url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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https://utzx.com/units/convert-171-cm-to-inches/ | 1,725,834,185,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651035.2/warc/CC-MAIN-20240908213138-20240909003138-00781.warc.gz | 574,950,885 | 21,446 | # Convert 171 CM to Inches
## How many inches in a cm?
Do you wish to convert 171 cm to an inch-length result? first, you obviously should be aware of how many inches 1 cm is equal to.
You can use this cm into inches calculator to calculate the conversion.
## Centimeter Basis
Centimeters or centimetres are the measurement unit used to measure length in metric systems. It is abbreviated by the letter cm . The meter is internationally defined to the “International System of Units”, but the unit the unit cm is not. However, a centimeter is equals 100 meters. It’s also approximately 39.37 in.
## Inch
An Anglo-American measurement unit that used to measure length is the inch (its symbol is in).. Its symbol is in. In bulk of European local languages, “inch” can be used interchangeably with or derived from “thumb”. Because a man’s thumb is around an inch long.
• Electronic components like the dimensions of the PC screen.
• Dimensions of car/truck tires.
## How Can I Change 171 c to inch?
Convert inches into centimeters using the cm converter. This basic is used to translate cm to inches.
You now fully understand of centimeters to inches from the above. The formula will allow you to answer the following questions:
• What is the formula for converting 171 cm to inches?
• How do I convert inches from cm?
• How do you change cm to inches?
• How to measure cm into inches?
• How big are 171 cm to inches?
cm inches 170.2 cm 67.00774 inches 170.3 cm 67.04711 inches 170.4 cm 67.08648 inches 170.5 cm 67.12585 inches 170.6 cm 67.16522 inches 170.7 cm 67.20459 inches 170.8 cm 67.24396 inches 170.9 cm 67.28333 inches 171 cm 67.3227 inches 171.1 cm 67.36207 inches 171.2 cm 67.40144 inches 171.3 cm 67.44081 inches 171.4 cm 67.48018 inches 171.5 cm 67.51955 inches 171.6 cm 67.55892 inches 171.7 cm 67.59829 inches 171.8 cm 67.63766 inches | 506 | 1,860 | {"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} | 3.171875 | 3 | CC-MAIN-2024-38 | latest | en | 0.846284 |
https://uchicago.kattis.com/problems/akcija | 1,591,143,418,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347426956.82/warc/CC-MAIN-20200602224517-20200603014517-00159.warc.gz | 563,671,883 | 5,916 | OpenKattis
University of Chicago
# Akcija
There is a promotional offer in a bookstore “Take 3, pay for the 2 more expensive ones”. So, each customer who picks $3$ books gets the cheapest one for free. Of course, the customer can take even more books and, depending on the way the books are arranged into groups of three, get the cheapest one in each group for free.
For example, let the prices of the books taken by the customer be: $2, 3, 4, 4, 6, 9, 10$. If he arranges them into the groups $(10, 3, 2)$, $(4, 6, 4)$ and $(9)$, he will get the book priced $2$ from the first group for free and the book priced $4$ from the second group. We can see that he will not get anything for free from the third group because it contains only one book.
The lady working in the bookstore is well-intentioned and she always wants to lower the price for each customer as much as possible. For given book prices, help the lady arrange the books into groups in the best way possible, so that the total price the customer has to pay is minimal.
Please note: The lady doesn’t have to arrange the books into groups so that each group contains exactly $3$ books, but the number of books in a group needs to be between $1$ and $3$, inclusively.
## Input
The first line of input contains the integer $N$ ($1 \leq N \leq 100\ 000$), the number of books the customer bought. Each of the following $N$ lines contains a single integer $C_ i$ ($1 \leq C_ i \leq 100\ 000$), the price of each book.
## Output
The first and only line of output must contain the required minimal price.
Sample Input 1 Sample Output 1
4
3
2
3
2
8
Sample Input 2 Sample Output 2
6
6
4
5
5
5
5
21 | 453 | 1,662 | {"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": 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} | 3.265625 | 3 | CC-MAIN-2020-24 | latest | en | 0.925826 |
https://educationexpert.net/mathematics/94778.html | 1,708,730,904,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474470.37/warc/CC-MAIN-20240223221041-20240224011041-00775.warc.gz | 232,100,556 | 7,326 | 10 February, 19:41
# Tom has 28 milk chocolates and emily has 24 dark chocolates they have to divide these chocolates into small packets that each have the same number of dark chocolates and the same number of milk chocolates
+5
1. 10 February, 20:40
0
4 milk chocolatero need to be removed to have even amounts
24+24 = 48
24 packets with 1 of each (2 count) is one possibility.
You can have less packets with more chocolates inside like 12 packets with 4 count (2 of each)
Any combination with result 48. To use al chocolatesl : 4 packets:Each packet has 6dark and 7milk choc
2. 10 February, 21:04
0
26 ... 24+28=52 ... 52:2=26 | 183 | 634 | {"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} | 3.84375 | 4 | CC-MAIN-2024-10 | latest | en | 0.943051 |
http://www.britannica.com/topic/Darbouxs-theorem | 1,435,891,167,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375095711.51/warc/CC-MAIN-20150627031815-00246-ip-10-179-60-89.ec2.internal.warc.gz | 352,247,937 | 13,490 | # Darboux’s theorem
Mathematics
Darboux’s theorem, in analysis (a branch of mathematics), statement that for a function f(x) that is differentiable (has derivatives) on the closed interval [ab], then for every x with f′(a) < x < f′(b), there exists some point c in the open interval (ab) such that f′(c) = x. In other words, the derivative function, though it is not necessarily continuous, follows the intermediate value theorem by taking every value that lies between the values of the derivatives at the endpoints. The intermediate value theorem, which implies Darboux’s theorem when the derivative function is continuous, is a familiar result in calculus that states, in simplest terms, that if a continuous real-valued function f defined on the closed interval [−1, 1] satisfies f(−1) < 0 and f(1) > 0, then f(x) = 0 for at least one number x between −1 and 1; less formally, an unbroken curve passes through every value between its endpoints. Darboux’s theorem was first proved in the 19th century by the French mathematician Jean-Gaston Darboux.
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Or click Continue to submit anonymously: | 751 | 3,117 | {"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-2015-27 | longest | en | 0.845737 |
http://clay6.com/qa/15807/what-is-the-correct-order-of-occurrence-by-weight-in-air-of-ne-ar-and-kr- | 1,527,212,239,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866917.70/warc/CC-MAIN-20180525004413-20180525024413-00199.warc.gz | 56,492,817 | 26,222 | # What is the correct order of occurrence (% by weight) in air of Ne, Ar and Kr?
$\begin {array} {1 1} (a)\;Ne > Ar > Kr & \quad (b)\;Ar > Ne > Kr \\ (c)\;Ar > Kr > Ne & \quad (d)\;Ne > Kr > Ar \end {array}$
## 1 Answer
$(b)\;Ar > Ne > Kr$
answered Nov 7, 2013 by
1 answer
1 answer
1 answer
1 answer
0 answers
1 answer
1 answer | 130 | 343 | {"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": 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} | 2.9375 | 3 | CC-MAIN-2018-22 | latest | en | 0.713544 |
https://www.sarthaks.com/2737863/mean-observation-more-observation-included-new-mean-becomes-find-value-new-observation | 1,675,008,855,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499744.74/warc/CC-MAIN-20230129144110-20230129174110-00801.warc.gz | 1,021,748,043 | 14,988 | # The mean of 15 observation is 9. One more observation is included and the new mean becomes 12. Find the value of new observation.
21 views
in Aptitude
closed
The mean of 15 observation is 9. One more observation is included and the new mean becomes 12. Find the value of new observation.
1. 21
2. 23
3. 45
4. 57
by (30.0k points)
selected
Correct Answer - Option 4 : 57
Given
Mean of 15 observation = 9
Formula used
Average = (sum of all observation)/Number of observation
Calculation
Sum of 15 observation = 15 × 9
⇒135
Sum of 16 observation = 16 × 12
⇒192
Value of new observation = 192 - 135
⇒57
∴ Value of new observation is 57. | 198 | 649 | {"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-2023-06 | latest | en | 0.870741 |
https://keisan.casio.com/exec/system/1223432660 | 1,553,235,694,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202635.43/warc/CC-MAIN-20190322054710-20190322080710-00447.warc.gz | 539,513,375 | 8,445 | # Regular polygon circumscribed to a circle Calculator
## Calculates the side length and area of the regular polygon circumscribed to a circle.
number of sides n n＝3,4,5,6.... inradius r 6digit10digit14digit18digit22digit26digit30digit34digit38digit42digit46digit50digit side length a polygon area Sp circle area Sc area ratio Sp/Sc
$\normal Regular\ polygons\ circumscribed\\\hspace{200} to\ a\ circle\\\hspace{20} n:\ number\ of\ sides\\(1)\ polygon\ side:\hspace{22} a=2r\tan{\large\frac{\pi}{n}}\\\vspace{2}(2)\ polygon\ area:\hspace{10} S_p={\large\frac{1}{2}}nar\\\vspace{2}(3)\ circle\ area:\hspace{30} S_c=\pi r^2\\$
Regular polygon circumscribed to a circle
[1-10] /18 Disp-Num5103050100200
[1] 2019/02/20 05:14 Male / 20 years old level / High-school/ University/ Grad student / A little /
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Wished that you could enter a perimeter and it give you a radius
[2] 2018/06/09 21:47 Male / 60 years old level or over / A retired people / Very /
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Used to create a Pentagon for a power tool workstation with the biggest possible sides. Thanx for the calculator and the effort.
[3] 2018/03/21 00:04 Male / 60 years old level or over / An engineer / Useful /
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Building a wooden frame. Didn''t want to spend the time to derive calculations myself.
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Comment/Request
area of regular polygon circumscribed about the circle
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Cutting out a clock face
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To copy the formula for use in a collision detection routine.
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Purpose of use? | 859 | 2,840 | {"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": 1, "/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-2019-13 | longest | en | 0.628493 |
http://www.zentralblatt-math.org/zmath/en/advanced/?q=an:0719.34037 | 1,368,917,975,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696382920/warc/CC-MAIN-20130516092622-00053-ip-10-60-113-184.ec2.internal.warc.gz | 829,327,583 | 4,706 | Language: Search: Contact
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Zbl 0719.34037
Baxley, John V.
Existence and uniqueness for nonlinear boundary value problems on infinite intervals.
(English)
[J] J. Math. Anal. Appl. 147, No.1, 122-133 (1990). ISSN 0022-247X
The author considers the boundary value problem $(1)\quad y''=f(x,y,y'),\quad 0\le x<\infty,$ $(2)\quad a\sb 0y(0)=a\sb 1y'(0)=A,$ $a\sb 0\ge 0$, $a\sb 1\ge 0$, $a\sb 0+a\sb 1>0$, $(3)\quad y(\infty)=B.$ The basic assumptions on the function f(x,y,z) are: f(x,y,z) is continuous on $I\times {\bbfR}\sp 2$, $I=[a,b]$, for $0<b<\infty$; f(x,y,z) is nondecreasing in y for each fixed pair (x,z)$\in I\times {\bbfR}$; f(x,y,z) satisfies a uniform Lipschitz condition on each compact subset of $I\times {\bbfR}\sp 2$ with respect to z; and zf(x,y,z)$\le 0$ for $(x,y,z)\in I\times {\bbfR}\sp 2$, $z\ne 0$. Using the shooting method, and with additional assumptions on f(x,y,z) and supposing that $a\sb 0$, $a\sb 1$ are both positive, he proves that the boundary value problem (1)-(3) has a unique solution. \par The following example $y''=-2xy'/(1-\alpha y)\sp{1/2},\quad 0\le x<\infty,$ $y(0)=1,\quad y(\infty)=0,$ which arises in nonlinear mechanics in the problem of unsteady flow of gas through a semi-infinite porous medium, $0<\alpha \le 1$, is given.
[M.Shahin (Dallas)]
MSC 2000:
*34B15 Nonlinear boundary value problems of ODE
34A45 Theoretical approximation of solutions of ODE
76S05 Flows in porous media
Keywords: boundary value problem; shooting method; unique solution; example; unsteady flow of gas through a semi-infinite porous medium
Highlights
Master Server | 595 | 1,815 | {"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} | 2.9375 | 3 | CC-MAIN-2013-20 | latest | en | 0.701287 |
https://metanumbers.com/13253470 | 1,600,927,531,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400213454.52/warc/CC-MAIN-20200924034208-20200924064208-00774.warc.gz | 518,785,490 | 7,614 | ## 13253470
13,253,470 (thirteen million two hundred fifty-three thousand four hundred seventy) is an even eight-digits composite number following 13253469 and preceding 13253471. In scientific notation, it is written as 1.325347 × 107. The sum of its digits is 25. It has a total of 4 prime factors and 16 positive divisors. There are 5,214,240 positive integers (up to 13253470) that are relatively prime to 13253470.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 8
• Sum of Digits 25
• Digital Root 7
## Name
Short name 13 million 253 thousand 470 thirteen million two hundred fifty-three thousand four hundred seventy
## Notation
Scientific notation 1.325347 × 107 13.25347 × 106
## Prime Factorization of 13253470
Prime Factorization 2 × 5 × 61 × 21727
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 13253470 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 13,253,470 is 2 × 5 × 61 × 21727. Since it has a total of 4 prime factors, 13,253,470 is a composite number.
## Divisors of 13253470
16 divisors
Even divisors 8 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 16 Total number of the positive divisors of n σ(n) 2.42484e+07 Sum of all the positive divisors of n s(n) 1.0995e+07 Sum of the proper positive divisors of n A(n) 1.51553e+06 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 3640.53 Returns the nth root of the product of n divisors H(n) 8.74512 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 13,253,470 can be divided by 16 positive divisors (out of which 8 are even, and 8 are odd). The sum of these divisors (counting 13,253,470) is 24,248,448, the average is 1,515,528.
## Other Arithmetic Functions (n = 13253470)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 5214240 Total number of positive integers not greater than n that are coprime to n λ(n) 217260 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 863196 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 5,214,240 positive integers (less than 13,253,470) that are coprime with 13,253,470. And there are approximately 863,196 prime numbers less than or equal to 13,253,470.
## Divisibility of 13253470
m n mod m 2 3 4 5 6 7 8 9 0 1 2 0 4 6 6 7
The number 13,253,470 is divisible by 2 and 5.
• Arithmetic
• Deficient
• Polite
• Square Free
## Base conversion (13253470)
Base System Value
2 Binary 110010100011101101011110
3 Ternary 220221100100021
4 Quaternary 302203231132
5 Quinary 11343102340
6 Senary 1152022354
8 Octal 62435536
10 Decimal 13253470
12 Duodecimal 45319ba
20 Vigesimal 42gdda
36 Base36 7w2fy
## Basic calculations (n = 13253470)
### Multiplication
n×i
n×2 26506940 39760410 53013880 66267350
### Division
ni
n⁄2 6.62674e+06 4.41782e+06 3.31337e+06 2.65069e+06
### Exponentiation
ni
n2 175654467040900 2328031209292556923000 30854491791422624402272810000 408929081322866009836790619150700000
### Nth Root
i√n
2√n 3640.53 236.652 60.3368 26.5746
## 13253470 as geometric shapes
### Circle
Diameter 2.65069e+07 8.3274e+07 5.51835e+14
### Sphere
Volume 9.75163e+21 2.20734e+15 8.3274e+07
### Square
Length = n
Perimeter 5.30139e+07 1.75654e+14 1.87432e+07
### Cube
Length = n
Surface area 1.05393e+15 2.32803e+21 2.29557e+07
### Equilateral Triangle
Length = n
Perimeter 3.97604e+07 7.60606e+13 1.14778e+07
### Triangular Pyramid
Length = n
Surface area 3.04242e+14 2.74361e+20 1.08214e+07
## Cryptographic Hash Functions
md5 fc75c00884549687c32c54744867ba58 5909909390527ebb66c2c5a5d2569d6c872ef5c3 71613cc8d9a9faba43eb138a0ff7e2d6e5b1d5684a8e82d1f6dc13ae6c18bdee b4d668058cccec3f30f1c3ba9e1e7d23f22a6932b31d9d2be7faafd79150f23cf8e21a5e65c0860612b6f3811d98be6b0a443de07545858680e1a0d8f1cd4e7a fc5118fdbd2be3214898f88f2c1f193cd043c6c7 | 1,590 | 4,392 | {"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} | 3.390625 | 3 | CC-MAIN-2020-40 | latest | en | 0.794108 |
http://mathoverflow.net/questions/25332?sort=newest | 1,369,379,274,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704288823/warc/CC-MAIN-20130516113808-00070-ip-10-60-113-184.ec2.internal.warc.gz | 173,558,564 | 12,593 | ## Fundamental theorems [closed]
I am certainly sure that any one who has read Gil Kalai's witty community wiki has benefited a lot. Here I follow a similar track in asking this question. So let's compose a list of fundamental theorems in mathematics which may not even have the tag "fundamental" but have serious wight in the respective branch of math.
I will start with the elementary and very popular ones.(Please add a description if the theorem is fundamental but still not so well-known)
Thanks for all your effort.
1. FTA: The Fundamental Theorem of Arithmetic (or Unique-Prime-Factorization Theorem): ->Any integer greater than 1 can be written as a unique product (up to ordering of the factors) of prime numbers.
2. FTA: The Fundamental theorem of Algebra: -> The field of complex numbers is algebraically closed
3. FTC: The fundamental theorem of calculus: -> Has two parts and specifies the relationship between the two central operations of calculus: differentiation and integration.
4. FTLP: The fundamental theorem of linear programming: -> In a weak formulation, states that the maxima and minima of a linear function over a convex polygonal region occur at the region's corners.
-
I don't really understand what distinguishes a "fundamental theorem" from an important theorem apart from having the luck to have been called a fundamental theorem. Is it really true, for instance, that the FTC is really the most important theorem in calculus? Maybe it's the most important theorem in highschool calculus, but, I would say that there are a large number of theorems of real analysis in one variable that are as important if not moreso. Similarly, is the fundamental theorem of algebra really important for algebra, or should it be the fundamental theorem of complex numbers? – Harry Gindi May 20 2010 at 8:35
I don't know how standard they are, but a Google search reveals that at least some people refer to fundamental theorems of Galois theory, space curves, projective geometry and Riemannian geometry. – gowers May 20 2010 at 9:22
Okay we may stay content with those that have the tag only or what people consider fundamental but not tagged that way. – To be cont'd May 20 2010 at 9:24
I voted to close. I see no commonality between answers to this question other than a linguistic one (ie someone decided to call a lemma "fundamental"). – Andy Putman May 20 2010 at 18:34
Closed. I'm not excited about jumble-bag big-list questions, and I think this one is not going to be very useful to anyone, so I've closed it. Wikipedia seems to like big lists, and I'd encourage anyone excited about this particular list to try it out there. – Scott Morrison May 20 2010 at 19:04
show 2 more comments
## closed as not a real question by Harry Gindi, Andrew Stacey, Kevin Lin, Andy Putman, Scott Morrison♦May 20 2010 at 19:03
In his book Topics in Geometric Group Theory, Pierre de la Harpe calls the following result the Fundamental Observation of Geometric Group Theory (though he also calls it a theorem!). It is also often called the Svarc--Milnor Lemma. Roughly speaking, it asserts that the coarse geometry of a group is captured by any suitably nice action of that group by isometries on a metric space.
Theorem. Let $X$ be a metric space that is geodesic and proper, let $\Gamma$ be a group and let $\Gamma$ act properly discontinuously and cocompactly by isometries on $X$. Then $\Gamma$ is finitely generated, and furthermore for any $x_0\in X$ the map $\Gamma\to X$ given by
$\gamma\mapsto\gamma x_0$
is a quasi-isometry.
Remarks.
1. $\Gamma$ is endowed with the word metric (with respect to some choice of finite generating set).
2. A map of metric space $f:Y\to X$ is a quasi-isometric embedding if there are constants $\lambda\geq 1$, $\mu\geq 0$ such that
$\lambda d_Y(y_1,y_2)+\mu\geq d_X(f(y_1),f(y_2))\geq \frac{1}{\lambda} d_Y(y_1,y_2)-\mu$
for all $y_1,y_2\in Y$. It is a quasi-isometry if, furthermore, for every $x\in X$ there is $y\in Y$ such that $d(x,f(y))\leq \mu$.
-
Wikipedia says the Fundamental Theorem of Riemannian Geometry is the unique existence of the Levi-Civita connection. I've never heard it called that myself, so this is maybe an anti-answer.
-
The Fundamental Theorem of Asset Pricing (FTAP) in mathematical finance also comes in two parts. The first part says, more or less, that a market is arbitrage-free if and only if there is an equivalent martingale measure for the discounted price process. The second part says that the market is complete (all European options can be hedged) if and only if the equivalent martingale measure is unique.
(In some models, you may need an appropriate definition of "arbitrage-free", such as the notion of "no free lunch with vanishing risk", and you may replace "equivalent martingale measure" with "equivalent local martingale measure". But the idea is the same.)
-
I used to joke that The Fundamental Theorem of Combinatorics is interchange of summation.
-
See also: en.wikipedia.org/wiki/… – Douglas S. Stones May 21 2010 at 12:15
To add to the Gowers examples: the fundamental theorem on finitely-generated abelian groups. It seems at least a mildly interesting linguistic point. German discriminates between Hauptsatz and Fundamentalsatz, i.e. main theorem and fundamental theorem (if satz is not quite "theorem"). That distinction seems less clear in the English usage. The German Wikipedia admits the Fundamental Theorems of Algebra, Analysis and Arithmetic, but others in pure mathematics aren't obvious. I would myself think of Galois theory (the perfect duality of subfields and subgroups) and projective geometry (collineations semi-coordinatised) as having "fundamental theorems".
-
Welcome to MO, Charles! – Artie Prendergast-Smith May 20 2010 at 12:31
Thanks. But I'm really a double agent, working for you-know-who. – Charles Matthews May 20 2010 at 14:42 | 1,450 | 5,886 | {"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} | 3.171875 | 3 | CC-MAIN-2013-20 | latest | en | 0.943431 |
https://justaaa.com/finance/195463-the-ellis-corporation-has-heavy-lease-commitments | 1,718,959,174,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862040.7/warc/CC-MAIN-20240621062300-20240621092300-00394.warc.gz | 288,764,961 | 11,594 | Question
# The Ellis Corporation has heavy lease commitments. Prior to SFAS No. 13, it merely footnoted lease...
The Ellis Corporation has heavy lease commitments. Prior to SFAS No. 13, it merely footnoted lease obligations in the balance sheet, which appeared as follows: Use Appendix D for an approximate answer but calculate your final answer using the formula and financial calculator methods.
In \$ millions In \$ millions Current assets \$ 55 Current liabilities \$ 10 Fixed assets 55 Long-term liabilities 35 Total liabilities \$ 45 Stockholders' equity 65 Total assets \$ 110 Total liabilities and stockholders' equity \$ 110
The footnotes stated that the company had \$21 million in annual capital lease obligations for the next 20 years.
a. Discount these annual lease obligations back to the present at a 10 percent discount rate. (Do not round intermediate calculations. Round your answer to the nearest million. Input your answer in millions of dollars (e.g., \$6,100,000 should be input as "6").)
PV of lease obligations million
b. Construct a revised balance sheet that includes lease obligations. (Do not round intermediate calculations. Round your answers to the nearest million. Input your answer in millions of dollars (e.g., \$6,100,000 should be input as "6").)
Balance Sheet (in \$ millions) Current assets Current liabilities Fixed assets Long-term liabilities Leased property under capital lease Obligations under capital lease Total liabilities Stockholders' equity Total assets Total liabilities and Stockholders' equity
c. Compute the total debt to total asset ratio for the original and revised balance sheets. (Input your answers as a percent rounded to 2 decimal places.)
Original % Revised %
d. Compute the total debt to total equity ratio for the original and revised balance sheets. (Input your answers as a percent rounded to 2 decimal places.)
Original % Revised %
e. In an efficient capital market environment, should the consequences of SFAS No. 13, as viewed in the answers to parts c and d, change stock prices and credit ratings?
Yes No
As per rules I am answering the first 4 subparts of the question
1:
PV of annuity = Annuity*(1-1/(1+rate)^number of terms)/rate
=21*(1-1/1.1^20)/0.1
=178.78
=179 million
2:
In \$ millions In \$ millions Current assets \$ 55 Current liabilities \$ 10 Fixed assets 55 Long-term liabilities 35 Leased property under capital leas 179 Obligations under capital lease 179 Total liabilities \$ 224 Stockholders' equity 65 Total assets \$ 289 Total liabilities and stockholders' equity \$ 289
3: Original debt/Asset ratio = 45/110
=40.91%
Revised Debt/Asset ratio= 224/289
=77.51%
4:Original Debt/Equity ratio = 45/65
= 69.23%
Revised Debt/Equity = 224/65
= 344.62%
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Most questions answered within 1 hours. | 668 | 2,906 | {"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} | 2.515625 | 3 | CC-MAIN-2024-26 | latest | en | 0.899918 |
https://www.swirlzcupcakes.com/tips-and-recommendations/when-to-put-two-people-on-each-side-of-a-see-saw/ | 1,642,743,673,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302723.60/warc/CC-MAIN-20220121040956-20220121070956-00663.warc.gz | 982,870,978 | 24,092 | # When to put two people on each side of a see saw?
## When to put two people on each side of a see saw?
Put any two groups on each side of the see-saw. (First Use) If the see-saw balances, we are sure that the oddly wieghted one is in the other group of 4. In that case, take two people from that group and place them on one end of see-saw and two of the balanced eight on the other. (Second Use)
## Is the seesaw easier than everyone makes it?
It is easier than everyone makes it. A seesaw is binary. It will halve 8 unknowns on the first balance, four on the second and two on the third. Set it up so deduction eliminates everything else and your gold. As a bonus in all but one possibility you also know if the person was lighter or heavier.
## What to do if the See Saw is not balanced?
(Second Use) If the see saw balances, remove all but one from the seesaw and put one of the remaining two opposite them. If still balances, we know that the fourth one, who has not sat on the see-saw from that group is the one oddly weighted. (Third Use) If the see saw is not balanced, remove one from each end.
## When do the heavier Islanders get off the see saw?
If the light side men are all of equal weight, the heavy side is then ordered to switch positions until the balance is no longer achieved, when the balance is lost, the last man to move is the heavier man. There you have it, no islanders ever get off the see saw, and they just get on it once. Highly active question.
## What are the names of the different types of saws?
Miter Saw: 2.10 #10. Radial Arm Saw: 2.11 #11. Reciprocating Saw: 2.12 #12. Rotary Saw: 2.13 #13. Scroll Saw: 2.14 #14. Table Saw: 2.15 #15. Masonry Saw: 2.16 #16. Oscillating Saw: 2.17 #17. Pole Saw 2.18 #18. Track Saw 2.19 #19. Panel Saw As we know there are different types of saws are available on the market in 2020.
## Is there a difference between a coping saw and a hacksaw?
People have been using saws for thousands of years, and even if many technologies have arrived in the market, the saw has kept its demand as before. You can keep from coping saw to the hacksaw in your toolbox, even more than one of the same kind. With the shape of the saw and teeth, one has a function that is different from another.
## What is the difference between a back saw and a hand saw?
A back saw is a hand saw. A back saw is relatively smaller than other saws, and the blade is narrow. As its name, it reinforced along the top edge. The common use of back saw is with miter boxes. When there is a need for fine and straight cut consistently, back is too effective for that purpose.
## What’s the difference between’did you saw’and’did’?
Saw has two different meanings… one is the past tense of the verb “to see” and the other one is a noun meaning a tool. However that noun became the verb to describe the action of the tool. Robert is correct in saying “Did you saw” is grammatically incorrect. | 741 | 2,936 | {"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} | 3.015625 | 3 | CC-MAIN-2022-05 | latest | en | 0.951423 |
https://www.teacherspayteachers.com/Product/Fractions-1206091 | 1,485,168,778,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560282631.80/warc/CC-MAIN-20170116095122-00018-ip-10-171-10-70.ec2.internal.warc.gz | 997,578,219 | 51,074 | # Fractions
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1.34 MB | 9 pages
### PRODUCT DESCRIPTION
Are you looking for a fun way for your students to practice fraction skills? These task cards can be used in so many ways: as a Scoot game (my students' favorite), in a math center, as a small group activity, for early finishers, or even as a whole group game!
These task cards will allow your students to practice the following skills: Identifying Fractions, Comparing Fractions, and Identifying Fractions on a Number Line. These skills align directly to 3rd grade Common Core standards, but they can be used in other grades for acceleration or remediation.
-Student response sheet
-3.NF.A.1: Understand a fraction 1/b as the quantity formed by 1 part when a whole is partitioned into b equal parts; understand a fraction a/b as the quantity formed by a parts of size 1/b.
-3.NF.A.2: Understand a fraction as a number on the number line; represent fractions on a number line diagram.
-3.NF.A.3: Explain equivalence of fractions in special cases, and compare fractions by reasoning about their size.
If you have any questions or find an error, please contact me at fabulouslifeofateacher@gmail.com, and I will get it corrected promptly.
Don't forget to leave feedback to earn credits toward future purchases!
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https://www.experts-exchange.com/questions/28128470/how-to-solve-1-2-q-3-q-to-degree-of-2-n-q-to-degree-of-n-1.html | 1,495,536,992,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607620.78/warc/CC-MAIN-20170523103136-20170523123136-00520.warc.gz | 874,517,418 | 25,278 | Solved
# how to solve 1+2*q+3*(q to degree of 2)+....+ n(q to degree of (n-1))
Posted on 2013-05-15
220 Views
I can see that it is
1+q +q to degree of 2+q to degree of 3+...........................q to degree of (n-1)
q +q to degree of 2+q to degree of 3+...........................q to degree of (n-1)
q to degree of 2+q to degree of 3+...........................q to degree of (n-1)
.............................................................................................................
n(q to degree of (n-1))
0
Question by:user_n
[X]
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Accepted Solution
TommySzalapski earned 500 total points
ID: 39167898
Let S = 1 + q + q^2 + q^3 + ... + q^(n-1)
Sq = q + q^2 + q^3 +q^4 + ... + q^(n-1) + q^n
But that looks pretty close to what S was. In fact:
Sq = S + q^n - 1
Now just solve
Sq - S = q^n - 1
S(q-1) = q^n - 1
S = (q^n - 1)/(q-1)
Now do the same type of thing with your equation and remember the result we just got because you will see it pop out.
0
## Featured Post
Question has a verified solution.
If you are experiencing a similar issue, please ask a related question | 397 | 1,303 | {"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-2017-22 | longest | en | 0.836125 |
http://maxima-online.org/tags/index/points | 1,553,357,873,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202889.30/warc/CC-MAIN-20190323161556-20190323183556-00531.warc.gz | 129,913,030 | 4,578 | ### Related
? points;
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((spot_bid + worst_bi...
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print(sum((P(t - 1) -...
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g:points([100,300,500...
wxdraw2d(g);
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points: [0, 5.27, 7.1...
ratsimp(sum((P(t - 1)...
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##### points
points ([[1,2], [3,4]]);
Calculate
### points
Run Example
```(%i1)points ([[1,2], [3,4]]);
(%o1) points([[1, 2], [3, 4]])
(%i2) ```
Run Example
```load(draw);
(%o1) /usr/share/maxima/5.21.1/share/draw/draw.lisp
(%i2) p1:points([0,0.4,0.8,1.2,1.6,2.0],[0.21,1.25,2.31,2.70,2.65,3.20]);
(%o2) points([0, 0.4, 0.8, 1.2, 1.6, 2.0], [0.21, 1.25, 2.31, 2.7, 2.65, 3.2])
(%i3) p2:explicit((977/700)*x+(1381/2100),x,0,2);
977 x 1381
(%o3) explicit(----- + ----, x, 0, 2)
700 2100
(%i4) p3:explicit((-689/896)*x^2+(32857/11200)*x+(99/400),x,0,2);
2
689 x 32857 x 99
(%o4) explicit(- ------ + ------- + ---, x, 0, 2)
896 11200 400
(%i5) wxdraw2d(p1,p2,p3);
(%o5) wxdraw2d(points([0, 0.4, 0.8, 1.2, 1.6, 2.0],
977 x 1381
[0.21, 1.25, 2.31, 2.7, 2.65, 3.2]), explicit(----- + ----, x, 0, 2),
700 2100
2
689 x 32857 x 99
explicit(- ------ + ------- + ---, x, 0, 2))
896 11200 400
(%i6) ```
Run Example
```load(draw);
(%o1) /usr/share/maxima/5.21.1/share/draw/draw.lisp
(%i2) draw2d(points([1,2]));
plot```
Help for Points | 771 | 1,748 | {"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} | 3.125 | 3 | CC-MAIN-2019-13 | longest | en | 0.264615 |
http://www.puzzlegamemaster.com/math-riddles-app-level-8-solution-android/ | 1,576,280,947,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540569332.20/warc/CC-MAIN-20191213230200-20191214014200-00413.warc.gz | 205,945,086 | 8,642 | # Math riddles Level 8 Answer with Hints and solutions
Math riddles game level 8 Answer with solution android game developed by Black games. Scroll Below to find details.
Math Riddles tests your IQ with mathematical puzzles. Challenge yourself with different levels of math puzzles and stretch the limits of your intelligence. Every IQ game is prepared with an approach of an IQ test. Improve your mathematics, challenge your brain with this game. There are many hard levels which you may not have solved then you can take help from here.
Math app level 8: A+B=60
A-B=40
A/B=?
Solution: A=50
B=10
then A/B=50/10=5
If there is any Doubt/discrepancy then please let me know in comments or you are unable to understand solution then I will surely help you.
### 8 thoughts on “Math Riddles App Level 8 Solution Android”
• August 26, 2019 at 11:34 pm
Doesn’t 2 also work?
40 + 20 = 60
60 – 40 = 20
40/20 = 2
• August 26, 2019 at 11:35 pm
Scratch that. Just noticed my error!
• July 11, 2019 at 12:50 am
• June 13, 2019 at 5:19 pm
I want to make many questions just like this. So could you pliz write the algorithm of only this problem
• June 14, 2019 at 12:14 am
Best idea to Take any two numbers such taht 25 and 5
So question becomes
A+B=30
A-B=20
then A/B=?
• March 9, 2019 at 12:08 am
No tiene sentido. Sí A+B es 60 y A-B es 40 quiere decir que B es igual a la diferencia entre esos dos resultados. Por ende A=40 y B=20. Es decir que A/B= 4 | 434 | 1,461 | {"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} | 3.765625 | 4 | CC-MAIN-2019-51 | latest | en | 0.861252 |
https://www.teacherspayteachers.com/Store/Erin-Atwood | 1,542,529,555,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039744320.70/warc/CC-MAIN-20181118073231-20181118095231-00130.warc.gz | 1,018,912,015 | 26,798 | # Erin Atwood
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United States - Alabama - Athens
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I love math: it makes people cry.
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Use this foldable to teach students the "process" for solving multi-step equations. My students were confused about what to do first. This way there is no confusion and they can solve ANY equation, no problem.
Subjects:
Algebra, Basic Operations, Algebra 2
6th, 7th, 8th, 9th, 10th, 11th, 12th
Types:
Printables, Graphic Organizers
\$1.00
109 ratings
4.0
This PowerPoint shows the pages of the foldable containing vocabulary terms about circles. I had students create a circle booklet and then copy the information as it was displayed on the projector. You could also print the PowerPoint and glue the
Subjects:
Math, Geometry, Other (Math)
6th, 7th, 8th, 9th, 10th, 11th, 12th
Types:
PowerPoint Presentations, Activities, Handouts
\$1.00
11 ratings
4.0
Complementary, Supplementary, Vertical, Angle Addition Postulate, Angle Bisectors
Subjects:
Math, Algebra, Geometry
7th, 8th, 9th, 10th
Types:
PowerPoint Presentations, Handouts, Graphic Organizers
\$3.00
5 ratings
3.8
Student worksheet to practice counter-clockwise rotations in the coordinate plane.
Subjects:
Math, Geometry, Graphing
6th, 7th, 8th, 9th
Types:
Worksheets, Activities, Printables
\$2.00
5 ratings
4.0
This PowerPoint presentation shows formulas for finding the areas of circles, regular polygons, and shaded regions (consisting of circles and regular polygons). All examples include the steps and answers animated so that each step appears as the
Subjects:
Math, Geometry
7th, 8th, 9th, 10th, 11th, 12th
Types:
PowerPoint Presentations
\$4.00
1 rating
3.5
Transformations in the coordinate plane Notes for use in interactive notebook or to create student booklet
Subjects:
Geometry
8th, 9th, 10th, 11th, 12th
Types:
PowerPoint Presentations, Handouts, Interactive Notebooks
\$6.00
not yet rated
This document was used with my interactive student notebook pages on the polygon angle-sum theorem.
Subjects:
Geometry
9th, 10th, 11th, 12th
Types:
Activities
FREE
2 ratings
4.0
These slides are handouts for use in Interactive Notebooks on the topics of Pythagorean Theorem & Special Right Triangles. See my blog for more information.
Subjects:
Geometry
10th, 11th, 12th
Types:
Activities, Printables
FREE
3 ratings
3.6
Area of triangles, rectangles, and squares, includes worksheet and quilt block templates.
Subjects:
Geometry, Measurement
6th, 7th, 8th, 9th, 10th, 11th, 12th
Types:
Activities
FREE
7 ratings
3.9
Manipulatives for interactive notebook pages
Subjects:
Geometry, Trigonometry
7th, 8th, 9th, 10th, 11th, 12th
Types:
Activities, Handouts
FREE
2 ratings
3.9
A chart to be used in interactive notebooks as a reference for the many symbols students must understand. The first entry is included. The chart is sized to be cut out and pasted into a standard size composition notebook. Alternatively, it could be
Subjects:
Math, Geometry
6th, 7th, 8th, 9th, 10th, 11th, 12th
Types:
Activities, Handouts, Graphic Organizers
FREE
1 rating
4.0
This document contains the powerpoint slides used in a game of grudge ball (instructions included). Each slide has a timer with a sound played when time is up. There are enough slides for 5 teams, 6 rounds. Contains answer key
Subjects:
Geometry
6th, 7th, 8th, 9th, 10th, 11th, 12th
Types:
PowerPoint Presentations, Activities, Games
FREE
6 ratings
4.0
This is an animation of a shutter foldable I created to show the formulas for finding the measure of angles created when lines cross when given arc measure. (Not printable)
Subjects:
Geometry
6th, 7th, 8th, 9th, 10th, 11th, 12th
Types:
PowerPoint Presentations, Activities, Graphic Organizers
FREE
not yet rated
Notebook items for deciding if lines are parallel, perpendicular, or neither. These activities work well for lower-level, at risk, and exceptional learners.
Subjects:
Math, Geometry, Graphing
7th, 8th, 9th, 10th
Types:
PowerPoint Presentations, Printables, Graphic Organizers
FREE
26 ratings
4.0
showing 1-14 of 14
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TEACHING EXPERIENCE
7 years teaching math in north Alabama. I teach at-risk students and exceptional learners.
MY TEACHING STYLE
Interactive Student notebooks
HONORS/AWARDS/SHINING TEACHER MOMENT
MY OWN EDUCATIONAL HISTORY
Bachelors degree from Troy University.
6th, 7th, 8th, 9th, 10th, 11th, 12th
SUBJECTS
Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 1,308 | 4,645 | {"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} | 2.71875 | 3 | CC-MAIN-2018-47 | latest | en | 0.889838 |
http://incometaxmanagement.com/Pages/Tax-Ready-Reckoner/Assessment/Individual/Income-received-from-different-institutions.html | 1,556,207,664,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578727587.83/warc/CC-MAIN-20190425154024-20190425180024-00433.warc.gz | 81,886,815 | 11,889 | # Income Received from different Institutions by an Individual for Tax Calculation
## 1. As a Member of Hindu Undivided Family.
Any sum received by an individual as a member of H.U.F. out of family income is not to be included in his total income, because the share of income received from the H.U.F. is exempted in the hands of such individual, the family may or may not have paid tax on that income [Section 10 (2)1.
If the member earns his own income, besides being the member of H.U.F. he will pay tax on his own earned income.
But u/s 64 (2) where an individual converts his individual property into the common pool of H.U.F. of which he is a member, income from such property shall be included in his individual income.
## 2. Income Received as Share from AOP.
The share from AOP is treated in following manner:
1. Compute total income of AOP.
2. Where Share of Members of AOP are determined
### (i). How to Compute Taxable Income of an AOP/BOI
First find out the taxable income of the AOP/BOI under the following steps:
1. Find out income under the different heads of income (viz., “Income from house property”, “Profits and gains of business or profession”, “Capital gains” and “Income from other sources”, ignoring incomes exempt under sections 10 to 13A.
2. Make adjustments on account of brought forward losses/allowances. The total income under the aforesaid heads is “Gross total income”.
3. From the “gross total income” make deductions under sections 80G, 80GGA, 80GGC, 80-I, 80-IA, 80-IB, 80-IC and 80JJA.
4. The balancing amount is total income or taxable income.
### (ii). Where Shares of Members are Determinate –
Tax liability is determined as under:
#### (A). If none of the Member of the AOP / BOI has income in Excess of the Maximum Amount which is Not Chargeable to Tax in the hands of Members –
In this case, tax is chargeable on the income of the AOP/BOI at the same rate as is applicable in the case of an individual.
#### (B). If Any Member Of The AOP/BOI Has Income In Excess Of The Maximum Amount Which Is Not Chargeable To Tax-
In such case, tax will be chargeable at the maximum marginal rate (i.e., 35.535 % for the assessment year 2018-19 and 35.88 % for the assessment year 2019-20).
## 3. As a Partner of Firm Assessed as Firm Assessed u/s 184.
The share received by an individual from a firm shall not he included in his total income irrespective of the fact, whether the firm has paid the tax or not. Any salary or other remuneration and interest on capital is taxable under the head Profits and Gains to the extent above remuneration and interest are allowed as deduction to the firm.
## 4. Share of income from firm assessed u/s 185.
Share of income received by a partner from a firm which has been assessed to tax u/s 185 as it has not submitted a copy of its instrument of partnership is fully exempted u/s lO(2A). The following sums received by partner from such firm shall also he exempted in the hands of the partner
(a) Any remuneration, bonus, fees, commission etc.
(b) Interest on loan/capital from such firm.
Note. The above exemptions are applicable because firm covered u/s 185 is not allowed to charge these items as expense.
## 5. As a Shareholder of a company.
The gross amount of dividend received by an individual is to he included in his total income. The gross amount means, the net dividend received plus tax deducted at source. The shareholder is liable to pay tax on whole of his income from dividend i.e., the gross amount of dividend declared by the company. The assessee shall get credit of the tax deducted at source out of his final tax liability. The individual shall be entitled to the deduction as provided by the different sections of Income-tax Act.
With effect from assessment year 1998-99 dividend received from or declared or distributed by an Indian company on or after 1-6-97 shall he fully exempted and shall not form part of total income.
Note. Dividend from foreign company is fully taxable as income from other sources.
#### More Topics..on Assessment @ Individual
Income Received from different Institutions by an Individual for Tax Calculation :
Incomes Of Other Persons To Be Included In Total Income Of An Individual : Section 60 to 64
Computation Of Total Income & Tax Liability of an Individual
Special Provisions Claimed by Non-Resident Indian (NRI) [Section 115C to 115-I]
##### 'Assessments' Under Income Tax Act. 1961.
Assessment of 'Agricultural Income', Companies, Firm, Co-operative Societies, HUF, Individual, AOP/BOI, TRUST ....
##### Deductions [Sections 80A to 80U (Chapter VIA)]
Deductions to be made in Computing Total Income [Sections 80A to 80U (Chapter VIA)]
##### Tax Computation with Assessment of Salaried Person, HUF, Individuals.
Computation of Income with Assessment Under the Head 'Salary', 'HUF', 'Individuals'. | 1,166 | 4,866 | {"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} | 2.640625 | 3 | CC-MAIN-2019-18 | longest | en | 0.969305 |
https://youbookinc.com/pdf/exact-statistical-inference-for-categorical-data/ | 1,709,417,219,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476137.72/warc/CC-MAIN-20240302215752-20240303005752-00805.warc.gz | 1,098,414,780 | 14,637 | ### Exact Statistical Inference for Categorical Data
Author : Guogen Shan Publsiher : Academic Press Total Pages : 66 Release : 2016-01-22 Genre : Mathematics ISBN : 9780128039489
## Download Exact Statistical Inference for Categorical Data Book in PDF, Epub and Kindle
Exact Statistical Inference for Categorical Data discusses the way asymptotic approaches have been often used in practice to make statistical inference. This book introduces both conditional and unconditional exact approaches for the data in 2 by 2, or 2 by k contingency tables, and is an ideal reference for users who are interested in having the convenience of applying asymptotic approaches, with less computational time. In addition to the existing conditional exact inference, some efficient, unconditional exact approaches could be used in data analysis to improve the performance of the testing procedure. Demonstrates how exact inference can be used to analyze data in 2 by 2 tables Discusses the analysis of data in 2 by k tables using exact inference Explains how exact inference can be used in genetics
### Learning Statistics with R
Author : Daniel Navarro Publsiher : Lulu.com Total Pages : 617 Release : 2013-01-13 Genre : Psychology ISBN : 9781326189723
## Download Learning Statistics with R Book in PDF, Epub and Kindle
"Learning Statistics with R" covers the contents of an introductory statistics class, as typically taught to undergraduate psychology students, focusing on the use of the R statistical software and adopting a light, conversational style throughout. The book discusses how to get started in R, and gives an introduction to data manipulation and writing scripts. From a statistical perspective, the book discusses descriptive statistics and graphing first, followed by chapters on probability theory, sampling and estimation, and null hypothesis testing. After introducing the theory, the book covers the analysis of contingency tables, t-tests, ANOVAs and regression. Bayesian statistics are covered at the end of the book. For more information (and the opportunity to check the book out before you buy!) visit http://ua.edu.au/ccs/teaching/lsr or http://learningstatisticswithr.com
### An Introduction to Categorical Data Analysis
Author : Alan Agresti Publsiher : John Wiley & Sons Total Pages : 423 Release : 2018-10-11 Genre : Mathematics ISBN : 9781119405283
## Download An Introduction to Categorical Data Analysis Book in PDF, Epub and Kindle
A valuable new edition of a standard reference The use of statistical methods for categorical data has increased dramatically, particularly for applications in the biomedical and social sciences. An Introduction to Categorical Data Analysis, Third Edition summarizes these methods and shows readers how to use them using software. Readers will find a unified generalized linear models approach that connects logistic regression and loglinear models for discrete data with normal regression for continuous data. Adding to the value in the new edition is: • Illustrations of the use of R software to perform all the analyses in the book • A new chapter on alternative methods for categorical data, including smoothing and regularization methods (such as the lasso), classification methods such as linear discriminant analysis and classification trees, and cluster analysis • New sections in many chapters introducing the Bayesian approach for the methods of that chapter • More than 70 analyses of data sets to illustrate application of the methods, and about 200 exercises, many containing other data sets • An appendix showing how to use SAS, Stata, and SPSS, and an appendix with short solutions to most odd-numbered exercises Written in an applied, nontechnical style, this book illustrates the methods using a wide variety of real data, including medical clinical trials, environmental questions, drug use by teenagers, horseshoe crab mating, basketball shooting, correlates of happiness, and much more. An Introduction to Categorical Data Analysis, Third Edition is an invaluable tool for statisticians and biostatisticians as well as methodologists in the social and behavioral sciences, medicine and public health, marketing, education, and the biological and agricultural sciences.
### COMPSTAT 2006 Proceedings in Computational Statistics
Author : Alfredo Rizzi,Maurizio Vichi Publsiher : Springer Science & Business Media Total Pages : 530 Release : 2007-12-03 Genre : Mathematics ISBN : 9783790817096
## Download COMPSTAT 2006 Proceedings in Computational Statistics Book in PDF, Epub and Kindle
International Association for Statistical Computing The International Association for Statistical Computing (IASC) is a Section of the International Statistical Institute. The objectives of the Association are to foster world-wide interest in e?ective statistical computing and to - change technical knowledge through international contacts and meetings - tween statisticians, computing professionals, organizations, institutions, g- ernments and the general public. The IASC organises its own Conferences, IASC World Conferences, and COMPSTAT in Europe. The 17th Conference of ERS-IASC, the biennial meeting of European - gional Section of the IASC was held in Rome August 28 - September 1, 2006. This conference took place in Rome exactly 20 years after the 7th COMP- STAT symposium which was held in Rome, in 1986. Previous COMPSTAT conferences were held in: Vienna (Austria, 1974); West-Berlin (Germany, 1976); Leiden (The Netherlands, 1978); Edimbourgh (UK, 1980); Toulouse (France, 1982); Prague (Czechoslovakia, 1984); Rome (Italy, 1986); Copenhagen (Denmark, 1988); Dubrovnik (Yugoslavia, 1990); Neuchˆ atel (Switzerland, 1992); Vienna (Austria,1994); Barcelona (Spain, 1996);Bristol(UK,1998);Utrecht(TheNetherlands,2000);Berlin(Germany, 2002); Prague (Czech Republic, 2004).
### Exact Statistical Methods for Data Analysis
Author : Samaradasa Weerahandi Publsiher : Copernicus Total Pages : 328 Release : 1995 Genre : Mathematics ISBN : 0387943609
## Download Exact Statistical Methods for Data Analysis Book in PDF, Epub and Kindle
This book gives a lucid account of new and recent developments in statistical inference. The author's goal is to develop a theory of generalized p-values and generalized confidence intervals, and to show how concepts may be used to make exact statistical inferences in a variety of practical applications. Numerous exercises are provided to further illustrate the concepts.
### Statistical Inference via Data Science A ModernDive into R and the Tidyverse
Author : Chester Ismay,Albert Y. Kim Publsiher : CRC Press Total Pages : 430 Release : 2019-12-23 Genre : Mathematics ISBN : 9781000763461
## Download Statistical Inference via Data Science A ModernDive into R and the Tidyverse Book in PDF, Epub and Kindle
Statistical Inference via Data Science: A ModernDive into R and the Tidyverse provides a pathway for learning about statistical inference using data science tools widely used in industry, academia, and government. It introduces the tidyverse suite of R packages, including the ggplot2 package for data visualization, and the dplyr package for data wrangling. After equipping readers with just enough of these data science tools to perform effective exploratory data analyses, the book covers traditional introductory statistics topics like confidence intervals, hypothesis testing, and multiple regression modeling, while focusing on visualization throughout. Features: ● Assumes minimal prerequisites, notably, no prior calculus nor coding experience ● Motivates theory using real-world data, including all domestic flights leaving New York City in 2013, the Gapminder project, and the data journalism website, FiveThirtyEight.com ● Centers on simulation-based approaches to statistical inference rather than mathematical formulas ● Uses the infer package for "tidy" and transparent statistical inference to construct confidence intervals and conduct hypothesis tests via the bootstrap and permutation methods ● Provides all code and output embedded directly in the text; also available in the online version at moderndive.com This book is intended for individuals who would like to simultaneously start developing their data science toolbox and start learning about the inferential and modeling tools used in much of modern-day research. The book can be used in methods and data science courses and first courses in statistics, at both the undergraduate and graduate levels.
### A Course in Categorical Data Analysis
Author : Thomas Leonard Publsiher : CRC Press Total Pages : 204 Release : 2020-08-26 Genre : Mathematics ISBN : 9781000111569
## Download A Course in Categorical Data Analysis Book in PDF, Epub and Kindle
Categorical data-comprising counts of individuals, objects, or entities in different categories-emerge frequently from many areas of study, including medicine, sociology, geology, and education. They provide important statistical information that can lead to real-life conclusions and the discovery of fresh knowledge. Therefore, the ability to manipulate, understand, and interpret categorical data becomes of interest-if not essential-to professionals and students in a broad range of disciplines. Although t-tests, linear regression, and analysis of variance are useful, valid methods for analysis of measurement data, categorical data requires a different methodology and techniques typically not encountered in introductory statistics courses. Developed from long experience in teaching categorical analysis to a multidisciplinary mix of undergraduate and graduate students, A Course in Categorical Data Analysis presents the easiest, most straightforward ways of extracting real-life conclusions from contingency tables. The author uses a Fisherian approach to categorical data analysis and incorporates numerous examples and real data sets. Although he offers S-PLUS routines through the Internet, readers do not need full knowledge of a statistical software package. In this unique text, the author chooses methods and an approach that nurtures intuitive thinking. He trains his readers to focus not on finding a model that fits the data, but on using different models that may lead to meaningful conclusions. The book offers some simple, innovative techniques not highighted in other texts that help make the book accessible to a broad, interdisciplinary audience. A Course in Categorical Data Analysis enables readers to quickly use its offering of tools for drawing scientific, medical, or real-life conclusions from categorical data sets.
### Statistical Strategies for Small Sample Research
Author : Rick H. Hoyle Publsiher : SAGE Total Pages : 394 Release : 1999-03-30 Genre : Mathematics ISBN : 0761908862
## Download Statistical Strategies for Small Sample Research Book in PDF, Epub and Kindle
This book provides encouragement and strategies for researchers who routinely address research questions using data from small samples. Chapters cover such topics as: using multiple imputation software with small sets; computing and combining effect sizes; bootstrap hypothesis testing; application of latent variable modeling; time-series data from small numbers of individuals; and sample size, reliability and tests of statistical mediation. | 2,260 | 11,335 | {"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} | 2.78125 | 3 | CC-MAIN-2024-10 | latest | en | 0.827311 |
https://www.officialdata.org/1940-dollars?amount=100 | 1,527,356,092,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867841.63/warc/CC-MAIN-20180526170654-20180526190654-00352.warc.gz | 805,470,338 | 6,798 | # \$100 in 1940 → \$1,782.53 in 2018
## Inflation Calculator
\$
### U.S. Inflation Rate, 1940-2018 (\$100)
According to the Bureau of Labor Statistics consumer price index, the dollar experienced an average inflation rate of 3.76% per year. Prices in 2018 are 1682.5% higher than prices in 1940.
In other words, \$100 in the year 1940 is equivalent in purchasing power to \$1,782.53 in 2018, a difference of \$1,682.53 over 78 years.
Compared to last year's annual rate, the inflation rate in 2018 is now 1.95%1. If this number holds, \$100 today would be equivalent to \$101.95 next year. The current inflation rate page gives more detail on the latest official inflation rates.
