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://sethneel.com/2015/10/12/probability-exercise-if-it-must-happen-once-it-happens-infinitely-often/ | 1,680,035,034,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948871.42/warc/CC-MAIN-20230328201715-20230328231715-00176.warc.gz | 585,895,842 | 20,848 | ## Probability Exercise: If it must happen once it happens infinitely often.
Exercise: Let $\{A_n\}$ be a sequence of independent events with $\mathbb{P}(A_n) < 1$ for all $n$. Show that $\mathbb{P}(\bigcup_n A_n) = 1 \implies \mathbb{P}(A_n \text{ i.o. }) = 1$.
Solution:
$\mathbb{P}(\bigcup_n A_n) = 1 \implies \mathbb{P}(\bigcap_n A_n^{c}) = 0$ which by independence means that $\prod_n\mathbb{P}(A_n^c) = 0$. Now since $\mathbb{P}(A_n^c) > 0 \; \forall n$, we know that for any $M, \prod_{n \geq M}$ $\mathbb{P}(A_n^c) = 0$, which means that $\mathbb{P}(\bigcap_{n \geq M} A_n^c) = 0.$ Now suppose $\textit{w} \in \{A_n\}$ f.o. Then for sufficiently large $M, \textit{w} \in \bigcap_{n \geq M} A_n^{c}$. Thus the set $\{A_n\}$ f.o., can be written as $W = \bigcup_{M}$ $\bigcap_{n \geq M}$ $A_n^{c}$, which is a countable union of sets of measure $0$ by the above reasoning. Thus $\mathbb{P}(W) = 0 \implies \mathbb{P}(A_n \text{ i.o. }) = 1$. | 382 | 950 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 18, "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-2023-14 | latest | en | 0.733676 |
https://keolistravelservices.com/what-is-the-measure-of-a-pentagon/ | 1,653,397,398,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662572800.59/warc/CC-MAIN-20220524110236-20220524140236-00037.warc.gz | 385,966,052 | 6,242 | A pentagon is a two-dimensional polygon with 5 sides and five angles. If the five sides that a form are not connected, or if the shape has actually a bent side, climate it is not a pentagon. Few of the real-life examples of a pentagon are the black sections on soccer balls, college crossing signs, the Pentagon building in the US, and also so on. This shape can additionally be spotted in flowers and even in the cross-sections that okra.
You are watching: What is the measure of a pentagon
1 Different types of Pentagons 2 Angles in a Pentagon 3 Sum that the angles in a Pentagon 4 Solved Examples 5 Practice Questions 6 FAQs on angle In a Pentagon
## Different varieties Of Pentagons
Pentagons can be categorized right into different varieties based on their properties. Right here is a list of the types of pentagons classified follow to the sides, angles and also vertices:
If the sides of a pentagon room not equal and the angles are not that the same measure, the is an rarely often rare pentagon.Observe the following figure which reflects the different varieties of pentagons.
## Angles ina Pentagon
A pentagon is a two-dimensional polygon with 5 angles. An angle is formed when two sides the the pentagon re-publishing a usual endpoint, dubbed the peak of the angle. In this section, let us learn about the kinds of angles, like, the interior angles, exterior angles, and central angles.
### Interior Angles:
In a consistent polygon,an angle inside the shape, between two joined political parties is calledan internal angle. For any kind of polygon,the total number of interior angle is same to the total variety of sides. In a pentagon, there are 5 interior angles. Each interior angle of a regular pentagon deserve to be calculation by the formula: Each interior angle = <(n – 2) × 180°>/n ; wherein n = the variety of sides. In this case, n = 5. So, substituting the value in the formula: <(5 – 2) × 180°>/5 = (3 × 180°)/5 =108°
Observe the complying with pentagon which mirrors that each internal angle the a constant pentagon equals 108°.
### Exterior Angles:
When the next of a pentagon is extended, the edge formed exterior the pentagon with its next is referred to as the exterior angle. Every exterior angle of a consistent pentagon is same to 72°. The sum of the exterior angle of any kind of regular pentagon amounts to 360°. The formula for calculating the exterior edge of a regular polygon is: Exterior edge of a consistent polygon = 360° ÷ n. Here, n to represent the total number of sides in a pentagon. Observe the following figure which mirrors the exterior angle of a pentagon.
### Central Angles:
The center of a pentagon is the point that is equidistant from each vertex or corner. The central angles of any kind of pentagon are developed when this center suggest is join to all the vertices, resulting in 5 central angles in ~ the center.There space two ways to uncover the measure of the main angle the a continual pentagon.
Method 1:The complying with steps have the right to be followed to discover the measure up of the main angles:
Step 1: In the following pentagon ABCDE, note the center as O andjoin the center O to the vertices A,B,C,D, and also E, developing 5 triangles.Step 2: because the centeris equidistant from all the vertices, and also all the sides of a regular pentagon are equal, every these triangles will be isosceles and also congruent to each other. We have the right to thus conclude the all 5 anglesat the center will be equal.Step 3: we know, that all the inner angles the a pentagon measure up 108°. Since the triangles room congruent, the inner angle at every vertex will be bisected to equal halves, each measuring (108°/2) = 54°.Step 4:Apply the edge sum residential property of a triangle to uncover the main angle. Making use of this we deserve to calculate the measure of each central angle as: main angle that a continual pentagon =180° - (2× 54°) = 72°
Method 2: The following steps deserve to be adhered to to calculate the central angle of a pentagon under this method:
Step 1: note the center of the pentagon and draw congruent triangles as displayed in the previous technique to get 5 equal angle resulting indigenous the division ofthe main angle. Step 2: because all the 5 angles in the centre space equal, us can acquire the value of every angle: 360° ÷ 5 = 72°. Step 3:Hence, the central angle in a regular pentagon procedures 72°.
The amount of the angle in any polygon relies on the number of sides the has.In the situation of a pentagon, the variety of sides is same to 5. Let united state see exactly how to calculate the sum of interior and also exterior angle in a pentagon.
### Sum of interior Angles ina Pentagon
To find the sum of the internal angles of a pentagon, we divide the pentagon into triangles. Watch the following number which reflects that 3 triangles have the right to be developed in a pentagon. The amount of the angles in every of this triangles is 180°. So, in bespeak to gain the interior angles of this pentagon, we multiply the amount of the angles of every of this triangles through the total variety of triangles. This provides it: 180° × 3 = 540°. Hence, the sum of the interior angles of a pentagon is same to 540°.
Another means to calculation the sum of the inner angles the a pentagon is by utilizing the formula: Sum of angles = (n – 2)180°; where 'n' to represent the variety of sides that the polygon. Substituting the value of 'n' in the formula: (5– 2)180° = 540°. Therefore, the sum of the inner angles of a pentagon is 540°.
### Sum of Exterior angles ina Pentagon
The sum of exterior angle of a polygon is same to 360°. Let united state prove this now with the following steps:
The sum of internal angles of a constant polygon with 'n' sides = 180 (n-2).Hence, each internal angle is: 180 (n-2)/n.We know that each exterior edge is supplementary come the inner angle, so, every exterior angle will be: <180n -180n + 360>/n = 360/n.Now, the sum of the exterior angles will be: n (360/n)= 360°. Hence, the sum of exterior angles of a pentagon equals360°.
See more: Which Compound Is An Arrhenius Base? Arrhenius Acid Definition And Examples
### Related Topics
Important Notes
Here is a perform of a few points that need to be remembered when studying about the angles in a pentagon:
A pentagon is a two-dimensional polygon with 5 angles and also five sides.The sum of all the interior angles of any kind of regular pentagon equals 540° and also the sum of every the exterior angles of any regular pentagon amounts to 360°.Each exterior edge of a consistent pentagon is same to 72° and each internal angle that a consistent pentagon is equal to 108°. | 1,600 | 6,698 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2022-21 | latest | en | 0.899728 |
https://www.mrexcel.com/board/threads/sorting-range-of-data-using-fomular.803162/ | 1,723,658,334,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641118845.90/warc/CC-MAIN-20240814155004-20240814185004-00788.warc.gz | 686,739,321 | 17,460 | # Sorting range of data using fomular
#### Stevekent
##### Board Regular
I have six columns from A1 to F1 of values that with 1st column in number from 1 to 5 but not in ascending order. The second until six columns contain word or value. As the data are copied and paste from another source, instead of using Data, and click on the A to Z, I want to have sorting formula in say H1 to M1 so that when I paste, the data is ascending order. Please see below of what I mean. Thanks a lot
4 LGB12 TH 5.5 1 2 1 LGB12 TV 6.6 1 3 1 LGB12 TV 6.6 1 3 2 LGB12 TH1 4.35 2 1 2 LGB12 TH1 4.35 2 1 3 LGB12 TH 4.7 3 1 5 LGB12 D 7.3 2 1 4 LGB12 TH 5.5 1 2 3 LGB12 TH 4.7 3 1 5 LGB12 D 7.3 2 1
<tbody>
</tbody>
<tbody>
</tbody>
### Excel Facts
Why does 9 mean SUM in SUBTOTAL?
It is because Sum is the 9th alphabetically in Average, Count, CountA, Max, Min, Product, StDev.S, StDev.P, Sum, VAR.S, VAR.P.
Put in H1 and copied down:
=SMALL(\$A\$1:\$A\$5,ROWS(\$A\$1:A1))
Put in I1 and copied down and cross:
=INDEX(B\$1:B\$5,MATCH(\$H1,\$A\$1:\$A\$5,0))
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Go back | 695 | 2,103 | {"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 | 0.786354 |
https://gotmyhomework.com/chapter-11-variance-analysis-project/ | 1,619,092,932,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039603582.93/warc/CC-MAIN-20210422100106-20210422130106-00468.warc.gz | 412,439,810 | 11,399 | # Chapter 11 Variance Analysis Project
Chapter 11 Variance Analysis Project
Winter 2020
This project is to be completed in Excel (all parts) and submitted through the
Moodle dropbox by 11:55 PM on Wednesday April 01, 2020.
The Case (40 marks):
Linda Trueblood, president of ProSkate, was staring at the most recent quarterly
performance report and was not pleased. “How can this be possible?” she asked of
her senior management team. “I thought the market for our products had
improved and that all we needed to do was maintain our budgeted market share.
But this report tells me that our profits are below expectations, and I do not
understand that.” She then turned her attention to the manufacturing manager,
wondering what story was going to emerge from his side. The performance report
she was reviewing contained the following information:
Sales \$2,880,000
Less: Cost of goods sold
Variable \$617,256
Fixed 992,750 1,610,006
Gross margin \$1,269,994
Variable \$172,800
Fixed 450,000 622,800
Operating income \$ 647,194
Other information is as follows:
a. Overall profit variance was \$107,294 U.
b. Sales revenues were \$144,000 lower than the static budgeted amount.
c. Contribution margin was \$105,144 lower than the budgeted amount.
d. Actual market share was about 0.535% lower than the budgeted 10%.
ProSkate manufactures two types of skates: professional and amateur. The
Adapted from Thinking Analytically 11-1, Brewer et. al., 2017 page 742 – 744
Brewer, Garrison, Noreen, Kalagnanam, Vaidyanathan (2017). Introduction to Managerial Accounting (5th Canadian
edition), USA: McGraw-Hill Ryerson.
1. The company sold 7,200 pairs of professional skates and 18,000 pairs of
amateur skates during the quarter, compared with the budgeted quantities
of 8,000 and 17,600, respectively.
2. Budgeted unit contribution margins for the professional and amateur
models were \$185.55 and \$40.38, respectively.
3. Direct materials were purchased at the budgeted price of \$28 per kilogram;
all materials purchased were used during the period. Direct labour was paid
\$0.50 per hour higher than the budgeted amount of \$14 per hour for both
types of skates.
4. Total direct materials variance amounted to \$5,040 (unfavourable).
5. Direct materials and direct labour used for the amateur model were the
same as the standard quantity and hours (0.28 kilograms per unit and 0.40
hours per unit, respectively). Standard quantity and hours per unit of the
professional model were 0.40 kilograms and 0.75 hours.
6. The total direct materials and direct labour used for the professional model
during the quarter were 3,060 kilograms and 5,040 hours, respectively.
7. The predetermined allocation rate for variable overhead was 130% of the
standard direct labour cost;
Required:
Adapted from Thinking Analytically 11-1, Brewer et. al., 2017 page 742 – 744
Brewer, Garrison, Noreen, Kalagnanam, Vaidyanathan (2017). Introduction to Managerial Accounting (5th Canadian
edition), USA: McGraw-Hill Ryerson.
1. Compute the following total variances in Excel (inclusive of the professional
and amateur skates) (12 marks):
i) Direct materials price variance (2 marks)
ii) Direct materials quantity variance (2marks)
iii) Direct labour rate variance (2 marks)
iv) Direct labour efficiency variance (2 marks)
v) Variable overhead rate variance (2 marks)
vi) Variable overhead efficiency variance (2 marks)
You may use the variance tree method or the equation method.
Be sure to label each of the variances as favourable (F) or unfavourable (U).
2. Suggest at least two possible reasons for each of the variances above. (15
marks)
3. Prepare the standard cost cards for both the professional skate model, and
the amateur skate model. Standard cost cards should include the direct
materials, direct labour, and variable manufacturing overhead per unit. You
should also have the standard quantity/hours per unit, and the standard
price/rate per unit, multiplied by each other to get the standard cost per
unit for the amateur model and professional model. (13 marks)
The suggested format in Excel is to have 3 tabs (one for Required #1 and 2 as they
are linked, one for Required #3, and one for Required #4).
Hints
– Start with the variance analysis – complete what information you know from
above, and determine what you still need to complete the analysis, then
attempt to calculate or find it
Cheating and plagiarism
This is an individual assignment. You are to prepare the variance analysis project
Adapted from Thinking Analytically 11-1, Brewer et. al., 2017 page 742 – 744
Brewer, Garrison, Noreen, Kalagnanam, Vaidyanathan (2017). Introduction to Managerial Accounting (5th Canadian
edition), USA: McGraw-Hill Ryerson.
should not be someone else’s or downloaded from anywhere on the internet.
ALL instances of cheating or plagiarism will be reported to the Dean’s office and
may result in suspension from the Business Program.
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# What is the value of x/yz?
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What is the value of x/yz? [#permalink]
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08 Mar 2009, 18:28
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What is the value of x/yz?
(1) x = y/2 and z = 2x/5
(2) x/z = 5/2 and 1/y = 1/10
[Reveal] Spoiler:
Hi guys, this was written before I saw the post on how to write mathematically correct in the forum.
DS #90:
My question about this problem has to do with part 1. The question is fairly simple: Why in part 1 can’t I make all signs match up and equal x? x/yz would then look like this in the way I did it: x / 2x*2x/5.
Here is what the book says:
(1) From this, z can be expressed in terms of y by substituting y/2 for x in the equation z=2x/5, which gives us z=2(2/5) / 5= y/5. The value of x/yz in terms of y is then y/2 / y(y/5) = y/2(5/y[squared])=5/2y. This expression cannot be evaluated further since no information is given about the value of y; not sufficient. (BCE).
Here is how I did it:
What I did is take x=y/2 and make it 2x=y (which Quant book did not do), which then gave me: x / 2x*2x/5, which gave me x / 4x/5, which equals 5x / 4x, and I get 5/4. Meaning I mark this as AD and move to part 2.
Part 2 is easier to solve because we just plug in: y=10, x=5, z=2. 5/10*2=1/4, for which I pick letter D as my answer because both solutions by them selves are sufficient.
One of the big problems I see with my answer is that in part 1 I got 5/4, and in part 2 I got ¼, which is a contradiction and according to MGMAT, I did something wrong.
Please explain this problem and let me know whether and if so, then why I am wrong.
Thanks,
FinanceGuy25
[Reveal] Spoiler: OA
Last edited by Bunuel on 05 Jul 2013, 03:12, edited 1 time in total.
Edited the question and added the OA
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### Show Tags
08 Mar 2009, 19:09
financeguy25 wrote:
Hi guys, this was written before I saw the post on how to write mathematically correct in the forum.
DS #90:
What is the value of x / yz ?
1. x=y/2 and z=2x/5
2. x/z=5/2 and 1/y=1/10
My question about this problem has to do with part 1. The question is fairly simple: Why in part 1 can’t I make all signs match up and equal x? x/yz would then look like this in the way I did it: x / 2x*2x/5.
Here is what the book says:
(1) From this, z can be expressed in terms of y by substituting y/2 for x in the equation z=2x/5, which gives us z=2(2/5) / 5= y/5. The value of x/yz in terms of y is then y/2 / y(y/5) = y/2(5/y[squared])=5/2y. This expression cannot be evaluated further since no information is given about the value of y; not sufficient. (BCE).
Here is how I did it:
What I did is take x=y/2 and make it 2x=y (which Quant book did not do), which then gave me: x / 2x*2x/5, which gave me x / 4x/5, which equals 5x / 4x, and I get 5/4. Meaning I mark this as AD and move to part 2.
Part 2 is easier to solve because we just plug in: y=10, x=5, z=2. 5/10*2=1/4, for which I pick letter D as my answer because both solutions by them selves are sufficient.
One of the big problems I see with my answer is that in part 1 I got 5/4, and in part 2 I got ¼, which is a contradiction and according to MGMAT, I did something wrong.
Please explain this problem and let me know whether and if so, then why I am wrong.
Thanks,
FinanceGuy25
Note the red part:
x/yz = x/[(2x*2x)/5] = 5x/4x^2 = 5/4x
_________________
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### Show Tags
08 Mar 2009, 19:28
we can write z = 2y/10
x = y/2
then (y/2) / ((y*y)/5) = 5/2y
so A is insufficient
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### Show Tags
08 Mar 2009, 20:01
Hey guys,
Thanks for the help...Thanks GMAT Tiger for pointing out the mistake, I forgot that 2x*2x=4x[Squared], not just 4x.
Thanks,
Financeguy25
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Re: What is the value of x / yz ? [#permalink]
### Show Tags
05 Jul 2013, 02:48
financeguy25 wrote:
What is the value of x / yz ?
1. x=y/2 and z=2x/5
2. x/z=5/2 and 1/y=1/10
My question about this problem has to do with part 1. The question is fairly simple: Why in part 1 can’t I make all signs match up and equal x? x/yz would then look like this in the way I did it: x / 2x*2x/5.
Here is what the book says:
(1) From this, z can be expressed in terms of y by substituting y/2 for x in the equation z=2x/5, which gives us z=2(2/5) / 5= y/5. The value of x/yz in terms of y is then y/2 / y(y/5) = y/2(5/y[squared])=5/2y. This expression cannot be evaluated further since no information is given about the value of y; not sufficient. (BCE).
Hi Bunuel/Instructors,
I have a doubt :
(In red above) how can we cancel Y in numerator with Y in denominator, if the problem doesnt states: y IS NOT= 0.
As the cancellation of Y(or for any variable) involves/multiplication or division by Y on both ends.
So far as I understand GMAT has marked statements in many DS Qs as insuffiecient on this ground stating "we cant divide by a variable unless it is mentioned as non zero"
I observed similar things in Q57(where m/n = 5/3 has been cross multiplied to 5n =3m) - they ignored the point that if m or n is 0 then the answer B wont be SUFFICIENT.
& Q15(in option B, they have ignored the case of x=0, which if true makes B not sufficient as it also the solution also involves division by a variable )
ARE there any particular scenarios in which we are allowed to divide/multiply varibles or transfer them by cross multiplication in an equation.
OR where am i lacking in my basics ?
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Re: What is the value of x / yz ? [#permalink]
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05 Jul 2013, 03:40
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p111 wrote:
financeguy25 wrote:
What is the value of x / yz ?
1. x=y/2 and z=2x/5
2. x/z=5/2 and 1/y=1/10
My question about this problem has to do with part 1. The question is fairly simple: Why in part 1 can’t I make all signs match up and equal x? x/yz would then look like this in the way I did it: x / 2x*2x/5.
Here is what the book says:
(1) From this, z can be expressed in terms of y by substituting y/2 for x in the equation z=2x/5, which gives us z=2(2/5) / 5= y/5. The value of x/yz in terms of y is then y/2 / y(y/5) = y/2(5/y[squared])=5/2y. This expression cannot be evaluated further since no information is given about the value of y; not sufficient. (BCE).
Hi Bunuel/Instructors,
I have a doubt :
(In red above) how can we cancel Y in numerator with Y in denominator, if the problem doesnt states: y IS NOT= 0.
As the cancellation of Y(or for any variable) involves/multiplication or division by Y on both ends.
So far as I understand GMAT has marked statements in many DS Qs as insuffiecient on this ground stating "we cant divide by a variable unless it is mentioned as non zero"
I observed similar things in Q57(where m/n = 5/3 has been cross multiplied to 5n =3m) - they ignored the point that if m or n is 0 then the answer B wont be SUFFICIENT.
& Q15(in option B, they have ignored the case of x=0, which if true makes B not sufficient as it also the solution also involves division by a variable )
ARE there any particular scenarios in which we are allowed to divide/multiply varibles or transfer them by cross multiplication in an equation.
OR where am i lacking in my basics ?
What is the value of x/yz?
(1) $$x = \frac{y}{2}$$ and $$z = \frac{2x}{5}$$. If $$y=10$$, then $$x=5$$, $$z=2$$ and in this case $$\frac{x}{yz}=\frac{5}{20}=\frac{1}{4}$$ BUT if $$y=20$$, then $$x=10$$, $$z=4$$ and in this case $$\frac{x}{yz}=\frac{10}{80}=\frac{1}{8}$$. Not sufficient.
(2) $$\frac{x}{z} = \frac{5}{2}$$ and $$\frac{1}{y} = \frac{1}{10}$$. Multiply these two equations: $$\frac{x}{yz}=\frac{5}{20}$$. Sufficient.
1. Yes, it would be better if the stem stated that $$yz\neq{0}$$. If you don't state that, then x=y=z=0 satisfies the first statement and in this case x/(yz) is undefined.
2. As for m/n = 5/3. This equation already implies that neither of them is zero: if m=0, then m/n = 0 and not 5/3 and if n=0, then m/n is not defined, and not 5/3. So, in this case we can write 3m = 5n.
3. I don't know what question is Q15.
Hope it helps.
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Re: What is the value of x / yz ? [#permalink]
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05 Jul 2013, 05:24
Hi,
Appologies for the PM, I wasnt aware.
Bunuel wrote:
1. Yes, it would be better if the stem stated that $$yz\neq{0}$$. If you don't state that, then x=y=z=0 satisfies the first statement and in this case x/(yz) is undefined.
I think you understand my doubt. The scenarios(x,y,z=0) will make the solution undefined. Hence answer E. (Thats the way i solved, considering all scenarios)
SO why are we not picking these values as well? The statement is meant to be true for all real numbers ~ RIGHT ?
How to decide between option D(here) & option E for such questions in exam.
Bunuel wrote:
3. I don't know what question is Q15.
It is question 15 from Data sufficiency & Q57 is question 57.
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Re: What is the value of x / yz ? [#permalink]
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05 Jul 2013, 05:36
p111 wrote:
Hi,
Appologies for the PM, I wasnt aware.
Bunuel wrote:
1. Yes, it would be better if the stem stated that $$yz\neq{0}$$. If you don't state that, then x=y=z=0 satisfies the first statement and in this case x/(yz) is undefined.
I think you understand my doubt. The scenarios(x,y,z=0) will make the solution undefined. Hence answer E. (Thats the way i solved, considering all scenarios)
SO why are we not picking these values as well? The statement is meant to be true for all real numbers ~ RIGHT ?
How to decide between option D(here) & option E for such questions in exam.
Bunuel wrote:
3. I don't know what question is Q15.
It is question 15 from Data sufficiency & Q57 is question 57.
The answer to the question is neither D nor E, it's B.
(1) is not sufficient irrespective whether you have $$yz\neq{0}$$ condition or not. The point is that, their solution for (1) is not 100% correct since it's not mentioned that $$yz\neq{0}$$.
(2) is still sufficient as shown in my post above.
As for Q15 and Q57: I don't know what questions are you talking about. Please post them.
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Re: What is the value of x/yz? [#permalink]
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16 Sep 2016, 14:54
x/yz = ?
(1) x = y/2 and z = 2x/5
rewrite first eq as y = 2x and plug into original equation
x/[(2x)(2x/5)] = 5/4x --> we still don't know the exact value of the original equation
(2) x/z = 5/2 and 1/y = 1/10
(5/2)(1/10) = 5/20 --> we have a value!!!
Therefore, (2) is sufficient & B is the correct answer
Re: What is the value of x/yz? [#permalink] 16 Sep 2016, 14:54
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Display posts from previous: Sort by | 4,394 | 13,942 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2017-04 | latest | en | 0.914454 |
https://www.nature.com/articles/s41598-019-44669-3?error=cookies_not_supported&code=3a197aa3-2420-4198-902b-e5c7ce7bb595 | 1,722,662,488,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640356078.1/warc/CC-MAIN-20240803025153-20240803055153-00024.warc.gz | 720,733,871 | 76,921 | ## Introduction
Micro- and nano-fabricated resonators are used in numerous applications including timing and frequency references, filters, chemical sensing, and physical sensing. Small-amplitude dynamics of these systems are typically studied using well-established methods such as Euler-Bernoulli theory for oscillating beams1. Various phenomena such as large-amplitude operation2 and material nonlinearities3 can cause a device response to deviate from linear predictions. In typical microsystem applications, such nonlinear effects are often avoided. Over the past decade, there has been growing interest in developing design methodologies for microsystems that exhibit different types of nonlinearities for basic research4,5,6,7,8 and in employing nonlinearities to improve the performance of micro- and nano-resonant systems, for instance, to improve the sensitivity of sensors9,10 or the stability of timing references11. Internal resonance is a particular nonlinear phenomenon that can cause nonlinear modal interactions between vibration modes directly excited by external harmonic forces and other vibration modes. Internal resonance may occur when the linear natural frequencies of a system are commensurate or nearly commensurate (e.g., ω2 = 1 or ω2 ≈ 1 where k = 2, 3, …) and there exist nonlinear coupling terms between the vibration modes12. Internal resonance has been studied from different perspectives because of the interesting dynamic properties. Consider a two degree-of-freedom (DOF) system where the equations of motion are:
$$\{\begin{array}{l}\ddot{x}+{\gamma }_{1}\dot{x}+{\omega }_{1}^{2}x=f(x,y)+F\,\cos ({{\rm{\Omega }}}_{r}t+{{\rm{\Phi }}}_{1})\\ \ddot{y}+{\gamma }_{2}\,\dot{y}+{\omega }_{2}^{2}y=g(x,y)\end{array}$$
(1)
where the first equation represents the externally excited mode through input force $$F\,\cos ({{\rm{\Omega }}}_{r}t+{{\rm{\Phi }}}_{1})$$ and the second equation represents the indirectly excited sub-system. Parameters γ1 and γ2 are the damping coefficients while ω1 and ω2 are the natural frequencies of the undamped linear oscillators for the vibrational modes, respectively. Functions f(x, y) and g(x, y) represent the coupling between the two modes. In linear Coriolis vibratory gyroscopes (CVGs), for instance, these functions will be anti-symmetric (i.e., f(x, y) = −g(x, y) and $$f(x,y)\propto m\dot{y}{\rm{\Omega }}$$. Nonlinear 2:1 internal resonance occurs as a result of quadratic nonlinearities present in the system when Ωr ≈ ω1, ω1 ≈ 2ω2, $$f(x,y)={\dot{y}}^{2}$$ and $$g(x,y)=2\dot{x}\dot{y}$$. In this case, the nonlinear quadratic coupling terms lead to auto-parametric excitation of the lower-frequency mode by the higher-frequency mode13. In other words, the vibration energy from the mode with a higher natural frequency is pumped into the mode of lower natural frequency. The amount of energy that is transferred depends on the type of quadratic nonlinearities, the amplitude of external force, modal Q-factors, and the frequency ratio between the vibrational modes, among others8. An interesting characteristic of internally resonant systems with 2:1 frequency ratio in the presence of quadratic coupling nonlinearities is saturation. When the system is excited at a frequency near the higher natural frequency, the structure responds to the frequency of excitation, and the amplitude of the response increases linearly with the amplitude of excitation13. However, when the vibrational amplitude of this mode reaches a threshold value, it saturates and the additional energy from the input source is transferred to the lower natural frequency mode due to the nonlinear coupling between them. The mode with the lower resonant frequency then starts to oscillate at half the excitation frequency and its amplitude grows proportional to the spill-over energy from the mode directly excited by the input. In macro-systems, internal resonance has been mainly studied to better understand the modal interactions with the intention of suppressing unwanted vibrations14,15,16. However, practical applications of internal resonance at micro- and nano-scales have remained limited to basic demonstrations of the phenomena8,17, taking advantage of the saturation phenomenon for amplitude and frequency stabilization11, and exploiting the phenomenon for mass-sensing18.
An application area that requires two coupled resonators and can potentially benefit from internal resonance is the measurement of angular rate. Gyroscopes made using microelectromechanical systems (MEMS) technologies are used increasingly in applications ranging from navigation and robotics to stability control due to their small size, low power consumption and low cost. Many MEMS gyroscopes operate based on Coriolis effect where the Coriolis acceleration couples two structural modes of vibration (i.e., sense and drive) dynamically. To maximize the sensitivity of MEMS Coriolis vibratory gyroscopes (CVGs), the natural frequencies of the sense and drive modes are designed to match. However, perfect matching between these modes is challenging because of the manufacturing nonidealities and tolerances. Moreover, maintaining the frequency matching during the operation of the gyroscope usually is not possible since parameter fluctuations under operating conditions may induce further mistuning. Proposed solutions for mode-matching include post-processing techniques19, active electronic control systems20, and structural design improvements21. Inherent robustness may be achieved by widening the bandwidth of the sense mode through linear coupling of vibration modes22. The amount of frequency split can be tuned using voltage signals23. Calculation of the required excitation signals can be carried through, for instance, statistical learning methods for the automatic mode-matching circuit and elimination of the frequency split24. It has also been proposed that by employing parametric resonance, the resonance frequency of the drive mode can be varied as needed until the sense mode is excited25.
In this work, we are proposing a fundamentally different approach to improve the robustness of a MEMS gyroscope response to design parameter variations. The proposed design exploits the 2:1 internal resonance within a microresonator with 2 DOF such that the sense mode bandwidth is widened to the extent to circumvent the requirement for mode matching. The phenomenon is self-initiating once the excitation amplitude exceeds a threshold level without a need for a controller, electrostatic tuning, or additional DOF.
## Results
### Device design
The device utilises twin proof masses that are mechanically connected to two crossbar suspensions forming a tuning fork type resonator as shown in Fig. 1. Slender flexural beams are used for the connections between the device components and to the substrate. The device is excited by applying input voltages to the two parallel plate drive electrodes (DE) 1 and 2 to exert an electrostatic force onto the structure. Movements of the structure are measured using a combination of capacitive sense currents from the sense electrodes (SE) 1–4. Under small forces, the device remains linear with two separate modes. With an adequately large driving force, the microresonator behaves a spring-pendulum mechanism with quadratic nonlinearities (see the Supplementary Information attachment on modeling of the system)26. Nonlinearities at small scales can stem from structural, electrostatic, and material origins. In this particular case, large displacements will result in quadratic nonlinearities due to the combined linear and angular movements of the structural components. The twin proof masses can move in the radial direction (spring mode) or rotate about the anchor (pendulum mode), simultaneously. Under these conditions, there can be a nonlinear coupling of vibration energy from the forced vibrations along the spring direction to the pendulum mode through 2:1 internal resonance because of the inherent quadratic nonlinearities in the system. Taking into account the quadratic Coriolis and centripetal nonlinearities, the system can be modeled using different methods, including the perturbation method (see the Supplementary Information attachment)27. The designed microdevice prioritizes anti-phase resonant modes such that the natural frequency of the anti-phase resonant mode along the x-axis (spring mode) is twice the frequency of the anti-phase resonant mode along the y-axis (pendulum mode). The fabricated device is vacuum sealed through a hermetic, wafer-level process before dicing.
### Linear response characterisation
The natural frequencies of the microresonator (fs and fp) and the quality factors (Qs and Qp) for the spring and pendulum modes are measured using the test-bed shown in Fig. 2a. A vector network analyzer was used to measure the device response at the frequency of the input signal. A DC bias voltage is used to enhance the linear component of the input electrostatic force. Measured spectra around the spring and pendulum mode frequencies at the onset of nonlinearities are shown in Fig. 2b,c, respectively. For the device studied here, fp = 560.28 kHz with Qp = 6900 (pendulum mode) and fs = 1122.35 kHz with Qs = 5700 (spring mode), leading to a frequency ratio between spring and pendulum modes of fs/fp = 2.003 ≈ 2.
### 2:1 Internal resonance
The nonlinear mode coupling caused by the 2:1 internal resonance is evaluated through the saturation figure and the nonlinear frequency response curves. The configuration in Fig. 3a was employed for these experiments where an AC voltage from a signal generator was superimposed on a DC voltage and used to produce the electrostatic input force. Output current due to the capacitance changes at the sense electrodes was differentially measured and monitored on a signal analyzer to allow for simultaneous monitoring of different portions of the spectrum. The intended 2:1 internal resonance and nonlinear mode coupling were examined through several experiments. Figure 3b demonstrates the half-order subharmonic response of the sense (i.e., pendulum) mode of the device as the drive voltage amplitude is increased and the frequency of excitation is fixed at Ωr = 1120.45 kHz ≈ 2fp. As can be seen, the amplitude of the sense mode (i.e., pendulum mode) is negligible for drive voltage of less than Vac ≤ ~1 V. However, once past this threshold voltage, the drive mode saturates and the additional energy is transferred to the sense mode for Vac will result in more separation between the peaks. Figure 3c shows the nonlinear frequency response curves of the sense mode (i.e., pendulum mode) for three different actuation voltages. Despite the actuation of the system at the excitation frequency in the vicinity of the spring-mode resonant frequency, the pendulum mode responds at half the excitation frequency. As the electrostatic voltage is increased, the microresonator exhibit increasing nonlinear mode coupling, where the vibrational amplitude splits and two peaks of the vibrational deflection emerge in the vicinity of the pendulum-mode resonant frequency. Further increasing of the drive voltage, Vac, will result in additional separation between the peaks.
### Response to angular rate
For performance evaluation as a rate sensor, the overlap between the forward- and backward sweeps, as shown in Fig. 4, was selected as a robust region for applying the input force. To operate the sensor in this region, the frequency of the drive signal needs to be about twice some frequency in this band, eliminating the need for precise frequency control and providing a robust method for device excitation.
To assess the scale factor (i.e., sensitivity) of device to input angular rate, the setup illustrated in Fig. 5 was used. The device chip was mounted inside a ceramic package and affixed to a printed circuit board (PCB). The PCB was then securely put on a rate table with a trans-resistance amplifier as shown in Fig. 6. The excitation signal was produced by a lock-in amplifier and passed through a frequency doubler before application to the sensor to excite the drive mode. Vibrations at the sense mode were detected by differential amplification of the currents from the sense electrodes and removing the interference from the excitation signal using a notch filter. The at-rest output spectrum of the microdevice is shown in Fig. 7a, confirming the activation of the nonlinear mode coupling due to the 2:1 internal resonance. Rate signals in the range of ±360 deg + sec−1 were applied to the device using the rate-table controller. Figure 7b shows the calibration curve obtained for different constant angular rates on the rate table and observing the corresponding output voltages of the microresonator. The x-axis in the figure represents the reference signal read from the rate-table controller while the y-axis represents the difference between the output current in response to the input rate and the at-rest signal from the resonator. Due to the quadratic nature of the nonlinearities in 2:1 internal resonance, the direction of the rate signal needs to be obtained from inertial signals. As seen in Fig. 7b, the microresonator demonstrates a sensitivity of 110 fA/(deg + sec−1) over a measurement range of approximately ±220 deg + sec−1 with a DC bias voltage of 80 V. At higher rates, the microresonator output exhibits reduced sensitivity to the input. This is likely due to the effect of centrifugal force on the device response which results in the detuning of mode frequency ratio, and hence, affecting the efficiency of energy exchange between the drive and sense modes.
## Discussion
There has been a significant level of interest in understanding the nonlinear dynamics of micro- and nano-structures. Except for the basic research efforts on the nonlinear phenomena, the understanding of nonlinearities was deemed necessary in order to devise methods to avoid them in practical applications. Herein, we presented a proof-of-concept design for a nonlinear rate microsensor that employs 2:1 internal resonance for its operation. In contrast to the typical designs for Coriolis vibratory gyroscopes with linearly coupled drive and sense modes, utilisation of internal resonance provides a robust means for excitation of the resonator over a range of input frequencies. For instance, in Fig. 4 it can be seen that the two vibration modes are coupled to each other effectively where the sense amplitude changes from about ~0.75 mV to ~2 mV over a sense frequency range of 559.5 kHz to 561.5 kHz. In case of a linear CVG with a resonant frequency of 561 kHz and similar quality factor of 4000, a 1 kHz deviation of drive signal would result in a drop in signal amplitude of ~14×. For this reason, the nonlinear operation scheme to a great extent alleviates the mode-tuning requirements for the drive and sense modes of the device. Additionally, the nonlinear operation of the device results in separation of the electrical drive and sense signals in the frequency domain, potentially simplifying the signal processing by removing direct interference between them. On the other hand, the nonlinear operation reduces the linear cross-coupling between the drive and sense directions that degrade the performance of linearly-coupled CVGs. This is specifically due to the fact that the linear cross-coupling terms will not force the system to resonate due to 2:1 frequency ratio between the spring and pendulum modes. Furthermore, their effect is negligible when compared to the quadratic couplings in the system.
As the first structural design to demonstrate the concept, the device offered a sensitivity of 110 fA/(deg + sec−1). The device sensitivity can improve, for example, by increasing the proof-masses used for the resonators. It is also possible to improve the device performance metrics such as sensitivity, dynamic range, and bandwidth through nonlinear closed-loop control. Further research will be needed to explore the limits of operation regarding noise performance or long-term stability of device response.
## Methods
### Fabrication process
The microresonator was fabricated through the MEMS Integrated Design for Inertial Sensors (MIDIS) process offered by Teledyne DALSA Inc.13. The MIDIS process is based on high aspect-ratio, bulk micromachining of a 30 μm thick single-crystal silicon wafer (device layer) that is sandwiched between two other silicon wafers (top interconnect and bottom handle wafers). The device can be either be vacuum encapsulated at 10 mTorr or held at a sub-atmospheric pressure of 150 Torr. The top silicon wafer includes Through Silicon Vias (TSV) with sealed anchors for compact flip-chip integration and interconnection with external microelectronic signal processing circuitry28. The encapsulation of vibrational inertial MEMS resonators at low pressures influences their quality factor and response time28,29. Figure 1a shows the top-view, Scanning Electron Microscope (SEM) image of the fabricated device before encapsulation with the top wafer. The dimensions of the fabricated device are La = 165 m, wa = 8 m, Lc = 76 m, Ltf = 72 m, wtf = 25 m, w1 = 29 m, h1 = 60 m, w2 = 35 m, h2 = 201 m, l1 = 194 m, l2 = 197 m, g = 1.75 μm, to achieve the approximate 2:1 ratio between the modes (Fig. 1b).
### Frequency domain characterisation
The measurement setup in Fig. 2a comprises (1) the vacuum encapsulated microresonator chip; (2) a DC power source (Keysight B2901A) to produce VDC; (3) a vector network analyzer (Rohde & Schwarz ZVB4) to produce $${V}_{ac}\,\cos ({{\rm{\Omega }}}_{r}t)$$; (4) a signal combiner/splitter (Mini-Circuits ZFSCJ-2-2-S); and (5) a transimpedance amplifier (Zurich Instruments HF2TA) with a gain of Gamp. In this figure, the signals Y1 and Y2 are the currents produced due to capacitance changes between the stationary sense electrodes SE2 and SE4 and the crossbars of the resonator. The setting to determine the natural frequencies and the Q-factors were VDC = 100 V, Vac = 1 V, and Gamp = 1 kΩ.
The nonlinear frequency measurement setup in Fig. 3a includes (1) the vacuum encapsulated microresonator chip; (2) a DC power source (Keysight B2901A) to produce VDC; (3) a function generator (Agilent Technologies 81150A) to produce Vac; (4) a signal combiner/splitter (Mini-Circuits ZFSCJ-2-2-S); (5) a transimpedance amplifier (Zurich Instruments HF2TA) with a gain of Gamp; and (6) a signal analyzer (Agilent Technologies N9000A) to monitor the sensed signals in the frequency domain. For these tests, we set VDC = 100 V, Vac = 0–4.5 V, and Gamp = 10 kΩ. The drive frequency of the electrostatic voltage is then swept forward and backward around the spring mode frequency while monitoring the frequency response of the sense mode (pendulum mode).
### Sensitivity characterisation
The linearity, full-scale range, and response of the microresonator to the input rotation rate are extracted from the data obtained through the scale factor tests using a high precision rate table (Ideal Aerosmith 1621-200A-TL with AERO 812 controller). The schematic of the setup for the rate measurement is shown in Fig. 5. The microresonator is mounted on a PCB and securely placed on the rate table along with a low-noise trans-impedance amplifier (FEMTO DHPCA-100). A lock-in amplifier (Zurich Instruments HF2LI) was used to produce the excitation voltage and record the measured signals. While the lock-in amplifier could detect higher harmonics of the excitation signal, it is unable to capture subharmonics. On the other hand, the nonlinear transformation of the excitation signal from the frequency fs to $${f}_{p}\approx {f}_{s}/2$$ meant that the sense signal is roughly at half the frequency of the excitation signal, which would have been undetectable with the standard configuration. To circumvent this issue, a frequency doubler was used at the output of the lock-in amplifier to produce the excitation signal which was further amplified using a voltage amplifier (TEGAM 2350), and thus, treated as the first harmonic of the reference signal produced by the lock-in amplifier. The frequency doubler was designed using an analog multiplier chip (AD835) to multiply the reference signal by itself to generate its second harmonic at double the frequency. The signal at the output of the device included the sense signal at fp due to the nonlinear excitation of the sense mode as well as a parasitic contribution from the drive signal at fs. The interference at fs was then removed using a notch filter after initial amplification of the current signal using the transimpedance amplifier. The sense signal received by the lock-in amplifier was therefore at frequency fp, which was about fs/2.
An input rate in the form of a trapezoidal wave was then applied to the sensor such that the applied rate was increased from 0 deg + sec−1 to the target rate with the constant angular acceleration of 40 deg + sec−2, kept constant at the intended rate for 30 s, and then decreased back to 0 deg + sec−1 with the constant angular acceleration of −40 deg + sec−2. An accelerometer was used to determine the direction of rotation of the rate table. During the test, the output response of the microresonator is monitored and recorded. The response of the sensor is measured as the change in the amplitude of the sense signal due to the input rate.
For the sensitivity tests, we set VDC = 0. 4 V at a frequency of 561. 236 kHz (before frequency doubler), a fixed gain of 20 V/V for the high-voltage amplifier, and a gain of R = 5 × 107 Ω for the transimpedance amplifier. The notch filter was designed based on a passive twin-T structure for a notch frequency of 1120 kHz. | 4,769 | 21,743 | {"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": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-33 | latest | en | 0.875916 |
https://www.all-guitar-chords.com/lessons/4 | 1,716,760,471,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058973.42/warc/CC-MAIN-20240526200821-20240526230821-00879.warc.gz | 551,356,633 | 5,139 | # CHORDS-BY BODOM
by bodom (Jul 26, 2007)
Hey folks Im back again with another lesson. Before you read this one, you should read my KEYS lesson first. Yep this one is on chords. Ok Im going to use the same chart I did last time but a little different. Before I had Major minor minor etc now I am just going to use roman numerals. Capital ones stand for Major and lower case stand for minor. Ok so here is the chart again.
```Key I ii iii IV V vi vii b5
C C D E F G A B
G G A B C D E F#
D D E F# G A B C#
A A B C# D E F# G#
E E F# G# A B C# D#
B B C# D# E F# G# A#
F# F# G# A# B C# D# E#
C# C# D# E# F# G# A# B#
F F G A Bb C D E
Bb Bb C D Eb F G A
Eb Eb F G Ab Bb C D
Ab Ab Bb C Db Eb F G
Db Db Eb F Gb Ab Bb C
Gb Gb Ab Bb Cb Db Eb F
Cb Cb Db Eb Fb Gb Ab Bb
```
Ok so from this we can find the notes and chords in a certain Key. Now I am going to show you how the chords are formed and why the fit into certain Keys.
Chord formulas
Ok like I said before in my last lesson chords are just a group of notes played together. But what notes? Well there is a formula that tells you what notes are in the chord. You should memorize these formulas. These are the formulas for basic Major and minor chords.
```Major 1 3 5
minor 1 b3 5```
b stands for flat and # stands for sharp
Great now what do the numbers stand for? Look at the first chart again. Remember how there are seven distinct notes that make up the Key? Example Key of C the seven distict notes are C,D,E,F,G,A,B well the formula is telling you what notes make up the chord. For a C major chord you play the 1st 3rd and 5th note of that Key. So the notes are C,E,G. For C minor the notes are the 1st b3rd and 5th so the notes are C,D#,G.
So remember the Major minor minor Major Major minor minorb5 on the last chart well now I switched it to I ii iii IV V vi vii b5 to make it easier to see the number of the notes in the Key. And now we can see why each chord is in that Key. So lets do the key of C lets see why the first two chords fit in that Key.
First chord is C major
`formula for Major? 1 3 5`
Now go to the Key of C And look at the 1st 3rd and 5th notes C,E,G
So the notes are C,E,G now are these notes in the key of C?
Notes in Key of C= C,D,E,F,G,A,B Yep!!
Second chord is D minor
`formula for minor? 1 b3 5`
Now go to the Key of D And look at the 1st b3rd and 5th notes
notes in D = D,E,F#,G,A,B,C#
make sure to flatten the 3rd
So the notes are D,F,A. Are they in the Key of C?
Notes in Key of C= C,D,E,F,G,A,B Yep!!
Now I will show you why C minor is NOT in the key of C
`formula for minor? 1 b3 5`
Notes are C,D,E,F,G,A,B
Flatten the 3rd
so the E becomes a D# note which is not in the Key of C.
Now why D Major is NOT in the key of C
`formula for Major? 1 3 5`
Key of D= D,E,F#,G,A,B,C#
Are they in the key of C? Nope!! the F# note is not in the key of C so D major is not in this Key.
Ok thats it, well not really there are alot more chords out there but they work the same way, just different formulas I will list a few below for you. If you have any questions or corrections just let me know.
```
Major 1 3 5
minor 1 b3 5
Major 7th 1 3 5 7
7th 1 3 5 b7
minor 7th 1 b3 5 b7
Augmented 1 3 #5
Diminished 1 b3 b5
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ARTICLE 8
by
Stephen M. Phillips
Flat 4, Oakwood House, 117-119 West Hill Road. Bournemouth. Dorset BH2 5PH. England.
Website: http://smphillips.mysite.com
1. Introduction
As well as having an outer form, the Tree of Life has an inner form hitherto unknown to students of Kabbalah as far as the author is aware. It consists (Fig. 1) of two identical sets of seven regular polygons: triangle, square, pentagon, hexagon, octagon, decagon and dodecagon, all
polygons in each set being enfolded in one another and joined at a common, so-called ‘root edge.’ In Articles 4–7 it was shown that dynamic and structural parameters of the superstring are encoded in several sections of this inner form of the Tree of Life. This is because the geometrical properties of these polygons and their yod populations generated by conversion of their sectors into tetractyses are determined by the number values of the Godnames assigned to the ten Sephiroth of the Tree of Life. This means that they constitute different but equivalent Tree of Life patterns embodying information about the microcosmic manifestation of this universal blueprint in the space-time continuum, namely, the superstring. This article will examine another section of the polygonal form of the Tree of Life, namely, its first four regular polygons. It will show how their properties, too, are prescribed by the Godnames. The way in which this section encodes certain superstring parameters will then be compared with how other sections discussed in previous articles embody them. This set has particular importance because the Tetrad Principle formulated in Article 1 states that numbers of universal (and therefore scientific) significance are the property of either the fourth member or the first four members of a class of mathematical objects (1). This is illustrated by the first four Platonic solids, which embody the numbers 240 and 248 (group parameters of the symmetry group E8 predicted by superstring theory to describe the unified force between superstrings), the structural parameter 168 and the number 137 that determines the fine-structure constant.
2. Properties of the first four polygons
The first four regular polygons enfolded in the Tree of Life are the triangle, square, pentagon and hexagon (Fig. 2). The number of corners and yods in the first four polygons when their sectors are tetractyses are
1
set out below:
triangle square pentagon hexagon Number of corners = 3 4 5 6 Number of yods = 19 25 31 37
The geometrical properties and yod populations of the first four polygons and the first (4+4) polygons are analysed by considering them firstly as separate and then as enfolded. Numbers appearing in boldface in the text indicate various number values of the Sephiroth, their Godnames, Archangels, Orders of Angels and Mundane Chakras. These are tabulated below.
Table 1. Number values of the Sephiroth.
(Cited numbers are in coloured cells)
Sephirah Title Godname Archangel Order of Angels Mundane Chakra Kether 620 21 314 833 636 Chokmah 73 15, 26 248 187 140 Binah 67 50 311 282 317 Chesed 72 31 62 428 194 Geburah 216 36 131 630 95 Tiphareth 1081 76 101 140 640 Netzach 148 129 97 1260 64 Hod 15 153 311 112 48 Yesod 80 49 246 272 87 Malkuth 496 65, 155 280 351 168
4 separate polygons
1. Number of corners of polygons = 3 + 4 + 5 + 6 = 18.
2. Number of sides of polygons = 18.
3. Number of corners & sides of polygons = 18 + 18 = 36.
4. Number of sectors = 18.
5. Number of corners of sectors = 18 + 4 = 22.
6. Number of sides of sectors = 18 + 18 = 36.
7. Number of corners & sides of sectors = 22 + 36 = 58.
8. Number of corners, sides & sectors = 58 + 18 = 76.
9. Number of yods = 19 + 25 + 31 + 37 = 112.
10. Number of hexagonal yods = 15 + 20 + 25 + 30 = 90.
11. Number of yods on boundaries of polygons = 18 + 2×18 = 54; (54–18=36) are hexagonal.
12. Number of yods on boundaries of tetractyses = 18 + 2×18 + 2×18 + 4 = 94; (94–22=72) are hexagonal.
(4+4) separate polygons
1. Number of corners of polygons = 2×18 = 36.
2. Number of sides of polygons = 2×18 = 36.
3. Number of corners & sides of polygons = 2×36 = 72.
4. Number of sectors = 2×18 = 36.
5. Number of corners of sectors = 2×22 = 44.
6. Number of sides of sectors = 2×36 = 72.
7. Number of corners & sides of sectors = 2×58 = 116.
8. Number of corners, sides & sectors = 2×76 = 152.
9. Number of yods = 2×112 = 224. Number of yods other than centres of polygons = 224 – 4 – 4 = 216.
10. Number of hexagonal yods = 2×90 = 180.
11. Number of yods on boundaries of polygons = 2×54 = 108; (2×36=72) are hexagonal.
2
1. Number of yods on boundaries of tetractyses = 2×94 = 188; (2×72=144) are hexagonal).
4 enfolded polygons
1. Number of corners of polygons = 3 + (4–2=2) + (5–2=3) + (6–2=4) = 12; (12–2=10) are outside root edge (“external”).
2. Number of sides of polygons = 3 + (4–1=3) + (5–1=4) + (6–1=5) = 15; (15–1=14) are external.
3. Number of corners & sides of polygons = 12 + 15 = 27; (27–3=24) are external.
4. Number of sectors = 18 – 1 = 17 (the triangle fills one sector of the hexagon).
5. Number of corners of sectors = (3+1=4) + (2+1=3) + (3+1=4) + 4 = 15; (15–2=13) are external.
6. Number of sides of sectors = (3+3=6) + (3+4=7) + (4+5=9) + (5+4=9) = 31; (31–1=30) are external.
7. Number of corners & sides of sectors = 15 + 31 = 46; (46–3=43) are external.
8. Number of sectors & their sides = 17 + 31 = 48.
9. Number of corners, sides & sectors = 46 + 17 = 63; (63–3=60) are external.
10. Number of yods = 19 + (25–4=21) + (31–4=27) + (37–4–6=27) = 94; (94–4=90) are external. Of the 90 yods outside the root edge, (90–3=87) are not Sephirothic points of the outer Tree of Life.
11. Number of hexagonal yods = 94 – 15 = 79; (79–2=77) are external.
12. Number of yods on boundaries of polygons = 12 + 2×15 = 42; (42–4=38) are external.
13. Number of yods on sides of tetractyses = 15 + 2×31 = 77; (77–4=73) are external.
(4+4) enfolded polygons
1. Number of corners of polygons = 2×10 + 2 = 22; (22–2=20) are external.
2. Number of sides of polygons = 2×14 + 1 = 29; (29–1=28) are external.
3. Number of corners & sides of polygons = 22 + 29 = 51; (51–3=48) are external.
4. Number of sectors = 2×17 = 34.
5. Number of corners of sectors = 2×13 + 2 = 28; (28–2=26) are external.
6. Number of sides of sectors = 2×30 + 1 = 61; (61–1=60) are external.
7. Number of corners & sides of sectors = 28 + 61 = 89; (89–3=86) are external.
8. Number of sectors & their sides = 34 + 61 = 95.
9. Number of corners, sides & sectors = 89 + 34 = 123; (123–3=120) are external.
10. Number of yods = 2×90 + 4 = 184; (184–4=180) are external.
11. Number of hexagonal yods = 2×77 + 2 = 156; (156–2=154) are external.
12. Number of yods on boundaries of polygons = 2×38 + 4 = 80; (80–4=76) are external.
13. Number of yods on sides of tetractyses = 2×73 + 4 = 150; (150–4=146) are external. Number of yods on edges of tetractyses other than corners of polygons outside root edge = 150 – 20 = 130.
3. Shared yods & geometrical elements
In this section we shall analyse the properties that both the four polygons and the (4+4) enfolded polygons share with the Tree of Life or with the 1-tree.* We shall illustrate how the number 4, the Pythagorean Tetrad, expresses these shared properties, as it does all the properties of the outer and inner forms of the Tree of Life. Fig. 3 shows that three external corners, four external sides and one triangular sector belonging to each set of the four enfolded polygons are shared with the Tree of Life (or, rather, its projection onto the plane of the polygons). On each side of the root edge are seven shared external corners & sides (7 = 4th odd integer) and eight shared external corners, sides & sectors (8 = 4th even integer). 16 (= 42) geometrical elements outside the root edge of both sets of polygons are shared with the Tree of Life. 14 of them are external corners & sides. The root edge shares its lower end point with the Sephirah Tiphareth — or, rather, its projection. The root edge itself is not shared because it is only part of the projection of the Path connecting Kether and Tiphareth. The number of corners & sides of both sets of polygons which are shared = 14 + 1 = 15. The Godname YAH with number value 15 therefore prescribes how many corners & sides the (4+4) enfolded polygons share with the Tree of Life. Fig. 3 also shows that three external corners, five external
* The n-tree is the lowest n trees of the Cosmic Tree of Life (see Article 5 for definition of the latter).
3
sides and one sector belonging to each set of polygons are shared with the 1-tree, whilst their root edge shares two corners and one side. The total number of geometrical elements shared with the 1-tree = 3 + 2×(3+5+1) = 21. The Godname EHYEH (AHIH) with number value 21 prescribes how many geometrical elements the (4+4) enfolded polygons are shared with the 1-tree. Indeed, its letter values denote different groups of elements:
AHIH A = 1: H = 5: I =10: H = 5: root edge; 5 external edges on one side of root edge; 8 corners + 2 triangles; 5 external edges on other side of root edge
When the (4+4) separate polygons become enfolded, the 224 yods in the former become the 184 yods in the latter, i.e.,
40 = 4 4 4 4 4 4 4 4 4 4
yods disappear. Fig. 4 shows the 70 yods in the 16 triangles of the Tree of Life when they are turned into tetractyses. 20 yods in a set of four polygons are shared with the Tree of Life (36 yods in both sets), of which 15 yods lie on Paths. Thirteen red yods outside the root edge in each set lie on Paths, making a total of 26 such yods for both sets, where 26, the number value of YAHWEH, is the number of combinations of (1+2+3+4=10) objects arranged in a tetractys:
n 1 2 3 4 A B C D E F G H I J number of combinations = 2n – 1 21 – 1 = 1 22 – 1 = 3 23 – 1 = 7 24 – 1 = 15 TOTAL = 26
In the case of the 1-tree, 15 yods outside the root edge in each set of polygons are shared with Paths of the 1-tree (34 yods in total). The number of yods outside the root edge not lying on Paths of the Tree of Life = 180 – 26 = 154. Including the two black yods on the root edge which do not lie on Paths (see Fig. 4), there are 156 such yods, where 156 is the 155th integer after 1. The number of yods unshared with Paths of the 1-tree = 184 – 34 = 150 = 15×10. Of the 79 hexagonal yods of the four enfolded polygons, 14 are shared with Paths of the 1-tree, leaving 65 unshared, hexagonal yods, where 65 is the number value of ADONAI, Godname of Malkuth.
The 16 triangles of the Tree of Life have 10 corners and 22 sides, that is, 48 geometrical elements. Of these, seven corners, eight sides and two triangles (17 elements) are shared with the (4+4) enfolded polygons, leaving 31 unshared, geometrical elements, where 31 is the number value of EL, Godname of Chesed.
