problem Topics: Types of problems that may appear on the exam…
1. 1-D Kinematics – be able to apply any of the six basic kinematics formulas to solve for any variable.
2. 2-D Kinematics
a. Given a vector with a provided direction (in degrees relative to some known direction) and magnitude, resolve the vector into its X and Y components.
b. Given the X and Y components of a vector, provide the vector’s magnitude and direction (in degrees relative to some known direction).
c. Projectile Motion Problems (acceleration in the y dimension, constant velocity in the x dimension). Possible variations…
i. Symmetric problem (initial and final positions are at the same elevation) -- be able to apply the Range Formula.
ii. Asymmetric problem (initial and final positions are at different elevations)
iii. Half of a symmetric problem (flight either begins or ends with horizontal velocity)
d. River Problems – be able to identify component and resultant vectors. Use two provided vectors to find the third. Provided vectors will be orthogonal. Solutions may not be orthogonal.
i. Given two component vectors, find the resultant (magnitude and direction)
ii. Given a resultant vector and one component, find the other component vector (magnitude and direction)
3. 1-D Newton’s Laws – Apply Newton’s laws to solve problems. The general strategy is to write two equations for net force – one using Newton’s 2nd Law (the object’s mass times it acceleration), and one using the vector sum of all forces acting on the object in question.
a. Simple problems relating to acceleration, net force, and friction.
b. Vertically Accelerating Objects:
i. Supported by tension
ii. Supported by normal force (e.g. elevator problems)
c. Find tensions and acceleration in systems of blocks connected by strings. Some blocks slide on a horizontal surface, and at least one block hangs vertically. There may or may not be friction between the surface and the sliding block(s).
4. 2-D Newton’s Laws – Resolve forces into useful directions (x and y components or perpendicular and parallel components). Then apply Newton’s Laws.
a. Static equilibrium – a non-moving mass is supported by strings. Find tensions.
b. One or more blocks on an incline in a “tug-of-war” with another block that is hanging vertically (facilitated by a pulley). Find acceleration and tension.
5. Work and Energy – Be able to apply all of the formulas, except efficiency…
a. Work
b. Kinetic Energy
c. Work-Energy Theorem
d. Power
e. Potential Energies (gravitational and spring)
f. Conservation of energy without non-conservative work
g. Conservation of energy with non-conservative work
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Get Full Access to Essential Calculus (Available Titles Cengagenow) - 1 Edition - Chapter 1 - Problem 1.1.10
Get Full Access to Essential Calculus (Available Titles Cengagenow) - 1 Edition - Chapter 1 - Problem 1.1.10
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# Solved: Sketch a rough graph of the number of hours of
ISBN: 9780495014423 247
## Solution for problem 1.1.10 Chapter 1
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Problem 1.1.10
Sketch a rough graph of the number of hours of daylight as a function of the time of year.
Step-by-Step Solution:
Step 1 of 3
Eco 110: Exam #2 Study Guide Chapter Six 1. Give an example of a price floor and a price ceiling. 2. Does a price ceiling or price floor cause a shortage 3. What two groups are responsible for carrying the burden of tax 4. What determines how the burden of tax is divided between the two groups Why 5. Why do economists oppose price controls 6. What is the difference between a binding price ceiling and a non- binding price ceiling 7. In a market with a binding price ceiling, an increase in the ceiling will __________ the quantity supplied, ______ the quantity demanded, and reduce the _______. a. Increase, decrease, surplus b. Decrease, increase, surplus c. Increase, decrease, shortage d. Decrease, increase, shortage 8. Which of the following would increase quantity supplied, decrease quantity demanded, and increase the price that consumers pay a. The imposition of a binding price floor b. The removal of binding price floor c. The passage of a tax levied on producers d. The repeal of a tax levied on producers 9. When a good is taxed, the burden of the tax falls mainly on consumers if… a. The tax is levied on consumers b. The tax is levied on producers c. Supply is inelastic, and demand is elastic d. Supply is elastic, and demand is inelastic 10. When the government imposes a binding price floor, it causes… a. The supply curve t
Step 2 of 3
Step 3 of 3
##### ISBN: 9780495014423
Since the solution to 1.1.10 from 1 chapter was answered, more than 240 students have viewed the full step-by-step answer. This full solution covers the following key subjects: . This expansive textbook survival guide covers 13 chapters, and 4396 solutions. The full step-by-step solution to problem: 1.1.10 from chapter: 1 was answered by , our top Calculus solution expert on 01/17/18, 03:06PM. This textbook survival guide was created for the textbook: Essential Calculus (Available Titles CengageNOW), edition: 1. Essential Calculus (Available Titles CengageNOW) was written by and is associated to the ISBN: 9780495014423. The answer to “Sketch a rough graph of the number of hours of daylight as a function of the time of year.” is broken down into a number of easy to follow steps, and 19 words.
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Solved: Sketch a rough graph of the number of hours of
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How to determine the required size of a septic tank
# How to determine the required size of a septic tank
In this article, we will explain how to determine the required size of a septic tank. The average residential septic tank size required for a normal daily sewage wastewater flow is provided in a table of septic tank sizes (in gallons). This article also explains how to calculate septic tank volume based on a septic tank inside dimensions measured in feet.
##### Required Septic Tank Size for Daily Water Usage Volume in Gallons
Average Sewage WastewaterFlow in Gallons/Day Minimum Septic Tank Size in Gallons(Effective Capacity Needed) 0-500 900 601-700 1200 801-900 1500 1001-1240 1900 2001-2500 3200 4501-5000 5800
Septic Tank in Gallons (Based on Number of Bedrooms)
0 bedrooms = 750 gal.
1-3 bedrooms = 1000 gal.
4 bedrooms = 1200 gal.
5 bedrooms = 1500 gal.
6 bedrooms = 1750 gal.
Calculating Septic Tank Capacity in Gallons Round Septic Tanks: 3.14 x radius squared x depth (all in feet) = cubic capacity. Cubic capacity x 7.5 = gallons capacity.
Rectangular Septic Tanks: Length x Width x Depth in feet x 7.5 = gallons
Rectangular Septic Tanks (alternative method 1): Length x width in inches / 231 = gallons per inch of septic tank depth. Multiply this number by septic tank depth in inches
Rectangular Septic Tanks (alternative method 2): Length x Width x Depth in feet / .1337 = gallons
##### Example for Computing Septic Tank Size, Capacity or Volume in Gallons
One gallon of water = .1337 cubic feet volume. For a rectangular septic tank, multiply depth in feet by width by length. Divide this by .1337 to solve for the number of gallons in
the rectangular septic tank.
Example: How many gallons is held in a 4 foot deep by 5 foot wide by 8 foot long septic tank? 4 x 5 x 8 = 160 cubic feet. So, using the conversion factor to convert cubic feet to
gallons, 160 / .1337 = 1196 – or about a 1200-gallon tank. One cubic foot of volume contains liquid in the amount of 7.481 gallons. Therefore, a second approach to calculating septic
tank size or capacity in gallons is to multiply the septic tank volume in cubic feet by this constant which we can round up to 7.5 gallons/cubic foot.
Source: http://engineeringfeed.com
Continue Reading about the "How to determine the required size of a septic tank" on the next page below
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# Root of 27300
#### [Root of twenty-seven thousand three hundred]
square root
165.2271
cube root
30.1107
fourth root
12.8541
fifth root
7.7132
In mathematics extracting a root is known as the determination of the unknown "x" in the equation $y=x^n$ The outcome of the extraction of the root is considered as a mathematical root. In the case of "n is equivalent to 2", one talks about a square root or sometimes a second root also, another possibility could be that n is equivalent to 3 by this time one would declare it a cube root or simply third root. Considering n beeing greater than 3, the root is declared as the fourth root, fifth root and so on.
In maths, the square root of 27300 is represented as this: $$\sqrt[]{27300}=165.22711641858$$
Moreover it is legit to write every root down as a power: $$\sqrt[n]{x}=x^\frac{1}{n}$$
The square root of 27300 is 165.22711641858. The cube root of 27300 is 30.110702110076. The fourth root of 27300 is 12.854070033207 and the fifth root is 7.7131633575377.
Look Up
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Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :
# JEE Main Maths Integral Calculus Previous Year Questions With Solutions
Calculus is the study of change, it basically analyses things that change and is a significant part of Mathematics. Calculus is a branch of mathematics focused on limits, integrals, derivatives, functions and infinite series. Integral calculus and Differential calculus are the two main branches of this topic. This article covers indefinite integral, definite integral, integration using partial fractions, integration by parts and properties of a definite integral. The integral calculus questions from the previous years of JEE Main are present on this page, along with the detailed solution for each question. About 2-4 questions are asked from this topic in JEE Examination.
## JEE Main Maths Integral Calculus Previous Year Questions With Solutions
Question 1: ∫{[sin8x − cos8x] / [1 − 2 sin2x cos2x]} dx = _________.
Solution:
∫{[sin8x − cos8x] / [1 − 2 sin2x cos2x]} dx
= ∫{[(sin4x + cos4x) * (sin4x − cos4x)] / [(sin2x + cos2x)2 − 2 sin2x cos2x]} dx
= ∫(sin4x – cos4x) dx
= ∫[sin2x + cos2x] * [sin2x – cos2x] dx
= ∫(sin2x + cos2x) dx
= ∫−cos2xdx
= [−sin2x / 2] + c
Question 2: ∫x2dx / (a + bx)2 = ___________.
Solution:
Put a + bx = t
⇒ x = [t − a] / [b] and dx = dt / [b]
I =∫([t − a] / b)2 * [1 / t2] * [dt / b]
= [1 / b2]∫(1 − (2a / t) + [a2 * t−2]) dt
= [1 / b2] * [(t − 2a log t) − (a2 / t)]
= [1 / b2] [(x + a / b) − [2a / b] * log (a + bx) − [a2 / b] * [1 / (a + bx)]
Question 3: ∫[x5 / √(1 + x3)] dx = ________.
Solution:
Put 1 + x3 = t2
⇒ 3x2 dx = 2tdt and x3 = t2 − 1
So, ∫[x5 / √(1 + x3)] dx = ∫{[x2 * x3] / √(1 + x3)} dx
= [2 / 3] ∫{[(t2 − 1) * t] dt / [t]}
= [2 / 3] ∫(t2 − 1) dt
= [2 / 3] [(t3 / 3) − t] + c
= [2 / 3] [{(1 + x3)3/2 / 3} − {(1 + x3)½}]+ c
Question 4: ∫tan32x sec2x dx = __________.
Solution:
∫tan32x sec2x dx = ∫[(sec2 2x − 1) sec2x * tan2x] dx
=∫sec32x tan2x dx −∫sec2x tan2x dx ……. (i)
Now, we take ∫sec32x tan2x dx
Put sec 2x = t
⇒ sec 2x tan 2x = dt/2, then it reduces to
[1 / 2] ∫t2 dt = t3 / 6
= [sec32x] / [6]
From (i), ∫sec32x tan2x dx −∫sec2x tan2x dx
= [sec3 2x / 6] − [sec2x / 2] + c
Trick: Let sec 2x = t, then sec 2x tan 2x dx = [1 / 2] dt
[1 / 2] ∫(t2 − 1) dt = [1 / 6]t3 − [1 / 2]t + c
= [sec3 2x / 6] − [sec 2x / 2] + c
Question 5: If ∫{[4ex + 6e−x] / [9ex − 4e−x]} dx = Ax + B log (9e2x − 4) + C, then find A, B and C.
Solution:
I =∫{[4ex + 6e−x] / [9ex − 4e−x]} dx
= [4 / 9]∫[9e2xdx] / [9e2x − 4] + 6∫[dx] / [9e2x − 4]
∫dx / [9e2x − 4] = [1 / 8] log (9e2x − 4) − [1 / 4] log3 − [1 / 4] x + constant
I = [35 / 36] log (9e2x − 4) − [3 / 2] x − [3 / 2] log3 + constant
Comparing with the given integral, we get
A = −3 / 2, B = 35 / 36, C = [−3 / 2] log3 + constant
Question 6: ∫x cos2x dx = ______.
Solution:
x cos2x dx = [1 / 2] ∫x (1 + cos2x) dx
= [x2 / 4] + [1 / 2] [(x sin2x) / (2) −∫(sin2x / 2) dx] + c
= [x2 / 4] + (x sin2x / 4) + (cos2x / 8) + c
Question 7: ∫x / [1 + x4] dx = ________.
Solution:
Put t = x2 ⇒ dt = 2x dx, therefore,
∫x / [1 + x4] dx = [1 / 2] ∫1 / [1 + t2] dt
= [1 / 2] tan−1 t + c
= [1 / 2] tan−1 x2 + c
Question 8: ∫[1 + x2] / √[1 − x2] dx = ________.
Solution:
Put x = sinθ ⇒ dx = cosθ dθ, then it reduces to
∫(1 + sin2θ) dθ = θ + [1 / 2]∫(1 − cos2θ) dθ
= [3θ / 2] − [1 / 2] sinθ * √[1 − sin2θ] + c
= [3 / 2] sin−1 x − [1 / 2]x √[1 − x2] + c
Question 9: ∫√(1 + sin [x / 2]) dx = _________.
Solution:
∫√(1 + sin [x / 2]) dx = ∫√(sin2 [x / 4] + cos2 [x / 4] + 2 sin [x / 4] cos [x / 4]) dx
= ∫(sin [x / 4] + cos [x / 4]) dx
= 4 (sin [x / 4] – cos [x / 4]) + c
Question 10: ∫[sinx] / [sin (x − α)] dx = ________.
Solution:
∫[sinx] / [sin (x − α)] dx =
∫[sin (x − α + α)] / [sin (x − α)] dx
= ∫{[(sin (x − α) cosα + cos (x − α) sinα] / [sin (x − α)]} dx
= ∫cosα dx +∫sinα * cot (x − α) dx
= x cosα + sinα * log sin (x − α) + c
Question 11: ∫(log x)2 dx = _______.
Solution:
∫(log x)2 dx
Put log x = t
⇒ et = x
⇒ dx = et dt, then it reduces to
∫t2 * et dt = t2 * et − 2t * et + 2et + c
= x (log x)2 − 2x log x + 2x + c
Question 12: ∫[cos2θ] * log ([cosθ + sinθ] / [cosθ − sinθ]) dθ = ___________.
Solution:
We know that
log ([cosθ + sinθ] / [cosθ − sinθ]) = log ([1 + tanθ] / [1 − tanθ]) = log tan (π / 4 + θ) ∫secθ dθ = log tan (π / 4 + θ / 2)
∫sec2θ dθ = [1 / 2] * log tan (π / 4 + θ)
2 sec2θ = [d / dθ] log tan (Ï€ / 4 + θ) …….(i)
Integrating the given expression by parts, we get
I = [1 / 2] sin2θ log tan (π / 4 + θ) − [1 / 2] ∫[sin2θ * 2 sec2θ] dθ by (i)
= [1 / 2] sin2θ log tan (π / 4 + θ) −∫tan2θ dθ
= [1 / 2] sin2θ log tan (π / 4 + θ) − [1 / 2] log sec2θ
Question 13: ∫dx / (sinx + sin2x) = _________.
Solution:
I = ∫dx / [sinx (1 + 2cosx)]
= ∫[sinx dx] / [sin2x * (1 + 2 cosx)]
= ∫sinx dx / {(1 − cosx) * (1 + cosx) * (1 + 2 cosx)}
Now differential coefficient of cosx is − sinx, which is given in numerator and hence, we make the substitution cosx = t ⇒ −sinx dx = dt
I = −∫dt / [(1 − t) (1 + t) (1 + 2t)]
We split the integrand into partial fractions
I = −∫[{1 / 6 (1 − t)} − {1 / 2 (1 + t)} + {4 / 3 (1 + 2t)}] dt
[1 / 6] log (1 − cosx) + [1 / 2] log (1 + cosx) − [2 / 3] log (1 + 2 cosx)
Question 14: For which of the following values of m, the area of the region bounded by the curve y = x − x2 and the line y = mx equals 9 / 2?
Solution:
The equation of curve is y = x − x2
⇒ x2 − x = −y
(x − [1 / 2])2 = − (y − [1 / 4])
This is a parabola whose vertex is (12, 14)
Hence, point of intersection of the curve and the line x − x2 = mx
⇒ x (1 − x − m) = 0 i.e., x =0 or x = 1 − m
[9 / 2] = ∫1−m0 (x − x2 − mx) dx
= ([x2 / 2] − [x3 / 3] − [mx2 / 2])1−m0
= [(1 − m)] * [(1 − m)2 / 2] − [(1 − m)3 / 3] = [(1 − m)3/ 6]
(1 − m)3 = [6 * 9] / [2] = 27
⇒ 1 − m = 271/3 = 3
⇒ m = −2
Also, (1 − m)3 − 33 = 0
(1 − m −3) [(1 − m)2 + 9 + (1 − m)3] = 0
(1 − m)2 + 3 (1 − m) + 9 = 0
m2 − 2m + 1 − 3m + 3 + 9 = 0
m2 − 5m + 13 = 0
m = (5 ± √[25 − 52]) / 2 i.e., m is imaginary
Hence, m = −2.
Question 15: Let f : R → R and g : R → R be continuous functions, then the value of the integral ∫π/2−π/2 [f (x) + f (−x)] [g (x) − g (−x)] dx = ___________.
Solution:
Let h(x) = {f (x) + f (−x)} {g (x) − g (−x)}
h (−x) = {f (−x) + f (x)} {g (−x) − g (x)}
= −{f (−x) + f (x)} {g (x) − g (−x)}
= − h (x)
Therefore, ∫π/2−π/2 h (x) dx = 0.
