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https://www.hackmath.net/en/example/5587?tag_id=95,124 | 1,561,544,930,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000266.39/warc/CC-MAIN-20190626094111-20190626120111-00187.warc.gz | 756,715,790 | 6,976 | # Conical area
A right angled triangle has sides a=12 and b=19 in right angle. The hypotenuse is c. If the triangle rotates on the c side as axis, find the volume and surface area of conical area created by this rotation.
Result
V = 2422.438
S = 988.1
#### Solution:
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Carla is 5 years old and Jim is 13 years younger than Peter. One year ago, Peter’s age was twice the sum of Carla’s and Jim’s age. Find the present age of each one of them. | 1,001 | 3,685 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2019-26 | latest | en | 0.921608 |
https://justaaa.com/accounting/186388-the-range-of-105-100-79-86-101-94-98 | 1,723,434,360,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641028735.71/warc/CC-MAIN-20240812030550-20240812060550-00239.warc.gz | 250,058,053 | 10,117 | Question
the range of 105 , 100 , 79 , 86 , 101 , 94 , 98...
the range of 105 , 100 , 79 , 86 , 101 , 94 , 98 , 87 and 83 is :
Range is the difference between the lowest and the highest value from the given data.
The given set of numbers:- 105, 100, 79, 86, 101, 94, 98, 87 and 83
Firstly we should arrange the given numbers in ascending order which will make the calculation of lowest and highest value easy.
79, 83, 86, 87, 94, 98, 100, 101, 105
The lowest value = 79
The highest value = 105
Range = Highest value - Lowest value
= 105 - 79 = 26
Therefore the range of 105 , 100 , 79 , 86 , 101 , 94 , 98 , 87 and 83 is 26. | 222 | 635 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-33 | latest | en | 0.715691 |
https://math.stackexchange.com/questions/1824674/on-the-proof-of-lucas-theorem | 1,582,156,196,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144429.5/warc/CC-MAIN-20200219214816-20200220004816-00474.warc.gz | 475,495,012 | 31,488 | # On the proof of Lucas' theorem
Lucas theorem states that
Let $m,n$ be two natural numbers, $p$ be a prime. Suppose that $m, n$ admit the following base $p$ representation $$m=m_0+m_1p+\cdots+m_sp^s,\qquad n=n_0+n_1p+\cdots+n_sp^s$$ with $0\le m_i, n_i \le p-1$, then $$\binom{n}{m}\equiv\prod_{i=0}^{s}\binom{n_i}{m_i}\ (\bmod p)$$
I have read the proof of Lucas' Theorem on Wikipedia and in this paper. There are some steps that I don't fully understand. It is follows
\begin{align*} \sum^{n}_{m=0}\binom{n}{m}X^m&=(1+X)^n=\prod_{i=0}^{s}\left( (1+X)^{p^i}\right)^{n_i}\\ &\equiv \prod_{i=0}^{s}(1+X^{p^i})^{n_i}=\prod_{i=0}^{s}\left( \sum_{m_i=0}^{n_i}\binom{n_i}{m_i}X^{m_ip^i}\right)\ (\bmod p)\\ &=\prod_{i=0}^{s}\left(\sum_{m_i=0}^{p-1}\binom{n_i}{m_i}X^{m_ip^i} \right)\\ &=\sum_{m=0}^{n}\left(\prod_{i=0}^{s}\binom{n_i}{m_i} \right)X^m\ (\bmod p) \end{align*}
How can we get \begin{align*} &\prod_{i=0}^{s}\left( \sum_{m_i=0}^{n_i}\binom{n_i}{m_i}X^{m_ip^i}\right) \\ &=\prod_{i=0}^{s}\left(\sum_{m_i=0}^{p-1}\binom{n_i}{m_i}X^{m_ip^i} \right)\\ &=\sum_{m=0}^{n}\left(\prod_{i=0}^{s}\binom{n_i}{m_i} \right)X^m \end{align*}
I don't know how can we change the index from $n_i$ to $p-1$ in the first equation and how can we get the second equation.
• $m_i$ has two different meanings: In some places it's a constant depending on $m$; in other places it's a variable being summed over, and may as well be called $k$. – mr_e_man Jan 28 at 2:24
For the first line, observe that $${n_i\choose k}=0$$ for $k>n_i$. So, \begin{align*} \sum_{m_i=0}^{n_i}\binom{n_i}{m_i}X^{m_ip^i} =\sum_{m_i=0}^{p-1}\binom{n_i}{m_i}X^{m_ip^i} \end{align*} , as the extra terms are basically zero.
For the second line, notice that $m$ can be uniquely written in the form $$m_0'+m_1'p+\ldots+m_s'p^s$$
Thus, the coefficient of $X^m$ in \begin{align*} \prod_{i=0}^{s}\left(\sum_{m_i=0}^{p-1}\binom{n_i}{m_i}X^{m_ip^i} \right) \end{align*} will be the product of the coefficients of $X^{m_ip^i}$. | 841 | 1,983 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2020-10 | latest | en | 0.755594 |
https://nrich.maths.org/public/leg.php?code=106&cl=2&cldcmpid=975 | 1,563,213,383,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195523840.34/warc/CC-MAIN-20190715175205-20190715201205-00292.warc.gz | 499,925,248 | 19,674 | # Search by Topic
#### Resources tagged with Cubes & cuboids similar to More Christmas Boxes:
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### There are 52 results
Broad Topics > 3D Geometry, Shape and Space > Cubes & cuboids
### More Christmas Boxes
##### Age 7 to 11 Challenge Level:
What size square should you cut out of each corner of a 10 x 10 grid to make the box that would hold the greatest number of cubes?
### All Wrapped Up
##### Age 7 to 11 Challenge Level:
What is the largest cuboid you can wrap in an A3 sheet of paper?
### Christmas Presents
##### Age 7 to 11 Challenge Level:
We need to wrap up this cube-shaped present, remembering that we can have no overlaps. What shapes can you find to use?
### Thinking 3D
##### Age 7 to 14
How can we as teachers begin to introduce 3D ideas to young children? Where do they start? How can we lay the foundations for a later enthusiasm for working in three dimensions?
### Double Your Popcorn, Double Your Pleasure
##### Age 7 to 11 Challenge Level:
We went to the cinema and decided to buy some bags of popcorn so we asked about the prices. Investigate how much popcorn each bag holds so find out which we might have bought.
### Four Layers
##### Age 5 to 11 Challenge Level:
Can you create more models that follow these rules?
### The Third Dimension
##### Age 5 to 11 Challenge Level:
Here are four cubes joined together. How many other arrangements of four cubes can you find? Can you draw them on dotty paper?
### Changing Areas, Changing Volumes
##### Age 11 to 14 Challenge Level:
How can you change the surface area of a cuboid but keep its volume the same? How can you change the volume but keep the surface area the same?
### Three Sets of Cubes, Two Surfaces
##### Age 7 to 11 Challenge Level:
How many models can you find which obey these rules?
### Cuboid-in-a-box
##### Age 7 to 11 Challenge Level:
What is the smallest cuboid that you can put in this box so that you cannot fit another that's the same into it?
### Next Size Up
##### Age 7 to 11 Challenge Level:
The challenge for you is to make a string of six (or more!) graded cubes.
### Castles in the Middle
##### Age 7 to 14 Challenge Level:
This task depends on groups working collaboratively, discussing and reasoning to agree a final product.
##### Age 7 to 11 Challenge Level:
If you had 36 cubes, what different cuboids could you make?
### Cubes
##### Age 7 to 11 Challenge Level:
How many faces can you see when you arrange these three cubes in different ways?
### Triple Cubes
##### Age 5 to 11 Challenge Level:
This challenge involves eight three-cube models made from interlocking cubes. Investigate different ways of putting the models together then compare your constructions.
### Cubic Conundrum
##### Age 7 to 16 Challenge Level:
Which of the following cubes can be made from these nets?
### Take Ten
##### Age 11 to 14 Challenge Level:
Is it possible to remove ten unit cubes from a 3 by 3 by 3 cube so that the surface area of the remaining solid is the same as the surface area of the original?
### Cereal Packets
##### Age 7 to 11 Challenge Level:
How can you put five cereal packets together to make different shapes if you must put them face-to-face?
### Plutarch's Boxes
##### Age 11 to 14 Challenge Level:
According to Plutarch, the Greeks found all the rectangles with integer sides, whose areas are equal to their perimeters. Can you find them? What rectangular boxes, with integer sides, have. . . .
### Construct-o-straws
##### Age 7 to 11 Challenge Level:
Make a cube out of straws and have a go at this practical challenge.
### Cubist Cuts
##### Age 11 to 14 Challenge Level:
A 3x3x3 cube may be reduced to unit cubes in six saw cuts. If after every cut you can rearrange the pieces before cutting straight through, can you do it in fewer?
### Three Cubed
##### Age 7 to 11 Challenge Level:
Can you make a 3x3 cube with these shapes made from small cubes?
### Little Boxes
##### Age 7 to 11 Challenge Level:
How many different cuboids can you make when you use four CDs or DVDs? How about using five, then six?
### Making Cuboids
##### Age 7 to 11 Challenge Level:
Let's say you can only use two different lengths - 2 units and 4 units. Using just these 2 lengths as the edges how many different cuboids can you make?
### Nine Colours
##### Age 11 to 16 Challenge Level:
Can you use small coloured cubes to make a 3 by 3 by 3 cube so that each face of the bigger cube contains one of each colour?
### Making Maths: Link-a-cube
##### Age 7 to 11 Challenge Level:
Make a cube with three strips of paper. Colour three faces or use the numbers 1 to 6 to make a die.
### A Puzzling Cube
##### Age 7 to 11 Challenge Level:
Here are the six faces of a cube - in no particular order. Here are three views of the cube. Can you deduce where the faces are in relation to each other and record them on the net of this cube?
### Green Cube, Yellow Cube
##### Age 7 to 11 Challenge Level:
How can you paint the faces of these eight cubes so they can be put together to make a 2 x 2 x 2 cube that is green all over AND a 2 x 2 x 2 cube that is yellow all over?
### Holes
##### Age 5 to 11 Challenge Level:
I've made some cubes and some cubes with holes in. This challenge invites you to explore the difference in the number of small cubes I've used. Can you see any patterns?
### Christmas Boxes
##### Age 11 to 14 Challenge Level:
Find all the ways to cut out a 'net' of six squares that can be folded into a cube.
### Dicey
##### Age 7 to 11 Challenge Level:
A game has a special dice with a colour spot on each face. These three pictures show different views of the same dice. What colour is opposite blue?
### Icosian Game
##### Age 11 to 14 Challenge Level:
This problem is about investigating whether it is possible to start at one vertex of a platonic solid and visit every other vertex once only returning to the vertex you started at.
### Boxed In
##### Age 11 to 14 Challenge Level:
A box has faces with areas 3, 12 and 25 square centimetres. What is the volume of the box?
### Tic Tac Toe
##### Age 11 to 14 Challenge Level:
In the game of Noughts and Crosses there are 8 distinct winning lines. How many distinct winning lines are there in a game played on a 3 by 3 by 3 board, with 27 cells?
### Cube Drilling
##### Age 7 to 11 Challenge Level:
Imagine a 4 by 4 by 4 cube. If you and a friend drill holes in some of the small cubes in the ways described, how many will not have holes drilled through them?
### Drilling Many Cubes
##### Age 7 to 14 Challenge Level:
A useful visualising exercise which offers opportunities for discussion and generalising, and which could be used for thinking about the formulae needed for generating the results on a spreadsheet.
### All in the Mind
##### Age 11 to 14 Challenge Level:
Imagine you are suspending a cube from one vertex and allowing it to hang freely. What shape does the surface of the water make around the cube?
### Cuboids
##### Age 11 to 14 Challenge Level:
Find a cuboid (with edges of integer values) that has a surface area of exactly 100 square units. Is there more than one? Can you find them all?
### Painted Faces
##### Age 7 to 11 Challenge Level:
Imagine a 3 by 3 by 3 cube made of 9 small cubes. Each face of the large cube is painted a different colour. How many small cubes will have two painted faces? Where are they?
### Troublesome Dice
##### Age 11 to 14 Challenge Level:
When dice land edge-up, we usually roll again. But what if we didn't...?
### Solids
##### Age 11 to 16 Challenge Level:
A task which depends on members of the group working collaboratively to reach a single goal.
### Always, Sometimes or Never? Shape
##### Age 7 to 11 Challenge Level:
Are these statements always true, sometimes true or never true?
### The Solid
##### Age 11 to 16 Challenge Level:
A task which depends on members of the group working collaboratively to reach a single goal.
### Sending a Parcel
##### Age 11 to 14 Challenge Level:
What is the greatest volume you can get for a rectangular (cuboid) parcel if the maximum combined length and girth are 2 metres?
### Dice, Routes and Pathways
##### Age 5 to 14
This article for teachers discusses examples of problems in which there is no obvious method but in which children can be encouraged to think deeply about the context and extend their ability to. . . .
### Counting Triangles
##### Age 11 to 14 Challenge Level:
Triangles are formed by joining the vertices of a skeletal cube. How many different types of triangle are there? How many triangles altogether?
### Painting Cubes
##### Age 11 to 14 Challenge Level:
Imagine you have six different colours of paint. You paint a cube using a different colour for each of the six faces. How many different cubes can be painted using the same set of six colours?
### How Many Dice?
##### Age 11 to 14 Challenge Level:
A standard die has the numbers 1, 2 and 3 are opposite 6, 5 and 4 respectively so that opposite faces add to 7? If you make standard dice by writing 1, 2, 3, 4, 5, 6 on blank cubes you will find. . . .
### Paper Folding - Models of the Platonic Solids
##### Age 11 to 16
A description of how to make the five Platonic solids out of paper.
### Which Solid?
##### Age 7 to 16 Challenge Level:
This task develops spatial reasoning skills. By framing and asking questions a member of the team has to find out what mathematical object they have chosen. | 2,281 | 9,538 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2019-30 | longest | en | 0.905522 |
https://sonichours.com/how-tall-is-1-76-meters-in-feet/ | 1,716,403,697,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058560.36/warc/CC-MAIN-20240522163251-20240522193251-00367.warc.gz | 470,451,645 | 20,057 | General
# How Tall Is 1.76 Meters In Feet
There are two ways to convert the measurement of height. The first way is by using the inch. It is a standard unit of length, one-twelfth of a foot. It is equal to 25.34 millimeters. The second way is by using the centimeter. Generally, the measurement of a meter is represented in meters. Likewise, the measurements of a foot are represented in centimeters.
Lastly, if you want to convert 1.76 meters to feet, just multiply the length by 12 inches. This way, you’ll get 5.77 feet. However, if you’d like to convert your height in centimeters, then you’ll need to multiply the length by four to get 176 feet. In this way, you’ll get a more accurate measurement. Using the standard method, you can calculate the length of your foot in either units.
The same rule holds for the centimeter. One meter is equal to about 3.28 feet. For instance, if you’re 5 feet and 8 inches tall, you’ll be at 17.3 inches. So, you’d have to multiply 1.76 meters by 12 inches. Those who are 5 feet and eighteen inches tall would be at five feet and nine and a half feet. By the way, a meter equals 3.84 inches, which is close to 5.77 ft.
If you want to know how tall you are, you can look up the conversion formulas online. You can even find a calculator for the conversion. It’s easier than you think! And remember that if you know your own height, you can also find out how much you need to buy. In this case, a meter is equivalent to five feet and seven inches. The numerator is four sixteenths of an inch, which makes the conversion ratio of 1.76 meters to feet quite accurate.
In a metric system, a meter is equivalent to about three feet and twelve inches. It is not necessary to know how tall you are to calculate the conversion. It’s enough to know your height in inches. Then, multiply that number by the number of centimeters. Then, add two to the numerator. Then, subtract three from the numerator. The result is a meter.
In the English language, one meter is equivalent to about three feet and twelve inches. A meter equals five feet and nine inches, while five meters is equal to five feet and nine inches. Thus, you are 1.6 meters tall, and you are a half-meter short. You can divide the numbers by three to get the conversion from a meter to a foot. Once you have the height of your choice, you can measure your height in centimeters.
In other words, you can convert the measurement of 1.76 meters to feet by using the centimeter. A meter is equal to a foot, which is 12 inches. Similarly, one meter is equivalent to 5.77 feet. In the U.S., the average male is about five feet and nine inches. A metric is a tenth of a kilometer. That is about three-tenths of a pound.
In the United States, the average adult male is 5.77 feet tall. The same is true in the United Kingdom. For instance, the average American man is five feet and nine inches high. Therefore, it is roughly equivalent to 176 centimeters. If you are a woman, you should be aware of your proportional ratio of one foot to a metric foot. It is easy to convert the two measurements and use the metric system for the conversion.
To convert a meter to a foot, multiply the number of inches in a meter by three. Then, take that value and divide it by the number of inches in a foot. Then, you can convert the length of a person in meters to feet by multiplying it by three. It is easy to determine how tall a person is. A foot is 12 inches long, whereas a meter is three.
Visit the rest of the site for more useful articles! | 841 | 3,529 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-22 | latest | en | 0.935989 |
https://www.enotes.com/homework-help/teacher-buying-pencils-315-students-pencils-sold-197081 | 1,485,161,500,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560282202.61/warc/CC-MAIN-20170116095122-00374-ip-10-171-10-70.ec2.internal.warc.gz | 894,140,672 | 13,568 | # Teacher is buying pencils for 315 students. The pencils are sold in boxes of tens. How can she use rounding to decide how many pencils to buy?How do I get the answer?
hala718 | High School Teacher | (Level 1) Educator Emeritus
Posted on
There are 315 students ,
then the teachers needed 315 pencils.
Since the pencils comes in 10- pecils cases, then she can buy 32 cases .
Then, each student will have 1 pencil and there will be 5 extra pencils.
Then, we will round the pencils needed from 315 to 320
If she buy 1 less case. then there will be 31 cases with 310 pencils and 5 studens will not get any.
Then , the teacher needs to buy 32 cases
william1941 | College Teacher | (Level 3) Valedictorian
Posted on
The teacher has to buy pencils for 315 students. If she has more than 315 pencils, it is fine as there will be a few extra pencils but if she has less than 315 pencils, a few students will not be able to do their work.
Now pencils are sold in boxes that contain 10 pencils each. So the teacher rounds 315 to the next higher number that is divisible by 10. Here it is 320. Therefore she needs to buy 32 boxes of pencils to satisfy the requirement of all the students.
nuke34 | eNotes Newbie
Posted on
She should round up the # of students to 320 in order to have enough for everyone.10 pencils times 32 boxes =320 pencils...
neela | High School Teacher | (Level 3) Valedictorian
Posted on
The problem is telling about the utility and application of rounding.
The number of students = 315.
The number of pencils in a bos = 10.
Since each pencil box has 10 pencils, teacher has to purchase a number nearest 10 to 315 . There are two nearest 10's: 310 or 320 . But with 310 pencils , it is short of 5 pencils for 315 students, if she gives one pencil to each.
With 320 pencils, she can give one each to 315 students and retain extra 5 pencils.
So the number boxes of pencils each containing 10 pencils to be purchased by the teacher = 320/10 = 32 | 505 | 1,979 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2017-04 | longest | en | 0.940049 |
http://mathhelpforum.com/math-topics/51048-units-converts.html | 1,481,334,375,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542932.99/warc/CC-MAIN-20161202170902-00011-ip-10-31-129-80.ec2.internal.warc.gz | 177,314,838 | 10,449 | 1. ## units converts
convert to cubic feet:
64.2 inches x 30.0 inches x 26.0 inches
2. Originally Posted by tangerine_tomato
convert to cubic feet:
64.2 inches x 30.0 inches x 26.0 inches
Multiply each term by $\frac{1~ft}{12~in}$
Then when you multiply everything together, the units will be in $ft^3$
--Chris
3. Originally Posted by tangerine_tomato
convert to cubic feet:
64.2 inches x 30.0 inches x 26.0 inches
Originally Posted by Chris L T521
Multiply each term by $\frac{1~ft}{12~in}$
Then when you multiply everything together, the units will be in $ft^3$
--Chris
u can do that $\frac{64.2in}{1} *\frac{1ft}{12in}$ and that would cancel inches out blah blah....then repeaton on other 2 dimensions then multiply to geth ur answer in feet cubed
however this may also help u in the future $\frac{1ft^3}{1}*\frac{12in}{1ft}*\frac{12in}{1ft}* \frac{12in}{1ft}$thus canceling feet cubed so you now have $1ft^3= 1728in^3$
thus $64.2in*30.0in*26.0in =50076in^3 * \frac{1ft^3}{1728in^3}$giving you $28.979...ft^3$
we can check your answer by doing the above method chris showed (probobly safer but much much slower.....(as u can record how man inches cubed are in a feet for future use)
so $\frac{64.2in}{1}\frac{1ft}{12in}=5.35ft$
$\frac{30in}{1}\frac{1ft}{12in}=2.5ft$
$\frac{26in}{1}\frac{1ft}{12in}=2.1111...ft$
and finaly
$5.35ft*2.5ft*2.1111..ft=28.979...ft^3$
hope that gives u help in the future..shortcuts are always nice(just can get messy if u make mistakes) | 510 | 1,487 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2016-50 | longest | en | 0.745342 |
http://slideplayer.com/slide/3615942/ | 1,506,324,473,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690340.48/warc/CC-MAIN-20170925055211-20170925075211-00072.warc.gz | 304,871,370 | 37,466 | # Applied Quantitative Methods MBA course Montenegro
## Presentation on theme: "Applied Quantitative Methods MBA course Montenegro"— Presentation transcript:
Applied Quantitative Methods MBA course Montenegro
Peter Balogh PhD
6. Measures of dispersion
In the previous part of the presentation we considered several measures of the typical, or average value. The mean is widely regarded as the most important descriptive statistic. When references are made to the average time or the average weight or the average cost it is generally the mean that has been calculated. Knowledge of the mean, the median and the mode will increase our understanding of the data but will not provide a sufficient understanding of the differences in the data.
6. Measures of dispersion
In many applications it is the differences that are of particular interest to us. In market research, for example, we are interested not only in the typical values but also in whether opinions or behaviours are fairly consistent or vary considerably. A niche market is defined by difference. Quality control, whether in the manufacturing or the service sector, is concerned with difference from the expected.
6. Measures of dispersion
In this part I introduce ways of measuring this variability, or dispersion, and then consider ways of comparing different distributions. Measures of dispersion can be absolute (considering only one set of data at a time and giving an answer in the original units e.g. £'s, minutes, years), or relative (giving the answer as a percentage or proportion and allowing direct comparison between distributions).
6.1 The standard deviation
The standard deviation is the most widely used measure of dispersion, since it is directly related to the mean. If you choose the mean as the most appropriate measure of central location, then the standard deviation would be the natural choice for a measure of dispersion. Unlike the mean, the standard deviation is not so well known and does not have the same intuitive meaning. The standard deviation measures differences from the mean - a larger value indicating a larger measure of overall variation. The standard deviation will also be in the same units as the mean (£‘s, minutes, years) and a change of units (e.g. from £’s to dollars, or metres to centimetres) will change the value.
6.1 The standard deviation
The application of computer packages will generally make the determination of the standard deviation a relatively straightforward procedure, but it is worth checking what version of the formula is being used (the divisor can be n or n - 1). I will continue to follow the practice of showing the calculations by hand, as you may still need to do them. Such calculations do have the additional advantage of showing how the standard deviation is related to the mean. The standard deviation is particularly important in the development of statistical theory, since most statistical theory is based on distributions described by their mean and standard deviation.
Untabulated data We have already seen how to calculate the mean from simple data. We will need this calculation of the mean before we calculate the standard deviation. We can again use the first 10 observations on the number of cars entering a car park in 10-minute intervals: The mean of this data is 19.6 cars. The differences about the mean are shown diagrammatically in Figure 6.2. To the left of the mean the differences are negative and to the right of the mean the differences are positive. It can be seen, for example, that the observation 9 is 10.6 units below the mean, a deviation of The sum of these differences is zero - check this by adding all the deviations. This summing of deviations to zero illustrates the physical interpretation of the mean as being the centre of gravity with the observations as a number of "weights in balance'. A
Untabulated data To calculate the standard deviation we follow six steps: Compute the mean Calculate the differences from the mean Square these differences Sum the squared differences Average the squared differences to find variance: Square root variance to find standard deviation.
6.1.2 Tabulated discrete data
Table 6.2, showing the number of working days lost by employees in the last quarter, typifies the tabulation of discrete data.
6.1.2 Tabulated discrete data
We need to allow for the fact that 410 employees lost no days, 430 lost one day and so on by including frequency in our calculations. In this example there are 1440 employees in total and we need to include 1440 squared differences. The formula for the standard deviation becomes
6.1.2 Tabulated discrete data
6.1.3 Tabulated (grouped) data
When data is presented as a grouped frequency distribution we must determine whether it is discrete or continuous (as this will affect the way we view the range of values) and determine the mid-points. Once the mid-points have been determined we proceed as before using mid-point values for x and frequencies, as shown in Table 6.4. The approach shown clearly illustrates how the standard deviation summarizes differences, but would be extremely tedious to perform by hand.
6.1.3 Tabulated (grouped) data
Some algebraic manipulation of the formula given in Section 6.1.2, will provide a simplified formula that is easier to work with for both calculations by hand and the construction of spreadsheets. The simplified formula is usually presented as follows: The formula does lose its intuitive appeal but is easier to use. Formula of this kind can be presented in a variety of ways. Using a formula presented in different ways should not be a problem. What you do need to be sure about are the stages required in the calculations (e.g. what columns to add] and the assumptions being made (e.g. is n or [n - 1) being used as the divisor?). The use of this simplified formula is illustrated in Table 6.5.
6.1.3 Tabulated (grouped) data
6.1.4 The variance The variance is the squared value of the standard deviation, and therefore is calculated easily once the standard deviation is known. It is sometimes used as a descriptive measure of dispersion or variability rather than the standard deviation, but its importance lies in more advanced statistical theory. As we will see, you can add variances but you cannot add standard deviations. Variance is mentioned here for completeness.
6.2 Other measures of dispersion
While the standard deviation is the most widely used measure of dispersion, it is not the only one. As we saw when looking at measures of location, different measures (mean, median and mode) are appropriate for different situations and the same is true for measures of dispersion. Furthermore, some of the measures of dispersion are specifically linked to certain measures of location and it would not make sense to mix and match the statistics.
6.2.1 The range The range is the most easily understood measure of dispersion as it is the difference between the highest and lowest values. If we were again concerned with the 10 observations: the range would be 22 cars (31 - 9). It is, however, a rather crude measure of spread, being dependent on the two most extreme observations. It is also highly unstable as new data is added. If this measure is to be used, it may well be better to quote the highest and lowest figure, rather than the difference.
6.2.1 The range The range has, however, found a number of specialist applications, particularly in quality control (range charts). When dealing with data presented as a frequency distribution we will not always know exactly the highest and lowest values, only the group they lie in. If the groups are open-ended (e.g. 60 and more), then any values used will merely be based on assumptions that we have made about the widths of the groups. In such cases there seems little point in quoting either the range or the extreme values.
6.2.2 The quartile deviation
If we are able to quote a half-way value, the median, then we can also quote quarter-way values, the quartiles. These are order statistics like the median and can be determined in the same way. With untabulated data or tabulated discrete data it will merely be a case of counting through the ordered data set until we are a quarter of the way through and three quarters of the way through and noting the values; this will give the first quartile and third quartile, respectively. When working with tabulated continuous data, further calculations are necessary. Consider for example the data given in Table 6.6 (see Table 5.6 for the determination of the median).
6.2.2 The quartile deviation
The lower quartile (referred to as Q1), will correspond to the value one-quarter of the way through the data, the 11th ordered value: and the upper quartile (referred to as Q3) to the value three-quarters of the way through the data, the 33rd ordered value:
The graphical method To estimate any of the order statistics graphically, we plot cumulative frequency against the value to which it refers, as shown in Figure 6.4. The value of the lower quartile is £12 and the value of the upper quartile is £25 (to an accuracy of the nearest £1 which the scale of this graph allows).
Calculation of the quartiles
We can adapt the median formula (see Section 5.1.3) as follows: where O is the order value of interest, l is the lower boundary of corresponding group, i is the width of this group, F is the cumulative frequency up to this group, and f is the frequency in this group.
Calculation of the quartiles
The lower quartile will lie in the group '£10 but under £15' and can be calculated thus: The upper quartile will lie in the group '£20 but under £30' and can be calculated thus:
Calculation of the quartiles
The quartile range is the difference between the quartiles: and the quartile deviation (or semi-interquartile range) is the average difference:
Calculation of the quartiles
As with the range, the quartile deviation may be misleading. If the majority of the data is towards the lower end of the range, for example, then the third quartile will be considerably further above the median than the first quartile is below it, and when we average the difference of the two numbers we will disguise this difference. This is likely to be the case with a country's personal income distribution. In such circumstances, it would be preferable to quote the actual values of the two quartiles, rather than the quartile deviation.
6.2.3 Percentiles The formula given in Section for an order value, O, can be used to find the value at any position in a grouped frequency distribution of continuous data. For data sets that are not skewed to one side or the other, the statistics we have calculated so far will usually be sufficient, but heavily skewed data sets will need further statistics to fully describe them. Examples would include some income distributions, wealth distributions and times taken to complete a complex task. In such cases, we may want to use the 95th percentile, i.e. the value below which 95% of the data lies. Any other value between 1 and 99 could also be calculated. An example of such a calculation is shown in Table 6.7.
6.2.3 Percentiles For this wealth distribution, the first quartile and the median are both zero. The third quartile is £ None of these statistics adequately describes the distribution. To calculate the 95th percentile, we find 95% of the total frequency, here 0.95 x = 25365 and this is the item whose value we require.
6.2.3 Percentiles It will be in the group labelled 'under £ ' which has a frequency of 800 and a width of (i.e ). Using the formula, we have:
6.2.4 Back to raw data So far this chapter has taken us from individual numbers (raw data) through ordered data to grouped data, looking at the methods used to find the measures of dispersion. The previous chapter did the same for measures of location. However, the idea of grouping the data developed when calculation had to be done by hand, or at least using slide-rules and calculators. It was the only practical method when large amounts of data were being analysed. Now we have computers and suitable software, which can deal with huge amounts of data very quickly and easily, without having to make assumptions about an even spread of data within each group, or guessing what the highest or lowest value was.
6.2.4 Back to raw data Add to this that most data starts life as individual bits of raw data, and you can see that most of the descriptive statistics we have been discussing can be found very easily, provided someone has recorded them electronically. An example using Excel is shown as Figure 6.5. An example of the output from SPSS is shown as Figure 6.6. If you are trying to describe secondary data for which you only have tabulated data, then, of course, you have to go back to the methods we have been discussing.
6.3 Relative measures of dispersion
All of the measures of dispersion described earlier in this chapter have dealt with a single set of data. In practice, it is often important to compare two or more sets of data, maybe from different areas, or data collected at different times. In Part 4 we look at formal methods of comparing the difference between sample observations, but the measures described in this section will enable some initial comparisons to be made. The advantage of using relative measures is that they do not depend on the units of measurement of the data.
6.3.1 Coefficient of variation
This measure calculates the standard deviation from a set of observations as a percentage of the arithmetic mean: Thus the higher the result, the more variability there is in the set of observations.
6.3.1 Coefficient of variation
If, for example, we collected data on personal incomes for two different years, and the results showed a coefficient of variation of 89.4% for the first year, and 94.2% for the second year, then we could say that the amount of dispersion in personal income data had increased between the two years. Even if there has been a high level of inflation between the two years, this will not affect the coefficient of variation, although it will have meant that the average and standard deviation for the second year are much higher, in absolute terms, than the first year.
6.3.2 Coefficient of skewness
Skewness of a set of data relates to the shape of the histogram which could be drawn from the data. The type of skewness present in the data can be described by just looking at the histogram, but it is also possible to calculate a measure of skewness so that different sets of data can be compared. Three basic histogram shapes are shown in Figure 6.7, and a formula for calculating skewness is shown below.
6.3.2 Coefficient of skewness
A typical example of the use of the coefficient of skewness is in the analysis of income data. If the coefficient is calculated for gross income before tax, then the coefficient gives a large positive result since the majority of income earners receive relatively low incomes, while a small proportion of income earners receive high incomes. When the coefficient is calculated for the same group of earners using their after tax income, then, although a positive result is still obtained, its size has decreased.
6.3.2 Coefficient of skewness
These results are typical of a progressive tax system, such as that in the UK. Using such calculations it is possible to show that the distribution of personal incomes in the UK has changed over time. A discussion of whether or not this change in the distribution of personal incomes is good or bad will depend on your economic and political views; the statistics highlight that the change has occurred.
6.4 Variability in sample data
We would expect the results of a survey to identify differences in opinions, income and a range of other factors. The extent of these differences can be summarized by an appropriate measure of dispersion (standard deviation, quartile deviation, range). Market researchers, in particular, seek to explain differences in attitudes and actions of distinct groups within a population. It is known, for example, that the propensity to buy frozen foods varies between different groups of people.
6.4 Variability in sample data
As a producer of frozen foods you might be particularly interested in those most likely to buy your products. Supermarkets of the same size can have very different turnover figures and a manager of a supermarket may wish to identify those factors most likely to explain the differences in turnover. A number of clustering algorithms have been developed in recent years that seek to explain differences in sample data.
6.4 Variability in sample data
As an example, consider the following algorithm or procedure that seeks to explain the differences in the selling prices of houses: 1 Calculate the mean and a measure of dispersion for all the observations in your sample. In this example we could calculate the average price and the range of prices (Figure 6.8).
6.4 Variability in sample data
It can be seen from the range that there is considerable variability in price relative to the average price. Usually the standard deviation would be preferred to the range as a measure of dispersion for this type of data. 2 Decide which factors explain most of the difference (range) in price, for example, location, house-type, number of bedrooms. If location is considered particularly important, we can divide the sample on that basis and calculate the chosen descriptive statistics (Figure 6.9).
6.4 Variability in sample data
In this case we have chosen to segment the sample by location, areas X and Y. The smaller range within the two new groups indicates that there is less variability of house prices within areas. We could have divided the sample by some other factor and compared the reduction in the range. 3 Divide the new groups and again calculate the descriptive statistics. We could divide the sample a second time on the basis of house-type (Figure 6.10). 4 The procedure can be continued in many ways with many splitting criteria. A more sophisticated version of this procedure is known as the automatic interactive detection technique.
Case 2: using measures of difference and performance
Managers are likely to meet a number of measures of difference and increasingly also various measures of performance (benchmarking, for instance, has become an important management tool, where targets are determined using the performance of the 'best' organizations on certain measures). Managers need to be able to respond to this type of information with insight and confidence.
Case 2: using measures of difference and performance
It is important for managers to clarify what these measures mean in business terms and what the underlying assumptions are. In the same way that you don't need to be an accountant to use accounting information, you don't need to be a statistician to use statistical information. Managers should look for a business understanding in the information they are given and develop responses that allow their organization to interpret and apply such information. Knowing the assumptions will reveal some of the thinking of those that devised them. Management is a process that involves a judgement as to what is appropriate and when. | 3,957 | 19,310 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2017-39 | longest | en | 0.92496 |
https://spmaddmaths.blog.onlinetuition.com.my/2020/08/6-4-equations-of-straight-lines-part-1.html | 1,723,228,275,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640768597.52/warc/CC-MAIN-20240809173246-20240809203246-00348.warc.gz | 435,306,139 | 11,888 | \
# 7.4.1 Equations of Straight Lines (Part 1)
7.4.1 Axes Intercepts and Gradient
(A) Formula for gradient:
1. Gradient of the line joining (x1, yl) and (x2, y2) is:
2.
Gradient of the line with knowing x–intercept and y–intercept
is:
3.
The gradient of the straight line joining P and Q is equal to the tangent of angle θ, where θ is the angle made by the straight line PQ and the positive direction of the x-axis.
(B) Collinear points
The gradient of a straight line is always constant i.e.the gradient of AB is equal to the gradient of BC.
Example 1:
The gradient of the line passing through point (k, 1 – k) and point (–3k, –3) is 5. Find the value of k.
Solution:
Example 2:
Based on the diagram below, find the gradient of the line.
Solution: | 216 | 758 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-33 | latest | en | 0.822198 |
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1. 1. Mapwork
2. 2. Calculations you must be able to do <ul><li>Map code </li></ul><ul><li>Distance </li></ul><ul><li>Gradient </li></ul><ul><li>Area </li></ul><ul><li>Speed/distance/time </li></ul><ul><li>Co-ordinates / grid reference </li></ul><ul><li>Bearing & direction </li></ul><ul><li>Magnetic bearing and magnetic declination </li></ul><ul><li>Vertical exaggeration & cross section </li></ul><ul><li>Word scale conversion </li></ul><ul><li>General </li></ul>
3. 3. 1) Map code <ul><li>The map code indicates the position of a map on a grid of longitude and latitude </li></ul><ul><li>South Africa is divided into a series of grid blocks by using lines of latitudes and longitudes (like when playing ‘Battle ship’) </li></ul><ul><li>E.g. 2634CB </li></ul><ul><li>26 = latitude </li></ul><ul><li>34 = longitude </li></ul><ul><li>C = Big block </li></ul><ul><li>B = Small block in the big block </li></ul>
4. 5. 26˚ 34˚ ? 2634CB 19˚ 20˚ 21˚ 22˚ 23˚ 24˚ 25˚ 26˚ 27˚ 28˚ 29˚ 30˚ 31˚ 32˚ 33˚ 34˚ 35˚ 36˚ 37˚ 38˚ 39˚ 40˚ 41˚
5. 6. Each grid block can be divided into lines of 60 seconds latitudes and longitudes Divide the grid block in two half's (30’ latitude and 30’ longitude) Each quarter now become a ‘Big block’ numbered A to D 24˚ 25˚ 34˚ 35˚ Divide each big block in half’s (15’ and 45’ latitude and longitude) Number each ‘small block’ in the big block A to D Finally find now big block C and small block B 30’ 30’ 15’ 15’ 45’ 45’ A B C D A B C D A B C D A B C D A B C D
6. 7. 2) Distance <ul><li>Always measure the map distance in millimeters!!!!! </li></ul><ul><li>NOW you can convert the map distance into the real distance by using the following formula </li></ul>Distance = Map distance x scale 1000( m) OR 1 000 000( km )
7. 8. 124mm 1: 50 000
8. 9. Distance = map distance x scale 1000 000 What is the real distance in km ? = 124mm x 50 000 1000 000 = 6,2km Remember, if you had to give the answer in meters you must divide by a 1000 !!!
9. 10. 3) Gradient <ul><li>Gradient is the steepness of a slope </li></ul><ul><li>You can easily see on a topographic map whether the slope is steep or gentle </li></ul><ul><li>When the contour lines </li></ul><ul><li>are close together the slope is </li></ul><ul><li>steep and when contour lines </li></ul><ul><li>are far apart the slope is gentle </li></ul>
10. 11. <ul><li>Gradient is shown in a fraction format e.g. 1:8 </li></ul><ul><li>1 represents the vertical distance (1m change in height) </li></ul><ul><li>8 represents the horizontal distance (8m change with distance) </li></ul>1m 8m
11. 12. Gradient = Height difference (m) . Distance (m) Calculate the gradient between X and Y.
12. 13. 72mm X Y
13. 14. <ul><li>Height difference between X and Y </li></ul><ul><li>X = 2140m </li></ul><ul><li>Y = 2320m </li></ul><ul><li>DH = 2320m </li></ul><ul><li> - 2140m </li></ul><ul><li>180m </li></ul>To calculate the distance in meters use the normal formula for distance (D) D = map distance x scale 1000 = 72mm x 50 000 1000 = 3600m
14. 15. Gradient = Height difference (m) Distance (m) = 180m 3600m = 180m ÷ 180m 3600m ÷ 180m = 1 . 20 = 1: 20 1m 20m
15. 16. 4) Area <ul><li>To calculate the area of a given place you need to do TWO distance calculations!!! </li></ul><ul><li>Convert each side in to meters or kilometers before you apply it to the area formula </li></ul>Calculate the map area in km² Click here for topographic map
16. 17. Area = Length x Breadth Length Breadth L = 155mm x 50 000 1000 000 = 7,75km B = 155mm x 50 000 1000 000 = 7,75km Area = L ength x B readth = 7.75km x 7.75km = 60.063km²
17. 18. 5) Speed / distance / time Distance Speed Time
18. 19. <ul><li>If you walk from V to W at an average speed of 5km/h. How long will it take to cover the distance? </li></ul><ul><li>Fist, calculate the distance in km between the tow points </li></ul><ul><li>Secondly, apply it to the formula </li></ul>D = 93mm x 50 000 1000 000 = 4,65km Time = Distance Speed = 0.93 x 60 = 0,93h = 4,65km 5km/h = 55 minutes 48seconds Click here for topographic map
19. 20. 6) Co-ordinates (Grid reference) <ul><li>Always give the latitude first followed by the longitude </li></ul><ul><li>Co-ordinates are given in degrees, minutes and seconds </li></ul><ul><li>E.g. 23˚38’41”S and 31˚16’52”E </li></ul>
20. 21. <ul><li>Remember, between each degree (°) it is divided into 60 minutes and between each minute (‘) it is divided into 60 seconds (“) </li></ul><ul><li>Always start with the latitude and then the longitude </li></ul><ul><li>Write down the latitude degrees, minutes and seconds South </li></ul><ul><li>Then write down the longitude degrees, minutes and seconds East </li></ul><ul><li>E.g. 29°17’33” South & 24°35’22” East </li></ul>
21. 22. 29°S 24°E 30’ 45’ 15’ 30’ 45’ 15’
22. 23. 29°26' 24°38' Latitude: Measure (mm) from 29°26’ to the point (A) Measure (mm) the distance from 29°26’ to the next latitude (B) Divide A by B x 60 Longitude: Measure (mm) from 24°38’ to the point (C) Measure (mm) the distance from 24°38’ to the next latitude (D) Divide C by D x 60 C D B A
23. 24. 7) Bearing & Direction <ul><li>Bearing refers to the degrease between two points. </li></ul><ul><li>0˚ is always at the ‘top’ or true north </li></ul><ul><li>Always measure the degrease in a clockwise direction from 0˚ </li></ul><ul><li>Direction refers to the compass direction </li></ul>
24. 25. A B Bearing: 240˚ Direction: SW
25. 26. Magnetic declination <ul><li>The difference between true north and magnetic north </li></ul><ul><li>Magnetic declination changes from year to year </li></ul><ul><li>Change can be to the east (angle becomes smaller) or to the west (angle becomes greater </li></ul><ul><li>Change to east you subtract </li></ul><ul><li>Change to west you add </li></ul>
26. 28. 138° Step 1: Calculate the difference in years 2010 – 2002 = 8 years Step 2: Calculate the magnetic change 8years x 7’ W = 56’W Step 3: Determine the total change 24°34’W + 56’W = 25°30’W Step 4: Calculate the magnetic bearing True bearing + magnetic declination 138° + 25°30’ = 163°30’
27. 29. CROSS SECTION <ul><li>A cross section is a side view of a feature on the land surface </li></ul>Vertical scale: 1cm represents 20m Horizontal scale 1: 50 000
28. 30. trig beacon 7 . Draw a cross section from the settlement TO 1362,1 1340m 1320m 1300m 1280m 7 1360m Always make sure that your starting point is on the LEFT hand side!! Otherwise you will draw a mirror image. 1360m
29. 31. 1280 1362,1 1360 1340 1320 1300 1280 Place a piece of paper on the line joining the settlement and the trig beacon Mark the settlement , trig beacon and every contour line that touches your paper Find the height of each contour line and write it down below the appropriate marking. Don’t forget the heights of the settlement and trig beacon 1362,1 1340m 1320m 1300m 1280m 7
30. 32. Now you are going to draw the cross section Draw the horizontal axis (slightly longer than the line between the settlement and the trig beacon) Transfer the markings from the piece of paper onto the horizontal axis On the vertical axis make a mark every 20cm’s Indicate the heights on the vertical axis, starting with the lowest altitude (1280m) from horizontal axis next to the first marking Each marking’s height increase by 20m upwards till you get to the highest altitude from the horizontal axis 1280m 1300m 1320m 1340m 1360m 1360m 1380m Indicate each intersection point between the vertical and horizontal axis Finally, connect each point
31. 33. 1362,1 1340m 1320m 1300m 1280m 7
32. 34. A cross section from the settlement to trig beacon 7 Horizontal scale 1: 50 000 Vertical scale 1cm r.p. 20m Footpath Trig beacon 7 Settlement
33. 35. Vertical exaggeration <ul><li>Tells you by how much the vertical axis (height) has been stretched out (distorted) </li></ul><ul><li>The ‘bigger’ the vertical scale, the more stretched your cross section </li></ul><ul><li>E.g. 1cm rp 20cm will be more stretched as 5mm rp 20cm </li></ul>If you stretch this object out (upwards) will it still show the real shape?
34. 36. <ul><li>Lets see the difference </li></ul>VS = 1:2000 and HS 1:50 000 VE = VS HS = 1/2000 1/50000 = 1 . X 50 000 2000 1 = 25 times Everything on this cross section appears 25 times higher. Poor representation! VS = 1:4000 and HS 1:50 000 VE = VS HS = 1/4000 1/50000 = 1 . X 50 000 4000 1 = 12,5 times Everything on this cross section appears 12,5 times higher. Better representation! 1cm represents 20m = 1:2000 5mm represents 20m = 1:4000
35. 37. 10) Converting a word scale into a fraction scale <ul><li>When will you use this? </li></ul><ul><li>A cross section can have a scale for the vertical axis </li></ul><ul><li>E.g. A vertical scale of 1cm represents 20m </li></ul><ul><li>You need to make sure that all the units are the same. </li></ul><ul><li>Convert the ‘larger’ unit into the smaller unit </li></ul><ul><li>Therefore, 20m will be converted into cm’s </li></ul>
36. 38. 1cm represents 20m 1cm represents 20m x 100 1cm represents 2000cm 1:2000 5mm represents 20m 5mm represents 20m x 1000 5mm represents 20000mm 5 5 1mm represents 4000mm 1:4000 Units must be the same Nee to be 1 There is 1000mm in 1m There is 100cm in 1 meter
37. 39. Convex and concave slopes Convex slope Concave slope
38. 40. Landforms due to inclined strata Cuesta???? Homoclinal ridge???? Hogsback???? CUESTA WHY?? Dip slope scarp slope <25˚
39. 41. Cuesta and its contour lines
40. 42. Back to AREA V W Back to SPEED
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Easy calculations for mapwork.
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0 | 3,518 | 10,251 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2021-39 | latest | en | 0.51074 |
https://thunderbitz.com/coin-change-2-leetcode-solution/ | 1,656,262,346,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103271763.15/warc/CC-MAIN-20220626161834-20220626191834-00725.warc.gz | 624,250,540 | 32,426 | # 518. Coin Change 2 LeetCode Solution | Easy Approach
Share:
Coin Change 2 You are given an integer array `coins` representing coins of different denominations and an integer `amount` representing a total amount of money.
Return the number of combinations that make up that amount. If that amount of money cannot be made up by any combination of the coins, return `0`.
You may assume that you have an infinite number of each kind of coin.
The answer is guaranteed to fit into a signed 32-bit integer.
Example 1:
```Input: amount = 5, coins = [1,2,5]
Output: 4
Explanation: there are four ways to make up the amount:
5=5
5=2+2+1
5=2+1+1+1
5=1+1+1+1+1
```
Example 2:
```Input: amount = 3, coins = [2]
Output: 0
Explanation: the amount of 3 cannot be made up just with coins of 2.
```
Example 3:
```Input: amount = 10, coins = [10]
Output: 1
```
Constraints:
• `1 <= coins.length <= 300`
• `1 <= coins[i] <= 5000`
• All the values of `coins` are unique.
• `0 <= amount <= 5000`
Time: O(n)
Space: O(n)
### C++
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### Python
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#### Watch Tutorial
Checkout more Solutions here | 372 | 1,193 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2022-27 | latest | en | 0.689791 |
https://findthefactors.com/tag/tetris/ | 1,674,866,618,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499468.22/warc/CC-MAIN-20230127231443-20230128021443-00225.warc.gz | 277,721,083 | 14,603 | # 1274 Imagining With Sain Smart Jr.’s Tetris Puzzle
My granddaughter recently received a Sain Smart Jr. Tetris Puzzle for her birthday. I got to watch as she and her younger sister had lots of fun creating pictures using the colorful pieces. Here are some of their creations:
As they played they experimented and learned what they liked and what they didn’t like. In the process, they learned some mathematics and may not have even realized it.
Yeah, they could also make all the Tetris puzzle pieces fit in the frame. It seems to help if you save some of the five wood grained pieces for last.
From what I could tell, this was a very fun and educational toy.
Now here I’ll share some information about the number 1274:
• 1274 is a composite number.
• Prime factorization: 1274 = 2 × 7 × 7 × 13, which can be written 1274 = 2 × 7² × 13
• The exponents in the prime factorization are 1, 2, and 1. Adding one to each and multiplying we get (1 + 1)(2 + 1)(1 + 1) = 2 × 3 × 2 = 12. Therefore 1274 has exactly 12 factors.
• Factors of 1274: 1, 2, 7, 13, 14, 26, 49, 91, 98, 182, 637, 1274
• Factor pairs: 1274 = 1 × 1274, 2 × 637, 7 × 182, 13 × 98, 14 × 91, or 26 × 49,
• Taking the factor pair with the largest square number factor, we get √1274 = (√49)(√26) = 7√26 ≈ 35.69314
35² + 7² = 1274
1274 is the hypotenuse of a Pythagorean triple:
490-1176-1274 which is (5-12-13) times 98.
It is also 2(35)( 7), 35² – 7², 35² + 7² | 473 | 1,429 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2023-06 | latest | en | 0.942741 |
http://mathhelpforum.com/discrete-math/110548-permutation-problems.html | 1,511,570,265,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934809160.77/warc/CC-MAIN-20171124234011-20171125014011-00546.warc.gz | 191,643,125 | 13,204 | 1. ## Permutation problems
,
I'm having a hard time solving these problems. How do you solve them?
1) A family consisting of an old man, 6 adults and 4 children, is to be seated in a row for dinner. The children which to occupy two seats at each end and the old man refuses to have a child on either side of him. In how many ways can the seating arrangement be made for the dinner?
2) Ten guests are to be seated in a row. Three of them are to be seated togather. Of the remaining seven, two do not wish to sit next to each other. Find the number of possible arrangements.
3) The letters of the word ZENITH are written in all possible orders. If all the words are written in a dictionary, what is the rank of the word ZENITH?
2. Originally Posted by saberteeth
,
I'm having a hard time solving these problems. How do you solve them?
1) A family consisting of an old man, 6 adults and 4 children, is to be seated in a row for dinner. The children which to occupy two seats at each end and the old man refuses to have a child on either side of him. In how many ways can the seating arrangement be made for the dinner?
2) Ten guests are to be seated in a row. Three of them are to be seated togather. Of the remaining seven, two do not wish to sit next to each other. Find the number of possible arrangements.
3) The letters of the word ZENITH are written in all possible orders. If all the words are written in a dictionary, what is the rank of the word ZENITH?
You need to show that you have done some work on these.
3. Originally Posted by Plato
You need to show that you have done some work on these.
Thanks for your reply Plato. I have spend hours trying to solve them but, i couldn't even figure out the starting point. This is why i am unable to show my workings
4. Originally Posted by saberteeth
I have spend hours trying to solve them but, i couldn't even figure out the starting point. This is why i am unable to show my workings.
I find that incredible. How can anyone spend hours not knowing where to start?
Maybe you need more help then we can give.
Originally Posted by saberteeth
3) The letters of the word ZENITH are written in all possible orders. If all the words are written in a dictionary, what is the rank of the word ZENITH?
Can you at least tell us how many ways “ZENITH” can be rearranged?
What is the last arrangement in a dictionary order?
5. Originally Posted by Plato
I find that incredible. How can anyone spend hours not knowing where to start?
Maybe you need more help then we can give.
Can you at least tell us how many ways “ZENITH” can be rearranged?
What is the last arrangement in a dictionary order?
That would be 6!
6. Originally Posted by saberteeth
That would be 6!
Yes it is. Now answer these.
What is the first arrangement in a dictionary order?
What is the last arrangement in a dictionary order?
7. Originally Posted by Plato
Yes it is. Now answer these.
What is the first arrangement in a dictionary order?
What is the last arrangement in a dictionary order?
The first arrangement is EHINTZ and the last one is ZTNIHE
8. Originally Posted by saberteeth
The first arrangement is EHINTZ and the last one is ZTNIHE
Now there are 720 in the list.
ZEHINT is #696.
What number is ZENITH?
You can list the other 23.
9. Originally Posted by Plato
Now there are 720 in the list.
ZEHINT is #696.
What number is ZENITH?
You can list the other 23.
This is the part that i am unable to understand. How did you know the order for ZEHINT is #696? Did you subtract 4! from all the possible arrangements? If yes, then what is the central idea behind finding the number of a particular arrangement?
All this may sound stupid but, this is why i need your help as i am self-studying college math..
10. Well I will try to help you, even though I think self-study of mathematics is not a good plan.
There are 720 ways to arrange the letters “ZENITH”.
There are 600 of those arrangements that do not begin with a Z.
Therefore, in a dictionary ordering the arrangement “ZEHINT” is #601
There are 24 arrangements that begin with a ZE.
There are 12 arrangements that begin with a ZEH or ZEI.
So the first that begins with a ZEN #613.
There are 6 arrangements that begin with a ZEN.
Can you finish?
11. Originally Posted by Plato
Well I will try to help you, even though I think self-study of mathematics is not a good plan.
There are 720 ways to arrange the letters “ZENITH”.
There are 600 of those arrangements that do not begin with a Z.
Therefore, in a dictionary ordering the arrangement “ZEHINT” is #601
There are 24 arrangements that begin with a ZE.
There are 12 arrangements that begin with a ZEH or ZEI.
So the first that begins with a ZEN #613.
There are 6 arrangements that begin with a ZEN.
Can you finish?
ZENHIT # 613
ZENHTI # 614
ZENIHT # 615
ZENITH # 616
The only thing i did not understand is the statement "There are 12 arrangements that begin with a ZEH or ZEI" How did you get 12? Can you please elaborate?
12. Ok, this took me few hours and i figured it out with the help of your steps. Thanks!
,
,
,
,
,
,
,
,
,
,
,
# the letters of the word ZENITH are written in all possible orders. How many words are possible if all these words are written as in a dictionery ?
Click on a term to search for related topics. | 1,302 | 5,294 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2017-47 | longest | en | 0.9723 |
2017forms5.blogspot.com | 1,539,718,668,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510866.52/warc/CC-MAIN-20181016180631-20181016202131-00117.warc.gz | 2,375,846 | 17,061 | ## Area:
Calculating area:
To work out the area of a square or rectangle, multiply its height by its width.
If the height and width are in cm, the area is shown in cm². If the height and width are in m, the area is shown in m².
A square with sides of 5 m has an area of 25 m², because 5 × 5 = 25.
Area 1
Area 2
Area 3
Area 4
## Perimetre:
Calculating perimeter:
The perimeter is the distance all the way around the outside of a 2D shape.
To work out the perimeter, add up the lengths of all the sides
The perimeter of this shape is 10 + 10 + 6 + 6 = 32
Perimeter 1
Perimeter 2
Perimeter 3
### Units of measurement (converting length units)
Basic conversion rules:
1) If you need to convert from a LARGER unit to a SMALLER unit, MULTIPLY.
For example: 5km into metres: 5 x 1000= 5.000 metres
2) If you need to convert from a SMALLER unit to a LARGER unit, DIVIDE.
For example: 10cm into metres: 10 / 100= 0,1 metres
## miércoles, 29 de agosto de 2018
### Lines, Polygons and Polyhedrons
Lines, Polygons and Polyhedrons
Lines, segments, rays, parallel
Lines and intersections
Intersections
Identifying polygons
Matching Polygons
Polyhedrons
Geometry facts
## lunes, 30 de julio de 2018
### Generators
Generating Electricity
Electricity is the energy we used the most nowadays. But what is the process behind the generation of this type of energy?
### Scientists and their contribution: Energy
Scientists and their contribution: Energy
Here you have a file with brief biographies of individuals who have made significant contributions to energy and science.
Lehigh University
Allient Energy Kids
## jueves, 5 de julio de 2018
### Linear Perspective
Linear Perspective
Would you like your drawings look more realistic? Have you got problems when illustrating landscapes? Look at the following video and learn some concepts about perspective.
Divisibility
## martes, 26 de junio de 2018
### Sharing the Planet: Energy
Sharing the Planet: Energy.
Our third unit of inquiry is called Energy. Its central idea is "Humans use, control and transform energy to make their lives easier".
The key concepts to work are Form, Change, Responsibility.
Related Concepts: Energy transformation, energy conservation, efficiency, power
Kids & Energy
Easy Science for Kids
EIA
Space Environment
Dr Binocs
Energy Turtlediary
Non-renewable energies
Renewable energies | 604 | 2,385 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-43 | longest | en | 0.822323 |
http://hitchhikersgui.de/Center_of_population | 1,537,743,918,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267159820.67/warc/CC-MAIN-20180923212605-20180923233005-00515.warc.gz | 114,516,407 | 13,123 | # Center of population
The point on earth closest to everyone in the world on average is in the north of South Asia, with a mean distance of 5,000 kilometers (3,000 mi). Its antipodal point is correspondingly the farthest point from everyone on earth, and is located in the South Pacific near Easter Island, with a mean distance of 15,000 kilometers (9,300 mi). The data used by this figure is lumped at the country level, and is therefore lacking precision.
In demographics, the centre of population (or population center) of a region is a geographical point that describes a centrepoint of the region's population. There are several different ways of defining such a "centre point", leading to different geographical locations; these are often confused.[1]
## Definitions
Three commonly used (but different) center points are:
1. the mean centre, also known as the centroid or centre of gravity;
2. the median centre, which is the intersection of the median longitude and median latitude;
3. the geometric median, also known as Weber point, Fermat–Weber point, or point of minimum aggregate travel.
A further complication is caused by the curved shape of the Earth. Different centre points are obtained depending on whether the centre is computed in three-dimensional space, or restricted to the curved surface, or computed using a flat map projection.
### Mean centre
The mean center, or centroid, is the point on which a rigid, weightless map would balance perfectly, if the population members are represented as points of equal mass.
Mathematically, the centroid is the point to which the population has the smallest possible sum of squared distances. It is easily found by taking the arithmetic mean of each coordinate. If defined in the three-dimensional space, the centroid of points on the Earth's surface is actually inside the Earth. This point could then be projected back to the surface. Alternatively, one could define the centroid directly on a flat map projection; this is, for example, the definition that the US Census Bureau uses.
Contrary to a common misconception, the centroid does not minimize the average distance to the population. That property belongs to the geometric median.
### Median centre
The median centre is the intersection of two perpendicular lines, each of which divides the population into two equal halves.[2] Typically these two lines are chosen to be a parallel (a line of latitude) and a meridian (a line of longitude). In that case, this center is easily found by taking separately the medians of the population's latitude and longitude coordinates. Tukey called this the cross median.[3]
### Geometric median
The geometric median is the point to which the population has the smallest possible sum of distances (or equivalently, the smallest average distance). Because of this property, it is also known as the point of minimum aggregate travel. Unfortunately, there is no direct closed-form expression for the geometric median; it is typically computed using iterative methods.
## Determination
In practical computation, decisions are also made on the granularity (coarseness) of the population data, depending on population density patterns or other factors. For instance, the centre of population of all the cities in a country may be different from the center of population of all the states (or provinces, or other subdivisions) in the same country. Different methods may yield different results.
