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https://labs.da.org/wordpress/alindroth/2019/01/30/a-solution-to-exercise-42/ | 1,586,045,523,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370526982.53/warc/CC-MAIN-20200404231315-20200405021315-00255.warc.gz | 533,292,762 | 9,213 | A Solution to Exercise 42
I mentioned in my last post that we had discussed exercise 42 from chapter 2 in our last meeting and that Dr. Prudhom gave me a hint on a tactic I could take to solve it. I tried using that method, but I found it to be actually much more difficult than I had anticipated—maybe I went about it in the wrong way, but the algebra was incredibly messy and felt much more convoluted than necessary. So I came back to the original statement of the exercise and tried to approach it from a group-theoretic perspective.
The first thing I did was to look up some examples of dihedral groups online, using a very handy website I found called GroupProps, which is basically “Wikipedia for groups.” You can find pages on a huge variety of groups that describe their properties and definitions. I looked up the dihedral groups, and specifically, the dihedral groups with orders that are multiples of 4 (i.e. the symmetry groups of regular polygons with an even number of vertices). I arrived at this restriction because I noted that the exercise implies that a rotation by 180° is an element of the group, and this is only the case in these specific dihedral groups. After all, if you rotate a triangle by 180° about its center, you don’t end up with a symmetry.
I noticed that in their presentations of these groups, GroupProps lists each reflection as a product of some “base” reflection with some number of rotations. That is, there are two basic, essential elements to the group, from which every other element can be constructed. (These are called a generating set of the group.) This led me to believe, although I have not been able to prove it, that any reflection in the dihedral group of order 2can be written as the product of a single reflection x and some number of rotations by 360°/k degrees. (Call one rotation of 360/k° a.)
This observation lets me restate the problem (working in the dihedral group of order 2k) as follows:
I can then manipulate this expression into this form without changing the value of either side of the equation:
Note that I multiplied both sides by the inverse of x^2. But x^2 is the identity, so its inverse is also the identity, so I actually didn’t change the value of either side of the equation! This is important. (Just as a note, this works specifically because x is its own inverse, so I could have just replaced x with x^-1 without further justification. But this makes it even clearer that this step is justified.)
Once I have it in this form, I can use a second observation I gleaned from the GroupProps website, specifically, that
(Again, I don’t know how to prove it, but it makes intuitive sense if you play around with some concrete examples to see it in action.)
Edit (Jan. 31): I did find a way to prove it, and it’s actually really easy and I don’t quite see how I missed it. There’s absolutely no trick to it, all you have to do is write out the multiplication:
The last equality holds because the xs cancel with the x^-1 s that are right next to them.
That fact lets me rewrite the equation as follows, again, without changing the value of either side:
Now I just have to solve the modular congruence n-m=m-n mod 2k, which I’m comfortable doing thanks to my GOA Number Theory course last semester. I get that either n=m mod 2k, or n-m=m-n=k mod 2k. But the first possibility is actually not a possibility, because we assumed that the two rotations were distinct, and this would imply that they are the same rotation. So we have that n-m=m-n=k mod 2k.
Recall that I never changed the value of either side of the equation through all of my manipulations. Considering just the left hand side of all of the equations, this means that I have
My notation was different from the book’s notation, but this equality is exactly what I was asked to prove. I included the book’s notation under my equation for clarity.
I’m not sure that I took the most direct method, and I know that there are a lot of “holes” in my reasoning, insofar as there are facts that I used that I haven’t seen proven. But I think that this method does work regardless.
One response to “A Solution to Exercise 42”
1. Tina Bessias
Congratulations on finding this solution, Aram! It’s fun to see this post so soon after hearing you talk about the problem in our meeting yesterday.
One blogging suggestion: how about including a link to the Group-Props site? | 1,001 | 4,411 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2020-16 | latest | en | 0.972301 |
https://www.beatthegmat.com/search.php?st=0&sk=t&sd=d&sr=posts&author_id=89765-45.html | 1,600,910,772,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400213006.47/warc/CC-MAIN-20200924002749-20200924032749-00652.warc.gz | 741,920,928 | 9,939 | ## Search found 232 matches
#### Number System
##### Number System
A Number when divided by 225 gives a remainder of 32.What will be the remainder when same number is divided by 15?
a.4
b.3
c.2
d.1
by ruplun
Thu Jan 09, 2020 10:09 am
Forum: Problem Solving
Topic: Number System
Replies: 2
Views: 228
#### Number System
##### Number System
TWO numbers when divided by a certain divisor gives remainders 35 and 20 respectively and when their sum is divided by the same divisor, the remainder is 15.
by ruplun
Thu Jan 09, 2020 10:07 am
Forum: Problem Solving
Topic: Number System
Replies: 0
Views: 115
#### Twelve years ago and again five years ago, there were extend
###### Critical Reasoning
Twelve years ago and again five years ago, there were extended periods when the Darfir Republic's currency,the pundra, was weak: its value was unusually low relative to the world's most stable currencies. Both times a weak pundra made Darfir's manufactured products a bargain on world markets, and D...
by ruplun
Sat Jun 08, 2013 8:50 pm
Forum: Critical Reasoning
Topic: Twelve years ago and again five years ago, there were extend
Replies: 8
Views: 5617
###### Sentence Correction
The article explained that there is evidence that the dolphins suffer in captivity, living only a average of 12 years, with some even committing suicide. a) evidence that the dolphins suffer in captivity, living only a average b) ample proof that dolphins living in captivity only survive an average ...
by ruplun
Sun Mar 31, 2013 7:27 am
Forum: Sentence Correction
Replies: 1
Views: 608
#### will it be is or are
##### will it be is or are
Thirteen percent of Cleveland’s teens are pregnant.
I feel the verb should be is "is".
Do confirm...
by ruplun
Wed Mar 06, 2013 11:49 pm
Forum: Sentence Correction
Topic: will it be is or are
Replies: 1
Views: 801
#### Query
##### Query
Am posting GMAT Verbal topics but fail to get answer ..its happening for the past one year.Is this forum closed?
This is urgent ....
by ruplun
Tue Mar 05, 2013 3:33 am
Topic: Query
Replies: 4
Views: 1566
#### gmat aspirant from Kolkata
##### gmat aspirant from Kolkata
Hi
Can we for a group study in kolkata? Do let me know
by ruplun
Tue Jan 15, 2013 8:51 pm
Forum: Let's Meet Up
Topic: gmat aspirant from Kolkata
Replies: 1
Views: 952
#### Allen's study
According to the passage, which of the following was true of most villages in seventeenth-century England? (A) The resident squire had significant authority. (B) Church members were selected on the basis of their social status within the community. (C) Low population density restricted agricultural ...
by ruplun
Mon Dec 24, 2012 2:13 am
Topic: Allen's study
Replies: 8
Views: 5522
#### Partoria, large trucks
##### Partoria, large trucks
In the nation of Partoria, large trucks currently account for 6 percent of miles driven on Partoria’s roads but are involved in 12 percent of all highway fatalities. The very largest trucks—those with three trailers—had less than a third of the accident rate of single- and double-trailer truck...
by ruplun
Mon Dec 17, 2012 11:44 pm
Forum: Critical Reasoning
Topic: Partoria, large trucks
Replies: 3
Views: 1094
#### Borania people
##### Borania people
In Borania many people who want to quit smoking wear nicotine skin patches, which deliver small doses of nicotine through the skin. Beginning next month, these patches can be purchased without a doctor's prescription. Although nonprescription patches will be no more effective than those obtained by ...
by ruplun
Mon Dec 17, 2012 10:05 pm
Forum: Critical Reasoning
Topic: Borania people
Replies: 0
Views: 833
#### Columnist: People should avoid using a certain artificial
###### Critical Reasoning
I fail to understand the reasoning behind (B).I feel the correct answer to be B...please elaborate .
The logic being :If we dont eat the food that this artificial fat has vitamins , then the problem is solved.
by ruplun
Mon Oct 08, 2012 2:53 am
Forum: Critical Reasoning
Topic: Columnist: People should avoid using a certain artificial
Replies: 8
Views: 5160
#### Veritas books
##### Veritas books
Please can u tell me how are Veritas books ... I plan to buy them 4 preparation , do u think it will be helpful..Please suggest.
by ruplun
Wed Jun 13, 2012 7:29 pm
Topic: Veritas books
Replies: 1
Views: 1750
#### MGMAT CR
###### Critical Reasoning
Antoine: The alarming fact is that among children aged 19 years and younger, the number taking antipsychotic medicines soared 73 percent in the last four years. That is greater than the increase in the number of adults taking antipsychotic medicines during the same period. Lucy: But the use of anti...
by ruplun
Wed Jun 06, 2012 7:54 am
Forum: Critical Reasoning
Topic: MGMAT CR
Replies: 27
Views: 4119
#### ds
##### ds
[/img]
by ruplun
Fri Mar 09, 2012 8:35 am
Forum: Data Sufficiency
Topic: ds
Replies: 1
Views: 609
#### consecutive
##### consecutive
If m and n are two consecutive positive integers, is m \gt n ? 1. m-1 and n+1 are consecutive positive integers 2. m is an even integer (a)let m=3 , n =4 then m-1 =2 and n+1 =5 , 2 and 5 not consecutive integers.Hence insuff let m=6, n =7 then m-1 = 5 and n+1=8.insuff (b)m=2,n=3 m=6,n=5.insufficient...
by ruplun
Fri Mar 09, 2012 8:25 am
Forum: Data Sufficiency
Topic: consecutive
Replies: 2
Views: 618 | 1,499 | 5,387 | {"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-2020-40 | latest | en | 0.907334 |
http://techie-buzz.com/tech-news/google-scientific-calculator.html | 1,582,126,908,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144165.4/warc/CC-MAIN-20200219153707-20200219183707-00275.warc.gz | 144,243,398 | 7,409 | Math nerds, heads up. Google has just added a full featured scientific calculator to Google search, thereby allowing you to perform complex scientific and trigonometrical calculations using only your mouse. Head over to Google.com, type in a simple calculation and you should see the 34 button calculator popping in, an example is shown below:
This is so very cool, far better than how Google used to show mathematical results previously. Moreover, you don’t have to type the exact expression in the search box over and over again. Once the scientific calculator shows up, you can just key in numbers directly in the calculator and perform mathematical calculations on the fly. Now that this has happened, I won’t mind unpinning Windows calculator from the taskbar. (related: use Google Chrome as a calculator)
Google’s advanced calculator supports trigonometrical, logarithmic, factorial and basic arithmetic functions e.g Pi(π) and factorial of a number (!). Classical Google specific calculation techniques have gone nowhere, Google has only aggrandized the functionality by amalgamating all the core methods into a newer calculator UI that just works.
For example, you can type “square root of [number]” (without quotes) to find the square root of any real number through Google search. That still works, only difference is that now you get the result displayed in the calculator itself.
However, here is a little catch.
The new calculator does not support square root of negative integers. At least, not yet. Try finding the square root of -1 or any other negative integer in the new scientific calculator of Google and you’ll get an error. (high school tip: square root of a negative number is the same as square root of a positive number, just add the imaginary “i” to it.)
But wait, you can still apply the old technique to get the square root of a negative number from Google search. Here is how:
The best part : Google’s improved calculator supports voice commands, so you can use your microphone to type in numbers and the calculator will “listen”. It might take you some time to get accustomed to but once you master the operation, you will love it even more.
The new scientific calculator is by no means limited to PC’s or desktop computers; it works across all mobile devices e.g iOS, Android and so forth. Here is how the calculator looks when viewed from Android (just rotate your phone to get the full view)
Although separate from the new calculator UI, I am more excited about how Google has revamped the display of unit conversions. Previously, Google used to display the result of a conversion in the search result page, which has now been organized under a neat tabular format.