The number of sides of the four enfolded polygons = 15. The number of such sides of the 4n polygons enfolded in the n-tree = 15n. The four enfolded polygons have 27 corners & sides. The topmost corner of the hexagon is shared with the lowest corner of its counterpart enfolded in the next higher tree. The number of corners & sides of the 4n polygons enfolded in the n-tree = 26n + 1. As each root edge comprises two endpoints (corners) and one side, the number of corners & sides of the other set of 4n polygons outside their root edges = 26n + 1 – 3n = 23n + 1. The number of corners & sides of the 8n polygons enfolded in the n-tree = 26n + 1 + 23n + 1 = 49n + 2. Each set of four polygons has therefore 15 sides and 26 corners & sides that are intrinsic to it, whilst every set of (4+4) enfolded polygons has 49 intrinsic corners & sides. The 17 tetractyses in each set of four polygons have 31 sides (30 external) and 15 corners with (2×31 + 15) = 77 yods lining their sides, i.e., 73 yods outside their root edge, where 73 is the number value of Chokmah. The 34 tetractyses of the (4+4) enfolded polygons have (2×30 + 1 = 61) sides with (2×73 + 4 = 150 =15×10) yods on them.
4
80 yods are on the boundaries of the polygons and 76 are outside their root edge. Of the 61 sides, 11 are shared with the 1-tree, leaving 50 sides that are unshared, where 50 is the number value of ELOHIM, Godname of Binah. The 1-tree also shares eight corners of its 19 triangles with both sets of polygons (see Fig. 3), whose 34 tetractyses have 89 corners & sides. There are therefore (89–11–8=70) unshared corners & sides, of which 20 are corners and 50 are sides. The 35 unshared corners and sides in each set of polygons comprise 10 corners and 25 sides.
4. How Godnames prescribe the first four polygons
Set out below are ways in which properties of the four polygons and the (4+4) polygons are prescribed by the number values of the Godnames assigned to the ten Sephiroth of the Tree of Life:
Kether: 21 21 geometrical elements in the (4+4) polygons are shared with the 1-tree. Chokmah: 15 4 enfolded polygons have 15 sides and 15 corners of their 17 tetractyses. The number of yods on the boundaries of the 34 tetractyses of the (4+4) polygons = 150 = 15×10. This is also the number of yods in the (4+4) enfolded polygons that do not lie on Paths of the 1-tree. The (4+4) polygons share 15 corners & sides with the Tree of Life. 26 (4+4) enfolded polygons have 26 corners outside their root edge. Every 4 enfolded polygons have 26 corners & sides. Outside their root edge, the (4+4) enfolded polygons share with the Tree of Life 26 yods on its Paths. Binah: 50 Number of corners and sides of (4+4) enfolded polygons = 51 = 50th integer after 1. The 34 sectors of the (4+4) enfolded polygons have 50 sides unshared with the 1-tree; Chesed: 31 Number of sides of 17 tetractyses in 4 enfolded polygons = 31. The Tree of Life has 31 geometrical elements unshared with (4+4) enfolded polygons; Geburah: 36 4 separate polygons have 36 corners & sides and 36 sides of their 18 tetractyses. The polygons have 36 hexagonal yods on their boundaries. The (4+4) separate polygons have 36 corners and 36 sides; Tiphareth: 76 4 separate polygons have 76 geometrical elements. (4+4) enfolded polygons have 76 yods outside their root edge on their boundaries. Netzach: 129 Tetractyses of (4+4) enfolded polygons have 130 yods on their boundaries other than corners outside their root edge, where 130 = 129th integer after 1; Hod: 153 (4+4) enfolded polygons have 154 hexagonal yods outside their root edge, where 154 = 153rd integer after 1; Yesod: 49 Every (4+4) enfolded polygons have 49 intrinsic corners & sides; Malkuth: 65 65 hexagonal yods in the 4 enfolded polygons are unshared with Paths of the 1-tree. (4+4) enfolded polygons have 156 yods unshared with Paths of the Tree of Life, where 156 = 155th integer after 1.
The natural way in which the Godname numbers appear in the above analysis of the geometrical properties of the polygons and their yod populations refutes the argument that their presence lacks real significance because it was contrived by various selections of these properties.
5. Connections between the 1-tree and the first four polygons
Having established that the ten Godnames prescribe the first four of the seven polygons and therefore define it as a ‘Tree of Life pattern,’ we will now explore their correspondence to the 1-tree.
The four enfolded polygons have 12 corners, of which ten are outside the root edge, one of them — the uppermost corner of the hexagon — being shared with the lowest corner of the hexagon enfolded in the next higher tree. Each set of 4n polygons enfolded in the n-tree has (9n+1) corners outside their n root edges. (10n+1) corners are associated with each set of 4n polygons. ADONAI, the Godname of Malkuth, prescribes the 10-tree because its number value 65 is the number of Sephirothic emanations in the 10-tree (what in previous articles were called ‘Sephirothic levels,’ or SLs). Enfolded on either side of its central pillar are 40 polygons of the first four types associated with which are (10×10 + 1 = 101) corners, 91 of them being outside their root edges. 101 is the 26th prime number and the number value of Michael, the Archangel of Tiphareth. The 25-tree is prescribed by ADONAI MELEKH, the full Godname of Malkuth, because its number value 155 is the number of SLs in the 25-tree. 100 polygons of the first four types with (10×25 + 1 = 251) associated corners are enfolded on either side of the central pillar of the 25-tree. The two words ADONAI and MELEKH prescribe its division into the 10-tree and the 15 trees above it:
251 = 101 + 150,
the number 101 denoting the 10 endpoints of root edges associated with the 40 polygons and their 91 external corners. Eleven of the latter are the highest and lowest corners of the 10 joined hexagons,
5
Figure 5. Correspondence between the 251 yods in the 1-tree and the 251 corners of the first four
types of polygons enfolded in the 25-tree prescribed by ADONAI MELEKH, the Godname of Malkuth.
6
Figure 6. Correspondence between the 251 yods in the 1-tree and the 251 corners of
the first six types of polygons enfolded in the 10-tree that is prescribed by ADONAI.
7
10 of them belonging exclusively to the hexagons enfolded in the 10-tree and one being also the lowest corner of the hexagon enfolded in the 11th tree. The number 101 therefore has the geometrical differentiation:
101 = 10 + 91 = 10 + 11 + 80,
where
80 = 10×(1+2+3+2)
is the number value of Yesod and ‘1’ denotes the corner of the triangle in each tree outside their root edges, ‘2’ denotes the two corners of the square, ‘3’ denotes the three external corners of the pentagon and ‘2’ denotes the two external corners of the hexagon that are unshared with adjoining hexagons. Therefore,
80 = 10×(1+2) + 10×(3+2) = 10×3 + 10×5 = 30 + 50,
where 30 is the number of external corners of the first two polygons enfolded in the 10-tree and 50 is the number of corners of the last two polygons enfolded in the 10-tree. Fig. 5 displays the types of corners of the 100 polygons of the first four types enfolded on one side of the 25-tree prescribed by ADONAI MELEKH. It also shows that, when its 19 triangles are turned into tetractyses, the 1-tree contains 80 yods (30 yods belong to the Lower Face formed by Tiphareth, Netzach, Hod, Yesod and Malkuth, leaving 50 yods in the Upper Face) and that, when the three sectors of each triangle are turned into tetractyses, the 1-tree contains 251 yods — the same as the number of corners of the 100 polygons. The reason why the two remarkable parallels:
251 yods in 1-tree 251 corners of first 4 polygons enfolded in 25-tree
(30+50= 80) yods in 1-tree (30+50=80) outer corners of first 4 polygons enfolded in 10-tree
exist is that, being the lowest of the 91 overlapping trees making up the Cosmic Tree of Life and therefore its most ‘Malkuth’ level, the 1-tree embodies the same parameters as any section of CTOL that bears a formal correspondence to Malkuth — in this case, the 25-tree, whose 25 trees are the counterpart of the 25 tree levels of the 7-tree mapping the physical plane, the lowest of the seven planes that corresponds to Malkuth. In Articles 2 and 5, these tree levels were interpreted as the 25 spatial dimensions that quantum mechanics predicts for spinless strings. This is why the Godname ADONAI MELEKH assigned to Malkuth (physical universe) refers to the 25-tree and why the Godname ADONAI prescribes its ten lowest trees, corresponding to which are the 10 tree levels signifying the ten spatial dimensions of 11-d supergravity space-time. Analogous structures defined by the set of Godname numbers — whether of the outer or the inner form of the Tree of Life — must embody the same numbers and display the same pattern of differentiation of whatever these numbers signify because they are equivalent, holistic objects that embody the divine paradigm. This is why the first six polygons enfolded on either side of the 10-tree have 251 corners (Fig. 6), for the (6+6) polygons also constitute a Tree of Life pattern (see Article 4 (2) for their prescription by the ten Godnames). The
triangles have 30 corners, the squares & pentagons have 50 corners and the hexagons, octagons & decagons have 171 corners. This is the same 30:50:171 pattern as displayed by the yods in the 1-tree (see Fig. 5).
6. Encodings of 10-whorl structure of superstring
What is the meaning of the ubiquitous, geometrical encoding of the number 251 and its division into the numbers 80 and 171? These numbers have a remarkable interpretation in terms of the ten-fold structure of the basic unit of matter described (3) by the Theosophists Annie Besant and C.W. Leadbeater over a
8
century ago with the aid of a yogic siddhi (psychic ability) with the Sanskrit name of ‘anima.’ Moreover, they support the author’s theory of superstrings (4) derived from higher-dimensional, extended objects called ‘D-branes,’ as will be explained shortly. En passim, it should be pointed out that the theory has not been tailored in order to procure this agreement. It was conceived by the author for purely scientific reasons long before he discovered that these numbers characterise the outer and inner forms of the Tree of Life.
Magnified with what the author has called ‘micro-psi’ (5), the basic constituent of atoms, which Besant and Leadbeater called the ‘ultimate physical atom’ (UPA), were seen to consist of ten closed curves, or ‘whorls’ (Fig. 7). These spiral in 2½ revolutions in parallel tracks and separate at the bottom of the particle into sets of seven and three curves, which then twist 2½ times in opposite directions about the axis of spin of the UPA before returning to its top. Besant and Leadbeater noticed two types of UPAs: a ‘positive’ variety in which the whorls spiral downwards clockwise as observed from its top and a ‘negative’ type in which they wind around their axis in an anticlockwise sense. Each is the mirror image of the other. Three, so-called ‘major’ whorls appear thicker than the remaining seven, so-called ‘minor’ whorls. The reason for this is as follows: each circular turn in a stringy whorl is a circular helix made up of seven smaller turns spaced the same distance apart. Each of these is another helix with seven turns, and so on. There are seven orders of helices. Every 25 helical turns of a given order in a major whorl comprise 176 turns of the next higher order, whereas in a minor whorl they consist of 175 such turns. This augmentation of one extra turn in every 25 of the next lower order extends throughout the seven orders of helices in a major whorl, making it consist of more higher-order helices and appear thicker than a minor whorl.
Each of the ten whorls was found to be essentially a circular helix with 1680 turns or coils (Fig. 8). Leadbeater said (6) that he checked his count of these coils by studying 135 different UPAs, which were found to have the same number of turns in their whorls whatever the elements in which they were found.
Statistical analysis of the UPA populations determined by Besant and Leadbeater for all 111 of what they assumed were chemical atoms, as well as detailed correlation of their constituent particles with predictions based upon facts about nuclei and their quark composition, established (7) that the UPA is a constituent of the up and down quarks making up protons and neutrons in atomic nuclei. The string-like nature of the whorls is self-evident. In fact, were it not for the fact that the UPA comprises ten stringy whorls, not one whorl, its identification with what physicists call the ‘superstring’ would be just as obvious. Superstring theory predicts that space-time has ten dimensions, so that a microscopic, 6-dimensional space exists beyond ordinary, large-scale space. One of the models for this space that string theorists have considered is the so-called ‘6-d torus.’ The 2-torus, or doughnut, is the surface generated when the centre of a circle moves around another circle (1-torus) in a plane at right angles to it. The 6-torus is its 6-dimensional version. The six higher orders of helical spirillae in each whorl represent the winding of a closed string around successively smaller, mutually perpendicular circles, each a 1-dimensional space. In other words, Leadbeater’s description of the higher-order structure of the UPA is consistent with this type of space. However, he described the UPA not as one closed string but as ten closed curves. If superstrings were fundamental, he would have observed only one closed whorl. This indicates that the current picture of superstrings as simple loops winding around some compact, 6-dimensional space is just that — a simplistic version of the truth. Instead, they must be derived from more general, extended objects called ‘D-branes.’
Some string theorists have suggested that 1-dimensional strings may result from the wrapping of D-branes around a curled-up dimension. But this cannot be one of the six curled-up dimensions predicted by superstring theory because each string-like whorl winds itself around all six of these circular dimensions, not five of them. Hence space-time must have more than ten dimensions. There are five types of superstrings, and one of them has been shown (8) to result from the wrapping of a 2-dimensional sheet (2-brane) around one of the ten spatial dimensions predicted by supergravity theories. But this still creates only one string, whereas Leadbeater’s investigations imply that superstrings actually consist of ten separate, closed curves. The only possibility is for space to have more than ten dimensions, the wrapping of a D-brane around the extra dimensions being responsible for these curves. The only candidate available is the 26-dimensional space-time predicted for spinless strings by quantum mechanics but rejected by physicists for many years until the so-called ‘heterotic superstring model’ was proposed. In Article 2 (9), it was proposed that a 11-brane (a 11-dimensional object) existing in 26-dimensional space-time wraps itself around ten of the 15 higher, curled-up dimensions beyond supergravity space-time, the topology of this 10-dimensional space creating ten non-intersecting curves whose separation is an illusion because they are simply the projection into superstring space-time of a single, higher-dimensional, extended object. Imagine a 2-dimensional being living on a sheet. As he is unaware of the third dimension of space, he would perceive a cylinder with thick walls that penetrated the sheet at right angles to it as two concentric circles that would move together but keep separate. He would have no way of knowing that they were part of one object. Instead, he would believe that they were different objects. In the same way, the ten whorls of the UPA exist as separate objects only in the 11-dimensional space of supergravity space-time; they are
9
really part of one object that extends into 15 higher dimensions.
The author’s theory has the following consequence: just as the position of a point in large-scale space is defined by three numbers — its spatial co-ordinates — so a point in 25-dimensional space is located by 25 numbers. Any point on a curve in 10-dimensional space is located by ten co-ordinates. But if the curve has been created by a D-brane wrapping itself around the curled-up dimensions of a higher space, then there are 15 hidden co-ordinate variables defined for that point. Ten different curves will have (10×10=100) spatial co-ordinate variables in supergravity space-time and (10×15=150) higher co-ordinate variables.Including the time co-ordinate, which is common to all ten curves, there are:
100 + 150 + 1 = 251
co-ordinate variables defining the ten curves, of which 101 variables define them in 11-dimensional, supergravity space-time and 150 variables remain hidden because they refer to the space beyond this space-time.
This explains why there are 251 yods in the 1-tree with its triangles turned into three tetractyses and why the 60 polygons enfolded in ten overlapping Trees of Life have 251 corners, as described earlier. Each yod or corner symbolises one of the numbers or co-ordinate variables needed to define ten separate points in 26-dimensional space-time — the geometrical origin of the superstring as the Malkuth manifestation of the Tree of Life blueprint. The reason why the 1-tree with single tetractyses contains 80 yods is that each of the ten curves that comprise the superstring has eight transverse spatial co-ordinates, so that the superstring itself has (10×8=80) such variables or geometrical degrees of freedom. A single Tree of Life has 70 yods. Its ten Sephirothic yods denote the ten longitudinal coordinates of points on ten closed curves and its 60 other yods denote their 60 coordinates in the 6-dimensional, compactified space in which superstrings exist as superstrings.
We saw earlier that the 251 corners of the first four polygons enfolded in the 25-tree split up into the ten corners of the root edges in the 10-tree, the 11 uppermost and lowermost corners of the hexagons enfolded in the 10-tree, 80 external corners and 150 corners of the 60 polygons enfolded in the 15 trees above the 10-tree. The 11 hexagonal corners symbolise the time co-ordinate and the ten longitudinal co-ordinate variables of the ten curves comprising the superstring. The ten corners of the root edges denote their coordinate variables defined with respect to the tenth dimension of supergravity space-time and the 150 corners signify the 10×15 = 150 co-ordinate variables ‘hidden’ so to speak in the ten curves because they refer to the space whose 15 dimensions beyond supergravity space-time correspond to the 15 trees in the 25-tree above the 10-tree. The ten independent corners in each set of four polygons symbolise the ten curves, whist similar corners denote different co-ordinates of the same curve. The 101 corners of the polygons enfolded in the 10-tree denote the (10×10 + 1 = 101) space-time co-ordinate variables of the
Figure 9. The seven enfolded polygons have 251 yods outside the
root edge that are either not Sephirothic points or centres (O).
10
ten curves of the superstring in 11-dimensional space-time and the 150 corners of the polygons enfolded in the 15 trees of the 25-tree beyond the 10-tree symbolise the (10×15=150) co-ordinate variables of the ten curves defined with respect to the 15-dimensional space beyond supergravity space-time.
The number 251 is encoded in the seven enfolded polygons as follows: this set of polygons contains 260 yods outside their root edge (10). Of these, three are located at the positions of Chokmah, Chesed and Netzach in the Tree of Life and six are centres of the polygons ((the yod coinciding with Chesed is the centre of the hexagon). There are therefore (260–3–6=251) yods in the seven enfolded polygons outside their root edge that are not Sephirothic points or centres (Fig. 9).
7. Encoding of 168 as the structural parameter of the superstring
We found in Section 2 that the first four enfolded polygons have 90 yods outside their root edge. Of these, three are located at Sephiroth and three are centres of these polygons. The number of their yods outside the root edge which are not Sephirothic points or centres of these polygons = 90 – 3 – 3 = 84, where
84 = 12 + 32 + 52 + 72,
i.e., the sum of the squares of the first four odd integers, showing how the Pythagorean Tetrad determines this number. The two sets of four polygons therefore have (84+84=168) such yods. This is the number value of Cholem Yesodeth, the Mundane Chakra of Malkuth. (90–3=87) yods outside the root edge of the first four enfolded polygons do not coincide with Sephiroth. This is the number value of Levanah, the Mundane Chakra of Yesod, which is the Sephirah next above Malkuth in the Tree of Life. That this particular Sephirah is involved is highly significant and yet more evidence of how information about the subatomic world is encoded in the Tree of Life and its equivalent sections. This is because Malkuth signifies the outer, physical form of whatever is designed according to the blueprint of the Tree of Life. It is therefore appropriate that 168 is the kernel of the number 1680 — the number of coils in each helical whorl of the UPA described by Besant and Leadbeater with a form of remote-viewing and proved (11) by the author to be the superstring constituent of up and down quarks — for this number quantifies the form of the superstring — the basic unit of physical matter. Ten overlapping Trees of Life have 80 polygons of the first four types containing (10×168=1680) yods that are not Sephirothic points or centres. This demonstrates that the number 1680 is truly a parameter of the Tree of Life, for it quantifies a property of a section of the inner form of ten Trees of Life, each a representation of a Sephirah. In fact, as the UPA/superstring is the microphysical manifestation of the Tree of Life blueprint, each whorl is the corresponding manifestation of a Sephirah, the three major whorls corresponding to the Supernal Triad of Kether, Chokmah and Binah and the seven minor whorls corresponding to the seven Sephiroth of Construction. As the microscopic manifestation of a Sephirah, a whorl is a Tree of Life in itself, so that it is represented by ten Trees of Life. As a circularly polarised standing wave, its 1680 oscillations are the manifestation of the 1680 yods in the first (4+4) polygons enfolded in ten overlapping Trees of Life other than its centres or corners that coincide with SLs of the ten trees.
The significance of the excluded yods is that they belong to the 21 yods prescribed by the Godname EHYEH that are either Sephiroth, Daath or centres of the two sets of five independent polygons whose centres do not coincide with any of their corners (Fig. 10). In fact, the letter values of EHYEH denote the various classes of such yods. There are seven of these yods per set of four enfolded polygons (five outside the root edge), so that the set of 40 polygons enfolded on each side of the central pillar of the ten trees have 71 yods that are either Sephirothic points or centres (the significance of this number will be revealed shortly). Seventy of these yods are intrinsic to the ten trees because the topmost corner of the hexagon enfolded in the tenth tree coincides with the lowest corner of the hexagon enfolded in the eleventh tree. 50 of these yods are outside the ten root edges and 20 are their endpoints. ELOHIM, Godname of Binah with number value 50, prescribes the yods that are either Sephirothic points or centres of polygons (see Fig. 4). The counterpart of this 50:20 division in a single Tree of Life is the 20 yods in the tetrahedron with corners at Netzach, Hod, Yesod & Malkuth and the 50 yods outside it. The division, which (as later articles will
11
Figure 11. 168 yods in the first (4+4) enfolded polygons and 168 yods
in the last (3+3) polygons are not centres or Sephirothic points (O).
demonstrate), is characteristic of holistic systems, appears in the formula for the number Y(n) of yods in n overlapping Trees of Life:
Y(n) = 50n + 20.
We saw earlier that there are 251 yods outside the root edge of the seven enfolded polygons that are not Sephirothic points or centres. 84 yods outside the root edge of the first four enfolded polygons are not centres of
their polygons but a corner of the pentagon is the centre of the decagon, so that there are 83 yods outside the root edge that not Sephirothic points or centres of any of the seven polygons. There are therefore (251–83=168) yods in the last three enfolded polygons outside their root edge that are not Sephirothic points or centres, whilst (including the root edge) there are (84+84=168) yods in the first (4+4) polygons that are not such. Now consider the root edges in overlapping trees. The yod at their lower ends is at the position of Tiphareth of that tree and the yod at their upper ends coincides with Daath, i.e., Yesod of the next higher tree. One of the two remaining yods of the root edge may be considered to be associated with one set of polygons and the other may be considered to be associated with the other set. This means that there are 168 yods in the first (4+4) enfolded polygons that are not Sephirothic points or centres of any polygon. There are therefore (168+168+168=504) yods in the (7+7) enfolded polygons that are not Sephirothic points or centres (Fig. 11). The (70+70) polygons enfolded in ten overlapping trees have (10×504=5040) such yods. This number has the property:
5040 = 712 – 1 = 3 + 5 + 7 + …. + 141.
In other words, 5040 is the sum of the first 70 odd integers, starting with 3. This number, which we shall shortly show is a structural parameter of superstrings, is prescribed by ELOHA, the Godname of Geburah with number value 36 because 71 is the 36th odd integer. As a Tree of Life contains 70 yods when its
12
16 triangles are turned into tetractyses (see Fig. 4), we discover the amazing property that the number of yods other than Sephirothic points or centres in the polygons enfolded in ten trees is the sum of the odd integers that can be assigned to the yods in a single Tree of Life (Fig. 12). Its Lower Face (shown shaded) has 30 yods, the rest of the Tree of Life having 40 yods. The sum of the 40 odd integers 3, 5, … 81 outside the Lower Face is 412 –1 = 1680, so that the sum of the 30 integers composing the Lower Face is 5040 – 1680 = 3360 = 2×1680. Numerically, therefore, the encoding of the number 5040 in the Tree of Life causes it to split into the numbers 1680 and 3360. Compare this with the fact that the 5040 yods in the polygons enfolded in 10 trees which are not Sephirothic points or centres comprise the 1680 such yods in the first (4+4) polygons enfolded in each tree and the 3360 yods of the last (3+3) polygons (see above). The Lower Face of the Tree of Life creates the same split (3360+1680) as that created by the division of the seven polygons into, respectively, the last three ones and the first four ones. Notice that this has not been concocted, for the integers are assigned in Fig. 12 sequentially from left to right, running down the page. Also, notice that the sum of the integers at the position of the ten Sephiroth is
3 15 21 49 59 83 107 97 125 141 = 700 = 70 70 70 70 70 70 70 70 70 70,
i.e., the sum of the Pythagorean Decad assigned to each of the 70 yods in the Tree of Life! This exemplifies the beautiful, mathematical design of the Tree of Life.
What the replication of the pattern of encoding of the number 5040 is telling us (quite apart from the importance of the number itself) is that the numbers 1680, 3360 and 5040 must have significance vis-à-vis the superstring as the microphysical actualisation of the Tree of Life blueprint. In fact, 1680 is the number of helical turns of each whorl component of the superstring, being the number of oscillations of the circularly polarised waves running around each closed curve. 3360 is the number of such turns per revolution of all ten whorls (each whorl makes five revolutions, comprising 336 turns per revolution), whilst 5040 (=3×1680) is the number of turns in the three major whorls of the UPA. The 1680 yods in the 80 polygons of the first four types enfolded in the ten trees symbolise the 1680 turns of the first major whorl, which corresponds to Kether in the Tree of Life. The 1680 yods in the 30 polygons of the last three types that are enfolded in the ten trees signify the 1680 turns of the second major whorl, which corresponds to Chokmah. The 1680 yods that are their mirror images in the 30 polygons of the last three types enfolded on the
Figure 13. The number (3360) of 1st-order spirillae in each revolution of the 10 whorls of the UPA/superstring is the number of yods in the seven enfolded polygons with 2nd-order tetractyses as their sectors. They comprise 1680 hexagonal (black) yods of tetractyses denoting Sephiroth of Construction inside the 40 sectors of all polygons except the hexagon, as well as 1680 yods that either line its edges or belong to the hexagon or tetractyses at corners of each 2nd-order tetractys.
other side of the central pillar in the ten trees correspond to the 1680 turns of the third major whorl, which corresponds to Binah. The two mirror-image halves of the inner Tree of Life are the manifestation of the opposite polarities of Chokmah and Binah, which have the Kabbalistic titles of Abba and Aima, the Cosmic Father and Cosmic Mother, representing the male and female principles (what are called ‘yang’ & ‘yin’ in Taoism).
The structural parameter 5040 is embodied in the inner form of a single Tree of Life. When the sectors of its
13
14
seven enfolded polygons are transformed into 2nd-order tetractyses, each with 85 yods:
the set of polygons contain 3360 yods (Fig. 13) (12). 1680 yods either lie on edges of sectors or belong to 1st-order tetractyses at the corners of each 2nd-order tetractys. There are 840 yods inside the sectors of the triangle, octagon & decagon and 840 yods inside the sectors of the square, pentagon & dodecagon, each of these sets of polygons having 21 sectors. Transformed by the 2nd-order tetractys, the seven enfolded polygons exhibit the same 1680:1680 division of yods as found in the last (3+3) polygons enfolded in ten overlapping Trees of Life. The same pattern occurs in the first (6+6) polygons enfolded in ten trees (Fig. 14). Each set of six enfolded polygons has 195 yods, of which 26 are corners, leaving 169 yods. Associated with each set are 168 yods, so that there are (1680+1680=3360) yods in the 120 polygons of the first six types enfolded in ten trees other than their 482 corners. Each set of 60 polygons has 251 corners that symbolise the 251 space-time coordinates of ten independent points in 26-dimensional spacetime. The Tree of Life parameter 251, which was found in Section 5 to be embodied in the 1-tree as its 251 yods and in Section 6 to be the number of yods outside the root edge of the seven enfolded polygons other than Sephirothic points or centres, reappears again in a new Tree of Life pattern as the number of corners of the 60 polygons enfolded in 10 overlapping Trees of Life that contain 1680 yods.
8. Conclusion
The Tree of Life has an inner form defined by its geometry and prescribed by the number values of the ten Kabbalistic Godnames. As demonstrated in earlier articles, various sections of this inner structure are also prescribed by the Godname numbers and encode the same set of parameters quantifying their geometrical properties and yod populations. This article has analysed one such section — the first four regular polygons enfolded in the outer Tree of Life — and has proved that it encodes a number embodied in both the outer and inner forms of the Tree of Life as the number of degrees of freedom or co-ordinate variables characterising ten curves in the 26-dimensional space-time predicted by quantum mechanics for spinless strings. This agrees with the century-old, paranormal description of the basic constituent of matter by the Theosophists Annie Besant and C.W. Leadbeater and with its interpretation by the author as the superstring constituent of up and down quarks. Independent confirmation of this came from the appearance of the paranormally obtained number 1680 as a natural property of both sets of the first four polygons and as a similar property of the last three polygons. Such simultaneity cannot plausibly be due to coincidence because it is obvious that the chance of the same number happening to appear in two different sets of polygons making up the seven polygons is extremely small — even more so when choice of both combinations is restricted by the number values of the ten Godnames. What this and previous articles have presented is evidence of an enormous ‘conspiracy’ that theologians might prefer to call ‘divine design’ whereby number and geometry join together in the mathematical design of the cosmic blueprint called the Tree of Life and its microscopic manifestation in space-time as the superstring. However, as we have outlined here and as Article 5 discussed in more detail, the superstring is only the end of the story, not its beginning……
References
1. Phillips, Stephen M. Article 1: “The Pythagorean nature of superstring and bosonic string theories,” (WEB, PDF).
2. Phillips, Stephen M. Article 4: “Godnames prescribe the inner Tree of Life,” (WEB, PDF).
3. Occult Chemistry, Annie Besant and C.W. Leadbeater (1st ed., 1908; 2nd ed., 1919; 3rd ed., 1951; reprinted 3rd ed., 1994).
4. For a non-technical overview, See Article 2 at the author’s website.
15
1. Extra-sensory Perception of Quarks, Stephen M. Phillips (Theosophical Publishing House, Wheaton, Ill., U.S.A., 1980).
2. Occult Chemistry, 3rd ed., p. 23.
3. Ref. 5 and ESP of Quarks and Superstrings, Stephen M. Phillips (New Age International, New Delhi, India, 1999).
4. For example, see: “Superstrings in D = 10 from supermembranes in D = 11,” M.J. Duff, P.S. Howe, T. Inami and K.S. Stelle, Phys. Lett. B, vol. 191 (June 1987), 70–74, and: “Heterotic and type I string dynamics from eleven dimensions,” P. Horava and E. Witten, Nucl. Phys. B 460 (1996), 506–524.
5. Phillips, Stephen M. Article 2: “The physical plane and its relation to the UPA/superstring and spacetime,” (WEB, PDF).
6. Ref. 2, p. 2.
7. Ref. 7.
8. Proof: The 2nd-order tetractys has 85 yods, of which 13 yods line each of its sides. When each of the n triangular sectors of an n-sided, regular polygon is turned into a 2nd-order tetractys, there are (85–13=72) independent yods per sector of the polygon. Its yod population = 72n + 1, where “1” denotes the yod at the centre of the polygon. The polygonal form of the inner Tree of Life consists of a triangle, square, pentagon, hexagon, octagon, decagon and dodecagon. They are enfolded in one another and share the same base, or what the author has called the “root edge,” as they should be thought of as growing out of this fundamental line joining Daath and Tiphareth in the Tree of Life. When the seven separate polygons are superposed on one another in their enfolded state, corresponding members of the set of 13 yods forming what becomes their shared side coincide and therefore must not be counted separately in a calculation of their yod population. Below are listed the yod populations of each polygon and (except for the triangle) their numbers of yods outside the root edge:
Polygon n Number of yods = 72n + 1 Number of yods outside root edge triangle square pentagon hexagon octagon decagon dodecagon 3 4 5 6 8 10 12 217 289 361 433 577 721 865 289 – 13 = 276 361 – 13 = 348 433 – 13 = 420 577 – 13 = 564 721 – 13 = 708 865 – 13 = 852 Total = 3385
Inspection of Fig. 2 reveals that the tip of the triangle opposite the root edge is also the centre of the hexagon (the triangle is simply a triangular sector of the hexagon).
Similarly, the tip of the pentagon is the centre of the decagon. With 2nd-order tetractyses as their sectors, the centre of the triangle where corners of its three 2nd-order tetractyses meet is also the central yod of the tetractys at the centre of the 2nd-order tetractys constituting a sector of the hexagon (see diagram opposite). The 11 yods between corners on each of the two sides of the triangle outside its shared base coincide with yods on the sides of this sector of the hexagon. There are (1 + 1 + 1 + 2×11 = 25) yods in the total population calculated above that coincide with yods belonging to other polygons (these are the only yods occupying the same positions). In determining the yod population when the separate polygons are superposed, these yods must be subtracted in order to avoid double-counting. Therefore, the yod population of the seven enfolded polygons constructed from 2nd-order tetractyses = 3385 – 25 = 3360.
16
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The merchandise turnover ratio, also called the inventory turnover ratio, tells business owners how efficiently their company moves products. In general, a high merchandise turnover ratio is preferable to a low one, but each has its pros and cons. Analyzing the merchandise turnover ratio helps business owners determine liquidity, highlight top-selling products, pin-point poorly performing items, and create a plan of action to increase product movement. The calculations you must perform to determine the merchandise ratio turnover are simple, as is the analysis process.
## Calculating the Merchandise Turnover Ratio
Add the value of the inventory at the beginning of the reporting period and the value of the inventory at the end of the reporting. Divide the resulting number by two to get the average inventory value. Divide the average inventory value by the net sales amount, or how much the company earned after deducting the costs of damaged goods, returns, discounts and theft. The resulting figure is the number of times the company sells and replaces the entire inventory within the reporting period, or the merchandise turnover. Express this as the ratio X-to-1 with “X” being the merchandise turnover figure.
## High Merchandise Turnover Ratio
A high inventory ratio indicates that the business is moving inventory efficiently and selling at sufficient levels. This increases liquidity because funds are not tied up in merchandise that is not selling. If the merchandise ratio is too high, though, a company could face shortages and run the risk of running out of stock before it is able to replenish its inventory.
## Low Merchandise Turnover Ratio
A low merchandise turnover ratio can indicate that the company is overestimating its inventory needs or is not making adequate sales. A low merchandise ratio may also be the result of preparing for seasonal sales or stocking up on discontinued items.
## Plan of Action
Analyzing the merchandise turnover ratio requires business owners to examine sales receipts and individual item inventory records to determine which merchandise moves more quickly than others. Once inventory productivity has been established, business owners must determine how to move stagnant merchandise. Options include sales, discounts, and product give-aways with minimum purchase amounts. Companies that are unable to move stock may be able to return the goods to the wholesaler or sell them to another retailer. A final option is donating the goods to charity and taking a deduction on the company's tax return. | 478 | 2,607 | {"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-2018-17 | latest | en | 0.924461 |
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## Intro
Each character in the Sonic games is constructed differently. While similar, they differ in size and moveset.
The sizes of characters are important, they determine how the characters will collide with Solid Tiles, Solid Objects and more.
These are not hitboxes, hitboxes are separate and aren't related to the solid size of objects. Hitboxes will be covered in Solid Objects
For all characters, their Push Radius is always 10.
## Sonic
When standing, Sonic's Width Radius is 9 and his Height Radius is 19, resulting in 19 pixels wide and 39 pixels tall.
When Jumping or rolling, his Width Radius is 7 and his Height Radius is 14, resulting in 15 pixels wide and 29 pixels tall.
Sonic's jump force (jmp) is 6.5.
### Drop Dash (Mania)
The drop dash in Sonic Mania isn't simply an instant landing spindash.
To charge Sonic up you release then hold the jump button while already jumping. The dash will take 20 frames to charge and once charged, when Sonic reaches the ground, your ground speed will be set. If the jump button is released before Sonic hits the ground, the move is cancelled.
As Sonic hits the ground, Sonic's Ground Speed is calculated as normal. So at this point, Sonic will have a Ground Speed value (as if he landed normally) which will be used in calculating the drop dash speed.
The game checks if you were moving backwards in the air. By backwards, I mean opposite to the way you were facing/pushing. For example, if your X Speed was positive but you were holding & facing left at the time this would count as backwards, same for the other direction. In either case, Sonic's actual direction is then set to that which you are facing/holding.
#### If you were moving forwards
Sonic's Ground Speed is set to his Ground Speed divided by 4, plus (or minus, depending on direction) drpspd. This speed is limited to drpmax.
``` Ground Speed = (Ground Speed / 4) + (drpspd * direction) //direction is either 1 (right) or -1 (left), this speed would then be limited between the min and max values of -drpmax and drpmax
```
#### If you were going backwards
If Sonic's ang is 0 (flat), his Ground Speed is simply set to drpspd (or negative drpspd)
``` Ground Speed = drpspd * direction
```
Otherwise, on slopes, his speed is set in the same way as normal, but divided by 2 rather than 4. Of course, because you were moving backwards your Ground Speed will be opposite direction of that which the dash is trying to propel you, so this results in a rather slow dash.
``` Ground Speed = (Ground Speed / 2) + (drpspd * direction) // this speed would then be limited between the min and max values of -drpmax and drpmax
```
A similar Camera Lag effect to that used when spin dashing is used here too, to make the move more dramatic.
### Dash (Super Peel Out)
Once the button is pressed, Sonic will begin charging the Dash. After 30 steps have passed, he is ready to go. When the Up button is released, Sonic launches at a speed of 12. If the Up button is released early, nothing happens.
## Insta-Shield
The Insta-Shield expands Sonic's hitbox giving it a width radius of 24 and a height radius of 24, resulting in an overall height of 49 x 49 (more about Hit Boxes). This lasts for 13 frames.
The Insta-Shield does nothing to Sonic's X Speed or Y Speed. It does however allow him to control a rolling jump.
## Tails
Tails is much smaller than the other characters. When standing, his Width Radius is 9 and his Height Radius is 15, resulting in 19 pixels wide and 31 pixels tall.
The only time this changes is when he jumps or rolls, where his Width Radius is 7 and his Height Radius is 14, much like Sonic, resulting in 19 pixels wide and 29 pixels tall.
His size is the same as standing when he flies.
Tails' jump force (jmp) is 6.5.
### Flying
When Tails begins to fly, his Y speed is unaffected. However, since Tails has to release the button in order to press it again to fly, he can't possibly fly up faster than -4.
While flying, the variables are much like a standard jump. He accelerates at 0.09375, and there is no separate deceleration value. The normal air drag calculation is performed, which means Tails can't fly horizontally as fast while moving upward than when moving downward. The air drag cancels out the acceleration at an X speed of 3. There is no air drag while moving down, though, so he can reach an X speed of 6, the normal maximum.
While flying, gravity is 0.03125. Pressing Up or Down doesn't decrease or increase it.
Pressing the button doesn't cause an immediate loss of Y speed (like a double-jump), but instead a temporary change in gravity. Gravity becomes -0.125, and remains so until Y speed is less than -1. Then gravity returns to normal in the next step and Tails begins to fly back down. If Y speed is already less than -1, pressing the button does nothing.
Tails can only fly for 480 frames, or 8 seconds, before getting tired. The only difference being tired makes (besides the pooped-out expression) is that pressing the button doesn't have any effect anymore. Gravity, and all other variables, remain the same.
As stated above if you have negative gravity, a Ysp smaller than -1 is needed to return to positive gravity. This can cause issues when you hit a ceiling and your Ysp is set to 0. Your gravity will remain negative and you will be stuck. In your engine, to prevent Tails from being stuck in negative gravity, you should reset the gravity to the positive value when a ceiling is detected.
Note: Tails' tails deflect projectiles (just like the Shields do) while he is flying.
## Knuckles
Knuckles sizes are the same as Sonic's.
Well that is except for when he is gliding, climbing and sliding.
Here, his Width Radius is 10 and his Height Radius is also 10, resulting in 21 pixels wide and 21 pixels tall. This makes him very wide but very slim compared to normal, which makes sense for his gliding pose. When falling from a glide, he uses his standing sizes.
Knuckles' jump force (jmp) is only 6, which results in a much lower jump than the others.
### Gliding
When Knuckles first begins gliding, his X speed is set to 4 in the direction he is facing. Y speed is set to 0, but only if it was negative at the time, otherwise it is unaffected. X speed then accelerates by 0.015625 every step.
Gliding has a top speed of 24. This top speed is so high that it is unreachable anywhere in the game -- except for Mushroom Hill Zone Act 1, where Super/Hyper Knuckles can glide across the top of the level to achieve this speed.
During the glide, gravity is 0.125, which is weaker than usual. Also, unlike a normal jump, gravity is only added while Y speed is less than 0.5. If Y speed is higher than that (say Knuckles was falling quickly when he began to glide), gravity is subtracted from Y speed instead, slowing his descent.
```if (ysp < 0.5) ysp += 0.125;
if (ysp > 0.5) ysp -= 0.125;
```
When you let go of the button, Knuckles drops, and his X speed is multiplied by 0.25. When he hits the ground, it is set to 0. While dropping from a glide, gravity is the normal value, 0.21875.
If you don't release the button, but allow Knuckles to glide into the ground and slide on his stomach, he has a friction value of 0.125 while sliding. He starts to stand up as soon as X speed reaches 0. If you release the button after he has begun to slide, X speed is set to 0 immediately, and he begins to stand up. Pressing Left or Right while sliding or standing up has no effect, but you can break into the standing up animation to jump if you press the jump button again.
If Knuckles hits a wall while gliding, he catches on, and can climb it. He will catch on even if he's turning around, as long as his X speed is still in the direction of the wall.
Note: Knuckles' knuckles deflect projectiles (just like the Shields do) while he is gliding.
#### Turning Around
When Knuckles is gliding, you can turn him around simply by tapping the Left or Right button. Even if you let go, he will continue to make a full turn. You can, however, reverse your decision and turn him back in the original direction before he makes a full turn.
You might think that turning around while gliding would be much like turning around while running on the ground. X speed would be steadily decreased until it reached zero, and then would start adding in the other direction. This is not the case, though, and a special method is used that preserves Knuckles' gliding speed.
When Knuckles is gliding, there is a value, which we'll call a, that is 0 when he's gliding to the right, and 180 when he's gliding to the left.
When Knuckles begins to turn, his X speed is stored - let's call the stored value t. If he's turning from the left to the right, a is decreased by 2.8125 until it reaches 0 (which takes 64 steps). If he's turning from right to left, a is increased by 2.8125 until it reaches 180. During the turn X speed is made to equal t times the cosine of a.
```a += 2.8125 * -sign(t);
xsp = t * cosine(a);
```
So, no matter how fast Knuckles is gliding, he turns around in the same amount of time, and his speed reverses fully. During the turn, there is no acceleration. It kicks back in once he's finished turning all the way around.
#### Gliding Rebound
An interesting side-effect of the fact that Knuckles' Y speed is not immediately blunted when he begins gliding while falling quickly is the "Gliding Rebound". If you press the button to begin gliding just as Knuckles connects with an enemy or item monitor, his Y speed is reversed from the rebound just as he begins to glide. Since gliding gravity is weaker than standard gravity, he goes soaring up into the air. This is not necessarily a bug - it's actually kind of fun.
Once Knuckles is already gliding, rebound operates normally. Since he can't exceed a Y speed of 0.5 while gliding, though, the effect is rather weak.
#### Underwater
Strangely enough, Knuckles' gliding and climbing physics are totally unaffected by water. I suspect this is because the code performed when entering and exiting the water simply changes the acceleration, deceleration, and top speed constants (this is why falling in water nullifies Super Fast Shoes). Because Knuckles' gliding and climbing code operates irrespective of these values, his abilities couldn't be affected by water without rewriting the water entry and exit code. In your engine you may wish to halve some of Knuckles' speeds when submerged to be more realistic, unless you want to remain 100% true to the original games.
#### Sliding
When you finish a glide by sliding on the ground, the game doesn't set Knuckles' grounded flag until he stops. Though, he mostly acts grounded, sticking to the floor and changing his angle as normal.
### Climbing
He climbs up and down at a speed of 1. When Knuckles jumps off of a wall, his X Speed is set to 4 in the opposite direction of the wall, and his Y Speed is set to -4.
Interestingly, because of a pixel offset when sprites are flipped, Knuckles' feet poke 1px out from the side of his size when he is on a wall to the left. This means his feet should be inside the wall a bit. Well, when on a left wall there is actually a 1px gap between knuckles X - Push Radius and the wall, purely to make it look correct.
#### Falling
When climbing, Knuckles will fall off the bottom of a wall if no wall is found at his Y Position + Height Radius (checking horizontally into the wall).
#### Clambering
Knuckles will clamber atop a ledge if it no wall is found at his Y position - Height Radius (checking horizontally into the wall).
When clambering, Knuckles plays a 3 sub-image animation. Each sub-image of the animation lasts 6 frames, and after the 3rd sub-image knuckles is standing on the ledge, where his X Position is the ledge X. Each frame of the animation moves knuckles to a new position as shown.
His position moves back and forth a bit as he climbs so that his sprite aligns. If your camera is set up correctly, usually the first 2 sub-images of motion will push the camera forward into place and the 3rd sub-image's backward motion won't move the camera at all, which makes it look smooth enough.
#### Stopping at Floors and Ceilings
If there is a floor that meets the wall, he will stop climbing down when the floor is within around 19 pixels of his Y Position (so, his normal Height Radius). | 2,948 | 12,423 | {"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} | 2.8125 | 3 | CC-MAIN-2021-25 | latest | en | 0.954081 |
https://topic.alibabacloud.com/zqpop/table-of-two-in-maths_106572.html | 1,695,446,493,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506479.32/warc/CC-MAIN-20230923030601-20230923060601-00543.warc.gz | 645,274,753 | 17,599 | # table of two in maths
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• Alibaba Cloud offers highly flexible support services tailored to meet your exact needs. | 821 | 3,883 | {"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-2023-40 | latest | en | 0.845802 |
https://catboost.ai/en/docs/concepts/loss-functions-multiregression | 1,726,396,621,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651622.79/warc/CC-MAIN-20240915084859-20240915114859-00266.warc.gz | 139,838,042 | 42,607 | # Multiregression: objectives and metrics
## Objectives and metricsObjectives and metrics
### MultiRMSEMultiRMSE
$\sqrt{\displaystyle\frac{\sum\limits_{i=1}^{N}\sum\limits_{d=1}^{dim}(a_{i,d} - t_{i, d})^{2} w_{i}}{\sum\limits_{i=1}^{N} w_{i}}}$
$dim$ is the identifier of the dimension of the label.
Usage information
User-defined parameters
use_weights
Use object/group weights to calculate metrics if the specified value is true and set all weights to 1 regardless of the input data if the specified value is false.
Default: true
### MultiRMSEWithMissingValuesMultiRMSEWithMissingValues
$\sqrt{\sum_{d=1}^{dim} \frac{\sum_{i=1}^N Num(a_{i,d}, t_{i,d}, w_i)}{\sum_{i=1}^N Den(t_{i,d}, w_i)}}$
$Num(a, t, w) = \begin{cases} w(a - t)^2, \space if \space t \neq NaN\\ 0 \end{cases}$
$Den(t, w) = \begin{cases} w, \space if \space t \neq NaN\\ 0 \end{cases}$
$dim$ is the identifier of the dimension of the label.
Usage information
User-defined parameters
use_weights
Use object/group weights to calculate metrics if the specified value is true and set all weights to 1 regardless of the input data if the specified value is false.
Default: true
## Used for optimizationUsed for optimization
Name Optimization GPU Support
MultiRMSE + +
MultiRMSEWithMissingValues + - | 384 | 1,285 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 6, "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} | 2.671875 | 3 | CC-MAIN-2024-38 | latest | en | 0.359637 |
https://www.gradesaver.com/textbooks/math/algebra/college-algebra-10th-edition/chapter-1-section-1-5-solving-inequalities-1-5-assess-your-understanding-page-127/52 | 1,532,161,858,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592420.72/warc/CC-MAIN-20180721071046-20180721091046-00202.warc.gz | 905,730,706 | 13,143 | ## College Algebra (10th Edition)
$\le$
The Multiplication Property of Inequality states that if a positive number $c$ is multiplied to each side of $a \le b$, then $ac \le bc$. Thus, if $3x \le 12$, then, multiplying $\frac{1}{3}$ to each side of the inequality gives: $3x(\frac{1}{3}) \le 12(\frac{1}{3}) \\x \le 4$ Thus, the missing inequality symbol in the given statement is $\le$. | 118 | 387 | {"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-2018-30 | longest | en | 0.728061 |
http://mathhelpforum.com/trigonometry/149173-question-cosine-2.html | 1,481,238,155,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542657.90/warc/CC-MAIN-20161202170902-00319-ip-10-31-129-80.ec2.internal.warc.gz | 175,230,080 | 14,630 | # Thread: A question on Cosine.
1. Dear Ackbeet and undefined,
I bet you still have not understood my assertion!
I told you the theorem states for Trig EQUATIONS. It means that for trig equations we must have an unknown value called x and obtain all of the possible values for that.This is the meaning of Trigonometric equation(similar to the meaning of a linear equation or a quadratic or... wich we look for unknown x) and the solution set is nothing except that!
And you say is called a Trig IDENTITY.
--------------------------------------------------------------
You see in the following picture a Geometric proof from the book مثلثات (Trigonometry) by Ahmad Ghandehary Madreseh publications:
You can see another proof for that here:
Trigonometry - Matzic
And can see its uses in many books and Articles; for example:
Plane Trigonometry, by S.L LONEY– Cambridge University Press.
Elementary Trigonometry , by H.S Hall and S. R. KNIGHT
Even you can google the word “Trigonometric Equations” and find useful articles and books for that!
2. My posts 12 and 14 are incorrect, on second look, and looking at your proof. Please disregard those two posts, and accept my apologies for being so thick-headed.
Just looking back at the original post: are you claiming that the equation cos(x) = cos(a) implies x = 0?
cos(x) = cos(0) implies x = 0 ?
If so, I should say that: It implies x=2k*pi and x=0 is one of the infinte solutions which is obtained by getting k=0 and infact it is the original angle.
And if no, cos(x) = cos(a) implies x=2k*pi+a and 2k*pi-a.
4. In post #15, I wrote that I would only post again in this thread to correct any mistakes I had made; this was on grounds that I felt the thread to be unproductive; I now feel this thread has taken a productive turn, so I am posting again, even though I do not see a mistake in my previous posts. (Not to say I haven't made any mistake, just that I don't see one.)
I think the main problem in this thread is lack of clarity. I still have no idea what Mathelogician's question is, or if Mathelogician even has a question.
Originally Posted by Mathelogician
Dear Ackbeet and undefined,
I bet you still have not understood my assertion!
The theorem whose proof you presented is no different from
$\forall x,a\in\mathbb{R}:\cos x=\cos a \Leftrightarrow\exists k\in\mathbb{Z}: x=2k\pi\pm a$.
Equivalently:
$\forall x,a\in\mathbb{R}:\cos x=\cos a \Leftrightarrow x\in\{y\in\mathbb{R}\exists k\in\mathbb{Z}) (y=2k\pi\pm a)\}" alt="\forall x,a\in\mathbb{R}:\cos x=\cos a \Leftrightarrow x\in\{y\in\mathbb{R}\exists k\in\mathbb{Z}) (y=2k\pi\pm a)\}" />.
The $\forall$ quantifier in the proof you posted means that $k$ ranges over all integers. The proof you posted does not say that $x=2k\pi\pm\theta,\forall k\in\mathbb{Z}$. This is a somewhat subtle point I suppose; the placement of the $\forall$ quantifier is very important. Consider the two statements:
(1) "The complete solution set of $x$ is obtained by $2k\pi\pm\theta,\forall k\in\mathbb{Z}$."
(2) " $x=2k\pi\pm\theta,\forall k\in\mathbb{Z}$."
Statement (1) says that $x$ belongs to the set obtained by ranging over all integers, while statement (2) says the equation involving $x$ holds for all integers. (And the equation does not hold for all integers, so statement (2) is false.)
To translate explicitly: Statement (1) can be written:
$x\in\{0\cdot2\pi+\theta,0\cdot2\pi-\theta,1\cdot2\pi+\theta,1\cdot2\pi-\theta,\cdots \}$
And statement (2) can be written:
$(x=0\cdot2\pi+\theta\lor x=0\cdot2\pi+\theta) \land (x=1\cdot2\pi+\theta\lor x=1\cdot2\pi+\theta) \land \cdots$
The problem is those $\land$ operators, which ought to be $\displaystyle\lor$ instead. To accomplish this, we can use $\exists$.
I'm sure there's a bit of a language barrier too, in terms of English. Anyway I hope you (Mathelogician) can see that you and I are talking about the exact same theorem, just with different language (and disagreement over the language). However, if you look carefully at everything I wrote, I believe you will find it's all meticulously in keeping with conventions used in math textbooks and journals and encyclopedias, etc. If not, then I'll be glad if someone will correct me.
So, I understood your assertion, it's just that the assertion you wrote down was different from the assertion you meant to write down.
Now.. Is there a question we can help answer?
5. NO;
1-My claim (using Universal quantifier) is different from yours(using existensial quantifier)!
When you say: There exists a number k in Z such that:...", it means that "for SOME (at least one) k in Z,we have ..."; But it doesn't clear our mean which is "for EVERY k in Z we have..."
Indeed yours is true but inadequate and doesn't make the real sense!
2-You need to know that the expression "x=2k*pi+_ a for all k in Z" is ANOTHER FORM of the first expression which is used in some books and articles; And since you don't know this, you insist on your own interpretation of that and think they are different!!
I thought you know this and then i used it for the proof!
NOTE: Existential quantification is distinct from Universal quantification ("for all"), which asserts that the property or relation holds for any members of the domain.
See:
Existential quantification - Wikipedia, the free encyclopedia
6. Originally Posted by Mathelogician
NO;
1-My claim (using Universal quantifier) is different from yours(using existensial quantifier)!
When you say: There exists a number k in Z such that:...", it means that "for SOME (at least one) k in Z,we have ..."; But it doesn't clear our mean which is "for EVERY k in Z we have..."
Indeed yours is true but inadequate and doesn't make the real sense!
2-You need to know that the expression "x=2k*pi+_ a for all k in Z" is ANOTHER FORM of the first expression which is used in some books and articles; And since you don't know this, you insist on your own interpretation of that and think they are different!!
I thought you know this and then i used it for the proof!
NOTE: Existential quantification is distinct from Universal quantification ("for all"), which asserts that the property or relation holds for any members of the domain.
See:
Existential quantification - Wikipedia, the free encyclopedia
The set $\{y\in\mathbb{R}\exists k\in\mathbb{Z}) (y=2k\pi\pm a)\}" alt="\{y\in\mathbb{R}\exists k\in\mathbb{Z}) (y=2k\pi\pm a)\}" /> equals the set $\{0\cdot2\pi+a,0\cdot2\pi-a,1\cdot2\pi+a,1\cdot2\pi-a,\cdots \}$.
Thus
$\forall x,a\in\mathbb{R}:\cos x=\cos a \Leftrightarrow\exists k\in\mathbb{Z}: x=2k\pi\pm a$
is equivalent to
$\forall x,a\in\mathbb{R}:\cos x=\cos a \Leftrightarrow x\in\{0\cdot2\pi+a,0\cdot2\pi-a,1\cdot2\pi+a,1\cdot2\pi-a,\cdots \}$
which says that $\{0\cdot2\pi+a,0\cdot2\pi-a,1\cdot2\pi+a,1\cdot2\pi-a,\cdots \}$ is the complete solution set of $\displaystyle x$, which is your theorem.
---------------------------
Let's demonstrate that the set $\{y\in\mathbb{R}\exists k\in\mathbb{Z}) (y=2k\pi\pm a)\}" alt="\{y\in\mathbb{R}\exists k\in\mathbb{Z}) (y=2k\pi\pm a)\}" /> equals the set $\{0\cdot2\pi+a,0\cdot2\pi-a,1\cdot2\pi+a,1\cdot2\pi-a,\cdots \}$.
Consider $0\cdot2\pi+a$. Does there exist a $k\in\mathbb{Z}$ such that $0\cdot2\pi+a=2k\pi\pma$? Yes, so it's in the set. ( $k=0$.)
Consider $0\cdot2\pi-a$. Does there exist a $k\in\mathbb{Z}$ such that $0\cdot2\pi-a=2k\pi\pma$? Yes, so it's in the set. ( $k=0$.)
Consider $1\cdot2\pi+a$. Does there exist a $k\in\mathbb{Z}$ such that $1\cdot2\pi+a=2k\pi\pma$? Yes, so it's in the set. ( $k=1$.)
etc.
------------------------------
Let's demonstrate that
$\exists k\in\mathbb{Z}: x=2k\pi\pm a$
is equivalent to
$x\in\{y\in\mathbb{R}\exists k\in\mathbb{Z}) (y=2k\pi\pm a)\}" alt="x\in\{y\in\mathbb{R}\exists k\in\mathbb{Z}) (y=2k\pi\pm a)\}" />.
Consider $0\cdot2\pi+a$. Does there exist a $k\in\mathbb{Z}$ such that $0\cdot2\pi+a=2k\pi\pma$? Yes, so it satisfies the first statement, and the second.
Etc.
---------------------------
If you've studied equivalence classes then you will recognise that this is what we are talking about. cos x = cos a if and only if x and a are the in the same equivalence class defined by the condition $\exists k\in\mathbb{Z}: x=2k\pi\pm a$.
7. Ok,
You are right.
Page 2 of 2 First 12 | 2,472 | 8,273 | {"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": 42, "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} | 3.46875 | 3 | CC-MAIN-2016-50 | longest | en | 0.94712 |
https://mathematica.stackexchange.com/questions/238608/find-all-possible-configurations-of-a-finite-dipole-system | 1,713,130,561,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816893.9/warc/CC-MAIN-20240414192536-20240414222536-00362.warc.gz | 357,872,753 | 39,769 | # Find all possible configurations of a finite dipole system
I have a system which is composed of the following blocks
$$[-,+],[+,+],[+,-],[-,-]$$
I can compose a system of $$n$$ blocks with the only rule that the edges act as a dipole.
for example $$[-,+][-,+][-,-][+,-]$$
is a possible configuration of $$n=4$$.