Practical uses for finding the center of population include locating possible sites for forward capitals, such as Brasília, Astana or Austin. Practical selection of a new site for a capital is a complex problem that depends also on population density patterns and transportation networks.
## World
It is important to use a method that does not depend on a two-dimensional projection when dealing with the entire world. As described in INED (Institut national d'études démographiques),[4] the solution methodology deals only with the globe. As a result, the answer is independent of which map projection is used or where it is centered. As described above, the exact location of the center of population will depend on both the granularity of the population data used, and the distance metric. With geodesic distances as the metric, and a granularity of 1,000 kilometers (600 mi), meaning that two population centers within 1000 km of each other are treated as part of a larger common population center of intermediate location, the world's center of population is found to lie somewhere north of South Asia[5] with an average distance of 5,200 kilometers (3,200 mi) to all humans.[4] The data used in the reference support this result to a precision of only a few hundred kilometers, hence the exact location is not known. Another analysis utilising city level population data found that the world's center of population is located close to Almaty, Kazakhstan.[6]
## By country
### Australia
Australia's population centroid is in central New South Wales. By 1996 it had moved only a little to the north-west since 1911.[7]
In Canada, a 1986 study placed the point of minimum aggregate travel just north of Toronto in the city of Richmond Hill, and moving westward at a rate of approximately 2 metres per day.[8]
### China
China's population centroid has wandered within southern Henan from 1952 to 2005. Incidentally, the two end point dates are remarkably close to each other.[9] China also plots its economic centroid or center of economy/GDP, which has also wandered, and is generally located at the eastern Henan borders.
### Estonia
The center of population of Estonia was on the northwestern shore of Lake Võrtsjärv in 1913 and moved an average of 6 km northwest with every decade until the 1970s. The higher immigration rates during the late Soviet occupation to mostly Tallinn and Northeastern Estonia resulted the center of population moving faster towards north and continuing urbanization has seen it move northwest towards Tallinn since the 1990s. The center of population according to the 2011 census was in Jüri, just 6 km southeast from the border of Tallinn.[10]
### Finland
In Finland, the point of minimum aggregate travel is located in the former municipality of Hauho.[11] It is moving slightly to the south-west-west every year because people are moving out of the periphery areas of northern and eastern Finland.
### Germany
In Germany, the centroid of the population is located in Spangenberg, Hesse, close to Kassel.[12]
### Great Britain
The centre of population in Great Britain did not move much in the 20th century. In 1901, it was in Rodsley, Derbyshire and in 1911 in Longford. In 1971 it was at Newhall, South Derbyshire and in 2000, it was in Appleby Parva, Leicestershire.[13][14][15][need quotation to verify]
### Ireland
The centre of population of the entire island of Ireland is located near Kilcock, County Kildare. This is significantly further east than the Geographical centre of Ireland, reflecting the disproportionately large cities of the east of the island (Belfast and Dublin).[16] The centre of population of the Republic of Ireland is located southwest of Edenderry, County Offaly.[17]
### Japan
The centroid of population of Japan is in Gifu Prefecture, almost directly north of Nagoya city, and has been moving east-southeast for the past few decades.[18] More recently, the only large regions in Japan with significant population growth have been in Greater Nagoya and Greater Tokyo.
### New Zealand
New Zealand's median centre of population over time
In June 2008, New Zealand's median centre of population was located near Taharoa, around 100 km (65 mi) southwest of Hamilton on the North Island's west coast.[19] In 1900 it was near Nelson and has been moving steadily north ever since.[20]
### Sweden
The demographical center of Sweden (using the median center definition) is Hjortkvarn in Hallsberg Municipality, Örebro county. Between the 1989 and 2007 census the point moved a few kilometres to the south, due to a decreasing population in northern Sweden and immigration to the south.[21]
### Russia
The center of population in the Russian Federation is calculated by A. K. Gogolev to be at 56°34′N 53°30′E / 56.567°N 53.500°E, 42 km (26 mi) south of Izhevsk.[22]
### United States
The mean center of the United States population (using the centroid definition) has been calculated for each U.S. Census since 1790. Over the last two centuries, it has progressed westward and, since 1930, southwesterly, reflecting population drift. For example, in 2010, the mean center was located near Plato, Missouri, in the south-central part of the state, whereas, in 1790, it was in Kent County, Maryland, 47 miles (76 km) east-northeast of the then-new federal capital, Washington, D.C.
## Sources
• Bellone F. and Cunningham R. (1993). "All Roads Lead to... Laxton, Digby and Longford." Statistics Canada 1991 Census Short Articles Series.
## References
1. ^ Kumler, Mark P.; Goodchild, Michael F. (1992). "The population center of Canada – Just north of Toronto?!?". In Janelle, Donald G. Geographical snapshots of North America: commemorating the 27th Congress of the International Geographical Union and Assembly. pp. 275–279.
2. ^ "Centers of Population Computation for the United States, 1950-2010" (PDF). Washington, DC: Geography Division, U.S. Census Bureau. March 2011.
3. ^ Tukey, John (1977). Exploratory Data Analysis. Addison-Wesley. p. 668. ISBN 9780201076165.
4. ^ a b Claude Grasland and Malika Madelin (May 2001). "The unequal distribution of population and wealth in the world" (PDF). Population Et SociétéS. Institut national d'études démographiques. 368: 1–4. ISSN 0184-7783.
5. ^ exact phrase in the paper is "at the crossroads between China, India, Pakistan and Tajikistan"
6. ^ "Center of World Population". City Extremes. 2017. Retrieved August 21, 2017.
7. ^ "Figure 15: Shifts in the Australian Population Centroid*, 1911–1996". Parliament of Australia Parliamentary Library. Archived from the original on 19 August 2000. Retrieved 7 January 2009.
8. ^ "The Population Center of Canada – Just North of Toronto?!?" (PDF). Retrieved 21 April 2012.
9. ^ http://sourcedb.igsnrr.cas.cn/zw/lw/201007/P020100706529106697457.pdf
10. ^ Haav, Mihkel (2010) - "Eesti dünaamika 1913-1999"
11. ^ Uusirauma.fi Kaupunkilehti Uusi Rauma 03.08.2009 Päivän kysymys? Missä Rauman keskipiste? (in Finnish)
12. ^ Dradio.de Archived 24 October 2007 at the Wayback Machine. (in German)
13. ^ "News Item:". University of Leeds. Retrieved 25 November 2007.
14. ^ "Population Centre". Appleby Magna & Appleby Parva. Archived from the original on 23 November 2007. Retrieved 25 November 2007.
15. ^ "Coffee Break: The movable Midlands; ANSWERS TO CORRESPONDENTS". The Daily Mail. London. 7 February 2002. p. 64.
16. ^ http://imgur.com/q439KW0
17. ^ http://imgur.com/prTTQOK
18. ^ "Our Country's Center of Population (我が国の人口重心)". Stat.go.jp. Retrieved 21 April 2012.
19. ^ "Subnational Population Estimates: At 30 June 2008 -- Commentary". Statistics New Zealand. Retrieved 11 November 2014.
20. ^ http://nzbooks.org.nz/tag/bridget-williams-books/
21. ^ "Sweden's demographic centre, SCB.se, 2008-03-18". Scb.se. 18 March 2008. Archived from the original on 29 March 2012. Retrieved 21 April 2012.
22. ^ Сайт "Встарь, или Как жили люди"
23. ^ https://www.ptt.cc/man/Geography/D1F0/DD1E/M.1105123514.A.427.html | 2,713 | 11,457 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2018-39 | latest | en | 0.935283 |
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Lesson Transcript
Instructor: Eric Istre
Eric has taught high school mathematics for more than 20 years and has a master's degree in educational administration.
This lesson will explain the law of syllogism and provide several examples showing when it can be used to reach a valid conclusion and when it cannot.
## Definition of the Law of Syllogism
In recent years, a satellite television provider made some humorous ads in which a person with cable television eventually had some negative consequences. For example, one went like this:
When you pay too much for cable, you throw things.
When you throw things, people think you have anger issues.
When people think you have anger issues, your schedule clears up.
When your schedule clears up, you grow a scraggly beard.
When you grow a scraggly beard, you start taking in stray animals.
When you start taking in stray animals, you can't stop taking in stray animals.
So, what we end up with is: When you pay too much for cable, you can't stop taking in stray animals.
This sure seems like a strange conclusion, doesn't it? Reaching a conclusion is not always easy. Sometimes there is no correct answer. Even when there is one, it can be hard to know what it is. At the end of the lesson, I'll explain why there is a problem with the conclusion of these ads. But for now, let's consider conclusions in mathematics. Thankfully, when it comes to mathematics, the conclusion is not always so uncertain.
The law of syllogism, also called reasoning by transitivity, is a valid argument form of deductive reasoning that follows a set pattern. It is similar to the transitive property of equality, which reads:
if a = b and b = c then, a = c.
There are also three parts involved in the law of syllogism, and each of these parts is called a conditional statement. A conditional statement has a hypothesis, which follows after the word if, and it has a conclusion, which follows after the word then. A letter is used to represent each phrase of the conditional statement.
Let me introduce the pattern, and then we can look at some examples.
Statement 1: If p, then q.
Statement 2: If q, then r.
Statement 3: If p, then r.
Statements 1 and 2 are called the premises of the argument. If they are true, then statement 3 must be the valid conclusion.
## Examples
Now that we know what syllogism is, let's test our knowledge with some examples.
First, an example with a valid conclusion:
Statement 1: If it continues to rain (p), then the soccer field will become wet and muddy (q). This becomes if p, then q.
Statement 2: If the soccer field becomes wet and muddy (q), then the game will be canceled (r). This becomes if q, then r.
Statement 3: If it continues to rain (p), then the game will be canceled (r). This final statement is the conclusion, and becomes if p, then r.
This follows the pattern for the law of syllogism; therefore, it is a valid conclusion.
Now, let's try an example with an invalid conclusion:
Statement 1: If the bank robber steals the money (p), then the sheriff will track him down (q). This is If p, then q.
Statement 2: If the sheriff tracks him down (q), then the bank robber will be arrested (r). This is If q, then r.
Statement 3: If the bank robber steals the money (p), then the bank robber will be rich (s). This is If p, then s.
Instead of building on statement 2, this final statement simply offers another possibility of statement 1. This does not follow the law of syllogism pattern, so statement 3 is an invalid conclusion.
Now, we'll do look at an example that we'll call valid conclusion possible:
Statement 1: If the truck runs over some nails (p), then a tire will go flat (q). If p, then q.
Statement 2: If a tire goes flat (q), then the deliveries will not be made on time (r). If q, then r.
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# Multiplication
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Multiplication is the binary mathematical operation of scaling one number or quantity by another (multiplying). It is one of the four basic operations in elementary arithmetic (with addition, subtraction and division). The result of this operation is called product and the multiplied numbers are called factors, or the multiplier and the multiplicand.
The product mn of a positive whole number m times another quantity n agrees with the result of successively adding n to itself a total of m times. For example, 2 multiplied by 3 (often said as "3 times 2") gives the same result as 2 taken 3 times, i.e., of addding 3 copies of 2 : 2 × 3 = 2 + 2 + 2. Similarly, multiplied by 2 is the same value obtained by adding the number pi to itself. Multiplication does not give the result of a repeated addition when the multiplier is not a whole number. We can calculate the product
using the rules for manipulating radicals and products, but we cannot view this computation as a repeatedly adding . Instead of being viewed as a special type of repeated addition, multiplication should be viewed as a basic numerical operation, separate from and as important as addition.
Multiplication can be visualised as counting objects arranged in a rectangle (for natural numbers) or as finding the area of a rectangle whose sides have given lengths (for numbers generally). The inverse of multiplication is division: as 2 times 3 equals to 6, so 6 divided by 3 equals to 2.
Multiplication can be defined outside of the context of real numbers. There are natural multiplication operations on the complex numbers, matrices, tensors, sequences, and functions. The common thread that allows all of these operations to be called "multiplication" is the existence of an additional "addition" operation and a collection of common axioms satisfied by the addition and multiplication operations in each context. The most general context in which a multiplication operation exists, encompassing all of the above examples, is that of the abstract ring encountered in abstract algebra.
## Properties
Commutativity
Multiplication of numbers is commutative, meaning a × b = b × a. Multiplication is not always commutative in other contexts. For example, multiplication of matrices is in general not a commutative operation:
but reversing the order of multiplication, we have
a different result.
Associativity
Multiplication is associative, meaning a × (b × c) = (a × b) × c.
Distributivity
Multiplication is distributive over addition, meaning a × (x + y) = a × x + a × y.
## Products of sequences
### Capital pi notation
The product of a sequence can be written using capital Greek letter Π (Pi). Unicode position U+220F (∏) contains a symbol for the product of a sequence, distinct from U+03A0 (Π), the letter. The meaning of this notation is given by:
where i is an index of multiplication, m is its lower bound and n is its upper bound. Example:
If m = n, the value of the product just equals to xm. If m > n, the product is the empty product, with the value 1. | 822 | 3,751 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 7, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2018-47 | latest | en | 0.936531 |
https://math.stackexchange.com/questions/351856/why-does-the-series-sum-limits-n-2-infty-frac-cosn-pi-3n-converge?noredirect=1 | 1,569,155,835,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514575513.97/warc/CC-MAIN-20190922114839-20190922140839-00143.warc.gz | 577,913,717 | 33,757 | # Why does the series $\sum\limits_{n=2}^\infty\frac{\cos(n\pi/3)}{n}$ converge?
Why does this series $$\sum\limits_{n=2}^\infty\frac{\cos(n\pi/3)}{n}$$ converge? Can't you use a limit comparison with $1/n$?
• Can you please edit the title so it is clear what you are asking? Is it $\sum_{n=1}^{\infty} \cos \dfrac{n\pi}{3n}$ – Aryabhata Apr 5 '13 at 6:04
• Careful, your first statement is not correct. The cosine term will oscillate as $n$ gets large and never approach a single value. – Jared Apr 5 '13 at 6:04
• @Jared you are right, will correct it. Is it an alternating series? – Billy Thompson Apr 5 '13 at 6:07
• Yes, this sounds like the way to go. It's not a strictly alternating series, but the sign change in cosine is what causes convergence. You may even be able to evaluate this explicitly using telescoping series, because we know the values of $\cos(\frac{n\pi}{3})$ for all integral $n$. – Jared Apr 5 '13 at 6:08
• @Jared is there any other way to evaluate it? – Billy Thompson Apr 5 '13 at 6:10
First of all your conclusion is wrong since $\lim_{n \to \infty} \cos(n \pi/3)$ doesn't exist.
The convergence of $$\sum_{n=1}^N \dfrac{\cos(n\pi/3)}{n}$$ can be concluded based on Abel partial summation (The result is termed as generalized alternating test or Dirichlet test). We will prove the generalized statement first.
Consider the sum $S_N = \displaystyle \sum_{n=1}^N a(n)b(n)$. Let $A(n) = \displaystyle \sum_{n=1}^N a(n)$. If $b(n) \downarrow 0$ and $A(n)$ is bounded, then the series $\displaystyle \sum_{n=1}^{\infty} a(n)b(n)$ converges.
First note that from Abel summation, we have that \begin{align*}\sum_{n=1}^N a(n) b(n) &= \sum_{n=1}^N b(n)(A(n)-A(n-1))\\&= \sum_{n=1}^{N} b(n) A(n) - \sum_{n=1}^N b(n)A(n-1)\\ &= \sum_{n=1}^{N} b(n) A(n) - \sum_{n=0}^{N-1} b(n+1)A(n) \\&= b(N) A(N) - b(1)A(0) + \sum_{n=1}^{N-1} A(n) (b(n)-b(n+1))\end{align*} Now if $A(n)$ is bounded i.e. $\vert A(n) \vert \leq M$ and $b(n)$ is decreasing, then we have that $$\sum_{n=1}^{N-1} \left \vert A(n) \right \vert (b(n)-b(n+1)) \leq \sum_{n=1}^{N-1} M (b(n)-b(n+1))\\ = M (b(1) - b(N)) \leq Mb(1)$$ Hence, we have that $\displaystyle \sum_{n=1}^{N-1} \left \vert A(n) \right \vert (b(n)-b(n+1))$ converges and hence $$\displaystyle \sum_{n=1}^{N-1} A(n) (b(n)-b(n+1))$$ converges absolutely. Now since $$\sum_{n=1}^N a(n) b(n) = b(N) A(N) + \sum_{n=1}^{N-1} A(n) (b(n)-b(n+1))$$ we have that $\displaystyle \sum_{n=1}^N a(n)b(n)$ converges.
In your case, $a(n) = \cos(n \pi/3)$. Hence, $$A(N) = \displaystyle \sum_{n=1}^N a(n) = - \dfrac12 - \cos\left(\dfrac{\pi}3(N+2)\right)$$which is clearly bounded.
Also, $b(n) = \dfrac1{n}$ is a monotone decreasing sequence converging to $0$.
Hence, we have that $$\sum_{n=1}^N \dfrac{\cos(n\pi/3)}{n}$$ converges.
Look at some of my earlier answers for similar questions.
For what real numbers $a$ does the series $\sum \frac{\sin(ka)}{\log(k)}$ converge or diverge?
Give a demonstration that $\sum\limits_{n=1}^\infty\frac{\sin(n)}{n}$ converges.
If the partial sums of a $a_n$ are bounded, then $\sum{}_{n=1}^\infty a_n e^{-nt}$ converges for all $t > 0$
If you are interested in evaluating the series, here is a way out. We have for $\vert z \vert \leq 1$ and $z \neq 1$, $$\sum_{n=1}^{\infty} \dfrac{z^n}n = - \log(1-z)$$ Setting $z = e^{i \pi/3}$, we get that $$\sum_{n=1}^{\infty} \dfrac{e^{in \pi/3}}n = - \log(1-e^{i \pi/3})$$ Hence, \begin{align} \sum_{n=1}^{\infty} \dfrac{\cos(n \pi/3)}n & = \text{Real part of}\left(\sum_{n=1}^{\infty} \dfrac{e^{in \pi/3}}n \right)\\ & = \text{Real part of} \left(- \log(1-e^{i \pi/3}) \right)\\ & = - \log(\vert 1-e^{i \pi/3} \vert) = 0 \end{align} Hence, $$\sum_{n=2}^{\infty} \dfrac{\cos(n \pi/3)}n = - \dfrac{\cos(\pi/3)}1 = - \dfrac12$$
• I think it is time to write a general answer for generalized alternating test and close tons of similar questions as abstract duplicates. – user17762 Apr 5 '13 at 6:13
• I support the motion. – Did Apr 5 '13 at 7:10
• I'm a little confused about something. On the one hand, you show that $\sum a_n b_n$ should converge absolutely. But $\sum \cos (n \pi 3) n^{-1}$ does not converge absolutely, as the nth term is at least $\frac{1}{2n}$ in absolute value. So I feel I must be missing something? – davidlowryduda Apr 5 '13 at 8:14
• @mixedmath Yes, you are absolutely right. $\sum a_n b_n$ doesn't converge absolutely. I have changed it. The proof remains un affected. What we have is $\sum A(n)(b(n) - b(n+1))$ converges absolutely and this is what we want. This doesn't mean that $\sum a(n)b(n)$ converges absolutely. Thanks for pointing this out. – user17762 Apr 5 '13 at 15:16
• Awesome. Thanks, and great writeup! – davidlowryduda Apr 5 '13 at 16:03
Note that $$\cos(n\pi/3) = 1/2, \ -1/2, \ -1, \ -1/2, \ 1/2, \ 1, \ 1/2, \ -1/2, \ -1, \ \cdots$$ so your series is just 3 alternating (and convergent) series inter-weaved. Exercise: Prove that if $\sum a_n, \sum b_n$ are both convergent, then the sequence $$a_1, a_1+b_1, a_1+b_1+a_2, a_1+b_1+a_2+b_2, \cdots$$ is convergent. Applying that twice proves your series converges. | 1,930 | 5,090 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2019-39 | latest | en | 0.735644 |
https://math.answers.com/Q/A_square_plus_b_equals_7_and_a_plus_b_square_equals_11_what_are_the_values_of_a_and_b | 1,712,925,295,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296815919.75/warc/CC-MAIN-20240412101354-20240412131354-00121.warc.gz | 340,958,702 | 47,494 | 0
# A square plus b equals 7 and a plus b square equals 11 what are the values of a and b?
Updated: 12/17/2022
Wiki User
14y ago
A^2 + b = 7 A + B^2 =11 -------------- A=2 and B=34(2^2) + 3 = 7 2 + 9(3^2) = 11
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14y ago
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Q: A square plus b equals 7 and a plus b square equals 11 what are the values of a and b?
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### 12 plus 10 plus 11 equals?
12 + 10 + 11 equals 33
### If the sides of a square are lengthened by 6cm the area becomes 121cm2 find the length of a side of the original square?
5cm because the square root of 121 is 11 and 5 plus 6 equals 11.
it equals 28
### Does 1 plus 1 equals 11 I am 6 And i want to know?
No it equals 2 10 plus 1 equals 11
### What are the values of a and b given that y plus 4x equals 11 is the perpendicular bisector equation of the line joining a 2 to 6 b?
Their values work out as: a = -2 and b = 4
8+11= 8+11
### Squareroot 12 plus squareroot 75 plus squareroot 48?
Two times the square root of three plus five times the square root of three plus four times the square root of three equals eleven times the square root of three.
-3
### How do you add 2x plus 3y equals 18 in 5x-y equals 11?
How do you add 2x plus 3y equals 18 in 5x-y equals 11
### What is the answer to C over 3 plus 4 equals 11?
The answer is 21 because if 21 is divided by 3 it equals 7 plus 4 equals 11. | 468 | 1,419 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-18 | latest | en | 0.784308 |
https://www.gradesaver.com/textbooks/math/algebra/elementary-and-intermediate-algebra-concepts-and-applications-6th-edition/chapter-4-polynomials-connecting-the-concepts-exercises-page-243/8 | 1,531,849,571,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589757.30/warc/CC-MAIN-20180717164437-20180717184437-00531.warc.gz | 888,330,188 | 13,093 | ## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$\dfrac{1}{c^8}$
Use the rule $a^{-m} =\dfrac{1}{a^m}$ to obtain: $c^{-8}=\dfrac{1}{c^8}$ | 63 | 168 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2018-30 | latest | en | 0.657631 |
https://sudokuprimer.com/solving-containers.php | 1,591,522,540,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348526471.98/warc/CC-MAIN-20200607075929-20200607105929-00591.warc.gz | 531,809,152 | 3,707 | Solving Sudoku Containers
# Tool: Solving containers
Looking at an entire row, column or box
Next technique: solving sections >>
## Introduction
Sudoku puzzles cannot have the same number twice in any row, column or box (container). You can take advantage of this by using deduction to determine (1) which numbers are missing, and (2) where the missing numbers go.
## Details
When you see a row, column or box that is only missing one number you know what that number is and you can fill it in. This is the most basic use of this technique. You only need one container to illustrate this, as in this row example.
This row is only missing one number and it is a 3. You can fill it in.
You can do the same thing with multiple numbers missing in a container. But as the number of empty cells increases the ability to find numbers decreases, and you need numbers in other adjacent containers to help you know which numbers are missing. Here is a row with two numbers missing.
It is pretty easy to see that this row is missing a 1 and a 3. The cells in column one and four are the empty ones. So far no rocket science here. In order to fill these two numbers in you must know where at least one of them goes. If there is a 1 or a 3 in either column one or four (or in box one or two) you would know how to complete the row.
Now let's look at a formula.
## Formula
This is kinda like math but don't get nervous. It's not too hard to grasp. To solve any number of empty cells in a container where n is the number of empty cells, you need to have n-1 known numbers outside of that container to solve one of the empty cells. Stick with me - here's an example.
Look at row one in the above image. It has four empty cells. The missing numbers are 1, 3, 5 and 7. So n is 4 and n-1 is 3. That means we need n-1 (three) known numbers to solve one of the n (four) empty cells in row one. Once one cell is solved n becomes 3 and n-1 becomes 2. And so on until we have only one cell left to solve. At that point n is 1 and n-1 is 0. Or in other words we need to find zero numbers to solve the last cell (because it's the only number left for that row).
Look at column seven. It has the numbers 1, 5, 7 and 9 in it and they are all outside of row one. 1, 5 and 7 happen to be three (n-1) of the four (n) numbers we're missing. So that tells us row one column seven is a 3. Good job!
After the 3 is filled in we have three (n) empty cells left so we need two (n-1) known numbers to solve the next empty cell. You'll notice that column one has two of the three needed numbers in it (a 1 and a 5), and therefore a 7 goes in to row one column one.
Each time you fill in a number n and n-1 go down one number. With this knowledge you can now finish solving row one.
This video illustrates the concept of solving containers.
## Skill Levels
This technique can be used for any level of difficulty. If you're only missing one or two numbers it can be easy to use.
There are times when you only have one or two known numbers in a given container. This can help you find numbers. But it is tricky to do if there aren't many numbers yet. This is a more advanced tool when there are fewer numbers in a container.
Next technique: solving sections >>
## More Sudoku Information
Sudoku Techniques
Sudoku Patterns
Awesome Sudoku T-Shirts
Sudoku Myths
Free Blank Puzzle Grids | 826 | 3,370 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2020-24 | latest | en | 0.947423 |
https://www.examfocus.com/psat-problem-solving-practice-test-4/ | 1,721,446,999,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514981.25/warc/CC-MAIN-20240720021925-20240720051925-00896.warc.gz | 660,080,243 | 15,806 | # PSAT Problem Solving Practice Test 4
Question – 1
1. Which of the following is the cube of 3/25? Indicate the correct option.
• A. 0.0144
• B. 0.001728
• C. 0.0001728
• D. 0.00001728
• E. 0.00144
• Answer Explanation:(3/25)^3= 0.12^3=0.001728Option B is correct.
Question – 2
2. What part of the share of Amanda of 1/15 does Peter have if Peter has 3/40? Indicate the correct option.
• A. 120/3
• B. 3/120
• C. 1/45
• D. 9/8
• E. 8/9
• Answer Explanation:Let the required number be x1/15*x=3/40x=3/40*15= 45/40 = 9/8Peter has 9/8 part of Amanda’s share.Option D is true.
Question – 3
3. The average of n numbers is n. If each number is multiplied by n, which of the following will be the new average? Indicate the correct option.(Note:[x^2=x*x] )
• A. 2n
• B. n
• C. n^3
• D. n^2
• E. n/2
• Answer Explanation:Average = Sum of observations/number of observationsn = Sum/nSum = n*nEach observation is multiplied by n. Hence, the sum of the observations is equal to the old sum multiplied by n.New sum = Old sum*n = n*n*nAverage = n*n*n/n=n^2Option D is true.
Question – 4
4. One apple has the same nutrients as 4 biscuits and 3 biscuits have the same nutritional value as 4 bananas. What is the ratio of the number of apples to the number of bananas having the same nutrients? Indicate the correct option.
• A. 3:16
• B. 16:3
• C. 4:3
• D. 3:4
• E. 12:1
• Answer Explanation:1 apple = 4 biscuits3 biscuits = 4 bananas1 apple = 4 biscuits = 4*4/3 bananas = 16/3 bananas1 apple = 16/3 bananas3 apples = 16 bananasRequired ratio = 3:16Option A is true.
Question – 5
5. Peter bought oranges at \$5 per dozen. He sold half of his purchase at twice his investment and half of his purchase was rotten and could not be sold. How much was his profit percent? Indicate the correct option.
• A. 100%
• B. 50%
• C. 25%
• D. 10%
• E. 200%
• Answer Explanation:Let his purchase be of 10 dozen.His investment = \$5*10=\$50He sold 5 dozen at \$50*2=\$100He could not sell 5 dozen.Profit percent = (100-50)/50*100= 100%Option A is true.
Question – 6
6. A two-digit number is six times the sum of its digits and six less than three times the product of its digits. Which of the following is the number? Indicate the correct option.
• A. 54
• B. 45
• C. 36
• D. 63
• E. 72
• Answer Explanation:Let the two digit number be 10x+y, where x and y are the digits in the units and tens place.10x+y=6*(x+y) and 10x+y=3xy-610x+y=6x+6y10x-6x=6y-y4x=5yPut x=5y/4 in the second equation10(5y/4)+y=3(5y/4)y-650y/4+y=15y^2/4-650y+4y=15y^2-2415y^2-54y-24=05y^2-18y-8=05y^2-20y+2y-8=05y(y-4)+2(y-4)=0y=-2/5, y=4y=4 and x= 5*4/4=5The number is 54Option A is correct.
Question – 7
7. Some children are made to stand in a long row. Each child is given sweets equal to his position in the row. They have to walk up to some guests in a sequence and give their sweets to them. How many children shall have to give their sweets if at least 250 sweets have to be given to the guests? Indicate the correct option.
• A. 17
• B. 22
• C. 18
• D. 16
• E. 20
• Answer Explanation:The number of sweets given to the guests forms an AP.The first term of the AP, a, is 1 and the sum of n terms is at least 250 and the common difference d is 1.Sn=(n/2)[2a+(n-1)d]250<=(n/2)[2*1+(n-1)1]250<=(n/2)(n+1)250*2<=n(n+1)n(n+1)>=50022*23=506>=500The least possible value of n is 22.Hence, 22 children will have to give their sweets if at least 250 sweets have to be given.Option B is true.
Question – 8
8. What is the measure of each exterior angle of an equilateral triangle? Indicate the correct option.
• A. 90 degrees
• B. 120 degrees
• C. 80 degrees
• D. 180 degrees
• E. 300 degrees
• Answer Explanation:Each interior angle of an equilateral triangle is 60 degrees.Each exterior angle is 180-60=120 degrees.Option B is true.
Question – 9
9. Which of the following points lies in the third quadrant? Indicate the correct option.
• A. Abscissa is negative and ordinate is positive
• B. Abscissa is positive and ordinate is positive
• C. Abscissa is negative and ordinate is negative
• D. Abscissa is positive and ordinate is negative
• E. Abscissa is negative and ordinate is zero
• Answer Explanation:In the third quadrant, abscissa is negative and ordinate is negative.Only option C is true.
Question – 10
10. When a wire circumscribing a circle of area 616 sq.cm. is bent to form a square, how much lesser area does it enclose? Indicate the correct option.
• A. 14 sq.cm.
• B. 616 sq.cm.
• C. 484 sq.cm.
• D. 132 sq.cm.
• E. 88 sq.cm.
• Answer Explanation:Area of circle = 616 sq.cm.Radius of circle = Sqrt[616/pi]= sqrt[616*7/22]= sqrt[196]Radius = 14cmCircumference = 2*pi*radius= 2*22/7*14= 88Circumference of square = 88 cmSide of square = 88/4=22Area of square = 22*22= 484 sq.cm.Difference in area = 616 – 484 = 132 sq.cm.Option D is true.
Question – 11
11. Glory and Glen complete a piece of work in 8 days working together. In how many days would Glen complete the work if Glory is twice as efficient as Glen is? Indicate the correct option.
• A. 12
• B. 6
• C. 24
• D. 48
• E. 18
• Answer Explanation:Let the time taken by Glory be X and that taken by Glen be Y to complete the work.Glory is twice as fast as Glen and hence she takes half as much time as Glen does.Hence, 2X=YWork done by them in one day1/X+1/Y=1/81/(Y/2)+1/Y=1/82/Y+1/Y=1/83/Y=1/8Y=3*8=24Glen would complete it in 24 days.Option C is correct.
Question – 12
12. Robin’s takes 4 hours to water his garden working alone and Rubin takes 6 hours. They both take turns of one hour each to water the garden, starting with Robin. How much time will they take to water the garden? Indicate the correct option.
• A. 5 hours 20 minutes
• B. 4 hours 40 minutes
• C. 5 hours
• D. 4 hours
• E. 4 hours 50 minutes
• Answer Explanation:Garden watered in one hour by Robin = 1/4Garden watered in one hour by Rubin = 1/6Garden watered by them working together in two hours = 1/4+1/6= 5/12Garden watered in 4 hours = 2*5/12=10/12=5/6Remaining garden = 1-5/6=1/6Time taken by Robin to water 1/6 garden = 1/6*4=4/6 hours = 4/6*60=40 minutesOption B is correct.
Question – 13
13. Find the next number in the series 8, 24, 12, 36, 18, 54, 27, …. Indicate the correct option.
• A. 88
• B. 83
• C. 64
• D. 81
• E. 45
• Answer Explanation:8, 24, 12, 36, 18, 54, 278*3 = 2424/2 = 1212*3 = 3636/2 = 1818*3 = 5454/2 = 2727*3 = 81Option D is true.
Question – 14
14. If x+1/x=6, then which of the following is correct? Indicate the correct option.(Note:[x^2=x*x] )
• A. x^2+1/x^2=36
• B. x^2+1/x^2=34
• C. x^2+1/x^2=38
• D. x^2+1/x^2=40
• E. x^2+1/x^2=24
• Answer Explanation:x+1/x=6x^2+1/x^2+2*x*1/x=36x^2+1/x^2+2=36x^2+1/x^2=36-2=34Option B is correct..
Question – 15
15. 8th January 1993 was a Sunday. On which day does 8th January 1998 fall? Indicate the correct option.
• A. Sunday
• B. Monday
• C. Thursday
• D. Friday
• E. Saturday | 2,420 | 6,950 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-30 | latest | en | 0.871355 |
http://ecomputernotes.com/computer-graphics/simple-line-drawing-method/write-short-note-on-bresenhams-line-drawing-algorithm | 1,586,298,914,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371806302.78/warc/CC-MAIN-20200407214925-20200408005425-00215.warc.gz | 53,468,746 | 9,137 | # Bresenham’s Line Drawing Algorithm.
by Dinesh Thakur Category: Line Drawing Method
You know that DDA algorithm is an incremental scan conversion method which performs calculations at each step using the results from the preceding step. Here we are going to discover an accurate and efficient raster line generating algorithm, the Bresenham's line-drawing algorithm.
This algorithm was developed by Jack E. Bresenham in 1962 at IBM. As stated above, in this lecture, I will explain how to draw lines using the Bresenham's line-drawingalgorithm. And then show you the complete line drawing function. Before we begin on this topic, a revision of the concepts developed earlier like scan conversion methods and rasterization may be helpful. Once we finish this aspect, we will proceed towards exposition of items listed in the synopsis.
In particular, we will emphasize the following
(a) The working of Bresenham’s algorithm.
(b) Implementation of the Bresenham’s algorithm.
The working of Bresenham’s algorithm The following is an explanation of how the Bresenham's line-drawing algorithm works, rather than exact implementation.
Let’s take a look at this image. One thing to note here is that it is impossible to draw the true line that we want because of the pixel spacing. Putting it in other words, there's no enough precision for drawing true lines on a PC monitor especially when dealing with low resolutions. The Bresenham's line-drawing algorithm is based on drawing an approximation of the true line.
The true line is indicated in bright color, and its approximation is indicated in black pixels. In this example the starting point of the line is located exactly at 0, 0 and the ending point of the line is located exactly at 9, 6. Now let discuss the way in which this algorithm works. First it decides which axis is the major axis and which is the minor axis.
The major axis is longer than the minor axis. On this picture illustrated above the major axis is the X axis. Each iteration progresses the current value of the major axis (starting from the original position), by exactly one pixel. Then it decides which pixel on the minor axis is appropriate for the current pixel of the major axis. Now how can you approximate the right pixel on the minor axis that matches the pixel on the major axis?
That’s what Bresenham's line-drawing algorithm is all about. And it does so by checking which pixel's center is closer to the true line. Now you take a closer look at the picture. The center of each pixel is marked with a dot. The algorithm takes the coordinates of that dot and compares it to the true line. If the span from the center of the pixel to the true line is less or equal to 0.5, the pixel is drawn at that location. That span is more generally known as the error term.
You might think of using floating variables but you can see that the whole algorithm is done in straight integer math with no multiplication or division in the main loops(no fixed point math either). Now how is it possible? Basically, during each iteration through the main drawing loop the error term is tossed around to identify the right pixel as close as possible to the true line. Let's consider these two deltas between the length and height of the line: dx = x1 - x0; dy = y1 - y0;
This is a matter of precision and since we're working with integers you will need to scale the deltas by 2 generating two new values: dx2 = dx*2; dy2 = dy*2; These are the values that will be used to change the error term. Why do you scale the deltas? That’s because the error term must be first initialized to 0.5 and that cannot be done using an integer. Finally the scaled values must be subtracted by either dx or dy (the original, non-scaled delta values) depending on what the major axis is (either x or y).
The implementation of Bresenham’s algorithm
The function given below handles all lines and implements the complete Bresenham's algorithm.
function line(x0, x1, y0, y1)
boolean steep := abs(y1 - y0) > abs(x1 - x0)
if steep then
swap(x0, y0)
swap(x1, y1)
if x0 > x1 then
swap(x0, x1)
swap(y0, y1)
int deltax := x1 - x0
int deltay := abs(y1 - y0)
real error := 0
real deltaerr := deltay / deltax
int y := y0
if y0 < y1 then ystep := 1 else ystep := -1
for x from x0 to x1
if steep then plot(y,x) else plot(x,y)
error := error + deltaerr
if error ? 0.5
y := y + ystep
error := error - 1.0
Note:-To draw lines with a steeper slope, we take advantage of the fact that a steep line can be reflected across the line y=x to obtain a line with a small slope. The effect is to switch the x and y variables throughout, including switching the parameters to plot.
Related Articles on Computer Graphics
Dinesh Thakur holds an B.C.A, MCSE, MCDBA, CCNA, CCNP, A+, SCJP certifications. Dinesh authors the hugely popular blog. Where he writes how-to guides around Computer fundamental , computer software, Computer programming, and web apps. For any type of query or something that you think is missing, please feel free to Contact us.
Related Articles | 1,184 | 5,067 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2020-16 | latest | en | 0.927808 |
https://turningtooneanother.net/2020/01/05/what-is-number-statement/ | 1,685,384,411,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644907.31/warc/CC-MAIN-20230529173312-20230529203312-00142.warc.gz | 692,961,314 | 8,926 | # What is number statement?
## What is number statement?
In mathematics education, a number sentence is an equation or inequality expressed using numbers and mathematical symbols. The term is used in primary level mathematics teaching in the US, Canada, UK, Australia, New Zealand and South Africa.
## What is an example of number sentence?
Examples of number sentences include: 32 + 57 =? 5 x 6 = 10 x? They will usually comprise of addition, subtraction, multiplication or division – or a combination of all four!
What is a math statement?
Brielfy a mathematical statement is a sentence which is either true or false. For example “The square root of 4 is 5″ is a mathematical statement (which is, of course, false).
What is a true number statement?
Truth Values of a Number Sentence: A number sentence that is an equation is said to be true if both numerical expressions evaluate to the same number; it is said to be false otherwise. For example, 3 < 4, 6 + 8 > 15 > 12, and (15 + 3)2 < 1000 – 32 are all true number sentences, while the sentence 9 > 3(4) is false.
### How do you use the word number in a sentence?
Number sentence example
1. You have my number , don’t you?
2. Soon they came into the main road where a number of the king’s men were waiting.
3. I told you not to give my number to anyone.
4. It has a large number of landlocked nations without ports to access the international markets, both for imports and exports.
### Does or mean multiply?
When you do an Or operation you do a join of all your result sets which can be seen as an addition. The Or and And should be seen as operations on your (sub)result set(s) and not as multiplications or additions.
What are the types of numbers?
What does it look like?
Type of Number Example
Prime Number P=2,3,5,7,11,13,17,…
Composite Number 4,6,8,9,10,12,…
Whole Numbers W=0,1,2,3,4,…
Integers Z=…,−3,−2,−1,0,1,2,3,… | 489 | 1,895 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2023-23 | latest | en | 0.925012 |
https://epolylearning.com/mcq-question-answers/Aptitude/Time-and-Work | 1,725,929,261,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651164.37/warc/CC-MAIN-20240909233606-20240910023606-00252.warc.gz | 209,141,561 | 10,749 | # Time and Work MCQs
Welcome to our comprehensive collection of Multiple Choice Questions (MCQs) on Time and Work, a fundamental topic in the field of Aptitude. Whether you're preparing for competitive exams, honing your problem-solving skills, or simply looking to enhance your abilities in this field, our Time and Work MCQs are designed to help you grasp the core concepts and excel in solving problems.
In this section, you'll find a wide range of Time and Work mcq questions that explore various aspects of Time and Work problems. Each MCQ is crafted to challenge your understanding of Time and Work principles, enabling you to refine your problem-solving techniques. Whether you're a student aiming to ace Aptitude tests, a job seeker preparing for interviews, or someone simply interested in sharpening their skills, our Time and Work MCQs are your pathway to success in mastering this essential Aptitude topic.
Note: Each of the following question comes with multiple answer choices. Select the most appropriate option and test your understanding of Time and Work. You can click on an option to test your knowledge before viewing the solution for a MCQ. Happy learning!
So, are you ready to put your Time and Work knowledge to the test? Let's get started with our carefully curated MCQs!
### Time and Work MCQs | Page 1 of 3
Q1.
A can do a work in 15 days and B in 20 days. If they work on it together for 4 days, then the fraction of the work that is left is :
Q2.
A can lay railway track between two given stations in 16 days and B can do the same job in 12 days. With help of C, they did the job in 4 days only. Then, C alone can do the job in:
d.
10
Q3.
A, B and C can do a piece of work in 20, 30 and 60 days respectively. In how many days can A do the work if he is assisted by B and C on every third day?
Q4.
A is thrice as good as workman as B and therefore is able to finish a job in 60 days less than B. Working together, they can do it in:
Q5.
A alone can do a piece of work in 6 days and B alone in 8 days. A and B undertook to do it for Rs. 3200. With the help of C, they completed the work in 3 days. How much is to be paid to C?
Q6.
If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be: | 584 | 2,347 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-38 | latest | en | 0.946239 |
http://forum.allaboutcircuits.com/threads/help-3-phase-induction-motor-problem.103238/ | 1,484,833,437,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280668.34/warc/CC-MAIN-20170116095120-00477-ip-10-171-10-70.ec2.internal.warc.gz | 106,379,732 | 16,656 | Help: 3 Phase Induction Motor Problem
Discussion in 'Homework Help' started by ParagM, Nov 6, 2014.
1. ParagM Thread Starter New Member
Nov 6, 2014
2
1
I will be helpful if someone solve this problem,
A 4 pole, 1440 rpm, 3 phase induction motor is operated from a per phase voltage of 240 V, 50 Hz and driving a constant toque load. Calculate the following at f=25 Hz, airgap flux = constant = 4.8.
i. Supply voltage per phase
ii. Slip
iii. Slip frequency
iv. Percentage rotor loss
I have calculated synchronous speed Ns=120f1/p =1500
slip s = Ns-N / Ns = 0.04 & slip freq fr = s f1 = 2
segecious likes this.
2. WBahn Moderator
Mar 31, 2012
18,085
4,917
I'm just guessing here, but I suspect your instructor believes it would be helpful if YOU at least attempted to solve YOUR homework.
Sep 20, 2014
52
1
Bahn if you don't want to help him then why are you trolling his topic? He has obviously attempted the problem.
Apr 5, 2008
15,799
2,385
Hello,
When he made an attemp, he should show it here.
That way we can see where he needs assistance.
Bertus
Sep 20, 2014
52
1
Bertus, did you read this? I know you are smarter than me.
I have calculated synchronous speed Ns=120f1/p =1500
slip s = Ns-N / Ns = 0.04 & slip freq fr = s f1 = 2
Sounds a lot like an attempt.
Sep 20, 2014
52
1
Paragm how did you get Synchronous speed?
7. WBahn Moderator
Mar 31, 2012
18,085
4,917
It's an attempt at half of the problem. For the other half, he explicitly asked to be given the answer without any further effort.
Sep 20, 2014
52
1
Bahn you are a smart man and I'm sure a great person, but I think you are a little too quick to judge people. You do not know how long Parag spent trying to figure out what to do next for this 3 phase power problem. When I was in engineering school, it would sometimes take me hours upon hours to solve problems and I'd still get the wrong answer. When I read your posts here, I interpret them as judgmental, almost as if you are trying to scare people away from engineering because they aren't trying hard enough. Sometimes people get stuck and they need help even if its just a small inch to set them in the right direction....Also I noticed you joined in 2012, when I was in college and asked questions on this board in 2008 I never ran into a person like you.
9. WBahn Moderator
Mar 31, 2012
18,085
4,917
You're right, I have no idea how long Parag spent trying to work the other half because he has given no indication or evidence of trying at all. If he has spent hours trying, then it shouldn't be hard to post SOMETHING to show for it instead of just asking for someone to supply the answer to him. If you've read very much of my stuff on these boards at all, then you will have seen numerous examples of when I have been more than willing to work with someone for page after page after page to walk them through their struggles, going as far back to first principles as necessary. But I do ask that the person make (and show) their efforts to work their own problems so that we can guide them in their struggles based upon what they are doing right and what they are doing wrong.
As for you never running into a person like me in 2008.... so? What's your point?
Sep 20, 2014
52
1
When I was here in 2008 nobody would throw things back in my face like i see you doing with this guy, and other guys here like in the laplace topic. I see you coming down on people hard for deriving extremely complex equations, not everyone is a genius like you are.
11. panic mode Senior Member
Oct 10, 2011
1,328
305
WBahn is doing the best thing any student in trouble could hope for, you just don't see it...
12. WBahn Moderator
Mar 31, 2012
18,085
4,917
I'm not going to come down hard on someone for deriving an extremely complex equation -- you need to review that thread if you think that is what is happening. I will come down on someone for refusing, after several attempts, to start asking if their results make any sense compared to easily seen limiting cases. I will also come down on people for failing to use units, particularly when not doing so has already resulted in them either making a mistake or failing to catch a mistake that tracking units would have caught. I just guess that's baggage that comes from seeing someone die because they couldn't be bothered to track their units (and having come within about three feet of being that person instead).
Sep 20, 2014
52
1
I apologize Bahn.
14. ParagM Thread Starter New Member
Nov 6, 2014
2
1
Thanks vu2nan & others for devoting your valuable time!
This question is from university paper & I have given problem statement as it is.
I think, the question is incomplete.
Still, any kind of help will be appreciated. | 1,228 | 4,716 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2017-04 | latest | en | 0.957596 |
https://physics-network.org/what-is-linear-magnification-and-its-formula/ | 1,685,731,407,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648850.88/warc/CC-MAIN-20230602172755-20230602202755-00507.warc.gz | 514,051,014 | 23,990 | # What is linear magnification and its formula?
The linear magnification of a mirror can be defined as the ratio of the height of the image to the height of the object. m=h′h. Where, m is the magnification, h is the height of the object and h′is the height of the image formed.
## What is linear magnification Physics 11?
Linear magnification produced by a spherical mirror is the ratio of the size of the image formed by the mirror so that of the size of the object.
## What is magnification and linear magnification?
10.3. Magnification (linear) is the numerical ratio between the image size and the object size. This term is also found in the lens equation (Equations [10.1] and [10.2], left sides) and is therefore directly linked with image formation: [10.5]
## What is magnification in physics definition?
Magnification is defined as the ratio of the height of image to the height of object. m=hiho.
## What is the SI unit of linear magnification?
Magnification is a ratio of lengths, hence it has no units.
m=U−V=U−V.
## What is linear magnification Class 12?
Linear magnification produced by a spherical mirror is defined as the ratio of the size of the image formed by the mirror to the size of the object. It is denoted by m.
## What is the importance of linear magnification?
The main importance is of linear magnification. Is that a negative value of linear magnification will denote that the image is inverted and a positive value will denote that it doesn’t erect image. Also longitude and magnification will denote the factor by which an image increases in size.
## What is linear magnification of concave mirror?
Solution : The linear magnification of a concave mirror is `m =(h_(2))/(h_(1))= “size of image”/”size of object” = v/u`
Clearly, `m gt 1`, when image is enlarged, and `mlt1`, when image is smaller in size than the object.
## What are linear and longitudinal magnification?
Linear magnification is the ratio of the image length to the object length measured in the plane that is perpendicular to the optical axis. In longitudinal magnification, the image length is measured along the optical axis.
## What are the 4 types of magnification?
There are four ways of creating magnification: Increase the size of the object Decrease the viewing distance Transverse magnification Telescopic magnification.
## What is the difference between linear magnification and angular magnification?
The two basic types of magnification are: Linear magnification: It refers to the ratio of image length to object length. Angular magnification: It refers to the ratio of the tangents of the angles subtended by an object and its image from a given point.
## What is magnification formula?
Magnification (m) = h / h’ Here, h is the height of the object and h’ is the height of the object. Besides, it can also be related to the object distance and image distance. So, it can be written as: m = -v / u.
## What is magnification and its types?
Magnification is the process of enlarging the apparent size, not physical size, of something. This enlargement is quantified by a calculated number also called “magnification”. When this number is less than one, it refers to a reduction in size, sometimes called minification or de-magnification.
## What is a magnification definition?
1 : the act of magnifying. 2a : the state of being magnified. b : the apparent enlargement of an object by an optical instrument that is the ratio of the dimensions of an image formed by the instrument to the corresponding dimensions of the object. — called also power.
## What is the value of linear magnification of a plane mirror?
The magnification produced by a plane mirror is +1.
## What is the SI unit of lens?
The S.I unit of power of lens is dioptre denoted by D. 1 D=1 m1=1 m−1. Thus, 1 Dioptre is the power of the lens having focal length of 1 m.
## What is the unit of refractive index?
The refractive index has no unit.
## What is mirror formula?
The relation between focal length of mirror, distance of the object and distance of the image is known as mirror formula. It is given by. u1+v1=f1.
## What is the linear magnification of a spherical mirror?
The linear magnification for a spherical mirror is the ratio of the size of the image to the size of the object and is denoted by m .
## What is a lens in physics class 12?
A lens is a uniform transparent medium bounded between two spherical or one spherical and one plane surface. Convex Lens. A lens which is thinner at edges and thicker at middle is called a convex or converging lens. Concave Lens.
## What is longitudinal magnification?
Longitudinal magnification denotes the factor by which an image increases in size, as measured along the optical axis. Angular magnification is equal to the ratio of the tangents of the angles subtended by an object and its image when measured from a given point in…
## What is linear magnification of plane mirror concave and convex mirror?
Concave mirror can produce image of magnification -1 and convex mirror always made diminished image so its magnification is always less than 1 and plane mirror always produce image of same size so it will produce image if magnification of 1.
## Is the linear magnification of a lens constant?
On the other hand, if your object is larger, then the image becomes larger (higher). M=HiHo is the formula for magnification. Hence, these two factors are independent of the magnification: the magnification of a lens is constant. But linear magnification is not constant for the lens. | 1,247 | 5,566 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2023-23 | latest | en | 0.929944 |
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Which of the following is closest in value to 9^9 – 9^2 ?
A.9^9
B.9^8
C.9^7
D.9^6
E.9^5
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24 Feb 2009, 11:35
Its A-9^9
see ,
9^9-9^2=9^2((9^7) - 1)
~9^9
because 9^7 is too large compared to 1
janet1511 wrote:
Which of the following is closest in value to 9^9 – 9^2 ?
A.9^9
B.9^8
C.9^7
D.9^6
E.9^5
--
edited a small typo
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Last edited by nitya34 on 25 Feb 2009, 22:09, edited 2 times in total.
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24 Feb 2009, 13:15
Agree with A
subtracting 81 makes it closer to 9 ^ 9 among the answer choices
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24 Feb 2009, 17:42
nitya34 wrote:
Its A-9^9
see ,
9^9-9^2=9^2((9^4.5) - 1)
~9^9
because 9^2 is too small compared to 9^9
Right approach but wrong equation
9^9-9^2 = 9^2(9^7 - 1)
~ 9^2*9^7 = 9^9
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24 Feb 2009, 18:47
agreed with the answer
9^2(9^7-1)
approx = 9^9
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25 Feb 2009, 22:03
thanks
I corrected it
xyz21 wrote:
nitya34 wrote:
Its A-9^9
see ,
9^9-9^2=9^2((9^4.5) - 1)
~9^9
because 9^2 is too small compared to 9^9
Right approach but wrong equation
9^9-9^2 = 9^2(9^7 - 1)
~ 9^2*9^7 = 9^9
_________________
http://gmatclub.com/forum/math-polygons-87336.html
http://gmatclub.com/forum/competition-for-the-best-gmat-error-log-template-86232.html
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25 Feb 2009, 22:46
1
This post received
KUDOS
nitya34 wrote:
thanks
I corrected it
xyz21 wrote:
nitya34 wrote:
Its A-9^9
see ,
9^9-9^2=9^2((9^4.5) - 1)
~9^9
because 9^2 is too small compared to 9^9
Right approach but wrong equation
9^9-9^2 = 9^2(9^7 - 1)
~ 9^2*9^7 = 9^9
happens to all of us
Re: PS [#permalink] 25 Feb 2009, 22:46
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,250 | 3,530 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2018-09 | latest | en | 0.824061 |
https://stats.stackexchange.com/questions/6024/effectively-fitting-this-kind-of-model-y-c-1-x-3-x-4-c-2-x-1-x-9 | 1,716,550,642,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058709.9/warc/CC-MAIN-20240524091115-20240524121115-00191.warc.gz | 483,901,553 | 42,731 | Effectively fitting this kind of model: $y = c_1 (x_3 - x_4) + c_2 (x_1 - x_9)$
Given observations of $\{y, x_1, x_2, \cdots, x_n\}$, we can always do a linear regression and get all the coefficients $\{c_i\}$ for the model
$$y = c_0 + c_1 x_1 + \cdots + c_n x_n.$$
However, this may not be the best answer. Let me explain it.
When we are doing a regression, we have estimates $\{d_i\}$ for the standard deviations of the coefficients $\{c_i\}$ and it may turn out, in my particular problem, that most of these coefficients have low $t$-values.
On the other hand, in my problem, I already know the underlying model is more like
$$y = \Sigma_i c_i (x_{m_i} - x_{n_i})$$
such as
$$y = c_1 (x_3 - x_4) + c_2 (x_1 - x_9)$$
and the problem is I don't know $\{m_i\}$ and $\{n_i\}$.
That is, in my strange case, if I already know it is of the form $y = c_1 (x_3 - x_4) + c_2 (x_1 - x_9)$, then when I find $c_1$ and $c_2$, I will find them to have high $t$-values.
Nevertheless I don't know $(x_3 - x_4)$ and $(x_1 - x_9)$ are the "special" combinations. And if I just solve $y = c_1 x_1 + \cdots + c_9 x_9$, I will find all $\{c_i\}$ to have low $t$-values.
(The reason for this strange phenomenon is, my $\{x_i\}$ have significant correlations with each other.)
It seems that I can solve the model
$$y = c_{1,2} (x_1 - x_2) + c_{1,3} (x_1 - x_3) + \cdots + c_{4,6} (x_4 - x_6) + \cdots$$
and find all $\{c_{i,j}\}$ with high $t$-values. But then there will be $36$ coefficients $\{c_{i,j}\}$ instead of $9$. I wonder if there are faster methods?
Thank you.
• Seems to me there might be some better way of formulating or parameterising the problem, but it's hard to tell without knowing a bit more.. could you give us some idea what the $x_i$s are and how you know the underlying model is of that form? Jan 6, 2011 at 12:33
(This response picks up where @AVB, who has provided useful comments, left off by suggesting we need to figure out which differences $X_i - X_j$ ought to be included among the independent variables.)
The big question here is what is an effective method to identify the model. Later we can worry about faster methods. (But regression is so fast that you could process dozens of variables for millions of records in a matter of seconds.)
To make sure I'm not going astray, and to illustrate the procedure, I simulated a dataset like yours, only a little simpler. It consists of 60 independent draws from a common multivariate normal distribution with five unit-variance variables $Z_1, Z_2, Z_3, Z_4,$ and $Y$. The first two variables are independent of the second two and have correlation coefficient 0.9. The second two variables have correlation coefficient -0.9. The correlations between $Z_i$ and $Y$ are 0.5, 0.5, 0.5, and -0.5. Then--this changes nothing essential but makes the data a little more interesting--I rescaled the variables, thus: $X_1 = Z_1, X_2 = 2 Z_2, X_3 = 3 Z_3, X_4 = 4 Z_4$.
Let's begin by establishing that this simulation emulates the stated problem. Here is a scatterplot matrix.
The full regression of $Y$ against the $X_i$ is highly significant ($F(4, 55) = 15.28,\ p < 0.0001$) but all four t-values equal 1.24 ($p = 0.222$), which is not significant at all. The estimated coefficients are 0.26, 0.13, 0.088, and -0.066 (rounded to two sig figs).
Here is my proposal: systematically combine variables in pairs (six pairs in this case, 36 pairs for nine variables), one pair at a time. Regress a pair along with all remaining variables, seeking highly significant results for the pairs.
What is a "pair"? It is the linear combination suggested by the estimated coefficients. In this case, they are
\eqalign{ X_{12} =& X_1 / 0.26 &+ X_2 / 0.13 \cr X_{13} =& X_1 / 0.26 &+ X_3 / 0.088 \cr X_{14} =& X_1 / 0.26 &- X_4 / 0.066 \cr X_{23} =& X_2 / 0.13 &+ X_3 / 0.088 \cr X_{24} =& X_2 / 0.13 &- X_4 / 0.066 \cr X_{34} =& X_3 / 0.088 &- X_4 / 0.066 \text{.} }
In general, with $\hat{\beta}_i$ representing the estimated coefficient of $X_i$ in this full regression, the pairs are defined by
$$X_{ij} = X_i / \hat{\beta}_i + X_j / \hat{\beta}_j\text{.}$$
This is so systematic that it's straightforward to script.
The "identification regressions" are the model
$$Y \sim X_{12} + X_3 + X_4$$
along with the five additional permutations thereof, one for each pair.
You are looking for results where $X_{ij}$ becomes significant: ignore the significance of the remaining $X_k$. To see what's going on, I will list the results of all six identification regressions for the simulation. As a shorthand, I list the variables followed by a vector of their t-values only:
\eqalign{ X_{12}, X_3, X_4:&\ (5.50, 1.24, -1.24) \cr X_{13}, X_2, X_4:&\ (1.36, 4.94, -1.13) \cr X_{14}, X_2, X_3:&\ (1.31, 5.16, 1.17) \cr X_{23}, X_1, X_4:&\ (1.64, 3.10, -1.09) \cr X_{24}, X_1, X_3:&\ (1.50, 4.15, 1.07) \cr X_{34}, X_1, X_2:&\ (5.56, 1.25, 1.25) }
As you can see from the first component of each vector (the t-value for the pair), precisely two disjoint pairs exhibit significant t-statistics: $X_{12}$, with $t = 5.50\ (p \lt 0.001)$, and $X_{34}$, with $t = 5.56\ (p \lt 0.001)$. The model thus identified is
$$Y \sim X_{12} + X_{34}\text{.}$$
(In general, we would also include--provisionally--any remaining $X_i$ not participating in any of the pairs. There aren't any in this case.)
The regression results are
\eqalign{ \hat{\beta_{12}} &= 0.027\ (t = 5.54,\ p \lt 0.001) \cr \hat{\beta_{34}} &= 0.0055\ (t = 5.58,\ p \lt 0.001), \cr F(2, 57) &= 30.92\ (p \lt 0.0001). }
Translating back to the original $X_i$, the model is
\eqalign{ Y &= 0.027(X_1 / 0.26 + X_2 / 0.13) + 0.0055(X_3 / 0.088 - X_4 / 0.066) \cr &= 0.103 X_1 + 0.206 X_2 + 0.0629 X_3 - 0.0839 X_4 \cr &= 0.103 (Z_1 + Z_2) + 0.021 (Z_3 - Z_4) \text{.} }
(The last line shows how this all relates to form of the original question.) That's exactly the form used in the simulation: $Z_1$ and $Z_2$ enter with the same coefficient and $Z_3$ and $Z_4$ enter with opposite coefficients. This method got the right answer.