So when I am scratching my head over a math problem inundated with half a dozen different units, I can use the dropdown menu on the unit conversion page and convert the values in real time. Speed, mass, length, volume, area. temperature, time – Google has got you covered. | 618 | 2,982 | {"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-2020-10 | latest | en | 0.885093 |
https://imathworks.com/physics/physics-classical-quantum-coin-toss/ | 1,721,610,372,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517805.92/warc/CC-MAIN-20240722003438-20240722033438-00219.warc.gz | 267,517,217 | 7,568 | # [Physics] Classical/Quantum Coin Toss
classical-mechanicsprobabilityquantum mechanics
I am having a brainfreeze moment and have confused myself, help appreciated!
Quantum Coin: Superposition Heads AND Tails.
Classical Mechanics: Deterministic (in principle, if not in practice) if I repeat the same experiment I get the same result.
Quantum Mechanics: Non-deterministic no way that I can predict if I get heads or tails.
Now think of some physical implementation of a quantum coin perhaps I send some electron to mirror, afterwards it is on a superposition on both sides of the mirror. Perhaps reflected (heads) with probability 0.9 and transmitted (tails) with probability 0.1.
My question is does a classical analogy exist here? It can't be both deterministic and agree with the probabilities predicted by quantum mechanics right? Is the problem just that I should not be applying classical physics at all here? Does this question even make any sense?
3) Quantum correlations are stronger than classical correlations, this is because, in quantum mechanics, we work with probabilites complex amplitudes $\psi$, instead of working directly with probabilites $p$ (The relation is $p = |\psi|^2$). Some experiment results cannot be explained by classical correlations.
4) If you consider, for instance, a superposition 1-spin quantum state like $\psi = |+_z>$ + $|-_z>$, a measurement of the spin on the $z$ axis will gives you always $+1$ OR $-1$. So, from the point of view of the measurement, it is an OR, it is not a AND. You will have 50% probability to measure +1, and 50% probability to measure -1. | 366 | 1,613 | {"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} | 2.625 | 3 | CC-MAIN-2024-30 | latest | en | 0.909653 |
https://www.developerfaqs.com/4384/for-each-element-of-the-list-find-closest-date-from-a-different-list | 1,571,106,164,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986655735.13/warc/CC-MAIN-20191015005905-20191015033405-00220.warc.gz | 812,122,689 | 8,968 | # For each element of the list find closest date from a different list
• A+
Category：Languages
I have 2 lists:
``l1 = [ '09/12/2017', '10/24/2017' ] l2 = [ '09/15/2017', '10/26/2017', '12/22/2017' ] ``
For every ticker in l1 I want to find the closest element from l2 after it, so the output should be
``l3 = [ '09/15/2017', '10/26/2017' ] ``
The right way seems to be to explicitely iterate in parallel over both lists in reverse order, but I was hoping for a more "pythonic" solution..
EDIT: I do want an optimal complexity solution, which (assuming the lists are sorted), I think is O(max(len(l1), len(l2))).
You could use a list comprehension in combination with `min` method by passing a `lambda` expression.
``from datetime import datetime l1 = [ '09/12/2017', '10/24/2017' ] l2 = [ '09/15/2017', '10/26/2017', '12/22/2017' ] l1 = [min(l2, key=lambda d: abs(datetime.strptime(d, "%m/%d/%Y") - datetime.strptime(item, "%m/%d/%Y"))) for item in l1] ``
Output
``['09/15/2017', '10/26/2017'] ``
If you want a more efficient solution you can write your own `insert` sort algorithm.
``def insertSortIndexItem(lst, item_to_insert): index = 0 while index < len(lst) and item_to_insert > lst[index]: index = index + 1 return lst[index] l2 = sorted(l2, key=lambda d: datetime.strptime(d, "%m/%d/%Y")) l1 = [insertSortIndexItem(l2, item) for item in l1] `` | 442 | 1,376 | {"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-2019-43 | latest | en | 0.741853 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Exercises%3A_General_Chemistry/Exercises%3A_Gray/Homework_05 | 1,723,624,016,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641104812.87/warc/CC-MAIN-20240814061604-20240814091604-00810.warc.gz | 121,704,881 | 31,193 | # Homework 5: Crystal Field Splitting
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These homework problems are suggested and will not be turned in for review. However, answers will be available for them the following week by your class TAs. For more homework feel free to go to the Homework page.
## Q14
The Co2+ ion in aqueous solution is octahedrally coordinated and paramagnetic, with three unpaired electrons. Which one or ones of the following statements follow from this observation:
1. Co(H2O)42+ is square planar
2. Co(H2O)42+ is tetrahedral
3. Co(H2O)62+ has a Δ0 that is larger than the electron-pairing energy;
4. the d levels are split in energy and filled as follows: (t2g)5(eg)2
5. the d levels are split in energy and filled as follows: (t2g)6(eg)1
## Q15
The coordination compound potassium hexafluorochromate(III) is paramagnetic. What is the formula for this compound? What is the configuration of the Cr d electrons?
## Q16
How many unpaired electrons are there in Cr3+, Cr2+, Mn2+, Fe2+, Co3+, Co2+ in
1. a strong octahedral ligand field and
2. a very weak octahedral field?
## Q19
What is the d-orbital electronic configuration of Cr(NH3)63+? How many unpaired electrons are present? If six Br- groups were substituted for the six NH3 groups to give CrBr63-, would you expect Δ0 to increase or decrease?
## Q21
For each of the following, sketch the d-orbita1 energy levels and the distribution of d electrons among them:
1. Ni(CN)42- (square planar)
2. Ti(H2O)62+ (octahedral)
3. NiCl42- (tetrahedra1)
4. CoF63- (high-spin complex)
5. Co(NH3)63+ (low—spin complex)
## Q23
Pt(II) can occur in the complex ion PtCl42-.
1. What is the geometry of this ion? In the valence bond theory, what Pt orbitals are used in making bonds to the Cl- ions?
2. What is the systematic name for the sodium salt of this ion?
3. Using crystal field theory, draw the d-electron configuration for this ion. ls the ion paramagnetic or diamagnetic?
4. Pt(II) can be oxidized to Pt(IV). Draw the d-electron configuration for the chloride complex ion of Pt(IV). Explain the difference between this configuration and that of Pt(II), Is the Pt(IV) chloride complex ion paramagnetic or diamagnetic?
## Q24
A solution is prepared that is 0.025M in tetraamminecopper(II), Cu(NH3)42+. What will be the concentration of Cu2+ hydrated copper ion if the ammonia concentration is 0.10, 0.50, 1.00, and 3.00 M respectively? What ammonia concentration is needed to keep the Cu2+ concentration less than 10-15 M?
## Q27
The ion Co(NH3)63+ is very stable, with Kf = 2.3 x 1034. If the hydrolysis constant for the ammonium ion, Kb, is 5 x 10-10, show that the equilibrium in the reaction
Co(NH3)63+ + 6H+ ↔ Co3+ + 6NH4+
lies far to the right. Then why does Co(NH3)63+ remain intact in hot concentrated sulfuric acid?
## Q29
What is the solubility of Cu(OH)2 in pure water? In buffer at pH 6? Copper(II) forms a complex with NH3, Cu(NH3)42+, with Kf = 1.0 x 1012. What concentration of ammonia must be maintained in a solution to dissolve 0.10 mole of Cu(OH)2 per liter of solution?
## Q31
Predict the electron configuration of an octahedral d4 complex with
1. strong field ligands and
2. weak field ligands, and state the number of unpaired electrons
## Q32
Predict which of the following complexes absorbs light of the shorter wavelength and explain your reasoning: [Co(H2O)6]3+ or [Co(en)3]3+.
## Q33
Compare the magnetic properties of [Fe(H2O)6]2+ and [Fe(CN)6]4-.
## Q34
What change in magnetic properties (if any) can be expected when NO2- ligands in an octahedral complex are replaced by Cl- ligands in a d6 complex?
## Q35
Draw the orbital splitting diagram for the following complex and give its electron configuration: tetrahedral CoCl42-.
## Q36
The complex ion PdCl42- is diamagnetic. Propose a structure for PdCl42-.
## Q37
Explain the following differences in color:
1. For [Cr(H2O)6]Cl3 and [Cr(NH3)6]Cl3, one complex is violet while the other yellow Match the expected color with each complex and explain your identifications.
2. For [Co(H2O)6]2+ and [CoCl4]2-, one complex is blue while the other reddish. Correlate a color with each complex and explain your identifications.
3. One of the following solids is yellow, and the other is green: Fe(NO3)2·6H2O versus K4[Fe(CN)6·3H2O]. Indicate which is which and explain your reasoning.
Homework 5: Crystal Field Splitting is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts. | 3,036 | 8,812 | {"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": 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} | 2.796875 | 3 | CC-MAIN-2024-33 | latest | en | 0.200623 |
https://www.convertunits.com/from/siriometer/to/picometer | 1,643,188,270,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304928.27/warc/CC-MAIN-20220126071320-20220126101320-00204.warc.gz | 742,112,360 | 17,138 | ## ››Convert siriometre to picometre
siriometer picometer
How many siriometer in 1 picometer? The answer is 6.684587153547E-30.
We assume you are converting between siriometre and picometre.
You can view more details on each measurement unit:
siriometer or picometer
The SI base unit for length is the metre.
1 metre is equal to 6.684587153547E-18 siriometer, or 1000000000000 picometer.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between siriometers and picometers.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of siriometer to picometer
1 siriometer to picometer = 1.4959787E+29 picometer
2 siriometer to picometer = 2.9919574E+29 picometer
3 siriometer to picometer = 4.4879361E+29 picometer
4 siriometer to picometer = 5.9839148E+29 picometer
5 siriometer to picometer = 7.4798935E+29 picometer
6 siriometer to picometer = 8.9758722E+29 picometer
7 siriometer to picometer = 1.04718509E+30 picometer
8 siriometer to picometer = 1.19678296E+30 picometer
9 siriometer to picometer = 1.34638083E+30 picometer
10 siriometer to picometer = 1.4959787E+30 picometer
## ››Want other units?
You can do the reverse unit conversion from picometer to siriometer, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Picometre
The SI prefix "pico" represents a factor of 10-12, or in exponential notation, 1E-12.
So 1 picometre = 10-12 metre.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 576 | 1,977 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-05 | latest | en | 0.76047 |
http://www.tesla-institute.com/index.php/plc-areas-of-application | 1,611,068,247,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519395.23/warc/CC-MAIN-20210119135001-20210119165001-00703.warc.gz | 182,795,993 | 10,262 | ## PLC - Areas of application
Every machine or system has a controller. Dependent on the technology type used, controllers can be separated into hydraulic, pneumatic, electronic and electrical controllers. Often, a mixture of different technologies is applied. Moreover, differentiation is created between hard-wired programmable (e.g. wiring of electro-mechanical or electronic components) and PLCs. The initial is utilized principally in cases, where any reprogramming by the user is out of the query and the task size guarantees the development of a special controller.
Characteristic applications for such controllers can be found in cars, video cameras, and automatic washing machines. Nevertheless, if the task size does not guarantee the development of a special controller or if the user is to include the facility of setting timers and counters, or of making easy or independent program changes, then a universal controller use, where the program is written to a memory of electronic, is the ideal option? The PLCstands for such a universal controller. It can be applied for different applications and, through the program installed in its memory, offers the user with an easy means of changing, expanding and optimizing control processes.
The creative task of a PLC engaged the input signals interconnection along with a specified program and, if "true", to switch the corresponding output. Boolean algebra forms the basis of mathematical for this operation, which recognizes accurately two defined statuses of one variable: "0" and "1". Consequently, an output can only think these two statuses. For example, a linked motor could thus be either switched on or off, i.e. controlled.