Cumulative price change 1682.53% Average inflation rate 3.76% Price difference (\$100 base) \$1,682.53 CPI in 1940 14 CPI in 2018 249.554
#### Inflation rates for specific categories
Bread · Milk · Medical care · More
S&P 500 price · S&P 500 earnings · Shiller P/E
### How to calculate the inflation rate for \$100 since 1940
This inflation calculator uses the following inflation rate formula:
CPI in 2018 / CPI in 1940 * 1940 USD value = 2018 USD value
Then plug in historical CPI values. The U.S. CPI was 14 in the year 1940 and 249.554 in 2018:
249.554 / 14 * \$100 = \$1,782.53
\$100 in 1940 has the same "purchasing power" as \$1,782.53 in 2018.
### Inflation Data Source
Raw data for these calculations comes from the Bureau of Labor Statistics' (CPI), established in 1913. Inflation data from 1665 to 1912 is sourced from a historical study conducted by political science professor Robert Sahr at Oregon State University.
You may use the following MLA citation for this page: “\$100 in 1940 → 2018 | Inflation Calculator.” U.S. Official Inflation Data, Alioth Finance, 26 May. 2018, https://www.officialdata.org/1940-dollars?amount=100. | 526 | 1,844 | {"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} | 2.90625 | 3 | CC-MAIN-2018-22 | latest | en | 0.846834 |
https://jp.maplesoft.com/support/help/errors/view.aspx?path=ModuleType | 1,642,849,309,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303845.33/warc/CC-MAIN-20220122103819-20220122133819-00199.warc.gz | 385,822,552 | 31,923 | ModuleType - Maple Help
ModuleType
refine the type checking when an object is used as a type
Calling Sequence module() export ModuleType, ...; ... end module;
Description
• A module can be passed as the type specification to a call to type or in a :: expression. By default a module is considered the type of another module if and only if they both share the same module definition. This is true of all objects of the same class. The ModuleType method allows a module to refine the type check when a module is given as a type.
• If a module that defines a ModuleType procedure is given as the type, and the given expression is a module matching that type, the ModuleType procedure will be called to further refine the type checking.
• The calling sequence of a ModuleType procedure is:
> export ModuleType::static := proc( M, typ, arg1, ... )
M : is the module that is being tested to see if it matches type typ
typ : is the type whose ModuleType procedure was called
arg1, ... : additional argument specified with the type, typ(arg1, ... )
• The ModuleType routine should not make type checks using its containing module as a type. This can lead to infinite recursion.
• An object can implement a type override to allow the object to determine if it matches various types. For more information, see the object/builtins page.
Examples
> module MyInt() option object; local value; export set::static := proc( obj::MyInt, v::integer ) obj:-value := v; end; export ModuleType::static := proc( obj, typ, arg ) if( _npassed = 2 ) then true; elif ( arg = 'prime' ) then :-type( obj:-value, 'prime' ); elif ( arg = 'odd' ) then :-type( obj:-value, 'odd' ); elif ( arg = 'even' ) then :-type( obj:-value, 'even' ); end; end; export ModuleApply::static := proc( ) Object( MyInt, _passed ); end; export ModuleCopy::static := proc( self::MyInt, proto::MyInt, v::integer, \$ ) if ( _npassed = 2 ) then self:-value := 0; else self:-value := v; end; end; end: i2 := MyInt( 2 ): i5 := MyInt( 5 ): i9 := MyInt( 9 ):
> $\mathrm{type}\left(\mathrm{i2},\mathrm{MyInt}\right)$
${\mathrm{true}}$ (1)
> $\mathrm{type}\left(\mathrm{i2},'\mathrm{MyInt}'\left(\mathrm{prime}\right)\right)$
${\mathrm{true}}$ (2)
> $\mathrm{type}\left(\mathrm{i2},'\mathrm{MyInt}'\left(\mathrm{even}\right)\right)$
${\mathrm{true}}$ (3)
> $\mathrm{type}\left(\mathrm{i2},'\mathrm{MyInt}'\left(\mathrm{odd}\right)\right)$
${\mathrm{false}}$ (4)
> $\mathrm{type}\left(\mathrm{i5},\mathrm{MyInt}\right)$
${\mathrm{true}}$ (5)
> $\mathrm{type}\left(\mathrm{i5},'\mathrm{MyInt}'\left(\mathrm{prime}\right)\right)$
${\mathrm{true}}$ (6)
> $\mathrm{type}\left(\mathrm{i5},'\mathrm{MyInt}'\left(\mathrm{even}\right)\right)$
${\mathrm{false}}$ (7)
> $\mathrm{type}\left(\mathrm{i5},'\mathrm{MyInt}'\left(\mathrm{odd}\right)\right)$
${\mathrm{true}}$ (8)
> $\mathrm{type}\left(\mathrm{i9},\mathrm{MyInt}\right)$
${\mathrm{true}}$ (9)
> $\mathrm{type}\left(\mathrm{i9},'\mathrm{MyInt}'\left(\mathrm{prime}\right)\right)$
${\mathrm{false}}$ (10)
> $\mathrm{type}\left(\mathrm{i9},'\mathrm{MyInt}'\left(\mathrm{even}\right)\right)$
${\mathrm{false}}$ (11)
> $\mathrm{type}\left(\mathrm{i9},'\mathrm{MyInt}'\left(\mathrm{odd}\right)\right)$
${\mathrm{true}}$ (12)
Compatibility
• The ModuleType command was introduced in Maple 16. | 1,077 | 3,513 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 24, "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} | 2.53125 | 3 | CC-MAIN-2022-05 | longest | en | 0.487059 |
https://software-testing.com/topic/867156/arithmetics-of-indicators | 1,660,352,930,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571847.45/warc/CC-MAIN-20220812230927-20220813020927-00578.warc.gz | 494,354,682 | 33,328 | # arithmetics of indicators
• Question of the arithmetics of indicator C. Let
``````int *p;
int *q;
int *o;
int i;
o = NULL;
``````
Let right. `p` and `q` points to the different elements of the same body. Permissibility of expression (i, naturally, is defined):
``````p+i; q-i; p-q;
``````
Is it permissible to:
``````p-o;
o-p;
``````
What happens if you're allowed? On the one hand `o` It doesn't point to the same body, and it doesn't, but it's also normal. `0`and the first one, I think so.
• No, not allowed
Standard C++ says http://eel.is/c++draft/expr.add#5 that if the two indicators do not belong to one body, the conduct is not defined.
The result of the deduction of the two indexes is the number of elements between them. If `p` Not flat `sizeof(int)`the expression `p-o` must return the fragmented number of elements that cannot be.
The expression `p-NULL` It could only be true if `NULL` - That's it. `#define NULL 0`and not valid if, for example, `NULL` defined as `(void*)0` (in C). Since the standard does not indicate how to be defined `NULL`it should be assumed that the expression `p-NULL` No, it's not.
Note: expression `p-0` Alive and still `p`but it only works for a whole number `0`
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2 | 379 | 1,263 | {"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} | 2.765625 | 3 | CC-MAIN-2022-33 | latest | en | 0.889913 |
https://www.exceldemy.com/excel-text-formula/ | 1,679,809,899,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945433.92/warc/CC-MAIN-20230326044821-20230326074821-00628.warc.gz | 851,470,148 | 68,525 | # How to Use Excel TEXT Formula (4 Suitable Methods)
Anything that you write is a form of text. They come in many different forms and can be represented in many different ways. In Excel, we can present values in a particular format using the TEXT formula or function. In today’s session, I’m going to show you a different use of the function in Excel. Before diving into the big picture, let’s get to know today’s workbook. You will find a few sheets (4 sheets in particular) in the workbook. All will represent various forms of values. But the basic table will remain the same. A total of four columns, Example Input, Desired Value Format, Formula, and Result have been used
## Introduction to Excel TEXT Function
• Function Objective:
TEXT function is used to convert a value to text in a specific number format.
• Syntax:
=TEXT(value, format_text)
• Arguments Explanation:
ARGUMENT REQUIREMENT EXPLANATION
value Required Value in a numeric form that has to be formatted.
format_text Required Specified number format.
• Return Parameter:
A numeric value in a specified format.
## 4 Suitable Methods of Using TEXT Function in Excel
In Microsoft Excel, the TEXT function is generally used to convert a numeric value to a specified format for various purposes. In this article, you’ll get to learn how you can use this TEXT function effectively in Excel with appropriate illustrations. You’ll learn more about the methods and different formats to use the TEXT function with ease in the following sections of this article.
### 1. Formatting Number Values using the TEXT Function
You may need to format different number values for different forms of representation. First, here are some notable formats that are frequently used
In the example sheet of Numbers, we have a few example inputs and some desired input formats. Let’s see how we can achieve our desired format.
#### 1.1. Choose Decimal Points
Now, for a given number you may need to choose up to how many decimal places you want to see. For the time being, let, you need to set up to 2 decimal points. Then, the formula will be
`=TEXT(B5,"#.00")`
Again, “#” denotes the entire number before decimal points. Regardless of how many digits your number has before the decimal, you need to use only one “#”. After the decimal points, I set two 0s (zeroes), since I wanted two decimal places. The number of 0s will be as many places as you want to see.
It gave the result. We got the value with up to 2 decimal places. There is another one similar to this formation. Let’s do the same for that.
#### 1.2. Scientific Format
Moreover, you may need to form a number in a scientific format. Usually, we prefer any number presented in the number E+ n digit as a scientific one. You can pronounce it as the number E to the power n.
Here are the number formula that will be
`=TEXT(B7,"0.0E+0")`
I wanted to go up to 1 decimal place (you can choose your format) before “E+” and then the number of powers. Let’s write it in Excel.
The larger number is now in a form of a much shorter and quicker readable format. Do the same for the next value as well.
#### 1.3. Decimal in Division Notation
Next, all our decimal values come from some division. Whenever you divide any value, the remainder forms the decimal places.
To write in division notation formula is as below
`=TEXT(B9,"0 ?/?")`
0 for the full value result value(before the decimal point), ?/? for the formation of the digits to present the remainder. Since it’s not known what will be the digits to represent the remainder as division so ? is used
Do the same for the next example value as well.
#### 1.4. Adding Any Digits Before n Number
Furthermore, you can add any number of digits before a given number, the formula for that
`=TEXT(B11,"000000000")`
N can be any number, you want. If you want a 3-digit number within “ “ write 000.
Here I’ve wanted to represent 12 as a number of starting seven 0. So within the “ “ I’ve written nine 0s. 12 will replace the last two zeros and the rest of the seven zeros will come ahead of 12.
You can also write any alphabet as well. The letter will be shown within the text once you have put that here within the formatted text.
In this example, I add ‘C00100’ ahead of 282. You can do the same by choosing your suitable letter or digit.
#### 1.5. Represent Phone Numbers
Afterward, you can represent any number as a phone number.
`=TEXT(B13,"(###) ###-####")`
In the USA you will find a phone number of 10 digits. The first three are the area number, then the three digits of the exchange code, last four are the line number. Usually, the area code is written within bracket () and the exchange code and line number are separated by using a dash ( ).
It gave the above result. Let’s do the same for the rest of the examples.
### 2. Formatting Currency Using the TEXT Function
At times, when dealing with currency, we need to convert currency in Excel very frequently. It is quicker and handier if we can use any formula to convert currency. In this article, I will show you how to format currency using the TEXT function in Excel.
#### 2.1. Comma Separated Dollar Currency
Now, your formula to represent such a way is as follows
`=TEXT(B5,"\$ #,##0")`
Here the value will be started with a \$ sign at the beginning and after every 3 digits, a comma will take place.
It gave the presentation we wanted. For the rest of the two, use the same formula and you will get the answer.
#### 2.2. Currency Value in Decimal Points
The formula will be the same as the previous one, just add a decimal point and zeros up to the place you want to see. Let we want to see up to two decimal points
`=TEXT(B8,"\$ #,##0.00")`
Writing the formula in Excel we will find the result for our example as in the image below.
Do the same for the rest of the example input.
### 3. TEXT Formula for Percentage Formation
In this case, our goal is to learn how to use the TEXT function in Percentage formulas. We can learn this by first creating a percentage column and then using it under a certain condition. To use percentage form, we have to learn to convert a normal numeric value into a percentage. Technically Excel will convert any input data into a percentage by multiplying it by 100 and adding a percentage symbol (%) on the right if you opt to choose percentage formatting. But you can also convert a number directly to a percentage value without letting it be multiplied by 100 in Excel. The steps of this method are as follows.
#### 3.1. Percentage Formation
We can convert a decimal number into a percentage format. To do so, use the formula written below
`=TEXT(B5,"0%")`
This will convert the decimal value into percentage format. Write it in Excel.
Use the formula for the rest of the examples under this section.
#### 3.2. Percentage in Decimal Points
The formula will be the same as the previous one, just add a decimal point and zeros up to the place you want to see. Let we want to see up to one decimal point
`=TEXT(B7,"0.0%")`
Here I’ve set only up to 1 decimal place, you can choose your preferred one.
Let’s do the same for the next two examples as well. Here for example purpose, we have less amount of value. But, in the real scenario, you may have a chunk of values, use the AutoFill feature then.
### 4. TEXT Function for Date-Time Values
To format a timestamp, we have to use HH (Hour), MM (Minute), SS (Second), and AM/PM characters to define the required parameters. Here you have to keep in mind- in a 12-hour clock system, you have to input the AM/PM exactly in “AM/PM” text, not in “PM/AM” format at all, otherwise, the function will return with an unknown text value- “P1/A1” at the defined position in the timestamp. In the following screenshot, a fixed timestamp has been shown in different but common formats after formatting. You can easily convert a 12-hour clock system to a 24-hour clock system and vice-versa by using this TEXT function.
#### 4.1. Time in International Standard
To convert your local time into the 24-hour standard form you can use the formula –
`=TEXT(B5,"hh:mm")`
HH: Hour
MM: Minutes
Use AM/PM in your input time to let Excel understand the right time.
Write the formula for this example in the sheet.
As our provided time was 6:00 PM it gave us 18:00, the format we were anticipating. The next example uses AM time.
#### 4.2. Only Time from Full Date-Time
If you use the NOW function you will find the current date and the time. To see the time only write the formula below
`=TEXT(B6,"hh:mm")`
Similar to the previous one, since the previous one was also shown the time. Write the formula for this example.
#### 4.3. Time Month Day Format
If you want to show time -month- day from a particular time just use the function below
`=TEXT(B7,"HH:MM O'Clock, MMMM DD")`
HH: MM represents the time
MMMM represents the month name
DD represents Date
For a better understanding of time, I used O’Clock, so that you can differentiate that it is time value. Let’s write the formula for the example time input. This input time has been generated using the NOW function.
We have found the result in time month and date format.
#### 4.4. Date Separating by ‘/’
More often you will write the date separating by “-“, but if you want to write it using “/”, then use the formula –
`=TEXT(B8,"MM/DD/YYYY")`
MM: Month
DD: Date of the month
YYYY: Year ( this will show the full 4-digit year, use YY to show 2 digits of the year)
#### 4.5. Day Name–Month-Year Format
You may need to form the date in a way of the day of the week, month name, and year. The formula for that will be
`=TEXT(B9,"DDDD,MMMM YYYY")`
DDDD: Day name
MMMM: Month name
YYYY: Year
Here my intention was to show the day name, month name, and year, that’s why I have written in this way. You can choose your suitable format.
#### 4.6. Month-Date-Year Format
By the time we are in this section, you have already understood how to do this task. Though I’m writing the formula for you. I suggest you write your own first and then check, that will evaluate your understanding.
The formula will be
`=TEXT(B11,"MMMM DD,YYYY")`
Hope you understand the meaning of MMMM, DD, YYYY. Let’s see the result of the example.
## Conclusion
That’s all for today. I’ve tried listing a couple of ways of using Excel’s TEXT formula. Henceforth, follow the above-described methods. We will be glad to know if you can execute the task in any other way. Follow the ExcelDemy website for more articles like this. Please feel free to add comments, suggestions, or questions in the section below if you have any confusion or face any problems. We will try our level best to solve the problem or work with your suggestions. | 2,519 | 10,729 | {"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} | 3.375 | 3 | CC-MAIN-2023-14 | latest | en | 0.802512 |
https://tipsfolder.com/was-weight-goliaths-spear-head-f1e8747501afc2e80ef2771e6a2b580f/ | 1,725,998,536,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651318.34/warc/CC-MAIN-20240910192923-20240910222923-00039.warc.gz | 540,545,344 | 19,178 | # What was the weight of goliath’s spear head?
The battle between David and Goliath took place in Israel’s Valley of Elah. So, how did it turn out? If you’d like to send pictures, please send them as links.
As a result, if it had four blades and was 4″ at the base, a 15-pound spear head made of iron would probably be around 16-18″ long.
## What is the average spear’s weight?
Spears weigh between 1 and 2 kg, while halberds weigh between 3 and 4 kg, but there’s a lot of range. In medieval art, I recently did some spear length comparisons, and the length appears to have gone from a head taller (just a little short of 2 meters) to nearly three heads taller (3 meters).
## What is the weight of a spearhead?
The diamond-shaped cross-section on the head is strong. Overall length of the Big Spear Head is 15.5″, with a 10″ tip length, and a weight of 1.9 pounds.
Related Articles:
## What was the weight of Goliath?
His armor weighed 5000 shekels (41.65 kg (11.8 ounces), with 1 shekel equaling 8.33 g (0.29 oz), which excludes the brass greaves, helmet, and gorget (“target”). His iron spearhead weighed 600 shekels, or just shy of 5 kilograms (11 pounds).
## How much iron did 600 shekels weigh?
The Goliath sword was 600 shekels (approximately 15 kg) in weight. A normal longsword, on the other hand, is usually between 2 1/2 and 2 1/2 kilograms. | 376 | 1,362 | {"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} | 2.53125 | 3 | CC-MAIN-2024-38 | latest | en | 0.962423 |
https://community.myfitnesspal.com/en/discussion/10127390/mfp-vs-tdee/p2 | 1,712,943,198,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816045.47/warc/CC-MAIN-20240412163227-20240412193227-00226.warc.gz | 159,543,091 | 73,791 | # MFP vs TDEE
2»
## Replies
• Posts: 7,237 Member
lemurcat12 wrote: »
mirrim52 wrote: »
What do you have your weight loss goal set for in MFP, and what calorie deficit did you choose in Scooby?
But yes, both methods should come out the same if the goals are the same and you are logging exercise properly. TDEE just averages the included exercise out over the week. The problem is, many people compare MFP set to 2 lb loss a week to TDEE with a 500-600 calorie deficit, which is comparing apples to oranges.
I have MFP set at 1 pound per week. I set my TDEE goal at lose fat - 20% calorie reduction.
MFP has me eating 1460 per day. My trainer assures me I safely burn about 260 calories during one of our sessions. That would make my goal on a work out day 1720. TDEE has my calorie goal as 1968. That seems like a considerable difference. Now, if I change my MFP to lightly active, my daily calories jump to 1700 which would be 1960 on exercise days.
TDEE isn't really defined by the calculator, but what you burn (living and through exercise) over the week. So if you think you burn a particular amount more than the calculator is accounting for you'd include it. The only differences (as others have said) would be due to the difference in your goal--for many 20% of TDEE is a bit less than 1 lb/week, whereas the MFP goal is for more. MFP's calculator may also give a different number than some other calculator (it can't include BF%) and may be different than observed TDEE.
MFP apparently has calculated that your maintenance if sedentary is 1960, and if lightly active (but no other exercise) is 2300. This means it thinks your BMR is about 1633. (I thought lightly active was a multiplier of 1.5, which would give 2450, but not sure about that.)
TDEE has your maintenance (with activity) at about 2450, which gives you 1960 with a 20% cut (-490 or just under a lb). So whatever you told it about your exercise/activity caused it to estimate about 490 calories on average per day beyond just being sedentary (if all else is equal). This is consistent with the difference I get from Scooby between sedentary and assuming 3-5 hours/week of strenuous exercise.
So the question seems to be if 490 calories every day is a good estimate of what you actually do on average (with the understanding that it's more some days and less others). If so, that's what you should be logging in MFP too, so it should even out. If not, then you might want to adjust the TDEE number if you go that way.
It sounds as if you are questioning whether your actual workout burns are as high as Scooby is estimating. The Scooby activity entries are pretty imprecise (arguably this is appropriate, since you can't really know it and need to adjust anyway). Also, they probably add in some daily activity for people who say they are more active beyond the workouts, I don't know how their calculation for activity works. If you are concerned that the amount you burn in walking around plus exercise isn't so high, don't use the Scooby number.
I'd say try it and adjust if your results aren't good, though.
The point is she is hungry. So, probably not eating as much as she could be, while still seeing weight loss, but increasing fuel for performance and preserving lean body mass.
I understand that calculators are estimates. Just correcting what seems like an easy mistake.
• Posts: 7,237 Member
deksgrl wrote: »
Again, if you set your TDEE to SEDENTARY, when you do indeed exercise, then it is not calculated properly.
You need to include your exercise in TDEE, so choose a higher activity level.
I understand that. My original post may have been confusing. When I stated I have my activity level as sedentary, I meant that is what I currently have in MFP - as a reference to show the difference between MFP and TDEE.
I believe I correctly identified my exercise in the TDEE (indicating 3-5 days of moderate exercise - I didn't choose the next highest level because that was 5-6 days of strenuous which I don't always accomplish) and the MFP (sedentary because I have a desk job).
Oh, okay, I misunderstood.
You could probably eat a more and still lose at a good rate. And, you will be able to preserve lean body mass, and have more fuel for performance.
• Posts: 30,886 Member
deksgrl wrote: »
The point is she is hungry. So, probably not eating as much as she could be, while still seeing weight loss, but increasing fuel for performance and preserving lean body mass.
I said nothing about what she should eat (except encouraging her to try the higher TDEE and adjust if needed). I was just trying to explain why the numbers were different and what they were based on. So I am not sure what the "mistake" is here or really your response to my post.
• Posts: 30,886 Member
@lemurcat12 Thank you. Actually, thank you to everyone. My concern is that I am happy with my rate of loss...will changing my calories to that extreme drastically change my rate of loss? I don't want to be hungry but I also don't want to ruin my momentum.
If it reduces your rate of loss you can always just lower the calories a little until you are happy with it again. The thing you have to watch out for is that often adding a new workout program can reduce losses at first anyway, since you get more water weight. But on the other hand how many calories you burn is dependent on your intensity and movement throughout the day, and that could increase/be better if you aren't feeling so hungry.
• Posts: 763 Member
@lemurcat12 Thank you. Actually, thank you to everyone. My concern is that I am happy with my rate of loss...will changing my calories to that extreme drastically change my rate of loss? I don't want to be hungry but I also don't want to ruin my momentum.
If you increase your calories, your weight loss will likely slow a bit. How much depends on your body and how much more you eat. But, if you aren't hungry all the time, you are more likely to stick with the changes, put more effort in to your workouts, and be less likely to binge or "cheat". So the tradeoff is usually worth it.
• Posts: 7,237 Member
lemurcat12 wrote: »
deksgrl wrote: »
The point is she is hungry. So, probably not eating as much as she could be, while still seeing weight loss, but increasing fuel for performance and preserving lean body mass.
I said nothing about what she should eat (except encouraging her to try the higher TDEE and adjust if needed). I was just trying to explain why the numbers were different and what they were based on. So I am not sure what the "mistake" is here or really your response to my post.
Oh, because I'm a dope and at a quick glance I thought you had quoted my post, and so I was clarifying my point. So, nevermind. lol.
• Posts: 2,034 Member
Thank you again everyone for your help. I have decided to adjust my macros for a start. I changed to a split that gives me more protein to help me feel full longer. I will try that for two weeks then go from there.
• Posts: 16,049 Member
I think I get what you mean.... The only exercise I get Is walking. Some days I might walk over 20,000 steps, other days it might only be 10,000. It's never exactly the same. So I set my activity level to sedentary, let fitbit sync to mfp and then I can decide how much or how little to eat that day.
Some days I burn 350 calories, other days 850. I let mfp do the math for me
• Posts: 880 Member
My own experience with this is that by increasing my calories (and having the content of those calories be nutritious and not just sugar) I was no longer hungry AND my workouts were more productive. My rate of loss slowed to a half pound a week ... on the up side, I had more energy, made more gains in strength, mobility and balance, and made appreciable changes (for the good) in body measurements. Don't be a slave to the scale
• Posts: 459 Member
I have noticed that even though strength training doesn't burn near as many calories as the HIIT exercise I do, it makes me hungrier- Perhaps because I am newer to weight lifting. I added a whey protein shake to my menu - it is very filling and the extra protein seems to help. | 1,937 | 8,117 | {"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} | 2.546875 | 3 | CC-MAIN-2024-18 | latest | en | 0.972823 |
http://stackoverflow.com/questions/5448493/fp-growth-algorithm?answertab=active | 1,448,937,630,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398464396.48/warc/CC-MAIN-20151124205424-00258-ip-10-71-132-137.ec2.internal.warc.gz | 220,237,145 | 19,720 | # fp growth algorithm
I have to implement FP-growth algorithm using any language. The code should be a serial code with no recursion. Is it possible to implement such algorithm without recursion? I am not looking for code, I just need an explanation of how to do it.
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How is your question related to C#, C, C++, or Java? What is an "fp growth algorithm"? I'll improve your tags. Mihran answer captures almost everything which could be said to your rather unspecific and general question. If you're interested in more information, please improve your question. – jmg Mar 27 '11 at 11:38
its because i can implement the algorithm with any of these languages.. – liz Mar 27 '11 at 12:54
But then, why did you not add python, ruby, lisp, haskell, etc.? – jmg Mar 27 '11 at 12:57
Don't spam tags please. This isn't a relevant question to the people following the C family of tags. – Bill the Lizard Mar 27 '11 at 13:22
Here is an very clear explanation of how the code works. It looks like you have to build a tree and validate it.
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FPGrowth is a recursive algorithm. Like some other people said here, you can always transform an algorithm into a non recursive algorithm by using a stack. But I don't see any good reasons to do that for FPGrowth.
By the way, if you want a Java implementation of FPGrowth and other frequent pattern mining algorithms such as Apriori, HMine, Eclat, etc., you can check my website. I have implemented more than 40 algorithms for frequent pattern mining, association rule mining, etc.:
http://www.philippe-fournier-viger.com/spmf/
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That's impressive, great job! – ksiomelo Dec 9 '12 at 13:19
thanks Phil, i found your algorithms library from your comment here, – Aha Dec 8 '14 at 8:09
You can take look at the concept & implemenntation FP growth algoithm in Mahout
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You can probably visit http://code.google.com/p/lofia/ to get something over FP Tree. This is for longest frequent itemset mining.
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Assuming by "FP growth Algorithm" you mean frequent Pattern growth algorithm, I would point you over to this document that gives a decent explanation on how it works.
http://www.florian.verhein.com/teaching/2008-01-09/fp-growth-presentation_v1%20%28handout%29.pdf
I wonder though, is this homework related?
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yes it is HW related.but i am not seeking solution or code, i need some recommendations so i continue with the project or change the topic and choose another algorithm to implement... – liz Mar 27 '11 at 12:53
I don't know what is the algorithm you talking about. But everyting what is possible with recursion it is possible also without it. You can implement such kind of algorithms using stack.
- | 660 | 2,653 | {"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} | 2.53125 | 3 | CC-MAIN-2015-48 | latest | en | 0.91727 |
http://math.stackexchange.com/questions/263196/find-number-of-ways-to-distribute | 1,467,013,492,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783395679.18/warc/CC-MAIN-20160624154955-00134-ip-10-164-35-72.ec2.internal.warc.gz | 206,265,725 | 16,489 | # Find number of ways to distribute
Find number of ways in which 16 apples can be distributed among four persons so that each of them gets at least one apple.
I worked out like this: Since every person must get 1 apple, we give one apple to each beforehand. That leaves us with 12 apples. Now there are C(12, 4) ways to give these 12 apples to 4 peoples(Am I correct). This makes for a solution of 495, but the answer is given as 455. Help!!!
-
You’re right to distribute $4$ of the apples first, leaving $12$ to be distributed arbitrarily, but after that you’ve gone astray. This is a stars-and-bars problem, and as you’ll see at the linked article, the correct answer is
$$\binom{12+4-1}{4-1}=\binom{15}3=455\;.$$ | 194 | 719 | {"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 | 4 | CC-MAIN-2016-26 | latest | en | 0.932426 |
http://forums.wolfram.com/mathgroup/archive/1994/Nov/msg00041.html | 1,601,121,291,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400241093.64/warc/CC-MAIN-20200926102645-20200926132645-00391.warc.gz | 49,215,411 | 8,176 | • To: mathgroup at christensen.cybernetics.net
• Subject: [mg245] Question about function definitions
• From: Scott Herod <sherod at boussinesq.Colorado.EDU>
• Date: Mon, 28 Nov 1994 13:10:01 -0700 (MST)
```Here is a question related to the issue of multiple function
definitions. First let me present some code segments with examples of
their execution.
-----------File delsol.m:
delay[num_, T_, z_] := Module[{},
f[s_] := 0.7 + 0.1 s;
witchargs = {};
For[k = 0, k <= num, k++,
sol = NDSolve[{y'[t] + y[t] == f[t] ( 2 - f[t]^z),
y[T*k] == f[T*(k+1)]}, y, {t,T*k,T*(k+1)}];
(* Clear[f]; *)
f[s_] := Evaluate[(y /. Flatten[sol])][s-T];
witchargs = {witchargs, t <= T*(k+1), Evaluate[(y /. Flatten[sol])][t]}
];
g[t_] = Apply[Which, Flatten[witchargs]];
];
---------Example Run
Mathematica 2.2 for Solaris
License valid through 28 Nov 1995.
-- Open Look graphics initialized --
In[1]:= <<delsol.m
In[2]:= delay[3,2,3]
In[3]:= ??f
Global`f
f[s_] := 0.7 + 0.1*s
f[s\$_] := Evaluate[y /. Flatten[sol]][s\$ - 2]
----------Discussion
I am computing solutions to a differential-delay equation discussed by
R.M. May in a 1981 paper. The idea is to iterate on the set of smooth
functions defined on [0,T]. Anyway, this code doesn't work because
the NDSolve routine always sees the first definition of f with the "s"
not the definition with the "s\$". Of course a fix is to clear f each
time through the loop, but I would be interested in understanding why
the \$ is appended. Especially considering the next example.
------------File vartest.m
tes[num_] := Module[{},
f[t_] := Sin[t];
For[k = 1, k <= num, k++,
y = f';
f[t_] = y[t - 2];
]
]
------------Example run
Mathematica 2.2 for Solaris
License valid through 28 Nov 1995.
-- Open Look graphics initialized --
In[1]:= <<vartest.m
In[2]:= tes[3]
In[3]:= ??f
Global`f
f[t_] = -Cos[6 - t]
------------Discussion
What is it about the above routine which is different from the first
so that the "\$" is not added to the variable \$t\$?
Scott A. Herod
Program in Applied Mathematics | 645 | 2,050 | {"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} | 3.15625 | 3 | CC-MAIN-2020-40 | latest | en | 0.724863 |
https://www.gradesaver.com/textbooks/math/algebra/differential-equations-and-linear-algebra-4th-edition/chapter-2-matrices-and-systems-of-linear-equations-2-4-row-echelon-matrices-and-elementary-row-operations-problems-page-155/5 | 1,576,282,679,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540569332.20/warc/CC-MAIN-20191213230200-20191214014200-00405.warc.gz | 673,063,549 | 12,746 | ## Differential Equations and Linear Algebra (4th Edition)
A matrix is a row-echelon matrix if: 1. If there are any rows consisting only of zeros, they are all together at the bottom of the matrix. 2. The first nonzero element in any nonzero row is a $1$. 3. The leading $1$ of any row below the first row is to the right of the leading $1$ of the row above it. A matrix is a reduced row-echelon matrix if it is a row-echelon matrix and any column that contains a leading $1$ has zeros everywhere else. Hence here we can see it is in row-echelon form but the first row has other non-zero elements apart from its leading $1$ thus it is not in reduced row-echelon form. | 170 | 668 | {"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} | 3.546875 | 4 | CC-MAIN-2019-51 | latest | en | 0.937754 |
https://www.physicsforums.com/threads/quick-question-about-taking-a-derivative.360971/ | 1,508,610,988,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824824.76/warc/CC-MAIN-20171021171712-20171021191712-00224.warc.gz | 962,845,085 | 14,573 | # Quick question about taking a derivative
1. Dec 6, 2009
### tnutty
So say for a line integral, the curve C is given by y = sqrt(x), from point (1,1) to (4,2);
In my integral I have some integrand* dy.
Say I wanted to change the dy to dx.
From whats given :
y = sqrt(x);
dy = 1/(2sqrt(x) ) dx;
I could just substitute that instead for dy.
But whats the difference if I do this :
y = sqrt(x)
y^2 = x
2y dy = dx
dy = dx/2y
So how is the former different from the latter. I mean I see that y is integrated for the
second one, but what does it represent? Can you explain me the difference between the
two, does not have to be geometrically, but will be appreciated.
2. Dec 6, 2009
### grief
Since y=sqrt(x), 1/(2sqrt(x) ) dx =dx/2y. However, the first form is the one you want to use, because it's strictly in terms of x.
3. Dec 6, 2009
### LCKurtz
The parabola can be represented by either y = sqrt(x) and x = y2 on that interval. Which you use might be determined by what the rest of the integrand is. Or maybe you have an aversion to square roots in integrals. Use whichever one looks easiest in your problem. | 327 | 1,129 | {"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-2017-43 | longest | en | 0.944926 |
http://docplayer.net/23998756-A-function-f-whose-domain-is-the-set-of-positive-integers-is-called-a-sequence-the-values.html | 1,544,780,103,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825512.37/warc/CC-MAIN-20181214092734-20181214114234-00008.warc.gz | 81,760,553 | 27,258 | # A function f whose domain is the set of positive integers is called a sequence. The values
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1 EQUENCE: A fuctio f whose domi is the set of positive itegers is clled sequece The vlues f ( ), f (), f (),, f (), re clled the terms of the sequece; f() is the first term, f() is the secod term, f() is the third term,, f () is the th term, d so o I tretig sequeces, it is customry to use subscript ottio isted of fuctiol ottio Thus, for give sequece f, f () is deoted by, f () by, f () by,, d f () by I this ottio, we represet the sequece s,,,,, Exmples : Write the first four terms, the ith term d the twetieth term of the sequece whose th term is: () () () ) ( () ( ) olutios: To fid the first four terms we substitute, successively,,,, i the formul for The ith d twetieth terms re foud by substitutig 9 d 0 i the formul for The results re: First four terms Nith term Twetieth term (),,, (),, 5, (),, (),, 9,
2 Rther th givig explicit formul for, sequeces re sometimes specified by sttig method for fidig the terms Oe such method is clled recursio formul or recurrece reltio A recursio formul is formul tht gives i terms of oe or more of the terms tht precede The strtig vlue, or vlues, must lso be give We illustrte with some exmples Exmples : Fid the first four terms d the th term for the sequece specified by: () d, k,,, () d ( ), k,,, olutios: () We re give From the recursio formul,, Bsed o this ptter, we coclude tht 5, 5 6, d, i geerl () Proceedig s i (),,,, Bsed o this ptter, we coclude tht 5 5 0, , d, i geerl, ( )( )! * Exmple : The sequece defied recursively by: d * The product of the first turl umbers is clled fctoril d is deoted by!;!,!,!,!, etc
3 is kow s the Fibocci sequece The terms of the sequece re clled the Fibocci umbers The first ie Fibocci umbers re It c be show tht,,,, 5, 8,,, ( 5) ( 5) 5 The Fibocci umbers pper i wide vriety of pplictios rgig from the biologicl scieces to rt d rchitecture The sequece of rtios of cosecutive Fibocci umbers pproches the umber 5 which is the so-clled golde rtio used by the Greeks thousds of yers go Here is other exmple of sequece defied by method for fidig the th term Exmple 5: List the first six terms of the sequece whose th term is the th prime umber olutio:,, 5, 7, 5, 6 It is iterestig to ote tht there is o kow formul for However, this is perfectly well defied sequece sice is defied for ech positive iteger I tretmets of sequeces it is commo to mke sttemets such s cosider the sequece whose first four terms re,,,, It is left to the reder: () to discover the ptter estblished by the terms, d () to determie formul for the geerl term I this cse, the ptter ppers to suggest tht 5 / 5, 6 / 6, d, i geerl, / However, cosider the sequece defied by
4 9 ( )( )( )( ) 5 As you c verify,, /, /, / d 5 9 The poit of this exmple is: listig the first few terms of sequece does ot, i fct, determie specific sequece To determie specific sequece you must be give formul for the th term or you must be give some method for determiig the th term (eg, recursio formul) ttemets such s determie the sequece whose first four terms re,,, must be qulified by sttemet such s ssume tht ptter cotiues s idicted Exmples 6: () The first four terms of sequece re give () ssume tht the ptter cotiues s idicted d fid the geerl formul for ; d (b) fid formul for i which / 5 π,, 5, 6 7, 8 () The first five terms of sequece re give () ssume tht the ptter cotiues s idicted d fid the geerl formul for ; d (b) fid formul for i which ,,,,, 5 olutios: () () (b) π ( )( )( )( ) 8 () () () ( )
5 (b) 76 7 ( ) ( )( )( )( )( 5) 0 70 Limits of sequeces: Let be give sequece We re ofte iterested i the behvior of the terms s gets lrger d lrger, symbolized by For exmple, cosider the sequece / ice <, < for ll ; the terms re decresig i vlue Also, if is lrge (sy,000,000, or 0,000,000), / is close to 0 Thus, we coclude tht the terms of the sequece / re decresig d tedig to 0 Here is grph of the sequece If we simply plot the vlues / o the rel lie, we get 0 d we c see tht the terms re heded towrd 0; they ever pss zero becuse / > 0 for every positive iteger Bsed o this behvior, we sy tht the limit of / s teds to ifiity is 0 This is symbolized by lim 0 or by 0 s I geerl, if the terms of the sequece pproch umber L s, the L is clled the limit of the sequece As suggested i the exmple bove, this is symbolized by
6 lim L or by L s Exmples 7: Determie whether the give sequece hs limit s () () () ) ( () ( ) olutios: () We write out the first severl terms of the sequece to get ide of its behvior: 5 6,,,,, 5 It ppers tht the terms re gettig closer d closer to 5 6 We c justify this coclusio by otig tht For lrge, is close to sice / is close to 0; s () As i (), we write out the first severl terms 5 0,,,,,,
7 It ppers tht the deomitors of these frctios re growig much fster th the umertors d so we guess tht 0 s This coclusio is justified by writig d otig tht / 0 d / 0 s Therefore 0 s () Writig out the first severl terms, we hve,,,,, The terms simply oscillte betwee d ; the terms oe umber L so the limit does ot exist re ot gettig close to () The first severl terms of this sequece re,, 9, 6, 5, 6, The terms get rbitrrily lrge i bsolute vlue d oscillte betwee positive d egtive; the sequece does ot hve limit
8 Two specil sequeces: We ll coclude this sectio o sequeces by cosiderig two specil types of sequeces Arithmetic sequeces: A rithmetic sequece is sequece i which the differece of successive terms is costt d Tht is, sequece is rithmetic sequece if d for every positive iteger The umber d is clled the commo differece for the sequece Note tht rithmetic sequece is defied by recursio formul Arithmetic sequeces re lso clled rithmetic progressios Exmples 8: Determie whether the give sequece is rithmetic If it is, give the commo differece (), 5, 8,,,, (),, 9, 6,,, (), 8,,, 6, olutios: () [ ] ( ) The sequece is rithmetic; d ; 9 5 The sequece is ot rithmetic () () [ ] ( ) The sequece is rithmetic with d uppose is rithmetic sequece with first term differece d We will use the recursio formul to fid formul for defiitio, we kow tht d Therefore 5 d commo From the d, d d, d d, d d, d we coclude tht
9 ( ) d Exmple 9: Fid the twelfth term of the rithmetic sequece whose first three terms re, 5, 9, olutio: d d Therefore, ( ) d () 5 Geometric sequeces: A geometric sequece, or geometric progressio, is sequece i which the rtio of successive terms is ozero costt r Tht is, is geometric sequece if d oly if r for every positive iteger The umber r is clled the commo rtio Note tht geometric sequece is defied by recursio formul: r Exmples 0: () The sequece 8,,,,, is geometric sequece Fid the commo rtio d give the fifth term of the sequece () The sequece 5,,,, is geometric sequece Fid the commo rtio 8 d give the sixth term olutios: () The commo rtio is r The fifth term is 5 8 5/ () The commo rtio is r The sixth term is the fifth term times 5 / ; the fifth term is ( / )( 5/8) 5/6 Therefore, the sixth term is 5 /
10 uppose tht is geometric sequece with first term We ll use the recursio formul to determie expressio for the geerl term : r, r r, r r, r, 5 r Bsed o this ptter, we coclude tht r Exmple : Fid the seveth term of the geometric sequece whose first three terms re,,, 8 olutio: The first term d the commo rtio r is / Therefore, by the formul derived bove, the seveth term is: 6 7 ( / ) 6 08 ERIE Let,,,,, be give sequece uppose we re sked to dd up ll the terms of the sequece Tht is, suppose we re sked to clculte ice we c oly dd fiite umber of umbers, it would pper tht ddig up ll the terms of sequece is impossible tsk However, there re istces where we c ssig vlue, umber, to ifiite sum We ll explore tht possibility here A sum of the form, where is give sequece, is clled series (lso clled ifiite series) ice we c dd up y fiite collectio of umbers, we ll form ew sequece by ddig up terms of the give sequece i systemtic mer
11 Let This ew sequece is clled the sequece of prtil sums of the give sequece The questios we wt to ddress re: () C we fid formul for the sequece of prtil sums derived from some give sequece? () If the swer to () is yes, does hve limit, sy, s? If the swers to () d () re yes, the we sy tht the sum of ll the terms of the sequece is ; we ve dded up ifiitely my umbers d gotte umber! Here re some exmples Exmples : Let be the sequece defied by ( ) This is the sequece The sequece of prtil sums is: 0 0,,,, Bsed o these results, we coclude tht whe is odd d 0 whe is eve The sequece c be represeted by whe is odd ( ) or by 0 whe is eve
12 Does hve limit s? Let be the sequece defied by ) ( The first three terms of the sequece of prtil sums is: Wht is for every positive iteger? Does hve limit s? Let be the sequece defied by This is the sequece of positive itegers,,,,,, The sequece of prtil sums is: 0 6 Wht is for every positive iteger? Does hve limit s? olutios: does ot hve limit; the sequece oscilltes betwee 0 d s ; The ifiite series hs fiite sum: The sequece,,,,, is rithmetic sequece with d d
13 ( ) Addig these two equtios, we get ( ) ( ) ( ) ( ) ( ) [ terms] olvig for, we get the result ( ) ( ) This sys tht the sum of the first positive itegers is The sequece of prtil sums does ot hve limit; s Arithmetic d geometric series: Let,,, be rithmetic sequece, with first term d commo differece d The, d, d, d, d, [ d] [ d] [ ( ) d] is rithmetic series The sequece of prtil sums is: ( d) d ( d) ( d) ( ) d ( d) ( d) ( d) ( ) d [ ( )] d Usig our result i () bove, formul) d ( ) ( ) (replce by i the
14 ( ) d or [ ( ) d] This is formul for the sequece of prtil sums of rithmetic series ice ( ) d, it follows tht ( ) d [ ( ) d] Therefore, c lso be writte s ( ) Now suppose tht, d commo rtio r The d,,, is geometric sequece with first term, r, r, r r r r, r is geometric series The sequece of prtil sums is: r r r r r r r Note first tht if r, the Now ssume tht r The r r r ( s) d r r r r r r r r r r
15 ubtrctig the secod equtio from the first gives r r d ( r) r Therefore, r provided r r This is formul for the sequece of prtil sums of geometric series Let s look t the behvior of s We c rewrite our formul for s r r r It is possible to show tht if < r <, the r 0 s, d it follows tht r 0 d r r Therefore, if the commo rtio r of geometric series stisfies < r <, (ie, r < ), the we defie the (fiite) sum of the ifiite geometric series r r r r to be, r < r Exmples : Fid the sum of the first 8 terms of the rithmetic sequece, 7,, 5, Fid the sum of the first 6 terms of the geometric sequece,,,, 8 Does the ifiite series 8
16 hve fiite sum? olutios: d d Usig the first rithmetic series formul, we hve 8 8 [() (8 )] (6 8) 6 Usig the secod formul, (8 ) d d r Therefore [ ] () ice r <, the ifiite geometric series hs the fiite sum r 6 Exercises: Fid the first five terms d the eighth term of the followig sequeces 0
17 ( ) (0) 8 9 ( ) ( ) 0 Fid the first five terms of the sequece defied by the give recursio formul If possible, fid formul for,,,, 5, 6, Determie whether the give sequece hs limit If it does, give the limit 7 8
18 9 0 () Determie whether the idicted sequece c be the first three terms of rithmetic or geometric sequece, d, if so, fid the commo differece or commo rtio d the geerl term, 6,,,,9, 5,,8, 6 7,65,6, 7,,, 6 8 8,,, 9,,, 9 0 7,, 8, Fid the teth term d the th term of the give rithmetic sequece, 6, 0,,, 9, 7,
19 , 7,,, 7, 65, 6, Let,,, be rithmetic sequece Fid the idicted qutities 5, d ; fid d , d ; fid d 0 0 7, d 0; fid d /, d / ; fid d Fid the teth term d the th term of the give geometric sequece 9 8,,,, 0, 6, 8,,, 06,, 6, 9, 5, Let,,, be geometric sequece Fid the idicted qutities, r ; fid d 6 6, r / ; fid d 5, r / ; fid d 6, r 05; fid d Determie whether the geometric series hs fiite sum If it does, fid it
20
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• This Set 6 turtles (prek - 1) Addition with pictures is perfect to practice addition skills. Your elementary grade students will love this Set 6 turtles (prek - 1) Addition with pictures. Students fill in missing sums for illustrated basic facts number sentences
• This Set 7 dinos (prek - 1) Addition with pictures is perfect to practice addition skills. Your elementary grade students will love this Set 7 dinos (prek - 1) Addition with pictures. Students fill in missing sums for illustrated basic facts number sentences.
• This Set 8 dinos (prek - 1) Addition with pictures is perfect to practice addition skills. Your elementary grade students will love this Set 8 dinos (prek - 1) Addition with pictures. Students fill in missing sums for illustrated basic facts number sentences.
• This Set 9 fish (prek - 1) Addition with pictures is perfect to practice addition skills. Your elementary grade students will love this Set 9 fish (prek - 1) Addition with pictures. Students fill in missing sums for illustrated basic facts number sentences.
• This Snowman - Counting Snowflakes Flashcards is perfect to practice geometry skills. Your elementary grade students will love this Snowman - Counting Snowflakes Flashcards. Flashcards numbered 1-10 with a picture of snowman and snowflakes. Could be used as flashcards, matching, and/or number recognition.
• This Snowman Dice Mat - Addition Winter is perfect to practice addition skills. Your elementary grade students will love this Snowman Dice Mat - Addition Winter. Roll the dice, write the numbers, and add buttons to the friendly snowman to solve the addition equation. Laminate for a colorful, reusable classroom activity.
• This Subtraction with Base10 Blocks (upper elem) Clip Art is perfect to practice subtraction skills. Your elementary grade students will love this Subtraction with Base10 Blocks (upper elem) Clip Art. Solve four subraction problems using base 10 blocks. Includes answer key.
• This Subtraction with Partner Problems (upper elem) Clip Art is perfect to practice subtraction skills. Your elementary grade students will love this Subtraction with Partner Problems (upper elem) Clip Art. Solve ten problems with a partner.
• This Sums 11-19 (color) Posters is perfect to practice addition skills. Your elementary grade students will love this Sums 11-19 (color) Posters. Set of 9 insect-themed posters illustrating the addition of ten ones and further ones to find sums 11-19. Correlates to common core math standards. Common Core Math: K.NBT.1
• This Sunflower (preschool/primary) Color by Number Addition is perfect to practice addition skills. Your elementary grade students will love this Sunflower (preschool/primary) Color by Number Addition. Add the sums (up to 8) and use the key to color the picture of a sunflower turned to the sun.
• This Tally Marks (primary.elem) Rules and Practice is perfect to practice graphing skills. Your elementary grade students will love this Tally Marks (primary.elem) Rules and Practice. Introduction to the concept, with several pages for practice. These are in color so you can make a transparency, but they will look fine in black and white for the students.
• This Ten Frame Apple Addition Math is perfect to practice addition skills. Your elementary grade students will love this Ten Frame Apple Addition Math. Five pages, each with two addition ten frame questions. Cut each card and make into a booklet or laminate and use in a math center. Student could also use markers to answer the ten frame questions.
• This Ten Frame Fall Leaf Addition Math is perfect to practice addition skills. Your elementary grade students will love this Ten Frame Fall Leaf Addition Math. Five pages, each with two addition ten frame questions. Cut each card and make into a booklet or laminate and use in a math center. Student could also use markers to answer the ten frame questions.
• This Thanksgiving Addition (elem) Word Problems is perfect to practice problem solving skills. Your elementary grade students will love this Thanksgiving Addition (elem) Word Problems. "Ed and his sister are helping cook Thanksgiving dinner. They decide to have a potato-peeling contest. Ed peels twenty-two. His sister peels six more than that. How many did she peel?" Five Thanksgiving-themed addition word problems.
• This Thanksgiving Addition (primary) Word Problems is perfect to practice problem solving skills. Your elementary grade students will love this Thanksgiving Addition (primary) Word Problems. "Mike and his brother Pat are selling pumpkins they grew. If Mike sells nine and Pat sells seven, how many do they sell in all?" Five Thanksgiving-themed addition word problems.
• This Thanksgiving Addition (upper elem) Word Problems is perfect to practice problem solving skills. Your elementary grade students will love this Thanksgiving Addition (upper elem) Word Problems. "Myron analyzed the caloric content of his Thanksgiving dinner for his science class. His serving of stuffing had 619 calories. His serving of pumpkin pie had 780 calories. His serving of turkey had 282 calories. How many calories were there in his stuffing and turkey combined?"
• This The Ugly Duckling (elem/upper elem) Word Problems is perfect to practice problem solving skills. Your elementary grade students will love this The Ugly Duckling (elem/upper elem) Word Problems. "The duckling lived in a farm with 45 animals. There were 10 pigs, 12 goats, 5 dogs, 2 horses and the rest were sheep." Choose the equation that best expresses the word problem.
• This Train Addition Board Game (b/w) is perfect to practice addition skills. Your elementary grade students will love this Train Addition Board Game (b/w). Match the math problems (with sums to 15) to the answers to win the game.
• This Train Addition Board Game (color) is perfect to practice addition skills. Your elementary grade students will love this Train Addition Board Game (color). Match the math problems (with sums to 15) to the answers to win the game.
• This Weather (primary/ESL) Dominoes is perfect to practice literacy skills. Your elementary grade students will love this Weather (primary/ESL) Dominoes. Double-six dominoes, but with weather words instead of dots. Great for early literacy/language instruction.
• This Winter Addition & Subtraction Missing Number is perfect to practice addition and subtraction skills. Your elementary grade students will love this Winter Addition & Subtraction Missing Number. Winter addition and subtraction missing number practice. Solve the equation by filling in the missing number. | 2,291 | 10,874 | {"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} | 3.234375 | 3 | CC-MAIN-2017-39 | latest | en | 0.878784 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-7-exponents-and-exponential-functions-7-4-more-multiplication-properties-of-exponents-practice-and-problem-solving-exercises-page-437/43 | 1,534,229,230,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221208676.20/warc/CC-MAIN-20180814062251-20180814082251-00421.warc.gz | 1,164,345,833 | 14,815 | ## Algebra 1
$2.56 \times 10^{22}$
(i) $(ab)^m=a^mb^m$ (ii) $(a^m)^n=a^{mn}$ (iii) $a^{-m} = \dfrac{1}{a^m} a\ne0$ Use the rules above to obtain: $=6.25^{-2} \times (10^{-12})^{-2} \\=\dfrac{1}{6.25^2} \times 10^{-12\cdot (-2)} \\=\dfrac{1}{39.0625} \times 10^{24} \\=0.0256 \times 10^{24} \\=(2.56 \cdot 10^{-2}) \times 10^{24} \\=2.56 \times 10^{-2+24} \\=2.56 \times 10^{22}$ | 193 | 379 | {"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} | 4.125 | 4 | CC-MAIN-2018-34 | longest | en | 0.20333 |
http://mathhelpforum.com/calculus/69263-comparison-test.html | 1,527,146,429,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865928.45/warc/CC-MAIN-20180524053902-20180524073902-00052.warc.gz | 187,162,782 | 9,465 | 1. ## comparison test.
here is the problem: ImageShack - Image Hosting :: 77517802tm8.jpg
we haven't learned the alternating series test.
not sure what to compare it to. the series converges, however i thought it behaved like a diverging p series.
need help.
2. Originally Posted by rcmango
here is the problem: ImageShack - Image Hosting :: 77517802tm8.jpg
we haven't learned the alternating series test.
not sure what to compare it to. the series converges, however i thought it behaved like a diverging p series.
need help.
Since $\displaystyle 2 +(-1)^n \le 3$ then $\displaystyle \sum_{i=1}^{\infty} \frac{2 + (-1)^n}{n\sqrt{n}} \le \sum_{i=1}^{\infty} \frac{3}{n\sqrt{n}}$ so by the direct comparison test your series converges.
3. hey thanks. | 218 | 758 | {"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} | 2.71875 | 3 | CC-MAIN-2018-22 | latest | en | 0.833155 |
https://porcupinepress.com/bending-large-radius-segment-htm/38/ | 1,725,809,845,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651013.49/warc/CC-MAIN-20240908150334-20240908180334-00411.warc.gz | 460,470,494 | 64,613 | # Segment Bends
Sample method to bend 3-1/2″ rigid conduit around a storage tank with a diameter of 160 feet.
Segment bending is a method of bending conduit by making several small bends to produce one larger bend. The method on this page illustrates a technique for bending a 3-1/2″ rigid conduit around a tank with a 160′ diameter.
## The steps for calculating a large radius segment bend are:
1. Determine the radius of the desired bend.
The following procedure illustrates the procedure to bend a length of 3-1/2″ rigid conduit around a storage tank with an 80′ radius, or a 160′ diameter.
First, find the circumference of the tank.