I want to share a cool observation in this regard. First, here's the scatterplot matrix for the model.
Notice how $X_{12}$ and $X_{34}$ look uncorrelated. Furthermore, $Y$ is only weakly correlated with these variables. Doesn't look like much of a relationship, does it? Now consider an alternative set of pairs, $X_{13}$ and $X_{24}$. The regression of $Y$ on these is still highly significant ($F(2, 57) = 16.61\ (p \lt 0.0001).$ Moreover, the coefficient of $X_{24}$ is significant ($t = 2.39,\ p = 0.020$) even though that of $X_{13}$ is not ($t = 0.24,\ p = 0.812$). But look at the scatterplot matrix!
Clearly $X_{13}$ and $X_{24}$ are strongly correlated. But, even though this is the wrong model, $Y$ is also visibly correlated with these two variables, much more so than in the preceding scatterplot matrix!
The lesson here is that mere bivariate plots can be deceiving in a multiple regression setting: to analyze the relationship between any candidate independent variable (such as $X_{12}$) and the independent variable ($Y$), we must make sure to "factor out" all other independent variables. (This is done by regressing $Y$ on all other independent variables and, separately, regressing $X_{12}$ on all the others. Then one looks at a scatterplot of the residuals of the first regression against the residuals of the second regression. It's a theorem that the slope in this bivariate regression equals the coefficient of $X_{12}$ in the full multivariate regression of $Y$ against all the variables.)
This insight shows why we might want to systematically perform the "identification regressions" I have proposed, rather than using graphical methods or attempting to combine many of the pairs in one model. Each identification regression assesses the strength of the contribution of a proposed linear combination of variables (a "pair") in the context of all the remaining independent variables.
Note that although correlated variables were involved, correlation is not an essential feature of the problem or of the solution. Even where you don't expect the original variables $X_i$ to be strongly correlated, you could expect a model to have (unknown) linear constraints among the variables. That is the important issue to cope with. The presence of correlation only means that it can be problematic to identify such pairs solely by inspecting the original regression results.
Following the procedure I have proposed does not guarantee you will find a unique solution. It's conceivable, for instance, that you will find so many highly significant pairs that they are linearly dependent, forcing you to select among them by some other criterion. Nevertheless, the results you get ought to limit the sets of pairs you need to examine; they can be obtained with a straightforward procedure without intervention; and--if this simulation is any guide--they have a good chance of producing effective results.
You can not use the 36-coefficient model, and not because it's going to be slow. Speed is the least of your worries here.
The real trouble is that you've taken an already under-determined problem (because of the correlations), and converted it into a problem which is severely under-determined for any data, because of linear dependencies. Simply put, $x_1-x_2=(x_1-x_3)-(x_2-x_3)$, so you can only determine 2 out of the 3 coefficients for these terms in the best case. The only way to fix this will be to prescribe some artificial regularization condition, like having minimal $\sum c_i^2$, or whatever may be right in your case.
It seems to me that you might be better off if you start by analyzing the correlation matrix and first figuring out which terms of the form $c_i-c_j$ should really appear in your problem. | 2,899 | 9,797 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-22 | latest | en | 0.902802 |
http://math.stackexchange.com/questions/400265/sum-k-1n-mk-where-mk-is-defined-by-2mk-k | 1,469,372,773,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824109.37/warc/CC-MAIN-20160723071024-00063-ip-10-185-27-174.ec2.internal.warc.gz | 154,887,782 | 21,654 | # $\sum_{k=1}^n m(k)$, where $m(k)$ is defined by $2^{m(k)} || k$.
I'm looking at the sum: $$f(n) = \sum_{k=1}^n m(k),$$ where $m(k)$ is defined by $2^{m(k)} || k$, i.e. $2^{m(k)}$ is the largest power of $2$ that divides $k$. For example, we have $f(8) = 0+1+0+2+0+1+0+3 = 7$. Here's a table of $n$, $m(n)$, and $f(n)$ for a few small $n$:
n m(n) f(n)
1 0 0
2 1 1
3 0 1
4 2 3
5 0 3
6 1 4
7 0 4
8 3 7
The values $m(n)$ in the table are the last $m(k)$-value for the given $n$, i.e. the value of $m(k)$ when $k=n$. I haven't spent much time trying to figure this out, mostly because it's a funny little problem and I thought I'd share it. It seems to me that we have $f(2^j) = 2^j-1 \forall j$ and $f(n)\sim n$.
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Hmm... since when does $2^3$ divide $4$? :-) – Peter Košinár May 23 '13 at 14:52
All your $m$'s are one too high. – N. S. May 23 '13 at 14:57
OK -- the notation should be fixed now and consistent with the answers. – Douglas B. Staple May 23 '13 at 15:15
Let $m(k)$ be the exponent of $2$ occurring in $k$, i.e. $2^{m(k)}\|k$. One quickly deduces $$\tag1f(2n+1)=f(2n)=n+f(n)$$ as only the even numbers contribute and $2k$ contributes one more to $2n$ and $2n+1$ than $k$ does to $n$, i.e. $m(2k)=m(k)+1$. Since $f(1)=m(1)=0$, we find $f(3)=f(2)=1$, $f(5)=f(4)=3$, and so on. The sequence goes $$0, 1, 1, 3, 3, 4, 4, 7, 7, 8, 8, 10, 10, 11, 11, 15, 15, 16, 16, 18, 18, 19, 19, 22, 22, 23, 23, 25, 25, 26, 26,\ldots$$ And according to OEIS, this equals $m(n!)$ and also $n$ minus the number of $1$s in the binary expansion of $n$. One easily verifies that these two discriptions indeed follow the recursion formula $(1)$.
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There is a name for your "m" function, it is called the 2-adic valuation, and is usually denoted by $\nu_2$.
It is easy to show that
$$f(2^j)=2^{j}-1$$
Indeed, for $1 \leq k \leq j-1$ there are exactly $2^{j-k-1}$ numbers between $1$ and $2^j$ with $\nu_2(n)=k$, namely
$$\{ 2^k\cdot 1, 2^k \cdot 3, 2^k \cdot 5,..., 2^k\cdot (2^{j-k}-1) \}$$
When $k=n$ there is exactly $1$ number with $\nu_2(n)=k$.
Then, by rearranging the terms of your sum into groups of equal terms you get
$$f(2^j)=j+\sum_{k=1}^{j-1} k 2^{j-k-1}$$
It is easy to show that this sum is exactly $2^j-1$.
Edit With the new fix, you need to add a 1 to every single term, thus the sum increases by exactly $2^j$.
-
As a bit of interesting trivia concerning $f(n)$ we see that the Mellin-Perron summation formula applies, giving $$f(n) = \frac{1}{2} v_2(n) + \frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} L(s) n^s \frac{ds}{s} \quad\text{where}\quad L(s) = \sum_{n\ge 1} \frac{v_2(n)}{n^s}.$$ But we have $$L(s) = \sum_{m\ge 1} \frac{v_2(2m)}{(2m)^s} = \frac{1}{2^s} \sum_{m\ge 1} \frac{1+v_2(m)}{m^s} = \frac{1}{2^s} (\zeta(s) + L(s)),$$ so that $$L(s) = \frac{\zeta(s)}{2^s-1},$$ giving $$f(n) = \frac{1}{2} v_2(n) + \frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} \frac{\zeta(s)}{2^s-1} n^s \frac{ds}{s}$$
We can apply the Cauchy Residue Theorem to this integral, treating first the poles at $s=1$ and $s=0$, which yields $$f(n) \sim \frac{1}{2} v_2(n) + n - \frac{1}{2}\log_2 n - \frac{1}{4} - \frac{1}{2}\log_2 \pi.$$ The above approximation is excellent, giving for example $f(200) \sim 196.6023238$ when the correct value is $197$, or $f(411) \sim 405.5827546$ when the correct value is $405.$ Finally, $f(1000)$ is $994$ when the approximation gives $995.4413598.$
The complete asymptotic expansion is obtained by including all the poles on the imaginary axis at $\frac{2\pi i k}{\log 2}$, with $k$ a nonzero integer. These poles are all simple. We have $$\operatorname{Res}\left(\frac{1}{2^s-1}; s = \frac{2\pi i k}{\log 2}\right) = \frac{1}{\log 2}.$$ Now introducing $$\rho_k = \frac{2\pi i k}{\log 2}$$ we finally obtain $$f(n) \sim \frac{1}{2} v_2(n) + n - \frac{1}{2}\log_2 n - \frac{1}{4} - \frac{1}{2}\log_2 \pi + \frac{1}{\log 2} \sum_{k\in \mathbb{Z}\setminus 0} \frac{\zeta(\rho_k)}{\rho_k} e^{2\pi i k\log_2 n},$$ where the sum is a Fourier series in $\log_2 n$ that describes the fluctuations.
Taking the terms up to $|k|=7$, we obtain $f(200) \sim 197.0498970.$ With $34$ terms, we obtain $f(411)\sim 405.3130370$ and $f(1000) \sim 993.9869087.$
-
One simplification of the sum is $f(n)=\sum\limits_{k=1}^\infty \left\lfloor\frac{n}{2^k}\right\rfloor$
(obviously, only terms for $k\leq \log_2(n)$ are going to be non-zero). This can be evaluated very easily on the computer, since each consecutive term corresponds to dividing the previous one by 2 and ignoring any remainder.
-
This is interesting; the point is to count the number of integers $\leq n$ divisible by $2^k$ for each $k\geq 1$, which is exactly $f(n)$. This also immediately gives $f(2^j) = 2^j-1$ as the sum of a geometric series. – Douglas B. Staple May 24 '13 at 17:36
Answer updated to reflect the modification in question. – Peter Košinár May 24 '13 at 19:47 | 1,986 | 4,950 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2016-30 | latest | en | 0.882919 |
https://nrich.maths.org/public/leg.php?code=3&cl=3&cldcmpid=2018 | 1,474,990,746,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661123.53/warc/CC-MAIN-20160924173741-00257-ip-10-143-35-109.ec2.internal.warc.gz | 893,884,093 | 9,272 | # Search by Topic
#### Resources tagged with Integers similar to What an Odd Fact(or):
Filter by: Content type:
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Challenge level:
### There are 39 results
Broad Topics > Numbers and the Number System > Integers
### AB Search
##### Stage: 3 Challenge Level:
The five digit number A679B, in base ten, is divisible by 72. What are the values of A and B?
### Slippy Numbers
##### Stage: 3 Challenge Level:
The number 10112359550561797752808988764044943820224719 is called a 'slippy number' because, when the last digit 9 is moved to the front, the new number produced is the slippy number multiplied by 9.
### Two Much
##### Stage: 3 Challenge Level:
Explain why the arithmetic sequence 1, 14, 27, 40, ... contains many terms of the form 222...2 where only the digit 2 appears.
### As Easy as 1,2,3
##### Stage: 3 Challenge Level:
When I type a sequence of letters my calculator gives the product of all the numbers in the corresponding memories. What numbers should I store so that when I type 'ONE' it returns 1, and when I type. . . .
### Double Digit
##### Stage: 3 Challenge Level:
Choose two digits and arrange them to make two double-digit numbers. Now add your double-digit numbers. Now add your single digit numbers. Divide your double-digit answer by your single-digit answer. . . .
### Seven Up
##### Stage: 3 Challenge Level:
The number 27 is special because it is three times the sum of its digits 27 = 3 (2 + 7). Find some two digit numbers that are SEVEN times the sum of their digits (seven-up numbers)?
### Phew I'm Factored
##### Stage: 4 Challenge Level:
Explore the factors of the numbers which are written as 10101 in different number bases. Prove that the numbers 10201, 11011 and 10101 are composite in any base.
### Arrange the Digits
##### Stage: 3 Challenge Level:
Can you arrange the digits 1,2,3,4,5,6,7,8,9 into three 3-digit numbers such that their total is close to 1500?
### The Patent Solution
##### Stage: 3 Challenge Level:
A combination mechanism for a safe comprises thirty-two tumblers numbered from one to thirty-two in such a way that the numbers in each wheel total 132... Could you open the safe?
### Times Right
##### Stage: 3 and 4 Challenge Level:
Using the digits 1, 2, 3, 4, 5, 6, 7 and 8, mulitply a two two digit numbers are multiplied to give a four digit number, so that the expression is correct. How many different solutions can you find?
### Mini-max
##### Stage: 3 Challenge Level:
Consider all two digit numbers (10, 11, . . . ,99). In writing down all these numbers, which digits occur least often, and which occur most often ? What about three digit numbers, four digit numbers. . . .
### Even Up
##### Stage: 3 Challenge Level:
Consider all of the five digit numbers which we can form using only the digits 2, 4, 6 and 8. If these numbers are arranged in ascending order, what is the 512th number?
### 1 Step 2 Step
##### Stage: 3 Challenge Level:
Liam's house has a staircase with 12 steps. He can go down the steps one at a time or two at time. In how many different ways can Liam go down the 12 steps?
### Sissa's Reward
##### Stage: 3 Challenge Level:
Sissa cleverly asked the King for a reward that sounded quite modest but turned out to be rather large...
### Six Times Five
##### Stage: 3 Challenge Level:
How many six digit numbers are there which DO NOT contain a 5?
### Not a Polite Question
##### Stage: 3 Challenge Level:
When asked how old she was, the teacher replied: My age in years is not prime but odd and when reversed and added to my age you have a perfect square...
### Base Puzzle
##### Stage: 3 Challenge Level:
This investigation is about happy numbers in the World of the Octopus where all numbers are written in base 8 .Octi the octopus counts.
### Lesser Digits
##### Stage: 3 Challenge Level:
How many positive integers less than or equal to 4000 can be written down without using the digits 7, 8 or 9?
### Aba
##### Stage: 3 Challenge Level:
In the following sum the letters A, B, C, D, E and F stand for six distinct digits. Find all the ways of replacing the letters with digits so that the arithmetic is correct.
### Whole Numbers Only
##### Stage: 3 Challenge Level:
Can you work out how many of each kind of pencil this student bought?
### Score
##### Stage: 3 Challenge Level:
There are exactly 3 ways to add 4 odd numbers to get 10. Find all the ways of adding 8 odd numbers to get 20. To be sure of getting all the solutions you will need to be systematic. What about. . . .
### Latin Numbers
##### Stage: 4 Challenge Level:
Let N be a six digit number with distinct digits. Find the number N given that the numbers N, 2N, 3N, 4N, 5N, 6N, when written underneath each other, form a latin square (that is each row and each. . . .
### Eight Dominoes
##### Stage: 2, 3 and 4 Challenge Level:
Using the 8 dominoes make a square where each of the columns and rows adds up to 8
### Dalmatians
##### Stage: 4 and 5 Challenge Level:
Investigate the sequences obtained by starting with any positive 2 digit number (10a+b) and repeatedly using the rule 10a+b maps to 10b-a to get the next number in the sequence.
### Never Prime
##### Stage: 4 Challenge Level:
If a two digit number has its digits reversed and the smaller of the two numbers is subtracted from the larger, prove the difference can never be prime.
### Ordered Sums
##### Stage: 4 Challenge Level:
Let a(n) be the number of ways of expressing the integer n as an ordered sum of 1's and 2's. Let b(n) be the number of ways of expressing n as an ordered sum of integers greater than 1. (i) Calculate. . . .
### Pairs
##### Stage: 3 Challenge Level:
Ann thought of 5 numbers and told Bob all the sums that could be made by adding the numbers in pairs. The list of sums is 6, 7, 8, 8, 9, 9, 10,10, 11, 12. Help Bob to find out which numbers Ann was. . . .
### Always Perfect
##### Stage: 4 Challenge Level:
Show that if you add 1 to the product of four consecutive numbers the answer is ALWAYS a perfect square.
### Coffee
##### Stage: 4 Challenge Level:
To make 11 kilograms of this blend of coffee costs £15 per kilogram. The blend uses more Brazilian, Kenyan and Mocha coffee... How many kilograms of each type of coffee are used?
### Upsetting Pitagoras
##### Stage: 4 and 5 Challenge Level:
Find the smallest integer solution to the equation 1/x^2 + 1/y^2 = 1/z^2
### What Are Numbers?
##### Stage: 2, 3, 4 and 5
Ranging from kindergarten mathematics to the fringe of research this informal article paints the big picture of number in a non technical way suitable for primary teachers and older students.
### Fracmax
##### Stage: 4 Challenge Level:
Find the maximum value of 1/p + 1/q + 1/r where this sum is less than 1 and p, q, and r are positive integers.
### Whole Number Dynamics I
##### Stage: 4 and 5
The first of five articles concentrating on whole number dynamics, ideas of general dynamical systems are introduced and seen in concrete cases.
### In Particular
##### Stage: 4 Challenge Level:
Write 100 as the sum of two positive integers, one divisible by 7 and the other divisible by 11. Then find formulas giving all the solutions to 7x + 11y = 100 where x and y are integers.
### Diophantine N-tuples
##### Stage: 4 Challenge Level:
Take any whole number q. Calculate q^2 - 1. Factorize q^2-1 to give two factors a and b (not necessarily q+1 and q-1). Put c = a + b + 2q . Then you will find that ab+1 , bc+1 and ca+1 are all. . . .
### Our Ages
##### Stage: 4 Challenge Level:
I am exactly n times my daughter's age. In m years I shall be exactly (n-1) times her age. In m2 years I shall be exactly (n-2) times her age. After that I shall never again be an exact multiple of. . . .
### Whole Number Dynamics IV
##### Stage: 4 and 5
Start with any whole number N, write N as a multiple of 10 plus a remainder R and produce a new whole number N'. Repeat. What happens?
### Euler's Squares
##### Stage: 4 Challenge Level:
Euler found four whole numbers such that the sum of any two of the numbers is a perfect square. Three of the numbers that he found are a = 18530, b=65570, c=45986. Find the fourth number, x. You. . . .
### Rudolff's Problem
##### Stage: 4 Challenge Level:
A group of 20 people pay a total of £20 to see an exhibition. The admission price is £3 for men, £2 for women and 50p for children. How many men, women and children are there in the group? | 2,177 | 8,454 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2016-40 | longest | en | 0.831136 |
http://www.safalniveshak.com/latticework-of-mental-models-permutation-and-combination/ | 1,484,935,894,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280850.30/warc/CC-MAIN-20170116095120-00034-ip-10-171-10-70.ec2.internal.warc.gz | 675,805,126 | 17,641 | # Latticework of Mental Models: Permutation and Combination
Here’s an interesting trivia.
If you wear different tie on the same shirt, most people will think that you’re wearing a different shirt. That’s an interesting way to multiply your options without really buying new cloths (except few new ties).
“That’s not a trivia, that’s a Jugaad.”, you might want to say. Anyways, That brings me to an equally interesting mindbender.
If you have 2 shirts (white, blue), 3 pants (black, gray and brown) and 3 different ties (pink, orange, red), in how many different ways can you get dressed? Assumption here is that getting dressed requires you to wear all three i.e. a shirt, a pant and a tie.
Using the multiplication principle we can say that there are total 2 x 3 x 3 = 18 ways to get dressed. Of course some of the dress combinations will look outright funny but our concern here is to find out all possible ways to get dressed. Moreover, today we are getting into Maths discipline and most mathematicians don’t really have whole lot of fashion sense anyways.
So that’s the simplest example of using the idea of combinations in real life. Now let’s say, for some strange reason, we were also considering the order in which you put on the cloths, i.e. it matters to us if one puts on the shirt first instead of tie.
Imagine wearing a tie first and then squeezing the shirt inside the tie, funny right? I told you mathematicians don’t care much about the dressing etiquettes 🙂
Okay, back to the same question again. In how many ways can you get dressed if the order of dressing matters?
For each of those 18 combinations, there are six ways to dress. So for a given combination of pant, shirt and tie you could go pant first, followed by the shirt and then tie. Or you could wear the shirt first, followed by the tie and finally the pant. Or you could do shirt, pant and tie. And so on.
Here the order (or the arrangement) of the objects matter. So permutation is just a fancy pants (no pun intended) definition of all possible ways of doing something. Simply put, permutations or rearrangements mean the different ways we can order or arrange a number of objects.
Combinations means the different ways we can choose a number of different objects from a group of objects where no order is involved, just the number of ways of choosing them.
For that matter a combination lock should really be called a permutation lock. The order you put the numbers in matters. A true combination lock would accept both 10-17-23 and 23-17-10 as correct.
In most simple terms, permutations and combinations (P&C) is all about counting the possible outcomes.
I am not sure about the current curriculums in schools but I learnt about P&C after 10th standard. Perhaps they are teaching these mathematical principles to kids in junior classes now.
Mathematicians get a kick out of converting simple numbers into complicated equations containing english and greek letters. So this is how a typical maths text book would describe the idea of permutations –
If one event can happen in ‘n‘ different ways, and a second event can happen independent of the first in ‘m’ different ways, the two events can happen in n x m (n multiplied by m) different ways.
To add little more clarity to the above definitions, it would help to think it this way – Permutations are for lists (order matters) and combinations are for groups (order doesn’t matter).
## The Factorial (!)
Let me take help from Peter Bevelin, author of Seeking Wisdom, for explaining the idea of factorial. Bevelin writes –
We have 3 hats to choose from – one black, one white and one brown. In how many ways can we arrange them if the order white, black and brown is different from the order black, white and brown? This is the same as asking how many permutations there are with three hats, taken three at a time. We can arrange the hats in 6 ways:
1. Black-White-Brown
2. Black-Brown-White
3. White- Black-Brown
4. White-Brown-Black
5. Brown-White-Black
6. Brown-Black-White.
Another way to think about this: We have three boxes in a row where we put a different hat in each box. We can fill the first box in three ways, since we can choose between all three hats. We can then fill the second box in two ways, since we now can choose between only two hats. We can fill the third box in only one way, since we have only one hat left. This means we can fill the box in 3 x 2 x1= 6 ways.
Another way to write this is 3! [another fancy pant word] If we have n (6) boxes and can choose from all of them, there are n (6) choices. Then we are left with n-1 (5) choices for box number two, n-2 (4) choices for box number three and so on. The number of permutations of n boxes is n!. What n! [pronounced as n-Factorial] means is the product of all numbers from 1 to n.
Suppose we have a dinner in our home with 12 people sitting around a table. How many seating arrangements are possible? The first person that enters the room can choose between twelve chairs, the second between eleven chairs and so on, meaning there are 12! or 479,001,600 different seating arrangements.
The number of ways we can arrange ‘r’ objects from a group of ‘n‘ objects is called a permutation of ‘n’ objects taken ‘r’ at a time and is defined as
p(n,r) = n! / (n-r)!
A safe has 100 digits. To open the safe a burglar needs to pick the correct 3 different numbers. Is it likely? The number of permutations or ways of arranging 3 digits from 100 digits is 970,200 (100! / (100-3)!). If every permutation takes the burglar 5 seconds, all permutations are tried in 5 6 days assuming a 24-hour working day.
In how many ways can we combine 2 flavors of ice cream if we can choose from strawberry (S), vanilla (V), and chocolate (C) without repeated flavors? We can combine them in 3 ways: SV, SC, VC. VS and SV are a combination of the same ice creams. The order doesn’t matter. Vanilla on the top is the same as vanilla on the bottom.
The number of ways we can select ‘r‘objects from a group of ‘n’ objects is called a combination of n objects taken r at a time and is defined as
C(n,r) = n! / r! (n-r)!
The number of ways we can select 3 people taken from a group of 10 people is 120 i.e.
(10! / (3!) (10-3)!)
## P&C In Investing
The study of permutations applies to investing in a broad sense because a good understanding of probability is sometimes necessary to make rational financial choices.
Infact, if you don’t understand P&C properly, you would have a tough time understanding probability. So P&C forms the foundation for understanding the concept of probability.
Charlie Munger, in his famous talk at USC Business School in 1994 entitled A Lesson on Elementary Worldly Wisdom, said –
Obviously, you’ve got to be able to handle numbers and quantities—basic arithmetic. And the great useful model, after compound interest, is the elementary math of permutations and combinations. And that was taught in my day in the sophomore year in high school…It’s very simple algebra. It was all worked out in the course of about one year between Pascal and Fermat. They worked it out casually in a series of letters.
It’s not that hard to learn. What is hard is to get so you use it routinely almost everyday of your life. The Fermat/Pascal system is dramatically consonant with the way that the world works. And it’s fundamental truth. So you simply have to have the technique.
If you don’t get this elementary, but mildly unnatural, mathematics of elementary probability into your repertoire, then you go through a long life like a one legged man in an asskicking contest. You’re giving a huge advantage to everybody else.
One of the advantages of a fellow like Buffett, whom I’ve worked with all these years, is that he automatically thinks in terms of decision trees and the elementary math of permutations and combinations.
## Conclusion
For me, preparing tea is a combination problem. All I know is that I need to throw in the ingredients (tea, water, milk, sugar) in a pan and heat it for 5-10 minutes. But for my wife it’s a permutation problem. She insists that milk be added at the last. No wonder, nobody wants to drink my tea.
Okay now that you’re equipped with the knowledge of P&C, riddle me this. If my wife, my two kids and I go to a movie theatre, in how many different ways can four of us be seated on our four allotted seats?
Did you say 24?
That could be correct except that the real world hardly presents itself like a well defined mathematics problem. In real world, there’s always a catch.
The catch in my case is that my kids are identical twins. So for all practical purposes, assuming the kids are wearing the same attire, half of the seating arrangements will look identical. And the correct answer would be 12.
Talking about probability, the odds of having an identical twins is approximately 1 in 300 i.e. 0.33 percent. My wife and I consider ourselves very lucky 🙂
One goal to learn different mental models from multiple disciplines is to learn how problems can be transformed. Remember that painting of the old lady and young woman?
Do you see both? Once you can see both of them it’s easy to switch between them. That’s the beauty of multidisciplinary thinking. The more models you have available, faster you can turn problems into each other.
It’s completely fine to use one model to understand the idea, and another to work out the details. But life becomes difficult when we think there’s only one way to approach it.
Beware! These mental models and multidisciplinary ideas do no good sitting inside your head like artifacts in a museum for they need to be taken out and played with.
I hope these mental models help you cut through the optical illusion around you and you start seeing many young and old ladies – metaphorically 🙂
Take care and keep learning.
Anshul Khare worked for 12+ years as a Software Architect. He is an avid learner in various disciplines like psychology, philosophy, and spirituality with special interests in human behaviour and value investing. You can connect with Anshul on Twitter.
Very well described Anshul, As you rightly said we learnt P&C after 10th Standard but today the concept get cleared finally for me. We smile while walking along when finally interpret different mathematical models. I read the below book couple of years back and it helped me immensely to understand many mathematical concept.
Thank You…
• Anshul Khare says:
2. Great information and sharing, was much helpful to me.
3. Darren says:
Hi … I am still trying to figure how P&C can figure in investing or even normal daily thinking….. other than figuring out ways to sit…
• Anshul Khare says: | 2,421 | 10,693 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2017-04 | longest | en | 0.927333 |
https://ncertmcq.com/tag/maths-rd-sharma-solutions-class-8/page/3/ | 1,709,576,179,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476464.74/warc/CC-MAIN-20240304165127-20240304195127-00799.warc.gz | 409,764,195 | 26,534 | ## RD Sharma Class 8 Solutions Chapter 15 Understanding Shapes I (Polygons) Ex 15.1
These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 15 Understanding Shapes I Ex 15.1
Question 1.
Draw rough diagrams to illustrate the following:
(i) Open curve
(ii) Closed curve
Solution:
Question 2.
Classify the following curves as open or closed.
Solution:
Open curves : (i), (iv) and (v) are open curves.
(ii) , (iii), and (vi) are closed curves.
Question 3.
Draw a polygon and shade its interior. Also draw its diagonals, if any.
Solution:
In the given polygon, the shaded portion is its interior region AC and BD are the diagonals of polygon ABCD.
Question 4.
Illustrate, if possible, each one of the following with a rough diagram:
(i) A closed curve that is not a polygon.
(ii) An open curve made up entirely of line segments.
(iii) A polygon with two sides.
Solution:
(i) Close curve but not a polygon.
(ii) An open curve made up entirely of line segments.
(iii) A polygon with two sides. It is not possible. At least three sides are necessary
Question 5.
Following are some figures : Classify each of these figures on the basis of the following:
(i) Simple curve
(ii) Simple closed curve
(iii) Polygon
(iv) Convex polygon
(v) Concave polygon
(vi) Not a curve
Solution:
(i) It is a simple closed curve and a concave polygon.
(ii) It is a simple closed curve and convex polygon.
(iii) It is neither a curve nor polygon.
(iv) it is neither a curve not a polygon.
(v) It is a simple closed curve but not a polygon.
(vi) It is a simple closed curve but not a polygon.
(vii) It is a simple closed curve but not a polygon.
(viii) It is a simple closed curve but not a polygon.
Question 6.
How many diagonals does each of the following have ?
(ii) A regular hexagon
(iii) A triangle.
Solution:
Here n = 4
Question 7.
What is a regular polygon ? State the name of a regular polygon of:
(i) 3 sides
(ii) 4 sides
(iii) 6 sides.
Solution:
A regular polygon is a polygon which has all its sides equal and so all angles are equal,
(i) 3 sides : It is an equilateral triangle.
(ii) 4 sides : It is a square.
(iii) 6 sides : It is a hexagon.
Hope given RD Sharma Class 8 Solutions Chapter 15 Understanding Shapes I Ex 15.1 are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RD Sharma Class 8 Solutions Chapter 20 Mensuration I (Area of a Trapezium and a Polygon) Ex 20.1
These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1
Other Exercises
Question 1.
A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m² ?
Solution:
Area of floor = 1080 m²
Base of parallelogram shaped tile (b) = 24 cm
and corresponding height (h) = 10 cm
Area of one tile = b x h = 24 x 10 = 240 cm²
Question 2.
A plot is in the form of a rectangle ABCD having semi-circle on BC as shown in Fig. If AB = 60 m and BC = 28 m, find the area of the plot.
Solution:
Length of rectangular portion (l) = 60 m
and breadth (b) = 28 m
Area of the rectangular plot = l x b = 60 x 28 m² = 1680 m²
Radius of semicircular portion (r) = $$\frac { b }{ 2 }$$ = $$\frac { 28 }{ 2 }$$ = 14 m
Area = $$\frac { 1 }{ 2 }$$ πr²
= $$\frac { 1 }{ 2 }$$ x $$\frac { 22 }{ 7 }$$ x 14 x 14 m²
= 308 m²
Total area of the plot = 1680 + 308 = 1988 m²
Question 3.
A playground has the shape of a rectangle, with two semi-circles on its smaller sides as diameters, added to its outside. If the sides of the rectangle are 36 m and 24.5 m, find the area of the playground. (Take π = $$\frac { 22 }{ 7 }$$).
Solution:
Length of rectangular portion (l) = 36 m
and breadth (b) = 24.5 m
= $$\frac { 22 }{ 7 }$$ x 150.0625 m²
= 471.625 m²
Total area of the playground = 471.625 + 882 = 1353.625 m²
Question 4.
A rectangular piece is 20 m long and 15 m wide. From its four corners, quadrants of radii 3.5 have been cut. Find the area of the remaining part.
Solution:
Length of rectangular piece (l) = 20 m
Area of rectangular piece = l x b = 20 x 15 = 300 m²
Area of the remaining portion = 300 – 38.5 m² = 261.5 m²
Question 5.
The inside perimeter of a running track (shown in Fig.) is 400 m. The length of each of the straight portion is 90 m and the ends are semi-circles. If track is everywhere 14 m wide, find the area of the track. Also, find the length of the outer running track.
Solution:
Inner perimeter = 400 m.
Length (l) = 90 m.
Perimeter of two semicircles = 400 – 2 x 90 = 400 – 180 = 220 m
Question 6.
Find the area of the Figure in square cm, correct to one place of decimal. (Take π = $$\frac { 22 }{ 7 }$$)
Solution:
Length of square (a) = 10 cm.
Area = a² = (10)² = 100 cm²
Base of the right triangle AED = 8 cm
and height = 6 cm
Question 7.
The diameter of a wheel of a bus is 90 cm which makes 315 revolutions per minute. Determine its speed in kilometres per hour. (Take π = $$\frac { 22 }{ 7 }$$)
Solution:
Diameter of the wheel (d) = 90 cm.
Question 8.
The area of a rhombus is 240 cm² and one of the diagonal is 16 cm. Find another diagonal.
Solution:
Area of rhombus = 240 cm²
Length of one diagonal (d1) = 16 cm
Second diagonal (d2)
Question 9.
The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Solution:
In rhombus, diagonal (d1) = 7.5 cm
and diagonal (d2) = 12 cm
Question 10.
The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.
Solution:
diagonal AC = 24 m
and perpendicular BL = 13 m
and perpendicular DM on AC = 8 m
Area of the field ABED = $$\frac { 1 }{ 2 }$$ x AC x (BL + DM)
= $$\frac { 1 }{ 2 }$$ x 24 x (13 + 8) m²
= 12 x 21 = 252 m²
Question 11.
Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.
Solution:
Side of rhombus (b) = 6 cm
Altitude (h) = 4 cm
Question 12.
The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m² is Rs 4.
Solution:
Number of rhombus shaped tiles = 300
Diagonals of each tile = 45 cm and 130 cm
Rate of polishing the tiles = Rs 4 per m²
Total cost = 202.5 x 4 = Rs 810
Question 13.
A rectangular grassy plot is 112 m long and 78 broad. It has a gravel path 2.5 m wide all around it on the side. Find the area of the path and the cost of constructing it at Rs 4.50 per square metre.
Solution:
Length of rectangular plot (l) = 112 m
and breadth (b) = 78 m
Width of path = 2.5 m
Inner length = 112 – 2 x 2.5 = 112 – 5 = 107 m
and inner breadth = 78 – 2 x 2.5 = 78 – 5 = 73 m
Area of path = outer area – inner area
= (112 x 78 – 107 x 73) m² = 8736 – 7811 = 925 m²
Rate of constructing = Rs 4.50 per m²
Total cost = 925 x Rs 4.50 = Rs 4162.50
Question 14.
Find the area of a rhombus, each side of which measures 20 cm and one of whose diagonals is 24 cm.
Solution:
Side of rhombus = 20 cm.
One diagonal (d1) = 24 cm
Diagonals of a rhombus bisect each other at right angle
AB = 20 cm
and OA = $$\frac { 1 }{ 2 }$$ AC = $$\frac { 1 }{ 2 }$$ x 24 cm = 12 cm
In right-angled ∆AOB,
AB² = AO² + BO² (Pythagoras theorem)
⇒ (20)² = (12)² + BO²
⇒ 400 = 144 + BO²
⇒ BO² = 400 – 144 = 256 = (16)²
⇒ BO = 16 cm
and diagonal BD = 2 x BO = 2 x 16 = 32 cm
Now area of rhombus ABCD
Question 15.
The length of a side of a square field is 4 m. What will be the altitude of the rhombus if the area of the rhombus is equal to the square field and one of its diagonal is L m ?
Solution:
Side of square = 4 m
Area of square = (a)² = 4 x 4 =16 m²
Diagonals of a rhombus bisect each other at right angles.
In right ∆AOB
AB² = QA² + BO² (Pythagoras theorem)
= (8)² + (1)² = 64 + 1 = 65
AB = √65 m.
Now, length of perpendicular AL (h)
Question 16.
Find the area of the field in the form of a rhombus, if the length of each side be 14 cm and the altitude be 16 cm.
Solution:
Length of each side of rhombus = 14 cm.
Length of altitude = 16 cm
Area = Base x altitude = 14 x 16 cm² = 224 cm²
Question 17.
The cost of fencing a square field at 60 paise per metre is Rs 1,200. Find the cost of reaping the field at the rate of 50 paise per 100 sq. metres.
Solution:
Cost of fencing the square field = Rs 1,200
Rate = 60 paise per m.
Question 18.
In exchange of a square plot one of whose sides is 84 m, a man wants to buy a rectangular plot 144 m long and of the same area as of the square plot. Find the width of the rectangular plot.
Solution:
Side of a square plot = 84 m
Area = (a)² = (84)² = 84 x 84 m² = 7056 m²
Area of rectangular field = 7056 m²
Length (l) = 144 m
Question 19.
The area of a rhombus is 84 m². If its perimeter is 40 m, then find its altitude.
Solution:
Area of rhombus = 84 m²
Perimeter = 40 m
Question 20.
A garden is in the form of a rhombus whose side is 30 metres and the corresponding altitude is 16 m. Find the cost of levelling the garden at the rate of Rs 2 per m².
Solution:
Side of rhombus garden (b) = 30 m.
Altitude (h) = 16 m
Area = Base x altitude = 30 x 16 = 480 m²
Rate of levelling the garden = Rs 2 per m²
Total cost = Rs 480 x 2 = Rs 960
Question 21.
A field in the form of a rhombus has each side of length 64 m and altitude 16 m. What is the side of a square field which has the same area as that of a rhombus ?
Solution:
Length of side of rhombus (b) = 64 m
and altitude (h) = 16 m
Area = b x h = 64 x 16 m² = 1024 m²
Now area of square = 1024 m²
Side of the square = √Area = √1024 m = 32 m
Question 22.
The area of a rhombus is equal to the area of a triangle whose base and the corresponding altitude are 24.8 cm and 16.5 cm respectively. If one of the diagonals of the rhombus is 22 cm, find the length of the other diagonal.
Solution:
Base of triangle (b) = 24.8 cm
and altitude (h) = 16.5 cm
Area of triangle = $$\frac { 1 }{ 2 }$$ x base x height
= $$\frac { 1 }{ 2 }$$ x bh= $$\frac { 1 }{ 2 }$$ x 24.8 x 16.5 cm² = 204.6 cm²
Area of rhombus = 204.6 cm²
Length of one diagonal (d1 = 22 cm
Second diagonal (d2)
Hope given RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1 are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5
These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5
Other Exercises
Question 1.
Mr. Cherian purchased a boat for Rs 16,000. If the total cost of the boat is depreciating at the rate of 5% per annum, calculate its value after 2 years.
Solution:
Cost of boat = Rs 16,000
Rate of depreciating = 5% p.a.
Period = 2 years
Value of boat after 2 years
Question 2.
The value of a machine depreciates at the rate of 10% per annum. What will be its value 2 years hence, if the present value is Rs 1,0,000 ? Also, find the total depreciation during this period.
Solution:
Present value of machine = Rs 1,00,000
Rate of depreciation = 10% p.a.
Period (n) = 2 years
Value of machine after 2 years
Question 3.
Pritam bought a plot of land for Rs 6,40,000. Its value is increasing by 5% of its previous value after every six months. What will be the value of the plot after 2 years ?
Solution:
Present value of plot = Rs 6,40,000
Increase = 5% per half year
Period (n) = 2 years or 4 half years
Question 4.
Mohan purchased a house for Rs 30,000 and its value is depreciating at the rate of 25% per year. Find the value of the house after 3 years.
Solution:
Present value of the house (P) = Rs 30,000
Rate of depreciation = 25% p.a.
Period (n) = 3 years
Value of house after 3 years
Question 5.
The value of a machine depreciates at the rate of 10% per annum. It was purchased 3 years ago. If its present value is Rs 43,740, find its purchased price.
Solution:
Let the purchase price of machine = Rs P
Rate of depreciation = 10% p.a.
Period (n) = 3 years.
and present value = Rs 43,740
Question 6.
The value of a refrigerator which was purchased 2 years ago, depreciates at 12% per annum. If its present value is Rs 9,680, for how much was it purchased ?
Solution:
Let the refrigerator was purchased for = Rs P
Rate of depreciation (R) = 12% p.a.
Period (n) = 2 years
and present value (A) = Rs 9,680
Question 7.
The cost of a TV set was quoted Rs 17,000 at the beginning of 1999. In the beginning of2000, the price was hiked by 5%. Because of decrease in demand the cost was reduced by 4% in the beginning of 2001. What is the cost of the TV set in 2001 ?
Solution:
List price of TV set in 1999 = Rs 17,000
Rate of hike in 2000 = 5%
Rate of decrease in 2001 = 4%
Price of TV set in 2001
Question 8.
Ashish started the business with an initial investment of Rs 5,00,000. In the first year, he incurred a loss of 4%. However, during the second year he earned a profit of 5% which in third year, rose to 10%. Calculate the net profit for the entire period of 3 years.
Solution:
Initial investment = Rs 5,00,000
In the first year, rate of loss = 4%
In the second year, rate of gain = 5%
and in the third year, rate of gain = 10%
Investment after 3 years
Hope given RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5 are helpful to complete your math homework.
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## RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6
These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6
Other Exercises
Question 1.
Write the following squares of bionomials as trinomials :
Solution:
Using the formulas
(a + b)2 = a2 + 2ab + b2 and (a – b)2 = a2 – 2ab + b2
(i) (a + 2)2 = (a)2 + 2 x a x 2 + (2)2
{(a + b)2 = a2 + 2ab + b2}
= a2 + 4a + 4
(ii) (8a + 3b)2 = (8a)2 + 2 x 8a * 3b + (3b)2 = 642 + 48ab + 9 b2
(iii) (2m+ 1)2 = (2m)2 + 2 x 2m x1 + (1)2
= 4m2 + 4m + 1
Question 2.
Find the product of the following binomials :
Solution:
Question 3.
Using the formula for squaring a binomial, evaluate the following :
(i) (102)2
(ii) (99)2
(iii) (1001)2
(iv) (999)2
(v) (703)
2
Solution:
(i) (102)2 = (100 + 2)2
= (100)2 + 2 x 100 x 2 + (2)2
{(a + b)2 = a2 + 2ab + b2}
= 10000 + 400 + 4 = 10404
(ii) (99)2 = (100 – 1)2
= (100)2 – 2 x 100 X 1 +(1)2
{(a – b)2 = a2 – 2ab + b2}
= 10000 -200+1
= 10001 -200 =9801
(iii) (1001 )2 = (1000 + 1)2
{(a + b)2 = a2 + 2ab + b2}
= (1000)2 + 2 x 1000 x 1 + (1)2
= 1000000 + 2000 + 1 = 1002001
(iv) (999)2 = (1000 – 1)2
{(a – b)2 = a2 – 2ab + b2}
= (1000)2 – 2 x 1000 x 1 + (1)2
= 1000000 – 2000 + 1
= 1000001 -2000 = 998001
Question 4.
Simplify the following using the formula:
(a – b) (a + b) = a2 – b2 :
(i) (82)2 (18)2
(ii) (467)2 (33)2
(iii) (79)2 (69)2
(iv) 197 x 203
(v) 113 x 87
(vi) 95 x 105
(vii) 1.8 x 2.2
(viii) 9.8 x 10.2
Solution:
(i) (82)2 – (18)2 = (82 + 18) (82 – 18)
{(a + b)(a- b) = a2 – b2} = 100 x 64 = 6400
(ii) (467)2 – (33)2 = (467 + 33) (467 – 33)
= 500 x 434 = 217000
(ii) (79)2 – (69)2 = (79 + 69) (79 – 69)
148 x 10= 1480
(iv) 197 x 203 = (200 – 3) (200 + 3)
= (200)2 – (3)2
= 40000-9 = 39991
(v) 113 x 87 = (100 + 13) (100- 13)
= (100)2 – (13)2
= 10000- 169 = 9831
(vi) 95 x 105 = (100 – 5) (100 + 5)
= (100)2 – (5)2
= 10000 – 25 = 9975
(vii) 8 x 2.2 = (2.0 – 0.2) (2.0 + 0.2)
= (2.0)2 – (0.2)2
= 4.00 – 0.04 = 3.96
(viii)9.8 x 10.2 = (10.0 – 0.2) (10.0 + 0.2)
(10.0)2 – (0.2)2
= 100.00 – 0.04 = 99.96
Question 5.
Simplify the following using the identities :
Solution:
Question 6.
Find the value of x, if
(i) 4x = (52)2 – (48)2
(ii) 14x = (47)2 – (33)2
(iii) 5x = (50)2 – (40)2
Solution:
(i) 4x = (52)2 – (48)2
4x = (52 + 48) (52 – 48)
Question 7.
If x + $$\frac { 1 }{ x }$$= 20, find the value of x2+ $$\frac { 1 }{ { x }^{ 2 } }$$
Solution:
Question 8.
If x – $$\frac { 1 }{ x }$$ = 3, find the values of x2 + $$\frac { 1 }{ { x }^{ 2 } }$$ and x4 + $$\frac { 1 }{ { x }^{ 4 } }$$
Solution:
Question 9.
If x2 – $$\frac { 1 }{ { x }^{ 2 } }$$= 18, find the values of x+ $$\frac { 1 }{ x }$$ and x– $$\frac { 1 }{ x }$$
Solution:
Question 10.
Ifx+y = 4 and xy = 2, find the value of x2+y2.
Solution:
x + y = 4
Squaring on both sides,
(x + y)2 = (4)2
⇒ x2 +y2 + 2xy = 16
⇒ x2+y2 + 2 x 2 = 16 (∵ xy = 2)
⇒ x2 + y2 + 4 = 16
⇒ x2+y2 = 16 – 4= 12 ‘
∴ x2+y2 = 12
Question 11.
If x-y = 7 and xy = 9, find the value of x2+y2.
Solution:
x-y = 7
Squaring on both sides,
(x-y)2 = (7)2
⇒ x2+y2-2xy = 49
⇒ x2 + y2 – 2 x 9 = 49 (∵ xy = 9)
⇒ x2 +y2 – 18 = 49
⇒ x2 + y2 = 49 + 18 = 67
∴ x2+y2 = 67
Question 12.
If 3x + 5y = 11 and xy = 2, find the value of 9x2 + 25y2
Solution:
3 x + 5y = 11, xy = 2
Squaring on both sides,
(3x + 5y)2 = (11)2
⇒ (3x)2 + (5y)2 + 2 x 3x x 5y = 121
⇒ 9x2 + 25y2 + 30 x 7 = 121
⇒ 9x2 + 25y2+ 30 x 2 = 121 (∵ xy = 2)
⇒ 9x2 + 25y2 + 60 = 121
⇒ 9x2 + 25y2 = 121 – 60 = 61
∴ 9x2 + 25y2 = 61
Question 13.
Find the values of the following expressions :
(i)16x2 + 24x + 9, when X’ = $$\frac { 7 }{ 45 }$$
(ii) 64x2 + 81y2 + 144xy when x = 11 and y = $$\frac { 4 }{ 3 }$$
(iii) 81x2 + 16y2-72xy, whenx= $$\frac { 2 }{ 3 }$$ andy= $$\frac { 3 }{ 4 }$$
Solution:
Question 14.
If x + $$\frac { 1 }{ x }$$ = 9, find the values of x+ $$\frac { 1 }{ { x }^{ 4 } }$$.
Solution:
Question 15.
If x + $$\frac { 1 }{ x }$$ = 12, find the values of x– $$\frac { 1 }{ x }$$.
Solution:
Question 16.
If 2x + 3y = 14 and 2x – 3y = 2, find the value of xy.
Solution:
2x + 3y = 14, 2x – 3y= 2
We know that
(a + b)2 – (a – b)2 = 4ab
∴ (2x + 3y)2 – (2x – 3y)2 = 4 x 2x x 3y = 24xy
⇒ (14)2 – (2)2 = 24xy
⇒ 24xj= 196-4= 192
⇒ xy = $$\frac { 192 }{ 24 }$$ = 8
∴ xy = 8
Question 17.
If x2 + y2 = 29 and xy = 2, find the value of
(i) x+y
(ii) x-y
(iii) x4 +y4
Solution:
x2 + y2 = 29, xy = 2
(i) (x + y)2 = x2 + y2 + 2xy
= 29 + 2×2 = 29+ 4 = 33
∴ x + y= ±√33
(ii) (x – y)2 = x2 + y2 – 2xy
= 29- 2×2 = 29- 4 = 25
∴ x-y= ±√25= ±5
(iii) x2 + y2 = 29
Squaring on both sides,
(x2 + y2)2 = (29)2
⇒ (x2)2 + (y2)2 + 2x2y2 = 841
⇒ x4 +y + 2 (xy)2 = 841
⇒ x4 + y + 2 (2)2 = 841 (∵ xy = 2)
⇒ x4 + y + 2×4 = 841
⇒ x4 + y + 8 = 841
⇒ x4 + y = 841 – 8 = 833
∴ x4 +y = 833
Question 18.
What must be added to each of the following expressions to make it a whole square ?’
(i) 4x2 – 12x + 7
(ii) 4x2 – 20x + 20
Solution:
(i) 4x2 – 12x + 7 = (2x)2 – 2x 2x x 3 + 7
In order to complete the square,
we have to add 32 – 7 = 9 – 7 = 2
∴ (2x)2 – 2 x 2x x 3 + (3)2
= (2x-3)2
∴ Number to be added = 2
(ii) 4x2 – 20x + 20
⇒ (2x)2 – 2 x 2x x 5 + 20
In order to complete the square,
we have to add (5)2 – 20 = 25 – 20 = 5
∴ (2x)2 – 2 x 2x x 5 + (5)2
= (2x – 5)2
∴ Number to be added = 5
Question 19.
Simplify :
(i) (x-y) (x + y) (x2 + y2) (x4 + y4)
(ii) (2x – 1) (2x + 1) (4x2 + 1) (16x4 + 1)
(iii) (7m – 8m)2 + (7m + 8m)2
(iv) (2.5p -5q)2 – (1.5p – 2.5q)2
(v) (m2 – n2m)2 + 2m3n2
Solution:
(i) (x – y) (x + y) (x2 + y2) (x4 +y)
= (x2 – y2) (x2 + y) (x4 + y4)
= [(x2)2 – (y2)2] (x4+y4)
= (x4-y4) (x4+y4)
= (x4)2 – (y4)2 = x8 – y8
(ii) (2x – 1) (2x + 1) (4x2 + 1) (16x4 + 1)
= [(2x)2 – (1)2] (4x2 + 1) (16x4 + 1)
= (4x2 – 1) (4x2 + 1) (16x4 + 1)
= [(4x2)2-(1)2] (16x4+ 1)
= (16x4-1) (16x4+ 1)
= (16x4)2– (1)2 = 256x8 – 1
(iii) (7m – 8m)2 + (7m + 8n)2
= (7m)2 + (8n)2 – 2 x 7m x 8n + (7m)2 + (8n)2 + 2 x 7m x 8n
= 49m2 + 64m2 – 112mn + 49m2 + 64m2 + 112mn
= 98 m2 + 128n2
(iv) (2.5p – 1.5q)2 – (1.5p – 2.5q)2
= (2.5p)2 + (1.5q)2 – 2 x 2.5p x 1.5q
= [(1.5p)2 + (1.5q)2 – 2 x 1.5 p x 2.5q]
= (6.25p2 + 2.25q2 – 7.5 pq) – (2.25p2 + 6.25q2-7.5pq)
= 6.25p2 + 2.25q2 – 7.5pq – 2.25p2 – 6.25q2 + 7.5pq
= 6.25p2 – 2.25p2 + 2.25g2 – 6.25q2
= 4.00P2 – 4.00q2
= 4p2 – 4q2 = 4 (p2 – q2)
(v) (m2 – n2m)2 + 2m3M2
= (m2)2 + (n2m)2 -2 x m2 x n2m + 2;m3m2
= m4 + n4m2 – 2m3n2 + 2m3n2
= m4 + n4m2 = m4 + m2n4
Question 20.
Show that :
Solution:
Hope given RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.6 are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RD Sharma Class 8 Solutions Chapter 19 Visualising Shapes Ex 19.2
These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 19 Visualising Shapes Ex 19.2
Other Exercises
Question 1.
Which among the following are nets for a cube ?
Solution:
Nets for a cube are (ii), (iv) and (vi)
Question 2.
Name the polyhedron that can be made by folding each net:
Solution:
(i) This net is for a square
(ii) This net is for triangular prism.
(iii) This net is for triangular prism.
(iv) This net is for hexagonal prism.
(v) This net is for hexagon pyramid.
(vi) This net is for cuboid.
Question 3.
Dice are cubes where the numbers on the opposite faces must total 7. Which of the following are dice ?
Solution:
Figure (i) shows the net of cube or dice.
Question 4.
Draw nets for each of the following polyhedrons:
Solution:
(i) Net for cube is given below :
(ii) Net of a triangular prism is as under :
(iii) Net of hexagonal prism is as under :
(iv) The net for pentagonal pyramid is as under:
Question 5.
Match the following figures:
Solution:
(a) (iv)
(b) (i)
(c) (ii)
(d) (iii)
Hope given RD Sharma Class 8 Solutions Chapter 19 Visualising Shapes Ex 19.2 are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RD Sharma Class 8 Solutions Chapter 19 Visualising Shapes Ex 19.1
These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 19 Visualising Shapes Ex 19.1
Other Exercises
Question 1.
What is the least number of planes that can enclose a solid ? What is the name of the solid ?
Solution:
The least number of planes that can enclose a solid is called a Tetrahedron.
Question 2.
Can a polyhedron have for its faces :
(i) three triangles ?
(ii) four triangles ?
(iii) a square and four triangles ?
Solution:
(i) No, polyhedron has three faces.
(ii) Yes, tetrahedron has four triangles as its faces.
(iii) Yes, a square pyramid has a square as its base and four triangles as its faces.
Question 3.
Is it possible to have a polyhedron with any given number of faces ?
Solution:
Yes, it is possible if the number of faces is 4 or more.
Question 4.
Is a square prism same as a cube ?
Solution:
Yes, a square prism is a cube.
Question 5.
Can a polyhedron have 10 faces, 20 edges and 15 vertices ?
Solution:
No, it is not possible as By Euler’s formula
F + V = E + 2
⇒ 10 + 15 = 20 + 2
⇒ 25 = 22
Which is not possible
Question 6.
Verify Euler’s formula for each of the following polyhedrons :
Solution:
(i) In this polyhedron,
Number of faces (F) = 7
Number of edges (E) = 15
Number of vertices (V) = 10
According to Euler’s formula,
F + V = E + 2
⇒ 7 + 10 = 15 + 2
⇒ 17 = 17
Which is true.
(ii) In this polyhedron,
Number of faces (F) = 9
Number of edges (E) = 16
Number of vertices (V) = 9
According to Euler’s formula,
F + V = E + 2
⇒ 9 + 9 = 16 + 2
⇒ 18 = 18
Which is true.
(iii) In this polyhedron,
Number of faces (F) = 9
Number of edges (E) =18
Number of vertices (V) = 11
According to Euler’s formula,
F + V = E + 2
⇒ 9 + 11 = 18 + 2
⇒ 20 = 20
Which is true.
(iv) In this polyhedron,
Number of faces (F) = 5
Number of edges (E) = 8
Number of vertices (V) = 5
According to Euler’s formula,
F + V = E + 2
⇒ 5 + 5 = 8 + 2
⇒ 10 = 10
Which is true.
(v) In the given polyhedron,
Number of faces (F) = 9
Number of edges (E) = 16
Number of vertices (V) = 9
According to Euler’s formula,
F + V = E + 2
⇒ 9 + 9 = 16 + 2
⇒ 18 = 18
Which is true.
Question 7.
Using Euler’s formula, find the unknown:
Solution:
We know that Euler’s formula is
F + V = E + 2
(i) F + 6 = 12 + 2
⇒ F + 6 = 14
⇒ F = 14 – 6 = 8
Faces = 8
(ii) F + V = E + 2
⇒ 5 + V = 9 + 2
⇒ 5 + V = 11
⇒ V = 11 – 5 = 6
Vertices = 6
(iii) F + V = E + 2
⇒ 20 + 12 = E + 2
⇒ 32 = E + 2
⇒ E = 32 – 2 = 30
Edges = 30
Hope given RD Sharma Class 8 Solutions Chapter 19 Visualising Shapes Ex 19.1 are helpful to complete your math homework.
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## RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.5
These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.5
Other Exercises
Question 1.
Construct a quadrilateral ABCD given that AB = 4 cm, BC = 3 cm, ∠A = 75°, ∠B = 80° and ∠C = 120°.
Solution:
Steps of construction :
(i) Draw a line segment AB = 4 cm.
(ii) At A draw a ray AX making an angle of 75°.
(iii) At B draw another ray BY making an angle of 80° and cut off BC = 3 cm.
(iv) At C, draw another ray CZ making an angle of 120° which intersects AX at D.
Then ABCD is the required quadrilateral.
Question 2.
Construct a quadrilateral ABCD where AB = 5.5 cm, BC = 3.7 cm, ∠A = 60°, ∠B = 105° and ∠D = 90°.
Solution:
∠A = 60°, ∠B = 105° and ∠D = 90°
But ∠A + ∠B + ∠C + ∠D = 360° (Sum of angles of a quadrilateral)
⇒ 60° + 105° + ∠C + 90° = 360°
⇒ 255° + ∠C = 360°
⇒ ∠C = 360° – 255° = 105°
Steps of construction :
(i) Draw a line segment AB = 5.5 cm.
(ii) At A, draw a ray AX making an angle of
(iii) At B, draw another ray BY making an angle of 105° and cut off BC = 3.7 cm.
(iv) At C, draw a ray CZ making an angle of 105° which intersects AX at D.
Then ABCD is the required quadrilateral.
Question 3.
Construct a quadrilateral PQRS where PQ = 3.5 cm, QR = 6.5 cm, ∠P = ∠R = 105° and ∠S = 75°.
Solution:
∠P = 105°, ∠R = 105° and ∠S = 75°
But ∠P + ∠Q + ∠R + ∠S = 360° (Sum of angles of a quadrilateral)
⇒ 105° + ∠Q + 105° + 75° = 360°
⇒ 285° + ∠Q = 360°
⇒ ∠Q = 360° – 285° = 75°
Steps of construction :
(i) Draw a line segment PQ = 3.5 cm.
(ii) At P, draw a ray PX making an angle of 105°.
(iii) At Q, draw another ray QY, making an angle of 75° and cut off QR = 6.5 cm.
(iv) At R, draw a ray RZ making an angle of 105° which intersects PX at S.
Then PQRS is the required quadrilateral.
Question 4.
Construct a quadrilateral ABCD when BC = 5.5 cm, CD = 4.1 cm, ∠A = 70°, ∠B = 110° and ∠D = 85°.
Solution:
∠A = 70°, ∠B = 110°, ∠D = 85°
But ∠A + ∠B + ∠C + ∠D = 360° (Sum of angles of a quadrilateral)
⇒ 70° + 110° + ∠C + 85° = 360°
⇒ 265° + ∠C = 360°
⇒ ∠C = 360° – 265° = 95°
Steps of construction:
(i) Draw a line segment BC = 5.5 cm.
(ii) At B, draw a ray BX making an angle of 110°.
(iii) At C, draw another ray CY making an angle of 95° and cut off CD = 4.1 cm.
(iv) At D, draw a ray DZ making an angle of 85° which intersects BX at A.
Then ABCD is the required quadrilateral.
Question 5.
Construct a quadrilateral ABCD, where ∠A = 65°, ∠B = 105°, ∠C = 75°, BC = 5.7 cm and CD = 6.8 cm.
Solution:
∠A = 65°, ∠B = 105°, ∠C = 75°
But ∠A + ∠B + ∠C + ∠D = 360° (Sum of angles of a quadrilateral)
⇒ 65° + 105° + 75° + ∠D = 360°
⇒ 245° + ∠D = 360°
⇒ ∠D = 360° – 245° = 115°
Steps of construction:
(i) Draw a line segment BC = 5.7 cm.
(ii) At B, draw a ray BX making an angle of
(iii) At C draw a another ray CY making an angle of 75° and cut off CD = 6.8 cm.
(iv) At D, draw a ray DZ making an angle of 115° which intersects BX at A.
Then ABCD is the required quadrilateral.