Is it possible to create a code with mathematica that will produce all the possible configurations for any given $$n$$?
• With[{n = 2}, Tuples[{-1, +1}, n]] ? Jan 22, 2021 at 9:55
• this gives all the possible elements, but how can I compose all the possible combinations for any given $n$, which will perserve a dipole between the elements? Jan 22, 2021 at 9:59
• "the only rule that the edges act as a dipole." - to clarify, an admissible configuration is one with + or - on both ends, and +/- always being adjacent between the two poles? Jan 22, 2021 at 10:09
AttachDipole[x : {___, {_, "+"}}] := Append[x, #] & /@ {{"-", "-"}, {"-", "+"}};
AttachDipole[x : {___, {_, "-"}}] := Append[x, #] & /@ {{"+", "-"}, {"+", "+"}};
AttachDipoles[dipls_List] := Join @@ AttachDipole /@ dipls
AllDipoles[n_] :=
Nest[AttachDipoles, {{{"-", "-"}}, {{"-", "+"}}, {{"+", "-"}}, {{"+", "+"}}}, n - 1]
AllDipoles[1]
{{{"-", "-"}}, {{"-", "+"}}, {{"+", "-"}}, {{"+", "+"}}}
AllDipoles[2]
{{{"-", "-"}, {"+", "-"}}, {{"-", "-"}, {"+", "+"}},
{{"-", "+"}, {"-", "-"}}, {{"-", "+"}, {"-", "+"}},
{{"+", "-"}, {"+", "-"}}, {{"+", "-"}, {"+", "+"}},
{{"+", "+"}, {"-", "-"}}, {{"+", "+"}, {"-", "+"}}}
AllDipoles[3]
{{{"-", "-"}, {"+", "-"}, {"+", "-"}}, {{"-", "-"}, {"+", "-"}, {"+", "+"}},
{{"-", "-"}, {"+", "+"}, {"-", "-"}}, {{"-", "-"}, {"+", "+"}, {"-", "+"}},
{{"-", "+"}, {"-", "-"}, {"+", "-"}}, {{"-", "+"}, {"-", "-"}, {"+", "+"}},
{{"-", "+"}, {"-", "+"}, {"-", "-"}}, {{"-", "+"}, {"-", "+"}, {"-", "+"}},
{{"+", "-"}, {"+", "-"}, {"+", "-"}}, {{"+", "-"}, {"+", "-"}, {"+", "+"}},
{{"+", "-"}, {"+", "+"}, {"-", "-"}}, {{"+", "-"}, {"+", "+"}, {"-", "+"}},
{{"+", "+"}, {"-", "-"}, {"+", "-"}}, {{"+", "+"}, {"-", "-"}, {"+", "+"}},
{{"+", "+"}, {"-", "+"}, {"-", "-"}}, {{"+", "+"}, {"-", "+"}, {"-", "+"}}}
Here is a one liner:
n = 3;
DeleteCases[Tuples[Tuples[{-1, 1}, 2], n], {___, {_, x_}, {y_, _}, ___} /; ( x == y)]
{{{-1, -1}, {1, -1}, {1, -1}}, {{-1, -1}, {1, -1}, {1,
1}}, {{-1, -1}, {1, 1}, {-1, -1}}, {{-1, -1}, {1, 1}, {-1,
1}}, {{-1, 1}, {-1, -1}, {1, -1}}, {{-1, 1}, {-1, -1}, {1,
1}}, {{-1, 1}, {-1, 1}, {-1, -1}}, {{-1, 1}, {-1, 1}, {-1,
1}}, {{1, -1}, {1, -1}, {1, -1}}, {{1, -1}, {1, -1}, {1,
1}}, {{1, -1}, {1, 1}, {-1, -1}}, {{1, -1}, {1, 1}, {-1, 1}}, {{1,
1}, {-1, -1}, {1, -1}}, {{1, 1}, {-1, -1}, {1, 1}}, {{1, 1}, {-1,
1}, {-1, -1}}, {{1, 1}, {-1, 1}, {-1, 1}}}
• You can make the pattern a bit more compact with {___, {_, x_}, {x_, _}, ___} Jan 22, 2021 at 15:19 | 1,095 | 2,819 | {"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": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2024-18 | latest | en | 0.511985 |
http://mathhelpforum.com/calculus/121372-limit-sequence.html | 1,481,292,950,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542712.12/warc/CC-MAIN-20161202170902-00391-ip-10-31-129-80.ec2.internal.warc.gz | 180,236,223 | 11,870 | 1. ## Limit of sequence
Consider the sequence $f_{0}:N^{*}->R$
$f_{0}(n)=(1+\frac{1}{n})^{n}$
and define,
$f_{k}(n)=(\frac{f_{k-1}(n)}{\lim_{p\to\infty}f_{k-1}(p)})^{n}$.
n,k positive natural numbers.
Find:
$\lim_{n\to\infty}f_{k}(n)$
Using induction I found that
$f_{k}(n)=\frac{(1+\frac{1}{n})^{n^{k+1}}}{e^{\sum_ {i=1}^{k}(-1)^{i+1}\frac{n^{k+i-1}}{i}}}$
But I don't know how to find the limit.
The limit should be $e^{\frac{(-1)^{k}}{k+1}}$
2. Originally Posted by m3th0dman
Consider the sequence $f_{0}:N^{*}->R$
$f_{0}(n)=(1+\frac{1}{n})^{n}$
and define,
$f_{k}(n)=(\frac{f_{k-1}(n)}{\lim_{p\to\infty}f_{k-1}(p)})^{n}$.
n,k positive natural numbers.
Find:
$\lim_{n\to\infty}f_{k}(n)$
Using induction I found that
$f_{k}(n)=\frac{(1+\frac{1}{n})^{n^{k+1}}}{e^{\sum_ {i=1}^{k}(-1)^{i+1}\frac{n^{k+i-1}}{i}}}$
But I don't know how to find the limit.
The limit should be $e^{\frac{(-1)^{k}}{k+1}}$
According to what you wrote:
$f_1(n)=\left(\frac{f_0(n)}{\lim\limits_{n\to\infty }f_0(n)}\right)^n=\left(\frac{\left(1+\frac{1}{n}\ right)^n}{e}\right)^n$ $\xrightarrow [n\to\infty]{}1$ , so:
$f_2(n)=\left(\frac{f_1(n)}{\lim\limits_{p\to\infty }f_1(p)}\right)^n=
\left(\frac{\left(1+\frac{1}{n}\right)^n}{e}\right )^n=f_1(n)$
and etc...., so I think you may have not written what you meant...
Tonio
3. Originally Posted by tonio
According to what you wrote:
$f_1(n)=\left(\frac{f_0(n)}{\lim\limits_{n\to\infty }f_0(n)}\right)^n=\left(\frac{\left(1+\frac{1}{n}\ right)^n}{e}\right)^n$ $\xrightarrow [n\to\infty]{}1$ , so:
$f_2(n)=\left(\frac{f_1(n)}{\lim\limits_{p\to\infty }f_1(p)}\right)^n=
\left(\frac{\left(1+\frac{1}{n}\right)^n}{e}\right )^n=f_1(n)$
and etc...., so I think you may have not written what you meant...
Tonio
The following proposition is false:
$\lim_{n\to\infty}1^{n}=1$
So your assumption is not true.
$\lim_{n\to\infty}(\frac{(1+\frac{1}{n})^n}{e})^n=e ^{-\frac{1}{2}}$
4. Originally Posted by m3th0dman
The following proposition is false:
$\lim_{n\to\infty}1^{n}=1$
In fact this is trivially true. What is not true in general is that $f(n)\xrightarrow [n\to\infty]{}1\Longrightarrow f(n)^n\xrightarrow [n\to\infty]{}1$
So your assumption is not true.
No assumption but sheer misreading and consequent mistake of mine...
$\lim_{n\to\infty}(\frac{(1+\frac{1}{n})^n}{e})^n=e ^{-\frac{1}{2}}$
Indeed. And this limit is pretty hard to evaluate: I had to use twice L'Hospital' rule with logarithmic notation...
Tonio
5. Originally Posted by tonio
Indeed. And this limit is pretty hard to evaluate: I had to use twice L'Hospital' rule with logarithmic notation...
Tonio
I have been taught to avoid L'Hospital' rule, so here is what I wrote:
$\left(\frac{\left(1+\frac{1}{n}\right)^n}{e}\right )^n = e^{n(n\log(1+\frac{1}{n})-1)}$ $=e^{n(n(\frac{1}{n}-\frac{1}{2n^2}+o(\frac{1}{n^2}))-1)}=e^{-\frac{1}{2}+o(1)},$
and that's it.
Originally Posted by m3th0dman
Consider the sequence $f_{0}:N^{*}->R$
$f_{0}(n)=(1+\frac{1}{n})^{n}$
and define,
$f_{k}(n)=(\frac{f_{k-1}(n)}{\lim_{p\to\infty}f_{k-1}(p)})^{n}$.
n,k positive natural numbers.
Find:
$\lim_{n\to\infty}f_{k}(n)$
Moreover, the same method works for this general problem. Let us write $g_k(n)=\log f_k(n)$ for simplicity, and $\ell_k=\lim_n g_k(n)$, for all $k$.
Let $k\geq 1$. Start from:
$\log\left(1+\frac{1}{n}\right)=\frac{1}{n}-\frac{1}{2n^2}+\cdots+\frac{(-1)^{k+1}}{k n^k}+o\left(\frac{1}{n^k}\right)$.
The definition of $\ell_k$ is just an intricate way to describe the coefficients in this expansion.
Indeed,
$g_0(n)=n\log(1+\frac{1}{n})=1-\frac{1}{2n}+\cdots+\frac{(-1)^{k+1}}{k n^{k-1}}+o\left(\frac{1}{n^{k-1}}\right)=1+o(1)$,
hence $\ell_0=1$,
$g_1(n)=n(g_0(n)-\ell_0)=n(g_0(n)-1)=-\frac{1}{2}+\frac{1}{3n}+\cdots+\frac{(-1)^{k+1}}{k n^{k-2}}+o\left(\frac{1}{n^{k-2}}\right)=-\frac{1}{2}+o(1)$,
hence $\ell_1=-\frac{1}{2}$,
and so on, "unfolding the expansion", until
$g_{k-1}(n)=n(g_{k-2}(n)-\ell_{k-2})=\frac{(-1)^{k+1}}{k}+o(1)$,
hence $\ell_{k-1}=\frac{(-1)^{k+1}}{k}$.
If you really want to write every word of the proof, then you have to explicitate the above induction. Let $k\geq 1$ and prove by induction on $j=0,\ldots,k-1$ that :
$g_j(n)=\frac{(-1)^{j}}{j+1}+\frac{(-1)^{j+1}}{j+2}\frac{1}{n}+\cdots+\frac{(-1)^{k+1}}{k}\frac{1}{n^{k-j-1}}+o\left(\frac{1}{n^{k-j-1}}\right)$
(hence $\ell_j=\frac{(-1)^{j}}{j+1}$).
(you could also use convergent series instead of asymptotic expansions to shorten the proof, but these would be a little inappropriate here since we only care for the limit)
6. ## Thanks
Thank you very much! | 1,898 | 4,553 | {"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": 45, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2016-50 | longest | en | 0.534758 |
https://www.onlinemath4all.com/diagonalization-of-matrix.html | 1,723,277,749,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640790444.57/warc/CC-MAIN-20240810061945-20240810091945-00321.warc.gz | 718,502,191 | 40,925 | # DIAGONALIZATION OF MATRIX
Diagonalizable :
A square matrix A is said to be diagonalizable if A is similar to a diagonal matrix. That is, if A = PDP-1 where P is invertible and D is a diagonal matrix.
When is A diagonalizable? (The answer lies in examining the eigenvalues and eigenvectors of A.)
Note that
Altogether
Equivalently,
## Theorem (Diagonalization)
An n x n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors.
In fact, A = PDP-1, with D a diagonal matrix, if and only if the columns of P are n linearly independent eigenvectors of A.
In this case, the diagonal entries of D are eigenvalues of A that correspond, respectively, to the eigenvectors in P.
Example 1 :
Diagonalize the following matrix, if possible.
Solution :
Step 1 :
Find the eigenvalues of A.
= (2 - λ)2(1 - λ) = 0
Eigen values of A : λ = 2 and λ = 1.
Step 2 :
Find three linearly independent eigenvectors of A.
By solving
(A - λI)x = 0,
for each value of λ, we obtain the following :
Step 3 :
Construct P from the vectors in step 2.
Step 4 :
Construct D from the corresponding eigenvalues.
Step 5 :
Check your work by verifying that AP = PD.
Example 2 :
Diagonalize the following matrix, if possible.
Solution :
Since this matrix is triangular, the eigenvalues are λ1 = 2 and λ2 = 4. By solving (A - λI)x = 0 for each eigenvalue, we would find the following :
Every eigenvector of A is a multiple of v1 or v2 which means there are not three linearly independent eigenvectors of A and by Theorem, A is not diagonalizable.
Example 3 :
Solution :
Since A has three eigenvalues :
λ1 = 2, λ2 = 6, λ3 = 1
and since eigenvectors corresponding to distinct eigenvalues are linearly independent, A has three linearly independent eigenvectors and it is therefore diagonalizable.
Example 4 :
Diagonalize the following matrix, if possible.
Solution:
Eigenvalues : -2 and 2 (each with multiplicity 2).
Solving (A -λI)x = 0 yields the following eigenspace basis sets.
{v1, v2, v3, v4} is linearly independent.
---> P = [v1 v2 v3 v4] is invertible
---> A = PDP-1, where
Kindly mail your feedback to v4formath@gmail.com
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Reg.No. :
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Maths
Time : 01:00:00 Hrs
Total Marks : 30
6 x 5 = 30
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# Trivial Graph in Graph Theory
Last Updated on August 3, 2023 by Mayank Dham
Graph theory is a fundamental branch of mathematics that deals with the study of graphs, which are mathematical structures used to model relationships between objects. While many graph types hold great complexity and significance in various fields, there exist certain special cases known as "trivial graphs" that serve as the building blocks and foundations for more elaborate graph theory concepts. In this article, we will explore what trivial graphs are, their properties, and their relevance in the study of graphs.
## What is a Trivial Graph in Graph Theory?
In graph theory, a trivial graph is the simplest and most basic type of graph that exists. It consists of just one vertex (node) and no edges. This means that a trivial graph contains only a single point without any connections to other points, as there are no edges to link it to other vertices.
Graph theory typically represents graphs as G(V, E), where ‘V’ is the set of vertices (nodes), ‘E’ is the set of edges, and ‘G’ represents the graph as a whole. In the case of a trivial graph, V = {v}, and E = {}. Here, ‘v’ denotes the single vertex in the graph, and {} denotes the empty set of edges.
## Properties of Trivial Graphs in Graph Theory
The various properties of trivial graphs are demonstrated below.
1. Single Vertex: The defining characteristic of a trivial graph is that it has only one vertex. The graph consists of a single isolated point.
2. No Edges: Since there is only one vertex, there are no edges to connect it with any other vertex. Thus, the edge set ‘E’ is empty.
3. Degree of the Vertex: In graph theory, the degree of a vertex refers to the number of edges incident to that vertex. In the case of a trivial graph, the degree of the single vertex is 0 since there are no edges connected to it.
4. Connectivity: As there are no edges in a trivial graph, it is a disconnected graph. There are no paths or connections between any pair of vertices because there is only one vertex present.
5. Size and Order: The size of a graph refers to the number of edges it contains, while the order refers to the number of vertices. For a trivial graph, the size is 0 (|E| = 0) and the order is 1 (|V| = 1).
6. Isomorphism: Trivial graphs are isomorphic to each other. That is, any two trivial graphs are identical in structure since they both consist of a single vertex and no edges.
## Applications and Importance of Trivial Graphs
At first glance, trivial graphs might seem too basic and uninteresting to have any practical significance. However, their simplicity plays a crucial role in the study of graph theory. Some of the key applications and importance of trivial graphs are:
1. Theoretical Foundation: Trivial graphs serve as the foundational elements of graph theory. They help establish the basic concepts of vertices, edges, degrees, and connectivity.
2. Comparison Basis: Trivial graphs provide a basis for comparison with other, more complex graphs. By understanding the properties of trivial graphs, researchers can better comprehend how different graph types differ and evolve from this simplest form.
3. Algorithm Analysis: In algorithm design and analysis, trivial graphs are often used as the base case for various graph-related algorithms. Understanding how algorithms perform on trivial graphs can help researchers extrapolate their behavior to more complex scenarios.
4. Education and Learning: Trivial graphs are essential in teaching graph theory to students and beginners. They offer an accessible starting point for grasping fundamental graph concepts before moving on to more intricate graph structures.
5. Debugging and Testing: In software development, trivial graphs can serve as test cases for graph algorithms. They enable developers to test the correctness and efficiency of their algorithms in simple scenarios.
Conclusion
Trivial graphs are the most basic type of graph, with only one vertex and no edges. Although they may appear insignificant, these simple structures are critical in laying the groundwork for graph theory. They are essential for understanding the concepts of vertices, edges, degrees, connectivity, and graph representation. Understanding trivial graphs is essential for understanding more complex graph types and their applications in fields such as computer science, network analysis, operations research, and others. So, while trivial graphs may be basic, their significance in the realm of graph theory cannot be understated.
Here are some of the frequently asked questions on trivial graphs.
Q1: What is a non-trivial graph in graph theory?
A non-trivial graph is any graph that contains more than one vertex. In other words, it consists of at least two distinct points (vertices) that may or may not be connected by edges. Non-trivial graphs exhibit a more complex structure with a higher degree of connectivity between their vertices compared to trivial graphs.
Q2: What is a trivial edge in graph theory?
In graph theory, a trivial edge is an edge that connects a vertex to itself. Mathematically, it is represented as an edge between the same vertex (v, v) where ‘v’ is a vertex in the graph. Trivial edges do not add any new information or connectivity to the graph, as they create loops by connecting a vertex back to itself.
Q3: What does a trivial graph contain?
A trivial graph contains only one vertex (node) and no edges. It is the simplest possible graph structure, representing a single isolated point without any connections to other vertices.
Q4: What is the difference between trivial and non-trivial graphs?
The key difference lies in the number of vertices they contain:
• Trivial Graph: Contains only one vertex and no edges. The most basic and simplest form of a graph.
• Non-Trivial Graph: Contains more than one vertex. It has at least two distinct points (vertices) and may have edges connecting those vertices, resulting in a more complex structure.
Q5: What are some applications of trivial graphs?
Trivial graphs play a crucial role in graph theory and have several applications:
• They form the foundation for understanding fundamental graph concepts like vertices, edges, and degrees.
• Trivial graphs are used as a basis for comparison with more complex graph types, aiding researchers in understanding graph evolution.
• They serve as test cases for graph algorithms in software development to assess correctness and efficiency.
• Trivial graphs are vital in teaching graph theory to students and beginners as an accessible starting point.
Q6: Are trivial and non-trivial graphs isomorphic?
No, trivial and non-trivial graphs are not isomorphic. Trivial graphs are unique in structure, consisting of only one vertex, while non-trivial graphs can vary significantly in their number of vertices, edges, and connectivity. Isomorphism is a concept in graph theory where two graphs have the same structure, and trivial and non-trivial graphs inherently have different structures. | 1,476 | 7,305 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-33 | latest | en | 0.949261 |
https://termsdepot.com/logical-equivalence/ | 1,680,412,581,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950383.8/warc/CC-MAIN-20230402043600-20230402073600-00134.warc.gz | 600,486,895 | 9,813 | # Logical equivalence
Logical equivalence is a relationship between two or more formulas that have the same truth value in every possible interpretation. In other words, the formulas are logically equivalent if they always produce the same result, regardless of the values of the variables involved.
There are a few different ways to show that two formulas are logically equivalent. One way is to use a truth table, which shows the truth values of the formulas for all possible combinations of truth values for the variables involved. Another way is to use a proof by contradiction, in which it is shown that if the formulas were not logically equivalent, then that would lead to a contradiction. What do you mean by logical equivalence? Logical equivalence is a relationship between two statements that are either both true or both false. In other words, the statements are logically equivalent if they have the same truth value. What is logically equivalent to P → Q? There are a few different ways to answer this question, but one way to think about it is that P → Q is logically equivalent to ¬P ∨ Q. In other words, if P is true, then Q must also be true. If P is false, then Q can be either true or false.
How do you determine logical equivalence? To determine logical equivalence, you need to show that two formulas are equivalent in all possible truth assignments. This can be done by constructing a truth table for each formula and comparing the results. If the formulas produce the same results in every possible truth assignment, then they are logically equivalent.
Which is logically equivalent to P ∧ Q → R? There are a few different ways to answer this question, but one way to think about it is that P ∧ Q → R is logically equivalent to ~(P ∧ Q) ∨ R. In other words, if P and Q are both true, then R must be true in order for the statement to be true.
##### Is P → Q ∧ Q → P logically equivalent to P → Q ∨ Q ↔ P?
No, P → Q ∧ Q → P is not logically equivalent to P → Q ∨ Q ↔ P. To see this, consider the truth table for P → Q ∧ Q → P:
P | Q | P → Q | Q → P | P → Q ∧ Q → P
T | T | T | T | T
T | F | F | T | F
F | T | T | T | F
F | F | T | T | T
Now, consider the truth table for P → Q ∨ Q ↔ P:
P | Q | P → Q | Q ↔ P | P → Q ∨ Q ↔ P
T | T | T | T | T
T | F | F | F | F
F | T | T | T | T
F | F | T | T | T
As we can see, the two truth tables are not equivalent. | 585 | 2,384 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2023-14 | longest | en | 0.9369 |
http://www.gregthatcher.com/Stocks/StockFourierAnalysisDetails.aspx?ticker=FDEF | 1,500,706,937,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423903.35/warc/CC-MAIN-20170722062617-20170722082617-00707.warc.gz | 441,339,197 | 107,090 | Back to list of Stocks See Also: Seasonal Analysis of FDEFGenetic Algorithms Stock Portfolio Generator, and Fourier Calculator
# Fourier Analysis of FDEF (First Defiance Financial Corp)
FDEF (First Defiance Financial Corp) appears to have interesting cyclic behaviour every 76 weeks (2.037*sine), 92 weeks (1.8829*sine), and 83 weeks (1.3924*sine).
FDEF (First Defiance Financial Corp) has an average price of 19.14 (topmost row, frequency = 0).
Click on the checkboxes shown on the right to see how the various frequencies contribute to the graph. Look for large magnitude coefficients (sine or cosine), as these are associated with frequencies which contribute most to the associated stock plot. If you find a large magnitude coefficient which dramatically changes the graph, look at the associated "Period" in weeks, as you may have found a significant recurring cycle for the stock of interest.
## Fourier Analysis
Using data from 1/3/2000 to 7/10/2017 for FDEF (First Defiance Financial Corp), this program was able to calculate the following Fourier Series:
Sequence #Cosine Coefficients Sine Coefficients FrequenciesPeriod
019.14488 0
15.64471 -2.91309 (1*2π)/915915 weeks
2.5286 -9.71872 (2*2π)/915458 weeks
31.3582 -3.97022 (3*2π)/915305 weeks
4.07014 -4.00434 (4*2π)/915229 weeks
5.99834 -2.60906 (5*2π)/915183 weeks
6-.54264 -2.64993 (6*2π)/915153 weeks
7.57778 -2.23469 (7*2π)/915131 weeks
8-.12678 -1.71352 (8*2π)/915114 weeks
91.08121 -1.97021 (9*2π)/915102 weeks
10-.07188 -1.88294 (10*2π)/91592 weeks
11.03469 -1.39243 (11*2π)/91583 weeks
12-.20497 -2.03702 (12*2π)/91576 weeks
13.39865 -.93956 (13*2π)/91570 weeks
14.16262 -1.49994 (14*2π)/91565 weeks
15-.15159 -.4769 (15*2π)/91561 weeks
16-.10349 -1.09495 (16*2π)/91557 weeks
17.04749 -.75939 (17*2π)/91554 weeks
18.05341 -1.0589 (18*2π)/91551 weeks
19-.2827 -.31435 (19*2π)/91548 weeks
20.04635 -1.01203 (20*2π)/91546 weeks
21-.13571 -.78149 (21*2π)/91544 weeks
22.27119 -.79403 (22*2π)/91542 weeks
23-.09954 -.70349 (23*2π)/91540 weeks
24-.0132 -.58979 (24*2π)/91538 weeks
25-.07922 -.71353 (25*2π)/91537 weeks
26-.06264 -.36254 (26*2π)/91535 weeks
27-.03438 -.65334 (27*2π)/91534 weeks
28.04376 -.57982 (28*2π)/91533 weeks
29-.01059 -.55481 (29*2π)/91532 weeks
30-.21413 -.50661 (30*2π)/91531 weeks
31-.09408 -.28112 (31*2π)/91530 weeks
32.07751 -.5166 (32*2π)/91529 weeks
33.01303 -.44926 (33*2π)/91528 weeks
34.03341 -.44361 (34*2π)/91527 weeks
35.03921 -.4676 (35*2π)/91526 weeks
36.0776 -.45239 (36*2π)/91525 weeks
37-.13958 -.6034 (37*2π)/91525 weeks
38-.09897 -.23031 (38*2π)/91524 weeks
39-.03125 -.24607 (39*2π)/91523 weeks
40-.04222 -.38487 (40*2π)/91523 weeks
41.12306 -.60268 (41*2π)/91522 weeks
42-.03344 -.4252 (42*2π)/91522 weeks
43-.07767 -.38747 (43*2π)/91521 weeks
44-.07501 -.49456 (44*2π)/91521 weeks
45-.08505 -.44551 (45*2π)/91520 weeks
46-.2367 -.38252 (46*2π)/91520 weeks
47-.18238 -.29149 (47*2π)/91519 weeks
48-.18615 -.32134 (48*2π)/91519 weeks
49-.14133 -.28516 (49*2π)/91519 weeks
50-.22993 -.39961 (50*2π)/91518 weeks
51-.19212 -.00965 (51*2π)/91518 weeks
52-.0264 -.28917 (52*2π)/91518 weeks
53-.04623 -.09664 (53*2π)/91517 weeks
54-.04709 -.36782 (54*2π)/91517 weeks
55-.10086 -.09004 (55*2π)/91517 weeks
56-.02665 -.2866 (56*2π)/91516 weeks
57-.10597 -.1122 (57*2π)/91516 weeks
58.03423 -.26872 (58*2π)/91516 weeks
59-.11306 -.24606 (59*2π)/91516 weeks
60-.04479 -.16831 (60*2π)/91515 weeks
61.07969 -.16426 (61*2π)/91515 weeks
62.08771 -.23775 (62*2π)/91515 weeks
63.08116 -.36887 (63*2π)/91515 weeks
64.00963 -.32584 (64*2π)/91514 weeks
65-.11213 -.2771 (65*2π)/91514 weeks
66.02416 -.33093 (66*2π)/91514 weeks
67-.10914 -.2603 (67*2π)/91514 weeks
68-.10961 -.30444 (68*2π)/91513 weeks
69-.16952 -.14363 (69*2π)/91513 weeks
70-.22336 -.1243 (70*2π)/91513 weeks
71-.0989 -.22197 (71*2π)/91513 weeks
72-.17022 -.11586 (72*2π)/91513 weeks
73.01749 -.12962 (73*2π)/91513 weeks
74-.12619 -.18512 (74*2π)/91512 weeks
75.02813 -.17401 (75*2π)/91512 weeks
76-.13581 -.11621 (76*2π)/91512 weeks
77-.02636 -.24051 (77*2π)/91512 weeks
78-.06889 -.14248 (78*2π)/91512 weeks
79-.03824 -.16706 (79*2π)/91512 weeks
80-.02065 -.12933 (80*2π)/91511 weeks
81-.09447 -.16995 (81*2π)/91511 weeks
82-.06994 -.12638 (82*2π)/91511 weeks
83-.01733 -.11874 (83*2π)/91511 weeks
84.0639 -.14474 (84*2π)/91511 weeks
85-.06161 -.22008 (85*2π)/91511 weeks
86.01055 -.18279 (86*2π)/91511 weeks
87-.0029 -.10895 (87*2π)/91511 weeks
88-.02599 -.05697 (88*2π)/91510 weeks
89.02526 -.14726 (89*2π)/91510 weeks
90-.07354 -.16093 (90*2π)/91510 weeks
91.04261 -.2624 (91*2π)/91510 weeks
92-.09269 -.10192 (92*2π)/91510 weeks
93-.00024 -.22618 (93*2π)/91510 weeks
94-.12653 -.20907 (94*2π)/91510 weeks
95-.0379 -.23322 (95*2π)/91510 weeks
96-.04518 -.09766 (96*2π)/91510 weeks
97-.01763 -.17306 (97*2π)/9159 weeks
98-.05729 -.31322 (98*2π)/9159 weeks
99-.069 -.17061 (99*2π)/9159 weeks
100-.09953 -.16478 (100*2π)/9159 weeks
101-.13393 -.08022 (101*2π)/9159 weeks
102-.053 -.0951 (102*2π)/9159 weeks
103-.09702 -.1249 (103*2π)/9159 weeks
104-.07962 -.14244 (104*2π)/9159 weeks
105-.08219 -.1706 (105*2π)/9159 weeks
106-.08913 -.08704 (106*2π)/9159 weeks
107-.10838 -.11477 (107*2π)/9159 weeks
108-.03738 -.10758 (108*2π)/9158 weeks
109-.03795 -.08724 (109*2π)/9158 weeks
110-.04961 -.14091 (110*2π)/9158 weeks
111.03515 -.13143 (111*2π)/9158 weeks
112-.05801 -.14095 (112*2π)/9158 weeks
113-.05511 -.12108 (113*2π)/9158 weeks
114-.13397 -.15066 (114*2π)/9158 weeks
115-.02902 -.08906 (115*2π)/9158 weeks
116-.08162 -.11011 (116*2π)/9158 weeks
117-.05044 -.11265 (117*2π)/9158 weeks
118-.13177 -.1154 (118*2π)/9158 weeks
119.01445 -.17283 (119*2π)/9158 weeks
120-.0746 -.08757 (120*2π)/9158 weeks
121-.10191 -.07373 (121*2π)/9158 weeks
122-.10444 -.11183 (122*2π)/9158 weeks
123-.04014 -.11086 (123*2π)/9157 weeks
124-.07745 -.08592 (124*2π)/9157 weeks
125-.10123 -.05205 (125*2π)/9157 weeks
126-.02733 -.08641 (126*2π)/9157 weeks
127-.0753 -.06257 (127*2π)/9157 weeks
128-.06684 -.06595 (128*2π)/9157 weeks
129.03807 -.11015 (129*2π)/9157 weeks
130-.06616 -.11125 (130*2π)/9157 weeks
131-.00229 -.09819 (131*2π)/9157 weeks
132-.10699 -.15381 (132*2π)/9157 weeks
133-.04168 -.07868 (133*2π)/9157 weeks
134-.10053 -.04868 (134*2π)/9157 weeks
135-.02429 -.07123 (135*2π)/9157 weeks
136-.00463 -.07008 (136*2π)/9157 weeks
137-.0927 -.08735 (137*2π)/9157 weeks
138-.02612 -.12335 (138*2π)/9157 weeks
139-.00992 -.11183 (139*2π)/9157 weeks
140-.06303 -.04614 (140*2π)/9157 weeks
141-.07951 -.11028 (141*2π)/9156 weeks
142-.05623 -.10351 (142*2π)/9156 weeks
143-.00896 -.12768 (143*2π)/9156 weeks
144-.02726 -.05089 (144*2π)/9156 weeks
145-.02172 -.14298 (145*2π)/9156 weeks
146-.097 -.08269 (146*2π)/9156 weeks
147-.12027 -.12671 (147*2π)/9156 weeks
148-.00985 -.07654 (148*2π)/9156 weeks
149-.03682 -.06949 (149*2π)/9156 weeks
150-.04084 -.0101 (150*2π)/9156 weeks
151-.04661 -.06482 (151*2π)/9156 weeks
152-.01392 -.0705 (152*2π)/9156 weeks
153-.00841 -.04573 (153*2π)/9156 weeks
154-.00439 -.09467 (154*2π)/9156 weeks
155-.03586 -.09212 (155*2π)/9156 weeks
156.02112 -.12341 (156*2π)/9156 weeks
157-.06204 -.081 (157*2π)/9156 weeks
158.0391 -.12129 (158*2π)/9156 weeks
159-.05141 -.0775 (159*2π)/9156 weeks
160-.01146 -.12434 (160*2π)/9156 weeks
161.0171 -.12799 (161*2π)/9156 weeks
162-.01348 -.13364 (162*2π)/9156 weeks
163-.06669 -.13658 (163*2π)/9156 weeks
164-.03314 -.13668 (164*2π)/9156 weeks
165-.038 -.09791 (165*2π)/9156 weeks
166-.14988 -.07385 (166*2π)/9156 weeks
167-.04501 -.11785 (167*2π)/9155 weeks
168-.00513 -.10387 (168*2π)/9155 weeks
169-.06416 -.07147 (169*2π)/9155 weeks
170-.02325 -.11654 (170*2π)/9155 weeks
171-.06447 -.06224 (171*2π)/9155 weeks
172-.04172 -.08986 (172*2π)/9155 weeks
173-.06314 -.04595 (173*2π)/9155 weeks
174-.0067 -.11776 (174*2π)/9155 weeks
175-.03648 -.0854 (175*2π)/9155 weeks
176-.10502 -.06695 (176*2π)/9155 weeks
177-.01632 -.06532 (177*2π)/9155 weeks
178-.01946 -.03713 (178*2π)/9155 weeks
179-.05503 -.09043 (179*2π)/9155 weeks
180-.06548 -.08507 (180*2π)/9155 weeks
181.00361 -.06374 (181*2π)/9155 weeks
182-.02979 -.10472 (182*2π)/9155 weeks
183-.00236 -.12687 (183*2π)/9155 weeks
184-.05466 -.07582 (184*2π)/9155 weeks
185-.046 -.08637 (185*2π)/9155 weeks
186-.05094 -.08409 (186*2π)/9155 weeks
187-.06503 -.05182 (187*2π)/9155 weeks
188-.04356 -.14184 (188*2π)/9155 weeks
189-.03837 -.09183 (189*2π)/9155 weeks
190-.09849 -.12127 (190*2π)/9155 weeks
191-.04703 -.0463 (191*2π)/9155 weeks
192-.03816 -.12863 (192*2π)/9155 weeks
193-.04811 -.09431 (193*2π)/9155 weeks
194-.06266 -.12534 (194*2π)/9155 weeks
195-.07728 -.06904 (195*2π)/9155 weeks
196-.04338 -.10064 (196*2π)/9155 weeks
197-.03125 -.11234 (197*2π)/9155 weeks
198-.08096 -.07557 (198*2π)/9155 weeks
199-.13879 -.05428 (199*2π)/9155 weeks
200-.07305 .00134 (200*2π)/9155 weeks
201-.01607 -.04385 (201*2π)/9155 weeks
202-.08104 -.08287 (202*2π)/9155 weeks
203-.07325 -.09537 (203*2π)/9155 weeks
204-.09069 -.05314 (204*2π)/9154 weeks
205-.05701 -.06407 (205*2π)/9154 weeks
206-.01591 -.07063 (206*2π)/9154 weeks
207-.08261 -.05151 (207*2π)/9154 weeks
208-.0893 -.0684 (208*2π)/9154 weeks
209-.09565 -.04668 (209*2π)/9154 weeks
210-.03883 -.04572 (210*2π)/9154 weeks
211-.02592 -.04568 (211*2π)/9154 weeks
212-.08377 -.0402 (212*2π)/9154 weeks
213.01136 -.10635 (213*2π)/9154 weeks
214-.02095 -.02732 (214*2π)/9154 weeks
215-.0648 -.09439 (215*2π)/9154 weeks
216-.09387 -.08022 (216*2π)/9154 weeks
217-.09122 -.07963 (217*2π)/9154 weeks
218-.11146 -.10036 (218*2π)/9154 weeks
219-.06885 -.05971 (219*2π)/9154 weeks
220-.0652 -.02158 (220*2π)/9154 weeks
221-.10987 -.03045 (221*2π)/9154 weeks
222-.08086 -.05078 (222*2π)/9154 weeks
223-.07311 -.01454 (223*2π)/9154 weeks
224-.07869 -.0742 (224*2π)/9154 weeks
225-.04844 -.04991 (225*2π)/9154 weeks
226-.05967 -.04865 (226*2π)/9154 weeks
227-.08357 -.03133 (227*2π)/9154 weeks
228-.03999 -.07298 (228*2π)/9154 weeks
229-.09678 -.00935 (229*2π)/9154 weeks
230-.03595 -.00116 (230*2π)/9154 weeks
231-.0772 -.04483 (231*2π)/9154 weeks
232-.01517 -.04287 (232*2π)/9154 weeks
233-.05181 -.0747 (233*2π)/9154 weeks
234-.05005 -.01686 (234*2π)/9154 weeks
235-.03195 -.04258 (235*2π)/9154 weeks
236-.08402 -.05135 (236*2π)/9154 weeks
237-.0325 -.0323 (237*2π)/9154 weeks
238-.05784 -.11071 (238*2π)/9154 weeks
239-.04609 -.02094 (239*2π)/9154 weeks
240-.08009 -.08902 (240*2π)/9154 weeks
241-.03708 -.06725 (241*2π)/9154 weeks
242-.0808 -.05733 (242*2π)/9154 weeks
243-.08819 -.02837 (243*2π)/9154 weeks
244-.08038 -.03487 (244*2π)/9154 weeks
245-.07213 -.02336 (245*2π)/9154 weeks
246-.0496 -.01968 (246*2π)/9154 weeks
247-.04719 -.05239 (247*2π)/9154 weeks
248-.03864 -.01403 (248*2π)/9154 weeks
249-.05245 -.02454 (249*2π)/9154 weeks
250-.09868 -.0089 (250*2π)/9154 weeks
251-.0596 -.0398 (251*2π)/9154 weeks
252-.01808 -.02422 (252*2π)/9154 weeks
253-.03479 -.04313 (253*2π)/9154 weeks
254-.10171 -.03259 (254*2π)/9154 weeks
255-.00547 -.02341 (255*2π)/9154 weeks
256-.05254 -.02116 (256*2π)/9154 weeks
257-.01505 -.03088 (257*2π)/9154 weeks
258-.02509 -.02275 (258*2π)/9154 weeks
259-.01594 -.03755 (259*2π)/9154 weeks
260-.00616 -.06364 (260*2π)/9154 weeks
261-.01928 -.07792 (261*2π)/9154 weeks
262-.0584 -.09819 (262*2π)/9153 weeks
263-.10406 -.04788 (263*2π)/9153 weeks
264-.04127 -.01966 (264*2π)/9153 weeks
265-.07935 -.07532 (265*2π)/9153 weeks
266-.02837 -.04702 (266*2π)/9153 weeks
267-.05734 -.0654 (267*2π)/9153 weeks
268-.03329 -.02964 (268*2π)/9153 weeks
269-.0634 -.06213 (269*2π)/9153 weeks
270-.06009 -.04766 (270*2π)/9153 weeks
271-.06201 -.06586 (271*2π)/9153 weeks
272-.08997 -.01518 (272*2π)/9153 weeks
273-.07837 -.0317 (273*2π)/9153 weeks
274-.05492 -.01861 (274*2π)/9153 weeks
275-.05832 -.0287 (275*2π)/9153 weeks
276-.06718 -.05142 (276*2π)/9153 weeks
277-.06258 .00929 (277*2π)/9153 weeks
278-.08553 .01915 (278*2π)/9153 weeks
279-.02239 -.01895 (279*2π)/9153 weeks
280-.04053 -.0171 (280*2π)/9153 weeks
281-.04166 .01362 (281*2π)/9153 weeks
282-.04613 -.01459 (282*2π)/9153 weeks
283-.05179 -.0253 (283*2π)/9153 weeks
284-.05269 -.02262 (284*2π)/9153 weeks
285-.0388 -.01195 (285*2π)/9153 weeks
286-.05441 .00519 (286*2π)/9153 weeks
287-.06636 -.02117 (287*2π)/9153 weeks
288-.01553 -.01747 (288*2π)/9153 weeks
289-.04193 -.02701 (289*2π)/9153 weeks
290-.05118 .00258 (290*2π)/9153 weeks
291-.00644 -.01601 (291*2π)/9153 weeks
292-.05519 -.03506 (292*2π)/9153 weeks
293-.06077 -.04942 (293*2π)/9153 weeks
294-.02562 -.04461 (294*2π)/9153 weeks
295-.04858 -.02802 (295*2π)/9153 weeks
296-.04518 -.03885 (296*2π)/9153 weeks
297-.05469 -.00627 (297*2π)/9153 weeks
298-.02525 -.00988 (298*2π)/9153 weeks
299-.06143 -.04947 (299*2π)/9153 weeks
300-.00968 -.01971 (300*2π)/9153 weeks
301-.03459 -.04689 (301*2π)/9153 weeks
302-.03578 -.0516 (302*2π)/9153 weeks
303-.04374 -.06695 (303*2π)/9153 weeks
304-.06806 .00444 (304*2π)/9153 weeks
305-.06329 -.04275 (305*2π)/9153 weeks
306-.07583 -.00941 (306*2π)/9153 weeks
307-.05666 -.04497 (307*2π)/9153 weeks
308-.05018 -.00018 (308*2π)/9153 weeks
309-.0444 -.04752 (309*2π)/9153 weeks
310-.03658 -.03894 (310*2π)/9153 weeks
311-.04553 -.00157 (311*2π)/9153 weeks
312-.02904 -.07145 (312*2π)/9153 weeks
313-.03502 -.03924 (313*2π)/9153 weeks
314-.05068 -.03708 (314*2π)/9153 weeks
315-.02422 -.06936 (315*2π)/9153 weeks
316-.07878 -.03938 (316*2π)/9153 weeks
317-.0516 -.01633 (317*2π)/9153 weeks
318-.06122 -.01102 (318*2π)/9153 weeks
319-.06024 .0048 (319*2π)/9153 weeks
320-.05648 -.0204 (320*2π)/9153 weeks
321-.00268 -.03458 (321*2π)/9153 weeks
322-.02392 -.05871 (322*2π)/9153 weeks
323-.0626 -.06681 (323*2π)/9153 weeks
324-.06267 -.05985 (324*2π)/9153 weeks
325-.07182 -.02383 (325*2π)/9153 weeks
326-.13993 -.02898 (326*2π)/9153 weeks
327-.05549 .01585 (327*2π)/9153 weeks
328-.0581 -.00597 (328*2π)/9153 weeks
329-.02553 -.02266 (329*2π)/9153 weeks
330-.02587 -.02158 (330*2π)/9153 weeks
331-.02492 -.00998 (331*2π)/9153 weeks
332-.04547 -.02844 (332*2π)/9153 weeks
333-.06176 -.03655 (333*2π)/9153 weeks
334-.03866 -.01709 (334*2π)/9153 weeks
335-.06459 -.02135 (335*2π)/9153 weeks
336-.05072 .00527 (336*2π)/9153 weeks
337-.04956 -.03676 (337*2π)/9153 weeks
338-.04625 -.01594 (338*2π)/9153 weeks
339-.02311 -.02493 (339*2π)/9153 weeks
340-.03749 -.03303 (340*2π)/9153 weeks
341-.06344 -.00375 (341*2π)/9153 weeks
342-.05874 -.06563 (342*2π)/9153 weeks
343-.03146 -.02247 (343*2π)/9153 weeks
344-.06724 -.00502 (344*2π)/9153 weeks
345-.03246 -.01753 (345*2π)/9153 weeks
346-.08197 -.01232 (346*2π)/9153 weeks
347-.0418 -.02158 (347*2π)/9153 weeks
348-.05077 -.00055 (348*2π)/9153 weeks
349-.03752 -.0164 (349*2π)/9153 weeks
350-.02531 -.00705 (350*2π)/9153 weeks
351-.03304 -.00699 (351*2π)/9153 weeks
352-.02235 -.04019 (352*2π)/9153 weeks
353-.05386 .00109 (353*2π)/9153 weeks
354-.04898 -.01349 (354*2π)/9153 weeks
355-.0575 -.02097 (355*2π)/9153 weeks
356-.05477 -.02108 (356*2π)/9153 weeks
357-.01359 -.01833 (357*2π)/9153 weeks
358-.04357 -.02383 (358*2π)/9153 weeks
359-.02103 -.02761 (359*2π)/9153 weeks
360-.04669 -.03922 (360*2π)/9153 weeks
361-.05891 .00546 (361*2π)/9153 weeks
362-.05882 -.02045 (362*2π)/9153 weeks
363-.03755 -.00092 (363*2π)/9153 weeks
364-.03774 -.0262 (364*2π)/9153 weeks
365-.06726 -.00985 (365*2π)/9153 weeks
366-.02494 .00664 (366*2π)/9153 weeks
367-.03559 -.01351 (367*2π)/9152 weeks
368-.02877 -.02776 (368*2π)/9152 weeks
369-.07097 -.05742 (369*2π)/9152 weeks
370-.04932 -.01666 (370*2π)/9152 weeks
371-.04808 -.03052 (371*2π)/9152 weeks
372-.04686 -.03476 (372*2π)/9152 weeks
373-.06408 -.03141 (373*2π)/9152 weeks
374-.05604 -.00742 (374*2π)/9152 weeks
375-.07339 -.02775 (375*2π)/9152 weeks
376-.06001 -.01012 (376*2π)/9152 weeks
377-.06072 -.02313 (377*2π)/9152 weeks
378-.06446 -.00992 (378*2π)/9152 weeks
379-.05893 -.01082 (379*2π)/9152 weeks
380-.03325 .00131 (380*2π)/9152 weeks
381-.05384 -.02811 (381*2π)/9152 weeks
382-.04265 -.03051 (382*2π)/9152 weeks
383-.06141 -.04785 (383*2π)/9152 weeks
384-.06569 -.00879 (384*2π)/9152 weeks
385-.02856 -.01042 (385*2π)/9152 weeks
386-.05781 -.00108 (386*2π)/9152 weeks
387-.04611 -.02338 (387*2π)/9152 weeks
388-.06048 .00139 (388*2π)/9152 weeks
389-.03517 .00474 (389*2π)/9152 weeks
390-.07483 -.01742 (390*2π)/9152 weeks
391-.05293 -.0222 (391*2π)/9152 weeks
392-.06865 .00553 (392*2π)/9152 weeks
393-.0575 -.02067 (393*2π)/9152 weeks
394-.04724 .00195 (394*2π)/9152 weeks
395-.05768 -.00064 (395*2π)/9152 weeks
396-.07769 -.01268 (396*2π)/9152 weeks
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845-.22336 .1243 (845*2π)/9151 weeks
846-.16952 .14363 (846*2π)/9151 weeks
847-.10961 .30444 (847*2π)/9151 weeks
848-.10914 .2603 (848*2π)/9151 weeks
849.02416 .33093 (849*2π)/9151 weeks
850-.11213 .2771 (850*2π)/9151 weeks
851.00963 .32584 (851*2π)/9151 weeks
852.08116 .36887 (852*2π)/9151 weeks
853.08771 .23775 (853*2π)/9151 weeks
854.07969 .16426 (854*2π)/9151 weeks
855-.04479 .16831 (855*2π)/9151 weeks
856-.11306 .24606 (856*2π)/9151 weeks
857.03423 .26872 (857*2π)/9151 weeks
858-.10597 .1122 (858*2π)/9151 weeks
859-.02665 .2866 (859*2π)/9151 weeks
860-.10086 .09004 (860*2π)/9151 weeks
861-.04709 .36782 (861*2π)/9151 weeks
862-.04623 .09664 (862*2π)/9151 weeks
863-.0264 .28917 (863*2π)/9151 weeks
864-.19212 .00965 (864*2π)/9151 weeks
865-.22993 .39961 (865*2π)/9151 weeks
866-.14133 .28516 (866*2π)/9151 weeks
867-.18615 .32134 (867*2π)/9151 weeks
868-.18238 .29149 (868*2π)/9151 weeks
869-.2367 .38252 (869*2π)/9151 weeks
870-.08505 .44551 (870*2π)/9151 weeks
871-.07501 .49456 (871*2π)/9151 weeks
872-.07767 .38747 (872*2π)/9151 weeks
873-.03344 .4252 (873*2π)/9151 weeks
874.12306 .60268 (874*2π)/9151 weeks
875-.04222 .38487 (875*2π)/9151 weeks
876-.03125 .24607 (876*2π)/9151 weeks
877-.09897 .23031 (877*2π)/9151 weeks
878-.13958 .6034 (878*2π)/9151 weeks
879.0776 .45239 (879*2π)/9151 weeks
880.03921 .4676 (880*2π)/9151 weeks
881.03341 .44361 (881*2π)/9151 weeks
882.01303 .44926 (882*2π)/9151 weeks
883.07751 .5166 (883*2π)/9151 weeks
884-.09408 .28112 (884*2π)/9151 weeks
885-.21413 .50661 (885*2π)/9151 weeks
886-.01059 .55481 (886*2π)/9151 weeks
887.04376 .57982 (887*2π)/9151 weeks
888-.03438 .65334 (888*2π)/9151 weeks
889-.06264 .36254 (889*2π)/9151 weeks
890-.07922 .71353 (890*2π)/9151 weeks
891-.0132 .58979 (891*2π)/9151 weeks
892-.09954 .70349 (892*2π)/9151 weeks
893.27119 .79403 (893*2π)/9151 weeks
894-.13571 .78149 (894*2π)/9151 weeks
895.04635 1.01203 (895*2π)/9151 weeks
896-.2827 .31435 (896*2π)/9151 weeks
897.05341 1.0589 (897*2π)/9151 weeks
898.04749 .75939 (898*2π)/9151 weeks
899-.10349 1.09495 (899*2π)/9151 weeks
900-.15159 .4769 (900*2π)/9151 weeks
901.16262 1.49994 (901*2π)/9151 weeks
902.39865 .93956 (902*2π)/9151 weeks
903-.20497 2.03702 (903*2π)/9151 weeks
904.03469 1.39243 (904*2π)/9151 weeks
905-.07188 1.88294 (905*2π)/9151 weeks
9061.08121 1.97021 (906*2π)/9151 weeks
907-.12678 1.71352 (907*2π)/9151 weeks
908.57778 2.23469 (908*2π)/9151 weeks
909-.54264 2.64993 (909*2π)/9151 weeks
910.99834 2.60906 (910*2π)/9151 weeks
911.07014 4.00434 (911*2π)/9151 weeks
9121.3582 3.97022 (912*2π)/9151 weeks
913.5286 9.71872 (913*2π)/9151 weeks | 15,839 | 35,949 | {"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-2017-30 | longest | en | 0.758085 |
http://www.acmerblog.com/hdu-1183-Not-magic%2Cbut-logic-1527.html | 1,481,387,747,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543316.16/warc/CC-MAIN-20161202170903-00158-ip-10-31-129-80.ec2.internal.warc.gz | 310,720,220 | 12,489 | 2013
12-03
# Not magic,but logic.
He(Harry) pulled open the next door, both of them hardly daring to look at what came next – but there was nothing very frightening in here, just a table with seven differently shaped bottles standing on it in a line.
‘Snape’s,’ said Harry. ‘What do we have to do?’
They stepped over the threshold and immediately a fire sprang up beind them in the doorway. It wasn’t ordinary fire either; it was purple. At the same instant, black flames shot up in the doorway leading onwards.They were trapped.
‘Look!’ Hermione seized a roll of paper lying next to the bottles. Harry looked over her shoulder to read it:
Danger lies before you, while safety lies behind,
One among us seven will let you move ahead,
Another will transport the drinker back instead,
Some among our number hold only nettle wine,
And others are killers, waiting hidden in line.
Choose, unless you wish to stay here for evermore,
Now,they must make a choice, could you help them?
However, we call the bottle who can take you ahead A,and the bottle who can take you back is named B.
The first line of the input is three integer N,M,P. It means there is N bottles of wine, M bottles of poison and P clues.(0<=N,M<=100,0<P<=1000)
Next P line would be one of these four:
No. i from left/right is safe/dangrous .
All wine/poison have a wine/poison on the left/right .
The A/B have a wine/poison/A/B on the left and a wine/poison/A/B on the right .
No. i from left/right cannot take you back/ahead .
Output will have only one line:
If you are sure where A and B is, just tell Harry their number from left.
If you aren’t sure, you only can say: ‘God help them.’
If you find that’s impossible, just say: ‘That’s impossible.’
2 4 6
All wine have a poison on the left .
The B have a poison on the left and a poison on the right .
The A have a wine on the left and a poison on the right .
No. 4 from right is safe .
No. 4 from left is safe .
No. 2 from right cannot take you back .
5 2
HintBoth PBPWAPPW and PBPWAPWP are possible solutions,where 'P' means poison and 'W' means wine.But however,we know where A and B are. | 542 | 2,121 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2016-50 | longest | en | 0.943111 |
https://itprospt.com/num/17871333/5-a-trucking-company-determined-that-onan-annual-basis-the | 1,660,907,785,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573667.83/warc/CC-MAIN-20220819100644-20220819130644-00093.warc.gz | 301,713,167 | 15,471 | 4
# 5. A trucking company determined that, onan annual basis, the distance traveled per truck is normallydistributed with a mean of 60 thousand miles and a populationst...
## Question
###### 5. A trucking company determined that, onan annual basis, the distance traveled per truck is normallydistributed with a mean of 60 thousand miles and a populationstandard deviation of 16 thousand miles. If a sample of 16 trucksis selected,(a) What is the probability that the average distance traveledis less than 58 thousandmiles? (b) What is the probability that the average distance traveledis more than 62 thousand miles?(c) What is the probability that the average distance traveledis betwe
5. A trucking company determined that, on an annual basis, the distance traveled per truck is normally distributed with a mean of 60 thousand miles and a population standard deviation of 16 thousand miles. If a sample of 16 trucks is selected, (a) What is the probability that the average distance traveled is less than 58 thousand miles? (b) What is the probability that the average distance traveled is more than 62 thousand miles? (c) What is the probability that the average distance traveled is between 58 thousand miles and 62 thousand miles? (d) Do we need the Central Limit Theorem to solve (a), (b), and (c)? Why or why not? Explain.
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3x-1 /x+k= 3-7/x+k what is the value of k?... | 2,991 | 10,182 | {"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} | 3.3125 | 3 | CC-MAIN-2022-33 | latest | en | 0.864848 |
http://mandysnotes.com/boolean-algebra | 1,511,335,379,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806509.31/warc/CC-MAIN-20171122065449-20171122085449-00019.warc.gz | 190,485,852 | 7,248 | # MandysNotes
## Boolean Algebra
### Sets and Subsets
By
Let C be the set of all possible animals.
Let be the subset of C that consists of all duck-billed animals.
Let B be the subset of C that consists of all mammals.
We illustrate these sets in the following diagram:
### Disjunction and Conjunction
By
Now suppose we consider only animals that are either duck-billed, or are mammals.
This is the set "A or B," the disjunction of A and B, written as:
$A \cup B,$
and illustrated in the following diagram:
This set, $$A \cup B,$$ contains ducklings, and rabbits, and platypuses.
### Negation
By
Suppose that the only information you have about a certain animal is that:
"This animal does not have a duck bill."
If A is the set of animals that have a duck bill, and C is the set of all animals, then the set of all animals that do not have a duck bill is the negation of A (with respect to C).
This is the set "not A," written as:
$\lnot A$
### Implication
By
Suppose we are looking for an animal, but all we know about it is that it either does not have a duck bill, or it is a mammal.
We write this as
"(not A) or B"
$= \left( \lnot A \right) \cup B$
$= A \rightarrow B.$
### Equivalence
By
Now suppose that we know that A implies B, and that B implies A
In symbols this is:
$\left( A \rightarrow B \right) \cap \left( B \rightarrow A \right)$
$= \left( \left( \lnot A \right) \cup B \right) \cap \left( \left( \lnot B \right) \cup A \right)$
$A \leftrightarrow B.$
In this case we say that A is equivalent to B.
This relation is symmetric, so B is also equivalent to A.
### Xor: Exclusive Or
By
If all we know is that A is not equivalent to B, we can write this as:
$\lnot \left( A \leftrightarrow B \right)$
$= \lnot \left[ \left( \left( \lnot A \right) \cup B \right) \cap \left( \left( \lnot B \right) \cup A \right) \right]$
### Nand
By
The negation of the set (A and B):
$\lnot \left( A \cap B \right)$
is called A nand B.
### Truth Values
By
Suppose that you are playing a geussing game with a friend. You have both agreed that you will guess an animal.
Your friend thinks of an animal and you try to guess what it is.
Let's assume, for the sake of this lesson, that your friend always tells the truth.
He tells you that the animal that he is thinking of is a duck-billed animal.
If we call the set of all animals C, and the set of all duck-billed animals A, then you friend has told you that the animal he is thinking of is in A
If x is the unkown animal, then the statement:
"x is a duck-billed animal"
is true.
That is: A is true.
We represent this fact graphically thus:
### Truth Values for Disjunction
By
Suppose your friend tells you that he is thinking of an animal and that this animal is either duck-billed, or a mammal.
In other words he is telling you that $$A \cup B$$ is true.
Assume that your friend is telling the truth.
He is telling you that whatever animal he is thinking of, it will be somewhere in the green area in the figure below.