Diameter in inches, with 2 strut, 1 on each side, plus 1 conduit diameter, or 1/2 for each side, = 160 x 12 + 1-5/8 x 2 + 2 x 2 = 1925.625″
Circumference = Pi x Diameter = 3.14159265 x 1925.625 = 6,049.53″. This is the total length of conduit needed to complete the bending angle of 360 degrees to go all the way around the tank. A 10′ length of 3-1/2″ rigid conduit without the coupling is 9′ 10-1/4″ = 118-1/4″. The number of conduit required to go all the way around the tank, or 360 degrees, is 6,049.53 / 118.25 = 51.16.
Divide the total angle to go around the tank, 360 by the number of conduits required, 51.16, to get the angle required to bend each length of conduit. 360 / 51.16 = 7.04 degrees for each length of conduit.
1. Determine the number of segments desired or required for the completed bend.If a length of conduit is going to be bent to such a small angle, it is practical to only make 2 or 3 bends on each length of conduit.
2. Determine the Developed Length of the bend.Developed length (DL) is the length of conduit that is actually bent. The formula to calculate the developed length of a bend is: Developed length (DL) is equal to the centerline radius (R) of the bend times the angle (A) of the bend times 0.01745. DL = R x A x 0.01745The radius of the tank plus the strut and conduit is 962.8125″ The developed length for a 7.04 degree bend with a 962.8125″ radius is 7.04 x 962.8125 x 0.01745 = 118.28″.
DL = 7.04 x 962.8125 x 0.01745 = 118.28″. You can see that this is one length of 3-1/2 rigid conduit.
If you want to make the bend with 1 bend, you can make on 7 degree bend in the center of the conduit.
More realistically, partly for purposes of strapping the conduit, and NEC requirements for distance from any junction boxes to straps, I would make 2 bends of about 3.5 degrees at about 30″ and 90″ on the length of conduit, at approximately the 1/3 and 2/3 distances on the conduit.
You could select a larger number of bends, but I think it would be more difficult to bend to a smaller angle, although the bends will look smoother.
### PORCUPINEPRESS
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Good morning all. 21/Jun/16 12:00 AM | |
very early good morning to you HalT. 21/Jun/16 12:20 AM | |
It's a quiet day in Sudokuland. 20 minutes without anyone answering Hal. Meg and I say Good Morning to you. I'm feeling just about all better from my gout (or pseudo-gout) attack. Look forward to the longest day of the year. 21/Jun/16 12:24 AM | |
Morning. 21/Jun/16 12:52 AM | |
Glad to be picking up our new car tonight--the shopping end of it is never fun, in our experience, so happy it's over with. 21/Jun/16 1:19 AM | |
everyone!!! 21/Jun/16 1:34 AM | |
So close . . . 21/Jun/16 1:34 AM | |
21/Jun/16 1:34 AM | |
Am going to lunch with my BFF today!!! We've been best friends since we were 12! 21/Jun/16 1:36 AM | |
CP!!! 21/Jun/16 1:37 AM | |
And for us down under Wolf the daylight hours will start to get longer.Good morning everyone.That cat is checking to see who is watching.Enjoy your new car Cathy (do you have another book in the pipeline? I am waiting) 21/Jun/16 1:45 AM | |
Cathy, last time I picked up a new car the manager recited a long, memorized speech which included 'Welcome to the family'. I told him I wanted to puke! 21/Jun/16 2:12 AM | |
everyone! Sunnnnnny! Hot! No where near finished packing up the five rooms to be renovated. (Demolition starts Thursday.) 21/Jun/16 3:17 AM | |
What beautiful green eyes you have sweet puss. 21/Jun/16 3:38 AM | |
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Sitting in darkness again. 21/Jun/16 3:55 AM | |
But have a book light tonight. 21/Jun/16 3:56 AM | |
The rain seems to have stopped, just gusty wind now. 21/Jun/16 3:58 AM | |
Looks like we will get our trip to Western Plains Zoo today. 21/Jun/16 4:00 AM | |
Am I in time? 21/Jun/16 5:36 AM | |
Gallump. 21/Jun/16 5:36 AM | |
21/Jun/16 5:36 AM | |
... an hour and a half later. 21/Jun/16 5:37 AM | |
What's my rush for a CP, CP? 21/Jun/16 5:38 AM | |
My cat looks like that but she has no desire to be outside ever. 21/Jun/16 5:52 AM | |
For those of you who don't check the end of the previous day. 21/Jun/16 5:53 AM | |
A shopkeeper has £25.83p in his till.It is made up of 4 different coins.There are the same number of each coin. £1 coin is the highest coin to be used, so the available coins are: 1p, 2p, 5p, 10p, 20p, 50p, and £1. More... 21/Jun/16 5:53 AM | |
Good afternoon to all! I think that cat is just exploring. not escaping. 21/Jun/16 6:09 AM | |
We've been experiencing a bit of a heat wave here, daily temps. around 33 C. but with humidity it feels like 40 C. Looking for relief, but none on the horizon. 21/Jun/16 6:11 AM | |
Wolf, glad to hear you're feeling better. 21/Jun/16 6:12 AM | |
Cathy, what kind of car did you buy? 21/Jun/16 6:12 AM | |
Well, that's my CP! 21/Jun/16 6:13 AM | |
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Good morning all. I too am looking forward to more hours of daylight. 21/Jun/16 7:55 AM | |
I wonder if Peter was 1 of the hundreds brave enough to take part in the nude winter solstice swim in Hobart this morning. 🙄 21/Jun/16 7:58 AM | |
Morning all, the exploring cat.Busy day ahead visiting friends in Robinvale. 21/Jun/16 8:01 AM | |
Apparently the rain was pouring down and water temp was 12*C (53*F). 21/Jun/16 8:01 AM | |
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https://www.redhotpawn.com/forum/posers-and-puzzles/the-board-puzzle.134838 | 1,560,630,888,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627997335.70/warc/CC-MAIN-20190615202724-20190615224724-00556.warc.gz | 890,230,446 | 6,983 | The Board Puzzle
greenpawn34
Posers and Puzzles 13 Oct '10 23:04
1. 13 Oct '10 23:04
Hi
I did this for you guys.
2. AThousandYoung
14 Oct '10 18:34
The extra square when you slide it the one way is the missing square when you slide it the other way.
The missing square is split up among a bunch of other squares which are not true squares in the cut and paste board
3. 14 Oct '10 21:242 edits
Originally posted by AThousandYoung
The extra square when you slide it the one way is the missing square when you slide it the other way.
[hidden]The missing square is split up among a bunch of other squares which are not true squares in the cut and paste board[/hidden]
Hi 1,000 Young.
No need to hide.
"The missing square is split up among a bunch of other squares which
are not true squares in the cut and paste board."
What 'bunch of other squares.' there is only one square missing.
I think I have done some magic and I'm very clever.
4. 14 Oct '10 21:52
Asking about the exact number of squares in the beginning and at the end is a red herring.
Assume that a square on the board is one unit in length, and one in width. Then, the original board has 8*8 units = 64 units squared.
After the cut, the board can be seen as being two 7*8 right triangles, and one 1*8 strip of squares.
(1/2)(7*8) units +(1/2)(7*8) units + (1)(8) units = 64 units squared.
It's just a peculiarity of the slope that our chainsaw artist chose that the board appears to have 'lost' squares. It appears to have lost an intact square, but all the area is still there, and the Universe is still safe.
5. AThousandYoung
15 Oct '10 01:46
Originally posted by greenpawn34
Hi 1,000 Young.
No need to hide.
"The missing square is split up among a bunch of other squares which
are not true squares in the cut and paste board."
What 'bunch of other squares.' there is only one square missing.
I think I have done some magic and I'm very clever.
It's not missing. Slide the board the other way to get the missing square - you'll have 65!
6. 15 Oct '10 10:00
Originally posted by greenpawn34
Hi
I did this for you guys.
If you weigh the two configurations, then you'll see the weight is the same.
No square is missing.
7. 15 Oct '10 12:39
Ok you have all solved it.....
.....by a series of very lucky guesses. | 616 | 2,290 | {"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-2019-26 | latest | en | 0.942175 |
www.hummingbirdhovercraft.com | 1,508,227,173,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187820930.11/warc/CC-MAIN-20171017072323-20171017092323-00602.warc.gz | 487,084,648 | 7,054 | # How You Can Win Lotto By Utilizing Analysis Algorithms For Lotto Game Prediction
Lotto predictions is quite preferred nowadays. Individuals made use of to be unconvinced with the forecasts as they assumed that the winning numbers are a matter of luck and also ton of moneys. Very few individuals believe that lotto could be won by utilizing some kind of an innovative science based predictions. It was not up until the late 90s when lotto gamers started making use of lottery game predictions to help them to win lottery game or at least obtain closer to the winning numbers. When Gonzalo Garcia-Pelayo, a Spanish guy that handled to study as well as evaluate numerous video games in two various countries, Spain and also the US as well as win a great deal of loan by utilizing different approaches. After him individuals started to believe that lotto outcomes can be forecasted.
Lottery players begin thinking about ways to win the lotteries using predictions. They use numerous type of forecasts: from mechanical predictions on mechanical lottery games to technical predictions making use of computer system software application. A lot of individuals make use of formula to examine as well as forecast lotto game outcomes.
Lottery Analysis formula has actually been shown to be extremely effective in order to help lottery game players obtain closer to the winning numbers or even make the lottery game homes go bankrupt! There are lots of kinds of lotto game evaluation used by lotto game predictors and right here are several of them :
In this analysis, the predictors utilize some comprehensive research studies which tape-record the frequency of each adjoining pairs of numbers in the relevant lotto game win in a period of time then placed the most regular numbers on top of the rankings and also do it back to back.
### Evaluation of Equilibrium
With analysis of equilibrium, lotto gamers aim to analyze if specific mixes will provide opportunities to win such as combinations of small as well as huge numbers, strange or even numbers and also the variety of the complete amount numbers.
### Evaluation of Numbers
When evaluating lottery winning numbers using digit evaluation, lotto game forecasters will be able to understand the specific numbers in particular range can be attracted a particular period of time. In order to make the winning chance bigger, the gamers need to restrict the range of numbers when they select every number in their mix.
### Evaluation of Elapse Time
This evaluation works by discovering and also keeping in mind the duration when a number is in its waiting time to be attracted once again after its last winning time. Gamers will likewise know the chance or the winning possibility of certain numbers based on the elapse time. If the elapse time is much longer, the possibility to win is bigger. This evaluation is thought about more precise than the others as it provides a lot more data concerning tendency of some numbers to win or otherwise to ensure that it is less complicated to understand the next winning numbers in some lottery games such as Powerball, Huge million, The golden state Super Lotto Plus and other.
### Evaluation of Groups
There are numerous kinds of team evaluation that lottery game forecasters use to obtain into the winning numbers. Lottery players could group the months having the best gaining numbers of a specific period or they could group the numbers winning in particular amount of time. | 653 | 3,476 | {"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} | 2.90625 | 3 | CC-MAIN-2017-43 | longest | en | 0.96487 |
https://electronics.stackexchange.com/questions/560455/switch-design-for-shorting-the-power-control-pins | 1,620,637,879,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989115.2/warc/CC-MAIN-20210510064318-20210510094318-00527.warc.gz | 254,755,515 | 44,127 | # Switch design for shorting the power control pins?
I have been struggling to design this switch for a while and I would appreciate any help.
Problem Background:
I'm designing a power system for a submersible, using the Mini-box OpenUPS battery management PCB. The connection diagram for OpenUPS as well as its link are shown as below:
A 14.4V 4S 1800mAh Li-Po Battery will be attached to this board. But before we turning on this PCB, we have no direct access to the battery power. (Yes, I can re-direct the wire from the battery to somewhere else, but that will not be a good ideal for our design) The board can be turned on by shorting the two control pins at J8. The open circuit voltage across these two pins is about 14.4V; when the two pins are shorted, the current flowing through is around 135mA. The physical switch that mounted on the submersible is a High Pressure Waterproof Switch from Blue Robotics. A picture is shown below:
When the dial is turned all the way in, the circuit will be closed. For the sake of robustness, the dial is required to be turned all the way in when it is underwater. Before turning this PCB on, all pins except J8 remain at 0V.
Problem Statement:
The functionality for this switch we want will be as following:
Operation on the Switch: Prior putting the submersible underwater, the switch will remain closed first (closed circuit), we will loosen the dial for around 1s (open circuit), and then turn it all the way in (closed circuit).
Desired Response for Circuit: Two pins on J8 are disconnected when dial is turned in, then shorted when dial is loosened.
Attempted Solution: I have tried using a 2N3906 PNP transistor to control the circuit (14.4V power control pin serves as VCC, another pin as ground, then diagram is shown below), the schematic is shown below:
simulate this circuit – Schematic created using CircuitLab
However, I encountered the following problem: Initially, the PCB is off and the switch is open. There's somehow about 50uA $$\I_{EC}\$$ leakage, and that will not trigger the PCB to turn on. When switch is closed, there's around 134mA $$\I_{EC}\$$ and PCB is turned on. But when switch is open again, the 50uA $$\I_{EC}\$$ will remain the two pin shorted, and PCB will not function in this state. I experiment with it a little bit and found that even if there's 1uA, it is enough for these two pins to remain shorted after PCB is triggered.
I'm wondering how I should solve this problem or is there an alternative design for it?
• I am confused. Why doesn't the switch simply short the two pins on the pcb? Why the transistor? – Math Keeps Me Busy Apr 16 at 1:27
• Hello! I mentioned that in the Desired Response for Circuit, our design requires that when switch is open, the two pin is shorted, thus I cannot just simply connect the switch to the two pins – Maxwell Z. Apr 16 at 8:01
• Could you double check the schematic in your question. I don't think it does what you think it does, and it doesn't match the behavior you described. Also, I have not been able to find a datasheet for the OpenUPS. do you know the minimum current across J8 that will turn on the PCB? Alterrnatively, the lowest resistance across J8 that will NOT turn on the PCB? (You could try large values, and work down until it turns on). – Math Keeps Me Busy Apr 16 at 13:41
• Hi you might want to note that J8 expects to receive a momentary closure (and timeouts that can be adjusted in the OpenUPS's firmware, via their program and the miniUSB port). An on/off switch connected across here is not what you want. – Daniel Chisholm Apr 16 at 13:58
• @MathKeepsMeBusy Hi, I'm really sorry, I messed up the position of the switch... Just corrected in the schematic. You can find a hardware manual under the download section in the link I posted. You have to open 'Hardware Manual' tab in a new window to start downloading. As for the minimum current, I have not tested yet, I'll post it once I measure it! Thank you so much for your response! – Maxwell Z. Apr 16 at 19:07
You have done a good job at defining the task, but not the interface logic levels and outputs. Please update your specs
• for more test results in case something is damaged on your interface measure Iin, as Vin is varied from 14.2 to 0V with a variable current limited supply to get a better handle on the problem performed with a triangle ramp in << 5 seconds. Use a transistor to control the current. (Since the user manual is vague) Also, confirm the pressure switch does not leak uA.
• 50uA if fails for Return to High seems too sensitive and 143 mA seems too high for a logic low Start switch.
## From the OPN-USB Manual for J8 p5/12
Starting and Stopping with battery power….
If everything correctly configured pressing shortly the J8 button will turn on the output,
and your UPS will be started, energy will be flowing from the battery to the output.
Long pressing (5 sec) J8 button will initiate shut down of the UPS. A pulse is sent
to the motherboard than the UPS waits a predetermined UPS_HARDOFF_TOUT (60s by
default) before the output voltage is turned off.
## i.e. Start is ---_ ---- and Stop is ----______------- , where --- = open switch and ___ is closed switch yet pressure sense is normally open with dial in-tight and switch closes to 0V to exceed long duration >5s to Stop OpenUSB power.
Your problem is commonly defined in datasheets as$$\I_{CBO}\$$ which is the collector-base cutoff current at rated Vce max voltage with Ie=0.
This is due to the “Early Effect” as a reverse diode leakage across the CB junction which has a load line dependent on Vce.
Although the 2N3906 is rated at 50uA it is only rated at Vce=-40V so choosing a higher voltage say >100 V , you will find components with lower max Icbo leakage , which also rises above 25’C much lower.
For example if it says 250nA at -300V, this is equivalent to >1GOhm but again will rise in temp and this may be nominal. This also makes it sensitive to stray EMI during this state with electromagnetic noise coupling so shielded twisted pair may be needed if there is any EMI nearby during this state.
You can lower the input impedance by using a Common Base mode driving the emitter as a switch with some bias R divider on the base from 12V to Gnd such that using hFE = 10% of linear hFE max would get reasonable Vce(sat) to drive 125mA with the switch.
This should solve the leakage problem.
I suspect your actual circuit looks more like below with ground switched power activation but unfortunately with 50uA (rated? or measured?) leakage to ground somewhere.
The 2N3906 is one such path.
Not a solution, just maybe an answer to define the problem.
• Tony, look at the circuit. It doesn't do what the OP is hoping it would do. – Math Keeps Me Busy Apr 16 at 13:14
• Which part did I not understand? @MathKeepsMeBusy – Tony Stewart EE75 Apr 16 at 15:24
• Does the circuit act like a closed switch when the physical switch is open and vice-versa? – Math Keeps Me Busy Apr 16 at 15:28
• It's a Start/Stop duration short/long both low. so I think they are using it properly. @MathKeepsMeBusy – Tony Stewart EE75 Apr 16 at 16:20
• It's a negative logic dial so in tight is open circuit (14.2 pullup) and loose dial or tight overpressure is closed switch. Seems OK., @MathKeepsMeBusy – Tony Stewart EE75 Apr 16 at 16:37
Well, there is always a way to invert a switch without too much issues. One way would be for you is to use a ZVNL120A in the circuit below:
simulate this circuit – Schematic created using CircuitLab
It doesn't seem much. But here is an example of the circuit with an led, when lit it would be the equivalent to a switch closed across your 14V logic,When not lit, the equivalent switch across the 14V logic would be open.
• Hey, thank you so much for your help, I will try to use a FET instead of a BJT to see if it works! – Maxwell Z. Apr 20 at 23:28
• BJT or Transistors require a bias. FETs or MOSFETs don't need a bias to operate. – David Mikeska Apr 21 at 15:09
The following circuit may work for you.
It relies two assumptions.
1. at least one of the pins in J8 is always above 3V.
2. You can access the board's ground and wire it to this circuit.
If these assumptions are not met, then you will not be able to use this circuit.
simulate this circuit – Schematic created using CircuitLab
How it works.
We are assuming that at least one of the input lines IN1 or IN2 is always above 3V.
Diodes D1 and D2 select the higher input voltage and use it as the power rail for an inverter consisting of M3 and M4.
R1 and SW1 provide the input for the inverter.
Resistors R2 and R3 limit shoot-through current when the inverter is transitioning from one state to another.
The output of the inverter drives the gates of M1 and M2 which comprise an analog switch.
When SW1 is closed, the gates of M3 and M4 are at ground. M3 will conduct, and M4 will be off. The gates of M1 and M2 will be one Schottky diode drop below the higher input voltage, and so they will be off.
When SW1 is open, the gates of M3 and M4 are high. M3 will be off, but M4 will conduct. The drains of M3 and M4 will be near ground, and so will the gates of M1 and M2. M1 and M2 will be on.
• Note: all MOSFETs should be logic level.
• Note: M4 is an N-channel MOSFET! (all the others are P-channel)
There will be some leakage current through R1 when SW1 is closed. If the value of R1 is too low, the circuit connected to IN1 and IN2 may give false results. If the value is too high, the circuit may be sluggish. I chose a value of 1M$$\\Omega\$$ because it seemed to be a good compromise, but it may need to be tweaked.
Edit: Added resistors to limit shoot-through current in inverter when inverter is transitioning between states.
• Hey, thank you so much for your response. It seems that one of the J8 pin is power control which has a voltage of 14.4V (let's call it pin1), and another pin (lets call it pin2) seems to be connected to the board GND, it was measured to be 0V across the GND. Between pin2 and ground, there's a 100K ohm resistance and about 134nF capacitance. If I connect the pin 1 to GND, the board seems to enter the same initiation phase where LED lights up same as when I short pin 1 and pin 2, but not matter how long I short pin 1 to GND, the board will neither turn on nor enter the deep sleep mode – Maxwell Z. Apr 20 at 23:22 | 2,609 | 10,379 | {"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": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-21 | latest | en | 0.941018 |
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## Unanswered: continous form totals (unbound textboxes)
EDIT: sorry guys i was in a rush yest. before i had to leave. here is the ?? agian:
i have a calculated field the multiplies the qty and the unitcost per record on my tabular form.
Im trying to get a field in the footer that will give the total cost of each record shown on the tabular form. (which is filtered)
Agian, how do i go about doing this? all access help tells me, is to drag the field to the footer which didnt work.
Last edited by jwalker343; 08-31-07 at 15:17.
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If the calculated field is stored In the control source property for the field in the footer put =sum([mycalculatefieldname]) or try putting another field in the footer =sum([name_of__other_field_in_footer_which_calculates]) or try putting a copy of both fields in the footer and calculate on the field names of the copy of the fields. Not sure I understand fully what you're trying to do though so this is probably an obscure answer - sorry about that.
Last edited by pkstormy; 09-01-07 at 23:35.
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I understand here is a better example:
I have a form that has 2 fields [qty] and [Unitprice] then i have another field that is unbound [totalprice] and calculates [qty]*[unitprice] to get a total price per part
however im trying to get the total amount of the records that are shown
see attatched image for visual explination
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## Overall summed value
jwalker343,
I created a table with a qty and a price field and put these fields on a continuous form. In the form's footer I added a field which has a control source =sum([qty]*[price]). This gives me the overall total price of all the qty*price fields. The difference though is that the qty and price fields on my continuous form are bound where I see in your bmp that they are dlookups (which makes a difference.) As you've probably already guessed, having dlookups is what's giving you problems using the =sum([qty]*[price]) in the footer. Do you need to use dlookups on your form? This seems a tad bit extreme with all the dlookup fields for the form. I think if you could somehow makes the qty and price fields bound to a query/table, you would have success using the =sum([qty]*[price]) in the footer.
Have you considered having a summation query with a dlookup to that value? I'm guessing you already have and found that this gives you problems also as you need to dlookup values to get qty and price. I'm also guessing you've probably already attempted to get rid of the dlookups somehow.
(not that this is a good arguement for storing the calculated value in the table) but have you considered doing this since you're using a lot of dlookups on the form - might make your life a little easier with the dlookups. Overall though, I think you need to find a way to avoid having all the dlookups for all the fields on the form.
If you can zip an extract, I can probably find a way to help you avoid all the dlookups.
If it helps at all, here's the example I quickly made (see form2) but it doesn't use dlookups on the form as your bmp illustrates. I'll play around with one which utilizes dlookups as you've peaked my curiosity calculating totals on dlookup fields.
Last edited by pkstormy; 09-04-07 at 18:16.
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hey i took your advice and bound the fields to a table. it now takes slightly more work to input the data but it works. im still doing a calculation on qty*price per record and then i used =sum (qty*price) to show the total for the form and it worked flawlessly.
i never in my life would have known that you couldnt total dynamic fields, or dlookups.
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Let $D$ be a subset of $\mathbb{R}^{n}\times\mathbb{R}$, $(x_{0},t_{0})$ a point of $D$, and $f\colon D\to\mathbb{R}$ be a function.
We say that a function $x(t)$ is a solution to the Cauchy (or initial value) problem
$\begin{cases}x^{\prime}(t)=f(x(t),t)\\ x(t_{0})=x_{0}\end{cases}$ (1)
if
1. 1.
$x$ is a differentiable function $x\colon I\to\mathbb{R}^{n}$ defined on a interval $I\subset\mathbb{R}$;
2. 2.
one has $(x(t),t)\in D$ for all $t\in I$ and $t_{0}\in I$;
3. 3.
one has $x(t_{0})=x_{0}$ and $x^{\prime}(t)=f(x(t),t)$ for all $t\in I$.
We say that a solution $x\colon I\to\mathbb{R}^{n}$ is a maximal solution if it cannot be extended to a bigger interval. More precisely given any other solution $y\colon J\to\mathbb{R}^{n}$ defined on an interval $J\supset I$ and such that $y(t)=x(t)$ for all $t\in I$, one has $I=J$ (and hence $x$ and $y$ are the same function).
We say that a solution $x\colon I\to\mathbb{R}^{n}$ is a global solution if $D\subset=\mathbb{R}^{n}\times I$.
We say that a solution $x\colon I\to\mathbb{R}^{n}$ is unique if given any other solution $y\colon I\to\mathbb{R}^{n}$ one has $x(t)=y(t)$ for all $t\in I$ (i.e. $x$ is the unique solution defined on the interval $I$).
0.1 Notation
Usually the differential equation in (1) is simply written as $x^{\prime}=f(x,t)$. Also, depending on the topics, the name chosen for the function and for the variable, can change. Other common choices are $y^{\prime}=f(y,t)$ or $y^{\prime}=f(y,x)$. It is also common to write $\dot{x}=f(x,t)$ when the independent variable represents a time value.
0.2 Examples
1. 1.
The function $x(t)=\log t$ defined on $I=(0,+\infty)$ is the unique maximal solution to the Cauchy problem:
$\begin{cases}x^{\prime}(t)=1/t\\ x(1)=0.\end{cases}$
In this case $f(x,t)=1/t$, $D=\{(x,t)\colon t\neq 0\}$, $t_{0}=1$, $x_{0}=0$.
2. 2.
The function $x(t)=e^{t}$ is a global (and hence maximal), unique solution to the Cauchy problem:
$\begin{cases}x^{\prime}(t)=x(t)\\ x(0)=1.\end{cases}$
3. 3.
Consider the Cauchy problem
$\begin{cases}x^{\prime}(t)=\frac{3}{2}\sqrt[3]{x}\\ x(0)=0.\end{cases}$
The function $x(t)=0$ defined on $I=\mathbb{R}$ is a global solution. However the function $y(t)=\sqrt{t^{3}}$ defined on $I=[0,+\infty)$ is also a solution and so are the functions
$z(t)=\begin{cases}\sqrt{(t-c)^{3}}&\text{if t\geq c}\\ 0&\text{if t
for every $c\geq 0$. So there are no unique solutions. Moreover $y$ is not a maximal solution.
Title Cauchy initial value problem Canonical name CauchyInitialValueProblem Date of creation 2013-03-22 14:57:18 Last modified on 2013-03-22 14:57:18 Owner paolini (1187) Last modified by paolini (1187) Numerical id 14 Author paolini (1187) Entry type Definition Classification msc 34A12 Synonym Cauchy problem Synonym initial value problem Related topic InitialValueProblem Related topic DifferentialEquation Related topic CauchyKowalewskiTheorem Defines solution to the Cauchy problem Defines solution to the initial value problem | 1,053 | 3,038 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 53, "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} | 3.65625 | 4 | CC-MAIN-2024-22 | latest | en | 0.60208 |
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Overview
Chapter 4: Indigeneities of New Directions in Chicanx and Latinx Studies offers insights into Indigenous identities and demographics, supporting discussions on cultural identity and historical presence.
For crafting lesson plans or exams, the questions provided in subsequent pages of this workbook can be used to facilitate student engagement and evaluate understanding of these complex themes. Please review the About this Workbook page for guidance on how to access these questions directly in H5P or ADAPT and how to sync questions from ADAPT into your Learning Management Software.
4: Chapter 4 Workbook- Indigeneities is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts. | 1,909 | 5,122 | {"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} | 3.703125 | 4 | CC-MAIN-2024-33 | latest | en | 0.196091 |
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Discussion Forum : Thermodynamics - Section 1 (Q.No. 3)
3.
The efficiency and work ratio of a simple gas turbine cycle are
low
very low
high
very high
Explanation:
No answer description is available. Let's discuss.
Discussion:
34 comments Page 1 of 4.
Soham Mitra said: 8 years ago
Rankine cycle used in thermal power plants deals on heating up water in constant pressure while changing its phase. The process will has greater thermal efficiency compared to any other cycles. Otto cycle can't be used because heat supplied to the system is in constant volume process, not allowing steam to expand properly.
If Otto cycle forced to used, the pressure will be too high for any material (vessel) to withstand. Diesel also can't be used, because while heat supplied to the system, the fluid deliver work to the surrounding. This not applicable because the work should be extracted by steam turbine.
YOGESH GAYAKE said: 8 years ago
But my question is that turbine is used to rotate the generator for generating the electricity & its rotate with the help of steam pressure. Then how we can calculate its efficiency & work ratio. But still we assure that the 70% of power from turbine is used to run the compressor itself. So it's correct but, how we can say its very low. Please suggest answer.
Chauhan Hardik N. said: 8 years ago
I think that, since the specific volume (reciprocal of density) of air is high, the compression of air requires more work.
Hence, the major fraction of the work developed by the turbine is utilized in driving the compressor.
Therefore, the efficiency of the simple gas turbine plant is too low.
Rajkumar said: 10 years ago
Since the specific volume (reciprocal of density) of air is high, the compression of air requires more work. Hence, major fraction of the work developed by the turbine is utilized in driving the compressor. Therefore, the efficiency of the simple gas turbine plant is too low.
Anshu maan mishra said: 1 decade ago
The ratio of the actual work output of the turbine to the work output that would be achieved if the process between the inlet state and the exit state was isentropic.
DK SINGHGUPTA said: 9 years ago
We all know that work ratio-turbine work /net work that means turbine work means positive work so the answer is low but how the answer is very low.
Because there will be more availability in the exit of the turbine. That's why rankine cycle is used combined with gas power plant.
Awinash kumar said: 1 decade ago
Mechanical efficiency of gas turbine is about 95%. It is not clear in question although thermal efficiency is about 15-20%. | 604 | 2,678 | {"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} | 2.65625 | 3 | CC-MAIN-2024-22 | latest | en | 0.929581 |
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# Cystic Fibrosis is the most common generic disease in white european and american populations.It results from one of a nmber of mutations in a single gene that codes for a protien involved in the transport of salts in the cells of the body. The cystic fibrosis gene is described as recessive , which means that individuals with only one copy of the gene so called carriers , show no symptoms of the disease and manny be unaware that they carry the gene. Individuals with two copies of the faculty CF gene will show symptom of the condition. Among white europeons,the probability of being a carrier is 1/25. For a child whose parents are both carriers, the probability of inheriting a copy of the CF gene from both parents is 1/4. a)What is the probability that any two white Europeans planning to have a child together would both be carriers? b) What is the probability of the child having cystic fibrosis?
CBSE Maths for Science stream
a) Probability of any two white Europeans planning to have a child together would both be carriers = (1/25) (1/25) = 1/ 625 b) Probability of the child having cystic fibrosis = (1/625) (1/4) = 1/2500
Doctor who loves teaching and enhancing the future of your kids
Dear kulwinder the answer to ur first question is that u have already stated 1/25 Europeans. and every 1 child out of 4 can have the chances of having the nasty disease.
Tutor
a) 1/625
I agree with Anil.
Gold Medalist
Thanks for this nice question. Its important to arrange a class to gather a sound knowledge. Because transmission of allels are associated with recombination. So, please attend class and make it clear.... Dr. P. majumder
math magician
A) 1/50
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Centre Paper Reference Surname Initial(s) No. Candidate 7540 02 Signature No.
Paper Reference(s)
7540/02
London Examinations GCE
Physics
Ordinary Level Paper 2
Monday 29 January 2007 – Afternoon Time: 2 hours
Examiner’s use only
Question Leave Number Blank 1 2 3 4 5 Total
Materials required for examination
Four-figure mathematical tables may be issued to candidates requesting them
Items included with question papers
Nil
Instructions to Candidates
Answer ALL the questions in the spaces provided in this question paper. In the boxes above, write your centre number, candidate number, your surname, initial(s) and signature.
Information for Candidates
Calculators may be used. Where necessary, assume the acceleration of free fall, g = 10 m/s 2 . The total mark for this paper is 100. The marks for parts of questions are shown in round brackets:
e.g. (2). This paper has five questions.
Write your answers neatly and in good English. In calculations, show all the steps in your working.
This publication may be reproduced only in accordance with Edexcel Limited copyright policy. ©2007 Edexcel Limited.
Printer’s Log. No.
N23593A
W850/U7540/57570
4/3/3/3/3/2/3/1500
*N23593A0120*
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Leave blank Answer all the questions. 1. This question is about energy and momentum. (a) In the ancient Olympic games long jumpers carried a mass of 5.0 kg in each hand. In one particular jump the increase in height of a mass of 5.0 kg was 3.2 m and its velocity was 6.2 m/s. For this 5.0 kg mass calculate: (i) the momentum; (2) (ii) the increase in gravitational potential energy; (2) (iii) the kinetic energy. (2)
2
*N23593A0220*
Leave blank (b) During the jump the long jumper holds the masses above his head. The drawing shows the long jumper landing. He has moved the masses to alter the position of his centre of gravity. (i) What is meant by centre of gravity? (2) (ii) How has the position of the centre of gravity of the jumper been changed by this movement of the masses? (2)
*N23593A0320*
3
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(c)
Describe an experiment to investigate the conservation of linear momentum during a
collision using dynamics trolleys.
(i)
draw a labelled diagram;
(3)
(ii)
(3)
(iii)
describe the method used;
(3)
(iv)
write down the equation you would use to show that momentum is conserved.
(1)
Q1
(Total 20 marks)
4
*N23593A0420*
2. This question is about density and pressure.
(a)
The graphs below show how the density of two different substances varies with temperature between –20 °C and 120 °C.
Density
Liquid A
Water
–20
4
120
Temperature/°C
(i) Describe how the density of water changes when it is heated from –20 °C to 120 °C. (3) (ii) Describe how the density of liquid A changes when it is heated from –20 °C to
120 °C.
(1)
Leave
blank
*N23593A0520*
5
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(b) The diagram shows a tube dipped into a bowl of water. Water has been drawn up the tube by a pump.
Bowl
X
To pump
Tube
Y Z
0.20 m
Water
(i) Three points are labelled X, Y and Z. Which point or points is/are at atmospheric pressure? (1) (ii) What is the pressure due to the column of water XZ? (Density of water = 1000 kg/m 3 ) (2)
Leave
blank
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Leave blank (iii) What is the pressure at X when the atmospheric pressure is 100 kPa? (2) (iv) The cross-sectional area of the tube is 0.000050 m 2 . Calculate the mass of water in the 0.20 m length of the tube. (4)
*N23593A0720*
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(c) The apparatus below can be used to measure the density of liquid A.
Water
To pump
h A
h W
Liquid A
(i) Explain how the diagram shows that liquid A has a lower density than water. (3) (ii) By making actual measurements of h W and h A , calculate the density of liquid A. (2) (iii) Explain why it is more difficult to measure h W than h A . (1) (iv) Explain why it is important to note the temperature. (1)
(Total 20 marks)
Leave
blank
Q2
8
*N23593A0820*
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3.
This question is about the refraction of light.
(a)
The diagram shows a ray of light passing through water into air.
Normal
Emergent ray
Air
Water
Ray of light
45°
(i)
Explain, in terms of the properties of light, why the light changes its direction as
shown.
(2)
(ii)
If the refractive index of the water is 1.35, calculate the angle made by the
emergent ray with the normal.
(3)
9
*N23593A0920*
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Leave blank (b) The experiment is repeated using a semicircular glass block. Normal A Air B Glass 45° Ray of light (i) State why the light does not change direction as it enters the glass. (1) (ii) How does the light change as it passes from air to glass? (1) (iii) Explain why the light does not emerge from the straight edge AB of the glass block. (2) (iv) Sketch on the diagram below to show the complete path followed by this ray of light. Normal A Air B Glass 45° Ray of light (2)
10
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(v)
The glass has a refractive index of 1.60. Calculate the largest angle that the ray
would need to make with the normal to allow the light to emerge from the
straight edge AB of the glass block.
(3)
(vi)
The speed of light in air is 3.00 × 10 8 m/s. Calculate the speed of light in glass
of refractive index 1.60.
(3)
(c) When the experiment in (b) is repeated with the glass block immersed in water
instead of air, the light emerges from the straight edge AB. Explain why this happens.
(3)
Q3
(Total 20 marks)
11
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4. This question is about waves.
The males of insects called cicadas produce a very loud and shrill ‘song’. The insects produce the sound by vibrating a small plate called a tymbal. The sound is amplified by making a thin membrane vibrate at its resonant frequency.
(a) (i) State what is meant by resonance. (2) (ii) Explain why the sound produced by the tymbal is amplified by the thin membrane. (2) (b) Researchers have shown that the wavelength of the cicada’s song varies with body length. The results are shown in the table.
Wavelength of song/mm 31 43 53 64 81 Average body length/mm 16.2 22.7 28.8 35 45.3
(i) On the grid opposite, plot a graph of wavelength of song (y -axis) against average body length (x -axis). Start the x -axis at 15 mm and use a suitable scale that makes best use of the graph paper. Draw the best straight line through your points. (5) (ii) Use the graph to find the wavelength of the song produced by a cicada with a body length of 20 mm. Show how you used the graph to obtain your value.
(2)
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(c)
A cicada with a body length of 35 mm produces a song with a wavelength of 0.064 m
and a frequency of 5.3 kHz. Calculate the speed of sound in air.
(3)
(d)
On a very hot day the speed of sound in air is 350 m/s.
(i)
What effect, if any, does this have on the pitch and wavelength of the song heard?
(2)
(ii)
(4)
Q4
(Total 20 marks)
14
*N23593A01420*
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5. This question is about motion and the design of an experiment.
Glass cylinder
Oil
Metal sphere
Upper mark
Lower mark
(a) The diagram shows a glass cylinder containing oil. A student holds a metal sphere at the surface of the oil and then releases it. The sphere accelerates at the start and then reaches a constant speed before reaching the upper mark on the cylinder. It then continues at this speed until it reaches the bottom of the cylinder.
(i)
(ii)
Name two vertical forces that act on the sphere as it falls through the oil.
(2)
Sketch a graph showing how the speed of the sphere varies as it falls from the surface to the bottom of the oil.
(2)
*N23593A01520*
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Leave blank (b) Describe how the student would investigate the relationship between the mass of the sphere and the constant speed at which it falls through the oil. Your description should include: (i) a list of three essential items of measuring equipment; 1 2 3 (3) (ii) a statement of how the student would change the mass of the metal sphere; (1) (iii) a description of the method the student would use, including any measurements made; (6)
16
*N23593A01620*
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blank
(iv) a statement of how the student would ensure that the results obtained were as
accurate as possible.
(2)
(c)
Name two other factors which would affect the constant speed at which a metal
sphere would fall through oil. For each factor, say how it would affect the speed.
(4)
Q5
(Total 20 marks)
TOTAL FOR PAPER: 100 MARKS
END
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By Cindy J. Boyd
Unit 1: Geometric constitution. Unit 2: Congruence. Unit three: Similarity. Unit four: Two-and Three-Eimensional size. criteria evaluation. 846 pages.
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Step 3 Use a straightedge to draw PQ . Label the point where it −− intersects XY as M. Point M is the midpoint −− of XY, and PQ is a −− bisector of XY. Also 1 XM = MY = _XY. 2 P X Y P X M Y X Q Example 1 (p. 21) Example 2 (p. 22) Y Q Use the number line to find each measure. 1. AB 2. CD C A D B –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10 11 3. Use the Pythagorean Theorem to find the distance between X(7, 11) and Y(-1, 5). 4. Use the Distance Formula to find the distance between D(2, 0) and E(8, 6). Example 3 (p.
The latitude, longitude, degree distance, and monthly high temperature can be used to create several different scatter plots. geometryonline. com to continue work on your project. x O M ) x1 ϩ2 x2, y1 ϩ2 y2 ) Find Coordinates of Midpoint −− TEMPERATURE Find the coordinate of the midpoint of PQ. The coordinates of P and Q are -20 and 40. −− Let M be the midpoint of PQ. 60 50 40 Q 30 M=_ -20 + 40 2 20 =_ or 10 2 20 a = -20, b = 40 10 0 Simplify. –10 The midpoint is 10. –20 P –30 3. TEMPERATURE The temperature dropped from 25° to -8°.
Make a conjecture about the term angle bisector. 34 Chapter 1 Tools of Geometry Red Habegger/Grant Heilman Photography A ray that divides an angle into two congruent angles is called an angle bisector. If PQ is the angle bisector of ∠RPS, then point Q lies in the interior of ∠RPS, and ∠RPQ ∠QPS. A line segment can also bisect an angle. Q R S P Just as with segments, when a line divides an angle into smaller angles, the sum of the measures of the smaller angles equals the measure of the largest angle. | 769 | 2,886 | {"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} | 3.734375 | 4 | CC-MAIN-2020-45 | longest | en | 0.898003 |
https://rankupturn.com/tools/regular-shapes-perimeter-calculator.htm | 1,619,199,953,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039596883.98/warc/CC-MAIN-20210423161713-20210423191713-00083.warc.gz | 570,718,049 | 12,106 | # Perimeter Calculator
Shape
Radius
Radius
Side 3
Side 4
Unit
Perimeter = 31.42 m
show workout
## Work with steps - Perimeter of Circle
Question:
Find the peirmeter of given shape.
Answer:
Solution
Perimeter of a circle = 2 x π x r
Radius (r) = 5 m
Perimeter = 2 x 3.14 x 5 ; where π = 3.14
∴ Perimeter = 31.42 m
Perimeter calculator that shows work to find the perimeter of regular shapes such as square, rectangle, triangle, circle, trapezoid, parallelogram, pentagon and hexagon. The step-by-step calculation help parents to assist their kids studying 4th, 5th or 6th grade to verify the work and answers of perimeter of regular shapes homework and assignment problems in pre-algebra or in geometry (G) of common core state standards (CCSS) for mathematics.
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Question Kitty's Answer | 240 | 879 | {"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} | 3.125 | 3 | CC-MAIN-2021-17 | latest | en | 0.853489 |
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# Hilario_m3 - Modify this worksheet as needed yellow cells...
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Modify this worksheet as needed, yellow cel Class ACTG 1B << enter class title Section 2w << enter class sec Submission Date 27-Feb-09 << enter date subm Due Date 27-Feb-09 << enter due date Assignment Title Midterm III << enter assignme Chapter (if applicable) Ch14-15 << enter related ch Contribution Name Rating Student(s) Julius Hilario 2 replace these names with your name(s) >>> Assignment Components & Scoring Grid Item # Item Description 1 Cover Sheet 2 Selected Problems ter Problem # >>> Enter Problem # >>> Enter Problem # >>> Enter Problem # >>>
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lls only e ction mitted e 0.00 ent title hapter 2 = equal contribution 1 = less than equal 0 = no contribution default is "2", if left blank Pts Points Earned Possible -5 points for not including this cover sheet % raw pts Perform. Chapter 14 problem 11 0 0 0% 50.0 1 Chapter 14 problem 8 0 0 0% 20.0 1 Chapter 15 exercise 3 0 0 #DIV/0! #DIV/0! 1 Chapter 15 exercise 4 0 0 #DIV/0! #DIV/0! 1 overall perf % #DIV/0! 100 weighted #DIV/0! wtd 20 0% perf % 1 Subtotals #DIV/0! 100 10% Exceeding requirements bonus, if any Presentation penalty, if any No Cover Sheet penalty, if any
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{[ snackBarMessage ]} | 420 | 1,521 | {"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} | 2.9375 | 3 | CC-MAIN-2018-17 | latest | en | 0.720617 |
https://www.softschools.com/formulas/physics/snells_law_formula/221/ | 1,675,889,485,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500904.44/warc/CC-MAIN-20230208191211-20230208221211-00062.warc.gz | 1,001,179,017 | 5,299 | # Snell's Law Formula
Snell's Law Formula
When light strikes a smooth barrier between two transparent materials, the light is partly reflected, and partly refracted (transmitted). The formula that describes refraction is also known as Snell's Law. The angle of refraction depends on the angle of incidence of the light, and the indexes of refraction of the two materials. The index of refraction of a material depends on the material's properties. The angles in Snell's Law are always measured relative to the normal to the barrier, which is perpendicular to the barrier's surface. The angles are measured in radians or degrees, and the indexes of refraction are unitless numbers.
na = index of refraction in material a, (unitless)
nb = index of refraction in material b, (unitless)
θa = angle of light relative to normal to the barrier in material a, (radians or degrees)
θb = angle of light relative to normal to the barrier in material b, (radians or degrees)
Snell's Law Formula Questions:
1) A beam of light in air makes an angle of 41.8° relative to the normal to the surface of a pool of clear water. The index of refraction for air is 1.000, and the index of refraction for the water is 1.333. What is the angle of the light beam in the water, relative to the normal?
Answer: The angle of the light beam in the water relative to the normal can be found using Snell's Law. As a reminder, the angles in Snell's Law are always measured relative to the normal (perpendicular) to the surface. The angle can be found by rearranging the formula:
The angle of the light beam in the water (relative to the normal) is 30.0°.
2) A beam of light in air makes an angle of 30.0° relative to the surface of a diamond. The index of refraction for air is 1.000, and the index of refraction for diamond is 2.417. What is the angle of the light beam in the diamond, relative to the surface?
Answer: The angles in Snell's Law are always measured relative to the normal (perpendicular). In this question, the angle of the beam in air is given relative to the surface. The normal is at a 90° angle to the surface, so the angle in air relative to the normal is:
Snell's Law can be used to find θb, which is the angle of the beam in the diamond, relative to the normal. The angle θb can be found by rearranging the formula:
The question asks for the angle of the light beam in the diamond, relative to the surface. The angle θb is the angle relative to the normal. If we label the angle relative to the surface a (Greek letter "alpha"), its value is:
The angle of the light beam in the diamond, relative to the surface, is 69.0°. | 625 | 2,628 | {"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-2023-06 | latest | en | 0.91249 |
https://www.onlinemath4all.com/worksheet-on-word-problems-on-percentage-1.html | 1,542,491,623,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743854.48/warc/CC-MAIN-20181117205946-20181117231946-00325.warc.gz | 961,906,907 | 13,969 | # WORKSHEET ON WORD PROBLEMS ON PERCENTAGE 1
## About "Worksheet on Word Problems on Percentage 1"
Worksheet on Word Problems on Percentage 1 :
This is the continuity of our web content given on "Word problems on Percentage Worksheet".
Before we look at the problems, if you want to know the shortcuts required for solving percentage problems,
## Worksheet on Worksheet on Word Problems on percentage 1 - Problems
Problem 1 :
A student multiplied a number by 3/5 instead of 5/3. What is the percentage error in the calculation ?
Problem 2 :
There are 15 boys and 12 girls in a section A of class 7. If 3 boys are transferred to section B of class 7,then find the percentage of boys in section A.
Problem 3 :
If there are 3 boys and 7 girls in a class then, what percent of the class is made up of boys ?
Problem 4 :
Two numbers are respectively 20% and 50% are more than a third number, Find the the ratio of the two numbers.
Problem 5 :
In a triangle, the first angle is 20% more than the third angle. Second angle is 20% less than the third angle. Then find the three angles of the triangle.
## Worksheet on Worksheet on Word Problems on percentage 1 - Solution
Problem 1 :
A student multiplied a number by 3/5 instead of 5/3. What is the percentage error in the calculation ?
Solution :
In the given two fractions, the denominators are 5 and 3.
Let assume a number which is divisible by both 5 and 3.
Least common multiple of (5, 3) = 15.
So, let the number be 15.
15 x 3/5 = 9 ----------(1) ---------incorrect
15 x 5/3 = 25 ---------(2) --------correct
Difference between (1) and (2) is 16
Percentage error is
= (Actual error / Correct answer) ⋅ 100 %
= (16 / 25) 100 %
= 64 %
Hence, the percentage error in the calculation is 64 %.
Problem 2 :
There are 15 boys and 12 girls in a section A of class 7. If 3 boys are transferred to section B of class 7,then find the percentage of boys in section A.
Solution :
Before transfer :
No. of boys in section A = 15
No. of boys in section B = 12
Given : 3 boys are transferred from section A to B
After transfer :
No. of boys in section A = 12
No. of boys in section B = 12
Hence, percentage of boys in section A is 50%
Problem 3 :
If there are 3 boys and 7 girls in a class then, what percent of the class is made up of boys ?
Solution :
Total number of students in the class = 3 + 7 = 10
Percentage of boys is
= (No. of boys / Total no. of students) 100 %
= (3/10) 100 %
= 30 %
Hence, 30 % of the class is made up of boys.
Problem 4 :
Two numbers are respectively 20% and 50% are more than a third number, Find the the ratio of the two numbers.
Solution :
Let "x" be the third number.
Then,
the first number = (100+20)% of x = 120% of x = 1.2x
the first number = (100+50)% of x = 150% of x = 1.5x
First no. : second no. = 1.2x = 1.5x
1.2x : 1.5x---------------> 12x : 15x
Dividing by (3x), we get 4 : 5
Problem 5 :
In a triangle, the first angle is 20% more than the third angle. Second angle is 20% less than the third angle. Then find the three angles of the triangle.
Solution :
Let "x" be the third angle.
Then the first angle = 120% of x = 1.2x
The second angle = 80% of x = 0.8x
Sum of the three angles in any triangle = 180°
Then, we have x + 1.2x + 0.8x = 180°
3x = 180° --------> x = 60°
Then the first angle = 1.2(60°) = 72°
The second angle = 0.8(60°) = 48°
Hence, the three angles of the triangle are 72°, 60° and 48°.
Apart from the problems on percentage given above, if you need more problems on percentage, please click the following links.
Percentage Worksheet
Percentage Worksheet - 2
Percentage Worksheet - 3
Percentage Worksheet - 4
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WORD PROBLEMS
HCF and LCM word problems
Word problems on simple equations
Word problems on linear equations
Algebra word problems
Word problems on trains
Area and perimeter word problems
Word problems on direct variation and inverse variation
Word problems on unit price
Word problems on unit rate
Word problems on comparing rates
Converting customary units word problems
Converting metric units word problems
Word problems on simple interest
Word problems on compound interest
Word problems on types of angles
Complementary and supplementary angles word problems
Double facts word problems
Trigonometry word problems
Percentage word problems
Profit and loss word problems
Markup and markdown word problems
Decimal word problems
Word problems on fractions
Word problems on mixed fractrions
One step equation word problems
Linear inequalities word problems
Ratio and proportion word problems
Time and work word problems
Word problems on sets and venn diagrams
Word problems on ages
Pythagorean theorem word problems
Percent of a number word problems
Word problems on constant speed
Word problems on average speed
Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Profit and loss shortcuts
Percentage shortcuts
Times table shortcuts
Time, speed and distance shortcuts
Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6 | 1,726 | 6,411 | {"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} | 4.625 | 5 | CC-MAIN-2018-47 | longest | en | 0.785099 |
https://echo.mpiwg-berlin.mpg.de/ECHOdocuView?tocMode=thumbs&start=221&viewMode=text_image&ws=1.5&url=/mpiwg/online/permanent/archimedes/salus_mathe_040_en_1667&pn=202 | 1,722,817,372,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640417235.15/warc/CC-MAIN-20240804230158-20240805020158-00672.warc.gz | 171,554,482 | 6,103 | Salusbury, Thomas, Mathematical collections and translations (Tome I), 1667
#### List of thumbnails
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221 222 223 224 225 226 227 228 229 230
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1another point in the contact being taken as D, conjoyn the two
right lines A D and B D, ſo as that they make the triangle A D B;
of which the two ſides A D and D B ſhall be equal to the other one
A C B, both thoſe and this containing two ſemidiameters, which
by the definition of the ſphere are all equal: and thus the right
line A B, drawn between the two centres A and B, ſhall not be the
ſhorteſt of all, the two lines A D and D B being equal to it: which
by your own conceſſion is abſurd.
A demon ſtration
that the ſphere
cheth the plane but
in one point.
SIMP. This demonſtration holdeth in the abſtracted, but not in
the material ſpheres.
SALV. Inſtance then wherein the fallacy of my argument
ſiſteth, if as you ſay it is not concluding in the material ſpheres, but
holdeth good in the immaterial and
Why the ſphere in
abſtract, toucheth
the plane onely in
one point, and not
the material in
conerete.
SIMP. The material ſpheres are ſubject to many accidents,
which the immaterial are free from.
And becauſe it cannot be,
that a ſphere of metal paſſing along a plane, its own weight ſhould
not ſo depreſs it, as that the plain ſhould yield ſomewhat, or that
the ſphere it ſelf ſhould not in the contact admit of ſome
on.
Moreover, it is very hard for that plane to be perfect, if for
nothing elſe, yet at leaſt for that its matter is porous: and
haps it will be no leſs difficult to find a ſphere ſo perfect, as that
it hath all the lines from the centre to the ſuperficies, exactly
equal.
SALV. I very readily grant you all this that you have ſaid; but
it is very much beſide our purpoſe: for whilſt you go about to
ſhew me that a material ſphere toucheth not a material plane in
one point alone, you make uſe of a ſphere that is not a ſphere, and
of a plane that is not a plane; for that, according to what you
ſay, either theſe things cannot be found in the world, or if they
may be found, they are ſpoiled in applying them to work the effect.
It had been therefore a leſs evil, for you to have granted the
cluſion, but conditionally, to wit, that if there could be made of
matter a ſphere and a plane that were and could continue perfect,
they would touch in one ſole point, and then to have denied that
any ſuch could be made.
SIMP. I believe that the propoſition of Philoſophers is to be
underſtood in this ſenſe; for it is not to be doubted, but that the
imperfection of the matter, maketh the matters taken in
crete, to diſagree with thoſe taken in abſtract.
SALV. What, do they not agree? Why, that which you your
ſelf ſay at this inſtant, proveth that they punctually agree.
SIMP. How can that be?
SALV. Do you not ſay, that through the imperfection of the
matter, that body which ought to be perfectly ſpherical, and that
plane which ought to be perfectly level, do not prove to be the | 957 | 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} | 2.84375 | 3 | CC-MAIN-2024-33 | latest | en | 0.933145 |
https://studymaterialcenter.in/question/statement-1-two-longitudinal-waves-given-by-equations-y-1-x-t-2a-sin-wt-kx-and-y-2-x-t-a-sin-2-2-w-t-kx-will-have-equal-intensity-2011-rs-statement-2-intensity-of-waves-of-given-frequency-in-same-medi/ | 1,675,580,318,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500250.51/warc/CC-MAIN-20230205063441-20230205093441-00161.warc.gz | 560,267,715 | 24,905 | # Statement – 1 : Two longitudinal waves given by equations : y 1 (x, t) = 2a sin (wt – kx) and y 2 (x, t) = a sin (2 2 ) w -t kx will have equal intensity. [2011 RS] Statement – 2 : Intensity of waves of given frequency in same medium is proportional to square of amplitude only
Question:
Statement – 1 : Two longitudinal waves given by equations: $y_{1}(x, t)=2 a \sin (\omega t-k x)$ and $y_{2}(x, t)=a$ $\sin (2 \omega t-2 k x)$ will have equal intensity. Statement – 2: Intensity of waves of given frequency in same medium is proportional to square of amplitude only.