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# Thread: Writing Polynomial functions
1. ## Writing Polynomial functions
hello I just started on these tonight and i understand all the terms but this one is tricky because it has to go through a specific point and i dont quite understand how to do it yet.
Any help is appreciated:
The polynomial of degree 4, ${P}{\left({x}\right)}$ has a root of multiplicity 2 at ${x}={4}$ and roots of multiplicity 1 at ${x}={0}$ and ${x}=-{3}$. It goes through the point ${\left({5},{24}\right)}$.
Find a formula for ${P}{\left({x}\right)}$.
My train of thought
P(x)=
(X-4)^2 Multiplicity of 2 and zeros at 4
(X) Multiplicity of 1 and zero at 0
(X+3) Multiplicity of 1 and zero at -3
When i graph it, i get a 4th degree polynominal that goes through (-3,0)(0,0) and bounces off (4,0) but does not go through (5,24) instead it goes through (5,40).
2. Originally Posted by xsavethesporksx
hello I just started on these tonight and i understand all the terms but this one is tricky because it has to go through a specific point and i dont quite understand how to do it yet.
Any help is appreciated:
The polynomial of degree 4, ${P}{\left({x}\right)}$ has a root of multiplicity 2 at ${x}={4}$ and roots of multiplicity 1 at ${x}={0}$ and ${x}=-{3}$. It goes through the point ${\left({5},{24}\right)}$.
Find a formula for ${P}{\left({x}\right)}$.
My train of thought
P(x)=
(X-4)^2 Multiplicity of 2 and zeros at 4
(X) Multiplicity of 1 and zero at 0
(X+3) Multiplicity of 1 and zero at -3
When i graph it, i get a 4th degree polynominal that goes through (-3,0)(0,0) and bounces off (4,0) but does not go through (5,24) instead it goes through (5,40).
$\displaystyle P(x) = kx(x+3)(x-4)^2$
$\displaystyle P(5) = 24$
solve for $\displaystyle k$
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# The perimeter of a quadrant of a circle of radius $\dfrac{7}{2}cm$ is :(a) $3.5cm$ (b) $5.5cm$ (c) $7.5cm$ (d) $12.5cm$
Last updated date: 20th Jun 2024
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Hint:As we know that Circumference of circle is $2\pi r$ . Perimeter implies that the circumference of a circle and quadrant of a circle means one-fourth part of the circle. Using this concept we will get the required solution .
As We learn that Perimeter of the circle is $2\pi r$.
But in this question we have to find the perimeter of a quadrant of circle means that one-fourth of the perimeter of circle
Here we see that the blue colored part is the quadrant of the circle .
Perimeter of quadrant of circle is $\dfrac{{2\pi r}}{4}$
where we take $\pi = \dfrac{{22}}{7}$ and Radius of circle is given $r = \dfrac{7}{2}$ ;
By putting these value ,
= $\dfrac{{2 \times \dfrac{{22}}{7} \times \dfrac{7}{2}}}{4}$
When we solve we get
= $\dfrac{{22}}{4}$$cm$
In decimal form
= $5.5cm$
So, the correct answer is “Option B”.
Note:Students should also remembered that Area of Circle is = $\pi {r^2}$ And the area of quadrant is $\dfrac{{\pi {r^2}}}{4}$.Perimeter of semi-circle is $\pi r$ and Area of semi-circle is $\dfrac{{\pi {r^2}}}{2}$.
Perimeter of Square is $4 \times a$ and Area of Square is ${a^2}$ where a is the side of the square.
Perimeter of rectangle is $2\left( {l + b} \right)$ and Area of rectangle is $l \times b$ where $l =$ Length of Rectangle and $b =$ breadth of Rectangle.
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0
# Is 17 quarts bigger than 4 gallons?
Wiki User
2013-03-07 01:29:44
1 gallon = 4 quarts
2 gallons = 8 quarts
3 gallons = 12 quarts
4 gallons = 16 quarts
It does appear to be.
Wiki User
2013-03-07 01:29:44
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# 1616
## 1,616 is an even composite number composed of two prime numbers multiplied together.
What does the number 1616 look like?
This visualization shows the relationship between its 2 prime factors (large circles) and 10 divisors.
1616 is an even composite number. It is composed of two distinct prime numbers multiplied together. It has a total of ten divisors.
## Prime factorization of 1616:
### 24 × 101
(2 × 2 × 2 × 2 × 101)
See below for interesting mathematical facts about the number 1616 from the Numbermatics database.
### Names of 1616
• Cardinal: 1616 can be written as One thousand, six hundred sixteen.
### Scientific notation
• Scientific notation: 1.616 × 103
### Factors of 1616
• Number of distinct prime factors ω(n): 2
• Total number of prime factors Ω(n): 5
• Sum of prime factors: 103
### Divisors of 1616
• Number of divisors d(n): 10
• Complete list of divisors:
• Sum of all divisors σ(n): 3162
• Sum of proper divisors (its aliquot sum) s(n): 1546
• 1616 is a deficient number, because the sum of its proper divisors (1546) is less than itself. Its deficiency is 70
### Bases of 1616
• Binary: 110010100002
• Base-36: 18W
### Squares and roots of 1616
• 1616 squared (16162) is 2611456
• 1616 cubed (16163) is 4220112896
• The square root of 1616 is 40.1995024845
• The cube root of 1616 is 11.7349286169
### Scales and comparisons
How big is 1616?
• 1,616 seconds is equal to 26 minutes, 56 seconds.
• To count from 1 to 1,616 would take you about twenty-six minutes.
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 1616 cubic inches would be around 1 feet tall.
### Recreational maths with 1616
• 1616 backwards is 6161
• The number of decimal digits it has is: 4
• The sum of 1616's digits is 14
• More coming soon!
MLA style:
"Number 1616 - Facts about the integer". Numbermatics.com. 2022. Web. 4 July 2022.
APA style:
Numbermatics. (2022). Number 1616 - Facts about the integer. Retrieved 4 July 2022, from https://numbermatics.com/n/1616/
Chicago style:
Numbermatics. 2022. "Number 1616 - Facts about the integer". https://numbermatics.com/n/1616/
The information we have on file for 1616 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!
Keywords: Divisors of 1616, math, Factors of 1616, curriculum, school, college, exams, university, Prime factorization of 1616, STEM, science, technology, engineering, physics, economics, calculator, one thousand, six hundred sixteen.
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# Digital Logic
## Description
Digital systems are based on two logic levels which are usually represented as 1's and 0's.We need different ways to combine different logical signals or conditions to provide a logical result,
LOGIC GATES
Overview on Digital Logic:
Digital (or Boolean) logic is that the description of binary values through printed circuit board (PCB) technology that uses circuits and digital logic gates to line up the implementation of basic computer operations. Digital logic is an ordinary part of electrical engineering (EE) and design/style courses. Furthermore, it is the support for digital computing and provides a basic understanding on how circuits and hardware communicate within a computer. Combinations of logic gates used in digital logic, form circuits which will perform specific tasks within larger circuits or systems. The process of manufacturing complex circuits using combinations of basic devices is named Combinational Logic.
Logic gate is an elementary building block of a digital circuit . Most logic gates have two inputs and one output. At any given moment, every terminal is in one of the two binary conditions represented by different voltage levels : low (0) or high (1).
e.g., consider the logical statement: "If I move the switch on the wall up, the light will turn on" At first glance, this seems to be a correct statement. However looking at a few other factors, it is realized that there'smore to it than this. In this example, a more complete statement would be: "I f I move the switch on the wall up and the light bulb is good and the power is on, the light will turn on."
If these two statements are logical expressions and use logical terminology, first statement can be reduced to:
Light = Switch
This means nothing more than that the light will follow the action of the switch, so that when the switch is up/on/true/1, the lightwill also be on/true/1. Conversely, if the switchis down/off/false/0, the lightwill also be off/false/0.
For second version of the statement, a slightly more complex expression is combine the separate variables of Switch, Bulb, and Power in this expression. The symbol normally used is a dot,which is the same symbol used for multiplication in some mathematical expressions.
Using this symbol, three-variable expression becomes.
Light = Switch • Bulb• Power
Light = Switch and Bulb and Power
Normally, symbols are used rather than words to designate the and function that are used to
Generally logic state of a terminal can, and does, change often, as the circuit processes data. In most logic gates, the low state is approximately zero volts (0 V), while high state is approximately five volts positive (+5V).
There are seven basic logic gates :
1. AND
2. OR
3. XOR
4. NOT
5. NOR
6.NAND
7.XNOR
Using combinations of logic gates, complex operations can be performed. In theory, there is no limit to the number of gates that can be arrayed together in a single device. But in practice, there is a limit to the number of gates that can be packed into a given physical space. Arrays of logic gates are found in digital integrated circuits (!Cs).As IC technology advances, the required physical volume for each individual logic gate decreases and digital devices of the same or smaller size become capable of performing ever-morecomplicated operations at ever-increasing speed.
UNIVERSAL GATES
Universal gates are the ones which can be used for implementing any gate like AND, OR and NOT or any combination of these basic gates;NAND and NOR gates are universal gates; but there are some rules that need to be followed when implementing NAND or NOR based gates.
BOOLEAN EXPRESSIONS
Digital circuits use ON-OFF devices to implement operations of a system of logic called two-valued using Boolean expression. The statements may take the form of algebraic expressions, logic block diagrams, or truth tables, aswell as circuits. Boolean expression is composed of variables and terms. The simplification of Boolean expressions can lead to more effective computer programs, algorithms and circuits.
FORMS OF BOOLEAN EXPRESSIONS.
(1) Sum of Product form (SOP)
e.g. W = ( X Y Z ) + ( X Y Z ) + ( X Y Z)
Each term in such an expression is called minterm.
(2) Product of Sum form (POS)
e.g. S = (P + Q + R) (P + Q + R) (P + Q + R)
Each term in such an expression is called maxterm.
Minterm and Maxterm
n variables can be combined to form 2" minterms or maxterms.
Each minterm is obtained from an AND term of n variables with each variable being primed, if corresponding bit of binary number is 0 and unprimed,bit is 1.
Each maxterm is obtained from an OR term of n variables with each variable being unprimed, if corresponding bit is a 0 and primed bit is 1.
Maxterm is complement of its corresponding minterm and vice versa.
MINIMlZATION OF BOOLEAN EXPRESSION.
(1) Convert equation from POS form to SOP form.
(2) Remove parenthesis if any in the expression.
(3) If there are two or more identical terms, keep only one of them and drop the other.
(4) Ifa variable and its complements are present in a term, reduce it to 0. (A. A= 0).
(5) Group two terms of which one contains a variable and other its complement, except for which both are identical. They can be reduced to a single term and in the reduced term, above variable will be absent
(C + c = 1)
e.g. (A . B . C ) + (A. B. C) = A. B ( C + C) = A . B
(6) If there are two terms which are identical except that one contains an extra variable, reduce them into a single term by dropping the larger one.
e.g. B.C + A . B . C = B . C (1 + A ) = B . C (1) = B
• Digital Logic Questions can be used by any candidate who is preparing for UGC NET Computer Science
• This Digital Logic MCQ section can help in testing your analytical skills thereby giving you an edge over other students
• Any student who wants to prepare for DOEACC A Level, DOEACC B Level, and DOEACC C level can also use these Objective Type Questions Answer.
• This section contains Combinational Circuits MCQ for the preparation of various competitive exams like the UGC NET Computer Science, JRF, CSIR, GATE Computer Science.
• All candidates who have to appear for the Kendriya Vidyalaya Entrance exam can also refer to this Digital Logic quiz questions with answers
• You can also get access to the Digital Logic MCQ ebook.
• Knowledge Representaion Questions can be used in the preparation of JRF, CSIR, and various other exams.
• This Digital Logic Multiple Choice Questions Answers section can also be used for the preparation of various competitive exams like UGC NET, GATE, PSU, IES, and many more.
• Knowledge Representaion Objective Type Questions Answer can be used to gain a credit score in various undergraduate and postgraduate courses like BSc, MSc and MCA
## Syllabus for Digital Logic Practice Questions:
Boolean algebra Rules, Combinational and sequential logic circuits, Definition of Minimization, Introduction of Number representations and computer arithmetic (fixed and floating point) A main component of digital logic consists of 5 different logic gates: AND OR XOR NAND NOR These basic logic gates are utilized in conjunction with each other to create elaborate engineering designs that deliver various computing outcomes. There are some ways that a number of logic gates can be combined to perform a selected task. They may all work, but some combinations will perform the task that better than others. For example, a circuit designer might want to design a combinational logic circuit that uses the minimum number of gates, or performs the specified task within the least time, or at the minimum cost.
## Significance of this section for Digital Electronics Practice Questions:
This section introduces practice questions on the fundamentals of digital logic, quiz on Digital Logic Quiz and the entire of digital electronics Practice questions depends on just seven forms of logic gates. You will be able to learn all Digital logic related problems by solving Digital logic Practice Test. You will get chance to evaluate and rate your performance level by participating in Digital logic Practice Test. Enhance your Digital Logic knowledge by practicing GATE MCQ for Digital Electronics Practice Questions. You can prepare Digital Logic Quiz for enhancing knowledge in Digital Logic. This Section contains updated MCQ on SQL Quiz on Digital Electronics Practice Questions for competitive exams. You will be able to know the correct answers of GATE CSE- Digital Logic Quiz along with brief explanations of the correct answers on one click. You can use these online MCQs on Digital logic Practice Test in improving knowledge for GATE CS exam.
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# Show that any simple set is the union of a finite number of mutually disjoint canonical intervals
Show that any simple set is the union of a finite number of mutually disjoint canonical intervals.
The set of canonical intervals: $$\mathcal{I}=\{[a,b) \quad | \quad a,b \in \mathbb{R} \quad \text{and} \quad a < b \}$$
Simple sets: A subset $$S$$ of $$\mathbb{R}$$ is said to be simple if it’s the union of a finite number of canonical intervals.
-proof-
Let $$s_1$$ be a simple set $$\implies s_1=\cup_{i=1}^{n} I_i$$ where each $$I_i=[a_i,b_i)$$
Now we do relabelling - we re-label the intervals according to their left-end points in this manner -
Let $$a_1$$ be the minimum taken overall $$A=\{a_i\}_{i=1}^{n}$$
define $$a_2 = a_1$$ if $$|A| else define $$a_2 = \min\{A-a_1\}$$ ..etc
Continuing in this manner we then construct $$a_1 \leq a_2 \leq a_3 \leq .. \leq a_n$$
First, $$a_1 \leq a_2$$
case1: if $$b_1> b_2 \implies I_1 \supset I_2$$ so re-write $$s_1 = [a_1,a_2) \cup [a_2,b_1)=[a_1,b_1)$$
case 2: if $$b_1=b_2 \implies s_1=[a_1,b_1) \cup [a_2=b_1, b_2)=[a_1,b_2)$$
case3: if $$b_2> b_1$$ then $$s_1=[a_1,a_2) \cup [a_2,b_1) \cup [b_1,b_2)=[a_1,b_2)$$
case4: if $$a_2 > b_1$$ then $$s_1 = [a_1,b_1) \cup [a_2,b_2)$$ where $$[a_1,b_1) \cap [a_2,b_2) = \emptyset$$
case5: if $$a_2=b_1$$ then $$s_1 = [a_1,b_1) \cup [a_2=b_1,b_2)=[a_1,b_2)$$
then we consider the case where $$a_2 \leq a_3$$ we continue this process $$(n-1)$$ times (where everytime we re-write $$s_1$$ in a manner where we we don't write an overlapped part more than once until the final case $$a_{n-1} \leq a_n$$ we will then see that we have constructed s_1\$ which is a finite union of pairwise disjoint canonical intervals
I am not sure if my attempt is correct but if there is an easier way, please share an idea with me
An approach like yours can be made to work, but there is a completely different approach that is easier, if perhaps less natural.
Define a relation $$\sim$$ on $$S$$ as follows: for any $$x,y\in S$$,
$$x\sim y\text{ iff }[\min\{x,y\},\max\{x,y\}]\subseteq S\;,$$
i.e., $$x\sim y$$ if and only if the closed interval between $$x$$ and $$y$$ is contained in $$S$$. It’s not hard to show that $$\sim$$ is an equivalence relation on $$S$$. Let $$\mathscr{C}$$ be the set of $$\sim$$-equivalence classes; the members of $$\mathscr{C}$$ are certainly pairwise disjoint subsets of $$S$$ whose union is $$S$$, so we’ll be done if we can show that they are canonical sets and that $$\mathscr{C}$$ is finite.
Let $$C\in\mathscr{C}$$. $$S$$ is a bounded set, so $$C$$ is bounded, and we can let $$a=\inf C$$ and $$b=\sup C$$; now we need only show that $$C=[a,b)$$. Certainly $$C\subseteq[a,b]$$; I’ll show next that $$(a,b)\subseteq C$$. Suppose that $$x\in(a,b)$$; then $$a, so there is a $$c_0\in C$$ such that $$a\le c_0. Similarly, $$x, so there is a $$c_1\in C$$ such that $$x. Then $$c_0\sim c_1$$, so $$[c_0,c_1]\subseteq S$$, and since $$c_0, it’s clear that $$[c_0,x]\subseteq S$$ as well. But then $$x\sim c_0\in C$$, so $$x\in C$$, and since $$x\in(a,b)$$ was arbitrary, it follows that $$(a,b)\subseteq C$$. Now we must show that $$a\in C$$ and $$b\notin C$$.