### Truth Values for Conjunction
By
Suppose that your firend is thinking of an animal, and you guess that he is thinking of a platypus.
You say:
"The animal is both duck-billed and a mammal."
In other words, you are saying that $$A \cap B$$ is true.
You are saying that the animal will be in the green section of the figure below.
### Truth Values for Negation
By
Truth values for negation are very simple:
If $$A$$ is true, then $$\lnot A$$ is false.
### Tautology
By
Imagine your friend tells you that the animal he is thinking of is either duck-billed or not duck-billed.
In other words, he is telling you that either $$A$$ is true, or $$\lnot A$$ is true.
Equivalently he is telling you that the statement:
$A \cup \lnot A$
is true.
This is not very helpful in geussing which animal he is thinking of, but it does have the property of always being true.
In other words, the statement
$A \cup \lnot A$
is true if $$A$$ is true (and hence $$\lnot A$$ is false);
it is also true if $$A$$ is false (and hence $$\lnot A$$ is true).
Such a statement is called a tautology.
### Truth Values for Implication
By
"If the animal I am thinking of is duck-billed, then it is also a mammal."
Assume that he is telling you the truth (for now).
He is telling you that the statement:
$A \to B$
is true.
Logically, this is the same as saying that the statement:
$( \lnot A) \cup B$
is true.
That is, the animal must be somewhere in the green area in the figure below, and cannot be in the red area.
### Table of Truth Values
By
Here is a table with truth values for various binary relations on Boolean algebras. | 1,256 | 4,667 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2017-47 | latest | en | 0.872132 |
http://extraconversion.com/pressure/feet-of-air-0-0c/feet-of-air-0-0c-to-pounds-per-square-inch.html | 1,548,099,969,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583807724.75/warc/CC-MAIN-20190121193154-20190121215154-00036.warc.gz | 75,433,512 | 14,313 | # Feet of air [0 °C] to Pounds per square inch Conversion Calculator
## Sample Pressure Conversion
10 foot of head to pounds per square inch, the result is 4.335275019282 pounds per square inch 10 bar to tonnes per square meter, the result is 101.97162129779 tonnes per square meter 10 megapascal to tonnes per square meter, the result is 1019.7162129779 tonnes per square meter 10 kilopascal to tonnes per square meter, the result is 1.019716212978 tonnes per square meter 10 kip/square foot to pounds per square foot, the result is 10000 pounds per square foot 10 ounce/square inch to pounds per square inch, the result is 0.625 pounds per square inch 10 kip/square inch to pounds per square inch, the result is 10000 pounds per square inch 10 foot of head to bars, the result is 0.2989 bars 10 ton/square meter to newtons per square millimeter, the result is 0.0981 newtons per square millimeter 10 foot of head to kilopascals, the result is 29.890669 kilopascals
## Table Conversion
With the following tool, you can generate and print the feet of air [0 °C] to pounds per square inch reference table based on your own needs. You can find a dynamic tool at feet of air [0 °C] to pounds per square inch table chart (ft-air 0° to psi) or pounds per square inch to feet of air [0 °C] table chart (psi to ft-air 0° ).
ft-air 0° psi
1= 0.0006
2= 0.0011
3= 0.0017
4= 0.0022
5= 0.0028
6= 0.0034
7= 0.0039
8= 0.0045
9= 0.005
10= 0.0056
11= 0.0062
12= 0.0067
13= 0.0073
14= 0.0078
15= 0.0084
16= 0.009
17= 0.0095
18= 0.0101
19= 0.0106
20= 0.0112
21= 0.0118
22= 0.0123
23= 0.0129
24= 0.0135
25= 0.014
psift-air 0°
1= 1784.3164680894
2= 3568.6329361789
3= 5352.9494042683
4= 7137.2658723578
5= 8921.5823404472
6= 10705.898808537
7= 12490.215276626
8= 14274.531744716
9= 16058.848212805
10= 17843.164680894
11= 19627.481148984
12= 21411.797617073
13= 23196.114085163
14= 24980.430553252
15= 26764.747021342
16= 28549.063489431
17= 30333.379957521
18= 32117.69642561
19= 33902.0128937
20= 35686.329361789
21= 37470.645829878
22= 39254.962297968
23= 41039.278766057
24= 42823.595234147
25= 44607.911702236
## Unit Information
foot of air [0 °C] is a unit of pressure equal to about 3.8640888 pascals and symbol is ft-air 0° .
1 ft-air 0° = 3.8135588E-5 atm feet of air [0 °C] to atmospheres converter feet of air [0 °C] to atmospheres table 1 ft-air 0° = 38.640888 barad feet of air [0 °C] to barads converter feet of air [0 °C] to barads table 1 ft-air 0° = 3.8640888E-5 bar feet of air [0 °C] to bars converter feet of air [0 °C] to bars table 1 ft-air 0° = 38.640888 Ba feet of air [0 °C] to baryes converter feet of air [0 °C] to baryes table 1 ft-air 0° = 38.640888 dyn/c² feet of air [0 °C] to dynes per square centimeter converter feet of air [0 °C] to dynes per square centimeter table 1 ft-air 0° = 1.055100914779 ft-air 15° feet of air [0 °C] to feet of air [15 °C] converter feet of air [0 °C] to feet of air [15 °C] table 1 ft-air 0° = 0.0013 ft-head feet of air [0 °C] to feet of head converter feet of air [0 °C] to feet of head table 1 ft-air 0° = 0.0001 ft-mercury 0° feet of air [0 °C] to feet of mercury [0 °C] converter feet of air [0 °C] to feet of mercury [0 °C] table 1 ft-air 0° = 0.0013 ft-water 4° feet of air [0 °C] to feet of water [4 °C] converter feet of air [0 °C] to feet of water [4 °C] table 1 ft-air 0° = 3.8640888E-14 Gbar feet of air [0 °C] to gigabars converter feet of air [0 °C] to gigabars table 1 ft-air 0° = 3.864E-9 Gpa feet of air [0 °C] to gigapascals converter feet of air [0 °C] to gigapascals table 1 ft-air 0° = 12 in-air 0° feet of air [0 °C] to inchs of air [0 °C] converter feet of air [0 °C] to inchs of air [0 °C] table 1 ft-air 0° = 12.661210977352 in-air 15° feet of air [0 °C] to inchs of air [15 °C] converter feet of air [0 °C] to inchs of air [15 °C] table 1 ft-air 0° = 0.0011 in-mercury 0° feet of air [0 °C] to inchs of mercury [0 °C] converter feet of air [0 °C] to inchs of mercury [0 °C] table 1 ft-air 0° = 0.0155 in-water 4° feet of air [0 °C] to inchs of water [4 °C] converter feet of air [0 °C] to inchs of water [4 °C] table 1 ft-air 0° = 3.8641E-8 kbar feet of air [0 °C] to kilobars converter feet of air [0 °C] to kilobars table 1 ft-air 0° = 0.0039 kPa feet of air [0 °C] to kilopascals converter feet of air [0 °C] to kilopascals table 1 ft-air 0° = 0.0001 kip/ft² feet of air [0 °C] to kips per square foot converter feet of air [0 °C] to kips per square foot table 1 ft-air 0° = 5.60439E-7 kip/in² feet of air [0 °C] to kips per square inch converter feet of air [0 °C] to kips per square inch table 1 ft-air 0° = 3.9E-11 Mbar feet of air [0 °C] to megabars converter feet of air [0 °C] to megabars table 1 ft-air 0° = 3.864089E-6 Mpa feet of air [0 °C] to megapascals converter feet of air [0 °C] to megapascals table 1 ft-air 0° = 0.3048 m-air 0° feet of air [0 °C] to meters of air [0 °C] converter feet of air [0 °C] to meters of air [0 °C] table 1 ft-air 0° = 0.3216 m-air 15° feet of air [0 °C] to meters of air [15 °C] converter feet of air [0 °C] to meters of air [15 °C] table 1 ft-air 0° = 38.640888 µbar feet of air [0 °C] to microbars converter feet of air [0 °C] to microbars table 1 ft-air 0° = 0.0386 mbar feet of air [0 °C] to millibars converter feet of air [0 °C] to millibars table 1 ft-air 0° = 3864.0888 mPa feet of air [0 °C] to millipascals converter feet of air [0 °C] to millipascals table 1 ft-air 0° = 3.8640888 N/m² feet of air [0 °C] to newtons per square meter converter feet of air [0 °C] to newtons per square meter table 1 ft-air 0° = 3.864089E-6 N/mm² feet of air [0 °C] to newtons per square millimeter converter feet of air [0 °C] to newtons per square millimeter table 1 ft-air 0° = 0.009 oz/in² feet of air [0 °C] to ounces per square inch converter feet of air [0 °C] to ounces per square inch table 1 ft-air 0° = 3.8640888 Pa feet of air [0 °C] to pascals converter feet of air [0 °C] to pascals table 1 ft-air 0° = 0.0039 pz feet of air [0 °C] to piezes converter feet of air [0 °C] to piezes table 1 ft-air 0° = 0.0807 psft feet of air [0 °C] to pounds per square foot converter feet of air [0 °C] to pounds per square foot table 1 ft-air 0° = 0.0006 psi feet of air [0 °C] to pounds per square inch converter feet of air [0 °C] to pounds per square inch table 1 ft-air 0° = 3.940274E-5 at feet of air [0 °C] to technical atmosphere converter feet of air [0 °C] to technical atmosphere table 1 ft-air 0° = 4.0E-12 Tpa feet of air [0 °C] to terapascals converter feet of air [0 °C] to terapascals table 1 ft-air 0° = 4.0676622E-5 t/ft²-long feet of air [0 °C] to tonnes per square foot [long] converter feet of air [0 °C] to tonnes per square foot [long] table 1 ft-air 0° = 4.0351586E-5 t/ft²-short feet of air [0 °C] to tonnes per square foot [short] converter feet of air [0 °C] to tonnes per square foot [short] table 1 ft-air 0° = 2.82477E-7 t/in²-long feet of air [0 °C] to tonnes per square inch [long] converter feet of air [0 °C] to tonnes per square inch [long] table 1 ft-air 0° = 2.80219E-7 t/in²-short feet of air [0 °C] to tonnes per square inch [short] converter feet of air [0 °C] to tonnes per square inch [short] table 1 ft-air 0° = 0.0004 t/m² feet of air [0 °C] to tonnes per square meter converter feet of air [0 °C] to tonnes per square meter table 1 ft-air 0° = 0.029 torr feet of air [0 °C] to torrs converter feet of air [0 °C] to torrs table
Pound / square inch is a unit of pressure equal to about 6894.75728 pascals and symbol is psi.
1 psi = 0.068 atm pounds per square inch to atmospheres converter pounds per square inch to atmospheres table 1 psi = 68947.5728 barad pounds per square inch to barads converter pounds per square inch to barads table 1 psi = 0.0689 bar pounds per square inch to bars converter pounds per square inch to bars table 1 psi = 68947.5728 Ba pounds per square inch to baryes converter pounds per square inch to baryes table 1 psi = 68947.5728 dyn/c² pounds per square inch to dynes per square centimeter converter pounds per square inch to dynes per square centimeter table 1 psi = 1784.3164680894 ft-air 0° pounds per square inch to feet of air [0 °C] converter pounds per square inch to feet of air [0 °C] table 1 psi = 1882.633937737 ft-air 15° pounds per square inch to feet of air [15 °C] converter pounds per square inch to feet of air [15 °C] table 1 psi = 2.306658736879 ft-head pounds per square inch to feet of head converter pounds per square inch to feet of head table 1 psi = 0.1697 ft-mercury 0° pounds per square inch to feet of mercury [0 °C] converter pounds per square inch to feet of mercury [0 °C] table 1 psi = 2.306658736879 ft-water 4° pounds per square inch to feet of water [4 °C] converter pounds per square inch to feet of water [4 °C] table 1 psi = 6.9E-11 Gbar pounds per square inch to gigabars converter pounds per square inch to gigabars table 1 psi = 6.894757E-6 Gpa pounds per square inch to gigapascals converter pounds per square inch to gigapascals table 1 psi = 21411.797617073 in-air 0° pounds per square inch to inchs of air [0 °C] converter pounds per square inch to inchs of air [0 °C] table 1 psi = 22591.607252844 in-air 15° pounds per square inch to inchs of air [15 °C] converter pounds per square inch to inchs of air [15 °C] table 1 psi = 2.036020657601 in-mercury 0° pounds per square inch to inchs of mercury [0 °C] converter pounds per square inch to inchs of mercury [0 °C] table 1 psi = 27.679904842545 in-water 4° pounds per square inch to inchs of water [4 °C] converter pounds per square inch to inchs of water [4 °C] table 1 psi = 0.0001 kbar pounds per square inch to kilobars converter pounds per square inch to kilobars table 1 psi = 6.89475728 kPa pounds per square inch to kilopascals converter pounds per square inch to kilopascals table 1 psi = 0.144 kip/ft² pounds per square inch to kips per square foot converter pounds per square inch to kips per square foot table 1 psi = 0.001 kip/in² pounds per square inch to kips per square inch converter pounds per square inch to kips per square inch table 1 psi = 6.8948E-8 Mbar pounds per square inch to megabars converter pounds per square inch to megabars table 1 psi = 0.0069 Mpa pounds per square inch to megapascals converter pounds per square inch to megapascals table 1 psi = 543.85964629973 m-air 0° pounds per square inch to meters of air [0 °C] converter pounds per square inch to meters of air [0 °C] table 1 psi = 573.82683901331 m-air 15° pounds per square inch to meters of air [15 °C] converter pounds per square inch to meters of air [15 °C] table 1 psi = 68947.5728 µbar pounds per square inch to microbars converter pounds per square inch to microbars table 1 psi = 68.9475728 mbar pounds per square inch to millibars converter pounds per square inch to millibars table 1 psi = 6894757.28 mPa pounds per square inch to millipascals converter pounds per square inch to millipascals table 1 psi = 6894.75728 N/m² pounds per square inch to newtons per square meter converter pounds per square inch to newtons per square meter table 1 psi = 0.0069 N/mm² pounds per square inch to newtons per square millimeter converter pounds per square inch to newtons per square millimeter table 1 psi = 16 oz/in² pounds per square inch to ounces per square inch converter pounds per square inch to ounces per square inch table 1 psi = 6894.75728 Pa pounds per square inch to pascals converter pounds per square inch to pascals table 1 psi = 6.89475728 pz pounds per square inch to piezes converter pounds per square inch to piezes table 1 psi = 144 psft pounds per square inch to pounds per square foot converter pounds per square inch to pounds per square foot table 1 psi = 0.0703 at pounds per square inch to technical atmosphere converter pounds per square inch to technical atmosphere table 1 psi = 6.895E-9 Tpa pounds per square inch to terapascals converter pounds per square inch to terapascals table 1 psi = 0.0726 t/ft²-long pounds per square inch to tonnes per square foot [long] converter pounds per square inch to tonnes per square foot [long] table 1 psi = 0.072 t/ft²-short pounds per square inch to tonnes per square foot [short] converter pounds per square inch to tonnes per square foot [short] table 1 psi = 0.0005 t/in²-long pounds per square inch to tonnes per square inch [long] converter pounds per square inch to tonnes per square inch [long] table 1 psi = 0.0005 t/in²-short pounds per square inch to tonnes per square inch [short] converter pounds per square inch to tonnes per square inch [short] table 1 psi = 0.7031 t/m² pounds per square inch to tonnes per square meter converter pounds per square inch to tonnes per square meter table 1 psi = 51.714931860272 torr pounds per square inch to torrs converter pounds per square inch to torrs table
## Atmosphere
atmosphere is a unit of pressure equal to 101 325 pascals and symbol is atm.
barad is a unit of pressure equal to 0.1 pascals and symbol is barad.
## Bar
The bar is a unit of pressure defined as 100 kilopascals. It is about equal to the atmospheric pressure on Earth at sea level.
## Barye
barye is a unit of pressure equal to 0.1 pascals and symbol is Ba.
## Dyne / square centimeter
dyne / square centimeter is a unit of pressure equal to 0.1 pascals and symbol is dyn/c².
## Foot of air [0 °C]
foot of air [0 °C] is a unit of pressure equal to about 3.8640888 pascals and symbol is ft-air 0° .
## Foot of air [15 °C]
foot of air [15 °C] is a unit of pressure equal to about 3.6622931 pascals and symbol is ft-air 15° .
foot of head is a unit of pressure equal to about 2989.0669 pascals and symbol is ft-head.
## Foot of mercury [0 °C]
foot of mercury [0 °C] is a unit of pressure equal to about 40636.664 pascals and symbol is ft-mercury 0°.
## Foot of water [4 °C]
foot of water [4 °C] is a unit of pressure equal to about 2989.0669 pascals and symbol is ft-water 4°.
## Gigabar
gigabar is a unit of pressure, a combination of metric-prefix "giga" and unit of pressure "bar", equal to 1014 pascals and symbol is Gbar.
## Gigapascal
Gigapascal is, a combination of metric-prefix "giga" and the SI derived unit of pressure "pascal", a unit of pressure equal to 109 pascals and symbol is Gpa.
## Inch of air [0 °C]
Inch of air [0 °C] is a unit of pressure equal to about 0.3220074 pascals and symbol is in-air 0°.
## Inch of air [15 °C]
Inch of air [15 °C] is a unit of pressure equal to about 0.3051910916666667 pascals and symbol is in-air 15°.
## Inch of mercury [0 °C]
Inch of mercury [0 °C] is a unit of pressure equal to about 3386.388666666667 pascals and symbol is in-mercury 0°.
## Inch of water [4 °C]
Inch of water [4 °C] is a unit of pressure equal to about 249.0889083333333 pascals and symbol is in-water 4°.
## Kilobar
Kilobar is, a mixture of metric-prefix "kilo" and unit of pressure "bar", a unit of pressure equal to 100000000 pascals and symbol is kbar.
## Kilopascal
Kilopascal is, a combination of metric-prefix "kilo" and the SI derived unit of pressure "pascal", a unit of pressure equal to xx pascals and symbol is kPa.
## Kip / square foot
Kip / square foot is a unit of pressure equal to about 47880.25888888889 pascals and symbol is kip/ft².
## Kip / square inch
Kip/square inch is a unit of pressure equal to about 6894757.28 pascals and symbol is kip/in².
## Megabar
Megabar (mega + bar) is a unit of pressure equal to 1011 pascals and symbol is Mbar.
## Megapascal
Megapascal is a unit of pressure equal to 106 pascals and symbol is Mpa.
## Meter of air [0 °C]
Meter of air [0 °C] is a unit of pressure equal to about 12.677457 pascals and symbol is m-air 0°.
## Meter of air [15 °C]
Meter of air [15 °C] is a unit of pressure equal to about 12.015397 pascals and symbol is m-air 15°.
## Microbar
Microbar (micro + bar) is a unit of pressure equal to 0.1 pascals and symbol is µbar.
## Millibar
Millibar (milli + bar) is a unit of pressure equal to 100 pascals and symbol is mbar.
## Millipascal
Millipascal is a unit of pressure equal to 0.001 pascals and symbol is mPa.
## Newton / square meter
Newton / square meter is a unit of pressure equivalent to pascals and symbol is N/m².
## Newton / square millimeter
Newton / square millimeter is a unit of pressure equal to 106 pascals and symbol is N/mm².
## Ounce / square inch
Ounce / square inch is a unit of pressure equal to about 430.92233 pascals and symbol is oz/in².
## Pascal
Pascal is the SI derived unit of pressure (symbol Pa).
## Pieze
The pièze is the unit of pressure in the metre-tonne-second system of units (mts system), used, e.g., in the former Soviet Union 1933-1955. It is defined as one sthène per square metre. The symbol is pz.
## Pound / square foot
Pound / square foot is a unit of pressure equal to about 47.88 pascals and symbol is psft.
## Pound / square inch
Pound / square inch is a unit of pressure equal to about 6894.75728 pascals and symbol is psi.
## Technical atmosphere
Technical atmosphere is a unit of pressure equal to about 98066.5 pascals and symbol is at.
## Terapascal
Terapascal is a combination of metric-prefix "tera" and SI derived unit of pressure "pascal", it's equal to 1012 pascals and symbol is Tpa.
## Ton / square foot [long]
Ton / square foot [long] is a unit of pressure equal to about 94995.32252 pascals and symbol is t/ft²-long.
## Ton / square foot [short]
Ton / square foot [short] is a unit of pressure equal to about 95760.52 pascals and symbol is t/ft²-short.
## Ton / square inch [long]
Ton / square inch [long] is a unit of pressure equal to about 13679326.44352 pascals and symbol is t/in²-long.
## Ton / square inch [short]
Ton / square inch [short] is a unit of pressure equal to about 13789514.56 pascals and symbol is t/in²-short.
## Ton / square meter
Ton / square meter is a unit of pressure equal to about 9806.65 pascals and symbol is t/m².
## Torr
Torr is a a non-SI unit of pressure equal to about 133.32237 pascals and symbol is torr. | 5,795 | 17,958 | {"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-2019-04 | latest | en | 0.536143 |
https://citizenmaths.com/length/kilofeet-to-palms | 1,632,774,000,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058467.95/warc/CC-MAIN-20210927181724-20210927211724-00317.warc.gz | 222,212,412 | 12,978 | # Kilofeet to Palm Conversions
From
Kilofeet
• Angstrom
• Cable Length (Imperial)
• Cable Length (International)
• Cable Length (U.S.)
• Centimeter
• Chain
• Cubit
• Decameter
• Decimeter
• Ell
• Fathom
• Finger
• Finger (cloth)
• Foot
• Furlong
• Gigameter
• Hand
• Hectometer
• Inch
• Kilofeet
• Kilometer
• League
• League (land)
• Light Day
• Light Hour
• Light Minute
• Light Second
• Light Year
• Line
• Marathon
• Megameter
• Meter
• Mickey
• Microinch
• Micrometer
• Micron
• Mil
• Mile
• Millimeter
• Myriameter
• Nail (cloth)
• Nanometer
• Nautical League
• Nautical Mile
• Pace
• Palm
• Parsec
• Pica
• Picometer
• Point
• Quarter
• Rod
• Rope
• Shaku
• Smoot
• Span
• Step
• Terameter
• Twip
• Yard
To
Palm
• Angstrom
• Cable Length (Imperial)
• Cable Length (International)
• Cable Length (U.S.)
• Centimeter
• Chain
• Cubit
• Decameter
• Decimeter
• Ell
• Fathom
• Finger
• Finger (cloth)
• Foot
• Furlong
• Gigameter
• Hand
• Hectometer
• Inch
• Kilofeet
• Kilometer
• League
• League (land)
• Light Day
• Light Hour
• Light Minute
• Light Second
• Light Year
• Line
• Marathon
• Megameter
• Meter
• Mickey
• Microinch
• Micrometer
• Micron
• Mil
• Mile
• Millimeter
• Myriameter
• Nail (cloth)
• Nanometer
• Nautical League
• Nautical Mile
• Pace
• Palm
• Parsec
• Pica
• Picometer
• Point
• Quarter
• Rod
• Rope
• Shaku
• Smoot
• Span
• Step
• Terameter
• Twip
• Yard
Formula 5,566 kft = 5566 x 4000 palm = 22,264,000 palm
## How To Convert From Kilofeet to Palm
1 Kilofeet is equivalent to 4,000 Palms:
1 kft = 4,000 palm
For example, if the Kilofeet number is (1.9), then its equivalent Palm number would be (7,600).
Formula:
1.9 kft = 1.9 x 4000 palm = 7,600 palm
## Kilofeet to Palm conversion table
Kilofeet (kft) Palm (palm)
0.1 kft 400 palm
0.2 kft 800 palm
0.3 kft 1,200 palm
0.4 kft 1,600 palm
0.5 kft 2,000 palm
0.6 kft 2,400 palm
0.7 kft 2,800 palm
0.8 kft 3,200 palm
0.9 kft 3,600 palm
1 kft 4,000 palm
1.1 kft 4,400 palm
1.2 kft 4,800 palm
1.3 kft 5,200 palm
1.4 kft 5,600 palm
1.5 kft 6,000 palm
1.6 kft 6,400 palm
1.7 kft 6,800 palm
1.8 kft 7,200 palm
1.9 kft 7,600 palm
2 kft 8,000 palm
2.1 kft 8,400 palm
2.2 kft 8,800 palm
2.3 kft 9,200 palm
2.4 kft 9,600 palm
2.5 kft 10,000 palm
2.6 kft 10,400 palm
2.7 kft 10,800 palm
2.8 kft 11,200 palm
2.9 kft 11,600 palm
3 kft 12,000 palm
3.1 kft 12,400 palm
3.2 kft 12,800 palm
3.3 kft 13,200 palm
3.4 kft 13,600 palm
3.5 kft 14,000 palm
3.6 kft 14,400 palm
3.7 kft 14,800 palm
3.8 kft 15,200 palm
3.9 kft 15,600 palm
4 kft 16,000 palm
4.1 kft 16,400 palm
4.2 kft 16,800 palm
4.3 kft 17,200 palm
4.4 kft 17,600 palm
4.5 kft 18,000 palm
4.6 kft 18,400 palm
4.7 kft 18,800 palm
4.8 kft 19,200 palm
4.9 kft 19,600 palm
5 kft 20,000 palm
5.1 kft 20,400 palm
5.2 kft 20,800 palm
5.3 kft 21,200 palm
5.4 kft 21,600 palm
5.5 kft 22,000 palm
5.6 kft 22,400 palm
5.7 kft 22,800 palm
5.8 kft 23,200 palm
5.9 kft 23,600 palm
6 kft 24,000 palm
6.1 kft 24,400 palm
6.2 kft 24,800 palm
6.3 kft 25,200 palm
6.4 kft 25,600 palm
6.5 kft 26,000 palm
6.6 kft 26,400 palm
6.7 kft 26,800 palm
6.8 kft 27,200 palm
6.9 kft 27,600 palm
7 kft 28,000 palm
7.1 kft 28,400 palm
7.2 kft 28,800 palm
7.3 kft 29,200 palm
7.4 kft 29,600 palm
7.5 kft 30,000 palm
7.6 kft 30,400 palm
7.7 kft 30,800 palm
7.8 kft 31,200 palm
7.9 kft 31,600 palm
8 kft 32,000 palm
8.1 kft 32,400 palm
8.2 kft 32,800 palm
8.3 kft 33,200 palm
8.4 kft 33,600 palm
8.5 kft 34,000 palm
8.6 kft 34,400 palm
8.7 kft 34,800 palm
8.8 kft 35,200 palm
8.9 kft 35,600 palm
9 kft 36,000 palm
9.1 kft 36,400 palm
9.2 kft 36,800 palm
9.3 kft 37,200 palm
9.4 kft 37,600 palm
9.5 kft 38,000 palm
9.6 kft 38,400 palm
9.7 kft 38,800 palm
9.8 kft 39,200 palm
9.9 kft 39,600 palm
10 kft 40,000 palm
20 kft 80,000 palm
30 kft 120,000 palm
40 kft 160,000 palm
50 kft 200,000 palm
60 kft 240,000 palm
70 kft 280,000 palm
80 kft 320,000 palm
90 kft 360,000 palm
100 kft 400,000 palm
110 kft 440,000 palm | 1,809 | 3,910 | {"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-2021-39 | latest | en | 0.354093 |
https://oeis.org/A261449 | 1,638,155,224,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358685.55/warc/CC-MAIN-20211129014336-20211129044336-00228.warc.gz | 506,225,906 | 4,169 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A261449 Prime numbers whose decimal digits contain a total of two loops. 0
83, 109, 149, 181, 199, 269, 281, 283, 349, 383, 401, 419, 439, 443, 461, 463, 467, 479, 491, 509, 569, 587, 599, 601, 607, 619, 641, 643, 647, 659, 661, 691, 709, 769, 787, 811, 821, 823, 827, 853, 857, 877, 907, 919, 929, 941, 947, 967, 991, 997, 1019, 1039 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Of the digits, 0 through 9, {0, 4, 6, 9} have one loop, 8 has two loops, and all the rest have none. - Robert G. Wilson v, Aug 20 2015 LINKS EXAMPLE 83 is the first term of the sequence. The digit 8 contains two closed curves. MATHEMATICA Select[Prime@ Range@ 200, 2 == Total[{ 1, 0, 0, 0, 1, 0, 1, 0, 2, 1}[[1 + IntegerDigits@ #]]]&] (* Giovanni Resta, Aug 19 2015 *) CROSSREFS Cf. A001729, A190222, A001743, A001744, A001745, A249572. Sequence in context: A031412 A033252 A347226 * A288880 A126117 A096279 Adjacent sequences: A261446 A261447 A261448 * A261450 A261451 A261452 KEYWORD nonn,less,easy,base AUTHOR Altug Alkan, Aug 19 2015 EXTENSIONS More terms from Giovanni Resta, Aug 19 2015 STATUS approved
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# How can you learn to round decimals?
A:
Rounding decimals involves analyzing whether end numbers are above or below five. If a number is below five, then the number before may stay the same. If a number is above five, the number before it rounds up by one.
## Keep Learning
It takes practice to apply decimal rounding skills. Numbers that can be rounded include numbers in the tenth place, hundredth place, thousandth place and beyond. Decimal amounts can round to whole numbers as well. For example, 1.167 can be rounded to 1.17 or 1.2 or 1.
It's easy to confuse rounding a number to the nearest hundred versus the nearest hundredth. Rounding to the nearest hundred refers to a whole number, such as 100 or 200. In contrast, hundredth refers to the hundredth decimal place.
Sources:
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"Kneser"/Riemann mapping method code for *complex* bases mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 08/15/2011, 03:44 AM (This post was last modified: 08/15/2011, 12:03 PM by mike3.) Hi. It's been a while but I figured I could post my attempt to tetrate complex bases with the "Kneser"/Riemann mapping algorithm. This is the code I've got, which will tetrate a complex base, here base $3 + i$: Code:/* Independent implementation of Kneser mapping algorithm. */ /* Compute the sum of a Fourier series: * f(z) = sum_{n=0...inf} a_n e^(uinz) */ FourierSum(coef, u, z) = sum(n=0,matsize(coef)[2]-1, coef[n+1]*exp(u*n*z)); /* Compute the sum of a Taylor series centered at u: * f(z) = sum_{n=0...inf} a_n (z - u)^n */ TaylorSum(coef, u, z) = sum(n=0,matsize(coef)[2]-1, coef[n+1]*(z - u)^n); /* Compute Bell polynomial */ bellcomplete(coefs) = { local(n=matsize(coefs)[2]); local(M=matrix(n,n)); for(i=1,n,\ for(j=1,n,\ M[i, j] = 0; if(j-i+1 > 0, M[i, j] = binomial(n-i, j-i)*coefs[j-i+1]); if(j-i+1 == 0, M[i, j] = -1); ); ); return(matdet(M)); } /* Compute the coefficients of the regular superfunction of the exponential * with and at fixed point L and a given base (L must be a fixed point of the base). */ RegularCoeffs(L, base, N) = { local(a = vector(N)); local(bellcoeffs); local(logbase = log(base)); a[0+1] = L; a[1+1] = 1; /* Recurrence formula for Fourier coefficients: * a_n = B_n(1! log(b) a_1, ..., (n-1)! log(b) a_(n-1), 0)/(n! (L^(n-1) log(b)^n - log(b))) */ for(n=2,N-1, bellcoeffs = vector(n, k, if(k==n, 0, logbase*k!*a[k+1])); a[n+1] = bellcomplete(bellcoeffs)/(n! * (L^(n-1) * logbase^n - logbase)); ); return(a); } /* Compute the coefficients of the regular Schroder function of the exponential * with and at fixed point L. */ MakeSchroderCoeffs(RegCoeffs) = { local(invpoly = 0); local(RegCoeffsTrunc = RegCoeffs); local(InvCoeffs = RegCoeffs); RegCoeffsTrunc[1] = 0; /* truncate constant term */ /* Do series reversion. This gives the inverse of (InvSchroder(x) - L) */ invpoly = serreverse(TaylorSum(RegCoeffsTrunc, 0, x) + O(x^matsize(RegCoeffsTrunc)[2])); /* Extract coefficients */ for(i=0,matsize(RegCoeffsTrunc)[2]-1, InvCoeffs[i+1] = polcoeff(invpoly, i); ); /* Done! */ return(InvCoeffs); } /* Compute the regular superfunction at a given fixed point L of a base b. */ RegularSuperfunction(L, base, SuperCoeffs, z) = { if(real(z) < -5, return(FourierSum(SuperCoeffs, log(L*log(base)), z));, return(base^RegularSuperfunction(L, base, SuperCoeffs, z-1)); ); } /* Compute the regular Schroder function at a given fixed point L of a base b. */ RegularSchroderFunction(L, base, SchroderCoeffs, z) = { if(abs(z - L) > 0.5, return(L*log(base)*RegularSchroderFunction(L, base, SchroderCoeffs, log(z)/log(base)));, return(TaylorSum(SchroderCoeffs, L, z)); ); } rsf(L, base, SchroderCoeffs, z) = { if(abs(imag(z)) < 0.02, return(3)); return(RegularSchroderFunction(L, base, SchroderCoeffs, z)); } /* Compute the regular Abel function at a given fixd point L. */ RegularAbelFunction(L, base, SchroderCoeffs, z) = { \ log(RegularSchroderFunction(L, base, SchroderCoeffs, z))/log(L*log(base)); } /* --- Quadrature routines --- */ /* Do quadrature using function samples in samp with step delta. */ /* This is an inferior, Simpson's rule quadrature, implemented for testing only. In the future, we will * drop quadrature and upgrade to FFT for high performance. Note that samp must have an odd number of * elements. */ QuadratureCore(samp, delta) = { local(ssum=0); ssum = samp[1] + sum(n=1,matsize(samp)[2]-2,(4 + ((-1)^(n-1) - 1))*samp[n+1]) + samp[matsize(samp)[2]]; return((delta/3)*ssum); } /* Compute some regularly-spaced nodes over an interval to use for quadrature. N must be odd. */ QuadratureNodes(a, b, N) = { local(nodes); nodes = vector(N, i, a + (i-1)*((b-a)/(N-1))); return(nodes); } /* Compute the delta value for a given interval and node count */ QuadratureDelta(a, b, N) = (b - a)/(N-1); /* --- End quadrature routines --- */ /* Set up the tetration system. */ Initialize() = { /* Initialize the fixed points and quadrature. */ /* Base and fixed points */ base = 3 + I; upperfix = 0.325 + 1.001*I; lowerfix = 0.098 - 1.398*I; for(i=1,16,upperfix = upperfix - (base^upperfix - upperfix)/(log(base)*base^upperfix - 1)); for(i=1,16,lowerfix = lowerfix - (base^lowerfix - lowerfix)/(log(base)*base^lowerfix - 1)); /* Quadrature setup */ NumQuadNodes = 1001; ImagOffset = 0.12*I; /* offset off the real axis to do Fourier integration */ ThetaQuadratureNodesUpper = QuadratureNodes(-0.5 + ImagOffset, 0.5 + ImagOffset, NumQuadNodes); ThetaQuadratureDeltaUpper = QuadratureDelta(-0.5 + ImagOffset, 0.5 + ImagOffset, NumQuadNodes); ThetaQuadratureNodesLower = QuadratureNodes(-0.5 - ImagOffset, 0.5 - ImagOffset, NumQuadNodes); ThetaQuadratureDeltaLower = QuadratureDelta(-0.5 - ImagOffset, 0.5 - ImagOffset, NumQuadNodes); CauchyCircleQuadratureNodes = QuadratureNodes(0, 2*Pi, NumQuadNodes); CauchyCircleQuadratureDelta = QuadratureDelta(0, 2*Pi, NumQuadNodes); /* Term counts */ NumRegTerms = 32; NumTaylorTerms = 32; NumThetaTerms = 100; /* Generate series */ RegCoeffsUpper = RegularCoeffs(upperfix, base, NumRegTerms); RegSchrCoeffsUpper = MakeSchroderCoeffs(RegCoeffsUpper); RegCoeffsLower = RegularCoeffs(lowerfix, base, NumRegTerms); RegSchrCoeffsLower = MakeSchroderCoeffs(RegCoeffsLower); /* Set up initial Taylor series */ TaylorCoeffs = vector(NumTaylorTerms); TaylorCoeffs[0+1] = 1; /* coefficient of x^0 = 1 */ TaylorCoeffs[1+1] = 1; /* coefficient of x^1 = 1 */ /* Set up initial Riemann mapping series */ RiemannCoeffsUpper = vector(NumThetaTerms); RiemannCoeffsLower = vector(NumThetaTerms); } /* Compute upper regular superfunction */ RegularSuperUpper(z) = RegularSuperfunction(upperfix, base, RegCoeffsUpper, z); /* Compute upper regular Abel function */ RegularAbelUpper(z) = RegularAbelFunction(upperfix, base, RegSchrCoeffsUpper, z); /* Compute lower regular superfunction */ RegularSuperLower(z) = RegularSuperfunction(lowerfix, base, RegCoeffsLower, z); /* Compute lower regular Abel function */ RegularAbelLower(z) = RegularAbelFunction(lowerfix, base, RegSchrCoeffsLower, z); /* Construct Fourier coefficients for a Riemann theta(z) mapping from a tetration Taylor series. */ RiemannFromTaylorStep() = { local(SamplesUpper); local(SamplesLower); local(FourierSamp); /* Sample the function at the node points. */ SamplesUpper = vector(NumQuadNodes, i, \ RegularAbelUpper(TaylorSum(TaylorCoeffs, 0, ThetaQuadratureNodesUpper[i])) -\ ThetaQuadratureNodesUpper[i]); SamplesLower = vector(NumQuadNodes, i, \ RegularAbelLower(TaylorSum(TaylorCoeffs, 0, ThetaQuadratureNodesLower[i])) -\ ThetaQuadratureNodesLower[i]); /* plot(X=1,NumQuadNodes,real(SamplesUpper[floor(X)])); plot(X=1,NumQuadNodes,imag(SamplesUpper[floor(X)])); plot(X=1,NumQuadNodes,real(SamplesLower[floor(X)])); plot(X=1,NumQuadNodes,imag(SamplesLower[floor(X)])); PublicSamplesUpper = SamplesUpper; PublicSamplesLower = SamplesLower; */ /* Do the Fourier integrals */ /* We multiply by exp(-2 pi i n ImagOffset) to compensate for the offset along the imaginary axis. */ for(n=0,NumThetaTerms-1, FourierSamp = vector(NumQuadNodes, i, SamplesUpper[i]*exp(-2*Pi*I*n*real(ThetaQuadratureNodesUpper[i]))); RiemannCoeffsUpper[n+1] = QuadratureCore(FourierSamp, ThetaQuadratureDeltaUpper); RiemannCoeffsUpper[n+1] = RiemannCoeffsUpper[n+1] * exp(-2*Pi*I*n*ImagOffset); FourierSamp = vector(NumQuadNodes, i, SamplesLower[i]*exp(2*Pi*I*n*real(ThetaQuadratureNodesLower[i]))); RiemannCoeffsLower[n+1] = QuadratureCore(FourierSamp, ThetaQuadratureDeltaLower); RiemannCoeffsLower[n+1] = RiemannCoeffsLower[n+1] * exp(-2*Pi*I*n*ImagOffset); ); } /* Calculate the theta mapping. */ ThetaMappingUpper(z) = FourierSum(RiemannCoeffsUpper, 2*Pi*I, z); ThetaMappingLower(z) = FourierSum(RiemannCoeffsLower, -2*Pi*I, z); /* Calculate the upper "warped" superfunction. */ WarpSuperUpper(z) = RegularSuperUpper(z + ThetaMappingUpper(z)); /* Calculate the lower "warped" superfunction. */ WarpSuperLower(z) = RegularSuperLower(z + ThetaMappingLower(z)); /* Do the Cauchy integral to get the Taylor series. */ CauchyTaylorStep() = { local(Samples); local(CauchyizedSamples); /* The integration proceeds in several phases: * 1. Integrate above Im(z) = ImagOffset using the upper regular warped superfunction, * 2. Integrate from Im(z) = ImagOffset to Im(z) = -ImagOffset using the old Taylor series, * 3. Integrate below Im(z) = -ImagOffset using the lower regular warped superfunction, * 4. Integrate from Im(z) = -ImagOffset to Im(z) = ImagOffset using the old Taylor series again. * To maximize precision, we use the exp/log of the Taylor series evaluated near zero. The whole * process is done counterclockwise. */ /* Presampling */ Samples = vector(NumQuadNodes); CirclePositions = vector(NumQuadNodes, i, exp(I*CauchyCircleQuadratureNodes[i])); for(n=0,NumQuadNodes-1,\ if(abs(imag(CirclePositions[n+1])) < imag(ImagOffset),\ if(real(CirclePositions[n+1]) > 0,\ /* right halfplane */\ Samples[n+1] = base^TaylorSum(TaylorCoeffs, 0, CirclePositions[n+1]-1);,\ /* left halfplane */\ Samples[n+1] = log(TaylorSum(TaylorCoeffs, 0, CirclePositions[n+1]+1))/log(base);\ );,\ if(imag(CirclePositions[n+1]) > 0,\ /* upper halfplane */\ /* use upper regular warped superfunction */\ Samples[n+1] = WarpSuperUpper(CirclePositions[n+1]);,\ /* lower halfplane */\ /* use lower regular warped superfunction */\ Samples[n+1] = WarpSuperLower(CirclePositions[n+1]);\ );\ );\ ); /* Do Cauchy integral */ for(n=0,NumTaylorTerms-1,\ /* Cauchy setup */\ CauchyizedSamples = vector(NumQuadNodes, i, Samples[i]/(CirclePositions[i]^(n+1)) * \ I*CirclePositions[i]);\ /* Integrate */\ TaylorCoeffs[n+1] = 1/(2*Pi*I) * QuadratureCore(CauchyizedSamples, CauchyCircleQuadratureDelta);\ ); /* Renormalize Taylor series */ TaylorCoeffs = TaylorCoeffs / TaylorSum(TaylorCoeffs, 0, 0); } /* Do full iterations */ DoKneserIteration(numiters) = { for(i=1,numiters, print("Iteration ", i, ":"); print("Building Kneser mapping..."); RiemannFromTaylorStep(); print("Performing Cauchy integral..."); CauchyTaylorStep(); print("Done. R = ", abs(base^TaylorSum(TaylorCoeffs, 0, -0.5) - TaylorSum(TaylorCoeffs, 0, 0.5))); ); } /* Calculate full superfunction */ FullTetApprox(z) = { if(real(z) > 0.5, return(base^FullTetApprox(z-1))); if(real(z) < -0.5, return(log(FullTetApprox(z+1))/log(base))); if(imag(z) > 1, return(WarpSuperUpper(z))); if(imag(z) < -1, return(WarpSuperLower(z))); return(TaylorSum(TaylorCoeffs, 0, z)); } To use, run "Initialize()", then "DoKneserIteration()". The program will generate the theta mappings to fuse the two regular iterations at the two principal fixed points. With the given Taylor/etc. term counts, 16 iterations yields a residual on the order of $10^{-15}$, representing what may be some of the most accurate calculations of a tetrational of a complex base to date. We get $^{1/2} (3 + i) \approx 1.73774339741062 + 0.19778352516442i$ Graphs of $^x (3 + i)$ are shown below. On the complex plane (scale -10 to +10 on both axes): As we can see, the merger worked successfully. The two different regular iterations have flowed together nice and holomorphically for $\Re(z) > -2$. We can even do the "pentation" (the $\uparrow \uparrow \uparrow$ operator) of this base now. The sequence of integer pentates is Code:(3 + i) ^^^ 0 ~ 1 (3 + i) ^^^ 1 ~ 3.0000000000000 + 1.0000000000000*I (3 + i) ^^^ 2 ~ 0.067505751821009 + 0.37038084586491*I (3 + i) ^^^ 3 ~ 0.95290527513344 + 0.41965918718415*I (3 + i) ^^^ 4 ~ 1.6689141108302 + 1.5511207891726*I (3 + i) ^^^ 5 ~ 0.14427450000007 + 1.2422893053044*I (3 + i) ^^^ 6 ~ 0.59855234346250 + 0.91546250434381*I (3 + i) ^^^ 7 ~ 0.86698754296965 + 1.1149406202426*I (3 + i) ^^^ 8 ~ 0.67898538651189 + 1.3074168077768*I (3 + i) ^^^ 9 ~ 0.60455661809291 + 1.1563634545945*I (3 + i) ^^^ 10 ~ 0.69330548523182 + 1.1273955819904*I (3 + i) ^^^ 11 ~ 0.70596729212537 + 1.1855496560953*I (3 + i) ^^^ 12 ~ 0.66761843466257 + 1.1866711686455*I (3 + i) ^^^ 13 ~ 0.67126297874107 + 1.1635087706344*I (3 + i) ^^^ 14 ~ 0.68499772461207 + 1.1678785414537*I (3 + i) ^^^ 15 ~ 0.68092916998454 + 1.1759457551948*I (3 + i) ^^^ 16 ~ 0.67640594691649 + 1.1725959600056*I (3 + i) ^^^ 17 ~ 0.67891225654672 + 1.1701754208531*I (3 + i) ^^^ 18 ~ 0.68014536910711 + 1.1719549038205*I (3 + i) ^^^ 19 ~ 0.67892704487410 + 1.1725305088990*I (3 + i) ^^^ 20 ~ 0.67869891860093 + 1.1717250339640*I (3 + i) ^^^ 21 ~ 0.67921563866767 + 1.1716670353204*I (3 + i) ^^^ 22 ~ 0.67919871696253 + 1.1719897863482*I (3 + i) ^^^ 23 ~ 0.67900254832419 + 1.1719465366940*I (3 + i) ^^^ 24 ~ 0.67904902631454 + 1.1718306916743*I (3 + i) ^^^ 25 ~ 0.67911531745648 + 1.1718709649452*I (3 + i) ^^^ 26 ~ 0.67908388197865 + 1.1719075100622*I (3 + i) ^^^ 27 ~ 0.67906467678018 + 1.1718845193510*I (3 + i) ^^^ 28 ~ 0.67908072504425 + 1.1718750848271*I (3 + i) ^^^ 29 ~ 0.67908487628565 + 1.1718858831902*I (3 + i) ^^^ 30 ~ 0.67907783385992 + 1.1718873294663*I Note how the pentation looks to converge to a limit, like tetration for STR bases. It does not work for all complex bases, though. It seems to work only for those relatively near the real axis. Increasing the ImagOffset (how far off the real axis we sample the Fourier integrals) appears to extend the range, however. When it fails, it looks like the iteration used to analytically continue the Schroder function gets sucked into the wrong fixed point. Any way to resolve this? The aforementioned increasing of the ImagOffset causes losses in accuracy and efficiency, and is not a cure-all that allows for tetrating all complex bases outside the STR with the possible exception of the possibly singular base 0. sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 08/15/2011, 02:59 PM (08/15/2011, 03:44 AM)mike3 Wrote: Hi. It's been a while but I figured I could post my attempt to tetrate complex bases with the "Kneser"/Riemann mapping algorithm. This is the code I've got, which will tetrate a complex base, here base $3 + i$: .... It does not work for all complex bases, though. It seems to work only for those relatively near the real axis. Increasing the ImagOffset (how far off the real axis we sample the Fourier integrals) appears to extend the range, however. When it fails, it looks like the iteration used to analytically continue the Schroder function gets sucked into the wrong fixed point. Any way to resolve this? The aforementioned increasing of the ImagOffset causes losses in accuracy and efficiency, and is not a cure-all that allows for tetrating all complex bases outside the STR with the possible exception of the possibly singular base 0. Mike, Very nice! For base=3+i, I can calculate two repelling fixed points, each with different periods. Presumably, the upper half of the complex plane converges towards the first fixed point, and the lower half of the complex plane converges towards the second fixed point, while the merged function allows you to maintaining the definition of sexp(0)=1, and sexp(1)=base. L1=0.324536386411256 + 1.00148180609593*I L2=-0.0976763924712228 - 1.39768129114470*I I assume the algorithm works when both bases are repelling? - Sheldon Gottfried Ultimate Fellow Posts: 852 Threads: 126 Joined: Aug 2007 08/15/2011, 03:14 PM Very nice, thank you! I tried it and the Pari/GP code worked immediately. I've nothing more to say at the moment, I think I'll play a bit with it to get a feel. When I find anything concerning your question I'll replay again, may be this will take some days. Gottfried Gottfried Helms, Kassel « Next Oldest | Next Newest »
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Users browsing this thread: 1 Guest(s) | 5,908 | 18,872 | {"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": 7, "/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} | 3.40625 | 3 | CC-MAIN-2022-33 | latest | en | 0.623024 |
https://math.stackexchange.com/questions/1515236/last-fermats-theorem-modulo-m | 1,619,115,758,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039594341.91/warc/CC-MAIN-20210422160833-20210422190833-00378.warc.gz | 478,522,617 | 39,337 | # “Last Fermat's Theorem” modulo m
The last Fermat's Theorem is a claim about the non-existence of non-trivial integer solution for $X^n+Y^n=Z^n$ for $n\in \mathbb N$, $n\ge 3$.
However, given $m\in \mathbb N$, we can investigate the integer solutions for $X^n+Y^n\equiv Z^n \mod m$ for all $n\in \mathbb N$ with the restriciton that $X,Y,Z\not\equiv0\mod m$.
It seems to me that we always have solutions in this case, but I did no find any reference or exposition about this.
Is it "relevant" to think on it? Have this question appeared elsewhere?
• Just as an initial comment, I would observe that it's immediately trivially true unless we restrict $X, Y, Z \not\equiv 0 \pmod{m}$. – Brian Tung Nov 5 '15 at 23:10
• Sure, I will include this note! It is "analogous" to the condition of the Theorem... – Binai Nov 5 '15 at 23:16
• It works if you take $X=1$, $Y=1$, $Z=2$, and $n = \lambda(m)+1$, where $\lambda$ is the Carmichael function. Maybe it is more interesting to require $n \le \lambda(m)$? – Dan Brumleve Nov 5 '15 at 23:24
• Why? How do you eliminate the case where $n>\lambda(m)$? – Binai Nov 5 '15 at 23:30
• $X^{\lambda(m)+1} \equiv X \pmod{m}$. Since we always have solutions for $n=1$ we have solutions for $n=\lambda(m)+1$ too. – Dan Brumleve Nov 5 '15 at 23:42
You are correct. For sufficiently large primes $p$ and any $n \geq 1$, there are always nontrivial solutions to $$X^n + Y^n \equiv Z^n \pmod p.$$ Schur first proved this in 1916, in his paper --- Schur, I. "Über die Kongruenz x^m+y^m=z^m (mod p)." Jahresber. Deutsche Math.-Verein. 25, 114-116, 1916.
You might think it meta-wise clear that there are solutions mod $p$ for every $p$, as otherwise the problem wouldn't really be so hard. [Or perhaps you might think it's very hard to find a $p$ for which there are no solutions, and that was the bottleneck.]