This function has coined the name PLC: Programmable logic controller, i.e. the behavior ofinput/output is related to that of a pneumatic switching valve or electromagnetic relay controller; the program is saved in a memory of electronic. However, the PLC tasks have quickly multiplied: the functions of timer and counter, setting and resetting of memory, mathematical computing operations all stand for functions, which can be implemented by practically any of PLCs nowadays.
The requirement to be met by PLC‘s continued to grow up in line with their speedily spreading usage and the automation technology development. Visualization is the representation statuses of machine for instance the control program being executed, through display or monitor. Also controlling, i.e. the facility to intervene in control processes or, alternatively, to make such intervention by unauthorized persons impossible. It also became required to interconnect and harmonize individual systems controlled via PLC by means of automation technology. Therefore a master computer makes easy the means to issue higherlevel commands for program processing to some PLC systems.
The networking of some PLCs as well as that of a master computer and PLC is affected through special communication interfaces. To this effect, a lot of the more current PLCs are well-matched with open, standardized bus systems, for instance Profibus to EN 50170.
End of the seventies, binary inputs and outputs were finally extended with the analogue inputs and outputs addition, since many of today‘s technical applications need analogue processing such as speed setting, force measurement, servo-pneumatic positioning systems. At the same time, the analogue signals acquisition or output allows an actual/set point value comparison and as a result the automatic control engineering realization functions, a task, which broadly exceeds the scope suggested by the name as programmable logic controller.
The PLCs presently on offer in the market place have been modified to customer demands to such an extent that it has become possible to buy a highly suitable PLC for virtually any application. As such, miniatures PLCs are currently available with a minimum number of inputs/outputs beginning from just a few hundred Pounds. Also available are larger PLC swith 28 or 256 inputs/outputs. A lot of PLCs can be extended by means of additional input/output, positioning, communication and analogue modules. Special PLCs are available for shipping or mining, safety technology tasks. Yet further PLCs are capable to process numerous programs concurrently or multitasking. Lastly, PLCs are coupled with other automation components, accordingly creating significantly wider areas of application.
Top | 821 | 4,449 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-04 | longest | en | 0.905921 |
https://kindsonthegenius.com/blog/2018/03/what-are-non-parametric-tests.html | 1,550,748,131,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247504594.59/warc/CC-MAIN-20190221111943-20190221133943-00224.warc.gz | 601,648,589 | 11,147 | # What are Non-Parametric Tests?
Non-parametric tests which are also called distribution-free tests are applied when the distribution of the population is not known. In other words, non-parametric tests makes no assumption about the probability distributions of the observations.
Common non-Parametric Tests
Kolmogorov-Smirnov Test: This test is used to examine wether a sample is taken from a given distribution. In case of two samples, it tests if the two samples are taken from the same distribution.
Mann-Withney U (also called Wilcoxon rank sum test): Just like the Kolmogorov-Smirnov test, this is used to test wether two samples are drawn from the same distribution.
Wilcoxon signed-rank test: This is used to test wether matched pair samples are drawn from populations with different mean ranks.
Sign Test: Used to test whether matched pair samples are taken from populations with equal medians.
Kruskal-Wallis one-way ANOVA on ranks test: This is used to test if two samples are taken from the same distribution. It compares two samples which could be of equal or different sample sizes.
Chi-Square Tests:This test is used to determine whether there is significant difference in between the expected frequencies and the observed frequencies in one or more categories.
Categories of Non-Parametric Tests
Non-Parametric tests could be categorized into three different categories as shown below:
• Goodness of Fit Tests: In this categories, the distribution of the variable being analyzed is the same as hypothetical
• Tests for Independence: Here, the claim is that the rows and columns of variables being tested are independent.
• Tests for Homogeneity: In tests for homogeneity, the variables being analyzed are distributed equally
Table 1.0 summarized the three categories of non-parametric tests
Table 1.0: Summary of Non-Parametric Tests | 389 | 1,863 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2019-09 | longest | en | 0.908343 |
http://www.rodsmith.org.uk/alfred-korzybski/science-sanity%20-%200341.htm | 1,620,304,837,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988753.97/warc/CC-MAIN-20210506114045-20210506144045-00208.warc.gz | 95,807,038 | 3,262 | SCIENCE AND SANITY - online book
An Introduction To Non-aristotelian Systems And General Semantics.
ON FUNCTION 141 It is a genuine and fundamental semantic impasse. These static statements are very harmful, and yet they cannot be abolished, for the present. There are even weighty reasons why, without the formulation and application of oo-valued semantics, it is not possible (1933) to abolish them. What can be done under such structural circumstances ? Give up hope, or endeavour to invent methods which cover the discrepancy in a satisfactory (1933) way? The analysis of the psycho-logics of the mathematical propositional function and A semantics gives us a most satisfactory structural solution, necessitating, among others, a four-dimensional theory of propositions. We see (1933) that we can make definite and static statements, and yet make them semantically harmless. Here we have an example of abolishing one of the old A tacitly-assumed 'infinities'. The old 'general' statements were supposed to be true for 'all time'; in quantitative language it would mean for 'infinite numbers of years'. When we use the date, we reject the fanciful tacit A 'infinity' of years of validity, and limit the validity of our statement by the date we affix to it. Any reader who becomes accustomed to the use of this structural device will see what a tremendous semantic difference it makes psycho-logically. But the above does not exhaust the question structurally. We have seen that when we speak about oo-valued processes, and stages of processes, we use variables in our statements, and so our statements are not propositions but propositional functions which are not true or false, but are ambiguous. But, by assigning single values to the variables, we make propositions, which might be true or false; and so investigation and agreement become possible, as we then have something definite to talk about. A fundamental structural issue arises in this connection; namely, that in doing this (assigning single values to the variables), our attitude has automatically changed to an extensional one. By using our statements with a date, we deal with definite issues, on record, which we can study, analyse, evaluate., and so we make our statements of an extensional character, with all cards on the table, so to say, at a given date. Under such extensional and limited conditions, our statements then become, eventually, propositions, and, therefore, true or false, depending on the amount of information the maker of the statements possesses. We see that this criterion, though difficult, is feasible, and makes agreement possible. A structural remark concerning the -system may not be amiss here. In the -system the 'universal' proposition (which is usually a propositional function) always implies existence. In A 'logic', when it is | 579 | 2,836 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-21 | latest | en | 0.902833 |
http://www.wyzant.com/resources/answers/9827/how_would_i_work_out_the_answer_to_this_algebraic_equation | 1,405,124,658,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1404776429991.81/warc/CC-MAIN-20140707234029-00075-ip-10-180-212-248.ec2.internal.warc.gz | 653,864,640 | 11,224 | Search 75,797 tutors
0 0
## How would I work out the answer to this algebraic equation?
8x+6=4x+26
step by step how to would be nice
8x+6=4x+26
First, you want to get all the variables you are trying to solve for on one side
so subtract 4x from both side
8x+6=4x+26
-4x -4x
That leaves you with
4x+6=26 (because 4x-4x=0 so that term is eliminated)
Next, subtract six from both sides
4x+6=26
-6 -6
That leaves 4X=20 (again, 6-6=0 so that term is eliminated)
Now divide the whole thing by 4 and get X=5
8x + 6 = 4x + 26
Remember that you are solving for x
Therefore, place all the figures with x on one side and those without x
on the other side.
During the reshuffling, the signs are bound to change.
For instance, bringing 4x to the place of 8x changes the sign to -4x.
Likewise, transferring 6 to the position of 26 changes the sign to -6.
New equation: 8x - 4x = 26 - 6
4x = 20
x = 20/4
x = 5 | 314 | 926 | {"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-2014-23 | latest | en | 0.92093 |
https://www.physicsforums.com/threads/thrust-pressure.16299/ | 1,597,110,695,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738723.55/warc/CC-MAIN-20200810235513-20200811025513-00270.warc.gz | 802,205,043 | 15,902 | # Thrust & Pressure
## Main Question or Discussion Point
Hi there.
I am trying to create a realistic physics simulation for testing thrust based hovering vehicle designs and i am stuck.
I understand basic physics like moment of inertia and forces and the like but thrust is something i am having trouble with.
Currently if i place a vertical thruster at each corner of a rectangular craft, when the thrust force overcomes the weight of the craft it just kind of floats streight up forever.
As it is, the thrust force is just a variable being incremented and added to the rest of the force calculations.
The thrusters are supposed to be propeller style things expelling air at high speed.
The main problem is that i expected the craft to reach an equallibrium for a given amount of thrust rather than just floating on out into space. There is a huge hole in my knowledge of physics here and if you can provide any links to good resources or have any comments of your own, i would be most grateful for any help.
Related General Engineering News on Phys.org
russ_watters
Mentor
Originally posted by Nemos
Currently if i place a vertical thruster at each corner of a rectangular craft, when the thrust force overcomes the weight of the craft it just kind of floats streight up forever.
As it is, the thrust force is just a variable being incremented and added to the rest of the force calculations.
The thrusters are supposed to be propeller style things expelling air at high speed.
The main problem is that i expected the craft to reach an equallibrium for a given amount of thrust rather than just floating on out into space.
There is an equilibrium altitude, the problem is that the atmosphere is (depening on your definition) about 450,000 feet high, so the pressure/density gradient isn't very steep on the scale of a tabletop experiment. It may well be that a 1.01 thrust:weight ratio at sea level will take your craft to 1,000 feet before it drops to 1.00. The loss of thrust with altitude, however, is different for every engine.
enigma
Staff Emeritus
Gold Member
You will need to model an altitude control system.
If you don't mind oscillation about the desired point, something as simple as:
Thrust = (desired-actual)/desired * Full_Thrust
may do the trick.
Hi there.
Thanks for you input chaps.
I will try some of this out tonight now that i have some idea of what should be going on.
Thanks again.:0)
Cliff_J | 534 | 2,431 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2020-34 | latest | en | 0.934127 |
https://2022.help.altair.com/2022.1/activate/personal/en_us/topics/reference/oml_language/ControlSystem/c2d.htm | 1,716,192,707,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058254.21/warc/CC-MAIN-20240520080523-20240520110523-00100.warc.gz | 57,987,356 | 11,841 | c2d
Computes a discrete-time state-space model from a continuous-time state-space model.
Syntax
SYSD = c2d(SYSC, Ts)
SYSD = c2d(SYSC, Ts, METHOD)
SYSD = c2d(SYSC, Ts, 'prewarp', W0)
Inputs
SYSC
Continuous-time state-space model.
Ts
Sampling time in seconds.
METHOD
A string to specify the method used to convert the SYSC model to the SYSD model.
zoh
Zero-Order-Hold method (default method).
prewarp
Pre-warp method. This method requires W0.
tustin
Tustin approximation method.
bilin
Bilinear transformation.
W0
The pre-warp frequency.
Outputs
SYSD
A discrete-time state-space model.
Example
Default method:
A = [10.0 5;
5 10 ];
B = [0.0 -10;
10 0.0 ];
C = [-10 0.0;
0.0 10 ];
D = [10 0.0;
0.0 -10 ];
ResSys1 = c2d (ss (A, B, C, D), 2, 'zoh')
ResSys1 = object [
Scaled: 0
TimeUnit: seconds
ts: 2
a: [Matrix] 2 x 2
5.34324e+12 5.34324e+12
5.34324e+12 5.34324e+12
b: [Matrix] 2 x 2
3.56216e+12 -3.56216e+12
3.56216e+12 -3.56216e+12
c: [Matrix] 2 x 2
-10 0
0 10
d: [Matrix] 2 x 2
10 0
0 -10
e: [Matrix] 0 x 0
type: StateSpaceModel
]
Using the pre-warp method:
ResSys2 = c2d (ss (A, B, C, D), 2, 'prewarp',0.5) %For the same state space matrices given above
ResSys2 = object [
Scaled: 0
TimeUnit: seconds
ts: 2
a: [Matrix] 2 x 2
-1.28904 0.15908
0.15908 -1.28904
b: [Matrix] 2 x 2
1.17581 2.13639
-2.13639 -1.17581
c: [Matrix] 2 x 2
2.13639 -1.17581
1.17581 -2.13639
d: [Matrix] 2 x 2
1.30930 -15.79057
-15.79057 -18.69070
e: [Matrix] 0 x 0
type: StateSpaceModel
]
Computes a discrete-time state-space model from a continuous-time state-space model.