Suppose that $$b\in C$$. Then $$b\in S$$, so there is an $$i\in\{1,\ldots,n\}$$ such that $$b\in I_i=[a_i,b_i)$$. Then $$b, so choose any $$x\in(b,b_i)$$; $$x\in S$$, and $$[b,x]\subseteq I_i\subseteq S$$, so $$b\sim x$$, and therefore $$x\in C$$, contradicting the definition of $$b$$ as $$\sup C$$. This shows that $$b\notin C$$.
Now suppose that $$a\notin C$$; $$a=\inf C$$, so there is a strictly decreasing sequence $$\langle x_k:k\in\Bbb Z^+\rangle$$ in $$C$$ converging to $$a$$. Each of the points $$x_k$$ belongs to one of the sets $$I_i$$ with $$i\in\{1,\ldots,n\}$$, and there are only finitely many of these sets $$I_i$$, so there must be some $$i\in\{1,\ldots,n\}$$ such that $$A=\{k\in\Bbb Z^+:x_k\in I_i\}$$ is infinite. Now $$\{x_k:k\in A\}\subseteq[a_i,b_i)$$, so
$$a_i=\inf[a_i,b_i)\le\inf\{x_k:k\in A\}=a\;.$$
Pick any $$k\in A$$; then $$[a,x_k]\subseteq[a_i,x_k]\subseteq S$$, so $$a\sim x_k\in C$$, and therefore $$a\in C$$. This completes the proof that $$C=[a,b)$$ and hence that the partition of $$S$$ into $$\sim$$-equivalence classes is a partition of $$S$$ into canonical intervals.
Finally, it’s easy to check that for each $$C\in\mathscr{C}$$ there is an $$i(C)\in\{1,\ldots,n\}$$ such that $$I_i\subseteq C$$ and that the map $$\mathscr{C}\to\{1,\ldots,n\}:C\mapsto i(C)$$ is injective, so that $$|\mathscr{C}|\le n$$.
• Wow. Your proof looks genius. Thanks. I will learn it :) Apr 18, 2020 at 15:37
• @Dreamer123: You’re welcome. It’s a technique that is often useful when one is dealing with intervals. Apr 18, 2020 at 15:40
I think a variation on your technique might be a bit more rigorous and a bit easier.
The statement to prove is this: given a simple set $$S = \bigcup_{i=1}^{n} I_i$$, the set $$S$$ can be rewritten as a disjoint union of canonical intervals.
Here's a proof by induction on $$n$$.
The basis step is $$n=1$$, and so $$S=I_1$$ is a union of just one canonical interval.
Suppose that the statement is true for $$n-1$$, and so in particular the set $$S' = \bigcup_{i=1}^{n-1} I_i$$ can be written as a disjoint union of canonical intervals, $$S' = \bigcup_{j=1}^m J_j$$, for some $$m$$ and some canonical intervals $$J_j = [a_j,b_j)$$ such that $$(*) \,\, - \infty = b_0 \,\,<\,\, a_1 \,\,<\,\, b_1 \,\,\le\,\, a_2 \,\,<\,\, b_2 \,\,\le\,\, a_3 < \cdots < b_{m-1} \,\, \le \,\,a_m \,\,<\,\, b_m \,\,<\,\, a_{m+1} = +\infty$$ By convention I'm setting $$b_{0}=-\infty$$ and $$a_{m+1}=+\infty$$.
So $$S = S' \cup I_n$$ can be rewritten $$(\#) \qquad S = \bigcup_{j=1}^m J_j \cup \underbrace{I_n}_{=[a',b')}$$
We now look at where $$a'$$ and $$b'$$ fit into the sequence $$(*)$$, and using that information we identify exactly how to alter the terms of the union $$(\#)$$ in order to get a disjoint union such that the value of that union is still equal to $$S$$.
Let $$j \in \{0,1,...,m,m+1\}$$ be the minimal value such that $$b_{j-1} \le a'$$. It follows that $$a_j,a' \in [b_{j-1},b_j)$$.
Let $$k \in \{0,1,...,m,m+1\}$$ be the maximal value such that $$b' \le a_k$$. It follows that $$b_{k-1},b' \in (a_{k-1},a_k]$$.
Noting that $$b_{j-1} \le a' < b' \le a_k$$, it follows that $$j \le k$$.
Case 1: If $$j=k$$ then the union $$(\#)$$ is already disjoint and we are done.
Case 2: If $$j then, the union $$(\#)$$ becomes disjoint, without changing the value of the union, by removing the terms $$[a_j,b_j)$$, ... ,$$[a_{k-1},b_{k-1})$$ and inserting a new term $$\bigl[\min\{a_j,a'\},\max\{b_{k-1},b'\}\bigr)$$
• Wow. Thanks. Your proof is elegant :) Apr 18, 2020 at 15:35
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# average velocity
• Oct 6th 2008, 07:37 PM
jojoferni244
average velocity
how would you do this problem. i know the equation for av. velocity is
s2-s1/t2-t1. I keep getting -9.8
At a time t seconds after it is thrown up in the air, a tomato is at a height (in meters) of f(t)=−4https://webwork3.math.luc.edu/webwor...144/char3A.png9t^2+60t+2 m.
What is the average velocity of the tomato during the first 5 seconds?
• Oct 6th 2008, 07:45 PM
RNAS
Assuming what you typed is correct, the tomato went ker-splat!
I got 35.5 m/s. Take $\displaystyle \frac {f(5) - f(0)} {5 - 0}$.
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language agnostic - what is the best algorithm to use for this problem
Equilibrium index of a sequence is an index such that the sum of elements at lower indexes is equal to the sum of elements at higher indexes. For example, in a sequence A:
``A[0]=-7 A[1]=1 A[2]=5 A[3]=2 A[4]=-4 A[5]=3 A[6]=0``
3 is an equilibrium index, because:
``A[0]+A[1]+A[2]=A[4]+A[5]+A[6]``
6 is also an equilibrium index, because:
``A[0]+A[1]+A[2]+A[3]+A[4]+A[5]=0``
(sum of zero elements is zero) 7 is not an equilibrium index, because it is not a valid index of sequence A.
If you still have doubts, this is a precise definition: the integer k is an equilibrium index of a sequence if and only if and .
Assume the sum of zero elements is equal zero. Write a function
``int equi(int[] A);``
that given a sequence, returns its equilibrium index (any) or -1 if no equilibrium indexes exist. Assume that the sequence may be very long.
1. Calculate the total sum of all of the elements in `A`
2. For every index `i`, calculate the sum of the elements from `A[0]` to `A[i - 1]`, until the sum is equal to `(totalSum - A[i]) / 2`.
Note that the sum of elements from `A[0]` to `A[i - 1]` can be tracked as a running total, which means that the complexity of the whole algorithm is O(n). Implementing as code is left as an exercise for the reader.
Here's a solution that uses `O(n)` memory. Compute `S[i] = A[0] + A[1] + ... + A[i]`. Then the sum of a subsequence `[i, j]` is `Sum(i, j) = S[j] - S[i - 1]` (`S[x < 0] = 0`).
So for each `i` from `0` to `A.Length - 1` check if `Sum(0, i - 1) = Sum(i + 1, A.Length - 1)`.
In fact, if you're allowed to destroy the given array, you don't even need `S`, you can do it all in `A`.
Pseudocode - worst case is 2 passes through A.
``````R = sum(A)
L = e = 0
for i = 0 .. A.size
L+=e
R-=(e=A[i])
return i if L==R
end
return NULL
``````
a = (-7, 1, 5, 2, -4, 3, 0)
sumleft = 0
sumright = 0
for i in range(len(a)):
``````for j in range(i+1,len(a)):
sumright += a[j]
if sumright == sumleft:
print i
sumleft += a[i]
sumright = 0
``````
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# Sqrt(Cos(X))*cos(300x)+sqrt(Abs(X))-0.7)*(4-x*x)^0.01,sqrt(6-x^2),-sqrt(6-x^2)from -4.5 To 4.5
### Surviving the Storm: Solar Powered Generators for Dependable Home Energy
Sqrt(Cos(X))cos(300x)+sqrt(Abs(X))-0.7)(4-x*x)^0.01 is an equation that can be used to generate a graph of a function of x. This equation is a combination of several trigonometric functions and square root functions that can be used to generate a graph from -4.5 to 4.5. Examining the graph generated by this equation can give us a better understanding of the behavior of the equation.
## What is Sqrt(Cos(X))cos(300x)+sqrt(Abs(X))-0.7)(4-x*x)^0.01?
Sqrt(Cos(X))cos(300x)+sqrt(Abs(X))-0.7)(4-x*x)^0.01 is an equation that combines several trigonometric functions and square root functions. The equation includes a cosine function, a square root function, and an absolute value function. The equation also includes a power function and a constant. These components of the equation are combined to generate a graph of a function of x.
## Examining the Graph from -4.5 to 4.5
The graph generated by this equation is a combination of several trigonometric functions and square root functions. The graph starts at -4.5 and increases steadily until it reaches a peak at 0. It then decreases steadily until it reaches a second peak at 4.5. The graph then decreases steadily until it reaches the starting point at -4.5. The graph has several smaller peaks and valleys throughout its range, which can be used to better understand the behavior of the equation.
The equation also produces a second graph, which is a combination of a square root function and a constant. This graph starts at -4.5 and increases steadily until it reaches a peak at 0. It then decreases steadily until it reaches a second peak at 4.5. The graph then decreases steadily until it reaches the starting point at -4.5. The graph has several smaller peaks and valleys throughout its range, which can be used to better understand the behavior of the equation.
Finally, the equation produces a third graph, which is a combination of a square root function and a constant. This graph starts at -4.5 and decreases steadily until it reaches a peak at 0. It then increases steadily until it reaches a second peak at 4.5. The graph then decreases
In mathematics, an equation is a statement of an equality involving two expressions. The expression on the left side is known as the equation’s left-hand side and the expression on the right side is known as the equation’s right-hand side. An equation can be written in a variety of ways, including the “sqrt(cos(x))*cos(300x)+sqrt(abs(x))-0.7)*(4-x*x)^0.01”, “sqrt(6-x^2)”, and “-sqrt(6-x^2)” forms, as is seen in the equation,
sqrt(cos(x))*cos(300x)+sqrt(abs(x))-0.7)*(4-x*x)^0.01,sqrt(6-x^2),-sqrt(6-x^2), from -4.5 to 4.5.
This equation describes a mathematical relationship between x and the three expressions, sqrt(cos(x))*cos(300x)+sqrt(Abs(x))-0.7)*(4-x*x)^0.01, sqrt(6-x^2) , and -sqrt(6-x^2). The equation is a representation of a continuous function, which covers the range from -4.5 to 4.5.
The first expression, sqrt(cos(x))*cos(300x)+sqrt(Abs(x))-0.7)*(4-x*x)^0.01, is a polynomial equation in which the independent variable x is multiplied by a certain number of terms and then added to some constant value. It is a function that is continuous over the range from -4.5 to 4.5.
The second expression, sqrt(6-x^2), is a square root equation, involving the square root of 6 minus x squared. This expression is also continuous over the range from -4.5 to 4.5.
The third expression, -sqrt(6-x^2), is the same as the second expression but with a negative sign, representing the negative value of the function at certain values of the independent variable x. This expression is also continuous over the range from -4.5 to 4.5.
This equation can be used to represent real-world scenarios involving the calculation of certain factors, such as the cost of a product or the volume of a material. This equation can also be used to calculate the value of certain phenomena, such as the motion of a wave or the height of a jump.
Overall, this equation represents a relationship between the independent variable x and the three expressions given. It is a continuous function with a range of -4.5 to 4.5. This equation can be used to calculate certain values or factors in a variety of real-world scenarios, making it a very useful mathematical tool.
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# Search by Topic
#### Resources tagged with Sine, cosine, tangent similar to Ladder and Cube:
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### There are 36 results
Broad Topics > Pythagoras and Trigonometry > Sine, cosine, tangent
### Squ-areas
##### Age 14 to 16 Challenge Level:
Three squares are drawn on the sides of a triangle ABC. Their areas are respectively 18 000, 20 000 and 26 000 square centimetres. If the outer vertices of the squares are joined, three more. . . .
### Inscribed in a Circle
##### Age 14 to 16 Challenge Level:
The area of a square inscribed in a circle with a unit radius is, satisfyingly, 2. What is the area of a regular hexagon inscribed in a circle with a unit radius?
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Straight lines are drawn from each corner of a square to the mid points of the opposite sides. Express the area of the octagon that is formed at the centre as a fraction of the area of the square.
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Describe how to construct three circles which have areas in the ratio 1:2:3.
### Lying and Cheating
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Follow the instructions and you can take a rectangle, cut it into 4 pieces, discard two small triangles, put together the remaining two pieces and end up with a rectangle the same size. Try it!
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Six circular discs are packed in different-shaped boxes so that the discs touch their neighbours and the sides of the box. Can you put the boxes in order according to the areas of their bases?
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The sides of a triangle are 25, 39 and 40 units of length. Find the diameter of the circumscribed circle.
### Where Is the Dot?
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A dot starts at the point (1,0) and turns anticlockwise. Can you estimate the height of the dot after it has turned through 45 degrees? Can you calculate its height?
### Dodecawhat
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Follow instructions to fold sheets of A4 paper into pentagons and assemble them to form a dodecahedron. Calculate the error in the angle of the not perfectly regular pentagons you make.
### At a Glance
##### Age 14 to 16 Challenge Level:
The area of a regular pentagon looks about twice as a big as the pentangle star drawn within it. Is it?
### Raising the Roof
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How far should the roof overhang to shade windows from the mid-day sun?
##### Age 14 to 16 Challenge Level:
If you were to set the X weight to 2 what do you think the angle might be?
##### Age 11 to 18
Logo helps us to understand gradients of lines and why Muggles Magic is not magic but mathematics. See the problem Muggles magic.
### Circle Box
##### Age 14 to 16 Challenge Level:
It is obvious that we can fit four circles of diameter 1 unit in a square of side 2 without overlapping. What is the smallest square into which we can fit 3 circles of diameter 1 unit?
### History of Trigonometry - Part 2
##### Age 11 to 18
The second of three articles on the History of Trigonometry.
### History of Trigonometry - Part 3
##### Age 11 to 18
The third of three articles on the History of Trigonometry.
##### Age 14 to 16 Challenge Level:
In this problem we are faced with an apparently easy area problem, but it has gone horribly wrong! What happened?
### Round and Round a Circle
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Can you explain what is happening and account for the values being displayed?
### Trigonometric Protractor
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An environment that simulates a protractor carrying a right- angled triangle of unit hypotenuse.
### A Scale for the Solar System
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The Earth is further from the Sun than Venus, but how much further? Twice as far? Ten times?
### 8 Methods for Three by One
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This problem in geometry has been solved in no less than EIGHT ways by a pair of students. How would you solve it? How many of their solutions can you follow? How are they the same or different?. . . .
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The coke machine in college takes 50 pence pieces. It also takes a certain foreign coin of traditional design...
### Far Horizon
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An observer is on top of a lighthouse. How far from the foot of the lighthouse is the horizon that the observer can see?
### Farhan's Poor Square
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From the measurements and the clue given find the area of the square that is not covered by the triangle and the circle.
### Moving Squares
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How can you represent the curvature of a cylinder on a flat piece of paper?
### Sine and Cosine
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The sine of an angle is equal to the cosine of its complement. Can you explain why and does this rule extend beyond angles of 90 degrees?
### Screen Shot
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A moveable screen slides along a mirrored corridor towards a centrally placed light source. A ray of light from that source is directed towards a wall of the corridor, which it strikes at 45 degrees. . . .
### Orbiting Billiard Balls
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What angle is needed for a ball to do a circuit of the billiard table and then pass through its original position?
### Figure of Eight
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On a nine-point pegboard a band is stretched over 4 pegs in a "figure of 8" arrangement. How many different "figure of 8" arrangements can be made ?
### Sine and Cosine for Connected Angles
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The length AM can be calculated using trigonometry in two different ways. Create this pair of equivalent calculations for different peg boards, notice a general result, and account for it.
### The History of Trigonometry- Part 1
##### Age 11 to 18
The first of three articles on the History of Trigonometry. This takes us from the Egyptians to early work on trigonometry in China.
### Round and Round
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Prove that the shaded area of the semicircle is equal to the area of the inner circle.
### Cosines Rule
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Three points A, B and C lie in this order on a line, and P is any point in the plane. Use the Cosine Rule to prove the following statement.
### Making Maths: Clinometer
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Make a clinometer and use it to help you estimate the heights of tall objects.
### Eight Ratios
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Two perpendicular lines lie across each other and the end points are joined to form a quadrilateral. Eight ratios are defined, three are given but five need to be found.
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# Question Video: Divisibility of Numbers Mathematics
Which of these are divisible by both 2 and 3? [A] 6, 54, 36 [B] 15, 9, 12 [C] 4, 6, 48 [D] 5, 54, 9 [E] 5, 8, 36.
01:33
### Video Transcript
Which of these are divisible by both two and three? Option A) six, 54, and 36; option B) 15, nine, and 12; option C) four, six, and 48; option D) five, 54, and nine; and option E) five, eight, and 36.
As only even numbers are divisible by two, we can immediately rule out options B, D, and E as they contain odd numbers. Option B contains the odd numbers 15 and nine, option D contains the odd numbers five and nine, and option E contains the odd number five. None of these odd numbers are divisible by two.
Which of the remaining sets of numbers — set A or set C — are all divisible by three? Six, 54, 36, and 48 are all divisible by three as they’re in the three times table. However, four is not divisible by three. Therefore, set C cannot be correct.
This leaves the correct answer set A. The numbers six, 54, and 36 are divisible by both two and three.