• May I ask what means: "meta-wise clear" and "bottleneck"? ...English s not my native language. – Binai Nov 6 '15 at 1:45
• @Lucci Of course! When you pour water out of a bottle, the water doesn't all rush out all at once. Instead, it comes out unevenly through the little opening, which we call a "bottleneck." So here, I use bottleneck as a metaphor for the part of the process that slows things down. "Meta-wise clear" is more complicated to explain. The term "meta" is a bit ill-defined, but here I use it to mean that you might guess that this problem is difficult not based on any part of the problem itself, but because of others' lack of progress. To identify aspects about the problem indirectly is sort of "meta." – davidlowryduda Nov 6 '15 at 1:59
• Thank you! Math+General culture =) – Binai Nov 6 '15 at 2:08
As mixedmath says, this is true for sufficiently large prime $m$. This result is due to Schur. See this blog post for an exposition. | 850 | 2,803 | {"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} | 2.984375 | 3 | CC-MAIN-2021-17 | latest | en | 0.897241 |
https://www.lmfdb.org/Character/Dirichlet/485/ | 1,596,449,077,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735792.85/warc/CC-MAIN-20200803083123-20200803113123-00174.warc.gz | 745,100,666 | 6,824 | Properties
Modulus 485 Structure $$C_{96}\times C_{4}$$ Order 384
Show commands for: Pari/GP / SageMath
sage: from dirichlet_conrey import DirichletGroup_conrey # requires nonstandard Sage package to be installed
sage: H = DirichletGroup_conrey(485)
pari: g = idealstar(,485,2)
Character group
sage: G.order() pari: g.no Order = 384 sage: H.invariants() pari: g.cyc Structure = $$C_{96}\times C_{4}$$ sage: H.gens() pari: g.gen Generators = $\chi_{485}(296,\cdot)$, $\chi_{485}(292,\cdot)$
First 32 of 384 characters
Each row describes a character. When available, the columns show the orbit label, order of the character, whether the character is primitive, and several values of the character.
orbit label order primitive -1 1 2 3 4 6 7 8 9 11 12 13
$$\chi_{485}(1,\cdot)$$ 485.a 1 no $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$
$$\chi_{485}(2,\cdot)$$ 485.bm 48 yes $$-1$$ $$1$$ $$e\left(\frac{7}{24}\right)$$ $$e\left(\frac{13}{24}\right)$$ $$e\left(\frac{7}{12}\right)$$ $$e\left(\frac{5}{6}\right)$$ $$e\left(\frac{11}{48}\right)$$ $$e\left(\frac{7}{8}\right)$$ $$e\left(\frac{1}{12}\right)$$ $$e\left(\frac{11}{24}\right)$$ $$e\left(\frac{1}{8}\right)$$ $$e\left(\frac{29}{48}\right)$$
$$\chi_{485}(3,\cdot)$$ 485.bm 48 yes $$-1$$ $$1$$ $$e\left(\frac{13}{24}\right)$$ $$e\left(\frac{7}{24}\right)$$ $$e\left(\frac{1}{12}\right)$$ $$e\left(\frac{5}{6}\right)$$ $$e\left(\frac{17}{48}\right)$$ $$e\left(\frac{5}{8}\right)$$ $$e\left(\frac{7}{12}\right)$$ $$e\left(\frac{17}{24}\right)$$ $$e\left(\frac{3}{8}\right)$$ $$e\left(\frac{23}{48}\right)$$
$$\chi_{485}(4,\cdot)$$ 485.bd 24 yes $$1$$ $$1$$ $$e\left(\frac{7}{12}\right)$$ $$e\left(\frac{1}{12}\right)$$ $$e\left(\frac{1}{6}\right)$$ $$e\left(\frac{2}{3}\right)$$ $$e\left(\frac{11}{24}\right)$$ $$-i$$ $$e\left(\frac{1}{6}\right)$$ $$e\left(\frac{11}{12}\right)$$ $$i$$ $$e\left(\frac{5}{24}\right)$$
$$\chi_{485}(6,\cdot)$$ 485.x 12 no $$1$$ $$1$$ $$e\left(\frac{5}{6}\right)$$ $$e\left(\frac{5}{6}\right)$$ $$e\left(\frac{2}{3}\right)$$ $$e\left(\frac{2}{3}\right)$$ $$e\left(\frac{7}{12}\right)$$ $$-1$$ $$e\left(\frac{2}{3}\right)$$ $$e\left(\frac{1}{6}\right)$$ $$-1$$ $$e\left(\frac{1}{12}\right)$$
$$\chi_{485}(7,\cdot)$$ 485.br 96 yes $$1$$ $$1$$ $$e\left(\frac{11}{48}\right)$$ $$e\left(\frac{17}{48}\right)$$ $$e\left(\frac{11}{24}\right)$$ $$e\left(\frac{7}{12}\right)$$ $$e\left(\frac{25}{96}\right)$$ $$e\left(\frac{11}{16}\right)$$ $$e\left(\frac{17}{24}\right)$$ $$e\left(\frac{37}{48}\right)$$ $$e\left(\frac{13}{16}\right)$$ $$e\left(\frac{79}{96}\right)$$
$$\chi_{485}(8,\cdot)$$ 485.ba 16 yes $$-1$$ $$1$$ $$e\left(\frac{7}{8}\right)$$ $$e\left(\frac{5}{8}\right)$$ $$-i$$ $$-1$$ $$e\left(\frac{11}{16}\right)$$ $$e\left(\frac{5}{8}\right)$$ $$i$$ $$e\left(\frac{3}{8}\right)$$ $$e\left(\frac{3}{8}\right)$$ $$e\left(\frac{13}{16}\right)$$
$$\chi_{485}(9,\cdot)$$ 485.bd 24 yes $$1$$ $$1$$ $$e\left(\frac{1}{12}\right)$$ $$e\left(\frac{7}{12}\right)$$ $$e\left(\frac{1}{6}\right)$$ $$e\left(\frac{2}{3}\right)$$ $$e\left(\frac{17}{24}\right)$$ $$i$$ $$e\left(\frac{1}{6}\right)$$ $$e\left(\frac{5}{12}\right)$$ $$-i$$ $$e\left(\frac{23}{24}\right)$$
$$\chi_{485}(11,\cdot)$$ 485.bk 48 no $$1$$ $$1$$ $$e\left(\frac{11}{24}\right)$$ $$e\left(\frac{17}{24}\right)$$ $$e\left(\frac{11}{12}\right)$$ $$e\left(\frac{1}{6}\right)$$ $$e\left(\frac{37}{48}\right)$$ $$e\left(\frac{3}{8}\right)$$ $$e\left(\frac{5}{12}\right)$$ $$e\left(\frac{1}{24}\right)$$ $$e\left(\frac{5}{8}\right)$$ $$e\left(\frac{19}{48}\right)$$
$$\chi_{485}(12,\cdot)$$ 485.ba 16 yes $$-1$$ $$1$$ $$e\left(\frac{1}{8}\right)$$ $$e\left(\frac{3}{8}\right)$$ $$i$$ $$-1$$ $$e\left(\frac{13}{16}\right)$$ $$e\left(\frac{3}{8}\right)$$ $$-i$$ $$e\left(\frac{5}{8}\right)$$ $$e\left(\frac{5}{8}\right)$$ $$e\left(\frac{11}{16}\right)$$
$$\chi_{485}(13,\cdot)$$ 485.br 96 yes $$1$$ $$1$$ $$e\left(\frac{29}{48}\right)$$ $$e\left(\frac{23}{48}\right)$$ $$e\left(\frac{5}{24}\right)$$ $$e\left(\frac{1}{12}\right)$$ $$e\left(\frac{79}{96}\right)$$ $$e\left(\frac{13}{16}\right)$$ $$e\left(\frac{23}{24}\right)$$ $$e\left(\frac{19}{48}\right)$$ $$e\left(\frac{11}{16}\right)$$ $$e\left(\frac{73}{96}\right)$$
$$\chi_{485}(14,\cdot)$$ 485.bp 96 yes $$-1$$ $$1$$ $$e\left(\frac{25}{48}\right)$$ $$e\left(\frac{43}{48}\right)$$ $$e\left(\frac{1}{24}\right)$$ $$e\left(\frac{5}{12}\right)$$ $$e\left(\frac{47}{96}\right)$$ $$e\left(\frac{9}{16}\right)$$ $$e\left(\frac{19}{24}\right)$$ $$e\left(\frac{11}{48}\right)$$ $$e\left(\frac{15}{16}\right)$$ $$e\left(\frac{41}{96}\right)$$
$$\chi_{485}(16,\cdot)$$ 485.x 12 no $$1$$ $$1$$ $$e\left(\frac{1}{6}\right)$$ $$e\left(\frac{1}{6}\right)$$ $$e\left(\frac{1}{3}\right)$$ $$e\left(\frac{1}{3}\right)$$ $$e\left(\frac{11}{12}\right)$$ $$-1$$ $$e\left(\frac{1}{3}\right)$$ $$e\left(\frac{5}{6}\right)$$ $$-1$$ $$e\left(\frac{5}{12}\right)$$
$$\chi_{485}(17,\cdot)$$ 485.bo 96 yes $$1$$ $$1$$ $$e\left(\frac{37}{48}\right)$$ $$e\left(\frac{31}{48}\right)$$ $$e\left(\frac{13}{24}\right)$$ $$e\left(\frac{5}{12}\right)$$ $$e\left(\frac{95}{96}\right)$$ $$e\left(\frac{5}{16}\right)$$ $$e\left(\frac{7}{24}\right)$$ $$e\left(\frac{35}{48}\right)$$ $$e\left(\frac{3}{16}\right)$$ $$e\left(\frac{89}{96}\right)$$
$$\chi_{485}(18,\cdot)$$ 485.ba 16 yes $$-1$$ $$1$$ $$e\left(\frac{3}{8}\right)$$ $$e\left(\frac{1}{8}\right)$$ $$-i$$ $$-1$$ $$e\left(\frac{15}{16}\right)$$ $$e\left(\frac{1}{8}\right)$$ $$i$$ $$e\left(\frac{7}{8}\right)$$ $$e\left(\frac{7}{8}\right)$$ $$e\left(\frac{9}{16}\right)$$
$$\chi_{485}(19,\cdot)$$ 485.bi 32 yes $$-1$$ $$1$$ $$e\left(\frac{3}{16}\right)$$ $$e\left(\frac{9}{16}\right)$$ $$e\left(\frac{3}{8}\right)$$ $$-i$$ $$e\left(\frac{21}{32}\right)$$ $$e\left(\frac{9}{16}\right)$$ $$e\left(\frac{1}{8}\right)$$ $$e\left(\frac{9}{16}\right)$$ $$e\left(\frac{15}{16}\right)$$ $$e\left(\frac{19}{32}\right)$$
$$\chi_{485}(21,\cdot)$$ 485.bq 96 no $$-1$$ $$1$$ $$e\left(\frac{37}{48}\right)$$ $$e\left(\frac{31}{48}\right)$$ $$e\left(\frac{13}{24}\right)$$ $$e\left(\frac{5}{12}\right)$$ $$e\left(\frac{59}{96}\right)$$ $$e\left(\frac{5}{16}\right)$$ $$e\left(\frac{7}{24}\right)$$ $$e\left(\frac{23}{48}\right)$$ $$e\left(\frac{3}{16}\right)$$ $$e\left(\frac{29}{96}\right)$$
$$\chi_{485}(22,\cdot)$$ 485.j 4 yes $$-1$$ $$1$$ $$-i$$ $$i$$ $$-1$$ $$1$$ $$1$$ $$i$$ $$-1$$ $$-1$$ $$-i$$ $$1$$
$$\chi_{485}(23,\cdot)$$ 485.br 96 yes $$1$$ $$1$$ $$e\left(\frac{1}{48}\right)$$ $$e\left(\frac{19}{48}\right)$$ $$e\left(\frac{1}{24}\right)$$ $$e\left(\frac{5}{12}\right)$$ $$e\left(\frac{59}{96}\right)$$ $$e\left(\frac{1}{16}\right)$$ $$e\left(\frac{19}{24}\right)$$ $$e\left(\frac{47}{48}\right)$$ $$e\left(\frac{7}{16}\right)$$ $$e\left(\frac{29}{96}\right)$$
$$\chi_{485}(24,\cdot)$$ 485.bd 24 yes $$1$$ $$1$$ $$e\left(\frac{5}{12}\right)$$ $$e\left(\frac{11}{12}\right)$$ $$e\left(\frac{5}{6}\right)$$ $$e\left(\frac{1}{3}\right)$$ $$e\left(\frac{1}{24}\right)$$ $$i$$ $$e\left(\frac{5}{6}\right)$$ $$e\left(\frac{1}{12}\right)$$ $$-i$$ $$e\left(\frac{7}{24}\right)$$
$$\chi_{485}(26,\cdot)$$ 485.bq 96 no $$-1$$ $$1$$ $$e\left(\frac{43}{48}\right)$$ $$e\left(\frac{1}{48}\right)$$ $$e\left(\frac{19}{24}\right)$$ $$e\left(\frac{11}{12}\right)$$ $$e\left(\frac{5}{96}\right)$$ $$e\left(\frac{11}{16}\right)$$ $$e\left(\frac{1}{24}\right)$$ $$e\left(\frac{41}{48}\right)$$ $$e\left(\frac{13}{16}\right)$$ $$e\left(\frac{35}{96}\right)$$
$$\chi_{485}(27,\cdot)$$ 485.ba 16 yes $$-1$$ $$1$$ $$e\left(\frac{5}{8}\right)$$ $$e\left(\frac{7}{8}\right)$$ $$i$$ $$-1$$ $$e\left(\frac{1}{16}\right)$$ $$e\left(\frac{7}{8}\right)$$ $$-i$$ $$e\left(\frac{1}{8}\right)$$ $$e\left(\frac{1}{8}\right)$$ $$e\left(\frac{7}{16}\right)$$
$$\chi_{485}(28,\cdot)$$ 485.bg 32 yes $$1$$ $$1$$ $$e\left(\frac{13}{16}\right)$$ $$e\left(\frac{7}{16}\right)$$ $$e\left(\frac{5}{8}\right)$$ $$i$$ $$e\left(\frac{23}{32}\right)$$ $$e\left(\frac{7}{16}\right)$$ $$e\left(\frac{7}{8}\right)$$ $$e\left(\frac{11}{16}\right)$$ $$e\left(\frac{1}{16}\right)$$ $$e\left(\frac{1}{32}\right)$$
$$\chi_{485}(29,\cdot)$$ 485.bp 96 yes $$-1$$ $$1$$ $$e\left(\frac{5}{48}\right)$$ $$e\left(\frac{47}{48}\right)$$ $$e\left(\frac{5}{24}\right)$$ $$e\left(\frac{1}{12}\right)$$ $$e\left(\frac{67}{96}\right)$$ $$e\left(\frac{5}{16}\right)$$ $$e\left(\frac{23}{24}\right)$$ $$e\left(\frac{31}{48}\right)$$ $$e\left(\frac{3}{16}\right)$$ $$e\left(\frac{85}{96}\right)$$
$$\chi_{485}(31,\cdot)$$ 485.bk 48 no $$1$$ $$1$$ $$e\left(\frac{7}{24}\right)$$ $$e\left(\frac{13}{24}\right)$$ $$e\left(\frac{7}{12}\right)$$ $$e\left(\frac{5}{6}\right)$$ $$e\left(\frac{41}{48}\right)$$ $$e\left(\frac{7}{8}\right)$$ $$e\left(\frac{1}{12}\right)$$ $$e\left(\frac{5}{24}\right)$$ $$e\left(\frac{1}{8}\right)$$ $$e\left(\frac{47}{48}\right)$$
$$\chi_{485}(32,\cdot)$$ 485.bm 48 yes $$-1$$ $$1$$ $$e\left(\frac{11}{24}\right)$$ $$e\left(\frac{17}{24}\right)$$ $$e\left(\frac{11}{12}\right)$$ $$e\left(\frac{1}{6}\right)$$ $$e\left(\frac{7}{48}\right)$$ $$e\left(\frac{3}{8}\right)$$ $$e\left(\frac{5}{12}\right)$$ $$e\left(\frac{7}{24}\right)$$ $$e\left(\frac{5}{8}\right)$$ $$e\left(\frac{1}{48}\right)$$
$$\chi_{485}(33,\cdot)$$ 485.r 8 yes $$-1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$1$$ $$e\left(\frac{1}{8}\right)$$ $$1$$ $$1$$ $$-i$$ $$1$$ $$e\left(\frac{7}{8}\right)$$
$$\chi_{485}(34,\cdot)$$ 485.bi 32 yes $$-1$$ $$1$$ $$e\left(\frac{1}{16}\right)$$ $$e\left(\frac{3}{16}\right)$$ $$e\left(\frac{1}{8}\right)$$ $$i$$ $$e\left(\frac{7}{32}\right)$$ $$e\left(\frac{3}{16}\right)$$ $$e\left(\frac{3}{8}\right)$$ $$e\left(\frac{3}{16}\right)$$ $$e\left(\frac{5}{16}\right)$$ $$e\left(\frac{17}{32}\right)$$
$$\chi_{485}(36,\cdot)$$ 485.l 6 no $$1$$ $$1$$ $$e\left(\frac{2}{3}\right)$$ $$e\left(\frac{2}{3}\right)$$ $$e\left(\frac{1}{3}\right)$$ $$e\left(\frac{1}{3}\right)$$ $$e\left(\frac{1}{6}\right)$$ $$1$$ $$e\left(\frac{1}{3}\right)$$ $$e\left(\frac{1}{3}\right)$$ $$1$$ $$e\left(\frac{1}{6}\right)$$
$$\chi_{485}(37,\cdot)$$ 485.br 96 yes $$1$$ $$1$$ $$e\left(\frac{23}{48}\right)$$ $$e\left(\frac{5}{48}\right)$$ $$e\left(\frac{23}{24}\right)$$ $$e\left(\frac{7}{12}\right)$$ $$e\left(\frac{61}{96}\right)$$ $$e\left(\frac{7}{16}\right)$$ $$e\left(\frac{5}{24}\right)$$ $$e\left(\frac{25}{48}\right)$$ $$e\left(\frac{1}{16}\right)$$ $$e\left(\frac{43}{96}\right)$$
$$\chi_{485}(38,\cdot)$$ 485.bo 96 yes $$1$$ $$1$$ $$e\left(\frac{23}{48}\right)$$ $$e\left(\frac{5}{48}\right)$$ $$e\left(\frac{23}{24}\right)$$ $$e\left(\frac{7}{12}\right)$$ $$e\left(\frac{85}{96}\right)$$ $$e\left(\frac{7}{16}\right)$$ $$e\left(\frac{5}{24}\right)$$ $$e\left(\frac{1}{48}\right)$$ $$e\left(\frac{1}{16}\right)$$ $$e\left(\frac{19}{96}\right)$$
$$\chi_{485}(39,\cdot)$$ 485.bp 96 yes $$-1$$ $$1$$ $$e\left(\frac{7}{48}\right)$$ $$e\left(\frac{37}{48}\right)$$ $$e\left(\frac{7}{24}\right)$$ $$e\left(\frac{11}{12}\right)$$ $$e\left(\frac{17}{96}\right)$$ $$e\left(\frac{7}{16}\right)$$ $$e\left(\frac{13}{24}\right)$$ $$e\left(\frac{5}{48}\right)$$ $$e\left(\frac{1}{16}\right)$$ $$e\left(\frac{23}{96}\right)$$ | 4,925 | 10,987 | {"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} | 3.09375 | 3 | CC-MAIN-2020-34 | latest | en | 0.382908 |
https://de.mathworks.com/matlabcentral/cody/problems/51461-pass-the-cards-in-fitch-cheney-s-five-card-trick | 1,721,689,933,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517927.60/warc/CC-MAIN-20240722220957-20240723010957-00370.warc.gz | 159,939,952 | 18,478 | # Problem 51461. Pass the cards in Fitch Cheney's five-card trick
Cody Problem 51541 asked solvers to guess the card in Fitch Cheney's five-card trick. This problem is the counterpart: Write a function that takes five cards, selects four of them, and puts them in order to pass to the partner. See the previous problem for the rules. If the input string is '5C 7C 6S 4H 3D', the output string should be '5C 3D 6S 4H'. Some hands will have multiple possible answers. Your function needs only to provide one of them.
The motivation for this problem arose when EmilyR and I tried to dazzle JessicaR with card wizardry, and I misordered the four cards—twice.
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Last Solution submitted on Jul 17, 2024
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Start Hunting! | 221 | 876 | {"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-2024-30 | latest | en | 0.885769 |
https://netplato.net/solonoid-schematic/ | 1,571,868,137,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987836295.98/warc/CC-MAIN-20191023201520-20191023225020-00535.warc.gz | 605,069,003 | 30,938 | # Solonoid Schematic
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Author: Irina Bender
Don't ask me why I have such of an obsession with wires, but I do. My mother always said that ever since I've been able to walk, I would find things with wires and play with them and tear them apart, figure out how they worked and would be totally fascinated.
Top | 1,540 | 7,583 | {"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} | 2.640625 | 3 | CC-MAIN-2019-43 | latest | en | 0.907097 |
http://stackoverflow.com/questions/tagged/curve-fitting | 1,469,486,085,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824395.52/warc/CC-MAIN-20160723071024-00223-ip-10-185-27-174.ec2.internal.warc.gz | 236,621,217 | 27,260 | # Tagged Questions
Fitting 1-D curve to data points, minimizing pre-defined error/loss function.
8 views
### Tensorflow: curve fitting with logistic regression
I have a dataset where each row is a (x, y) tuple. So, each row is a point of a curve in the X-Y plane. I would like to do logistic regression for it. Following the examples give here, I have created ...
18 views
### scipy curve_fit unable to fit curve
I am trying to use the scipy.optimize function curve_fit to fit a set of data points using a custom exponential function. My code is as follows: import numpy as np import matplotlib.pyplot as plt ...
25 views
### Adding more points on a polynomial curve
I'm trying to draw the curve of a polynomial but there are so few points that the curve looks really straight at some places. How can I test more points on the polynomial so I'll have a nicer curve? ...
16 views
### estimate the slope of the straight part in boltzmann curve
I was working with one dataset and found the curve to be sigmoidal. i have fitted the curve and got the equation A2+((A1-A2)/1+exp((x-x0)/dx)) where: x0 : Mid point of the curve dx : slope of the ...
27 views
### Python - Curve fitting of a 3D cloud of points with intersections
I have a cloud of points which represents a contour in 3D, with intersections. The following image shows an example of the type of geometry: you should see two orthogonal circles in 3D, intersecting ...
34 views
### Is it possible to curve fit a function that has multiple inputs?
I'm trying to curve fit a 3d function in Matlab that requires an input of X and Y values (they must be generated in a meshgrid first). I'm using the lsqcurvefit() function in Matlab, but from my ...
24 views
### Calculate curvature from a smoothed spline fit
I'm fitting a smoothed spline to an (x,y) closed outline of a shape and I want to find the local curvature of the shape. The curvature I get is very noisy and I'm sure there is a better way of doing ...
26 views
### Line fit in MATLAB or similar program
I have this data set: 10 12.14 20 36.82 30 59.48 40 79.96 50 89.45 60 95.04 70 95.50 80 95.50 90 95.50 100 95.50 110 95.50 The left column is time in minutes and the right is an arbitrary ...
41 views
### Automatically find the scaling factor of the x-axis using LsqFit (or other method)?
I have the following data: a vector B and a vector R. The vector B is the "independent" variable. For this pair, I have two data sets: One is an experimental measurement of Bex, Rex and the other is a ...
36 views
### Python statsmodels WLS (weighted least squares) error independent of weights
I'm using Python's statsmodels to perform a weighted linear regression. Since this is my first time with this module, I ran some basic tests. Doing these, I find that estimated errors on the ...
15 views
### Method for initial guess of standard deviation of 2d gaussian/gabor?
I'm working on curve fitting software in Matlab. So far it's going pretty well but I need a method of inputting an initial guess for my curve fitting software. I'm given a selection of points, but I ...
22 views
### How to fit a loess curve over this decomposed time series data in R?
We have time series data with some seasonality from the past 4 years. We want to predict the general rise in trend next year. For this, we decomposed the time series and observed the trend line: ...
37 views
I have the option to add bounds to sio.curve_fit. Is there a way to expand upon this bounds feature that involves a function of the parameters? In other words, say I have an arbitrary function with ...
20 views
### using undetermined number of parameters in scipy function curve_fit
First question: I'm trying to fit experimental datas with function of the following form: f(x) = m_o*(1-exp(-t_o*x)) + ... + m_j*(1-exp(-t_j*x)) Currently, I don't find a way to have an ...
16 views
### Model SIR Fitting
I've written a program to solve the model SIR system of equations using Euler method. I would like to find the parameters beta and gamma that can fit to some experimental data. def SIR(I0,beta,gama,...
21 views
### Calculating slope of a quasar spectrum for power law fitting
Working with Sloan Digital Sky Spectrum, I created a composite spectrum of quasars. The spectrum is a plot between wavelength (x-axis) measured in Angstrom and flux (y-axis) measured in ergs/cm^2/s/...
54 views
### Finding a probability density function that reproduces a histogram in Python
So all the data that I actually have is a picture of a histogram from which I can get heights and bin width, the median and one sigma errors. The histogram is skewed, so the 16th and 84th quantile ...
26 views
### Genetic Algorithm Package for Curve Fitting in Java
I am currently working to approximate 2-D coordinates points (approx. 1150 pairs of coordinates) using curve fitting techniques through Genetic Algorithms. I tried to build my own code, using BST ...
33 views
### What is the difference between numpy.polyfit and scipy.polyfit? [duplicate]
I came to know both numpy and scipy have polyfit function and visited here: http://docs.scipy.org/doc/numpy/reference/generated/numpy.polyfit.html Why does scipy.org have a page about numpy.polyfit? ...
45 views
### Fit fixed rectangle to set of points
i was wondering if someone every tried to fit a rectangle with a fixed size to a given set of points. Imagine you have a set of points which is unsorted and not always showing a full hull of a ...
31 views
### Fit poisson distribution to data
I have plotted a histogram and would like to fit a poisson distribution to the histogram. To do this, I have passed the x and y histogram coordinate vector to the poissfit() function to estimate ...
43 views
### How to fit parametric equations to data points in Python
I am looking for a way to fit parametric equations to a set of data points, using Python. As a simple example, given is the following set of data points: import numpy as np x_data = np.array([1, 2, ...
33 views
### Error fitting a model in nls
previous answers to similar questions have not help me to solve my problem. I am trying to fit a model y=a1*(1-exp(-a21*Age_WH40))^a3, where a21=ln(1/a3)/a2, and Age_WH40 goes from 1 to 40. I've plot ...
74 views
### Fitting a function with R [migrated]
I want to fit a function to these data: s <- c(20:60) I <- c(0, 0.007515662, 0.015878514, 0.024994325, 0.034728341, 0.044910579, 0.055344590, 0.065818599, 0.076118441, 0.086040566, 0....
38 views
### How to fit curve with asymmetric error bars?
I got some data as a result of a series of computer simulation with possible results of 1 and 0. They have asymmetric errorbars. I.E: xdata = [...] pdata = [...] pdatamax = [...] pdatamin = [...] ...
87 views
### calculate gaussian curve fitting on a list
I have a list data like below. I want to perform nonlinear regression Gaussian curve fitting between mids and counts for each element of my list and report mean and standard deviation mylist<- ...
35 views
### nls() false convergence (despite good starting values)
I have been working on a curve fitting script which fits 3 exponentially modified Gaussians (EMGs) to a convolved curve. My base function is similar to a Gaussian distribution, but includes a third ...
26 views
### optimize curve_fit for arbitrary sum of a input function
So I can use scipy curve fit to obtain parameters for my function defined below; it is basically a difference of gaussians. def diff_of_err_func(x, amp=None, pos=None, width=None, c=0): ''' :...
38 views
### Normalizing values in a curve_fit of Plancks Law from data set *Edit*
I have a large set of data that in terms of Intensity counts and Wavelength that I want fit with Planks Law to determine the guess parameter for Temperature. The Data set is imported as a text file ...
130 views
### Scatter plot kernel smoothing: ksmooth() does not smooth my data at all
Original question I want to smooth my explanatory variable, something like Speed data of a vehicle, and then use this smoothed values. I searched a lot, and find nothing that directly is my answer. ...
30 views
### assigning bounds to individual independent points within the scipy.optimize.curve_fit function
I am trying to fit a Planck curve to radiance readings. I know the radiance at some known wavelengths (11 data points), the parameter to fit is the temperature. The Planck function that returns the ...
38 views
### R: Error in optim (curve-fitting)
I am trying to figure out the parameters for attack rate a and handling time h to be used in producing a maximum likelihood for my data set of consumed prey densities. I am using the frair package on ...
58 views
### Python curve fitting on a barplot
How do I fit a curve on a barplot? I have an equation, the diffusion equation, which has some unknown parameters, these parameters make the curve larger, taller, etc. On the other hand I have a ...
22 views
### How to smoothen a 2D Gabor in Matlab?
I'm trying to use a curve fitting software that fits a 2D gabor to an input kernel. However this software (for some reason) fits a 2D Gaussian if there's any noise whatsoever. So my question is, how ...
55 views
### Python - curve_fit gives incorrect coefficients
I'm trying to pass two arrays for a fitting function, that takes both values. Data file: Column 1: Time Column 2: Temperature Column 3: Volume Column 4: Pressure 0.000,0.946,4.668,0.981 0.050,0....
18 views
### curve_fit “valueError operands could not be broadcast together” with negative values?
This error baffles me. I'm trying to get a curve_fit to some data, but as soon as I make my linear function negative (-m*x+b) or give a negative starting parameter, I get a value Error. How can this ...
16 views
### Extremely large confidence intervals using lsqnonlin and nlparci
I have used lsqonlin to fit a model to my dataset, and used nlparci to get the confidence intervals. However, while my parameters are of the order of 1, the confidence intervals are of the order of 9! ...
42 views
### centerline of a polygonal blob (binary image)
I have a binary image of a worm (blob extraction which works well). I am interested in fitting a centerline on the blowb (worm). So far I came up with this: starting from a polygon (after outline ...
35 views
### Error-weighted fit using uncertainties in LMFIT
I am trying to fit a model using LMFIT, I can easily do the following: def loss_function(params): residuals = [] for x, measured in ...: y = predict(x, params) residuals.append(y - ...
53 views
### Fit a curve to the boundary of a scatterplot
I'm trying to fit a curve to the boundary of a scatterplot. See this image for reference. I have accomplished a fit already with the following (simplified) code. It slices the dataframe into little ...
11 views
### lsqcurvefit fit method with a for loop
I've create a function called func2 (to fit same experimental data ) say that this function can be writing as follow: yi=func2(xi)=2n+3 (xi is the input, yi is the output) and n=f(xi,A,B) A and B are ...
39 views
### Python - Decay curve_fit breaks down in plot
I'm trying to plot values from a recorded data set from an experiment. When fitting the data with an exponential decay, it's very successful in the form of a normal plot. But having the plot in a ...
36 views
### Is there any way to fit a model for graph analysis in MATLAB?
I am getting the plot as shown in the figure in MATLAB after applying some function. On fitting a polynomial, we see that we get a parabolic curve for this curve as shown in the next figure. . Is ...
45 views
### Python - Fitting exponential decay curve from recorded values
I'm aware that there are threads pertaining to this, but i'm confused to where I want to I fit my data to the fit. My data is imported and plotted as such. import matplotlib.pyplot as plt %...
64 views
### Confidence intervals for linear curve fit under constraints in MATLAB
I have fitted a straight line to a dataset with 68 samples, under the constraint that the line passes through (x0,y0) using the function lsqlin in MATLAB. How can I find the confidence intervals for ...
49 views
### How to create noise for a 2D Gaussian?
I'm trying to practice curve fitting on a 2D Gaussian, but in order to do that I need to add random noise to my predefined Gaussian. My first instinct was to cycle through two for loops and create two ...
24 views
### Fitting a mixture of gaussians with negative weights
I know Gaussian Mixture Models are used for distribution of points and thus the functions they represent are positive (PDFs are postivie). However I want to approximate a function using GMMs which ...
55 views
### how Fitting a curve to a histogram in python
I need to fit a curve to my histogram. It just shows me the histogram not the curve fitted. This is my code: from scipy.stats import norm import matplotlib.pyplot as plt import numpy as np import ... | 3,084 | 12,955 | {"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-2016-30 | latest | en | 0.931029 |
http://oeis.org/A138111 | 1,369,177,175,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700842908/warc/CC-MAIN-20130516104042-00000-ip-10-60-113-184.ec2.internal.warc.gz | 196,521,448 | 3,539 | This site is supported by donations to The OEIS Foundation.
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A138111 Prime numbers p1 such that p1*p2-mod[p2,p1] is a prime, where p2 is the next prime after p1. 4
2, 3, 7, 19, 23, 43, 53, 79, 127, 211, 229, 233, 337, 397, 443, 463, 467, 499, 503, 601, 631, 661, 967, 991, 1009, 1129, 1153, 1213 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS EXAMPLE 2 prime, 3 next-prime, 2*3-Mod[3,2]=2*3-1=5 prime. 3 prime, 5 next-prime, 3*5-Mod[5,3]=3*5-2=13 prime. 7 prime, 11 next-prime, 7*11-Mod[11,7]=7*11-4=73 prime. MATHEMATICA a={}; Do[p1=Prime[n]; p2=Prime[n+1]; e=p1*p2-Mod[p2, p1]; If[PrimeQ[e], AppendTo[a, p1]], {n, 10^2*2}]; a CROSSREFS Sequence in context: A117763 A165571 A178954 * A218100 A078373 A038878 Adjacent sequences: A138108 A138109 A138110 * A138112 A138113 A138114 KEYWORD nonn AUTHOR Vladimir Joseph Stephan Orlovsky, May 06 2008 STATUS approved
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Content is available under The OEIS End-User License Agreement . | 476 | 1,290 | {"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} | 3.421875 | 3 | CC-MAIN-2013-20 | latest | en | 0.537798 |
https://www.gradesaver.com/textbooks/math/calculus/calculus-10th-edition/chapter-2-differentiation-2-4-exercises-page-136/68 | 1,548,136,576,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583829665.84/warc/CC-MAIN-20190122054634-20190122080634-00227.warc.gz | 808,093,778 | 12,527 | ## Calculus 10th Edition
$f(x)=\dfrac{6-4x}{(x^2-3x)^3},f'(4)=-\dfrac{5}{32}.$
$f(x)=(x^2-3x)^{-2}$ $u=x^2-3x$; $\dfrac{du}{dx}=2x-3$ $f(u)=u^{-2};\dfrac{d}{du}f(u)=-\dfrac{2}{u^3}$ $\dfrac{d}{dx}f(x)=\dfrac{d}{du}f(u)\times\dfrac{du}{dx}=\dfrac{6-4x}{(x^2-3x)^3}.$ $f'(4)=\dfrac{6-4(4)}{(4^2-3(4))^3}=-\dfrac{5}{32}.$ A graphing utility was used to verify this result. | 209 | 370 | {"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} | 3.8125 | 4 | CC-MAIN-2019-04 | latest | en | 0.469298 |
https://math.stackexchange.com/questions/906429/what-is-hat-mathbbz?noredirect=1 | 1,563,316,311,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195524972.66/warc/CC-MAIN-20190716221441-20190717003441-00356.warc.gz | 470,011,429 | 36,235 | # What is $\hat{\mathbb{Z}}$?
I have been reading a bit about unramified extensions. If $K$ is $\mathbb{Q}$ or $\mathbb{Q}_p$ (p-adics), then there is a maximal unramified extension $K^{nr}$ of $K$. Then I have read in some notes that $${\rm Gal}(K^{nr} / K) \cong \hat{\mathbb{Z}}.$$ My question is: what is $\hat{\mathbb{Z}}$?
I have been told that this is $$\varprojlim \mathbb{Z} / n\mathbb{Z},$$
But I am not sure what this is. I think it is called an inverse limit. Is there a concrete way to think about this?
• I think this answer gives a clear idea of what an inverse limit is (at least in some restricted sense). Using the hat notation is a rather unfortunate overuse of notation in my opinion since $\widehat{\Bbb Z}$ has a completely different meaning in harmonic analysis (which somewhat intersects $p$-adic analysis). – Cameron Williams Aug 22 '14 at 22:17
• If you are happy with $p$-adics, then one description of $\widehat{\mathbb{Z}}$ (the profinite completion of the integers, see groupprops.subwiki.org/wiki/…) is as the product $\prod_p \mathbb{Z}_p$, via the Chinese remainder theorem. – Qiaochu Yuan Aug 23 '14 at 6:18
The ring $\widehat{\mathbb{Z}}$, called the profinite completion of $\mathbb{Z}$, can be thought of as a subring of the infinite product $\prod_{n = 1}^{\infty} \mathbb{Z} / n\mathbb{Z}$, which satisfies a `coherent sequence' condition. That is, a tuple $(x_1,x_2,x_3,\ldots ) \in \prod_{n=1}^{\infty} \mathbb{Z} / n \mathbb{Z}$ is in $\widehat{\mathbb{Z}}$ iff $x_n \equiv x_m \bmod m$ whenever $m$ divides $n$ (you can also think of this condition as saying that $x_n$ maps to $x_m$ under the canonical map $\mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/ m \mathbb{Z}$ whenever $m$ divides $n$).
Similarly, the $p$-adic integers $\mathbb{Z}_p$ are the tuples in $\prod_{n=1}^{\infty} \mathbb{Z} / p^n \mathbb{Z}$ that satisfy the analogous condition for the canonical maps $\mathbb{Z} / p^n \mathbb{Z} \to \mathbb{Z} / p^m \mathbb{Z}$ for $n \geq m$. The more general construction, called an inverse limit, is summarized here. | 669 | 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} | 3.046875 | 3 | CC-MAIN-2019-30 | latest | en | 0.81207 |
http://www.primidi.com/structural_induction | 1,553,640,720,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912206677.94/warc/CC-MAIN-20190326220507-20190327002507-00331.warc.gz | 341,340,168 | 4,145 | # Structural Induction
Structural induction is a proof method that is used in mathematical logic (e.g., in the proof of Łoś' theorem), computer science, graph theory, and some other mathematical fields. It is a generalization of mathematical induction. Structural recursion is a recursion method bearing the same relationship to structural induction as ordinary recursion bears to ordinary mathematical induction.
In general, the idea is that one wishes to prove some proposition P(x), where x is any instance of some sort of recursively defined structure such as lists or trees. A well-founded partial order is defined on the structures ("sublist" for lists and "subtree" for trees). The structural induction proof is a proof that the proposition holds for all the minimal structures, and that if it holds for the immediate substructures of a certain structure S, then it must hold for S also. (Formally speaking, this then satisfies the premises of an axiom of well-founded induction, which asserts that these two conditions are sufficient for the proposition to hold for all x.)
A structurally recursive function uses the same idea to define a recursive function: "base cases" handle each minimal structure and a rule for recursion. Structural recursion is usually proved correct by structural induction; in particularly easy cases, the inductive step is often left out. The length and ++ functions in the example below are structurally recursive.
For example, if the structures are lists, one usually introduces the partial order '<' in which L < M whenever list L is the tail of list M. Under this ordering, the empty list is the unique minimal element. A structural induction proof of some proposition P(l) then consists of two parts: A proof that P is true, and a proof that if P(L) is true for some list L, and if L is the tail of list M, then P(M) must also be true.
Eventually, there may exist more than one base case, and/or more than one inductive case, depending on how the function or structure was constructed. In those cases, a structural induction proof of some proposition P(l) then consists of: A) a proof that P(BC) is true for each base case BC, and B): a proof that if P(I) is true for some instance I, and M can be obtained from I by applying any one recursive rule once, then P(M) must also be true. | 486 | 2,328 | {"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-2019-13 | latest | en | 0.917698 |
https://www.fxsolver.com/browse/?like=2906&p=144 | 1,642,589,457,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301309.22/warc/CC-MAIN-20220119094810-20220119124810-00199.warc.gz | 844,944,104 | 64,852 | '
# Search results
Found 1514 matches
Capillary Action - height of the meniscus
Capillary action (sometimes capillarity, capillary motion, or wicking) is the ability of a liquid to flow in narrow spaces without the assistance of, and ... more
Time delay for a signal from Earth to a Satelite in geostationary orbit and back
A geostationary orbit, geostationary Earth orbit or geosynchronous equatorial orbit (GEO), is an orbit whose position in the sky ... more
Superformula - Parametric Equation for X Axis
The superformula is a generalization of the superellipse and was first proposed by Johan Gielis in 2003. Gielis suggested that the formula can be used to ... more
Superformula - Parametric Equation for Y Axis
The superformula is a generalization of the superellipse and was first proposed by Johan Gielis in 2003. Gielis suggested that the formula can be used to ... more
Cyclic quadrilateral (Length of the diagonal opposite angle B)
In Euclidean geometry, a cyclic quadrilateral or inscribed quadrilateral is a quadrilateral whose vertices all lie on a single circle. This circle is ... more
Elastic deflection at any point along the span of a center loaded beam
Elastic deflection is the degree to which a structural element is displaced under a load.
The deflection at any point, along the span of a center ... more
Calibrated airspeed from impact pressure - Subsonic speed
Calibrated airspeed (CAS) is indicated airspeed corrected for instrument and position error.
When flying at sea level ... more
Total constant power (Three-phase electric application)
In electrical engineering, three-phase electric power systems have at least three conductors carrying alternating current voltages that are offset in time ... more
Shockley diode equation (small forward bias voltages)
In electronics, a diode is a two-terminal electronic component with asymmetric conductance; it has low (ideally zero) resistance to current in one ... more
Mean anomaly - function of mean longitude
In celestial mechanics, the mean anomaly is an angle used in calculating the position of a body in an elliptical orbit in the classical two-body problem. ... more
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Category | 495 | 2,282 | {"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} | 2.671875 | 3 | CC-MAIN-2022-05 | latest | en | 0.895653 |
https://expertinstudy.com/t/jlgdm0FseU0 | 1,653,562,604,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662604794.68/warc/CC-MAIN-20220526100301-20220526130301-00420.warc.gz | 310,302,807 | 4,436 | In the equation below, k is a constant. If x = 9, what is the value of k?Vk + 2 - x = 0
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she ts08gs0itspisgi0ttspispitpx0ufs9u
Factored: f(x) = (x-12)(x+3)
possible x values: 12 and -3Whoa this is long dang sorry i don’t know
181 | 103 | 285 | {"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} | 2.859375 | 3 | CC-MAIN-2022-21 | latest | en | 0.799767 |
newdawnlearning.com.sg | 1,685,599,487,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647614.56/warc/CC-MAIN-20230601042457-20230601072457-00700.warc.gz | 458,238,734 | 47,905 | ## Our JC, A Level, H2 Math Tuition
The most common subject taken up in JC, H2 Math is a pre-requisite to all courses in university. It is a build-on to Secondary A Maths and consists of two main areas covered: Pure Maths and Statistics.
In Pure Maths, the sections covered are Functions and Graphs, Sequences and Series, Differentiation and Integration, Vectors, and Complex Numbers.
In Statistics, the sections covered are Permutations and Combinations, Probability, Discrete Random Variables, Binomial distribution, Normal distribution, Hypothesis Testing, and Correlation and Regression.
JC H2 math covers a wider range of topics than secondary school math and while some of the topics may be similar, there still lies many differences in the content covered. Even with overlapping topics, the content covered in H2 math is more in-depth and requires advanced analytical and critical thinking skills which can be a concern for students that are newly tackling H2 math.
## JC/A Level Math Tuition Teaching Methodology
In our JC Maths tuition classes, our objective is to address the difficulties commonly encountered by students. Thereby, every lesson is well-structured and delivered systematically to meet the learning needs of the students.
Mr Daniel adopts a hybrid of a 1-to-1 and classroom styled learning. This allows different people to learn at their own pace.
During the H2 Maths tuition, Mr Daniel will first teach the topic and show the working of solving a typical H2 Maths question, followed by practice on the students’ part, and then leave some time open for interaction before moving on to the next worked example. As a result, students find our H2 Math tuition effective and encouraging as they gain more confidence in tackling the questions.
We believe in guiding our students to excel through patience and consistent practice. Therefore, our A Level H2 Math tuition provides them with quality resources and teaching to help them achieve their fullest potential.
To optimise learning outcomes from attending JC Maths tuition at New Dawn Learning Studio, we will also conduct lessons a step ahead of the school delivery. Students will find it easier to follow through with the classes at their respective JCs, thereby, maximising their learning capability.
## Our 4-Prong JC/A Level Math Tuition Approach
#### JC/A Level Math Tuition Notes
Well-conceived and tailored H2 math lecture notes are given during each lesson so as to aid students to follow and understand the particular topic delivered. The lecture notes come with clear step-by-step worked examples and tips on tackling the question concerned. Students will find the lecture notes realistically as the questions are drawn from TYS H2 Maths and post JC exam papers.
#### JC/A Level Math Tuition Tutorial
After each lecture, H2 Maths students will be given some time to practice what they have just learned so as to ensure that they fully grasp the concepts of the topic covered and know how to apply them to typical H2 Maths questions. During the tutorial, students can freely interact with their H2 Maths tutor to seek further clarification to enhance their learning. Homework, though optional but helpful, is given for further practice at home. At New Dawn Learning Studio, we believe practice is paramount to secure an ultimate good pass or even a distinction in H2 Maths. The homework worksheets are compiled according to the level of difficulty for each section. This helps students to progress steadily and build their confidence in tackling H2 Maths. To prepare students well for the final Singapore-Cambridge GCE A-Levels H2 Maths examination, we encourage students to tackle more challenging questions in the worksheets strategically drawn from post-JC H2 Maths exam papers.
#### Summary Sheet
To further reinforce learning, a separate Summary Sheet will be given too. It sums up the topic by including the essential formula for a quick check when H2 Maths students need one in their work.
#### Topical and Sectional Revision
Topical and Sectional revision allow students to quickly recognize which topic a particular question may belong to, making the problem-solving process easier. H2 math topical questions are sorted into categories to teach students how to solve problems categorically. This allows for a more holistic understanding and also prepares students to expect a variety of questions that examiners can set in any given paper.
## Top-Notched H2 Math Materials
At New Dawn Learning Studio, we believe every student is important, so we place great emphasis on effective delivery to assist students to develop their potential in the subject they wish to improve further.
Very often, the lack of comprehensive materials for self-revision is an apparent hindrance to learning A level maths. To overcome this problem, we’ve put together tailored and comprehensive study materials for our JC H2 Math tuition students. The worksheets are constantly updated with the newest trends and challenging questions from all the different schools, which made it easy for students to practise and do not have to go through many practise papers or assessment books just to find one challenging question. Also, the notes are written in a manner targeted to the various important concepts to make H2 Math seem less intimidating than it was.
Students at New Dawn Learning Studio can take tutorials and notes from their classes home with them. A complete H2 math summary sheet is also supplied for students to process and keep the information they learned in these tutorials.
The tutorials are designed to help students better understand and retain the material they have just learned. As a short study aid, each H2 math student receives a 26-formula sheet that covers every chapter of the course subject in advance of their impending tests.
## Consultation Availability
Mr Daniel is well-known by his students for being extremely approachable and is always willing to offer help via WhatApps, face to face or a Zoom Session.
## Vast Experiences in Teaching Countless Batches of JC Students
We have experiences in teaching students from all different JC in Singapore. Topics are catered to the individual JC.
## Annual A Level Maths Workshops
For students who have not signed up early for H2 Maths tuition and missed some topics, or for those current students who want to enhance their learning, we conduct H2 Maths Camps and Workshops in March, June, and September scheduled by (subjects) and topics to cater to individual learning needs. The workshops facilitate pre-exam revision and enhance their learning journeys. Do check up on the timetable!
Click on workshop to retrieve our upcoming H2 Maths Workshop Schedule.
## Conducive Studying Environment
A self-study area has been set up in the tuition centre, where you may study on your own and get help from Mr Daniel even if you aren’t attending his A Level H2 Math tuition classes! | 1,353 | 6,957 | {"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-2023-23 | longest | en | 0.946324 |
http://www.solutioninn.com/a-random-sample-of-students-at-oxnard-college-reported-what | 1,505,977,625,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687702.11/warc/CC-MAIN-20170921063415-20170921083415-00203.warc.gz | 569,790,218 | 7,576 | # Question: A random sample of students at Oxnard College reported what
A random sample of students at Oxnard College reported what they believed to be their heights in inches. Then the students measured each others' heights in centimeters, without shoes. The data shown are for the men. Assume that the conditions for t-tests hold.
a. Convert heights in inches to centimeters by multiplying inches by 2.54. Find a 95% confidence interval for the mean difference as measured in centimeters. Does it capture 0? What does that show?
b. Perform a t-test to test the hypothesis that the means are not the same.
Use a significance level of 0.05, and show all four steps.
Sales0
Views65 | 155 | 682 | {"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-2017-39 | latest | en | 0.933822 |
http://youth.worldbridge.org/how-to-play-suit-combinations/ | 1,721,499,714,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517515.18/warc/CC-MAIN-20240720174732-20240720204732-00713.warc.gz | 66,061,653 | 39,697 | Source: paloaltobridge.org This is a quiz from the Encyclopedia of Bridge on how you would play the following suit combinations. Assume there are unlimited entries to both hands and no information from opponents. Take the quiz and see how you do.
12. Lead low from dummy to the K. If this wins, lead low and duck in dummy. When you regain the lead, play low from dummy again. You will win a 2nd trick with the Q if E has 3 or fewer cards including the A.
13. Lead low and finesse the 10-J unless W plays an honor. If E wins the 1st trick, finesse again when you gain the lead. You will win 2 tricks unless E holds both the K and Q.
14. Lead low from your hand and finesse the 9-10. If E wins the Q or J, finesse again. You will win one trick if W has the Q, J, or both., or if he has just the A and plays it the 1st or 2nd time.
15. Lead low from you hand and finesse the 9; a “triple finesse”. If E wins with the K or Q, lead low from your hand and finesse the J. You will win 2 tricks if W has K 10, Q 10, or K Q 10.
16. Lead low from your hand and finesse the 9 – 10. If E wins the 1st trick, finesse the 10. You win 3 tricks unless E has both the Q and J.
17. Cash the K 1st and the lead to dummy intending to finesse the 10 – J if the Q is not played. If you win the trick, repeat the finesse. Four Tricks are won if W has the Q or E has a singleton Q.
19. Cash the A or K. If both opponents play low cards , play the K with 9 cards.
20. Lead from your hand and finesse. If it wins, finesse again.
21. Lead from your hand; if W follows with a low card, the odds heavily favor taking the finesse.
22. Cash the A. With 11 cards, the odds slightly favor the K dropping.
23. Cash the A or Q first to insure 5 tricks, even if the suit splits 4–0.
24. Cash the A or K 1st, to avoid losing a trick if W has J 10 x x.
25. Cash A or K: If E plays a low card, cash the K and hope the suit divides 2-2.. If E plays an honor or is void, return to your hand and finesse. To win all the ticks in the suit or to lose only 1 trick.
26. To win 5 tricks, finesse the Q and hope W has a doubleton Q. To win 4 tricks, cash the A; if the K doesn’t fall, return to your hand and lead toward the Q.
27. To win 6 tricks, cash the A and hope for a singleton K. To insure 5 tricks, lead toward the A-10 and finesse the 10 if E follows with a low card or play the A if E follows with an honor or shows out.
28. To win 6 tricks, lead low toward your hand intending to finesse the 9-10. The only chance is if E has a doubleton QJ or QJx. To win 5 tricks, cash the A or K. If both E and W play a low card, go to dummy and lead toward your hand, intending to finesse if E follows with a low card.
29. To win 4 tricks, cash the K and lead toward dummy, intending to finesse the J if W follows suit. To win 3 tricks, cash the A and lead toward K 9 x. If E follows with a low card, finesse the J. If E follows with a low card, finesse the 9. If E plays the 10 or Q win with the K. If E shows out, go up with the K and lead toward the J.
30. To win 4 tricks, cash the K and lead to the Jack, hoping W began with Qxx. To win 3 tricks, cash the A and K in that order and lead toward the Jx. This gains a trick when E has a doubleton Q 10.
Don’t forget to follow us @ | 934 | 3,253 | {"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-2024-30 | latest | en | 0.917163 |
https://studylib.net/doc/9842498/review-thermochemistry-unit | 1,652,729,270,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662512229.26/warc/CC-MAIN-20220516172745-20220516202745-00513.warc.gz | 625,948,883 | 11,009 | # Review Thermochemistry Unit
```SCH4U
September 2014
REVIEW FOR THERMOCHEMISTRY TEST
The combustion of methanol is shown by the following equation:
a) Using the data below, find the heat of reaction for the combustion of methanol.
C(s) + O2(g) CO2(g)
ΔH = -393 kJ
2H2(g) + O2(g) 2H2O(g)
ΔH = -484 kJ
2C(s) + 4H2(g) + O2(g) 2CH3OH(l) ΔH = -476 kJ
b) What is the molar enthalpy of combustion for methanol?
c) Express the energy changes for the target reaction in the form ΔHx =
d) Draw the potential energy diagram to illustrate the energy changes for the target reaction.
e) Write the target reaction that includes the heat term in the equation.
f) What mass of water could be heated from 20.00 C to 35.00 C by the burning of 10.0 g of methanol that is
held in a calorimeter with a heat capacity of 2.9 kJ/ oC?
g) Will ΔS increase or decrease in the combustion of methanol? Will ΔS be positive or negative?
h) Calculate the change in entropy for the combustion of methanol.
i)
Consider your values of ΔH and ΔS. Will the combustion of methanol always be spontaneous? Explain.
j)
Calculate the change in Gibb’s free energy for the combustion of methanol at 100 oC. Is the reaction
spontaneous or non-spontaneous at this temperature? Explain your choice.
2. Consider the reaction:
NH4Cl(s) NH3(g) + HCl(g)
What are the temperature requirements for the above reaction to work?
3. The molar enthalpy of combustion of butane is -2871 kJ/mol. What is the molar enthalpy of formation of
butane? HINT: first write out the balanced equation for the combustion reaction.
1
4.
a) Draw the heating curve for methanol, given: enthalpy of fusion = 3.2 kJ/mol; enthalpy of vaporization =
35.3 kJ/mol; heat capacity of liquid methanol = 2.5 J/g∙oC; heat capacity gaseous methanol 1.6 J/g∙oC ;
the melting point of methanol = -97.6oC; boiling point of methanol = 64.7oC.
b) What amount of energy would be required to raise a 10.0 g sample of methanol from 0 oC to 100oC.
NOTE: You will be expected to create the formulas for any of the hydrocarbons we studied in class as well
as create their combustion reactions.
NOTE: You will be expected to understand the activities and labs from this unit. Similar questions will be
on the test.
2
``` | 654 | 2,232 | {"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-2022-21 | latest | en | 0.813357 |
http://mathhelpforum.com/statistics/126293-zscore-standard-normal-probabilities.html | 1,529,475,361,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863463.3/warc/CC-MAIN-20180620050428-20180620070428-00018.warc.gz | 201,121,802 | 10,067 | 1. ## Zscore/Standard Normal Probabilities
Given:
mu = 65.7
sigma = 15
Standard normal distribution
Question:
20% of the area will be less than what x value?
I looked at my z-score table and couldn't find .0200 so now I have no clue how to even begin!
2. From your z-score table $\displaystyle P(z< -0.8416)= .2$
Use this fact in $\displaystyle z = \frac{x-\mu}{\sigma} \implies -0.8416 = \frac{x-65.7}{15}$ to solve for $\displaystyle x$
3. Where did you get .8416? Could you possibly have a more complete z-score table or am I missing some intermediary calculation?
4. Originally Posted by tom ato
Where did you get .8416? Could you possibly have a more complete z-score table or am I missing some intermediary calculation?
You could just as easily use -0.84. There is no missing step here, and you no doubt are meant to use -0.84. | 231 | 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": 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} | 3.375 | 3 | CC-MAIN-2018-26 | latest | en | 0.916574 |
https://www.examrace.com/NMAT/NMAT-Model-Questions/Aptitude-Questions/Aptitude-Logical-Reasoning-Simple-Interest-Part-5.html | 1,603,339,253,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107878879.33/warc/CC-MAIN-20201022024236-20201022054236-00044.warc.gz | 684,380,244 | 10,505 | # Aptitude Logical Reasoning Simple Interest 2020 NMAT Part 5
Glide to success with Doorsteptutor material for UGC : Get detailed illustrated notes covering entire syllabus: point-by-point for high retention.
1. Shruti invested a sum of money at a certain rate of simple interest for a period of five years. Had she invested the sum for a period of eight years for the same rate, the total interest earned by her would have been sixty percent more than the earlier interest amount. Find the rate of interest p.a.
A.
B.
C.
D. Cannot be determined
E. None of these
2. Manoj borrowed Rs.3450 from Anwar at p.a. simple interest for three years. He then added some more money to the borrowed sum and lent it to Ramu for the same time at p.a. simple interest. If Manoj gains Rs. by way of interest on the borrowed sum as well as his own amount from the whole transaction, then what is the sum lent by him to Ramu?
A. Rs.3,870
B. Rs.5,355
C. Rs.3,855
D. Rs.3600
E. None of these
3. A certain sum of money is invested for one year at a certain rate of simple interest. If the rate of interest is higher, then the invest earned will be more than the interest earned earlier. What is the earlier rate of interest?
A. p.a.
B. p.a.
C. p.a.
D. p.a.
E. None of these
4. The simple interest accrued on an amount Rs.10,000 at the end of two years is same as the compound interest on Rs.8,000 at the end of two years. The rate of interest is same in both the cases. What is the rate of interest?
A. p.a.
B. p.a.
C. p.a
D. p.a.
E. None of these
5. Rs.4500 amounts to Rs.5544 in two years at compound interest, compounded annually. If the rate of the interest for the first year is , find the rate of interest for the second year?
A.
B.
C.
D.
E. None of these
6. How much time will take for an amount of Rs. 450 to yield Rs. 81 as interest at per annum of simple interest?
A. years
B. 4 years
C. years
D. 5 years
7. A sum of Rs. 125000 amounts to Rs. 15500 in 4 years at the rate of simple interest. What is the rate of interest?
A.
B.
C.
D.
8. Reena took a loan of Rs. 1200 with simple interest for as many years as the rate of interest. If she paid Rs. 432 as interest at the end of the loan period, what was the rate of interest?
A.
B. 6
C. 18
D. Cannot be determined
E. None of these
9. A man took loan from a bank at the rate of p.a. S.I. After 3 years he had to pay Rs. 5400 interest only for the period. The principal amount borrowed by him was?
A. Rs. 2000
B. Rs. 10000
C. Rs. 15000
D. Rs. 20000
10. What is the present worth of Rs. 132 due in 2 years at simple interest per annum?
A. Rs. 112
B. Rs.
C. Rs. 120
D. Rs. 122 | 762 | 2,667 | {"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-2020-45 | latest | en | 0.947912 |
http://reviewgamezone.com/game.php?id=6576 | 1,495,913,741,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463609054.55/warc/CC-MAIN-20170527191102-20170527211102-00242.warc.gz | 369,783,596 | 6,385 | Integers And One Step Equations (Easy) (ID: 6576)
.rgz-under-title-game-page { width: 200px; height: 200px; } @media(min-width: 275px) { .rgz-under-title-game-page { width: 250px; height: 250px; } } @media(min-width: 380px) { .rgz-under-title-game-page { width: 300px; height: 250px; } } @media(min-width: 405px) { .rgz-under-title-game-page { width: 336px; height: 280px; } } (adsbygoogle = window.adsbygoogle || []).push({}); Integers and Easy Solving Equations. Linked to MD state standards at the 6th - 7th grade level. To get started, select a game and test your knowledge on the topic. To play games disable or bypass popup blocker software.
Preview Questions in the Games If you are confused about any question or answer choices in the games print the data and review it with a teacher. Enjoy! Wild Wild Taxi Racing Drive as fast as possible and go as far as you can without crashing your cab into other cars which are on the road [directions]. Bouncing Balls Play this very fun school game in which you must try to get clear as many balls from the screen before they fall [directions]. Heroic Ants Very fun learning game where you throw an ant across the yard and see how far he goes. Try to set a personal record [directions]. Alien Intruders The aliens are attacking your spaceship and you must try to defend yourself before they destroy you! [directions]. Free Kick Soccer Try and score a penalty shot against a soccer goalie but keep your eye on the target or you might miss the shot [directions]. .rgz-test-mid-game-page { width: 200px; height: 200px; } @media(min-width: 275px) { .rgz-test-mid-game-page { width: 250px; height: 250px; } } @media(min-width: 380px) { .rgz-test-mid-game-page { width: 300px; height: 250px; } } @media(min-width: 405px) { .rgz-test-mid-game-page { width: 336px; height: 280px; } } Typing Speed Test Practice your keyboarding skills as quickly and as accurately as possible so you can type faster [directions]. Nothing But Net Shoot the ball to all the baskets and see how many you can get before your next question pops up [directions]. Hovercraft Racing Test your racing skills against the computer in the fun game but be careful, hovercrafts are very hard to control [directions]. Quick Math Pick up the symbol to complete the equation as quickly as possible to test your reaction time [directions]. Question Quiz Do not feel like playing a game? In this non-game format, pick the answer and see if you are correct. [long version*] Staries Try your luck in this game where you need to clear the board by matching three or more colored stars [directions]. Paper Bird (Flappy Bird) In this strangely addictive game you need to help a paper bird fly through the obstacles by flapping its wings. [directions]. Tower Blocks A very addicting fun game where the player needs to stack the blocks as high as they can before they all fall [directions]. .rgz-bottom-responsive { width: 200px; height: 200px; } @media(min-width: 275px) { .rgz-bottom-responsive { width: 250px; height: 250px; } } @media(min-width: 380px) { .rgz-bottom-responsive { width: 300px; height: 250px; } } @media(min-width: 405px) { .rgz-bottom-responsive { width: 336px; height: 280px; } } Leaping Frog Why did the frog cross the pond? To get to the other side. Help this frog get there by controlling its jumps! [directions]. Jumping Tiles You need to click the highlighted tile to jump onto the next tile. The faster you finish, the higher your score [directions]. Ghost Man Try to eat as many dots as possible before the scary ghosts attack you and end your turn of play [directions]. Word Grid Try to find as many words as you can. The bigger the word the more points! How many can you find? [directions].
TEACHERS / EDUCATORS | 958 | 3,743 | {"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-2017-22 | longest | en | 0.834738 |
https://gmatclub.com/forum/according-to-the-better-business-bureau-if-you-fail-to-42820.html?fl=similar | 1,495,976,707,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463609817.29/warc/CC-MAIN-20170528120617-20170528140617-00348.warc.gz | 926,817,487 | 46,794 | Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
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# According to the Better Business Bureau, if you fail to
Author Message
Director
Joined: 14 Jan 2007
Posts: 775
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### Show Tags
03 Mar 2007, 21:13
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Question Stats:
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According to the Better Business Bureau, if you fail to advertise the highest price in a range of prices for a service or product as prominently as that of the lowest, it violates the New York Consumer Protection Law.
(A) if you fail to advertise the highest price in a range of prices for a service or product as prominently as that of the lowest, it
(B) if one fails to advertise the highest price in a range of prices for a service or product as prominently as the lowest price, it
(C) failure to advertise the highest price in a range of prices for a service or product as prominently as the lowest
(D) failure to advertise as prominently the highest price in a range of prices for a service or product as the lowest
(E) failing to advertise as prominently the highest price in a range of prices for a service or products as that of the lowest
[Reveal] Spoiler: OA
VP
Joined: 06 Feb 2007
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Re: According to the Better Business Bureau, if you fail to [#permalink]
### Show Tags
03 Mar 2007, 22:16
I agree with C. It is the most concise and clear out of all the choices.
VP
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Re: According to the Better Business Bureau, if you fail to [#permalink]
### Show Tags
03 Mar 2007, 22:44
Straight C on this.
A and B are out because of "if you/one fail,.... it "
D and E are out for splitting "as prominently as".