SYSD = c2d(SYS, Ts, METHOD) computes the discrete-time approximation of the continuous-time systems models SYS, with sampling period T's using METHOD.
Based on the SLICOT library functions ab04md and expm. | 723 | 1,796 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-22 | latest | en | 0.567846 |
https://www.mycoursehelp.com/QA/calculus-hw/47840/1 | 1,632,041,562,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056752.16/warc/CC-MAIN-20210919065755-20210919095755-00468.warc.gz | 955,440,735 | 8,770 | ### Create an Account
Home / Questions / Calculus hw
# Calculus hw
1-Consider the given function. Evaluate the Riemann sum for 2 = x = 14, with six subintervals, taking the sample points to be left endpoints. L6 = 2-If f(x) = 3x2 - 2x, 0 = x = 3, evaluate the Riemann sum with n = 6, taking the sample points to be right endpoints. R6 = 3-The graph of a function g is given. a = 4 Consider the integral I when answering the following questions. (a) Estimate the integral I with six subintervals using right endpoints. R6 = (b) Estimate the integral I with six subintervals using left endpoints.
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1-Consider the given function. Evaluate the Riemann sum for 2 = x = 14, with six subintervals, taking the sample points to be left endpoints. L6 = 2-If f(x) = 3x2 - 2x, 0 = x = 3, evaluate the Riemann sum with n = 6, taking the sample points to be right endpoints. R6 = 3-The graph of a function g is given. a = 4 Consider the integral I when answering the following questions. (a) Estimate the integral I with six subintervals using right endpoints. R6 = (b) Estimate the integral I with six subintervals using left endpoints. L6 = (c) Estimate the integral I with six subintervals using midpoints. M6 = 4-The table gives the values of a function obtained from an experiment. Use them to estimate 9f(x) dx3using three equal subintervals with right endpoints, left endpoints, and midpoints. x3456789f(x)-3.4-2.2-0.60.20.71.51.7(a) Estimate 9f(x) dx3using three equal subintervals with right endpoints. R3 = If the function is known to be an increasing function, can you say whether your estimate is less than or greater than the exact value of the integral? less than greater than one cannot say (b) Estimate 9f(x) dx3using three equal subintervals with left endpoints. L3 = If the function is known to be an increasing function, can you say whether your estimate is less than or greater than the exact value of the integral? less than greater than one cannot say (c) Estimate 9f(x) dx3using three equal subintervals with midpoints. M3 = If the function is known to be an increasing function, can you say whether your estimate is less than or greater than the exact value of the integral? less than greater than one cannot say 5-Express the limit as a definite integral on the given interval. lim n ? n[3 - 5(xi*)2 + 6(xi*)5] ?x, [4, 7]i = 17dx4 6- (a) 6g(x) dx0 (b) 18g(x) dx6 (c) 21g(x) dx0 7-If 7f(x) dx = 81and 7f(x) dx =...
Apr 29 2020 View more View Less | 836 | 2,615 | {"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} | 4.125 | 4 | CC-MAIN-2021-39 | latest | en | 0.818079 |
https://www.featuredanswer.com/aa12320933/physics-help-please-lateral-acceleration-66f2 | 1,660,439,970,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571989.67/warc/CC-MAIN-20220813232744-20220814022744-00169.warc.gz | 664,581,562 | 44,168 | ## Featured Answer is a community of 2,593,952 amazing learners
We're a place where learners ask for help for their tasks and share their knowledge.
# Physics help please! Lateral Acceleration?
The "cornering performance" of an automobile is evaluated on a skid pad, where the maximum speed that a car can maintain around a circular path on a dry, flat surface is measured. Then the centripetal acceleration, also called the lateral acceleration, is calculated as a multiple of the free-fall acceleration g. The main factors affecting the performance of the car are its tire characteristics and the suspension system. A certain car can negotiate a skidpad of radius 62.2 m at 86.0 km/h. Calculate its maximum lateral acceleration.
In the following work, 'ac' is centripetal acceleration, 'v' is the tangential velocity and 'r' is the radius of the curvature of the road.
First I will convert the units of velocity:
86.0 km/h = 23.9 m/s
ac = v^2/r
ac = (23.9)^2/(62.2)
ac = 9.18 m/s^2
Reference: Physics, AP Physics. | 253 | 1,025 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2022-33 | latest | en | 0.929126 |
http://list.seqfan.eu/pipermail/seqfan/2017-August/017908.html | 1,638,078,392,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358469.34/warc/CC-MAIN-20211128043743-20211128073743-00296.warc.gz | 43,727,010 | 2,610 | # [seqfan] Re: Just a quick (but hard?) funny sequence idea
Chris Thompson cet1 at cam.ac.uk
Wed Aug 30 17:59:00 CEST 2017
```On Aug 27 2017, Gaurav Verma wrote:
>Here are the first 10 terms of the sequence:
>
>a(1) = 3
>a(2) = 3
>a(3) = 878
>a(4) = 11404
>a(5) = 11404
>a(6) = 595413
>a(7) = 1797640
>a(8) = 98274734
>a(9) = 298419478
>a(10) = work in progress, will update soon
The value for a(9) seems to be wrong: pi^298419478 = 31415926 95426383...
matches only 8 digits of pi.
It's easy to see that each term of the sequence is of the form q+1 where
p/q is a rather good approximation of log_10(pi). So good that is has to
be a convergent? A proof of that eludes me at the moment, but certainly
the values a(1) to a(8) correspond to convergents.
Assuming that conjecture, it is straightforward to compute more terms
using the continued fraction for log_10(pi) = [0,2,87,4,1,1,1,4,52,...]
a(1) to a(8) as above
a(9) = 198347106
a(10) = 8128636028
a(11) = 75041122922
a(12) = 922797637351
a(13) = 11598859508648
a(14) = 28036830572808
a(15) = 1213341301344107
a(16) = 21996765548122104
a(17) = 71928417857731452
a(18) = 240751079727999028
a(19) = 5127701092145711019
a(20) = 81320964235147379208
a(21) = 1224942164619356399124
a(22) = 7268332023480991015532
a(23) = 26242236697890514923907
a(24) = 1042421135892139605940710
a(25) = a(24)
a(26) = 44876593316757784085298300
a(27) = 1837855483715284868285348842
a(28) = 4146393986688580399663359468
a(29) = 14747720463039036730368089028
[Computations done with bc(1) using 100 decimal digits.]
If the conjecture is wrong, at least these values are upper limits for
each a(n).
--
Chris Thompson
Email: cet1 at cam.ac.uk
``` | 668 | 1,684 | {"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-2021-49 | latest | en | 0.716949 |
http://www.uvm.edu/~dhowell/StatPages/Resampling/random_sampling_and_random_assig.html | 1,432,700,325,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207928869.6/warc/CC-MAIN-20150521113208-00042-ip-10-180-206-219.ec2.internal.warc.gz | 812,025,844 | 3,977 | # Parametric and Resampling Statistics (cont):
## Random Sampling and Random Assignment
The major assumption behind traditional parametric procedures--more fundamental than normality and homogeneity of variance--is the assumption that we have randomly sampled from some population (usually a normal one). Of course virtually no study you are likely to run will employ true random sampling, but leave that aside for the moment. To see why this assumption is so critical, consider an example in which we draw two samples, calculate the sample means and variances, and use those as estimates of the corresponding population parameters. For example, we might draw a random sample of anorexic girls (potentially) given one treatment, and a random sample of anorexic girls given another treatment, and use our statistical test to draw inferences about the parameters of the corresponding populations from which the girls were randomly sampled. We would probably like to show that our favorite treatment leads to greater weight gain than the competing treatment, and thus the mean of the population of all girls given our favorite treatment is greater than the mean of the other population. But statistically, it makes no sense to say that the sample means are estimates of the corresponding population parameters unless the samples are drawn randomly from that (those) populations(s). (Using the 12 middle school girls in your third period living-arts class is not going to give you a believable estimate of U. S. (let alone world) weights of pre-adolescent girls.) That is why the assumption of random sampling is so critical. In the extreme, if we don't sample randomly, we can't say anything meaningful about the parameters, so why bother? That is part of the argument put forth by the resampling camp.
Of course, those of us who have been involved in statistics for any length of time recognize this assumption, but we rarely give it much thought. We assume that our sample, though not really random, is a pretty good example of what we would have if we had the resources to draw truly random samples, and we go merrily on our way, confident in the belief that the samples we actually have are "good enough" for the purpose. That is where the parametric folks and the resampling folks have a parting of the ways.
The parametric people are not necessarily wrong in thinking that on occasion nonrandom sampling is good enough. If we are measuring something that would not be expected to vary systematically among participants, such as the effect of specific stimulus variations on visual illusions, then a convenience sample may give acceptable results. But keep in mind that any inferences we draw are not statistical inferences, but logical inferences. Without random sampling we cannot make a statistical inference about the mean of a larger population. But on nonstatistical grounds it may make good sense to assume that we have learned something about how people in general process visual information. But using that kind of argument to brush aside some of the criticisms of parametric tests doesn't diminish the fact that the resampling approach legitimately differs in its underlying philosophy.
The resampling approach, and for now I mean the randomization test approach, and not bootstrapping, really looks at the problem differently. In the first place, people in that area don't give a "population" the centrality that we are used to assigning to it in parametric statistics. They don't speak as if they sit around fondly imagining those lovely bell-shaped distributions with numbers streaming out of them, that we often see in introductory textbooks. In fact, they hardly appear to think about populations at all. And they certainly don't think about drawing random samples from those imaginary populations. Those people are as qualified as you could wish as statisticians, but they don't worry too much about estimating parameters, for which you really do need random samples. They just want to know the likelihood of the sample data falling as they did if treatments were equally effective. And for that, they don't absolutely need to think of populations.
In the history of statistics, the procedures with which we are most familiar were developed on the assumption of random sampling. And they were developed with the expectation that we are trying to estimate the corresponding population mean, variance, or whatever. This idea of "estimation" is central to the whole history of traditional statistics--we estimate population means so that we can (hopefully) conclude that they are different and that the treatments have different effects.
But that is not what the randomization test folks are trying to do. They start with the assumption that samples are probably not drawn randomly, and assume that we have no valid basis (or need) for estimating population parameters. This, I think, is the best reason to think of these procedures as nonparametric procedures, though there are other reasons to call them that. But if we can't estimate population parameters, and thus have no legitimate basis for retaining or rejecting a null hypothesis about those parameters, what basis do we have for constructing any statistical test. It turns out that we have legitimate alternative ways for testing our hypothesis, though I'm not sure that we should even be calling it a null hypothesis.