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Math 126 Week 5 Discussion
# Math 126 Week 5 Discussion
398.3k points
Week 5 Discussion
1. Select a topic of interest to you and record the topic in your posting, for example: What is the average number of hours people watch TV every week? Make sure the question you ask will be answered with a number, rather than answers with words.
2. Write a hypothesis of what you expect your research to reveal. Example: Adults 21 years and over watch an average of 2.5 hours of TV per day.
3. Sample at least fifteen people and record their data in a simple table or chart; study the examples from Section 12-3.
4. You can gather your data at work, on the phone, or via some other method. This is your Sampling Design. Which of the four sampling techniques best describes your design?
5. Explain in moderate detail the method you used to gather your data. In statistics this venture is called the Methodology.
6. Make sure you break your sample into classes or groups, such as males/females, or ages, or time of day, etc.
7. Calculate the mean, median, and mode for your data as a whole.
8. Now calculate the mean, median, and mode of each of your classes or groups.
9. Indicate which measure of central tendency best describes your data and why. Then compare your results for each class or group, and point out any interesting results or unusual outcomes between the classes or groups. This is called a comparative analysis using our results to explain interesting outcomes or differences (i.e., between men and women).
Comment on at least two of your classmates postings. Make sure you comment on their hypothesis (topic), their design, and whether you agree or do not agree with their best measure of central tendency
Math 126
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#### Oh Snap! This Answer is Locked
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Surround your text in *italics* or **bold**, to write a math equation use, for example, $x^2+2x+1=0$ or $$\beta^2-1=0$$
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## Sunday, July 10, 2011
### Blogging my way through CLRS section 3.1 [part 5]
Part 4 here.
I wrote an entire blog post explaining the answers to 2.3 but Blogger decided to eat it. I don't want to redo those answers so here is 3.1:
For now on I will title my posts with the section number as well to help Google.
Question 3.1-1: Let $f(n)$ and $g(n)$be asymptotically non-negative functions. Using the basic definition of $\theta$-notation, prove that $\max(f(n) , g(n)) \in \theta(f(n) + g(n))$ .
CLRS defines $\theta$ as $\theta(g(n))= \{ f(n) :$ there exists some positive constants $c_1, c_2$, and $n_0,$ such that $0 \leq c_1g(n) \leq f(n) \leq c_2g(n)$ for all $n \geq n_0\}$ Essentially we must prove that there exists some $c_1$ and $c_2$ such that $c_1 \times (f(n) + g(n)) \leq \max(f(n), g(n)) \leq c_2 \times (f(n) + g(n))$ There are a variety of ways to do this but I will choose the easiest way I could think of. Based on the above equation we know that $\max(f(n), g(n)) \leq f(n) + g(n)$ (as f(n) and g(n) must both me non-negative) and we further know that $\max(f(n), g(n))$ can't be more than twice f(n)+g(n). What we have then are the following inequalities: $$\max(f(n), g(n)) \leq c_1 \times (f(n) + g(n))$$ and $$c_2 \times (f(n) + g(n)) \leq 2 \times \max(f(n), g(n))$$ Solving for $c_1$ we get 1 and for $c_2$ we get $\frac {1} {2}$
Question 3.1-2: Show for any real constants $a$ and $b$ where $b \gt 0$ that $(n+a)^b \in \theta(n^b)$
Because $a$ is a constant and the definition of $\theta$ is true after some $n_0$ adding $a$ to $n$ does not affect the definition and we simplify to $n^b \in \theta(n^b)$ which is trivially true
Question 3.1-3: Explain why the statement "The running time of $A$ is at least $O(n^2)$," is meaningless.
I'm a little uncertain of this answer but I think this is what CLRS is getting at when we say a function $f(n)$ has a running time of $O(g(n))$ what we really mean is that $f(n)$ has an asymptotic upper bound of $g(n)$. This means that $f(n) \leq g(n)$ after some $n_0$. To say a function has a running time of at least g(n) seems to be saying that $f(n) \leq g(n) \And f(n) \geq g(n)$ which is a contradiction.
Question 3.1-4: Is $2^{n+1} = O(2^n)$? Is $2^{2n} = O(2^n)$?
$2^{n+1} = 2 \times 2^n$. which means that $2^{n+1} \leq c_1 \times 2^n$ after $n_0$ so we have our answer that $2^{n+1} \in o(2^n)$ Alternatively we could say that the two functions only differ by a constant coefficient and therefore the answer is yes.
There is no constant such that $2^{2n} = c \times 2^n$ and thefore $2^{2n} \notin O(2^n)$
Question 3.1-5: Prove that for any two functions $f(n)$ and $g(n)$, we have $f(n) \in \theta(g(n)) \iff f(n) \in O(g(n)) \And f(n) \in \Omega(g(n))$
This is an "if an only if" problem so we must prove this in two parts:
Firstly, if $f(n) \in O(g(n))$ then there exists some $c_1$ and $n_0$ such that $f(n) \leq c_1 \times g(n)$ after some $n_0$. Further if $f(n) \in Omega(g(n))$ then there exists some $c_2$ and $n_0$ such that $f(n) \geq c_2 \times g(n)$ after some $n_0$.
If we combine the above two statements (which come from the definitions of $\Omega$ and O) than we know that there exists some $c_1, c_2, and n_0,$ such that $c_1g(n) \leq f(n) \leq c_2g(n)$ for all $n \geq n_0\}$
We could do the same thing backward for the other direction: If $f(n) \in \theta(g(n))$ then we could split the above inequality and show that each of the individual statements are true.
Question 3.1-6: Prove that the running time of an algorithm is $\theta(g(n)) \iff$ its worst-case running time is $O(g(n))$ and its best case running time $\Omega(g(n))$.
I'm going to try for an intuitive proof here instead of a mathematical one. If the worst case is asymptotically bound above in the worst case by a certain function and is asymptotically bound from below in the best case which means that the function is tightly bound by both those functions. f(n) never goes below some constant times g(n) and never goes above some constant times g(n). This is what we get from the above definition of $\theta(g(n)))$ A mathematical follows from question 3.1-5.
Question 3.1-7: Prove that $o(g(n)) \cap \omega(g(n)) = \varnothing$
little o and little omega are defined as follows: $o(g(n)) = \{ f(n) : \forall c > 0 \exists n_0 \text{such that } 0 \leq f(n) \leq c \times g(n) \forall n \gt n_0$ and $\omega(g(n)) = \{ f(n) : \forall c > 0 \exists n_0 \text{such that } 0 \leq c \times g(n) \leq f(n) \forall n \gt n_0$
In other words
$$f(n) \in o(g(n)) \iff \lim_{n \to \infty} \frac {f(n)} {g(n)} = 0$$ and $$f(n) \in \omega(g(n)) \iff \lim_{n \to \infty} \frac {f(n)} {g(n)} = \infty$$
It is obvious that these can not be true at the same time. This would require that $0 = \infty$
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# The angle A of the triangle ABC is equal to 76; the outer angle at the vertex B is equal to 94
The angle A of the triangle ABC is equal to 76; the outer angle at the vertex B is equal to 94 degrees. Find the corner C.
Let’s solve a geometric problem. But first, let’s make a short condition:
Angle A – 76 °;
Outside angle at apex В – 94 °;
Angle C -?
So, remember that the outer and inner angles at one vertex in a triangle always add up to 180 °. Thus, we calculate the degree measure of the angle B:
180 – 94 = 86 °;
Now we find the sum of angle A and angle B:
86 + 76 = 162 °;
Since the sum of all angles in a triangle is always 180 °, we can easily find the degree measure of the last angle in a triangle:
180 – 162 = 18 ° – angle С;
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# Introduction to Backtracking
Backtracking is a powerful algorithmic technique that allows us to efficiently explore all possible solutions to a problem. It is particularly useful when dealing with optimization or combinatorial problems. In this tutorial, we will delve deep into the concept of backtracking algorithms, understanding their working principles, and exploring some common backtracking problems.
## What is Backtracking?
At its core, backtracking is a systematic way of searching for solutions to a problem by incrementally building candidates and backtracking when a candidate is found to be invalid. It utilizes a depth-first search approach, exploring a branch of the solution space and undoing the choices that lead to dead ends.
## Basic Structure of a Backtracking Algorithm
Let's take a look at the basic structure of a backtracking algorithm:
1. Define the problem: First, we need to precisely define the problem and understand its constraints.
2. Identify the decision variables: Determine the variables that define the state of the problem and the solution space.
3. Define the constraints: Establish the conditions that must be satisfied for a solution to be valid.
4. Define the objective function: If applicable, specify an objective function that needs to be optimized.
5. Implement the backtracking algorithm: Develop a recursive function that systematically explores the solution space.
6. Apply pruning techniques: Introduce techniques like bounding and pruning to optimize the algorithm's performance.
7. Test and validate the algorithm: Evaluate the algorithm on various test cases to ensure its correctness and efficiency.
## Common Backtracking Problems
Now that we understand the basics of backtracking, let's explore some common backtracking problems and their solutions.
### Problem 1: N-Queens
The N-Queens problem is a classical backtracking problem that involves placing N queens on an N x N chessboard in such a way that no two queens threaten each other. Here's how we can solve it using backtracking:
``````def solve_n_queens(n, queens, result):
if len(queens) == n:
result.append(queens.copy())
return
for i in range(n):
if is_valid_move(queens, i):
queens.append(i)
solve_n_queens(n, queens, result)
queens.pop()
def is_valid_move(queens, col):
row = len(queens)
for i in range(row):
if queens[i] == col or abs(queens[i] - col) == abs(i - row):
return False
return True
``````
### Problem 2: Subset Sum
Given a set of positive integers and a target sum, the Subset Sum problem requires finding all possible subsets that sum up to the target. Here's a backtracking solution:
``````def subset_sum(numbers, target_sum, subset, result, start_index):
if sum(subset) == target_sum:
result.append(subset.copy())
return
for i in range(start_index, len(numbers)):
subset.append(numbers[i])
subset_sum(numbers, target_sum, subset, result, i + 1)
subset.pop()
``````
These are just two examples of the wide variety of problems that can be solved using backtracking algorithms. By applying the principles we've discussed and tailoring our approach to the specific requirements of the problem, we can tackle numerous optimization and combinatorial problems.
## Conclusion
Backtracking algorithms offer an efficient and systematic way of exploring all possible solutions to a problem. By understanding their underlying principles and applying them to various problems, programmers can devise elegant and optimized solutions. In this tutorial, we covered the basics of backtracking and explored two common backtracking problems. Armed with this knowledge, you are now equipped to tackle more complex challenges using this powerful algorithmic technique.
Now that the blog post is complete, you can convert it to HTML format using a markdown-to-HTML converter tool.
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12
Q:
# A train covers 180 km distance in 5 hours. Another train covers the same distance in 1 hour less. What is the difference in the distances covered by these trains in one hour if they are moving in the same direction?
A) 15 kms B) 9 kms C) 6 kms D) 18 kms
Explanation:
The first train covers 180 kms in 5 hrs
=> Speed = 180/5 = 36 kmph
Now the second train covers the same distance in 1 hour less than the first train => 4 hrs
=> Speed of the second train = 180/4 = 45 kmph
Now, required difference in distance in 1 hour = 45 - 36 = 9 kms.
Q:
The ratio between the speeds of two trains is 2 : 5. If the first train runs 350 km in 5 h, then the difference between the speeds (in km/h) of both the trains is:
A) 350 B) 105 C) 165 D) 180
Explanation:
2 267
Q:
Given that the lengths of the paths of a ball thrown with different speeds by two boy sare the same, if they take 0.6 seconds and 1 second respectively to cover the said length, what is the average speed of travel for the first throw, if the same for the second is 96 km/h?
A) 160 km/h B) 200 km/h C) 150 km/h D) 100 km/h
Explanation:
1 312
Q:
A man divided his journey into three parts of distances of 18 km, 20 km and 27 km. He travelled the distances at the speeds of 6 km/h, 5 km/h and 9 kn/h,respectively. What was his average speed during the entire journey?
A) 6.5 km/h B) 4.5 km/h C) 7.5 km/h D) 5.5 km/h
Explanation:
2 579
Q:
The distance between two railway stations is 1176 km. To cover this distance, an express train takes 5 hours less than a passenger train while the average speed of the passenger train is 70 km/h less than that of the express train. The time taken by the passenger train to complete the travel is:
A) 12 hours B) 23 hours C) 17 hours D) 18 hours
Explanation:
2 656
Q:
The distance between the places H and O is D units. The average speed that gets a person from H to O in a stipulated time is S units. He takes 20 minutes more time than usual if he travels at 60 km/h, and reaches 44 minutes early if he travels at 75 km/h. The sum of the numerical values of D and S is:
A) 376 B) 358 C) 344 D) 384
Explanation:
0 413
Q:
A train that is running at the speed of 72 km/h crossesan electric pole in 36 seconds. The length of the train (in metres) is:
A) 720 m B) 460 m C) 360 m D) 620 m
Explanation:
1 455
Q:
An athlete crosses a distance of 900 m in 10 minutes. What is his speed in km per hour?
A) 5.4 km/h B) 6.9 km/h C) 3.6 km/h D) 4.8 km/h
Explanation:
2 3484
Q:
Rakesh walking at the speed of 8 km/h crosses a bridge in 30 minutes. What is the length of the bridge in kilometres?
A) 5 km B) 2 km C) 3 km D) 4 km
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# Algebra
Translate the sentence into an equation.
The sum of a number times
6
and
8
is
7
.
Use the variable
b
for the unknown number.
1. 👍 0
2. 👎 1
3. 👁 1,476
1. 6b + 8 = 7
1. 👍 0
2. 👎 1
👩🏫
Ms. Sue
2. The sum of a number times 5 and 16 is at most -15
1. 👍 0
2. 👎 1
3. ;kln
1. 👍 0
2. 👎 1
4. The sum of 6 times a number and 4 equals 8
1. 👍 0
2. 👎 0
5. Translate the sentence into an equation.
The sum of 5 times a number and 3 is8.
Use the variable for the unknown number.
1. 👍 0
2. 👎 0
## Similar Questions
1. ### Math
1. I am thinking of a number. It is increased by 7. The sum is 21. What is my number? A. n=28*** B. n=14 C. n=7 D. n= 147 2. I am thinking of a number. It is increased by 9. The product is 27. What is my number? A. n = 36 B. n =
2. ### algebra
Translate the sentence into an inequality. the sum of a number times 3 snd 27 is at most -19. Use the variable c for the unknown number.
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4. ### algebra
Translate this sentence into an equation. 41 is the sum of 16 and Malik's age . use variable m represent Malik's age
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translate sentence to inequality The sum of a number times 9 and 30 is at least 17
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Translate the sentence into an inequality. The sum of a number times 10 and 27 is at least −20..... How do I write this????
3. ### match
the sum of two times a number and 5 is 11? equation form.
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Translate the sentence into an inequality. Ten times the sum of a number and 27 is at least 26.
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Write an inequality and solve. Two times the sum of a number and four is no more than three times the sum of the number and seven decreased by four.
2. ### Math
Translate the phrase into an expression: 8 times the sum of 7 and y
3. ### math
how do i translate: the sum of a number times 5 and 29 is at most 23 into an inequality sentence number
4. ### Math
Five times the sum of a number and 17 is at most 25 translate ?
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Topic: Integral
Replies: 5 Last Post: Jul 22, 2011 3:00 PM
Messages: [ Previous | Next ]
Axel Vogt Posts: 1,068 Registered: 5/5/07
Re: Integral
Posted: Jul 22, 2011 10:01 AM
On 21.07.2011 02:42, [email protected] wrote:
> Hello all,
>
> I'm trying to evaluate the following integral
>
> S(a,b) = integrate(0,2pi) sin(a*cos(t))*sin(b*sin(t)) /
> (sin(t)*cos(t)) dt
>
> I get S(a,b) = 16pi*a*b*J(2,sqrt(a^2+b^2))/(a^2+b^2)
>
> where J(k,z) is the bessel function of the first kind, order k.
> However funny numerics started occuring and i think i have traced it
> back to this integral. It would appear (in mathematica and maple)
> that numerically this appears reasonable for a,b<1. but for a,b>1 the
> numerical integration and my result diverge quite significantly...too
> much to perhaps be numerical error in the integration.
>
> My steps are as follows:
>
> cos(a*z+b/z) = -J(0,2sqrt(ab)) + sum(k=0,oo) (-1)^k ((a*z)^(2k) +
> (b/z)^(2k))*J(2k,2sqrt(ab))/(sqrt(ab))^(2k)
>
> I derived this by using binomial theorem in the series for cos - also
> numerically it appears bang on. Using this, convering the integral to
> that over the unit circle, yeilds my result.
>
> Alternatively one may use similar laurent series for sin(z+1/z),
> multiply them together, get the residue and get the same result.
>
> what have i missed?
For a=1/2/2, b=1/2 I get 0.765149580251623 for the integral and
0.765143786388934 for your formula, using Maple with 15 Digits.
So it seems that it is not correct for a,b<1 as well.
Date Subject Author
7/20/11 Rancid Moth
7/21/11 Dann Corbit
7/22/11 Rancid Moth
7/22/11 Axel Vogt
7/22/11 Sam2718
7/22/11 Axel Vogt
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Question
# A rational number equivalent to $\frac{5}{7}$is ..................
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Solution
## Rational numbers equivalent to $\frac{5}{7}$Equivalent rational numbers are rational numbers written in the form of different numerators and denominators but have the same value.We can get an equivalent rational number by multiplying the numerator and denominator by the same number.We have $\frac{5}{7}$Multiply the numerator and denominator by the same number that is$2$, we get$\frac{5}{7}=\frac{5×2}{7×2}=\frac{10}{14}$Thus, $\frac{5}{7}=\frac{10}{14}$Hence, A rational number equivalent to $\frac{5}{7}$is $\frac{10}{14}$
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Victorian Curriculum Year 10A - 2020 Edition
5.09 Circles
Lesson
Graphs of equations of the form $\left(x-h\right)^2+\left(y-k\right)^2=r^2$(xh)2+(yk)2=r2 (where $h$h, $k$k, and $r$r are any number and $r\ne0$r0) are called circles.