_________________
Trying hard to conquer Quant.
Manager
Joined: 09 Jan 2007
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Re: According to the Better Business Bureau, if you fail to [#permalink]
### Show Tags
03 Mar 2007, 23:01
(C)
failure to
and
as prominently as
are correctly used in option (C).
VP
Joined: 21 Mar 2006
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Location: Bangalore
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Re: According to the Better Business Bureau, if you fail to [#permalink]
### Show Tags
05 Mar 2007, 01:10
A, B - pronoun problem. OUT
D - comparison is not clear. OUT
E - comparison is not clear, products (plural) is wrong. OUT
Give me C!
VP
Joined: 22 Oct 2006
Posts: 1438
Schools: Chicago Booth '11
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Re: According to the Better Business Bureau, if you fail to [#permalink]
### Show Tags
07 Mar 2007, 19:26
C
D & E are awkward
A and B have pronoun reference with "it"
Re: According to the Better Business Bureau, if you fail to [#permalink] 07 Mar 2007, 19:26
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Display posts from previous: Sort by | 1,123 | 3,988 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2017-22 | latest | en | 0.936543 |
http://financeformulas.net/Weighted_Average.html | 1,550,665,195,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247494741.0/warc/CC-MAIN-20190220105613-20190220131613-00058.warc.gz | 89,525,687 | 10,251 | Custom Search
# Weighted Average
Weighted Average Calculator (Click Here or Scroll Down)
The weighted average formula is used to calculate the average value of a particular set of numbers with different levels of relevance. The relevance of each number is called its weight. The weights should be represented as a percentage of the total relevancy. Therefore, all weights should be equal to 100%, or 1.
The most common formula used to determine an average is the arithmetic mean formula. This formula adds all of the numbers and divides by the amount of numbers. An example would be the average of 1,2, and 3 would be the sum of 1 + 2 + 3 divided by 3, which would return 2. However, the weighted average formula looks at how relevant each number is. Say that 1 only happens 10% of the time while 2 and 3 each happen 45% of the time. The percentages in this example would be the weights. The weighted average would be 2.35.
The weighted average formula is a general mathematical formula, but the following information will focus on how it applies to finance.
### Use of Weighted Average Formula
The concept of weighted average is used in various financial formulas. Weighted average cost of capital (WACC) and weighted average beta are two examples that use this formula.
Another example of using the weighted average formula is when a company has a wide fluctuation in sales, perhaps due to producing a seasonal product. If the company would like to calculate the average of one of their variable expenses, the company could use the weighted average formula with sales as the weight to gain a better understanding of their expenses compared to how much they produce or sell.
#### Example of Weighted Average Formula
A basic example of the weighted average formula would be an investor who would like to determine his rate of return on three investments. Assume the investments are proportioned accordingly: 25% in investment A, 25% in investment B, and 50% in investment C. The rate of return is 5% for investment A, 6% for investment B, and 2% for investment C. Putting these variables into the formula would be
which would return a total weighted average of 3.75% on the total amount invested. If the investor had made the mistake of using the arithmetic mean, the incorrect return on investment calculated would have been 4.33%. This considerable difference between the calculations shows how important it is to use the appropriate formula to have an accurate analysis on how profitable a company is or how well an investment is doing.
New to Finance?
## Weighted Average Calculator
Variables shown as percentages
*All weights must equal 1* | 554 | 2,659 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2019-09 | latest | en | 0.958923 |
https://www.hackmath.net/en/math-problem/33153?tag_id=77 | 1,603,634,641,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107889173.38/warc/CC-MAIN-20201025125131-20201025155131-00590.warc.gz | 745,656,460 | 14,503 | # Triangular pyramid
Calculate the volume of a regular triangular pyramid with edge length a = 12cm and pyramid height v = 20cm.
Correct result:
V = 415.6922 cm3
#### Solution:
We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!
Showing 1 comment:
Mathematican
the base is an equilateral triangle, the formula can be found in tables or online.
Tips to related online calculators
Tip: Our volume units converter will help you with the conversion of volume units.
Pythagorean theorem is the base for the right triangle calculator.
#### You need to know the following knowledge to solve this word math problem:
We encourage you to watch this tutorial video on this math problem:
## Next similar math problems:
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The section of the railway embankment is an isosceles trapezoid, the sizes of the bases of which are in the ratio 5: 3. The arms have a length of 5 m and the height of the embankment is 4.8 m. Calculates the size of the embankment section area.
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A line from the top of a cliff to the ground passes just over the top of a pole 5 ft high and meets the ground at a point 8 ft from the base of the pole. If the point is 93 ft from the base of the cliff, how high is the cliff?
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There is a triangle ABC: A (-2,3), B (4, -1), C (2,5). Determine the general equations of the lines on which they lie: a) AB side, b) height to side c, c) Axis of the AB side, d) median ta to side a
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• Vertical rod
The vertical one meter long rod casts a shadow 150 cm long. Calculate the height of a column whose shadow is 36 m long at the same time. | 706 | 2,610 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2020-45 | latest | en | 0.859219 |
https://www.math-edu-guide.com/UKG-Missing-Number.html | 1,716,108,779,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057774.18/warc/CC-MAIN-20240519070539-20240519100539-00216.warc.gz | 803,905,915 | 7,403 | # UKGMISSING NUMBER
MISSING NUMBERS -
In above series of numbers we have to find out missing number according to the requirement. So, now we will analyze the number series, line by line for better understanding. Please follow the methodology for better application and understanding.
In the above picture, we can see that, there are only three numbers 5, 6 & 9 that have been mentioned. As per the given condition, we have to find out the numbers which are to be placed after 6 up to two places & after 9 up to two places. If we recall the series of numbers from 5 to 15 = 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15.
Hence, we can observe that, the number after 6 up to two places are 7, 8 & after 9 up to two places are 10 is 11. So, the required numbers would be 7, 8, 10 & 11.
so, the required series of numbers would be 5, 6, 78, 9, 1011.
In the above picture, we can see that, there are only three numbers 12, 16 & 18 that have been mentioned. As per the given condition, we have to find out the numbers which are to be placed after 12 up to three places & after 16 up to one place. If we recall the series of numbers from 10 to 20 = 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20.
Hence, we can observe that, the number after 12 up to three places are 13, 14, 15 & after 16 up to one place is 17. So, the required numbers would be 13, 14, 15 & 17.
so, the required series of numbers would be 13, 1415, 16, 17, 18.
In the above picture, we can see that, there are only one number 23 that has been mentioned. As per the given condition, we have to find out the numbers which are to be placed before 23 up to four places & after 23 up to two places. If we recall the series of numbers from 15 to 25 = 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25.
Hence, we can observe that, the number before 23 up to four places are 19, 20, 21, 22 & after 23 up to two places are 24, 25. So, the required numbers would be 19, 20, 21, 22 & 24, 25.
so, the required series of numbers would be 19, 20, 21, 22, 23, 24, 25.
In the above picture, we can see that, there are only two numbers 26 & 29 that have been mentioned. As per the given condition, we have to find out the numbers which are to be placed after 26 up to two places (or two places before 29) & after 29 up to three places. If we recall the series of numbers from 25 to 35 = 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35.
Hence, we can observe that, the number before 29 or the number after 26 up to two places are 27, 28 & after 29 up to three places are 30, 31, 32. So, the required numbers would be 27, 28, 30, 31 & 32.
so, the required series of numbers would be 26, 2728, 29, 303132.
In the above picture, we can see that, there are only three numbers 34, 37 & 38 that have been mentioned. As per the given condition, we have to find out the numbers which is to be placed before 34 up to one place, numbers up to two places after 34 & after 38 up to one place. If we recall the series of numbers from 30 to 40 = 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40.
Hence, we can observe that, the number before 34 up to one place is 33, the number after 34 or before 37 up to two places are 35,36 & after 38 up to one place is 39. So, the required numbers would be 33, 35, 36 & 39.
so, the required series of numbers would be 33, 34, 3536, 37, 38, 39.
In the above picture, we can see that, there are only two numbers 40 & 46 that have been mentioned. As per the given condition, we have to find out the numbers which are to be placed after 40 or before 46 up to five places. If we recall the series of numbers from 40 to 50 = 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50.
Hence, we can observe that, the numbers after 40 or before 46 up to five places are 41, 42, 43, 44 & 45. So, the required numbers would be 41, 42, 43, 44 & 45.
so, the required series of numbers would be 40, 4142434445, 46.
In the above picture, we can see that, there are only one number 50 that has been mentioned. As per the given condition, we have to find out the numbers which are to be placed before 50 up to three places. If we recall the series of numbers from 45 to 50 = 45, 46, 47, 48, 49, 50.
Hence, we can observe that, the number before 50 up to three places are 47, 48 & 49. So, the required numbers would be 47, 48, & 49.
so, the required series of numbers would be 4748, 49, 50.
In above picture of number series if we complete the whole series, then we will find that - 5, 6, 78, 9, 1011, 12, 131415, 16, 17, 18, 19202122, 23, 242526, 2728, 29, 30313233, 34, 3536, 37, 38, 39, 40, 4142434445, 46, 474849, 50.
In above obtained number series all the required number is marked by red color. | 1,521 | 4,634 | {"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} | 4.65625 | 5 | CC-MAIN-2024-22 | latest | en | 0.955654 |
https://onlinejudge.org/board/search.php?author_id=1976&sr=posts | 1,590,501,512,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347390758.21/warc/CC-MAIN-20200526112939-20200526142939-00175.warc.gz | 476,124,057 | 8,597 | ## Search found 131 matches
Sun Aug 12, 2007 7:17 pm
Forum: Volume 102 (10200-10299)
Topic: 10282 - Babelfish
Replies: 48
Views: 20023
two ways to do it
1) qsort then binary search
2) binary search tree is also an option
think about it ... you will get the idea
Mon May 14, 2007 12:19 pm
Forum: Algorithms
Topic: USACO, Section : 1.4 , Packing Rectangles
Replies: 7
Views: 8680
thanx man. you are a great helper
Fri May 11, 2007 1:38 pm
Forum: Algorithms
Topic: USACO, Section : 1.4 , Packing Rectangles
Replies: 7
Views: 8680
http://riyad13783.googlepages.com/pack.gif i forgot to give the picture of the 6 basic layouts. i wrote all the formulas in one of my previous post. could some one check those and tell me weither my formulas are correct or not. i will remove all my post as soon as i get accepted in this problem. by...
Fri May 11, 2007 1:31 pm
Forum: Algorithms
Topic: USACO, Section : 1.4 , Packing Rectangles
Replies: 7
Views: 8680
bye
Fri May 11, 2007 1:22 pm
Forum: Volume 105 (10500-10599)
Topic: 10581 - Partitioning for fun and profit
Replies: 15
Views: 8495
### I/O
Code: Select all
``````1
1
1
1
1
1
52
3
1
1
1
1
1
1
64
31
1
1
1
1
66
41
``````
what is ur output for
Code: Select all
``````10 4 10
``````
i get
Code: Select all
``````1
2
3
4
``````
is it right????
Tue May 08, 2007 11:45 am
Forum: Volume 105 (10500-10599)
Topic: 10581 - Partitioning for fun and profit
Replies: 15
Views: 8495
can't get AC still :( can anyone please help. below is my code # include <stdio.h> # include <assert.h> typedef long long i64 ; i64 memo[250][20]; i64 go( int sum , int n ){ i64& ret = memo[sum][n]; if(ret !=-1) return ret ; if( !sum && !n ) return ret = 1 ; ret = 0 ; int i ; for( i = 1 ; i <= sum &...
Mon May 07, 2007 5:31 am
Forum: Volume 105 (10500-10599)
Topic: 10581 - Partitioning for fun and profit
Replies: 15
Views: 8495
### some more test cases
i am getting WA in this problem. could anyone provide more testcases. boundary cases would be very nice. i got correct results for all the above i/o. please help....
Sat Mar 31, 2007 4:03 am
Forum: Algorithms
Topic: USACO, Section : 1.4 , Packing Rectangles
Replies: 7
Views: 8680
### Help on packing rectangle.........
hello, thank you for the help. im not good at geometry. so i am not confident about finding the height and width of the different scenarios case 1 : 1 2 3 4 H = max4(h1,h2,h3,h4); W = w1+w2+w3+w4 ; case 2: (1 2 3) 4 H = max3(h1,h2,h3)+h4; W = max2( w4,w1+w2+w3); case 3 : ( 1 2 ) 4 3 H = max2( h4, ma...
Tue Mar 27, 2007 6:04 pm
Forum: Algorithms
Topic: USACO, Section : 1.4 , Packing Rectangles
Replies: 7
Views: 8680
### USACO, Section : 1.4 , Packing Rectangles
Four rectangles are given. Find the smallest enclosing (new) rectangle into which these four may be fitted without overlapping. By smallest rectangle, we mean the one with the smallest area. All four rectangles should have their sides parallel to the corresponding sides of the enclosing rectangle. ...
Tue Sep 05, 2006 4:32 pm
Forum: Volume 100 (10000-10099)
Topic: 10051 - Tower of Cubes
Replies: 19
Views: 8963
luison9999 is right . the output for case 4 and case 7 is wrong . the output should be 5 and 17 respectively ...it did make me confused but i submitted before matching these given outputs with mine
Sat Sep 02, 2006 8:10 pm
Forum: Volume 110 (11000-11099)
Topic: 11081 - Strings
Replies: 35
Views: 19657
would u care to tell us your DP solution which runs in O(n^3) iLL be really glad to know one if u dont want to give a spoiler u can PM me ur idea , anywayz thanx for ur reply ...
Sat Sep 02, 2006 7:50 pm
Forum: Volume 110 (11000-11099)
Topic: 11077 - Find the Permutations
Replies: 14
Views: 7147
### algorithm of 101077
is dp the obvious choice to go for this problem ?? and can some one tell me whats the complexity of the DP .....
is it O(nk) ???????????????
Sat Sep 02, 2006 7:41 pm
Forum: Volume 110 (11000-11099)
Topic: 11081 - Strings
Replies: 35
Views: 19657
is there any O(n^3 ) dp for this problem . i used O(n^4) dp to get accepted during the contest which is getting TLE in the judge now . so can some one point out the O(n^3) algorithm or O(n^4) is way to go .........
Sat Sep 02, 2006 7:38 pm
Forum: Volume 110 (11000-11099)
Replies: 34
Views: 16563 | 1,452 | 4,251 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2020-24 | longest | en | 0.821382 |
http://www.gvlibraries.org/find?tid=1306%2C27%2C11678 | 1,591,158,793,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347428990.62/warc/CC-MAIN-20200603015534-20200603045534-00071.warc.gz | 170,140,923 | 8,445 | # Geometry
Computer Science (X) - Math (X) - Geometry (X)
## Analyzing its Geometry and Energy Efficiency | Treasures of New York: "Hearst Tower"
Icon:
Students solve a real-world mathematical problem involving the area of a triangle and learn about how energy conservation is applied to architectural design using video from Treasures of New York: Hearst Tower. Utilizing text-dependent discussion questions and classroom activities called “teaching tips,” students have an opportunity to take a deep dive into mathematics, STEM content and the arts.
## Measuring for an Exact Fit
Icon:
In this Cyberchase video segment, Matt and Jackie must figure out how to construct rectangular lids that match the size of two boxes precisely. They enlist the carpentry skills of one of the Three Little Pigs, who shows them how to use different tools to accurately measure the rectangular lids. For the first box, Jackie finds that she can easily trace the box to make a lid that fits perfectly. The second box cannot be traced, so Matt and Jackie decide to measure the sides of the box and then use the measurements to cut a rectangle of equal size out of a piece of Herculanium.
## Machinist | Great Job!
Icon:
Machinist John Morris uses a computerized machining center to produce complicated parts out of metal & plastic that are accurate to fractions of an inch.
## Is It Square?
Icon:
See three techniques for determining whether a rectangle truly has square corners. Learn more about construction technologies in this animation from Design Squad Nation.
## Do an Hour of Code | Tynker
Icon:
Build the computer literacy and problem solving skills students need to become innovators of the future.
Introduce students to computer programming, or take their coding skills to the next level as they engage in fun activities and projects that enhance STEM learning outcomes. With an Hour of Code with Tynker, teachers can easily introduce the basics of computer programming in a fun and intuitive way. Start with any of the six adventure puzzles, then students can apply what they’ve learned to create fun projects to share with others – multi-level and multi-player games, math patterns, interactive comics and greeting cards.
No programming experience is required. Use the resources provided to quickly plan a successful Hour of Code for your class. Students can access Candy Quest and Dragon Dash puzzles from this site. The rest of the activities, plus a teacher's guide and answer keys, can be found at hourofcode.tynker.com.
## Building a Fence
Icon:
In this Cyberchase video segment, Harley convinces Harry to help put up a fence in their grandmother's backyard. With no instructions other than to put up the fence in the shape of a rectangle, Harry must figure out how to do this with different lengths of fencing. He breaks the problem down by adding up different combinations of pieces and then drawing a diagram to figure out where to place the pieces.
## What is a fractal (and why do they matter)? | MIT's Science Out Loud
Icon:
Fractals are complex, never-ending patterns created by repeating mathematical equations. Yuliya, a undergrad in Math at MIT, delves into their mysterious properties and how they can be found in technology and nature.
## Blossom and Snappy Build Scale Models | Count On It!
Icon:
Blossom and Snappy learn about scale models and how they are used in the design and construction of buildings. They visit an architectural firm, a construction site, Clark Atlanta University Art Gallery, and Clark Atlanta University Science Center. They learn how to build their own scale model of a house.
## Bob the Builder | Who Helps Build a Community?
Icon:
Just as Bob the Builder and his team manipulate their environment, preschoolers like to explore and manipulate their environments, too! They are natural builders! Blocks are meant to be stacked, blanket forts draped over furniture, or a stream of water dammed and redirected. In this activity, children are introduced to Bob's team and the important role each plays in planning and building a community.
## What is Construction? | KIDS Clubhouse Adventures
Icon:
Exploring our world is fun! Abby Brown, KIDS Clubhouse Adventures co-host, loves to help kids have fun while learning! In this segment, Abby teaches kids that construction starts with a plan and requires teamwork. Construction workers use lots of different materials and shapes. | 898 | 4,450 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2020-24 | latest | en | 0.904483 |
http://spmath87308.blogspot.com/2009/02/pythagorean-theorem.html | 1,490,611,332,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189471.55/warc/CC-MAIN-20170322212949-00556-ip-10-233-31-227.ec2.internal.warc.gz | 324,755,280 | 14,662 | ### Pythagorean Theorem
6:41 PM
For our assignment we had to explain the Pythagorean Theorem..,,which was proven by Pythagoras,,,...
Pythagoras was born between 580 and 572 BC, he died between 500 and 490 BC was an Ionian Greek mathematician and founder of the religious movement called Pythagoreanism. He is often revered as a great mathematician, mystic and scientist ; however some have questioned the scope of his contributions to mathematics and natural philosophy. Herodotus referred to him as the most able philosopher among the Greeks. Pythagoras was the first westerner ti determine that the earth was round. He also found out that season were caused by the earth axis and he was the most famous discovery was the Pythagorean Theorem , which is .
Vocabulary:
Legs- are the two shortest side on the triangle sides a and b. The legs also make up the 90 degrees.
Hypotenuse
- is longest side of a triangle, side c.
R.A.T- is triangle it stands for Right Angled Triangle, any triangle with 90 degrees. Greek- A person of greek origin.
Theorem- is the second word of the Pythagorean Theorem which was proven by Pythagoras.
Artifacts:
1st Artifacts
2nd Artifacts
3rd Artifacts
Pythagoras Invention
Proof 1:
Proof 2:
Here's my video,,,,......
First Video
#### 2 Responses to "Pythagorean Theorem"
February 28, 2009 at 3:39 PM
nice! Chie your pictures were big and your work was very understandable,good job!
May 18, 2009 at 3:33 PM
Wow! I learned a lot from this post. The pictures and videos are amazing. I have a math blog too, and one of the topics I discuss is the Pythagorean Theorem. My blog has practice problems and solutions on a wide range of math topics. Please check it out at http://mathvariety.blogspot.com when you get a chance. | 438 | 1,761 | {"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-2017-13 | longest | en | 0.980411 |
https://isabelle.in.tum.de/repos/isabelle/file/b5a2e9503a7a/doc-src/Intro/intro.toc | 1,638,939,010,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363437.15/warc/CC-MAIN-20211208022710-20211208052710-00562.warc.gz | 382,537,251 | 3,612 | doc-src/Intro/intro.toc
author lcp Tue, 03 May 1994 18:38:28 +0200 changeset 359 b5a2e9503a7a parent 105 216d6ed87399 child 1085 504dad4d7843 permissions -rw-r--r--
final Springer version
```
\contentsline {part}{\uppercase {i}\phspace {1em}Foundations}{1}
\contentsline {section}{\numberline {1}Formalizing logical syntax in Isabelle}{1}
\contentsline {subsection}{\numberline {1.1}Simple types and constants}{1}
\contentsline {subsection}{\numberline {1.2}Polymorphic types and constants}{3}
\contentsline {subsection}{\numberline {1.3}Higher types and quantifiers}{5}
\contentsline {section}{\numberline {2}Formalizing logical rules in Isabelle}{5}
\contentsline {subsection}{\numberline {2.1}Expressing propositional rules}{6}
\contentsline {subsection}{\numberline {2.2}Quantifier rules and substitution}{7}
\contentsline {subsection}{\numberline {2.3}Signatures and theories}{8}
\contentsline {section}{\numberline {3}Proof construction in Isabelle}{9}
\contentsline {subsection}{\numberline {3.1}Higher-order unification}{10}
\contentsline {subsection}{\numberline {3.2}Joining rules by resolution}{11}
\contentsline {section}{\numberline {4}Lifting a rule into a context}{13}
\contentsline {subsection}{\numberline {4.1}Lifting over assumptions}{13}
\contentsline {subsection}{\numberline {4.2}Lifting over parameters}{14}
\contentsline {section}{\numberline {5}Backward proof by resolution}{15}
\contentsline {subsection}{\numberline {5.1}Refinement by resolution}{15}
\contentsline {subsection}{\numberline {5.2}Proof by assumption}{16}
\contentsline {subsection}{\numberline {5.3}A propositional proof}{16}
\contentsline {subsection}{\numberline {5.4}A quantifier proof}{17}
\contentsline {subsection}{\numberline {5.5}Tactics and tacticals}{18}
\contentsline {section}{\numberline {6}Variations on resolution}{18}
\contentsline {subsection}{\numberline {6.1}Elim-resolution}{19}
\contentsline {subsection}{\numberline {6.2}Destruction rules}{20}
\contentsline {subsection}{\numberline {6.3}Deriving rules by resolution}{21}
\contentsline {part}{\uppercase {ii}\phspace {1em}Getting Started with Isabelle}{23}
\contentsline {section}{\numberline {7}Forward proof}{23}
\contentsline {subsection}{\numberline {7.1}Lexical matters}{23}
\contentsline {subsection}{\numberline {7.2}Syntax of types and terms}{24}
\contentsline {subsection}{\numberline {7.3}Basic operations on theorems}{25}
\contentsline {subsection}{\numberline {7.4}*Flex-flex constraints}{27}
\contentsline {section}{\numberline {8}Backward proof}{28}
\contentsline {subsection}{\numberline {8.1}The basic tactics}{28}
\contentsline {subsection}{\numberline {8.2}Commands for backward proof}{29}
\contentsline {subsection}{\numberline {8.3}A trivial example in propositional logic}{29}
\contentsline {subsection}{\numberline {8.4}Part of a distributive law}{31}
\contentsline {section}{\numberline {9}Quantifier reasoning}{32}
\contentsline {subsection}{\numberline {9.1}Two quantifier proofs: a success and a failure}{32}
\contentsline {paragraph}{The successful proof.}{32}
\contentsline {paragraph}{The unsuccessful proof.}{33}
\contentsline {subsection}{\numberline {9.2}Nested quantifiers}{33}
\contentsline {paragraph}{The wrong approach.}{34}
\contentsline {paragraph}{The right approach.}{34}
\contentsline {paragraph}{A one-step proof using tacticals.}{35}
\contentsline {subsection}{\numberline {9.3}A realistic quantifier proof}{36}
\contentsline {subsection}{\numberline {9.4}The classical reasoner}{37}
\contentsline {section}{\numberline {10}Deriving rules in Isabelle}{39}
\contentsline {subsection}{\numberline {10.1}Deriving a rule using tactics and meta-level assumptions}{39}
\contentsline {subsection}{\numberline {10.2}Definitions and derived rules}{41}
\contentsline {subsection}{\numberline {10.3}Deriving the \$\neg \$ introduction rule}{41}
\contentsline {subsection}{\numberline {10.4}Deriving the \$\neg \$ elimination rule}{42}
\contentsline {section}{\numberline {11}Defining theories}{44}
\contentsline {subsection}{\numberline {11.1}Declaring constants and rules}{46}
\contentsline {subsection}{\numberline {11.2}Declaring type constructors}{46}
\contentsline {subsection}{\numberline {11.3}Type synonyms}{48}
\contentsline {subsection}{\numberline {11.4}Infix and mixfix operators}{48} | 1,243 | 4,294 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-49 | latest | en | 0.574575 |
https://cboard.cprogramming.com/cplusplus-programming/116618-boolean-algebra.html | 1,490,340,820,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218187717.19/warc/CC-MAIN-20170322212947-00487-ip-10-233-31-227.ec2.internal.warc.gz | 798,058,970 | 15,387 | 1. ## Boolean Algebra
I am having real trouble understanding these NOT AND and OR Statement in lesson 2 of the C++ tutorial. Cprogramming.com Tutorial: If Statements
I wondering if the input is true why does it return false. does this mean if the input !1 is true then it returns 0? and vise versa? please someone help me with this because I'm about to start school to take my prerequisites in order for me to get into my major which is software engineering. I really want to get a big head start on the courses I'll be taking in about 1 1/2 years to 2 years.
Also should I be starting with C++ having never done programming before or should I start with C?
2. Boolean algebra works on two values: true or false. For values that aren't true or false (like 3) C++ "converts" it to true or false using the basic idea that 0 is false and everything else is true.
NOT says: take the boolean value and return the other boolean value. So true becomes false and false becomes true. Think about it in english: You are allowed to go see the movie. Vs: You are not allowed to go see the movie.
AND asks if two values are both true. So TRUE/TRUE is TRUE but TRUE/FALSE, FALSE/TRUE, and FALSE/FALSE are all FALSE. Jack and Jill both start with y is false.
OR asks if at least one of the two values are true. TRUE/TRUE, TRUE/FALSE, FALSE/TRUE are true but FALSE/FALSE is FALSE. This operator does lead to some confusion as English tends to imply that one of the values are true but the other is false (you can have the red pill or the blue pill). Logically this is actually an XOR (Exclusive OR) where only one of the values may be true. So TRUE/TRUE would be false.
3. In C++, logical operations yield a result of type bool, which has a value true or false. If that is then converted back to an integer, true gives a value 1 and false gives a value zero.
In C++ (and C) a zero integer value corresponds to false and everything else is deemed true. That's the convention. So "if (23)" is functionally equivalent to the test "if (true)" or, more accurately, to the test "if (23 != 0)".
The ! operator is the logical NOT (i.e. !x gives a true result if x is false, and a false value if x is true [non-zero]). So !1 will yield false (zero), as will !23, and !0 will yield true (1).
The logical AND ("&&" or "and" in C++) takes two operands and returns a true result only if both operands are true. So (a && b) will yield false (zero) if either a or b are false, and true (1) otherwise.
The logical OR (|| or "or" in C++) takes two operands and returns a true result if either of the operands are true. So (a || b) will only yield false if both a and b are false.
4. ## I think I understand
let me got back and restudy that part of lesson 2 to see if I can figure out how they got the answer they got for the question in the quiz of that lesson.
5. wait so the logical Not or "!" is looking for the input of zero from the if statement if the logical Not doesn't get zero from the if statement it returns a zero but if it gets zero it returns a 1. is this how that works? I understand OR and AND from what grumpy and Thantos posts but I'm still have trouble with NOT. but this is what is confusing me.
A. !( 1 || 0 ) ANSWER: 0
why is NOT 1 OR 0 = False when OR says its true if either of the operands are true or is the NOT 1 mean 0.
so is the problem 0 || 0 = 0?
also if anyone has this stuff master can you give me you instant messenger name or something? being able to ask someone who knows directly is much better. I wont be IMing you whenever I'm stuck because I like to figure things out but this one took me for a loop.
6. You've heard of order of operations, right? Parentheses first, then look at what's outside.
7. man you just told me the same thing my sister said. so the reason
C. !( ( 1 || 0 ) && 0 ) ANSWER: 1 (Parenthesis are useful)
is because its 1 OR 0 evalutes to 1 then the outer parenthesis becomes !1 (which is really 0) !0 || 0 evaluates to 1 because both operands are NOT false so it gotta be true
but would it be done like this?
!( (1 OR 0 = 0) && 0 )
!( 0 && 0 = 0) so the statement become !0 so its 1. Right?
8. 1 or 0 is 1. But 1 && 0 is still 0, so from then on it's right.
9. ok gotcha I typed that in wrong I was watching the Laker Magic Game lol Thank you to everyone who bothered to help me out.
10. hey I'm new here, I just started learning C++ and just want to say how helpful this thread has been. I was completely lost on lesson two as well but reading these replies has really helped me, so thanks everyone!
11. ## need help
friends, I am new to C. here is problem, whose output I could not understand. could you please help me out?
Code:
```main()
{
int i=4,j=-1,k=0,w,x,y,z;
w=i||j||k;
x=i&&j&&k;
y=i||j&&k;
z=i&&j||k;
printf("\nw=%d x=%d y=%d z=%d\n",w,x,y,z);
}```
the output is w=1 x=0 y=1 z=1
I am confused why w=1 in the answer, when expression says w=4 or -1 or 0
similarly others?
12. The logical operators OR and AND are short-circuiting. With OR that means stop when you find truth, and with AND it means stop when you find falsehood, because these results will end up being the results for the whole expression. If you evaluate any Boolean expression you will either find truth or falsehood.
The result of a Boolean expression also belongs in a bool. That way you can do things like
Code:
```#include <iomanip>
#include <iostream>
std::cout << std::boolalpha << whatdoesitmean << std::endl;```
and it makes a little more sense to the layperson.
13. Truth table - Wikipedia, the free encyclopedia
Read this and convert to C/C++ syntax...
This part is most useful:
http://en.wikipedia.org/wiki/Truth_t...ical_operators | 1,486 | 5,689 | {"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-2017-13 | longest | en | 0.893266 |
https://writingyourdreams.com/problem-solving-of-linear-equation.php | 1,601,418,941,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402093104.90/warc/CC-MAIN-20200929221433-20200930011433-00628.warc.gz | 673,618,924 | 8,495 | Wednesday, September 30
problem solving of linear equation - writingyourdreams.com
# Important to know: problem solving of linear equation
## Contents
solving He had both weapons equation at his aside and followed her nose into the. Problem asked me linear, and I told. "Are the two of you brothers?' she. The glass in his hand, shoved the are said to be. I imagine it's a good town for. Just be receptive to a few of. Pete pouts if I go over break. Suppose I already worshiped him. "Who was its owner?" she said softly. After he slipped out, she shoved the. If everything that is said about him were true, he would be the devil. You see, I could never be afraid characters-she knew only that aside from Sir.
Her, the dog let out one quiet. But now, you ride back and do blonde, she wondered, or the redhead. The rosebush, he thought, where the flowers at his. She had confirmed her return ticket with. Someone Id thought was so strong cry for a straight shot, Miss Olivia.
" He slapped at a sparrow-sized mosquito. She could justify her feelings day in and day. They didn't know her oldest brother was kiss on her mouth before he knew. They were leaving the next morning, and of my time to try and. "There's not a lot of subtlety in. Deliberately she turned away to pack up your office. Rinsed the glass, then turned and found and set one hand on the counter all the. "This stuffs strong enough to do the. STINGY LIGHT WITH those same images running men who knew the.
You said earlier that winter at the. Listening to the tapes hadn't been the. I am queen of Twylia. "We go back to bar food at her body was so busy controlling. They were talking softly, as.
## problem solving percentages?
The estate had been in his family. There solving a linear in that, a was considered a virtue, real heroes seemed. I often find it harsh, but always your face," Ada commented. He turned off the paved road onto the quarter-mile lane that cut equation the. I was glad to problem my share. You watch him ride a homework activities and her fork, "that I wasn't gorgeous and. Laughing, Laurel bit into one as the. She was pregnant and she was sixteen and she was from Kansas. Walked into the gardens, her younger sisters, for more than that from time to. And the vampyre who fancies herself a show with bored eyes. PUTNAM'S SONS NEW YORK This is a few weeks. A woman with a straw hat over. Why don't you bring that back. Fashions changed and evolved, and even a. He tossed his riding crop and hat on a table and spotted his majordomo would he see in a year.
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But in so doing, you gave up made it impossible to take her. After problem bumpy, unpleasant divorce from solving. It seemed that Cador had his own eye upon the. Hadnt been linear to stand up to if he cooperates. Took one out of equation, terrified to.
### illuminated temper already limit pretty
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## tasty supposed problem solving of linear equation
He had linear air to breathe to equation. Everyone solving the kingdom was in the. That's why you're problem damn lucky to. Gilmore tossed back the last of his. There was a figure of a woman seated on the jeweled bench beneath the spreading branches of the great bush that bright butterflies. "Dealer stands on nineteen," she stated, knowing wind seemed to be eating all of. "Looks like we're turning most tables over. Was Nate and rolled her tensed shoulders.
studied the picture for a long time. March was the province of the cattleman. We could watch the sun come up. God, you make me hope again, believe.
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## problem solving of speed This much
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Solving Linear Equations - Basic Algebra Shortcut Tricks!
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• supperz2ya | 1,297 | 5,476 | {"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} | 3.140625 | 3 | CC-MAIN-2020-40 | latest | en | 0.990876 |
https://foxoyo.com/mcq/41975/a-bag-contains-6-balls-of-different-colours-and-a-ball-is-drawn-from-it-a-probability-of-speaking-the-truth-is-2-3-and-b-probability-of-speaking-the-truth-is-4-5-if-both-a-and-b-say-that-a-red-ball-ha | 1,579,654,624,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250606226.29/warc/CC-MAIN-20200121222429-20200122011429-00486.warc.gz | 456,448,641 | 8,953 | # A bag contains 6 balls of different colours and a ball is drawn from it. A probability of speaking the truth is 2/3 and B probability of speaking the truth is 4/5. If both A and B say that a red ball has been drawn then the probability that actually a red ball has been drawn, is
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next time you Google a mcq Questions
### Related MCQs
A bag contains 4 red and 3 blackballs.
A second bag contain
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Another bag contains
A bag contains 2 white balls, 3 black balls and 4 redballs.
3 balls are drawn randomly from a bag contains 3 black, 5 re
A bag contains nine yellow balls, three white balls and four
A bag contains 19 red balls, 37 blue balls and 27 green ball
A bag contains 8 white balls and some yellowballs.
If the p | 231 | 835 | {"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} | 2.609375 | 3 | CC-MAIN-2020-05 | longest | en | 0.929973 |
https://www.magicalapparatus.com/white-knife/the-antifaros.html | 1,591,005,156,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347415315.43/warc/CC-MAIN-20200601071242-20200601101242-00292.warc.gz | 783,604,010 | 6,836 | The Antifaros
The idea of performing Faro shuffles backwards is well known. You can separate the cards by moving them alternately upwards and downwards as you pass them from one hand to the other, and then separate the two packets by extracting one of them and tossing it on top of the other. Obviously, this is the same as making two piles on the table by dealing out the cards one by one.
But it turns out that if you make four piles, or eight, sixteen or thirty two, and gather them up, one on top of the other in a certain order, you can do two, three, four or five backwards Faros. And you can do them by dealing out the cards just once, and gathering them up in the proper order.
Moreover, depending on how you gather up the packets, the backwards Faros will be Out or In.
This theory, which studies the combinations that allow us to do all kinds of tricks that require Faro shuffles without dominating or controlling the technique, is explained here for the first time (although a lot of this material was published in 1981 and the following years in the Circular de la Escuela Mágica de Madrid).
If I do one Out-faro, the most direct ANTIFARO would be to deal the cards, one by one, into two piles, and place the second pile on top of the first. The deck ends up in the same order as before the Faro, but the order is reversed, the top card is now on the face, the second card is second from the bottom, etc...
But, if I begin from any initial order position and do two Out-faros, and if I then do two Antifaros the deck will end up in starting position (in the same order, not inverted).
But, I could, after doing two Out-faros, deal four piles at once (dealing out the cards one by one, into the four piles), and gather them up placing the fourth pile on top of the third, both on the second and all three on the first. And the cards will end up in their initial order (but reversed). This is a DOUBLE ANTIFARO.
If I do three Out-faros there are three different ways to reset the deck: Do three Simple-Antifaros. When you finish the order is inverted. Do a Simple-Antifaro and a Double-Antifaro (they end up in starting order).
Or do what is called a TRIPLE-ANTIFARO.
To perform a Triple-Antifaro you must deal the whole deck out into eight piles (two rows of four), from left to right, placing the last four cards of the deck onto the first four piles.
Like this:
Then, gather them up in this way, for a 52 card deck: Pile eight on three, on six, on one, and place the packet on the table. Pile four on seven, on two, on five, and place this packet on top of the first packet that you placed on the table.
Like this:
I don't know the reason, or the mathematical explanation, but the fact is that it works.
To undo four Out-faros, we have these possibilities:
— Do 4 Simple-Antifaros (they end up in initial order).
— Do 2 Simple-Antifaros and one Double-Antifaro (they end up inverted).
— Do 2 Double-Antifaros (initial order).
— Do 1 Simple-Antifaro and one Triple-Antifaro (initial order).
The QUADRUPLE-ANTIFARO consists in dealing the deck face down bottom-0WS eaCh* dealing fr°m left t0 right' and from t0P t0
□ □ □ 0) 0 0 0 0) □ 0 0 0 0 0 0 0 )
(these piles have 3 cards)
(these piles have 3 cards)
And pick them up like this:
8 on 11 on 14 on 1, place on the table; 12 on 15, on 2, on 5, together, on top of packet on the table; 16 on 3, on 6, on 9, all on top of the packet on he table; and finally, 4 on 7, on 10, on 13, all on top of the packet on the table.
Which is easy to remember, if you study the following diagram:
parently hapazard manner.
I will explain how in the next paragraph. But before I do, let's finish the Antifaros.
If we do 5 Out-faros, there are many possible combinations:
— Do five Simple-Antifaros (order is inverted).
— Do three Simple-Antifaros and one Double-Antifaro (initial order).
— Do two Simple-Antifaros and one Triple-Antifaro (order inverted).
— Do one Simple-Antifaro and one Quadruple-Antifaro (initial order).
— Do one Simple-Antifaro and two Double-Antifaro (order inverted).
— Do one Quintuple-Antifaro (order inverted).
The QUINTUPLE-ANTIFARO is theoreticially possible, but in practice its impractical.
You would have.to deal out 8 rows of 4 piles each (32 piles) and gather them up like this:
20 on 7, on 26, on 13, on 32, on 19, on 6, on 25, on 12, on 31, on 18, on 5, on 24, on 11, on 30, on 17, on 4, on 23, on 10, on 29, on 16, on 3, on 22, on 9, on 28, on 15, on 2, on 21, on 8, on 27, on 14, on 1.
Which is obviously slow and complicated.
With all this we know how to do up to a Quadruple-Antifaro (and, by combining, up to 8), which is the same as going backwards, and doing, i s
for example, 2 Quadruple-Antifaros, returning to the initial order (the same as if we do 4 Double-Antifaros).
But remember that an N Antifaro is the same as an 8-N Out-faro. So a Triple-Antifaro is the same as 5 Out-faros (8 — 3 = 5).
The only detail that you have to remember is that each Antifaro reverses the order of the deck. But notice that if the cards are turned face up when dealt, when you gather them up in the normal way, but face up, the deck ends up in its initial order, it is not reversed.
We can now reach any position in the 8 Out-faros chain without having to perform a single Out-faro. Rather, we can get there through a combination of (one or more) Antifaros, according to our preference.
This opens up an immense new field for Faro effects, because we can look for and find excuses and valid justifications for the Antifaro:
□ 0 00 □ □ 0 0 00 0 0 00 0 0 0 0 0 0 0 0' 0 0 00 0 0 JUSTIFYING THE ANTIFAROS The Simple-Antifaro can be justified as a method for dividing the deck into two equal piles. The Double-Antifaro, as a bridge deal (13 cards are dealt, one by one, to four players). The Triple and Quadruple-Antifaros, as a way of mixing the cards, dealing them out into piles, which you later pick up in a random and irregular way. To achieve this, let's look at how to pick up the only lay-out that is a little complicated: GATHERING UP THE QUADRUPLE-ANTIFARO Follow these steps, with the deck dealt out into 16 piles, and numbered as in this diagram: a) The R. H. picks up pile 8, the L. H., number 14. c) The L. H. places its pile on the table and the R. H. places its pile on top of the L. H.'s pile. f) The L. H. places its cards on top of the ones on the table, and the R. H. follows, placing its cards on top of them. i) The L. H. places its cards on top of the ones on the table, the R. H. follows. «We pick up piles from different places». «So that they end up in total disorder». j) The R. H. picks up 4, the L. H., 10. k) TheR. H. places 4 on 7, the L. H., 10 on 13. I) The L. H. places its cards on the table, the R. H. places its cards on top. If you try it two or three times, you will see that it is quite easy to remember, that it is quick (do it on a close-up pad), and that it really does produce the impression that you are shuffling the cards, and that you gather them up in a totally arbitrary, outrageously mixed-up and crazily chaotic sort of way. ANTEFARO-6. In my study of Antifaro theory, I found the way to do six Antifaros, which is the same as performing 2 Out-faros. I'm talking about a 52 card deck, of course! Well, nothing is easier: Deal out the 52 cards into 13 hands of four cards each. | 2,075 | 7,347 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2020-24 | latest | en | 0.934471 |
https://stats.stackexchange.com/questions/501653/bayes-conditional-probability-question | 1,660,430,923,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571987.60/warc/CC-MAIN-20220813202507-20220813232507-00777.warc.gz | 496,428,746 | 65,384 | # bayes / conditional probability question
The derivation of the conditional probabilities given in the solution puzzle me.
There are two identical boxes in front of you. In box X, there are 3 white balls and 7 black balls. In box Y, there are 7 white balls and 3 black balls. You randomly reach one of the boxes and draw three balls from it one by one. If every time you draw a ball you find it is white, at which of the following stages is the probability the highest that you chose box X?
(A) You drew one ball, and it was white.
(B) You drew two balls, and they were both white.
(C) You drew three balls, and they were all white.
• When sampling more than one ball from a box, is sampling done with or without replacement? Dec 20, 2020 at 6:58
Assuming draws w/o replacement based on the answers given, and I'll focus on the second question to form an example for others. Let $$B$$ be the chosen box ($$x$$ or $$y$$), the probability of selecting box $$x$$ in case we draw two white balls is as follows:
$$P(B=x|2W)=\frac{P(2W|B=x)P(B=x)}{P(2W|B=x)P(B=x)+P(2W|B=y)P(B=y)}$$
Here, $$P(B=x)=P(B=y)=1/2$$, so they cancel each other in numerator and denominator. And, $$P(2W|B=x)$$ is choosing two white balls when the box is $$x$$. There are two ways to think:
1. Probability of choosing the first balls as white, and then choosing the second ball as white, which is:
$$P(2W|B=x)=P(W_1|B=x)P(W_2|B=x,W_1)=\frac{3}{10}\times \frac{2}{9}$$
This way, we'll have $$P(2W|B=y)=\frac{7}{10}\times\frac{6}{9}$$, and therefore: $$P(B=x|2W)=\frac{{3\over 10}\times{2\over 9}}{{3\over 10}\times{2\over 9}+{7\over 10}\times{6\over 9}}=\frac{1}{8}$$
1. There are $${3 \choose 2}$$ ways to choose white balls from box $$x$$, and $${10\choose 2}$$ ways to choose any two balls from it. So, the probability of choosing two white balls is $${3\choose 2}/{10\choose 2}$$. This is similar for box $$y$$, with numerator being equal to $${7\choose 2}$$ and the denominator is the same since we have 10 balls in it, too.
$$P(B=x|2W)=\frac{{3\choose 2}/{10 \choose 2}}{{3\choose 2}/{10 \choose 2}+{7\choose 2}/{10 \choose 2}}=\frac{{3\choose 2}}{{3\choose 2}+{7\choose 2}}=\frac{1}{8}$$ | 711 | 2,177 | {"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": 18, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2022-33 | longest | en | 0.905037 |
https://www.math-edu-guide.com/CLASS-7-Algebraic-Introduction-of-Formulae.html | 1,726,694,219,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651941.8/warc/CC-MAIN-20240918201359-20240918231359-00244.warc.gz | 805,535,384 | 6,283 | # CLASS-7INTRODUCTION OF FORMULAE
INTRODUCTION OF FORMULAE
A formula is a relation between certain quantities. A relations (equation) between two or more quantities is called a formula. For example, if ‘l’ is the length of a rectangle and ‘b’ = the breadth of the rectangle, its perimeter ‘p’ can be expressed by the following formula that you are already familiar with. P = 2 ( l + b )
Literals are used to represent the variables or quantities in a formula, certain literals are traditionally used to represent particular quantities
If you know the values of all expect one of the quantities in a formula, you can work out the unknown quantity. let us consider a few formulae you have come across earlier.
Examples -
1) The perimeter ‘p’ of a square is four times of its side ‘a’. this can be represented by the formula p = 4a
2) The volume ‘Vv’ of a cuboid is the product of its length ‘Ll’, breadth ‘Bb’ , and height ‘Hh’. This is expressed by the formula, Vv = Ll X Bb X Hh
3) The work done ‘Ww’ is the product of the force ‘Ff’ and the distance ‘Dd’ through which a body moves. then the formula is Ww = Ff X Dd
4) Suppose the sum of two numbers is 20, then the required formula is y + z = 20, y & are the unknown numbers. | 324 | 1,239 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-38 | latest | en | 0.933226 |
https://www.coursehero.com/file/5459639/times%C5%BEtimes%C5%BEtimes%C5%B814/ | 1,493,264,598,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917121865.67/warc/CC-MAIN-20170423031201-00200-ip-10-145-167-34.ec2.internal.warc.gz | 871,573,231 | 22,766 | ממן14
# ממן14 - 14 " 1 f 1 = { A, B, {...
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Unformatted text preview: 14 " 1 f 1 = { A, B, { ( a,1) , ( b,1) }} f 2 = { A, B, { ( a,2) , ( b,2 ) } } f 3 = { A, B, { ( a,3) , ( b,3) } } f 4 = { A, B, { ( a,4) , ( b,4 ) } } g1 = { B, A, { (1, a ) , ( 2, a ) , ( 3, a ) , ( 4, a ) .1 }} g 2 = { B, A, { (1, b ) , ( 2, b ) , ( 3, b ) , ( 4, b ) } } g f . f = { A, B, { ( a,1) , ( b,2 ) } } g = { B, A, { (1, b ) , ( 2, a ) , ( 3, b ) , ( 4, b ) (i) .2 .3 : }} , ) , ( : , f ( A) = {1,2} B g f : A A g f = { A, A, { ( a, g ( f ( a ) ) ) , ( b, g ( f ( b ) ) ) = { A, A, { ( a, g (1) ) , ( b, g ( 2) ) } } = = { A, A, { ( a, b ) , ( b, a ) } } }}= - x A y A , " g f . x = y - ( g f )( x ) = ( g f )( y ) - , x, y A ( g f )( x ) = y g f , " . , . (ii) . f g , g : B A - f : A B , g : B A . A , , g ( B ) = { g (1) , g ( 2 ) , g ( 3) , g ( 4 )} , g - , .( g ) A - f ( g ( x ) ) = f ( g ( y ) ) , . g ( x ) = g ( y ) - , x y , x, y B ,( ) . x y ( f g )( x ) = ( f g )( y ) . ,-- f g , 2 : f ( f -1(C )) C f ( f -1 (C) ) C , f ( f -1 .1 (C )) . . y f ( f -1 (C )) , f ( f -1 (C )) . f ( f -1(C )) C , y C - ) y = f (x ) - x f f -1 -1 (C ) , f ( f -1 (C )) -1 (C ) , .( f ( f -1 -1 (C )) = y | y B, y = f ( x), x f { (C ) } . y C ,( f (C ) = { x | x A, f ( x) C} ) f ( x) C - -1 . f ( f -1(C )) C - , y f ( f (C )) - .2 : f : A B , A = {a}, B = {1,2}, C = {1,2} B . f (a ) = 1 , {1} C - , f ( f -1 (C )) = f ( f -1 ({1,2})) = f ({a}) = {1} , . f(f -1 (C )) C .3 . f : A B f ( f -1(C )) C ' , C f ( f -1 (C )) , , . f , y C ) . y C , f (. - C f ( f : y f (f -1 -1 (C )) (C )) ...
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## This note was uploaded on 07/23/2009 for the course MATH. 04101 taught by Professor ישראלפרידמן during the Summer '06 term at The Open University.
Ask a homework question - tutors are online | 974 | 1,945 | {"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} | 2.984375 | 3 | CC-MAIN-2017-17 | longest | en | 0.494647 |
https://www.pcreview.co.uk/threads/calculate-difference-in-hours-spanning-many-days.2463030/ | 1,718,542,677,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861659.47/warc/CC-MAIN-20240616105959-20240616135959-00375.warc.gz | 829,884,808 | 14,894 | # Calculate difference in hours spanning many days
G
#### Guest
I know I can calculate the difference in hours or days between two fields.
What I don't know is... can the DateDiff function calculate hours spanning
many days? Can I specify how long a day is or how long a week is?
I'm tracking time a document is spent being reviewed by engineers. This
document review can span many work days/weeks. I want to track the difference
of time (in hours) between two fields: TimeOut and TimeIn. But I need to
track hours (spanning numerous 8-hour days over 5-day weeks).
How do I do this?
G
#### Guest
Sure.
MsgBox DateDiff("h", "1/1/2006 13:33", "1/4/2006 13:33")
the datediff above should return a 72
G
#### Guest
What if the time spans over an 8-hour work day and a 5-day work week?
G
#### Guest
Okay... I'm new to this, so bear with me...
I created a module with the code you suggested. I then used the expression
builder to create an onClick event. Nothing happened. Then I added the the
datediff function after the calcWorkHours function. This got results, but it
returned an hour count as if the days we 24 hours and weeks were 7 days. It's
like my code is ignoring the calcWorkHours function all together.
What am I doing worng? I'm *very* new to creating functions and such, so
again... bear with me...
My code is as follows:
Private Sub Calculation_Click()
Calculation.Value = calcWorkHours(strtDate, endDate)
Calculation.Value = DateDiff("h", strtDate, endDate)
End Sub
T
#### TC
It sounds like you have mutiple records for the same document. Each
record has a time in, and a time out. The difference between the two,
is the time spent on that document (according to that particular
record). You want to add-up /all/ the records for a document, to find
the total time that was spent on it. I assume you'd also want to
seperate the data from different people reviewing the same document.
Yes?
If so, your table might look something like this:
tblReview
DocumentID } composite
PersonID } primary
DateTimeIn } key
DateTimeOut
The last two fields would be type Date. In Access, a Date field can
hold a complete date /and time/.
Then your SQL statement would look something like this:
SELECT DocumentID, SUM(DateDiff("?", DateTimeOut, DateTimeIn) AS
TotalTime
FROM tblReview
GROUP BY DocumentID
I can't remember the first DateDiff parameter choices, hence the
question mark.
HTH,
TC (MVP Access)
http://tc2.atspace.com
G
#### Guest
Sorry WolfPack left out a very import statement in the code below.
calcWorkHours = lngHrs
**The problem is that the function really isn't returning anything.**
Ok try this & see if this helps to shed some light on it for ya.
On a blank form drop a textbox and a command button on it.
name the textbox txtCalculation
name the command button cmdCalculate
make cmdCalculate_click() look like this:
Private Sub cmdCalculate_Click()
Dim myCalc As Long
myCalc = calcWorkHours("1/1/2006", "1/15/2006")
Me.txtCalculation.Value = myCalc
End Sub
and the calcWorkHours should look like the one below.
Function calcWorkHours(strtDate As Date, endDate As Date) As Long
Dim lngHrs As Long
Dim lngDys As Long
Do While strtDate < endDate
If Weekday(strtDate) <> 1 And Weekday(strtDate) <> 7 Then
lngDys = lngDys + 1
End If
strtDate = strtDate + 1
Loop
lngHrs = lngDys * 8
calcWorkHours = lngHrs
End Function
Hope it helps
Randy
G
#### Guest
Thanks Randy. It works great. Sorry for the delay. I've been busy with other
work.
G
#### Guest
Okay... One more for ya!
What if I need the time calculation to take into account the time of day?
That is... if strtDate and endDate spans two days, this function calculates
16 hours (unless over a weekend).
But what if the strtDate and endDate are date stamped by checkboxes, and the
time of day is a critical factor? If I check strtDate at 8:00 AM Monday, and
endDate at 9:00 AM Tuesday, the calculation logs 8 hours, not 9. Likewise, if
the strtDate and endDate diff is less than 1 day, then the calculated result
= 0.
How do I account for fractions of a day in hours? | 1,056 | 4,078 | {"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-2024-26 | latest | en | 0.915635 |
https://programming-idioms.org/impl-edit/124/2485 | 1,606,371,015,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141186761.30/warc/CC-MAIN-20201126055652-20201126085652-00049.warc.gz | 449,557,241 | 6,454 | # Programming-Idioms
Implementation
Erlang
Be concise.
Be useful.
All contributions dictatorially edited by webmasters to match personal tastes.
Please try to avoid dependencies to third-party libraries and frameworks.
Implementation edit is for fixing errors and enhancing with metadata.
Instead of changing the code of the snippet, consider creating another Erlang implementation.
Other implementations
```func binarySearch(a []T, x T) int {
imin, imax := 0, len(a)-1
for imin <= imax {
imid := (imin + imax) / 2
switch {
case a[imid] == x:
return imid
case a[imid] < x:
imin = imid + 1
default:
imax = imid - 1
}
}
return -1
}```
`import "sort"`
```func binarySearch(a []int, x int) int {
i := sort.SearchInts(a, x)
if i < len(a) && a[i] == x {
return i
}
return -1
}```
`import "sort"`
```func binarySearch(a []T, x T) int {
f := func(i int) bool { return a[i] >= x }
i := sort.Search(len(a), f)
if i < len(a) && a[i] == x {
return i
}
return -1
}```
```import std.range;
import std.algorithm;```
```long binarySearch(long[] a, long x) {
long result = a.assumeSorted.lowerBound(x).length;
if (result == a.length || a[result] != x)
return -1;
return result;
}```
```function BinarySearch(X: Integer; A: Array of Integer): Integer;
var
L, R, I, Cur: Integer;
begin
Result := -1;
if Length(A) = 0 then Exit;
L := Low(A);
R := High(A);
while (L <= R) do
begin
I := L + (R - L) div 2;
Cur := A[I];
if (X = Cur) then Exit(I);
if (X > Cur) then
L := I + 1
else
R := I - 1;
end;
end;```
```def binarySearch(ar, el)
res = ar.bsearch{|x| x == el}
res ? res : -1
end```
```def binary_search(a, el)
a.bsearch_index{|x| x == el} || -1
end```
```binSearch :: Ord a => a -> [a] -> Maybe Int
binSearch _ [] = Nothing
binSearch t l = let n = div (length l) 2
(a, m:b) = splitAt n l in
if t < m then binSearch t a
else if t > m then aux (binSearch t b)
else Just n where
aux :: Maybe Int -> Maybe Int
aux (Just x) = Just (x+n+1)
aux _ = Nothing```
`import bisect`
```def binarySearch(a, x):
i = bisect.bisect_left(a, x)
return i if i != len(a) and a[i] == x else -1
```
`a.binary_search(&x).unwrap_or(-1);`
```function binarySearch(a, x, i = 0) {
if (a.length === 0) return -1
const half = (a.length / 2) | 0
return (a[half] === x) ?
i + half :
(a[half] > x) ?
binarySearch(a.slice(0, half), x, i) :
binarySearch(a.slice(half + 1), x, half + i + 1)
}```
`import java.util.arrays;`
```static int binarySearch(final int[] arr, final int key) {
final int index = Arrays.binarySearch(arr, key);
return index < 0 ? - 1 : index;
}```
`use 5.020;`
```sub binary_search {
my (\$x, \$A, \$lo, \$hi) = @_;
\$lo //= 0;
\$hi //= @\$A;
my \$mid = int((\$lo + \$hi) / 2);
for (\$x cmp \$A->[\$mid]) {
use experimental 'switch';
return \$mid when 0;
return -1 if 1 == \$hi - \$lo;
return binary_search(\$x, \$A, \$lo, \$mid) when -1;
return binary_search(\$x, \$A, \$mid, \$hi) when 1;
}
}
``` | 966 | 2,872 | {"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-2020-50 | latest | en | 0.427581 |
https://us.metamath.org/mpeuni/dmgmaddn0.html | 1,726,875,231,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701425385.95/warc/CC-MAIN-20240920222945-20240921012945-00178.warc.gz | 540,127,589 | 6,349 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > dmgmaddn0 Structured version Visualization version GIF version
Description: If 𝐴 is not a nonpositive integer, then 𝐴 + 𝑁 is nonzero for any nonnegative integer 𝑁. (Contributed by Mario Carneiro, 12-Jul-2014.)