Question 6.
Construct a quadrilateral PQRS in which PQ = 4 cm, QR = 5 cm, ∠P = 50°, ∠Q = 110° and ∠R = 70°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 4 cm.
(ii) At P, draw a ray PX making an angle of 50°.
(iii) At Q, draw another ray QY making an angle of 110° and cut off QR = 5 cm.
(iv) At R, draw a ray RZ making an angle of 70° which intersects PX at S.
Then PQRS is the required quadrilateral.
Hope given RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.5 are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.4
These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.4
Other Exercises
Question 1.
Construct a quadrilateral ABCD, in which AB = 6 cm, BC = 4 cm, CD = 4 cm, ZB = 95° and ∠C = 90°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4 cm.
(ii) At B, draw a ray BX making an angle of 95° and cut off BA = 6 cm.
(iii) At C, draw a ray CY making an angle of 90° and cut off CD = 4 cm.
Then ABCD is the required quadrilateral.
Question 2.
Construct a quadrilateral ABCD, where AB = 4.2 cm, BC = 3.6 cm, CD = 4.8 cm, ∠B = 30° and ∠C = 150°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.6 cm.
(ii) At B, draw a ray BX making an angle of 30° and cut of BA = 4.2 cm.
(iii) At C, draw another ray CY making an angle of 150° and cut off CD = 4.8 cm.
Then ABCD is the required quadrilateral.
Question 3.
Construct a quadrilateral PQRS, in which PQ = 3.5 cm, QR = 2.5 cm, RS = 4.1 cm, ∠Q = 75° and ∠R = 120°.
Solution:
Steps of construction :
(i) Draw a line segment QR = 2.5 cm.
(ii) At Q, draw a ray QX making an angle of 75° and cut off QP = 3.5 cm.
(iii) At R, draw another ray RY making an angle of 120° and cut off RS = 4.1 cm.
(iv) Join PS.
Then PQRS is the required quadrilateral.
Question 4.
Construct a quadrilateral ABCD given BC = 6.6 cm, CD = 4.4 cm, AD = 5.6 cm and ∠D = 100° and ∠C = 95°.
Solution:
Steps of construction :
(i) Draw a line segment CD = 4.4 cm.
(ii) At C, draw a ray CX making an angle of 95° and cut off CB = 6.6 cm
(iii) At D, draw another ray DY making an angle of 100° and cut off DA = 5.6 cm.
(iv) Join AB.
Then ABCD is the required quadrilateral.
Question 5.
Construct a quadrilateral ABCD in which AD = 3.5 cm, AB = 4.4 cm, BC = 4.7 cm, ∠A = 125° and ∠B = 120°.
Solution:
Steps of construction :
(i) Draw a line segment AB 4.4 cm.
(ii) At A, draw a ray AX making an angle of 125° and cut off AD = 3.5 cm.
(iii) At B, draw another ray BY making an angle of 120° and cut off BC = 4.7 cm.
(iv) Join CD.
Then ABCD is the required quadrilateral.
Question 6.
Construct a quadrilateral PQRS in which ∠Q = 45°, ∠R = 90°, QR = 5 cm, PQ = 9 cm and RS = 7 cm.
Solution:
Steps of construction :
This quadrilateral is not possible to construct as shown in the figure.
Question 7.
Construct a quadrilateral ABCD in which AB = BC = 3 cm, AD = 5 cm, ∠A = 90° and ∠B = 105°.
Solution:
Steps of construction :
(i) Draw a line segment AB = 3 cm.
(ii) At A, draw a ray AX making an angle of 90° and cut off AD = 5 cm.
(iii) At B, draw another ray BY making an angle of 105° and cut off BC = 3 cm.
(iv) Join CD.
Then ABCD is the required quadrilateral.
Question 8.
Construct a quadrilateral BDEF where DE = 4.5 cm, EF = 3.5 cm, FB = 6.5 cm and ∠F = 50° and ∠E = 100°
Solution:
Steps of construction :
(i) Draw a line segment EF = 3.5 cm.
(ii) At E, draw a ray EX making an angle of 100° and cut off ED = 4.5 cm.
(iii) At F, draw another ray FY making an angle of 45° and cut off FB = 6.5 cm.
(iv) Join DB.
Then BDEF is the required quadrilateral.
Hope given RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.4 are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.3
These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.3
Other Exercises
Question 1.
Construct a quadrilateral ABCD in which AB = 3.8 cm, BC = 3.4 cm, CD = 4.5 cm, AD = 5 cm and ∠B = 80°.
Solution:
Steps of construction :
(i) Draw a line segment AB = 3.8 cm.
(ii) At B, draw a ray BX making an angle of 80° and cut off BC = 3.4 cm.
(iii) With centre A and radius 5 cm and with centre C and radius 4.5 cm, draw arcs which intersect each other at D.
Question 2.
Construct a quadrilateral ABCD given that AB = 8 cm, BC = 8 cm, CD = 10 cm, AD = 10 cm and ∠A = 45°.
Solution:
Steps of construction :
(i) Draw a line segment AB = 8 cm.
(ii) At A, draw a ray AX making an angle of 45° and cut off BC = 8 cm.
(iii) With centre A and C and radius 10 cm, draw arcs intersecting each other at D.
Then ABCD is the required quadrilateral.
Question 3.
Construct a quadrilateral ABCD in which AB = 7.7 cm, BC = 6.8 cm, CD = 5.1 cm, AD = 3.6 cm and ∠C = 120°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6.8 cm.
(ii) At C, draw a ray CX making an angle of 120° and cut off CD = 5.1 cm.
(iii) With centre B and radius 7.7 cm and with centre D and radius 3.6 cm draw arcs which intersect each other at A.
Then ABCD is the required quadrilateral.
Question 4.
Construct a quadrilateral ABCD in which AB = BC = 3 cm, AD = CD = 5 cm and ∠B = 120°
Solution:
Steps of construction :
(i) Draw a line segment AB = 3 cm.
(ii) At B, draw a ray BX making an angle of 120° and cut off BC = 3 cm.
(iii) With centres A and C, and radius 5 cm, draw arcs intersecting each other at D.
Then ABCD is the required quadrilateral.
Question 5.
Construct a quadrilateral ABCD in which AB = 2.8 cm, BC = 3.1 cm, CD = 2.6 cm and DA = 3.3 cm and ∠A = 60°.
Solution:
Steps of construction :
(i) Draw a line segment AB = 2.8 cm.
(ii) At A draw a ray AX making an angle of 60° and cut off AD = 3.3 cm.
(iii) With centre B and radius 3.1 cm and with centre D and radius 2.6 cm, draw arc which intersect each other at C.
(iv) Join CB and CD.
Then ABCD is the required quadrilateral.
Question 6.
Construct a quadrilateral ABCD in which AB = BC = 6 cm, AD = DC = 4.5 cm and ∠B = 120°.
Solution:
Steps of construction:
The construction is not possible to draw as arcs of radius 4.5 cm from A and C, do not intersect at any point.
Hope given RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.3 are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.2
These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.2
Other Exercises
Question 1.
Construct a quadrilateral ABCD in which AB = 3.8 cm, BC = 3.0 cm, AD = 2.3 cm, AC = 4.5 cm and BD = 3.8 cm.
Solution:
Steps of construction :
(i) Draw a line segment AB = 3.8 cm.
(ii) With centre A and radius 2.3 cm and with centre B and radius 3.8 cm draw arcs intersecting each other at D.
(iv) Again with centre A and radius 4.5 cm and with centre B and radius 3 cm, draw arcs intersecting each other at C.
(v) Join AC and BC and also CD.
Then ABCD is the required quadrilateral.
Question 2.
Construct a quadrilateral ABCD in which BC = 7.5 cm, AC = AD = 6 cm, CD = 5 cm and BD = 10 cm.
Solution:
Steps of construction :
(i) Draw a line segment CD = 5 cm.
(ii) With centre C and D and radius 6 cm, draw line segments intersecting each other at A.
(iv) Again with centre C and radius 7.5 cm and with centre D and radius 10 cm, draw arcs intersecting each other at B.
(v) Join CB, CA, DA, DB and AB.
Then ABCD is the required quadrilateral.
Question 3.
Construct a quadrilateral ABCD, when AB = 3 cm, CD = 3 cm, DA = 7.5 cm, AC = 8 cm and BD = 4 cm.
Solution:
Steps of construction :
This quadrilateral is not possible as
BD = 4 cm, AB = 3 cm and AD = 7.5 cm
The sum of any two sides of a triangle is greater than the third side.
But BD + AD = 4 + 3 = 7 cm
Question 4.
Construct a quadrilateral ABCD given AD = 3.5 cm, BC = 2.5 cm, CD = 4.1 cm, AC = 7.3 cm and BD = 3.2 cm.
Solution:
Steps of construction :
(i) Draw a line segment CD = 4.1 cm.
(ii) With centre C and radius 7.3 cm and with centre D and radius 3.5 cm, draw arcs intersecting each other at A.
(iv) Again with centre C and radius 2.5 cm and with centre D and radius 3.2 cm, draw arcs intersecting each other at B.
(v) Join CB’, and DB’ and join AB’.
Then ABCD is the required quadrilateral.
Question 5.
Construct a quadrilateral ABCD given AD = 5 cm, AB = 5.5 cm, BC = 2.5 cm, AC = 7.1 cm and BD = 8 cm.
Solution:
Steps of construction:
(i) Draw a line segment AB = 5 cm.
(ii) With centre A and radius 7.1 cm and with centre B and radius 2.5 cm, draw arcs which intersect each other at C.
(iii) Join AC and BC.
(iv) Again with centre A and radius 5 cm and with centre B and radius 8 cm, draw arcs which intersect each other at D.
(v) Join AD and BD and CD.
Then ABCD is the required quadrilateral.
Question 6.
Construct a quadrilateral ABCD in which BC = 4 cm, CA = 5.6 cm, AD = 4.5 cm, CD = 5 cm and BD = 6.5 cm.
Solution:
Steps of construction:
(i) Draw a line segment CD = 5 cm.
(ii) With centre C and radius 5.6 cm and with centre D and radius 4.5 cm, draw arcs which intersect each other at A.
(iv) Again with centre C and radius 4 cm and with centre D and radius 6.5 cm, draw arcs which intersect each other at B.
(v) Join BC and BD and AB.
Then ABCD is the required quadrilateral.
Hope given RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.2 are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.1
These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.1
Other Exercises
Question 1.
Construct a quadrilateral ABCD in which AB = 4.4 cm, BC = 4 cm, CD = 6.4 cm, DA = 3.8 cm and BD = 6.6 cm.
Solution:
Steps of construction :
(i) Draw a line segment AB = 4.4 cm.
(ii) With centre A and radius 3.8 cm and with centre B and radius 6.6 cm, draw arcs intersecting each other at D.
(iii) With centre B and radius 4 cm, and with centre D and radius 6.4 cm, draw arcs intersecting each other at C on the other side of BD.
(iv) Join AD, BD, BC and DC.
The ABCD is the required quadrilateral.
Question 2.
Construct a quadrilateral ABCD such that AB = BC = 5.5 cm, CD = 4 cm, DA = 6.3 cm and AC = 9.4 cm. Measure BD.
Solution:
(i) Draw a line segment AC = 9.4 cm.
(ii) With centre A and C and radius 5.5 cm, draw arcs intersecting each other at B.
(iii) Join AB and CB.
(iv) Again with centre A and radius 6.3 cm, and with centre C and radius 4 cm, draw arcs intersecting each other at D below the line segment AC.
Then ABCD is the required quadrilateral. On measuring BD, it is 5 cm.
Question 3.
Construct a quadrilateral XYZW in which XY = 5 cm, YZ = 6 cm, ZW = 7 cm, WX = 3 cm and XZ = 9 cm.
Solution:
Steps of construction :
(i) Draw a line segment XZ = 9 cm.
(ii) With centre X and radius 3 cm and with centre Z and radius 7 cm, draw arcs intersecting each other at W.
(iii) Join XW and ZW.
(iv) Again with centre X and radius 5 cm and with centre Z and radius 6 cm, draw arcs, intersecting each other at Y below the line segment XZ.
(v) Join XY and ZY.
Then XYZW is the required quadrilateral.
Question 4.
Construct a parallelogram PQRS such that PQ = 5.2 cm, PR = 6.8 cm and QS = 8.2 cm.
Solution:
Steps of construction:
In a parallelogram, diagonals bisect each other. Now
(i) Draw a line segment PQ = 5.2 cm.
(ii) With centre P and radius 3.4 cm ($$\frac { 1 }{ 2 }$$ of PR) and with centre Q and radius 4.1 cm ($$\frac { 1 }{ 2 }$$ of QS) draw arcs intersecting each other at O.
(iii) Join PQ and QO and produced them to R and S respectively such that PO = OR and QO = OS.
(iv) Join PS, SR and RQ.
Then PQRS is the required parallelogram.
Question 5.
Construct a rhombus with side 6 cm and one diagonal 8 cm. Measure the other diagonal.
Solution:
Steps of construction :
Sides of a rhombus are equal.
(i) Draw a line segment AC = 8 cm.
(ii) With centres A and C and radius 6 cm, draw two arcs above the line segment AC and two below the line segment AC, intersecting each other at D and B respectively.
(iii) Join AB, AD, BC and CD.
Then ABCD is the required rhombus.
JoinBD.
On measuring BD, it is approximately 9 cm
Question 6.
Construct a kite ABCD in which AB = 4 cm, BC = 4.9 cm and AC = 7.2 cm.
Solution:
Steps of construction :
(i) Draw a line segment AC = 7.2 cm.
(ii) With centre A and radius 4 cm draw an arc.
(iii) With centre C and radius 4.9 cm, draw another arc which intersects the first arc at B and D.
(iv) Join AB, BC, CD and DA.
Then ABCD is the required kite.
Question 7.
Construct, if possible, a quadrilateral ABCD given, AB = 6 cm BC = 3.7 cm, CD = 5.7 cm, AD = 5.5 cm and BD = 6.1 cm. Give reasons for not being able to construct, if you cannot.
Solution:
Steps of construction :
(i) Draw a line segment BD = 6.1 cm.
(ii) With centre B and radius 6 cm and with centre D and radius 5.5 cm, draw arcs intersecting at A.
(iv) Again with centre B and radius 3.7 cm and with centre D and radius 5.7 cm, draw two arcs intersecting each other at C below the BD.
(v) Join BC and DC.
Then ABCD is the required quadrilateral.
Question 8.
Construct, if possible a quadrilateral ABCD in which AB = 6 cm, BC = 7 cm, CD = 3 cm, AD = 5.5. cm and AC = 11 cm. Give reasons for not being able to construct, if you cannot.
Solution:
Steps of construction:
It is not possible to construct this quadrilateral ABCD because
AD + DC = 5.5 cm + 3 cm = 8.5 cm
and AC = 11 cm | 14,801 | 41,530 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2024-10 | latest | en | 0.921197 |
https://www.solutionlogy.com/what-is-3x-times-5/ | 1,718,579,150,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861671.61/warc/CC-MAIN-20240616203247-20240616233247-00028.warc.gz | 869,417,648 | 38,930 | # What is 3x times -5?
What is 3x times -5?Question: What is 3x times 5? Solution: First we will remove the parentheses, we get 3×x×-5 Now, we will multiply the above monomials, we get 15x Hence, 3x times 5=15x | 76 | 211 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-26 | latest | en | 0.802018 |
https://howmanyzeros.com/zeros-in-936-2-billion | 1,643,156,205,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304876.16/warc/CC-MAIN-20220125220353-20220126010353-00143.warc.gz | 370,348,604 | 19,459 | Home » Number of Zeros » How many Zeros in 936.2 Billion?
# How many Zeros in 936.2 Billion?
If you have been wondering about how many zeros in 936.2 billion?, then you have come to the right post.
Apart from the answer to this question, we provide you with related information regarding the number of zeros in 936.2 billion.
Read on to learn all about the amount of 0s in 936.2bn, and make sure to check out our app right below.
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## How many Zeros in Nine Hundred And Thirty-Six Point Two Billion?
936.2 billion in figures equals 936200000000, or 936,200,000,000 when written by thousand separators, which makes it easier to count the occurrences of 0. Thus, we get:
Zeros in 936.2 Billion = 8
8 is the answer to 936.2 billion has how many zeros? Questions on our website similar to how many zeros in 936.2 billion, include, for example:
If you like to learn how many zeros for a numeral different from 936.2 billion, fill in our application at the beginning of this post. After inserting your number, our tool displays the amount of 0’s automatically.
In the next part of how many zeros in nine hundred and thirty-six point two billion, we show you how many 100 and 1000 there are in 936.2 billion, and other related information.
## 936.2 Billion has how many Zeros?
You already know the answer to the question 936.2 billion has how many zeros?, but we are left with telling you how many 10, 100, 1000 et cetera there are in nine hundred and thirty-six point two billion:
• How many tens in 936.2 billion? Answer: 93,620,000,000 tens.
• How many hundreds in 936.2 billion? Answer: 9,362,000,000 hundreds.
• How many thousands in 936.2 billion? Answer: 936,200,000 thousands.
• How many ten thousands in 936.2 billion? Answer: 93,620,000 ten thousands.
• How many hundred thousands in 936.2 billion? Answer: 9,362,000 hundred thousands.
• How many millions in 936.2 billion? Answer: 936,200 million.
In the concluding section ahead we wrap how many 0 in nine hundred and thirty-six point two billion up, and explain the use of our search form located in the sidebar and our menu.
## Zeros in 936.2 Billion
The image sums how many zeros in 936.2 billion? up:
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Thanks for visiting How many Zeros in 936.2 Billion. | 745 | 2,904 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2022-05 | latest | en | 0.93635 |
https://physics.stackexchange.com/questions/551951/why-does-surface-tension-act-towards-the-centre-of-an-air-bubble-in-water | 1,718,936,522,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862032.71/warc/CC-MAIN-20240620235751-20240621025751-00430.warc.gz | 402,306,730 | 39,690 | # Why does surface tension act towards the centre of an air bubble in water?
Surface tension acts towards the centre of an air bubble. However, since surface tension is the result of the cohesive forces between the water molecules, should not the water molecules at the interface with the air bubble be directed towards the surrounding water molecules and not the centre of the bubble?
## Intuition
To build intuition, let's replace the water surface by an elastic surface (rubber or any other easily stretchable material). This works because of the similar nature of surface forces of water and the elastic surfaces. So now, the water drop is converted to an air filled balloon. Now, from experience you'd already know that the balloon feels a force towards the center which is balanced by the air pressure inside. The case is similar for a water drop.
## Mathematical Analysis
First let's analyse the 2D analog of a sphere, circle. Imagine a stretched rubber band in the shape of a circle. In this case, as explained above, the forces are between the surrounding particles and thus along the tangential direction to the circle. Now let's analyse a very small section of that circle. See the figure below:
Image source
Now let's compute the component of tension in $$x$$ and $$y$$ directions:
\begin{align} T_x&=T\sin\theta_0+T\sin\theta_0= 2 T\sin\theta_0 \\ T_y&=T\sin\theta_0-T\sin\theta_0=0 \end{align}
As you can see that the vertical component of the tension cancels out and only the tension along the center remains. As evident, this is true for any segment, and thus all the parts feel a force towards the center.
Now let's translate our above understanding to the 3D version i.e. a sphere. I won't do the exact proof for you here because it's just an extension of the above proof and will be more insightful if you do it yourself. An image to help you out:
This is an image of a general surface. However, for the case of a sphere, you can assume $$R_1=R_2$$. Now as we did above, try taking the components of the tension in the $$x$$, $$y$$ and $$z$$ directions. Use symmetry to find out which components cancel and which ones add. You'll find out that only the forces acting towards the center of the sphere remain, while the others cancel out.
Note: Only the forces at the boundary matter. The forces inside the boundary and on the surface are internl forces if we consider the whole surface as our system, and thus they cancel out.
• Thank you for your detailed answer! I understand that the molecular forces between the water molecules building the surface of the air bubble are tangential to the spherical shape of the bubble. But should there not also be a force directed towards the molecules surrounding the molecules that build the surface around the air bubble? Commented May 14, 2020 at 18:13
• @Aliska Those forces cancel out if you take the surface as your system. Only the forces on the boundary remain
– user258881
Commented May 14, 2020 at 18:17 | 677 | 2,984 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 7, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-26 | latest | en | 0.925154 |
http://mathhelpforum.com/algebra/110825-inverse-funciton.html | 1,519,406,401,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814801.45/warc/CC-MAIN-20180223154626-20180223174626-00552.warc.gz | 209,303,026 | 10,371 | 1. ## Inverse funciton
The function f(x)= ax+b , f-1(x)=8x-3. Find a and b
I would be very happy if you explained how to do it.
Thank you
2. Originally Posted by Oasis1993
The function f(x)= ax+b , f-1(x)=8x-3. Find a and b
I would be very happy if you explained how to do it.
Thank you
let f(x)= ax+b =y (say)
then ax+b =y $\quad \Rightarrow \quad x= \frac{(y-b)}{a}$
$or\ x=\frac{1}{a}y-\frac{b}{a}$
$\therefore f^{-1} (y)=\frac{1}{a}y-\frac{b}{a} \quad \Rightarrow \quad (\because f(x)=y , taking inverse both side )$
$or\ \color {blue} f^{-1} (x)=\frac{1}{a}x-\frac{b}{a}$................(1)
but given $f^{-1}(x)=8x-3$................(2)
on comparison eq(1) and (2)
$\frac{1}{a}=8 \quad \Rightarrow a= \frac{1}{8}$
and
$\frac{b}{a}=3 \quad \Rightarrow b=3a={3\over 8}$
3. Thank you very much!
4. Just to put in my oar: 8x- 3 says "first multiply x by 8 and then subtract 3".
The "inverse" is the inverse of the original operarations ("divide by 8" rather than "multiply by 8" and "add 3" rather than "subtract 3") done in the opposite order.
So the inverse of "first multiply x by 8 and then subtract 3" is "first add 3 to x and then divide by 8".
$f(x)= \frac{x+ 3}{8}= \frac{1}{8}x+ \frac{3}{8}$ | 447 | 1,212 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2018-09 | latest | en | 0.700683 |
http://www.chegg.com/homework-help/computer-organization-and-design-4th-edition-chapter-3.6-solutions-9780123744937 | 1,472,657,357,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982290634.12/warc/CC-MAIN-20160823195810-00031-ip-10-153-172-175.ec2.internal.warc.gz | 366,537,066 | 16,693 | View more editions
# Computer Organization and Design The Hardware Software Interface (4th Edition)Solutions for Chapter 3.6
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Chapter: Problem:
SAMPLE SOLUTION
Chapter: Problem:
• Step 1 of 3
For better performance, a multiplication operation between two unsigned integers can be substituted by a number of shift and an add operation. Assume that A and B are both 8-bit unsigned integers, the process would involve following steps:-
1. Represent number B in multiples of 2 as . This means, find the nearest multiple of 2 below number B.
2. Now, bitwise left shift the number A, C times. Check if any more shift is required.
3. Add the results of shifts or to A.
• Step 2 of 3
To multiply a pair of hexadecimal numbers 0x24 and 0xC9 using above steps,
1. First convert the hexadecimal numbers to integers and binary. So,
2. Represent number B in multiples of 2 as . This means, find the nearest multiple of 2 below number B. In this case B is 36, which can be represented as
, so in this case C is 5 and D is 4.
3. Now, bitwise left shift the number A, C times. Here, bitwise left shift the number 201 (i.e. A) , 5 (i.e. C) times, to give:-
Then, shift 201, left to 2 places to give.
4. Finally, add the above two numbers to get the product.
Which is in fact correct, as .
Therefore, the single multiplication operation can be replaced with 2 shifts and one addition operation.
• Step 3 of 3
To multiply a pair of hexadecimal numbers 0x41 and 0x18 using above steps,
1. First convert the hexadecimal numbers to integers and binary. So,
2. Represent number B in multiples of 2 as . This means, find the nearest multiple of 2 below number B. In this case B is 65, which can be represented as
, so in this case C is 6 and D is 1.
3. Now, bitwise left shift the number A, C times. Here, bitwise left shift the number 24 (i.e. A) , 6 (i.e. C) times, to give:-
4. Finally, add the above shift result to 24 (i.e. A) to get the product.
Which is in fact correct, as .
Therefore, the single multiplication operation can be replaced with 1 shifts and one addition operation.
Corresponding Textbook
Computer Organization and Design The Hardware Software Interface | 4th Edition
9780123744937ISBN-13: 0123744938ISBN: Authors:
Alternate ISBN: 9780080922812 | 640 | 2,490 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2016-36 | latest | en | 0.841761 |
https://testbook.com/objective-questions/bn/mcq-on-discrete-time-fourier-transform-dtft--5eea6a0839140f30f369d787 | 1,680,253,727,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949598.87/warc/CC-MAIN-20230331082653-20230331112653-00037.warc.gz | 632,011,697 | 82,593 | # Discrete Time Fourier Transform (DTFT) MCQ Quiz in বাংলা - Objective Question with Answer for Discrete Time Fourier Transform (DTFT) - বিনামূল্যে ডাউনলোড করুন [PDF]
Last updated on Dec 31, 2022
পাওয়া Discrete Time Fourier Transform (DTFT) उत्तरे आणि तपशीलवार उपायांसह एकाधिक निवड प्रश्न (MCQ क्विझ). এই বিনামূল্যে ডাউনলোড করুন Discrete Time Fourier Transform (DTFT) MCQ কুইজ পিডিএফ এবং আপনার আসন্ন পরীক্ষার জন্য প্রস্তুত করুন যেমন ব্যাঙ্কিং, এসএসসি, রেলওয়ে, ইউপিএসসি, রাজ্য পিএসসি।
## Top Discrete Time Fourier Transform (DTFT) MCQ Objective Questions
#### Discrete Time Fourier Transform (DTFT) Question 1:
A real valued discrete time signal x[n] has Fourier transform X(e) that is zero for $$\frac{{3\pi }}{{14}} \le \omega \le \pi .$$ To make the non-zero portion of Fourier transform occupy the region |ω| < π the signal is subjected to decimation and interpolation. If L is the up sampling factor and N is the down sampling factor. The value of L and N are respectively.
1. 3, 14
2. 14, 3
3. 3π, 14π
4. 14π, 3π
Option 1 : 3, 14
#### Discrete Time Fourier Transform (DTFT) Question 1 Detailed Solution
Let the signal in frequency be represented by
To occupy the entire region from -π to π, x[n] must be down sample by factor 14/3
Since it is not possible to down sample by non-integer factor, up sample single first by 3 (= L).
Then down sample signal by 14 (= M).
Thus, L = 3
M = 14
#### Discrete Time Fourier Transform (DTFT) Question 2:
$${\left( {\frac{2}{3}} \right)^n}u\left[ {n - 3} \right]\mathop \leftrightarrow \limits^{FT} \frac{{A{e^{ - j6\pi f}}}}{{1 - \left( {\frac{2}{3}} \right){e^{ - j2\pi f}}}}$$. The value of $$A$$ is_________.
#### Discrete Time Fourier Transform (DTFT) Question 2 Detailed Solution
Concept:
$${a^n}u\left[ n \right]\mathop \leftrightarrow \limits^{FT} \frac{1}{{1 - \;a{e^{ - j2\pi f}}}}\\ {\left( a \right)^{n - {n_o}}}u\left[ {n - {n_o}} \right]\mathop \leftrightarrow \limits^{FT} \frac{{{e^{- j2\pi f{n_o}}}}}{{1 - \;a{e^{-j2\pi f}}}}$$
Analysis:
$${(\frac{2}{3})^n}u\left[ n \right]\mathop \leftrightarrow \limits^{FT} \frac{1}{{1 - \;(2/3){e^{ - j2\pi f}}}}$$
$${\left( \frac{2}{3} \right)^{n - {3}}}u\left[ {n - {3}} \right]\mathop \leftrightarrow \limits^{FT} \frac{{{e^{- j2\pi f{(3)}}}}}{{1 - \;(2/3){e^{-j2\pi f}}}}$$
Multiplying both the sides by (2/3)3
$${\left( {\frac{2}{3}} \right)^3}{\left( {\frac{2}{3}} \right)^{n - 3}}u\left[ {n - 3} \right]\mathop \leftrightarrow \limits^{FT} \frac{{{{\left( {\frac{2}{3}} \right)}^3}{e^{- j2\pi f\left( 3 \right)}} \times 1}}{{1 - \;\frac{2}{3}{e^{ - j2\pi f}}}}$$
∴ On comparing we get,
$$A=8/27=0.296$$
#### Discrete Time Fourier Transform (DTFT) Question 3:
The period ‘T0’ of a function $$\cos \left[ {\frac{\pi }{4}\left( {t - 1} \right)} \right]$$ is _____
1. T0 = 8 sec
2. T0 = 4 sec
3. $${T_0} = \frac{1}{8}sec$$
4. T0 = 6 sec
Option 1 : T0 = 8 sec
#### Discrete Time Fourier Transform (DTFT) Question 3 Detailed Solution
Concept:
A signal x(t) is a said to be periodic with a period T if x(t ± T) = x(t)
The time period for a signal in the form of A cos (ωt + ϕ) is $$T = \frac{{2\pi }}{\omega}$$
For the sum of the two continuous-time signals with period T1 and T2 to be periodic, there must exist non-zero integers a, b such that:
a × T1 = b × T2
Calculation:
Given function is, $$\cos \left[ {\frac{\pi }{4}\left( {t - 1} \right)} \right]$$
Time period $$T = \frac{{2\pi }}{{\frac{\pi }{4}}} = 8\;sec$$
#### Discrete Time Fourier Transform (DTFT) Question 4:
What is the Discrete Time Fourier Series of a periodic Impulse train of period N?
1. 1/N
2. 1
3. N
4. An impulse train
Option 2 : 1
#### Discrete Time Fourier Transform (DTFT) Question 4 Detailed Solution
The Discrete-Time Fourier transform of a signal of infinite duration x[n] is given by:
$$X\left( {e^{jω }} \right) = \mathop \sum \limits_{n = - \infty }^\infty x\left[ n \right]{e^{ - jω n}}$$
Given signal x[n] = δ (x - kN)
$$X\left( {{e^{jω }}} \right) = \mathop \sum \limits_{n = - \infty }^\infty \delta \left( {n - kN} \right){e^{ - jω n}}$$
X (e) = e-jωkN
In discrete-time system ω = 2 π / N
Substitute the value of ω in the above equation.
X (e) = e-j2πk
= 1
Therefore, the Discrete-Time Fourier Series of a periodic Impulse train of period N is 1
#### Discrete Time Fourier Transform (DTFT) Question 5:
The frequency response and the main lobe width for rectangular window are
1. $$\frac{{\sin \frac{{\omega N}}{2}}}{{\sin \frac{\omega }{2}}}\;and\frac{{4\pi }}{N}$$
2. $$\frac{{\sin \frac{{\omega N}}{2}}}{{\frac{\omega }{2}}}\;and\frac{\pi }{N}$$
3. $$\frac{{\sin \frac{\omega }{2}}}{{\sin \frac{{\omega N}}{2}}}\;and\frac{{2\pi }}{N}$$
4. $$\frac{{\sin \frac{{\omega N}}{4}}}{{\sin \frac{\omega }{2}}}\;and\frac{{8\pi }}{N}$$
Option 1 : $$\frac{{\sin \frac{{\omega N}}{2}}}{{\sin \frac{\omega }{2}}}\;and\frac{{4\pi }}{N}$$
#### Discrete Time Fourier Transform (DTFT) Question 5 Detailed Solution
The rectangular window is the most common windowing technique to design a finite impulse response (FIR) filter.
A rectangular window is defined as
$${w_R}\left( n \right) = \left\{ {\begin{array}{*{20}{c}} {1\;:0 \le n \le N - 1}\\ {0\;\;\;\;:otherwise} \end{array}} \right.$$
Where N = length of the FIR filter
The frequency response of wR(n) is wR(ω) which is calculated as:
wR(ω) = DTFT of wR(n), i.e.
$${w_R}\left( \omega \right) = \mathop \sum \limits_{n = - \infty }^\infty {w_R}\left( n \right){e^{ - j\omega n}}$$
For the given rectangular sequence, the Fourier transform will be a sinc sequence given by:
$${w_R}\left( \omega \right) \approx \frac{{\sin \left( {\frac{{\omega N}}{2}} \right)}}{{\sin \left( {\frac{\omega }{2}} \right)}}$$
The spectrum is as shown:
Main-lobe width = 2 × (zero-crossing points of sin c function), i.e.
$$= 2 \times \frac{{2\pi }}{N} = \frac{{4\pi }}{N}$$
Types of Window Approximate main lobe width 1. Rectangular $$\frac{{4\pi }}{N}$$ 2. Bartlett $$\frac{{8\pi }}{N}$$ 3. Hanning $$\frac{{8\pi }}{N}$$ 4. Hamming $$\frac{{8\pi }}{N}$$ 5. Blackman $$\frac{{12\pi }}{N}$$
#### Discrete Time Fourier Transform (DTFT) Question 6:
The value of the following summation $$\underset{n=0}{\overset{\infty }{\mathop \sum }}\,n{{\left( \frac{1}{3} \right)}^{n}}$$ is _______.(up to two decimal places)
#### Discrete Time Fourier Transform (DTFT) Question 6 Detailed Solution
Concept:
$$f\left( n \right)\overset{DTFT}{\mathop{\leftrightarrow }}\,F\left( \omega \right)$$
$$n~f\left( n \right)\leftrightarrow \frac{jdF\left( \omega \right)}{d\omega }$$
Also, the discrete-time Fourier Transform of a sequence f(n) is given by:
$$F\left( \omega \right) = \mathop \sum \limits_{ - \infty }^\infty f\left( n \right){e^{ - j\omega n}}$$
Now, the DTFT of ‘nf(n)’ will be:
$$j\frac{d~F\left( \omega \right)}{d\omega }= \mathop \sum \limits_{ - \infty }^\infty nf\left( n \right){e^{ - j\omega n}}$$
At, ω = 0
$$j\frac{dF\left( 0 \right)}{d\omega }=\mathop \sum \limits_{ - \infty }^\infty nf\left( n \right)$$ ---(1)
Calculation:
$$f\left( x \right)={{\left( \frac{1}{3} \right)}^{n}}u\left( n \right)$$
$$F\left( \omega \right)=\frac{1}{1-\left( \frac{1}{3} \right){{e}^{-j\omega }}}$$
$$n{{\left( \frac{1}{3} \right)}^{n}}u\left( n \right)\leftrightarrow j\frac{dF\left( \omega \right)}{d\omega }$$
$$j\frac{dF\left( \omega \right)}{d\omega }=j\left[ \frac{\left( \frac{1}{3} \right)\left( {{e}^{-j\omega }} \right)\left( -j \right)}{{{\left( 1-\left( \frac{1}{3} \right){{e}^{-j\omega }} \right)}^{2}}} \right]$$
$$=\frac{\frac{1}{3}{{e}^{-j\omega }}}{{{\left( 1-\frac{1}{3}{{e}^{-j\omega }} \right)}^{2}}}$$
From Equation (1),
$$\mathop \sum \limits_{ - \infty }^\infty\,nf\left( n \right)u\left( n \right)=\mathop \sum \limits_{ 0 }^\infty\,nf\left( n \right)$$
$$=\mathop \sum \limits_{ 0 }^\infty\,n{{\left( \frac{1}{3} \right)}^{n}}u\left( n \right)=j\frac{dF\left( 0 \right)}{d\omega }$$
$$f\left( n \right)={{\left( \frac{1}{3} \right)}^{n}}u\left( n \right)$$
$$j\frac{dF\left( 0 \right)}{d\omega }=\frac{\frac{1}{3}}{{{\left( 1-\frac{1}{3} \right)}^{2}}}$$
$$=\frac{1}{3\times {{\left( \frac{2}{3} \right)}^{2}}}=\frac{1}{3}\times \frac{9}{4}$$
$$=\frac{3}{4}=0.75$$
#### Discrete Time Fourier Transform (DTFT) Question 7:
The energy spectrum density value of $$x\left( n \right) = {\left( {\frac{1}{2}} \right)^n}u\left( n \right)\;at\;\omega = 0$$ is _____.
#### Discrete Time Fourier Transform (DTFT) Question 7 Detailed Solution
Since the given signal is aperiodic it has energy spectral density and not power spectral density.
$$X\left( {{e^{j\omega }}} \right) = \frac{1}{{1 - \frac{1}{2}{e^{ - j\omega }}}}$$
$${S_{xx}}\left( \omega \right) = {\left| {X\left( {{e^{j\omega }}} \right)} \right|^2}$$
$$= {\left[ {\frac{1}{{\sqrt {{{\left( {1 - \frac{1}{2}\cos \omega } \right)}^2} + {{\left( {\frac{1}{2}\sin \omega } \right)}^2}} }}} \right]^2}$$
$$= \frac{1}{{1 + {{\left( {\frac{1}{2}} \right)}^2} - 2\left( {\frac{1}{2}} \right)\cos \omega }}$$
At ω = 0
$$= \frac{1}{{1 + {{\left( {\frac{1}{2}} \right)}^2} - 1}}$$
= 4
#### Discrete Time Fourier Transform (DTFT) Question 8:
The impulse response of the causal LTI system that is characterised by the difference equation $$y\left[ n \right] - \frac{3}{4}y\left[ {n - 1} \right] + \frac{1}{8}y\left[ {n - 2} \right] = 2x\left[ n \right]$$ is
1. $$4{\left( {\frac{1}{4}} \right)^n4} u \left[ n \right] - 2{\left( {\frac{1}{2}} \right)^n} u \left[ n \right]$$
2. $$4{\left( {\frac{1}{2}} \right)^n} u \left[ n \right] - 2{\left( {\frac{1}{4}} \right)^n} u \left[ n \right]$$
3. $$4{\left( {\frac{3}{4}} \right)^n}u \left[ n \right] - 2{\left( {\frac{1}{8}} \right)^n} u\left[ n \right]$$
4. $$3{\left( {\frac{1}{4}} \right)^n} u \left[ n \right] - 2{\left( {\frac{1}{2}} \right)^n} u \left[ n \right]$$
Option 2 : $$4{\left( {\frac{1}{2}} \right)^n} u \left[ n \right] - 2{\left( {\frac{1}{4}} \right)^n} u \left[ n \right]$$
#### Discrete Time Fourier Transform (DTFT) Question 8 Detailed Solution
The frequency response
$$\begin{array}{l} H\left( {{e^{j\omega }}} \right) = \frac{{y\left( {{e^{j\omega }}} \right)}}{{x\left( {{e^{j\omega }}} \right)}}\\ = \frac{2}{{1 - \frac{3}{4}{e^{ - j\omega }} + \frac{1}{8}{e^{ - j2\omega }}}}\\ = \frac{2}{{\left( {1 - \frac{1}{2}{e^{ - j\omega }}} \right)\left( {1 - \frac{1}{4}{e^{ - j\omega }}} \right)}}\\ = \frac{4}{{1 - \frac{1}{2}{e^{ - j\omega }}}} - \frac{2}{{1 - \frac{1}{4}{e^{ - j\omega }}}} \end{array}$$
Taking inverse Fourier transform
$$h\left[ n \right] = 4{\left( {\frac{1}{2}} \right)^n}u\left[ n \right] - 2{\left( {\frac{1}{4}} \right)^n}u\left[ n \right]$$
#### Discrete Time Fourier Transform (DTFT) Question 9:
The signal $$sin (\sqrt{2 \pi t})$$ is
1. periodic with period $$T = \sqrt{2 \pi}$$
2. not periodic
3. periodic with period T = 2π
4. periodic with period T = 4π2
Option 2 : not periodic
#### Discrete Time Fourier Transform (DTFT) Question 9 Detailed Solution
Concept:
Periodic Signal:
A signal x(t) is said to be periodic if for some positive constant T0;
x(t) = x (t + T0) for all t
The smallest value of T0 that satisfies the periodicity condition of the above equation is called the fundamental period of x(t).
Periodic signal repeat with some fixed time period T and continue with -∞ to +∞
The sum of two periodic functions is often periodic, but not always.
Suppose that x(t) is periodic with fundamental period P, and g(t) is periodic with fundamental period Q. Then x(t) + g(t) will be periodic with a period R, if P and Q both have a common multiple R.
Example: sin x + sin πx
sin x is periodic with a period 2π
sin πx is periodic with a period 2
But since 2π and 2 have no common multiple, the sum of the two signals will not be periodic.
Given:
x(t) = $$sin (\sqrt{2 \pi t})$$
• For negative values of ‘t, sine-function will apply to imaginary values.
• For positive values of ‘t’, sine-function will apply on real-values.
• So, there will not be repetition in the waveform.
So We can draw waveform only in the RHS but we cannot draw waveform in the LHS.
So, sin( 2πt ) is non-periodic.
Important Points
• Any analog sinusoid or harmonic signal is always periodic. As the sinusoids are periodic and harmonic are expressed in terms of sinusoids only.
• Additional phase or Amplitude will not affect the periodicity of a signal.
• The Sum of two periodic discrete-time signals is always periodic.
• All periodic signals are power signals but vice versa may not be true.
• The sequence obtained by uniform sampling of a continuous-time signal may not be periodic.
#### Discrete Time Fourier Transform (DTFT) Question 10:
For certain sequences which are neither absolutely summable nor square summable, it is possible to have a Fourier Transform (FT) representation if we
1. take short time FT
2. evaluate FT only the real part of the sequence
3. allow DTFT to contain impulses
4. evaluate FT over a period time span | 4,851 | 12,932 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2023-14 | longest | en | 0.38459 |
https://socratic.org/questions/how-do-you-simplify-1-2-ln-4t-4-ln-2#262296 | 1,638,641,389,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362999.66/warc/CC-MAIN-20211204154554-20211204184554-00007.warc.gz | 576,618,485 | 6,129 | # How do you simplify 1/2 ln (4t^4) - ln 2?
May 4, 2016
This is asking you to remember the properties of logarithms. Here are the ones you need to know:
• $\setminus m a t h b f \left(\ln {a}^{b} = b \ln a\right)$
• $\setminus m a t h b f \left(c \ln a - c \ln b = c \ln \setminus \frac{a}{b}\right)$
So, we can start by getting that exponent out in front:
$\frac{1}{2} \ln 4 {t}^{4} - \ln 2$
$= \frac{1}{2} \ln {\left(2 {t}^{2}\right)}^{2} - \ln 2$
Be careful that you do the above step correctly. It would be incorrect to change $\ln 4 {t}^{4}$ to $4 \ln 4 t$, because the exponent only applied to the $t$ at that time.
(If you did, you would imply that the expression was $\ln {\left(4 t\right)}^{4} = \ln 256 {t}^{4} \ne \ln 4 {t}^{4}$.)
Now, the exponent applies to the quantity $2 {t}^{2}$, so we are justified in moving the power out to the front as a coefficient!
$= \cancel{\frac{1}{2}} \cdot \cancel{2} \ln 2 {t}^{2} - \ln 2$
$= \ln 2 {t}^{2} - \ln 2$
With the same coefficients $c = 1$ in front, we can now turn this into a fraction:
$= \ln \setminus \frac{\cancel{2} {t}^{2}}{\cancel{2}}$
$= \ln {t}^{2}$
$= \textcolor{b l u e}{2 \ln t}$ | 444 | 1,164 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 15, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2021-49 | latest | en | 0.742609 |
https://dividedby.org/991000-divided-by-40 | 1,713,663,464,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817699.6/warc/CC-MAIN-20240421005612-20240421035612-00077.warc.gz | 186,311,097 | 23,756 | Home » Divided by 40 » 991000 Divided by 40
# 991000 Divided by 40
Welcome to 991000 divided by 40, our post which explains the division of nine hundred ninety-one thousand by forty to you. 🙂
The number 991000 is called the numerator or dividend, and the number 40 is called the denominator or divisor.
The quotient (integer division) of of 991000 and 40, the ratio of 991000 and 40, as well as the fraction of 991000 and 40 all mean (almost) the same:
991000 divided by 40, often written as 991000/40.
Read on to find the result in various notations, along with its properties.
## Calculator
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24775
=
24775 Remainder 0
The Long Division Steps are explained here. Read them now!
## What is 991000 Divided by 40?
We provide you with the result of the division 991000 by 40 straightaway:
991000 divided by 40 = 24775
The result of 991000/40 is an integer, which is a number that can be written without decimal places.
• 991000 divided by 40 in decimal = 24775
• 991000 divided by 40 in fraction = 991000/40
• 991000 divided by 40 in percentage = 24775%
Note that you may use our state-of-the-art calculator above to obtain the quotient of any two integers or whole numbers, including 991000 and 40, of course.
Repetends, if any, are denoted in ().
The conversion is done automatically once the nominator, e.g. 991000, and the denominator, e.g. 40, have been inserted.
To start over overwrite the values of our calculator.
## What is the Quotient and Remainder of 991000 Divided by 40?
The quotient and remainder of 991000 divided by 40 = 24775 R 0
The quotient (integer division) of 991000/40 equals 24775; the remainder (“left over”) is 0.
991000 is the dividend, and 40 is the divisor.
In the next section of this post you can find the additional information in the context of nine hundred ninety-one thousand over forty, followed by the summary of our information.
Observe that you may also locate many calculations such as 991000 ÷ 40 using the search form in the sidebar.
The result page lists all entries which are relevant to your query.
Give the search box a go now, inserting, for instance, nine hundred ninety-one thousand divided by forty, or what’s 991000 over 40 in decimal, just to name a few potential search terms.
Further information, such as how to solve the division of nine hundred ninety-one thousand by forty, can be found in our article Divided by, along with links to further readings.
## Conclusion
To sum up, 991000/40 = 24775. It is a whole number with no fractional part.
As division with remainder the result of 991000 ÷ 40 = 24775 R 0.
You may want to check out What is a Long Division?
For questions and comments about the division of 991000 by 40 fill in the comment form at the bottom, or get in touch by email using a meaningful subject line.
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Thanks for visiting our article explaining the division of 991000 by 40.
For a list of our similar sites check out the sidebar of our home page. – Article written by Mark, last updated on December 10th, 2023 | 833 | 3,332 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2024-18 | latest | en | 0.929679 |
https://mattcoaty.com/2015/11/21/multi-digit-multiplication-strategies/ | 1,500,662,892,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423808.34/warc/CC-MAIN-20170721182450-20170721202450-00259.warc.gz | 676,188,587 | 39,297 | ## Multi-digit Multiplication Strategies
This past week my third grade class investigated different ways to multiply numbers. Before diving into this concept I asked the students their thoughts on multiplication. A few students explained to the class their view on the topic of multiplication.
• double or triple “hopping”
• using arrays
• “timesing”
• Increase the number by “a lot”
Most students were able to showcase examples of the above. Even though their vocabulary wasn’t exactly spot-on, students were able to come to the whiteboard and show their thinking.
I received different responses from the students when asking them about multi-digit multiplication. Actually, it was more of a lack of response. I feel like some of this is due to exposure. A few students raised their hands and asked to show their process to multiply multi-digit numbers. These students showcased their ability to use the traditional algorithm. The class reviewed this method with a few examples. Although students were finding the correct product they had trouble explaining the process. Students weren’t able to communicate why it worked or another method to find a solution.
On Tuesday my class started to explore the partial-products algorithm. Students were able to decompose individual products and find the sum. This made sense to students. Students were able to connect an area model with the partial-products method. They started to write number models right next to each partial-product.
Later in the week students were introduced to the lattice method. This method seemed “fun” for the students, but didn’t make as much sense as the partial-products method. Students were able draw the boxes and create diagonals to find the product. Some students had trouble with laying the boxes out before multiplying.
During the last day of the week students were asked to explain in written form how to multiply multi-digit numbers. Even though all of the students could use the traditional, partial-products and lattice methods, they were stuck for a bit. Soon, most students started to lean towards using the partial-products method to explain how and why this method works. I asked one student in particular why it made sense and she said “I can see it visually and in number form.” Although most students were able to use the other methods effectively they didn’t seem confident enough to explain why the strategies worked.
Students will be expected to multiply multi-digit numbers on the next unit assessment. The method to multiply these numbers will be determined by the student, but I’m wondering how many will gravitate towards the strategy (not just the process) that they understand. | 547 | 2,704 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2017-30 | longest | en | 0.976902 |
https://physics.stackexchange.com/questions/710906/if-the-escape-velocity-at-the-event-horizon-is-the-speed-of-light-does-it-mean-t | 1,656,864,334,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104244535.68/warc/CC-MAIN-20220703134535-20220703164535-00393.warc.gz | 493,497,804 | 67,448 | # If the escape velocity at the event horizon is the speed of light does it mean that slower bodies won't move away at all?
If we say that the escape velocity from a planet is say 10 km/s we think that a slower body will move away from that planet but will be eventually forced to fall back on the planet. In simple words we don't say the body won't move at all but it couldn't leave for ever the planet. What is confusing for me is the escape velocity at the black hole event horizon. If it is the speed of light does it mean that a slower body would leave the horizon but fall down again or that is impossible for that body to make a path at all even 1mm away from the event horizon?
• You can't throw a mass up in the first place. May 27 at 20:41
Your arguments make sense in pre-relativistic Physics, but they don't apply to actual black holes, which are described by General Relativity. In this answer, I'll try to briefly sketch the difference.
### "Black Stars" in Newtonian Gravity
In Newtonian gravity, it is not difficult to compute the escape velocity at a distance $$d$$ from the center of a spherical mass to be $$v = \sqrt{\frac{2GM}{d}},$$ which is even quoted at Wikipedia. In this framework of Classical Mechanics, we can then ask ourselves: what if a massive body was so compact that the escape velocity at its surface would be $$c$$? With some algebra, we figure out that the distance $$d$$ needed for this to happen is $$d = \frac{2 G M}{c^2}.$$
In this framework, your arguments are correct. Suppose you have a spherical body of mass $$M$$ and radius, say, $$R = 0.99 d$$. Then one could shoot a rock upwards at speeds below that of light, it would be able to surpass the position $$r = d$$ and would eventually fall back. The limitation would be simply that objects with speeds below that of light would need to fall back eventually.
However, this argumentation is based on pre-relativistic Physics. In particular, it employs Newtonian gravity, which is not an appropriate description of strong gravitational fields and is inconsistent with the Special Theory of Relativity.
### Black Holes in General Relativity
In General Relativity, the setup is completely different. $$c$$ is now an incredibly important parameter in the theory that not only describes the speed of light. It describes the speed of causality itself. What happens this time is that one tries to describe the gravitational field of a spherical object with mass $$M$$ and finds that is it done by means of a curved spacetime with metric given by $$\mathrm{d}s^2 = -\left(1 - \frac{2G M}{c^2 r}\right)c^2\mathrm{d}t^2 + \left(1 - \frac{2G M}{c^2 r}\right)^{-1}\mathrm{d}r^2 + r^2 \mathrm{d}\Omega^2.$$
In this setup, what happens at $$r = \frac{2GM}{c^2}$$ is far different. Previously, in pre-relativistic Physics, this equation determined a region of space. In GR, this equation determines a region of spacetime that does not behave as a spatial region, but rather it is what is called a null surface. In layperson's terms, it behaves a bit more as an instant in time than as a surface in space. In this framework, crossing this region is not like stepping through your doorstep, but rather like literally walking backwards or forwards in time. Nothing can escape the event horizon without literally moving backwards in time.
The previous concepts of escape velocity or even of force don't really make sense anymore. If you attempt to define the force needed to hover over $$r = \frac{2 G M}{c^2}$$ (i.e., to stay fixed at this radius without falling in), you'll find out that this force is infinite, which is greatly different from the Newtonian prediction. The reason is simply that the Newtonian theory fails to describe these strong fields, and one must instead resort to a different theory (namely, General Relativity).
### In Short
No, after falling in, it is impossible for any body to move even a millimeter past the event horizon. To get out of the event horizon, as little as it might be, would actually be equivalent to travelling to the past. The apparent difficulty in reconciling this with the notion of escape velocity is that escape velocity is a notion appropriate to pre-relativistic Physics, while black holes and the event horizon are described in a different framework—namely, General Relativity.
The people who first postulated the existence of an object like a star or planet massive enough that the escape velocity would be equal to that of light did apparently think in those terms (light struggling to leave the object but inevitably being drawn back in, or light beams hovering motionlessly in space right next to the object).
Then hundreds of years later when general relativity was understood to predict the existence of black holes and they (later) became subjects of study by mathematical physicists, those ways of imagining the workings of a black hole had to be abandoned and replaced with something entirely different.
My understanding of this is that once any object falls through the event horizon of an uncharged, nonspinning black hole its world line points in only one direction- towards the singularity- and so it gets carried in that direction in the same way that it gets carried forward in time. Since no world lines can bend outward to cross the EH on an exit path, it makes no sense to think of a photon for instance on the inside of a black hole trying to head out, slowing down, and falling inwards again.
This is a frightfully complicated business and I hope someone here can give you a more complete answer than this. | 1,239 | 5,573 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 13, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2022-27 | longest | en | 0.958571 |
http://www.solving-math-problems.com/solve-quadratic-equation.html | 1,516,319,559,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887660.30/warc/CC-MAIN-20180118230513-20180119010513-00494.warc.gz | 571,118,608 | 12,860 | by Natasha
(MD)
Second Degree Polynomial
Solve the following quadratic equation for x.
15 = 100x - 5x²
Use the quadratic formula, complete the square, solving by factoring, or use the Indian method.
Sep 09, 2013 quadratic equation by: Staff Answer Part I Solve the following quadratic equation for x. 15 = 100x - 5x² This problem can be solved easily using the quadratic formula The first step is to rewrite the equation in the standard format Standard Format of quadratic equation: - 5x² + 100x = 15 subtract 15 from each side of the equation - 5x² + 100x - 15= 15 - 15 - 5x² + 100x - 15= 0 divide each side of the equation by 5 (- 5x² + 100x - 15) / 5 = 0 / 5 - 5x² / 5 + 100x / 5 - 15 / 5 = 0 / 5 - x² + 20x - 3 = 0 ------------------------------------------------------
Sep 09, 2013 quadratic equation by: Staff ------------------------------------------------------ Part II Quadratic formula: - x² + 20x - 3 = 0 x = unknown a = -1 b = 20 c = -3 Substitute the values of the coefficients a, b, and c into the quadratic formula ------------------------------------------------------
Sep 09, 2013 quadratic equation by: Staff ------------------------------------------------------ Part III Complete the arithmetic to find the value of "x" the final answer is: ------------------------------------------------------
Sep 09, 2013 quadratic equation by: Staff ------------------------------------------------------ Part IV A graph of the quadratic function is shown below Thanks for writing. Staff www.solving-math-problems.com | 378 | 1,538 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2018-05 | longest | en | 0.715698 |
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# The arithmetic mean (average) of a set of 10 numbers is 10.
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The arithmetic mean (average) of a set of 10 numbers is 10. [#permalink] 26 Sep 2007, 07:07
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The arithmetic mean (average) of a set of 10 numbers is 10. Is the median value of the same set also equal to 10?
1) Exactly half of the numbers are less than 10.
2) The mode of the set of numbers is 10.
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I am sure that it can't be (A). Here is why :
Median = (5th + 6th)/2
From (1), we know that 5th < 10, but we know nothing about 6th.
Case 1 : 5th = 9, 6th = 10, median = 19/2
Case 2 : 5th = 9, 6th = 11, median = 10
Therefore (1) is insufficient.
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E
Although half of the numbers are less than 10 in wht order are they placed. it can be 1-5 or 2-6 so on.
mode 10 still not suff.
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Re: DS : Mean/Median/Mode [#permalink] 26 Sep 2007, 11:04
coldweather999 wrote:
The arithmetic mean (average) of a set of 10 numbers is 10. Is the median value of the same set also equal to 10?
1) Exactly half of the numbers are less than 10.
2) The mode of the set of numbers is 10.
I'm getting C. Knowing both Stat 1 and Stat 2, we can say that the median value will never be 10.
Median = (5th term + 6th term)/2
Stat 1:
set = 5, 5, 5, 5, 5, 15, 15, 15, 15, 15
In this case median = 10.
set = 4, 4, 4, 4, 9, 15, 15, 15, 15, 15
median is not equal to 10.
Insuff
Stat 2:
set = 10 times 10.
Median = 10
set = 3, 4, 4, 5, 9, 10, 10, 10, 20, 25
median is not equal to 10
Insuff.
Stat 1 & 2:
Since exactly half the values are under 10 then the 5th term has to be less than 10. Since the mode is 10, the 6th term has to be 10.
(10+number less than 10)/2 has to be less than 10.
Suff.
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I get the same - C - using the same rationale as GK_GMAT. What's the OA?
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Bluebird wrote:
I get the same - C - using the same rationale as GK_GMAT. What's the OA?
I also got (C) using the same logic as GK_GMAT...(i didn't assume the numbers to be whole numbers since it's not mentioned anywhere)...but the OA is (E)
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Re: DS : Mean/Median/Mode [#permalink] 26 Sep 2007, 13:02
coldweather999 wrote:
The arithmetic mean (average) of a set of 10 numbers is 10. Is the median value of the same set also equal to 10?
1) Exactly half of the numbers are less than 10.
2) The mode of the set of numbers is 10.
C for me.
(1)
1,2,3,4,5,15,16,17,18,19 gives you median of 10
1,2,3,4,8,15,16,17,18,22 gives you median more than 10
INSUFFICIENT
(2)
If all numbers are 10, then median =10
1,2,3,4,5,6,10,10, large, large gives median not equal to 10
INSUFFICIENT
Together, you can have
1,2,3,4,5,10,10,20,22,23
gives you median of not 10
Say a<10
You know that median = (10+a)/2
Since a is always less than 10, the median will never equals to 10.
SUFFICIENT
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why does the 6th term have to be 10 ?
It could be 11, and the 5th term could be 9, which give median of 10
Or, the 6th term could be 25, and the 5th term could be 3, therefore giving a median of 15
Just because the mode is 10, I dont think it means that the 6th term has to be 10
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pmenon wrote:
why does the 6th term have to be 10 ?
It could be 11, and the 5th term could be 9, which give median of 10
Or, the 6th term could be 25, and the 5th term could be 3, therefore giving a median of 15
Just because the mode is 10, I dont think it means that the 6th term has to be 10
The question said exactly half the terms (5 terms) are less than 10. The mode is 10. This means the sixth term must be 10.
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bkk145 wrote:
pmenon wrote:
why does the 6th term have to be 10 ?
It could be 11, and the 5th term could be 9, which give median of 10
Or, the 6th term could be 25, and the 5th term could be 3, therefore giving a median of 15
Just because the mode is 10, I dont think it means that the 6th term has to be 10
The question said exactly half the terms (5 terms) are less than 10. The mode is 10. This means the sixth term must be 10.
Mode refers to the value in a set that is most repeated, right ? If that is the case, how does it mean that the sixth term has to be 10 ?
As long as 10 is the most repeated number in the set, thats all that matters right ?
Forgive me ! lol
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St1:
The set could be {0,0,0,0,1,10,10,10,10,59}. Mean = 10, Median = 5.5
The set could be {0,0,0,0,0,20,20,20,20,20}. Mean = 10, Median = 10.
Insufficient.
St2:
The set could be {10,10,10,10,10,10,10,10,10,10}. Mode = Mean = Median = 10
The set could be {1,2,3,4,5,6,10,10,10,49}. Mode = 10. Mean = 10. Median = 5.5
Insufficient.
St1 and St2:
The set could be {1,2,3,4,5,10,10,10,10,45}. Mode = Mean = 10. Median = 7.5
The set could eb {0,0,0,0,0,10,10,10,10,60}. Mode = Mean = 10. Median = 5.
Insufficient.