The circle defined by $x^2+y^2=1$x2+y2=1. It has a centre at $\left(0,0\right)$(0,0) and a radius of $1$1 unit
### Transformations of circles
A circle can be vertically translated by increasing or decreasing the $y$y-values by a constant number. However, the $y$y-value together with the translation must be squared together. So to translate $x^2+y^2=1$x2+y2=1 up by $k$k units gives us $x^2+\left(y-k\right)^2=1$x2+(yk)2=1.
Vertically translating up by $2$2 ($x^2+\left(y-2\right)^2=1$x2+(y2)2=1) and down by $2$2 ($x^2+\left(y+2\right)^2=1$x2+(y+2)2=1)
Similarly, a circle can be horizontally translated by increasing or decreasing the $x$x-values by a constant number. However, the $x$x-value together with the translation must be squared together. That is, to translate $x^2+y^2=1$x2+y2=1 to the left by $h$h units we get $\left(x+h\right)^2+y^2=1$(x+h)2+y2=1.
Horizontally translating left by $2$2 ($\left(x+2\right)^2+y^2=1$(x+2)2+y2=1) and right by $2$2 ($\left(x-2\right)^2+y^2=1$(x2)2+y2=1)
Notice that the centre of the circle $x^2+y^2=1$x2+y2=1 is at $\left(0,0\right)$(0,0). Translating the circle will also translate the centre by the same amount. So the centre of $\left(x-h\right)^2+\left(y-k\right)^2=r^2$(xh)2+(yk)2=r2 is at $\left(h,k\right)$(h,k).
A circle can be scaled both vertically and horizontally by changing the value of $r$r. In fact, $r$r is the radius of the circle
Expanding by a scale factor of $2$2 ($x^2+y^2=4$x2+y2=4) and compressing by a scale factor of $2$2 ($x^2+y^2=\frac{1}{4}$x2+y2=14)
Summary
The graph of an equation of the form $\left(x-h\right)^2+\left(y-k\right)^2=r^2$(xh)2+(yk)2=r2 is a circle.
Circles have a centre at $\left(h,k\right)$(h,k) and a radius of $r$r
Circles can be transformed in the following ways (starting with the circle defined by $x^2+y^2=1$x2+y2=1):
• Vertically translated by $k$k units: $x^2+\left(y-k\right)^2=1$x2+(yk)2=1
• Horizontally translated by $h$h units: $\left(x-h\right)^2+y^2=1$(xh)2+y2=1
• Scaled (both vertically and horizontally) by a scale factor of $r$r: $x^2+y^2=r^2$x2+y2=r2
#### Practice questions
##### Question 1
Consider the circle with equation $\left(x-0.4\right)^2+\left(y+3.8\right)^2=2$(x0.4)2+(y+3.8)2=2.
1. What is the centre of the circle?
Give your answer in the form $\left(a,b\right)$(a,b).
2. What is the radius of the circle?
Give an exact answer.
##### Question 2
A circle has its centre at $\left(3,-2\right)$(3,2) and a radius of $4$4 units.
1. Plot the graph for the given circle.
2. Write the equation of the circle in general form.
##### Question 3
Consider the equation of a circle given by $x^2+4x+y^2+6y-3=0$x2+4x+y2+6y3=0.
1. Rewrite the equation of the circle in standard form.
2. What are the coordinates of the centre of this circle?
Give your answer in the form $\left(a,b\right)$(a,b).
3. What is the radius of this circle?
### Outcomes
#### VCMNA339
Explore the connection between algebraic and graphical representations of relations such as simple quadratic, reciprocal, circle and exponential, using digital technology as appropriate
#### VCMNA359 (10a)
Describe, interpret and sketch parabolas, hyperbolas, circles and exponential functions and their transformations.
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# Statistics and Probability Data Analysis and Recommendations
We are searching for a talented Statistics and Probability nerd. We have data that we need help getting our heads around. My statistics understanding is quite a bit rusty. I will need help evaluating the data in order to create logical responses to variable events.
We would like to find which contacts have the highest winnings (with std dev probability and margins, etc.) with all (or some) of the variables (Day of week(a), Time(b), State(c), High Score(D), Position(e), Health(f)). Additionally, we want to know which contacts have the highest probability of being assigned with the same set of variables (yes, contact (x) has the highest winnings, but contact (y) is more likely to be assigned with variables (a+c+D+f...or some other combination).
Help us create conditional logic formulas for the best contacts with variable values. Ex. State is TX. It is Tuesday. It is 8:45. The Health is FAIR. The Position is 1000. The High Score is 5500. Which contact should we choose? Which contact has the highest probability of being assigned? Which contact, when assigned, has the highest winnings? What is the standard deviation of the winnings with this condition?
Not Required, but If you have excel skills and can build calculations into excel, that will be top consideration.
WE ARE LOOKING FOR FREELANCERS THAT HAVE STRONG KNOWLEDGE OF STATISTICS FORMULAS AND WILL SHOW THEIR CONDITIONAL LOGIC AND METHODS IN ANALYSIS.
Please provide a simple explanation of how you will analyze this data:
Simple example (I would like something complete). I will find Standard Deviation of each data set independently, then I will utilize this formula (xyz) in order to find which contact is most probable to be assigned contact with variables (a and b).
I want to know that you understand statistics, not just excel. Standard Deviation. Covariance. Etc.
Om arbejdsgiveren:
( 16 bedømmelser ) Boulder, United States
Projekt ID: #14503669
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# math
posted by .
estimate each sum or difference by rounding to the greatest place value.
797 - 234
• math -
I did two of these problems for you. Now it's your turn.
• math -
I have no idea
## Similar Questions
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4th grade math homework. rounding sums and differences. paper say "round each number to its highest place. then estimate the sum or difference." example says 92 +28= then says round to 90 and 30 to make 120. further down numbers get …
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Estimate by rounding each term to the greatest place: 78.17-37.52 Need to show my work
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4. ### MATH
IS ABOUT ESTIMATE THE SUM ROUNDING TO THE THOUSANDS PLACE
Estimate each sum or difference by rounding to the greatest place value. 10,982+4,821= should the answer be... 11,000+5,000=16,000 or 10,000+5,000=15,000
Estimate each sum or difference by rounding to the greatst place value. 38,347+17,039 would it be 40,000+20,000=60,000
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364 plus 519 round each number the greatest place value to estimate the sum of:
9. ### math
esimate each sum or difference by rounding to the place value indicated. 37,097 - 20,364 ten thousand
10. ### Math
Estimate the sum or difference by rounding each number to the greatest place. 5,613 _. 1,392 876 +. 48
More Similar Questions
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# Thread: Trigonometry-maximum & minimum values
1. ## Trigonometry-maximum & minimum values
For acosA +(-) bsinA +(-) c
a= rcosA & b= rsinA ( a not = A^o)
Then a^2+b^2 = r^2-cos^2-A + r^2-sin^2-A = r^2 that= r is sqrt(a^2+b^2)..................................... .......(I could not understand this method. could you explain me about it)
2. I may not understand your question.
However, to start, r is the hypotenuse of a triangle formed by the axes and an angle.
$a = r cos(A)$ is true because cosine is the adjacent side (a) over the hypotenuse (r), so this is just solving for a. This is similar with solving for b as well, but it is the side opposite the angle, so we use sine instead of cosine.
Pythagorean theorem says that for a right triangle $a^2 + b^2 = c^2$, so when we substitute the above this gives us $r^2 cos^2(A) + r^2 sin^2(A)$.
If you elaborate more on your question, that would help me.
Patrick
3. ## Trigonometry maximum & minimum vales
Is there any use of circle with coordinates in the solution and my problem of not understanding is at starting of it after reading the entire proof.
4. I'm sorry, I don't understand what you are asking.
5. 'Then a^2+b^2 = r^2cos^2-A + r^2sin^2-A = r^2", a pythagoren theorem?
like this?
where x = a and y = b?
6. ## Trignometry maximum &minimum vales
"Extreme values of acosA +(-) bcosA +(-) c(anot=A^o) is
[c - sqrt(a^2+b^2) , c + sqrt(a^2+b^2)]" (I think now it is better) (I edited the qestion could you again read it)
7. same, i can't understand . . . . my problem i think is the same as yours, we both don't learn latexing.
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## Calculus (3rd Edition)
Substituting $n = 4$ in the equation for $a_{n}$ gives: $a_{4} = 4^{2}-4 = 12$
For this problem, simply plug in n = 4 into the equation given, $a_n = n^2 -n$
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# Joules To Watt Seconds Conversion
J
Want to convert from watt seconds to joules instead?
Disclaimer: We've spent hundreds of hours building and testing our calculators and conversion tools. However, we cannot be held liable for any damages or losses (monetary or otherwise) arising out of or in connection with their use. Full disclaimer.
## How to convert joules to watt seconds (J to Ws)
The formula for converting joules to watt seconds is: Ws = J × 1. To calculate the joule value in watt seconds first substitute the joule value into the preceding formula, and then perform the calculation. If we wanted to calculate 1 joule in watt seconds we follow these steps:
Ws = J × 1
Ws = 1 × 1
Ws = 1
In other words, 1 joule is equal to 1 watt second.
## Example Conversion
Let's take a look at an example. The step-by-step process to convert 9 joules to watt seconds is:
1. Understand the conversion formula: Ws = J × 1
2. Substitute the required value. In this case we substitute 9 for J so the formula becomes: Ws = 9 × 1
3. Calculate the result using the provided values. In our example the result is: 9 × 1 = 9 Ws
In summary, 9 joules is equal to 9 watt seconds.
## Converting watt seconds to joules
In order to convert the other way around i.e. watt seconds to joules, you would use the following formula: J = Ws × 1. To convert watt seconds to joules first substitute the watt second value into the above formula, and then execute the calculation. If we wanted to calculate 1 watt second in joules we follow these steps:
J = Ws × 1
J = 1 × 1
J = 1
Or in other words, 1 watt second is equal to 1 joule.
## Conversion Unit Definitions
### What is a Joule?
A joule (J) is the standard unit of energy in the International System of Units (SI). It is defined as the amount of work done when a force of one newton is applied over a distance of one meter in the direction of the force. In other words, one joule is equal to the energy transferred when one watt of power is dissipated for one second.
Mathematically, 1 joule is equal to 1 N × 1 m, or 1 kg × m2/s2 in terms of mass, length, and time.
Joules are a versatile unit of measurement used not only in physics but also in various fields including engineering, chemistry, and biology to quantify energy, work, and heat. For example, it's commonly used to express the energy content of foods, the capacity of batteries, and the energy output of electronic devices.
### What is a Watt Second?
A watt-second (Ws), also known as a joule (J), is a unit of energy in the International System of Units (SI). It represents one watt of power expended for one second. Mathematically, it can be defined as: 1 watt-second (Ws) = 1 watt (W) × 1second (s) = 1 joule (J).
Watt-seconds are used to measure the energy consumed or produced by electronic devices or electrical systems in a very short amount of time. For example, the energy released by a camera flash, the energy stored in capacitors, or the energy consumed by a small electronic circuit during its operation can be measured in watt-seconds.
In practical terms, a watt-second is a relatively small unit of energy. To provide some context, 1 watt-second is equivalent to the energy required to power a 1-watt device for 1 second.
## Joules To Watt Seconds Conversion Table
Below is a lookup table showing common joules to watt seconds conversion values.
Joules (J)Watt Second (Ws)
1 J1 Ws
2 J2 Ws
3 J3 Ws
4 J4 Ws
5 J5 Ws
6 J6 Ws
7 J7 Ws
8 J8 Ws
9 J9 Ws
10 J10 Ws
11 J11 Ws
12 J12 Ws
13 J13 Ws
## Other Common JouleConversions
Below is a table of common conversions from joules to other energy units.
ConversionResult
1 joule in ergs10000000 erg
1 joule in kilojoules0.001 kJ
1 joule in megajoules0.000001 MJ
1 joule in gigajoules0.000000001 GJ
1 joule in calories0.2388458966275 cal
1 joule in kilocalories0.0002388458966275 kcal
1 joule in watt hours0.00027777777777778 Wh
1 joule in kilowatt hours0.00000027777777777778 kWh
1 joule in megawatt hours0.00000000027777777777778 MWh
1 joule in btus0.00094708628903179 Btu
1 joule in electron volts6241509074461000000 eV
1 joule in kilo electron volts6241509074461000 keV
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# Area And Perimeter Of Irregular Shapes Pdf
By Clodoaldo V.
In and pdf
16.01.2021 at 10:52
File Name: area and perimeter of irregular shapes .zip
Size: 2463Kb
Published: 16.01.2021
Grade 6 geometry worksheet area and perimeter of irregular rectangles.
The app can even sum multiple area calculations together by way of drawing layers. To introduce the lesson, the teacher revises the previous lesson. Area of Irregular Polygons How to find the area of irregular polygons by following 4 simple steps: 1. Guides pupils to draw two rectangles of dimensions 7 cm by 3 cm and 1o cm by 15cm.
## Perimeter of Irregular Shapes
The worksheets are very varied, and include:. Each worksheet is randomly generated and thus unique. The answer key is automatically generated and is placed on the second page of the file.
You can generate the worksheets either in html or PDF format — both are easy to print. To get the worksheet in html format, push the button " View in browser " or " Make html worksheet ". Sometimes the generated worksheet is not exactly what you want. Just try again! To get a different worksheet using the same options:. Created by IXL, a trusted leader in math education, this workbook covers the must-know skills for third graders, from multiplication and fractions to geometry and measurement.
Your child will love the vivid colors, engaging problems, and fun graphics, while you'll feel confident knowing our carefully crafted material is setting them up for success in the classroom and beyond. Write an expression for the area of a two-part rectangle in two ways, thinking of one rectangle or two grades This is a challenging topic for 3rd grade in the Common Core Standards 3. It is essentially an illustration of the distributive property.
If you make an html worksheet, make sure the background colors are set to be printed in your browser's options. Draw a two-part rectangle to match the given number sentence for its total area grades This worksheet may or may not fit the page when it's randomly generated. For the PDF worksheet, you need to just remake the worksheet until it fits nicely. Choose at least one. Level 1: Fill in the expression. Level 2: Write the expression.
Level 3: Draw the rectangle. Level 4: Any of the above. Basic instructions for the worksheets Each worksheet is randomly generated and thus unique. To get a different worksheet using the same options: PDF format: come back to this page and push the button again. Html format: simply refresh the worksheet page in your browser window. A variety of problems about area and perimeter grades There are six different problem types.
Type Write an expression for the area of a two-part rectangle from image distributive property Columns needs set to 1 for the layout to work properly here.
## perimeter of irregular shapes
Comparing Numbers. Daily Math Review. Division Basic. Division Long Division. Hundreds Charts.
The worksheets are very varied, and include:. Each worksheet is randomly generated and thus unique. The answer key is automatically generated and is placed on the second page of the file. You can generate the worksheets either in html or PDF format — both are easy to print. To get the worksheet in html format, push the button " View in browser " or " Make html worksheet ". Sometimes the generated worksheet is not exactly what you want.
How to find area and perimeter of irregular shapes Page 2. Created by Danielle Miller, Hawk Ridge. Math Facilitator. Area.
## Related Post
Some irregular figures are made of rectangular or square regions. The areas of such irregular figures can be determined by calculating the areas of these rectangles and squares. To find the area of a figure which is a combination of rectangles and a squares, we calculate the area of each figure separately and then add them to find total area.
Geometry worksheet on finding the area and perimeter of irregular rectangular shapes. The students are given the These worksheets are pdf files. Have your students practice finding the areas of squares and rectangles with these printables. Find the area of these irregular shapes rectilinear figures by multiplying decimals to the tenths and hundredths place. Click to see how my students found the area of irregular shapes using the distributive property of multiplication.
A regular shape is a shape in which every side is the same length. If a shape is not regular then it is irregular.
### Areas of Irregular Shapes (Rectilinear Figures)
Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. Are you getting the free resources, updates, and special offers we send out every week in our teacher newsletter? Grade Level.
В полном недоумении Сьюзан посмотрела в окно кабинета на видневшийся внизу ТРАНСТЕКСТ. Она точно знала, что на такой пароль уходит меньше десяти минут. - Должно ведь быть какое-то объяснение. - Оно есть, - кивнул Стратмор. - Тебя оно не обрадует. - В ТРАНСТЕКСТЕ сбой.
Да, панк, - сказала Росио на плохом английском и тотчас снова перешла на испанский. - Mucha joyeria. Вся в украшениях. В одном ухе странная серьга, кажется, в виде черепа. - В Севилье есть панки и рокеры. Росио улыбнулась: - Todo bajo el sol. Чего только нет под солнцем.
Area and perimeter of irregular shapes. Grade 6 Geometry Worksheet. Find the perimeter and area. 1. 7. 9. 3. 9. 5. 2.
#### area of irregular shapes worksheet pdf
Сьюзан замерла. Мгновение спустя, как в одном из самых страшных детских кошмаров, перед ней возникло чье-то лицо. Зеленоватое, оно было похоже на призрак. Это было лицо демона, черты которого деформировали черные тени. Сьюзан отпрянула и попыталась бежать, но призрак схватил ее за руку.