This difference over the role of random sampling is a critical difference between the two approaches. But that is not all. The resampling people, in particular, care greatly about random assignment. The whole approach is based on the idea of random assignment of cases to conditions. That will appear to create problems later on, but take it as part of the underlying rationale. Both groups certainly think that random assignment to conditions is important, primarily because it rules out alternative explanations for any differences that are found. But the resampling camp goes further, and makes it the center point of their analysis. To put it very succinctly, a randomization test works on the logical principle that if cases were randomly assigned to treatments, and if treatments have absolutely no effect on scores, then a particular score is just as likely to have appeared under one condition than under any other. Notice that the principle of random assignment tells us that if the null hypothesis is true, we could validly shuffle the data and expect to get essentially the same results. This is why random assignment is fundamental to the statistical procedure employed. | 1,249 | 6,584 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2015-22 | latest | en | 0.929399 |
https://bigideasmathanswers.com/two-trains-passes-in-the-opposite-direction/ | 1,726,179,950,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651498.46/warc/CC-MAIN-20240912210501-20240913000501-00614.warc.gz | 109,866,210 | 42,746 | # Two Trains Passes in the Opposite Direction | Relative Speed of Two Trains Running in Opposite Direction
Learn about the concept of Two Trains Passing in Opposite Direction completely by referring to the entire article. Know How to Calculate Speed Time and Distance when Two Trains Run in Opposite Direction. Refer to the Formulas and Solved Examples on Two Trains Passes in Opposite Direction and get a good grip on it. Detailed Solutions provided for each and every problem makes it easy for you to understand the entire concept.
## How to find Relative Speed while Two Trains Running in Opposite Direction?
When Two Trains Passes through a Moving Object having a certain length in the Opposite Direction
Let us assume the Length of the faster train is l meters and the length of the slower train is m meters
Speed of faster train = x km/hr
Speed of slower train = y km/hr
Relative Speed = (x+y) km/hr
Time taken by faster train to cross the slower train = (l+m) m/(x+y) km/hr
Using this Simple Formula you can calculate the measures easily when they run on parallel tracks in the opposite direction.
### Solved Problems on Two Trains Running on Parallel Tracks in the Opposite Direction
1. Two trains of length 130 m and 100 m respectively are running at the speed of 52 km/hr and 40 km/hr on parallel tracks in opposite directions. In what time will they cross each other?
Solution:
Speed of faster train = 52 km/hr
Speed of slower train = 40 km/hr
Relative Speed of Trains = (52 km/hr – 40 km/hr)
= 12 km/hr
= 12*5/18
= 3.33 m/sec
Length of first train = 130 m
Length of Second Train = 100 m
Time taken by the two trains to cross each other = sum of the length of trains/relative speed of trains
= (130+100) m/12 km/hr
= 230 m/3.33 m/sec
= 69.06 sec
Therefore, Two Trains Crosses each other in 69.06 sec
2. Two trains 170 m and 145 m long are running on parallel tracks in the opposite directions with a speed of 50 km/hr and 40 km/hr. How long will it take to cross each other?
Solution:
Speed of faster train = 50 km/hr
Speed of slower train = 40 km/hr
Relative Speed of Trains = (50 km/hr +40 km/hr)
= 110 km/hr
= 110*5/18
= 30.5 m/sec
Length of first train = 170 m
Length of second train = 145 m
Time taken by two trains to cross each other = Sum of Length of Trains/Relative Speed of Trains
= (170+145) m/30.5 m/sec
= 315 m/30.5 m/sec
= 10.3 sec
3. Two trains travel in opposite directions at 50 km/hr and 30 km/hr respectively. A man sitting in the slower train passes the faster train in 12 s. The length of the faster train is?
Solution:
Speed of faster train = 50 km/hr
Speed of second train = 30 km/hr
Time taken to cross each other = 12 sec
Relative Speed of Trains = (50 Km/hr +30 Km/hr)
= 80 km/hr
Relative Speed of Trains in m/sec = 80*5/18
= 22.22 m/sec
Length of faster train = 22.22 m/sec * 12 sec
= 266.6 m
Therefore, the length of the faster train is 266.6 m | 787 | 2,935 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2024-38 | latest | en | 0.916733 |
http://milesquarefabricstudio.com/tag/area-model-multiplication-4th-grade-worksheet/ | 1,621,254,728,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991772.66/warc/CC-MAIN-20210517115207-20210517145207-00050.warc.gz | 30,849,434 | 18,236 | ## Milesquarefabricstudio Think Positive, Think In Math Grade 5
Home » Area Model Multiplication 4th Grade Worksheet
# Area Model Multiplication 4th Grade Worksheet
Math Grade 5 September 17, 2020.
Most volumes begin with an explanation of basic arithmetic operations namely: addition, subtraction, multiplication, and division. Reference tables are supplied to provide clues for quick mental arithmetic and mastery of math facts. When ready to be tested, the student can select a drill, which has Read More...
Math Grade 5 September 16, 2020.
If you are looking for printable worksheets for your preschool child, the array of choices can be a little intimidating. You may just be looking for a few pages to keep your child occupied with something more constructive than yet another half hour in front Read More...
## 2nd Grade Subtraction With Regrouping
Math Grade 5 September 16, 2020.
Patterns and sequencing and basic addition and subtraction should follow on from counting and number recognition. By the time your child is starting kindergarten or school, they should be able to count to 20 with ease, write numbers, do simple addition sums, and have some Read More...
##### Easy Math Problems For 2nd Graders
Math Grade 5 September 17, 2020.
It is important to work with your child to help establish an appropriate pace. Part of the benefit of interactive learning games is that parents can monitor their child has progress and see how well things are going. You may find yourself pleasantly surprised at Read More...
## Fun Second Grade Math Worksheets
Math Grade 5 September 17, 2020.
But another person used the stairs to reach the same floor. This person found it very easy and reach there with little effort. Compare this person to a student who knows all the basic concepts learned in elementary grades. To learn grade six or grade Read More...
Math Grade 5 September 17, 2020.
The 1st grade worksheets can also used by parents to bridge between kindergarten lessons & 2nd grade program. It happens on many occasions that children forget or feel unable to recollect the lessons learned at the previous grade. In such situations, 1st grade worksheets become Read More...
#### 2nd Grade Math Workbook Free
Math Grade 5 September 17, 2020.
Most volumes begin with an explanation of basic arithmetic operations namely: addition, subtraction, multiplication, and division. Reference tables are supplied to provide clues for quick mental arithmetic and mastery of math facts. When ready to be tested, the student can select a drill, which has Read More...
###### Comparing 3 Digit Numbers Worksheets 2nd Grade
Math Grade 5 September 17, 2020.
All my toys were one way or the other math related. I had puzzles, and tons of things Mom had me do as games on daily basis at home to get me ready for kindergarten! In fact, she continued guiding me towards being math friendly Read More... | 614 | 2,911 | {"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} | 2.890625 | 3 | CC-MAIN-2021-21 | latest | en | 0.940353 |
http://www.letime.net/b/e/ | 1,516,715,958,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891976.74/warc/CC-MAIN-20180123131643-20180123151643-00521.warc.gz | 507,033,839 | 2,792 | # EXAMPLE FROM MECHANICS OF THE FLUIDS
Law on the flows When one measures the speed to which a basin is emptied, one measures a variation of level of the basin. The reference mark is the bottom of the basin with like locates algebraic the zero. This water which is not any more in the basin left in a place located low only the algebraic value. Initially this water always exists, and is some share on a total reference mark. Thus the bottom of the basin is a balance and the formula of the flow will be the height starting factor of expo of less time on OJ which depends on the form of the evacuation and which varies. One should not lose sight of the fact that the final result is a balance of a value which should not in no case to be comparable to zero. Analyze of a step Formulate and analyzes particular case of the swirls. Drawn from the observation of the lock of Nourriguier, the swirls are transformed into solitary wave when they do not have enough place for dévelloper. This wave goes up the current. There is a high limit and a low limit in the appearance of a swirl. These limits depend the total head and on the flow. Cavitation (liquid/gas) can generate swirls which will go in the direction of the current. Always with the same law of the limits (pressure/speed). Experimental data measured on a flow of water: initial time of 0,01s since the zero are not defined, the height is 60 cm from where: H = 60; T = 0,01 S H = 55; T = 13,33 S H = 50; T = 16,58 S H = 45; T = 21,14 S H = 40; T = 27,08 S H = 35; T = 34,44 S H = 30; T = 42,27 S H = 25; T = 49,94 S H = 20; T = 59,32 S H = 15; T = 77,13 S H = 10; T = 119,27 S If one take the current formula simplified for reasons of concept (What is in addition advised by the books of analysis); the flow is equal to a constant, factor of the square root of the difference in pressure. Always for reasons of concept I thus release myself from the problems of dimension to thus compare only the function square root of the pressure height by a factor K (the exit being with a negligible pressure). If one takes the values referred to above and that one calculates the medium flow, one realizes that qelles that are dimensions of volume and property of flow which one applies to measurement there is a shift of factor four which pushes me to seek a new formula. It is true that the current formula gives an aspect resembling the real curve éffectuée by a flow, but in the application of my trade, I refined another proposal for a formula.In the final analysis the formula that I propose is an alternative of a naturally stable system. An alternative because it appears a constant. The result obtained being the level of water, there is the function of the flow in the function of the variation, these functions being naturally stable. Moreover if I vary the drain opening thus the flow, I can obtain a particular case which will correspond to the formula K factor of the square root the height if not the formula is the height of origin factor of expo of less time that OJ divides. This value OJ being it even a built variable of constant a factor of author: Andre pierre jocelyn Hypothesis on time | 754 | 3,158 | {"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-2018-05 | longest | en | 0.925883 |
http://www.gmathacks.com/integrated-reasoning/how-is-gmat-integrated-reasoning-scored.html | 1,430,900,661,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1430458431586.52/warc/CC-MAIN-20150501053351-00004-ip-10-235-10-82.ec2.internal.warc.gz | 414,517,203 | 5,412 | ### Bookshelf
Total GMAT Math Jeff's complete Quant guide, on sale now!
Total GMAT Verbal Everything you need to ace GMAT Verbal!
1,800 Practice Math Questions
GMAT Official Guide, with IR
OG Math | OG Verbal
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## How is GMAT Integrated Reasoning Scored?
###### May 13, 2012
You should follow me on Twitter. While you're at it, take a moment to subscribe to GMAT Hacks via RSS or Email.
Starting in early June, every administration of the GMAT will replace one of the two AWA essays with a 30-minute Integrated Reasoning (IR) section. You can read a few other articles on this site to learn more about the new format.
The GMAT IR section has nothing to do with your score between 200 and 800. That will still rely entirely on your Quant and Verbal performance. Instead, Integrated Reasoning is treated similar to the AWA--a separate section with a separate score.
Your performance on the Integrated Reasoning section will earn you a score between 1 and 8, inclusive. Only integer scores are possible, so it seems likely that the majority of test-takers will score a 3, 4, 5, or 6, with 7 and 8 reserved for the best performances and 1 and 2 kept for the weakest.
This separate score reinforces my belief that the Integrated Reasoning section will not be a crucial part of MBA admissions for some time yet. Admissions committees can use the Quant and Verbal scores like they always have, and take a "wait-and-see" approach to determine whether the IR score has additional value.
To the extent that MBA programs will look at your IR score in the next few admissions cycles, it is likely to be treated as little more than a box to be checked. Like an AWA score of 2.5 or a Quant score in the 30th percentile, an Integrated Reasoning score of 3, 4, or below probably won't impress anyone at a top program. But between a 6, 7, and an 8, there will be little difference. If you have a reasonably good score, application evaluators won't analyze it further.
Further on, admissions committees will probably discover that IR scores correlate highly with Quant and Verbal scores. Integrated Reasoning does reflect critical thinking skills--which is also a huge part of what the main GMAT sections evaluate, especially for test-takers scoring 650 and above. At that point, it's not about grammar rules and math formulas anymore. If you are well prepared for GMAT Quant and Verbal, you are likely to do well on IR, as well.