Ans E
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ywilfred wrote:
St1:
The set could be {0,0,0,0,1,10,10,10,10,59}. Mean = 10, Median = 5.5
The set could be {0,0,0,0,0,20,20,20,20,20}. Mean = 10, Median = 10.
Insufficient.
St2:
The set could be {10,10,10,10,10,10,10,10,10,10}. Mode = Mean = Median = 10
The set could be {1,2,3,4,5,6,10,10,10,49}. Mode = 10. Mean = 10. Median = 5.5
Insufficient.
St1 and St2:
The set could be {1,2,3,4,5,10,10,10,10,45}. Mode = Mean = 10. Median = 7.5
m
The set could eb {0,0,0,0,0,10,10,10,10,60}. Mode = Mean = 10. Median = 5.
Insufficient.
Ans E
how does it make C insufficient?
If we take both sets together 10 can never be median..question solved.
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rishi2377 wrote:
ywilfred wrote:
St1:
The set could be {0,0,0,0,1,10,10,10,10,59}. Mean = 10, Median = 5.5
The set could be {0,0,0,0,0,20,20,20,20,20}. Mean = 10, Median = 10.
Insufficient.
St2:
The set could be {10,10,10,10,10,10,10,10,10,10}. Mode = Mean = Median = 10
The set could be {1,2,3,4,5,6,10,10,10,49}. Mode = 10. Mean = 10. Median = 5.5
Insufficient.
St1 and St2:
The set could be {1,2,3,4,5,10,10,10,10,45}. Mode = Mean = 10. Median = 7.5
m
The set could eb {0,0,0,0,0,10,10,10,10,60}. Mode = Mean = 10. Median = 5.
Insufficient.
Ans E
how does it make C insufficient?
If we take both sets together 10 can never be median..question solved.
my bad.... it should be C. I thought the question was asking for the value of the median in which case the answer would be E.
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Re: DS : Mean/Median/Mode [#permalink] 10 May 2008, 23:09
****makes no sense so Deleted*****
Last edited by [email protected] on 11 May 2008, 05:54, edited 1 time in total.
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Re: DS : Mean/Median/Mode [#permalink] 11 May 2008, 05:43
[email protected] wrote:
Should be 'E'.
S1 + S2 not suff
case1 - set = {7,8,9,10,10,16} mean=median=mode=10 and 3 elements are smaller than 10.
case2- set={1,2,3,4,10,10,10,40} mean=10, mode=10, median=7 and 4 elements and smaller than 10.
C)
case1 - median will be (9 + 10)/2 = 9.5 <10
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Re: DS : Mean/Median/Mode [#permalink] 11 May 2008, 05:52
tarkumar wrote:
[email protected] wrote:
Should be 'E'.
S1 + S2 not suff
case1 - set = {7,8,9,10,10,16} mean=median=mode=10 and 3 elements are smaller than 10.
case2- set={1,2,3,4,10,10,10,40} mean=10, mode=10, median=7 and 4 elements and smaller than 10.
C)
case1 - median will be (9 + 10)/2 = 9.5 <10
Oops..wrong numbers....
yes it should be 'C'
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Re: [#permalink] 12 May 2008, 12:05
ywilfred wrote:
St1:
The set could be {0,0,0,0,1,10,10,10,10,59}. Mean = 10, Median = 5.5
The set could be {0,0,0,0,0,20,20,20,20,20}. Mean = 10, Median = 10.
Insufficient.
St2:
The set could be {10,10,10,10,10,10,10,10,10,10}. Mode = Mean = Median = 10
The set could be {1,2,3,4,5,6,10,10,10,49}. Mode = 10. Mean = 10. Median = 5.5
Insufficient.
St1 and St2:
The set could be {1,2,3,4,5,10,10,10,10,45}. Mode = Mean = 10. Median = 7.5
The set could eb {0,0,0,0,0,10,10,10,10,60}. Mode = Mean = 10. Median = 5.
Insufficient.
Ans E
Quick question.
In you example St1 and St2. Wouldn't the mode be 0 as it's used 5 times and 10 is used 4 times.
Re: [#permalink] 12 May 2008, 12:05
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• ### $450 Tuition Credit & Official CAT Packs FREE November 15, 2018 November 15, 2018 10:00 PM MST 11:00 PM MST EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) # In the circle above, PQ is parallel to diameter OR, and OR post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message SVP Joined: 30 Apr 2008 Posts: 1832 Location: Oklahoma City Schools: Hard Knocks In the circle above, PQ is parallel to diameter OR, and OR [#permalink] ### Show Tags 10 Jun 2008, 16:07 7 11 00:00 Difficulty: 75% (hard) Question Stats: 68% (01:48) correct 32% (02:26) wrong based on 246 sessions ### HideShow timer Statistics Attachment: GMATCircle.jpg [ 20.23 KiB | Viewed 23006 times ] In the circle above, PQ is parallel to diameter OR, and OR has length 18. What is the length of minor arc PQ? A) $$2\pi$$ B) $$\frac{9\pi}{4}$$ C) $$\frac{7\pi}{2}$$ D) $$\frac{9\pi}{2}$$ E) $$3\pi$$ OPEN DISCUSSION OF THIS QUESTION IS HERE: in-the-circle-above-pq-is-parallel-to-diameter-or-93977.html _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings Current Student Joined: 28 Dec 2004 Posts: 3251 Location: New York City Schools: Wharton'11 HBS'12 Re: PS - Geometry [#permalink] ### Show Tags 10 Jun 2008, 17:53 5 1 jallenmorris wrote: Attachment: GMATCircle.jpg In the circle above, PQ is parallel to diameter OR, and OR has length 18. What is the length of minor arc PQ? A) $$2\pi$$ B) $$\frac{9\pi}{4}$$ C) $$\frac{7\pi}{2}$$ D) $$\frac{9\pi}{2}$$ E) $$3\pi$$ hey you have to realize that the PQ and OR are parallel..having said that angle P will also be 35.. ok so now that you this..you draw another line from the center to P, to make a issocles triangle.. this angle will be 35 as well..now knowing this ..you yet again draw another line from the center to Q..and form an issoceles triangle..each angle will be 70 70.. so then the angle at the center O will be 40.. arc length will be 1/9*18pi..i.e 2pi SVP Joined: 30 Apr 2008 Posts: 1832 Location: Oklahoma City Schools: Hard Knocks Re: PS - Geometry [#permalink] ### Show Tags 10 Jun 2008, 20:02 5 1 jallenmorris wrote: Attachment: The attachment GMATCircle.jpg is no longer available In the circle above, PQ is parallel to diameter OR, and OR has length 18. What is the length of minor arc PQ? A) $$2\pi$$ B) $$\frac{9\pi}{4}$$ C) $$\frac{7\pi}{2}$$ D) $$\frac{9\pi}{2}$$ E) $$3\pi$$ Attachment: GMATCircleAnswer.jpg [ 20.99 KiB | Viewed 22861 times ] Thanks (kudos) for the explanation. In the image above, I realize now how to solve it from your explanation. Angle RPQ's measure is 35 degrees because of the rule that opposite angles created by a line through 2 parallel lines will be equal. (ORP and RPQ being the opposite angles created). Then the red line (seen in the image) is drawn creating an isosceles triangle out of PRC (C being the Center of the circle). Because 2 of the 3 sides are the same length, we know it's isosceles triangle and angle CRP = CPR. Step 2 draw the blue line. THis creates the 2nd isosceles triangle out of CPQ. Because angle RPQ is 35 degrees, and angle CPR is 35, we know that angle CPQ is 70 degrees, and because CPQ is isosceles, CQP is also 70 degrees. This makes PCQ = 40 degrees. Because the angle is 40 degrees, that means that the arc takes up 40/360 of the circumference of the circle. This is 1/9. The stem tells us the diameter is 18, and the formula for the circumference is d$$\pi$$, so we have 18$$\pi$$ * 1/9, or 2$$\pi$$. _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings Director Joined: 01 Jan 2008 Posts: 596 Re: PS - Geometry [#permalink] ### Show Tags 15 Jul 2008, 13:11 pi*r - ((2*2*35)/180)*pi*r = pi*r*2/9 = pi*9*2/9=2*pi -> A Intern Joined: 24 Jun 2008 Posts: 5 Re: PS - Geometry [#permalink] ### Show Tags 15 Jul 2008, 13:48 excellent! thank you. I couldn't solve it in 2 mins though. Manager Joined: 18 Jan 2008 Posts: 221 Schools: The School that shall not be named Re: PS - Geometry [#permalink] ### Show Tags 15 Jul 2008, 13:58 3 It's really not as complicated as J Allen's explanation.... Given: PQ is parallel to diameter OR 1) PRO = QPR = 35 2) Arc PRO = Arc QPR = 70 (35x2) Now, a circle = 360 degree Arc PQ = 360- Arc OR - Arc PO - Arc QR = 350-180-70-70=40 Here is the tricky part: circumference = 2*pi*r=pi*d (this is where I fell and got the wrong answer when I was practicing!) Circumference=18pi Arc PQ = 18pi*40/360=2pi Yup, totally doable under 2 mins....and you can get it correctly if you don't fall into the trap of the 2*pi*r thing like I did.... Fortunately 4*pi was not one of the options and it forced me to review my steps. _________________ 明日やろうばかやろう 24 wrote: Your East Coast bias is worse than ESPN. SVP Joined: 30 Apr 2008 Posts: 1832 Location: Oklahoma City Schools: Hard Knocks Re: PS - Geometry [#permalink] ### Show Tags 15 Jul 2008, 14:18 1 fatb, You mistake my long explanation as complicating the question. You forget that people come here to learn how to do these problems. Just because you can solve it in 30 seconds, or explain it in 3 lines doesn't mean it's easy for someone to understand your explanation. I go overboard on my explanations to make sure that ANYONE can understand the method I used so they can learn the concept. It doesn't do any good or improve this community, if all of the explanations are so short that no one can understand them. In order to help people learn, we have to simplify things...once they get it, there are certainly steps that can be done without writing it out, or consciously thinking about that step, but until people get to that point...they need all of the steps which it appears you think complicate the explanation. fatb wrote: It's really not as complicated as J Allen's explanation.... Given: PQ is parallel to diameter OR 1) PRO = QPR = 35 2) Arc PRO = Arc QPR = 70 (35x2) Now, a circle = 360 degree Arc PQ = 360- Arc OR - Arc PO - Arc QR = 350-180-70-70=40 Here is the tricky part: circumference = 2*pi*r=pi*d (this is where I fell and got the wrong answer when I was practicing!) Circumference=18pi Arc PQ = 18pi*40/360=2pi Yup, totally doable under 2 mins....and you can get it correctly if you don't fall into the trap of the 2*pi*r thing like I did.... Fortunately 4*pi was not one of the options and it forced me to review my steps. _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings Manager Joined: 18 Jan 2008 Posts: 221 Schools: The School that shall not be named Re: PS - Geometry [#permalink] ### Show Tags 15 Jul 2008, 18:11 I think the only underlying assumption that was made in my explanation was that people should know the following: the angle formed by any three points on a circle is the half of the measure of the arc subtended. _________________ 明日やろうばかやろう 24 wrote: Your East Coast bias is worse than ESPN. Director Joined: 14 Aug 2007 Posts: 680 Re: PS - Geometry [#permalink] ### Show Tags 16 Jul 2008, 18:18 1 heres how I did it. angle PRO = 35 so for arc PO central angle will be 70 and thus PO is 70/360 of the circumference which is 18*Pi = 7/2*pi Since lines PQ and OR are parallel, QR must also be 70/360 * 18 * pi = 7/2*pi The semicircle in the bottom part will be 1/2 *18 * pi = 9*pi length of arc PQ = circumference - lenght of arc PO - length of arc QR - length of semicircle = 18*pi - 7/2* pi - 7/2*pi - 9*pi = pi ( 18 - 16) = 2*pi Thus A Director Joined: 25 Oct 2008 Posts: 512 Location: Kolkata,India Re: PS - Geometry [#permalink] ### Show Tags 16 Aug 2009, 16:18 Guys major confusion! We know that minor arcs are always twice their inscribed angles right? So,minor arc PO=70=minor arc QR. minor arc PQ=180-70-70=40.So we have minor arc PQ=40. Same logic gives us Angle PCQ(C is the centre of the circle) as 20 deg.!!! Applying formula for length of an arc=2pi(radius)(angle)/360 =2pi(9)(20)/360=pi What did I do wrong? _________________ http://gmatclub.com/forum/countdown-beginshas-ended-85483-40.html#p649902 SVP Joined: 30 Apr 2008 Posts: 1832 Location: Oklahoma City Schools: Hard Knocks Re: PS - Geometry [#permalink] ### Show Tags 16 Aug 2009, 18:45 The problem you're having is that at first you use INSCRIBED angles, so you know 35 degrees, so minor arc PO = 70. This is correct. It also means that QR is 70 which is right. Then you use that 180-70-70 = 40 to get the degree measure of PQ. The problem you have here is that 40 is not the INSCRIBED angle. It is THE angle of the arc. The INSCRIBED angle is not the angle formed with center. The inscribed angle must be one end of the diameter to some point on the circle. We double that angle measurement to get the angle measured from point on the cirle to CENTER then to Point on the circle. We already have the 40 degree measurement for PQ. We don't need to find the INSCRIBED angle for this arc. Go to the formula you already have and use 40 instead of 20 and you get $$2pi$$ which is the correct answer. tejal777 wrote: Guys major confusion! We know that minor arcs are always twice their inscribed angles right? So,minor arc PO=70=minor arc QR. minor arc PQ=180-70-70=40.So we have minor arc PQ=40. Same logic gives us Angle PCQ(C is the centre of the circle) as 20 deg.!!! Applying formula for length of an arc=2pi(radius)(angle)/360 =2pi(9)(20)/360=pi What did I do wrong? _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings Intern Joined: 16 Aug 2009 Posts: 4 Re: PS - Geometry [#permalink] ### Show Tags 16 Aug 2009, 19:58 1 Tejal 777.. you are all most at it but minor mistake You found arc PQ = 40 Degree (in terms of length ) 360 degree = 2*Pi*r = 2*pi*9 40 degree = (2*pi*r*9/360 ) *40 = 2Pi Director Joined: 25 Oct 2008 Posts: 512 Location: Kolkata,India Re: PS - Geometry [#permalink] ### Show Tags 17 Aug 2009, 01:51 ok ok..i'm getting there..ALMOST understood.. So lemme see.. we need the measurement of minor arc PQ.I got the measurement of that but in terms of length which is 40.but the answer choices do not indicate it so we have to go furthur.Great. NOW,the formula we have, Length of an arc=(circumference) (INSCRIBED ANGLE)/360 If we put length=18 pi * 40/360 we are getting the answer..but what exactly did we do..?!how are we inserting 40 for the value of the inscribed angle? Waht a mess.. _________________ http://gmatclub.com/forum/countdown-beginshas-ended-85483-40.html#p649902 SVP Joined: 30 Apr 2008 Posts: 1832 Location: Oklahoma City Schools: Hard Knocks Re: PS - Geometry [#permalink] ### Show Tags 17 Aug 2009, 04:32 40 is not the value of an inscribed angle. In this problem, ORP is an example of an inscribed angle. I'll refer to the center as C, so it won't be confusing. Since we know ORP is 35 degrees. Arc OP is 70 degrees, which means that the angle of OCP (where C is the center) measures 70 degrees because it's double 35, and 35 is the measure of the inscribed angle. Because we know that PQ and OR are parallel, Arc OP and arc QR will be the same length. These 2 arcs take up the only space other than what we're trying to find, so you did the 180-70-70=40 for arc PQ. This is where the label of Inscribed was misapplied. 40 is the measure of arc PQ, not an inscribed angle. When we find the measure of an arc, we are always measuring the angle formed by the start of the arc (here P) to the center back out to the end point of the arc (here Q). This is NOT an inscribed angle. This is like a slice of pizza. Because we don't have an inscribed angle with value of 40, we do not need to figure out anything else with regard to an inscribed angle. We now have a slice of pizza that is 40/360 of the total size of the circle. So take $$\frac{2*pi*r*(angle-of-arc)}{360}$$ You're getting caught up in what the 40 degrees actually is. tejal777 wrote: ok ok..i'm getting there..ALMOST understood.. So lemme see.. we need the measurement of minor arc PQ.I got the measurement of that but in terms of length which is 40.but the answer choices do not indicate it so we have to go furthur.Great. NOW,the formula we have, Length of an arc=(circumference) (INSCRIBED ANGLE)/360 If we put length=18 pi * 40/360 we are getting the answer..but what exactly did we do..?!how are we inserting 40 for the value of the inscribed angle? Waht a mess.. _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$\$.
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Joined: 04 Sep 2010
Posts: 43
### Show Tags
30 May 2011, 19:07
1
A good question and uses very basic concepts though hides it well at first look of question:
a) An angle drawn from intersects of lines from circle's perimeter is half of angle drawn from intersects of lines from circles's center for the same ARC
b) straight line has angle of 180
c) Arch of circle will be angle from center/ 360 * 2pi* r
So assume circle center is C and draw two lines CP & CQ, so angle OCP is 70 (as it would double of angle ORP) ALSO similarly angle RCQ is 70 (Using concept a)
Using concept b) if OCP + RCQ = 140 then angle PCQ = 40 and it is at the center of circle
using concept c) 40/360 * 2* pi * 18/2 = 2 pi so answer is A
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### Show Tags
24 Sep 2011, 20:40
1
central angle = 2*inscribed angle
let C be the center of the circle.
given ANGLE PRC = 35
this angle PRC is nothing but inscribed angle of arc OP.
=> OP's central angle = 70
also given PQ || OR
=> ANGLE QPR = 35 (AS PRC is 35)
this angle QPR is nothing but inscribed angle of arc QR.
=> QR's central angle = 70.
Length of arc PQ = (PQ's Central angle/360)*(2*pi*r)
= ((180-70-70)/360)*(2*pi*9)
=2pi
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Re: In the circle above, PQ is parallel to diameter OR, and OR [#permalink]
### Show Tags
22 Jul 2013, 02:54
1
Good explanations. Just wanted to sum up that in such problems involving length of arc, the angle swiped by the arc can be found either by using CENTER ANGLE theorem, or by using the property of isosceles triangle, whereever 2 sides of a triangle is the radii.
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Posts: 50572
Re: In the circle above, PQ is parallel to diameter OR, and OR [#permalink]
### Show Tags
22 Jul 2013, 02:58
6
3
In the circle above, PQ is parallel to diameter OR, and OR has length 18. What is the length of minor arc PQ?
A. $$2\pi$$
B. $$\frac{9\pi}{4}$$
C. $$\frac{7\pi}{2}$$
D. $$\frac{9\pi}{2}$$
E. $$3\pi$$
The Central Angle Theorem states that the measure of inscribed angle is always half the measure of the central angle.
Let C be the center of the circle.
According to the central angle theorem above <PCO=2<PRO=70.
As PQ is parallel to OR, then <QPR=<PRO=35. Again, according to the central angle theorem above <QCR=2<QPR=70.
<PCQ=180-(<PCO+<QCR)=180-70-70=40.
Minor arc $$PQ=\frac{40}{360}*circumference=\frac{2\pi{r}}{9}=2\pi$$
OPEN DISCUSSION OF THIS QUESTION IS HERE: in-the-circle-above-pq-is-parallel-to-diameter-or-93977.html
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Re: In the circle above, PQ is parallel to diameter OR, and OR [#permalink]
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Re: In the circle above, PQ is parallel to diameter OR, and OR &nbs [#permalink] 02 Nov 2018, 13:04
Display posts from previous: Sort by | 4,953 | 17,352 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2018-47 | latest | en | 0.831519 |
https://www.enotes.com/topics/math/questions/sqrt-x-sqrt-y-y-0-y-1-9-find-particular-solution-764096 | 1,718,830,053,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861832.94/warc/CC-MAIN-20240619185738-20240619215738-00441.warc.gz | 670,939,980 | 17,583 | `sqrt(x) + sqrt(y)y' = 0 , y(1) = 9` Find the particular solution that satisfies the initial condition
For the given problem: `sqrt(x)+sqrt(y)y' =0,` we may rearrange this to
`sqrt(y)y' = -sqrt(x)`
Recall that `y'` is denoted as `(dy)/(dx)` then it becomes:
`sqrt(y)(dy)/(dx) = -sqrt(x)`
Apply the variable separable differential equation in a form of `f(y) dy = g(x) dx` .
`sqrt(y)(dy) = -sqrt(x)dx`
Apply direct integration using the Power Rule: `int u^n du = u^(n+1)/(n+1)` .
Note: `sqrt(x) = x^(1/2) and sqrt(y) = y^(1/2)` .
`int sqrt(y)(dy) = int -sqrt(x)dx`
`int y^(1/2) (dy) = int -x^(1/2)dx`
`y^(1/2+1)/(1/2+1)= -x^(1/2+1)/(1/2+1) +C`
`y^(3/2)/(3/2) = -x^(3/2)/(3/2)+C`
`y^(3/2)*2/3 = -x^(3/2)*2/3+C`
`2/3y^(3/2) = -2/3x^(3/2)+C`
The general solution of the differential equation is `2/3y^(3/2)= -2/3x^(3/2)+C` .
Using the given initial condition `y(1)=9` , we plug-in `x=1` and `y=9` to solve for C:
`2/3(9)^(3/2)= -2/3*1^(3/2)+C`
`2/3*27=-2/3*1+C`
`18=-2/3+C`
`C = 18+2/3`
`C = 56/3`
So,
`2/3y^(3/2)= -2/3x^(3/2)+56/3`
`y^(3/2)=(3/2)(-2/3x^(3/2)+56/3)`
`y^(3/2)=-x^(3/2)+28`
`(y^(3/2))^(2/3)=(-x^(3/2)+28)^(2/3)
`y =(-x^(3/2)+28)^(2/3)`
or
y = `root(3)((-x^(3/2)+28)^2)` | 591 | 1,210 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-26 | latest | en | 0.621987 |
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# GetWiki
### imaginary number
ARTICLE SUBJECTS
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imaginary number
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{{short description|Complex number defined by real number multiplied by imaginary unit "i"}}{{pp-pc1}}{{redirect2|Imaginary Number|Imaginary numbers|the 2013 EP by The Maine|Imaginary Numbers (EP)}}{| class="wikitable" style="float: right; margin-left: 1em; text-align: center;"
...}} (repeats the patternfrom blue area)
1=i−3 = i}}
1=i−2 = −1}}
1=i−1 = −i}}
{{math|1=i0 = 1}}
{{math|1=i1 = i}}
{{math|1=i2 = −1}}
{{math|1=i3 = −i}}
1=i4 = 1}}
1=i5 = i}}
1=i6 = −1}}
1=i'n = i'n(mod 4)}}
An imaginary number is a complex number that can be written as a real number multiplied by the imaginary unit {{mvar|i}},j is usually used in engineering contexts where i has other meanings (such as electrical current) which is defined by its property {{math|1=i2 = −1}}.BOOK,weblink Fundamentals of Waves and Oscillations, Uno Ingard, K., Cambridge University Press, 1988, 0-521-33957-X, 38, Chapter 2,
The square of an imaginary number {{mvar|bi}} is {{math|−b2}}. For example, {{math|5i}} is an imaginary number, and its square is {{math|−25}}. Zero is considered to be both real and imaginary.BOOK,weblink A Text Book of Mathematics Class XI, Sinha, K.C., Rastogi Publications, 2008, 978-81-7133-912-9, Second, 11.2,
Originally coined in the 17th century by René DescartesBOOK, Mathematical Analysis: Approximation and Discrete Processes, illustrated, Mariano, Giaquinta, Giuseppe, Modica, Springer Science & Business Media, 2004, 978-0-8176-4337-9, 121,weblink Extract of page 121 as a derogatory term and regarded as fictitious or useless, the concept gained wide acceptance following the work of Leonhard Euler and Carl Friedrich Gauss. An imaginary number {{math|bi}} can be added to a real number {{mvar|a}} to form a complex number of the form {{math|a + bi}}, where the real numbers {{mvar|a}} and {{mvar|b}} are called, respectively, the real part and the imaginary part of the complex number.BOOK, College Algebra: Enhanced Edition, 6th, Richard, Aufmann, Vernon C., Barker, Richard, Nation, Cengage Learning, 2009, 1-4390-4379-5, 66,weblink Both the real part and the imaginary part are defined as real numbers. Some authors use the term pure imaginary number to denote what is called here an imaginary number, and imaginary number to denote any complex number with non-zero imaginary part.BOOK, Plane Trigonometry: A New Approach, Johnston, C. L., Lazaris, Jeanne, Prentice Hall, 1991, 3rd, 247,
## History
, A history of non-euclidean geometry: evolution of the concept of a geometric space
, Boris Abramovich
, Rozenfeld
, Springer
, 1988
, 0-387-96458-4
, Chapter 10
, 382
In 1843 William Rowan Hamilton extended the idea of an axis of imaginary numbers in the plane to a four-dimensional space of quaternion imaginaries, in which three of the dimensions are analogous to the imaginary numbers in the complex field.With the development of quotient rings of polynomial rings, the concept behind an imaginary number became more substantial, but then one also finds other imaginary numbers such as the j of tessarines which has a square of {{math|+1}}. This idea first surfaced with the articles by James Cockle beginning in 1848.Cockle, James (1848) "On Certain Functions Resembling Quaternions and on a New Imaginary in Algebra", London-Dublin-Edinburgh Philosophical Magazine, series 3, 33:435–9 and Cockle (1849) "On a New Imaginary in Algebra", Philosophical Magazine 34:37–47
## Geometric interpretation
(File:Rotations on the complex plane.svg|thumb|90-degree rotations in the complex plane)Geometrically, imaginary numbers are found on the vertical axis of the complex number plane, allowing them to be presented perpendicular to the real axis. One way of viewing imaginary numbers is to consider a standard number line, positively increasing in magnitude to the right, and negatively increasing in magnitude to the left. At 0 on this {{mvar|x}}-axis, a {{mvar|y}}-axis can be drawn with "positive" direction going up; "positive" imaginary numbers then increase in magnitude upwards, and "negative" imaginary numbers increase in magnitude downwards. This vertical axis is often called the "imaginary axis" and is denoted {{math|iâ„}}, scriptstylemathbb{I}, or {{math|â„‘}}.In this representation, multiplication by {{math|–1}} corresponds to a rotation of 180 degrees about the origin. Multiplication by {{mvar|i}} corresponds to a 90-degree rotation in the "positive" direction (i.e., counterclockwise), and the equation {{math|1=i2 = −1}} is interpreted as saying that if we apply two 90-degree rotations about the origin, the net result is a single 180-degree rotation. Note that a 90-degree rotation in the "negative" direction (i.e. clockwise) also satisfies this interpretation. This reflects the fact that {{math|−i}} also solves the equation {{math|1=x2 = −1}}. In general, multiplying by a complex number is the same as rotating around the origin by the complex number's argument, followed by a scaling by its magnitude.
## Square roots of negative numbers
Care must be used when working with imaginary numbers expressed as the principal values of the square roots of negative numbers. For example:BOOK, An Imaginary Tale: The Story of "i" [the square root of minus one], Paul J., Nahin, Princeton University Press, 2010, 978-1-4008-3029-9, 12,weblink Extract of page 12
6=sqrt{36}=sqrt{(-4)(-9)} ne sqrt{-4}sqrt{-9} = (2i)(3i) = 6 i^2 = -6.
Sometimes this is written as:
-1 = i^2 = sqrt{-1}sqrt{-1} stackrel{text{ (fallacy) }}{=} sqrt{(-1)(-1)} = sqrt{1} = 1.
The fallacy occurs as the equality sqrt{xy} = sqrt{x}sqrt{y} does not hold when the variables are not suitably constrained. In this case the equality does not hold as the numbers are both negative. This can be demonstrated by,
sqrt{-x}sqrt{-y} = i sqrt{x} i sqrt{y} = i^2 sqrt{x} sqrt{y} = -sqrt{xy} neq sqrt{xy},
where both x and y are non-negative real numbers.
## Notes
{{reflist|group=note}}
{{reflist}}
## Bibliography
• BOOK, Paul, Nahin, An Imaginary Tale: the Story of the Square Root of −1, Princeton, Princeton University Press, 1998, 0-691-02795-1, , explains many applications of imaginary expressions.
{{Complex numbers}}{{Number systems}}
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# Pole in lake
115 replies to this topic
### #81 amanda
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Posted 09 June 2008 - 06:47 PM
I too came up with 12 feet.
1/3 is in the water and 8 feet is out of the water.
8 feet = 2/3
1/2 of that is 4 feet
So each 1/3 is 4 feet
4x3=12 feet
It doesn't matter how much is in the ground as long as you know how much is in and out of the water.
I see in your calculations where you've accounted for the 1/3 of the length that is in the water and the remaining 2/3 that are not in the ground. Where is the portion that is in the ground?
In US english, when you say something is "out of the water," it means that it is completely out of the lake or whatever body of water it is, and the space below it is not considered to be "out of the water." Also, when stated as: "One half of the pole is in the ground," it is understood to mean: "One half of the total length of the pole is in the ground." The second part of the statement, directly following the first part says, "another one third is covered by water " which is understood to mean "another one third of the total length of the pole is covered by water," however, the "of the total length of the pole" is not repeated because it is simply understood. Unless the OP specifies or clarifies, I think the answer should be 48.
What is not clear is if the pole is straight or if it changes shape like lost in space mentioned, a pole that bends
One half of the pole, (3/6ths) is in the ground.
Another one third of the pole, (2/6ths) is in the water, leaving only 1/6th remaining.
8 feet, our remaining 1/6th of the total length of the pole, is out of the water.
So if 8 feet is 1/6th of the pole, the total length of the pole is 8*6=48 feet.
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### #82 plortylox
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Posted 09 June 2008 - 07:26 PM
Surely it's 48 feet
his name isnt Surely, and got 48 feet myself
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### #83 Abby Normal
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Posted 09 June 2008 - 07:45 PM
I see in your calculations where you've accounted for the 1/3 of the length that is in the water and the remaining 2/3 that are not in the ground. Where is the portion that is in the ground?
In US english, when you say something is "out of the water," it means that it is completely out of the lake or whatever body of water it is, and the space below it is not considered to be "out of the water." Also, when stated as: "One half of the pole is in the ground," it is understood to mean: "One half of the total length of the pole is in the ground." The second part of the statement, directly following the first part says, "another one third is covered by water " which is understood to mean "another one third of the total length of the pole is covered by water," however, the "of the total length of the pole" is not repeated because it is simply understood. Unless the OP specifies or clarifies, I think the answer should be 48.
What is not clear is if the pole is straight or if it changes shape like lost in space mentioned, a pole that bends
One half of the pole, (3/6ths) is in the ground.
Another one third of the pole, (2/6ths) is in the water, leaving only 1/6th remaining.
8 feet, our remaining 1/6th of the total length of the pole, is out of the water.
So if 8 feet is 1/6th of the pole, the total length of the pole is 8*6=48 feet.
It stated that half was in the ground. If something is in the ground then it is NOT in water. And we know that 8 feet is not in the water.
I say the the pole is sticking in the lake vertically and 6 feet is in the ground (half) and 4 feet is in the water (1/3) and 2 feet sticking out of the water. Which would be 8 feet out of water.
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### #84 Gambit
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Posted 10 June 2008 - 07:09 AM
i got 24... i'm pretty sure this is right...
Edited by Gambit, 10 June 2008 - 07:11 AM.
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Posted 10 June 2008 - 02:08 PM
There is a pole in a lake. One half of the pole is in the ground, another one third is covered by water and eight feet is out of the water. What is the total length of the pole in feet?
Has anyone thought that maybe the pole is not vertical? It could be along a steep shore of the lake at a diagonal.
I apologize for the crude drawing (I'm not very good with laptop touchpads), but this would be a way it could be 12'.
4' in W, 2' in A, 6' in G.
#### Attached Images
Edited by MAD, 10 June 2008 - 02:09 PM.
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### #86 rl_socal
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Posted 10 June 2008 - 03:58 PM
Edited by rl_socal, 10 June 2008 - 04:06 PM.
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### #87 rl_socal
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Posted 10 June 2008 - 04:12 PM
Spoiler for amendment
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Posted 10 June 2008 - 11:28 PM
I thought of another possible answer: 8'
The pole could be horizontal with 1/2 of its cross-sectional area in the ground and another 1/3 of its cross-sectional area in water, but the remaining portion of the cross-section would still be 8' long.
It is obvious that there are multiple answers to this question and it sure would be nice if the question-poser would clarify and/or explain what his thought process was.
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### #89 Gambit
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Posted 11 June 2008 - 08:18 AM
yeah although there are two possible answers to this question, i think the q-poser was going for the straight forward ans of 48ft.
Tex i think its about time you give an ans...
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### #90 dnae
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Posted 28 June 2008 - 10:56 PM
There is a pole in a lake. One half of the pole is in the ground, another one third is covered by water and eight feet is out of the water. What is the total length of the pole in feet?
48ft
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# Algorithm Loops Assignment Help
Analysis associated with iterative programs along with good examples
1) O(1): Time complexity of a function is recognized as because O(1) in the event that this doesn’t include loop, recursion as well as call to any other non-constant time function.
// set of non-recursive and non-loop statements
For instance swap() function perform O(1) time complexity. A loop or even recursion which operates a constant number of times can also be regarded as O(1).
``` // Here c is constant for (int i = 1; i<= c; i++){ //some 0(1) expressions } ```
2) O(n): Time Complexity of a loop is recognized as because O(n) when the loop variables is actually incremented or decremented with a constant amount. For instance following functions possess O(n) time complexity.
``` // Here c is a positive integer constant for (int i = 1; i<= n; i += c) { // some 0(1) expressions for (int i = n; i > 0; i -= c) { // some 0(1) expressions } ```
3) O(nc): Time complexity associated with nested loops will be corresponding to the particular number of times the innermost statement will be executed. As an example the following sample loops possess O(n2) time complexity
``` for (int i = 1; i <=n; i += c) { for (int j = 1; j <=n; j += c) { //some 0(1) expressions } } for (int i = n; i > 0; i += c) { for (int j = i+1; j <=n; j += c) { //some 0(1) expressions } } ```
For instance Selection sort as well as Insertion Sort possess O(n2) time complexity.
4) O(Logn) Time Complexity of a loop is recognized as O(Logn) when the loop variables is actually split or increased with a constant amount.
``` for (int i = 1; i <=n; i *c=c) { //some 0(1) expressions } for (int i = n; i > 0; i /= c) { // some 0(1) expressions ```
For instance Binary Search possess O(Logn) time complexity.
5) O(LogLogn) Time Complexity of a loop is recognized as O(LogLogn) when the loop variables is actually decreased or increased exponentially with a constant amount.
``` // Here c is a constant greater than 1 for (int i = 2; i <=n; i = pow(i, c)) { //some 0(1) expressions } //Here fun is sqrt or cuberoot or any other constant root for (int i =n; i > 0 i = fun(i)) { //some 0(1) expressions } ``` | 639 | 2,262 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2018-39 | latest | en | 0.884996 |
https://www.physicsforums.com/threads/quick-question.237481/ | 1,511,062,217,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805265.10/warc/CC-MAIN-20171119023719-20171119043719-00128.warc.gz | 847,021,795 | 15,508 | # Quick question
1. May 27, 2008
### chukie
k(x) is a continuous function. k(x)=-1 and k(4)=2 then is this statement true:
lim x->3- k(x) = lim x->3+ k(x)
i realli hv no idea. could sumone help me please?
2. May 27, 2008
### Dick
What does continuous mean in terms of limits? k(x)=(-1) means the function is constant. k(4)=2 contradicts the previous statement. What's the real problem?
3. May 27, 2008
### chukie
sorry i typed the question wrong k(3)=-1
4. May 27, 2008
### Dick
S'ok. But the question is still odd, because the values of the function don't have anything to do with whether the two limits are equal. If a function is continuous at x=3, what can you say about it's left and right hand limits? Do you mean to say k(x) is only defined on the interval [3,4]? Or is it defined and continuous everywhere?
5. May 27, 2008
### chukie
k(x) is continuous for all real numbers
6. May 27, 2008
### Dick
Fine. Then the values of the function have nothing to do with the problem. What does being continuous tell you about limits?
7. May 27, 2008
### chukie
umm im not exactly sure but the left hand limit should equal the right hand limit?
8. May 27, 2008
### Dick
Pretty much. And they both should equal the value of the function at x=3. Kind of a silly question then, yes?
9. May 27, 2008
### chukie
yes lol but thank you so much for ur help. it helped cleared things up for me. | 418 | 1,411 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2017-47 | longest | en | 0.916771 |
https://socratic.org/questions/how-do-you-solve-x-3-2-1 | 1,638,773,301,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363290.39/warc/CC-MAIN-20211206042636-20211206072636-00537.warc.gz | 588,055,124 | 5,933 | # How do you solve x - 3 > 2?
Sep 2, 2015
x>5
#### Explanation:
You use this property: an inequality holds if and only if, for any number $x$, the inequality obtained by adding $x$ to both sides holds. This means that we can add (or subtract) any number we want from an inequality. In your case, the good idea is to add $3$ to both sides, since we obtain
$x - 3 + 3 > 2 + 3$
and therefore
$x > 5$ | 123 | 404 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2021-49 | latest | en | 0.921837 |
https://www.physicsforums.com/threads/propeller-problem.117872/ | 1,555,757,794,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578529606.64/warc/CC-MAIN-20190420100901-20190420122901-00294.warc.gz | 789,600,482 | 16,436 | # Propeller Problem
#### Kenchin
An airplane propeller is 2.08 m in length (from tip to tip) and has a mass of 117 kg. When the airplane's engine is first started, it applies a constant torque of 1590Nm to the propeller, which starts from rest.
Question I:
What is the average power output of the engine during the first 5.00 rev?
Question II:
What is the instantaneous power output of the motor at the instant that the propeller has turned through 5.00 rev?
I've already solved for the angular accelleration (after 5 revolutions) which is alpha, angular speed omega (after 5 revolutions), and work after 5 revolutions W, moment of inertia 42.18kg*m^2.
For the last two parts I've tried to solve using P=torque+angular velocity .... that turned out to be wrong. Then I tried using P=Change in work/change in time but that failed. So now I'm a little at a loss. Is there any suggestions where to try next?
I figured it out, my methods were correct...... my ending units were wrong! @_@
Last edited:
Related Introductory Physics Homework News on Phys.org
#### Andrew Mason
Homework Helper
Kenchin said:
An airplane propeller is 2.08 m in length (from tip to tip) and has a mass of 117 kg. When the airplane's engine is first started, it applies a constant torque of 1590Nm to the propeller, which starts from rest.
Question I:
What is the average power output of the engine during the first 5.00 rev?
Question II:
What is the instantaneous power output of the motor at the instant that the propeller has turned through 5.00 rev?
I've already solved for the angular accelleration (after 5 revolutions) which is alpha, angular speed omega (after 5 revolutions), and work after 5 revolutions W, moment of inertia 42.18kg*m^2.
For the last two parts I've tried to solve using P=torque+angular velocity .... that turned out to be wrong. Then I tried using P=Change in work/change in time but that failed. So now I'm a little at a loss. Is there any suggestions where to try next?
I figured it out, my methods were correct...... my ending units were wrong! @_@
Energy is torque x angle (force x distance).
$$\tau\Delta\theta = \text{Work}$$
So $$P_{avg} = \Delta E/\Delta t = \tau\Delta\theta/\Delta t$$
All you have to do is figure out how long it takes to move the propeller 5 revolutions with that torque: Use $\theta = \frac{1}{2}\alpha t^2$ and $\alpha = \tau/I$ to find the time in terms of angle and torque (and I).
To find instantaneous power, use:
$$P = \tau\omega = \tau\alpha\Delta t$$
You have to assume that in the first 5 revolutions, the resistance to motion is only the moment of inertia of the propeller, not the propulsion of air by the propeller.
AM
"Propeller Problem"
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1. Given that $A$ is regular and $(A \cup B)$ is regular, does it follow that $B$ is necessarily regular? Justify your answer.
2. Given two finite automata $M1, M2$, outline an algorithm to decide if $L(M1) \subset L(M2)$. (note: strict subset)
edited | 328 views
a) A is regular , A U B is regular , then B is not necessary regular
example :- A = (a+b)* B = anbn n>=0 A U B is (a+b)* while B is not regular.
b ) we have two machine M 1 and M 2
draw a DFA using M1 and M2 where start state is, say, p0q0 (where p0 is start state in M1 and q0 is start state in M2)
∂(p0q0, 0) = ∂(p0, 0) U ∂(q0, 0)
if L(M1) ⊆ L (M2)
Then final state of M1 will come together with final state of M2, while Final state of M2 can come alone.
i.e All inputs of M1 is also in machine M2 , and there may be different inputs in M2.
answered by Veteran (54.9k points) 96 246 476
selected
this should be done on the minimal DFA? Also how to ensure strict subset?
DFA need not be minimal.
To ensure strict subset, there should be atleast one final state of M2 that appears alone.
M1 and M2 even need not to be DFA, they may be NFA , resulting M1 x M2 will be DFA (that may be not minimized).
+1 vote
a) A is regular , A U B is regular , then B is not necessary regular
example :- A = (a+b)* B = Any non regular language A U B is (a+b)* , while B is not regular.
B)
Algorithm ->
Using DFA for M1 & M2 , create product DFA. Make two copies of it.
Suppose A is final state for M1 & B is final state of M2. Assume that C is some non final state for M1 & D is some non final state for M2.
To check for subset ->
Now make A as final state whenever it is associated with some non final state for M2. If A is associated with final state of M2, then make that product state non final. This way we will calculate M1-M2. This DFA should accept Empty language otherwise L(M1) is not subset of L(M2)
To check for proper subset ->
Make B as final state, whenever it is associated with some non final state of M1. If A is associated with final state of M1, then make that product state non final. This way we will calculate M2-M1. This DFA should accept non Empty language otherwise L(M1) is not proper subset of L(M2)
if both tests pass then we can say L(M1) is proper subset of L(M2).
answered by Veteran (45.7k points) 171 533 843
Intersection of M2' and M1 should have no final states AND Intersection of M2 and M1' should have at least one final state.
answered ago by Junior (605 points) 1 2 11 | 739 | 2,526 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2017-43 | longest | en | 0.865777 |
https://afteracademy.com/blog/odd-even-linked-list/ | 1,721,567,416,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517701.96/warc/CC-MAIN-20240721121510-20240721151510-00501.warc.gz | 69,742,078 | 13,851 | Difficulty: Medium
#### Understanding The Problem
Problem Description
Given a singly linked list, write a program to group all odd nodes together followed by the even nodes.
Problem Note
• Please note here we are talking about the node number and not the value in the nodes.
• You should try to do it in place.
• The program should run in O(n) time complexity.
• The relative order inside both the even and odd groups should remain as it was in the input.
• The first node is considered odd, the second node even, and so on.
Example 1
``````Input: 1->2->3->4->5->NULL
Output: 1->3->5->2->4->NULL
Explanation: The first group is of elements at odd position (1,3,5) in the linked list and then the ones at the even position(2,4)``````
Example 2
``````Input: 2->1->3->5->6->4->7->NULL
Output: 2->3->6->7->1->5->4->NULL
Explanation: The first group is of elements at odd position (2,3,6,7) in the linked list and then the ones at the even position(1,5,4)``````
You should have a clear understanding of a to solve the problem.
You may try to solve this .
#### Solution
The problem could have various possible solutions in which you can use an auxiliary space to store the numbers in order and recreate a linked list but the question explicitly demands the time complexity to be O(n) and to be solved in-place.
So the basic idea is that we can put all the odd positioned nodes in a linked list and all the even positioned nodes in another linked list, and we can simply join them end to end.
In short, we will have two pointers. The first will point to the head and the second will point to the next of the head. Now in each iteration, each of the pointers will skip one element add the even positioned and odd positioned nodes respectively. This will allow us to use the same linked list and save us from using auxiliary space.
Look at the following diagram of ``` example 1 ``` to get a clear picture of how the linked list is being manipulated.
Solution Steps
1. Take two pointers oddPointer and evenPointer
2. Now, Do below process until evenPointer becomes null
• Delete one node in the odd list and move two steps forward.
• Delete one node in even list and move two steps forward.
3. Now, merge odd and even lists.
• This can be done by ``` oddPointer.next = even_head ```
• oddPointer points to the last node in the odd list
• We need to add the even list at the end of the odd list.
4. Return the resultant list.
Pseudo Code
``````ListNode oddEvenList(ListNode head) {
while(evenPointer != null and evenPointer.next != null){
oddPointer.next = oddPointer.next.next
oddPointer = oddPointer.next
evenPointer.next = evenPointer.next.next
evenPointer = evenPointer.next
}
oddPointer.next = even
}``````
Complexity Analysis
Time Complexity: O(n) (How?)
Space Complexity: O(1)
Critical Ideas To Think
• What do ``` oddPointer ``` and ``` evenPointer ``` represent?
• Why did we initialize ``` evenPointer ``` with ``` head.next ``` ?
• We created an ``` even ``` variable to store the head of ``` evenPointer. ``` How does this help to solve the problem?
• The even list has n/2 nodes. Why does the solution have O(1) space complexity? Answer: Since we are not creating any new nodes here, space complexity is O(1). we are just creating three new pointers overall irrespective of the input. So the space remains constant.
• Can we conclude that we are just adding the even nodes at the end of the linked list every time until the original list is over?
#### Suggested Problems To Solve
• Split the linked list into parts
• Segregate even and odd nodes in a Linked List
• Sum of alternate nodes in a linked list
• Count nodes in a circular linked list.
If you have any more approaches or you find an error/bug in the above solutions, please comment down below.
Happy Coding!
Enjoy Algorithms! | 911 | 3,821 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-30 | latest | en | 0.903874 |
https://sightkitchen.com/how-many-teaspoons-are-in-9-tablespoons/ | 1,723,112,684,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640726723.42/warc/CC-MAIN-20240808093647-20240808123647-00743.warc.gz | 417,573,592 | 53,208 | # How Many Teaspoons are in 9 Tablespoons?
There are nine tablespoons in a cup, so there would be 36 teaspoons in 9 tablespoons. However, this is not an exact answer because there are different sizes of teaspoons. A tablespoon is three teaspoons, a metric tablespoon is 5 mL, and a US teaspoon is 4.9 mL.
So using those measurements, there would be about two metric tablespoons or 4 US teaspoons in 9 tablespoons.
If you’re anything like me, you’ve probably found yourself in a situation where you need to know how many teaspoons are in 9 tablespoons. Well, fear not! I have the answer for you.
There are 36 teaspoons in 9 tablespoons. That means that each tablespoon is equal to 4 teaspoons. So, if you ever find yourself in a pinch and need to know how to convert tablespoons to teaspoons, now you know!
## How Many Teaspoons Are in 6 Tablespoons
There are six teaspoons in a tablespoon. 1 tablespoon = 3 teaspoons. Therefore, 6 tablespoons = 18 teaspoons.
Credit: www.masterclass.com
## How Many Teaspoons Does It Take to Get 9 Tablespoons?
A tablespoon is equivalent to three teaspoons, so it would take nine teaspoons to get nine tablespoons.
## Do 9 Tablespoons Equal 1/2 Cup?
There is a lot of confusion when it comes to measurements, especially when it comes to tablespoons and cups. So, do nine tablespoons equal 1/2 cup? The answer is no. 9 tablespoons is equivalent to 6 fluid ounces, or 3/4 cup.
This is because there are 16 tablespoons in 1 cup. So, if you have nine tablespoons, that would be less than half a cup.
## What is 1 Teaspoon Equal to in Tablespoons?
A teaspoon is a standard unit of measurement in many kitchens and recipes. But what exactly is a teaspoon, and how do you know how much it is? A teaspoon is a small spoon used to measure small amounts of liquid or dry ingredients.
A tablespoon is a larger spoon used to measure larger amounts of liquid or dry ingredients. So, one tablespoon is equal to three teaspoons. One teaspoon equals 1/6 of a fluid ounce (fl oz), or 4.93 milliliters (ml).
So if you need to convert from teaspoons to tablespoons, you must multiply the number of teaspoons by 3. For example, three teaspoons would be equal to 1 tablespoon. Remember that these conversions are for kitchen measurements and may not be accurate for other measurements, such as those used in medicine or baking.
When in doubt, it’s always best to consult an expert or reference guide before proceeding with your recipe!
## What are the Equivalent Values of 9 Teaspoons?
One tablespoon is equivalent to three teaspoons. Therefore, nine teaspoons are equal to three tablespoons or one-fourth cup.
## Conclusion
There are nine tablespoons in a cup, so there would be 36 teaspoons in 9 tablespoons.
Scroll to Top | 629 | 2,758 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-33 | latest | en | 0.94046 |
https://www.hackmath.net/en/example/542?tag_id=101,36 | 1,566,551,154,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027318243.40/warc/CC-MAIN-20190823083811-20190823105811-00163.warc.gz | 844,819,128 | 7,109 | # Motion problem
From Levíc to Košíc go car at speed 81 km/h.
From Košíc to Levíc go another car at speed 69 km/h.
How many minutes before the meeting will be cars 27 km away?
Result
t = 10.8 min
#### Solution:
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
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#### To solve this example are needed these knowledge from mathematics:
Need help calculate sum, simplify or multiply fractions? Try our fraction calculator. Do you have a linear equation or system of equations and looking for its solution? Or do you have quadratic equation? Do you want to convert length units?
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17. Two squares
Two squares whose sides are in the ratio 5:2 have sum of its perimeters 73 cm. Calculate the sum of area this two squares. | 946 | 3,777 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2019-35 | latest | en | 0.954782 |
https://www.digitalocean.com/community/tutorials/level-order-traversal-in-a-binary-tree | 1,716,799,790,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059039.52/warc/CC-MAIN-20240527083011-20240527113011-00172.warc.gz | 623,504,000 | 50,729 | # Level Order Traversal in a Binary Tree
Published on August 3, 2022
Vijaykrishna Ram
While we believe that this content benefits our community, we have not yet thoroughly reviewed it. If you have any suggestions for improvements, please let us know by clicking the “report an issue“ button at the bottom of the tutorial.
Level Order Traversal is one of the methods for traversing across a Binary Tree. In this article, we shall look at how we can implement this algorithm in C/C++.
But before that, let us have our concepts covered.
## Building the concepts
A Binary Tree is a data structure where every node has at-most two children. The topmost node is called the Root node.
There are 4 common ways of traversing the nodes of a Binary Tree, namely:
• In order Traversal
• Pre Order Traversal
• Post Order Traversal
• Level Order Traversal
Let’s understand what a level in a Binary Tree means.
A level is the number of parent nodes corresponding to a given a node of the tree. It is basically the number of ancestors from that node until the root node.
So, for the root node (topmost node), it’s level is 0, since it has no parents. If it has children, both of them will have a level of 1, since it has only one ancestor until the root node, which is the root node itself.
We also need to understand the notion of height in a Binary Tree. This is simply the length of the path from the root to the deepest node in the tree.
In this case, the height will be the length from the deepest node (40 or 50, since they have the maximum level) to the root. So the height of the tree is 2.
Now that we have our concepts covered, let’s understand how we can implement Level Order Traversal.
## Level Order Traversal
A Level Order Traversal is a traversal which always traverses based on the level of the tree.
So, this traversal first traverses the nodes corresponding to Level 0, and then Level 1, and so on, from the root node.
In the example Binary Tree above, the level order traversal will be:
(Root) 10 -> 20 -> 30 -> 40 -> 50
To do this, we need to do 2 things.
1. We must first find the height of the tree
2. We need to find a way to print the nodes corresponding to every level.
### Find the height of the Tree
We will find the height of the tree first. To do this, the logic is simple.
Since the height of the tree is defined as the largest path from the root to a leaf. So we can recursively compute the height of the left and right sub-trees, and find the maximum height of the sub-tree. The height of the tree will then simply be the height of the sub-tree + 1.
C- style Pseudo Code:
``````// Find height of a tree, defined by the root node
int tree_height(Node* root) {
if (root == NULL)
return 0;
else {
// Find the height of left, right subtrees
left_height = tree_height(root->left);
right_height = tree_height(root->right);
// Find max(subtree_height) + 1 to get the height of the tree
return max(left_height, right_height) + 1;
}
``````
Now that we have the height, we must print nodes for every level. To do this, we will use a `for` loop to iterate all levels until the height, and print nodes at every level.
``````void print_tree_level_order(Node* root) {
int height = tree_height(root);
for (int i=0; i<height; i++) {
// Print the ith level
print_level(root, i);
}
}
``````
Observe that we need another function to print the `i`th level of the tree.
Here again, we have a similar logic. But this time, after printing the root node, we change the root node to it’s left and right children and print both sub-trees.
This will continue until we reach a leaf node, that is when the auxiliary root will be `NULL` at the next step. (Since leaf_node->left = `NULL` and leaf_node->right = `NULL`)
``````void print_level(Node* root, int level_no) {
// Prints the nodes in the tree
// having a level = level_no
// We have a auxiliary root node
// for printing the root of every
// sub-tree
if (!root)
return;
if (level_no == 0) {
// We are at the top of a sub-tree
// So print the auxiliary root node
printf("%d -> ", root->value);
}
else {
// Make the auxiliary root node to
// be the left and right nodes for
// the sub-trees and decrease level by 1, since
// you are moving from top to bottom
print_level(root->left, level_no - 1);
print_level(root->right, level_no - 1);
}
}
``````
Now, we have finally completed the Level Order Traversal!
I will provide the complete program below, which also has a section to construct the Binary Tree using insertion.
## Complete C/C++ Code
While this is originally a C program, the same can be compiled on C++ as well.
``````/**
Code for https://journaldev.com
File Name: level_order.c
Purpose: Find the Level Order Traversal of a Binary Tree
@author Vijay Ramachandran
@date 28/01/2020
*/
#include <stdio.h>
#include <stdlib.h>
typedef struct Node Node;
// Define the Tree Node here
struct Node {
int value;
// Pointers to the left and right children
Node* left, *right;
};
Node* init_tree(int data) {
// Creates the tree and returns the
// root node
Node* root = (Node*) malloc (sizeof(Node));
root->left = root->right = NULL;
root->value = data;
return root;
}
Node* create_node(int data) {
// Creates a new node
Node* node = (Node*) malloc (sizeof(Node));
node->value = data;
node->left = node->right = NULL;
return node;
}
void free_tree(Node* root) {
// Deallocates memory corresponding
// to every node in the tree.
Node* temp = root;
if (!temp)
return;
free_tree(temp->left);
free_tree(temp->right);
if (!temp->left && !temp->right) {
free(temp);
return;
}
}
int tree_height(Node* root) {
// Get the height of the tree
if (!root)
return 0;
else {
// Find the height of both subtrees
// and use the larger one
int left_height = tree_height(root->left);
int right_height = tree_height(root->right);
if (left_height >= right_height)
return left_height + 1;
else
return right_height + 1;
}
}
void print_level(Node* root, int level_no) {
// Prints the nodes in the tree
// having a level = level_no
// We have a auxiliary root node
// for printing the root of every
// subtree
if (!root)
return;
if (level_no == 0) {
// We are at the top of a subtree
// So print the auxiliary root node
printf("%d -> ", root->value);
}
else {
// Make the auxiliary root node to
// be the left and right nodes for
// the subtrees and decrease level by 1, since
// you are moving from top to bottom
print_level(root->left, level_no - 1);
print_level(root->right, level_no - 1);
}
}
void print_tree_level_order(Node* root) {
if (!root)
return;
int height = tree_height(root);
for (int i=0; i<height; i++) {
printf("Level %d: ", i);
print_level(root, i);
printf("\n");
}
printf("\n\n-----Complete Level Order Traversal:-----\n");
for (int i=0; i<height; i++) {
print_level(root, i);
}
printf("\n");
}
int main() {
// Program to demonstrate Level Order Traversal
// Create the root node having a value of 10
Node* root = init_tree(10);
// Insert nodes onto the tree
root->left = create_node(20);
root->right = create_node(30);
root->left->left = create_node(40);
root->left->right = create_node(50);
// Level Order Traversal
print_tree_level_order(root);
// Free the tree!
free_tree(root);
return 0;
}
``````
Output
``````Level 0: 10 ->
Level 1: 20 -> 30 ->
Level 2: 40 -> 50 ->
-----Complete Level Order Traversal:-----
10 -> 20 -> 30 -> 40 -> 50 ->
``````
You can also download this through a Github gist that I created for this purpose. (Contains code for insertion as well)
## Conclusion
Hopefully you have a better understanding of how Level Order Traversal can be implemented in C/C++. If you have any questions, feel free to ask them in the comments section below!
Thanks for learning with the DigitalOcean Community. Check out our offerings for compute, storage, networking, and managed databases.
Vijaykrishna Ram
author
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DigitalOcean makes it simple to launch in the cloud and scale up as you grow — whether you're running one virtual machine or ten thousand. | 2,141 | 8,453 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-22 | longest | en | 0.922761 |
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# Test: Ohm’S Law
## 5 Questions MCQ Test General Science(Prelims) by IRS Divey Sethi | Test: Ohm’S Law
Description
This mock test of Test: Ohm’S Law for JEE helps you for every JEE entrance exam. This contains 5 Multiple Choice Questions for JEE Test: Ohm’S Law (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Ohm’S Law quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Test: Ohm’S Law exercise for a better result in the exam. You can find other Test: Ohm’S Law extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1
### V-I graph of which material shows the straight line
Solution:
Materials which obeys the ohm’s law show straight line in the V-I graph. Since silver is the only ohmic material in given options, it shows straight line curve.
QUESTION: 2
### The following fig. shows I-V graph for a given metallic wire at two temperatures T1and T2.Then,
Solution:
The slope of the graph = I / V
In case of T2, I/V is less. Hence, V/I (=R) is more.
Value of resistance increases with increasing temperature.
Hence, T2 >T
QUESTION: 3
### How many electrons are flowing per second through a section of a conductor corresponding to current of 1A?
Solution:
I=Q/t
1 Ampere = 1 Coulomb /1 Secon
Charge on 1 electron = 1.6 x 10^-19
Coulomb
By unitary method,
If 1.6 x 10^-19 Coulomb / Second (Ampere) = Current by 1 electron
then, 1( Coulomb / Second) or (Ampere) = 1 / (1.6 x 10^-19) electrons
i.e, 6.25 x 10^18 electrons.
QUESTION: 4
1 ohm is equal to
Solution:
1 Ohm is defined as the resistance of a conductor with a potential difference of 1 volt applied to the ends through which 1-ampere current flows.
Ohms is the SI unit of electrical resistance.
QUESTION: 5
Which of the following is an ohmic conductor?
Solution:
Ohmic conductors are those which follows ohm's Law
Constantan is an copper nickle alloy which follows ohm's law
An electrolyte is a chemical that produces an electrically conducting solution and hence conducts electrical current.
But Electrolyte can be Ohmic as well as non-ohmic conductor
Transistor is semi-conductor device and does not follow ohm's Law
thermionic valves - is vacuum tube (electronic tube) , uses ion emission.
So option C Constanton s an ohmic conductor. | 635 | 2,391 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2021-25 | latest | en | 0.890731 |
https://80southstreet.net/what-is-half-of-3/ | 1,674,873,524,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499470.19/warc/CC-MAIN-20230128023233-20230128053233-00442.warc.gz | 101,370,929 | 13,586 | # What Is Half Of 3
Categories :
## How do you write 3 divided by 2?
We can write 3 divided by 2 as 3 / 2.
## What is 3 and a half as an improper fraction?
31/2 as improper fraction
Hence, the improper fraction is 7/2.
## How do you solve 72 divided by 2?
Using a calculator, if you typed in 72 divided by 2, you’d get 36. You could also express 72/2 as a mixed fraction: 36 0/2.