Сьюзан казалось, что она сходит с ума. Она уже готова была выскочить из комнаты, когда Стратмор наконец повернул рубильник и вырубил электропитание. В одно мгновение в шифровалке установилась полная тишина. Сирены захлебнулись, мониторы Третьего узла погасли. Тело Грега Хейла растворилось в темноте, и Сьюзан, инстинктивно поджав ноги, прикрылась пиджаком Стратмора.
Sara M.
Calculate the area and perimeter of the following shapes. 4 cm. 4 m. 2 cm 2 m. 2 km. 6 cm. 4 m. 1 km. 3 km. 3 km. 3 cm. 2 m. 2 km. Area= ______ cm².
Sebastian R.
Acres of skin human experiments at holmesburg prison pdf free list of asian countries and their capitals pdf
Percy J.
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# Quiz Discussion
What is the probability of getting at least one six in a single throw of three unbiased dice?s
Course Name: Quantitative Aptitude
• 1]
91/216
• 2]
1/216
• 3]
200/216
• 4]
17/216
##### Solution
No Solution Present Yet
#### Top 5 Similar Quiz - Based On AI&ML
Quiz Recommendation System API Link - https://fresherbell-quiz-api.herokuapp.com/fresherbell_quiz_api
# Quiz
1
Discuss
A speaks truth in 75% cases and B in 80% of the cases. In what percentage of the cases are they likely to contradict each other, narrating the same incident?
• 1]
5%
• 2]
15%
• 3]
35%
• 4]
45%
##### Solution
2
Discuss
The students in a class are seated, according to their marks in the previous examination. Once, it so happens that four of the students got equal marks and therefore the same rank. To decide their seating arrangement, the teacher wants to write down all possible arrangements one in each of separate bits of paper in order to choose one of these by lots. How many bits of paper are required?
• 1]
24
• 2]
12
• 3]
36
• 4]
48
##### Solution
3
Discuss
One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is not a face card (Jack, Queen and King only)?
• 1]
5/13
• 2]
10/13
• 3]
1/13
• 4]
1/26
##### Solution
4
Discuss
In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?
• 1]
1/10
• 2]
2/5
• 3]
2/7
• 4]
5/7
##### Solution
5
Discuss
If four dice are thrown together, then what is the probability that the sum of the numbers appearing on them is 25 ?
• 1]
0
• 2]
1/2
• 3]
1
• 4]
1/1296
##### Solution
6
Discuss
How many natural numbers can be made with digits 0, 7, 8 which are greater than 0 and less than a million?
• 1]
496
• 2]
728
• 3]
486
• 4]
1084
##### Solution
7
Discuss
How many 5 digit even numbers with distinct digits can be formed using the digits 1, 2, 5, 5, 4?
• 1]
16
• 2]
24
• 3]
36
• 4]
48
##### Solution
8
Discuss
A teacher has to choose the maximum different groups of three students from a total of six students. Of these groups, in how many groups there will be included in a particular student?
• 1]
12
• 2]
10
• 3]
8
• 4]
6
##### Solution
9
Discuss
In how many different ways can the letters of the word 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions?
• 1]
32
• 2]
48
• 3]
36
• 4]
60
##### Solution
10
Discuss
A number lock on a suitcase has 3 wheels each labeled with 10 digits from 0 to 9. If the opening of the lock is a particular sequence of three digits with no repeats, how many such sequences will be possible?
• 1]
720
• 2]
760
• 3]
780
• 4]
680
# Quiz
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# Sal’s Shipping with Functions
First post.
Saw the part about coming back “after having been introduced to functions”
Assumed that functions were somehow required.
Finished project realizing no functions were needed at all.
Decided to try incorporating functions into the solution.
This is the result I came up with:
``````# Set shipping weight
weight = 41.5
# Ground Shipping cost calculator
def calc_GS_cost(weight):
if weight <= 2:
GS_cost = weight * 1.5 + 20
elif weight <= 6:
GS_cost = weight * 3 + 20
elif weight <= 10:
GS_cost = weight * 4 + 20
else:
GS_cost = weight * 4.75 + 20
return GS_cost
GSP_cost = 125
# Drone Shipping cost calculator
def calc_DS_cost(weight):
if weight <= 2:
DS_cost = weight * 4.5
elif weight <= 6:
DS_cost = weight * 9
elif weight <= 10:
DS_cost = weight * 12
else:
DS_cost = weight * 14.25
return DS_cost
# Lowest shipping cost function
def lowest_ship_cost(weight):
GS_cost = calc_GS_cost(weight)
DS_cost = calc_DS_cost(weight)
if GS_cost < GSP_cost and GS_cost < DS_cost:
print("Ground Shipping is cheapest at", GS_cost, "dollars for", weight, "lbs.")
elif GSP_cost < GS_cost and GSP_cost < DS_cost:
print("Ground Shipping Premium is cheapest at", GSP_cost, "dollars for", weight, "lbs.")
elif DS_cost < GSP_cost and DS_cost < GS_cost:
print("Drone Shipping is cheapest at", DS_cost, "dollars for", weight, "lbs.")
else:
print("Undetermined")
lowest_ship_cost(weight)
``````
Haven’t managed to find a proper solution to handle instances where different shipping methods costs are the same.
What do you mean by
Haven’t managed to find a proper solution to handle instances where different shipping methods costs are the same.
Also note the DRY principle (don’t repeat yourself) in factoring can be applied here.
If you know something is going to repeat, you can anticipate it, example:
``````def calc_GS_cost(weight):
if weight <= 2:
GS_cost += weight * 1.5
elif weight <= 6:
GS_cost += weight * 3
elif weight <= 10:
GS_cost += weight * 4
else:
GS_cost += weight * 4.75
return GS_cost
``````
But even better
``````def calc_GS_cost(weight):
if weight <= 2:
cost_multiplier = 1.5
elif weight <= 6:
cost_multiplier = 3
elif weight <= 10:
cost_multiplier = 4
else:
cost_multiplier = 4.75
return weight * cost_multiplier + flat_fee
``````
This way you save yourself 6 operations from your original (the multiplications and additions). In terms of this program it doesn’t matter, especially as addition and multiplication are rather fast for CPUs to process, but this factoring pattern is useful when the type of operation you do might not be as efficient (particularly when scaled to a larger size).
3 Likes
Thanks for your advice. Wasn’t expecting such a thoughtful response so quickly.
I really admire your solution. It’s simple and intuitive; the way of coding that I’m trying to improve on.
The project was designed so that it’s very hard to input a weight that would result in more than 1 shipping option with the same shipping cost. (E.g. when weight is 3 and 1/3 lbs) I was just overthinking how to handle such an occurrence.
But, If I had to add extra code to handle this occurrence, I would just add extra if statements to the last function like so:
``````# Lowest shipping cost function
def lowest_ship_cost(weight):
GS_cost = calc_GS_cost(weight)
DS_cost = calc_DS_cost(weight)
if GS_cost == GSP_cost and GS_cost < DS_cost:
print("Both Ground Shipping and Ground Shipping Premium is cheapest at", GS_cost, "dollars for", weight, "lbs.")
elif GS_cost == DS_cost and GS_cost < GSP_cost:
print("Both Ground Shipping and Drone Shipping is cheapest at", GS_cost, "dollars for", weight, "lbs.")
elif DS_cost == GSP_cost and DS_cost < GS_cost:
print("Both Ground Shipping Premium and Drone Shipping is cheapest at", DS_cost, "dollars for", weight, "lbs.")
elif GS_cost < GSP_cost and GS_cost < DS_cost:
print("Ground Shipping is cheapest at", GS_cost, "dollars for", weight, "lbs.")
elif GSP_cost < GS_cost and GSP_cost < DS_cost:
print("Ground Shipping Premium is cheapest at", GSP_cost, "dollars for", weight, "lbs.")
elif DS_cost < GSP_cost and DS_cost < GS_cost:
print("Drone Shipping is cheapest at", DS_cost, "dollars for", weight, "lbs.")
else:
print("Undetermined")
``````
But this looks quite clunky and the else statement should probably be removed as it serves no purpose anymore.
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|
The sum of first m terms of an A.P. is 4m2 − m.
Question:
The sum of first $m$ terms of an A.P. is $4 m^{2}-m$. If its nth term is 107 . find the value of $n$. Also, find the 21 st term of this A.P.
Solution:
$S_{m}=4 m^{2}-m$
We know
$a_{m}=S_{m}-S_{m-1}$
$\therefore a_{m}=4 m^{2}-m-4(m-1)^{2}+(m-1)$
$a_{m}=8 m-5$
Now,
$a_{n}=107$
$\Rightarrow 8 n-5=107$
$\Rightarrow 8 n=112$
$\Rightarrow n=14$
$a_{21}=8(21)-5=163$
|
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# 222 - Topic 05 - Limits Symbolically
## 2390 days ago by Professor222
var('x,y') def f(x,y): if x+y<>0: return sin(3*(x^2+y^2+5*pi))/(x^2+y^2) else: return 3
x0=0 y0=0 # Something was wrong with the f defn above and eventually sage would "forget" the defn. Gee whiz. # f(x,y) = sin(3*(x^2+y^2+5*pi))/(x^2+y^2) f(x,y) = x*y^2/(x^2+y^4)
f(pi,3)
9*pi/(pi^2 + 81) 9*pi/(pi^2 + 81)
var('t') x1=x0+t y1=y0 x2=x0 y2=y0+t x3=x0+t*cos(10*t) y3=y0+t*sin(10*t) x4=x0+t^2 y4=y0-t^5 x5=x0-t y5=y0-t/2 # If the limit exists,it must agree with the limit along any of given curves. L = limit(f(x=x5,y=y5),t=0) L
0 0
0 0
G = plot3d(f,(x,-1,1),(y,-1,1)) G += parametric_plot3d([x1,y1,f(x=x1,y=y1)],(t,0.01,1),color='red',thickness=10) G += parametric_plot3d([x2,y2,f(x=x2,y=y2)],(t,0.01,1),color='yellow',thickness=10) G += parametric_plot3d([x3,y3,f(x=x3,y=y3)],(t,0.01,1),color='orange',thickness=10) G += parametric_plot3d([x4,y4,f(x=x4,y=y4)],(t,0.01,1),color='green',thickness=10) G += parametric_plot3d([x5,y5,f(x=x5,y=y5)],(t,0.01,1),color='purple',thickness=10) G += point3d([x0,y0,L],color='white',size=25) G.show()
# The above stuff combined into an interactive cell var('t,x,y') @interact def _(x0=(0),y0=(0),f = (x*y^2/(x^2+y^4)),in3d=checkbox(default=false)): x1=x0+t y1=y0 x2=x0 y2=y0+t x3=x0+t*cos(10*t) y3=y0+t*sin(10*t) x4=x0+t^2 y4=y0-t^5 x5=x0-t y5=y0-t/2 L = limit(f(x=x5,y=y5),t=0) G = plot3d(f,(x,-1,1),(y,-1,1),color='lightblue') G += parametric_plot3d([x1,y1,f(x=x1,y=y1)],(t,0.01,1),color='red',thickness=10) G += parametric_plot3d([x2,y2,f(x=x2,y=y2)],(t,0.01,1),color='yellow',thickness=10) G += parametric_plot3d([x3,y3,f(x=x3,y=y3)],(t,0.01,1),color='orange',thickness=10) G += parametric_plot3d([x4,y4,f(x=x4,y=y4)],(t,0.01,1),color='green',thickness=10) G += parametric_plot3d([x5,y5,f(x=x5,y=y5)],(t,0.01,1),color='purple',thickness=10) G += point3d([x0,y0,L],color='white',size=25) if in3d: G.show(stereo="redcyan") else: G.show()
x0 y0 f in3d
## Click to the left again to hide and once more to show the dynamic interactive window
var('x,y') f = x*y/(x^2+y^2) limit(limit(f,y=0),x=0)
0 0
f(x,y) = sin(x*y)/(x+y)^2
var('z') limit(f(x=z,y=z),z=0)
1/4 1/4
limit(limit(f,y=0),x=0)
(x, y) |--> 0 (x, y) |--> 0
limit(f(x=t,y=2*t),t=0)
2/9 2/9
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1. ## math league
Of the integers between 10^3 and 10^4 that have no repeated digit, how many have digits that increase from left to right?
2. ## Re: math league
Originally Posted by victorwen28
Of the integers between 10^3 and 10^4 that have no repeated digit, how many have digits that increase from left to right?
Any subset of four nonzero digits fits that description.
So $\displaystyle \binom{9}{4}=~?$
3. ## Re: math league
126
{1234, 1235, 1236, 1237, 1238, 1239, 1245, 1246, 1247, 1248, 1249, 1256, 1257, 1258, 1259, 1267, 1268, 1269, 1278, 1279, 1289, 1345, 1346, 1347, 1348, 1349, 1356, 1357, 1358, 1359, 1367, 1368, 1369, 1378, 1379, 1389, 1456, 1457, 1458, 1459, 1467, 1468, 1469, 1478, 1479, 1489, 1567, 1568, 1569, 1578, 1579, 1589, 1678, 1679, 1689, 1789, 2345, 2346, 2347, 2348, 2349, 2356, 2357, 2358, 2359, 2367, 2368, 2369, 2378, 2379, 2389, 2456, 2457, 2458, 2459, 2467, 2468, 2469, 2478, 2479, 2489, 2567, 2568, 2569, 2578, 2579, 2589, 2678, 2679, 2689, 2789, 3456, 3457, 3458, 3459, 3467, 3468, 3469, 3478, 3479, 3489, 3567, 3568, 3569, 3578, 3579, 3589, 3678, 3679, 3689, 3789, 4567, 4568, 4569, 4578, 4579, 4589, 4678, 4679, 4689, 4789, 5678, 5679, 5689, 5789, 6789}
4. ## Re: math league
But there is a condition that the digits that have to increase from left to right.
5. ## Re: math league
Is there a way to do it without listing all the possibilities?
6. ## Re: math league
Originally Posted by MaxJasper
126
{1234, 1235, 1236, 1237, 1238, 1239, 1245, 1246, 1247, 1248, 1249, 1256, 1257, 1258, 1259, 1267, 1268, 1269, 1278, 1279, 1289, 1345, 1346, 1347, 1348, 1349, 1356, 1357, 1358, 1359, 1367, 1368, 1369, 1378, 1379, 1389, 1456, 1457, 1458, 1459, 1467, 1468, 1469, 1478, 1479, 1489, 1567, 1568, 1569, 1578, 1579, 1589, 1678, 1679, 1689, 1789, 2345, 2346, 2347, 2348, 2349, 2356, 2357, 2358, 2359, 2367, 2368, 2369, 2378, 2379, 2389, 2456, 2457, 2458, 2459, 2467, 2468, 2469, 2478, 2479, 2489, 2567, 2568, 2569, 2578, 2579, 2589, 2678, 2679, 2689, 2789, 3456, 3457, 3458, 3459, 3467, 3468, 3469, 3478, 3479, 3489, 3567, 3568, 3569, 3578, 3579, 3589, 3678, 3679, 3689, 3789, 4567, 4568, 4569, 4578, 4579, 4589, 4678, 4679, 4689, 4789, 5678, 5679, 5689, 5789, 6789}
Originally Posted by victorwen28
But there is a condition that the digits that have to increase from left to right.
Example: Consider the set $\displaystyle \{2,4,1,6\}$, how many ways can you arrange that set into a four-digit number of increasing digits, (in the literature that is known as a strictly sorted integer).
So there is an one-to-one correspondent between the strictly sorted four digit integers and the four elements subsets of $\displaystyle \{1,2,3,4,5,6,7,8,9\}$
7. ## Re: math league
So do you have to count in the end?
8. ## Re: math league
Originally Posted by victorwen28
So do you have to count in the end?
If you simply want the answer the please GO AWAY. You are a parasite.
If you have read questions, post them.
But if you are simply after an answer, log off.
9. ## Re: math league
I want to understand how to do it, but I am not clear of your approach.
10. ## Re: math league
Originally Posted by victorwen28
I want to understand how to do it, but I am not clear of your approach.
O.K.
If I give your the set $\displaystyle \{7,3,6,1\}$ how many way can you arrange those four into a strictly sorted integer?
And what is it?
11. ## Re: math league
(1,3,6,7)
only one way.
12. ## Re: math league
Originally Posted by victorwen28
(1,3,6,7)
only one way.
Correct!
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# How Many Zeros In A Billion
How Many Zeros In A Billion. On the short scale, it’s one followed by nine zeroes. One million has six zeros (1,000,000), while one billion has nine zeros (1,000,000,000).
A billion hours ago our ancestors were living in the stone age. On the short scale, it’s one followed by nine zeroes. It is also known as a thousand millions as 1,000 × 1000,000 = 1000,000,000.
### One Million Million, Or 10 12 (Ten To The Twelfth Power), As Defined On The Long Scale.
For numbers with lots of zeroes, it can be hard to get a real sense of scale. There are total 6 zeros in a million. It depends on where you are.we in canada and the u.s.
### A Billion Hours Ago Our Ancestors Were Living In The Stone Age.
These terms were first used by genevieve guitel, the french mathematician. Say it has 9 zeros.but it is 1,000 million in europe and a billion has 12 zeros and a trillion has 15 zeros. If you apply for that big number of money, you can see why billionaires just love making more cash.
### It Means That It Has Nine Zeroes.
You must have noticed that the zeros are increasing in a pattern. In the united states—as well as around the world in science and finance—a billion is 1,000 million, which is written as a one followed by nine zeros. In the united states—as well as around the world in science and finance—a billion is 1,000 million, which is written as a one followed by nine zeros.
### So, To Multiply 100 By One Billion You Just Need To Add 9 Zeros To The Right Of 100.
One billion typically has 9 zeros and is written as 1000000000. Every time three more zeros are added to the number. The numbers in between them have different names.