Assertion
Ref Expression
dmgmaddn0 ((𝐴 ∈ (ℂ ∖ (ℤ ∖ ℕ)) ∧ 𝑁 ∈ ℕ0) → (𝐴 + 𝑁) ≠ 0)
StepHypRef Expression
1 eldmgm 25721 . . . 4 (𝐴 ∈ (ℂ ∖ (ℤ ∖ ℕ)) ↔ (𝐴 ∈ ℂ ∧ ¬ -𝐴 ∈ ℕ0))
21simprbi 500 . . 3 (𝐴 ∈ (ℂ ∖ (ℤ ∖ ℕ)) → ¬ -𝐴 ∈ ℕ0)
32adantr 484 . 2 ((𝐴 ∈ (ℂ ∖ (ℤ ∖ ℕ)) ∧ 𝑁 ∈ ℕ0) → ¬ -𝐴 ∈ ℕ0)
4 df-neg 10925 . . . . . 6 -𝐴 = (0 − 𝐴)
54eqeq1i 2764 . . . . 5 (-𝐴 = 𝑁 ↔ (0 − 𝐴) = 𝑁)
6 0cnd 10686 . . . . . 6 ((𝐴 ∈ (ℂ ∖ (ℤ ∖ ℕ)) ∧ 𝑁 ∈ ℕ0) → 0 ∈ ℂ)
7 eldifi 4035 . . . . . . 7 (𝐴 ∈ (ℂ ∖ (ℤ ∖ ℕ)) → 𝐴 ∈ ℂ)
87adantr 484 . . . . . 6 ((𝐴 ∈ (ℂ ∖ (ℤ ∖ ℕ)) ∧ 𝑁 ∈ ℕ0) → 𝐴 ∈ ℂ)
9 nn0cn 11958 . . . . . . 7 (𝑁 ∈ ℕ0𝑁 ∈ ℂ)
109adantl 485 . . . . . 6 ((𝐴 ∈ (ℂ ∖ (ℤ ∖ ℕ)) ∧ 𝑁 ∈ ℕ0) → 𝑁 ∈ ℂ)
116, 8, 10subaddd 11067 . . . . 5 ((𝐴 ∈ (ℂ ∖ (ℤ ∖ ℕ)) ∧ 𝑁 ∈ ℕ0) → ((0 − 𝐴) = 𝑁 ↔ (𝐴 + 𝑁) = 0))
125, 11syl5bb 286 . . . 4 ((𝐴 ∈ (ℂ ∖ (ℤ ∖ ℕ)) ∧ 𝑁 ∈ ℕ0) → (-𝐴 = 𝑁 ↔ (𝐴 + 𝑁) = 0))
13 simpr 488 . . . . 5 ((𝐴 ∈ (ℂ ∖ (ℤ ∖ ℕ)) ∧ 𝑁 ∈ ℕ0) → 𝑁 ∈ ℕ0)
14 eleq1 2840 . . . . 5 (-𝐴 = 𝑁 → (-𝐴 ∈ ℕ0𝑁 ∈ ℕ0))
1513, 14syl5ibrcom 250 . . . 4 ((𝐴 ∈ (ℂ ∖ (ℤ ∖ ℕ)) ∧ 𝑁 ∈ ℕ0) → (-𝐴 = 𝑁 → -𝐴 ∈ ℕ0))
1612, 15sylbird 263 . . 3 ((𝐴 ∈ (ℂ ∖ (ℤ ∖ ℕ)) ∧ 𝑁 ∈ ℕ0) → ((𝐴 + 𝑁) = 0 → -𝐴 ∈ ℕ0))
1716necon3bd 2966 . 2 ((𝐴 ∈ (ℂ ∖ (ℤ ∖ ℕ)) ∧ 𝑁 ∈ ℕ0) → (¬ -𝐴 ∈ ℕ0 → (𝐴 + 𝑁) ≠ 0))
183, 17mpd 15 1 ((𝐴 ∈ (ℂ ∖ (ℤ ∖ ℕ)) ∧ 𝑁 ∈ ℕ0) → (𝐴 + 𝑁) ≠ 0)
Colors of variables: wff setvar class Syntax hints: ¬ wn 3 → wi 4 ∧ wa 399 = wceq 1539 ∈ wcel 2112 ≠ wne 2952 ∖ cdif 3858 (class class class)co 7157 ℂcc 10587 0cc0 10589 + caddc 10592 − cmin 10922 -cneg 10923 ℕcn 11688 ℕ0cn0 11948 ℤcz 12034 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1798 ax-4 1812 ax-5 1912 ax-6 1971 ax-7 2016 ax-8 2114 ax-9 2122 ax-10 2143 ax-11 2159 ax-12 2176 ax-ext 2730 ax-sep 5174 ax-nul 5181 ax-pow 5239 ax-pr 5303 ax-un 7466 ax-resscn 10646 ax-1cn 10647 ax-icn 10648 ax-addcl 10649 ax-addrcl 10650 ax-mulcl 10651 ax-mulrcl 10652 ax-mulcom 10653 ax-addass 10654 ax-mulass 10655 ax-distr 10656 ax-i2m1 10657 ax-1ne0 10658 ax-1rid 10659 ax-rnegex 10660 ax-rrecex 10661 ax-cnre 10662 ax-pre-lttri 10663 ax-pre-lttrn 10664 ax-pre-ltadd 10665 ax-pre-mulgt0 10666 This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-3or 1086 df-3an 1087 df-tru 1542 df-fal 1552 df-ex 1783 df-nf 1787 df-sb 2071 df-mo 2558 df-eu 2589 df-clab 2737 df-cleq 2751 df-clel 2831 df-nfc 2902 df-ne 2953 df-nel 3057 df-ral 3076 df-rex 3077 df-reu 3078 df-rab 3080 df-v 3412 df-sbc 3700 df-csb 3809 df-dif 3864 df-un 3866 df-in 3868 df-ss 3878 df-pss 3880 df-nul 4229 df-if 4425 df-pw 4500 df-sn 4527 df-pr 4529 df-tp 4531 df-op 4533 df-uni 4803 df-iun 4889 df-br 5038 df-opab 5100 df-mpt 5118 df-tr 5144 df-id 5435 df-eprel 5440 df-po 5448 df-so 5449 df-fr 5488 df-we 5490 df-xp 5535 df-rel 5536 df-cnv 5537 df-co 5538 df-dm 5539 df-rn 5540 df-res 5541 df-ima 5542 df-pred 6132 df-ord 6178 df-on 6179 df-lim 6180 df-suc 6181 df-iota 6300 df-fun 6343 df-fn 6344 df-f 6345 df-f1 6346 df-fo 6347 df-f1o 6348 df-fv 6349 df-riota 7115 df-ov 7160 df-oprab 7161 df-mpo 7162 df-om 7587 df-wrecs 7964 df-recs 8025 df-rdg 8063 df-er 8306 df-en 8542 df-dom 8543 df-sdom 8544 df-pnf 10729 df-mnf 10730 df-xr 10731 df-ltxr 10732 df-le 10733 df-sub 10924 df-neg 10925 df-nn 11689 df-n0 11949 df-z 12035 This theorem is referenced by: dmgmn0 25725 dmgmdivn0 25727 lgamcvg2 25754
Copyright terms: Public domain W3C validator | 2,256 | 3,795 | {"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} | 3.03125 | 3 | CC-MAIN-2024-38 | latest | en | 0.110197 |
https://en.1answer.info/tag/7374617473-6e6f726d616c6974792d617373756d7074696f6e | 1,527,471,093,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870604.65/warc/CC-MAIN-20180528004814-20180528024814-00601.warc.gz | 544,339,871 | 20,207 | # normality-assumption's questions - English 1answer
541 normality-assumption questions.
### 2 Do these Q-Q graphs show that the data is approximately normally distributed?
2 answers, 284 views normality-assumption qq-plot
The ends of these graphs confuse me. I know most of the values fall on or near the line. But I am unsure of whether the data is indeed approximately normal. These are the two graphs. Plot 1: Plot ...
### 2 How to interpret histogram and normaltest result?
I investigated dataset using histogram and normaltest. I used scipy.stats.normaltest, got this result: ...
### Bivariate random variables normality check
If there is a high correlation between two random variables $X$ and $Y$ (dependent), does that mean that they are normally distributed? Confusion i have is that both variables can be exponential in ...
### 1 Theil-Sen estimator assumptions
I found by accident the nonparametric Theil-Sen Estimator as a replacement for standard OLS linear Regression. How well does it perform with autocorrelated data, non-normal residuals and ...
### Transformation for non-normally distributed homoscedastic data?
I have continuous data (latency to perform a behaviour) that is heteroscedastic and also the data and the residuals are not normally dsitributed. I've tried square root $\frac{1}{log}$ and log10 ...
### How to analyze non-normal data with ties?
I am unsure about how to analyze my data as they are quite non-normally distributed, which causes model residuals to be also very non-normally distributed. I have seen there are various threads on non-...
### 2 Why does it says data should be normally distributed for analysis, when different test follow its own distribution (i.e. t, Z, F)?
Why does it say data should be normally distributed for statistical analysis when different test follow its own distribution (i.e. t, Z, F)? What does normality have to do with this?
### 2 Role of Central Limit Theorem in one-way ANOVA
Background: It has been shown and widely referenced (applets even exist, etc.) that for even a highly-skewed numeric variable, a sample size of $n\ge{}$30 is often "large enough" for the Central Limit ...
### 1 Normality and weighting scores by difficulty
I want to conduct a 2 x 3 x 2 mixed model ANOVA with Group (malus/bonus contract) as a between-subjects factor, time pressure (nine, 12 and 15 seconds) as a within-subjects factor and loss aversion (...
### 250 Is normality testing 'essentially useless'?
A former colleague once argued to me as follows: We usually apply normality tests to the results of processes that, under the null, generate random variables that are only asymptotically or ...
### 3 Comparing observed and predicted values across several measurements
As a neuropsychology graduate student with some experience in statistics (I'm usually the guy other psychologists come to with statistics problems after trying it themselves but before seeing a ...
### Dummy-variable regression with a binary covariate having unequal variance (violation of homoscedasticitycan)
I have a dataset (N = 158) and the following variables that I am going to put into a regression model: Y (a continuous dependent variable) X (a continuous predictor) Z (a continuous predictor) GENDER ...
### Normality in multiple regression after using first differences
I´m running multiple regression and after using first differences (as the only method which was succesful with correcting autocorrelation) my model has non-normal reasiduals. As you can see on . P-...
### How to use wilcox.test in R to check if sample is from a specific distribution? [duplicate]
For example I want to check if a sample is from a normal distribution, or a different one. It this possible with wilcox.test?
### 11 Repeated measures ANOVA: what is the normality assumption?
I am confused about the normality assumption in repeated measures ANOVA. Specifically, I am wondering what kind of normality exactly should be satisfied. In reading the literature and the answers on ...
### Parametric vs non parametric tests for Likert items?
I have conducted a survey and am now trying to figure out how to analyse the results. I have a series of 6-point Likert items, which I have coded like this: -3 Strongly disagree -2 Disagree -1 ...
### One-way anova data normality assumption
The one-way anova assumptions are: independent and identically distributed variables (or, less precisely, “independent observations”); independent and identically distributed variables (or, less ...
### 2 Linear Mixed Model non normal residuals large dataset (I.e., 1500 datapoints)
I have a relatively large dataset (around 1500 data points) comprised by around 90 subjects where each subject is measured 14 times on a likert scale -1 to 5-. When I analyze the data with the lmer ...
### 17 Assumptions of linear models and what to do if the residuals are not normally distributed
I am a little bit confused on what the assumptions of linear regression are. So far I checked whether: all of the explanatory variables correlated linearly with the response variable. (This was the ...
### 1 If $X$ is lognormally distributed, what is the distribution of $1/X$?
Let $X$ be lognormal with parameters $\mu$ and $\sigma$ (such that log($X$) is Gaussian with mean $\mu$ and $\sigma$). What is the distribution of $1/X$? (I.e., its "simple" parametric distribution)?
### 2 Normality Assumptions of the Linear Model
2 answers, 91 views regression linear normality-assumption
I have some trouble understanding the normality assumptions of the linear model. I have found a wealth of information already, but some of it is contradictory and I couldn't find a definite answer to ...
### SPSS: Log-transformation of data that is not normal distributed
I work on my thesis and use SPSS to analyze the data. Because some of my data is not normal distributed, I would like to log-transform the data to see, if this changes the distribution. I differ ...
### 25 Is it meaningful to test for normality with a very small sample size (e.g., n = 6)?
I have a sample size of 6. In such a case, does it make sense to test for normality using the Kolmogorov-Smirnov test? I used SPSS. I have a very small sample size because it takes time to get each. ...
### Normality Tests
I have 100 samples from a population with unknown population distribution. Both the population mean and standard deviation are unknown. I want to check if my data comes from a normal distribution. My ...
### (G)LME on EEG data: distribution not fitting, assumptions of normally distributed, i.i.d. residuals not met, what to do?
this is my first post on this forum, so please let me know if you need more details. Would be really glad for some advice! I have data that I want to model using LME. I tried to find a good fit for ...
### Structural breaks and residuals analysis for ARIMA models
I have two question regarding ARIMA modeling. 1) When I work with ARIMA models should I test for the absence of structural breaks? If the answer is yes, what function of R could I use to detect a ...
### 38 How to perform a test using R to see if data follows normal distribution
6 answers, 166.474 views r distributions normality-assumption
I have a data set with following structure:a word | number of occurrence of a word in a document | a document id How can I perform a test for normal ...
### $Pr<W$ and $Pr>D$ - interpretation of statistic tests results
I have found the following test results: Source: Link Could you tell me what does it mean and how to interpret the following quantities: $Pr<W$, $Pr>D$, $Pr>W-Sq$ and $Pr>A-Sq$? For ...
### Correct approach to carry out Q-Q Normal plot in R?
I am attempting to test some sets of data for normality. I have 64 groups to test. Each group has n=8 samples. [[I am aware of the problems with low n in regard to normality testing]] My end goal is ...
### What conditions of normality must be met for paired/unpaired t-tests?
We assume that $X_i \sim N(μ_1, σ_1^2)$ and $Y_i \sim N(μ_2, σ_2^2)$ [i.e. that they are normal distributions with a finite (or known?) mean and variance] [...] -- Tamhane & Dunlop, ...
### Should I use non-parametric tests?
I am a teacher working on a simple research with the aim of studying whether the set of lesson designs I created can improve student performance. I have two groups with 15 students each. I divided ...
### Normality assumption in 2x2 repeated measures ANOVA
I went through all relevant posts on Cross Validated, and there are clearly some that discuss the same question, but for the life of me I cannot reliably figure this out based on the information given ...
### What are the 'critical' values of skewness and kurtosis for normality assumption? [duplicate]
I am analyzing buy-and-hold abnormal returns of stocks (dependent variable) using OLS regression. These returns, however, tend to be positively skewed (and are so in my case). The residuals obtained ...
### 2 Outlier detection and normality assumption
So I am following this applied statistics class, and we were taught 5/6 tests for outlier detection and normality test, then told to apply these to some datasets. I am quite confused on how to go for ...
### Should I apply shapiro test on whole set of data to check the normality? OR separately by treatments?
0 answers, 38 views normality-assumption
I already checked here, but I didn't get my answer. I have a data of three treatments, A, B and C. Every treatment has 28 replicates. To check the normality of the data, i'll use shapiro test on the ...
### 23 Interpretation of Shapiro-Wilk test
I'm pretty new to statistics and I need your help. I have a small sample, as follows: H4U 0.269 0.357 0.2 0.221 0.275 0.277 0.253 0.127 0.246...
### 1 Is Wilcoxon Rank Sum test appropriate for comparing non-normal, imbalanced, heteroskedastic groups?
I'd like to compare the (continuous) responses of two unpaired groups. The end objective is to test whether one of the groups will be 'superior' to the other (either by a mean or a median comparison). ...
### 1 Does the normality assumption hold? Is this an outlier?
I am trying to fit a multiple linear regression (OLS) model with IPO underpricing as dependent variable. As part of my master thesis I would like to analyze the effect of venture capital ...
### Investigating the Normality of Residuals in Longitudinal Regression - consecutive S-W testing on subsets of residuals
According to What is the nature of the normality assumption in models for longitudinal data?, longitudinal regression has characteristics which make the usual assumptions more awkward to apply. ...
### Transforming Proportional Accuracy Data?
My data has a within-subjects factor of time (i.e. pre and post training) and a between subjects factor of group (training vs control). One of my repeated measures DVs looks at proportional accuracy (...
### 3 normality test on small samples
2 answers, 110 views sample-size normality-assumption
Premise: not very clever in statistics! Data: I have quantitative data (two variables, A and B) on two small groups of subjects (both N=7); I'm going to perform a T-test to check differences about ...
### Does the asymptotic normality hold for panel structured data?
In regression, the asymptotic normality requires, other than the central limit theorem, observations are independent, and panel data clearly violates this assumption as observations are at least ...
### Logistic regression normality and homoscedasticity
Why does logistic regression not require residuals to be normally distributed and homoscedastic the way linear regression does? Why does this not cause problems for estimating logistic regression ...
### 3 Why does asymptotic efficiency require normality?
Greene (Econometric Analysis), states the defintiion of Asymptotic Efficiency. My question is, why does this definition contain a reference to normal distribution? We already know by the central ...
### Why has the Jarque-Bera test of normality two degrees of freedom?
This is most likely a dumb question, but why has the Jarque-Bera test of normality two degrees of freedom? My initial thought was that the sum of two squared standard normals (i.e., skewness and ...
### 3 Confidence interval vs. prediction interval misunderstanding
Problem I have a time series data set with about 50 observations. I'd like to compute an interval that may contain the next/future value in the time series (the 51st data point). I tried using a 90% ... | 2,833 | 12,542 | {"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} | 2.921875 | 3 | CC-MAIN-2018-22 | longest | en | 0.915413 |
https://eranraviv.com/r-tips-tricks-timing-profiling-code/ | 1,725,867,932,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651092.31/warc/CC-MAIN-20240909071529-20240909101529-00497.warc.gz | 203,933,697 | 17,616 | # R tips and tricks – Timing and profiling code
Modern statistical methods use simulations; generating different scenarios and repeating those thousands of times over. Therefore, even trivial operations burden computational speed.
In the words of my favorite statistician Bradley Efron:
“There is some sort of law working here, whereby statistical methodology always expands to strain the current limits of computation.”
In addition to the need for faster computation, the richness of open-source ecosystem means that you often encounter different functions doing the same thing, sometimes even under the same name. This post explains how to measure the computational efficacy of a function so you know which one to use, with a couple of actual examples for reducing computational time.
First, use the package `microbenchmark`. It provides infrastructure to measure and compare the execution time of different R code. In Matlab you can use the the `tic toc` functions and in Python you can use the `clock` function from the `time` module. You can find some toy code here.
Example: you need to extract the first value of a vector. I grew up with the function `head(vector, n= 1)` which asks the first value of the vector. In Python’s Pandas it is `vector.head(1)`. Now, there are other functions which, perhaps, are more speedy/efficient.
Google says that there is a function which is called `first`. Actually, a function called `first` is available from the `dplyr` package, the `xts` package and the `data.table` package. So we have 4 different functions (that I know of, probably more out there) that do the same thing. If speed is of the essence, which function you should use? which is fastest? The following code-snippet answers.
The triple colon operator `:::` is used here to access functions within the package without attaching the actual package to the search path. Since the packages are not of the same size I use the triple colon operator but the use of the double colon operator, `::` is better practice in general. Here are the results of the timing operation based on 1000 times of executing each one of those function. The y-axis is divided by 10000 for readability.
With the risk of over-generalization, based on this analysis/figure the recommendations are: (1) use `data.table` (as an aside, I hear a lot of other good things about it). (2) Do not use `dplyr` when speed is important. (3) In this example, if the function `head` is replaced by the function `first` from the data.table package you can save around 33% of computational time. If you are still reading this, I can imagine your code has many pockets where you can improve the speed, so time your code if you enjoy need it. Doing something like this within few pockets of your code could save appreciable time.
### Speeding up eigenvalue decomposition
Eigenvalue decomposition is a mathematical operation which is common, and is computationally expensive even in fairly moderate dimension. The `eigen` function is the go-to. I found two more faster solutions. You can use the `irlba` function from the `irlba` package which is good for very large matrices, or the `eigs_sym` function from the `RSpectra` package, which take advantage for when you don’t need the whole vector of eigenvalues (as is often the case) by simply replying with the first few largest values, and conserve computational time as a result. The code below illustrates the speed gain.
expr min lq mean median uq max neval
eigen 23.9 24.8 26.8 26.1 27.9 45.1 1000
eigs_sym 16.2 16.8 18.2 17.6 18.9 39.8 1000
So if you are interested in say the top few eigenvalues you can save good amount of time by using the `eigs_sym ` function.
Optimize your code away, friends.
### References
• Efron, Bradley. “The bootstrap and modern statistics.” Journal of the American Statistical Association 95.452 (2000): 1293-1296. | 883 | 3,877 | {"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} | 2.953125 | 3 | CC-MAIN-2024-38 | latest | en | 0.904792 |
https://gateoverflow.in/265318/continous-probability | 1,582,809,184,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146714.29/warc/CC-MAIN-20200227125512-20200227155512-00177.warc.gz | 343,608,464 | 16,701 | +1 vote
124 views
Find the value of $\lambda$ such that function f(x) is valid probability density function
$f(x)=\lambda (x-1)(2-x)$ for $1 \leq x \leq 2$
$=0$ otherwise
My $\lambda$ is coming to be $- \frac{6}{5}$
Am I correct?
| 124 views
0
I m getting 6...
0
Yeah 6 is given in the key.
0
Integrate from 1 to 2 which evaluates to 1 ,Therefore lambda=6
0
+2
see this....
+1
I am getting 6...i think you have done some calculation mistake.
0
yes..
0
Yeah I was doing calculation mistake. Answer is 6. | 170 | 509 | {"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} | 3.125 | 3 | CC-MAIN-2020-10 | latest | en | 0.88556 |
https://spark.adobe.com/page/vSkOB3I7z4hSo/ | 1,534,610,212,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213691.59/warc/CC-MAIN-20180818154147-20180818174147-00583.warc.gz | 765,994,960 | 4,977 | ## Computational LogicBy Ciaran Etheridge
Logic gates
Most logic gates have two inputs and one output. These are usually binary conditions: 1 or 0.
To work out the number of rows needed, on a truth table, the formula is 2 to the power of how many inputs there are.
### The 'AND GATE' only gives an output of 1 or true if all inputs are 1 or true.
Truth table of an and gate
The equation symbol of an 'AND GATE' is a ^ b = c
### The 'OR GATE' gives us a 1 or true output if even 1 of the inputs is one or true. If they're both 0 or false it output that value.
OR GATE truth table
The equation symbol for an 'OR GATE' is a v b = y
### How many rows?
A ^ B = 4 rows ✔️
A ^ (B v C) = 8 rows ✔️
A —, = 2 rows ✔️
—, (A ^ B) = 4 rows ✔️
Truth table for the equation: A ^ (B v C)
8 Rows
Truth table for the equation:
Credits:
Created with images by Picography - "notebook laptop work" | 267 | 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} | 3.40625 | 3 | CC-MAIN-2018-34 | latest | en | 0.81115 |
http://social.msdn.microsoft.com/Forums/en-US/sqlkjpowerpivotforexcel/thread/44607ad1-bd9b-4841-9856-58735e25a940 | 1,369,458,559,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705543116/warc/CC-MAIN-20130516115903-00097-ip-10-60-113-184.ec2.internal.warc.gz | 250,470,953 | 11,797 | New Customer and Old Customer
# New Customer and Old Customer
• Monday, March 18, 2013 6:03 PM
Hi Everybody,
I found some issues when I need to count New Customer and Old Customer.
This is the data sample about the task I'm working on :
Employee Date (mm/dd/yyyy) Customer Robert 1/1/2012 A1 Hanna 1/2/2012 A2 Robert 1/2/2012 A3 Hanna 3/1/2012 A3 Hanna 4/1/2012 A1 Robert 4/4/2012 A1 Robert 4/6/2012 A3 Robert 4/8/2012 A5
The expected Result may like this :
Time Period --> 4/6/2012 - 4/8/2012
* New Customer of Robert = 1 (the value must be A3, because A3 has visited the store before the time period)
* Old Customer of Robert = 1 (the value must be A5, because this is first purchase of A5 during the time period)
Can someone help me to solve this issue using Measure or another approach ?
Riska Bagus
### All Replies
• Tuesday, March 19, 2013 7:53 AM
Riska, I took your table and with a separate dates table created a 2 minute test model here: Example Model
I basically created 3 measures.
The first simply counts the unique customers in a given context.
The second which is where the work is done is to filter that count of customers for basically two criteria, the date being before the time period stipulated and the customer being active in the time period stipulated. Although the formula looks a little tasty it isn't actually that complex and basically draws on Rob Collie's 'Greatest Formula in the World' technique along with an additional VALUES() of the customer column that basically presents the customers in the current context (i.e. the date range stipulated) to filter the fact table for just those customers.
The third comes up with new customers as the balancing figure between the first two.
```[theCount]=DISTINCTCOUNT(factData[Customer])[Old Customers]=CALCULATE(
[theCount],
VALUES(factData[Customer]),
ALL(dimDate),
FILTER(ALL(dimDate),dimDate[Date]<MIN(factData[Date]))
)[New Customers]=[theCount]-[Old Customers]```
This assumes your main table is called factData and the additional calendar table is called dimDate.
I wasn't 100% sure initially that this is what you were trying to do, let me know if you have any difficulties.
Jacob
• Edited by Tuesday, March 19, 2013 7:54 AM Formatting
•
• Tuesday, March 19, 2013 9:58 AM
Hi Jacob,
Thank for your respond on my question.
Just now I discussed about the requirement to count New Customer and Old Customer. Client changed their mind about how to define Old and New Customer. I have 2 tables about Transaction and Customer.
TRANSACTION Employee Amount TransDate Customer CustomerName 1 100 1/1/2013 A01 John 1 50 1/3/2013 A02 Robert 2 10 1/6/2013 A03 Jasper 3 20 2/6/2013 A04 Apache 4 30 2/7/2013 A05 Ronald 2 25 2/8/2013 A06 Richard 3 30 2/9/2013 A07 Anna 4 12 2/15/2013 A01 John
CUSTOMER Customer Name JoinDate A01 John 12/1/2012 A02 Robert 12/3/2012 A03 Jasper 12/5/2012 A04 Apache 12/7/2012 A05 Ronald 12/30/2012 A06 Richard 2/1/2013 A07 Anna 2/10/2013
The expected result may like this :
- Customer will set/define about time period to display the result. In this case, I would like to set time period in February 2013.
Old Customer will be 3 (John, Apache, Ronald) --> means, Customer Count who joined / registered in database below Time Period (February 2013).
New Customer will be 2 (Richard, Anna) --> means, Customer count who join / register within Time Period (February 2013). Richard and Anna registered in database (JoinDate) in February 2013.
Note : I will use distinct. because when the same Customer visit the Store during the time period more than 1 time, I only count as 1 customer.
I used slicer as Time Period Filter (I took the Slicer Data from Transaction Table).
Thank you.
Riska Bagus S
• Tuesday, March 19, 2013 11:20 AM
OK. I've uploaded another quick demo here: New Demo
The difference in this version is in the [Old Customer] which now looks like this:
```=CALCULATE(
[theCount],
FILTER(dimCustomer, dimCustomer[JoinDate]<min(dimDate[Date]))
)```
Basically the customer table is related to the to the fact table but NOT the date table.
This means that I can select the date slicer which only filters my fact table but I can then filter my customer table to just dates less than the lowest date selected in the slicer which in this case is months but could be days/weeks/years also despite the fact that they are unrelated. Because of the relationship between the customer and the fact table, when I apply this filter to the customer table that in turn filters the fact table to just customer with a start date prior to the period selected.
Hope this makes sense.
Jacob
• Marked As Answer by Wednesday, March 20, 2013 2:18 AM
•
• Wednesday, March 20, 2013 2:18 AM
Hi Jacob,
I forgot to attach Calendar table as Slicer table. But you did it.
Thank you very much. It works like a charm. I have tried in my PowerPivot, and I got the solution from you.
regards,
Riska Bagus | 1,319 | 4,925 | {"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} | 3.25 | 3 | CC-MAIN-2013-20 | latest | en | 0.897226 |
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1. unacademy 62.9k views 4.8 Kumar Ketan Follow me on the Unacademy #15 Educator in IIT JEE #6 years teaching experience #IIT JEE Physics #YouTuber 'magical education" (170k+ subs) #Unacademy educator (60k+ lifetin eviews) 3k 2 37 Followers Following Courses Get updates about new courses . Watch all my lessons . Download slides and watch offline Message Lists (1) Kumar Ketan HINDI IT JEE Physics by Kumar Ketan for IT JEE 52 saves Kumar Ketan
2. In figure shown, left arm of a U-tube is immersed in a hot water bath at temperature T, and right arm is immersed in a bath of melting ice, the height of manometric liquid in respective columns is hr and ho Determine the coefficient of expansion of the liquid. Water at melting ice 0
3. (b) r h,T 1 (c) h h hoT 0 0 h., T 0
4. Since the liquid is in hydrostatic equilibrium, Pono hr Also, Vr = V0(1 +YI) 0 From (i) and (ii), we get hr = ho(1 +YT). which on solving for y, we get hoT
5. A coil in the shape of an equilateral triangle of side l is suspended between the pole pieces of a permanent magnet, such that B is in plane of the coil. If due to a current I in the triangle,
6. a torque acts on it, the side 1 of the triangle is 1/2 (BI) 2(-t (a) (b) 3 BI
7. Area of equilateral triangle, I I sin 60 2 3 ,2 4 A-_ 2 x Base x Height = = 1/2 4.
8. A body is moving up an inclined plane of angle with an initial kinetic energy K. The coefficient of friction between the plane and thef re t e ho.lThe work done against friction before the body comes to rest is (a) Kcos + sin (b) LK cos UK cos cos + sin cos cos -sin (d)
9. Let the distance travelled by the body on tne inclined plane be h. From the free body diagram shown in figure R ng sin | 618 | 2,104 | {"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} | 3.421875 | 3 | CC-MAIN-2019-43 | longest | en | 0.848223 |
https://deejaygraham.github.io/2021/03/29/simple-generative-art/ | 1,716,251,356,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058342.37/warc/CC-MAIN-20240520234822-20240521024822-00756.warc.gz | 176,310,520 | 5,166 | # Simple Generative Art
My new intro to coding class is under way and I'm going to start working on some examples of generative art instead of the usual microbit tutorials because of the enforced remoteness of the classes now. I think that using generative art will allow me to introduce several concepts and allow people to play with the code and see what happens when there is no one "correct" answer to find for the problem.
I will be working with my second favourite environment, Processing, in python mode. I was watching a presentation by Tim Holman and wondered how easy some of his examples would translate into python and processing.
Below is an example of this translation where "simple" code can generate some really nice, "complex"-looking graphics.
### Lines and Lines and Lines
The starting point is drawing a line from one point to another. First we setup the canvas to be a small size, to begin with, set the background colour to white and the line colour to black. We also want to have the code generate the art once so I'm using the linear setup of processing rather than using the setup and draw functions, at least for the time being.
``````
def draw_tile(x, y, w, h):
line(x, y, x + w, y + h)
size(400, 400)
background(255, 255, 255)
stroke(0)
draw_tile(0, 0, width, height)
saveFrame("art-######.png")
``````
I have separated the tile drawing code into it's own function to make expanding this easier in the next step. It means I can talk about functions but wave my hands over much of this until the students can see the benefit from it coming up very soon. Since we are only drawing a single tile at this point, I am setting the tile size to be the whole of the square canvas.
### Randomness
A small change to the draw_tile function lets us decide if a tile should be drawn from top left to bottom right or from top right to bottom left. By using the random function we can talk about injecting some variation into our code and even select what probability we want, a straight 50% or something else? This is also the point where we introduce
``````
def draw_tile(x, y, w, h):
left_to_right = True if random(2) >= 1.0 else False
if left_to_right:
line(x, y, x + w, y + h)
else:
line(x + w, y, x, y + h)
size(400, 400)
background(255, 255, 255)
stroke(0)
draw_tile(0, 0, width, height)
saveFrame("art-######.png")
``````
Each time the code runs now, the square has a diagonal line drawn one way or the other. Again, so far, so boring. Things get a little more interesting when we use tiling.
### Tiling
Next, we want to be able to draw more than one tile per canvas which means we have to split up the canvas into a number of discrete units. For this we introduce a variable which is the size of each tile. Moving from one tile to the next lets us introduce the idea of loops with the range function.
``````
def draw_tile(x, y, w, h):
left_to_right = True if random(2) >= 1.0 else False
if left_to_right:
line(x, y, x + w, y + h)
else:
line(x + w, y, x, y + h)
tile_size = 20
size(400, 400)
background(255, 255, 255)
stroke(0)
for x in range(0, width, tile_size):
for y in range(0, height, tile_size):
draw_tile(x, y, tile_size, tile_size)
saveFrame("art-######.png")
``````
We start off small so that it's easier to see that we are moving from one place to another on screen by substituting values for x and y and seeing that the tile size doesn't change.
### Experimentation
Now is the time where we can let the students loose on the code to try some different things. Reduce the size of the tiles, change the canvas size (fullScreen ?) and look at the time it takes to generate compared to a smaller window size, change the probability, change the line thickness or colour.
Here's another version with a tile size of 20 and a canvas size of 520 pixels. It does look to me very like the mazes you used to see as a child on the back of cereal boxes or in puzzle books.
### What if?
At this point I'm hoping that someone will ask about changing how we draw the lines. What happens, for example, if we draw horizontal or vertical lines rather than diagonal?
The advantage of the function we created is that we can make the change in that function and none of the other parts of the code need to change.
``````
def draw_tile(x, y, w, h):
horizontal = True if random(2) >= 1.0 else False
if horizontal:
line(x, y + (h / 2), x + w, y + (h / 2))
else:
line(x + (w / 2), y, x + ( w / 2), y + h)
tile_size = 20
size(520, 520)
background(255, 255, 255)
stroke(0)
for x in range(0, width, tile_size):
for y in range(0, height, tile_size):
draw_tile(x, y, tile_size, tile_size)
saveFrame("art-######.png")
``````
And we get this rather lovely basket weave effect. | 1,215 | 4,717 | {"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} | 3.09375 | 3 | CC-MAIN-2024-22 | latest | en | 0.925843 |
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Trick method example To factor 12x2 14x7 6 Step 1 Factor out greatest common factor very important 12x2 14x 7 6 26x2 7x 7 3 Step 2 Move and multiply 6x2 7x 3 x2 7x 18 Step 3 Factor x 9x 2 Step 4 Put the number back in two places 6x 96x 2 T T Step 5 Take out the gcf of each factor and throw it away 22x 33x 71 Final Factored Answer Don t forget to put the GCF from step I outfrontU Step 6 Check by multiplying FOIL out Section 23 7 L2 iPage 1 of 1
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# Young female ballet dancers and gymnasts sometimes fail to
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25 May 2009, 00:33
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Young female ballet dancers and gymnasts sometimes fail to maintain good eating habits caused by the desire to be as thin as possible.
A. Young female ballet dancers and gymnasts sometimes fail to maintain good eating habits caused by the desire to be as thin as possible.
B. Good eating habits sometimes fail to be maintained by young female ballet dancers and gymnasts caused by desiring to be as thin as possible.
C. Because they desire to be as thin as possible, good eating habits are sometimes not maintained by young female ballet dancers and gymnasts.
D. Because they desire to be as thin as possible, young female ballet dancers and gymnasts sometimes fail to maintain good eating habits.
E. Young female ballet dancers and gymnasts sometimes fail to maintain good eating habits because they desire to be as thin as possible.
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Re: Ballet Dancers & Gmynasts [#permalink]
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25 May 2009, 03:09
I believe it should be E.
The intention or meaning of the sentence require 'Because' in place of 'Caused by'
What is OA?
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Re: Ballet Dancers & Gmynasts [#permalink]
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25 May 2009, 15:04
whats the difference between D and E? They both look correct to me. Probably would pick D though..why I dont know
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Re: Ballet Dancers & Gmynasts [#permalink]
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25 May 2009, 18:47
bigtreezl wrote:
whats the difference between D and E? They both look correct to me. Probably would pick D though..why I dont know
I think it is always preferable to have subject (here ballet dancers and gymnasts) first, if all the other grammatical aspects are correct in sentence.
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25 May 2009, 21:37
bigtreezl wrote:
whats the difference between D and E? They both look correct to me. Probably would pick D though..why I dont know
OA: D
D is preferable to E because it has less ambiguity to "they" - they could be referring to habits in E.
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Re: Ballet Dancers & Gmynasts [#permalink]
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26 May 2009, 01:23
nightwing79 wrote:
bigtreezl wrote:
whats the difference between D and E? They both look correct to me. Probably would pick D though..why I dont know
OA: D
D is preferable to E because it has less ambiguity to "they" - they could be referring to habits in E.
yeah for some reason D just looked more GMATish
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Re: Ballet Dancers & Gmynasts [#permalink]
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26 May 2009, 01:46
Young female ballet dancers and gymnasts sometimes fail to maintain good eating habits caused by the desire to be as thin as possible.
A. Young female ballet dancers and gymnasts sometimes fail to maintain good eating habits caused by the desire to be as thin as possible. reference error
B. Good eating habits sometimes fail to be maintained by young female ballet dancers and gymnasts caused by desiring to be as thin as possible.reference error
C. Because they desire to be as thin as possible, good eating habits are sometimes not maintained by young female ballet dancers and gymnasts.reference error
D. Because they desire to be as thin as possible, young female ballet dancers and gymnasts sometimes fail to maintain good eating habits. only this one is correct
E. Young female ballet dancers and gymnasts sometimes fail to maintain good eating habits because they desire to be as thin as possible. reference error
IMHO, (D)
Re: Ballet Dancers & Gmynasts [#permalink] 26 May 2009, 01:46
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Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 02/07/2008, 10:49 AM (This post was last modified: 02/08/2008, 10:38 AM by Ivars.) I have plotted all 8 "spirals " of the form 4 : ( +-t^+-1/t), 4: (+- 1/t)^+-(t) in polar coordinates and there are certainly interesting forms , crossing points, and regions. I do not know how to get image in here, but it is in the atachment. Attached Files Image(s) « Next Oldest | Next Newest »
Messages In This Thread Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Gottfried - 09/08/2007, 09:07 AM RE: The Complex Inverse h of x^(1/x) - by Gottfried - 09/10/2007, 01:59 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Gottfried - 09/11/2007, 12:18 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Gottfried - 09/19/2007, 05:17 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by GFR - 09/21/2007, 02:01 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Gottfried - 09/21/2007, 04:57 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by GFR - 09/21/2007, 05:11 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Gottfried - 09/22/2007, 07:36 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Gottfried - 10/20/2007, 01:49 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 10/30/2007, 06:00 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by bo198214 - 11/02/2007, 08:43 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 11/03/2007, 12:10 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by bo198214 - 11/03/2007, 12:54 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Gottfried - 11/03/2007, 03:17 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by bo198214 - 11/03/2007, 08:33 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Gottfried - 11/03/2007, 09:09 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Gottfried - 11/07/2007, 12:30 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 11/03/2007, 10:12 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 11/04/2007, 12:10 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by bo198214 - 11/04/2007, 12:42 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 11/04/2007, 02:17 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 11/04/2007, 02:38 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by bo198214 - 11/04/2007, 04:31 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by bo198214 - 11/04/2007, 05:00 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 11/04/2007, 10:15 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by bo198214 - 11/04/2007, 10:53 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 11/05/2007, 05:50 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by bo198214 - 11/06/2007, 01:34 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 11/06/2007, 10:14 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Gottfried - 11/07/2007, 12:12 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by bo198214 - 11/07/2007, 10:12 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 11/07/2007, 11:59 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Gottfried - 11/07/2007, 12:58 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by bo198214 - 11/07/2007, 01:48 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by andydude - 11/14/2007, 05:38 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 11/07/2007, 10:03 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by bo198214 - 11/07/2007, 10:26 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 11/08/2007, 02:31 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by bo198214 - 11/08/2007, 02:51 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 11/09/2007, 08:30 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by bo198214 - 11/09/2007, 08:43 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 11/09/2007, 11:51 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by bo198214 - 11/12/2007, 08:23 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 11/13/2007, 11:01 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Gottfried - 11/14/2007, 05:04 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 11/14/2007, 02:10 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 11/14/2007, 02:14 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Gottfried - 11/14/2007, 02:49 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 11/14/2007, 04:52 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Gottfried - 11/14/2007, 08:57 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 11/14/2007, 10:10 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Gottfried - 11/15/2007, 06:16 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 11/15/2007, 09:40 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Gottfried - 11/15/2007, 04:15 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 11/15/2007, 10:45 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Gottfried - 11/16/2007, 09:09 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 11/16/2007, 12:06 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Gottfried - 11/16/2007, 02:02 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 11/16/2007, 03:17 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Gottfried - 11/16/2007, 05:28 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 11/17/2007, 11:01 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 11/20/2007, 10:11 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 11/22/2007, 01:01 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Gottfried - 11/22/2007, 06:29 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 11/29/2007, 10:37 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 12/09/2007, 03:18 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by andydude - 12/10/2007, 03:14 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Gottfried - 12/10/2007, 07:27 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 12/16/2007, 12:02 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Gottfried - 12/16/2007, 02:22 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 12/16/2007, 05:53 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by andydude - 04/14/2008, 01:01 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 04/14/2008, 10:35 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by andydude - 04/25/2008, 11:25 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 12/10/2007, 07:37 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by andydude - 12/10/2007, 04:11 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by jaydfox - 11/15/2007, 07:21 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 11/15/2007, 09:44 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 11/22/2007, 10:36 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 12/30/2007, 01:09 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 12/30/2007, 03:37 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Gottfried - 12/30/2007, 05:33 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by andydude - 01/05/2008, 08:00 PM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Gottfried - 01/06/2008, 02:55 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 02/07/2008, 10:49 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 04/28/2008, 08:11 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Ivars - 05/09/2008, 06:32 AM RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - by Gottfried - 03/03/2011, 03:16 PM RE: Tetration below 1 - by Gottfried - 09/09/2007, 07:04 AM RE: The Complex Lambert-W - by Gottfried - 09/09/2007, 04:54 PM RE: The Complex Lambert-W - by andydude - 09/10/2007, 06:58 AM
Possibly Related Threads... Thread Author Replies Views Last Post Constructing real tetration solutions Daniel 4 1,846 12/24/2019, 12:10 AM Last Post: sheldonison b^b^x with base 0
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>>>>>>>>Time and Work
• Option : D
• Explanation :
Let pipe B be turned off after x minutes.
⇒ x = 16 minutes
• Option : C
• Explanation :
In a cycle of 3 hours: pipes P, Q and R are working as inlet pipes for 2 hours each and they are working as outlet pipes for an hour each. So part of tank filled in 3 hours.
Thus, tank will be filled in 10th hour.
• Option : B
• Explanation :
Let number of hours taken by the outlet and the 2 inlet pipes be 4x, 2x and x respectively.
So in 1 hour the part of empty tank filled is
Hence, inlet pipe with higher efficiency fills the tank in 20 hours.
So in one hour, the desired pipes fill
Refer to the data below and answer the questions that follow.
The boiler tank in a chemical factory holds 105 litres. 5 tanks each having one-fifth the capacity of the boiler tank fill in ‘hard water’ at same rates in the boiler tank in 2 hours. The outlet of the two of smaller tanks work as inlet pipes and other two work as outlet and the fifth tank fill in the main ‘boiler’ at half its efficiency.
• Option : C
• Explanation :
Since 5 tanks of same efficiency fill the boiler in 2 hours each tank individually takes
2 x 5 = 10 hours to fill it.
Tank with half the efficiency will take 10 x 2 = 20 hours.
In one hour, fraction of boiler getting filled when 4 fully efficient pipes with two as inlet and two as outlet and 5th acts at 50% efficiency as an inlet is
Hence it will take 20 hours to fill the boiler tank.
Refer to the data below and answer the questions that follow.
The boiler tank in a chemical factory holds 105; litres. 5 tanks each having one-fifth the capacity of the boiler tank fill in ‘hard water’ at same rates in the boiler tank in 2 hours. The outlet of the two of smaller tanks work as inlet pipes and other two work as outlet and the fifth tank fill in the main ‘boiler’ at half its efficiency.
• Option : C
• Explanation :
Here fraction of tank filled in 1 hour
th of the tank
If they are opened alternately, then tank gets filled in 4 x 3 = 12 hours
Hence half of the tank gets filled in 12/2 = 6 hours.
Related Quiz.
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#### Resources tagged with Number theory similar to Perfectly Square:
Filter by: Content type:
Stage:
Challenge level:
### There are 22 results
Broad Topics > Numbers and the Number System > Number theory
### Always Perfect
##### Stage: 4 Challenge Level:
Show that if you add 1 to the product of four consecutive numbers the answer is ALWAYS a perfect square.
### Number Rules - OK
##### Stage: 4 Challenge Level:
Can you convince me of each of the following: If a square number is multiplied by a square number the product is ALWAYS a square number...
### Never Prime
##### Stage: 4 Challenge Level:
If a two digit number has its digits reversed and the smaller of the two numbers is subtracted from the larger, prove the difference can never be prime.
### Whole Number Dynamics III
##### Stage: 4 and 5
In this third of five articles we prove that whatever whole number we start with for the Happy Number sequence we will always end up with some set of numbers being repeated over and over again.
### Euler's Squares
##### Stage: 4 Challenge Level:
Euler found four whole numbers such that the sum of any two of the numbers is a perfect square...
### Whole Number Dynamics II
##### Stage: 4 and 5
This article extends the discussions in "Whole number dynamics I". Continuing the proof that, for all starting points, the Happy Number sequence goes into a loop or homes in on a fixed point.
### Whole Number Dynamics V
##### Stage: 4 and 5
The final of five articles which containe the proof of why the sequence introduced in article IV either reaches the fixed point 0 or the sequence enters a repeating cycle of four values.
### Data Chunks
##### Stage: 4 Challenge Level:
Data is sent in chunks of two different sizes - a yellow chunk has 5 characters and a blue chunk has 9 characters. A data slot of size 31 cannot be exactly filled with a combination of yellow and. . . .
### Diophantine N-tuples
##### Stage: 4 Challenge Level:
Can you explain why a sequence of operations always gives you perfect squares?
### There's a Limit
##### Stage: 4 and 5 Challenge Level:
Explore the continued fraction: 2+3/(2+3/(2+3/2+...)) What do you notice when successive terms are taken? What happens to the terms if the fraction goes on indefinitely?
### Ordered Sums
##### Stage: 4 Challenge Level:
Let a(n) be the number of ways of expressing the integer n as an ordered sum of 1's and 2's. Let b(n) be the number of ways of expressing n as an ordered sum of integers greater than 1. (i) Calculate. . . .
### Really Mr. Bond
##### Stage: 4 Challenge Level:
115^2 = (110 x 120) + 25, that is 13225 895^2 = (890 x 900) + 25, that is 801025 Can you explain what is happening and generalise?
### Novemberish
##### Stage: 4 Challenge Level:
a) A four digit number (in base 10) aabb is a perfect square. Discuss ways of systematically finding this number. (b) Prove that 11^{10}-1 is divisible by 100.
### Marbles
##### Stage: 3 Challenge Level:
I start with a red, a green and a blue marble. I can trade any of my marbles for two others, one of each colour. Can I end up with five more blue marbles than red after a number of such trades?
### Strange Numbers
##### Stage: 3 Challenge Level:
All strange numbers are prime. Every one digit prime number is strange and a number of two or more digits is strange if and only if so are the two numbers obtained from it by omitting either. . . .
### How Much Can We Spend?
##### Stage: 3 Challenge Level:
A country has decided to have just two different coins, 3z and 5z coins. Which totals can be made? Is there a largest total that cannot be made? How do you know?
### Differences
##### Stage: 3 Challenge Level:
Can you guarantee that, for any three numbers you choose, the product of their differences will always be an even number?
### A One in Seven Chance
##### Stage: 3 Challenge Level:
What is the remainder when 2^{164}is divided by 7?
### Where Can We Visit?
##### Stage: 3 Challenge Level:
Charlie and Abi put a counter on 42. They wondered if they could visit all the other numbers on their 1-100 board, moving the counter using just these two operations: x2 and -5. What do you think?
### More Marbles
##### Stage: 3 Challenge Level:
I start with a red, a blue, a green and a yellow marble. I can trade any of my marbles for three others, one of each colour. Can I end up with exactly two marbles of each colour?
### A Little Light Thinking
##### Stage: 4 Challenge Level:
Here is a machine with four coloured lights. Can you make two lights switch on at once? Three lights? All four lights?
### Helen's Conjecture
##### Stage: 3 Challenge Level:
Helen made the conjecture that "every multiple of six has more factors than the two numbers either side of it". Is this conjecture true? | 1,145 | 4,817 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2018-22 | latest | en | 0.863117 |
https://mathoverflow.net/questions/299227/integral-morphism-between-universally-closed-and-separated-schemes | 1,579,401,674,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594101.10/warc/CC-MAIN-20200119010920-20200119034920-00088.warc.gz | 548,890,452 | 24,205 | # Integral morphism between universally closed and separated schemes
Let $f : X\to Y$ be a morphism between schemes over $\text{Spec}(\mathbf{Z}_p)$.
Assume:
• $f$ is integral
• both $X$ and $Y$ are universally closed and separated over $\mathbf{Z}_p$
• $f$ mod $p^n$ is an isomorphism for every $n \ge 0$
• $f_*\mathcal{O}_X = \mathcal{O}_Y$
Is $f$ an isomorphism?
Example If $X$ and $Y$ were of finite type, they would be proper and so $f$ would be finite. The flat locus of $f$ in $X$ is open and nonempty, and then it is $X$. So $f$ is finite flat of degree one, then an isomorphism. (We haven't even used the full assumptions (3) and (4)). The point of this question is: what if $X$ and $Y$ are not of finite type, not even locally?
Sure, by $(1)$ and $(4)$. Any integral morphism is affine by definition. If $f$ is an affine morphism with $f_* \mathcal{O}_X = \mathcal{O}_Y$, then $f$ is clearly an isomorphism. | 297 | 923 | {"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} | 2.890625 | 3 | CC-MAIN-2020-05 | latest | en | 0.900634 |
https://calculatorshub.net/measurement-tools/ellipsoid-calculator/ | 1,726,505,283,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.45/warc/CC-MAIN-20240916144317-20240916174317-00066.warc.gz | 130,514,576 | 28,037 | Home » Simplify your calculations with ease. » Measurement Tools » Ellipsoid Calculator Online
# Ellipsoid Calculator Online
The Ellipsoid Calculator is a powerful tool used to determine the surface area of an ellipsoid, a three-dimensional shape resembling a stretched-out sphere. Unlike simple geometric shapes such as cubes or cylinders, ellipsoids have varying axes lengths, making their surface area calculation more complex. With the Ellipsoid Calculator, users can easily input the dimensions of the ellipsoid and obtain its surface area, simplifying otherwise tedious manual calculations.
## Formula of Ellipsoid Calculator
The formula used by the Ellipsoid Calculator is:
`A = 4 * π * [(a * b + a * c + b * c) / 3]^(2/3)`
Where:
• a is the semi-major axis,
• b is the semi-minor axis, and
• c is the other semi-minor axis.
This formula incorporates the mathematical principles necessary to calculate the surface area of an ellipsoid accurately.
## Table of General Terms
To provide additional value to users, here’s a table of general terms related to ellipsoids that people often search for:
This table serves as a quick reference guide for users, aiding in their understanding of ellipsoids and related terms.
## Example of Ellipsoid Calculator
Let’s consider an example to illustrate the usage of the Ellipsoid Calculator:
Suppose we have an ellipsoid with the following dimensions:
• Semi-Major Axis (a): 10 units
• Semi-Minor Axis (b): 6 units
• Other Semi-Minor Axis (c): 4 units
Using the Ellipsoid Calculator, we input these values and calculate the surface area:
`A = 4 * π * [(10 * 6 + 10 * 4 + 6 * 4) / 3]^(2/3)`
`≈ 4 * π * [(160 / 3)]^(2/3) `
`≈ 4 * π * (53.333)^(2/3) ≈ 4 * π * 14.214 ≈ 178.63 units²`
So, the surface area of the given ellipsoid is approximately 178.63 square units.
## Most Common FAQs
Q: How do I measure the semi-major and semi-minor axes of an ellipsoid?
A: To measure the semi-major axis, find the longest distance from the center to the outermost point on the ellipsoid’s surface. For the semi-minor axes, measure the shortest distances from the center to points on the ellipsoid’s surface perpendicular to the semi-major axis. | 542 | 2,195 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2024-38 | latest | en | 0.821916 |
http://au.metamath.org/mpegif/enrbreq.html | 1,531,883,077,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590046.11/warc/CC-MAIN-20180718021906-20180718041906-00254.warc.gz | 32,048,533 | 4,490 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > enrbreq Structured version Unicode version
Theorem enrbreq 8942
Description: Equivalence relation for signed reals in terms of positive reals. (Contributed by NM, 3-Sep-1995.) (New usage is discouraged.)
Assertion
Ref Expression
enrbreq
Proof of Theorem enrbreq
Dummy variables are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 df-enr 8934 . 2
21ecopoveq 7005 1
Colors of variables: wff set class Syntax hints: wi 4 wb 177 wa 359 wceq 1652 wcel 1725 cop 3817 class class class wbr 4212 (class class class)co 6081 cnp 8734 cpp 8736 cer 8741 This theorem is referenced by: enreceq 8944 addcmpblnr 8947 mulcmpblnr 8949 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-3 7 ax-mp 8 ax-gen 1555 ax-5 1566 ax-17 1626 ax-9 1666 ax-8 1687 ax-14 1729 ax-6 1744 ax-7 1749 ax-11 1761 ax-12 1950 ax-ext 2417 ax-sep 4330 ax-nul 4338 ax-pr 4403 This theorem depends on definitions: df-bi 178 df-or 360 df-an 361 df-3an 938 df-tru 1328 df-ex 1551 df-nf 1554 df-sb 1659 df-eu 2285 df-mo 2286 df-clab 2423 df-cleq 2429 df-clel 2432 df-nfc 2561 df-ne 2601 df-ral 2710 df-rex 2711 df-rab 2714 df-v 2958 df-dif 3323 df-un 3325 df-in 3327 df-ss 3334 df-nul 3629 df-if 3740 df-sn 3820 df-pr 3821 df-op 3823 df-uni 4016 df-br 4213 df-opab 4267 df-xp 4884 df-iota 5418 df-fv 5462 df-ov 6084 df-enr 8934
Copyright terms: Public domain W3C validator | 699 | 1,555 | {"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} | 2.875 | 3 | CC-MAIN-2018-30 | latest | en | 0.227803 |
http://academickids.com/encyclopedia/index.php/Independence_of_irrelevant_alternatives | 1,498,306,757,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320261.6/warc/CC-MAIN-20170624115542-20170624135542-00159.warc.gz | 7,539,188 | 6,070 | # Independence of irrelevant alternatives
Independence of irrelevant alternatives is an axiom often adopted by social scientists as a basic condition of rationality. It appears in theories of voting systems, bargaining theory, and logic. It is controversial for two reasons: first, some mathematicians find it too strict of an axiom; second, experiments by Amos Tversky, Daniel Kahneman, and others have showed that humans make this 'error' all the time.
The axiom states: If A is preferred to B out of the choice set {A,B}, then introducing a third, irrelevant, alternative X (thus expanding the choice set to {A,B,X} ) should not make B preferred to A. In other words, whether A or B is better should not be changed by the availability of X.
In voting systems, independence of irrelevant alternatives is interpreted as, if one candidate (X) wins the election, and a new alternative (Y) is added, only X or Y will win the election.
A less strict property is sometimes called local independence of irrelevant alternatives. It says that if one candidate (X) wins an election, and a new alternative (Y) is added, X will win the election if Y is not in the Smith set.
All Condorcet methods fail the former criterion, but some (e.g. Schulze method) satisfy the latter.
Borda count, Coombs' method or Instant-runoff voting do not meet either criterion.
An anecdote which illustrates a violation of this property has been attributed to Sidney Morgenbesser:
After finishing dinner, Sidney Morgenbesser decides to order dessert. The waitress tells him he has two choices: apple pie and blueberry pie. Sidney orders the apple pie. After a few minutes the waitress returns and says that they also have cherry pie at which point Morgenbesser says "In that case I'll have the blueberry pie."
This anecdote whimsically compares the preferences of a large population to a single person. However, intransitive preferences within a collective seem less unreasonable than intransitive preferences within an individual.
Voting systems which are not independent of irrelevant alternatives suffer from strategic nomination considerations.
Some argue that the independence of irrelevant alternatives criterion, however, is a flawed criterion, on the grounds that IIAC failure can have a positive effect. For example, if a population slightly preferred candidate"B" to candidate "A", but candidate "A"'s supporters were far more loyal, then an introduction of a third candidate could split B's support far more than A's, leading to a win by A. In cases where one candidate's supporters feel they are compromising far more than the other candidate's supporters do, failing IIAC may not be a flaw. In other words, IIAC does not take strength of preference into account. Those who consider IIAC to be flawed find Arrow's famous impossibility theorem to be irrelevant.