## How do you measure half of 1 3?
Half of 1/3 cup is even trickier. 1/3 cup equals 5 tablespoons plus 1 teaspoon so, half of 1/3 cup would be 2 tablespoons plus 2 teaspoons.
## How do you write 3 and a half as a decimal?
Answer: 3 1/2 as a decimal is 3.5.
## How do you write 27 divided by 3?
Using a calculator, if you typed in 27 divided by 3, you’d get 9. You could also express 27/3 as a mixed fraction: 9 0/3.
## What is the half of 3 in fraction?
Answer: Half of 3 is 3/2 as a fraction and 1.5 as a decimal.
## What is half of 1/3 in math?
So half of 1/3 is 1/6.
## What is 1 third of a whole?
If the whole is divided into three equal parts, each part is a THIRD. A third is obtained by dividing a whole (1) by 3. If a whole is divided into four equal parts, each part is a FOURTH. A fourth is obtained by dividing a whole (1) by 4.
## What would 3 be as a fraction?
What is 3% as a fraction? 3% is equal to 3/10 as a fraction.
## How do you do 3 divided by 4?
The answer to 3 divided by 4 is . 75 or 3/4. Because we are dividing a smaller number (3) by a larger number (4), we know the answer is going to be less than one and can be expressed either as a decimal or as a fraction. In decimal form, the answer is 0.75.
## Can 2 be divided by 3?
Answer: 2 divided by 3 as a fraction is 2/3.
The divisor is represented by the denominator, i.e., 3. | 554 | 1,759 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2023-06 | latest | en | 0.973677 |
https://zxi.mytechroad.com/blog/category/greedy/page/9/ | 1,581,936,848,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875141806.26/warc/CC-MAIN-20200217085334-20200217115334-00338.warc.gz | 987,719,903 | 25,204 | # Posts published in “Greedy”
Given an integer array with even length, where different numbers in this array represent different kinds of candies. Each number means one candy of the corresponding kind. You need to distribute these candies equally in number to brother and sister. Return the maximum number of kinds of candies the sister could gain.
Example 1:
Example 2:
Note:
1. The length of the given array is in range [2, 10,000], and will be even.
2. The number in given array is in range [-100,000, 100,000].
Solution 1: Greedy
Give all unique candies to sisters until they have n/2 candies.
Time complexity: O(n)
Space complexity: O(n)
C++ Hashset
C++ Bitset
Given a string S, you are allowed to convert it to a palindrome by adding characters in front of it. Find and return the shortest palindrome you can find by performing this transformation.
For example:
Given "aacecaaa", return "aaacecaaa".
Given "abcd", return "dcbabcd".
Solution 1: Brute Force
Time complexity: O(n^2)
Space complexity: O(n)
C++ w/ copy
C++ w/o copy
Problem:
A string S of lowercase letters is given. We want to partition this string into as many parts as possible so that each letter appears in at most one part, and return a list of integers representing the size of these parts.
Example 1:
Solution 0: Brute Force
Time complexity: O(n^2)
Space complexity: O(1)
C++
Python
Solution 1: Greedy
Time complexity: O(n)
Space complexity: O(26/128)
C++
Java
Python3
Problem:
Given two arrays of length m and n with digits 0-9 representing two numbers. Create the maximum number of length k <= m + n from digits of the two. The relative order of the digits from the same array must be preserved. Return an array of the k digits. You should try to optimize your time and space complexity.
Example 1:
nums1 = [3, 4, 6, 5]
nums2 = [9, 1, 2, 5, 8, 3]
k = 5
return [9, 8, 6, 5, 3]
Example 2:
nums1 = [6, 7]
nums2 = [6, 0, 4]
k = 5
return [6, 7, 6, 0, 4]
Example 3:
nums1 = [3, 9]
nums2 = [8, 9]
k = 3
return [9, 8, 9]
# Solution:
Time complexity: O(k * (n1+n2)^2)
Space complexity: O(n1+n2)
## Java
Given a char array representing tasks CPU need to do. It contains capital letters A to Z where different letters represent different tasks.Tasks could be done without original order. Each task could be done in one interval. For each interval, CPU could finish one task or just be idle.
However, there is a non-negative cooling interval n that means between two same tasks, there must be at least n intervals that CPU are doing different tasks or just be idle.
You need to return the least number of intervals the CPU will take to finish all the given tasks.
Example 1:
Note:
1. The number of tasks is in the range [1, 10000].
2. The integer n is in the range [0, 100].
Idea:
Counting
Time complexity: O(n)
Space complexity: O(1)
Mission News Theme by Compete Themes. | 815 | 2,895 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2020-10 | longest | en | 0.710057 |
http://openstudy.com/updates/50ab1b73e4b06b5e49334ca5 | 1,444,373,779,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443737916324.63/warc/CC-MAIN-20151001221836-00135-ip-10-137-6-227.ec2.internal.warc.gz | 230,216,601 | 12,979 | ## needhelp07 2 years ago Problem : Find the derivatives of the function using the limit of definition y=x^4
1. irkiz
still the same type of question? try going http://www.math.hmc.edu/calculus/tutorials/limit_definition/ they show you examples. just apply your question to how they solve it.
2. irkiz
It cant be that you were taught 2 other similar questions but still cant solve it.
3. needhelp07
im confused about the exponent 4
4. mark_o.
do you know how to do (a+b)^4?
5. needhelp07
no
6. mark_o.
its the binomial theorem or pascal triangle
7. mark_o.
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 etc....
8. mark_o.
for example just look at the pascal triangle for the coefficients (a+b)^2= 1a^2 + 2ab +1b^2= a^2 +2ab+b^2 can you guess whats (a+b)^3? and for (a+b)^4?
9. mark_o.
note it can be like (x+y)^3 ? or (x+y)^4 ?
10. mark_o.
so power of 3 we get the coefficients of 1 3 3 1 therefore the answer for (x+y)^3= x^3 +3x^2y +3 xy^2 + y^3 can you guess for (x+y)^4 ?
11. mark_o.
the reason i want you to learn this is when you go solved for the derivative of y=x^4 you will get to find (x+y)^4
12. needhelp07
13. mark_o.
im here now to teach you the easier way of doing it
14. mark_o.
ok for the power of 4 the coefficents inthe pascal triangle are 1 4 6 4 1 therefore (a+b)^4 = a^4 + 4a^3 b +6 a^2 b^2 +4 a b^3 + b^4 or (x+y)^4= x^4 +4x^3 y + 6 x^2 y^2 + 4x y^3 + y^4 did you see it ?
15. needhelp07
yes
16. mark_o.
now for practice y=x^2 $y+\Delta y=x +\Delta x$ $\Delta y=x+\Delta x - y$ $\Delta y= (x+\Delta x)^{2}-x ^{2}$
17. mark_o.
can you try solving that first ?
18. needhelp07
wait
19. needhelp07
i try solving it
20. needhelp07
21. mark_o.
$\Delta y=f(x+\Delta x) - y$ $\Delta y=(x+\Delta x)^{2}- x ^{2}$ $\Delta y= x ^{2 }+ 2x \Delta x +\Delta x ^{2} -x ^{2}$ add or subtract we get $\Delta y= 2x \Delta x+\Delta x ^{2}$ now the derivative is $y'=f'(x)= \lim _{\Delta x ->0}\frac{ \Delta y }{ \Delta x }=$ $= \lim _{\Delta x->0}(\frac{ 2x \Delta x+\Delta x ^{2} }{ \Delta x })$ try to cancell something in there
22. needhelp07
23. mark_o.
ok good can you try it if its y=x^4 ? do the same process
24. needhelp07
okay
25. needhelp07
26. mark_o.
hmm to make it easier lets make delta x=h so that delta y=(x+h)^4 - x^4 try solving that first
27. needhelp07
I cant follow my professor's teaching. Good to know that u are here to help me .thank you so much. you are a great help! i need to do more assignments about this so i will go out now .
28. mark_o.
29. needhelp07
haha im wrong
30. mark_o.
do the process i gave you there, and you will arrive at the correct answer ... :D
31. needhelp07
i wish i have a tutor like u hahaha .
32. mark_o.
remember this ?. (x+y)^4= x^4 +4x^3 y + 6 x^2 y^2 + 4x y^3 + y^4 make y an h here so that (x+h)^4= x^4 +4x^3 h + 6 x^2 h^2 + 4x h^3 + h^4
33. needhelp07
anyway thx a lot
34. needhelp07
goodbye!
35. mark_o.
ok just read it through here again and learn from it good luck and have fun now :D
36. needhelp07
ok
37. mark_o.
ok just messege me here and if you have prob i will go and give you an advise or correct if you have something done incorrectly ... :D | 1,129 | 3,181 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2015-40 | longest | en | 0.636109 |
https://responsivedesigntool.com/qa/quick-answer-how-do-you-measure-an-ounce-of-dry-ingredients.html | 1,603,594,796,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107885126.36/warc/CC-MAIN-20201025012538-20201025042538-00076.warc.gz | 501,139,574 | 8,774 | # Quick Answer: How Do You Measure An Ounce Of Dry Ingredients?
## How do I measure 1 oz?
Ounces are usually used to measure liquids when baking.
Two ounces are equivalent to 1 tablespoon.
A half tablespoon is equal to 1 ounce.
Some liquid measuring cups are marked in ounces as well as quarter cup, half cup and full cup marks..
## Is a heaped tablespoon an ounce?
A rounded tablespoon means there is as much of the product you are measuring above the top edge of the spoon as there is in the “bowl” of the spoon. A heaped tablespoon means as much as you can get on the spoon without it falling off. … A stick of butter is equal to half a cup, 4 ounces, 8 tablespoons or 113 grams.
## Does 8 tbsp equal 1 cup?
1/4 cup = 4 tablespoons. 1/3 cup = 5 tablespoons plus 1 teaspoon. 3/8 cup = 6 tablespoons. 1/2 cup = 8 tablespoons.
## How much is an ounce of dry ingredients?
Two dry tablespoons make up 1 dry ounce.
## How can I measure an ounce without a scale?
You can also use your hand to measure food portions of meat and produce. For example, a single 3-ounce serving of chicken, beef, or fish is roughly the size of your palm. A 1-cup serving of fruit or vegetables is roughly the size of your closed fist. A single serving of cheese is about the size of your thumb.
## How do you measure 1 oz of powder?
Divide the weight of 28.35 g (1 oz.) by the weight of the powder in 1 tbsp. to convert ounces to tablespoons. In this example, 1 oz. corresponds to 28.35 / 10.67 = 2.6 tbsp.
## How much is an oz of liquid?
Fluid ounceUS customary units0.9607599 US fl ozConversions (US)1 US fl oz in …… is equal to …SI units29.57353 ml11 more rows
## What is 2 ounces equal to in cups?
How big is 2 ounces? What is 2 ounces in cups?…Convert 2 Ounces to Cups.fl ozcups2.000.252.010.251252.020.25252.030.2537596 more rows
## Does 4 ounces equal 1 cup?
Fluid Ounces (oz) to Cups Conversion 1 Fluid ounce (oz) is equal to 0.125 cup. To convert fluid oz to cups, multiply the fluid oz value by 0.125 or divide by 8. For example, to convert 4 fluid oz to cups, multiply 4 by 0.125, that makes 0.5 cup is 4 fl oz.
## What does 1 oz of cheese look like?
1 oz = A Pair of Dice That’s the proper serving size for a serving of cheese. You can estimate your portions knowing that one ounce of cheese is about the size of a pair of dice.
## How many tablespoons are in an ounce of taco seasoning?
2 TablespoonsA package of Taco Seasoning is equivalent to 1 ounce/2 Tablespoons. The serving size for a 1 oz package is 6 servings. This Taco Seasoning is 8 Tablespoons/1 cup total which I guesstimate to be 48 total servings.
## How many tablespoons is an ounce of cocoa powder?
3.84 tbspOne ounce of cocoa powder converted to tablespoon equals to 3.84 tbsp. How many tablespoons of cocoa powder are in 1 ounce? The answer is: The change of 1 oz ( ounce ) unit in a cocoa powder measure equals = into 3.84 tbsp ( tablespoon ) as per the equivalent measure and for the same cocoa powder type.
## How many teaspoons are in a dry ounce?
Dry Measures3 teaspoons1 tablespoon1/2 ounce2 tablespoons1/8 cup1 fluid ounce4 tablespoons1/4 cup2 fluid ounces5 1/3 tablespoons1/3 cup2.6 fluid ounces8 tablespoons1/2 cup4 ounces4 more rows
## How many dry ounces equal a cup?
In case of dry measures, a cup of dry flour (APF) weighs 4.5 ounces, not 8 ounces.
## Does 2 tbsp equal 1 oz?
1 US fluid ounce (oz) is equal to 2 tablespoons (tbsp). To convert fluid ounces to tablespoons, multiply the fluid ounce value by 2. For example, to convert 2 fluid oz to tbsp, multiply 2 by 2, that makes 4 tbsp is 2 fl oz.
## What makes up 1 ounce?
Ounce, unit of weight in the avoirdupois system, equal to 1/16 pound (437 1/2 grains), and in the troy and apothecaries’ systems, equal to 480 grains, or 1/12 pound. The avoirdupois ounce is equal to 28.35 grams and the troy and apothecaries’ ounce to 31.103 grams.
## How many tablespoons are in an ounce of powder?
Ounce to US tablespoon Conversion Chart – Baking powderounces to US tablespoons of Baking powder1 ounce=1.97 ( 2 ) US tablespoons2 ounces=3.94 ( 4 ) US tablespoons4 ounces=7.89 ( 8 ) US tablespoons5 ounces=9.86 ( 9 3/4 ) US tablespoons19 more rows
## Is 1 oz the same as 1 fl oz?
More information from the unit converter How many oz in 1 fl oz? The answer is 1. We assume you are converting between ounce [US, liquid] and US fluid ounce. You can view more details on each measurement unit: oz or fl oz The SI derived unit for volume is the cubic meter.
## What is 1 oz of flour in tablespoons?
3.63 tbspHow many tablespoons of all purpose flour (APF) are in 1 ounce? The answer is: The change of 1 oz ( ounce ) unit in a all purpose flour (APF) measure equals = into 3.63 tbsp ( tablespoon ) as per the equivalent measure and for the same all purpose flour (APF) type. | 1,322 | 4,822 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2020-45 | latest | en | 0.91457 |
http://gateoverflow.in/50582/isro2011-19 | 1,506,048,656,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818688158.27/warc/CC-MAIN-20170922022225-20170922042225-00235.warc.gz | 124,488,700 | 20,518 | 1.8k views
If node A has three siblings and B is parent of A, what is the degree of A?
1. 0
2. 3
3. 4
4. 5
asked in DS | 1.8k views
Now coming to question if node $A$ has three siblings and $B$ is the parent of $A$. This is the only information given.
It is not given whether node $A$ is an internal node or leaf node and also total number of node in the graph is not given.
So given information is not sufficient to decide what should be the degree of node $A$.
So the degree of node $A$ can be any of the above.
selected by
mahol hai sir. :)
is it mentioned anywhere that that graph has only 4 nodes ?? if not they why are u assuming A as leaf node ??
No other info is given other than <If node A has three siblings and B is parent of A>.we can not add extra information by our own assumption.
i am not adding anything . question is ambiguous i guess. i can't say answer is 0 or 4 or 3
Answer should be 4, because if we consider node A at leaf level , anyhow degree will be 1(consider undirected graph) but no option with degree 1. Now (consider A no-leaf node), so we can consider it strict 4-ary tree then every node have 0 or 4 childs, then we can say that degree of A=4
If A has 4 children den degree will be 5 not 4 bcz of link connnected to B...
I don't think one can solve this with the given data - they missed copying the remaining part of question. Should be Mark to All.
yes you are right shikha,
now ya i also think same Arjun sir(Missing Data)
Node "A" has 3 siblings and B is parent of A.
A is child of B.
There is no children for A
So, A must me leaf node
Degree of leaf node is 0
A/c to given condition B is parent of A and A has 3 Siblings.Degree of A would be 0
Without any assumptions Answer is matching with option A
how can someone answer this question, probably 1 could be answer but nothing is mentioned that it's a graph or a tree and worst thing is that 1 is not given in options.
ISRO always gives ambiguous questions!
it will be zero offcourse
In tree the number of subtrees of a node is called the degree of the node.So degree of leaf node is 0.
In graph degree of a node is the total no node connected with it.(If it is graph then ans is 1)
ans:0(a) | 598 | 2,202 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2017-39 | latest | en | 0.944261 |
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Measuring Length
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Estimating Length
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Estimating Length
Advance your students from relative measurements to learning about standardized units with this lesson that teaches them about inches and feet by using common classroom objects.
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This math lesson is filled with fun hands-on non-standard measurement activities! The lesson begins with an interactive story then has students explore measurement using several different objects as their tools!
Math
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Estimating Measurements of Mass and Volume Using Metric Units
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Estimating Measurements of Mass and Volume Using Metric Units
Students will become more familiar with common metric measurements by matching everyday objects with the metric mass and volume units they would use to measure them.
Math
Lesson Plan
Line Plots: Representing the Length of Classroom Items
Lesson Plan
Line Plots: Representing the Length of Classroom Items
In this lesson, your students will measure the lengths of items and then make a line plot to show the measurement data. They will get hands-on by measuring and surveying the class.
Math
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How Big Is It?
Lesson Plan
How Big Is It?
Your young scientists will have tons of fun visiting measurement stations and using tools to measure various objects. They'll even be able to create posters at the end of the activity.
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The Smaller They Come
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The Smaller They Come
Help your students use rulers in the classroom with this lesson! Encourage them to measure items around them using centimeters to understand shapes and sizes.
Math
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Order Up!
Lesson Plan
Order Up!
It’s time to get measuring! In this lesson students determine the lengths of different towers and then put them in order from shortest to longest.
Math
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Making Measurements for Line Plots
Lesson Plan
Making Measurements for Line Plots
In this hands-on lesson, your students will get to sharpen their measuring skills as they measure lengths of items in your classroom to the nearest quarter inch. They'll get to then use their data to create line plots.
Math
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Footy Fun!
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Footy Fun!
Looking for a fun way to learn about measurement? Inch on over to this engaging lesson. Students will love using their own feet to determine the lengths of different objects. | 488 | 2,426 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2020-50 | latest | en | 0.919433 |
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# Chapter 7 Risk and Return Prob and Soln - Chapter 7...
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Chapter 7 - Solutions Questions and Problems BASIC 7.1 Returns: Describe the difference between a total holding period return and an expected return. LO 2, LO3 Solution: The holding period return is the total return over some investment or “holding” period. It consists of a capital appreciation component and an income component. The holding period return reflects past performance. The expected return is a return that is based on the probability-weighted average of the possible returns from an investment. It describes a possible return (or even a return that may not be possible) for a yet- to-occur investment period. 7.3 Expected returns: You have chosen biology as your college major because you would like to be a medical doctor. However, you find that the probability of being accepted into medical school is about 10 percent. If you are accepted into medical school, then your starting salary when you graduate will be \$300,000 per year. However, if you are not accepted, then you would choose to work in a zoo, where you will earn \$40,000 per year. Without considering the additional educational years or the time value of money, what is your expected starting salary as well as the standard deviation of that starting salary? LO 3 Solution: E(salary) = 0.9(\$40,000) + (0.1) (\$300,000) = \$66,000 | 389 | 1,670 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2020-45 | latest | en | 0.923221 |
http://slidegur.com/doc/85876/section-8---2-similar-polygons | 1,521,825,420,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648404.94/warc/CC-MAIN-20180323161421-20180323181421-00471.warc.gz | 251,871,248 | 7,882 | ### Section 8 * 2 Similar Polygons
```Section 8 – 2
Similar Polygons
Objectives:
To identify similar polygons
To apply similar polygons
Similar Polygons:
Two polygons whose corresponding angles are
congruent and corresponding sides are
proportional.
Similarity Ratio:
The ratio of the lengths of the corresponding
sides.
Example 1
Understanding Similarity
ABCD ~ EFGH. Complete each statement.
∠E =
=
∠B =
=
Example 2 Determining Similarity
A) Determine whether the triangles are
similar. If they are, write the similarity statement
and give the similarity ratio.
B) Sketch ∆XYZ and ∆MNP with ∠X ≅∠M,
∠Y≅∠N, and ∠Z≅∠P. Also, XY = 12, YZ = 14,
ZX = 16, MN = 18, NP =21, and PM = 24.
Can you conclude that the two triangles are
similar? Explain.
C) Determine whether the parallelograms are
similar? Explain.
Example 3
Using Similar Figures
A)
LMNO ~ QRST. Find the value of x.
B)
LMNO ~ QRST. Find SR to the nearest tenth.
C)
If ∆ABC ~ ∆YXZ, find the value of x.
Homework:
8 – 2 Ditto
Section 8 – 2
Continued…
Objectives:
To apply similar polygons
Example 4 Real-World Connection
A) A map has the dimensions 12 in. tall by 16
in. wide. You want to reduce the map to fit on a
4 in. by 6 in. notecard. What are the dimensions
of the largest possible complete map that you
can fit on the index card?
B) You want to enlarge a 3 in. by 5 in. photo.
If you have photo paper that is 12 in. by 16 in.,
what are the dimensions of the largest complete
enlargement you can make?
C) A painting is 24 in. wide and 36 in. long.
The length of a postcard reduction of the
painting is 6 inches. How wide is the postcard?
Homework:
Textbook Page 425 – 426; #2 – 18 Even
``` | 499 | 1,658 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2018-13 | latest | en | 0.846764 |
https://www.shaalaa.com/question-bank-solutions/find-the-centre-and-radius-of-the-following-x2-y2-6x-8y-24-0-different-forms-of-equation-of-a-circle_171342 | 1,618,281,375,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038071212.27/warc/CC-MAIN-20210413000853-20210413030853-00256.warc.gz | 1,108,216,434 | 9,376 | # Find the centre and radius of the following: x2 + y2 − 6x − 8y − 24 = 0 - Mathematics and Statistics
Sum
Find the centre and radius of the following:
x2 + y2 − 6x − 8y − 24 = 0
#### Solution
Given equation of the circle is
x2 + y2 − 6x − 8y − 24 = 0
Comparing this equation with
x2 + y2 + 2gx + 2fy + c = 0, we get
2g = − 6, 2f = − 8 and c = − 24
∴ g = − 3, f = − 4 and c = − 24
∴ Centre of the circle = (−g, −f) = (3, 4)
and radius of the circle = sqrt("g"^2 + "f"^2 - "c")
= sqrt((- 3)^2 + (- 4)^2 - (- 24))
= sqrt(9 + 16 + 24)
= sqrt(49)
= 7.
Is there an error in this question or solution? | 254 | 612 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2021-17 | latest | en | 0.682817 |
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posted by .
What is the LCM AND GCF OF 369 AND 3075
• math -
369 = 3*3*41
3075 = 3*5*5*41
So, GCF = 3*41 = 123
LCM = 3*3*5*5*41 = 9225 = 369*3075/GCF
### Answer This Question
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Factors and multiples
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## Check these out next
Chapter 1
Factors
Highest Common Factor
Prime factors
multiples
Least common multiples
Chapter 1
Factors
Highest Common Factor
Prime factors
multiples
Least common multiples
## More Related Content
### Similar to FACTORS MULTIPLES Grade 7.ppt (20)
1. 1. http://www.education707.com 1 CHAPTER FACTORS AND MULTIPLES Aim The aim of this lesson is to solve problems involving factors and multiples. objective At the end of this lesson, students will be able to understand and solve problems involving Factors Prime factors Highest common factor H.C.F Multiples Least common multiple L.C.M
2. 2. http://www.education707.com 2 Factors and multiples Factors Defn. Whole numbers which multiply to give a certain number are called the factors of that number. Ex 1. Factors of 24 are 1 , 24 2 , 12 , 3 , 8 , 4 , 6 , Ex 2. Factors of 36 are 1 , 36 2 , 18 , 3 , 12 , 4 , 9 , 6 , Ex 3. Evaluation Find the factors of 8 , 12 , 9 , 20 and 84
3. 3. http://www.education707.com 3 Factors and multiples Prime numbers Defn. Prime numbers are numbers which have exactly two factors, namely 1 and the number itself. Note : 1 is not a prime number because it has only one factor, i.e 1 itself. e.g 2 , 3, 5, 7 , 11 , 13 , 17 , 19 , ….
4. 4. http://www.education707.com 4 12 ÷ 2 = 6 12 ÷ 3 = 4 12 ÷ 6 = 2 12 can be divided by 2, 4 and 6 so 12 is not a prime number. 7 ÷ 7 = 1 7 ÷ 1 = 7 7 can only be divided by itself and 1 to give an integer answer. 7 is a prime number Prime numbers
5. 5. http://www.education707.com 5 Copy the table below and circle the prime numbers between 1 and 40. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 Ex Soln. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
6. 6. http://www.education707.com 6 Prime factors Defn. Prime factors are those factors of a number which are prime numbers. 30 = 2 × 3 × 5 24 = 2 × 2 × 2 × 3 = 23 × 3 Index notation 60 = 2 × 2 × 3 × 5 = 22 × 3 × 5 Index notation 30 2 15 3 5 5 1 24 2 12 2 6 2 3 3 1 Ex. Find the prime factorisation of (a) 30 (b) 24 (c) 60 (a) (b) (c) Soln.
7. 7. http://www.education707.com 7 Highest Common Factor H.C.F H.C.F Defn. The H.C.F of two or more numbers is the largest common factor among the common factors of the numbers. Ex 1. Find the H.C.F of 12 and 18. Soln. 12 = 2 × 2 × 3 18 = 2 × 3 × 3 H.C.F = 2 × 3 = 6 Select common factors
8. 8. http://www.education707.com 8 Highest Common Factor H.C.F Ex 2. Find the H.C.F of 24 , 32 , 36. Soln. 24 = 2 × 2 × 2 × 3 32 = 2 × 2 × 2 × 2 × 2 H.C.F = 2 × 2 = 4 36 = 2 × 2 × 3 × 3 Step 2. Select factors common to 24, 32 and 36 Step 1. Find the prime factorisation of 24, 32 and 36 .
9. 9. http://www.education707.com 9 Multiples Multiples of 2 are 2 , 4 , 6 , 8 , 10 , 12 , ……… Multiples of 3 are 3 , 6 , 9 , 12 , 15 , 18 , …….. Ex 1. Ex 2. Evaluation List the multiples of 4. List the multiples of 5. Ex 3. Ex 4.
10. 10. http://www.education707.com 10 Least Common Multiple L.C.M L.C.M Defn. The L.C.M of two or more numbers is the smallest common multiple of the numbers. Ex 1. Find the L.C.M of 12 and 18. Soln. 12 = 2 × 2 × 3 18 = 2 × 3 × 3 L.C.M = 2 × 2 × 3 × 3 = 36 Find the prime factorisation of 12 and 18.
### Editor's Notes
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• http://www.education707.com | 1,397 | 3,563 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2023-14 | latest | en | 0.795046 |
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For more course tutorials visitwww.bus308.comBUS 308 Week 1 DQ 1 Language BUS 308 Week 1 DQ 2 Probability BUS 308 Week 1 Quiz BUS 308 Week 1 Problem Set BUS 308 Week 1 Quiz (New) BUS 308 Week 2 DQ 1 Hypotheses
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BUS 308 Week 1 DQ 1 Language
• BUS 308 Week 1 DQ 1 Language
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• BUS 308 Week 1 Quiz (New)
• BUS 308 Week 2 DQ 1 Hypotheses
• BUS 308 Week 2 DQ 2 Variation
• BUS 308 Week 2 Quiz BUS 308 Week 2 Quiz (New)
• BUS 308 Week 2 Problem Set
• Discussion 1-1/Language Numbers and measurements are the language of business. Organizations look at results in many ways: expenses, quality levels, efficiencies, time, costs, etc. What measures does your department keep track of?
### BUS 308 Course Experience Tradition / bus308.com
BUS 308 Week 1 DQ 2 Probability
BUS 308 Week 1 Problem Set
• Things vary in life – virtually nothing (except physical standards such as the speed of light) we interact with is constant over time. Much of this variation follows somewhat predictable patterns that can be examined using probability. An example of a subjective probability is: “Cops usually do not patrol this road, so I can get away with speeding.”
• For assistance with these calculations, see the Recommended Resources for Week One. Measurement issues. Data, even numerically code variables, can be one of 4 levels – nominal, ordinal, interval, or ratio. It is important to identify which level a variable is, as this impacts the kind of analysis we can do with the data. For example, descriptive statistics such as means can only be done on interval or ratio level data
### BUS 308 Course Experience Tradition / bus308.com
BUS 308 Week 1 Quiz (New)
BUS 308 Week 1 Quiz
• 1.Question :In statistical notation, M is to μ as s is to σ.
• 2.Question :A parameter refers to a sample characteristic. Question
• 3.Question :Data on the city from which members of a board of directors come represent interval data. Question
• 1. Question :Data on the city from which members of a board of directors come represent interval data.
• 2. Question : Inferential statistics infer the characteristics of samples.
• 3. Question :The mode is which of the following?
### BUS 308 Course Experience Tradition / bus308.com
BUS 308 Week 2 DQ 1 Hypotheses
BUS 308 Week 2 DQ 2 Variation
• Discussion 2-1/Hypotheses What is a hypothesis test? Why do we need to use them to make decisions about relating sample results to the population; why can’t we just make our decisions by the sample value?
• Variation exists in virtually all parts of our lives. We often see variation in results in what we spend (utility costs each month, food costs, business supplies, etc.). Consider the measures and data you use (in either your personal or job activities). When are differences (between one time period and another, between different production lines, etc.) between average or actual results important?
### BUS 308 Course Experience Tradition / bus308.com
BUS 308 Week 2 Problem Set
BUS 308 Week 2 Quiz (New)
• Problem Set Week TwoComplete the problems below and submit your work in an Excel document. Be sure to show all of your work and clearly label all calculations. All statistical calculations will use the Included in the Week Two tab of theEmployee Salary Data Set are 2 one-sample t-tests comparing male and female average salaries to the overall sample mean.
• 1.Question :What is the relationship between the power of a statistical test and decision errors?
• 2.Question :The desired sample depends on all of these factors except?
• 3.Question :What question does the z test answer?
• 4.Question :The desired sample size depends only the size of the population to be tested.
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BUS 308 Week 2 Quiz
BUS 308 Week 3 DQ 1 ANOVA
• 1.Question :How is the sum of squares unlike either the standard deviation or the variance?
• 2.Question :If sums of squares statistics are calculated for shoppers at three different retail outlets, what statistic will indicate the variability among those at each outlet?
• 3.Question :Which is the symbol used for the test statistic in ANOVA?
• In many ways, comparing multiple sample means is simply an extension of what we covered last week. Just as we had 3 versions of the t-test (1 sample, 2 sample (with and without equal variance), and paired; we have several versions of ANOVA – single factor, factorial (called 2-factor with replication in Excel), and within-subjects (2-factor without replication in Excel).
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BUS 308 Week 3 DQ 2 Effect Size
BUS 308 Week 3 Problem Set
• Several statistical tests have a way to measure effect size. What is this, and when might you want to use it in looking at results from these tests on job related data?
• ASSIGNMENT WEEK 3 Complete the problems below and submit your work in an Excel document. Be sure to show all of your work and clearly label all calculations. All statistical calculations will use the (Note: Questions 1- 4 have additional elements to respond to below the analysis results.) Last week, we found that the average performance ratings do not differ between males and females in the population.
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BUS 308 Week 4 DQ 1 Confidence Intervals
BUS 308 Week 4 DQ 2 Chi-Square Tests
• Discussion 4-1/Confidence Intervals Many people do not “like” or “trust” single point estimates for things they need measured. Looking back at the data examples you have provided in the previous discussion questions on this issue, how might adding confidence intervals help managers accept the results better? Why?
• Discussion 4-2/Chi-Square Tests Chi-square tests are great to show if distributions differ or if two variables interact in producing outcomes. What are some examples of variables that you might want to check using the chi-square tests? What would these results tell you?
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BUS 308 Week 4 Problem Set
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• ASSIGNMENT WEEK 4 Let’s look at some other factors that might influence pay. Complete the problems below and submit your work in an Excel document. Be sure to show all of your work and clearly label all calculations. All statistical calculations will use the Using our sample data, construct a 95% confidence interval for the population's mean salary for each gender. Interpret the results
• 1. Question :The goodness of fit test null hypothesis states that the sample data does not match an expected distribution.
• 2. Question :Statistical significance in the Chi-square test means the population distribution (expected) is not the source of the sample (observed) data.
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BUS 308 Week 4 Quiz
BUS 308 Week 5 DQ 1 Correlation
• 1.Question :With reference to problem 1, what statistic determines the correlation of experience with productivity, controlling for age in experience?
• 2.Question :In a problem where interest rates and growth of the economy are used to predict consumer spending, which of the following will increase prediction error?
• Discussion 5-1/Correlation What results in your departments seem to be correlated or related to other activities? How could you verify this? Create a null and alternate hypothesis for one of these issues. What are the managerial implications of a correlation between these variables?
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BUS 308 Week 5 DQ 2 Regression
BUS 308 Week 5 Final Paper (2 Papers)
• Discussion 5-2/Regression At times we can generate a regression equation to explain outcomes. For example, an employee’s salary can often be explained by their pay grade, appraisal rating, education level, etc. What variables might explain or predict an outcome in your department or life? If you generated a regression equation, how would you interpret it and the residuals from it?
• This tutorial contains 2 Different Papers The final paper provides you with an opportunity to integrate and reflect on what you have learned during the class.The question to address is: “What have you learned about statistics?” In developing your responses, consider – at a minimum – and discuss the application of each of the course elements in analyzing and making decisions about data (counts and/or measurements).
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BUS 308 Week 5 Problem Set
• ASSIGNMENT WEEK 5 Create a correlation table for the variables in our (Use analysis ToolPak or StatPlus:mac LE function Correlation). Reviewing the data levels from week 1, what variables can be used in a Pearson’s Correlation Table (which is what Excel produces)?
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www.bus308.com | 2,217 | 9,768 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2017-17 | longest | en | 0.78388 |
https://groupprops.subwiki.org/w/index.php?title=Degree_of_irreducible_representation_divides_order_of_group&diff=24342&oldid=24341 | 1,571,617,279,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987750110.78/warc/CC-MAIN-20191020233245-20191021020745-00109.warc.gz | 507,299,675 | 10,154 | # Difference between revisions of "Degree of irreducible representation divides order of group"
This article gives the statement, and possibly proof, of a constraint on numerical invariants that can be associated with a finite group
This article states a result of the form that one natural number divides another. Specifically, the (degree of a linear representation) of a/an/the (irreducible linear representation) divides the (order) of a/an/the (group).
View other divisor relations |View congruence conditions
## Statement
Let $G$ be a finite group and $\rho$ an irreducible representation of $G$ over an algebraically closed field of characteristic zero. Then, the degree of $\rho$ divides the order of $G$.
## Related facts
### Other facts about degrees of irreducible representations
Further information: Degrees of irreducible representations
### Breakdown for a field that is not algebraically closed
Let $G$ be the cyclic group of order three and $\R$ be the field. Then, there are two irreducible representations of $G$ over $\R$: the trivial representation, and a two-dimensional representation given by the action by rotation by multiples of $2\pi/3$. The two-dimensional representation has degree $2$, and this does not divide the order of the group, which is $3$.
We still have the following results:
## Proof
### Introduction of some algebraic integers
Further information: Convolution algebra on conjugacy classes
Using the convolution algebra on conjugacy classes, we can show that for any representation $\rho$ with character $\chi$, and any conjugacy class $c$, the number:
$\omega(c,\rho) = \frac{|c|\chi(c)}{\chi(1)}$
are algebraic integers. Note that $\chi(1)$ is the degree of $\rho$.
### A little formula
We know that if $\rho$ is irreducible:
$\sum_c |c| \chi(c) \overline{\chi(c)} = |G|$
by the orthonormality of the irreducible characters.
Dividing by $\chi(1)$, we get:
$\sum_c \omega(c,\rho) \overline{\chi(c)} = |G|/\chi(1)$
Note that since the characters are algebraic integers and so are the values taken by $\omega$, the overall left-hand side is an algebraic integer. Thus $|G|/\chi(1)$ is an algebraic integer.
But since it is a rational number, it must be a rational integer, or in other words, the degree of $\rho$ divides the order of $G$. | 545 | 2,303 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 26, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2019-43 | longest | en | 0.857795 |
https://quizzes.buzz/quizzes/105126-35201-9023-3-5511-2701/15182 | 1,604,013,772,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107905965.68/warc/CC-MAIN-20201029214439-20201030004439-00617.warc.gz | 476,907,073 | 13,045 | ## Simplification & Approximation Mcqs
MCQ: 105.126 * 35.201 – 90.23 * 3 + 55.11 * 27.01 = ____________?
1. 4890
2. 40000
3. 271
4. 5996
This above question "105.126 * 35.201 – 90.23 * 3 + 55.11 * 27.01 = ____________?" taken from a category of Mathematics Quizzes, if you learn all question of this topic click on it "Mathematics Quizzes". It takes five to ten minutes to complete this free Quantitative MCQs paper 1 test. You will see 4 or 5 option of each question. You must choose / think only one option and then press on answer key for check right answer. Practice "Quantitative MCQs paper 1" MCQs and share with your friends, brothers, sisters.
## Simplification & Approximation Mcqs
MCQ: (0.9 * 0.9 – 0.8 * 0.8)/1.7 = _________?
MCQ: Free notebooks were distributed equally among children of a class. The number of notebooks each child got was one-eighth of the number of children. Had the number of children been half, each child would have got 16 notebooks. Total how many notebooks were distributed?
MCQ: 7 2/5 of 110 / ? = 844
MCQ: ? % of 400 + 40% of 160 = 17% of 400
MCQ: In a regular week, there are 5 working days and for each day, the working hours are 8. A man gets Rs. 2.40 per hour for regular work and Rs. 3.20 per hours for overtime. If he earns Rs. 432 in 4 weeks, then how many hours does he work for______?
MCQ: 6 of 5/8 / 5/8 – 1/8 = ______?
MCQ: 32% of 425 – ?% of 250 = 36
MCQ: 31.2 * 14.5 * 9.6 = ________?
MCQ: [8 + {(21 * √9/441 – 1 /5 of 60% – 1/5) * 625 + 7} ÷ 4] = _________?
MCQ: Positive integers indicated by x and y satisfy 3 1/x * y 2/5 = 13 3/4, the fractions being in their lowest terms, then x = ? and y = ____________?
MCQ: 105 * 99 * 299 = _________?
MCQ: 16.66% of 33.33% of 66.66% of ? = 30
MCQ: 64309 – 8703 + 798 – 437 = ________?
MCQ: (815 / ?) / 100 * 28 * 3 – (815 * 14 * 6)/100 = 1680
MCQ: The value of (2.75)³ – (2.00)³ – (0.75)³ is__________?
MCQ: 109.13 * 34.14 – 108.12 * 33 = ________?
MCQ: 25 2/5 + 8 4/5 + 4/5 = _________?
MCQ: (98 / 7 / 2) / ? = 14
MCQ: 1/3 of 2/6 of 1/7 of 2/28 of 2140 = ________?
MCQ: 3/7 of 4/9 of 2/5 of 7560 = (?)²
MCQ: 43% of 784 = ________?
MCQ: 7 + 1/2 of [8 – 4 / 2 * 3 – 1 + 1 + (4 + 2) – 2 * 2] = ________?
MCQ: (49² * 18²) / 324 = ________?
MCQ: Jameel gets on the elevator at the 11th floor of a building and rides up at the rate of 57 floors per minute. At the same time, Nasir gets on an elevator at the 51st floor of the same building and rides down at the rate of 63 floors per minute. If they continue travelling at these rates, then at which floor will their paths cross?
MCQ: 183.21 + 2.9928 * 66.84 = _______?
MCQ: 6 2/3 * 9 3/5 * 2 1/3 + 1 1/3 = __________?
MCQ: [50 – {20 + 60% of (26/13 + 24/12 + 21) – 2} + 8] = 625/?
MCQ: 9000 + 16 2/3 % of ? = 10500
MCQ: 8 2/3 + 6 4/5 = _____?
MCQ: 52 * 40.1 = ________? | 1,151 | 2,888 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2020-45 | longest | en | 0.84072 |
https://math.stackexchange.com/questions/914216/evaluating-an-indefinite-integral-involving-x-x-lfloor-x-rfloor | 1,579,692,984,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250606975.49/warc/CC-MAIN-20200122101729-20200122130729-00498.warc.gz | 540,383,867 | 32,388 | # Evaluating an Indefinite Integral involving $\{x\} = x - \lfloor x \rfloor$
This is probably a pretty simple question, but I just want to check something that I'm not completely sure about. I want to evaluate $\int{\{x\}}^ndx$, with $\{x\} = x - \lfloor x \rfloor$. Instead of using substitution or anything like that, I instead simply looked for a function that measures the area under $\{x\}^n$. The function (as seen here for $n = 4$), is just slices of $y = x^n$ with $x \in[0, 1]$. So the area from $0$ to $x$ is given by
$$\lfloor x \rfloor\int{x^ndx} + \{x\}\int{x^ndx} = (\lfloor x \rfloor + \{x\})\int{x^ndx} = x\int{x^ndx} = \frac{x^{n+2}}{n+1} + C.$$
But since the area under $\{x\}^n$ is always positive, and $x^{n+2}$ is negative when $n+2$ is odd, the final antiderivative should actually be
$$\int{\{x\}^ndx} = \frac{|x^{n+2}|}{n+1} + C, n \neq -1$$
Is it wrong to do this? Should I be evaluating indefinite integrals solely using the tools created to evaluate them (such as substitution, integration by parts, etc.) or are geometric proofs for the value of indefinite integrals equally valid?
Also I have a strange feeling in my gut that this is really wrong for some reason, like instead of geometrically finding a function that measures the area under $\{x\}^n$ I should have been looking for a function whose slope is $\{x\}^n$, so can someone clarify whether the antiderivative I got is even correct or not?
Because $\{n+x\}^n=x^n$ where $x \in [0,1), n \in \mathbb Z$, this function is periodic and is defined by its values in $[0,1)$. So if $x \ge 0,\ n \ne -1$:
\begin{align} \int_0^x\{t\}^n\ dt&=\int_0^{\lfloor x \rfloor+\{x\}}\{t\}^n\ dt\\ &=\int_0^{\lfloor x \rfloor}\{t\}^n \ dt+\int_{\lfloor x \rfloor}^{\lfloor x \rfloor+\{x\}}\{t\}^n\ dt\\ &=\int_0^{\lfloor x \rfloor}\{t\}^n \ dt+\int_0^{\{x\}}\{t\}^n\ dt\\ &=\lfloor x\rfloor\int_0^1t^n \ dt +\int_0^{\{x\}}t^n\ dt\\ &=\frac{\lfloor x \rfloor+\{x\}^{n+1}}{n+1} \end{align}
Important fact:
The indefinite integral $\int f(x) \ dx$ is defined to be a function $F(x)$ such that $F'(x) = f(x)$. This will exist by the Fundamental Theorem of Calculus if and only if $f$ is a continuous function.
Hence, in this case, since $\{x\}^n$ is not a continuous function from $\mathbb R$ to $\mathbb R$, the indefinite integral is not defined in this region! In general, it isn't a good idea to think of the indefinite integral as the area under the curve.
However, due to the properties of this function, it is still integrable, and it is still possible to define a definite integral. In particular, since the function is continuous on $[n,\ n+1)$ for $n \in \mathbb Z$, it will have an indefinite integral (or an antiderivative) in each of these regions. Given $x \in \mathbb R$, it is possible to integrate this function from $0$ to $x$ to get the area under the curve as shown above. However, this will be a definite integral and not an indefinite one.
• Okay, I understand how you get that. Where did I go wrong in my process? – user3002473 Aug 30 '14 at 20:39
• I think the key cause is that in trying to integrate from $0$ to $x$, you have used $x$ as a dummy variable - i.e. $\int_0^x \{x\}^n dx$. As such, your $x$ is appearing in two different varieties, which is always gonna cause confusion! I'm not sure I see how you got that the integral should be $\lfloor x \rfloor\int{x^ndx} + \{x\}\int{x^ndx}$ – Mathmo123 Aug 30 '14 at 20:41
• OH! I finally realized exactly what you mean! Okay, awesome, that makes total sense to me know. Thanks for your help! :) – user3002473 Aug 30 '14 at 20:46
• See my edit - it contains an import distinction – Mathmo123 Aug 30 '14 at 20:52 | 1,187 | 3,667 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2020-05 | latest | en | 0.793448 |
http://theinfolist.com/php/SummaryGet.php?FindGo=Number-theoretic_function | 1,618,990,126,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039526421.82/warc/CC-MAIN-20210421065303-20210421095303-00256.warc.gz | 88,079,244 | 13,332 | TheInfoList
In number theory, an arithmetic, arithmetical, or number-theoretic function is for most authors any function ''f''(''n'') whose domain is the positive integers and whose range is a subset of the complex numbers. Hardy & Wright include in their definition the requirement that an arithmetical function "expresses some arithmetical property of ''n''". An example of an arithmetic function is the divisor function whose value at a positive integer ''n'' is equal to the number of divisors of ''n''. There is a larger class of number-theoretic functions that do not fit the above definition, for example, the prime-counting functions. This article provides links to functions of both classes. Arithmetic functions are often extremely irregular (see table), but some of them have series expansions in terms of Ramanujan's sum.
An arithmetic function ''a'' is * completely additive if ''a''(''mn'') = ''a''(''m'') + ''a''(''n'') for all natural numbers ''m'' and ''n''; * completely multiplicative if ''a''(''mn'') = ''a''(''m'')''a''(''n'') for all natural numbers ''m'' and ''n''; Two whole numbers ''m'' and ''n'' are called coprime if their greatest common divisor is 1, that is, if there is no prime number that divides both of them. Then an arithmetic function ''a'' is * additive if ''a''(''mn'') = ''a''(''m'') + ''a''(''n'') for all coprime natural numbers ''m'' and ''n''; * multiplicative if ''a''(''mn'') = ''a''(''m'')''a''(''n'') for all coprime natural numbers ''m'' and ''n''.
Notation
$\sum_p f\left(p\right)$ and $\prod_p f\left(p\right)$ mean that the sum or product is over all prime numbers: :$\sum_p f\left(p\right) = f\left(2\right) + f\left(3\right) + f\left(5\right) + \cdots$ :$\prod_p f\left(p\right)= f\left(2\right)f\left(3\right)f\left(5\right)\cdots.$ Similarly, $\sum_ f\left(p^k\right)$ and $\prod_ f\left(p^k\right)$ mean that the sum or product is over all prime powers with strictly positive exponent (so ''k'' = 0 is not included): :$\sum_ f\left(p^k\right) = \sum_p\sum_ f\left(p^k\right) = f\left(2\right) + f\left(3\right) + f\left(4\right) +f\left(5\right) +f\left(7\right)+f\left(8\right)+f\left(9\right)+\cdots$ $\sum_ f\left(d\right)$ and $\prod_ f\left(d\right)$ mean that the sum or product is over all positive divisors of ''n'', including 1 and ''n''. For example, if ''n'' = 12, :$\prod_ f\left(d\right) = f\left(1\right)f\left(2\right) f\left(3\right) f\left(4\right) f\left(6\right) f\left(12\right).\$ The notations can be combined: $\sum_ f\left(p\right)$ and $\prod_ f\left(p\right)$ mean that the sum or product is over all prime divisors of ''n''. For example, if ''n'' = 18, :$\sum_ f\left(p\right) = f\left(2\right) + f\left(3\right),\$ and similarly $\sum_ f\left(p^k\right)$ and $\prod_ f\left(p^k\right)$ mean that the sum or product is over all prime powers dividing ''n''. For example, if ''n'' = 24, :$\prod_ f\left(p^k\right) = f\left(2\right) f\left(3\right) f\left(4\right) f\left(8\right).\$
Ω(''n''), ''ω''(''n''), ''ν''''p''(''n'') – prime power decomposition
The fundamental theorem of arithmetic states that any positive integer ''n'' can be represented uniquely as a product of powers of primes: $n = p_1^\cdots p_k^$ where ''p''1 < ''p''2 < ... < ''p''''k'' are primes and the ''aj'' are positive integers. (1 is given by the empty product.) It is often convenient to write this as an infinite product over all the primes, where all but a finite number have a zero exponent. Define the ''p''-adic valuation ν''p''(''n'') to be the exponent of the highest power of the prime ''p'' that divides ''n''. That is, if ''p'' is one of the ''p''''i'' then ''ν''''p''(''n'') = ''a''''i'', otherwise it is zero. Then :$n=\prod_p p^.$ In terms of the above the prime omega functions ω and Ω are defined by :''ω''(''n'') = ''k'', :Ω(''n'') = ''a''1 + ''a''2 + ... + ''a''''k''. To avoid repetition, whenever possible formulas for the functions listed in this article are given in terms of ''n'' and the corresponding ''p''''i'', ''a''''i'', ω, and Ω.
Multiplicative functions
σ''k''(''n''), τ(''n''), ''d''(''n'') – divisor sums
σ''k''(''n'') is the sum of the ''k''th powers of the positive divisors of ''n'', including 1 and ''n'', where ''k'' is a complex number. σ1(''n''), the sum of the (positive) divisors of ''n'', is usually denoted by σ(''n''). Since a positive number to the zero power is one, σ0(''n'') is therefore the number of (positive) divisors of ''n''; it is usually denoted by ''d''(''n'') or τ(''n'') (for the German ''Teiler'' = divisors). :$\sigma_k\left(n\right) = \prod_^ \frac = \prod_^ \left\left(1 + p_i^k + p_i^ + \cdots + p_i^\right\right).$ Setting ''k'' = 0 in the second product gives :$\tau\left(n\right) = d\left(n\right) = \left(1 + a_\right)\left(1+a_\right)\cdots\left(1+a_\right).$
φ(''n'') – Euler totient function
φ(''n''), the Euler totient function, is the number of positive integers not greater than ''n'' that are coprime to ''n''. :$\varphi\left(n\right) = n \prod_ \left\left(1-\frac\right\right) =n \left\left(\frac\right\right)\left\left(\frac\right\right) \cdots \left\left(\frac\right\right) .$
J''k''(''n'') – Jordan totient function
J''k''(''n''), the Jordan totient function, is the number of ''k''-tuples of positive integers all less than or equal to ''n'' that form a coprime (''k'' + 1)-tuple together with ''n''. It is a generalization of Euler's totient, . :$J_k\left(n\right) = n^k \prod_ \left\left(1-\frac\right\right) =n^k \left\left(\frac\right\right)\left\left(\frac\right\right) \cdots \left\left(\frac\right\right) .$
μ(''n'') – Möbius function
μ(''n''), the Möbius function, is important because of the Möbius inversion formula. See Dirichlet convolution, below. :$\mu\left(n\right)=\begin \left(-1\right)^=\left(-1\right)^ &\text\; \omega\left(n\right) = \Omega\left(n\right)\\ 0&\text\;\omega\left(n\right) \ne \Omega\left(n\right).\end$ This implies that μ(1) = 1. (Because Ω(1) = ω(1) = 0.)
τ(''n'') – Ramanujan tau function
τ(''n''), the Ramanujan tau function, is defined by its generating function identity: :$\sum_\tau\left(n\right)q^n=q\prod_\left(1-q^n\right)^.$ Although it is hard to say exactly what "arithmetical property of ''n''" it "expresses", (''τ''(''n'') is (2π)−12 times the ''n''th Fourier coefficient in the q-expansion of the modular discriminant function) it is included among the arithmetical functions because it is multiplicative and it occurs in identities involving certain σ''k''(''n'') and ''r''''k''(''n'') functions (because these are also coefficients in the expansion of modular forms).
''c''''q''(''n'') – Ramanujan's sum
''c''''q''(''n''), Ramanujan's sum, is the sum of the ''n''th powers of the primitive ''q''th roots of unity: :$c_q\left(n\right)= \sum_ e^ .$ Even though it is defined as a sum of complex numbers (irrational for most values of ''q''), it is an integer. For a fixed value of ''n'' it is multiplicative in ''q'': :If ''q'' and ''r'' are coprime, then $c_q\left(n\right)c_r\left(n\right)=c_\left(n\right).$
''ψ''(''n'') - Dedekind psi function
The Dedekind psi function, used in the theory of modular functions, is defined by the formula :$\psi\left(n\right) = n \prod_\left\left(1+\frac\right\right).$
Completely multiplicative functions
λ(''n'') – Liouville function
''λ''(''n''), the Liouville function, is defined by :$\lambda \left(n\right) = \left(-1\right)^.$
''χ''(''n'') – characters
All Dirichlet characters ''χ''(''n'') are completely multiplicative. Two characters have special notations: The principal character (mod ''n'') is denoted by ''χ''0(''a'') (or ''χ''1(''a'')). It is defined as :$\chi_0\left(a\right) = \begin 1 & \text \gcd\left(a,n\right) = 1, \\ 0 & \text \gcd\left(a,n\right) \ne 1. \end$ The quadratic character (mod ''n'') is denoted by the Jacobi symbol for odd ''n'' (it is not defined for even ''n''.): :$\Bigg\left(\frac\Bigg\right) = \left\left(\frac\right\right)^\left\left(\frac\right\right)^\cdots \left\left(\frac\right\right)^.$ In this formula $\left(\tfrac\right)$ is the Legendre symbol, defined for all integers ''a'' and all odd primes ''p'' by :$\left\left(\frac\right\right) = \begin \;\;\,0 & \text a \equiv 0 \pmod p, \\+1 & \texta \not\equiv 0\pmod p \textx, \;a\equiv x^2\pmod p \\-1 & \text x. \end$ Following the normal convention for the empty product, $\left\left(\frac\right\right) = 1.$
''ω''(''n'') – distinct prime divisors
ω(''n''), defined above as the number of distinct primes dividing ''n'', is additive (see Prime omega function).
Ω(''n'') – prime divisors
Ω(''n''), defined above as the number of prime factors of ''n'' counted with multiplicities, is completely additive (see Prime omega function).
''ν''''p''(''n'') – ''p''-adic valuation of an integer ''n''
For a fixed prime ''p'', ''ν''''p''(''n''), defined above as the exponent of the largest power of ''p'' dividing ''n'', is completely additive.
(''x''), Π(''x''), ''θ''(''x''), ''ψ''(''x'') – prime count functions
These important functions (which are not arithmetic functions) are defined for non-negative real arguments, and are used in the various statements and proofs of the prime number theorem. They are summation functions (see the main section just below) of arithmetic functions which are neither multiplicative nor additive. (''x''), the prime counting function, is the number of primes not exceeding ''x''. It is the summation function of the characteristic function of the prime numbers. :$\pi\left(x\right) = \sum_1$ A related function counts prime powers with weight 1 for primes, 1/2 for their squares, 1/3 for cubes, ... It is the summation function of the arithmetic function which takes the value 1/''k'' on integers which are the k-th power of some prime number, and the value 0 on other integers. :$\Pi\left(x\right) = \sum_\frac.$ ''θ''(''x'') and ''ψ''(''x''), the Chebyshev functions, are defined as sums of the natural logarithms of the primes not exceeding ''x''. :$\vartheta\left(x\right)=\sum_ \log p,$ :$\psi\left(x\right) = \sum_ \log p.$ The Chebyshev function ''ψ''(''x'') is the summation function of the von Mangoldt function just below.
Λ(''n'') – von Mangoldt function
Λ(''n''), the von Mangoldt function, is 0 unless the argument ''n'' is a prime power , in which case it is the natural log of the prime ''p'': :$\Lambda\left(n\right) = \begin\log p &\text n = 2,3,4,5,7,8,9,11,13,16,\ldots=p^k \text\\ 0&\text n=1,6,10,12,14,15,18,20,21,\dots \;\;\;\;\text. \end$
''p''(''n'') – partition function
''p''(''n''), the partition function, is the number of ways of representing ''n'' as a sum of positive integers, where two representations with the same summands in a different order are not counted as being different: :$p\left(n\right) = |\left\|.$
λ(''n'') – Carmichael function
''λ''(''n''), the Carmichael function, is the smallest positive number such that $a^\equiv 1 \pmod$ for all ''a'' coprime to ''n''. Equivalently, it is the least common multiple of the orders of the elements of the multiplicative group of integers modulo ''n''. For powers of odd primes and for 2 and 4, ''λ''(''n'') is equal to the Euler totient function of ''n''; for powers of 2 greater than 4 it is equal to one half of the Euler totient function of ''n'': :$\lambda\left(n\right) = \begin \;\;\phi\left(n\right) &\textn = 2,3,4,5,7,9,11,13,17,19,23,25,27,\dots\\ \tfrac12\phi\left(n\right)&\textn=8,16,32,64,\dots \end$ and for general ''n'' it is the least common multiple of λ of each of the prime power factors of ''n'': :$\lambda\left(p_1^p_2^ \dots p_^\right) = \operatornamelambda\left(p_1^\right),\;\lambda\left(p_2^\right),\dots,\lambda\left(p_^\right)$
''h''(''n'') – Class number
''h''(''n''), the class number function, is the order of the ideal class group of an algebraic extension of the rationals with discriminant ''n''. The notation is ambiguous, as there are in general many extensions with the same discriminant. See quadratic field and cyclotomic field for classical examples.
''r''''k''(''n'') – Sum of ''k'' squares
''r''''k''(''n'') is the number of ways ''n'' can be represented as the sum of ''k'' squares, where representations that differ only in the order of the summands or in the signs of the square roots are counted as different. :$r_k\left(n\right) = |\|$
''D''(''n'') – Arithmetic derivative
Using the Heaviside notation for the derivative, ''D''(''n'') is a function such that : $D\left(n\right) = 1$ if ''n'' prime, and : $D\left(mn\right) = m D\left(n\right) + D\left(m\right) n$ (Product rule)
Summation functions
Given an arithmetic function ''a''(''n''), its summation function ''A''(''x'') is defined by :$A\left(x\right) := \sum_ a\left(n\right) .$ ''A'' can be regarded as a function of a real variable. Given a positive integer ''m'', ''A'' is constant along open intervals ''m'' < ''x'' < ''m'' + 1, and has a jump discontinuity at each integer for which ''a''(''m'') ≠ 0. Since such functions are often represented by series and integrals, to achieve pointwise convergence it is usual to define the value at the discontinuities as the average of the values to the left and right: :$A_0\left(m\right) := \frac12\left\left(\sum_ a\left(n\right) +\sum_ a\left(n\right)\right\right) = A\left(m\right) - \frac12 a\left(m\right) .$ Individual values of arithmetic functions may fluctuate wildly – as in most of the above examples. Summation functions "smooth out" these fluctuations. In some cases it may be possible to find asymptotic behaviour for the summation function for large ''x''. A classical example of this phenomenon is given by the divisor summatory function, the summation function of ''d''(''n''), the number of divisors of ''n'': :$\liminf_d\left(n\right) = 2$ :$\limsup_\frac = \log 2$ :$\lim_\frac = 1.$ An average order of an arithmetic function is some simpler or better-understood function which has the same summation function asymptotically, and hence takes the same values "on average". We say that ''g'' is an ''average order'' of ''f'' if :$\sum_ f\left(n\right) \sim \sum_ g\left(n\right)$ as ''x'' tends to infinity. The example above shows that ''d''(''n'') has the average order log(''n'').