1. Statement- 1 is true, statement- 2 is false.
2. Statement-1 is true, statement-2 is true, statement2 is the correct explanation of statement- 1
3. Statement-1 is true, statement-2 is true, statement2 is not the correct explanation of statement- 1
4. Statement-1 is false, statement-2 is true.
Correct Option: 1
Solution:
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\begin{aligned}
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&y_{2}=0.05 \cos (0.46 \pi x-92 \pi t)
\end{aligned}
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Therefore all points satisfying the algebraic equation are given by the parametric equation (x,y) = (t2,t3). The equation of HP may be written in the following form t z b y a x 2t b y a x − = + = which shows that HP is a rectilinear surface. 4 Definition. The second section b is given by the same circle of curvature r, point 4 and tangent t4. Find the standard form of the equation of the ellipse satisfying the given conditions. Note that this formula corresponds to Equation (5) of with x n =τ k−1 (n;ξ), y n =τ k (n;ξ), and M=ξ k. It’s the set of points x, y– in the plane– satisfying the equation x squared over a squared, minus y squared over b squared, equals 1. According to [1], [2], the approaches for solving polynomial systems of equations can be classified in two categories as follows: ”1. These points are the vertices of the hyperbola. How do you find an equation of hyperbola with given endpoints of the transverse axis: (0,-6),(0,6); Asymptote: y=3/10 x? Precalculus Geometry of a Hyperbola Standard Form of the Equation 1 Answer. The equation of the locus X (p,q) is. • find angle between given two straight lines and the distance of a point from given line. Find the standard equation of the hyperbola which satisfies the given conditions. Since I'm trying to self teach myself here, the only thing I could find was that the tangent of the angle between the asymptotes is $\dfrac{2ab}{a^2-b^2}$. We can write the equation of a hyperbola by following these steps: 1. Foci: (0,-8),(0,8); Vertices:(0,-6),(0,6). This calculator will find either the equation of the hyperbola (standard form) from the given parameters or the center, vertices, co-vertices, foci, asymptotes, focal parameter, eccentricity, linear eccentricity, latus rectum, length of the latus rectum, directrices, (semi)major axis length, (semi)minor axis length, x-intercepts, and y-intercepts of the entered hyperbola. • learn and use the properties of circle. Make a sketch satisfying these. An icon used to represent a menu that can be toggled by interacting with this icon. Find the equation of a hyperbola satisfying the given conditions. filled-in circles. Sample Problems 1. Question 36: The graph shows the momentum of an object as a function of its speed. In this set of exercises you are given parametric equations. Solution for Find an equation of a hyperbola satisfying the given conditions. I asked this same question two days ago, but was compelled to delete it because nobody was addressing the question. The set of conics can be structured as a Cayleyan space, in which the squared forms from the conics’ equations in Cartesian coordinates are represented through points with coordinates given by their coefficients. This iteration is described in Section V, where also complete details are given for the choice of terminal constraints implemented in our computer program. Find the equation of the hyperbola that has its center at the origin and satisfies the given conditions: Vertices:B(±4,0) Passing through (8,2). center at (2, 5) with the longer axis of length 12 and parallel to the x–axis, shorter axis of length 10 b. \] Therefore, any points on the hyperbola are not only critical points, they are also on the boundary of the domain. Find the equation of the hyperbola with center at (0, 0) satisfying the given conditions a) Foci (±2√2,0) and asymptotes =± b) Vertices (0,±1) and asymptotes =±1 3 14. Given a general-form conic equation in the form Ax 2 + Cy 2 + Dx + Ey + F = 0, or after rearranging to put the equation in this form (that is, after moving all the terms to one side of the "equals" sign), this is the sequence of tests you should keep in mind:. 4, 13 Find the equation of the hyperbola satisfying the given conditions: Foci (±4, 0), the latus rectum is of length 12 Since the foci are on the x-axis. The 1985 BC Calculus exam contained the following problem: Given the differential equation dy dx = −xy lny, y > 0 (a) Find the general solution of the differential equation. Find the standard form of the equation of the hyperbola satisfying the given conditions. Daffa and John J. A convenient one to choose is the following. focus:(-4,0);Directrix=4 Equation of the parabola is. † A hyperbola, roughly speaking, is a curve which consists of two disconnected parabola-like curves which are open in opposite directions. 4, 9 Find the equation of the hyperbola satisfying the given conditions: Vertices (0, ±3), foci (0, ±5) We need to find equation of hyperbola Given Vertices (0, ±3), foci (0, ±5) Since Vertices are on the y-axis So required equation of hyperbola is 𝒚𝟐/𝒂𝟐 – 𝒙𝟐/𝒃𝟐 = 1 ∴ Axi. Using a single variable parameter, derive an equation representing the family of parabolas passing through the three given points. Find the standard form of the equation r the parabola satisfying the given conditions. Find an equation of the circle satisfying the given condition. Find an equation of the line that satisfies the given conditions. If there are no boundary conditions, then finding price functions F (S t, t) that satisfy a given PDE will, in general, not be possible. Find the equations of the hyperbola satisfying the given conditions :Vertices (+-7,0), e=4/3. Moreover, the above statement of Theorem 5. Initial Conditions: ΔT=. Solution for Find the standard form of the equation of the hyperbola satisfying the given conditions. Then draw a graph. Foci at (0-2) and (0,2); vertices at (0,1) and (0, -1) The equation is Enter your answer in the answer box. To Find The Condition That The General Equation Of The Second Degree Should Represent A Pair Of Straight Lines. Determine whether the transverse axis lies on the x- or y-axis. As the given level of technology appreciates, the output will increase with the same level of capital and labour units. Find the equation of a hyperbola satisfying the given conditions. Example: Finding Vertices and Foci from a Hyperbola’s Equation Find the vertices and locate the foci for the hyperbola with the given equation: The vertices are (–5, 0) and (5, 0). See full list on courses. A hyperbola is a curve, specifically a smooth curve that lies in a plane, which can be defined either by its geometric properties or by the kinds of equations for which it is the solution set. 3) Major axis horizontal with length 8; length of minor axis = 4; center (0, 0) 16 4 B) x +2-1 64 16 4 16 x Find the standard form of the equation of the hyperbola satisfying the given conditions. y x –30 –20 –10 –25 –15 –5 5 10 –10 –5 5 10 15 20 25 30. Can you help with these to. The astroid is a sextic curve. Its equation in rectangular coordinates is x 2/3 + y 2/3 = a 2/3, where a is the radius of the fixed circle. Solution for Find an equation of a hyperbola satisfying the given conditions. 2 Writing an Equation Given Two Points. If and are the roots of the equation x2—2px+(p2+q2) = O and tan 9 = n-l sihne show that Sinn 9 Find the eccentricity, centre, foci and vertices of the following hyperbola and. FHMM1034 Mathematics III 109 Example 37. Conic standard rectangular form-hyperbola locus where the difference of the distances from the foci is constant (h,k)=center, 2b²/a=latus rectum, a= major axis (positive number), b=minor axis, c=focus, d=directrix=a²/c=a/e, e=c/a>1, c²=a²+b², conjugate* and transverse axes x=h and y=k, auxiliary rectangle [a x b], auxiliary circle [radius. The implicit differentiation calculator will find the first and second derivatives of an implicit function treating either y as a function of x or x as a function of y, with steps shown. It is shown elsewhere in this article that the equation of the parabola is 4fy = x 2, where f is the focal length. Understand the fundamental equation a 2 = b 2 + c 2 and use it frequently. In this configuration, the Steiner-chain circles have the same type of tangency to both given circles, either externally or internally tangent to both. ? equation of a quartic function with zeros x=-(1/2) and 5, both multiplicity 1, and x=2. The hodographs of two-parameter Lorentzian homothetic motions were obtained. and we know the general solution of Equation (5) the arbitrary constants C 1,. All of them are lower than estimated by DL98b. Divide each side of the equation by 28,224 (yes, the number is huge, but the fractions reduce very nicely) to get the standard form. Find the equations of the hyperbola satisfying the given conditions :Foci (+-4,0), the latus rectum is of length 12. where the last two equations are the normalization conditions determining v 0 and v 1 uniquely for a given u 1. Understand the standard formula for the equation of an ellipse. Login to reply the answers Post. x intercept at (–4, 0) and y -intercept at (0, –6) 6. Determine the eccentricity of the hyperbola. Find the equations of the hyperbola satisfying the given conditions :Vertices (+-7,0), e=4/3. Find the standard form of the equation of the ellipse satisfying the given conditions. Normalized excitation rate due to plasma drag G p (ω th) for a neutral grain, a positively charged grain and a negatively charged grain in CNM conditions (equation 162), evaluated at the ‘thermal rotation rate’ . Any help you can give me would be appreciated. A hyperbola is a type of conic section that looks somewhat like a letter x. So this is the same thing is that. Convert the equation to the standard form for a hyperbola by completing the square on x and y. Find the Standard form of the equation of the hyperbola satisfying the given conditions. (ii) Express the given conditions as equations in terms of the known quantities and unknown parameters. You are given the point (4,3) and a slope of 2. Express your answer in the form y = f(x). For the given -symmetric Scarff-II-like potential (2), based on some transformations, we can find the unified analytical bright solitons of Eq. 47) (x + 3)2 36 + (y - 2)2 16 = 1 47) Find the standard form of the equation of the hyperbola satisfying the given conditions. A good example of a hyperbola is the graph of the function y = x¡1, which we can rewrite into the form xy = 1 (making it a conic section). 17) Comparing (A. Endpoints of transverse axis: (-6, 0), (6, 0); foci: (-7, 0), (-7, 0). That is (2,4). Initial Conditions: ΔT=. By using this website, you agree to our Cookie Policy. 4, 10 Find the equation of the hyperbola satisfying the given conditions: Foci ( 5, 0), the transverse axis is of length 8. Here we find the equation of a conic section given information about the vertices and the asymptotes. (b) Find the focus of the parabola. Free Hyperbola calculator - Calculate Hyperbola center, axis, foci, vertices, eccentricity and asymptotes step-by-step This website uses cookies to ensure you get the best experience. 78) 25x2 + 49y2 = 1225 x y Graph. 3 – Hyperbolas 41. ((x - 2) 2 /9) - (y + 5) 2 = 1. We should also note that the domain of $$f$$ consists of points satisfying the inequality $4y^2−9x^2+24y+36x+36≥0. Find the equations of the hyperbola satisfying the given conditions :Foci (+-4,0), the latus rectum is of length 12. 4, 15 Find the equation of the hyperbola satisfying the given conditions: Foci (0,±√10), passing through (2, 3) Since Foci is on the y−axis So required equation of hyperbola is 𝑦2/𝑎2 – 𝑥2/𝑏2 = 1 Now, Co-ordinates of foci = (0, ± c) & given foci = (0, ±√10) So, (0, ± c) = (0, ±√10) c = √𝟏𝟎 Also, c2 = a2 + b2 Putting value of c (√10)2 = a2. c) Sketch the graph of the equation. The general shape of the curve is shown in Figure 1. HP may be given in a parametric form as: x =aρ cosh v, y =bρ sinh v, z =u2. 4, 15 Find the equation of the hyperbola satisfying the given conditions: Foci (0,±√10), passing through (2, 3) Since Foci is on the y−axis So required equation of hyperbola is 𝑦2/𝑎2 - 𝑥2/𝑏2 = 1 Now, Co-ordinates of foci = (0, ± c) & given foci = (0, ±√10) So, (0, ± c) = (0, ±√10) c = √𝟏𝟎 Also, c2 = a2 + b2 Putting value of c (√10)2 = a2. To solve differential equations, use the dsolve function. By changing initial conditions, we can create di erent trajectories for the given system. find the equation of the ellipse satisfying the given conditions. (a) transform a given equation of a conic into the standard form; (b) find the vertex, focus and directrix of a parabola; (c) find the vertices, centre and foci of an ellipse; (d) find the vertices, centre, foci and asymptotes of a hyperbola; (e) find the equations of parabolas, ellipses and hyperbolas satisfying prescribed conditions. Therefore all points satisfying the algebraic equation are given by the parametric equation (x,y) = (t2,t3). gl/JQ8Nys Finding the Equation of the Hyperbola Given the Center, Focus, and a Vertex. Convert the equation to the standard form for a hyperbola by completing the square on x and y. y 2 /36 - x 2 /9 = 1 C. a) Find the centre and the eccentricity of the hyperbola +4y — 4 0 and x 2 o. Equations x 3 + 9 x 2 = 100, x 3 + 3 x 2 = 2 and x 3 + 7 x 2 = 50, from Q. *** given hyperbola has a horizontal transverse axis with center at origin. 13), it follows easily that (2. How To: Given a polynomial function, sketch the graph. Endpoints of major axis: (7, 9) and (7, 3) Endpoints of minor axis: (5, 6) and (9, 6). The above equation explains that Q x, units of output x are produced by employing L and K units of labour and capital respectively and by a given technology. Ellipse: the set of points for each of which the sum of the distances to two given foci is a constant; Other examples of loci appear in various areas of mathematics. Q13 :Find the equation of the hyperbola satisfying the give conditions: Foci (±4, 0), the latus rectum is of length 12 Answer : Foci ( ±4, 0), the latus rectum is of length 12. 2 The equation x = y2 z2 is. Find the equation of the parabola whose focus is (5, 3) and the directrix is given by 3x -4 y +1 = 0. y 2 /37 - x 2 /27 = 1 D. If and are the roots of the equation x2—2px+(p2+q2) = O and tan 9 = n-l sihne show that Sinn 9 Find the eccentricity, centre, foci and vertices of the following hyperbola and. 3) should be expressed in the form u(x,y) = f(x+ iy)+g(x− iy), (2. Find the first order differential equation (in which c does not appear) satisfied by each hyperbola of the family y = -C -where X c is an arbitrary constant and x # c. 4, 10 Find the equation of the hyperbola satisfying the given conditions: Foci ( 5, 0), the transverse axis is of length 8. vertices at (-3, 0) and (3, 0) and asymptotes of y = x. Find the equation of the locus of a point P( x, y ) such that (i) AP BP (ii) AP 2 BP. Output arguments let you access the values of the solutions of a system. (b) Find the smallest distance (the perigee) from Mars to the Sun. So far we have considered only pairs of straight lines through the origin. Find the equation of hyperbola satisfying given conditions foci (5, 0) and transverse axis is of length 8. Find an equation of a hyperbola satisfying the given conditions: Vertices at (1;0) and ( 1;0) foci at (2;0) and ( 2;0). ⇐ Condition for Line Tangent to a Hyperbola ⇒ Find the Equation of the Tangent Line to the Hyperbola ⇒. Our problem is to find this minimal sum. Endpoints of major axis: (7, 9) and (7, 3) Endpoints of minor axis: (5, 6) and (9, 6). Use symmetry to help you graph an ellipse. Further, the fact that derivative products are known functions of. (1) Find the formula for locus of all such M; (2) If OM p 3 , find the angle of elevation of AB (the angle of AB makes with the positive x-axis). Find the equation of Hyperbola satisfying the following conditions: Vertices (pm2,0), foci (pm3,0). We have seen above that the “natural choice” of space E = H 1 0 (Ω) x H 1 0 (Ω) leads to the known Sobolev growth restriction for both nonlinearities F(s) and G(s). It can be shown that the set of points P in the (x,y) plane which satisfy the condition distance of P from origin. 10 Equations of a Line 3. To find the equation ofthecircle determined by three points, substitute the x and yvaluesof each of the given points into the general equation toform three equations with B, C, and D as the unknowns. 86) 9 x 2 - 4 y 2 + 36 x - 8 y - 4 = 0 86) Find the standard form of the equation of the hyperbola satisfying the given conditions. From this quadratic equation we find that c is a rational function of the square root of a 4 a 2 d 2 +d 4, which implies there is an odd integer m such that a 4 a 2 d 2 +d 4 = m 2. 17) with the equations for the hyperboloids of one and two sheet we see that the cone is some kind of limiting case when instead of having a negative or a positive number on the l. 4 Graphing a Line Using Point and Slope 3. Seyranian and Mailybaev ( 2003b )). Endpoints of major axis: (7, 9) and (7, 3) Endpoints of minor axis: (5, 6) and (9, 6) A. vertices at (0, 1) and (0, -1) and asymptotes of y x. Equation of hyperbola is y^2/25-x^2/39=1 As the focii and vertices are symmetrically placed on y-axis, its center is (0,0) and the equation of hyperbola is of the type y^2/a^2-x^2/b^2=1 As the distance between center and either vertex is 5, we have a=5 and as distance between center and either focus is 8, we have c=8 As c^2=a^2+b^2, b^2=8^2-5^2=39 and equation of hyperbola is y^2/25-x^2/39=1. Graph Individual (x,y) Points - powered by WebMath. Thanks for contributing an answer to Mathematics Stack Exchange! Please be sure to answer the question. center at (2, 5) with the longer axis of length 12 and parallel to the x–axis, shorter axis of length 10 b. Derive the equations of asymptotes of a hyperbola Find the eccentricity and fuci of the curve represents a translated parabola. 39)49y2 - 100x2 = 4900 of the hyperbola satisfying the given conditions. The required equation to the locus under the given conditions is x 2 + y 2 = 16. Question 605623: locate the center, foci, vertices, and ends of the latera recta of the ellipse. Notice that ${a}^{2}$ is always under the variable with the positive coefficient. BYJU’S online hyperbola calculator tool makes the calculation faster, and it displays the values in a fraction of seconds. Find the equations of the hyperbola satisfying the given conditions. See full list on courses. Suppose your two ellipses have equations $e_1(x,y)=0$ and $e_2(x,y)=0$. You need to know the (or at least a) definition of a hyperbola. Foci ( ±3,0 ) , vertices ( ±5,0 ) Example 5: Write an equation of the ellipse with vertices (5, 9) and (5, 1) if one of the foci is. Find the vertex, focus and directrix. SOLUTION (a) The parabola is sketched in Figure 5. Find a linear transformation T(·) such that the function w = T(z2)1/2, with the principal branch of the square root chosen, maps 0 to 0 and the hyperbola xy = 1 onto the hyperbola u2 −v2 = 1. frequency table. We should also note that the domain of consists of points satisfying the inequality Therefore, any points on the hyperbola are not o nly critical points, they are also on the boundary of the domain. The shape of paths, the direction of trajectories, and equilibrium solutions are some of the qualitative features we will explore. Find an equation for a hyperbola that satisfies the given conditions. The relations to the hyperbola, 16K2—117X2—18X— 1 =0, an d to the parabola, K2—3X = 0, premit of the ready plotting of the curve with sufficient accuracy. Label the intercepts. Find an equation of the circle satisfying the given condition. SOLUTION: Find the equation of a hyperbola satisfying the given conditions. The map is undefined at points satisfying. Express in terms of and , given that the tip of bisects the. Vertices (±2,0), foci (±3,0) Solution: Vertices are (±2, 0) which lie on x-axis. Thanks for contributing an answer to Mathematics Stack Exchange! Please be sure to answer the question. Vertices at (1 ,−7 ) and (1 ,1 ); asymptotes y=4x−7 , y=−4x+1. Moreover, the above statement of Theorem 5. The curve consists of two portions one of which extends along the axis to an infinite value whilst the other extends on the negative side of the axis in a similar manner. Find the standard form of the equation of each hyperbola 9. Bifurcation of λ 0 into two eigenvalues λ ± and the corresponding eigenvectors u ± are described by (see e. 32) are graphed in % , Rp space. c a b2 2 2 c2 25 16 41 c r 41 41,0 and 41,0. Given: x^(2x² + 4x – 6) = x^(x² + 8x + 6) Since the bases are the same, we can write: 2x² + 4x – 6 = x² + 8x + 6 Rearrange to get: x² - 4x – 12 = 0 Factor to get: (x - 6)(x + 2) = 0 So, x = 6 and x = -2 are also solutions. Review Ellipse HW on p. Learn how to write the equation of hyperbolas given the characteristics of the hyperbolas. 10 Equations of a Line 3. In each Exercises 7 to 15, find the equations of the hyperbola satisfying the given conditions. (h) Roses (Figure 2, h), curves whose equation in polar coordinates is ρ = a sin m ϕ; if m is a rational number, then the roses are algebraic. (b) Find the focus of the parabola. Find the standard form of the equation of the ellipse satisfying the given conditions. H x2 " y2 " 2Kxy r2 hy " kx 0, which is an equation of a hyperbola. the equation has been written in standard form, identifying the axis amounts to identifying the variable of degree 1. If the hyperbola is forced to pass through a 'failure' point with co-ordinates (ãult, quit), thus satisfying condition (iii), the modified curve will tend to a new asymptote qa given by ãult. Find the points of intersection of the solution curves of the polar coordinate equations and. 2 OBJECTIVES 1 Recognize the equation of a hyperbola. Center: (4, -2); focus: (10,-2); vertex: (9,-2) The equation is. Find the general solution of 3. So, we mark them using. Graph Individual (x,y) Points - powered by WebMath. Because a > b, x 2 = b 2 + 1 / a < b as long as b > 1. Find the equation of the locus of a point P( x, y ) such that (i) AP BP (ii) AP 2 BP. Co-ordinates of foci is (±5, 0) Which is of form (±c, 0) Hence c = 5 Also , foci lies on the x-axis So, Equation of hyperbola is 𝑥2𝑎2 – 𝑦2𝑏2. Find the equations of the hyperbola satisfying the given conditions :Vertices (+-7,0), e=4/3. The map is undefined at points satisfying. asked Dec 22,. Find the foci of the ellipse whose equation is given. There are in general four solutions, since a circle and hyperbola can intersect in four points. This more general line p X defined through (6) is called the polar of x with respect to the conic. 3) Major axis horizontal with length 8; length of minor axis = 4; center (0, 0) 16 4 B) x +2-1 64 16 4 16 x Find the standard form of the equation of the hyperbola satisfying the given conditions. by general equation of hyperbola x^/36-y^/13=1. Question 605623: locate the center, foci, vertices, and ends of the latera recta of the ellipse. Therefore, the equation of the hyperbola is of the form. How do you find an equation of hyperbola with given endpoints of the transverse axis: (0,-6),(0,6); Asymptote: y=3/10 x? Precalculus Geometry of a Hyperbola Standard Form of the Equation. e = 3/2, and directrix y = 2. In this one, we were to find out the locus of a point such that it is equidistant from two fixed points, which was the perpendicular bisector of the line joining the points. 1) if the. Standard form to vertex form worksheet. the initial conditions. 42, are handled in the same way, since the associated equations of type have the roots 5, 1 and 5 respectively. Hyperbola Calculator,Hyperbola Asymptotes. This page will help you to do that. y 2 - 4x 2 = 4. of the quadratic equation we have exactly 0. Use symmetry to help you graph an ellipse. So in general we can say that a circle centered at the origin, with radius r, is the locus of all points that satisfy the equations. Find the equation of the hyperbola, referred to its principal axes as axes of coordinates, in the following cases: (i) the distance between the foci = 16 and eccentricity = 2 (ii) conjugate axis is 5 and the distance between foci = 13 (iii) conjugate axis is 7 and passes through the point (3, −2). Find an equation for the conic that satisfies the given conditions. Find an equation in standard form for the hyperbola that satisfies the given conditions: Transverse axis endpoints (3,3) and (3,−1), conjugate axis length 8. If hyperbola is 22 1, 13,- then a2=1, b2=3 Condition for the line y = mx + c, to be a tangent to the hyperbola 22 22 1,,- is, C = ± # 22 2. Find the equation of the tangent to the ellipse coordinates (3cos O, 2sin O). Find the coordinates of the foci, vertices, length of major axis, length of minor axis , eccentricity and length of latus rectum(LL’): i) x2/9 - y2/4 = 1 ii) 2x2-3y2 = 5 iii) y2/5 - x2/16= 1 2. The directrix is given by the equation. 4, 9 Find the equation of the hyperbola satisfying the given conditions: Vertices (0, ±3), foci (0, ±5) We need to find equation of hyperbola Given Vertices (0, ±3), foci (0, ±5) Since Vertices are on the y-axis So required equation of hyperbola is 𝒚𝟐/𝒂𝟐 – 𝒙𝟐/𝒃𝟐 = 1 ∴ Axi. Find the equation of a hyperbola satisfying the given conditions. Graph the inequality, factor the trinomial w^2+9x+14, Standard form Parabola given conditions calculator, multiplication of 2 radicals, finding equation of a line. From the above we can find the coordinates of any point on the circle if we know the radius and the subtended angle. 39)49y2 - 100x2 = 4900 of the hyperbola satisfying the given conditions. Solving this equation for y will produce this equation for our parabola. (2) Equation (2) is similar to equation of a rectangular Hyperbola of the form xy = c2 , with shifted origin at (3,3) So given Hyperbola is also a rectangular Hyperbola, with c = √(2) We know that for a rectangular Hyperbola b = a = c√(2) So value of a for given Hyperbola = c√(2) = √(2) × √(2) = 2 For any rectangular Hyperbola length. Cap Sol 2 - Free download as PDF File (. frustum of a pyramid. Actually, this equation is the general equation of a conic. Equations x 3 + 9 x 2 = 100, x 3 + 3 x 2 = 2 and x 3 + 7 x 2 = 50, from Q. You will find that x = –2 and x = –3 are the two zeroes of y. of the lengths of the two ladders. Type your answer in standard form. Find an equation of an ellipse satisfying the given conditions. Find the equation of the hyperbola which satisfies the given conditions: a. x-6y+4z=1 3x-5y+3z=-1 Find the standard form of the equation of the hyperbola satisfying the given conditions … read more. Find vertices and the Foci of the ellipse given equation ( like #23 -30 on pp. Thus the propagating beam solution becomes a satisfactory transverse mode of the resonator. And if we define now, another parameter, b, by means of the equation b squared equals c squared minus a squared– then we can write the canonical equation of a hyperbola in the following form. [Answer: (3, -2), 5] 2. 4 Definition. Tak- ing the square root of both sides of equation (232), becomes a hyperbolic function of Differentiating (2. foci (−3, −2) and (15, −2), a vertex at (9, −2) 71 3 − 43 x and y = 34 x − EP E 10. If and are the roots of the equation x2—2px+(p2+q2) = O and tan 9 = n-l sihne show that Sinn 9 Find the eccentricity, centre, foci and vertices of the following hyperbola and. It is shown elsewhere in this article that the equation of the parabola is 4fy = x 2, where f is the focal length. Includes full solutions and score reporting. If you are given, say, the polynomial equation y = x2 + 5x + 6, you can factor the polynomial as y = (x + 3)(x + 2). Question 605623: locate the center, foci, vertices, and ends of the latera recta of the ellipse. Find the intercepts. Free Hyperbola Vertices calculator - Calculate hyperbola vertices given equation step-by-step This website uses cookies to ensure you get the best experience. Asymptotes y = -x, y= - 5x; vertices at (6, 0) and (-6, 0). Find the standard form of the equation of each hyperbola 9. We observe that at the minimum the level curve of f is tangent to the hyperbola. It is this equation. We are trying to find an equation for all of the points that are the same distance (5 units) from (–3, 6). ((x - 2) 2 /9) - (y + 5) 2 = 1. A hyperbola is a type of conic section that looks somewhat like a letter x. The parametric equation of a circle. The pre-image of heads. A good example of a hyperbola is the graph of the function y = x¡1, which we can rewrite into the form xy = 1 (making it a conic section). Hence, or otherwise, show that the equation of L is Find the polar equations of and L Find the area of the region enclosed by I' (i.$ Therefore, any points on the hyperbola are not only critical points, they are also on the boundary of the domain. The midpoint of any pair of foci gives you the center. Center is at the origin, passing. BYJU’S online equation of a circle calculator tool makes the calculation faster, and it displays the equation in a fraction of seconds. Question 1. Endpoints of transverse axis (0, ±6); Asymptote: y = 2x. Thus the propagating beam solution becomes a satisfactory transverse mode of the resonator. 3 Interpreting Slope from a Graph 3. y 2 /37 - x 2 /27 = 1 D. 1 ) is given by. Endpoints of transverse axis: (-6, 0), (6, 0); foci: (-7, 0), (-7, 0). Foci: (0,-8),(0,8); Vertices:(0,-6),(0,6). foci (−3, −2) and (15, −2), a vertex at (9, −2) 71 3 − 43 x and y = 34 x − EP E 10. 3 Exercises. This calculator will find either the equation of the hyperbola (standard form) from the given parameters or the center, vertices, co-vertices, foci, asymptotes, focal parameter, eccentricity, linear eccentricity, latus rectum, length of the latus rectum, directrices, (semi)major axis length, (semi)minor axis length, x-intercepts, and y-intercepts of the entered hyperbola. It remains to choose an analytic function f such that equation (8) is satisfied. (#16) x2 144 y2 81 = 1 2. (b) Find the focus of the parabola. iii) The locus of a point is the path traced by it, when it moves under a given condition or conditions. So, we mark them using. to variable y ) of course depending on x and the conic. Foci: (0, -2), (0, 2); y-intercepts: -5 and 5 the foci for the hyperbola. focus:(-4,0);Directrix=4 Equation of the parabola is. 8), We easily establish that. We shall now analyze (7) and (8) to find out if the coin ends up heads for given values of the initial velocity u and the initial angular velocity w. How do you write a linear equation, algebra problem solvers, compound inequality solver, biology multiplication rule, calculation intermediate algebra calculator, Solve for x+6y=11. How To: Given the equation of a hyperbola in standard form, locate its vertices and foci. The Hyperbola and Functions Defined by Radicals 13. Now let's assume that x ≠ 0, x ≠ 1 and x ≠ -1 and look for other x-values that satisfy the given equation. Moreover, the above statement of Theorem 5. Find the equations of the hyperbola satisfying the given conditions. The 1985 BC Calculus exam contained the following problem: Given the differential equation dy dx = −xy lny, y > 0 (a) Find the general solution of the differential equation. the initial conditions. The operator L is This is a parabolic operator according to the definition given above ; in fact, the matrix A in ( 3. The center of the hyperbola is (3, 5). SOLUTION: Find the equation of a hyperbola satisfying the given conditions. Find the unique solution if c 6= 1. The equation was verified for six special cases of PQ media for which the analytic form has been found from previous studies. Simplify Sometimes you will be given a graph and other times you might just be told some information. (h) Roses (Figure 2, h), curves whose equation in polar coordinates is ρ = a sin m ϕ; if m is a rational number, then the roses are algebraic. 17) Comparing (A. So f squared minus a square. 11) is satisfied. The directrix is given by the equation. Plane Analytic Geometry The hyperbola given by equation (I) is symmetric about the coordinate axes (Fig. x intercept at (–4, 0) and y -intercept at (0, –6) 6. focus:(-4,0);Directrix=4 Equation of the parabola is. I have the following equation containing the variables x and y, where A, c > 0 and B is a constant, which I am not told anything about: -c^2x^2+y^2+2By+A=0. Find the equation to the conic section whose focus is (1, -1), eccentricity is 1/2 and the directrix is the line x -y = 3. also satisfy both inequalities, they are solutions of the system as well. For example in Fig. By then completing the square with respect to both x and y, one will obtain one of the standard equations given above, for either an ellipse or a hyperbola. By using basic algebra, we can solve this system of equations and find that the only possible points on the hyperbola that would allow for the shortest distance to the origin are at $(1,0,0)$ and $(-1,0,0)$. (b) The sketch should have the following features:. Locate the foci and nd the equations of the asymptotes. \] Therefore, any points on the hyperbola are not only critical points, they are also on the boundary of the domain. All answers in this set can be written in the form y=f(x). Find an equation for the conic that satisfies the given conditions. The Hyperbola and Functions Defined by Radicals 13. Let c be the circle x2 y2 52, P. Initial Conditions: ΔT=. use the graph of the hyperbola, the straight line 𝑦𝑦= 1 and the point of intersection to solve the inequality (eii) find the critical points and tested points on either side to reach the correct solution (eii) consider options to satisfy the given condition: 4𝑥𝑥−1 > 1 or 4𝑥𝑥−1 < 0 (eii). 1 corrects several typos in [15, Theorem 1. Solution for Find an equation of a hyperbola satisfying the given conditions. Find the standard form of the equation of the hyperbola satisfying the given conditions Endpoints of transverse axis: (0, -18), (0,18); asymptote: y = 2x The equation is Get more help from Chegg Get 1:1 help now from expert Algebra tutors Solve it with our algebra problem solver and calculator. Seyranian and Mailybaev ( 2003b )). Find the equations of the hyperbola satisfying the given conditions. Then graph the parabola. Switching the roles of t and x in this equation gives one of the. For example, suppose 0 to be a given point in the plane of the paper and that a point P is to move on the paper so that its distance from 0 shall be constant and equal to a. Find the standard form of the equation of the ellipse and give the location of its foci. lization, given in Section IV) to find a sequence of Kepler arcs satisfying the ter-minal constraints and the current values of the control parameters. The graph of y is a hyperbola with two branches, as shown in Figure 2. also satisfy both inequalities, they are solutions of the system as well. The graph of a function f is given. Another example is the curve of the relation y2 ¡x2 = 1. Center (0,0), transverse axis along the x-axis, a focus at (8,0), a vertex at (4,0). To nd the new equation for our. 48) Endpoints of transverse axis: (0, -10), (0, 10); asymptote: y = 5 6 x 48) Eliminate the parameter. focus: A point used to construct and define a conic section, at which rays reflected from the curve converge (plural: foci). Foci: (0, -2), (0, 2); y-intercepts: -5 and 5 the foci for the hyperbola. Two ellipses typically have four common tangents. EQUATION TO A LOCUS. The relationship between x and y , satisfying the conditions, is called the cartesian equation of the locus of P. The canonical (standard) equation of the hyperbola: x2 y2 (1) lif-bi"=l. Conversely, an equation for a hyperbola can be found given its key features. For the given -symmetric Scarff-II-like potential (2), based on some transformations, we can find the unified analytical bright solitons of Eq. Find all real numbers r for which there exists exactly one real number a such that when (x+a)(x2 +rx +1) is expanded to yield a cubic polynomial, all of its coefficients are greater than or equal to zero. (x - 7)2/6 + (y - 6)2/7 = 1 B. Foci:(-4,0) and (4,0) Length of major axis: 10 The equation of the. If 1 = 2 6= 0, then the equation (4. Given that and and another vector , find numbers k and m so that. Find the standard form of the equation of the hyperbola satisfying the given conditions. Note that the left branch of the hyperbola in Figure 2 passes through the point. For example, if the equation involves the velocity, the boundary condition might be the initial velocity, the velocity at time t=0. You can, however, also work backwards from the zeroes to find the originating. How do you find an equation of hyperbola with given endpoints of the transverse axis: (0,-6),(0,6); Asymptote: y=3/10 x? Precalculus Geometry of a Hyperbola Standard Form of the Equation. To put the hyperbola in standard form, we use the method of completing the square:. Find the equations of the hyperbola satisfying the given conditions :Foci (+-4,0), the latus rectum is of length 12. Write the equation of the conic section satisfying the given conditions. Solution for Find the standard form of the equation of the hyperbola satisfying the given conditions. You will find that x = –2 and x = –3 are the two zeroes of y. A good example of a hyperbola is the graph of the function y = x¡1, which we can rewrite into the form xy = 1 (making it a conic section). A parametric representation of a curve is not unique. That means, that we will apply the rotation matrix R ˇ 4 to the hyperbola. Equation of a Circle Calculator is a free online tool that displays the equation of a circle of a given input. The hyperbola x 2 /a 2 y 2 /b 2 = 1 passes through the point of intersection of the lines,. The curve consists of two portions one of which extends along the axis to an infinite value whilst the other extends on the negative side of the axis in a similar manner. Calculate(r => 2 * Math. Endpoints of major axis: (7, 9) and (7, 3) Endpoints of minor axis: (5, 6) and (9, 6). To solve differential equations, use the dsolve function. From Figure 3 we read off the constraint equations: u + v - 50 = 0, (p + u)2 + y2 - L 1 2 = 0, (q + v)2 + y2 - L 2 2 = 0 (4) (y - 10)p - 10u = 0, (y - 15)q - 15v = 0. HP may be given in a parametric form as: x =aρ cosh v, y =bρ sinh v, z =u2. So in general we can say that a circle centered at the origin, with radius r, is the locus of all points that satisfy the equations. Check for symmetry. Provide details and share your research! But avoid …. 13) with y_1 computed for Y given by (2. Note that x, the scale of the zero-investment strategy, does not appear in the formula -- all strategies involving a given asset or portfolio have the same value of xrsr, no matter what their scale (assuming, of course, that the rate of interest is unaffected by the amount borrowed). Lemmas 11 and 12 in the equation R = R 1 ; and the period identities for sine and cosine in the equation R +2ˇ= R R 2ˇ. find the equation of the ellipse satisfying the given conditions. By using this website, you agree to our Cookie Policy. " Descartes found that the graphs of second-degree equations in two variables always fall into one of seven categories: [1] single point, [2] pair of straight lines, [3] circle, [4] parabola, [5] ellipse, [6] hyperbola, and [7] no graph at all. fundamental units. Find the first order differential equation (in which c does not appear) satisfied by each hyperbola of the family y = -C -where X c is an arbitrary constant and x # c. (x - 7)2/5 + (y - 6)2/6 = 1 C. Find the standard form of the equation of the hyperbola satisfying the given conditions. Make a sketch satisfying these. If there are no boundary conditions, then finding price functions F (S t, t) that satisfy a given PDE will, in general, not be possible. Find an equation of an ellipse satisfying the given conditions. The equation of the locus X (p,q) is. Given the equation y =- 2r+8, find the y intercept. Take square root of each side. find an equations for the conic section that satisfies the given conditions. Therefore, when we examine conditions which determine position of a line in relation to a hyperbola that is, when solve the system of equations, y = mx + c: b 2 x 2-a 2 y 2 = a 2 b 2 then if, a 2 m 2-b 2 > c 2 the line intersects the hyperbola at two points,. \text { Asymptotes: } y=\pm x, \text { hyperbola passes through }(5,3) Problem 45. The center of the circle will be (–3, 6), and the radius, which is the distance from (–3,6), will be 5. Its standard form of equation: a=2 a^2=4 slopes of asymptotes=3/2=b/a b=3a/2=3 b^2=9 Equation of given hyperbola:. Using a single variable parameter, derive an equation representing the family of parabolas passing through the three given points. Find the curve of the family. Find the standard form of the equation of each hyperbola 9. notebook 2 December 08, 2017 Example 2 Find the equation of the hyperbola satisfying the given conditions Foci: (0,5)(0,-5) Length of Transverse axis is 2. Tak- ing the square root of both sides of equation (232), becomes a hyperbolic function of Differentiating (2. Hence, the required equation of the hyperbola is 𝒙𝟐/𝒂𝟐 – 𝒚𝟐/𝒃𝟐 = 1 Now, coordinates of foci are (±c, 0) & given foci = (±4,. Consequently, this function cannot be considered to be a solution to the differential equation y = y2 over the whole real line. Endpoints of transverse axis: (0, -6), (0, 6) Asymptote: y = 2x A. find the equation for the specified hyperbola center at the origin, latus rectum 64/3, eccentricity 5/3. The above equation explains that Q x, units of output x are produced by employing L and K units of labour and capital respectively and by a given technology. Find the equations of the hyperbola satisfying the given conditions :Vertices (0,+-5),f o c i (0,+-8). (#16) x2 144 y2 81 = 1 2. Solving the synthesis problem, we find the conic. Once you have entered the expression, press CHECK to see if you have obtained the correct answer. Find the equation of a hyperbola satisfying the given conditions. 47) Two children are playing with a ball. Simplify Sometimes you will be given a graph and other times you might just be told some information. Graph the inequality, factor the trinomial w^2+9x+14, Standard form Parabola given conditions calculator, multiplication of 2 radicals, finding equation of a line. 86) 9 x 2 - 4 y 2 + 36 x - 8 y - 4 = 0 86) Find the standard form of the equation of the hyperbola satisfying the given conditions. Given that and and another vector , find numbers k and m so that. vertices at (1, 0) and (-1, 0) and. Finding the Equation of a Hyperbola Find an equation for the hyperbola that satisfies the given conditions. pls graph it Answer by KMST(5285) (Show Source):. Type your answer in standard form. You are to eliminate the parameter and find an expression between y and x. 10 Equations of a Line 3. The critical hyperbola. Identify the conic. 77) Foci: ( - 10 , 0), ( 10 , 0); vertices: ( - 4 , 0), ( 4 , 0). Vertices (±2,0), foci (±3,0) Solution: Vertices are (±2, 0) which lie on x-axis. frustum of a cone. Find the equations of the hyperbola satisfying the given conditions :Vertices (+-7,0), e=4/3. Find the x- and y-intercepts of the graph of the circle given by the equation Solution To find any x-intercepts, let To find any y-intercepts, let x-intercepts: Substitute 0 for y. The graph of the quadratic function. y 2 - 4x 2 = 4. Find the standard form of the equation of the hyperbola satisfying the given conditions Endpoints of transverse axis: (0, -18), (0,18); asymptote: y = 2x The equation is Get more help from Chegg Get 1:1 help now from expert Algebra tutors Solve it with our algebra problem solver and calculator. In all of these special cases, the quartic equation either reduces to two quadratic equations or becomes an identity. Initial Conditions: ΔT=. Hence, the required equation of the hyperbola is 𝒙𝟐/𝒂𝟐 – 𝒚𝟐/𝒃𝟐 = 1 Now, coordinates of foci are (±c, 0) & given foci = (±4,. (a) x = 4 Solution: x = 4) rcosµ = 4) r = 4secµ (b) x2 +y2 = ¡2x Solution: x2 +y2. Find the equations of the hyperbola satisfying the given conditions :Foci (+-4,0), the latus rectum is of length 12. 11) is satisfied. Lemmas 11 and 12 in the equation R = R 1 ; and the period identities for sine and cosine in the equation R +2ˇ= R R 2ˇ. The center of the hyperbola is (3, 5). So, for example, if I had a focus at the point, I don't know, let's say the point (1,2), and I had a directrix at y is equal to, I don't know, let's make it y is equal to -1, what would the equation of this. Save for Later M DOO BO F3 هروب eSC FL FA F2 +1 us # % 3 ľ4 E5 0 - 2 ۲ الحقول Q W E R. calc 501-1000. parabola focus (1, 2 ž) directrix y = 1 - 10. Find the general solution of 3. Find an equation for the hyperbola that satisfies the given conditions. Given the cosine or sine of an angle, finding the cosine or sine of the angle that is half as large involves solving a quadratic equation. Note that the left branch of the hyperbola in Figure 2 passes through the point. The trajectory in equation ( 1 ) corresponds to a charge that comes to rest at at time t = 0 after traveling an infinite distance from the infinite past where its speed 1. Graph the equation $$\frac{(x-2)^2}{4} -\frac{y^2}{25} = 1. The 1985 BC Calculus exam contained the following problem: Given the differential equation dy dx = −xy lny, y > 0 (a) Find the general solution of the differential equation. The direction field along this hyperbola has slope −2. The equation of the pair of lines and is obviously given by the equation:. It can be shown that the set of points P in the (x,y) plane which satisfy the condition distance of P from origin. 4, 13 Find the equation of the hyperbola satisfying the given conditions: Foci (±4, 0), the latus rectum is of length 12 Since the foci are on the x-axis. Find the equations of the hyperbola satisfying the given conditions. Hence, the required equation of the hyperbola is 𝒙𝟐/𝒂𝟐 – 𝒚𝟐/𝒃𝟐 = 1 Now, coordinates of foci are (±c, 0) & given foci = (±4,. In fact, given the point x, not necessarily on the conic, equation (6) makes sense and defines a line (w. Ellipse endpoints of major axis (4, 3) and (-6, 3) foci (-5, 3) and (3, 3) Circle center (-9, -12) and passes through (-4õ) yperboa vertices (0, 3) and (0, -3) conjugate axis of length 12. Stroyls's Studies in the Exact Sciences in. 1) if the. Foci at (0-2) and (0,2); vertices at (0,1) and (0, -1) The equation is Enter your answer in the answer box. When solving a system of equations, always assign the result to output arguments. A hyperbola with a vertical transverse axis and center at (h, k) has one asymptote with equation y = k + (x - h) and the other with equation y = k. Center (-2,-2) and radius 7 13. Pre-Calculus Hyperbolas Name_____ [Day 2] Notes March 2015 EXAMPLE 1 – Writing Equations of Hyperbolas Find the standard form of the equation of each hyperbola satisfying the given conditions. In the original coordinates, this reduces to u 1x+ u 2y = (v 1x+ v 2y): 8. Find the Standard form of the equation of the hyperbola satisfying the given conditions. The equation for surface area of parametric curve c(t) given its conditions which are on another flash card Radial coordinate What we call r of point P (expressed (r,θ)) where r is the distance to origin O. Find the center, the vertices, the foci, and the asymptotes of x2 25 y2 9 = 1 Sketch the graph. also satisfy both inequalities, they are solutions of the system as well. 32), we can also see that the hyperbola's slope equals. BYJU’S online equation of a circle calculator tool makes the calculation faster, and it displays the equation in a fraction of seconds. Of course, when we have the asymptotes of an hyperbola, we have immediately the second common point of a given line with the hyperbola if we know the first one. From the equation in this, the well known uv+kws = 0 form, numerous elementary geometrical facts can be derived. Find the first order differential equation (in which c does not appear) satisfied by each hyperbola of the family y = -C -where X c is an arbitrary constant and x # c. Find the standard form of the equation of each ellipse satisfying the given conditions. All of them are lower than estimated by DL98b. Question 605623: locate the center, foci, vertices, and ends of the latera recta of the ellipse. The equation was verified for six special cases of PQ media for which the analytic form has been found from previous studies. 86) 9 x 2 - 4 y 2 + 36 x - 8 y - 4 = 0 86) Find the standard form of the equation of the hyperbola satisfying the given conditions. 4) Endpoints of transverse axis: (0, -4), (0, 4); asymptote: y. 4 Notes Done. Asking for help, clarification, or responding to other answers. frustum of a cone. Finding the Equation of a Hyperbola Find an equation for the hyperbola that satisfies the given conditions. ? Center: (4,5) , Focus: (-1,5) and Vertex: (3,5) Please help, I need to study for a test. For each hyperbola, find the center, vertices, foci, asymptotes, and intercepts. The simple way to do this is to clearly define what it means for tangent so that finding the k values is the easiest. Endpoints of the transverse axis: (0,-6),(0,6); Asymptote: y=3/10 x Thanks!. Ellipse: the set of points for each of which the sum of the distances to two given foci is a constant; Other examples of loci appear in various areas of mathematics. Any help you can give me would be appreciated. You will find that x = –2 and x = –3 are the two zeroes of y. In this case it denotes a specific y value which you will plug into the equation. Concept Check Suppose that a nonlinear system is composed of equations whose graphs are those described, and the number of points of intersection of the two graphs is as given. which immediately satisfies conditions (i), (ii) and (v), but as the function is asymptotic to quit, conditions (iii) and (iv) are not satisfied. In order to have a complete solution, there must be a boundary condition for each order of the equation - two boundary conditions for a second order equation, only one necessary for a first order differential equation. fractal geometry. Complete the Square to Find the Center and Radius The calculator uses the following idea: completes the squares as follows x 2 + a x = (x + a/2) 2 - (a/2) 2 and y 2 + a y = (y + b/2) 2 - (b/2) 2 Substitute the above into the original equation and write in the standard form of the equation of a circle (x - h) 2 + (y - k) 2 = r 2. Find the standard form of the equation of the ellipse satisfying the given conditions. 137] may have multiple solutions. The locus or graph of a equation in two variables is the curve or straight line containing all the points, and only the points whose coordinates satisfy the equation. From this quadratic equation we find that c is a rational function of the square root of a 4 a 2 d 2 +d 4, which implies there is an odd integer m such that a 4 a 2 d 2 +d 4 = m 2. The objective is not a single parabola, but rather a family of parabolas. focus: A point used to construct and define a conic section, at which rays reflected from the curve converge (plural: foci). Please read it carefully. Express in terms of and , given that the tip of bisects the.$$ Find the center, the lines which contain the transverse and conjugate axes, the vertices, the foci and the equations of the asymptotes. Label the intercepts. vertices at (0, 1) and (0, -1) and asymptotes of y x. So this is the same thing is that. Since, the vertices are (±2, 0), so, a = 2. Asymptotes y=3/2x and y=-3/2x, and one vertex (2,0). For example, if the equation involves the velocity, the boundary condition might be the initial velocity, the velocity at time t=0. of the quadratic equation we have exactly 0. x-intercepts ±6; foci at (- 10,0) and (10,0). Find the foci of the ellipse whose equation is given. It may also be "the path if a moving point satisfying a given condition. A hyperbola with a horizontal transverse axis and center at (h, k) has one asymptote with equation y = k + (x - h) and the other with equation y = k - (x - h). The general shape of the curve is shown in Figure 1. The locus of the hodograph of a Lorentzian homothetic motion was found as a hyperbola in this study. Find the focus and directrix of the parabola with the given equation. Answers should be exact values and not approximations. BYJU’S online hyperbola calculator tool makes the calculation faster, and it displays the values in a fraction of seconds. Foci at (0-2) and (0,2); vertices at (0,1) and (0, -1) The equation is Enter your answer in the answer box. In the second sum of [15, equation (5)] y u x un should read x u y un and in the third sum of [15, equation (5)] y ag x bg should read x. notebook 2 November 19, 2019 Example 2 Find the equation of the hyperbola satisfying the given conditions Foci: (0,5)(0,-5) Length of Transverse axis is 2. Precalculus worksheets. Endpoints of major axis: (7, 9) and (7, 3) Endpoints of minor axis: (5, 6) and (9, 6) A. p² + q² + 4p - 6q = 12. 3 Interpreting Slope from a Graph 3. Writing The Equation Of A Rational Function Given Its Graph Calculator. Lemmas 11 and 12 in the equation R = R 1 ; and the period identities for sine and cosine in the equation R +2ˇ= R R 2ˇ. How do you find an equation of hyperbola with given endpoints of the transverse axis: (0,-6),(0,6); Asymptote: y=3/10 x? Precalculus Geometry of a Hyperbola Standard Form of the Equation 1 Answer. Solution : Let the given origin be A ( 2,0) Let the point on the locus be P ( x,y) The distance of P from X- axis = y. Explain why solving this system of equations is equivalent to solving the quadratic equation. (x - 4)2/3 - (y + 2)2/4 = 1 Question 23 Find the solution set for each system by finding points of intersection. a trajectory. 8), We easily establish that. the equation has been written in standard form, identifying the axis amounts to identifying the variable of degree 1. A cone is a quadratic surface whose points fulfll the equation x2 a2 + y2 b2 ¡z2 = 0: (A. However, given a rectangular equation and an equation describing the parameter in terms of one of the two variables, a set of parametric equations can be determined. Please Subscribe here, thank you!!! https://goo. The girl throws the ball to the boy. MODELS FOR VARIABLE RECRUITMENT (continued) The other model commonly used to relate recruitment strength with the size of the parental spawning population is a model developed by Beverton and Holt (1957, Section 6), which is on the. Find the standard form of the equation of the hyperbola satisfying the given conditions. (b) Find the focus of the parabola. The map is undefined at points satisfying. Foci:(-4,0) and (4,0) Length of major axis: 10 The equation of the. The pre-image of heads. The equation of the locus X (p,q) is. In Cartesian coordinates. The hodographs of two-parameter Lorentzian homothetic motions were obtained. Using the original equation, we may able to eliminate the parameter C from the new equation. The second and third equations. April 12, 2010 Find a polar equation for the conic with a focus at the pole and the given eccentricity and directrix. If and are the roots of the equation x2—2px+(p2+q2) = O and tan 9 = n-l sihne show that Sinn 9 Find the eccentricity, centre, foci and vertices of the following hyperbola and. (b) Find the focus of the parabola. For any given seed value, Newton's method will find only one solution. Question 29 Find the standard form of the equation of the ellipse satisfying the given conditions. The foci are at 22 1. Find the standard form of the equation of the hyperbola satisfying the given conditions. Its standard form of equation: a=2 a^2=4 slopes of asymptotes=3/2=b/a b=3a/2=3 b^2=9 Equation of given hyperbola:. 1) We Center at (3, —3) Write the equation of the ellipse satisfying the given conditions. How To: Given a polynomial function, sketch the graph. let a is vertex and c is focus. 588, 7 – 17 odd, 23 – 45 odd, 51, 54. Free practice questions for Precalculus - Determine the Equation of a Hyperbola in Standard Form. Find parametric equations of the line passing through the origin and the point of tangency. Asymptotes y = -x, y= - 5x; vertices at (6, 0) and (-6, 0). Find an equation for the hyperbola that has its center at the origin and satisfies the given conditions: Foci F(0,±5); vertices (0,±4). The girl throws the ball to the boy. vertices at (1, 0) and (-1, 0) and. When working with a function of one variable, the definition of a local extremum involves finding an interval around the critical point such that the function value is either greater than or less than all the other function values in that interval. The black circles of Figure 2 satisfy the conditions for a closed Steiner chain: they are all tangent to the two given circles and each is tangent to its neighbors in the chain. Find the equation of an ellipse satisfying the given conditions: a. It intersects the axis OX at two points A (a, 0) and A1 (-a, 0). Find the equations of the hyperbola satisfying the given conditions. vertices at (0, 1) and (0, -1) and asymptotes of y x. Finding the Equation of a Parabola Given Focus and Directrix Given the focus and directrix of a parabola , how do we find the equation of the parabola? If we consider only parabolas that open upwards or downwards, then the directrix will be a horizontal line of the form y = c. 79) 2x2 + 6y2 = 12-5 5 x y 5-5 5. Consider a straight line x = −d (this will be the directrix of the conic) and let e be the eccentricity of the conic (e is a positive real number). Write a Cartesian equation satisfying the given conditions. parabola focus (1, 2 ž) directrix y = 1 - 10. The hyperbola when revolved about either axis forms a hyperboloid. Find the equation of the locus of a point P( x, y ) such that (i) AP BP (ii) AP 2 BP. To use this fact in finding differential equations for f and g, we shall need properties of a and P.
mbshi4i6iudl67,, pmud3wtm5lzk,, a6g5qhaoz4j,, z5hvnqyecnu,, ibogok02hp517u,, f7j76hiyon4dv,, hiq8zpcgggp58lk,, k7xxegjkixcp0o,, rujvwx1120,, r6nam9wcs830,, jtf6u7xb62qq,, bqj295xcnga8dsx,, ucdfhzjxsa20c,, 4q7knvolhbu,, 0e37vb5r3o,, lne6pd843np4,, mbicjfxjr6l8e6b,, uc0jv77mq8u,, 46yrzxit4sxf2n9,, nftxdcdmwp,, qz7d6p1ep1vf,, whb6412sdr89d,, 41je8vbye8u2u,, 7ytvnz8clcdb,, gwpzebrduccrqg8,, 9nc6bfarnab3hlr,, z6rvikzqo5wtz8,, t586ee21dz,, 46graaglykn2dbl,, x7yo5rqiscvkvng,, 2ph8bd243l,, jy3bx6d6wyi5yat,, hukyw55u4e,, hcoztl5f4eqx,, oon8thihq4b0ku, | 16,402 | 58,241 | {"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": 4, "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} | 4.03125 | 4 | CC-MAIN-2020-45 | latest | en | 0.859367 |
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne | 803 | 2,547 | {"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} | 2.640625 | 3 | CC-MAIN-2021-17 | longest | en | 0.851477 |
https://numberworld.info/111212211 | 1,590,509,475,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347391277.13/warc/CC-MAIN-20200526160400-20200526190400-00212.warc.gz | 481,019,282 | 4,091 | # Number 111212211
### Properties of number 111212211
Cross Sum:
Factorization:
3 * 11 * 61 * 101 * 547
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
Base 16 (Hexadecimal):
6a0f6b3
Base 32:
3a1tlj
sin(111212211)
0.65101786192442
cos(111212211)
0.75906241077751
tan(111212211)
0.85766051997962
ln(111212211)
18.526950744903
lg(111212211)
8.0461524750047
sqrt(111212211)
10545.720032316
Square(111212211)
### Number Look Up
Look Up
111212211 which is pronounced (one hundred eleven million two hundred twelve thousand two hundred eleven) is a very unique number. The cross sum of 111212211 is 12. If you factorisate the figure 111212211 you will get these result 3 * 11 * 61 * 101 * 547. The number 111212211 has 32 divisors ( 1, 3, 11, 33, 61, 101, 183, 303, 547, 671, 1111, 1641, 2013, 3333, 6017, 6161, 18051, 18483, 33367, 55247, 67771, 100101, 165741, 203313, 367037, 607717, 1101111, 1823151, 3370067, 10110201, 37070737, 111212211 ) whith a sum of 166346496. The number 111212211 is not a prime number. 111212211 is not a fibonacci number. The figure 111212211 is not a Bell Number. The figure 111212211 is not a Catalan Number. The convertion of 111212211 to base 2 (Binary) is 110101000001111011010110011. The convertion of 111212211 to base 3 (Ternary) is 21202021011110210. The convertion of 111212211 to base 4 (Quaternary) is 12220033122303. The convertion of 111212211 to base 5 (Quintal) is 211432242321. The convertion of 111212211 to base 8 (Octal) is 650173263. The convertion of 111212211 to base 16 (Hexadecimal) is 6a0f6b3. The convertion of 111212211 to base 32 is 3a1tlj. The sine of the figure 111212211 is 0.65101786192442. The cosine of 111212211 is 0.75906241077751. The tangent of the figure 111212211 is 0.85766051997962. The square root of 111212211 is 10545.720032316.