About the author: Jeff Sackmann has written many GMAT preparation books, including the popular Total GMAT Math, Total GMAT Verbal, and GMAT 111. He has also created explanations for problems in The Official Guide, as well as 1,800 practice GMAT math questions.
Total GMAT Verbal The comprehensive guide to the GMAT Verbal section. Recognize, dissect, and master every question type you'll face on the test. Everything you need, all in one place, including 100+ realistic practice questions. Click to read more. | 705 | 3,050 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2015-18 | longest | en | 0.924235 |
https://arcomalaga.org/and-pdf/527-vector-questions-and-answers-class-11-pdf-785-984.php | 1,653,533,822,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662595559.80/warc/CC-MAIN-20220526004200-20220526034200-00060.warc.gz | 166,910,849 | 10,031 | # Vector Questions And Answers Class 11 Pdf
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You are browsing Grade 11 questions. View questions in All Grades. Grade 11 Grade
Vectors Worksheet Answers This worksheet pairs well with the second-grade curriculum, which focuses heavily on Earth and space science. With just a few clicks of your mouse, you can transform the hard work you do into an organized, yet quality-driven work flow. The vectors in problem 8 are linearly dependent and cannot form a basis of anything.
All the exercise of Chapter 2 — Physics and Mathematics questions with Solutions to help you to revise complete Syllabus and Score More marks. The magnitudes of these vectors are 3 m and 4 m respectively. Find the resultant.
## Grade 11 Physics Solution
Find vector AB. State and prove parallelogram law of vector addition and determine magnitude and direction of resultant vector. State the characteristics of vector product. Position vector 20 Define the following terms. Resultant vector ii. Composition of vectors. What does it represent?
## Eleventh Grade (Grade 11) Vectors Questions
What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of path skate? If the round trip takes 10 min, what is the a net displacement, b average velocity, and c average speed of the cyclist? Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.
These Physics Lab Manual may be freely downloadable and used as a reference book. Learning does not mean only gaining knowledge about facts and principles rather it is a path which is informed by scientific truths, verified experimentally. The laboratory is important for making the study complete, especially for a subject like Science and Physics. CBSE has included the practicals in secondary class intending to make students familiarised with the basic tools and techniques used in the labs. By performing the experiments, students will know the concept in a better way as they can now view the changes happening in front of their eyes.
Chapter List. The cross product of two vectors is non-commutative whereas the dot product of two vectors is always commutative. The cross product of two perpendicular vectors has maximum magnitude whereas the dot product of two mutually perpendicular vectors is always zero. The cross product of two parallel vectors is null or zero vector whereas, the dot product of two parallel vectors is always equal to product of their magnitudes. The necessary condition for addition of vectors is both are must be vector or the vectors having the same, nature only can be added. Hence when angle between two vectors is then the resultant of the two vectors is equal to either of the vector.
Therefor weight has both magnitude and direction, and is a vector. Page Question Title. Solution continued 3. Power is the rate.
## NEET Vectors Important Questions
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These solutions for Physics And Mathematics are extremely popular among Class 11 Science students for Physics Physics And Mathematics Solutions come handy for quickly completing your homework and preparing for exams. A vector is defined by its magnitude and direction, so a vector can be changed by changing its magnitude and direction. If we rotate it through an angle, its direction changes and we can say that the vector has changed. No, it is not possible to obtain zero by adding two vectors of unequal magnitudes.
Practice the multiple choice questions to test understanding of important topics in the chapters. Download latest questions with answers for Physics Scalars and Vectors in pdf free or read online in online reader free. Students should practice the multiple choice questions to gain more marks in NEET exams. Download in pdf free by clicking on links below. You should also carefully go through the syllabus for Physics Scalars and Vectors and download MCQs for each topics which you have studied.
Она смотрела на обмякшее тело коммандера и знала, о чем он думает. Рухнул не только его план пристроить черный ход к Цифровой крепости.
### Unit: Vectors (Prerequisite)
Под визг покрышек, в снопе искр Беккер резко свернул вправо и съехал с дороги. Колеса мотоцикла подпрыгнули, ударившись о бетонное ограждение, так что он едва сумел сохранить равновесие. Из-под колес взметнулся гравий. Мотоцикл начал подниматься по склону. Колеса неистово вращались на рыхлой земле.
А неприятности немалые. - Ты сама видишь. Впервые за последний час она позволила себе улыбнуться. - Этих слов я и ждала от. Он пожал плечами: - Как только мы получим ключ, я проинформирую директора. Сьюзан не могла не поразить идея глобального прорыва в области разведки, который нельзя было себе даже представить.
Сьюзан не могла не поразить идея глобального прорыва в области разведки, который нельзя было себе даже представить. И он попытался сделать это в одиночку. Похоже, он и на сей раз добьется своей цели. Ключ совсем. Танкадо мертв. Партнер Танкадо обнаружен.
Но как только шифр будет взломан… - Коммандер, а не лучше ли будет… - Мне нужен ключ! - отрезал. Сьюзан должна была признать, что, услышав о Цифровой крепости, она как ученый испытала определенный интерес, желание установить, как Танкадо удалось создать такую программу. Само ее существование противоречило основным правилам криптографии. Она посмотрела на шефа. - Вы уничтожите этот алгоритм сразу же после того, как мы с ним познакомимся.
Главный криптограф АНБ испробовала все - подмену букв, шифровальные квадраты, даже анаграммы.
Все когда-то бывает в первый раз, - бесстрастно ответил Бринкерхофф. Она встретила эти слова с явным неодобрением. - Я все проверяю дважды. - Ну… ты знаешь, как они говорят о компьютерах.
В тот момент Сьюзан поняла, за что уважает Тревора Стратмора. Все эти десять лет, в штиль и в бурю, он вел ее за. Уверенно и неуклонно. Не сбиваясь с курса.
Лучшее, что мог сделать директор, - не мешать ему работать и наблюдать за тем, как коммандер творит свое чудо. Стратмор разработал план… и план этот Фонтейн не имел ни малейшего намерения срывать. ГЛАВА 75 Пальцы Стратмора время от времени касались беретты, лежавшей у него на коленях.
Скажи, что это .
0 Response | 1,621 | 6,633 | {"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} | 3.796875 | 4 | CC-MAIN-2022-21 | latest | en | 0.902767 |
https://www.snapxam.com/solver?p=%5Cint%5Cleft%28%5Cleft%28-21%5Ccdot1%5Cright%29%5Ccdot%20x%2B%5Cfrac%7B1%7D%7Bx%7D%5Cright%29dx&method=9 | 1,719,138,532,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862466.81/warc/CC-MAIN-20240623100101-20240623130101-00770.warc.gz | 852,693,040 | 13,782 | 👉 Try now NerdPal! Our new math app on iOS and Android
# Integrate $\int\left(-21x+\frac{1}{x}\right)dx$
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## Final answer to the problem
$-\frac{21}{2}x^2+\ln\left|x\right|+C_0$
Got another answer? Verify it here!
## Step-by-step Solution
How should I solve this problem?
• Integrate by substitution
• Integrate by partial fractions
• Integrate by parts
• Integrate using tabular integration
• Integrate by trigonometric substitution
• Weierstrass Substitution
• Integrate using trigonometric identities
• Integrate using basic integrals
• Product of Binomials with Common Term
• FOIL Method
Can't find a method? Tell us so we can add it.
1
Expand the integral $\int\left(-21x+\frac{1}{x}\right)dx$ into $2$ integrals using the sum rule for integrals, to then solve each integral separately
$\int-21xdx+\int\frac{1}{x}dx$
Learn how to solve problems step by step online.
$\int-21xdx+\int\frac{1}{x}dx$
Learn how to solve problems step by step online. Integrate int(-21x+1/x)dx. Expand the integral \int\left(-21x+\frac{1}{x}\right)dx into 2 integrals using the sum rule for integrals, to then solve each integral separately. The integral \int-21xdx results in: -\frac{21}{2}x^2. The integral \int\frac{1}{x}dx results in: \ln\left|x\right|. Gather the results of all integrals.
## Final answer to the problem
$-\frac{21}{2}x^2+\ln\left|x\right|+C_0$
## Explore different ways to solve this problem
Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more
SnapXam A2
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acsch | 817 | 2,159 | {"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} | 4.28125 | 4 | CC-MAIN-2024-26 | latest | en | 0.591662 |
https://r4r.co.in/mcqs/learn.php?test_id=1971&subid=656&subcat=NCERT%20Class%2012&test=MCQ%20Questions%20for%20Class%2012%20Physics%20Chapter%204%20Moving%20Charges%20and%20Magnetism | 1,656,563,771,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103661137.41/warc/CC-MAIN-20220630031950-20220630061950-00185.warc.gz | 515,553,720 | 6,519 | # NCERT Class 12/MCQ Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Sample Test,Sample questions
1.12 cm
2.16 cm
3.11 cm
4.18 cm
1.0°
2.45°
3.90°
4.180°
1. sin θ
2.cos θ
3.cot θ
4.tan θ
## Question: `A current carrying power line carries current from west to east. Then the direction of the magnetic field 2 m above it is :`
1.west to east
2. south to north
3.north to south
4.None of the above
1.neutrons
2.deutrons
3.β particles
4.α partifcles
1. 2π
2.π/2
3.π/4
4.1/π
1.-1
2.1
3.0
4.infinity
## Question: `Circular loop of radius 0.0157 m carries a current 2 A. The magnetic field at the centre of the loop is :`
1.1.57 × 10-3Wb/m²
2.8.0 × 10-5 Wb/m²
3.2.0 × 10-3 Wb/m²
4.3.l4 × 10-1 Wb/m²
1.Gauss
2.Oersted
3.Ampere
4.Gilbert
## Question: `Isoclinic lines are the lines joining places with :`
1.equal dip
2.equal declination
3.equal dip and declination
4.None of the above
## Question: `same everywhere`
1. same everywhere
2. inversely proportional to the distance
3.directly proportional to the distance
4.None of the above
1. electric flux
2. magnetic flux
3.magnetic field
4.electric field
## Question: `The current sensitibility of a moving coil galanometer increases with decrease in:`
1.magnetic field
2.area of a coil
3.number of turns
4.None of the above
1.4F
2.2F
3.F
4.F/4
1.B/4
2.B/2
3.4B
4.2B
1.iron
2.steel
3.copper
4.aluminium
## Question: `The permeability of a paramagnetic substance is:`
1. very large
2. small but more than unity
3.less than unity
4.negative
## Question: `The radius of the trajectory of a charged particle in a uniform magnetic field is proportional to the:`
1.charge on the particle
2.energy of the particle
3.momentum of the particle
4.All the above
## Question: `The sensitivity of a tangent galvanometer can be increased by increasing:`
2.the external magnetic field
3. the number of turns of the coil
4.all the above
## Question: `The SI unit of magnetic dipole moment is’`
1.Ampere
2.Ampere metre²
3.Tesla
4.None of the above
## Question: `The strength of the magnetic field around an infinite current carrying conductor is :`
1.same everywhere
2.inversely proportional to the distance
3.directly proportional to the distance
4.None of the above
## Question: `To convert galvanometer into voltmeter one should connect :`
1.high resistance in series with galvanometer
2.low resistance in series with galvanometer
3.high resistance in parallel with galvanometer
4. low resistance in parallel with galvanometer
## Question: `What happens to the magnetic field at the centre of a circular current carrying coil if we double the radius of the coil keeping the current unchanged?`
1.halved
2.doubled
4.remains unchanged
## Question: `What is the angle of dip at the magnetic poles ?`
1. 30°
2.0°
3.45°
4.None of the above
## Question: `When charged particle enters-a uniform magnetic field, its K.E.:`
1. remains constant
2.increases
3.decreases
4.becomes zero
## Question: `When we double the radius of a coil keeping the current through it unchanged, what happens to the magnetic field directed along its axis at far off points?`
1.halved
2.doubled
4.remains unchanged
1.gauss
2.tesla
3.oersted
4.weber/metre²
## Question: `Which of the following shows that the earth behaves as a magnet?`
1.Repulsion between like poles .