Dirichlet convolution
Given an arithmetic function ''a''(''n''), let ''F''''a''(''s''), for complex ''s'', be the function defined by the corresponding Dirichlet series (where it converges): :$F_a\left(s\right) := \sum_^\infty \frac .$ ''F''''a''(''s'') is called a generating function of ''a''(''n''). The simplest such series, corresponding to the constant function ''a''(''n'') = 1 for all ''n'', is ''ς''(''s'') the Riemann zeta function. The generating function of the Möbius function is the inverse of the zeta function: :$\zeta\left(s\right)\,\sum_^\infty\frac=1, \;\;\mathfrak \,s >0.$ Consider two arithmetic functions ''a'' and ''b'' and their respective generating functions ''F''''a''(''s'') and ''F''''b''(''s''). The product ''F''''a''(''s'')''F''''b''(''s'') can be computed as follows: :$F_a\left(s\right)F_b\left(s\right) = \left\left( \sum_^\frac \right\right)\left\left( \sum_^\frac \right\right) .$ It is a straightforward exercise to show that if ''c''(''n'') is defined by :$c\left(n\right) := \sum_ a\left(i\right)b\left(j\right) = \sum_a\left(i\right)b\left\left(\frac\right\right) ,$ then :$F_c\left(s\right) =F_a\left(s\right) F_b\left(s\right).$ This function ''c'' is called the Dirichlet convolution of ''a'' and ''b'', and is denoted by $a*b$. A particularly important case is convolution with the constant function ''a''(''n'') = 1 for all ''n'', corresponding to multiplying the generating function by the zeta function: :$g\left(n\right) = \sum_f\left(d\right).$ Multiplying by the inverse of the zeta function gives the Möbius inversion formula: :$f\left(n\right) = \sum_\mu\left\left(\frac\right\right)g\left(d\right).$ If ''f'' is multiplicative, then so is ''g''. If ''f'' is completely multiplicative, then ''g'' is multiplicative, but may or may not be completely multiplicative.
Relations among the functions
There are a great many formulas connecting arithmetical functions with each other and with the functions of analysis, especially powers, roots, and the exponential and log functions. The page divisor sum identities contains many more generalized and related examples of identities involving arithmetic functions. Here are a few examples:
Dirichlet convolutions
:$\sum_\mu\left(\delta\right)= \sum_\lambda\left\left(\frac\right\right)|\mu\left(\delta\right)|= \begin 1 & \text n=1\\ 0 & \text n\ne1 \end$ where ''λ'' is the Liouville function. :$\sum_\varphi\left(\delta\right) = n.$ ::$\varphi\left(n\right) =\sum_\mu\left\left(\frac\right\right)\delta =n\sum_\frac.$ Möbius inversion :$\sum_ J_k\left(d\right) = n^k.$ ::$J_k\left(n\right) =\sum_\mu\left\left(\frac\right\right)\delta^k =n^k\sum_\frac.$ Möbius inversion :$\sum_\delta^sJ_r\left(\delta\right)J_s\left\left(\frac\right\right) = J_\left(n\right)$ :$\sum_\varphi\left(\delta\right)d\left\left(\frac\right\right)= \sigma\left(n\right).$ :$\sum_|\mu\left(\delta\right)|= 2^.$ ::$|\mu\left(n\right)|=\sum_\mu\left\left(\frac\right\right)2^.$ Möbius inversion :$\sum_2^= d\left(n^2\right).$ ::$2^=\sum_\mu\left\left(\frac\right\right)d\left(\delta^2\right).$ Möbius inversion :$\sum_d\left(\delta^2\right)= d^2\left(n\right).$ ::$d\left(n^2\right)=\sum_\mu\left\left(\frac\right\right)d^2\left(\delta\right).$ Möbius inversion :$\sum_d\left\left(\frac\right\right)2^= d^2\left(n\right).$ :$\sum_\lambda\left(\delta\right)=\begin &1\text n \text\\ &0\text n \text \end$ where λ is the Liouville function. :$\sum_\Lambda\left(\delta\right) = \log n.$ ::$\Lambda\left(n\right)=\sum_\mu\left\left(\frac\right\right)\log\left(\delta\right).$ Möbius inversion
Sums of squares
For all $k \ge 4,\;\;\; r_k\left(n\right) > 0.$ (Lagrange's four-square theorem). : :$r_2\left(n\right) = 4\sum_\left\left(\frac\right\right),$ where the Kronecker symbol has the values :$\left\left(\frac\right\right) = \begin +1&\textn\equiv 1 \pmod 4 \\ -1&\textn\equiv 3 \pmod 4\\ \;\;\;0&\textn\text.\\ \end$ There is a formula for r3 in the section on class numbers below. :$r_4\left(n\right) = 8 \sum_d = 8 \left(2+\left(-1\right)^n\right)\sum_d = \begin 8\sigma\left(n\right)&\text n \text\\ 24\sigma\left\left(\frac\right\right)&\text n \text \end,$ where ''ν'' = ''ν''2(''n''). :$r_6\left(n\right) = 16 \sum_ \chi\left\left(\frac\right\right)d^2 - 4\sum_ \chi\left(d\right)d^2,$ where $\chi\left(n\right) = \left\left(\frac\right\right).$Hardy & Wright, § 20.13 Define the function σ''k''*(''n'') as :$\sigma_k^*\left(n\right) = \left(-1\right)^\sum_\left(-1\right)^d d^k= \begin \sum_ d^k=\sigma_k\left(n\right)&\text n \text\\ \sum_d^k -\sum_d^k&\text n \text. \end$ That is, if ''n'' is odd, σ''k''*(''n'') is the sum of the ''k''th powers of the divisors of ''n'', that is, σ''k''(''n''), and if ''n'' is even it is the sum of the ''k''th powers of the even divisors of ''n'' minus the sum of the ''k''th powers of the odd divisors of ''n''. :$r_8\left(n\right) = 16\sigma_3^*\left(n\right).$ Adopt the convention that Ramanujan's ''τ''(''x'') = 0 if ''x'' is not an integer. :$r_\left(n\right) = \frac\sigma_^*\left(n\right) + \frac\left\$
Divisor sum convolutions
Here "convolution" does not mean "Dirichlet convolution" but instead refers to the formula for the coefficients of the product of two power series: :$\left\left(\sum_^\infty a_n x^n\right\right)\left\left(\sum_^\infty b_n x^n\right\right) = \sum_^\infty \sum_^\infty a_i b_j x^ = \sum_^\infty \left\left(\sum_^n a_i b_\right\right) x^n = \sum_^\infty c_n x^n .$ The sequence $c_n = \sum_^n a_i b_$ is called the convolution or the Cauchy product of the sequences ''a''''n'' and ''b''''n''.
These formulas may be proved analytically (see Eisenstein series) or by elementary methods. :$\sigma_3\left(n\right) = \frac\left\.$ Ramanujan, ''On Certain Arithmetical Functions'', Table IV; ''Papers'', p. 146 :$\sigma_5\left(n\right) = \frac\left\.$ Koblitz, ex. III.2.8 :$\begin \sigma_7\left(n\right) &=\frac\left\\\ &=\sigma_3\left(n\right) + 120\sum_\sigma_3\left(k\right)\sigma_3\left(n-k\right). \end$ :$\begin \sigma_9\left(n\right) &= \frac\left\\\ &= \frac\left\. \end$ :$\tau\left(n\right) = \frac\sigma_\left(n\right) + \frac\sigma_\left(n\right) - \frac\sum_\sigma_5\left(k\right)\sigma_5\left(n-k\right),$ where ''τ''(''n'') is Ramanujan's function. Since ''σ''''k''(''n'') (for natural number ''k'') and ''τ''(''n'') are integers, the above formulas can be used to prove congruences for the functions. See Ramanujan tau function for some examples. Extend the domain of the partition function by setting ''p''(0) = 1. :$p\left(n\right)=\frac\sum_\sigma\left(k\right)p\left(n-k\right).$ This recurrence can be used to compute ''p''(''n'').
Class number related
Peter Gustav Lejeune Dirichlet discovered formulas that relate the class number ''h'' of quadratic number fields to the Jacobi symbol. An integer ''D'' is called a fundamental discriminant if it is the discriminant of a quadratic number field. This is equivalent to ''D'' ≠ 1 and either a) ''D'' is squarefree and ''D'' ≡ 1 (mod 4) or b) ''D'' ≡ 0 (mod 4), ''D''/4 is squarefree, and ''D''/4 ≡ 2 or 3 (mod 4). Extend the Jacobi symbol to accept even numbers in the "denominator" by defining the Kronecker symbol: :$\left\left(\frac\right\right) = \begin \;\;\,0&\text a \text \\\left(-1\right)^&\texta \text \end$ Then if ''D'' < −4 is a fundamental discriminant :$\begin h\left(D\right) & = \frac \sum_^r\left\left(\frac\right\right)\\ & = \frac \sum_^\left\left(\frac\right\right). \end$ There is also a formula relating ''r''3 and ''h''. Again, let ''D'' be a fundamental discriminant, ''D'' < −4. Then :$r_3\left(|D|\right) = 12\left\left(1-\left\left(\frac\right\right)\right\right)h\left(D\right).$
Prime-count related
Let $H_n = 1 + \frac12 + \frac13 + \cdots +\frac$ be the ''n''th harmonic number. Then :$\sigma\left(n\right) \le H_n + e^\log H_n$ is true for every natural number ''n'' if and only if the Riemann hypothesis is true. The Riemann hypothesis is also equivalent to the statement that, for all ''n'' > 5040, :$\sigma\left(n\right) < e^\gamma n \log \log n$ (where γ is the Euler–Mascheroni constant). This is Robin's theorem. :$\sum_\nu_p\left(n\right) = \Omega\left(n\right).$ :$\psi\left(x\right)=\sum_\Lambda\left(n\right).$ :$\Pi\left(x\right)= \sum_\frac.$ :$e^=\prod_p.$ :$e^= \operatorname,2,\dots,\lfloor x\rfloor$
Menon's identity
In 1965 P Kesava Menon proved :$\sum_ \gcd\left(k-1,n\right) =\varphi\left(n\right)d\left(n\right).$ This has been generalized by a number of mathematicians. For example, B. Sury :$\sum_ \gcd\left(k_1-1,k_2,\dots,k_s,n\right) =\varphi\left(n\right)\sigma_\left(n\right).$ N. Rao :$\sum_ \gcd\left(k_1-a_1,k_2-a_2,\dots,k_s-a_s,n\right)^s =J_s\left(n\right)d\left(n\right),$ where ''a''1, ''a''2, ..., ''a''''s'' are integers, gcd(''a''1, ''a''2, ..., ''a''''s'', ''n'') = 1. László Fejes Tóth :$\sum_ \gcd\left(k^2-1,m_1\right)\gcd\left(k^2-1,m_2\right) =\varphi\left(n\right)\sum_ \varphi\left(\gcd\left(d_1, d_2\right)\right)2^,$ where ''m''1 and ''m''2 are odd, ''m'' = lcm(''m''1, ''m''2). In fact, if ''f'' is any arithmetical function :$\sum_ f\left(\gcd\left(k-1,n\right)\right) =\varphi\left(n\right)\sum_\frac,$ where * stands for Dirichlet convolution.
Miscellaneous
Let ''m'' and ''n'' be distinct, odd, and positive. Then the Jacobi symbol satisfies the law of quadratic reciprocity: :$\left\left(\frac\right\right) \left\left(\frac\right\right) = \left(-1\right)^.$ Let ''D''(''n'') be the arithmetic derivative. Then the logarithmic derivative : $\frac = \sum_ \frac$ Let ''λ''(''n'') be Liouville's function. Then :$|\lambda\left(n\right)|\mu\left(n\right) =\lambda\left(n\right)|\mu\left(n\right)| = \mu\left(n\right),$ and :$\lambda\left(n\right)\mu\left(n\right) = |\mu\left(n\right)| =\mu^2\left(n\right).$ Let ''λ''(''n'') be Carmichael's function. Then :$\lambda\left(n\right)\mid \phi\left(n\right).$ Further, :$\lambda\left(n\right)= \phi\left(n\right) \textn=\begin1,2, 4;\\ 3,5,7,9,11, \ldots \text p^k \textp\text;\\ 6,10,14,18,\ldots \text 2p^k\textp\text. \end$ See Multiplicative group of integers modulo n and Primitive root modulo n. :$2^ \le d\left(n\right) \le 2^.$ :$\frac<\frac < 1.$ :$\begin c_q\left(n\right) &=\frac\phi\left(q\right)\\ &=\sum_\mu\left\left(\frac\right\right)\delta. \end$ Note that $\phi\left(q\right) = \sum_\mu\left\left(\frac\right\right)\delta.$ :$c_q\left(1\right) = \mu\left(q\right).$ :$c_q\left(q\right) = \phi\left(q\right).$ :$\sum_d^\left(\delta\right) = \left\left(\sum_d\left(\delta\right)\right\right)^2.$ Compare this with 13 + 23 + 33 + ... + ''n''3 = (1 + 2 + 3 + ... + ''n'')2 :$d\left(uv\right) = \sum_\mu\left(\delta\right)d\left\left(\frac\right\right)d\left\left(\frac\right\right).$ :$\sigma_k\left(u\right)\sigma_k\left(v\right) = \sum_\delta^k\sigma_k\left\left(\frac\right\right).$ :$\tau\left(u\right)\tau\left(v\right) = \sum_\delta^\tau\left\left(\frac\right\right),$ where ''τ''(''n'') is Ramanujan's function. Apostol, ''Modular Functions ...'', ch. 6 eq. 3
First 100 values of some arithmetic functions
Notes
References
* * * * * * * * * * * * * * * * * *
* | 8,535 | 25,479 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 125, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2021-17 | latest | en | 0.834031 |
https://www.physicsforums.com/threads/unknown-resistor-in-circuit-find-current-through-it-tricky.665474/ | 1,721,299,280,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514828.10/warc/CC-MAIN-20240718095603-20240718125603-00268.warc.gz | 790,692,967 | 17,194 | # Unknown resistor in circuit - find current through it (tricky)
• Engineering
• Color_of_Cyan
In summary, the homework statement says that there is no solution if the value of the unknown resistor is unknown.
Color_of_Cyan
## Homework Statement
http://imageshack.us/a/img210/3330/homeworkprob15.jpg
Find I0 in the network.
(The middle resistor value was not given)
## Homework Equations
V = IR
Voltage Division:
(Voltage across series resistor) = [(resistance) / total series resistance)](total input V)
Current Division (for 2 parallel resistors):
(current across parallel resistor) = [(other resistor) / (sum of parallel resistors)](total incoming current)]
Parallel resistors = (1/R1 + 1/R2)-1
Series Resistors = R1 + R2
Delta Y conversion and back for resistors
## The Attempt at a Solution
I really do not know where I should start.
Other than all I can do being to simplify the 8 and 4 ohm resistors to 12 ohm. But then there's still that unknown resistor in the middle.
I don't see how it would help to use y to T or the other way around to help.
Last edited by a moderator:
If you redraw this, so that the unknown resistor is "vertical" you can see that it is parallel to some other/s.
Are you sure you aren't given the value of another parameter (e.g., voltage) of the circuit? Otherwise, you can only determine the current as an algebraic expression (in terms of R).
Obviously has no solution if unknown resistor remains unknown.
rude man said:
Obviously has no solution if unknown resistor remains unknown.
...Or if the total current or some other voltage drop is not provided.
Yep, this is exactly all that was given.
Darn... looks like my professor really dropped the ball then (he wrote this problem up himself). Thanks anyway guys.
(By the way, the given answer by him is I knot = 1A)
Last edited:
Well like NascentOxygen says, you could try to work out the algebraic expression for it... All you'd have to do at a later stage would be to insert the numbers...?
gneill said:
...Or if the total current or some other voltage drop is not provided.
I say, good point there, gneill!
## 1. What is a resistor and how does it affect current in a circuit?
A resistor is an electrical component that resists the flow of current through a circuit. It does this by converting electrical energy into heat. The higher the resistance of a resistor, the lower the current that can pass through it.
## 2. How do I calculate the current through an unknown resistor in a circuit?
To calculate the current through an unknown resistor in a circuit, you can use Ohm's Law, which states that current (I) is equal to voltage (V) divided by resistance (R), or I = V/R. Measure the voltage across the resistor and use the known value of the other resistor(s) in the circuit to calculate the current through the unknown resistor.
## 3. Are there any other methods for finding the current through an unknown resistor?
Yes, there are other methods for finding the current through an unknown resistor. One method is to use Kirchhoff's Laws, which state that the sum of currents entering a junction in a circuit is equal to the sum of currents leaving the junction. By applying Kirchhoff's Laws to a circuit with an unknown resistor, you can solve for the current through the resistor.
## 4. Can I use a multimeter to measure the current through an unknown resistor?
Yes, you can use a multimeter to measure the current through an unknown resistor. Set the multimeter to measure current and place it in series with the resistor in the circuit. The multimeter will measure the current passing through the circuit, including the unknown resistor.
## 5. How do I determine the resistance of an unknown resistor in a circuit?
To determine the resistance of an unknown resistor in a circuit, you can use a known voltage source and measure the current passing through the resistor using a multimeter. Then, use Ohm's Law (R = V/I) to calculate the resistance. Alternatively, you can use a circuit simulator tool or manually solve the circuit using Kirchhoff's Laws to determine the resistance of the unknown resistor.
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1K | 1,096 | 4,739 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2024-30 | latest | en | 0.946834 |
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Descriptive Statistics Data Analysis
This exercise requires the use of a descriptive statistics calculator. You can find this tool in some versions of Excel (as part of the Analysis ToolPak) or you can use one of the many free online descriptive calculators such as the Descriptive Statistics Calculator (Links to an external site.) by Calculator Soup.
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First, use either Excel or the Calculator Soup descriptive statistics calculator to calculate the descriptive statistics for the given data set. This is explained in Chapter 1 of your course text.
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Identify any mode(s) in the data set. Is there a mode at all? Is there more than one mode?
Calculate manually the interquartile range and the values of Q1 and Q3. (It is important to calculate this manually because the interquartile range and quartiles output from Calculator Soup might not be accurate.) Address the following when completing this component:
Test to see if there are any outliers in the set. If so, which number(s)?
Which method from Section 2.4 of the text did you first use to check for outliers?
Now try the other method from Section 2.4 of the text. Do you come to the same conclusion about outliers in the data set?
Explain which descriptive statistic you think best summarizes this set of numbers and why.
Choose three of the descriptive statistics that you feel best represent this data set. Why were they chosen?
The Descriptive Statistics Data Analysis assignment
Must be two to three double-spaced pages in length (not including title and references pages) and formatted according to APA Style as outlined in the Writing Center’s APA Style (Links to an external site.) resource.
Must include a separate title page with the following:
Title of paper
Student’s name
Course name and number
Instructor’s name
Date submitted
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Sorting algorithms are among the most fundamental algorithms of computer science. A sorting algorithm takes an array of data and rearranges it so that it is in ascending, or sometimes descending order according to some sort key.
In this animation of sorting the data consists of an array of integers--the integers themselves are the sort key. There are many different algorithms that can be used to sort an array. Some algorithms are conceptually very intuitive while others are more subtle. Among sorting algorithms, those that are most efficient tend to be the most subtle. The heap sort algorithm belongs to this category. The heap sort algorithm relies on the connection between an complete binary tree and the array of values that correspond to it. The values in the array at the bottom of the screen are the same values that are in the tree enumerated in level order.
The primary advantage of the heap sort is its efficiency. The execution time efficiency of the heap sort is O(n log n). The memory efficiency of the heap sort, unlike the other n log n sorts, is constant, O(1), because the heap sort algorithm is not recursive.
The heap sort algorithm has two major steps. The first major step involves transforming the complete tree into a heap. The second major step is to perform the actual sort by extracting the largest element from the root and transforming the remaining tree into a heap.
In this step, the complete tree is transformed into a heap. This process is accomplished by beginning at the top of the tree. Each subtree is transformed into a heap by an operation similar to the enqueue operation of a priority queue. When you watch the animation, notice that the nodes at the top of the tree turn from blue to red. As each subtree is transformed into a heap, the node that constitutes its root becomes red. The big-O of this transformation into a heap is O(n log n) because it is done on each of the nodes n, the reheaping process must travel down one branch of the tree (log n).
Once the tree has been transformed into a heap, the root node contains the largest value--a property that is true of all heaps. The root value is swapped with the value in the right-most node on the lowest level. As you watch the animation, notice that the later node turns back to blue. Also its value in the list at the bottom of the screen becomes blue. The blue nodes and values are those that are in sorted order. Excluding the right-most node at the bottom level from the tree, the remaining subtree is almost a heap. It is then changed back into a heap. This operation is similar to the dequeue operation of a priority queue. The big-O of the second step is O(n log n) because it must be performed on each node (n), the reheaping process being log n. | 571 | 2,817 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2015-40 | latest | en | 0.928048 |
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Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
(Page 2)
• The main use for the database is to identify a number sequence that you have come across, perhaps on the Internet, in your work, on a test, etc.
• This page and the next two or three will show some examples.
Identifying a Sequence - Description of Database
You discover what you think may be a new algorithm for putting a file of patient's medical records into alphabetical order. (Perhaps you are a computer scientist or someone working in information science.)
To handle a pile of 1, 2, 3, 4, ... records, your algorithm takes
0, 1, 3, 5, 9, 11, 14, 17, 25, ...
steps.
How can you check if someone has discovered this algorithm before? You decide to ask the OEIS if this sequence has appeared before in the scientific literature.
You go to the main web page, which gives you the following. (These responses have been slightly edited to save space.)
The On-Line Encyclopedia of Integer Sequences
Enter a sequence, word, or sequence number:
Hints
You replace the example by your sequence and click "Submit":
Enter a sequence, word, or sequence number:
Hints
This produces the following response.
Greetings from the On-Line Encyclopedia of Integer Sequences!
Search: seq:1,3,5,9,11,14,17,25
Displaying 1-1 of 1 result found. page 1
Sort: relevance | references | number | modified | created Format: long | short | data
A003071 Sorting numbers: maximal number of comparisons for sorting n elements by list merging. (Formerly M2443) +2013
0, 1, 3, 5, 9, 11, 14, 17, 25, 27, 30, 33, 38, 41, 45, 49, 65, 67, 70, 73, 78, 81, 85, 89, 98, 101, 105, 109, 115, 119, 124, 129, 161, 163, 166, 169, 174, 177, 181, 185, 194, 197, 201, 205, 211, 215, 220, 225, 242, 245, 249, 253, 259, 263, 268, 273, 283, 287, 292, 297, 304 (list; graph; refs; listen; history; edit; internal format)
OFFSET 1,3 COMMENTS Comment from Jeremy Gardiner, Dec 28 2008: The following sequences all appear to have the same parity: A003071, A029886, A061297, A092524, A093431, A102393, A104258, A122248, A128975. REFERENCES J.-P. Allouche and J. Shallit, The ring of k-regular sequences, Theoretical Computer Sci., 98 (1992), 163-197. D. E. Knuth, Art of Computer Programming, Vol. 3, Sections 5.2.4 and 5.3.1. N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence). LINKS J.-P. Allouche and J. Shallit, The ring of k-regular sequences, Theoretical Computer Sci., 98 (1992), 163-197. FORMULA Let n = 2^e_1 + 2^e_2 + ... + 2^e_t, e_1 > e_2 > ... > e_t >= 0, t >= 1. Then a(n) = 1 - 2^e_t + Sum_{1<=k<=t} (e_k+k-1)*2^e_k [Knuth, Problem 14, Section 5.2.4]. a(n) = a(n-1) + A061338(n) = a(n-1) + A006519(n) + A000120(n)-1 = n + A000337(A000523(n)) + a(n-2^A000523(n)). a(2^k) = k*2^k + 1 = A002064(k). - Henry Bottomley (se16(AT)btinternet.com), Apr 27 2001 G.f.: x/(1-x)^3 + 1/(1-x)^2*Sum(k>=1, (-1+(1-x)*2^(k-1))*x^2^k/(1-x^2^k)). - Ralf Stephan (ralf(AT)ark.in-berlin.de), Apr 17 2003 CROSSREFS Cf. A001855. KEYWORD nonn,easy,nice AUTHOR N. J. A. Sloane (njas(AT)research.att.com). EXTENSIONS More terms from David W. Wilson (davidwwilson(AT)comcast.net)
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And so you discover that your sequence is the number of steps needed for sorting by list merging, a well-known algorithm.
The entry directs you to 5.3.1 of Volume 3 of D. E. Knuth, The Art of Computer Programming, where you find your algorithm described. The entry even gives an explicit formula for the nth term.
You decide not to apply for a patent!
The OEIS's files are full of stories like that (although that one is imaginary). A frequent comment is
... your database saved me six months of work.
The OEIS has helped people from almost every country in the world, and from almost every field. From school children to high-school students to undergraduates to ... to professionals to retirees. Anyone who likes numbers will find something interesting here.
The database has been called the mathematical analogue of a fingerprint file [B. Cipra, Mathematicians get an on-line fingerprint file, Science, 265 (22 July, 1994), page 473], since it serves to identify number sequences.
Other comments, stories and anecdotes will be mentioned in subsequent demonstrations.
It is hoped that eventually the database will include every (interesting) number sequence that has ever been published.
Description of the Database
• The database is first of all a very large collection of number sequences.
• The example in the 5th demonstration will be taken from chemistry. Other examples have come from physics, computer science, etc., and all branches of mathematics, especially discrete mathematics, combinatorics, number theory, algebra, game theory, recreational mathematics, ...
• The typical entry for a sequence gives among other things:
• the beginning of the sequence
• its name or description
• any references to the scientific literature or links to relevant web sites
• any formulae, recurrences, generating functions, etc. that are known
• cross-references to other sequences
• a list of keywords
• the name of the person who submitted it
• A more complete description of the entries is available from the web page:
eishelp2.html
• Some details
• The sequences in the database have been collected over several decades. The database has grown from 2400 sequences in 1973 to 5500 in 1995 to nearly 200000 today.
• New sequences currently arrive at a rate of over 15000 per year.
• The web pages receive thousands of hits each day, from almost every country on earth.
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• Please see the OEIS Foundation web site for more information about the history of the OEIS and its conversion to a wiki — and to make donations!
• The OEIS is much more than just this database, however, as will become apparent in subsequent demonstrations.
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Last modified November 19 03:29 EST 2017. Contains 294912 sequences. | 1,954 | 7,051 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2017-47 | longest | en | 0.857836 |
https://mocktestpro.in/class/mcq-questions-for-class-11-maths-chapter-4-principle-of-mathematical-induction/ | 1,709,242,626,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474853.43/warc/CC-MAIN-20240229202522-20240229232522-00251.warc.gz | 404,819,056 | 21,442 | Practice Questions with Answers - Multiple Choice Questions
# MCQ Questions for Class 11 Maths Chapter 4 Principle of Mathematical Induction
## MCQ Questions for Class 11 Maths Chapter 4 Principle of Mathematical Induction
Students are advised to solve the Principle of Mathematical Induction Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Principle of Mathematical Induction Class 11 with answers will boost your confidence thereby helping you score well in the exam.
Explore numerous MCQ Questions of Principle of Mathematical Induction Class 11 with answers provided with detailed solutions by looking below.
Question 1.
For all n∈N, 3n5 + 5n³ + 7n is divisible by
(a) 5
(b) 15
(c) 10
(d) 3
Given number = 3n5 + 5n² + 7n
Let n = 1, 2, 3, 4, ……..
3n5 + 5n³ + 7n = 3 × 1² + 5 × 1³ + 7 × 1 = 3 + 5 + 7 = 15
3n5 + 5n³ + 7n = 3 × 25 + 5 × 2³ + 7 × 2 = 3 × 32 + 5 × 8 + 7 × 2 = 96 + 40 + 14 = 150 = 15 × 10
3n5 + 5n³ + 7n = 3 × 35 + 5 × 3³ + 7 × 3 = 3 × 243 + 5 × 27 + 7 × 3 = 729 + 135 + 21 = 885 = 15 × 59
Since, all these numbers are divisible by 15 for n = 1, 2, 3, …..
So, the given number is divisible by 15
Question 2.
{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. {1 – 1/(n + 1)} =
(a) 1/(n + 1) for all n ∈ N.
(b) 1/(n + 1) for all n ∈ R
(c) n/(n + 1) for all n ∈ N.
(d) n/(n + 1) for all n ∈ R
Answer: (a) 1/(n + 1) for all n ∈ N.
Let the given statement be P(n). Then,
P(n): {1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. {1 – 1/(n + 1)} = 1/(n + 1).
When n = 1, LHS = {1 – (1/2)} = ½ and RHS = 1/(1 + 1) = ½.
Therefore LHS = RHS.
Thus, P(1) is true.
Let P(k) be true. Then,
P(k): {1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] = 1/(k + 1)
Now, [{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] ∙ [1 – {1/(k + 2)}] = [1/(k + 1)] ∙ [{(k + 2 ) – 1}/(k + 2)}] = [1/(k + 1)] ∙ [(k + 1)/(k + 2)] = 1/(k + 2)
Therefore p(k + 1): [{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] = 1/(k + 2)
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.
Question 3.
For all n ∈ N, 32n + 7 is divisible by
(a) non of these
(b) 3
(c) 11
(d) 8
Given number = 32n + 7
Let n = 1, 2, 3, 4, ……..
32n + 7 = 3² + 7 = 9 + 7 = 16
32n + 7 = 34 + 7 = 81 + 7 = 88
32n + 7 = 36 + 7 = 729 + 7 = 736
Since, all these numbers are divisible by 8 for n = 1, 2, 3, …..
So, the given number is divisible by 8
Question 4.
The sum of the series 1 + 2 + 3 + 4 + 5 + ………..n is
(a) n(n + 1)
(b) (n + 1)/2
(c) n/2
(d) n(n + 1)/2
Given, series is series 1 + 2 + 3 + 4 + 5 + ………..n
Sum = n(n + 1)/2
Question 5.
The sum of the series 1² + 2² + 3² + ……….. n² is
(a) n(n + 1) (2n + 1)
(b) n(n + 1) (2n + 1)/2
(c) n(n + 1) (2n + 1)/3
(d) n(n + 1) (2n + 1)/6
Answer: (d) n(n + 1) (2n + 1)/6
Given, series is 1² + 2² + 3² + ……….. n²
Sum = n(n + 1)(2n + 1)/6
Question 6.
For all positive integers n, the number n(n² − 1) is divisible by:
(a) 36
(b) 24
(c) 6
(d) 16
Given,
number = n(n² − 1)
Let n = 1, 2, 3, 4….
n(n² – 1) = 1(1 – 1) = 0
n(n² – 1) = 2(4 – 1) = 2 × 3 = 6
n(n² – 1) = 3(9 – 1) = 3 × 8 = 24
n(n² – 1) = 4(16 – 1) = 4 × 15 = 60
Since all these numbers are divisible by 6 for n = 1, 2, 3,……..
So, the given number is divisible 6
Question 7.
If n is an odd positive integer, then aⁿ + bⁿ is divisible by :
(a) a² + b²
(b) a + b
(c) a – b
(d) none of these
Given number = aⁿ + bⁿ
Let n = 1, 3, 5, ……..
aⁿ + bⁿ = a + b
aⁿ + bⁿ = a³ + b³ = (a + b) × (a² + b² + ab) and so on.
Since, all these numbers are divisible by (a + b) for n = 1, 3, 5,…..
So, the given number is divisible by (a + b)
Question 8.
n(n + 1) (n + 5) is a multiple of ____ for all n ∈ N
(a) 2
(b) 3
(c) 5
(d) 7
Let P(n): n(n + 1)(n + 5) is a multiple of 3.
For n = 1, the given expression becomes (1 × 2 × 6) = 12, which is a multiple of 3.
So, the given statement is true for n = 1, i.e. P(1) is true.
Let P(k) be true. Then,
P(k): k(k + 1)(k + 5) is a multiple of 3
⇒ K(k + 1) (k + 5) = 3m for some natural number m, …… (i)
Now, (k + 1) (k + 2) (k + 6) = (k + 1) (k + 2)k + 6(k + 1) (k + 2)
= k(k + 1) (k + 2) + 6(k + 1) (k + 2)
= k(k + 1) (k + 5 – 3) + 6(k + 1) (k + 2)
= k(k + 1) (k + 5) – 3k(k + 1) + 6(k + 1) (k + 2)
= k(k + 1) (k + 5) + 3(k + 1) (k +4) [on simplification] = 3m + 3(k + 1 ) (k + 4) [using (i)] = 3[m + (k + 1) (k + 4)], which is a multiple of 3
⇒ P(k + 1) (k + 1 ) (k + 2) (k + 6) is a multiple of 3
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.
Question 9.
For any natural number n, 7ⁿ – 2ⁿ is divisible by
(a) 3
(b) 4
(c) 5
(d) 7
Given, 7ⁿ – 2ⁿ
Let n = 1
7ⁿ – 2ⁿ = 71 – 21 = 7 – 2 = 5
which is divisible by 5
Let n = 2
7ⁿ – 2ⁿ = 72 – 22 = 49 – 4 = 45
which is divisible by 5
Let n = 3
7ⁿ – 2ⁿ = 7³ – 2³ = 343 – 8 = 335
which is divisible by 5
Hence, for any natural number n, 7ⁿ – 2ⁿ is divisible by 5
Question 10.
The sum of the series 1³ + 2³ + 3³ + ………..n³ is
(a) {(n + 1)/2}²
(b) {n/2}²
(c) n(n + 1)/2
(d) {n(n + 1)/2}² | 2,488 | 5,138 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2024-10 | latest | en | 0.812128 |
https://sugarrushbelfast.com/formula-762 | 1,670,464,330,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711232.54/warc/CC-MAIN-20221208014204-20221208044204-00439.warc.gz | 596,808,736 | 4,635 | # System of three equations solver
One tool that can be used is System of three equations solver. We can help me with math work.
## The Best System of three equations solver
In this blog post, we will show you how to work with System of three equations solver. Some students prefer to start with the easiest problems first and work their way up to the more difficult ones. Others prefer to start with the most difficult problem and then work their way down to the easier ones. And still others prefer to work on all the problems at the same time. Ultimately, it is up to the individual student to decide what works best for them.
There are a few different ways that you can solve for the x intercept of a line or a curve. One way is to set the equation equal to zero and solve for x. Another way is to graph the equation and find the point where it crosses the x-axis.
In order to solve an equation by factoring, you need to find the factors of each side of the equation and set them equal to each other. For example, if you have the equation x^2+5x+6=0, you would need to find the factors of each side that add up to 0. In this case, the factors would be (x+2)(x+3)=0. Thus, you would set x+2=0 or x+3
There are a few different ways to solve limits, depending on the equation you are trying to solve. If you are given a limit equation with a rational expression, you can solve it by finding the lowest common denominator and then cancelling out any factors that are common to the numerator and denominator. If you are given a limit equation with an irrational expression, you can solve it by using methods like algebra or trigonometric identities.
The 3x3 system of equations solver is a tool that can be used to solve systems of three linear equations with three variables. This tool can be used to find the values of the variables that make all of the equations true.
When factoring, one is looking for numbers that multiply together to equal the number being factored. In this case, one is looking for two numbers that when multiplied together, equal x. There are an infinite number of values for x that can be found by factoring.
## Help with math
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Square root calculator and solver Linear congruence solver Solving systems of equations solver Word math problems Find answers to math word problems Linear equation solver 4 unknowns | 645 | 2,977 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2022-49 | latest | en | 0.958728 |
https://www.sinosplice.com/life/archives/2014/07/29/the-beauty-of-chinese-numbers | 1,695,447,507,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506479.32/warc/CC-MAIN-20230923030601-20230923060601-00108.warc.gz | 1,117,258,047 | 20,785 | # The Beauty of Chinese Numbers
The beauty of Chinese numbers is that they are consistent. You learn the rules, and they just work. Even if you try to get flippant and say 一十 for “10” instead of just regular old , no one’s going to get upset.
The consistent beauty of Chinese numbers is made all the more obvious by the relative skankiness of numbers in English. I noticed this because my daughter (now somewhere between 2½ and 3 years old) has pretty much mastered all the numbers to 100 in Chinese, but the teens in English continue to stump her. She can count to 20 no problem, but if you ask her to read a random two-digit number that starts with “1” in English, that’s where the trouble starts. If she’s speaking Chinese, she can read either Arabic numerals or Chinese characters all the way to 100, but doing it in English trips her up especially for the range 11-20. She’s actually much better in the 60-90 range, because they’re regular.
If English numbers were totally regular like Chinese, we’d see these little gems (pretty much all of which my daughter has tried to pull off at one time or another):
– 10 = Onety
– 11 = Onety-one
– 12 = Onety-two
– 13 = Onety-three
– 14 = Onety-four
– 15 = Onety-five
– 16 = Onety-six
– 17 = Onety-seven
– 18 = Onety-eight
– 19 = Onety-nine
– 20 = Twoty
– 30 = Threety
– 50 = Fivety
Somehow I missed it my whole life, but one of the things that makes the names of the teens so bizarre (aside from the inexplicable “eleven” and “twelve”), is that the digits are represented backwards, only for these 7 numbers. When you see “36” and read it “thirty-six,” the “thirty” corresponds to the 3 on the left, and the 6 to the “six” on the right. So you’re reading the number, digit by digit, left to right. But for numbers like 14, 15, and 16, not only do they sound like 40, 50, and 60, but the order of the syllables better matches 40, 50, and 60 as well. And both the “-ty” and the “-teen” from those numbers originally represented “10,” right? These pairs are essentially pronounced as if they were the same numbers! Confusing as hell. I feel for my daughter.
Fortunately kids don’t realize they have good reason to be frustrated, and just jeep on truckin’.
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##### John Pasden
John is a Shanghai-based linguist and entrepreneur, founder of AllSet Learning.
1. Spanish is the same as English until you get to sixteen, where it turns logical: dieciseis = ten-and-six, and so on. But the prefix for twenty always bothered me, “ven”? I should look it up…
• “But the prefix for twenty always bothered me, “ven”? I should look it up…”
In French it’s vingt, so I’m going to blame Latin.
Quickly thinking about it, the European languages all seem to be universally troublesome with numbers. French gets my vote for being the worst, what with 70 (unless you’re Belgian or Swiss) being sixty-ten (soixante dix), 71 being sixty-eleven (soixante et onze), eighty being four-twenty (quatre vingt), 90 being four-twenty-ten (quatre vingt dix)…
• Yeah, French is a good ’un, Chris, but my vote goes to Danish with 56 being “six-and-half-third” and 67 being “seven-and-three.” A lot of cultures of Eurasiafrica and the Americas use the base 20 numeral system. (Surprisingly, upon discovery by the Europeans, not all Australian nations found it necessary to be able to count up to 20.) Note in this context that French « doigt » can mean both finger (doigt de la main = “finger-doigt”) and toe (doigt de pied = “foot-doigt”), so there are altogether 20 doigts.
2. This is why China doesn’t have the concept of “teenagers”. All the baggage associated with it – simply assumed to be true by Americans – such as teenage rebellion and such, doesn’t exist. People aged 13-19 in China are simply people, without any special abilities attributed to them because of a quirk of language.
• True, Chinese doesn’t have ‘teenagers’, but China does have ‘少年’ and ‘青年’, and my Chinese friends and colleagues are just as prone to whingeing about the youth of today as anybody else. Whorfianism is still nonsense.
• That phase happens in the early 20s here. Well, at least in my city it does.
3. I never noticed the backwards teen numeral thing!
Not to mention the fact that we count by units of ten (ten ones = one ten, ten tens = one hundred, ten hundreds = one thousand) up until we get to ten thousand, and then for some reason we switch to counting by units of 1000 (one thousand thousand = one million, one thousand million = one billion, etc).
4. I learned to teach my Chinese friends to better distinguish between (for example) forty and fourteen by writing them in a sort of English pinyin:
14 = fǒr-teèn
40 = fòr-tee
Note that the second syllable in 40 is a neutral tone. I think a lot of English learners default to thinking that the “n” at the end of 14 is the big difference between the pair, when bigger factor is actually the point of emphasis: 2nd syllable in 14, 1st syllable in 40.
• Benjamin,
That’s a good point, and one that difference in emphasis is one I also take care to emphasize when I help my daughter get those numbers straight.
Using the tone marks is a cool trick! | 1,304 | 5,134 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2023-40 | longest | en | 0.958105 |
http://gluedideas.com/content-collection/british-encyclopedia/Involution_P1.html | 1,642,850,330,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303845.33/warc/CC-MAIN-20220122103819-20220122133819-00394.warc.gz | 29,462,120 | 3,118 | Home >> British Encyclopedia >> Inflammation to Language 1 >> Involution_P1
# Involution
## root, quantity, power, raised, negative and am
Page: 1 2
INVOLUTION. If a quantity be conti nually multiplied by itself, it is said to be involved or raised; and the power to which it is raised is expressed by the number of times the quantity- has been employed in the multiplication.
Thus, aXa, or a', is called the setond power of a; axaXa, or 0, the third pow. er, xa....(a), or an, the nth power.
If the quantity to be involved be nega. tive, the signs of the even powers will be positive, and the sig-ns of the odd power negative.
For-a x-a=as;-ax-aX - a - &c.
A simple quantity- is raised to any pow er, by multiplying the index of every fac tor in the quantity- by the exponent of the power, and prefixing the proper sign de termined by the last article.
'rhus, on raised to the nui power is am", Because .x1 v am x am ....to n factors, by the rule of multiplication, is amn ; also, 47Pn=-a bxa bx&e. to n factors, or a a....to n factors xbXb X b-to n factors =_-hnxbn ; and a' b3 e raised to the fifth power is am 105 c5. Also, raised to the nth rrjwer is* am n ; where the positive or negative sign is to be pre fixed; according as n is an even or odd number.
If the quantity to be involved be a frac tion, both the numerator and denomina tor must be raised to the proposed power. If the quantity proposed b e a compound one, the involution may either be repre sented by the proper index, or it may ac tually take place.
Let a+b be the quantity to be raised to any power.
b a+ b b f -a b-l-b2 a >7-bi2 or a2+2 a b+b2 the sq.or 2d power a -Fb a3+2 n2 b+a 62 + n2 b--1-2 a b2+ 63 or a3+3 a2 6+3 a 62+63 the 3d pr. a -1-b a4-1-3 a3 b±3a22-1 b3 a3 6+3 a2 b2-1-3 a D3-Fb4 or a4+4 a3 b-F6 a2.52--F4 a b34-b4 the fourth power.
If b be negative, or the quantity to be involved be a -b, wherever an odd pow er of b enters, the sign of the term must be negative.
Hence, as----44.-_-a4 - 4 a' b ± 6 a' b2 -4 a b3+64.
Evolx-rros or the extraction of roots, is the method of determining a quantity, which, raised to a proposed power, mill produce a given quantity.
Since the nth power of am is amn, the root of ems must be am ; i. r. to ex
tract any root of a single quantity, we must divide the index of that quantity by the index of the root required.
When the index of the quantity is not exactly divisible by the number which ex presses the root to be extracted, that root must be represented according to the no tation already pointed out Thus the square, cube, fourth, root of a' arc respectively represented by (a' + x91, (a' + (a' -I- 1 (a' -I- the same roots of 1 ,are represe nt e dby 31 (a"-i-x") of n.
If the root to be extracted be express ed by an odd number, the sign of th e root will be the same with the sign of the pro posed quantity.
If the root to be extracted be expressed by an even number, and the quantity pro posed be positive, the root may be either positive or negative. Because either a positive or negative quantity, raised to such a power, is positive.
If the root proposed to be extracted be expressed by an even number, and the sign of the proposed quantity be negative, the root cannot be extracted; because no quantity, raised to an even power, can produce a negative result. Such roots are galled impossible.
Any root of a product may be found by taking that root of each factor, and mul tiplying the roots, so taken, together.
2. I Thus, (a b)n = an x b because each of these quantities, raised to the pow er, is a 6.
I ? 2 In a=b, then an x an = and in the r * rfa n n same manner a X a =a .
Any root of a fraction may be found by taking that root both of the numerator and denominator. Thus, the cube root off, is L or and Cr to 2 =— Or 4 b n Xb--n.
To extract the square root of a compound pantie!".
a'+2 a (cr-}-1.
a' 2 a4-b)2 a 2 a 64-6' • • Sizice the square root of a'+2 a 6+62 is a±b, whatever be the values of a and b, we may obtain a general rule for the extraction of the square root, by observ ing in what manner a and b maybe deriv ed from (0+2 a 6+6'.
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Rules 1. Time yourself 2. Work your solution on a seperate
Author Message
CEO
Joined: 15 Aug 2003
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Rules 1. Time yourself 2. Work your solution on a seperate [#permalink]
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10 Mar 2004, 20:23
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
Rules
1. Time yourself
2. Work your solution on a seperate sheet of paper
If the square root of the product of three distinct positive
integers is equal to the largest of the three numbers, what is the
product of the two smaller numbers?
(1) The largest number is 12.
(2) The average (arithmetic mean) of the three numbers is 20/3
Director
Joined: 03 Jul 2003
Posts: 652
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10 Mar 2004, 20:58
Ans: D
Time more than 4 mins
Manager
Joined: 11 Oct 2003
Posts: 102
Location: USA
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10 Mar 2004, 21:58
I think it it is A..2 min
Manager
Joined: 28 Jan 2004
Posts: 203
Location: India
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10 Mar 2004, 23:50
I will go for A (only statement 1 can be used to answer).
Time taken 30 seconds.
Sol. Three integers be X,Y,Z and largest be Z.
sqrt(XYZ) = Z ,replace Z by 12
sqrt(XY*12) = 12
or sqrt(XY) = sqrt(12)
or XY = 12
Director
Joined: 13 Nov 2003
Posts: 789
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11 Mar 2004, 03:40
Agree with mantha and mdfrahim, think that the answer is A)
Time: 2,5 min.
SVP
Joined: 30 Oct 2003
Posts: 1790
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11 Mar 2004, 05:42
a) sqrt( a * b * 12 ) = 12
so a * b = 12
Sufficient
b) a+b+c = 20
Only number that satisfies this and a * b = c is c = 12
Sufficient
< 2 minutes
CEO
Joined: 15 Aug 2003
Posts: 3454
Re: [#18] 2 Min. Challenge : Product [#permalink]
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11 Mar 2004, 06:30
praetorian123 wrote:
Rules
1. Time yourself
2. Work your solution on a seperate sheet of paper
If the square root of the product of three distinct positive
integers is equal to the largest of the three numbers, what is the
product of the two smaller numbers?
(1) The largest number is 12.
(2) The average (arithmetic mean) of the three numbers is 20/3
D is correct
good work anand
Senior Manager
Joined: 23 Sep 2003
Posts: 293
Location: US
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11 Mar 2004, 06:51
I agree with Anandnk and Kpadma.
The only distinct positive integers that satisfy the requirement that their sum = 20, and that the largest integer = the product of the other 2 integers are 6, 2 and 12.
If you try other combinations, you'll find that they don't sum to 20. Example: a=3; b=4 ; ab = c = 12 but 3 + 4 + 12 <> 20. Same with 8, 2 and 16.
11 Mar 2004, 06:51
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## Introduction to Evaluation of Infix/Postfix Expression
This article describes the Evaluation of Infix/Postfix; what is Infix, Postfix, and Prefix. How to perform conversion among them.
### Expression
An expression is the combination of operands and operators, resulting in equations to return the answer. E.g., A+B, 2+4, C*2, 6/2, etc.
### Infix
Infix refers to the operator exits between the operands. It makes the expression in the form of A operator B., E.g., 2+3, (5+6)*5-9, A+B, where A and B are the operands and “+” is the operator.
### Postfix
Postfix (Reverse Polish) refers to the operator that exits after the pair of operands. It makes the expression in the form of an <operand><operand> <operator>. E.g., AB+, (4+5)*78 = 4 5 + 78*.
### Prefix
Prefix (also called Polish) refers to the existence of the operator always before the pair of operands. In other words, the operator always comes before the operator. It makes the expression in the form of an <operator> <operand><operand>. E.g., +AB, *CD, (4+5)*78 = *+4 5 7 8.
### Why need for Postfix Expression?
The computer reads the equation from left to right or right to left.
Consider the following example –
A + B * C + D
The computer first scans the whole expression and later evaluates the expression according to the high precedence of the operator. In this expression, the computer first evaluates the expression B * C, then again scans the rest of the expression and adds the result to A, then scans the rest expression and finally adds the result to D.
It is clearly understood that the computer scans the expression operator to find the high precedence to solve the expression, and this action is repetitive according to the expression length, which makes the procedure time taking.
This repetitive procedure of scanning makes the evaluation procedure slow and inefficient. That’s why Postfix evaluation comes into play.
The above-given expression can be solved easily through Postfix using Stack.
Following rules for precedence and associativity:
1. ()
2. ^ associativity R-L
3. / and * associativity L-R
4. + and – associativity L-R
Let’s take some examples:
Infix Postfix Prefix A + B * C + D ABC*+D+ ++A*BCD 4/5 45/ /45 2+3*5 235*+ +2*35 (6+2) * (5-1) 62 + 51 – * *+ 62 – 51
### Infix to Postfix manual conversion rule:
1. Parenthesize the Infix expression according to the operator’s high precedence value.
2. Start solving the expression from left to right and move each operator to its right parentheses.
3. Remove all parentheses.
### Prefix manual conversion rule:
1. Parenthesize the Infix expression according to the operator’s high precedence value.
2. Start solving the expression from left to right and move each operator to its left parentheses.
3. Remove all parentheses.
Note: Write the Infix expression in the parentheses if not already.
Example 1: Conversion of Infix equation to Postfix and Prefix.
Postfix
Expression: A + B * C + D
Step 1: ([A + B] * [C + D]) #write infix expression into parentheses as per precedence value.
Step 2: ([A B +] * [C D +]) #Move the operator to its right parentheses.
Step 3: AB + C D + * #Remove the brackets.
Prefix
Expression: A + B * C + D
Step 1: ([A + B] * [C + D]) #write infix expression into parentheses as per precedence value.
Step 2: ([+ A B] * [+ C D]) #Move the operator to its left parentheses.
Step 3: *+ AB + CD #Remove the brackets.
Example 2: Conversion of Infix equation to Postfix and Prefix.
Postfix
Expression: ( AX + [B * C] )
Step 1: (AX + [B * C]) #write infix expression into parentheses as per precedence value.
Step 2: (AX + [B C *]) #Move the operator to its right parentheses.
Step 3: AX B C *+ #Remove the brackets.
Prefix
Expression: ( AX + [B * C] )
Step 1: (AX + [B * C]) #write infix expression into parentheses as per precedence value.
Step 2: (+ AX [* B C]) #Move the operator to its left parentheses.
Step 3: + AX * B C #Remove the brackets.
Example 3: Conversion of Infix equation to Postfix and Prefix.
Postfix
Expression: ( [A + B] * [C + E] )
Step 1: ( [A + B] * [C + E] ) #write infix expression into parentheses as per precedence value.
Step 2: ([A B +] * [C E +]) #Move the operator to its right parentheses.
Step 3: A B + C E +* #Remove the brackets.
Prefix
Expression: ( [A + B] * [C + E] )
Step 1: ( [A + B] * [C + E] ) #write infix expression into parentheses as per precedence value.
Step 2: ( [+ A B] * [+ C E] ) #Move the operator to its left parentheses.
Step 3: *+ A B + C E #Remove the brackets.
Example 4: Conversion of Infix equation to Postfix and Prefix.
Postfix
Expression: ( [ AX + [B * CY] ] / [D - E] )
Step 1: ( [ AX + [B * CY] ] / [D - E] ) #write infix expression into parentheses as per precedence value.
Step 2: ( [ AX + [B CY *] ] / [D E – ] ) #Move the operator to its right parentheses.
Step 3: A X BCY * + D E – / #Remove the brackets.
Prefix
Expression: ( [ AX + [B * CY] ] / [D - E] )
Step 1: ( [ AX + [B * CY] ] / [D - E] ) #write infix expression into parentheses as per precedence value.
Step 2: ( [ + AX [* B CY] ] / [- D E] ) #Move the operator to its left parentheses.
Step 3: / + A X * BCY – D E #Remove the brackets.
Example 5: Conversion of Infix equation to Postfix and Prefix.
Postfix
Expression: ( AX * [ BX * [ [ [CY + AY ] + BY ] * CX ] ] )
#write infix expression into parentheses as per precedence value.
Step 1: ( AX * [ BX * [ [ [CY + AY ] + BY ] * CX ] ] )
#Move the operator to its right parentheses.
Step 2: ( AX * [ BX * [ [ [CY AY + ] BY +] CX *] ] )
#Remove the brackets.
Step 3: AX BX CY AY + BY + CX ***
Prefix
Expression: ( AX * [ BX * [ [ [CY + AY ] + BY ] * CX ] ] )
#write infix expression into parentheses as per precedence value.
Step 1: ( AX * [ BX * [ [ [CY + AY ] + BY ] * CX ] ] )
#Move the operator to its left parentheses.
Step 2: ( * AX[ * BX [ [ [ + CY AY] + BY] * CX ] ] )
#Remove the brackets.
Step 3: * AX * BX *++CY AY BY CX
### Conclusion
Evaluating infix and postfix expressions is a fundamental task in programming and mathematics. While infix notation is the standard notation we are familiar with, postfix notation offers simplicity and ease of evaluation advantages. Converting infix expressions to postfix notation simplifies the evaluation process and eliminates ambiguity. Understanding these notations and their evaluation methods is valuable for anyone working with mathematical expressions and programming languages.
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posted by .
A ball is thrown toward a cliff of height h with a speed of 27m/s and an angle of 60 degrees above horizontal. It lands on the edge of the cliff 3.0s later.
What is the height of the cliff?
What is the maximum height of the ball?
What is the ball's impact speed?
Please provide the Formulas and steps!
Thank You!
• Physics -
Vo = 27m/s @ 60o
Xo = 27*cos60 = 13.5 m/s.
Yo = 27*sin60 = 23.4 m/s.
Tr = (Y-Yo)/g = (0-23.4)/-9.8 = 2.39 s.=
Rise time.
hmax = Yo*t + 0.5g*t^2.
hmax = 23.4*2.39 - 4.9*(2.39)^2=27.84 m.
Tr + Tf = 3.0 s.
2.39 + Tf = 3
Tf = 3 - 2.39 = 0.61 s. = Fall time.
h = hmax - 0.5g*Tf^2.
h = 27.84 - 4.9*(0.61)^2 = 23.3 = Ht. of cliff.
V = Vo + gt = 0 + 9.8*0.61 = 5.98 m/s.=
Impact velocity of ball. | 318 | 745 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2017-26 | latest | en | 0.794332 |
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4-81.
This problem is a checkpoint for addition and subtraction of mixed numbers. It will be referred to as Checkpoint 4.
Compute each sum or difference. Simplify if possible.
a.$5 \frac { 1 } { 2 } + 4 \frac { 2 } { 3 }$
b.$1 \frac { 5 } { 6 } + 2 \frac { 1 } { 5 }$
1. $9 \frac { 1 } { 3 } - 4 \frac { 1 } { 5 }$
1. $10 - 8 \frac { 2 } { 3 }$
Check your answers by referring to the Checkpoint 4 materials located at the back of your book.
Ideally, at this point you are comfortable working with these types of problems and can solve them correctly. If you feel that you need more confidence when solving these types of problems, then review the Checkpoint 4 materials and try the practice problems provided. From this point on, you will be expected to do problems like these correctly and with confidence.
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Lacunary function
In analysis, a lacunary function, also known as a lacunary series, is an analytic function that cannot be analytically continued anywhere outside the circle of convergence within which it is defined by a power series. The word "lacunary" is derived from ("pl." lacunae), meaning gap, or vacancy.
The first known examples of lacunary functions involved Taylor series with large gaps, or lacunae, between non-zero coefficients "a""n". More recent investigations have also focused attention on Fourier series with similar gaps between the non-zero coefficients. So there is a slight ambiguity in modern usage of the term lacunary series, which may be used to refer to series of both the Taylor and Fourier types.
A simple example
Consider the lacunary function defined by a simple power series:
:$f\left(z\right) = sum_\left\{n=0\right\}^infty z^\left\{2^n\right\} = z + z^2 + z^4 + z^8 + cdots,$
The power series clearly converges uniformly on any open domain |"z"| < 1, by comparison with the familiar geometric series, which is absolutely convergent when |"z"| < 1. So "f"("z") is analytic on the open unit disk. Nevertheless "f"("z") has a singularity at every point on the unit circle, and cannot be analytically continued outside of the open unit disk, as the following argument demonstrates.
Clearly "f"("z") has a singularity at "z" = 1, because
:$f\left(1\right) = 1 + 1 + 1 + cdots,$
is a divergent series. But since
:$f\left(z^2\right) = f\left(z\right) - z qquad f\left(z^4\right) = f\left(z^2\right) - z^2 qquad f\left(z^8\right) = f\left(z^4\right) - z^4 cdots,$
we can see that "f"("z") has a singularity when "z"2 = 1 (that is, when "z" = −1), and also when "z"4 = 1 (that is, when "z" = ±"i"), and so forth. By induction, "f"("z") must have a singularity at every one of the 2"n"th roots of unity, and since these are dense on the unit circle, every point on the unit circle must be an essential singularity of "f"("z"). [(Whittaker and Watson, 1927, p. 98) This example apparently originated with Weierstrass.]
An elementary result
Evidently the argument advanced in the simple example can also be applied to show that series like
:$f\left(z\right) = sum_\left\{n=0\right\}^infty z^\left\{3^n\right\} = z + z^3 + z^9 + z^\left\{27\right\} + cdots qquad g\left(z\right) = sum_\left\{n=0\right\}^infty z^\left\{4^n\right\} = z + z^4 + z^\left\{16\right\} + z^\left\{64\right\} + cdots,$
also define lacunary functions. What is not so evident is that the gaps between the powers of "z" can expand much more slowly, and the resulting series will still define a lacunary function. To make this notion more precise some additional notation is needed.
We write
:$f\left(z\right) = sum_\left\{k=1\right\}^infty a_kz^\left\{lambda_k\right\} = sum_\left\{n=1\right\}^infty b_n z^n,$
where "b""n" = "a""k" when "n" = λ"k", and "b""n" = 0 otherwise. The stretches where the coefficients "b""n" in the second series are all zero are the "lacunae" in the coefficients. The monotonically increasing sequence of positive natural numbers {λ"k"} specifies the powers of "z" which are in the power series for "f"("z").
Now a theorem of Hadamard can be stated. [(Mandelbrojt and Miles, 1927)] If
:$lim_\left\{k oinfty\right\} frac\left\{lambda_k\right\}\left\{lambda_\left\{k-1 > 1 + delta ,$
where "δ" > 0 is an arbitrary positive constant, then "f"("z") is a lacunary function that cannot be continued outside its circle of convergence. In other words, the sequence {λ"k"} doesn't have to grow as fast as 2"k" for "f"("z") to be a lacunary function – it just has to grow as fast as some geometric progression (1 + δ)"k". A series for which λ"k" grows this quickly is said to contain Hadamard gaps. See Ostrowski-Hadamard gap theorem.
Lacunary trigonometric series
Mathematicians have also investigated the properties of lacunary trigonometric series
:$S\left(lambda_k, heta\right) = sum_\left\{k=1\right\}^infty a_k cos\left(lambda_k heta\right) qquad S\left(lambda_k, heta,omega\right) = sum_\left\{k=1\right\}^infty a_k cos\left(lambda_k heta + omega\right) ,$
for which the λ"k" are far apart. Here the coefficients "a""k" are real numbers. In this context, attention has been focused on criteria sufficient to guarantee convergence of the trigonometric series almost everywhere (that is, for almost every value of the angle "θ" and of the distortion factor "ω").
*Kolmogorov showed that if the sequence {λ"k"} contains Hadamard gaps, then the series "S"(λ"k", "θ", "ω") converges (diverges) almost everywhere when
::$sum_\left\{k=1\right\}^infty a_k^2,$
:converges (diverges).