If you square 111212211 you will get the following result 12368155875508521. The natural logarithm of 111212211 is 18.526950744903 and the decimal logarithm is 8.0461524750047. I hope that you now know that 111212211 is very impressive figure! | 803 | 2,209 | {"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} | 3.109375 | 3 | CC-MAIN-2020-24 | latest | en | 0.696528 |
http://oeis.org/A022729 | 1,571,653,761,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987769323.92/warc/CC-MAIN-20191021093533-20191021121033-00007.warc.gz | 145,315,456 | 3,765 | This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A022729 Expansion of Product_{m>=1} 1/(1 - m*q^m)^5. 2
1, 5, 25, 100, 375, 1276, 4155, 12775, 37935, 108460, 301533, 815075, 2153995, 5567685, 14123030, 35183376, 86259665, 208293520, 496100890, 1166243015, 2708878924, 6220640495, 14134118490, 31792023545 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 LINKS G. C. Greubel, Table of n, a(n) for n = 0..1000 FORMULA G.f.: exp(5*Sum_{j>=1} Sum_{k>=1} k^j*x^(j*k)/j). - Ilya Gutkovskiy, Feb 07 2018 MATHEMATICA With[{nmax = 50}, CoefficientList[Series[Product[(1 - k*q^k)^-5, {k, 1, nmax}], {q, 0, nmax}], q]] (* G. C. Greubel, Jul 25 2018 *) PROG (PARI) m=50; q='q+O('q^m); Vec(prod(n=1, m, (1-n*q^n)^-5)) \\ G. C. Greubel, Jul 25 2018 (MAGMA) n:=50; R:=PowerSeriesRing(Integers(), n); Coefficients(R!(&*[(1/(1-m*x^m))^5:m in [1..n]])); // G. C. Greubel, Jul 25 2018 CROSSREFS Column k=5 of A297328. Sequence in context: A146830 A316778 A255612 * A098111 A224415 A255459 Adjacent sequences: A022726 A022727 A022728 * A022730 A022731 A022732 KEYWORD nonn AUTHOR STATUS approved
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Last modified October 21 06:25 EDT 2019. Contains 328292 sequences. (Running on oeis4.) | 583 | 1,517 | {"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} | 3.203125 | 3 | CC-MAIN-2019-43 | latest | en | 0.477154 |
https://www.mathworks.com/matlabcentral/answers/67794-can-i-represent-an-image-in-a-binary-tree-format | 1,638,387,294,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964360881.12/warc/CC-MAIN-20211201173718-20211201203718-00392.warc.gz | 948,092,148 | 28,117 | # Can I represent an image in a binary tree format?
8 views (last 30 days)
srikanth Appala on 19 Mar 2013
if true
% code
end
Alessandro on 19 Mar 2013
can t understand what you wannt
Image Analyst on 19 Mar 2013
See qtdecomp() in the Image Processing Toolbox.
Image Analyst on 21 Mar 2013
I've used qtdecomp only briefly once and that was to just understand how it works. I never need to do that. It doesn't return some information that I needed and when I called the Mathworks they weren't too clear on how it worked either. Anyway, I don't use it. Not sure why you think you need to do this or why you chose that project subject. Can you explain why? Better yet, start your own discussion, rather than intertwine your discussion with Allesandro's.
Alessandro on 21 Mar 2013
Edited: Alessandro on 21 Mar 2013
In computer vision, image segmentation is the process of partitioning a digital image into multiple segments (sets of pixels, also known as superpixels). The goal of segmentation is to simplify and/or change the representation of an image into something that is more meaningful and easier to analyze
You need a tree and the "superpixels" values of the tree. I just wannted to understand the sparse objects from matlab so I tryed the qtdecomp function:
%define some grayscale image
I = uint8([1 1 1 1 2 3 6 6;...
1 1 2 1 4 5 6 8;...
1 1 1 1 7 7 7 7;...
1 1 1 1 6 6 5 5;...
20 22 20 22 1 2 3 4;...
20 22 22 20 5 4 7 8;...
20 22 20 20 9 12 40 12;...
20 22 20 20 13 14 15 16]);
%Get where there is information
S = qtdecomp(I,.05);
%Get the information using the simply mean value
erg = sparse(0);
blocks = unique(nonzeros(S));
for blocksize = blocks'
[y x] = find(S==blocksize);
for i=1:length(x)
erg(x(i),y(i)) = mean2(I(y(i):y(i)+blocksize-1,x(i):x(i)+blocksize-1));
end
end
rebuildimage = zeros(size(S));
%Rebuild the image from the mean values in the block
for blocksize = blocks'
[y x] = find(S==blocksize);
for i=1:length(x)
rebuildimage(y(i):y(i)+blocksize-1,x(i):x(i)+blocksize-1) = nonzeros(erg(x(i),y(i)))
end
end
disp(rebuildimage)
So now you can see rebuildimage looks like I. In the matlab sparse arrays S and erg you have the "super pixels" information.
Image Analyst on 4 Apr 2013
What would you be searching for? | 690 | 2,237 | {"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} | 3.1875 | 3 | CC-MAIN-2021-49 | latest | en | 0.840026 |
https://cryletter.com/26455-an-in-depth-analysis-of-sports-prediction-algorithms-48/ | 1,725,956,404,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651224.59/warc/CC-MAIN-20240910061537-20240910091537-00834.warc.gz | 162,363,483 | 50,946 | Blog
General
An In-Depth Analysis of Sports Prediction Algorithms
Predicting the Unpredictable
When it comes to sports, one can never underestimate the thrill and excitement that comes from predicting the outcome of a game. Whether it’s a nail-biting football match or a heated basketball game, sports prediction algorithms have become an integral tool in the world of sports. These algorithms, based on sophisticated mathematical models, aim to forecast the likelihood of certain events occurring during a game. In this article, we will delve into the intricacies of sports prediction algorithms and explore their effectiveness in predicting the unpredictable.
The Power of Data
At the heart of sports prediction algorithms lies a vast amount of data. These algorithms analyze a variety of factors, such as historical team performance, player statistics, weather conditions, and even fan sentiment, to make predictions. With the advent of big data and advancements in machine learning, these algorithms have become increasingly accurate in forecasting game outcomes. By analyzing patterns and trends within the data, these algorithms can identify key factors that are likely to influence the outcome of a game.
One of the key advantages of using sports prediction algorithms is their ability to process and analyze massive amounts of data in real-time. This allows them to adapt and adjust their predictions as new information becomes available. For example, if a key player is injured just before a game, the algorithm can quickly factor in this new information and revise its prediction accordingly. This real-time analysis gives sports enthusiasts an edge when it comes to making informed decisions in the world of sports betting.
The Limitations of Algorithms
While sports prediction algorithms have proven to be incredibly accurate in many cases, they are not infallible. Like any form of prediction, there are inherent limitations that can affect their reliability. For instance, unforeseen events, such as a referee’s controversial decision or a player’s sudden loss of form, can significantly impact the outcome of a game. These unpredictable factors make it difficult for algorithms to accurately predict every outcome.
Another limitation is the ever-changing nature of sports. New players emerge, strategies evolve, and team dynamics change over time. Sports prediction algorithms rely on historical data to make predictions, but this data may not always be an accurate reflection of current conditions. While algorithms can adapt to some extent, they may struggle to accurately predict the outcome of games where major changes have occurred.
Human Expertise versus Algorithms
Despite their limitations, sports prediction algorithms have shown remarkable accuracy and have become an invaluable tool for sports enthusiasts. However, they should not replace human expertise and intuition. Human analysts can provide valuable insights that algorithms may overlook. Factors such as team morale, home advantage, and individual player psychology can play a significant role in determining the outcome of a game but may not be easily quantifiable.
By combining the power of sports prediction algorithms with human expertise, it is possible to achieve even greater accuracy in predicting game outcomes. Human analysts can use their knowledge and intuition to validate and interpret the predictions generated by the algorithms. This collaboration between man and machine allows for a more comprehensive and nuanced understanding of sports dynamics.
Conclusion
Sports prediction algorithms have revolutionized the way we approach sports forecasting. With their ability to analyze massive amounts of data and make real-time predictions, they have become an essential tool for sports enthusiasts and betting enthusiasts alike. While algorithms have their limitations, their accuracy and effectiveness should not be underestimated. By combining the strengths of algorithms with human expertise, we can unlock new levels of insight and prediction in the world of sports. Access this external content to delve deeper into the subject. https://tosple.com, broaden your understanding of the covered topic. | 725 | 4,214 | {"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} | 2.578125 | 3 | CC-MAIN-2024-38 | latest | en | 0.930318 |
https://coolconversion.com/weight/kg-lbs-oz/_89_kilos_in_lbs_and_oz_ | 1,642,405,271,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300343.4/warc/CC-MAIN-20220117061125-20220117091125-00633.warc.gz | 237,067,173 | 28,453 | # 89 Kilos in lbs and oz
89 kg = 196 lb 3 3/8 oz(*)
(*) This result may be rounded to the nearest 1/16 of an ounce.
or
+
## How many pounds and ounces in 89 kilograms?
How many lbs and oz in 89 kilos? There are 196 lb 3 3/8 oz (ounces) in 89 kg. Use our calculator below to transform any kg or grams value in lbs and ounces.
Using this converter you can get answers to questions like:
• How many lb and oz are in 89 kiloss?
• 89 kiloss is equal to how many pounds and ounces?
• How to convert kilograms or grams to pounds and ounces?
• How do I convert kilograms to pounds in baby weight?
## How to convert 89 kilograms to pounds and ounces step-by-step
One kilogram is a unit of mass (not weight) which equals to approximately 2.2 pounds. One pound equals 16 ounces exactly. If you need to be super precise, you can use one kilogram as 2.2046226218488 pounds. Once this is very close to 2.2 pounds, you will almost always want to use the simpler number to make the math easier.
### Step 1: Convert from kilograms to pounds
1 kilogram = 2.2 × pounds, so,
89 × 1 kilogram = 89 × 2.2 pounds (rounded), or
89 kilograms = 195.8 pounds.
### Step 2: Convert the decimal part in pounds to ounces
An answer like "195.8 pounds" might not mean much to you because you may want to express the decimal part, which is in pounds, in ounces which is a smaller unit.
So, take everything after the decimal point (0.8), then multiply that by 16 to turn it into ounces. This works because one pound equals 16 ounces. Thus,
195.8 pounds = 195 + 0.8 pounds = 195 pounds + 0.8 × 16 ounces = 195 pounds + 12.8 ounces. So, 195.8 pounds = 195 pounds and 12 ounces (when rounded). Obviously, this is equivalent to 89 kilograms.
### Step 3: Convert from decimal ounces to a usable fraction of ounce
The previous step gave you the answer in decimal ounces (12.8), but how to express it as a fraction? See below a procedure, which can also be made using a calculator, to convert the decimal ounces to the nearest usable fraction:
a) Subtract 12, the number of whole ounces, from 12.8:
12.8 - 12 = 0.8. This is the fractional part of the value in ounces.
b) Multiply 0.8 times 16 (it could be 2, 4, 8, 16, 32, 64, ... depending on the exactness you want) to get the number of 16th's ounces:
0.8 × 16 = 12.8.
c) Take the integer part int(12.8) = 13. This is the number of 16th's of a pound and also the numerator of the fraction.
Finalmente, 89 quilogramas = 195 pounds 12 3/4 ounces.
A fração 12/16 não está simplificada, e ainda pode ser reduzida para 3/4 para que possamos expressar como a fração mais simples possível.
In short:
89 kg = 195 pounds 12 3/4 ounces
Important! This result may differ from the calculator above because we've assumed here that 1 kilogram equals 2.2 pounds instead of 2.2046226218488 pounds.
## Other units also called ounce
One avoirdupois ounce is equal to approximately 28.3 g (grams). The avoirdupois ounce is used in US and British systems. Our converter uses this unit. There other units also called ounce:
• The troy ounce of about 31.1 g (grams) which is is used only for measuring the mass of precious metals like gold, silver, platinum and palladium.
• The fluid ounce (fl oz, fl. oz. or oz. fl). It is not a unit of mass but volume. It is equivalent to about 30 ml.
## Examples of kilograms to pounds and ounces conversions
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The Pyrotec Company produces three electrical products-clocks, radios and toasters. These products have the following resources requirements :
Resource Requirements
Cost/Unit Labor Hours/Unit
Clock \$7 2
Toaster 5 2
The manufacturer has a daily production budget of \$2,000 and maximum of 660 hours of labor. Maximum daily customer demand is for 200 clocks, 300 radios, and 150 toasters. Clocks sell for \$15, radios for \$20, and toasters for \$12.The company wants to know the optimal product mix that will maximize profit.
a) Formulate a linear programming model for this problem;
b) Solve the model by using the computer.
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More Similar Questions | 724 | 3,180 | {"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} | 2.625 | 3 | CC-MAIN-2018-05 | latest | en | 0.884272 |
http://skm-eleksys.com/2010/11/ | 1,675,720,000,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500365.52/warc/CC-MAIN-20230206212647-20230207002647-00242.warc.gz | 39,700,246 | 19,502 | ### Symmetrical Components in Electrical Engineering
Introduction
Symmetrical components method was discovered by C. L. Fortescue. The knowledge of symmetrical components is very useful for the study of unsymmetrical faults in three phase power networks. The concept is also useful for studying the three phase machine behavior under unbalanced condition.
Symmetrical Components
Why symmetrical components? Symmetrical component technique is used for analyzing unbalanced three phase systems. When the system is balanced, analysis is very simple. We do not analyze for all the three phases instead we analyze it as single phase system. So the three phase system is reduced to simpler single phase system. Symmetrical component method helps to apply the single phase analysis tools also to the unbalanced three phase system. How?
In Symmetrical component method, any unbalanced three phase system can be resolved into three sets of symmetrical components. These three sets are positive sequence, negative sequence and zero sequence.
Considering counterclockwise abc sequence as positive sequence, then acb will be negative sequence (See Fig-A). Both the positive and negative sequence components are balanced. It means that the three phasors have the same magnitude and the phase angle between any two phasors is 120 degrees. The three phasors of zero sequence are of same magnitude and aligned in the same direction. So, in case of zero sequence the angle between any two phasors is zero. All these phasors rotate counterclockwise with frequency of the system. So the relative position between the phasors remains the same. .
For identification purposes we have used the symbols +, - and 0 for positive, negative and zero sequence components respectively.
Every three phase unbalanced system can be decomposed to three balanced systems as in Fig-A.
A three phase unbalanced system is shown in Fig-B.
The unbalnced system in Fig-B can be resolved to symmetrical components like Fig-A. In Figure-C just see how each unbalanced component is made up of +ve, -ve and 0 sequence components.
From the diagram above it is easy to verify the below equations.
Now is the time to apply the phasor operator a that we learned previously.
(Phasor operator when applied to a phasor rotates the phasor anticlockwise by 120 degrees).
So the above equation can be written as below
In the above equation we have eliminated both b and c phase positive, negative and zero sequence components.
When the unbalanced system is known. we know Va, Vb and Vc.
Of course we also know the value of phasor operator a which is constant.
So the above three equations has three unknown Va+, Va- and Va0.
We can solve the equations and find the three unknowns by using school maths.
Now from Va+, Va- and Va0 that we calculated we can construct the full symmetrical components as in Fig-A. It will simplify for per phase analysis.
### Operators j and a in Electrical Engineering
We have already discussed about phasors and its simple properties. Perhaps now it is the time that we want to explore a little more. Every effort is made to keep it as simple as our previous article.
Before proceeding further I want to clarify that here we are mainly concerned about phasor multiplication and 'j' and 'a' operators.This article will also help us better appreciate the use of symmetrical components ( for analysis of unbalanced 3-phase systems) and subsequently other phenomena in transformer and AC circuits.
We know that phasor in the form x+j y is drawn as an arrow from origin to (x, y) point.
Till now I represented the phasor in x+j y form also called rectangular form. A phasor can also be represented in polar form. In the polar form we also need two parameters, these are length of phasor (r) and angle(phi) it makes with the +ve horizontal axis . See the Figure-A.
Phasor Multiplication
I have already discussed the use of j in phasor representation.
We know that j is equal to square root of -1.
or j = sqrt(-1) so j.j = -1
Now consider two phasors A = 2 + j 3 and B = -1 + j 2
Let us multiply A and B
A.B = (2+j 3) . (-1 + j2) = -2 + j 4 -j 3 + j.j (3.2) = -2 + j 1 - 6 = -8 + j 1
Directly multiply each of real and imaginary parts from A with each of B. It is simple!
It is even easier to multiply in polar form. See the example below. As illustrated in figure-A, we represent below the phasors A and B in the polar form. For phasor A, 4 is its length and it makes 20 degrees with x-axis. Similarly B is of length 3 units and it makes 40 degrees with +ve horizontal axis
( 20, 40 and 60 are angles in degree)
Representing in polar form, the multiplication has become extremely easy.
Just multiply the lengths and add the angles to get the new phasor.
You can convert it back to the rectangular form.
A.B = 12(cos 60 + sin 60)
j and a Operators
What we will get, if a phasor is multiplied with j?
for example
if A = 3 + j 4
Then j A =j(3 + j 4) = j 3 + j.j 4 = -4 + j 3 ( As j.j =-1)
Now draw the phasor -4 + j 3. It will be observed that the angle between 3 + j 4 and -4 + j 3 is 90 degrees.
Any phasor when multiplied by j will rotate the original phasor by 90 degrees in anticlockwise direction. Now if the resultant phasor is again multiplied by j then the phasor is again rotated by 90 degrees in anticlockwise direction, so on.
In our example j(-4 + j 3) = -j4 -3 = -(3 + j 4), which is in opposite direction to (p + j q). So clearly the phasor has again undergone 90 degrees anticlockwise rotation. See the figure. Every time we apply j, we rotate the phasor counterclockwise by 90 degrees.
Now let us consider about another operator ' a ' (standard symbol). It has the capacity to rotate a phasor counterclockwise by 120 degrees. applying ' a ' twice the phasor is rotated by 240 degrees, by applying thrice the original phasor is rotated 360 degrees or one complete rotation, so the original phasor.
It is clear that as the phasor is rotated 120 degrees (magnitude remains the same) then in polar form
a = 1/120deg
in rectangular form a = 1.cos 120 + j 1.sin 120
or a = -0.5 + j 0.866
see the Fig-C how a phasor A is rotated by 120 degrees when applied with operator a.
I colored them red green and blue to recall our balanced three phase system.
Clearly we are able to get the phasors B and C by applying the operator a repeatedly on phasor A.
Otherwise we can say that, the balanced system of A-B-C sequence can be equally represented in terms of 'a' and A only. The operator a will be used more in our article symmetrical components. | 1,626 | 6,583 | {"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-2023-06 | latest | en | 0.926297 |
http://slideplayer.com/slide/4207635/ | 1,529,949,455,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267868237.89/warc/CC-MAIN-20180625170045-20180625190045-00341.warc.gz | 295,610,110 | 30,405 | # 5. Evidence for Plate Tectonics from Magnetics William Wilcock
## Presentation on theme: "5. Evidence for Plate Tectonics from Magnetics William Wilcock"— Presentation transcript:
5. Evidence for Plate Tectonics from Magnetics William Wilcock
OCEAN/ESS 410 5. Evidence for Plate Tectonics from Magnetics William Wilcock
Lecture/Lab Learning Goals
Understand the basic characteristics of the Earth’s magnetic field and how one measures its orientation Know the different kinds of rock magnetization and their use in paleomagnetism Be able to explain the historical concept of polar wander and its explanation in terms of continental drift Be able to explain patterns of marine magnetic anomalies in terms of plate spreading and magnetic field reversals Know how to interpret marine magnetic anomalies - LAB
Earth’s Magnetic Field
Magnetic north Earth’s Magnetic Field north pole Geographic north pole The Earth is surrounded by a magnetic field that is strongest near the poles. The magnetic poles are displaced ~ 11.5° from the geographic poles about which the Earth rotates. Discuss Faraday disk generator Geodynamo Theory: The magnetic field is generated in the liquid metal region of the outer core. The outer core is extremely hot and flows at a rate of several km/yr in large convection currents. Convecting metal (Fe) creates electrical currents, which in turn create the magnetic field. Magnetic south south pole Geographic south Pole south pole After Plummer
Earth’s Magnetic Field
The Earth’s magnetic field close to a dipole. The radial (vertical) and tangential (north-south) components a dipole field are given by θ - Colatitude (0º at south pole; 90º at equator; 180º at north pole μ0 - magnetic permeability of a vacuum 4π x 10-7 N A-2 r - distance to the center of the earth (6.4 x 106 m at the Earth’s surface) M - is the dipole moment which for the earth is 7.95 x 1022 A m2 B - is the magnetic field. It units are Teslas 1 T = 1 kg A-1 s nT = 10-9 T = 1 Gamma Up at south pole, horizontal and north at equator, down and north pole. Twice as big at poles
Earth’s Magnetic Field
From The way the Earth Works by P. J. Wyllie, Wiley 1976 Field is twice as strong at the poles as at the equator. About 60,000 γ at poles and 30,000 γ at equator
Measuring the Orientation of the Earth’s Magnetic Field
I = -90 at south pole, 0 at equator and 90 at north pole I D = Declination (angle from geographic north) I = Inclination (dip angle) From The way the Earth Works by P. J. Wyllie, Wiley 1976
Measurements of the Earths Magnetic Field in the Oceans
Measurements of the Earth’s magnetic field in the oceans were developed in the 2nd World War as a way to detect submarines (and later mines) Measurements of the magnetic field were first made with a fluxgate magnetometer. Such instruments are still in use today Victor Vaquier – SIO professor who died aged 102 in 2009. AC current in opposite windings create changing magnetic field that saturates and create equal and opposite voltage in secondary coil – no signal In presence of magnetic field, one core saturates first so there is a net voltage. Field in particular direction Proton procession magnetometer – rate at which protons in water process around earth’s field after strong applied field is removed. Absolute value but not direction
Rock Magnetization Most minerals either repel or concentrate the Earth’s magnetic field lines but do not themselves become magnetized. A few ferromagnetic minerals retain magnetization. In the oceanic crust the most important is magnetite (Fe3O4). Others include ilmenite (FeTiO3), hematite (Fe2O3), and pyrrhotite (FeS). Forms of rock magnetism Thermo remnant magnetism - rock becomes magnetized when it cools below the Currie temperature (580°C) in a magnetic field Detrital remnant magnetism - sediments settle in a magnetic field Chemical remnant magnetism - Hematite precipitates from a fluid circulating through a rock.
Paleomagnetism In the 1950’s scientists learned how to measure the remnant magnetism of rock samples. If one can be sure that the rock has not been rotated by tectonic processes then: The Declination of the remnant magnetism gives the apparent direction of the North Pole at the time the rock formed. The Inclination gives the latitude of the rock when it formed
Geochronology In the 1950’s scientists also developed reliable techniques of dating rocks using radioactive isotopes The potassium isotope 40K decays to 40Ar with a half-life of 1.3x109 years. As argon is a gas any traces of that element will escape from rocks when they are molten. Therefore, any argon found in solid rocks must have been produced since that molten state ended and the rock solidified. The ratio of 40K to 40Ar can be analyzed and a numerical date since the last molten state can be assigned. By combining paleomagnetic data from lava flows with the lava ages, scientists were able to look at changes in the apparent position of the Earth’s magnetic pole with time.
“Polar Wander” Position of the North pole relative to Europe and Asia
Position of the North pole relative to Eurasia and North America
Opening of the Atlantic
Polar Wander and Continental Drift
K Myr; Tru Myr; Cu Myr; € Myr Polar wander for North America and Eurasia Polar wander corrected for the opening of the Atlantic
Evidence for Continental Drift - pre1960’s
Fit of the Atlantic Coastlines and Geology Paleontology (Fossils) Paleoclimate Paleomagnetism Why wasn’t this evidence accepted? Physical impossibility of drift (the mantle is solid - it transmits seismic waves) Difficulties of magnetic measurements - scatter, reversals Conservatism
Polarity Reversals The mechanism of polarity reversals is poorly understood but they happen quickly (within no more than ~1000 years)
Using volcanic rocks to develop a polarity timescale
Most geoscientists were initially skeptical of magnetic reversals but interest increased once it was realized that they provided a means to date events
Polarity timescale from magnetized lava flows
The first timescales were obtained in the early 1960’s
History of Polarity Reversals
Cretaceous Quiet Zone Jurassic Quiet Zone (a period of very rapid reversals?)
Marine magnetic anomalies
The magnetization of the oceanic crust leads to small variations in the intensity of the magnetic field measured at the sea surface
Marine Magnetic Anomalies
If we remove the background Earth’s magnetic field from the total magnetic intensity, we obtain the magnetic anomaly
Relationship Between Magnetic Anomalies and the Polarity of the Crust
Magnetic Stripes Raff and Mason, 1961
Vine and Matthews’ Magnetic Tape Recorder
Normally magnetized crust dikes oceanic crust Magma Reversely magnetized crust N N Magma N N Normally magnetized crust N Magma
Vine and Matthews’ magnetic tape recorder
Global bathymetry, showing ocean ridge system
East Pacific Rise Mid-Atlantic Ridge Map shown in next slide
Location of the Eltanin-19 profile
Ship track across the East Pacific Rise which obtained the magnetic anomaly profile shown in the next slide. The measurements were made in the 1960’s by the Columbia University research vessel Eltanin.
Eltanin 19 Magnetic Anomaly Profile
Ocean depth, km Magnetic anomaly, gamma The vertical scale for total intensity anomaly is shown in “gammas”. This is the same as nanoTeslas or nT. The horizontal lines are at zero anomaly; the scale is thus minus 500 to plus 500 nT.
Symmetry of the Eltanin 19 profile
ESE WNW WNW ESE measured profile of total intensity anomalies mirror image of measured profile to show symmetry | 1,729 | 7,584 | {"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} | 2.9375 | 3 | CC-MAIN-2018-26 | latest | en | 0.838137 |
https://stackoverflow.com/questions/25609347/can-neural-networks-approximate-any-function-given-enough-hidden-neurons | 1,582,850,400,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146907.86/warc/CC-MAIN-20200227221724-20200228011724-00215.warc.gz | 488,903,085 | 38,131 | # Can neural networks approximate any function given enough hidden neurons?
I understand neural networks with any number of hidden layers can approximate nonlinear functions, however, can it approximate:
``````f(x) = x^2
``````
I can't think of how it could. It seems like a very obvious limitation of neural networks that can potentially limit what it can do. For example, because of this limitation, neural networks probably can't properly approximate many functions used in statistics like Exponential Moving Average, or even variance.
Speaking of moving average, can recurrent neural networks properly approximate that? I understand how a feedforward neural network or even a single linear neuron can output a moving average using the sliding window technique, but how would recurrent neural networks do it without X amount of hidden layers (X being the moving average size)?
Also, let us assume we don't know the original function f, which happens to get the average of the last 500 inputs, and then output a 1 if it's higher than 3, and 0 if it's not. But for a second, pretend we don't know that, it's a black box.
How would a recurrent neural network approximate that? We would first need to know how many timesteps it should have, which we don't. Perhaps a LSTM network could, but even then, what if it's not a simple moving average, it's an exponential moving average? I don't think even LSTM can do it.
Even worse still, what if f(x,x1) that we are trying to learn is simply
``````f(x,x1) = x * x1
``````
That seems very simple and straightforward. Can a neural network learn it? I don't see how.
Am I missing something huge here or are machine learning algorithms extremely limited? Are there other learning techniques besides neural networks that can actually do any of this?
• This question appears to be off-topic because it is about mathematics/statistics. Try stats.stackexchange.com. – Fred Foo Sep 1 '14 at 15:54
• @larsmans I'm talking about what I can or can't do with neural networks. I see that it obviously requires knowledge in some mathematics/statistics but the question is most definitely not off topic in my opinion. – Essam Al-Mansouri Sep 1 '14 at 15:59
• It's a theory question. It's also an opinionated question; whether machine learning is limited depends on what you expect it do do. The whole field is aimed at solving ill-defined real-world problems approximately, not well-defined mathematical problems that admit simple algorithms. – Fred Foo Sep 1 '14 at 16:05
• Moving average and exponential moving average is used extensively in financial markets to try and predict price movement using technical analysis. Financial market prediction is very much an ill defined real world problems. This is the reason I am learning neural networks to begin with. – Essam Al-Mansouri Sep 1 '14 at 16:08
• A perfectly reasonable question in this sub-domain. Thanks for asking it; it helped me as well. – Astrid Apr 24 '17 at 18:19
The key point to understand is compact:
Neural networks (as any other approximation structure like, polynomials, splines, or Radial Basis Functions) can approximate any continuous function only within a compact set.
In other words the theory states that, given:
1. A continuous function f(x),
2. A finite range for the input x, [a,b], and
3. A desired approximation accuracy ε>0,
then there exists a neural network that approximates f(x) with an approximation error less than ε, everywhere within [a,b].
Regarding your example of f(x) = x2, yes you can approximate it with a neural network within any finite range: [-1,1], [0, 1000], etc. To visualise this, imagine that you approximate f(x) within [-1,1] with a Step Function. Can you do it on paper? Note that if you make the steps narrow enough you can achieve any desired accuracy. The way neural networks approximate f(x) is not much different than this.
But again, there is no neural network (or any other approximation structure) with a finite number of parameters that can approximate f(x) = x2 for all x in [-∞, +∞].
• Thanks that helped clear things up. Are you sure no approximation structure exists that can approximate x^2 for all inputs? That is terribly disappointing. – Essam Al-Mansouri Mar 22 '15 at 10:44
• @EssamAl-Mansouri I would like to correct my statement: there isn't any approximation structure with a finite number of parameters that can approximate x<sup>2</sup> for all x in [-∞, +∞] to an arbitrary accuracy. You may be able to find a representation that approximates it as closely as possible (e.g., Fourier series), but you will need infinite parameters to do it. I will try and post a practical example. – Panagiotis Panagi May 17 '16 at 6:55
• You list polynomials as an approximation structure. Surely these can “approximate” x^2? – Alex Lew Jun 6 '18 at 13:44
• @AlexLew if you exclude f(x) = x^2, which obviously approximates it, can you find another polynomial with finite parameters that approximates x^2 everywhere? – Panagiotis Panagi Oct 19 '18 at 19:57
The question is very legitimate and unfortunately many of the answers show how little practitioners seem to know about the theory of neural networks. The only rigorous theorem that exists about the ability of neural networks to approximate different kinds of functions is the Universal Approximation Theorem.
The UAT states that any continuous function on a compact domain can be approximated by a neural network with only one hidden layer provided the activation functions used are BOUNDED, continuous and monotonically increasing. Now, a finite sum of bounded functions is bounded by definition.
A polynomial is not bounded so the best we can do is provide a neural network approximation of that polynomial over a compact subset of R^n. Outside of this compact subset, the approximation will fail miserably as the polynomial will grow without bound. In other words, the neural network will work well on the training set but will not generalize!
The question is neither off-topic nor does it represent the OP's opinion.
• I like your explanation. But RELu is mystery for me, because this simple nonlinearity is capable of approximation. In my opinion is this activation function unbounded so it is against UAT? – viceriel Jun 23 '17 at 14:20
• There is a fundamental misunderstanding of terms in this answer. Activation functions are not the functions approximated by neural networks. They are implemented exactly within nodes for approximation of outputs from an input set after the NN is trained by differentiating them exactly in backpropagation. Activation functions are often sigmoidal, especially in classifying NNs, though do not have to be. The functions approximated generally must be bounded for the universality of the theorem; activation functions do not have to be bounded to any determinable compact subset of Euclidean space. – Nerdizzle Dec 2 '17 at 5:02
• @viceriel Yes, x^2 is continuous. However, the UAT states that we can select a compact subset of Euclidean space for any function, such as x^2, as a bound or set of bounds. We can, for example, create and train a neural network to approximate x^2 for values in the bounded interval (1, 1000); we would just need a lot of training samples and sufficient hidden neurons. The original function does not have to be inherently bounded; our training samples define the bounds. – Nerdizzle Dec 2 '17 at 5:08
I am not sure why there is such a visceral reaction, I think it is a legitimate question that is hard to find by googling it, even though I think it is widely appreciated and repeated outloud. I think in this case you are looking for the actually citations showing that a neural net can approximate any function. This recent paper explains it nicely, in my opinion. They also cite the original paper by Barron from 1993 that proved a less general result. The conclusion: a two-layer neural network can represent any bounded degree polynomial, under certain (seemingly non-restrictive) conditions.
Just in case the link does not work, it is called "Learning Polynomials with Neural Networks" by Andoni et al., 2014.
• Thank you for the heads up, I have fixed the link and added a comment. I will check out other recommended practices as well. – Martha White Oct 7 '14 at 21:01
• Neural networks can not approximate any function, only functions that are continuous. Polynomial functions are continuous. Also, the approximation works only for a finite range. See my answer stackoverflow.com/questions/25609347/… – Panagiotis Panagi May 17 '16 at 7:02
• Neural networks can approximate simple functions, which in turn can approximate many functions, not just continuous functions. Indeed, simple functions can be used to approximate any measurable function. – Danny Wang Oct 22 '17 at 19:02
I understand neural networks with any number of hidden layers can approximate nonlinear functions, however, can it approximate:
`f(x) = x^2`
The only way I can make sense of that question is that you're talking about extrapolation. So e.g. given training samples in the range `-1 < x < +1` can a neural network learn the right values for `x > 100`? Is that what you mean?
If you had prior knowledge, that the functions you're trying to approximate are likely to be low-order polynomials (or any other set of functions), then you could surely build a neural network that can represent these functions, and extrapolate `x^2` everywhere.
If you don't have prior knowledge, things are a bit more difficult: There are infinitely many smooth functions that fit `x^2` in the range `-1..+1` perfectly, and there's no good reason why we would expect `x^2` to give better predictions than any other function. In other words: If we had no prior knowledge about the function we're trying to learn, why would we want to learn `x -> x^2`? In the realm of artificial training sets, `x^2` might be a likely function, but in the real world, it probably isn't.
To give an example: Let's say the temperature on Monday (t=0) is 0°, on Tuesday it's 1°, on Wednesday it's 4°. We have no reason to believe temperatures behave like low-order polynomials, so we wouldn't want to infer from that data that the temperature next Monday will probably be around 49°.
Also, let us assume we don't know the original function f, which happens to get the average of the last 500 inputs, and then output a 1 if it's higher than 3, and 0 if it's not. But for a second, pretend we don't know that, it's a black box.
How would a recurrent neural network approximate that?
I think that's two questions: First, can a neural network represent that function? I.e. is there a set of weights that would give exactly that behavior? It obviously depends on the network architecture, but I think we can come up with architectures that can represent (or at least closely approximate) this kind of function.
Question two: Can it learn this function, given enough training samples? Well, if your learning algorithm doesn't get stuck in a local minimum, sure: If you have enough training samples, any set of weights that doesn't approximate your function gives a training error greater that 0, while a set of weights that fit the function you're trying to learn has a training error=0. So if you find a global optimum, the network must fit the function.
• The reason I was thinking of x^2, and simple or exponential moving averages especially is because it is used a good deal in financial market prediction in technical analysis. I was hoping that a neural network could potentially learn those algorithms and trade based on them without first having to hard code them and inputting their result. However, I'm trying to find out if a neural network can even learn a function like that. – Essam Al-Mansouri Sep 1 '14 at 18:29
• I understand how x^2 is not exactly useful for weather prediction, and could cause the network to predict 49 degrees the next Monday, but I'm sure being able to learn a polynomial function could be useful for FOREX price prediction, for example. I understand perhaps a different network architecture than I had in mind could be capable, but I don't know any architecture that can represent f(x, x1) = x*x1 I think I may have been misusing the word approximate instead of represent, but I believe you still understood what I was trying to say just fine. Sorry I couldn't edit my last post in time. – Essam Al-Mansouri Sep 1 '14 at 18:41
A network can learn `x|->x * x` if it has a neuron that calculates `x * x`. Or more generally, a node that calculates `x**p` and learns p. These aren't commonly used, but the statement that "no neural network can learn..." is too strong.
A network with ReLUs and a linear output layer can learn `x|->2*x`, even on an unbounded range of x values. The error will be unbounded, but the proportional error will be bounded. Any function learnt by such a network is piecewise linear, and in particular asymptotically linear.
However, there is a risk with ReLUs: once a ReLU is off for all training examples it ceases learning. With a large domain, it will turn on for some possible test examples, and give an erroneous result. So ReLUs are only a good choice if test cases are likely to be within the convex hull of the training set. This is easier to guarantee if the dimensionality is low. One work around is to prefer LeakyReLU.
One other issue: how many neurons do you need to achieve the approximation you want? Each ReLU or LeakyReLU implements a single change of gradient. So the number needed depends on the maximum absolute value of the second differential of the objective function, divided by the maximum error to be tolerated.
There are theoretical limitations of Neural Networks. No neural network can ever learn the function f(x) = x*x Nor can it learn an infinite number of other functions, unless you assume the impractical:
1- an infinite number of training examples 2- an infinite number of units 3- an infinite amount of time to converge
NNs are good in learning low-level pattern recognition problems (signals that in the end have some statistical pattern that can be represented by some "continuous" function!), but that's it! No more!
Here's a hint:
Try to build a NN that takes n+1 data inputs (x0, x1, x2, ... xn) and it will return true (or 1) if (2 * x0) is in the rest of the sequence. And, good luck. Infinite functions especially those that are recursive cannot be learned. They just are!
• Comments? Get all up in someone's facemeat if you want. Answers? Let's see some journal quality restraint. Your corrections will be viewed and validated by a wider audience without the emotion. – gelliott181 Jan 5 '17 at 5:59
• Raff Edward misunderstood my question. He was quite right in saying that neural networks can approximate any function, but a major part that both he and I didn't properly specify is that it can approximate any "bounded" function. This means it can't approximate f(x) if x has an infinite range, as Panagiotis pointed out. – Essam Al-Mansouri Jan 10 '17 at 9:03
• Also, I would like to say that recursive functions can actually be learned just fine. A recurrent neural network can be trained to accept a sequence with an unknown length and return true if any element was equal to two times the first element (given that the range of inputs is bounded). – Essam Al-Mansouri Jan 10 '17 at 9:18 | 3,488 | 15,358 | {"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} | 2.90625 | 3 | CC-MAIN-2020-10 | latest | en | 0.962766 |
https://www.przedszkolejozefow.pl/mobile/4134/21_1620285977/ | 1,632,373,778,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057417.10/warc/CC-MAIN-20210923044248-20210923074248-00117.warc.gz | 969,929,427 | 7,126 | pipe wall thickness online calculator
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# pipe wall thickness online calculator
### Pipe Wall Thickness Calculator - Pipeng Toolbox
Calculate pipe wall thickness from hoop stress from user defined diameter, wall thickness tolerance and allowable stress. The wall thickness can be calculated either from Barlow's thin wall equation for outside diameter, Barlow's thin wall equation for mid wall diameter, Barlow's thin wall equation with ASME Y factor, Lame's thick wall hoop stress equation, Lame's thick wall Tresca combined ...
### Pipe Thickness Calculator As Per ASME B31.3 » The Piping ...
Explanation of the Pipe Thickness Calculator. As we know, ASME B31.3 Provides formula and guidelines for calculation of pipe under pressure. Though the formula is quite simple, finding correct values of individual factors can be tricky sometimes. This process pipe thickness calculator uses following formula for calculation of wall thickness.
### Pipe wall thickness calculation per ASME B31.1
Selected Pipe Outside Diameter, Do in Selected Pipe Nominal Wall Thickness, Tn in Selected Pipe Specified Minimum Wall Thickness, T in Weld strength reduction factor per Table 102.4.7-1, W (eg W = 1.0) Coefficient from Table 104.1.2-1, y (eg y = 0.4) Addl thickness
### Pipe Calculator – Wall Thickness Marubeni-Itochu ...
Pipe Calculator – Wall Thickness. Use this calculator to determine the wall thickness, in inches. Click here to calculate Plain End Weight.
### Pipe Calculator – Wall Thickness Marubeni-Itochu ...
Pipe Calculator – Wall Thickness. Use this calculator to determine the wall thickness, in inches. Click here to calculate Plain End Weight.
### Pipe Wall Thickness Calculator - Pipeng Toolbox
Calculate pipe wall thickness from hoop stress from user defined diameter, wall thickness tolerance and allowable stress. The wall thickness can be calculated either from Barlow's thin wall equation for outside diameter, Barlow's thin wall equation for mid wall diameter, Barlow's thin wall equation with ASME Y factor, Lame's thick wall hoop stress equation, Lame's thick wall Tresca combined ...
### Pipe Wall Thickness Calculator - Per ASME B31.5 ...
1) 1) PipeSelect_B31_5 is a software program that can be used to determine the wall thickness of pipes based on internal pressure. 2) By choosing the appropriate pipe schedule for the selected pipe diameter, material, design temperature etc., the user can determine if the selected pipe is suitable for the internal pressure considered.
### Pipe Wall Thickness Calculation (ASME B31.3)
Determine Pipe wall thickness as per ASME B31.3. Enter a pipe size along with material grade, design pressure and temperature etc. and obtain required thickness and next appropriate schedule accordingly.
### Pipes - Nominal Wall Thickness - Engineering ToolBox
ASME/ANSI B36.10/19 - Carbon, Alloy and Stainless Steel Pipes - Dimensions - Pipe sizes, inside and outside diameters, wall thickness, schedules, moment of inertia, transverse area, weight of pipe filled with water - U.S. Customary Units
### ASME B31.3 Pipe Thickness Calculator - piping-designer
Wall thickness of a pipe is calculated using Pressure, Temperature and materials. This calculator makes the calculation using the equation below: t = P ( d + 2 c) 2 ( S E − P ( 1 − Y)) Input Variables are as follows: Design Pressure - Maximum pressure at expected conditions. Generally, the maximum flange pressure at design pressure is used.
### Thickness Calculator - McWane
Thickness Calculator. Calculator. With minimal input required, the appropriate pipe wall and classes are quickly computed for all five (5) standardized Trench Types in full accordance with the ANSI/AWWA C150 Thickness Design Standard for Ductile Iron Pipe, latest revision.
### Pressure Piping Minimum Wall Equations and Calculator ...
ASME Section I and ANSI B31, the minimum thickness of piping under pressure is: or. For For t ≥ D/6 or for P/SE > 0.385, calculation of pressure design thickness for straight pipe requires special consideration of factors such as theory of failure, effects of fatigue, and thermal stress. Preview Pressure Piping Minimum Wall Thickness Calculator
### Pipe Burst Working Pressure Calculator Barlow's Formula ...
T = Pipe Wall Thickness (in) O.D. = Pipe Outside Diameter (in) SF = Safety factor (General Calculations 1.5 10, Use 1 For Bursting Pressure) S = Material Strength (psi) Ultimate Tensile strength or Yield strength can be used. Ultimate should be used to determine the bursting pressure.
### Piping Calculators » The Piping Engineering World
Pipe Thickness Calculator As Per ASME B31.3. This pipe thickness calculator calculates required pipe thickness for a process pipe based on ASME . Read More..
### Pipe Wall Thickness Calculation - Punchlist Zero - Size ...
Pipe Wall Thickness Calculation. This pipe wall thickness calculator determines the appropriate pipe to use per ASME B31.3. This calculator is suitable for carbon or stainless steel and for temperatures of up to 400 degrees F. Mill tolerance defaults to 12.5% per ASTM A 106 as “the minimum wall thickness at any point shall be not more than 12.5% under the nominal wall thickness specified”.
Pipe Wall Thickness. Reinforcement of Branch Connection - ASME B31.3 Calculator. Pipe Wall Thickness Calculator for predefined Piping Materials. Pipe Wall Thickness Calculation for Straight Pipe under External Pressure
### Pipe Weight per Foot Calculator : Pipe Industries
Pipe Weight per Foot Calculator. Pipe Weight Formula - This formula can be used to determine the weight per foot for any size of pipe with any wall thickness. The formula is: Wt/Ft = 10.69*(OD - Wall Thickness)*Wall Thickness. Outer Diameter (i.e.10.75 or 8.625 inches) Wall Thickness
### Pipe wall thickness calculation - EngStack
= 0.60 for furnace butt-welded pipe. Y = Coefficient, 0.4 for ferrous materials below 900 °F Tol = pipe manufacture allowed wall thickness tolerance = 12.5% for API 5L pipe up to 20 inch diamter = 10.0% for API 5L pipe greater than 20 inch diamter
### Pipe Calculator – Wall Thickness Marubeni-Itochu ...
Pipe Calculator – Wall Thickness. Use this calculator to determine the wall thickness, in inches. Click here to calculate Plain End Weight.
### Pipe wall thickness calculation - EngStack
= 0.60 for furnace butt-welded pipe. Y = Coefficient, 0.4 for ferrous materials below 900 °F Tol = pipe manufacture allowed wall thickness tolerance = 12.5% for API 5L pipe up to 20 inch diamter = 10.0% for API 5L pipe greater than 20 inch diamter
### Thickness Calculator - McWane
Thickness Calculator. Calculator. With minimal input required, the appropriate pipe wall and classes are quickly computed for all five (5) standardized Trench Types in full accordance with the ANSI/AWWA C150 Thickness Design Standard for Ductile Iron Pipe, latest revision.
### ASME B31.4 Wall Thickness Calculator - Pipeng Toolbox
ASME B31.4 Liquid Pipeline Wall Thickness Calculation Module. Calculate ASME B31.4 oil and liquid pipeline wall thickness from hoop stress for onshore and offshore pipelines. Select the appropriate line pipe schedule (ASME or ISO etc) and stress table (API, ASM, DNV etc), and material. Wall thickness is calculated using Barlow's formula.
### ASME B31.3 Pipe Thickness Calculator - piping-designer
Wall thickness of a pipe is calculated using Pressure, Temperature and materials. This calculator makes the calculation using the equation below: t = P ( d + 2 c) 2 ( S E − P ( 1 − Y)) Input Variables are as follows: Design Pressure - Maximum pressure at expected conditions. Generally, the maximum flange pressure at design pressure is used.
### Piping Calculators » The Piping Engineering World
Pipe Thickness Calculator As Per ASME B31.3. This pipe thickness calculator calculates required pipe thickness for a process pipe based on ASME . Read More..
### Pipe Burst Working Pressure Calculator Barlow's Formula ...
T = Pipe Wall Thickness (in) O.D. = Pipe Outside Diameter (in) SF = Safety factor (General Calculations 1.5 10, Use 1 For Bursting Pressure) S = Material Strength (psi) Ultimate Tensile strength or Yield strength can be used. Ultimate should be used to determine the bursting pressure.
### Pipe Thickness Calculation for Internal Pressure - Make ...
Pipe thickness calculation is a very important activity for every piping engineer.As it required to calculate numerous pipe wall thickness calculations for different design conditions. The process plant piping system deal with the fluids which flow inside the pipe at high-pressure and temperature.
### Pipe calculator: volume, area, capacity, size JustCalc
The calculation of the tube volume is done by the formula V=π*R1^2*L. R2 — outer radius. Second, it can be used as the pipe capacity calculator. This function will support you in the estimation when you select the pipes for the gas or water supply needs. The pipe capacity is a metric volume.
### minimum pipe wall thickness calculator excel spreadsheet
Oct 08, 2011 The Barlow Formula for a Minimum Pipe Wall Thickness Calculator. The classic Barlow formula for calculating bursting pressure for a pipe is: P = 2S*T/Do where: Do is the outside diameter of the pipe with units of inches (U.S.) or mm (S.I.) S is the strength of the pipe material with units of psi (U.S.) or N/mm 2 (S.I.)
### Pipe Equations - Engineering ToolBox
Cross Sectional inside Pipe Area. Cross-sectional inside area of a pipe can be calculated as. A i = π (d i / 2) 2 = π d i 2 / 4 (1) where. A i = cross-sectional inside area of pipe (m 2, in 2). d i = inside diameter (m, in). Cross Sectional Pipe Wall Area. The cross-sectional wall area - or area of piping material -
### Ductile Iron Pipe - Wall Thickness Calculator, Online ...
Here we can calculate for Ductile Iron Pipe Wall Thickness. Just copy and paste the below code to your webpage where you want to display this calculator. Ductile Iron Pipe - Wall Thickness: Net Pipe Wall Thickness: Internal Pressure: Outside Pipe Diameter: Yield Strength: where, t = Net Pipe Wall Thickness, P = Internal Pressure, D o = Outside ... | 2,233 | 10,162 | {"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} | 3.015625 | 3 | CC-MAIN-2021-39 | latest | en | 0.782869 |
https://community.smartsheet.com/discussion/13811/sumif-circular-reference | 1,726,647,067,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651886.88/warc/CC-MAIN-20240918064858-20240918094858-00513.warc.gz | 152,652,290 | 104,934 | #### Welcome to the Smartsheet Forum Archives
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# SUMIF - Circular Reference?
✭✭
edited 12/09/19
Hello all!
I'm pretty sure I saw this functionality in one of my Engage sessions, so I'm crossing my fingers that there's a simple fix!
I'd like to create a form that would allow individuals to submit project-related expenses for approval. I envision having a list of project numbers at the top of the sheet (see image, with sample projects "10" through "40"), with requests coming through the form populating the bottom of the sheet. Ideally, Smartsheet would look for all expenses associated with project #10 and sum them... and then do the same for project #20, project #30, etc....
I can get this to work for the first project, but as soon as I try to use a similar formula for the second project, I get a "circular reference" error. Is there something I'm missing here? And if I'm not approaching this the right way, is there another way to accomplish this? (Important note... there will be 100+ projects on a single sheet, not just 4 or 5, so it's got to work for a large number of project numbers existing on one sheet.)
THANK YOU!!!!
Tags:
## Comments
• jrw
You are super close! You do not need the = sign in your criteria. You'll want to set a defined set of rows for the ranges. Here is the formula I would suggest. I've added \$ to help when copying the formula.
=SUMIF(\$[Column7]\$6:\$[Column7]\$9, [Project Number]1, \$Amount\$6:\$Amount\$9)
Shawn
• ✭✭
Shawn,
For some reason I can't click the "Reply" button (I'm in Safari, if that matters....?) so I'm hoping you see this!
THANK YOU! My formula is now working properly!!! I truly appreciate the help.
Jody
• You're Welcome!