2.Attraction between unlike poles
3.Null points in the magnetic field of a bar magnet
4.No existence of isolated magnetic poles
## More MCQS
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R4Rin Top Tutorials are Core Java,Hibernate ,Spring,Sturts.The content on R4R.in website is done by expert team not only with the help of books but along with the strong professional knowledge in all context like coding,designing, marketing,etc! | 1,095 | 3,815 | {"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-2022-27 | latest | en | 0.731284 |
https://oeis.org/A296515 | 1,718,286,659,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861451.34/warc/CC-MAIN-20240613123217-20240613153217-00190.warc.gz | 395,035,883 | 5,488 | The OEIS mourns the passing of Jim Simons and is grateful to the Simons Foundation for its support of research in many branches of science, including the OEIS.
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A296515 Number of edges in a maximal planar graph with n vertices. 3
0, 0, 1, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99, 102, 105, 108, 111, 114, 117, 120, 123, 126, 129, 132, 135, 138, 141, 144, 147, 150, 153, 156, 159, 162, 165, 168, 171, 174 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,4 COMMENTS {a(n)} is a monotonic increasing sequence because a maximal planar graph of order n can be generated on n + 1 nodes. Therefore a(n) <= a(n + 1). Maximal planar graphs of order n > 5 are not unique. |E(G_2n)| = (2n - 1) + 2*Sum_{k=0..(floor(log_2(n - 1)))} floor((n - 1)/2^k) where |E(G_2n)| is the size of a minimal planar graph G of order 2n. Number of edges of a maximal 3-degenerate graph of order n (this class includes 3-trees). The intersection of this class and maximal planar graphs is the Apollonian networks (planar 3-trees); neither class contains the other. - Allan Bickle, Nov 14 2021 a(n) is the number of blocks to dig (in a staircase fashion) to get out of a hole of depth n in Minecraft. - Max R Anderson, Oct 19 2023 LINKS Stefano Spezia, Table of n, a(n) for n = 0..10000 Allan Bickle, Structural results on maximal k-degenerate graphs, Discuss. Math. Graph Theory 32 4 (2012), 659-676. Allan Bickle, A Survey of Maximal k-degenerate Graphs and k-Trees, Theory and Applications of Graphs 0 1 (2024) Article 5. Dawei He, Yan Wang, and Xingxing Yu, The Kelmans-Seymour conjecture I, arXiv:1511.05020 [math.CO], 2015. D. R. Lick and A. T. White, k-degenerate graphs, Canad. J. Math. 22 (1970), 1082-1096. Mathematics Stack Exchange, You dig a hole in Minecraft straight down n blocks, how many blocks must you dig to get out? K. Stephenson, Circle Packing: A Mathematical Tale, Notices Amer. Math. Soc. 50 (2003). Squid Tamar-Mattis, Planar Graphs and Wagner's and Kuratowski's Theorems, REU (2016). Eric Weisstein's World of Mathematics, Kuratowski Reduction Theorem, Polyhedral Graph David R. Wood, A Simple Proof of the Fary-Wagner Theorem, arXiv:cs/0505047 [math.CO], 2005. Index entries for linear recurrences with constant coefficients, signature (2,-1). FORMULA a(n) = floor(6/2^n) + 3n - 6 (see comments section of A008486). G.f.: x^2 + 3*x^3/(x - 1)^2. - R. J. Mathar, Apr 14 2018 E.g.f.: 6 + x*(x + 6)/2 + 3*exp(x)*(x - 2). - Stefano Spezia, Feb 13 2023 a(n) = 3*(n - 2) for n >= 3. - Max R Anderson, Oct 19 2023 MATHEMATICA CoefficientList[Series[(x^2 (1 + x + x^2))/(x - 1)^2, {x, 0, 60}], x] (* or *) LinearRecurrence[{2, -1}, {0, 0, 1, 3}, 60] (* Robert G. Wilson v, Mar 04 2018 *) CROSSREFS Cf. A008486, A008585, A242088. Sequence in context: A337244 A366847 A031193 * A008585 A008486 A135943 Adjacent sequences: A296512 A296513 A296514 * A296516 A296517 A296518 KEYWORD nonn,easy AUTHOR Joseph Wheat, Jonathan Sondow, Feb 28 2018 STATUS approved
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Last modified June 13 09:43 EDT 2024. Contains 373383 sequences. (Running on oeis4.) | 1,227 | 3,528 | {"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-2024-26 | latest | en | 0.764946 |
https://www.asrmeta.com/refractive-index-calculator-both-simple-and-complex/ | 1,722,666,396,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640361431.2/warc/CC-MAIN-20240803060126-20240803090126-00642.warc.gz | 523,232,005 | 26,082 | ## Inroduction to refractive index calculator
The refractive index describes the change in speed of light as it enters from air into the material under observation. Therefore, the refractive index is a material property and everything we see around in our daily lives is composed of different materials. All of the materials have a distinct refractive index and usually, this value of the refractive index varies depending upon the wavelength of light. This phenomenon is called dispersion of light, where the velocity of wave in a medium depends upon the frequency of the wave and these quantities can be related to the speed of light and wavelength, respectively. The refractive index (n) can be calculated using a simple formula that relates the speed of light in a material (v), and the speed of light in the air (c).
$n=\frac{c}{v}$
The speed of light in air is close to the speed of light in a vacuum, therefore, both of these values will yield almost the same result. The refractive index value from this formula will be in the range of 1 ≤ n ≤ ∞.
However, normally, all the materials show a complex refractive index value (N). This complex value of the refractive index is related to two properties of a material known as relative permittivity (εr) and relative permeability (μr). Permittivity and permeability define the ease of setting up electric and magnetic polarizations, respectively, in a material. The refractive index can be found in terms of these quantities as follows.
$N=\sqrt{\mu_r \epsilon_r}$
Both relative permittivity and relative permeability are also complex quantities and the refractive index calculator presented here also take that into account.
## Instructions on using refractive index calculator
1. Select the refractive index calculator depending upon your desired calculations. For simple calculations involving the speed of light in a material and speed of light in air, the default calculator (“Refractive index from speed of light”) will work. For calculations involving permittivity and permeability, select “Complex refractive index from permittivity and permeability”.
2. You will need to enter the real and complex parts of relative permittivity and relative permeability separately. The result will also be displayed in real and imaginary values, separately.
3. For small values of speed of light in a material, the refractive index will yield high values, which might not be accurate. REMEMBER, the speed of light you have to enter should be in meters per second (m/s).
4. For the cases where the imaginary values of relative permittivity and relative permeability cancel each other, the resultant imaginary part refractive index will have no value, as it should be.
5. The refractive index calculator calculates the refractive index at a particular wavelength. For calculation at different wavelength, reinsert the new values.
6. This calculator is a dynamic calculator, i.e., the results which change when new values are inserted or already entered values are changed. | 614 | 3,025 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-33 | latest | en | 0.888456 |
https://www.acmicpc.net/problem/6747 | 1,503,435,643,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886112682.87/warc/CC-MAIN-20170822201124-20170822221124-00659.warc.gz | 858,815,024 | 15,586 | 시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율
1 초 128 MB 0 0 0 0.000%
## 문제
There are N beads which of the same shape and size, but with different weights. N is an odd number and the beads are labeled as 1, 2, ..., N. Your task is to find the bead whose weight is median (the ((N+1)/2)th among all beads). The following comparison has been performed on some pairs of beads:
A scale is given to compare the weights of beads. We can determine which one is heavier than the other between two beads. As the result, we now know that some beads are heavier than others. We are going to remove some beads which cannot have the medium weight.
For example, the following results show which bead is heavier after M comparisons where M=4 and N=5.
From the above results, though we cannot determine exactly which is the median bead, we know that Bead 1 and Bead 4 can never have the median weight: Beads 2, 4, 5 are heavier than Bead 1, and Beads 1, 2, 3 are lighter than Bead 4. Therefore, we can remove these two beads.
Write a program to count the number of beads which cannot have the median weight.
## 입력
The first line of the input file contains a single integer t (1 ≤ t ≤ 11), the number of test cases, followed by the input data for each test case. The input for each test case will be as follows:
The first line of input data contains an integer N (1≤N≤99) denoting the number of beads, and M denoting the number of pairs of beads compared. In each of the next M lines, two numbers are given where the first bead is heavier than the second bead.
## 출력
There should be one line per test case. Print the number of beads which can never have the medium weight.
## 예제 입력
1
5 4
2 1
4 3
5 1
4 2
## 예제 출력
2 | 456 | 1,697 | {"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} | 3.140625 | 3 | CC-MAIN-2017-34 | longest | en | 0.950357 |
https://www.12000.org/my_notes/Murphy_ODE/reports/maple_2021_1_and_mma_12_3_1/KERNELsubsubsection96.htm | 1,716,847,294,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059045.34/warc/CC-MAIN-20240527205559-20240527235559-00225.warc.gz | 537,908,566 | 3,329 | ##### 4.2.46 $$y'(x)=f(x) y(x)+g(x) y(x)^k$$
ODE
$y'(x)=f(x) y(x)+g(x) y(x)^k$ ODE Classification
[_Bernoulli]
Book solution method
The Bernoulli ODE
Mathematica
cpu = 0.351682 (sec), leaf count = 67
$\left \{\left \{y(x)\to \left (\exp \left (-\left ((k-1) \int _1^xf(K[1])dK[1]\right )\right ) \left (-(k-1) \int _1^x\exp \left ((k-1) \int _1^{K[2]}f(K[1])dK[1]\right ) g(K[2])dK[2]+c_1\right )\right ){}^{\frac {1}{1-k}}\right \}\right \}$
Maple
cpu = 0.117 (sec), leaf count = 81
$\left [y \left (x \right ) = \left (\int \left (-k \,{\mathrm e}^{\int \left (f \left (x \right ) k -f \left (x \right )\right )d x} g \left (x \right )+{\mathrm e}^{\int \left (f \left (x \right ) k -f \left (x \right )\right )d x} g \left (x \right )\right )d x +\textit {\_C1} \right )^{-\frac {1}{k -1}} {\mathrm e}^{\frac {\left (\int f \left (x \right )d x \right ) k}{k -1}} {\mathrm e}^{\int -\frac {f \left (x \right )}{k -1}d x}\right ]$ Mathematica raw input
DSolve[y'[x] == f[x]*y[x] + g[x]*y[x]^k,y[x],x]
Mathematica raw output
{{y[x] -> ((C[1] - (-1 + k)*Inactive[Integrate][E^((-1 + k)*Inactive[Integrate][
f[K[1]], {K[1], 1, K[2]}])*g[K[2]], {K[2], 1, x}])/E^((-1 + k)*Inactive[Integrat
e][f[K[1]], {K[1], 1, x}]))^(1 - k)^(-1)}}
Maple raw input
dsolve(diff(y(x),x) = f(x)*y(x)+g(x)*y(x)^k, y(x))
Maple raw output
[y(x) = 1/((-k*Int(exp(Int(f(x),x)*k)/exp(Int(f(x),x))*g(x),x)+_C1+Int(exp(Int(f
(x),x)*k)/exp(Int(f(x),x))*g(x),x))^(1/(k-1)))*exp(1/(k-1)*Int(f(x),x)*k)/exp(1/
(k-1)*Int(f(x),x))] | 727 | 1,509 | {"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": 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} | 3.65625 | 4 | CC-MAIN-2024-22 | latest | en | 0.192155 |
http://www.netlib.org/lapack/lawn41/node47.html | 1,485,145,502,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560282110.46/warc/CC-MAIN-20170116095122-00004-ip-10-171-10-70.ec2.internal.warc.gz | 601,721,429 | 2,472 | Next: Input File for Testing Up: Testing the Nonsymmetric Eigenvalue Previous: Tests Performed on the Contents
### Input File for Testing the Nonsymmetric Eigenvalue Routines
An annotated example of an input file for testing the nonsymmetric eigenvalue routines is shown below.