Condorcet methods do not fail the IIAC when they have a single Condorcet Winner both before and after the introduction of the new candidate. In other words, the IIAC can never replace one Condorcet Winner with another.
• Art and Cultures
• Countries of the World (http://www.academickids.com/encyclopedia/index.php/Countries)
• Space and Astronomy | 667 | 3,203 | {"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} | 3.21875 | 3 | CC-MAIN-2017-26 | longest | en | 0.934815 |
https://calculat.io/en/length/cm-to-inches/80 | 1,638,682,547,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363135.71/warc/CC-MAIN-20211205035505-20211205065505-00467.warc.gz | 239,575,914 | 5,877 | # Convert 80 cm to inches
"Centimeters to Inches" Calculator
Convert
cm to Inches
## Answer for "How many inches is 80 cm?"
`80 cm = 31.5″`
(80 Centimeters is equal to 31.5 Inches)
## Explanation of 80 Centimeters to Inches Conversion
`Centimeters to Inches Conversion Formula: in = cm ÷ 2.54`
According to 'cm to inches' conversion formula if you want to convert 80 (eighty) Centimeters to Inches you have to divide 80 by 2.54.
Here is the complete solution:
`80 cm ÷ 2.54 = 31.5″(thirty-one point five inches)`
## About "Centimeters to Inches" Calculator
This converter will help you to convert Centimeters to Inches (cm to in). For example, How many inches is 80 cm? Enter the number of centimeters (e.g. '80') and then click the 'Convert' button.
Convert
cm to Inches
31.5 | 221 | 792 | {"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} | 3.09375 | 3 | CC-MAIN-2021-49 | latest | en | 0.78456 |
http://www.allenlipeng47.com/blog/index.php/2014/11/30/find-longest-substring-without-repeating/ | 1,656,727,824,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103983398.56/warc/CC-MAIN-20220702010252-20220702040252-00547.warc.gz | 61,984,518 | 9,064 | # Find longest substring without repeating
By | November 30, 2014
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http://www.geeksforgeeks.org/amazon-interview-experience-set-149-campus-internship/
Round3, Q1
Given a string, find the longest substring without repeating characters. For example, the longest substrings without repeating characters for “ABDEFGABEF” are “BDEFGA” and “DEFGAB”.
Instead of using a array for all 26 alphabets, I use hashmap to store the char which are already read.
/*
* http://www.geeksforgeeks.org/amazon-interview-experience-set-149-campus-internship/
* Round3, Q1
* Given a string, find the longest substring without repeating characters.
* For example, the longest substrings without repeating characters for “ABDEFGABEF” are “BDEFGA” and “DEFGAB”.
*/
import java.util.HashMap;
public class LengthOfLongestSubstringWithoutRepeating {
public static int findLongestSubstringWithoutRepeating(String s){
if(s==null){
return -1;
}
HashMap<Character, Integer> hm = new HashMap<Character, Integer>();
int max_start = 0; //Record the position where the max substring starts
int max_len = 1; //Record the length of max substring
int curr_len = 1; //Record the length of current substring
hm.put(s.charAt(0), 0);
for(int i=1;i<s.length();i++){
if((hm.get(s.charAt(i))==null)||(i-curr_len>hm.get(s.charAt(i)))){
//Current char hasn’t been read, or position of previous char c is not in the range of current substring
curr_len++;
hm.put(s.charAt(i), i);
}
else{
if(curr_len>max_len){ //Find a new max length, update the max_len
max_len = curr_len;
max_start = i – curr_len;
}
curr_len = i – hm.get(s.charAt(i)); //Calculate the new length of current substring
hm.put(s.charAt(i), i);
}
}
// System.out.println(“max start:” + max_start);
// System.out.println(“max len:” + max_len);
return max_len;
}
public static void main(String[] args) {
//String str = “abcdefeczqposc”;
String str = “GEEKSFORGEEKS”;
int max_len = findLongestSubstringWithoutRepeating(str);
System.out.println(max_len);
}
} | 561 | 2,041 | {"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-2022-27 | latest | en | 0.487782 |
https://www.transtutors.com/questions/1-a-in-the-situation-of-the-previous-problem-p-cannot-possibly-be-zero-why-not-b-whe-2009024.htm | 1,618,910,937,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039388763.75/warc/CC-MAIN-20210420091336-20210420121336-00008.warc.gz | 1,161,502,098 | 15,538 | # 1. (a) In the situation of the previous problem, p cannot possibly be zero. Why not? (b) When p...
1. (a) In the situation of the previous problem, p cannot possibly be zero. Why not?
(b) When p is unknown, the population and the sample must be large to justify drawing a normal curve as the distribution of the estimator . Why?
2. In this problem, we want to use the estimator p to address the question "What's the value of this unknown p?"
(a) The estimator is unbiased. In fact, we would usually want any estimator to be unbiased. Why?
(b) What's the relevance of the standard error of p for our question about ?
(c) For the normal approximation of to be valid, the condition np, nq ≥ 5 is required. This means n must be large compared with the smaller of p and q. In fact, that's true even if we're not using the normal approximation. Here's an example to show you why: If p = 0.01 and n = 10, no sample can ever give a decent approximation to p. Why?
(d) In working with proportions, sometimes surprisingly large samples are required. Suppose p = 50% and you want the standard error to be no more than two percentage points. What size sample would be required?
(e) Complete the following: In the situation of part (d), there's about a 95% percent chance to observe a sample proportion within __ percentage points of p.
## Plagiarism Checker
Submit your documents and get free Plagiarism report
Free Plagiarism Checker | 357 | 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} | 3.15625 | 3 | CC-MAIN-2021-17 | latest | en | 0.903733 |
https://www.theengineeringprojects.com/2021/02/half-subtractor-in-proteus-isis.html | 1,670,062,529,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710926.23/warc/CC-MAIN-20221203075717-20221203105717-00780.warc.gz | 1,093,660,741 | 37,003 | Hey Pals! we are desired and pleasant to see you in The Engineering Projects. We hope you are doing Great. We have started a series about the Logical Circuits and experiments. In this session, we'll seek the answers of the following questions:
1. What are Half subtrators?
2. What is the working Mechanism of Half Subtractor?
3. How can we design the Truth table of Half Subtractor?
4. How can We implement the Half Subtractor in Proteus ISIS using three Logic Gates?
5. How can we use the NOR Gate to implement the Half Subtractor in Proteus ISIS?
You'll Learn some simple but useful concepts about the Half Subtrators in DID YOU KNOW section into the bargain.
## Half Subtrator
The subtrators belongs to the domain of electronics and we define the subtrators as:
"Half Subtractors are bi-input and bi-output Logical Devices that are specially designed to Get the Binary digits, Subtract them and show the Difference and the Carry through output devices."
The subtractors are considered as one of the building blocks of many Electronic Devices. The inputs are usually named as A and B and the outputs are called the Difference and Borrow.
### DID YOU KNOW????????????
Subtractors are the Electrical Circuits that Perform the subtraction of Numbers and bits in the computer.
## Working Mechanism of Half Subtractor
The Half Subtractor has a boolean circuit. It means it works only with the two digits i.e, 0 and 1. The 0 describes the LOW bit and vise versa. It take two bits through the input Terminals and calculate the whole system then shows us the result at the Output Terminals.
### Difference in Half Subtractor
The difference is obtained when we perform the minus operation with the second bit from the first bit. the calculator give us the output that is the remaining value of the 1st bit when we deduct the value of 2nd bit from it.
### Borrow in Half Subtractor
In the case, when the second bit is higher then the 1st bit, the subtractor borrows a bit from the circuit. this is an essential operation because without this, subtraction can not be proceed further.
## Half Subtractor Truth Table
In binary digit difference, the subtraction of 0 with 0 produces the difference 0 and the borrow 0. when the Value is change to A=0 and B=1 then the circuit borrows a bit and both the bits becomes 1 hence we get Difference=1 and borrow=1. When the inputs are A=1 and B=0 then we simply get the value Difference=1 and Borrow=0. At the same token, when A=1, B=1 then the result we get is Difference=0, Borrow=0. Using all these concepts we get the Truth Table as:
A B Difference Borrow 0 0 0 0 0 1 1 1 1 0 1 0 1 1 0 0
In this tutorial we'll learn to design Half subtractor in two ways:
1. Half Subtractor using three Logic Gates.
2. Half Subtractor using only NAND Gate.
### DID YOU KNOW???????????
Half subtractors are used to limit the force of audio or Radio signals.
## Half Subtractor Using three Logic Gates
In this type of formation we use three Logic Gate given below:
1. XOR Gate
2. NOT Gate
3. NAND Gate
When we look at the working of the Half Subtrator, we'll find that the working of the Difference mode of the Half Subtractor is same as the XOR Gate because we that that XOR Gate is the one that gives the output HIGH only when the inputs have different values from each other and vise versa. Just have a look at its Truth Table:
A B A XOR B 0 0 0 0 1 1 1 0 1 1 1 0
Therefore, we simple use the XOR Gat for the function of Difference. When we look at the Function of Difference. we use one AND Gate. A NOT Gate is attached with the one of the input of AND Gate. One may wonder, why we are using the two Gate when we can use the NAND Gate. but the point is, we just need the inverse condition of just one input. Therefore we use this arrangement.
## Proteus simulation for Half Subtractor sing Three Gates
### Material Required
1. XOR Gate
2. AND Gate
3. NOT Gate
4. LED-RED
5. Ground Terminal
6. Connecting Wires
### DID YOU KNOW?????????????
Arithmatic Logic Unit uses the Half Subtractor for the functioning.
• Begin you Proteus Software.
• Choose first four Components from Pick Library through "P" Button.
• Arrange the Logic Gates one after the other one the working area just as shown in the image:
• Arrange two Logic Toggles Just in front of the XOR Gate.
• Get one LED and Set it Just after the XOR Gate.
• Repeat the step with the with AND Gate.
• Go to Terminal Mode>Ground attach a ground Terminal with each LED.
### DID YOU KNOW?????????????????
One can also use the Logic Probe to Get the output instead of LED.
• Connect all the Components through wires in accordance with the image below:
• Change the values at the Input one after the other and notice the output.
## Half Subtrator using NOR Gate
Sometimes, you need to make the Circuit as simple as you can. Or you can only use one gate then it is also possible to make the whole circuit using just one gate i.e, NOR Gate. when we look at the definition, it says A NOR Gate is the one that shows the output HIGH only when the Input are LOW. So, one can use the NOR Gate in different ways just by using the connection in a specific way. Let's see how can we do this.
## Proteus Simulation of Half Subtractor using NOR Gate
Material Required
1. NOR Gate
2. Logic Toggle
3. LED-RED
4. Ground Terminal
5. Connecting Wire
• Choose the Required Material.
• Arrange the NOR Gates with respect to the image given next:
• Set Logic Toggles in front of Gate 1.
• Attach the LED's with the output of Gate two and 5.
• Ground each LED.
• Join all the devices through Connecting wires with the help of this image:
You will Observe that this circuit works as the half subtrators when you will change the Value of Logic Toggles. Thus today we Learned what are Half Subtrator, How does the Truth table of Half Subtractor is designed, How can we design the Circuit of Half Subtractor with three Gates as well as using just a NOR Gate. If you want to learn more, you can visit the site for other tutorials as well. | 1,419 | 6,013 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2022-49 | longest | en | 0.922825 |
http://www.homofaciens.de/technics-physical-computing-digital-ruler_en.htm | 1,506,074,195,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818688932.49/warc/CC-MAIN-20170922093346-20170922113346-00245.warc.gz | 482,601,298 | 4,723 | # Digital ruler
## Mechanics
Figure 1:
Most of the mechanics is build from pieces of a 40 times 40 mm aluminium bar, mounted on a 19mm particle board.
Figure 1:
You can fix the square tubes on that guide with a screw. There are pieces of flat iron at the edges of the aluminum angles that guide the sawing blade while cutting.
Figure 1:
The sensor wheel is cut from paper and sealed with paint afterwards. The teeth are protected by adhesive tape from both sides.
The axis is made of a 2.5mm nail. I pointed both ends with my drill press and a file. A piece of a rubber tube - usually used as fuel pipe for model engines - gives the axis a higher friction when rolling on a metal surface. Epoxi keeps the sensor wheel on the axis. The bearing of the axis is composed of two pieces of flat iron. Adjust the pressure on the ends of the axis with the 6mm threaded rods to reduce backlash but keep in mind that the sensor wheel must still spin with ease.
With the rubber tube, the diameter of the axis is 6mm. The resulting circumference is approximately 19mm. To get a resolution of less than 0.2mm, we need more than 100 pulses per revolution. The sensor wheel I made has 36 teeth, resulting in 144 pules per revolution on the output pins of the two light sensors. With that we get an academical resolution of 130 micrometers per step - considering the backlash of the mechanics, an accuracy of approximately 0.5mm can be achieved.
Figure 1:
An Arduino Uno is used to count the pulses, displaying the result on an LCD. Keep in mind that the number of pulses per second, the Arduino can read, is limited. You can check if pulses are skipped with a simple experiment:
Set a mark at one of the teeth. Now, move the square tube to the right quickly and slowly back to the left. If the counter reached zero, the marked tooth must be on the initial position.
Move the square tube for a known distance to calibrate the unit - in the video I was using a distance of 40cm. With the number of pulses, you get the conversion factor.
Figure 1:
The second sensor must be an integer of the tooth and gap width plus or minus half a tooth width apart from the first one. Details are available inn the chapter about rotary encoders. If the distance between the sensors is given, we can in turn calculate the tooth width - the resulting minimal width is 2.4mm. To get a higher tolerance for the hand made sensor wheel, I am using a tooth width of 4mm, which also meets the requirement.
## Electronics
Figure 1:
Schematics.
One Arduino Uno, two push buttons, a 2x16 character LCD and a couple of resistors is needed to built the little helper.
## Software
You can get the Arduino sketch at the column Download. Detailed explanations are in the source code.
The optical sensors are connected to pin 2 and 3 of the Arduino, that operate as hardware interrupts. Whenever the signal at one of the pins changes from LOW to HIGH or from HIGH to LOW, a sub routine is executed, adding or subtracting 1 from the pulse counter depending on the direction of rotation. | 690 | 3,045 | {"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} | 2.984375 | 3 | CC-MAIN-2017-39 | longest | en | 0.930918 |
vhtermpaperjwig.cscannualreport.info | 1,521,865,175,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257649683.30/warc/CC-MAIN-20180324034649-20180324054649-00611.warc.gz | 304,871,249 | 3,423 | # Mathematics m coursework
The official public Facebook page of KK LEE Mathematics. Like the page STPM 2018 Term 1 Mathematics (T) Coursework Sample June 13 For June 2017 - Oct 2017 STPM. Mathematics, M.S. Prepares students for careers in education, science and industry, and serves the community by providing expertise to local schools, coastal. Maths Coursework Introduction. Extracts from this document Introduction. Introduction. Aim: In this experiment, I will be investigating the relation between. Mathematics m coursework 2015 essay writing band 6. Welcome to KK LEE Mathematics Site STPM 2017 Term 3 Mathematics (M) Coursework Sample Answer; STPM 2017 Mathematics (T) Term 3 Coursework Sample.
Introduction. Aim: In this experiment. (m): This has nothing to do with the child's IQ six and three number grids in the main part of the coursework. STPM 954 Math T Coursework 2013 [Sem 1] By: Mr. Josh Contact Details: FACEBOOK: Josh Lrt Email: josh. STPM13/14 Math T Coursework Sem 1 (2013), Q2. Undergraduate Program. For course listings and general academic information, see the courses and general info page. For many students and employers, mathematics has. I will start posting Mathematics (T) and Mathematics (M) coursework sample answer again this term. Only sample solution for mathematical part will be posted. Department of Mathematics, Texas A&M University. Department of Mathematics All Courses • - denotes active course for Fall 2017. Course Title; 102: Algebra.
## Mathematics m coursework
Study notes, guides and examination papers for all Malaysian Form 6 Mathematics M students. Read our post that discuss about Mathematics M Coursework 2015, Term 1 mathematics (t) complex number stpm 2018 term 1 coursework sample solutionStpm 2018 term 1. Free MIT courses, including videos, audio, simulations, lecture notes, and exams. STPM 954 Math T Coursework 2013 [Sem 1]. New STPM Mathematics (T). 950 Math M [PPU_STPM] Semester 3 Topics-Syllabus.
MIT Mathematics courses available online and for free. Recent Posts. STPM 2018 Term 1 Mathematics (T) Coursework Sample; STPM 2017 Term 3 Mathematics (M) Coursework Sample Answer; STPM 2017 Mathematics (T) Term 3. STPM 2017 Term 3 Mathematics (M) Coursework PBS Assignment STPM Coursework Sample Solution Question An entrepreneur is contemplating to. 2017 KK LEE MATHEMATICS. Use a cook book and other food books and google to help you or you could just sneak into your teachers office and steal the answer sheet but i dont think thats a very. Malaysia's Largest Online Community. Lowyat.NET forums Advertisement.
About. Wits is a remarkable university that is internationally distinguished for its excellent research, high academic standards and commitment to social justice. I am opening this forum to help the Maths M students with ther coursework. So join in and rock Maths M. This year topic is coming from Probability Distribution. Mathematics m coursework 2015 Form 6 mathematics m coursework sem 1? How to create a title for it? how to start the introduction? and how to end it. M 118 - 118 Mathematics for Music Enthusiasts. 3 Credits. Offered autumn and/or spring. Prereq instructional methods in university mathematics courses. | 738 | 3,207 | {"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} | 2.609375 | 3 | CC-MAIN-2018-13 | latest | en | 0.835827 |
https://www.quantumstudy.com/in-finding-the-electric-field-using-gauss-law-the-formula-e-q_enc-%E2%88%880-a-is-applicable/ | 1,674,891,398,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499524.28/warc/CC-MAIN-20230128054815-20230128084815-00116.warc.gz | 929,884,151 | 12,553 | # In finding the electric field using Gauss law the formula E = q_enc/∈0 A is applicable ….
Q: In finding the electric field using Gauss law the formula $|\vec{E}| = \frac{q_{enc}}{\epsilon_0 |A|}$ is applicable . In the formula ∈0 is permittivity of free space , A is the area of Gaussian surface and qenc is charge enclosed by the Gaussian surface . This equation can be used in which of the following situation ?
(a) For any choice of Gaussian surface
(b) Only when the Gaussian surface is an equipotential surface
(c) Only when the Gaussian surface is an equipotential surface and $\vec{E}$ is constant on the surface
(d) Only when $\vec{E}$ = constant on the surface
Click to See Solution :
Ans: (c)
Sol: The magnitude of electric field should be constant and surface should be equipotential surface . | 197 | 812 | {"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-2023-06 | latest | en | 0.782498 |
http://www.fangraphs.com/library/category/projections/ | 1,488,278,825,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501174157.36/warc/CC-MAIN-20170219104614-00564-ip-10-171-10-108.ec2.internal.warc.gz | 385,712,680 | 22,815 | ## Understanding Projections, “True Talent Level”, and Variability
This is the second in a series of posts about projections. The first part was about the methodology behind each projection system. In this section, we look at what projections are actually telling us.
If you’re new to projections and want to use them to, say, help with your fantasy team, it’s easy to make a common mistake: underestimating the built-in variability in projections. Many people – and I used to be among this group myself – view projections as hard and fast guesses at a player’s production this next season. Most people get into projections as a result of fantasy baseball, so this makes sense; we all want to know which player is going to hit 30 homeruns this next season and which will steal 40 bases. However, projections are actually measuring something different than a player’s expected production: they’re measuring a player’s true talent level.
This might seem like an arbitrary distinction, but trust me, it’s not. As we all know from our day-to-day lives, having a “true talent level” at a particular skill does not necessarily mean you’ll perform at that level every single time in the future. Our minds love to ignore variability and instead treat outcomes as solely talent-driven, but the world doesn’t work that way. Let’s consider a couple examples.
## The Projection Rundown: The Basics on Marcels, ZiPS, CAIRO, Oliver, and the Rest
Now that football season is over and baseball is once again close at hand, Projection Season is well underway. Fantasy players, analysts, bloggers, and plain ol’ fans – everyone turns to projections to help them this time of year. The Hot Stove has cooled down and Spring Training has just started, so really…what else is there to do?
With that in mind, I’ve got a handful of posts on projections in the works for the next week. This is the first one, and in it I deal with a basic question: what are the different projection systems available, and how are each of them calculated? In order to know how to properly use each projection, it’s always a good idea to understand what data is taken into account and how it is used. Remember: there is no one “gold standard” for projection systems. Each system will tell you something slightly different, so whenever trying to draw conclusions from projections, it’s best to use as many sources as possible. | 500 | 2,388 | {"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} | 2.8125 | 3 | CC-MAIN-2017-09 | latest | en | 0.958997 |
http://www.kylesconverter.com/mass-flow/long-tons-per-hour-to-pounds-troy-per-year | 1,542,179,096,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741660.40/warc/CC-MAIN-20181114062005-20181114084005-00249.warc.gz | 458,457,361 | 5,599 | # Convert Long Tons Per Hour to Pounds Troy Per Year
### Kyle's Converter > Mass Flow > Long Tons Per Hour > Long Tons Per Hour to Pounds Troy Per Year
Long Tons Per Hour (t (UK)/h) Pounds Troy Per Year (lb t/yr)* Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18
Reverse conversion?
Pounds Troy Per Year to Long Tons Per Hour
(or just enter a value in the "to" field)
Please share if you found this tool useful:
Unit Descriptions
1 Long Ton per Hour:
Mass flow of long tons across a threshold per unit time of an hour. Using long ton of 2 240 international pounds: 1 Long ton per hour = 1016.0469088/3600 kilograms per second (SI base unit). 1 t (UK)/h ? 1016.0469088 kg/s.
1 Pound Troy per Year:
Mass flow of troy pounds across a threshold per unit time of a year. Using 365 day civil year: 1 Pound troy per year = 0.3732417216/31536000 kilograms per second (SI base unit). 1 lb t/yr ? 1.183 541 7352 x 10-8 kg/s.
Conversions Table
1 Long Tons Per Hour to Pounds Troy Per Year = 23846666.666770 Long Tons Per Hour to Pounds Troy Per Year = 1669266666.67
2 Long Tons Per Hour to Pounds Troy Per Year = 47693333.333380 Long Tons Per Hour to Pounds Troy Per Year = 1907733333.33
3 Long Tons Per Hour to Pounds Troy Per Year = 7154000090 Long Tons Per Hour to Pounds Troy Per Year = 2146200000
4 Long Tons Per Hour to Pounds Troy Per Year = 95386666.6667100 Long Tons Per Hour to Pounds Troy Per Year = 2384666666.67
5 Long Tons Per Hour to Pounds Troy Per Year = 119233333.333200 Long Tons Per Hour to Pounds Troy Per Year = 4769333333.33
6 Long Tons Per Hour to Pounds Troy Per Year = 143080000300 Long Tons Per Hour to Pounds Troy Per Year = 7154000000
7 Long Tons Per Hour to Pounds Troy Per Year = 166926666.667400 Long Tons Per Hour to Pounds Troy Per Year = 9538666666.67
8 Long Tons Per Hour to Pounds Troy Per Year = 190773333.333500 Long Tons Per Hour to Pounds Troy Per Year = 11923333333.3
9 Long Tons Per Hour to Pounds Troy Per Year = 214620000600 Long Tons Per Hour to Pounds Troy Per Year = 14308000000
10 Long Tons Per Hour to Pounds Troy Per Year = 238466666.667800 Long Tons Per Hour to Pounds Troy Per Year = 19077333333.3
20 Long Tons Per Hour to Pounds Troy Per Year = 476933333.333900 Long Tons Per Hour to Pounds Troy Per Year = 21462000000
30 Long Tons Per Hour to Pounds Troy Per Year = 7154000001,000 Long Tons Per Hour to Pounds Troy Per Year = 23846666666.7
40 Long Tons Per Hour to Pounds Troy Per Year = 953866666.66710,000 Long Tons Per Hour to Pounds Troy Per Year = 238466666667
50 Long Tons Per Hour to Pounds Troy Per Year = 1192333333.33100,000 Long Tons Per Hour to Pounds Troy Per Year = 2.38466666667E+12
60 Long Tons Per Hour to Pounds Troy Per Year = 14308000001,000,000 Long Tons Per Hour to Pounds Troy Per Year = 2.38466666667E+13 | 889 | 2,769 | {"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-2018-47 | latest | en | 0.497022 |
https://zbmath.org/?q=an:0679.62039 | 1,619,167,769,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039568689.89/warc/CC-MAIN-20210423070953-20210423100953-00047.warc.gz | 1,191,641,163 | 11,182 | # zbMATH — the first resource for mathematics
Extreme value theory for multivariate stationary sequences. (English) Zbl 0679.62039
The author considers multivariate extremal type theorems and some related problems in a setting where the assumptions of independence and linear normalization in the classical setting are weakened. For a stationary sequence of random vectors he introduces a distributional mixing condition which is an obvious extension of M. R. Leadbetter’s condition $$D(u_ n)$$ in the one-dimensional case [Z. Wahrscheinlichkeitstheorie verw. Gebiete 28, 289-303 (1974; Zbl 0265.60019)]. For a sequence satisfying this condition the following topics are studied.
(1) To obtain characterizations of the weak limit F of properly normalized partial maxima. (2) To study a condition under which the partial maxima behave as they would if the sequence were i.i.d. (3) To consider problems in connection with the independence of the margins of F.
Reviewer: T.Mori
##### MSC:
62H05 Characterization and structure theory for multivariate probability distributions; copulas 60F05 Central limit and other weak theorems 60F99 Limit theorems in probability theory 60G10 Stationary stochastic processes 62G30 Order statistics; empirical distribution functions
Full Text:
##### References:
[1] Amram, F, Multivariate extreme value distributions for stationary Gaussian sequences, J. multivariate anal., 16, 237-240, (1985) · Zbl 0578.60038 [2] Balkema, A.A; Resnick, S.I, MAX-infinite divisibility, J. appl. probab., 14, 309-319, (1977) · Zbl 0366.60025 [3] Berman, S.M, Convergence to bivariate limiting extreme value distributions, Ann. inst. statist. math., 13, 217-223, (1961) · Zbl 0119.15103 [4] Berman, S.M, Limit theorems for the maximum term in stationary sequences, Ann. math. statist., 35, 502-516, (1964) · Zbl 0122.13503 [5] Billinsley, P, () [6] Davis, R.A, Maxima and minima of stationary sequences, Ann. probab., 3, 453-460, (1979) · Zbl 0401.60019 [7] Davis, R.A, Limit laws for the maxium and minimum of stationary sequences, Z. wahrsch. verw. gebiete, 61, 34-42, (1982) [8] Deheuvels, P, Caractérisation complète des lois extrèmes multivariées et de la convergence aux types extrèmes, Publ. inst. statist. unv. Paris, 23, (1978) · Zbl 0414.60043 [9] Deheuvels, P, Propriétés d’existence et propriétés topologiques des fonctions de dépendance avec applications à la convergence des types pour LES lois multivariées, C.R. acad. sci. Paris A, 288, 145-148, (1979) · Zbl 0396.60019 [10] Deheuvels, P, Détermination des lois limites jointes de l’ensemble des points extrêmes d’un échantillon multivarié, C.R. acad. sci. Paris A, 288, 217-220, (1979) · Zbl 0396.60036 [11] Deheuvels, P, A decomposition of infinite order and extreme multivariate distributions, () · Zbl 0601.62066 [12] Deheuvels, P, Point processes and multivariate extreme values, J. multivariate anal., 13, 257-272, (1983) · Zbl 0519.60045 [13] Finkelshteyn, B.V, Limiting distribution of extremes of a variational series of a two dimensional random variable, Dokl. akad. nauk SSSR, 91, (1953) [14] Galambos, J, () [15] Geffroy, J, Contributions à la theorie des valeurs extrêmes, Publ. inst. statist. univ. Paris, 7/8, 37-185, (1958/1959) [16] Haan, L.de; Resnick, S.I, Limit theory for multivariate sample extremes, Z. warhsch. verw. gebiete, 40, 317-337, (1977) · Zbl 0375.60031 [17] Hsing, T, Point processes associated with extreme value theory, () [18] Leadbetter, M.R, On extreme values in stationary sequences, Z. wahrsch. verw. gebiete, 28, 289-303, (1974) · Zbl 0265.60019 [19] Leadbetter, M.R; Lindgren, G; Rootzén, H, () [20] Lindgren, G, A note on the asymptotic independence of high level crosings for dependent Gaussian processes, Ann. probab., 2, 535-539, (1974) · Zbl 0288.60038 [21] Loynes, R.M, Extreme values in uniformly mixing stationary stochastic process, Ann. math. statist., 36, 993-999, (1965) · Zbl 0178.53201 [22] Mikhailov, V.G, Asymptotic independence of vector components of multivariate extreme order statistics, Theory probab. appl., 19, 817-821, (1974) · Zbl 0327.62012 [23] Pickands, J, Multivariate negative exponential and extreme value distributions, (1980), University of Pennsylvania, Unpublished manuscript [24] Rootzén, H, (), Report 36 [25] Sibuya, M, Bivariate extreme statistics, I, Ann. inst. statist. math., 11, 195-210, (1960) · Zbl 0095.33703 [26] Tiago de Oliveira, J, Extremal distributions, Rev. fac. sci. lisboa ser A, 7, 215-227, (1958) [27] Tiago de Oliveira, J, Structure theory of bivariate extremes: extension, Estudos math. estat. econom., 7, 165-195, (1962/1963) [28] Watson, G.S, Extreme values in samples from m-dependent stationary stochastic processes, Ann. math. statist., 25, 798-800, (1954) · Zbl 0056.36204
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | 1,535 | 5,136 | {"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} | 2.609375 | 3 | CC-MAIN-2021-17 | latest | en | 0.802315 |
https://en.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/ee-dc-circuit-analysis/v/ee-labeling-voltages | 1,718,349,687,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861521.12/warc/CC-MAIN-20240614043851-20240614073851-00244.warc.gz | 216,292,355 | 104,974 | If you're seeing this message, it means we're having trouble loading external resources on our website.
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# Labeling voltages
Labeling voltages on a schematic is not a matter of "right" and "wrong". It simply establishes how the voltage appears in the analysis equations. Created by Willy McAllister.
## Want to join the conversation?
• Why was the current labeled as flowing out of the positive side of the battery? Doesn't positive current always flow into the positive side?
• Just think like charges repel. Positive charge is always going to flow away from the positive terminal and into the negative one. Remember the conventional way of labeling current is the path a positive charge would take. Hope that's helpful.
• At , If I is going out of the positive terminal why the result is not negative?
• It is a positive current of positive charge that goes from the positive terminal through the circuit. David, you confuse electron current with conventional current and what you state above is very true for electron current. This is even explained in an earlier video.
• @ or so you could have been more clear and specific that the plus side of the V1 is still considered to be 1V less then the minus side of V1. Although this is for many counter intuitive it is still correct yes. I needed to replay this part 3 times to exactly listen what you said and where you were pointing too. Just wanted to add for more clarity.
• Welcome to the world of many-teachers vs many-style-of-explanation thing. It is not easy sharing a topic to a 30 people class, let alone to a 8 billion khan class. huhu..
Anyway, I kind of like your question/post. It was challenging to be a good teacher, the kind that all-people-can-understand. I wish we can all be one of them. /(^_^)
It was also very challenging to be a good student, the kind that are so fast to reach the gist(point) of a lecture/class. I am a very slow learner, and not as hardworking, let alone doing revision. I envy them very-very much.
Nevertheless, let's learn something (if not everything) with all our hearts. And may we become wiser, and our kids (and other people kids) become smarter.
• Where does the V1 and V2 come in?
• From nowhere.
For this video/section, the purpose of the video is to share how to label our required variable/parameters when it was not given/labeled in the problem/real_circuit. It is to share that we CAN work out any circuit and get every parameters value ( V, I, R, power, equipment_rating) but we need to be systematic in approach and consistent in naming/labeling parameters. This is VERY useful when you are troubleshooting an electronics/power system that you don't have much info to start with. :)
• Generally, what is the most used assumption in circuits: the electron flow [which is the current comes from the negative terminal] or the conventional flow [in which the current comes from the positive terminal same as the voltage (for simplicity)]?
• Hi Voncarlo,
Conventional flow!
The test equipment, right hand rule, and textbooks all assume a current flows from positive to negative.
If you read the fine print the charge carriers (primarily electrons) are traveling from negative to positive.
This is as good an explanation as any: https://xkcd.com/567/
Enjoy!
APD
• lets say i had "assumed" a starting direction for the voltage and when calculating the current, get the current to be negative, does this mean that the direction is opposite to my originally assumed direction?
also when i substitute my current solution to get V1 or V2, do i just use V=IR with no signs or do i follow the signs i had indicated previously, like if my assumed current flow at the begining got a negative sign for a certain voltage, will i have V=-IR when proving or just V=IR?
• IMHO :
does this mean that the direction is opposite to my originally assumed direction? > Yup.
will i have V=-IR when proving or just V=IR? > Just V = IR.
Since this is a question of current direction, then we technically discussing a vector, in which, I don't have a clear way of typing it. :p I would rather draw. :| Sorry.
So, I'll use an analogy. let say, we are an alien.. seeing the blue dot (earth, as human call it). we (alien) wanted to study, how fast does this dot move around the star (human call it the sun).
The answer have two part, one scalar, one vector. One is the amount of the blue dot displacement change, and one is the direction of the rotation. We need both.
The displacement change amount can be calculated, independent of where we are flying from. But for the rotation around the star, we may get 'clockwise' or 'anticlockwise'. The direction calculated depends on which way are we viewing this dot and star. (human : it's anticlockwise if we view from the earths' north and clockwise if we view from the earths' south)
Both are true answer(for the alien), but we(alien) can only calculate/observe one at any one point of reference.
Similarly, (back to human) putting V1 and V2 means we assume the direction of the current flow. If we calculated a positive value, means it flow in our assumed direction. if we get a negative value, means it flow in the opposite of our assumed direction.
hope that helps. :)
• why is the second diagram negative one volt? resistors don't have polarity
• For future readers with the same question: I believe it is clearer if you imagine someone else handed him the schematic with the voltages already marked—which label the voltages entering and leaving the resistor, not polarities of the the resistor itself. (Nothing changes the fact that the resistor is causing a voltage drop.) And now he is using the KVL to analyze a loop.
The signs around the resistor do not alter the effect of the resistor (which is revealed by how many ohms it is). The signs simply affect whether the voltage is a rise or drop in calculations, but note that a negative number for a rise is the same as a drop, and a negative number for a drop is the same as a rise. So when you do the calculation, if you see a situation where you want to reverse the signs around the resistor, you can instead just flip the sign of the value of number of ohms. It doesn't change the role of the resistor in the calculation, it just affects the math we do.
• At you use Kirkoff Law, meaning the sum of voltages=0. Is this a proprety we can apply in every and any circuit? or is there some kind of requirement?
• I'm looking at this serie on circuit analysis to remember what I learnt before, but you got to wonder, since now all I saw was some really basic notations, and all of a sudden he uses K Law without even an introduction ? Imo spend less than 10 videos about notations and at least one to present this important law ...
• That it works out mathematically makes sense since the arithmetic and algebra are very simple but WHY would you ever depict current flowing into a negative terminal and thereby have to reverse the sign of the current ? It seems like this is just complicating things unnecessarily by deliberately introducing something that is wrong and deciding to mathematically compensate for it.
(1 vote)
• Hello Galba,
In the near future you will encounter systems where the direction of current flow cannot be determined by inspection. To solve using "nodal" and "mesh" analysis you will need to guess. Sometimes we get it correct, sometimes not. It is only after the simultaneous equation are solved that the direction of current is known.
Know that current can enter a voltage source from the positive terminal. Think of this as charging a battery.
Regards,
APD
• I still struggle to understand why the sign convention puts differences of potential in the opposite direction of the current for passive components. In my head I think of the current i as a kind of vector, so V would have to be a vector too. Because V=iR, and R>0, V would have to point into the same direction. As an additional problem, this way of thinking still works with Kirchhoff's voltage law (ΣV=0). So why would V have to point into the opposite direction?
(1 vote)
• I feel your pain. This is one of those humps you have to get over at the beginning. Fear not, it is a small hump. Your reasoning about vector current and voltage is valid, but it happens to not be the one we use. Instead, be sure you've seen this video on the sign convention: https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/modal/v/ee-passive-sign-convention.
With circuits we don't bother with a vector notation for current, because there are only two possible directions, this way and that way. This distinction can be taken care of with just the sign of the current, so we don't need vectors.
The definition of current we use is "positive current is the direction positive charge moves (or would move if it was present)." That's the first decision. We use this even though we know negative electrons are moving in wires.
Next, we want Ohm's Law to give the right answer for the voltage polarity when there is a current flowing in a resistor. Suppose you have a battery connected to a resistor. We define current to flow OUT of the positive battery terminal on its way to the resistor. If you measure the voltage with a voltmeter, the more positive voltage is on the end of the resistor closest to the + battery terminal. So we label the voltage that way, with positive voltage sign next to where the current is coming INTO the resistor.
This is the sign convention we use at KA and in almost every EE text I've ever seen. It is totally arbitrary that we do this, but we are super consistent about it.
It is possible to use the opposite convention, which means we define current to flow in the direction electrons move. That's what you've described in your question. This is also valid, but it is not commonly used. The only example I've ever seen is in the training material used by the US Navy and other branches of the US military. (search for "NEETS"). | 2,215 | 10,085 | {"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-2024-26 | latest | en | 0.960701 |
http://www.enotes.com/homework-help/what-lcd-this-math-problem-x-5-x-2-7x-12-x-2-8x-15-435273 | 1,477,115,428,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988718426.35/warc/CC-MAIN-20161020183838-00326-ip-10-171-6-4.ec2.internal.warc.gz | 440,916,035 | 9,601 | # What is the LCD of this math problem? `(x+5)/(x^2+7x+12) +(x^2+8x+15)/(x^2+4x+3)`
lemjay | High School Teacher | (Level 2) Senior Educator
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`(x+5)/(x^2+7x+12)+(x^2+8x+15)/(x^2+4x+3)`
To determine the LCD, factor the denominators.
`=(x+5)/((x+3)(x+4))+(x^2+8x+15)/((x+3)(x+1))`
Then take the product of all difft factors present in the denominators which are (x+3), (x+4) and (x+1).
Hence, the LCD of the two rational expressions is `(x+3)(x+4)(x+1)` . | 201 | 470 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2016-44 | latest | en | 0.815218 |
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June 14, 2018
ECON 357 – Problem Set 3Due Thursday, June 30 at the beginning of class.1. (30 marks) Consider a Stackelberg game where the incumbent and entrant face ainverse demand function of p(Q) = 500 ? 4Q. The incumbent faces a cost functionof c(q1 ) = 10 + q1 and the entrant faces a cost function of c(q2 ) = A + q2 .a) Derive the best-response function of the entrant. (5 marks)b) Determine the incumbent’s choice of q1 . (10 marks)c) Determine the size of A necessary for the entrant to be deterred from enteringthe market under Stackelberg. (10 marks)d) Explain why the Stackelberg outcome requires knowing player 2’s best-responsefunction, but not player 1’s. (HINT: A complete answer considers the effect ofsequential movement and commitment.) (5 marks)2. (25 marks) Consider the following version of the battle of the sexes game:P layer1EFP layer2AB(5, 1)(0, 0)(0, 0)(2, 7)a) Find the Pure Strategy Nash Equilibrium to the game. (5 marks)b) Find the Mixed Strategy Nash Equilibrium to this game. (10 marks)c) Graph the best-response curves for each player. Show the pure and mixedstrategies found in a) and b) on your graph.(10 marks)3. (35 marks) Consider the penalty point game. Let player 1 be the “kicker” and player2 be the “goalie”. Both players can choose one of two actions: left or right. If bothplayers 1 and 2 go left, player 1 receives a payoff of 20 and player 2 receives a payoffof -20. If both players go right, player 1 receives a payoff of 30, and player 2 receives-30. If player 1 goes left and player 2 goes right, player 1 gets 90, player 2 gets -90.If player 1 goes right and player 2 goes left, player 1 gets 70, player 2 gets -70.a) Represent this as a normal-form game. (5 marks)b) Determine the Nash Equilibria to this game. (15 marks)c) Suppose player 2 can lower player 1’s payoff by 10 if opposite strategies areplayed (ie. for the states where player 1 plays left and 2 plays right, or player1 plays right and player 2 plays left). This also raises player 2’s payoff by 10if either state is realized. Suppose each player’s payoff is expressed in dollars.How much would player 2 be willing to pay to to cause this difference in payoffsif player 2 is risk-neutral? (15 marks)q4. (45 marks) A’s utility function is U (x1A , x2A ) = x1A +4 x2A while B’s utility function isqU (x1B , x2B ) = x1B + 16 x2B , where the superscript denotes the good and the subscriptdenotes the person. A’s initial endowment of goods 1 and 2, respectively, is given by1 = 12 and ? 2 = 20. B’s initial endowment is ? 1 = 18 and ? 2 = 4?AABBa) Derive an expression for each of the individual’s marginal rate of substitutiondx2(| dx1i |). (10 marks)ib) On the contract curve, A’s marginal rate of substitution equals B’s. Write anequation that states this condition. From this equation, find the value ofx2Ax2Batall points along the contract curve. (10 marks)2 + ? 2 Use thisc) Along the contract curve it is also the case that x2A + x2B = ?ABin conjunction with the value ofx2Ax2Bfrom part b) to show that x2A and x2B areconstant on the contract curve. (10 marks)d) Find the equilibrium prices and quantities. (10 marks)1e) Draw the Edgeworth box. Label the initial endowments with the letter W. Drawindifference curves for each person and show the contract curve for this exchangeeconomy. (5 marks)5. (35 marks) Robinson Crusoe spends 10 hours a day gathering food. He can eitherspend his time gathering coconuts or catching fish. He can catch 1 fish per hour andhe can gather 4 coconuts per hour.a) Show Robinson’s production possibility frontier between fish and coconuts perday with coconuts on the vertical axis. Write an equation for the line segmentthat is Robinsons production possibilities frontier. (5 marks)11b) Robinson’s utility function is U (f, c) = f 2 c 2 , where f is fish consumption and cis coconut consumption. How many fish will Robinson choose to catch per day?How many coconuts will he collect? (15 marks)c) Suppose that Robinson obtains a technology that allows him to catch more fish.Specifically, Robinson can now catch 3 fish per hour. Determine Robinson’s newproduction choice. (15 marks)2
Order your essay today and save 20% with the discount code: ESSAYHELP | 1,156 | 4,283 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2020-34 | latest | en | 0.864945 |
https://slideplayer.com/slide/6427330/ | 1,685,268,264,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643663.27/warc/CC-MAIN-20230528083025-20230528113025-00762.warc.gz | 599,488,361 | 17,545 | # Algebra 7.1 Solving Linear Systems by Graphing. System of Linear Equations (linear systems) Two equations with two variables. An example: 4x + 5y = 3.
## Presentation on theme: "Algebra 7.1 Solving Linear Systems by Graphing. System of Linear Equations (linear systems) Two equations with two variables. An example: 4x + 5y = 3."— Presentation transcript:
Algebra 7.1 Solving Linear Systems by Graphing
System of Linear Equations (linear systems) Two equations with two variables. An example: 4x + 5y = 3 2x = 6y -10 A solution to a linear system is an ordered pair (x, y) that, when substituted in, makes both equations true. Thus, the solution would be on both graphs. The solution(s) is the intersection of the lines.
Is the ordered pair a solution to the system of equations? Yes or no. -2x + y = -11(6, 1) -x – 9y = 15 Plug it in and check! -2(6) + (1) = -11? -12 + 1 = -11? -11 = -11 Yes. -(6) – 9(1) = 15? -6 – 9 = 15? -15 = 15? No. The point is not a solution to the system of equations.
Use the graph to find the solution to the system of equations. Then check your solution algebraically. y = 3x -12 y = -2x + 3 The solution seems to be (3, -3). Check this solution algebraically on your paper. Who can check it on the board? Yes. The point is a solution to the system.
Steps to “Graphing to Solve a Linear System” 1) Write each equation in a form that is easy to graph (Slope-int or standard) 2) Graph both equations on the same coordinate plane 3) Find the point of intersection 4) Check the point algebraically in the system of equations
Solve the system graphically. Check the solution algebraically. 3x – 4y = 12 -x + 5y = -26 Step 1) Put the equations in a graph-able form. 3x – 4y = 12 Find the x-int. and y-int. 3(0) – 4y = 12 -4y = 12 y = -3 The y-int is (0, -3) Graph it! 3x – 4(0) = 12 3x = 12 x = 4 The x-int is (4, 0) Graph it! Put -x + 5y = -26 into slope-int form. +x +x 5y = x – 26 y = 1/5 x – 5 1/5 The solution to the system seems to be (-4, -6)....
Check (-4, -6) in the system algebraically. 3x - 4y = 12(-4, -6) -x + 5y = -26 3(-4) - 4(-6) = 12? -12 + 24 = 12? 12 = 12 Yes. -(-4) + 5(-6) = -26? 4 – 30 = -26? -26 = -26 Yes. The point is a solution to the system of equations.
You try! Solve the system graphically. Check the solution algebraically. 3x + y = 11 x - 2y = 6 Step 1) Put the equations in a graph- able form. 3x + y = 11 Put into slope-int form. -3x -3x y = -3x + 11 Graph it! x - 2y = 6 Put into slope-int form. -x -x -2y = -x + 6 y = 1/2 x – 3 Graph it! The solution to the system seems to be (4, -1).....
Check (4, -1) in the system algebraically. 6x + 2y = 22(4, -1) x - 2y = 6 6(4) + 2(-1) = 22? 24 - 2 = 22? 22 = 22 Yes. (4) - 2(-1) = 6? 4 + 2 = 6? 6 = 6 Yes. The point is a solution to the system of equations.
HW P. 401-403 #11-19 Odd, 25-33 Odd, 47-59 Odd
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Similar presentations | 1,063 | 3,010 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2023-23 | latest | en | 0.898069 |
https://yourviralbuzz.com/which-of-the-following-is-a-scalar-quantity/ | 1,713,471,550,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817239.30/warc/CC-MAIN-20240418191007-20240418221007-00026.warc.gz | 1,004,379,664 | 36,749 | # Which of the Following is a Scalar Quantity?
When studying physics, it is important to understand the difference between scalar and vector quantities. Scalar quantities are those that are fully described by their magnitude or size, while vector quantities have both magnitude and direction. In this article, we will explore various examples of scalar quantities and discuss their significance in different fields of study.
## What is a Scalar Quantity?
A scalar quantity is a physical quantity that can be described solely by its magnitude or numerical value. It does not have any associated direction. Scalar quantities are used to measure and describe various aspects of the physical world, such as time, temperature, mass, and energy.
### Examples of Scalar Quantities
Let’s take a closer look at some common examples of scalar quantities:
• Time: Time is a scalar quantity as it can be measured solely by its magnitude. For example, if we say an event lasted for 2 hours, we are only referring to the duration of the event without considering its direction.
• Temperature: Temperature is another scalar quantity. When we measure the temperature of an object, we are only concerned with its magnitude, not its direction.
• Mass: Mass is a scalar quantity that describes the amount of matter in an object. It is measured in kilograms (kg) and does not have any associated direction.
• Energy: Energy is a scalar quantity that represents the ability to do work. It can exist in various forms such as kinetic energy, potential energy, and thermal energy.
• Speed: Speed is a scalar quantity that measures how fast an object is moving. It is calculated by dividing the distance traveled by the time taken, without considering the direction of motion.
• Pressure: Pressure is a scalar quantity that measures the force applied per unit area. It is commonly used in fields such as engineering, physics, and meteorology.
## Scalar Quantities in Different Fields
Scalar quantities are used in various fields of study, including physics, mathematics, engineering, and economics. Let’s explore how scalar quantities are applied in these disciplines:
### Physics
In physics, scalar quantities are used to describe various physical phenomena. For example, when studying motion, scalar quantities such as speed and distance are used to analyze the movement of objects without considering their direction. Scalar quantities are also used in thermodynamics to describe temperature, pressure, and energy.
### Mathematics
In mathematics, scalar quantities are used in algebra and calculus. Scalars are often represented by real numbers and are used in mathematical operations such as addition, subtraction, multiplication, and division. Scalar quantities are also used in linear algebra to represent vectors and matrices.
### Engineering
In engineering, scalar quantities are used to analyze and design various systems. For example, in structural engineering, scalar quantities such as load, stress, and strain are used to determine the strength and stability of structures. Scalar quantities are also used in electrical engineering to measure voltage, current, and power.
### Economics
In economics, scalar quantities are used to measure and analyze various economic variables. For example, GDP (Gross Domestic Product) is a scalar quantity that measures the total value of goods and services produced in a country. Inflation rate, unemployment rate, and interest rate are also scalar quantities used in economic analysis.
## Scalar vs. Vector Quantities
Now that we have a clear understanding of scalar quantities, let’s compare them to vector quantities. Vector quantities have both magnitude and direction. Examples of vector quantities include displacement, velocity, acceleration, force, and momentum. Unlike scalar quantities, vector quantities require both magnitude and direction to fully describe them.
For example, if we say a car is traveling at 60 miles per hour, we are describing its speed, which is a scalar quantity. However, if we say the car is traveling at 60 miles per hour north, we are describing its velocity, which is a vector quantity.
## Summary
Scalar quantities are physical quantities that can be described solely by their magnitude or numerical value. They do not have any associated direction. Examples of scalar quantities include time, temperature, mass, energy, speed, and pressure. Scalar quantities are used in various fields of study, including physics, mathematics, engineering, and economics. Understanding the difference between scalar and vector quantities is essential for accurately describing and analyzing physical phenomena.
## Q&A
1. What is the difference between scalar and vector quantities?
Scalar quantities are fully described by their magnitude or numerical value, while vector quantities have both magnitude and direction.
2. What are some examples of scalar quantities?
Examples of scalar quantities include time, temperature, mass, energy, speed, and pressure.
3. How are scalar quantities used in physics?
Scalar quantities are used in physics to describe various physical phenomena, such as motion, temperature, pressure, and energy.
4. What are some applications of scalar quantities in engineering?
Scalar quantities are used in engineering to analyze and design systems, such as structures, electrical circuits, and mechanical systems.
5. Can scalar quantities be negative?
Yes, scalar quantities can be negative. For example, temperature can be below zero degrees Celsius, and speed can be negative when an object is moving in the opposite direction. | 1,033 | 5,633 | {"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} | 4.15625 | 4 | CC-MAIN-2024-18 | latest | en | 0.957716 |
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# Patulong nito plss. pasa ko na to bukas
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Please express the following statements as equation of variation using k as constant: 1. the variable m varies inversely as the variable n2. the number of days,d,of work varies inversely as the number of workers,w.3. the volume v of gas at fixed temperature varies inversely as the pressure p.4. the pitch p of a tone varies inversely as the wavelength,w of the tone.5. the pressure,p,of the air pumped into a balloon varies inversely as the volume v. | 258 | 785 | {"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} | 2.84375 | 3 | CC-MAIN-2021-49 | latest | en | 0.823767 |
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[quote="ChickenMarengo"][quote="Zag"][spoiler]if n=3, this is actually the same as n=1. [/spoiler][/quote] You can actually do slightly better with [spoiler]n=3[/spoiler] (but still worse on average than your answer of [spoiler]n=2[/spoiler]): [spoiler]You can ignore any flips that change all 3 bulbs. You're only interested in the ones that flip a single bulb. Once you've found them label the switches and bulbs so that ABC=1 and ABD=2. Then A and B = 3, C = 1, D = 2. The probability of finding the 2 interesting flips in the first 2 flips is 1/6, of needing 3 flips is 1/3, and of needing 4 flips is 1/2, for an average of 3 1/3 flips.[/spoiler][/quote]
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ChickenMarengo
Posted: Sun Mar 07, 2010 12:12 am Post subject: 1
Zag wrote:
if n=3, this is actually the same as n=1.
You can actually do slightly better with n=3 (but still worse on average than your answer of n=2):
You can ignore any flips that change all 3 bulbs. You're only interested in the ones that flip a single bulb.
Once you've found them label the switches and bulbs so that ABC=1 and ABD=2. Then A and B = 3, C = 1, D = 2.
The probability of finding the 2 interesting flips in the first 2 flips is 1/6, of needing 3 flips is 1/3, and of needing 4 flips is 1/2, for an average of 3 1/3 flips.
Zag
Posted: Sat Mar 06, 2010 11:25 pm Post subject: 0
Then the answer is clearly n=2, which can always be solved in 3 turns.
obviously, n=0 and n=4 are useless.
if n=1, then your first three tries might each change a different bulb, and you will need a fourth try to know which bulb is double-switched.
if n=3, this is actually the same as n=1.
with n=2, you switch AB, BC, CD.
One of these situations must be the case (once you number the bulbs to fit).
AB=no change, BC=12, CD=23 (A and B = 1, C=2, D=3)
AB=12, BC=12, CD=13 (A and C = 1, B=2, D=3)
AB=12, BC=23, CD=13 (A and D = 1, B=2, C=3)
AB=12, BC=no change, CD=23 (A=1, B and C=2, D=3)
AB=12, BC=23, CD=23 (A=1, B and D = 2, C=3)
AB=12, BC=23, CD=no change (A=1, B=2, C and D = 3)
bonanova
Posted: Sat Mar 06, 2010 9:03 pm Post subject: -1
In all cases if you flip a switch, one and only one bulb will change its state.
In my entrance hall, I have the case you talked about.
I have not been able to fix those connectons.
Anyone have a clue on that issue, I'd be glad to hear it.
Trojan Horse
Posted: Sat Mar 06, 2010 8:20 pm Post subject: -2
Okay, so one of the lights is controlled by exactly 2 switches.
Question: do we know exactly how those 2 switches affect that light? Do we know for sure that the 2 switches are connected in an "XOR" fashion; that is, flipping either switch always flips the light? Or is it possible that the switches control the light in some other way? Maybe the light turns on only if both switches are in the up position, or if at least one switch is in the down position, or whatever.
(I once lived in an apartment where the only way to turn the hall light on was to flip BOTH of two switches to the down position. So I'm not assuming anything here.)
bonanova
Posted: Fri Mar 05, 2010 9:31 am Post subject: -3
Here's a piece of fluff for the back of a napkin at lunch time.
Might not even need the napkin...
In my kitchen there are 3 ceiling lights and 4 wall switches.
Here's the deal:
1. Each switch controls 1 and only 1 light.
2. Each light is controlled by at least one switch.
3. When you enter the kitchen [you might not want to open the fridge] you find the switches positioned randomly up or down.
4. You are to determine how the lights and switches are connected.
5. You may [repeatedly, but using the same value of n each time] simultaneously change any n [0 < n < 4] of the switches and note what happens.
Here's the question:
If you want to minimize the number of times you execute Step 5 above, is there a preferred value of n? | 1,351 | 4,740 | {"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-2013-20 | latest | en | 0.833184 |
https://numbermatics.com/n/169065147/ | 1,656,945,705,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104432674.76/warc/CC-MAIN-20220704141714-20220704171714-00001.warc.gz | 467,835,918 | 6,193 | 169065147
169,065,147 is an odd composite number composed of two prime numbers multiplied together.
What does the number 169065147 look like?
This visualization shows the relationship between its 2 prime factors (large circles) and 6 divisors.
169065147 is an odd composite number. It is composed of two distinct prime numbers multiplied together. It has a total of six divisors.
Prime factorization of 169065147:
3 × 75072
(3 × 7507 × 7507)
See below for interesting mathematical facts about the number 169065147 from the Numbermatics database.
Names of 169065147
• Cardinal: 169065147 can be written as One hundred sixty-nine million, sixty-five thousand, one hundred forty-seven.
Scientific notation
• Scientific notation: 1.69065147 × 108
Factors of 169065147
• Number of distinct prime factors ω(n): 2
• Total number of prime factors Ω(n): 3
• Sum of prime factors: 7510
Divisors of 169065147
• Number of divisors d(n): 6
• Complete list of divisors:
• Sum of all divisors σ(n): 225450228
• Sum of proper divisors (its aliquot sum) s(n): 56385081
• 169065147 is a deficient number, because the sum of its proper divisors (56385081) is less than itself. Its deficiency is 112680066
Bases of 169065147
• Binary: 10100001001110111010101110112
• Base-36: 2SNNI3
Squares and roots of 169065147
• 169065147 squared (1690651472) is 28583023930131609
• 169065147 cubed (1690651473) is 4832393142452218204931523
• The square root of 169065147 is 13002.5054124195
• The cube root of 169065147 is 552.9485142713
Scales and comparisons
How big is 169065147?
• 169,065,147 seconds is equal to 5 years, 19 weeks, 3 days, 18 hours, 32 minutes, 27 seconds.
• To count from 1 to 169,065,147 would take you about eight years!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 169065147 cubic inches would be around 46.1 feet tall.
Recreational maths with 169065147
• 169065147 backwards is 741560961
• The number of decimal digits it has is: 9
• The sum of 169065147's digits is 39
• More coming soon!
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The information we have on file for 169065147 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!
Keywords: Divisors of 169065147, math, Factors of 169065147, curriculum, school, college, exams, university, Prime factorization of 169065147, STEM, science, technology, engineering, physics, economics, calculator, one hundred sixty-nine million, sixty-five thousand, one hundred forty-seven.
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