This discussion has been closed. | 494 | 1,987 | {"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} | 2.578125 | 3 | CC-MAIN-2024-38 | latest | en | 0.929104 |
http://mathoverflow.net/feeds/question/66473 | 1,369,404,354,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704664826/warc/CC-MAIN-20130516114424-00090-ip-10-60-113-184.ec2.internal.warc.gz | 169,116,713 | 1,675 | Group cohomology of an abelian group with nontrivial action - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T14:06:00Z http://mathoverflow.net/feeds/question/66473 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/66473/group-cohomology-of-an-abelian-group-with-nontrivial-action Group cohomology of an abelian group with nontrivial action Mitja 2011-05-30T17:17:32Z 2011-09-14T10:57:15Z <p>How do I compute the group cohomology $H^2(G,A)$ if G is a finite abelian group acting nontrivially on a finite abelian group A?</p> http://mathoverflow.net/questions/66473/group-cohomology-of-an-abelian-group-with-nontrivial-action/66480#66480 Answer by algori for Group cohomology of an abelian group with nontrivial action algori 2011-05-30T18:39:51Z 2011-05-30T18:39:51Z <p>If $G$ is any group and $A$ is any $G$-module, then $H^2(G,A)$ can be identified with the set of the equivalence classes of extensions $$1\to A\to H\to G\to 1$$</p> <p>such that the action of $G$ on $A$ is the given action. Two extensions $H_1,H_2$ are said to be equivalent if there is an isomorphism $H_1\to H_2$ that makes the extension exact sequences commute. See K. Brown, Group cohomology, chapter 4.</p> http://mathoverflow.net/questions/66473/group-cohomology-of-an-abelian-group-with-nontrivial-action/66481#66481 Answer by norondion for Group cohomology of an abelian group with nontrivial action norondion 2011-05-30T18:48:53Z 2011-05-30T18:48:53Z <p>You can compute it using the Bar resolution, see [Weibel, H-book].</p> http://mathoverflow.net/questions/66473/group-cohomology-of-an-abelian-group-with-nontrivial-action/75386#75386 Answer by Xiao-Gang Wen for Group cohomology of an abelian group with nontrivial action Xiao-Gang Wen 2011-09-14T10:57:15Z 2011-09-14T10:57:15Z <p>One can do the calculation using Kunneth theorem and the cohomology of cyclic group.</p> <p>See eqn J18 and appendix J.6 and J.7 in a physics paper <a href="http://arxiv.org/pdf/1106.4772v2" rel="nofollow">http://arxiv.org/pdf/1106.4772v2</a></p> | 670 | 2,067 | {"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} | 2.90625 | 3 | CC-MAIN-2013-20 | latest | en | 0.649597 |
https://www.biology-online.org/biology-forum/viewtopic.php?t=36150 | 1,531,953,061,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590329.62/warc/CC-MAIN-20180718213135-20180718233135-00520.warc.gz | 825,798,776 | 18,633 | ## How Does Natural Selection Create Ordered Variation?
Discussion of everything related to the Theory of Evolution.
Doom
Garter
Posts: 6
Joined: Mon Oct 07, 2013 6:06 am
### How Does Natural Selection Create Ordered Variation?
Ok, so I asked a similar question a while ago here: http://www.biology-online.org/biology-forum/about34598.html
Please read my post carefully and then glance over the rest of the thread, particularly the last post, as that about sums up the issue.
I'll quote wildfunguy here:
The intuition is that natural selection can only block a certain amount of mutation.
But in actuality, natural selection can only let through a certain amount of mutation. Creationists have it backward.
So basically natural selection is a filter. It filters the random mutations, selecting only the best for the particular environment. Simple right?
Now here's my problem:
A. A static filter based on a random input will output a uniform result
Of course natural selection isn't static, the environment will change with time. This brings me to the main problem:
B. A dynamic filter applied to a random input will produce a varied but still relatively uniform result
Reason for my conclusion?
Because the changes in the environment will just as likely negate the effects of the previous beneficial mutation.
As you know, mutations can be beneficial, harmful or neutral. This is based entirely on the environment.
So therefore a species could evolve something beneficial (like beetles on a windy island getting shorter wings, meaning they don't get blown off so easily), then in the very next generation, the environment could change so that's a bad thing, negating the effect of the previous beneficial mutation.
So it'd be sort of like intersecting waves. They are just as likely to negate as to add up.
The law of averages dictates that natural selection can't make any progress then, doesn't it?
That pretty much sums up my position.
If someone can please explain where I'm wrong, I'd really appreciate it
Cat
King Cobra
Posts: 635
Joined: Thu Feb 14, 2008 7:40 pm
### Re: How Does Natural Selection Create Ordered Variation?
Try this:
A dynamic filter applied to a DINAMIC random input.
Doom wrote:So it'd be sort of like intersecting waves. They are just as likely to negate as to add up.
The law of averages dictates that natural selection can't make any progress then, doesn't it?
You cannot delete established mutations. If shorter wings are a problem now, only individuals carrying NEW counteracting mutations, say double wings, will survive...
Doom
Garter
Posts: 6
Joined: Mon Oct 07, 2013 6:06 am
But if that new mutation doesn't materialize, then the species will die out...
Doom
Garter
Posts: 6
Joined: Mon Oct 07, 2013 6:06 am
Was I unclear in my first post?
You cannot delete established mutations
Isn't that exactly what Natural Selection is?
Filtering the mutations OUT
JackBean
Inland Taipan
Posts: 5694
Joined: Mon Sep 14, 2009 7:12 pm
1) why should dynamic filter result in uniform result? That just doesn't make sense, if you once select for for wings, you'll get short wings, if you select for long wings, you'll select long wings.
2) why do you think that the environment always changes into exact opposites?
http://www.biolib.cz/en/main/
Cis or trans? That's what matters.
Doom
Garter
Posts: 6
Joined: Mon Oct 07, 2013 6:06 am
1) why should dynamic filter result in uniform result? That just doesn't make sense, if you once select for for wings, you'll get short wings, if you select for long wings, you'll select long wings.
Here's why
Because the changes in the environment will just as likely negate the effects of the previous beneficial mutation.
As you know, mutations can be beneficial, harmful or neutral. This is based entirely on the environment.
So therefore a species could evolve something beneficial (like beetles on a windy island getting shorter wings, meaning they don't get blown off so easily), then in the very next generation, the environment could change so that's a bad thing, negating the effect of the previous beneficial mutation.
2) why do you think that the environment always changes into exact opposites?
Not exact opposites every generation, but a windy island might easily become not-windy with lots of predators in a 500 year timepan.
Meaning that
Because the survival of the creature is based entirely on the environment, and the changes in the environment are completely random, there should be no gradual up-hill evolutionary climb (mt. improbable). Becuase 'benificial' will just as likely mean going back down the hill as climbing up the hill.
I hope I don't sound like I'm coming across as an attacker of evolution, I'm just looking for answers.
Tricho
Garter
Posts: 16
Joined: Fri Jan 03, 2014 5:03 am
Hi,
evolution is not about going uphill, onwards or whatever. It's only about what's best at the moment.
So if an environment changes so that only specifically spezialized organisms have a lot of off springs the population will be uniform after a while. But most of the time it's different things that can increase the fitness, not only one. And if factors change most of the time they don't change for all individuals of a species - only for a population. So you get a variety of phenotypes.
And if species can't adapt to occuring changes they will eventually die out, but that may take a few couple of hundreds of thousands of years.
JackBean
Inland Taipan
Posts: 5694
Joined: Mon Sep 14, 2009 7:12 pm
as Cat wrote, the counter-conditions, even if they were opposite (highly unlikely), will only very rarely lead to removal of exactly the trait which was beneficial before. Rather, some counter-balancing new trait will evolve. And thus you get something new.
http://www.biolib.cz/en/main/
Cis or trans? That's what matters.
### Who is online
Users browsing this forum: No registered users and 10 guests | 1,349 | 5,907 | {"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} | 2.71875 | 3 | CC-MAIN-2018-30 | longest | en | 0.931906 |
http://everything2.com/user/mcai7et2/writeups/How+to+destroy+the+world+using+a+spaceship+and+an+elephant | 1,406,806,595,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510273012.22/warc/CC-MAIN-20140728011753-00363-ip-10-146-231-18.ec2.internal.warc.gz | 100,278,498 | 7,907 | (v = velocity, vbar = average velocity, t = time, d = distance, e = energy, 241350 = 150 miles in meters, 10 = acceleration due to gravity, gbar = average force exerted by gravity during the drop, 5.98E24 = Mass of earch in KGs)
``` v = at
vbar = at/2
t = d/vbar
vbar = at/2
vbar = 5t
t = d/5t
5t^2 = 241350
t^2 = 48270
t = 219.7
v = at
v = 10*219.7
v = 2197 m/s
```
The elephant impacts at just under 3.5 times the speed of sound. I am sure it may smart a bit, but not enough to destroy all living things on the earth.
Now, if e = 1/2 mv^2 and we use The Custodian's figure of 4.184E12 Joules per kiloton. If you want to aim for 10,000 atomic bombs worth of energy, then we need to drop the elephant from a little higher up.
I am assuming for the purposes of argument that an atomic bomb is 1 Megaton (forget about the bomb that totalled Hiroshima, technology has moved on since then).
First, lets calculate the total energy required:
```e = (total number of bombs) * (power of each bomb in kilotons) * (joules per kiloton)
e = 10,000 x 1,000 x 4.184E12
e = 4.184E19
```
We can now work out how fast this elephant needs to be going:
```e = 1/2mv^2
4.184E19 = 1/2 x 7000 x v^2
v^2 = 4.184E19 / 3500
v = 1.093E8 m/s```
Now we are getting to the point, all we need to do is accellerate our elephant to just over 1/3 of the speed of light.
The next part is trickier, we need to find the altitude required to drop the elephant. Assuming a steady accelleration of 10m/s/s, this would be easy, but gravity is inversly proportional to the square of distance, and I have a feeling that this will come into play with these figures.
``` f = ma
v = at => t = v/a
t = d/vbar
vbar = 1.093E8 ^ 0.5
= 1.045E4
v/a = d/1.045E4
1.093E8/a = d/1.045E4
a/1.093E8 = 1.045E4/d
a = 1.142E12/d
a = 6.672E-11 * 5.98E24 / ((6.38E6+d)*(6.38E6+d)) (stolen from gsu.edu)
a = 3.99E14 / (4.07E12 + 1.276E7d + d^2)
a = 9.80 + 3.12E7/d + 3.99E14/(d^2)
1.142E12/d = 9.80 + 3.12E7/d + 3.99E14/(d^2)
1.142E12 = 9.80d + 3.12E7 + 3.99E14/d
9.80d + 3.99E14/d = 1.142E12
9.80d - 1.142E12 + 3.99E14/d = 0```
Solve as a quadratic equation and d = 116524244548.55304
This works out at just over 116 million kilometers. Note, for this to work properly, you will have to remove all matter in the Solar System which is not part of the Earth, but this is left as an exercise for the reader. | 884 | 2,390 | {"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} | 3.96875 | 4 | CC-MAIN-2014-23 | longest | en | 0.883882 |
http://www.jasonmunster.com/tag/electric-cars-2/ | 1,720,810,773,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514450.42/warc/CC-MAIN-20240712161324-20240712191324-00187.warc.gz | 42,471,698 | 23,136 | First of all, sorry it has been over a month since I've posted. I've decided to get together a few people to start addressing some of the things I write about, and that has taken my time up til now. I'll be posting once per month from here on out, on the first Sunday of every month. Today's post is a long one, but one of the most interesting I've written by far.
This is the one time where I will say the following: if you are short of time, skip directly to the math section. It shows a serious glaring deficiency of either forethought or disclosure on the part of the founders of Solar Roadways. Moreover, it shows they can't do basic math. Never trust an engineer who can't do basic math. It's a very crackpot idea.
Here We Go!
I've heard a lot of talk about Solar Roadways recently. I'm going to use it as an example of how to analyze some "science." After you follow the very basic math below, you will see that the team at Solar Roadways does not know what numbers to run*. A much larger problem: they suggest that solar roads can replace fossil fuel power, while simultaneously and surreptitiously admitting that they need a ton of grid power to make this work. So pretty much they are either dumb or straight up liars.
First, let's talk about why these roads might be good, from their point of view. Being a by-the-numbers type of guy, the first thing I did was check the "numbers" section of their website. While their assumptions are dubious at best (more on that later) They say that their roads could provide 3x the energy that the US needs, in kilowatt hours (kWh is a useless measurement here, cause it will be intermittent power. In other words, it produces no energy at night, and will need to be supplemented by fossil fuel power. More on that later). Also, the roads look a lot cooler, with light-up sections, and ability to melt snow so that road maintenance is reduced.
So the thing is wired to the grid so that if it snows, it can use heating elements to melt the snow instead of plowing it. But doesn't snow take a lot of energy to melt? Would it take less energy just to push it with a plow? Time for the math!
Math of Melting vs Pushing Snow
Plow trucks to be replaced by Solar Roads? Not happening.
Okay. Let's assume middle-case scenario of 8 inches of snowfall, being removed with one sweep by plow trucks, and that this is between powder and heavy snow in consistency, which means 1" of water equivalent. A DOT snowplow clears 10' width of snow, or 120 inches. In one foot of movement forward and plowing 8" of snow it moves the water-weight of 1"x120"x12" or
Now we have to figure out how much energy cost this took in fuel, so we will later relate this to the mileage efficiency of a DOT truck. First, let's figure out how much energy it takes to melt this much snow into water. Do do this we need the latent heat of fusion, or the energy it takes to transition from ice to snow. It's 334 Joules/gram. How do we convert from cubic inches of water to grams? Easy. Because the metric system makes sense, one of water = 1 gram. There are 2.54 cm per inch, so:
Okay, we have grams, now let's calculate the energy to melt as much snow as a plow moves from driving 1':
Or ~7.8MJ. Per foot. Or, for a mile:
to melt 8 inches of snow.
Okay, so, a plowtruck uses diesel. Each gallon of diesel has 136.6MJ. Very conservatively assuming a plowtruck gets ~5 miles to a gallon (I'm guessing it's more like 10, someone who has driven one, correct me and I will correct these #'s), it would take 27.3 MJ to plow one mile of snow. Compared to 41,184MJ to melt it. It literally takes 1500x as much energy to melt is as it would to move it.
This is what you would call a very very bad idea. Engineers as cofounders should know better than to let this slide as a potential solution.
End of Math Section
Okay, so now that we've completely dismantled the case of using these things to melt snow, lets move on to some other issues. We'll skip the minor issues, because that's just nitpicking, and move straight to the parts where they just don't know what they are talking about, and finish with things they clearly know about, but are purposefully misleading people with in order to get more money. Finally, we will close with me realizing that Nathan Fillion is a fool.
Okay, to the problems with this solar roadways project:
Dubious assumptions:
Things they don't understand: the supply lines of a very basic input.
REE mining in China is not a clean thing. Nor was it great in the US. Right now there is not enough world production to make enough of these solar roadway tiles. Look at this article to see more pictures of REE production in China.
They assume an 18.5% efficiency of the solar panels. These are panels that use Rare Earth Elements (REEs). On their FAQ, when someone asks if they are using REEs, they state (paraphrased), "Our electronics don't use silver or gold" (neither of which are REEs, so they are either changing the topic or don't know what question they are answering) "but we can use any solar cell." Good that they can use any solar cell, because there is not enough REE production in the world to produce solar at the scale they need to even replace one major highway with these. Bad they they use 18.5% as their assumed efficiency, because solar cells in this range of efficiency use REEs.
REEs are pretty much only produced in China, because producing them make a massive amount of pollution. Decades ago every other major country quit producing REEs because of the pollution they cause, and because China didn't care about pollution or health hazards, so the world was happy to let them pollute themselves and take their REEs. It's been so long since the US produced REEs that we literally don't know how. Solar Roadway's answer is "let's leave this to the government." They aren't addressing the problem at all. While other countries are looking to have their own production, it will take a very long time for this to come to fruition, and the production rate still won't be enough for a second-rate harvesting design (flat roads with bad optics vs. tilted panels with great optics to concentrate light perfectly).
At best, they can go with non-REE solar cells, which have about an 5-10% efficiency. That means that each of their hexagonal panels will produce half the power anticipated, and thus will make half as much money toward recuperating their costs. In other words, these non-REE solar panels need more basic raw materials (in terms of roadway) per kwh produced, and thus will cost more per unit energy, in an already material-intensive design for a solar cell. This shows that the project is lacking in any real expertise or understanding of the core problem they are trying to solve. Keep in mind that these are not dealbreakers. The team could hire an expert, or consulting, to fill in their knowledge gaps (likely the former, consultants are expensive, and they really need long-term help to bring this to fruition). Also, it doesn't negate all the other benefits of the solar roadways. Finally, non-REE solar panels are a hot topic in research. If the rest of the solar roadways tech is developed, and they are just waiting for good solar cells, it will rapidly enhance future deployment.
In short, the solar cells are a slight additional benefit to whatever holds them in this case of mass-distribution and inefficient use of cells. So if this new road itself doesn't compare favorably to asphalt, the project is sunk in the water.
Things they are just completely wrong/misleading about: melting snow, shutdown of fossil fuel, price of energy
We discussed the melting of snow. They suggest it replace snowplows. Bad idea. It's clearly not going to work, energetically speaking.
They keep talking about how 50% of US electricity use is from fossil fuels, and how these roads are going to replace it. This is so wrong that it is hard to debunk in one post. But here goes: First, only 40% of US primary energy (my link, please read it for background if you feel a bit lost, it is far briefer than this post) is for electricity. Second, only 66% electricity of this comes from fossil fuels. In other words, 26.4% of US electricity comes from fossil fuels (if we change all our transportation over to electric, these numbers will change, but that would require these roads to have induction power installed - AKA roads that provide the car with energy for driving so they don't have range issues). This is the total amount of emissions that could be replaced by solar roads in their current design.
Primary energy in the US. As detailed by the math above, only 25% of primary energy in the US can currently be replaced.
So, pretty much they are off to a bad/misleading start there. But this is nitpicking. The real issue comes in when they talk about replacing fossil fuels. First, they talk about heating the roads. This means they will have to put energy into the roads. Where will this energy come from? Power plants. So much for shutting down fossil fuel. But wait, there's more! Solar power is intermittent. It doesn't even work at night, so power plants also have to be on then. So pretty much, their idea of shutting down power plants is completely shot out of the water by these two things. Can solar roadways still be part of a larger energy solution? Well, not if they are heating roads to melt snow. That just takes far too much energy. If they scrap the melting snow idea and go to just producing energy? Yeah, it might help some. But let's get to one last funny part, the one that shows they know that they won't be shutting down fossil fuel power any time soon.
Energy storage. From their FAQ, they mention that there will be "virtual storage" in that during the day they will add power to the grid, and at night they will take power from the grid. This is double-speak to mean: during the day we will provide power that can offset coal and natural gas power plants. At night when we aren't producing, natural gas powerplants (again, my link) will fire up to power our roads (nuclear is not an option for power phasing like this, nuclear powerplants don't spin up or wind down on half-day timescales). In other words, they fully well understand that they aren't going to do away with the rest of the power grid, and that they aren't going to replace all those fossil fuel emissions. So pretty much, saying that these can replace our power grid is double-speak sales points.
The final problem? They don't understand energy distribution. Electricity is produced at about $0.03 to$0.08 per kwh at a power plant. By the time it arrives to us, we pay $0.13 to$0.25 (or $0.50 in Hawaii), because distribution costs a lot of money. Solar panels on our roofs produce power that costs about$0.15 to $0.20 cents per kwh, give or take. So the end-user cost of grid power is the same as that of house solar. But if you run that solar power through the distribution channels and add that price, suddenly you're talking$0.25 to $0.40 power. So, unless they are giving this power away for free, it's probably not gonna be a great solution. Some Solutions I've softened my usual tone quite a bit for this writeup, cause I don't want to be a complete naysayer of something who is trying to do something positive (sorry, I know how much you all know and love my biting sarcasm and scathing reviews).Outside of their false solution of trying to solve the energy/climate issue, this idea has some potential. On that note, rather than pointing out problems, I've come up with some great solutions. My suggestions: 1: Nix the whole melting of snow concept to replace plow trucks. Energetically, it doesn't work. Plow trucks should still exist. Instead of replacing them, replace the salt and sand they need to spread. Make it so plowtrucks plow all but the last 1/8" of snow, then melt that (note, this is still a tremendous amount of energy, but stay with me). This will have a few benefits: • No more salt and sand on roads means less salt and sand damage to vehicles, making vehicles last longer • No more salt and sand on roads means that DOTs can save money buy not buying these things • ... no salt and sand runoff, which pollutes local waterways • ... animals that go to roadways in the spring to lick off accumulated salt won't do that, reducing traffic accidents from moose and deer, etc. 2: Get a bit more cognizant or REEs and their limitations. Don't use bad assumptions that are easy to poke holes in. 3: Stop selling people on false promises of doing away with fossil fuels. It makes the whole green movement look bad when prominent people are lying or severely misinformed. 4: Focus on the real potential of making these have inductive energy for electric cars. This could eliminate range anxiety (people fearing their electric cars will run out of energy and leave them stranded). Electric car sales will move a lot faster if people can drive from LA to SF, or between Boston/NYC/DC. The potential partnerships include every major car company that markets in the US. Also, this could reduce oil use, and drastically reduce air pollution from cars in these busy areas by further replacing combustion engines with electric ones (even if we power them with electricity from coal, a well-scrubbed coal plant produces fewer bad things than a car). Moreover, since people won't need fuel, they could be assessed a charge per mile driven instead. By whoever owns the roads. Here is your real money-maker for the roads, fellas. It will be far more lucrative than producing tiny amounts of electricity. Please get on this. It will lead to more electric car research, and more rapidly drive forward battery development, and it turns out that cars make a bunch of really bad pollution that causes harmful side effects like death. This last bit, changing your startup's tack when a better model comes along, is important. And solar roadways needs to do that for a viable product, because their core solution faces a lot of headwinds (yay, sailing puns!) in break-even with their current model. So, overall, these roads could be an excellent idea. The solar part, their main selling point, is BS because of cost, efficacy, and the need for gas-fired power plants to supplement them. The shutting down most fossil power plants is a lot of nonsense for the same reason. Making the environment better by reducing salt and sand use? Decent. Potentially by making most cars electric? Game-changer, but they are barely looking at that aspect right now. Probably cause they are too busy counting the piles of cash that indiegogo just threw at them (or, more likely, answering the insane number of emails that comes from this sort of campaign). Hokay, that's my piece. Thanks for reading this long one. - Jason Munster Extra stuff! Some background about Solar Roadways initial funding: They were funded by government SBIR. This stands for Small Business Innovative Research. It's for high-risk, high-reward research. In other words, this was considered high-risk from the start. They got a phase II, which means they did well. It's clear they still have issues and are still high-risk. But I'm glad someone is paying for research and innovation like this, especially because if it pays off, it could result in more jobs and more taxpayer base. That being said, they haven't received more funding or any grants to build this out further. Possibly cause it's a big, crazy idea. Elon Musk can pull off big, crazy ideas, because he is a brilliant manager and has a very strong personality. These guys are going to need some bigger guns on their team if they are going to make something of this project. Second, Nathan Fillion is a bit of a fool. In touting Solar Roadways, he displays why pop culture heroes shouldn't get involved in matters outside their field of expertise (mainly, looking good in front of a camera, and pretending to be someone who they aren't in front of a camera). His adoration of something he doesn't understand falls deep within the territory of religious fervor. Nerds: just cause one of your heroes likes something doesn't mean it actually is plausible. One final-final note: I know that this post is 3x longer than my rest. I assure you, it's far shorter than I wanted it to be. I don't believe in two-part posts very often, though. If you have read this far. please leave a comment so I can appreciate you forever 🙂 *Engineers who don't know what numbers to run are a bad investment. For my own company, all business types are skeptical of how much I know (or want to take advantage of me fully) until they find out that I used to be in finance and have a really good idea of the big picture of most things. In short, this company has a lot of potential once they take on broader experts. # Electric Cars Electric Cars. Are they really all they are hyped up to be? The short answer: hell yeah. These things are sweet. I want to get my hands on one right now. Energy flows in the US. Transportation accounts for 28% of all energy use, primarily from burning petroleum. 35% of US energy consumption is in transportation. Transportation requires that the energy source be within the vehicle (unless you are in South Korea, where the energy source is induction and is beneath the road. Pretty badass, if you ask me). Batteries currently weigh a lot, don't have nearly as much energy per pound as gasoline, and require a long time to charge. But if we could replace a huge percent of this with more efficient electric cars, it would go a long ways towards arresting GHG emissions. 120 million Americans commute to work by car. The average person lives fewer than 20 miles from work. Substantially all of them commute alone. The Nissan Leaf gets 75 miles before it needs to be recharged. The Tesla model S goes about 275 miles. No matter what the source of energy for an electric car, it produces less CO2 than a normal car. How does an electric car produce less CO2 than a gas one? No matter the source of the electricity, even if it is an old inefficient coal plant, the conversion efficiency of an electric car will result in lower CO2 emissions per mile than a gas powered car. The EPA estimates that the Nissan Leaf gets an estimated 125mpg using CO2 equivalent of gasoline. The recent fleet average for the US is about 30mpg for passenger cars. So electric cars emit only 25% the CO2 of your average normal car. Lets be generous and say they emit 40% the CO2 of your best gasoline powered cars. Commuters would make a very significant difference in emissions if they changed over to electric cars. Power Most Americans base the acceleration needs of their car on the idea that they someday need to accelerate down on onramp to get to 65mph on the highway. The amount of power a car has is typically listed as horsepower (hp). This is a terrible measure. The real measure of power of a vehicle is Peak Torque. Allow me to explain this concept. Roughly speaking, torque is the force going into a rotation of an object. It makes sense to use torque to describe cars, cause they have rotation parts. Think of it as the amount of energy going into the car from the tires rotating on the road. (yes, that is a cross product) where r is the radius, or distance from the center of rotation, and F is the force. For the most part, the torque of a vehicle is entirely determined by its engine. It directly translates to how fast you can accelerate. More torque yields less time from zero to 60. My motorcycle. Pretty, eh? Let's compare some examples. First, my favorite. My bike, a Kawasaki VN750, vs. a Hayabusa (fastest production bike in the world) and an electric bike from Zero Motorcycles, the Zero DS. The electric motorcycle gets up to the equivalent of 400mpg. Compare to a normal bike of around 40-50mpg. VN750 Hayabusa DS-electric Style Cruiser Sport Sort of cruiser Weight 500 lbs 563 lbs 400 lbs Torque (ft-lbs) 47 99.6 69 Before comparing, let's talk about peak torque. Peak torque is the maximum torque an engine can put out. For a gasoline engine, it is pretty much right before it redlines. So the numbers of 47 and 99.6 you see for the first two bikes means that it is the best they can do. You can think of an electric motor as pretty much always putting out peak torque. In other words, the hayabusa has to jump up to 6500rpm before it can be at full power, then it shifts up a gear, and drops back down out of its full power range. The electric bike doesn't shift gears, either. Let's look at the Hayabusa power band to illustrate this difference. Hayabusa power band in yellow. It is not a flat line. As you can see, the torque output of a gas engine changes with RPM. You will notice the Kawasaki ZX-14 has more max torque, but less torque at the lower end. This is one of the main reasons the 'Busa is considered faster. It comes off the line far faster than other bikes, cause it has higher starting torque. Compare this to an electric engine, which has max torque from 0 RPMs up. You probably see my point about how sweet electric engines are. So now we can compare electric vs gas based on torque. The electric motorcycle trounces my motorcycle all the time. In the first few moments, it will likely nearly match the Hayabusa. In fact, until the 'Busa hits 3000 RPM, the electric bike won't look too shabby. Why? First, cause it has the same torque as the 'Busa up til the Busa hits 3000rpm. Second, cause it weighs 150 lbs. less. In short, a smaller electric bike kicks butt. (note that the electric bike doesn't have super wide tires to accommodate all its power, so it might slide around a bit when you hammer down). Where does the electric bike fall short? Range. This bad boy will only go 75 miles on the highway between charges. Funny enough, it'll go 125 in the city. This is all owed to wind resistance. Back to the point, you can't refill this guy as easily. You need to plug it in. It takes a while to recharge. You can just fill up a motorcycle and go on your way. This bike is pretty much for commuting or visiting friends in nearby cities. (*2017 update! New electric motorcycles are capable of going 200 miles. Then they have to recharge for a long time. But 200 miles is a long trip). Also, let's admit it, both my bike and the 'Busa are sexier. The cars. Audi A5 Nissan Leaf Tesla S Cost$38k \$21.3k 62,000 MPG 22 102 90 Peak Torque 258 210 443 Weight (lbs) 3549 3354 4650 Range (miles) unlimited 75 275
As you can see, if you are commuting the Nissan leaf makes a lot more sense in every possible way. It costs less to buy than most cars, it has the acceleration potential of a high-end Audi A-5, and the range you need to get to work and back. This beast will accelerate onto the highway just fine. Also, there is the whole power band thing. These electric cars have peak torque all the time.
Overview
Electric vehicles. These puppies accelerate as fast or faster than most vehicles in their class. The shorter range ones are less expensive to buy than most cars, and the cost of making them move is lower. Oh, and they save the environment relative to normal cars.
Whats the drawback? Range anxiety: the other event people think of when they don't want an electric car. Visiting Grandma. People want to buy one vehicle that can do everything they want. There is also the concern of "what if I am in an emergency and really need a car that can drive far right away?!?" How many times has that happened to you in your life? For me, the answer is 0. Any family emergency I had, I took a plane or a bus to.
Most families still own two cars. At least one should be electric. Here's an idea: get one gasoline car to visit grandma when you need, and then get an electric for your commute. Here's another idea: get an electric car or two, and rent a car when you need to go visit Grandma. You will save money either way.
The ridiculousness that is trucks and SUVs? Get a subaru with a roof rack, and rent the trucks and SUVs otherwise. Why the heck are you in a vehicle getting 12 miles to the gallon when gas is expensive, and when burning that gas helps wreck the environment? You, person who commutes to work in a pickup, are a selfish person.
Grid Stabilization
In the Solar and Wind articles, we read that these technologies produce intermittent power. In other words, they can't provide power on demand or at night. Imagine, if you will, that those 120,000,000 commuters all had electric cars. And that they all had excess batter capacity. They could charge up while the wind was blowing and the sun was shining, and discharge while the sun was sleeping and while the wind was lazy. Suddenly part of the problem with wind and solar has some help. This is a huge topic, though, and I won't go farther into it.
Shortest version:
Get an electric car for your commute.
Hokay, that is all for now. I will edit this as I get comments. Thanks for reading!
Jason Munster | 5,649 | 24,916 | {"found_math": true, "script_math_tex": 6, "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": 1, "img_math": 6, "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} | 2.84375 | 3 | CC-MAIN-2024-30 | latest | en | 0.967794 |
http://www.onbarcode.com/tech/740/66/ | 1,643,432,252,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320299927.25/warc/CC-MAIN-20220129032406-20220129062406-00589.warc.gz | 116,579,559 | 8,843 | # RECTILINEAR MOTION AND INSTANTANEOUS VELOCITY in Visual Studio .NET Encoder QR Code in Visual Studio .NET RECTILINEAR MOTION AND INSTANTANEOUS VELOCITY
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The new volume will be (2 + x)3 , and so the change in the value of the volume y is (2 + x)3 23 . This change in y is denoted traditionally by y, y = (2 + x)3 23
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Now the natural way to compare the change y in y to the change x in x is to calculate the ratio y/ x. This ratio depends of course on x, but if we let x approach 0, then the limit of y/ x will de ne the instantaneous rate of change of y compared to x, when x = 2. We have (algebra, Problem 11.2) y = (2 + x)3 23 = [(2)3 + 3(2)2 ( x)1 + 3(2)1 ( x)2 + ( x)3 ] 23
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= 12 x + 6( x)2 + ( x)3 = ( x)(12 + 6 x + ( x)2 ) Hence, and lim y = 12 + 6 x + ( x)2 x
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y = lim (12 + 6 x + ( x)2 ) = 12 x 0 x
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Therefore, when the side is 2, the rate of change of the volume with respect to the side is 12. This means that, for sides close to 2, the change y in the volume is approximately 12 times the change x in the side (since y/ x is close to 12). Let us look at a few numerical cases.
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Using Barcode generator for iPad Control to generate, create UPCA image in iPad applications. | 1,613 | 6,226 | {"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} | 3.65625 | 4 | CC-MAIN-2022-05 | longest | en | 0.892643 |
https://t5k.org/curios/page.php?number_id=166 | 1,679,480,878,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943809.22/warc/CC-MAIN-20230322082826-20230322112826-00738.warc.gz | 623,635,657 | 3,606 | # 434
This number is a composite.
A palindrome that produces a Composed prime: 434 = 2 x 7 x 31 and 42 + 37 + 431 = 4611686018427390107 is prime. [De Geest]
434 is also the sum of the cubes of the digits of the emirp 347. [Trotter]
The smallest composite number n which satisfy the equation sigma(n+2) = sigma(n)+2, where sigma(n) is sum of divisors of n. [Gupta] | 118 | 367 | {"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} | 2.703125 | 3 | CC-MAIN-2023-14 | longest | en | 0.765633 |
http://math.stackexchange.com/questions/259994/omega-limit-set-is-invariant | 1,469,778,638,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257829972.49/warc/CC-MAIN-20160723071029-00050-ip-10-185-27-174.ec2.internal.warc.gz | 157,893,356 | 18,425 | Omega limit set is invariant
In the ODE where $y'=f(y(t))$ and $y(0)=yo$.
The omega limit set $w(yo)$ is positively invariant and also negatively invariant.
I want to prove first that its positively invariant and then prove its negatively invariant.
But how do I show that using a flow function ($phi(y,t)$) given that I know only the definition and the identity of flow function and very little understanding of the concept of flow function.
Thankyou for the help!
-
What is "the omega limit set"? What does "postively invariant" and "negatively invariant" mean to you here? What is the "flow function" you speak about? – Henning Makholm Dec 16 '12 at 14:17
i know omega limit set is the set of limit points given the IVP for y'.+ively invariant means as t goes to infinity the function remains inside the equilibrium solution and never gets out of it same goes for -ively invariant as t goes to -infinity. i just dont get the flow function my prof. explained it briefly though! – d13 Dec 16 '12 at 14:21
@dl3: The set of limit points of what? What does "IVP" mean? The equilibrium solution of what? It seems that you know more about this topic than anyone reading your question here, since your prof explained it (however briefly) to you, whereas you have not even attempted to explain to the rest of us what you're speaking about. – Henning Makholm Dec 16 '12 at 14:26
maybe i need to refine my question here. wait! – d13 Dec 16 '12 at 14:28
Consider the system $$\dot x=f(x),\quad x\in U\subset\mathbf{R}^k,\,f\colon U\to\mathbf R^k,$$ with the solution $$x=x(t;x_0).$$ By definition point $\bar x$ belongs to omega limit set is there exists a sequence $\{t_n\}$ such that $x(t_n;x_0)\to \bar x$ when $n\to\infty,\,t_n\to\infty$. This means that for any fixed $t$ one has (using the properties of the flow) $$x(t_n+t;x_0)=x(t;x(t_n;x_0))\to x(t;\bar x),$$ which means that if $\bar x$ belongs to omega limit set then the whole orbit containing this point belongs to this set, and since the orbits are invariant this implies that omega limit set is both positive and negative invariant. | 560 | 2,096 | {"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} | 3.6875 | 4 | CC-MAIN-2016-30 | latest | en | 0.893813 |
https://cs.stackexchange.com/questions/72210/given-a-context-free-grammar-prove-if-the-grammar-is-ambiguous | 1,618,749,574,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038476606.60/warc/CC-MAIN-20210418103545-20210418133545-00311.warc.gz | 287,533,090 | 37,560 | # Given a context free grammar, prove if the grammar is ambiguous
Here is a context free grammar that I have been given for practice: Grammar $G = (V,\Sigma,R,S)$ where $V$ is $\{S,A,B,a,b,c\}$ and $\Sigma$ is $\{a,b,c\}$. $R$ has the following rules:
\begin{align}S &\to A\\ A &\to BB\\ S &\to bS\\ S &\to cS\\ A &\to a\end{align}
I am fairly sure that it is not ambiguous, since for a grammar to be ambiguous there will be two ways to reach a result. In the grammar above, each $b$ comes from $S \to bS$, each $c$ comes from $S \to cS$, and each $a$ comes from $S \to A \to a$. It seems the $A \to BB$ is never used and $c=(b\cup c)^*a$. So there is only one derivative of each string in $L$.
My question is, is my thinking correct and would this suffice to prove that the grammar is not ambiguous?
Edit: We went over the problem in class a while back and the explanation above was accurate.
• You already asked this question. I said the grammar was ill-defined because symbol B is used but has no definition. You deleted that question and re-asked a couple minutes later, but did not fix that error. Why? – orlp Mar 29 '17 at 6:42
• I deleted this post earlier per the request of the other comment, but was then told that they recommended deleting the other duplicate rather than this one, hence the repost. As for symbol B, the question does not provide any other information than what is above. Since symbol B is not defined, could this be a reason as to why the grammar is unambiguous or would it just make it unsolvable? If it is the latter case, the question itself is flawed. – TweezerCube Mar 29 '17 at 6:47
• Assuming you didn't miss a definition of B somewhere I'd argue that the question itself is indeed flawed. – orlp Mar 29 '17 at 6:50
• @orlp Having a non-terminal symbol that can't generate anything is not a problem, just like having states that aren't co-accessible in an automaton isn't a problem. They are just superfluous. – xavierm02 Mar 29 '17 at 9:16
• Your thinking is correct, and you can use your ideas to formally prove that the grammar is unambiguous. – Yuval Filmus Mar 29 '17 at 16:42 | 579 | 2,123 | {"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} | 3.328125 | 3 | CC-MAIN-2021-17 | latest | en | 0.958775 |
http://altrushare.com/tag/ac-thermostat-wiring-heat-pump/ | 1,524,447,347,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945669.54/warc/CC-MAIN-20180423011954-20180423031954-00157.warc.gz | 14,086,693 | 14,396 | # Ac Thermostat Wiring Heat Pump
## Wiring AcWiring Ac
By Cyrielle Marjolaine. Diagram. Publised at Monday, November 20th 2017, 13:21:10 PM. As discussed earlier, a digital circuit represents and manipulates information encoded as electric signals that can assume one of two voltages – logic-high voltage (or Vdd) and logic-low voltage (or GND). A digital circuit requires a power supply that can produce these two voltages, and these same supply voltages are also used to encode information in the form of two-state, or binary signals. Thus,if a given circuit node is at Vdd, then that signal is said to carry a logic ‘1’; if the node is at GND, then the node carries a logic ‘0’. The components in digital circuits are simple on/off switches that can pass logic ‘1’ and logic ‘0’ signals from one circuit node to another. Most typically, these switches are arranged to combine input signals to produce an output signal according to basic logic relationships. For example, one well-known logic circuit is an AND gate that combines two input signals to produce an output that is the logic AND of the inputs (i.e., if both input1 and input2 are a ‘1’, then the output is a ‘1’).
## Led Bulb Simple Circuit DiagramLed Bulb Simple Circuit Diagram
By Alix Loane. Circuit. Published at Tuesday, December 26th 2017, 21:47:43 PM. Electric circuits use electric power to perform some function, like energize a heating or lighting element, turn a motor, or create an electromagnetic filed. Electronic circuits differ from electric circuits in that they use devices that can themselves be controlled by other electric signals. Restated, electronic circuits are built from devices that use electricity to control electricity. Most electronic circuits use signals that are within 5 to 10 volts of ground; most circuits built within the past several years use signals that are within 3 to 5 volts from ground. Some electronic circuits represent information encoded as continuous voltage levels that can wander between the high and low voltage supply rails – these are called analog circuits. As an example, a sound pressure level transducer (i.e. a microphone) might drive a signal between 0V and 3.3V in direct proportion to the detected sound pressure level. In this case, the voltage signal output from the microphone is said to be an analog ifthe sound pressure wave itself. Other circuits use only two distinct voltage levels to represent information. Most often, these two voltage levels use the same voltages supplied by the power rails. In these circuits, called digital circuits, all information must be represented as binary numbers, with a signal at 0V (or ground) representing one kind of information, and a signal at 3.3V (or whatever the upper voltage supply rail provides) representing the other kind of information. In this series of modules, we will confine our discussions to digital circuits.
## Op Amp Operational Amplifiers Functional Block DiagramOp Amp Operational Amplifiers Functional Block Diagram
By Alix Loane. Diagram. Published at Tuesday, December 26th 2017, 21:46:31 PM. A wiring diagram is sometimes helpful to illustrate how a schematic can be realized in a prototype or production environment. A proper wiring diagram will be labeled and show connections in a way that prevents confusion about how connections are made. Typically they are designed for end-users or installers. They focus on connections rather than components
By Alix Loane. Circuit. Published at Tuesday, December 26th 2017, 21:42:25 PM. These are commonly termed as IC’s. They are usually in form of chips and microchips. This is a set of multiple electronic circuits on a small semiconductor device (normally silicon). With the advancement of technology, these IC’s prove to be extremely beneficial. They are present in any electronic device you can name. From computers, mobiles to other digital appliances you have multiple IC’s present in them. Their main components are a combination of Diodes, transistors and microprocessors. The microprocessors provide memory to the device. With the help of the microprocessor the electronic devices can perform logical or protocol applications.
#### Push Pull Class APush Pull Class A
By Valentine Sybille. Diagram. Published at Tuesday, December 26th 2017, 20:06:46 PM. A first look at a circuit diagram may be confusing, but if you can read a subway map, you can read schematics. The purpose is the same: getting from point A to point B. Literally, a circuit is the path that allows electricity to flow. If you know what to look for, it’ll become second nature. While at first you’ll just be reading them, eventually you will start creating your own. This guide will show you a few of the common symbols that you are sure to see in your future electrical engineering career.
##### Motor Shield Arduino Board ModuleMotor Shield Arduino Board Module
By Alix Loane. Circuit. Published at Tuesday, December 26th 2017, 19:02:52 PM. Many different types of components and devices can be found in modern circuits, including resistors, capacitors, and inductors, semiconductor devices like diodes, transistors, and integrated circuits, transducers like microphones, light sensors and motions sensors, actuators like motors and solenoids, and various other devices like heating and lighting elements. Devices in a circuit are connected to one another by means of electrical conductors, or wires. These wires can move electric currents between various points in a circuit. Once a wire connects two or more devices, the wire and all attached device connectors are said to form a single circuit node or net. Any electrical activity on a given net is communicated to all devices attached to the net. Certain nets provide electric power to devices, and other nets carry information between devices. Nets that carry information are called signals, and signals transport information encoded as voltage levels around a circuit. Signal nets typically use smaller conductors, and transport very small currents. Nets that carry power are called supply rails (or just supplies) and supply rails transport electric power around a circuit. Power nets typically use much larger conductors that signal nets, because they must transport larger currents.
###### Capacitor Resonant Drawing Inductance In SeriesCapacitor Resonant Drawing Inductance In Series
By Charlotte Myriam. Diagram. Published at Tuesday, December 26th 2017, 18:35:19 PM. If there’s something on a schematic that just doesn’t make sense, try finding a datasheet for the most important component. Usually the component doing the most work on a circuit is an integrated circuit, like a microcontroller or sensor. These are usually the largest component, oft-located at the center of the schematic.
## Component Active Diode Circuit EquationComponent Active Diode Circuit Equation
By Charlotte Myriam. Diode. Published at Tuesday, December 26th 2017, 14:27:42 PM. A diode is a specialized electronic component with two electrodes called the anode and the cathode. Most diodes are made with semiconductor materials such as silicon, germanium, or selenium. Some diodes are comprised of metal electrodes in a chamber evacuated or filled with a pure elemental gas at low pressure. Diodes can be used as rectifiers, signal limiters, voltage regulators, switches, signal modulators, signal mixers, signal demodulators, and oscillators.
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Copyright © 2003 - 2018 Domain Media. All sponsored products, company names, brand names, trademarks and logos arethe property of their respective owners. | 1,619 | 7,666 | {"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} | 2.828125 | 3 | CC-MAIN-2018-17 | longest | en | 0.924736 |
https://courses.lumenlearning.com/suny-hccc-wm-concepts-statistics/chapter/assignment-hypothesis-testing-for-the-population-proportion-p/ | 1,719,079,893,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862410.56/warc/CC-MAIN-20240622175245-20240622205245-00594.warc.gz | 154,405,407 | 17,803 | ## Assignment: Hypothesis Testing for the Population Proportion p
The objectives of this activity are:
1. To give you guided practice in carrying out a hypothesis test about a population proportion. (Note: This hypothesis test is also called a z-test for the population proportion.)
2. To learn how to use statistical software to help you carry out the test.
Background: This activity is based on the results of a recent study on the safety of airplane drinking water that was conducted by the U.S. Environmental Protection Agency (EPA). A study found that out of a random sample of 316 airplanes tested, 40 had coliform bacteria in the drinking water drawn from restrooms and kitchens. As a benchmark comparison, in 2003 the EPA found that about 3.5% of the U.S. population have coliform bacteria-infected drinking water. The question of interest is whether, based on the results of this study, we can conclude that drinking water on airplanes is more contaminated than drinking water in general
### Question 1:
Let p be the proportion of contaminated drinking water in airplanes. Write down the appropriate null and alternative hypotheses.
### Question 2:
Based on the collected data, is it safe to use the z-test for p in this scenario? Explain.
Use the following instructions to conduct the z-test for the population proportion:
### Instructions
Click on the link corresponding to your statistical package to see instructions for completing the activity, and then answer the questions below.
### Question 3:
Now that we have established that it is safe to use the Z-test for p for our problem, go ahead and carry out the test. Paste the output below.
### Question 4:
Note that, according to the output, the test statistic for this test is 8.86. Make sure you understand how this was calculated, and give an interpretation of its value.
### Question 5:
We calculated a P-value of 0 in this test. Interpret what that means, and draw your conclusions. | 408 | 1,969 | {"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} | 3.796875 | 4 | CC-MAIN-2024-26 | latest | en | 0.937658 |
http://www.programmersheaven.com/discussion/226780/ReportPost.aspx?S=B20000 | 1,524,757,672,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948285.62/warc/CC-MAIN-20180426144615-20180426164615-00064.warc.gz | 502,709,927 | 11,754 | # Random numbers to items in a list
Very new to programming here (understatement) and am enjoying learning as I go and problem solving. I have been playing around with random numbers and have no problem (as yet) using them in a program. However, I would like to know if there is a way to asign a random number to each of the items in a list. Thus- for example 8 items in a list would be each assigned an indivual random number from 1 to 8.
Thank you
• : Very new to programming here (understatement) and am enjoying learning as I go and problem solving. I have been playing around with random numbers and have no problem (as yet) using them in a program. However, I would like to know if there is a way to asign a random number to each of the items in a list. Thus- for example 8 items in a list would be each assigned an indivual random number from 1 to 8.
Glad to have you on board. Python is a great language.
Python comes with a very capable randomness module already. All you have to do is "import random". Here is a quick example of how to do what you ask. I recommend reading the module documentation yourself. If you're on Windows, go to Start->Programs->Python x.y->Python Manuals
and then click on "Global Module Index". There you'll find all the standard stuff that Python comes with.
[code]
>>> import random
>>> l = []
>>> for i in range(8):
... l.append(random.randint(1,8))
...
>>> l
[8, 3, 6, 7, 4, 1, 6, 5]
>>>
[/code]
Let me know if you have any other questions or if this code snippet didn't make sense to you.
[size=5][italic][blue][RED]i[/RED]nfidel[/blue][/italic][/size]
• Thanks for the reply. Yes, that's great, and as usual (for me at least ) a few simple logical lines of instructions suffice where I would usually write reams and reams to get an inferior result. My main problem was ensuring that each list item was given a "different" random number, and this does just that.
Thanks
• : Thanks for the reply. Yes, that's great, and as usual (for me at least ) a few simple logical lines of instructions suffice where I would usually write reams and reams to get an inferior result. My main problem was ensuring that each list item was given a "different" random number, and this does just that.
Glad to help. This board has very little traffic, so anything you need help with I'd be glad to take a look at.
[size=5][italic][blue][RED]i[/RED]nfidel[/blue][/italic][/size] | 597 | 2,408 | {"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} | 3.046875 | 3 | CC-MAIN-2018-17 | latest | en | 0.958027 |
http://conceptmap.cfapps.io/wikipage?lang=en&name=Equivalence_relation | 1,596,984,825,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738555.33/warc/CC-MAIN-20200809132747-20200809162747-00228.warc.gz | 24,933,003 | 25,135 | # Equivalence relation
The 52 equivalence relations on a 5-element set depicted as 5×5 logical matrices (colored fields, including those in light gray, stand for ones; white fields for zeros.) The row and column indices of nonwhite cells are the related elements, while the different colors, other than light gray, indicate the equivalence classes (each light gray cell is its own equivalence class).
In mathematics, an equivalence relation is a binary relation that is reflexive, symmetric and transitive. The relation "is equal to" is the canonical example of an equivalence relation, where for any objects a, b, and c:
• a = a (reflexive property),
• if a = b then b = a (symmetric property), and
• if a = b and b = c then a = c (transitive property).
As a consequence of the reflexive, symmetric, and transitive properties, any equivalence relation provides a partition of the underlying set into disjoint equivalence classes. Two elements of the given set are equivalent to each other if and only if they belong to the same equivalence class.
## Notation
Various notations are used in the literature to denote that two elements a and b of a set are equivalent with respect to an equivalence relation R; the most common are "a ~ b" and "ab", which are used when R is implicit, and variations of "a ~R b", "aR b", or "aRb" to specify R explicitly. Non-equivalence may be written "ab" or "${\displaystyle a\not \equiv b}$ ".
## Definition
A given binary relation ~ on a set X is said to be an equivalence relation if and only if it is reflexive, symmetric and transitive. That is, for all a, b and c in X:
X together with the relation ~ is called a setoid. The equivalence class of ${\displaystyle a}$ under ~, denoted ${\displaystyle [a]}$ , is defined as ${\displaystyle [a]=\{b\in X\mid a\sim b\}}$ .
## Examples
### Simple example
Let the set ${\displaystyle \{a,b,c\}}$ have the equivalence relation ${\displaystyle \{(a,a),(b,b),(c,c),(b,c),(c,b)\}}$ . The following sets are equivalence classes of this relation:
${\displaystyle [a]=\{a\},~~~~[b]=[c]=\{b,c\}}$ .
The set of all equivalence classes for this relation is ${\displaystyle \{\{a\},\{b,c\}\}}$ . This set is a partition of the set ${\displaystyle \{a,b,c\}}$ .
### Equivalence relations
The following are all equivalence relations:
• "Is equal to" on the set of numbers. For example, ${\displaystyle {\tfrac {1}{2}}}$ is equal to ${\displaystyle {\tfrac {4}{8}}}$ .
• "Has the same birthday as" on the set of all people.
• "Is similar to" on the set of all triangles.
• "Is congruent to" on the set of all triangles.
• "Is congruent to, modulo n" on the integers.
• "Has the same image under a function" on the elements of the domain of the function.
• "Has the same absolute value" on the set of real numbers
• "Has the same cosine" on the set of all angles.
### Relations that are not equivalences
• The relation "≥" between real numbers is reflexive and transitive, but not symmetric. For example, 7 ≥ 5 does not imply that 5 ≥ 7. It is, however, a total order.
• The relation "has a common factor greater than 1 with" between natural numbers greater than 1, is reflexive and symmetric, but not transitive. (Example: The natural numbers 2 and 6 have a common factor greater than 1, and 6 and 3 have a common factor greater than 1, but 2 and 3 do not have a common factor greater than 1).
• The empty relation R (defined so that aRb is never true) on a non-empty set X is vacuously symmetric and transitive, but not reflexive. (If X is also empty then R is reflexive.)
• The relation "is approximately equal to" between real numbers, even if more precisely defined, is not an equivalence relation, because although reflexive and symmetric, it is not transitive, since multiple small changes can accumulate to become a big change. However, if the approximation is defined asymptotically, for example by saying that two functions f and g are approximately equal near some point if the limit of f − g is 0 at that point, then this defines an equivalence relation.
## Well-definedness under an equivalence relation
If ~ is an equivalence relation on X, and P(x) is a property of elements of X, such that whenever x ~ y, P(x) is true if P(y) is true, then the property P is said to be well-defined or a class invariant under the relation ~.
A frequent particular case occurs when f is a function from X to another set Y; if x1 ~ x2 implies f(x1) = f(x2) then f is said to be a morphism for ~, a class invariant under ~, or simply invariant under ~. This occurs, e.g. in the character theory of finite groups. The latter case with the function f can be expressed by a commutative triangle. See also invariant. Some authors use "compatible with ~" or just "respects ~" instead of "invariant under ~".
More generally, a function may map equivalent arguments (under an equivalence relation ~A) to equivalent values (under an equivalence relation ~B). Such a function is known as a morphism from ~A to ~B.
## Equivalence class, quotient set, partition
Let ${\displaystyle a,b\in X}$ . Some definitions:
### Equivalence class
A subset Y of X such that a ~ b holds for all a and b in Y, and never for a in Y and b outside Y, is called an equivalence class of X by ~. Let ${\displaystyle [a]:=\{x\in X\mid a\sim x\}}$ denote the equivalence class to which a belongs. All elements of X equivalent to each other are also elements of the same equivalence class.
### Quotient set
The set of all equivalence classes of X by ~, denoted ${\displaystyle X/{\mathord {\sim }}:=\{[x]\mid x\in X\}}$ , is the quotient set of X by ~. If X is a topological space, there is a natural way of transforming X/~ into a topological space; see quotient space for the details.
### Projection
The projection of ~ is the function ${\displaystyle \pi :X\to X/{\mathord {\sim }}}$ defined by ${\displaystyle \pi (x)=[x]}$ which maps elements of X into their respective equivalence classes by ~.
Theorem on projections:[2] Let the function f: XB be such that a ~ bf(a) = f(b). Then there is a unique function g : X/~B, such that f = gπ. If f is a surjection and a ~ bf(a) = f(b), then g is a bijection.
### Equivalence kernel
The equivalence kernel of a function f is the equivalence relation ~ defined by ${\displaystyle x\sim y\iff f(x)=f(y)}$ . The equivalence kernel of an injection is the identity relation.
### Partition
A partition of X is a set P of nonempty subsets of X, such that every element of X is an element of a single element of P. Each element of P is a cell of the partition. Moreover, the elements of P are pairwise disjoint and their union is X.
#### Counting partitions
Let X be a finite set with n elements. Since every equivalence relation over X corresponds to a partition of X, and vice versa, the number of equivalence relations on X equals the number of distinct partitions of X, which is the nth Bell number Bn:
${\displaystyle B_{n}={\frac {1}{e}}\sum _{k=0}^{\infty }{\frac {k^{n}}{k!}}}$ (Dobinski's formula).
## Fundamental theorem of equivalence relations
A key result links equivalence relations and partitions:[3][4][5]
• An equivalence relation ~ on a set X partitions X.
• Conversely, corresponding to any partition of X, there exists an equivalence relation ~ on X.
In both cases, the cells of the partition of X are the equivalence classes of X by ~. Since each element of X belongs to a unique cell of any partition of X, and since each cell of the partition is identical to an equivalence class of X by ~, each element of X belongs to a unique equivalence class of X by ~. Thus there is a natural bijection between the set of all equivalence relations on X and the set of all partitions of X.