### One Trillion = 1012 = 1,000,000,000,000.
There are 9 zeros in a billion when we use the short scale, 1,000,000,000 (used by english speaking countries). Experts use this large value when they’re measuring something that exceeds millions because a billion is equal to one thousand million. 1 million is equal to 10 lakhs.
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Edit Article
# How to Calculate the Circumference of a Circle
Whether you're doing craft work, putting fencing around your hot tub, or just solving a math problem for school, knowing how to find the circumference of a circle will come in handy in a variety of circle-related problems.
### Method 1 Using the Diameter
1. 1
Write down the formula for finding the circumference of a circle using the diameter. The formula is simply this: C = πd. In this equation, "C" represents the circumference of the circle, and "d" represents its diameter. That is to say, you can find the circumference of a circle just by multiplying the diameter by pi. Plugging π into your calculator will give you its numerical value, which is a closer approximation of 3.14.
2. 2
Plug the given diameter into the equation and solve. Let's say you're solving this problem: you have a hot tub with a diameter of 8 feet, and you want to build a white fence that creates a 6-foot wide space around the tub. To find the circumference of the fence that has to be created, you should first find the diameter of the tub and the fence which will be 8 feet + 6 feet + 6 feet, which will account for the entire diameter of the tub and fence. The diameter is 8 + 6 + 6, or 20 feet. Now plug it into the formula, plug π into your calculator for its numerical value, and solve for the circumference:
• C = πd
• C = π x 20
• C = 62.8 feet
### Method 2 Using the Radius
1. 1
Write down the formula for finding the circumference of a circle using the radius. The radius is half as long as the diameter, so the diameter can be thought of as 2r. Keeping this in mind, you can write down the formula for finding the circumference of a circle given the radius: C = 2πr. In this formula, "r" represents the radius of the circle. Again, you can plug π into your calculator to get its numeral value, which is a closer approximation of 3.14.
2. 2
Plug the given radius into the equation and solve. For this example, let's say you're cutting out a decorative strip of paper to wrap around the edge of a pie you've just made. The radius of the pie is 5 inches (12.7 cm). To find the circumference that you need, just plug the radius into the equation:
• C = 2πr
• C = 2π x 5
• C = 10π
• C = 31.4 inches (79.8 cm).
## Community Q&A
Search
• What is a circumference?
wikiHow Contributor
Circumference is the distance around the perimeter of a circle. It is calculated by multiplying the distance across the center (diameter) by Pi (3.1416).
• What is the formula used to calculate diameter?
wikiHow Contributor
Diameter is just simply 2 times the radius. The circumference is diameter x pie OR 2 x radius x pie.
• What is the answer of the circumference of a circle?
wikiHow Contributor
The answer depends on your number. For example, pie or 3.14 is multiplied with your number which is the diameter or radius.
• If I know the area of a circle in meters, how can I find the circumference?
wikiHow Contributor
Take the area, divide by pi, take the square root of that answer. Now multiply that by 2pi and you have the circumference. Equation: Area (a) = πr^2, and circumference (c) = 2πr. So c = sqrt(a/π)*2π.
• How do I find the circumference when I only have the area?
wikiHow Contributor
Because the equation to find area is pi * r^2, the first step is to divide by pi or 3.14. Then you have to find the square root of the remaining number. The answer should be your radius. Now just fill in that number for r in the circumference equation to get the final answer.
• How do I find the area?
wikiHow Contributor
It's Pi times the radius, squared. Example: Radius = 2. Area? 2 * 3.14 = 6.28. 6.28 * 6.28 = 39.44. The area of a circle with the radius of 2 is 39.44 units.
• How do I change the mixed number into radius?
wikiHow Contributor
If your radius is a mixed number, turn the number into an improper fraction. To do this, simply multiply the whole number part by the denominator and add that number to the numerator. The denominator should remain the same throughout the process. You can then use the improper fraction in your formula.
• How can I find the diameter of the circle, once I know the circumference?
wikiHow Contributor
Circumference (C) = pi * diameter(d). To get the diameter of a circle take the circumference (C) and divide it by pi (3.14).
• How do I find the radius when I get the circumference?
wikiHow Contributor
The circumference = π x the diameter of the circle (Pi multiplied by the diameter of the circle). Simply divide the circumference by π and you will have the length of the diameter. The diameter is just the radius times two, so divide the diameter by two and you will have the radius of the circle!
• How do you find the radius of a semicircle when you have the perimeter?
wikiHow Contributor
This article tells us the perimeter (also called "circumference" when talking about a full circle) of a full circle is given by 2*pi*r. Since we need the perimeter of a semicircle instead of a full circle, divide this by 2 (since the arc length is half) but then add the diameter, which is two times the radius (to account for the straight part across the bottom). So, perimeter of semicircle = (2*pi*r)/2 + 2r = (2 + pi)r. Use algebra to solve for r = perimeter / (2 + pi), and you have your radius!
• How can the formulas for a circle be simplified?
• How to find the radius or diameter if I already have my circumference?
• What is the answer of the circle length of formula?
If this question (or a similar one) is answered twice in this section, please click here to let us know.
## Tips
• Consider buying an advanced calculator that already has π as one of the buttons. This will mean less typing for you and a more accurate answer because the π button produces an approximation to π that is much more accurate than 3.14.
• Remember: some worksheets will ask to replace pi with a subside, such as 3.14 or 22/7.
• To find circumference from diameter, just multiply pi by the diameter.
## Warnings
• If you are stuck, ask a friend, family member, or teacher. They will always help!
• Remember to always double-check your work because one mistake will set off all your data.
• Take your time. Remember the old adage—measure twice, cut once.
## Article Info
Featured Article
Categories: Featured Articles | Geometry
In other languages:
Français: calculer la circonférence d’un cercle, Italiano: Calcolare la Circonferenza di un Cerchio, Español: calcular la circunferencia, Deutsch: Einen Kreisumfang berechnen, Português: Calcular a Circunferência de um Círculo, Nederlands: De omtrek van een cirkel berekenen, Русский: вычислить длину окружности круга, 中文: 计算圆的周长, Bahasa Indonesia: Menghitung Keliling Lingkaran, Čeština: Jak vypočítat obvod kruhu, 日本語: 円の円周を計算する, ไทย: คำนวณเส้นรอบวงของวงกลม, हिन्दी: गोलाकार चीजों की परिधी ज्ञात करें, العربية: حساب محيط دائرة, Tiếng Việt: Tính Chu vi Hình tròn
Thanks to all authors for creating a page that has been read 6,334,632 times.
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# How Do You Calculate Discount
Filter Type:
### How do I calculate the discount points on a mortgage loan ...
(4 days ago) Once you have these two pieces of the puzzle, you can calculate the length of time it will take to “break even” on your discount point strategy. So let’s talk about the (A) and (B) values mentioned above: (A) — One point equals 1% of the loan amount. For instance, paying one discount point on a \$300,000 mortgage loan means you’ll pay ...
Category: Coupon,
### Step by Step Math Lesson on How To Calculate Discount …
(5 days ago) The rate of discount is usually given as a percent, but may also be given as a fraction. The phrases used for discounted items include, " off," "Save 50%," and "Get a 20% discount." Procedure: To calculate the discount, multiply the rate by the original price. To calculate the sale price, subtract the discount …
https://www.mathgoodies.com/lessons/percent/sale_price
Category: Coupon,
### Discount Calculator - Calculator Soup - Online Calculators
(3 days ago) Calculator Use. Calculate the list price, discount percentage or sale price given the other two values. You will also find the discount savings amount. Calculate Discount from List Price and Sale Price. The discount is list price minus the sale price then divided by the …
https://www.calculatorsoup.com/calculators/financial/discount-calculator.php
Category: Coupon,
### Calculating a 10 Percent Discount: How-to & Steps - …
(7 days ago) May 02, 2017 · To calculate a 10 percent discount, there are only two steps you need to follow. Step 1 is to convert your percentage to a decimal. To convert your 10 percent into a decimal, you divide by 100.
Category: Coupon,
### Prompt Payment: Discount Calculator
(6 days ago) Feb 20, 2020 · After you click Calculate, the system brings back the effective annual discount rate. For this example, with the current value of funds rate of 1% (.01), the result is .0283- larger than the .01 current value of funds rate. Taking the discount this vendor is offering will save the government money.
https://www.fiscal.treasury.gov/prompt-payment/calculator.html
Category: Coupon,
### How to Calculate a Percentage Discount - dummies
(7 days ago) You can use subtraction to calculate a percentage discount. In fact, when you hear the words discount or sale price, you should automatically think of subtraction. Here’s an example: Greg has his eye on a television with a listed price of \$2,100. The salesman offers him a 30% discount …
https://www.dummies.com/education/math/pre-algebra/how-to-calculate-a-percentage-discount/
Category: Coupon,
### How To Calculate a Discount | Indeed.com
(1 days ago) Mar 09, 2021 · Represent the discount percentage in decimal form. Calculators have a function for this, or you can simply move the decimal point two places to the left. Example: Shoe Mart wants to offer a 25% discount off winter boots. 25% in decimal …
Category: Coupon,
### Discount Calculator - Find Out the Sale Price
(2 days ago) Nov 05, 2015 · How to calculate discount and sale price? Just follow these few simple steps: Find the original price (for example \$90) Get the the discount percentage (for example 20%) Calculate the savings: 20% of \$90 = \$18. Subtract the savings from the original price to get the sale price: \$90 - \$18 = \$72. You're all set!
https://www.omnicalculator.com/finance/discount
Category: Coupon,
### Discount Calculator
(7 days ago) For example, if a good costs \$45, with a 10% discount, the final price would be calculated by subtracting 10% of \$45, from \$45, or equivalently, calculating 90% of \$45: 10% of \$45 = 0.10 × 45 = \$4.50. 90% of \$45 = 0.90 × 45 = \$40.50. In this example, you are saving 10%, or \$4.50. A fixed amount off of a price refers to subtracting whatever ...
https://www.calculator.net/discount-calculator.html
Category: Coupon,
### How To Calculate Percentage Discount (%) - How To Calculate
(4 days ago) Discount is the amount deducted from the usual cost of something. The discount amount can be displayed in actual amount or as a percentage of the original amount; also known as percentage discount. Percentage discount is a percentage of the original amount (price/cost) deducted. Given the original amount (price) and the discounted amount, you can easily calculate the percentage discount …
https://www.learntocalculate.com/how-to-calculate-percentage-discount/
Category: Coupon,
### Percentage Discount Calculator. Find Discounted Price ...
(1 days ago) Feb 27, 2020 · Here you go, that's the original price before the applied discount. How do I calculate discount in percentages? Reformulate the basic equation to: discount = 100 * (original_price - discounted_price) / original_price. Subtract the final price from the original price. Divide this number by the original price. Finally, multiply the result by 100 ...
https://www.omnicalculator.com/finance/percentage-discount
Category: Coupon,
### How to Calculate the Discount Factor or Discount Rate ...
(1 days ago) Mar 08, 2018 · To calculate the discount factor for a cash flow one year from now, divide 1 by the interest rate plus 1. For example, if the interest rate is 5 percent, the discount factor is 1 divided by 1.05, or 95 percent. For cash flows further in the future, the formula is 1/ (1+i)^n, where n equals how many years in the future you'll receive the cash flow.
https://www.sapling.com/6516198/calculate-discount-factor
Category: Coupon,
### Calculate Percentage Discount - Easy Excel Tutorial
(2 days ago) If you know the discounted price and the percentage discount, you can calculate the original price. Take a look at the previous screenshot. To calculate the discounted price, we multiplied the original price by (1 - Percentage Discount).
https://www.excel-easy.com/examples/discount.html
Category: Coupon,
### How to Calculate Bond Discount Rate: 14 Steps (with Pictures)
(5 days ago) May 26, 2011 · Calculate the bond discount rate. This tells your the percentage, or rate, at which you are discounting the bond. Divide the amount of the discount by the face value of the bond. Using the …
https://www.wikihow.com/Calculate-Bond-Discount-Rate
Category: Coupon,
### How to calculate discount rate or price in Excel?
(1 days ago) Calculate discount rate with formula in Excel. The following formula is to calculate the discount rate. 1. Type the original prices and sales prices into a worksheet as shown as below screenshot: 2. Select a blank cell, for instance, the Cell C2, type this formula = (B2-A2)/ABS (A2) (the Cell A2 indicates the original price, B2 stands the sales ...
https://www.extendoffice.com/documents/excel/1487-excel-calculate-discount-rate-price.html
Category: Coupon,
### Discounting Formula | Steps to Calculate Discounted Value ...
(3 days ago) Jul 22, 2019 · Formula to Calculate Discounted Values. Discounting refers to adjusting the future cash flows to calculate the present value of cash flows and adjusted for compounding where the discounting formula is one plus discount …
https://www.wallstreetmojo.com/discounting-formula/
Category: Coupon,
### Discount Rate Formula: Calculating Discount Rate [WACC/APV]
(6 days ago) Aug 16, 2019 · The definition of a discount rate depends the context, it's either defined as the interest rate used to calculate net present value or the interest rate charged by the Federal Reserve Bank. There are two discount rate formulas you can use to calculate discount rate, WACC (weighted average cost of capital) and APV (adjusted present value).
https://www.profitwell.com/recur/all/discount-rate-formula
Category: Coupon,
### How to Calculate Original Price After Discount | Bizfluent
(5 days ago) Jan 22, 2019 · If you're in the business of holding sales or offering discounts, at some point you're going to have to figure out the original price of the item before the discount was applied. Find this by deducting the percentage discount from 100 percent, then dividing the discounted price by the result.
https://bizfluent.com/how-7566601-calculate-original-price-after-discount.html
Category: Coupon,
### How do you calculate the discount rate? – Mvorganizing.org
(3 days ago) Apr 29, 2021 · How do you calculate the discount rate? To calculate the percentage discount between two prices, follow these steps: Subtract the post-discount price from the pre-discount price. Divide this new number by the pre-discount price. Multiply the resultant number by 100. Be proud of your mathematical abilities.
https://www.mvorganizing.org/how-do-you-calculate-the-discount-rate/
Category: Coupon,
### Excel formula: Get percentage discount | Exceljet
(1 days ago) Get percentage discount. Generic formula. = ( original_price - sale_price) / original_price. Summary. To calculate the percentage discount from an original price and a sale price, you can use a formula that divides the difference by the original price. In the example shown, the formula in E5, copied down, is: = …
https://exceljet.net/formula/get-percentage-discount
Category: Coupon,
### Discount Factor (Meaning, Formula) | How to Calculate?
(6 days ago) Jan 30, 2019 · The above example shows that the formula depends not only on the rate of discount and the tenure of the investment but also on how many times the rate compounding happens during a year. Example #2. Let us take an example where the discount factor is to be calculated from year 1 to year 5 with a discount …
https://www.wallstreetmojo.com/discount-factor-formula/
Category: Coupon,
### Java Program To Calculate Discount Of Product | Programs
(6 days ago) Apr 19, 2021 · Java program to calculate discount of a product. With the help of the following program, you can calculate the discount of a product instantly. The following code has been written in five different ways. By using standard values, using command line arguments, while loop, do while loop, user-defined method and creating a separate class. Find Missing ...
https://javatutoring.com/java-program-to-calculate-discount/
Category: Coupon,
### Accounting for sales discounts — AccountingTools
(7 days ago) Apr 12, 2021 · Then, when the customer actually takes the discount, you charge it against the allowance, thereby avoiding any further impact on the income statement in the later reporting period. Most businesses do not offer early payment discounts, so there is no need to create an allowance for sales discounts. Related Courses. Bookkeeping Guidebook
https://www.accountingtools.com/articles/what-is-the-accounting-for-sales-discounts.html
Category: Coupon,
### How to Calculate a Trade Discount | Bizfluent
(6 days ago) Nov 21, 2018 · If the discount is a percentage, you calculate the trade discount by converting the percentage to a decimal and multiplying that decimal by the listed price. If the reseller is purchasing \$1,000 worth of items at a 30-percent discount, the trade discount …
Category: Coupon,
### How Do I Calculate a Discount Rate Over Time Using Excel?
(6 days ago) May 04, 2021 · The discount rate is the interest rate used when calculating the net present value (NPV) of an investment. NPV is a core component of corporate budgeting and is a comprehensive way to calculate ...
Category: Coupon,
### How to Calculate Discount Rate in a DCF Analysis
(3 days ago) How to Calculate Discount Rate in Excel: Starting Assumptions. To calculate the Discount Rate in Excel, we need a few starting assumptions: The Cost of Debt here is based on Michael Hill’s Interest Expense / Average Debt Balance over the past fiscal year. That’s 2.69 / …
https://breakingintowallstreet.com/biws/how-to-calculate-discount-rate/
Category: Coupon,
### How do you calculate net discount rate? – Mvorganizing.org
(2 days ago) Feb 08, 2021 · How do you know what discount rate to use? How to calculate discount rate. There are two primary discount rate formulas – the weighted average cost of capital (WACC) and adjusted present value (APV). The WACC discount formula is: WACC = E/V x Ce + D/V x Cd x (1-T), and the APV discount formula is: APV = NPV + PV of the impact of financing.
https://www.mvorganizing.org/how-do-you-calculate-net-discount-rate/
Category: Coupon,
### How to calculate Discount price or Net Price on Casio ...
(1 days ago) Learn how to calculate Discount Price or Net price on Casio calculators with review of Discounted value as well.