*Zygmund showed under the same condition that "S"(λ"k", "θ", "ω") is not a Fourier series representing an integrable function when this sum of squares of the "a""k" is a divergent series. [(Fukuyama and Takahashi, 1999)]
A unified view
Greater insight into the underlying question that motivates the investigation of lacunary power series and lacunary trigonometric series can be gained by re-examining the simple example above. In that example we used the geometric series
:$g\left(z\right) = sum_\left\{n=1\right\}^infty z^n ,$
and the Weierstrass M-test to demonstrate that the simple example defines an analytic function on the open unit disk.
The geometric series itself defines an analytic function that converges everywhere on the "closed" unit disk except when "z" = 1, where "g"("z") has a simple pole. [This can be shown by applying Abel's test to the geometric series "g"("z"). It can also be understood directly, by recognizing that the geometric series is the Maclaurin series for "g"("z") = "z"/(1−"z").] And, since "z" = "e""iθ" for points on the unit circle, the geometric series becomes
:$g\left(z\right) = sum_\left\{n=1\right\}^infty e^\left\{in heta\right\} = sum_\left\{n=1\right\}^infty left\left(cos n heta + isin n heta ight\right) ,$
at a particular "z", |"z"| = 1. From this perspective, then, mathematicians who investigate lacunary series are asking the question: How much does the geometric series have to be distorted – by chopping big sections out, and by introducing coefficients "a""k" ≠ 1 – before the resulting mathematical object is transformed from a nice smooth meromorphic function into something that exhibits a primitive form of chaotic behavior?
*Analytic continuation
*Szolem Mandelbrojt
*Benoit Mandelbrot
*Mandelbrot set
Notes
References
*Katusi Fukuyama and Shigeru Takahashi, "Proceedings of the American Mathematical Society", vol. 127 #2 pp.599-608 (1999), "The Central Limit Theorem for Lacunary Series".
*Szolem Mandelbrojt and Edward Roy Cecil Miles, "The Rice Institute Pamphlet", vol. 14 #4 pp.261-284 (1927), "Lacunary Functions".
*E. T. Whittaker and G. N. Watson, "A Course in Modern Analysis", fourth edition, Cambridge University Press, 1927.
* [http://www.ams.org/proc/1999-127-02/S0002-9939-99-04541-4/S0002-9939-99-04541-4.pdf Fukuyama and Takahashi, 1999] A paper (PDF) entitled "The Central Limit Theorem for Lacunary Series", from the AMS.
* [http://hdl.handle.net/1911/8511 Mandelbrojt and Miles, 1927] A paper (PDF) entitled "Lacunary Functions", from Rice University.
* [http://mathworld.wolfram.com/LacunaryFunction.html MathWorld article on Lacunary Functions]
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### Look at other dictionaries:
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• space — 1. noun /speɪs/ a) The intervening contents of a volume. If it be only a Single Letter or two that drops, he thruſts the end of his Bodkin between every Letter of that Word, till he comes to a Space: and then perhaps by forcing thoſe Letters… … Wiktionary
• movement-image — by Tom Conley The movement image is the title of the first panel of a historical diptych, Cinema 1 and Cinema 2, that classifies modes of perception and production of film from its beginnings in 1895 up to 1985. In this work and its… … The Deleuze dictionary | 2,868 | 10,348 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 10, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2020-10 | latest | en | 0.916158 |
https://grahamshawcross.com/2012/02/03/a-dream-world/ | 1,686,187,203,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654031.92/warc/CC-MAIN-20230608003500-20230608033500-00525.warc.gz | 323,924,327 | 26,918 | # A Dream World:
A Celestial Geometer imagines a world made up entirely of the following diagram
He asks a handy systems analyst newly arrived in heaven, after a life of endless toil working on CAAD, to design a computer database which will contain all the information which this simple world contains. He tells the systems analyst that he wishes him to do this so that he, the Geometer, may ask questions of the database and so elucidate this strange world’s properties.
The by now eager systems analyst proceeds as follows (though unknown to him it matters little how he proceeds)
In his former life having been an avid hill walker the world he is now presented with reminds him of a map grid so he unconsciously, or less likely consciously, starts his task by labelling the intersections of his grid world
He now feels happier, this is now a more familiar world, and he makes a list of all the lines, defining a line by its end points
Line (0,2) to (2,2)
2 (0,1) to (2,1)
3 (0,0) to (2,0)
4 (0,0) to (0,2)
5 (1,0) to (1,2)
6 (2,0) to (2,2)
The systems analyst then proudly reports to the Geometer that his database contains all the information in this special world.
The Geometer asks him if the database really does hold all the information.
The analyst says that indeed it does and what is more the design is so neat that it fits inside the smallest computer.
The Geometer says in that case he will ask a question and the following conversation ensues:-
G – How many squares are there in this world of yours?
A – You will have to define what a square is
G – Everyone knows what a square is; it is a right angled equal sided quadrilateral
A – You will have to define …
No… Now I could have designed the database so that the inherent squareness of this world was more apparent.
It would just need a list of all the squares rather than a list of all the lines
Square 1 (0,0) (1,0) (1,1) (0,1) (0.0)
2 (1,0) (2,0) (2,1) (1,1) (1,0)
3 (1,1) (2,1) (2,2) (1,2) (1,1)
4 (0,1) (1,1) (1,1) (0,2) (0,1)
Of course I could probably make this more compact
G – Wouldn’t any description like this make it more difficult to answer the question “How many lines are there?”
A – Yes but I could write a program to find the lines from the squares or the squares from the lines but it would take some time for me to have these written and as I would be using my knowledge of squares to design the program it might be simpler just to add the list of squares to the original database, of course this might mean a larger computer.
G – Don’t forget the fifth square!
A – No, of course not
G – Perhaps it is time I asked another question?
How many ‘T’ shapes are there in this world; before you ask for a definition with tops and stalks both 2 units long?
A – Well again this would need a program but it would be quicker if we just added a further list to the database.
G – It seems to me that every time I ask a question of this database, which you told me, holds all the information about this simple world, you have to write an ad-hoc program to find the answer or you simply increase the size of the database by adding the answer to it. For your database really to hold all the information in this world wouldn’t you have had to build in all these possibilities?
A – Yes I would but to do this I would have needed to know not only all the questions you could ask but also all the answers or at least a way of finding them
G – This seems like an infinite regress. If we know the answers this thing can tell us them, if we don’t it can’t. It is not much use is it?
This was written in 1981 in response to the intention of the Scottish Special Housing Association, following the successful development of their House Design and Site Layout Programs, to develop a property database to describe in detail their 100,000 houses.
An important moral was that real things are sometimes best represented by themselves.
The story was used, with permission, in Peter Swinson’s 1981 paper Logic Programming: a computing tool for the architect of the future.
Bibliography
ARU research project A25/SSHA-DOE: House design: Application of computer graphics to architectural practice
Bijl, A., Shawcross, G 1974 Housing site layout system
Computer Aided Design: Volume 7 Number 1 January 1975
Swinson, P. S. G. 1981 Logic programming: a computing tool for the architect of the future
## About Graham Shawcross
Architect PhD student at Edinburgh University Interested in order, rhythm and pattern in Architectural Design
This entry was posted in Architecture, Databases, Housing and tagged , , . Bookmark the permalink. | 1,162 | 4,685 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2023-23 | latest | en | 0.920892 |
https://www.geeksforgeeks.org/maximum-possible-sub-array-sum-after-at-most-x-swaps/?ref=lbp | 1,632,499,189,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057558.23/warc/CC-MAIN-20210924140738-20210924170738-00669.warc.gz | 826,460,060 | 28,241 | Related Articles
# Maximum possible sub-array sum after at most X swaps
• Difficulty Level : Medium
• Last Updated : 13 Apr, 2021
Given an array arr[] of N integers and an integer X, the task is to find the maximum possible sub-array sum after applying at most X swaps.
Examples:
Input: arr[] = {5, -1, 2, 3, 4, -2, 5}, X = 2
Output: 19
Swap (arr[0], arr[1]) and (arr[5], arr[6]).
Now, the maximum sub-array sum will be (5 + 2 + 3 + 4 + 5) = 19
Input: arr[] = {-2, -3, -1, -10}, X = 10
Output: -1
Approach: For every possible sub-array, consider the elements which are not part of this sub-array as discarded. Now, while there are swaps left and the sum of the sub-array currently under consideration can be maximized i.e. the greatest element among the discarded elements can be swapped with the minimum element of the sub-array, keep updating the sum of the sub-array. When there are no swaps left or the sub-array sum cannot be further maximized, update the current maximum sub-array sum found so far which will be the required answer in the end.
Below is the implementation of the above approach:
## CPP
`// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the maximum``// sub-array sum after at most x swaps``int` `SubarraySum(``int` `a[], ``int` `n, ``int` `x)``{`` ``// To store the required answer`` ``int` `ans = -10000;` ` ``// For all possible intervals`` ``for` `(``int` `i = 0; i < n; i++) {`` ``for` `(``int` `j = i; j < n; j++) {` ` ``// Keep current ans as zero`` ``int` `curans = 0;` ` ``// To store the integers which are`` ``// not part of the sub-array`` ``// currently under consideration`` ``priority_queue<``int``, vector<``int``> > pq;` ` ``// To store elements which are`` ``// part of the sub-array`` ``// currently under consideration`` ``priority_queue<``int``, vector<``int``>, greater<``int``> > pq2;` ` ``// Create two sets`` ``for` `(``int` `k = 0; k < n; k++) {`` ``if` `(k >= i && k <= j) {`` ``curans += a[k];`` ``pq2.push(a[k]);`` ``}`` ``else`` ``pq.push(a[k]);`` ``}`` ``ans = max(ans, curans);` ` ``// Swap at most X elements`` ``for` `(``int` `k = 1; k <= x; k++) {`` ``if` `(pq.empty() || pq2.empty()`` ``|| pq2.top() >= pq.top())`` ``break``;` ` ``// Remove the minimum of`` ``// the taken elements`` ``curans -= pq2.top();`` ``pq2.pop();` ` ``// Add maximum of the`` ``// discarded elements`` ``curans += pq.top();`` ``pq.pop();` ` ``// Update the answer`` ``ans = max(ans, curans);`` ``}`` ``}`` ``}` ` ``// Return the maximized sub-array sum`` ``return` `ans;``}` `// Driver code``int` `main()``{`` ``int` `a[] = { 5, -1, 2, 3, 4, -2, 5 }, x = 2;`` ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]);` ` ``cout << SubarraySum(a, n, x);` ` ``return` `0;``}`
## Java
`// Java implementation of the approach``import` `java.io.*;``import` `java.util.*;``class` `GFG``{` ` ``// Function to return the maximum`` ``// sub-array sum after at most x swaps`` ``static` `int` `SubarraySum(``int``[] a, ``int` `n, ``int` `x)`` ``{` ` ``// To store the required answer`` ``int` `ans = -``10000``;` ` ``// For all possible intervals`` ``for` `(``int` `i = ``0``; i < n; i++)`` ``{`` ``for` `(``int` `j = i; j < n; j++)`` ``{` ` ``// Keep current ans as zero`` ``int` `curans = ``0``;` ` ``// To store the integers which are`` ``// not part of the sub-array`` ``// currently under consideration`` ``ArrayList pq = ``new` `ArrayList();` ` ``// To store elements which are`` ``// part of the sub-array`` ``// currently under consideration`` ``ArrayList pq2 = ``new` `ArrayList();` ` ``// Create two sets`` ``for` `(``int` `k = ``0``; k < n; k++) {`` ``if` `(k >= i && k <= j) {`` ``curans += a[k];`` ``pq2.add(a[k]);`` ``}`` ``else`` ``pq.add(a[k]);`` ``}` ` ``Collections.sort(pq);`` ``Collections.reverse(pq);`` ``Collections.sort(pq2);` ` ``ans = Math.max(ans, curans);` ` ``// Swap at most X elements`` ``for` `(``int` `k = ``1``; k <= x; k++) {`` ``if` `(pq.size() == ``0` `|| pq2.size() == ``0`` ``|| pq2.get(``0``) >= pq.get(``0``))`` ``break``;` ` ``// Remove the minimum of`` ``// the taken elements`` ``curans -= pq2.get(``0``);`` ``pq2.remove(``0``);` ` ``// Add maximum of the`` ``// discarded elements`` ``curans += pq.get(``0``);`` ``pq.remove(``0``);` ` ``// Update the answer`` ``ans = Math.max(ans, curans);`` ``}`` ``}`` ``}` ` ``// Return the maximized sub-array sum`` ``return` `ans;`` ``}` ` ``// Driver code.`` ``public` `static` `void` `main (String[] args)`` ``{` ` ``int``[] a = { ``5``, -``1``, ``2``, ``3``, ``4``, -``2``, ``5` `};`` ``int` `x = ``2``;`` ``int` `n = a.length;` ` ``System.out.println(SubarraySum(a, n, x));`` ``}``}` `// This code is contributed by avanitrachhadiya2155`
## Python3
`# Python3 implementation of the approach` `# Function to return the maximum``# sub-array sum after at most x swaps``def` `SubarraySum(a, n, x) :`` ` ` ``# To store the required answer`` ``ans ``=` `-``10000`` ` ` ``# For all possible intervals`` ``for` `i ``in` `range``(n) :`` ` ` ``for` `j ``in` `range``(i, n) :`` ` ` ``# Keep current ans as zero`` ``curans ``=` `0`` ` ` ``# To store the integers which are`` ``# not part of the sub-array`` ``# currently under consideration`` ``pq ``=` `[]`` ` ` ``# To store elements which are`` ``# part of the sub-array`` ``# currently under consideration`` ``pq2 ``=` `[]`` ` ` ``# Create two sets`` ``for` `k ``in` `range``(n) :`` ``if` `(k >``=` `i ``and` `k <``=` `j) :`` ``curans ``+``=` `a[k]`` ``pq2.append(a[k])`` ` ` ``else` `:`` ``pq.append(a[k])`` ` ` ``pq.sort()`` ``pq.reverse()`` ``pq2.sort()`` ` ` ``ans ``=` `max``(ans, curans)`` ` ` ``# Swap at most X elements`` ``for` `k ``in` `range``(``1``, x ``+` `1``) :`` ``if` `(``len``(pq) ``=``=` `0` `or` `len``(pq2) ``=``=` `0` `or` `pq2[``0``] >``=` `pq[``0``]) :`` ``break`` ` ` ``# Remove the minimum of`` ``# the taken elements`` ``curans ``-``=` `pq2[``0``]`` ``pq2.pop(``0``)`` ` ` ``# Add maximum of the`` ``# discarded elements`` ``curans ``+``=` `pq[``0``]`` ``pq.pop(``0``)`` ` ` ``# Update the answer`` ``ans ``=` `max``(ans, curans)`` ` ` ``# Return the maximized sub-array sum`` ``return` `ans`` ` ` ``# Driver code``a ``=` `[ ``5``, ``-``1``, ``2``, ``3``, ``4``, ``-``2``, ``5` `]``x ``=` `2``;``n ``=` `len``(a)``print``(SubarraySum(a, n, x))` `# This ccode is contributed by divyesh072019.`
## C#
`// C# implementation of the approach``using` `System;``using` `System.Collections.Generic;``class` `GFG``{` ` ``// Function to return the maximum`` ``// sub-array sum after at most x swaps`` ``static` `int` `SubarraySum(``int``[] a, ``int` `n, ``int` `x)`` ``{` ` ``// To store the required answer`` ``int` `ans = -10000;` ` ``// For all possible intervals`` ``for` `(``int` `i = 0; i < n; i++)`` ``{`` ``for` `(``int` `j = i; j < n; j++)`` ``{` ` ``// Keep current ans as zero`` ``int` `curans = 0;` ` ``// To store the integers which are`` ``// not part of the sub-array`` ``// currently under consideration`` ``List<``int``> pq = ``new` `List<``int``>();` ` ``// To store elements which are`` ``// part of the sub-array`` ``// currently under consideration`` ``List<``int``> pq2 = ``new` `List<``int``>();` ` ``// Create two sets`` ``for` `(``int` `k = 0; k < n; k++) {`` ``if` `(k >= i && k <= j) {`` ``curans += a[k];`` ``pq2.Add(a[k]);`` ``}`` ``else`` ``pq.Add(a[k]);`` ``}` ` ``pq.Sort();`` ``pq.Reverse();`` ``pq2.Sort();` ` ``ans = Math.Max(ans, curans);` ` ``// Swap at most X elements`` ``for` `(``int` `k = 1; k <= x; k++) {`` ``if` `(pq.Count == 0 || pq2.Count == 0`` ``|| pq2[0] >= pq[0])`` ``break``;` ` ``// Remove the minimum of`` ``// the taken elements`` ``curans -= pq2[0];`` ``pq2.RemoveAt(0);` ` ``// Add maximum of the`` ``// discarded elements`` ``curans += pq[0];`` ``pq.RemoveAt(0);` ` ``// Update the answer`` ``ans = Math.Max(ans, curans);`` ``}`` ``}`` ``}` ` ``// Return the maximized sub-array sum`` ``return` `ans;`` ``}` ` ``// Driver code.`` ``static` `void` `Main() {`` ``int``[] a = { 5, -1, 2, 3, 4, -2, 5 };`` ``int` `x = 2;`` ``int` `n = a.Length;`` ``Console.WriteLine(SubarraySum(a, n, x));`` ``}``}` `// This code is contributed by divyeshrabaiya07.`
## Javascript
``
Output:
`19`
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My Personal Notes arrow_drop_up | 3,477 | 10,326 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2021-39 | longest | en | 0.675395 |
https://www.perkins.org/resource/blindsquare-readiness-lesson-chalk-walk-and-talk/ | 1,685,912,818,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650264.9/warc/CC-MAIN-20230604193207-20230604223207-00785.warc.gz | 1,009,651,509 | 120,959 | Story
# BlindSquare Readiness Lesson: Chalk, Walk, and Talk
## Introduction
As a participant of the High Tech O&M conference I felt inspired to use technology with my students during lessons. I have a new 5th grader with low vision who excels in the area of technology. So, I jumped right into using BlindSquare in a “scavenger hunt” lesson. I set two destinations at the front of the school: the book drop and the third tree. Each spot had a prize waiting when my student got there. I thought this is going to be great. When we started to access BlindSquare she asked, “What does 2 o’clock mean?” And “About how far is a yard?” Then I realized what we had talked about at our conference, your student must have a good foundation of O&M skills to access BlindSquare. So, I needed to get back to teaching foundations to get this student ready. I needed to review clock and yards. Here are some ways you can teach these skills to a younger student.
## Teaching the clock
Materials: flexible measuring tape, side walk chalk, and 36 inch cane
## Method
• Draw a large circle on the ground with side walk chalk. (10 – 12 feet in diameter)
• Draw the numbers 12, 3, 6, and 9 around the clock edges (cones can be used for tactile markers)
• Have your student stand at the center of the clock while you stand on the 12. Each of you should be holding an end of the tape measure. Walk around to each number/cone and have your student turn to your voice announcing each stop. Switch places with your student. Stop at the approximate place of the additional numbers on the circle.
• You can also have your student walk to you so they travel in the direction of the clock number.
## Teaching yards
Measuring with a cane is a great way to teach yards, especially a 36 inch cane. You can also use terms like giant strides or you can measure the distance of a taller student from their foot to where there cane tip touches.
Start at the curb and measure using chalk or place cones at the top of the cane. A parking area with 10 yards from curb to curb would be a good place to measure.
## Reflection
My student and I worked on this lesson a couple of times. We began to talk about how many yards away a classroom was located or at what clock direction was the office from the gym. We dusted off the high contrast tactile APH clock and practiced setting times. Soon we will set a place outside and with few selected settings reach our destination.
By Heather Panico
SHARE THIS ARTICLE | 573 | 2,490 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2023-23 | longest | en | 0.949268 |
https://leancrew.com/all-this/2011/02/oil-can-what/ | 1,542,377,432,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743046.43/warc/CC-MAIN-20181116132301-20181116154301-00423.warc.gz | 662,271,849 | 6,286 | # Oil can what?
I’ve been working on a nonlinear finite element program for the past week or so, and to shake out the bugs, I run it on a few test problems. One of my favorites is an example of snap-through buckling. It’s a simple little toggle structure with surprisingly rich behavior.
The toggle looks like this:
Two bars of the same length, material, and cross-sectional area. They’re pinned together at the apex where the load is applied and also pinned to fixed points at the base.
The first step in the solution is to recognize the mirror symmetry of the problem and replace it with a simpler but equivalent problem. Notice that the connection point is constrained by the symmetry to move only vertically. The behavior, then, is like two single-bar systems back-to-back. The upper end of each bar is forced to move vertically along a wall. The load on each bar is half that of the original problem. Solving one of these problems (I’ll choose the left one) will give us the solution to the other and, therefore, to the original two-bar problem.
The initial angle of the bar, when it’s unloaded, is $\alpha$, and we’ll use the downward angular movement of the bar, $\theta$, as our deflection variable. (We could use another variable to track the deflection, like the vertical movement of the upper joint, but the math is simpler using the angle.)
Here’s a free-body diagram of the upper joint in the deflected position. $Q$ is the compressive force in the bar, and $R$ is the reaction force from the wall.
Vertical equilibrium requires
The horizontal equilibrium condition is
but we won’t be using this because the reaction of the wall is of no interest. Recall that in the original problem there is no wall; $R$ is just a stand-in for the horizontal component of the compressive force in the other bar.
The joint can move down because the bar shortens as the compressive force in it rises. The formula for the shortening, $\Delta L$, is
where $A$ is the cross-sectional area of the bar, $E$ is the modulus of elasticity, and $L$ is the original, unloaded length. This is a standard formula; you can look up its derivation in any strength of materials book.1
Because the right end of the bar runs down the wall, the horizontal projection of the bar doesn’t change:
so
and
Substituting this into the equilibrium equation gives
We can put this in nondimensional form:
One advantage of this form is that it allows us to plot an entire class of solutions in a single graph. We don’t need to know the values of any variable except the initial bar angle, $\alpha$. I’ve chosen an $\alpha$ of just under 0.4 radians, the small angle in a 5-12-13 triangle.2
Here’s a plot of the solution:
It was generated by this short set of Gnuplot commands:
set terminal aqua title "Snap-through" font "Helvetica,14" size 500,400
unset key
set xzeroaxis lt -1
set xrange [0:.9]
set mxtics 2
set mytics 2
set xlabel "Deflection angle (radians)"
set ylabel "Load (P/EA)"
alpha = atan(5./12)
plot 2*(1 - cos(alpha)/cos(alpha-x))*sin(alpha-x) with lines lw 3
The first thing to note is that the relationship between load and deflection is nonlinear, which is why I’m using this problem as a test case. The overall structural behavior is nonlinear despite the linear behavior of the material ($Q$ and $\Delta L$ have a linear relationship) because of structure’s geometry.
The most interesting part of this solution, though, is what happens at the peak. Imagine the toggle being loaded with a slowly increasing force. The state of the toggle will follow the curve from the origin up to the peak. What happens at the peak? According to the graph, any further increase in deflection must be accompanied by a decrease in load.
In many real world situtations, the load cannot decrease, and the toggle then jumps across to a position far to the right on the graph.
The value of $\theta$ at this point is about twice the initial angle, $\alpha$, so this means that the toggle has been inverted.
This jump through the horizontal position occurs very quickly, which is why the behavior is called “snap-through.”
Even in situations where the load can be decreased after the peak, like when you’re pushing on the toggle with your finger, it’s impossible to stay at one of the intermediate positions. Those positions are unstable, like a pencil balanced on an infinitely sharp tip.3
The snap-through toggle has practical applications in, for example, certain types of switches. The initial angle of the toggle has to be chosen to match the area and modulus of the bar to get a switch that doesn’t snap through inadvertently but is still easy enough to push by a normal person.
My favorite application of snap-through behavior is a three-dimensional version of a toggle: the shallow dome on the base of an oil can.
Like the toggle mechanism, the dome will suddenly snap through to an inverted position when enough force is applied. When that happens, the volume in the can is slightly reduced and a little oil is pushed out. Releasing the force allows the dome to pop back out and air enters through the tip to replace the lost oil. The characteristic “pocka-pocka-pocka” of an oil can comes from the repeated snap-through and spring-back of the base.
This is such a common use of snap-through that some engineering texts refer to the behavior as “oil-canning.”
This type of oil can is iconic. When Dorothy and the Scarecrow come across the rusted Tin Woodsman, they loosen him up with a few squirts from just such a can, both in the W.W. Denslow illustrations
and in the publicity stills for the movie.
For some reason, though, it looks like Dorothy used a different kind of oil can in the movie itself, the kind with a squeeze-operated pump handle off to the side.
Too bad. You can’t get a good “pocka-pocka” out of a squeeze handle.
1. Popov’s is my favorite, but any similar book will do. ↩︎
2. The 5-12-13 triangle needs more promotion. The 3-4-5 gets all the publicity. ↩︎
3. You can stabilize the toggle in an intermediate position by holding it between two fingers; in that case, the value on the plot would be the net force of your two fingers. ↩︎ | 1,400 | 6,161 | {"found_math": true, "script_math_tex": 15, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2018-47 | longest | en | 0.913747 |
https://math.stackexchange.com/questions/2657576/sum-limits-k-0-fracn2-2k-binomn2k-n-cdot2n-2 | 1,558,537,156,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256858.44/warc/CC-MAIN-20190522143218-20190522165218-00384.warc.gz | 556,500,779 | 32,894 | # $\sum\limits_{k=0}^{\frac{n}{2}} 2k\binom{n}{2k}=n\cdot2^{n-2}$
Let $n\geq 2$ be even. Show that $$\sum\limits_{k=0}^{\frac{n}{2}} 2k\binom{n}{2k}=n\cdot2^{n-2}$$ 1) In a combinatorial way. (hint: count pairs $(x,S)$ s.t. $x\in S\subset \{1,2,...,n\}$ where $|S|$ is even)
2) Using binomial theorem. (hint: derivate $(1+x)^n+(1-x)^n$ )
Progress:
1) From the hint I was able to derive that the LHS counts the number of subsets $S_i$ where $|S_i|$ is even for all $i$s, multiplied by its cardinality, though I'm not sure about that. How does one put it in a combinatorial proof?
2) I calculated the derivative, but nothing yet.
$$(1+x)^n+(1-x)^n=\sum_{k=0}^{n}\binom{n}{k}x^k+\sum_{k=0}^{n}\binom{n}{k}(-1)^{n-k}x^k$$
Using that since $n$ is even $(-1)^{n-k}=(-1)^k$. For even $k$ the terms cancel each other out so:
$$(1+x)^n+(1-x)^n=\sum_{k=0}^{n/2}2\binom{n}{2k}x^{2k}$$
Taking the derivative of both sides we get:
$$n(1+x)^{n-1}-n(1-x)^{n-1}=\sum_{k=0}^{n/2}4k\binom{n}{2k}x^{2k-1}$$
Plugging in $x=1$ yields $$n\cdot 2^{n-2}=\sum_{k=0}^{n/2}2k\binom{n}{2k}$$
Consider the set of pairs $(x,S)$ where $S$ is a subset of $[n]$ of even cardinality and $x$ is an element of S. If $S$ has cardinality $2k$ there are $\binom{n}{2k}$ choices and there are $2k$ ways to choose $x$ so we have \begin{eqnarray*} \sum\limits_{k=0}^{\frac{n}{2}} 2k\binom{n}{2k}. \end{eqnarray*} On the other hand we can choose $x$ in $n$ ways and then $S=\{x\} \cup S'$ where $S'$ is a subset of $[n] / \{x\}$, choose the first $n-2$ elements to be either in or not in $S'$ ($2^{n-2}$ ways) and then choose the final element to be in or not in $S'$ in order that the cardinality of $S'$ is odd, so that gives $n 2^{n-2}$.
For the second part, differentiate & set $x=1$.
For the second part consider $$\sum_{k=0}^{\frac {n}{2}} 2k. \binom {n}{2k}=n\sum_{k=1}^{\frac {n}{2}} \binom {n-1}{2k-1}$$ Let $n=2m$ Then our identity converts to $$2m\sum_{k=1}^{m} \binom {2m-1}{2k-1}$$
Now we have $$(1+x)^{2m-1}=\sum_{k=0}^{2m-1} \binom {2m-1}{k}.x^k$$ And $$(1-x)^{2m-1}=\sum_{k=0}^{2m-1} \binom {2m-1}{k}.(-x)^k$$
Hence we get $$(1+x)^{2m-1}- (1-x)^{2m-1}= 2\sum_{k=1}^{m} \binom {2m-1}{2k-1}x^k$$
Hence putting $x=1$ We have $$2^{2m-1}=2\sum_{k=1}^{m} \binom {2m-1}{2k-1}$$ Hence we get $$2^{2m-2}= \sum_{k=1}^{m} \binom {2m-1}{2k-1}$$
Thus the value of our identity becomes $$\sum_{k=0}^{\frac {n}{2}} 2k. \binom {n}{2k}=n. 2^{n-2}$$ | 1,092 | 2,421 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2019-22 | latest | en | 0.681223 |
https://undergroundmathematics.org/divisibility-and-induction/division-game/suggestion | 1,638,900,313,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363405.77/warc/CC-MAIN-20211207170825-20211207200825-00489.warc.gz | 647,775,608 | 5,463 | Scaffolded task
# Division game Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource
## Suggestion
This suggestion is in the form of a proof sorter activity: we’ve written out a proof of a certain property and then muddled the order, and your job is to put the various statements into the right order.
You might want to print and cut up the statements.
Claim: If $p$ is prime and $p$ divides $ab$, then $p$ divides $a$ or $p$ divides $b$.
Note that ‘or’ in maths is inclusive, so you should read the conclusion above as “$p$ divides $a$ or $p$ divides $b$ or both”.
So $p$ divides $b$, as required.
There are integers $m$ and $n$ such that $am + pn = 1$.
Assume that $p$ is prime and that $p$ divides $ab$.
The left-hand side is a multiple of $p$.
If $p$ divides $a$ then we are done, so assume that $p$ does not divide $a$. We want to show that $p$ must divide $b$.
So $abm + pbn = b$.
Since $p$ is prime, the highest common factor of $a$ and $p$ is $1$. | 294 | 1,045 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2021-49 | latest | en | 0.900705 |
https://jeopardylabs.com/play/7th-grade-inequalities-27 | 1,695,692,405,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510130.53/warc/CC-MAIN-20230926011608-20230926041608-00630.warc.gz | 374,980,828 | 10,954 | Inequality Symbol
Meanings
One-Step
Inequalities
Two-Step
Inequalities
Solve and
Graph
Word Problems
(Write an inequality and Solve)
100
What is the symbol for more than?
>
100
x + 7 < 8
x < 1
100
DAILY DOUBLE
3x - 1 ≥ 14
x ≥ 5
100
x > 9
open circle at 9, pointing right
100
Lisa is cooking muffins. The recipe calls for more than 7 cups of sugar. She has already put in 2 cups. How many more cups does she need to put in?
x + 2 > 7
x > 5
200
What is the symbol for "less than"
<
200
b - 9 ≥ -20
b ≥ -11
200
x/2 - 4 < 10
x < 28
200
DAILY DOUBLE
x ≤ 17
closed circle at 17, pointing left
200
After paying $5.00 for a salad, Taylor has no more than$27.00 left. How much money did she have before buying the salad?
x - 5 ≤ 27
x ≤ 32
300
DAILY DOUBLE
What is the symbol for "no more than?"
300
DAILY DOUBLE
4x > 24
x > 6
300
-5x + 3 > -22
x < 5
300
x -12 < -4
x < 8, closed circle at 8, pointing left
300
How many t-shirts can Banning buy with at least $40 if each t-shirt costs$8.00?
8x ≥ 40
x ≥ 5
400
What is the symbol for "at most"?
400
m/3 ≤ 12
m ≤ 36
400
DAILY DOUBLE
-6x-3>-15
What is x<2
400
-8x > 32
x < -4, open circle at -4, pointing left
400
DAILY DOUBLE
Last Friday Marissa had $22.50. Over the weekend she received some money for cleaning. She now has over$32.00. How much money did she get for cleaning?
x + 22.50 > 32
x > 9.50
500
DAILY DOUBLE
What is the symbol for "at least"?
500
-2x < 18
x > -9
500
9 < 2+(-x)
What is -7>x or x<-7
500
DAILY DOUBLE
1/2X ≥4
x ≥ 8, closed circle at 8 pointing right
500
Max's employer has at most $18.00 to spend on tape measures for his crew. How many can he buy with that amount if each tape measure costs$4.50?
4.5x ≤ 18
x ≤ 4
600
What is the symbol for "taller than"?
>
600
DAILY DOUBLE
k + 3.2 ≥ -7 1/2
k ≥ -10.7 or
k ≥ -10 7/10
600
DAILY DOUBLE
12 ≥ -2t - 10
What is -11 ≤ t or t ≥ -11?
600
-12 > c/-2 + 10
What is 44 < c or c > 44; open circle at 44, pointing right
600
DAILY DOUBLE
Keith has $500 in a savings account at the beginning of the summer. He wants to have at least$200 in the account by the end of the summer. He withdraws \$25 each week for food, clothes, and movie tickets. How many weeks can Keith withdraw money from his account?
500 - 25w ≥ 200
w ≤ 12
Click to zoom | 826 | 2,339 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2023-40 | longest | en | 0.909107 |
https://jehovajah.wordpress.com/2012/08/24/quaternion-8-clarifying-the-rotation-action/ | 1,500,640,749,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423774.37/warc/CC-MAIN-20170721122327-20170721142327-00248.warc.gz | 661,978,246 | 35,735 | # Quaternion 8: Clarifying the Rotation Action
Just as the unit couples or complex numbers form a circle around the origin so the unit quaternions form a sphere. The unit quaternions have to be understood as unit vectors around the origin, and this is somewhat confusing when someone keeps telling you quaternions are 4 dimensional! However this is due to a confusion about 4 dimensions that Hamilton never had, nor did any serious scientist or mathematician of the time. The question was what use could the extra facility used for evaluating the unit quaternions be usefully put to. Hamilton naturally favoured that it be used to record time, others wanted to use it to record relative position of one space to another.
The unit quaternions occupy a dual role as do the unit couples. Conjuncting them always produces a unit quaternion. This property is used to define a rotation operator acting on a general quaternion vector Q :uQu, just as for couples it is uCu. Because couples are also commutative uCu is the same as u2C which is always evaluated as -C if u =i. The confusion arises because Hamilton leaves the I,j,k as fundamental axes vectors while also using the procedural call I which is a call to evaluate i by rotating it to a defined axis through pi/2 radians. It in fact evaluates to another unit couple or quaternion. Now these are also valid axes but what is not clear is how they relate to the original axes in terms of rotation.
There are 2 scenarios: the original axes have rotated in space to new axial positions in the unit sphere. Or the unit sphere has oriented itself along different axial conventions relative to the original axial position. One is the axes rotating in space, the other is space rotating relative to the axes.
Hamilton actually chose to rotate a quaternion vector thus identifying the unit quaternions as bilateral operators on vectors. He did establish appropriate notation for the operators, but we all think of I 2 as = -1. In the quaternion group this really must be rewritten as IQi =-1 when Q is the unit quaternion (1,0,0,0) etc.
Now the absolutely crucial associativity rule for unit quaternions ijk=-1, without which I was lost in my exploration of polynomial rotations, and Hamilton was unhappy with his triples!
(ij)k =i(jk)
Which states ij can pre operate on k or k can post operate on ij, but this must give the same resultant . In addition I can pre operate on jk or jk can post operate on I and these must conform to the previous 2 and to each other. But how do we evaluate this associative rule?
Fundamentally we have to accept the formalism, because that encodes a restriction on the unit quaternions in space. The I unit quaternion can only act in the ij plane on the i axis and the j axis. You will need to concentrate here, because the unit quaternion in the i vector direction acts as an operator on the i vector and the j vector rotating the i vector to the j vector and the j vector to the -I vector, so 2 actions of the I unit quaternion is like acting on any I or j vector with -1.
Now the j unit quaternion acts only in the jk plane, that is the plane defined by the j vector and the k vector. So again concentrate carefully: the j unit quaternion acts on the j vector rotating it to the k vector, and acts on the k vector rotating it to the -j vector. So 2 applications of the j unit quaternion is like acting on j with -1.
So finally the k unit quaternion acts on the plane described by the k vector and the I vector. Thus it rotates the k vector to the I vector and the I ve tor to thr -k vector, and is like acting on the k vector with -1 when 2 actions are applied.
You of course realise that these unit quaternions act on the whole of their planes only , this they are cyclic in their planes and help to generate the general rotation of the sphere in combination .
Thus the unit quaternions denoted by ijk have a particular action on the vectors in space, and have to follow he formal sequence. This means that using quaternions involves quantized motion in a sequence, and it actually matters whether the sequenc starts pre or post the vector. Thus ijk can pre act on I or k vectors but not on j vectors. However it can post act on I and j vectors but not on k vectors, in each case the action of the associative quotients is equivalent to acting on the vector with -1
There is a subtle error here, and it was one Maxwell made and it is the reason why he gave quaternions the heave no. In this highlighted foregoing explanation i have not included the 4th calculation axis! I will proceed to highlight the correction using a notation change to make it clearer.
The 4 axes vectors are e,I,j,k and their four unit quaternions are 1,q1,q2,q3.
Q1 acts on e commutativity to take it to q1i, q1 then acts on i to take it to -1e. Finally q1acts on -1e to take it to – q1I . Thus we see that q1 acts in the ei plane, and it acts on the unit vector through its unit quaternion. Q1q1=-1 this is defined as an anticlockwise rotation. However q1actspre to q2 or q2 acts post q1. These operators only commute with the vector notation, not the quaternion.
Now q2 also acts on 1e but in a different plane, and not necessarily orthogonal until the third constraint is in place.
The first question is how to define anticlockwise. This can only be done by defining the face of the "clock". We usually use a hand rule. The first thing is to define the ei clock face with the left hand rule: thumb points down at face fingers curl anti clockwise. In thumb down position From that position I rotate the thumb to point at 2 pi/3 to the ei plane in the thumbs to right and positioned at an angle. The fingers curl away from me anticlockwise , the. Thumb points to the face of the ej plane. Extend the index finger to point at the direction of the -1e, then rotate the thumb to past the thumbs up position to point to the face of the ek plane anticlockwise curling away from you
So q2 rotates 1e to q2j, and the j to -1e and q2q2=-1,
Next q3 acts on 1e taking it to q3k and k to -1e thus q3 q3 =-1.
The different geometry is clear and explains the confusion due to being wedded to the Cartesian frame.
Now since q1q2q3=-1 we have to see that as a cross plane action. The planes intersect only in 1e and-1e, so q1q2 can can get from i to -j. How do we get to q3k? My notion that q1q2 can only act in the plane they define is clearly unfounded and limiting. This has been a long held assumption which is here shown to be wrong, and obscuring of the actual behaviours. The actual behaviours are not forces rotating the axes in a spherical surface but combinatorial Forms that interact with any other such form in a relative and comparative way that generates a resultant quantity which is a vector as well as a quaternion unit. Hamilton described this action as rotating a cone around its apex axis, which means that pre and post multiplying the quaternion vector is fundamentally different to acting on the calculation vector e. this vector is a special unit identity vector in the quaternion group and it is commutative. However, to help distinguish the geometrical action, I change it's name under each action . Thus the e,I ,j,k vectors are this e vector attached to each quaternion in its position. So instead of writing q2e I write q2j where j is a special quaternion identity vector. There is only one special quaternion identiyt vector, so giving it this infinite variety of names would be confusing, except for relativity. If I let it have multiple names then I can think of the myself as moving in space to these axis vectors, ie my axes moves relative to the fixed sphericall axes. However if I insist on it having only one name, then I am demanding that all axes, or quaternions move to it, that is spherical space moves relative to me.
So finally I can make the intuitive suggestion that the 3 quaternions q1q2q3 act as if on each other to rotate the ijk vectors around the e vector as an axis. By choosing the e vector carefully we can encode asymmetrical rotation around this axis. The e axis in fact is not suitable for a time dimension, as Hamilton wanted, it actually represents the axis of rotation.
One other aspect of the e axis is its scaling effect. All none unit quaternions scale, but the e axis scales every other quaternion in one go, thus it can be a powerful door into the microscopic or macroscopic worlds encoded by quaternions.
Their are many other unit quaternion associative actions that can now be calculated, but the point is a set of rotator actions have been identified, and linear combinations of these can be formed to describe any general, none destructive rotation of the axes vectors I,j k. These I have called polynomial rotations, and they have to be applied to the basis vectors of a 4d space a vector partitioning of a spherical space. How they could be applied is through a dot product, a cross product or a wedge product, or products involving determinants, or any other product rule we care to create.
Having understood the principles now I wish to change the notation to minimise this long lasting confusion.
There is one more important convention: when evaluating any association tha action is always away from the resultant. Thus ij has no meaning, and had no meaning until the associative rule gave it a context. However il acting on a vector a either in pre or post position has a meaning and a direction of action: away from the resultant. Ij acting on a in post position first gets a result from ja , and then proceeds from there . aij is not a acting on i but i acting on a in the post position and that is commutative for the element a a vector, but not if a is a quaternion. If all the items are quaternions as in ijk the. associative rule is understood to always precede from left to right Thus k acting on I is the same whether written ki or ik. The latter is confuse able with I acting on k, hence the convention | 2,229 | 9,942 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2017-30 | longest | en | 0.942789 |
http://clay6.com/qa/336/a-manufacturer-produces-three-products-which-he-sells-in-two-markets-annual | 1,524,805,285,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125949036.99/warc/CC-MAIN-20180427041028-20180427061028-00578.warc.gz | 68,237,004 | 27,753 | Home >> CBSE XII >> Math >> Matrices
# A manufacturer produces three products $$x, y, z$$ which he sells in two markets. Annual sales are indicated below: $\begin{array} { c c } \textbf{Market} & \textbf{Products} \\ I & 10,000 \quad 2,000 \quad 18,000 \\ II & 6,000 \quad 20,000 \quad 8,000 \end{array}$ If unit sale prices of x, y and z are Rs 2.50, Rs 1.50 and Rs 1.00, respectively, find the total revenue in each market
This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com
Toolbox:
• If A is an m-by-n matrix and B is an n-by-p matrix, then their matrix product AB is the m-by-p matrix whose entries are given by dot product of the corresponding row of A and the corresponding column of B: $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
step 1: (a)Matrix for the products x,y,z is $\begin{array} { c c } x & y & z \\ 10,000 & 2,000 & 18,000 \\ 6,000 & 20,000 & 8,000 \end{array}$
Matrix corresponding to sale price of each product$\begin{array}{1 1}x\\y\\z\end{array}\begin{bmatrix}2.50\\1.50\\1.00\end{bmatrix}$
The revenue collected by the market is given by$\begin{bmatrix}10000 & 2000 & 18000\\6000 & 20000 & 8000\end{bmatrix}\begin{bmatrix}2.50\\1.50\\1.00\end{bmatrix}$
Step 2: Multiply each row with the column
$\begin{bmatrix}25000+3000+18000\\15000+30000+8000\end{bmatrix}$
$\begin{bmatrix}46000\\53000\end{bmatrix}$
Thus Revenue in each market is Rs 46000 and 53000.
Total Revenue=46000+53000=Rs 99000
edited Mar 19, 2013 | 567 | 1,562 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2018-17 | latest | en | 0.649995 |
https://www.eduzip.com/ask/question/int-dfrac-dx-1-x-3-nbsp-nbsp-581626 | 1,623,696,822,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487613380.12/warc/CC-MAIN-20210614170602-20210614200602-00558.warc.gz | 705,006,385 | 8,839 | Mathematics
$\int { \dfrac { dx }{ 1+{ x }^{ 3 } } }$
SOLUTION
Let $I = \int \frac{1}{1+x^{3}}dx$
$=\int \frac{1}{(1+x)(x^{2}-x+1)}dx$ ___ (1)
Now lets use the partial decomposition Technique
assume $\frac{1}{(1+x)(x^{2}-x+1)} = \frac{A}{1+x}+\frac{Bx+c}{x^{2}-x+1}$ ___ (2)
$A(x^{2}-x+1)+(Bx+C)(x+1) = 1$ ___ (3)
A+x = -1 the eq'n becomes
$A = \frac{1}{3}$ __(4)
Now replace A into (3) we have
$x^{2}-x+1+3(Bx+C)(x+1) = 3$
Now equating the coefficients of $x^{2}$ and x, and
constants, we get, $B = \frac{-1}{3}$ __ (5)
$C = \frac{2}{3}$ __ (6)
$I = \int \frac{1}{3(1+x)} dx+\int \frac{-x+2}{3(x^{2}-x+1)}dx$
$= \frac{1}{3}ln(1+x)-\int \frac{2x-1}{6(x^{2}-x+1)}dx\int \frac{1}{2(x^{2}-x+1)}dx$
$= \frac{1}{3}ln(1+x)-\int \frac{d(x^{2}-x+1)}{6(x^{2}-x+1)}+\int \frac{1}{2(x-\frac{1}{2})^{2}+\frac{3}{4}}dx$
$= \frac{1}{3}ln (1+x) -\frac{1}{6}ln(x^{2}-x+1)+I_{1}$ __ (7)
where $I_{1} = \int \frac{1}{2(x-\frac{1}{2})^{2}+\frac{3}{4}}dx$ __ (8)
Let $x-\frac{1}{2} = \frac{\sqrt{3}}{2} tan(y)$ __ (9), $y = arc tan (\frac{2x-1}{\sqrt{3}})$ __ (10)
$dx = \frac{\sqrt{3}}{2}sec^{2}(y)dy$ ___ (11) substitute the values of (9) & (11) into 8,
we have
$I_{1} = \int \frac{2\sqrt{3}sec^{2}(y)}{6sec^{2}(y)}dy = \int \frac{1}{\sqrt{3}}dy = \frac{1}{\sqrt{3}}y$ __ (12)
Replacing y with value obtained in (10)
$I_{1} = \frac{1}{\sqrt{3}} arc \,tan (\frac{2x-1}{\sqrt{3}})$ __ (13)
So, $I = \frac{1}{6} ln \frac{(1+x)^{2}}{x^{2}-x+1}+\frac{1}{\sqrt{3}}arc\, tan (\frac{2x-1}{\sqrt{3}})$
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Subjective Medium Published on 17th 09, 2020
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http://cowaymalaysia.com.my/journal/ws0ayo.php?page=1%2F8-as-a-percent-2f11f9 | 1,611,780,692,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704832583.88/warc/CC-MAIN-20210127183317-20210127213317-00240.warc.gz | 22,208,457 | 6,714 | Here is the answer to the question: Express 1/8 as a percentage or how to convert the fraction 1/8 to percent. Como. (1 point) 47% 1.8% 57.1% 28% 2. Round to the nearest tenth of a percent if necessary. Anonymous. It is 12.5% and the nearest rounded percent is 13% since you always round up. 709% B. 50 percent C. 25 percent D. 0, Monosodium glutamate (MSG), a food-flavor enhancer, has been blamed for “Chinese Restaurant Syndrome,” the symptoms of which are headaches & chest pains. Most questions answered within 4 hours. If it's not what You are looking for, type in … 70.9% C. 79% D. 7.09% 3 is 1/3 is equivalent to 33 1/3% what percent is . What Is 37.5 Percent As A Fraction In Simplest Form? answered 02/20/14, Easy to Understand Science and Math Tutor, Parviz F. Multiply both numerator and denominator by 100. 3/25 = (3/25)*(4/4) = 12/100 = 12% Rename .709 as a percent. 47% B. You make road signs and need a sign that shows a distance in miles and kilometers (km). "%" is a fancy way of writing "/100".... How To Convert Fractions To Decimals, And Then To Percent? If one parent has dimples and is homosygous for the trait, what is the probability any of his or her offspring will NOT have dimples? 2)Write the decimal 0.079 as a percent. 70.9% C. 79% D. 7.09% 3 is 1/3 is equivalent to 33 1/3% what percent is, 1. 5 years ago. You can view more similar questions or ask a new question. 1/8 x 100 % = 12•5 %. 100 times 1 is 100 so we're left with 100 / 8. Use the fraction to percent calculator below to write any fraction as a percent. For Free. 1) Write the fraction 5/9 as a percent.Round to the nearest hundredth of a percent where necessary. To get a fraction to a percentage, we need to multiply the denominator to something to get it to 100. Below you can find the full step by step solution for you problem. Rename .709 as a percent. Since 1/4=25%, 1/8 has to equal half of that or 12.5%. sch. Rename 4/7 as a percent. A. answered 02/19/14, Veteran teacher; all mid. MSG has the following composition by mass: 35.51 percent C, 4.77 percent. 12.5 %. 1.8 is equivalent to 180%. ALso, whatever we multiply to the denominator we do to the numerator.Hence: You divide 1/8 and you get 0.125 and then you always move 2 places to the RIGHT to make it a percent. Please, input values in this format: a b/c or b/c. We do this to find an equivalent fraction having 100 as the denominator. (1)6 percent (2)11 percent (3)13 percent (4)88 percent (2)? Round to the nearest tenth of a percent if necessary. 1/8 = 12.5 %. 100 percent B. When we substitute the 12.5 for the ? Chemistry. 70.9% C. 79% D. 7.09% 3 is 1/3 is equvulant to 33 1/3% what percent is, Jungle, Inc., has a target debt−equity ratio of 0.72. 1/8 = 12.5%. Moto. Steps to convert decimal into percentage. if the ramp does 153 J of work,what is the efficient of the ramp? A)82.7 percent B)121.0 percent C)32.0 percent D)338.0 percent. Get a free answer to a quick problem. Response must be 50 to 100 words. Rename 4/7 as a percent. Here, 1/8 * 100 = 12.5. On the right side, the 100 on the top will cancel the 100 on the bottom of the fraction because 100/100 = 1 and 1 times anything doesn't change that thing's value. No packages or subscriptions, pay only for the time you need. 5 years ago. A. 5 years ago. math. 0 0. You divide 1/8 and you get 0.125 and then you always move 2 places to the RIGHT to make it a percent so it would be 12.5 12.5% 0 0. Rename .709 as a percent. Express 1.8 as a percent. 0 0. Round to the nearest tenth of a percent if necessary. Round your answers to 2 decimal places. What Happened With Justin Bieber's Parents? To get this total, 1/8 is actually 100 divided by 8 which totals 12.5. 709% B. you do 185 J of work pulling a cart up a ramp. 1/8 x 100% = 12.5%. A. (1 whole) 7/8=87.5%. 709% B. Write a, 1. 100 times 1/8. subjects; high sch. Here are some related questions which you might be interested in reading. (e.g., 32.16)) Required: (a) If, A woman gives birth to a son. 1.8% C. 57.1% D. 28% 2. What is the percent deviation (percent error) in this calculation? 1/8 expressed as percentage. Examples: Three tenths should be typed as 3/10. Percent means 'out of one hundred' so..... From here you can multiply both sides by 100. A. above, 1/ 8 = 0.125 = 12.5 % / Same as converting 1/8 of a dollar to cents, first convert it to decimal number, then Multiply by 100 ( shift decimal point twice to the right) and insert %, © 2005 - 2020 Wyzant, Inc. - All Rights Reserved, a Question Rename .709 as a percent. Its WACC is 11 percent, and the tax rate is 31 percent. Lv 6. Pam J. A student found that her mixture was 15 percent NH4cl 20 percent NaCl and 75 percent Sio2 Assuming her calculations are correct what did she most likely do incorrectly in her experiment? Sign in? Round to the nearest tenth of a percent if necessary. Just type in a decimal number: See it as a percent Maggie. 1.8% C. 57.1% D. 28% 2. Identify the part, whole, and percent in the following statement: Find 15% of 750. a. part = 750, whole = n, percent = 15 b. part = 15, whole = 750, percent = p c. part = n, whole = 750, percent = 15 d. none of these 2. 47% B. 70.9% C. 79%. In humans, having dimples is a dominant trait. (1 point) 709% 70.9% 79% 7.09% 3. So ?/100 times 100 = just ? 0 0. 1 0. A. If one-third is equivalent to 33thirty-three . 0 0. A. How Do You Deal With Someone You Love So Much Doing Drugs? A. What is the probability that her second child will be a girl? How do you think about the answers? You can sign in to vote the answer. 1. 1.8 as a percent - solution and the full explanation with calculations. One and one-half should be typed as 1 1/2. A. 709% B. ANONYMOUS. 6/8=75%. A. science; reading, Tiffany L. Didn't find the answer you were looking for? Lv 7. Choose an expert and meet online. If you are using a calculator, some of them will not be able to get this answer if you key in the percentage sign.So you can also try this way. A student finds the density of an ice cube to be 0.80g/cm3; it is actually 0.90 g/cm3. 47% B. (1 point) 0 percent 25 percent 50 percent 100 percent, So I have some math questions I'd like for some one to check :) 1. answered 02/19/14, Mathematics professor at Community Colleges. 1 decade ago. Ask a Question. 1.8% C. 57.1% D. 28% 2. 1.8% C. 57.1% D. 28% 2. 1/8 as a percent this fraction reads 1 divided by 8-- so divide 1 by eight and you get 0.125-- move decimal over two places to the right and you get 12.5% :) Upvote • 1 Downvote Rename 4/7 as a percent. We hope it will be very helpful for you and it will help you to understand the solving process. (Do not include the percent signs (%). A link to the app was sent to your phone. 47% B. 8/8=100%. We get the answer 12.5 percent. Start here or give us a call: (312) 646-6365, this fraction reads 1 divided by 8-- so divide 1 by eight and you get 0.125-- move decimal over two places to the right and you get 12.5% :). 5 years ago. Rename .709 as a percent. Round to the nearest tenth of a percent if necessary. 1 decade ago. Lv 7. So, 1/8 as percent is 12.5 %. Explain how you would determine the number of kilometers you would show on the sign if you know the number of miles? | 2,238 | 7,187 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2021-04 | latest | en | 0.935686 |
http://www.conquerclub.com/forum/viewtopic.php?f=62&t=98919&p=4146125 | 1,406,314,062,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997894473.81/warc/CC-MAIN-20140722025814-00000-ip-10-33-131-23.ec2.internal.warc.gz | 571,822,950 | 19,028 | ## Five Fives Game
Comment games, quizzes, trivia, elimination games, etc...
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### Re: Five Fives Game
I'm fine with more functions. The main reason why I dislike the ceiling/floor functions is that you can abuse them to essentially get any number you want out of a single 5 by just using repeated square rooting and random other functions.
Also, I can't figure out 373 without using the ceiling/floor functions.
maasman
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### Re: Five Fives Game
I got 373 without ceiling/floor already:
373 = (5!! + .5) x Γ5 + 5/5
373 = 15.5 x 24 + 1
373 = 372 + 1
373 = 373
Next target = 379
crispybits
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### Re: Five Fives Game
So what's the new rule? we can use any function except floor/ceiling?
From: Karl_R_Kroenen
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### Re: Five Fives Game
I think we've always been able to use any function, just now I've made a list of them that's as complete as I can find infor for - we still can use floor/ceiling but maasman and I are avoiding them if possible.
crispybits
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### Re: Five Fives Game
379 = (5+5!!/5)!! - √5 x √5
379 = (5+3)!! - 5
379 = 384 - 5
379 = 379
Last edited by maasman on Wed Apr 10, 2013 12:18 pm, edited 1 time in total.
maasman
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### Re: Five Fives Game
383 = σ(5!) + 5! + 5!! + 5 + Π5
383 = 240 + 120 + 15 + 5 + 3
383 = 383
(having done all that bloomin research I've gotta slip some of the "new" ones in there )
crispybits
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### Re: Five Fives Game
Because I'm lazy:
389 = (5+5!!/5)!! + √5 x √5
389 = (5+3)!! + 5
389 = 384 + 5
389 = 389
maasman
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### Re: Five Fives Game
397 = Γ5 x (5!! + (Π5 x .5)) + μ(5!!)
397 = 24 x (15 + (3 x .5)) + 1
397 = 24 x 16.5 + 1
397 = 396 + 1
397 = 397
crispybits
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### Re: Five Fives Game
401 = 5# x 5!! - 5 x 5 - Γ5
401 = 30 x 15 - 25 - 24
401 = 450 - 49
401 = 401
maasman
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### Re: Five Fives Game
409 = Γ5 x (5!! + σ(5!!)) + 5/5
409 = 24 x (15 + 2) + 1
409 = 24 x 17 + 1
409 = 408 + 1
409 = 409
crispybits
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### Re: Five Fives Game
419 = 5# x 5!! - 5# - 5/5
419 = 30 x 15 - 30 - 1
419 = 450 - 31
419 = 419
maasman
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Location: River Falls, USA
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### Re: Five Fives Game
Too easy to cheat this one
421 = 5# x 5!! - 5# + 5/5
421 = 30 x 15 - 30 + 1
421 = 450 - 29
421 = 421
Next target: 431
crispybits
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### Re: Five Fives Game
431 = 5# x 5!! - 5!! - σ(5!!)) - σ(5!!))
431 = 30 x 15 - 15 - 2 - 2
431 = 450 - 19
431 = 431
maasman
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### Re: Five Fives Game
433 = (5# - σ(5!!)) x 5!! + 5!! - σ(5!!)
433 = (30-2) x 15 + 15 - 2
433 = 28 x 15 + 13
433 = 420 + 13
433 = 433
crispybits
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### Re: Five Fives Game
439 = 5# x 5!! - 5!! + σ(5!!)) + σ(5!!))
439 = 30 x 15 - 15 + 2 + 2
439 = 450 - 11
439 = 439
maasman
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### Re: Five Fives Game
443 = Γ5 x (Γ5 - 5) - (5!! - σ(5!!))
443 = 24 x (24 - 5) - (15 - 2)
443 = 24 x 19 - 13
443 = 456 - 13
443 = 443
crispybits
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### Re: Five Fives Game
449 = 5# x 5!! + (√5 x √5)/5
449 = 30 x 15 - 5/5
449 = 450 - 1
449 = 449
maasman
Posts: 533
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### Re: Five Fives Game
457 = Γ5 x (Γ5 - 5) + 5/5
457 = 24 x (24 - 5) + 1
457 = 24 x 19 + 1
457 = 456 + 1
457 = 457
crispybits
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### Re: Five Fives Game
Still lazy.
461 = 5# x 5!! + 5!! - σ(5!!)) - σ(5!!))
461 = 30 x 15 + 15 - 2 - 2
461 = 450 + 11
461 = 461
maasman
Posts: 533
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Location: River Falls, USA
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### Re: Five Fives Game
Same here
463 = Γ5 x (Γ5 - 5) + 5 + σ(5!!)
463 = 24 x (24 - 5) + 5 + 2
463 = 24 x 19 + 7
463 = 456 + 7
463 = 463
crispybits
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### Re: Five Fives Game
Are you stuck on the next one maasman or you just taking a breather?
crispybits
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### Re: Five Fives Game
Just breather.
467 = 5# x 5!! + 5!! + φ(5) - σ(5!!))
467 = 30 x 15 + 15 + 4 - 2
467 = 450 + 17
467 = 467
maasman
Posts: 533
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### Re: Five Fives Game
479 = (5# + σ(5!!)) x 5!! - 5/5
479 = (30 + 2) x 15 - 1
479 = 32 x 15 - 1
479 = 480 - 1
479 = 479
crispybits
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### Re: Five Fives Game
487 = 5# x 5!! + Γ5 + 5!! - σ(5!!))
487 = 30 x 15 + 24 + 15 - 2
487 = 450 + 37
487 = 487
maasman
Posts: 533
Joined: Wed Apr 11, 2007 6:45 pm
Location: River Falls, USA
Medals: 70
### Re: Five Fives Game
Another cheat
491 = 5# x 5!! + Γ5 + 5!! + σ(5!!)
491 = 30 x 15 + 24 + 15 + 2
491 = 450 + 41
491 = 491
crispybits
Posts: 696
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