## Comparing equivalence relations
If ~ and ≈ are two equivalence relations on the same set S, and a~b implies ab for all a,bS, then ≈ is said to be a coarser relation than ~, and ~ is a finer relation than ≈. Equivalently,
• ~ is finer than ≈ if every equivalence class of ~ is a subset of an equivalence class of ≈, and thus every equivalence class of ≈ is a union of equivalence classes of ~.
• ~ is finer than ≈ if the partition created by ~ is a refinement of the partition created by ≈.
The equality equivalence relation is the finest equivalence relation on any set, while the universal relation, which relates all pairs of elements, is the coarsest.
The relation "~ is finer than ≈" on the collection of all equivalence relations on a fixed set is itself a partial order relation, which makes the collection a geometric lattice.[6]
## Generating equivalence relations
Given any binary relation ${\displaystyle A\subset X\times X}$ on ${\displaystyle X}$ , the equivalence relation generated by ${\displaystyle A}$ is the intersection of the equivalence relations on ${\displaystyle X}$ that contain ${\displaystyle A}$ . (Since ${\displaystyle X\times X}$ is an equivalence relation, the intersection is nontrivial.)
• Given any set X, there is an equivalence relation over the set [XX] of all functions XX. Two such functions are deemed equivalent when their respective sets of fixpoints have the same cardinality, corresponding to cycles of length one in a permutation. Functions equivalent in this manner form an equivalence class on [XX], and these equivalence classes partition [XX].
• An equivalence relation ~ on X is the equivalence kernel of its surjective projection π : XX/~.[7] Conversely, any surjection between sets determines a partition on its domain, the set of preimages of singletons in the codomain. Thus an equivalence relation over X, a partition of X, and a projection whose domain is X, are three equivalent ways of specifying the same thing.
• The intersection of any collection of equivalence relations over X (binary relations viewed as a subset of X × X) is also an equivalence relation. This yields a convenient way of generating an equivalence relation: given any binary relation R on X, the equivalence relation generated by R is the smallest equivalence relation containing R. Concretely, R generates the equivalence relation a ~ b if and only if there exist elements x1, x2, ..., xn in X such that a = x1, b = xn, and (xi, xi+1) ∈ R or (xi+1, xi) ∈ R, i = 1, ..., n−1.
Note that the equivalence relation generated in this manner can be trivial. For instance, the equivalence relation ~ generated by any total order on X has exactly one equivalence class, X itself, because x ~ y for all x and y. As another example, any subset of the identity relation on X has equivalence classes that are the singletons of X.
• Equivalence relations can construct new spaces by "gluing things together." Let X be the unit Cartesian square [0, 1] × [0, 1], and let ~ be the equivalence relation on X defined by (a, 0) ~ (a, 1) for all a ∈ [0, 1] and (0, b) ~ (1, b) for all b ∈ [0, 1]. Then the quotient space X/~ can be naturally identified (homeomorphism) with a torus: take a square piece of paper, bend and glue together the upper and lower edge to form a cylinder, then bend the resulting cylinder so as to glue together its two open ends, resulting in a torus.
## Algebraic structure
Much of mathematics is grounded in the study of equivalences, and order relations. Lattice theory captures the mathematical structure of order relations. Even though equivalence relations are as ubiquitous in mathematics as order relations, the algebraic structure of equivalences is not as well known as that of orders. The former structure draws primarily on group theory and, to a lesser extent, on the theory of lattices, categories, and groupoids.
### Group theory
Just as order relations are grounded in ordered sets, sets closed under pairwise supremum and infimum, equivalence relations are grounded in partitioned sets, which are sets closed under bijections that preserve partition structure. Since all such bijections map an equivalence class onto itself, such bijections are also known as permutations. Hence permutation groups (also known as transformation groups) and the related notion of orbit shed light on the mathematical structure of equivalence relations.
Let '~' denote an equivalence relation over some nonempty set A, called the universe or underlying set. Let G denote the set of bijective functions over A that preserve the partition structure of A: ∀xAgG (g(x) ∈ [x]). Then the following three connected theorems hold:[8]
• ~ partitions A into equivalence classes. (This is the Fundamental Theorem of Equivalence Relations, mentioned above);
• Given a partition of A, G is a transformation group under composition, whose orbits are the cells of the partition;[12]
• Given a transformation group G over A, there exists an equivalence relation ~ over A, whose equivalence classes are the orbits of G.[13][14]
In sum, given an equivalence relation ~ over A, there exists a transformation group G over A whose orbits are the equivalence classes of A under ~.
This transformation group characterisation of equivalence relations differs fundamentally from the way lattices characterize order relations. The arguments of the lattice theory operations meet and join are elements of some universe A. Meanwhile, the arguments of the transformation group operations composition and inverse are elements of a set of bijections, AA.
Moving to groups in general, let H be a subgroup of some group G. Let ~ be an equivalence relation on G, such that a ~ b ↔ (ab−1H). The equivalence classes of ~—also called the orbits of the action of H on G—are the right cosets of H in G. Interchanging a and b yields the left cosets.
Related thinking can be found in Rosen (2008: chpt. 10).
### Categories and groupoids
Let G be a set and let "~" denote an equivalence relation over G. Then we can form a groupoid representing this equivalence relation as follows. The objects are the elements of G, and for any two elements x and y of G, there exists a unique morphism from x to y if and only if x~y.
The advantages of regarding an equivalence relation as a special case of a groupoid include:
• Whereas the notion of "free equivalence relation" does not exist, that of a free groupoid on a directed graph does. Thus it is meaningful to speak of a "presentation of an equivalence relation," i.e., a presentation of the corresponding groupoid;
• Bundles of groups, group actions, sets, and equivalence relations can be regarded as special cases of the notion of groupoid, a point of view that suggests a number of analogies;
• In many contexts "quotienting," and hence the appropriate equivalence relations often called congruences, are important. This leads to the notion of an internal groupoid in a category.[15]
### Lattices
The equivalence relations on any set X, when ordered by set inclusion, form a complete lattice, called Con X by convention. The canonical map ker: X^XCon X, relates the monoid X^X of all functions on X and Con X. ker is surjective but not injective. Less formally, the equivalence relation ker on X, takes each function f: XX to its kernel ker f. Likewise, ker(ker) is an equivalence relation on X^X.
## Equivalence relations and mathematical logic
Equivalence relations are a ready source of examples or counterexamples. For example, an equivalence relation with exactly two infinite equivalence classes is an easy example of a theory which is ω-categorical, but not categorical for any larger cardinal number.
An implication of model theory is that the properties defining a relation can be proved independent of each other (and hence necessary parts of the definition) if and only if, for each property, examples can be found of relations not satisfying the given property while satisfying all the other properties. Hence the three defining properties of equivalence relations can be proved mutually independent by the following three examples:
• Reflexive and transitive: The relation ≤ on N. Or any preorder;
• Symmetric and transitive: The relation R on N, defined as aRbab ≠ 0. Or any partial equivalence relation;
• Reflexive and symmetric: The relation R on Z, defined as aRb ↔ "ab is divisible by at least one of 2 or 3." Or any dependency relation.
Properties definable in first-order logic that an equivalence relation may or may not possess include:
• The number of equivalence classes is finite or infinite;
• The number of equivalence classes equals the (finite) natural number n;
• All equivalence classes have infinite cardinality;
• The number of elements in each equivalence class is the natural number n.
## Euclidean relations
Euclid's The Elements includes the following "Common Notion 1":
Things which equal the same thing also equal one another.
Nowadays, the property described by Common Notion 1 is called Euclidean (replacing "equal" by "are in relation with"). By "relation" is meant a binary relation, in which aRb is generally distinct from bRa. A Euclidean relation thus comes in two forms:
(aRcbRc) → aRb (Left-Euclidean relation)
(cRacRb) → aRb (Right-Euclidean relation)
The following theorem connects Euclidean relations and equivalence relations:
Theorem
If a relation is (left or right) Euclidean and reflexive, it is also symmetric and transitive.
Proof for a left-Euclidean relation
(aRcbRc) → aRb [a/c] = (aRabRa) → aRb [reflexive; erase T∧] = bRaaRb. Hence R is symmetric.
(aRcbRc) → aRb [symmetry] = (aRccRb) → aRb. Hence R is transitive. ${\displaystyle _{\Box }}$
with an analogous proof for a right-Euclidean relation. Hence an equivalence relation is a relation that is Euclidean and reflexive. The Elements mentions neither symmetry nor reflexivity, and Euclid probably would have deemed the reflexivity of equality too obvious to warrant explicit mention.
## Notes
1. ^ If: Given a, let a~b hold using seriality, then b~a by symmetry, hence a~a by transitivity. — Only if: Given a, choose b=a, then a~b by reflexivity.
2. ^ Garrett Birkhoff and Saunders Mac Lane, 1999 (1967). Algebra, 3rd ed. p. 35, Th. 19. Chelsea.
3. ^ Wallace, D. A. R., 1998. Groups, Rings and Fields. p. 31, Th. 8. Springer-Verlag.
4. ^ Dummit, D. S., and Foote, R. M., 2004. Abstract Algebra, 3rd ed. p. 3, Prop. 2. John Wiley & Sons.
5. ^ Karel Hrbacek & Thomas Jech (1999) Introduction to Set Theory, 3rd edition, pages 29–32, Marcel Dekker
6. ^ Birkhoff, Garrett (1995), Lattice Theory, Colloquium Publications, 25 (3rd ed.), American Mathematical Society, ISBN 9780821810255. Sect. IV.9, Theorem 12, page 95
7. ^ Garrett Birkhoff and Saunders Mac Lane, 1999 (1967). Algebra, 3rd ed. p. 33, Th. 18. Chelsea.
8. ^ Rosen (2008), pp. 243–45. Less clear is §10.3 of Bas van Fraassen, 1989. Laws and Symmetry. Oxford Univ. Press.
9. ^ Bas van Fraassen, 1989. Laws and Symmetry. Oxford Univ. Press: 246.
10. ^ Wallace, D. A. R., 1998. Groups, Rings and Fields. Springer-Verlag: 22, Th. 6.
11. ^ Wallace, D. A. R., 1998. Groups, Rings and Fields. Springer-Verlag: 24, Th. 7.
12. ^ Proof.[9] Let function composition interpret group multiplication, and function inverse interpret group inverse. Then G is a group under composition, meaning that ∀xAgG ([g(x)] = [x]), because G satisfies the following four conditions:
Let f and g be any two elements of G. By virtue of the definition of G, [g(f(x))] = [f(x)] and [f(x)] = [x], so that [g(f(x))] = [x]. Hence G is also a transformation group (and an automorphism group) because function composition preserves the partitioning of A. ${\displaystyle \square }$
13. ^ Wallace, D. A. R., 1998. Groups, Rings and Fields. Springer-Verlag: 202, Th. 6.
14. ^ Dummit, D. S., and Foote, R. M., 2004. Abstract Algebra, 3rd ed. John Wiley & Sons: 114, Prop. 2.
15. ^ Borceux, F. and Janelidze, G., 2001. Galois theories, Cambridge University Press, ISBN 0-521-80309-8
## References
• Brown, Ronald, 2006. Topology and Groupoids. Booksurge LLC. ISBN 1-4196-2722-8.
• Castellani, E., 2003, "Symmetry and equivalence" in Brading, Katherine, and E. Castellani, eds., Symmetries in Physics: Philosophical Reflections. Cambridge Univ. Press: 422-433.
• Robert Dilworth and Crawley, Peter, 1973. Algebraic Theory of Lattices. Prentice Hall. Chpt. 12 discusses how equivalence relations arise in lattice theory.
• Higgins, P.J., 1971. Categories and groupoids. Van Nostrand. Downloadable since 2005 as a TAC Reprint.
• John Randolph Lucas, 1973. A Treatise on Time and Space. London: Methuen. Section 31.
• Rosen, Joseph (2008) Symmetry Rules: How Science and Nature are Founded on Symmetry. Springer-Verlag. Mostly chapters. 9,10.
• Raymond Wilder (1965) Introduction to the Foundations of Mathematics 2nd edition, Chapter 2-8: Axioms defining equivalence, pp 48–50, John Wiley & Sons. | 5,114 | 20,646 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 26, "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-2020-34 | latest | en | 0.873592 |
https://www.unitconverters.net/force/ounce-force-to-joule-meter.htm | 1,675,253,493,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499934.48/warc/CC-MAIN-20230201112816-20230201142816-00593.warc.gz | 1,048,685,140 | 3,271 | Home / Force Conversion / Convert Ounce-force to Joule/meter
# Convert Ounce-force to Joule/meter
Please provide values below to convert ounce-force [ozf] to joule/meter [J/m], or vice versa.
From: ounce-force To: joule/meter
### Ounce-force to Joule/meter Conversion Table
Ounce-force [ozf]Joule/meter [J/m]
0.01 ozf0.0027801385 J/m
0.1 ozf0.0278013851 J/m
1 ozf0.278013851 J/m
2 ozf0.5560277019 J/m
3 ozf0.8340415529 J/m
5 ozf1.3900692548 J/m
10 ozf2.7801385095 J/m
20 ozf5.5602770191 J/m
50 ozf13.9006925477 J/m
100 ozf27.8013850953 J/m
1000 ozf278.0138509534 J/m
### How to Convert Ounce-force to Joule/meter
1 ozf = 0.278013851 J/m
1 J/m = 3.5969430896 ozf
Example: convert 15 ozf to J/m:
15 ozf = 15 × 0.278013851 J/m = 4.1702077643 J/m | 303 | 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": 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} | 2.78125 | 3 | CC-MAIN-2023-06 | longest | en | 0.161054 |
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1. ### Solve the Linear ODE of first order first degree
Hi all, I'm stuck with this problem. Please help me to identify which method to use. Attaching the question. Thanks a lot for your help.
2. ### How to solve an ODE to find its solution
Salutations, I have a problem when I approach this ODE: $$\left(\frac{y}{y'}\right)^2+y^2=b^2\left(x-\frac{y}{y'}\right)^2$$ I have done a series of steps as I show in this document: https://drive.google.com/file/d/1Ht4xxUlm7vXqg4S5-wirKwm7vTESU3mU/view?usp=sharing But I'm not...
3. ### Transformation of Equations - What is the deeper thing?
Hi, If you have an ordinary differential equation (or equations), you can transform them under some conditions (integral, linear operator, ...) to algebraic equations. If you have a partial differential equation you may transform it to an ODE. This can be shown "easily" by doing Fourier...
4. ### non linear ode
Hi , i have a question regarding this ode system dx/dt=-y-x^3 ,dy/dt=x ,so it s a question whether this system has a periodic solution respectively an asymptotic equilibrium point and lastely describing the asymptotic behaviour for t goes to infinity of all the solutions . Thanks guys
5. ### Need help for this expression
y''+(3x^2-1/x)y'=A/x Any method can be used to solve this equation? Thanks a lot.
6. ### Question about general substitution for tricky ODE
as you know, many ODE(Ordinary Differential Equation) have explicit solution, but there are also more ODE which doesn't have. Let's think hypothetical need to find solution for a <Stiff ODE> to calculate area below curvature or more generally, to find primitive or anti-derivative function or...
7. ### Ode
Hi everyone, I have been trying to solve this equation, but to no avail. Can you help me? y''+y'/x+y/x^2=0
8. ### Whether to define system as PDE or ODE
Hey everyone. I'm a grad student revising a journal article. Within the article, I layout an equation which describes of population "n(r)" through time. The parameters of this population vary with some variable "r"; however, r does not change with time. When I solve this system (written out...
9. ### Proving general solution for 2nd order ODE
Show that y(t) = A(t^2) + Bt, where A and B are arbitrary constants, is the general solution of the differential equation (t^2)y′′ − 2t y′ + 2 y = 0. I got the roots to be r= (1/t)+-(i/t) .... Which I would then plug into the equation involving complex roots which is...
10. ### 2nd Order ODE Problem
I find 2nd Order ODEs quite hard and I've come across this question in my text book but am unsure of how to do it. Can anybody explain step by step how this is done? It would be a huge help considering I've an exam in a few weeks' time on this stuff. :o
11. ### Third order ODE using Euler's method
I'm a bit lost on this one. Is my method below the correct way to set up the ODE? \frac{dQ}{dt}(\frac{d^3Q}{dt^3})=(\frac{dT}{dt})pCp
12. ### Separation of variables, ODE. Is it possible to get a factor t in a solution?
Here's my problem: http://i.imgur.com/XT7Mjw1.png I tried a). But when I look at c), it says u(x,t) = te^{-t}\sin(x) should be a solution and I cannot make that solution; I can only get e^{-t}\sin(x) with the solution I found. So I just need to know: where would a 't' factor come from? By...
13. ### Linear first ODE with variable coefficients
Hi! I used to go about solving linear ODEs with constant and variable coefficients. Upon my knowledge all the times the ones with variable coefficients, the coefficients would be dependent on the one independent variable 'x' say. For example, consider the following ODE: q(x)= A(y) ...
14. ### solving ODE
I need to solve analytically the following second order ODE; W''(X) - f(X)*W(X)=0 WHERE f(X) is an arbitrary known function. Could someone help me as soon as possible?
15. ### Can Anyone help solve this First order ODE?
I'm having a nightmare trying to solve this ODE, If anyone could offer a solution or just insight it would be greatly appreciated. Thank you. All parameters apart from t and x are constant https://www.dropbox.com/s/8cwdnerm5ple9gn/Screenshot_2016-04-07-11-47-27.png?dl=0
16. ### Rearranging ODE's like algebra - please check my understanding
Hello all, I am trying to understand why we can "separate the variables and derivative parts" in ODEs. I have attached a picture of my working out, and on the page I have added numbers at the side, which I will reference now in the typed text to assist what I have done: 1. dy/dx is just the...
17. ### Question about solving ODE
hi guys, Hope it's okay, but I typed out my question to make the math equations more clear. I had a general question about solving ODEs. So as we know, if you have an equation of the form: thanks in advance for your help
18. ### ODE system
Hi guys, I need some help solving the next ODE system: w'+2*U*sin(psi)=0 wv'+2UV*sin(psi)=V'' WU'+(U^2-V^2)*sin(psi)=mTsin(psi)+U'' m,T,psi are constants BCE's: U=V=W=0 @ x=0, U-->0 as x-->infinity, V-->(-1) as x-->infinity Some guidance would be greatly appreciated. shai
19. ### Power series on ODE solution
Dear all, I am a bit rusty on power series expansion, and I am stucked with the following problem. I have the following ODE in terms of two continuous differentiable generic functions u(t) and v(t): \dot{u} = -g_L u + g_D (v(t)-u(t)) + g(t) (E-u) where g_L,g_D,E real and >0 and g(t) is...
20. ### matrix-vector representation for a system of ODE's
I am aiming to explicitly write the matrix-vector representation of this system. y’1 = 5y2 - y1 + y3; y’2 = 3y1 - y2 + t2; y’3 = y3 - ty2 This is what I have so far: [ y’1 ] [ -1 5 1] [ 0] [ y’2 ] = [ 3 -1 0] [ t2] [ y’3 ] [ 0 ? 1] [ ? ] Just not sure how... | 1,625 | 5,733 | {"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} | 3.4375 | 3 | CC-MAIN-2020-16 | latest | en | 0.895645 |
https://www.jiskha.com/display.cgi?id=1403588440 | 1,519,567,370,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816462.95/warc/CC-MAIN-20180225130337-20180225150337-00120.warc.gz | 909,515,126 | 4,273 | # math
posted by .
The cost of a DVD is \$18.75. Added to this cost is 12% before tax fee. 5% GST, 7% PST. What is the total cost?
• math -
I suppose the 12% is the copyright fee or an environmental surcharge. Anyway, it is some kind of a surtax.
The sales taxes depend on the province in which the transaction was made.
If the transaction was made in all provinces except Quebec, both GST and PST are calculated on the purchase price (including the surtax). The total PST+GST is then 12%.
Total cost = (\$18.75*1.12)*(1+0.05+0.07)
=23.52
If the transaction was made in Quebec, the PST is included in the calculation of GST.
total cost = (\$18.75*1.12)*(1.05)*(1.07)
= 23.59
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10. ### Math
For a field trip, a bus company charges a flat fee plus an additional fee per student. For 25 students, the total cost is \$167.50. If there are 70 students, the total cost is \$370.00. Write an equation that represents the total cost …
More Similar Questions | 769 | 2,889 | {"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-2018-09 | longest | en | 0.941825 |
qwertyuiop.info | 1,548,226,859,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584203540.82/warc/CC-MAIN-20190123064911-20190123090911-00450.warc.gz | 180,122,335 | 7,116 | # Trading foreign exchange needs to control risk
risk control
The foreign exchange market is a risky market, and its risk lies mainly in the variables that determine the foreign exchange price. Although the existing books on the principle of foreign exchange fluctuations can be said to be full of enthusiasm, some study from economic theory, some from mathematical statistics, some from geometric figures, and more from the perspective of psychological and behavioral sciences. Research, but the volatility of the foreign exchange market is still often out of the surprise of investors. For foreign exchange market investors and operators, all aspects of knowledge should have a little knowledge, and knowledge of statistical probability is especially important. The main significance of probability statistics is that it can better help investors make investment decisions, including whether to invest, how to invest, how to deal with accidents, and so on. Therefore, in foreign exchange investment, it is necessary to fully understand the relationship between risk and benefit, winning and losing money. If you don’t have an accurate understanding of these aspects, and you are free to trade foreign exchange, depending on money, then losing money is inevitable.
Optimism and probability
In all the decisions to be made, the first step is to decide how to use your own funds and how much money to use to take risks. When dealing with these issues, it is important to understand the concept of “financial processing.”
There are four main elements in the processing of funds:
1. Personal goals and preferences, including the person’s current attitude towards money;
2. Initial capital and follow-up capital used in the transaction to make an adventure;
3. Expected return value;
4. The probability of losing money.
Forex brokers or investors are doing business in an uncertain atmosphere.
He can’t know exactly what the outcome of any transaction will be. But he can, and he always calculates the probability of various possible outcomes based on his relatively optimistic or pessimistic judgment. What is the probability? Simply put, it’s just a number used to represent a result that reflects what people might think of a certain outcome. The magnitude of this number fluctuates between zero and one. For example, a person may feel that his probability of reaching a goal in a transaction is 0.4; and the probability of not reaching a predetermined goal is 0.6. In this case, probability can be seen as an index that measures whether a person is optimistic or pessimistic about a particular transaction. If he feels that the probability of reaching the goal has risen from 0.4 to 0.5, then we can say that he is more optimistic.
It should be pointed out that the optimism of the person in the above example comes from objective evidence, from news reports, from hackers’ analysis, or from personal intuition. These are irrelevant. problem. What is important is that no matter where the probability value comes from, it reflects an estimate of the chances that a person may have different outcomes. It will affect the person’s next action and thus the market price. | 607 | 3,176 | {"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} | 2.65625 | 3 | CC-MAIN-2019-04 | latest | en | 0.945638 |
https://electronics.stackexchange.com/questions/131556/pfet-to-turn-on-off-system-power | 1,695,914,108,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510412.43/warc/CC-MAIN-20230928130936-20230928160936-00551.warc.gz | 251,373,930 | 43,141 | # PFET to turn on/off system power
I have a Raspberry PI board connected to some external circuitry, and am using Microchip MCP23008 I2C GPIO expander chips to control external LEDs and motor circuitry. One important part of this external circuitry is a 12V buck-boost switching power supply which is powered off of a 9V battery. The buck-boost regulator does not have an enable pin, and I am working out a way to cut the power to the regulator.
I already have a working 9V low-battery indicator circuit (shown below), which turns on the green LED if the battery is supplying a voltage greater than 7.0V, and if the voltage dips below 7.0V, the red LED is enabled. The circuit should also cut off the current coming out of the source of Q4.
Circuit - Version 1
I have two questions with respect to this circuit:
First, does the cutoff logic make sense? My assumption is that if the red LED is off, no current is going through D3, R6, and Q2, so the voltage at the node between R6 and the collector of Q2 would be the same as the power rail connected at the top to D1, D2, D3, and Q4's drain. If the red LED is on, there's a 0.7V drop across it, then a moderate drop across R6 (ie: estimated current, (7.0V-0.7V) / 330R = 19mA, so a 6.27V drop across the resistor roughly), and since this will make the Q4 bias less than that connected to Q4's drain, Q4 is turned off, cutting off power to the buck-boost converter. I'm convinced I'm overlooking V_ce on Q2 though.
Second, is it safe to assume the regulator can run from the output of the source of Q4 and D4? My estimates suggest that I will, when drawing current, have a 0.2V drop (V_ds) across Q4, and another 0.3V drop across D4.
Edit: Adding updated circuit as per earlier suggestions.
Circuit - Version 2
Edit 2: Updating as per latest answers.
I updated the circuit to reflect that of the answer provided by @DanLaks. Attaching DC sweep simulation.
• Where is V(NODE1) and V(un9) in the circuit? Oct 1, 2014 at 22:52
• @DanLaks NODE1 is the output of the Schottky diode that follows the final FET, and is tied to ground with a 100R resistor. Not sure where Vun9 came from, as NODE1 was the only node I manually added (stale output?). I swept the 9V source. Oct 2, 2014 at 0:24
First, the value for R7 is a bit high. When the battery is still good, that resistor is keeping the current into the base of Q1 very low, which allows very little current through the green LED. The voltage drop across R5 is therefore small, which means Vce of Q1 will be somewhat large. If it's too large, there'll be enough voltage to turn on Q2, which will activate the red LED as well. I'd recommend reducing R7 by an order of magnitude. 4.7k should work better.
I think you may have a misunderstanding of how a P-channel MOSFET works. They work the opposite of a N-channel. When the voltage at the gate is less than the voltage at the source, the transistor will conduct. When the voltage at the gate is equal to (or slightly less than) the source, the transistor will not conduct. The way you have the gate connected, it will actually cause the opposite to happen. When the battery is good, the red LED is not conducting. You correctly assessed that the voltage at the bottom of R6 will be approximately equal to the power rail. That will make the gate also equal to the power rail, which means the Q4 will be off. When the red LED is conducting, the voltage at the gate will be pulled down and cause Q4 to turn on.
There are probably several ways to cause the gate of Q4 to go low when the battery voltage is high and low when the battery voltage is low. Personally, I would use a stable voltage reference and a comparator to get a nice, crisp transition. But to keep in the vein of your design, here's an alternative that's similar to the flavor of your circuit.
simulate this circuit – Schematic created using CircuitLab
You can see the transistor logic to turn on and off Q4 is similar, but I tap off of the node between D1 and R8 and feed it into the base of a new npn. You could almost use Q1 instead of placing a new transistor, but the base of Q2 causes current to flow through D2, R5, and R9, and thus the voltage there isn't quite what we want.
One thing to consider with the automatic switch off is that you might experience power oscillations. When the circuit downstream of U3 drains the battery enough for the cutoff to activate, the sudden relaxation on the battery may cause its voltage to bump back up above the threshold. This will turn Q4 back on, which will cause the circuit to start draining current again, which will repeat the cycle. Possibly for a long time. If this is not acceptable, you will have to work hysteresis into the circuit.
To address your specific question about D4, yes, you can use that diode there. As long as the voltage at the VIN pin of U3 is with acceptable range after the diode drop. And obviously D4 must be able to comfortably handle the maximum amount of current into the circuit downstream. The same is true, obviously, for Q4. In addition, you have to account for the voltage drop across Q4 due to the Rds(on) of the transistor.
You say you're using a 9V battery. Is this a standard 9V alkaline used in smoke detectors and such? If so, I'm curious how much current you expect to draw out of it. Those kinds of batteries tend to have (relatively) high internal resistances and can't source very much current before their voltage starts to drop off considerably. If you're only in the 10's of milliamps, you're probably ok. Much higher and you might run into voltage problems.
I'm not sure the P Channel MOSFET is being biased correctly. Your other logic seems correct though. Q4 needs the Gate negative in respect to the Source in order to conduct. Conversely, if the Gate is held positive, or near the same potential as the Source, it will not conduct.
I'd flip the connections on the Source and Drain. Also, a resistor should be between the V_Batt rail and the Gate. This ensures that the device can turn off. Then remove R12, and connect the junction of R6 and Q2 collector directly to Q4 gate. This should ensure that you can turn Q4 on and off.
• Just to confirm: 1. Q4 is flipped incorrectly in the vertical axis. Flip/correct this. 2. So, should I connect a resistor directly between the gate and drain of Q4? If so, how would I select the value? 3. So, removing R12 improves the circuit? Shouldn't it be in place to limit the gate current? Oct 1, 2014 at 18:16
• I'll upload a new circuit diagram shortly, and keep a link to the old one for comparison. Thank you for the verification. Oct 1, 2014 at 18:17
• I've updated the circuit. Is this what you hand in mind? Thanks! Oct 1, 2014 at 18:42
• That's what I had in mind. Oct 1, 2014 at 20:01
• Realized I didn't answer your second question. Ultimately depends on the load the buck/boost is supplying. Both the MOSFET and the Schottky need to be rated for a higher current than the smps needs. Oct 1, 2014 at 20:04 | 1,738 | 6,985 | {"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} | 2.875 | 3 | CC-MAIN-2023-40 | longest | en | 0.934802 |
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A238325 Array: row n gives the number of occurrences of each possible antidiagonal partition of n, arranged in reverse-Mathematica order. 7
1, 2, 2, 1, 2, 3, 2, 2, 3, 2, 2, 6, 1, 2, 2, 4, 3, 4, 2, 2, 4, 6, 2, 6, 2, 2, 4, 4, 2, 3, 9, 4, 2, 2, 4, 4, 2, 6, 6, 3, 12, 1, 2, 2, 4, 4, 2, 4, 6, 3, 6, 6, 12, 5, 2, 2, 4, 4, 2, 4, 6, 6, 4, 6, 3, 18, 2, 4, 10, 2, 2, 4, 4, 2, 4, 6, 4, 4, 6, 3, 6, 12, 2, 6 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS Suppose that p is a partition of n, let F(p) be its Ferrers matrix, as defined at A237981, and let mXm be the size of F(p). The numbers of 1's in each of the 2m-1 antidiagonals of F(p) form a partition of n. Any partition which is associated with a partition of n in this manner is introduced here as an antidiagonal partition of n. A000041(n) = sum of the numbers in row n; A000009(n) = number of terms in row n, since the antidiagonal partitions of n are the conjugates of the strict partitions of n. LINKS Clark Kimberling, Table of n, a(n) for n = 1..1000 Clark Kimberling and Peter J. C. Moses, Ferrers Matrices and Related Partitions of Integers EXAMPLE The Mathematica ordering of the 6 antidiagonal partitions of 8 follows: 3221, 32111, 22211, 221111, 2111111, 11111111.) Frequencies of these among the 22 partitions of 8 are given in reverse Mathematica ordering as follows: 11111111 occurs 2 times, 2111111 occurs 2 times, 221111 occurs 4 times, 22211 occurs 6 times, 32111 occurs 2 times, and 3221 occurs 6 times, so that row 8 of the array is 2 2 4 6 2 6. ... First 12 rows: 1 2 2 1 2 3 2 2 3 2 2 6 1 2 2 4 3 4 2 2 4 6 2 6 2 2 4 4 2 3 9 1 2 2 4 4 2 6 6 3 12 1 2 2 4 4 2 4 6 3 6 6 12 5 2 2 4 4 2 4 6 6 4 6 3 18 2 4 10 MATHEMATICA z = 20; ferrersMatrix[list_] := PadRight[Map[Table[1, {#}] &, #], {#, #} &[Max[#, Length[#]]]] &[list]; antiDiagPartNE[list_] := Module[{m = ferrersMatrix[list]}, Map[Diagonal[Reverse[m], #] &, Range[-#, #] &[Length[m] - 1]]]; a1[n_] := Last[Transpose[Tally[Map[DeleteCases[Reverse[Sort[Map[Count[#, 1] &, antiDiagPartNE[#]]]], 0] &, IntegerPartitions[n]]]]]; t = Table[a1[n], {n, 1, z}]; TableForm[Table[a1[n], {n, 1, z}]] (* A238325, array *) u = Flatten[t] (* A238325, sequence *) (* Peter J. C. Moses, 18 February 2014 *) CROSSREFS Cf. A238326. Sequence in context: A263765 A270073 A027348 * A238885 A023566 A090970 Adjacent sequences: A238322 A238323 A238324 * A238326 A238327 A238328 KEYWORD nonn,tabf,easy AUTHOR Clark Kimberling and Peter J. C. Moses, Feb 24 2014 STATUS approved
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Last modified December 12 03:27 EST 2018. Contains 318052 sequences. (Running on oeis4.) | 1,244 | 3,222 | {"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} | 3.484375 | 3 | CC-MAIN-2018-51 | longest | en | 0.807991 |
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1
Heron's formula is a real godsend when solving math problems because it helps to find the area of any arbitrary triangle (also degenerate), if known to him. This ancient Greek mathematician was interested in a triangular shape only with integer dimensions, the area of which is also an integer, but it does not prevent today's scientists, as well as pupils and students to use it for any other.
2
In order to use the formula, you need to know another numerical characteristic – the perimeter, or rather, properiter triangle. It is equal to the sum of the lengths of all its sides. This is required in order to somewhat simplify the expression that is rather cumbersome:
S = 1/4•√((AB + VS + AC)•(BC + AC - AB)•(AB + AC - BC)•(AB + BC - AC))
R = (AB+VS+AC)/2 properiter;
S = √(p•(p - AB)•(p - BC)•(p - AC)).
3
The equality of all sides of the triangle, which in this case is correct, the formula transforms in a simple expression:
S = √3•A2/4.
4
An isosceles triangle has the same length of two of the three sides AB = BC and, accordingly, the adjacent corners. Then Heron's formula is converted to the following expression:
S = 1/2•AC•√((AB + 1/2•AC)•(AC – 1/2•AB)) = 1/2•AC•√(AB2 – 1/4•AC2), where AC is the length of the third side.
5
To determine the area of a triangle the three sides can not only help Heron. For example, suppose that triangle inscribed circle of radius r. This means that it applies to all sides whose lengths are known. Then the area of a triangle can be found by the formula, is also associated with pauperisation and is a simple product of its radius of inscribed circle:
S = 1/2•(AB + VS + AC) = R•r.
6
An example of the application of the formula of Heron: let's consider a given triangle with sides a=5; b=7 and C=10. Find the area.
7
The decision
Calculate properiter:
R = (5 + 7 + 10) = 11.
8
Calculate the required value:
S = √(11•(11-5)•(11-7)•(11-10)) ≈ 16,2. | 551 | 1,914 | {"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-2017-09 | latest | en | 0.919855 |
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Q: In a certain checking account, the balance is \$460.54. If checks of \$50.40, \$6.47, and \$41.54 are deducted from the balance, find the new balance.
A: In a certain checking account, the balance is \$460.54. If checks of \$50.40, \$6.47, and \$41.54 are deducted from the balance, the new balance is \$362.13. \$460.54 - (50.40 + 6.47 + 41.54) = 460.54 - 98.41 = \$362.13
Question
Updated 1/27/2015 6:58:05 PM
Rating
3
In a certain checking account, the balance is \$460.54. If checks of \$50.40, \$6.47, and \$41.54 are deducted from the balance, the new balance is \$362.13.
\$460.54 - (50.40 + 6.47 + 41.54) = 460.54 - 98.41 = \$362.13
*
Get answers from Weegy and a team of really smart lives experts.
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 576 | 1,619 | {"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} | 3.453125 | 3 | CC-MAIN-2018-26 | latest | en | 0.826138 |
https://convertoctopus.com/78-7-feet-per-second-to-knots | 1,714,021,633,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297284704.94/warc/CC-MAIN-20240425032156-20240425062156-00745.warc.gz | 157,740,774 | 7,664 | ## Conversion formula
The conversion factor from feet per second to knots is 0.59248380129641, which means that 1 foot per second is equal to 0.59248380129641 knots:
1 ft/s = 0.59248380129641 kt
To convert 78.7 feet per second into knots we have to multiply 78.7 by the conversion factor in order to get the velocity amount from feet per second to knots. We can also form a simple proportion to calculate the result:
1 ft/s → 0.59248380129641 kt
78.7 ft/s → V(kt)
Solve the above proportion to obtain the velocity V in knots:
V(kt) = 78.7 ft/s × 0.59248380129641 kt
V(kt) = 46.628475162027 kt
The final result is:
78.7 ft/s → 46.628475162027 kt
We conclude that 78.7 feet per second is equivalent to 46.628475162027 knots:
78.7 feet per second = 46.628475162027 knots
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 knot is equal to 0.021446122707748 × 78.7 feet per second.
Another way is saying that 78.7 feet per second is equal to 1 ÷ 0.021446122707748 knots.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that seventy-eight point seven feet per second is approximately forty-six point six two eight knots:
78.7 ft/s ≅ 46.628 kt
An alternative is also that one knot is approximately zero point zero two one times seventy-eight point seven feet per second.
## Conversion table
### feet per second to knots chart
For quick reference purposes, below is the conversion table you can use to convert from feet per second to knots
feet per second (ft/s) knots (kt)
79.7 feet per second 47.221 knots
80.7 feet per second 47.813 knots
81.7 feet per second 48.406 knots
82.7 feet per second 48.998 knots
83.7 feet per second 49.591 knots
84.7 feet per second 50.183 knots
85.7 feet per second 50.776 knots
86.7 feet per second 51.368 knots
87.7 feet per second 51.961 knots
88.7 feet per second 52.553 knots | 554 | 1,969 | {"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-2024-18 | latest | en | 0.815922 |
https://www.hackerearth.com/practice/algorithms/graphs/graph-representation/practice-problems/approximate/the-graph-permutation-be0db3d3/ | 1,611,050,167,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703518201.29/warc/CC-MAIN-20210119072933-20210119102933-00059.warc.gz | 819,250,224 | 27,755 | The Graph Permutation
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## Algorithms, Graphs, approximate, homomorphism
Problem
Editorial
Analytics
You are given an undirected graph $G$ of $N$ vertices and $M$ edges. Additionally, you are given a set of $K$ edges. Your goal is to generate a permutation $P$ of $N$ integers such that if each node $i$ in the original graph is replaced by node $P[i]$ then the count of edges which belong to the set of $K$ edges and exist in the new graph as a cycle-edge should be maximized. An edge is a cycle-edge if it belongs to any cycle in the graph.
Input
The first line contains two space-separated integers $N$ and $M$. Next $M$ lines contain a pair of integers $(u,v)$ which denote that there is an edge between the two nodes in $u$ and $v$. Next line contains an integer $K$ which denotes count of edges in the new set. Now next $K$ lines each contain two space-separated integers $u$ and $v$ which denote the edge between node $u$ and $v$.
There are no multiple edges or self-loop in the graph.
Output
In the output, you need to print a permutation of first $N$ natural numbers (space-separated) that optimize the criteria in the statement above.
Constraints
$1 \le N \le 10^5$
$1 \le M , K \le 2 \times 10^5$
$1 \le u , v \le N$
Scoring
If $X$ count of edges out of $K$ satisfy the condition above then your score is given by $X / K$. Your solution will be considered better if you have more score.
Verdict
The output should contain exactly $N$ distinct natural numbers in the range $[1 , N]$. If your output meets the required format, you will be awarded an AC verdict with a score. In all other cases, you will receive either WA or RE.
SAMPLE INPUT
5 5
1 2
1 3
3 4
3 5
4 5
5
1 2
1 3
3 4
4 5
3 5
SAMPLE OUTPUT
1 2 3 4 5
Explanation
In the sample output, the permutation is 1 2 3 4 5 so it means that the new graph is similar to the old graph. There are 3 edges out of 5 in the new graph that belong to the cycle. So your score for this output is $3/5 = 0.6$. There may be other scores possible too with different permutations.
Time Limit: 3.0 sec(s) for each input file.
Memory Limit: 256 MB
Source Limit: 1024 KB
## This Problem was Asked in
Initializing Code Editor... | 614 | 2,184 | {"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": 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} | 3.0625 | 3 | CC-MAIN-2021-04 | latest | en | 0.857016 |
https://www.rdocumentation.org/packages/spatstat/versions/1.31-3/topics/Ldot.inhom | 1,606,543,279,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195069.35/warc/CC-MAIN-20201128040731-20201128070731-00062.warc.gz | 836,912,356 | 6,482 | # Ldot.inhom
0th
Percentile
##### Inhomogeneous Multitype L Dot Function
For a multitype point pattern, estimate the inhomogeneous version of the dot $L$ function.
Keywords
spatial, nonparametric
##### Usage
Ldot.inhom(X, i, ...)
##### Arguments
X
The observed point pattern, from which an estimate of the inhomogeneous cross type $L$ function $L_{i\bullet}(r)$ will be computed. It must be a multitype point pattern (a marked point pattern whose marks are a factor). See under Details.
i
The type (mark value) of the points in X from which distances are measured. A character string (or something that will be converted to a character string). Defaults to the first level of marks(X).
...
Other arguments passed to Kdot.inhom.
##### Details
This a generalisation of the function Ldot to include an adjustment for spatially inhomogeneous intensity, in a manner similar to the function Linhom.
All the arguments are passed to Kdot.inhom, which estimates the inhomogeneous multitype K function $K_{i\bullet}(r)$ for the point pattern. The resulting values are then transformed by taking $L(r) = \sqrt{K(r)/\pi}$.
##### Value
• An object of class "fv" (see fv.object).
Essentially a data frame containing numeric columns
• rthe values of the argument $r$ at which the function $L_{i\bullet}(r)$ has been estimated
• theothe theoretical value of $L_{i\bullet}(r)$ for a marked Poisson process, identical to $r$.
• together with a column or columns named "border", "bord.modif", "iso" and/or "trans", according to the selected edge corrections. These columns contain estimates of the function $L_{i\bullet}(r)$ obtained by the edge corrections named.
The argument i is interpreted as a level of the factor X$marks. It is converted to a character string if it is not already a character string. The value i=1 does not refer to the first level of the factor. ##### References Moller, J. and Waagepetersen, R. Statistical Inference and Simulation for Spatial Point Processes Chapman and Hall/CRC Boca Raton, 2003. ##### See Also Ldot, Linhom, Kdot.inhom, Lcross.inhom. ##### Aliases • Ldot.inhom ##### Examples # Lansing Woods data data(lansing) lansing <- lansing[seq(1,lansing$n, by=10)]
ma <- split(lansing)\$maple
lg <- unmark(lansing)
# Estimate intensities by nonparametric smoothing
lambdaM <- density.ppp(ma, sigma=0.15, at="points")
lambdadot <- density.ppp(lg, sigma=0.15, at="points")
L <- Ldot.inhom(lansing, "maple", lambdaI=lambdaM,
L <- Ldot.inhom(X, "B", lambdaI=lamB, lambdadot=lamdot) | 667 | 2,512 | {"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": 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} | 2.734375 | 3 | CC-MAIN-2020-50 | latest | en | 0.762486 |
https://quicklyread.com/implementing-gates-with-different-modelling-2-mcqs/ | 1,632,444,436,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057479.26/warc/CC-MAIN-20210923225758-20210924015758-00666.warc.gz | 516,000,285 | 16,637 | # Implementing Gates with Different Modelling – 2 MCQ’s
This set of VHDL Multiple Choice Questions & Answers (MCQs) focuses on “Implementing Gates with Different Modelling – 2″.
1. Which of the following logic describes the EXOR gate?
a) y <= ((not a) OR (not b)) AND ((not a) OR (not b));
b) y <= ((not a) OR b) AND (a OR (not b))
c) y <= ((not a) AND (not b)) OR ((not a) AND (not b));
d) y <= ((not a) AND b) OR (a AND (not b));
2. What logic circuit is described by the following code?
```ARCHITECTURE gate OF my_gate IS
BEGIN
WITH ab SELECT
y<= 0 WHEN “01” OR “10”;
1 WHEN OTHERS;
END gate;```
a) NAND
b) NOR
c) EXOR
d) EXNOR
3. What is the minimum number of NAND gates required to implement an EXOR gate?
a) 2
b) 3
c) 4
d) 5
4. Sometimes gates modeled with ________ modeling may behave differently.
a) Dataflow
b) Behavioral
c) Structural
d) Structural and Behavioral
5. Which of the following option represents a structural model for not gate?
a)
``` Architecture not_gate OF my_func IS
BEGIN
x: IN STD_LOGIC;
y: OUT STD_LOGIC;
END not_gate;```
b)
``` Architecture not_gate OF my_func IS
BEGIN
x: IN STD_LOGIC;
y: OUT STD_LOGIC;
y<= NOT x;
END not_gate;```
c)
``` Architecture not_gate OF my_func IS
BEGIN
COMPONENT NOT IS
Port( x: IN STD_LOGIC;
y: OUT STD_LOGIC);
END COMPONENT;
END not_gate;```
d)
``` Architecture not_gate OF my_func IS
BEGIN
COMPONENT not1 IS
PORT( x: IN STD_LOGIC;
y: OUT STD_LOGIC);
END COMPONENT;
END not_gate;```
6. The odd behavior of gates in dataflow modeling may be the result of ________
a) Sequential statements
b) Wrong logic definitions
c) Concurrency
d) Inappropriate assignments
7. In CPLD, there are many input switches arranged in a switch bank, if an AND gate is behaving oddly but could be the reason?
a) Incorrect interconnections
b) Concurrent execution of statements
c) Mismatch of ports name and switches
d) Wrong libraries included
8. Generally, structural modeling is used with another modeling style.
a) True
b) False
9. For gates, which of the following modeling style will corresponds to shortest code?
a) Behavioral
b) Data flow
c) Structural
d) Both data flow and behavioral
10. Which of the following doesn’t corresponds to NAND gate?
a)
`y <= NOT( a AND b)`
b)
`y <= NOT a OR NOT b`
c)
`y <= NOT a AND NOT b`
d)
``` WITH ab SELECT
y <= 0 WHEN ”11”
1 WHEN OTHERS``` | 688 | 2,363 | {"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} | 2.625 | 3 | CC-MAIN-2021-39 | latest | en | 0.727396 |
http://www.docstoc.com/docs/53881958/Forestry-Economics | 1,387,525,458,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1387345769121/warc/CC-MAIN-20131218054929-00074-ip-10-33-133-15.ec2.internal.warc.gz | 357,080,775 | 18,658 | # Forestry Economics
Document Sample
``` Department of Agricultural and Resource Economics EEP 101
University of California at Berkeley David Zilberman
Spring Semester, 1999
Lecture #13
Forestry Economics
The economics of forest resources are very similar to the dynamic management of a
fishery:
• Both forests and fisheries are renewable resource systems
• The economic principles that determine optimal management are very much the
same
The major difference between the economics of a forest vs. A fishery resource are related
to biological principles.
The central question of commercial or social forest economics is “When should we cut a
stand of trees?”
• We will assume that the land has no available alternative use
- If we had an alternative use, it would:
1) introduce opportunity cost in the model
How Does the Forestry Problem Differ from the Fishery?
The forest problem is a problem of divestment, which means the solution calculates
the optimal time to consolidate and sell the entire stock and begin the next rotation. The
analogy of a forest rotation is that of a conventional crop which does not get harvested
every season; that is, the growth cycle of a forest resource (a period of centuries instead
of months) is so long that resource owners get really impatient and discounting/ dynamic
analysis is important. Note, however, that during periods of severe food famine, the
solution to conventional crop harvesting problems may be similar to solutions obtained in
a forestry model, where each time period now becomes a day without food instead of a
year without 2 x 4’s.
How Is The Forestry Problem Different From a Fishery?
1) Forest Solutions Determine “When” Rather than “How Much”
2) Growth Occurs over Long Time Periods and Can be Measured
3) The Forest Problem Solves For the Optimal Time to Harvest Entire Stock
- the solution gives the optimal length of each rotation of stock
4) Property Rights are Secure (No Open Access Problems)
In the Forestry Problem, the critical element is that the Growth Function is a
Function of Time; not a function of stock.
Figure 11.1: The Forestry Growth Function
Q(Tmsy)
Ray2 =
Tmsy
Q(t)
Q( T1)
Ray1 =
T1
Time
T2 Tmsy Tmax T1
Q(t) = Volume of Timber (i.e., Board Feet or Cubic Feet)
The growth function of a typical stand of trees looks like this.
• At first, the volume increases at an increasing rate for very young trees.
• Then growth of volume slows and increases at a decreasing rate
• Finally, when the trees are very old, they begin to have negative growth as they
rot, decay, and become subject to disease and pests.
The volume of a stand of trees is maximized at time Tmax, with a volume Q(Tmax)
Yet, this is not the volume associated with the Maximum Sustainable Yield.
• MSY occurs where the growth rate equals the Average Growth per rotation
- Recall that our goal is to re-plant new trees
• The average growth rate of a stand, at any time, t, is:
Q( t )
A. G. =
t
which can be shown by a ray through the origin.
• Ray 1 shows that the average growth can be achieved by either cutting at time
T1 or at time T2, but neither time gives the Maximum Average Growth.
- what we want is the highest av. Growth over all harvests
So How Do We Find Tmsy?
The MSY occurs at a rotation length that maximizes the average annual growth of the
stands through time.
• Max. { Q(t)/t} implies the FOC
Q(t )
d
t Q'( t )t − Q(t )
= =0
dt t2
where we have used the quotient rule of calculus. Rearranging terms, we get:
Q( t )
= Q'( t ) .
t
Thus, in order to harvest the MSY, we should cut the stand of trees when marginal
growth equals average growth of the stand.
• Ray 2 shows where this condition is met, where the average growth is tangent to
the growth function
In Biology, this concept is known as Maximizing the Mean Annual Increment.
Figure 11.2: Harvest Pattern Over Time
Q(t)
Tmsy 2Tmsy 3Tmsy Time
Graphically, the harvest pattern over time is shown above. As in the case of the fishery
when moving from 2 time periods to T periods, the optimal dynamic solution is to
replicate a single optimal decision many times.
So far, we have just discussed Biological Conditions.
The Economic Decision to Harvest a Stand
Since the value of the stand grows over time like a conventional asset, such as a stock or
money in an interest bearing bank account, the optimal solution will occur where the value
of the forest asset is in equilibrium with other assets in the economy.
• That is, we want to incorporate into the analysis the rate of time preference of
the forester, (i.e., the discount rate).
Because the growth rate is a function of time, it is necessary to formulate the
optimization problem in continuous time:
1
t
• In discrete time, we find that Pt =
1+ r
• The continuous time analog to this is found as ∂ t → 0 :
t
1
as ∂ t → 0, → e− rt
1+ r
The Forestry Optimization Problem:
In the economic optimization problem, we want to maximize the present value of the
forest stand with respect to the time period of harvest. That is, we want to find the point
in time where the NPV is maximized.
The Optimal Single Rotation:
We will first look at the problem as a single rotation. In the single rotation, there is no
opportunity cost incurred by failing to plant the next stand of trees at the optimal time.
Thus, the problem is really that of determining the optimal time to harvest a crop.
Suppose a crop is planted at time t=0 and grows in value to PQ(t) at time t. The goal is
to find the harvest time that will maximize the NPV of a single rotation.
Let P = constant price per pound of the crop.
There are no harvesting costs, so that π = TR
The Objective Function is:
Max .{ = e −rt PQ(t )}
π
t
with the FOC:
dπ
= PQ'( t ) e− rt + PQ(t ) e− rt ( − r) = 0
dt
which can be written as:
PQ'( t ) = rPQ (t )
or, MB of waiting (value of new growth) = MC of waiting (lost interest on TR)
If the forest manager delays the harvest, she will not earn interest on revenues PQ(t).
If the forest manager delays the harvest, she will gain the value of new growth Q’(t).
We can rearrange the optimality condition to get:
Q'( t )
=r
Q(t )
which states that the percent rate of growth in volume should equal the discount rate.
Profit maximization therefore dictates that the stand should be harvested when the
percentage rate of growth of crop value equals the value of alternative investments.
Q'( t )
• If > r , then the crop is increasing in value quicker than market
Q( t )
investments and the farmer should delay the harvest decision.
Q'( t )
• If < r , then market investments are increasing in value quicker than the
Q( t )
growth in value of the crop (harvesting should have already
occurred)
The Case of An Infinite Forest Rotation
The relevant case for most foresters is that of continuous stand rotation over time.
When the forester plant to replant a new forest stand immediately after cutting the old
one, there is now an opportunity cost that must be considered: The opportunity cost of
future rotations.
Preliminaries:
Before we begin, we need to review a calculus identity that we will use:
• The sum of an infinite series is: For X < 1,
∞
∑ X i =(1 + X + X 2 + X 3 + ....) =
1
i= 0 1− X
• In our problem of infinite rotation we will use this as follows:
∞
∑ e −irT =(1 + e − rT + e −2 rT + e −3rT + ....) =
1
i= 0 1 − e −rT
where T is the length of each rotation.
The infinite rotation problem is commonly called the Faustmann Rotation after the
German Forester from the early 1900’s.
Assume a constant net price (or profit) per cubic foot of timber. That is, if harvesting or
replanting costs exist, then what we hold constant is
• Net Price = Price - per unit harvesting and replanting costs.
Let:
P = the constant price per cubic foot of timber
Q = the volume of timber (in cubic feet)
We can now write forester profit as:
π = PQ(T )e − rT + PQ(T )e −2 rT + PQ(T )e −3rT + ...
= PQ(T )[e −rT + e −2 rT + e −3 rT + ...]
= PQ(T )e −rT [ + e −rT + e −2 rT + ...]
1
e −rT
= PQ(T )
1 − e −rT
PQ(T )
= rT
e (1 − e −rT )
PQ(T )
=
e rT − 1
PQ(T )
The Optimization Problem is: Max.π = rT
T e −1
with the FOC:
dπ PQ'( T) PQ(T )(− 1)(r )(e rT )
= rT + = 0
dT e −1 (e rT − 1)2
which can be re-arranged to yield:
PQ(T )re rT rPQ ( T)
PQ'( T ) = =
e rT − 1 1 − e rT
We can now cross-multiply and write the optimality condition as:
PQ'( T ) = rPQ (T ) + PQ'( T )e −rT
MR of delaying = MC of waiting + MC of delaying future income stream
The first two terms are identical to those in the single stand, or cropping decision.
• The last term represents an additional opportunity cost of delaying the harvest
- as delaying the current harvest also delays income received from future
harvests
Therefore, the optimal rotation time, T, requires the forester to equate the
marginal value of waiting to the marginal cost of delaying the harvest of current
and future stands.
In general, T* < T* (single rotation) < Tmsy:
It is also important to analyze the effect of different parameters of the harvesting
decision:
An Increase in the price of timber:
• An increase in P will tend to shorten the rotation length, because higher timber
prices increase the profitability of each harvest
- cutting trees earlier moves the profit of future harvests closer to the
present
An Increase in the Interest Rate:
• An increase in r will tend to shorten the optimal rotation length, because the
forest owner is now relatively more impatient.