Category: Coupon,
### Percent Off Calculator
(7 days ago) For this calculator, a "stackable additional discount" means getting a further percent off of a product after a discount is applied. Using the same example, assume that the 20% discount is a discount applied by the store to the product. If you have a coupon for another 15% off, the 15% off would then be applied to the discounted price of \$223.20.
https://www.calculator.net/percent-off-calculator.html
Category: Coupon,
### Discount Points Calculator: How to Calculate Mortgage Points
(5 days ago) Discount points are a way of pre-paying interest on a mortgage. You pre-pay a lump sum of money and then obtain a lower interest rate for the duration of the loan. How Much Do They Cost? Points cost 1% of the balance of the loan. If a borrower buys 2 points on a \$200,000 home loan then the cost of points will be 2% of \$200,000, or \$4,000.
https://www.mortgagecalculator.org/calcs/discount-points.php
Category: Coupon,
### Calculate Discount in Excel | Discount Percentage
(3 days ago) Apr 19, 2016 · discounted value = (discount percentage * total value) / 100. For example, if you would like to know the discounted value of something that costs €3,000 and has a discount of 15%: (15 * 3000) / 100 = 450. To understand the formula in terms of cells, where the percentage is placed in cell A1 and the price in cell A2: (A1 * A2) / 100.
https://calcuworld.com/calculate-discount-in-excel/
Category: Coupon,
### Discounted Cash Flow DCF Formula - Calculate NPV | CFI
(2 days ago) These articles will teach you business valuation best practices and how to value a company using comparable company analysis, discounted cash flow (DCF) modeling, and precedent transactions, as used in investment banking, equity research, in each period divided by one plus the discount rate (WACC WACC WACC is a firm’s Weighted Average Cost of ...
https://corporatefinanceinstitute.com/resources/knowledge/valuation/dcf-formula-guide/
Category: Coupon,
### How to Calculate Discount Factor | GoCardless
(2 days ago) For example, to calculate discount factor for a cash flow one year in the future, you could simply divide 1 by the interest rate plus 1. For an interest rate of 5%, the discount factor would be 1 …
https://gocardless.com/guides/posts/how-to-calculate-discount-factor/
Category: Coupon,
### How Points Work on a Loan
(5 days ago) Feb 27, 2021 · How Points Work. Points are calculated as a percentage of your total loan amount, and one point is 1% of your loan. 1 Your lender might say that you can get a lower rate by paying points, and you need to decide whether the cost is worth it. For example, suppose that you’re getting a loan for \$100,000. One point is 1% of the loan value or \$1,000.
https://www.thebalance.com/discount-points-315671
Category: Coupon,
### How to Calculate Discounted Payback Period (DPP ...
(6 days ago) An initial investment of Rs.50000 is expected to generate Rs.10000 per year for 8 years. Calculate the discounted payback period of the investment if the discount rate is 11%. Given, Initial investment = Rs. 50000 Years(n) = 8 Rate(i) = 11 % CF = 10000 . To Find, Discounted Payback Period (DPP) Solution:
https://www.easycalculation.com/budget/learn-discounted-payback-period.php
Category: Coupon,
### mysql - SQL calculate discount when SELECT - Stack Overflow
(4 days ago) Mar 20, 2018 · So i want to get ALL data, and SUM the pricebefore - price than / price and * 100 so that i get the percentual discount and want to safe it AS discount. But i get all time this error: Query failed: ERROR: operator does not exist: character varying - character varying LINE 1: SELECT *, SUM((pricebefore - price ) / price * 100) AS disco...
https://stackoverflow.com/questions/49406694/sql-calculate-discount-when-select
Category: Coupon,
### How to Calculate Net Present Value (NPV) & Formula
(5 days ago) Jan 16, 2021 · Assume there is no salvage value at the end of the project and the required rate of return is 8%. The NPV of the project is calculated as follows: N P V = \$ 5 0 0 ( 1 + 0. 0 8) 1 + \$ 3 0 0 ( 1 + 0 ...
Category: Coupon,
### How to Calculate the Lease Liability and Right-of-Use (ROU ...
(3 days ago) In Example 2, the discount rate has increased from 6% to 7%. As a result, the daily discount rate for calculating the interest on the lease liability needs to be updated. Once you have determined the discount rate, you have all the inputs to complete the updated lease liability's present value calculation based on the modified terms.
Category: Coupon,
### How to calculate discount with percentage - YouTube
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Category: Watches,
### How to Calculate the Cost of Trade Credit
(2 days ago) Jan 04, 2020 · Below is a formula for calculating the cost of trade credit. You can also use this formula for calculating the cost if you don't take the trade discount. Let's say your company is offered terms of trade of 2/10, net 30 but is not able to take the 2% discount.
Category: Credit,
### Ford A-Plan, Z-Plan, X-Plan, D-Plan Pricing: How It's ...
(2 days ago) Dec 19, 2019 · Although you'll generally have no trouble finding deals for purchasing & leasing a Ford, Plan Pricing continues to be a source of confusion for a lot of consumers. Ford A, Z, X, and D-Plan each target specific groups of individuals. You don't have to be a Ford employee to get a discount, however- those with affiliated organizations can benefit ...
https://www.carsdirect.com/deals-articles/decoding-ford-a-plan-z-plan-x-plan-d-plan-pricing
Category: Coupon,
### How to Calculate Sales Tax | Definition, Formula, & Example
(1 days ago) Jul 12, 2018 · But before you start collecting, you need to know how to calculate sales tax. Figuring sales tax rates can be complicated due to tricky sales tax laws. Once you determine the rate at which you need to collect, calculating sales tax is relatively simple. Below, learn what sales tax is, which states have the tax, and how to find sales tax.
https://www.patriotsoftware.com/blog/accounting/how-to-calculate-sales-tax/
Category: Coupon,
### "How do you calculate the present value of an annuity ...
(6 days ago) Sep 24, 2021 · Related Posts “How do you calculate the present value of an annuity? Write a reflective report on a communication event. The event can involve yourself and a patient/carer, or yourself and another practitioner.
https://www.topassignmenthub.com/how-do-you-calculate-the-present-value-of-an-annuity-2/
Category: Coupon,
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# Probability CBSE Maths 12 Science MCQ Answers in English
Probability CBSE Maths 12 Science MCQ Answers in English to enable students to get Answers in a narrative video format for the specific question.
Expert Teacher provides Probability CBSE Maths 12 Science MCQ Answers through Video Answers in English language. This video solution will be useful for students to understand how to write an answer in exam in order to score more marks. This teacher uses a narrative style for a question from Probability not only to explain the proper method of answering question, but deriving right answer too.
Please find the question below and view the Answer in a narrative video format.
Question:
## Similar Questions from CBSE, 12th Science, Maths, Probability
Question 1 : Three persons A, B and C apply for a job of Manager in a Private company. Chance of their selection (A, B and C) are in the ratio 1:2:4. The probability that A, B, and C can introduce changes to improve profits of company are 0.8, 0.5 and 0.3 respectively, if the changes does not take place, find the probability that is due to the appointment of C. (View Answer Video)
Question 2 : Probability of solving specific problem independently by A and B are and respectively. If both try to solve the problem, independently, then find the probability that exactly one of them solves the problem. (View Answer Video)
Question 3 : Three cards are drawn without replacement from a pack of 52 cards. Find the probability that the cards drawn are king, queen and jack respectively.
Question 4 : If the mean and variance of a Binomial distribution are 9 and 6 respectively, find the number of trails. (View Answer Video)
Question 5 : A couple has 2 children. Find the probability that both are boys, if it is known that one of them is a boy. (View Answer Video)
### Application of Derivatives
Question 1 : The point on the curve which is nearest to the point (0, 5) is :
Question 2 : The line y=mx+1 is a tangent to the curve if the value of m is ________. (View Answer Video)
Question 3 : The function f is a differentiable function and satisfies the functional equation f(x) + f(y) = f(x + y) – xy – 1for every pair x, y of real numbers. If f(1) = 1, then the number of integers n ≠ 1 for which f(n) = n is_______________. (View Answer Video)
Question 5 : Find the maximum value of in the interval [1,3]. find the maximum value of the same function in [-3,-1]. (View Answer Video)
### Three Dimensional Geometry
Question 1 : Find the distance between the point (-1, -5, -10) and the point of intersection of line and plane
Question 2 : Find the equation of the plane passing through the point (-1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0. (View Answer Video)
Question 3 : The cartesian equation of a line is , write its vector form. (View Answer Video)
Question 4 : Find the equation of plane passing through the line of intersection of the planes and and passing through the point (3, -2, -1). Also, find the angle between the two given planes. (View Answer Video)
Question 5 : A plane makes intercepts -6, 3, 4 respectively on the co-ordinte axes. Find the length of the perpendicular from the origin on it. (View Answer Video)
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# Thread: Range of a specific value in quadratic equation to have 3 real roots
1. ## Range of a specific value in quadratic equation to have 3 real roots
Can anyone help me with the this?
Shenaya Perera
2. ## Re: Range of a specific value in quadratic equation to have 3 real roots
Hey pererashenaya7.
It means you have to find a value of k where you can get three solutions for x to satisfy f(x) = 0 which you will have to define.
3. ## Re: Range of a specific value in quadratic equation to have 3 real roots
Originally Posted by chiro
Hey pererashenaya7.
It means you have to find a value of k where you can get three solutions for x to satisfy f(x) = 0 which you will have to define.
Yes...I know that...:-)
For example if the discriminant of a quadratic equation is greater than or equal to 0 it has 2 real roots and if it is equal to zero it has one real root.
But in here we have to find the range of k it has three real roots.But how? And that's exactly where I've been stuck... :-)
Shenaya Perera
4. ## Re: Range of a specific value in quadratic equation to have 3 real roots
Let y= tan(x). Then the equation is $y^2- y- k= 0$. That has two distinct real roots if and only if $(-1)^2- 4(1)(-k)= 4k+ 1\ge 0$. Now, what about tan(x)? For a given value of y, how many roots does $tan(x)= k$ have?
5. ## Re: Range of a specific value in quadratic equation to have 3 real roots
Originally Posted by HallsofIvy
Let y= tan(x). Then the equation is $y^2- y- k= 0$. That has two distinct real roots if and only if $(-1)^2- 4(1)(-k)= 4k+ 1\ge 0$. Now, what about tan(x)? For a given value of y, how many roots does $tan(x)= k$ have?
Two real roots...
But the question asks for a range of k which the equation has three real roots??.That's where I've been stuck..
Shenaya Perera
6. ## Re: Range of a specific value in quadratic equation to have 3 real roots
Maybe I'm missing something in the question (see attached graph of $y=\tan^2{x}-\tan{x}$) ...
For $k < -\dfrac{1}{4}$, $\tan^2{x}-\tan{x} = k$ has no real solutions.
For $k \ge -\dfrac{1}{4}$, $\tan^2{x}-\tan{x}=k$ has an infinite number of real solutions.
If we're considering only a single period of the function ...
$k < -\dfrac{1}{4}$ yields no solutions, $k = -\dfrac{1}{4}$ yields a single solution, and $k > - \dfrac{1}{4}$ yields only two solutions.
Sorry, but I'm not seeing how the equation yields three real solutions unless there is a restricted interval for $x$ that was unstated in the original problem ... someone please enlighten me.
7. ## Re: Range of a specific value in quadratic equation to have 3 real roots
Originally Posted by skeeter
Maybe I'm missing something in the question (see attached graph of $y=\tan^2{x}-\tan{x}$) ...
For $k < -\dfrac{1}{4}$, $\tan^2{x}-\tan{x} = k$ has no real solutions.
For $k \ge -\dfrac{1}{4}$, $\tan^2{x}-\tan{x}=k$ has an infinite number of real solutions.
If we're considering only a single period of the function ...
$k < -\dfrac{1}{4}$ yields no solutions, $k = -\dfrac{1}{4}$ yields a single solution, and $k > - \dfrac{1}{4}$ yields only two solutions.
Sorry, but I'm not seeing how the equation yields three real solutions unless there is a restricted interval for $x$ that was unstated in the original problem ... someone please enlighten me.
No you aren't missing anything...I've posted the exact question..
I'm also stuck on the same case,I also don't see a way to find a range of k which this equation has 3 real roots..
Shenaya Perera
8. ## Re: Range of a specific value in quadratic equation to have 3 real roots
Originally Posted by pererashenaya7
No you aren't missing anything...I've posted the exact question..
I'm also stuck on the same case,I also don't see a way to find a range of k which this equation has 3 real roots..
Shenaya Perera
... then I advise you research the origin of this question or ask whoever assigned it for clarification.
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# What is the percentage by mass of phosphorus in a washing powder, in which a 0.085 g precipitate of magnesium pyrophosphate forms from a 2 g sample of powder? (Gravimetric analysis).
## The phosphorus in a 2 g sample of washing powder is precipitated as $M {g}_{2}$${P}_{2}$${O}_{7}$. The precipitate weighs 0.085 g.
Sep 10, 2017
Here's what I got.
#### Explanation:
The idea here is that the mass of phosphorus present in the washing powder will be equal to the mass of phosphorus present in the $\text{2-g}$ sample of magnesium pyrophosphate.
To find the mass of phosphorus present in the precipitate, start by calculating the percent composition of the salt. To do that, use the molar mass of magnesium pyrophosphate, ${\text{Mg"_2"P"_color(red)(2)"O}}_{7}$, and the molar mass of phosphorus.
(color(red)(2) * 30.974 color(red)(cancel(color(black)("g mol"^(-1)))))/(222.5533color(red)(cancel(color(black)("g mol"^(-1))))) * 100% = "27.835% P"
This means that every $\text{100 g}$ of magnesium pyrophosphate will contain $\text{27.835 g}$ of phosphorus.
You can thus say that your sample contains
0.085 color(red)(cancel(color(black)("g Mg"_2"P"_2"O"_7))) * "27.835 g P"/(100color(red)(cancel(color(black)("g Mg"_2"P"_2"O"_7)))) = "0.02366 g P"
Now, you know that $\text{2 g}$ of washing powder contain $\text{0.02366 g}$ of phosphorus. THis means that the mass of phosphorus present in $\text{100 g}$ of washing powder is equal to
100 color(red)(cancel(color(black)("g washing powder"))) * "0.02366 g P"/(2color(red)(cancel(color(black)("g washing powder")))) = "1.18 g P"
Therefore, the percent concentration by mass of phosphorus in the washing powder, i.e. the number of grams of phosphorus present for every $\text{100 g}$ of washing powder, will be equal to
$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{% concentration = 1.2% P}}}}$
I'll leave the answer rounded to two sig figs, but keep in mind that you have one significant figure for the mass of washing powder.
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# Assignment Operators
Assignment operators are used to assign values to variables. The most common assignment operator, by far, is the = operator.
OperatorExample
Assignment (=)x = 5
This operator
• Computes the value of the expression to its right
• Assigns it to the variable on its left
For example, we can assign the value 10 to a variable called x as follows:
x = 10
From that moment on, the variable x contains the value 10 and we can use it in our program.
On the right there can be any expression that evaluates to a value. For example, this would also be valid.
a = 10
x = 2 + 3 + a
## Compound Assignment Operators
In addition to the basic assignment operator =, many programming languages offer compound assignment operators, which combine assignment with another arithmetic operation.
OperatorExample
Addition Assignment (+=)x += 3 (equivalent to x = x + 3)
Subtraction Assignment (-=)x -= 2 (equivalent to x = x - 2)
Multiplication Assignment (*=)x *= 4 (equivalent to x = x * 4)
Division Assignment (/=)x /= 2 (equivalent to x = x / 2)
Modulus Assignment (%=)x %= 3 (equivalent to x = x % 3)
These operators are a shorthand way of writing common operations and can help make our code more concise and readable.
Let’s see an example in JavaScript using the += operator:
let x = 5;
x += 3; // equivalent to x = x + 3
console.log(x); // will print 8
In this case, the += operator adds 3 to the current value of the variable x and then assigns the result back to x. As a result, x is updated to 8.
## Examples of Assignment Operators in Different Languages
As we said, in practically all languages the assignment operator is the equal sign =.
Similarly, the assignment operators +=, -=, *=, and /= are practically a standard.
For example, this is how assignment operators are used in C++, C#, or Java,
// assignment operator
int x = 5;
// compound operators
x += 3; // Equivalent to x = x + 3
x -= 2; // Equivalent to x = x - 2
x *= 4; // Equivalent to x = x * 4
x /= 2; // Equivalent to x = x / 2
x %= 7; // Equivalent to x = x % 7
Whose usage is identical in JavaScript,
// assignment operator
let x = 5;
// compound operators
x += 3; // Equivalent to x = x + 3
x -= 2; // Equivalent to x = x - 2
x *= 4; // Equivalent to x = x * 4
x /= 2; // Equivalent to x = x / 2
x %= 7; // Equivalent to x = x % 7
Or, for example, in Python,
# assignment operator
x = 5
# compound operators
x += 3 # Equivalent to x = x + 3
x -= 2 # Equivalent to x = x - 2
x *= 4 # Equivalent to x = x * 4
x /= 2 # Equivalent to x = x / 2
x %= 7 # Equivalent to x = x % 7
As we can see, the use of the operators is identical. In the examples, only the way of declaring the variable changes, which has nothing to do with today’s topic, which is the operators.
## Best Practices and Cleanliness Tips Tips
In many languages (not all) assignment operations return true as a result of the assignment.
This leads to a very common mistake, if we get confused and put = instead of ==
if(a = 3) // this is always true
{ }
In this case, we are not comparing a with 3. We are storing 3 in a, and evaluating the result as a condition. Here we have two undesirable effects:
• We are storing a value that we did not want in a variable
• In addition, the conditional is always met, because = always returns true.
That is, be careful not to mistakenly put = instead of ==.
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