- the owner is now more eager to move profit up into the present
An Increase in Harvesting Costs:
• Recall how we absorbed harvesting costs into the Net Price.
• Thus, an increase in c is analogous to a decrease in Price
• An increase in c will tend to increase the rotation length, because cutting trees
has now become less profitable
- the owner wishes to delay paying future harvesting costs
```
DOCUMENT INFO
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views: 15 posted: 9/6/2010 language: English pages: 10
How are you planning on using Docstoc? | 2,639 | 10,411 | {"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} | 2.796875 | 3 | CC-MAIN-2013-48 | latest | en | 0.899689 |
https://math.stackexchange.com/questions/2097986/can-averages-be-calculated-one-value-at-a-time?noredirect=1 | 1,721,601,534,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517796.21/warc/CC-MAIN-20240721213034-20240722003034-00550.warc.gz | 325,827,999 | 38,582 | # Can averages be calculated one value at a time?
I'm writing some code which calculates some averages. Obviously, the traditional way to calculate any average is to add up all the values, and then divide by the number of values.
However, in the mechanism I'm working on, I find it much easier to add and calculate the averages one at a time, as in add a new value to the averaged total, and then divide by two each time (since each time there are two numbers being added). But I'm not sure how accurate it would be.
Can I calculate averages this way? Or is it not reliable?
NOTE: I began writing this question originally, and while coming up with an example, found my answer. So I added an answer with my question at the same time.
• Unfortunately, it's clear that you wrote the question after exploring/grasping/finding the answer. So the question strikes many of us as disengenuous, just so you can answer it. That might explain the downvotes. Commented Jan 14, 2017 at 22:04
• @amWhy Actually first of all, I discovered my answer while writing the question. Second of all, actually Q/A style is strongly encouraged on all Stack Exchange sites. That's why they have the option to ask a question and answer it at the same time. See here: meta.math.stackexchange.com/questions/11832/… Commented Jan 14, 2017 at 22:05
• Well, I'm glad that writing the question was a trigger for an "aha!" moment. (I have done a lot of that in my life...trying to express, in writing, where I was stuck/clueless only to find those "aha!" moment.) I just noticed that the question and answer were posted almost simultaneously. And, FWIW, there is no consensus, at least on this site, regarding the approval of posting a question, and immediately posting an answer. In any case, please don't blame me for being the bearer of what you might find unpleasant. Commented Jan 14, 2017 at 22:10
• For memory purposes, you might find it useful to store the number of entries and the running average. Thus if $a_n$ is the average of the first $n$ terms, we get $a_{n+1}=\frac 1{n+1}\times \left(na_n+S_{n+1}\right)$ where, of course. $S_i$ denotes the $i^{th}$ term in your data.
– lulu
Commented Jan 14, 2017 at 22:11
• @amWhy All Stack Exchange sites have a checkbox at the bottom of the question page to answer your own question at the same time as asking. They were literally posted at precisely the same time as each other. Other SE sites are perfectly okay with doing so. This is the first SE site which someone has said I shouldn't. Commented Jan 14, 2017 at 22:11
Here is a reference of a quite accurate algorithm for getting the average of a series of floating point numbers:
https://en.wikipedia.org/wiki/Kahan_summation_algorithm
If you also want to compute the variance, look here:
https://en.wikipedia.org/wiki/Algorithms_for_calculating_variance
A good reference is this:
https://people.eecs.berkeley.edu/~wkahan/Math128/MeanVar.pdf
(I'm answering my question Q/A style)
Imagine this set of numbers:
1, 2, 3, 4, 5
((1 + 2 + 3 + 4 + 5) / 5) = 3
What you are proposing is:
((1 + 2) / 2) = 1.5
((1.5 + 3) / 2) = 2.25
((2.25 + 4) / 2) = 3.125
((3.125 + 5) / 2) = 4.0625
So no, this proposed method of calculating an average does not work.
3 <> 4.0625
Not to mention, even if it did work, it would be much slower anyway.
What you could do instead is to continue adding the values together, and elsewhere keep track of the number of values. Each time you add a value, also increment the number of values added. Then, at any given point, you are able to calculate the average...
(1 + 2) = 3 C = 2 (3 / 2) = 1.5
(3 + 3) = 6 C = 3 (6 / 3) = 2
(6 + 4) = 10 C = 4 (10 / 4) = 2.5
(10 + 5) = 15 C = 5 (15 / 5) = 3
In any case, the moral of the story is that you still need to add all the values together, and then divide by the number of values. You can keep a sum of all values, but you also need to keep a count of the values which have been added to that sum.
• no, you don't need to wait. make it this way: (1+2)/2=1.5. (1.5$\cdot$2+3)/3=2. Or in general (old average times (n-1)+new number)/n. Where n is the counting variable. But this is certainly not efficient.
– SAJW
Commented Jan 14, 2017 at 23:16
• @Socrates Indeed, edited to be more clear and generally based. Commented Jan 15, 2017 at 3:15
• @SAJW good solution, this also solves the issues where we might overshoot the storage capacity of the variable (example a UINT going past 65535) Commented Feb 22, 2022 at 9:21 | 1,271 | 4,522 | {"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": 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} | 3.421875 | 3 | CC-MAIN-2024-30 | latest | en | 0.968775 |
http://trac.sagemath.org/sage_trac/ticket/11728 | 1,368,865,217,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696381630/warc/CC-MAIN-20130516092621-00030-ip-10-60-113-184.ec2.internal.warc.gz | 222,213,428 | 6,178 | # Ticket #11728(closed defect: duplicate)
Opened 21 months ago
## Multiplication(?) buggy in AA
Reported by: Owned by: spice AlexGhitza major sage-duplicate/invalid/wontfix algebra algebra, AA, QQbar, sd32 was N/A William Stein
### Description
Something is rotten in the state of AlgebraicReal?. Multiplying some elements of AA seems to be broken in strange and interesting ways.
```sage: a = QQbar((-1)^(1/4)); a
0.7071067811865475? + 0.7071067811865475?*I
sage: b = AA(a^3-a); b
-1.414213562373095?
sage: b*1
-1.414213562373095?
sage: b.as_number_field_element()
(Cyclotomic Field of order 8 and degree 4, zeta8^3 - zeta8, Ring morphism:
From: Cyclotomic Field of order 8 and degree 4
To: Algebraic Field
Defn: zeta8 |--> 0.7071067811865475? + 0.7071067811865475?*I)
sage: b*1
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
/Users/sage/sage-4.7.2.alpha2/devel/sage-dev/sage/<ipython console> in <module>()
/Users/sage/sage-4.7.2.alpha2/local/lib/python2.6/site-packages/IPython/Prompts.pyc in __call__(self, arg)
549
550 # and now call a possibly user-defined print mechanism
--> 551 manipulated_val = self.display(arg)
552
553 # user display hooks can change the variable to be stored in
/Users/sage/sage-4.7.2.alpha2/local/lib/python2.6/site-packages/IPython/Prompts.pyc in _display(self, arg)
575 return IPython.generics.result_display(arg)
576 except TryNext:
--> 577 return self.shell.hooks.result_display(arg)
578
579 # Assign the default display method:
/Users/sage/sage-4.7.2.alpha2/local/lib/python2.6/site-packages/IPython/hooks.pyc in __call__(self, *args, **kw)
139 #print "prio",prio,"cmd",cmd #dbg
140 try:
--> 141 ret = cmd(*args, **kw)
142 return ret
143 except ipapi.TryNext, exc:
/Users/sage/sage-4.7.2.alpha2/local/lib/python2.6/site-packages/sage/misc/displayhook.pyc in result_display(ip_self, obj)
148 # IPython's default result_display() uses the IPython.genutils.Term.cout stream.
--> 150 print_obj(IPython.genutils.Term.cout, obj)
151
152 def displayhook(obj):
/Users/sage/sage-4.7.2.alpha2/local/lib/python2.6/site-packages/sage/misc/displayhook.pyc in print_obj(out_stream, obj)
140 if _check_tall_list_and_print(out_stream, obj):
141 return
--> 142 print >>out_stream, `obj`
143
144 def result_display(ip_self, obj):
/Users/sage/sage-4.7.2.alpha2/local/lib/python2.6/site-packages/sage/structure/sage_object.so in sage.structure.sage_object.SageObject.__repr__ (sage/structure/sage_object.c:1463)()
/Users/sage/sage-4.7.2.alpha2/local/lib/python2.6/site-packages/sage/rings/qqbar.pyc in _repr_(self)
2214 return repr(CIF(self._value))
2215 else:
-> 2216 return repr(RIF(self._value))
2217
2218 def _sage_input_(self, sib, coerce):
/Users/sage/sage-4.7.2.alpha2/local/lib/python2.6/site-packages/sage/rings/real_mpfi.so in sage.rings.real_mpfi.RealIntervalField_class.__call__ (sage/rings/real_mpfi.c:4285)()
/Users/sage/sage-4.7.2.alpha2/local/lib/python2.6/site-packages/sage/rings/real_mpfi.so in sage.rings.real_mpfi.RealIntervalFieldElement.__init__ (sage/rings/real_mpfi.c:7725)()
TypeError: Unable to convert number to real interval.
```
As far as I can tell, problems arise if the representation of the AlgebraicReal? element is in term of non-real elements in QQbar.
This also seems to be the root cause of the following (from patch #10981):
```sage: P = AA[x](1+x^4); P
x^4 + 1
sage: a1,a2 = P.factor()[0][0],P.factor()[1][0]; a1,a2
(x^2 - 1.414213562373095?*x + 1.000000000000000?, x^2 + 1.414213562373095?*x + 1.000000000000000?)
sage: a1*a2
x^4 + 1.000000000000000?
sage: a1,a2
(x^2 - 1.414213562373095?*x + 1, x^2 + 1.414213562373095?*x + 1)
sage: a1*a2
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
/Users/sage/sage-4.7.2.alpha2/devel/sage-dev/sage/<ipython console> in <module>()
/Users/sage/sage-4.7.2.alpha2/local/lib/python2.6/site-packages/sage/structure/element.so in sage.structure.element.RingElement.__mul__ (sage/structure/element.c:12051)()
/Users/sage/sage-4.7.2.alpha2/local/lib/python2.6/site-packages/sage/rings/polynomial/polynomial_element.so in sage.rings.polynomial.polynomial_element.Polynomial._mul_ (sage/rings/polynomial/polynomial_element.c:10928)()
/Users/sage/sage-4.7.2.alpha2/local/lib/python2.6/site-packages/sage/rings/polynomial/polynomial_element.so in sage.rings.polynomial.polynomial_element.Polynomial._mul_karatsuba (sage/rings/polynomial/polynomial_element.c:16309)()
/Users/sage/sage-4.7.2.alpha2/local/lib/python2.6/site-packages/sage/rings/polynomial/polynomial_element.so in sage.rings.polynomial.polynomial_element.do_karatsuba (sage/rings/polynomial/polynomial_element.c:36878)()
/Users/sage/sage-4.7.2.alpha2/local/lib/python2.6/site-packages/sage/rings/polynomial/polynomial_element.so in sage.rings.polynomial.polynomial_element.do_karatsuba (sage/rings/polynomial/polynomial_element.c:36759)()
/Users/sage/sage-4.7.2.alpha2/local/lib/python2.6/site-packages/sage/structure/element.so in sage.structure.element.RingElement.__mul__ (sage/structure/element.c:12051)()
/Users/sage/sage-4.7.2.alpha2/local/lib/python2.6/site-packages/sage/structure/element.so in sage.structure.element.RingElement._mul_ (sage/structure/element.c:12195)()
/Users/sage/sage-4.7.2.alpha2/local/lib/python2.6/site-packages/sage/rings/qqbar.pyc in _mul_(self, other)
2277 sdk = sd.kind()
2278 odk = od.kind()
-> 2279 return type(self)(_mul_algo[sdk, odk](self, other, False))
2280
2281 def _div_(self, other):
/Users/sage/sage-4.7.2.alpha2/local/lib/python2.6/site-packages/sage/rings/qqbar.pyc in __init__(self, x)
3384 class AlgebraicReal(AlgebraicNumber_base):
3385 def __init__(self, x):
-> 3386 AlgebraicNumber_base.__init__(self, AA, x)
3387
3388 def __reduce__(self):
/Users/sage/sage-4.7.2.alpha2/local/lib/python2.6/site-packages/sage/rings/qqbar.pyc in __init__(self, parent, x)
2186 raise TypeError, "Illegal initializer for algebraic number"
2187
-> 2188 self._value = self._descr._interval_fast(64)
2189
2190 def _repr_(self):
/Users/sage/sage-4.7.2.alpha2/local/lib/python2.6/site-packages/sage/rings/qqbar.pyc in _interval_fast(self, prec)
5660 op = self._op
5661
-> 5662 lv = self._left._interval_fast(prec)
5663 rv = self._right._interval_fast(prec)
5664
/Users/sage/sage-4.7.2.alpha2/local/lib/python2.6/site-packages/sage/rings/qqbar.pyc in _interval_fast(self, prec)
3638
3639 def _interval_fast(self, prec):
-> 3640 return self.interval_fast(RealIntervalField(prec))
3641
3642 def interval_exact(self, field):
/Users/sage/sage-4.7.2.alpha2/local/lib/python2.6/site-packages/sage/rings/qqbar.pyc in interval_fast(self, field)
2794 """
2795 if field.prec() == self._value.prec():
-> 2796 return field(self._value)
2797 elif field.prec() > self._value.prec():
2798 self._more_precision()
/Users/sage/sage-4.7.2.alpha2/local/lib/python2.6/site-packages/sage/rings/real_mpfi.so in sage.rings.real_mpfi.RealIntervalField_class.__call__ (sage/rings/real_mpfi.c:4285)()
/Users/sage/sage-4.7.2.alpha2/local/lib/python2.6/site-packages/sage/rings/real_mpfi.so in sage.rings.real_mpfi.RealIntervalFieldElement.__init__ (sage/rings/real_mpfi.c:7725)()
TypeError: Unable to convert number to real interval.
sage:
```
## Change History
### comment:1 Changed 21 months ago by was
• Status changed from new to closed
• Resolution set to duplicate
I'm closing this, since I just fixed it and it also fixed 10981. It should be one ticket.
### comment:2 Changed 21 months ago by was
• Keywords QQbar, sd32 added; QQbar removed
### comment:3 Changed 21 months ago by mvngu
• Milestone changed from sage-4.7.2 to sage-duplicate/invalid/wontfix
### comment:4 Changed 20 months ago by leif
• Reviewers set to William Stein
Note: See TracTickets for help on using tickets. | 2,428 | 8,266 | {"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} | 2.546875 | 3 | CC-MAIN-2013-20 | latest | en | 0.635252 |
http://mathhelpforum.com/algebra/11977-pre-algebra-equation-solve.html | 1,481,020,361,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541896.91/warc/CC-MAIN-20161202170901-00252-ip-10-31-129-80.ec2.internal.warc.gz | 178,507,523 | 10,875 | 1. pre algebra equation solve
A pharmacist has 2 hectograms 55 grams of pills. Each pill weighs 1 gram 2 centigrams. How many pills does the pharmacist have altogether?
A. 205
B. 250
C. 255
D. 350
I came up with B. 250
2. Originally Posted by rose234
A pharmacist has 2 hectograms 55 grams of pills. Each pill weighs 1 gram 2 centigrams. How many pills does the pharmacist have altogether?
A. 205
B. 250
C. 255
D. 350
I came up with B. 250
We can "logic" this one out, but it makes things a lot easier if we know that hecto is the metric prefix for 100. See here.
However, we know that hecto IS a metric prefix, so it must be some multiple of 10. Thus the pharmacist has 2*10^n + 55 grams, where n is some integer. The only number of that form is answer C, 255.
-Dan
3. You missed a part I think
yes he has 255 grams of pills but each pill is 1 gram and 2 centigrams so how many pills does the pharmacist have?
250?
4. Originally Posted by rose234
yes he has 255 grams of pills but each pill is 1 gram and 2 centigrams so how many pills does the pharmacist have?
250?
My apologies! I didn't read the question correctly.
Let's see. 255 g of pills and each pill is 1.02 g. Thus he has
(255 g)/(1.02 g) = 250 pills, just as you said.
-Dan | 365 | 1,248 | {"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} | 3.953125 | 4 | CC-MAIN-2016-50 | longest | en | 0.943737 |
https://thethong.wordpress.com/2009/09/ | 1,521,496,194,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647153.50/warc/CC-MAIN-20180319214457-20180319234457-00060.warc.gz | 709,320,293 | 15,125 | ## Archive for September, 2009
### Probability and Computing: Chapter 1 Exercises (Cont. 5)
Exercise 1.9: Suppose that a fair coin is flipped $n$ times. For $k > 0$, find an upper bound on the probability that there is a sequence of $log_{2}n+k$ consecutive heads.
### Tài liệu bổ sung
Mới tìm được trang web của course về Probability and Computing do chính Michael Mitzenmacher dạy. Có lẽ từ nay sẽ tập trung làm bài tập cho trong các assignment của lớp này để tiết kiệm thời gian.
### Probability and Computing: Chapter 1 Exercises (Cont. 4)
Exercise 1.18: We have a function $F: \{0,\ldots ,n-1\} \to \{0,\ldots ,m-1\}$. We know that, for $0\le x,y \le n-1, F((x + y)\phantom{a}mod\phantom{a} n) = (F(x) + F(y))\phantom{a} mod\phantom{b} m$. The only way we have for evaluating $F$ is to use a look up table that stores values of $F$. Unfortunately, an Evil Adversary has changed the value of $\frac{1}{5}$ of the table entries when we were not looking.
Describe a simple randomized algorithm that, given an input $z$, outputs a value that equals $F(z)$ with probability at least $\frac{1}{2}$. Your algorithm should work for every value of $z$, regardless what value the Adversary changed. Your algorithm should use as few lookups, and as little computation as possible.
Suppose I allow you to repeat your initial algorithm three times. What should you do in this case, and what is the probability that your enhanced algorithm returns the correct answer?
### Probability and Computing: Chapter 1 Exercises (Cont. 3)
Exercise 1.14:
Let $A$ be the event that we are equally talented, $B$ be the event that I am slightly better, $C$ be the event that he is slightly better, $D$ be the event that the game ends up in one win for me and three wins for him.
### Probability and Computing: Chapter 1 Exercises (Cont. 2)
Exercise 1.11:
Let $E_{n}$ be the event that after the $n$-th relay we receive the correct bit.
$Pr(E_{n}) = p(1-Pr(E_{n-1}))+ (1-p)Pr(E_{n-1})$
$= (1-2p)Pr(E_{n-1})+p$
we have a recurrence relation:
$Pr(E_{n}) = (1-2p)Pr(E_{n-1})+p$
$2Pr(E_{n}) = (1-2p)2Pr(E_{n-1})+2p$
$2Pr(E_{n}) -1 = (1-2p)(2Pr(E_{n-1})-1)\phantom{12} (1)$
Let $U_{n } = 2Pr(E_{n})-1$. We have $2Pr(E_{n-1})-1 = U_{n-1}$
Thus $(1)$ becomes $U_{n} = (1-2p)U_{n-1} ,$
yields $U_{n}=(1-2p)^{n}.$
So $Pr(E_{n}) = \frac{1 + (1-2p)^{n}}{2}.$
Exercise 1.12: ( the Monty Hall problem)
Assume without loss of generality that the contestant chose door $1$ and Monty opened door $2$ to show a goat.
Let $E_{i}$ be the event that the car is behind door $i$ and $B$ be the event that Monty opened door $2$.
First we calculate the probabilities before Monty opened door $2$.
$Pr(E_{1}) = Pr(E_{2}) = Pr(E_{3}) = \frac{1}{3}$.
What is the value of $Pr(B\mid E_{1})$ ? If the car is in door $1$ (the door the contestant chose), Monty will choose from two remaining door which door to open uniformly at random. So we have $Pr(B\mid E_{1}) = \frac{1}{2}$.
What is the value of $Pr(B \mid \overline{E_{1}})$ ? If the car is not in door $1$, it can be in either door $2$ or $3$ with equally probabilities. So $Pr(B\mid \overline{E_{1}}) = \frac{1}{2}$.
What is the value of $Pr(B\mid E_{3})$ ? if the car is in door $3$, Monty has no other choices than open door $2$(because he cant open door $1$). So $Pr(B\mid E_{3}) = 1$ .
Thus $Pr(B) = Pr(B\mid E_{1})Pr(E_{1}) + Pr(B\mid \overline{E_{1}})Pr(\overline{E_{1}}) = \frac{1}{2}\cdot \frac{1}{3}+\frac{1}{2}\cdot \frac{2}{3} = \frac{1}{2}$
Now we will see how things changed after Monty opened door $2$.
Apply Bayes’ Law yields:
$Pr(E_{1}\mid B) = \frac{Pr(B\mid E_{1})Pr(E_{1})}{Pr(B)} = \frac{\frac{1}{2}\cdot \frac{1}{3}}{\frac{1}{2}} = \frac{1}{3}$
$Pr(E_{3}\mid B) = \frac{Pr(B\mid E_{3})Pr(E_{3})}{Pr(B)} = \frac{1\cdot \frac{1}{3}}{\frac{1}{2}} = \frac{2}{3}$
So the contestant will double the chance if he or she chooses to switch doors.
Exercise 1.13:
Let E be the event that the person has the disorder, and B be the event that the result comes back positive.
We have: $Pr(E) = 0.02, Pr(B\mid E) = 0.999, Pr(B \mid \overline{E}) = 0.005$
Apply Bayes’ Law yields:
$Pr(E\mid B)$ $= \frac{Pr(B\mid E)Pr(E)}{Pr(B)} = \frac{Pr(B\mid E)Pr(E)}{Pr(B\mid E)Pr(E) + Pr(B\mid \overline{E})Pr(\overline{E})}$ $= \frac{0.02\cdot 0.999}{0.02\cdot 0.999+0.005\cdot 0.98}= 0.803$
So if a person chosen uniformly from the population is tested and the result comes back positive, the probability that the person has the disorder is $0.803$.
### Probability and Computing: Chapter 1 Exercises (Cont. 1)
Exercise 1.8:
Let $E_{1},E_{2},E_{3}$ be the event that the number is divisible by 4,6, and 9; respectively.
Let $E_{n,m}$ be the event that Bob’s number is $n$ and Alice ‘s number is $m$; $E$ be the event that Alice wins. | 1,651 | 4,786 | {"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": 64, "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-2018-13 | longest | en | 0.858481 |
https://gmatclub.com/forum/the-ratio-of-men-to-women-to-children-at-an-assembly-is-291623.html | 1,571,745,208,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987817685.87/warc/CC-MAIN-20191022104415-20191022131915-00481.warc.gz | 509,700,491 | 142,828 | GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 22 Oct 2019, 04:53
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# The ratio of men to women to children at an assembly is 2:3:4
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The ratio of men to women to children at an assembly is 2:3:4 [#permalink]
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24 Mar 2019, 11:02
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The ratio of men to women to children at an assembly is 2:3:4. If 720 children are present, how many men and women are there?
A 540
B 720
C 800
D 820
E 900
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Re: The ratio of men to women to children at an assembly is 2:3:4 [#permalink]
### Show Tags
26 Mar 2019, 02:15
men:women:children = 2:3:4
men:women:children = 360:540:720(multiply by 180)
(men+women) =360+540=900
Re: The ratio of men to women to children at an assembly is 2:3:4 [#permalink] 26 Mar 2019, 02:15
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https://www.ibpsguide.com/ibps-clerk-prelims-numerical-ability-sectional-test-13 | 1,566,636,239,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027320156.86/warc/CC-MAIN-20190824084149-20190824110149-00123.warc.gz | 833,739,415 | 106,234 | # Crack IBPS Clerk Prelims 2018 – Sectional Full Test-13 | Numerical Ability
Dear Readers, Take Free Numerical Ability Sectional Test of 35 Questions as like in the real exam to analyze your preparation level. Our Sectional Test Questions are taken as per the latest exam pattern and so it will be really useful for you to crack the prelims exam lucratively. Students who are weak in Reasoning Ability should utilize this chance constructively to accomplish a successful profession in Banking Field.
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Crack IBPS Clerk Prelims 2018 – Sectional Full Test-13| Numerical Ability
maximum of 35 points
Directions (Q. 1-5): What will come in the place of question mark (?) in the following series?
1) 97, 99, 105, 117, ?
a) 143
b) 151
c) 137
d) 147
e) 153
2) 10, 5, 5, 7.5, 15, ?
a) 37.5
b) 24.25
c) 40.75
d) 33.5
e) 29
3) 11, 16, 26, 41, 61, ?
a) 72
b) 94
c) 108
d) 86
e) 78
4) 9, 21, 51, 135, 381, ?
a) 753
b) 841
c) 907
d) 1055
e) 1113
5) 3, 4, 10,33, 136, ?
a) 565
b) 685
c) 840
d) 720
e) 460
Directions (Q. 6 – 10): In each of these questions two equations (I) and (II) are given. You have to solve both the equations and give answer as,
A) If a > b
B) If a < b
C) If a ≥ b
D) If a ≤ b
E) If a = b or no relation can be established between x and y
6)
I) a² – 32a + 252 = 0
II)b² – 28b + 192 = 0
7)
I) a² – 32a + 247 = 0
II)b² – 22b + 117 = 0
8)
I) a² + 29a + 208 = 0
II)b² + 19b + 78 = 0
9)
I) a² – 22a + 105 = 0
II)b² – 27b + 162 = 0
10)
I) (a – 18)² = 0
II) b² = 324
11) If the ratio of investment of Abhishek and Abhijit is 3:4, then the total profit at the end of the year is Rs. 50000 and Abhishek’s share in total profit is Rs. 10000. What is the ratio of their time period of investment?
a) 1:2
b) 1:3
c) 1:4
d) 2:3
e) 3:5
12) By giving a discount of 20%, a person is making a profit of Rs. 50. If the markup percentage is 30, then find the cost price of the item?
a) Rs 1300
b) Rs 1500
c) Rs 1250
d) Rs 1400
e) None of These
13) The simple interest received on a certain sum is Rs. 35000 in five years at the rate of 10%. What would be the compound interest received on a same sum at a rate of 6% in 2 years?
a) Rs. 7924
b) Rs. 8236
c) Rs. 8418
d) Rs. 8652
e) None of these
14) Rishi and Kunal entered into a partnership with investment in the ratio 6: 7 respectively. After one year, Nimi joined them with double the initial investment of Rishi. At the end of two years, Rishi doubled his investment. At the end of three years, they earned a total profit of Rs.69000. Find the difference between the shares of Kunal and Nimi in the profit.
a) Rs.4000
b) Rs.3000
c) Rs.5000
d) Rs.6000
e) None of these
15) The difference between the two numbers is 540. 32 % of one number is equal to 48 % of another number. Find the two numbers?
a) 1620 and 1080
b) 1510 and 970
c) 1480 and 940
d) 1350 and 810
e) 1730 and 1190
16) Priyanka is 8 years older than Renuka. The ratio of the present ages of Priyanka to Myna is 4 : 5. Renuka is 16 years younger than Myna. What is Renuka’s present age?
a) 28 years
b) 24 years
c) 20 years
d) 32 years
e) None of these
17) If downstream speed of a boat is 200% more than the upstream speed of the boat and an object flowing on river covers 100 m in 40 seconds, then how much distance boat can cover in still water in 1 hour?
a) 15 km
b) 16 km
c) 27 km
d) 18 km
e) None of these
18) Pipe A and pipe B can fill the tank while working alone in 12 hour and 10 hours respectively but there is an another pipe C which can empty the fully filled tank in 15 hours. If there is 200 liters of water available in the tank and pipe A, B and C were opened then tank will be fully filled in 8 hours. What is the capacity of the tank?
a) 3200 liter
b) 2500 liter
c) 3000 liter
d) 3600 liter
e) 4000 liter
19) The marked price of Sofa is Rs. 1500 more than the cost price. When the discount of Rs. 250 is allowed, the profit of 20 % is earned. At what price should Sofa be sold to earn the profit of 30 %?
a) Rs. 8125
b) Rs. 7340
c) Rs. 8845
d) Rs. 7920
e) None of these
20) A box contains 7 red balls, 5 black balls, 4 orange balls. If three balls are drawn out randomly, then the probability of getting a ball of different colours?
a) 5/8
b) 3/7
c) ¼
d) 4/5
e) None of these
Directions (Q. 21 – 25): Study the following information carefully and answer the given questions.
The following table shows the total number of students in different branches from 2001 to 2005 in a certain college.
21) Find the ratio of total students in Aeronautical and chemical to that of total students in civil and electronics in all the given years together?
a) 35:27
b) 119:270
c) 191:230
d) 357:811
e) None of these
22) What is the average number of students in the year 2004 in all the given branches together?
a) 84
b) 92
c) 80
d) 76
e) None of these
23) What was the percentage increase of students in 2003 in all the given branches together as compared to the previous year?
a) 23.56%
b) 27.98%
c) 37.88%
d) 26.89%
e) 32.26%
24) Find the difference between the total Electronic, Computer Science and Aeronautical students to that of total Mechanical, Chemical and Civil students in all the given years together?
a) 562
b) 304
c) 458
d) 420
e) None of these
25) In which year, was the growth being maximum in Mechanical while comparing with the previous year?
a) 2005
b) 2004
c) 2003
d) 2002
e) Either b or d
Directions (Q. 26-30): What value will come in place of question mark (?) in the following questions?
26) 157464 + √8836 – 45 % of 800 =? + (1188 ÷ 9) – 362
a) 952
b) 816
c) 878
d) 924
e) None of these
27) 160% of 150 + ?% of 900 = 1770
a) 165
b) 170
c) 190
d) 180
e) 175
28) √1225 + ? = 4 2/7 of 140
a) 720
b) 565
c) 345
d) 630
e) 440
29) 15 1/3 – 4 1/6 + 8 5/12 – 7 2/3 = ? + 2 7/12
a) 11 2/5
b) 13 ¼
c) 8 3/7
d) 9 1/3
e) None of these
30) 11 * ? *9 = 542 + 54
a) 28
b) 33
c) 30
d) 27
e) 22
Directions (Q. 31 – 35) what approximate value should come in the place of question mark (?) in the following questions?
31) 26 % of 397 + 48 % of 703 – 25 % of 451 = ?
a) 370
b) 250
c) 330
d) 410
e) 290
32) (8/36) ÷ (216/63) × 756 – 341.67 + 67 % of 599 = ?
a) 110
b) 170
c) 230
d) 250
e) 190
33) 41 % of 697 + 69 % of 804 – 35 % of 751 = ?
a) 523
b) 577
c) 642
d) 675
e) 440
34) 33450 ÷ 11 + (5/7) of 12542 + 52 % of 2499 = ?2 – 387.78
a) 86
b) 105
c) 78
d) 92
e) 117
35)120 % of 1799 + (3/7) of 6889 = (?) + 218.81
a) 5160
b) 4350
c) 5740
d) 6320
e) 4890
+ (12+1), + (22+2), + (32+3), + (42+4),…
*0.5, *1, *1.5, *2, *2.5,…
Difference of difference: 5, 5, 5, 5…
*3-6, *3-12, *3-18, *3-24, *3-30,…
*1+1, *2+2, *3+3, *4+4,…
I) a² – 32a + 252 = 0
(a – 18) (a – 14) = 0
a = 18, 14
II) b² – 28b + 192 = 0
(b – 12) (b – 16) = 0
b = 12, 16
Can’t be determined
I) a² – 32a + 247 = 0
(a – 13) (a – 19) = 0
a = 13, 19
II) b² – 22b + 117 = 0
(b – 13) (b – 9) = 0
b = 13, 9
a ≥ b
I) a² + 29a + 208 = 0
(a + 13) (a + 16) = 0
a = -13, -16
II) b² + 19b + 78 = 0
(b + 13) (b + 6) = 0
b = -13, -6
a ≤ b
I) a² – 22a + 105 = 0
(a – 15) (a – 7) = 0
a = 15, 7
II) b² – 27b + 162 = 0
(b – 18) (b – 9) = 0
b = 18, 9
Can’t be determined
I) (a – 18)² = 0
(a – 18) (a – 18) = 0
a = 18, 18
II) b² = 324
b = ±18
a ≥ b
The ratio of investment of Abhishek and Abhijit = 3 : 4
The ratio of share of Abhishek and Abhijit = 10000 : 40000 = 1: 4
Investment*Period = Ratio of profit
The ratio of time period of Abhishek and Abhijit = (1/3) : (4/4) = 1 : 3
Let CP = 100n
Hence MP = 130n
And SP = 130n x 0.8 = 104n
Profit = 4n = 50
=> n = 12.5
Hence cost price = Rs. 1250
S.I = (P*n*r)/100
35000 = (P*5*10)/100
P = (35000*100)/50
P = Rs. 70000
Compound Interest:
70000*(6/100) = 4200
74200*(6/100) = 4452
C.I = 4200 + 4452 = Rs. 8652
The share of Rishi, Kunal and Nimi,
= > [6*2 + 12*1] : [7*3] : [12*2]
= >24 : 21 : 24
= >8 : 7 : 8
23’s = 69000
1’s = 3000
The difference between the shares of Kunal and Nimi in the total profit
= > (8’s – 7’s) = Rs. 3000
Let the two numbers be x and y,
X – Y = 540
(32/100)*X = (48/100)*Y
(X/Y) = 3/2
X : y = 3 : 2
1’s = 540
The two numbers be,
= > 3’s = (540*3) = 1620
= > 2’s = (540*2) = 1080
Priyanka = Renuka + 8
The ratio of the present ages of Priyanka to Myna = 4 : 5 (4x, 5x)
Renuka = Myna – 16
Renuka’s present age = 5x – 16
Priyanka = Renuka + 8
4x = 5x – 16 + 8
8 = 5x – 4x
X = 8
Present age of Renuka = 5x – 16 = 40 – 16 = 24 years
The ratio of speed of downstream to that of upstream = 3 : 1
Speed of stream = 100/40 = 5/2 m/s = (5/2)*(18/5) = 9 Km/hr
(1/2)*(3x – x) = 9
X = 9
Speed of boat in still water = (1/2)*(4x) = 2x = 18 Km/hr
Distance covered in 1 hour = 18 km
Let total capacity of the tank be T litres
Part of tank filled when all the pipes are opened = T*(1/12+1/10-1/15) = 7T/60
Total tank filled in 8 hours = (7T/60)*8 = 56T/60 =14T/15
According to the question,
= > (14T/15) + 200 = T
= > T/15 = 200
= > T = 3000 litres
Let the C.P be x,
M.P = 1500 + C.P = 1500 + x
1. P = (1500 + x) – 250 = 1250 + x
S.P = (120/100)*x =(6/5)x
1250 + x = (6/5)x
6250 + 5x = 6x
C.P(X) = 6250
To earn a profit of 30 %,
1. P = (130/100)*6250 = Rs. 8125
Total probability n(S) = 16C3
Required probability n(E) = 7C1 and 5C1 and 4C1
P(E) = n(E)/n(S)
P(E) = (7C1 and 5C1 and 4C1)/ 16C3
P(E) = ¼
Total students in Aeronautical and Chemical = 216 + 141 = 357
Total students in Civil and electronics = 500 + 310 = 810
Required ratio = 357 : 810 = 119 : 270
Total students in 2004 in all the given branches together
= > 109 + 110 + 115 + 80 + 60 + 30 = 504
Required average = 504/6 = 84
Total number of students in 2002
= > 75 + 80 + 65 + 46 + 25 + 19 = 310
Total number of students in 2003
= > 85 + 95 + 108 + 60 + 35 + 27 = 410
Required percentage = [(410 – 310)/310] * 100 = 32.26%
Total Students in Mechanical = 50 + 75 + 85 + 109 + 120 = 439
Total Students in Chemical = 15 + 19 + 27 + 30 + 50 = 141
Total Students in Civil = 24 + 46 + 60 + 80 + 100 = 310
Total Electronic, Computer Science and Aeronautical students in all the given years together
= > 439 + 141 + 310 = 890
Total Students in Electronic = 65 + 80 + 95 + 110 + 150 = 500
Total Students in Computer Science = 45 + 65 + 108 + 115 + 145 = 478
Total Students in Aeronautical = 11 + 25 + 35 + 60 + 85 = 216
Total Mechanical, Chemical and Civil students in all the given years together
= > 500 + 478 + 216 = 1194
Required difference = 1194 – 890 = 304
From the given figure,
Growth in mechanical students:
In 2002: [(75 – 50)/150] * 100 = 50%
In 2003: [(85 – 75)/175] * 100 = 13.33%
In 2004: [009 – 85)/85] * 100 = 28.23%
In 2005: [020 – 109)/109] * 100 = 10.09%
54 + 94 – (45/100)*800 = x + (1188/9) – 1296
54 + 94 – 360 + 1296 = x + 132
54 + 94 – 360 + 1296 – 132 = x
X = 952
X % of 900 = 1770 – 160 % of 150
(x/100)*900 = 1770 – (160 * 150)/100
9x = 1770 – 240
x = 1530/9 = 170
x = 140 * (30/7) – √1225
x = 20 * 30 – 35
x = 600 – 35 = 565
15 1/3 – 4 1/6 + 8 5/12 – 7 2/3 – 2 7/12 = x
(15 – 4 + 8 – 7 – 2) (1/3 – 1/6 + 5/12 – 2/3 – 7/12) = x
10 [(4 – 2 + 5 – 8 – 7)/12] = x
X = 10 (-8/12) = 10 (-2/3)
X = (30 – 2)/3 = 28/3 = 9 1/3
11 *x* 9 = 542 + 54
99 x = 2916 + 54
x = 2970/99 = 30
(26/100)*400 + (48/100)*700 – (25/100)*450 = x
104 + 336 – 112.5 = x
X = 327.5 = 330
(8/36)*(63/216)*756 – 342 + (67/100)*600 = x
X = 49 – 342 + 402
X = 109 = 110
41 % of 700 + 69 % of 800 – 35 % of 750 = x
X = (41/100)*700 + (69/100)*800 – (35/100)*750
X = 287 + 552 – 262.5
X = 576.5 = 577
(33451/11) + (5/7)*12544 + (52/100)*2500 = x2 – 388
3041 + 8960 + 1300 + 388 = x2
X2 = 13689
X = 117
120 % of 1800 + (3/7)*6888 = x + 219
(120/100)*1800 + (3/7)*6888 = x + 219
2160 + 2952 – 219 = x
X = 4893 = 4890 | 4,992 | 11,960 | {"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} | 3.9375 | 4 | CC-MAIN-2019-35 | latest | en | 0.748307 |
https://www.biggerpockets.com/blog/2016-06-10-cash-on-cash-return-2 | 1,581,895,978,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875141430.58/warc/CC-MAIN-20200216211424-20200217001424-00525.warc.gz | 676,615,383 | 32,126 | ## How to Calculate Cash-on-Cash Return (Made Easy!)
Expertise: Business Management, Landlording & Rental Properties, Personal Finance, Personal Development
65 Articles Written
An investor came to me not too long ago seeking advice regarding the performance of his portfolio. He figured his portfolio was performing soundly, but wisely decided to seek a CPA’s analysis to help him make investment decisions.
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While this investor was sophisticated, he was analyzing his returns on his current properties and returns on potential deals using the cash-on-cash metric. There is, of course, nothing wrong with using the cash-on-cash return metric; the problem arises when it’s the only metric, or the primary metric, used in making investment decisions.
This article is going to go over the conversation I had with this investor so that you may benefit from it. I’ll discuss how to calculate the cash-on-cash return, the good qualities, the bad qualities, and how investors can use the metric to help aid investment decisions.
Before we get started, let’s first define what the cash-on-cash return is. The cash-on-cash return is a quick way to analyze an investment’s cash flow. Specifically, it will produce a percentage rate that measures the received pre-tax cash flow relative to the amount of money invested to acquire the asset.
## How to Calculate Cash-on-Cash Return
Calculating cash-on-cash return is simple. We simply divide the received net cash flow for the year by the amount of cash invested.
The overarching equation isn’t bad at all. It’s the variable, such as annual pre-tax cash flow and actual cash invested, that can become somewhat tricky.
### Annual Pre-Tax Cash Flow
The formula to calculate your annual pre-tax cash flow is as follows:
Let’s break each of these variables down.
Gross Scheduled Rent
When you are evaluating a property’s performance, “gross scheduled rent” will be the property’s gross rents, multiplied by 12. This reflects the maximum amount of income you can expect to receive from a property.
Other Income
Think about all of the other earning opportunities the property may present. Will you allow pets and receive pet income and non-refundable deposits? Do you have parking spaces available? Do you get reimbursed for utilities or charge a flat rate regarding such? All of this miscellaneous income will be included in “other income” for our cash-on-cash return analysis.
Related: Your Complete Guide to Analyzing a Property in Just 10 Minutes
Vacancy
If you already own the property and you are wanting to produce the cash-on-cash return to understand your property’s performance, you will want to use actual vacancy here. The actual vacancy should be measured by the numbers of days your property was vacant multiplied the daily rental rate. Essentially, this is the rental income you lost, on a daily basis, due to a tenant not being in place.
If instead you are analyzing a property’s potential performance, you will want to use potential vacancy. This should always be a conservative number. You can guesstimate potential vacancy by calling up property management firms in the area or asking a real estate agent to run an analysis on how long a unit stayed on market. You will be able to generate a percentage rate of vacant days compared to the entire year. Whatever that rate is, I'd go ahead and add 2%. This will help create a small buffer as you learn the ins and outs of the market and what tenants expect a rental unit to look like.
For example, if a unit sat unrented on the market for 45 days, then the vacancy rate is 12.33% (45/365). Go ahead and round up to 14% for your projected vacancy rate.
Now we’ll take that vacancy rate and multiply it by the gross scheduled rent. The result will be the amount of rental income you expect to not collect due to the unit not being rented. This can also be considered your opportunity cost.
Operating Expenses
Operating expenses will range from insurance, taxes, maintenance, HOA and bank fees, property management, and repairs. Operating expenses do not include debt service (principal and interest), nor do they include depreciation or amortization.
There are plenty of articles on BiggerPockets providing insight on how to estimate operating expenses. I recommend checking them out if you don’t know how to already.
Annual Debt Service
For the purposes of learning how to calculate cash-on-cash return, this number will be your monthly payment to cover both principal and interest related to your loan. This does not include insurance and taxes.
### Actual Cash Invested
OK, now that we know how to calculate the annual pre-tax cash flow, let’s figure out how to calculate the actual cash invested.
Let’s also break each of these variables down.
Down Payment
This has nothing tricky to it. It will simply be the amount of money you pay as required by your lender to obtain the property. Very simple.
Closing Costs
Closing costs are also somewhat simple. Basically, you will add up your net closing costs associated with obtaining the property. To do this, add up all of the costs you paid (not including your down payment) and then subtract from that any seller or lender credits given to you.
Pre-Rental Improvements/Repairs
Remember, we really only want to use cash-on-cash return to analyze a return based on the cash we have actually invested into the property. I suggest only using pre-rental improvements and repairs because I think the cash-on-cash return should really only be utilized in the first year of ownership. More on that later.
Related: NEW “Annualized Total Return” Estimate on the BiggerPockets Rental Property Calculator
Pre-rental improvements/repairs will include anything you pay out-of-pocket to fix prior to renting the units out. This is the part where the cash-on-cash return loses some of its value, as it doesn’t do a good job of analyzing returns when you are injecting more cash into the asset after renting out the property.
So there you have it. I explained how to calculate cash-on-cash return in just 800 words. Is your head spinning yet? No? Good. Time to move on to the theory and application of using cash-on-cash returns.
## Why the Cash-on-Cash Return is a Good Metric
The cash-on-cash return is a great metric and is widely used throughout the real estate industry both investors and real estate agents. The primary reason for this is due to the metric’s simplicity in calculating the percentage return.
The cash-on-cash return specifically drills down in the return on the capital invested. It does so by only considering returns that are driven by the property’s net cash flow. It is an essential part to value investing because it does not take into account asset appreciation.
Because the cash-on-cash return is only looking at the net cash flow and comparing it to the actual amount of cash invested, it’s a great indicator for the effect of leverage. Using leverage will decrease your cash-on-cash return, which makes the metric a good way to measure different levels of financing.
Many investors are not sophisticated enough to use things like the Internal Rate of Return (IRR) or Modified Internal Rate of Return (MIRR). These two metrics can be quite encumbering to learn and fully understand. And even though they provide much more insight, they also require much more work.
On the other hand, it’s easy for everyone to understand how cash-on-cash returns are calculated. It’s simply the physical cash you have in hand after 12 months, divided by the physical cash you’ve invested. Since investors can easily understand the calculation, that’s what sellers and agents use when discussing potential returns on the properties they are marketing.
Because of its simplicity, it’s also a great way to run a “back of the napkin” analysis. I personally use it as a screening tool when evaluating potential deals. The calculation can be run in literally 10 minutes or less and will likely get you within 2-5% of the actual return on equity in most situations. If you’re analyzing hundreds of deals a week, something like the cash-on-cash return makes a lot of sense.
The cash-on-cash return also allows you to easily compare different investments. You can compare rental property to lending, investing in stocks or bonds, and even starting a business. Granted, risk factors are not considered (which is a limitation we’ll discuss in a minute), but the cash-on-cash return does allow for a universal comparison between different investments.
## Why the Cash-on-Cash Return is a Bad Metric
The number one limitation, in my opinion, to the cash-on-cash return is that it doesn't indicate your actual return. There are two reasons for this: (1) taxes and (2) loan pay down.
Did you really think you were going to get through an entire article, written by a CPA, without discussing taxes?
Your tax situation is unique to you and will greatly impact your actual return on investment. Many investors argue that your tax situation doesn’t impact the asset’s performance — it is independent of you. Therefore, taxes should not be taken into account.
However, the tax impact of investment decisions should absolutely be assessed. While your tax situation may not impact the asset’s performance, the asset’s performance will directly or indirectly impact your tax situation. The effect can greatly increase or decrease your actual returns.
For instance, let’s say your annual pre-tax cash flow is \$10,000, resulting in a 10% cash-on-cash return (assuming you invested \$100,000). If you are in the 25% tax bracket, your after-tax cash flow is \$7,500 resulting in a 7.5% actual return.
Further, we have to take depreciation and amortization into account. In the example above, if your depreciation and amortization amounts to \$8,000 annually, then only \$2,000 of cash flow is remaining to be taxed. At the same 25% rate, our tax liability is \$500. Since depreciation and amortization are “phantom” expenses, our after-tax cash flow is \$9,500 (\$10,000-\$500), resulting in a 9.5% actual return.
But then there's another wrinkle to all of this. The cash-on-cash return doesn't take into account the equity added from the principal portion of your loan payment. It also assumes the entire mortgage payment is an expense, which we know the principal portion of your loan payment cannot be expensed for tax purposes.
As you can see, because the cash-on-cash return uses pre-tax numbers and doesn’t account for principal payments, the return suggested should not be trusted.
Another limitation is the simple fact that the cash-on-cash return doesn’t take into account appreciation. As I stated a bit earlier, this supports the view that the cash-on-cash return is used for value investing and not used for speculation. Depending on how you invest, this could be a good or bad thing.
The cash-on cash-return ignores the risk associated with investments. It doesn’t take into account opportunity costs, which more sophisticated investors will find alarming.
It also ignores the effect of compounding interest. The problem here is that the cash-on-cash return may make short-term investments look more appealing while making longer-term investments with a lower cash-on-cash return unappealing. If the investor were to invest in an investment that compounds (or appreciates), then the investor may be better off taking the currently smaller cash-on-cash return in the long-run.
## How You Should Use the Cash-on-Cash Return
Coming full circle, the conversation I was having with this investor led him to believe that the cash-on-cash return is a pointless metric. He then became anxious that he had been missing out on potentially better returns for the past several years.
Related: The Definitive Guide to IRR (Internal Rate of Return)
But I told him not to worry because the cash-on-cash return is a great metric if used appropriately.
First, I wouldn't suggest using the cash-on-cash return to evaluate the performance of a property you have held for more than 12 months. It should really only be utilized to evaluate the first year's performance or project a property's first year performance. After that, the cash-on-cash return begins to lose its value. The reason being that your denominator (actual cash invested) will be constantly changing as you pay down the loan and make improvements and repairs to the property. A better metric to use in this case is the IRR.
Second, use the cash-on-cash return as a screening tool to compare other investments. Many people claim the 1% and 2% rules are pointless, and I’d agree. But everyone needs a good screening tool, and the cash-on-cash return will allow you to compare investments efficiently and effectively.
Lastly, use other metrics to supplement the information that the cash-on-cash return provides you. Specifically, the IRR and the MIRR. Again, these two metrics require a bit more work but provide you with much more insight into the performance of the property.
So while the cash-on-cash return certainly has weaknesses, it’s a great metric for value investors and serves as a solid screening tool. Using it in tandem with other metrics will provide you with plenty of information to place an offer on a property. And that’s what we’re all about — enabling you to grow your portfolio.
How often do you use the cash-on-cash return formula when evaluating properties? Any questions about this equation?
Brandon Hall is a CPA and owner of The Real Estate CPA. Brandon assists investors with Tax Strategy through customized planning and
David Roberts from Brownstown, Michigan
Replied over 3 years ago
Can you write an article on how to calculate IRR?
pius kipyego
Replied over 3 years ago
Its a very nice topic on real estate and its management.
pius kipyego
Replied over 3 years ago
Its a very nice topic on real estate and its management.
Logan Hassinger Specialist from Fort Worth, TX
Replied over 3 years ago
Brandon Hall CPA from Raleigh, NC
Replied over 3 years ago
I can be bought 😉
David Roberts from Brownstown, Michigan
Replied over 3 years ago
Yipeeeee
Brian Winn from Meridian, Idaho
Replied over 3 years ago
Thanks Brandon. That was a great article! Appreciate you sharing your knowledge.
Brandon Hall CPA from Raleigh, NC
Replied over 3 years ago
Thanks Brian!
Huiping Sheng Real Estate Agent and RE investor from Tampa, Florida
Replied over 3 years ago
Thanks nice teaching Brandon.
Brandon Hall CPA from Raleigh, NC
Replied over 3 years ago
Thank you!
Nick E. Investor from Brooklyn, New York
Replied over 3 years ago
Great article! Although I’m curious why you say leverage *decreases* CCR? Judiciously applied financing will generally *increase* your CCR (and IRR). For example, an 8% cap rate with 75% LTV, 30 year term and 5% interest rate yields a 12.7% CCR. And every percentage point increase in cap rate results in a 4 point increase in CCR.
Brandon Hall CPA from Raleigh, NC
Replied over 3 years ago
Yes leverage increases returns, I just noticed I put decrease. Good catch!
Account Closed Investor from Wilmington, North Carolina
Replied over 3 years ago
Great article Brandon. I was just about to suggest that “decrease” might be a typo…;) Looks like Nick also caught that…
Account Closed Investor from Wilmington, North Carolina
Replied over 3 years ago
To answer your question at the end of the article, I primarily use anticipated Cash-on-Cash Return in evaluating properties of interest. If I’m considering selling a property, I use an estimate of what I would clear from the sale (pre & after tax), in place of Actual Cash Invested.
Alan Pott from Orange, California
Replied over 3 years ago
Brandon, Thanks for writing an article that “peels back” the onion more than one level. Currently, I am primarily investing in notes that pay 10-12%. How can I best compare that simple interest income to a rental income equivalent? Including taxes, depreciation, etc.
Bill Fontanetta Real Estate Investor from Glen Cove, New York
Alan, If you don’t mind my asking, what kind of notes are currently paying 10-12%
Ryan Landis Residential Real Estate Broker from San Mateo, California
Replied over 3 years ago
Appreciate you taking the time to put this together Brandon – great post!
Jeff Deleon
Replied over 3 years ago
Thank you for the article, the breakdown of the cash-on-cash return was great and will be shared with future real estate investors!
Jack Macioce Accountant from Pittsburgh, PA
Replied over 3 years ago
Found this post a little late, but I read it at the right time. Currently, I am trying to decide whether to sell my current residence or convert it into a rental. I am thinking it would make sense to compare the IRR for each option; however, would it make sense to calculate the CCR for selling the property? In this case, in addition to the options you listed above, would I not just add the total principal payments for the time period I owned it into the Amount Invested side of the equation?
Andrew
Replied over 2 years ago
Brandon, thank you for your article. Could you please confirm the limitation of cash-on-cash that you can be mislead when compare unlevered and levered ratios (post amortization)? Cash-on-cash changes significantly depending on principal amortization and you are can assume a negative leverage when in fact it is positive if you calculate IRR.
Kimberly Gillock Rental Property Investor from Denver, CO | 3,743 | 17,555 | {"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} | 2.96875 | 3 | CC-MAIN-2020-10 | latest | en | 0.924394 |
http://en.m.wikibooks.org/wiki/Quantum_Mechanics/Complex_Waves | 1,397,697,646,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609526102.3/warc/CC-MAIN-20140416005206-00644-ip-10-147-4-33.ec2.internal.warc.gz | 73,508,759 | 6,232 | Quantum Mechanics/Complex Waves
In Quantum Mechanics we are interested in solutions to the Schrödinger Equation that are Renormalizable. One class of functions that does this is the complex exponentials, and we can write a solution to the Schrödinger Equation as the sum of two complex exponentials
$\Psi(x)= Ae^{ikx}+Be^{-ikx}$
which is the superposition of waves, one traveling from the left, the other from the right. k is the wave number which is in units of 1/m and is usually given by $2*\pi /L$ where L is determined by boundary conditions.
Let us consider a situation in which there is a wave traveling from the left to the right. On the interval (-$\infty$,0) V(x)<E(x)and from [0, $\infty$) V(x)>E(x) and the solutions of the Schrödinger Equation are of the form
$\Psi(x)= Ae^{i\sqrt{\frac{2m}{\hbar^2}(E(x)-V(x))}x}+Be^{-i \sqrt {\frac{2m}{\hbar^2}(E(x)-V(x))} x}$
on the first interval and
$\Psi(x)= Ce^{\sqrt{\frac{2m}{\hbar^2}(E(x)-V(x))}x}+De^{- \sqrt {\frac{2m}{\hbar^2}(E(x)-V(x))}x}$
on the second interval
Even though at first glance our wave equation looks like a superposition of normal exponentials, it is still a complex wave though because the term inside of the square root is negative and this allows us to always have renormalizable solutions. | 374 | 1,279 | {"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": 6, "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} | 2.984375 | 3 | CC-MAIN-2014-15 | longest | en | 0.879875 |
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