```NEP: Data file for testing Nonsymmetric Eigenvalue Problem routines
7 Number of values of N
0 1 2 3 5 10 16 Values of N (dimension)
5 Number of values of NB, NS, and MAXB
1 3 3 3 20 Values of NB (blocksize)
2 2 2 2 2 Values of NBMIN (minimum blocksize)
1 0 5 9 1 Values of NX (crossover point)
2 4 2 4 6 Values of NS (no. of shifts)
20 20 6 10 10 Values of MAXB (min. blocksize)
20.0 Threshold value
T Put T to test the error exits
1 Code to interpret the seed
NEP 21
```
The first line of the input file must contain the characters NEP in columns 1-3. Lines 2-11 are read using list-directed input and specify the following values:
line 2: The number of values of N line 3: The values of N, the matrix dimension line 4: The number of values of the parameters NB, NBMIN, NX, NS, and MAXB line 5: The values of NB, the blocksize line 6: The values of NBMIN, the minimum blocksize line 7: The values of NX, the crossover point line 8: The values of NS, the number of shifts line 9: The values of MAXB, the minimum blocksize line 10: The threshold value for the test ratios line 11: An integer code to interpret the random number seed = 0: Set the seed to a default value before each run = 1: Initialize the seed to a default value only before the first run = 2: Like 1, but use the seed values on the next line line 12: If line 9 was 2, four integer values for the random number seed
The remaining lines occur in sets of 1 or 2 and allow the user to specify the matrix types. Each line contains a 3-character identification in columns 1-3, which must be either NEP or SHS (CHS in complex, DHS in double precision, and ZHS in complex*16), and the number of matrix types must be the first nonblank item in columns 4-80. If the number of matrix types is at least 1 but is less than the maximum number of possible types, a second line will be read to get the numbers of the matrix types to be used. For example,
```NEP 21
```
requests all of the matrix types for the nonsymmetric eigenvalue problem, while
```NEP 4
9 10 11 12
```
requests only matrices of type 9, 10, 11, and 12.
Next: Input File for Testing Up: Testing the Nonsymmetric Eigenvalue Previous: Tests Performed on the Contents
Susan Blackford 2001-08-13 | 722 | 2,799 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2017-04 | latest | en | 0.645251 |
https://hextobinary.com/unit/area/from/sqmil/to/sqnauticalmile | 1,726,235,916,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651523.40/warc/CC-MAIN-20240913133933-20240913163933-00368.warc.gz | 264,488,817 | 19,120 | Square Mil to Square Nautical Mile Converter
Area
Square Mil
Square Nautical Mile
1 Square Mil = 1.8809839309625e-16 Square Nautical Miles
How many Square Nautical Miles are in a Square Mil?
The answer is one Square Mil is equal to 1.8809839309625e-16 Square Nautical Miles and that means we can also write it as 1 Square Mil = 1.8809839309625e-16 Square Nautical Miles. Feel free to use our online unit conversion calculator to convert the unit from Square Mil to Square Nautical Mile. Just simply enter value 1 in Square Mil and see the result in Square Nautical Mile.
How to Convert Square Mil to Square Nautical Mile (mil2 to nmi2)
By using our Square Mil to Square Nautical Mile conversion tool, you know that one Square Mil is equivalent to 1.8809839309625e-16 Square Nautical Mile. Hence, to convert Square Mil to Square Nautical Mile, we just need to multiply the number by 1.8809839309625e-16. We are going to use very simple Square Mil to Square Nautical Mile conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Square Mil} = 1 \times 1.8809839309625e-16 = \text{1.8809839309625e-16 Square Nautical Miles}$$
What is Square Mil Unit of Measure?
Square Mil is a unit of area measurement. In square mil, all sides are equal to 1 mil in length. Square mil is equal to 1/1000000 of square inches.
What is the symbol of Square Mil?
The symbol of Square Mil is mil2. This means you can also write one Square Mil as 1 mil2.
What is Square Nautical Mile Unit of Measure?
Square Nautical Mile is a unit of area measurement. In square nautical mile, all sides are equal to 1 nautical mile in length.
What is the symbol of Square Nautical Mile?
The symbol of Square Nautical Mile is nmi2. This means you can also write one Square Nautical Mile as 1 nmi2.
Square Mil to Square Nautical Mile Conversion Table
Square Mil [mil2]Square Nautical Mile [nmi2]
11.8809839309625e-16
23.761967861925e-16
35.6429517928875e-16
47.52393572385e-16
59.4049196548125e-16
61.1285903585775e-15
71.3166887516738e-15
81.50478714477e-15
91.6928855378663e-15
101.8809839309625e-15
1001.8809839309625e-14
10001.8809839309625e-13
Square Mil to Other Units Conversion Table
Square Mil [mil2]Output
1 square mil in ankanam is equal to9.6450617283951e-11
1 square mil in aana is equal to2.0290560831101e-11
1 square mil in acre is equal to1.5942236697094e-13
1 square mil in arpent is equal to1.8870415714807e-13
1 square mil in are is equal to6.4516e-12
1 square mil in barn is equal to6451600000000000000
1 square mil in bigha [assam] is equal to4.8225308641975e-13
1 square mil in bigha [west bengal] is equal to4.8225308641975e-13
1 square mil in bigha [uttar pradesh] is equal to2.5720164609053e-13
1 square mil in bigha [rajasthan] is equal to2.5507601265177e-13
1 square mil in bigha [bihar] is equal to2.5512286717283e-13
1 square mil in bigha [gujrat] is equal to3.9855626976839e-13
1 square mil in bigha [himachal pradesh] is equal to7.9711253953678e-13
1 square mil in bigha [nepal] is equal to9.525986892242e-14
1 square mil in biswa [uttar pradesh] is equal to5.1440329218107e-12
1 square mil in bovate is equal to1.0752666666667e-14
1 square mil in bunder is equal to6.4516e-14
1 square mil in caballeria is equal to1.4336888888889e-15
1 square mil in caballeria [cuba] is equal to4.8074515648286e-15
1 square mil in caballeria [spain] is equal to1.6129e-15
1 square mil in carreau is equal to5.0012403100775e-14
1 square mil in carucate is equal to1.3274897119342e-15
1 square mil in cawnie is equal to1.1947407407407e-13
1 square mil in cent is equal to1.5942236697094e-11
1 square mil in centiare is equal to6.4516e-10
1 square mil in circular foot is equal to8.8419332258162e-9
1 square mil in circular inch is equal to0.0000012732383948401
1 square mil in cong is equal to6.4516e-13
1 square mil in cover is equal to2.3912527798369e-13
1 square mil in cuerda is equal to1.6416284987277e-13
1 square mil in chatak is equal to1.5432098765432e-10
1 square mil in decimal is equal to1.5942236697094e-11
1 square mil in dekare is equal to6.4516042561233e-13
1 square mil in dismil is equal to1.5942236697094e-11
1 square mil in dhur [tripura] is equal to1.929012345679e-9
1 square mil in dhur [nepal] is equal to3.8103947568968e-11
1 square mil in dunam is equal to6.4516e-13
1 square mil in drone is equal to2.5117348251029e-14
1 square mil in fanega is equal to1.0033592534992e-13
1 square mil in farthingdale is equal to6.3750988142292e-13
1 square mil in feddan is equal to1.5477899705916e-13
1 square mil in ganda is equal to8.0375514403292e-12
1 square mil in gaj is equal to7.7160493827161e-10
1 square mil in gajam is equal to7.7160493827161e-10
1 square mil in guntha is equal to6.3769003162943e-12
1 square mil in ghumaon is equal to1.5942250790736e-13
1 square mil in ground is equal to2.8935185185185e-12
1 square mil in hacienda is equal to7.2004464285714e-18
1 square mil in hectare is equal to6.4516e-14
1 square mil in hide is equal to1.3274897119342e-15
1 square mil in hout is equal to4.5393417556031e-13
1 square mil in hundred is equal to1.3274897119342e-17
1 square mil in jerib is equal to3.1913807189542e-13
1 square mil in jutro is equal to1.1210425716768e-13
1 square mil in katha [bangladesh] is equal to9.6450617283951e-12
1 square mil in kanal is equal to1.2753800632589e-12
1 square mil in kani is equal to4.0187757201646e-13
1 square mil in kara is equal to3.2150205761317e-11
1 square mil in kappland is equal to4.1822896408661e-12
1 square mil in killa is equal to1.5942250790736e-13
1 square mil in kranta is equal to9.6450617283951e-11
1 square mil in kuli is equal to4.8225308641975e-11
1 square mil in kuncham is equal to1.5942250790736e-12
1 square mil in lecha is equal to4.8225308641975e-11
1 square mil in labor is equal to8.9999981353839e-16
1 square mil in legua is equal to3.5999992541535e-17
1 square mil in manzana [argentina] is equal to6.4516e-14
1 square mil in manzana [costa rica] is equal to9.2311302396923e-14
1 square mil in marla is equal to2.5507601265177e-11
1 square mil in morgen [germany] is equal to2.58064e-13
1 square mil in morgen [south africa] is equal to7.5307575580717e-14
1 square mil in mu is equal to9.677399951613e-13
1 square mil in murabba is equal to6.3768946788374e-15
1 square mil in mutthi is equal to5.1440329218107e-11
1 square mil in ngarn is equal to1.6129e-12
1 square mil in nali is equal to3.2150205761317e-12
1 square mil in oxgang is equal to1.0752666666667e-14
1 square mil in paisa is equal to8.1164614825204e-11
1 square mil in perche is equal to1.8870415714807e-11
1 square mil in parappu is equal to2.550757871535e-12
1 square mil in pyong is equal to1.95148215366e-10
1 square mil in rai is equal to4.03225e-13
1 square mil in rood is equal to6.3769003162943e-13
1 square mil in ropani is equal to1.2681600519438e-12
1 square mil in satak is equal to1.5942236697094e-11
1 square mil in section is equal to2.4909766860524e-16
1 square mil in sitio is equal to3.5842222222222e-17
1 square mil in square is equal to6.9444444444444e-11
1 square mil in square angstrom is equal to64516000000
1 square mil in square astronomical units is equal to2.882813900904e-32
1 square mil in square attometer is equal to6.4516e+26
1 square mil in square bicron is equal to645160000000000
1 square mil in square centimeter is equal to0.0000064516
1 square mil in square chain is equal to1.5942185484941e-12
1 square mil in square cubit is equal to3.0864197530864e-9
1 square mil in square decimeter is equal to6.4516e-8
1 square mil in square dekameter is equal to6.4516e-12
1 square mil in square digit is equal to0.0000017777777777778
1 square mil in square exameter is equal to6.4516e-46
1 square mil in square fathom is equal to1.929012345679e-10
1 square mil in square femtometer is equal to645160000000000000000
1 square mil in square fermi is equal to645160000000000000000
1 square mil in square feet is equal to6.9444444444444e-9
1 square mil in square furlong is equal to1.5942236697094e-14
1 square mil in square gigameter is equal to6.4516e-28
1 square mil in square hectometer is equal to6.4516e-14
1 square mil in square inch is equal to0.000001
1 square mil in square league is equal to2.767740830046e-17
1 square mil in square light year is equal to7.2080493955751e-42
1 square mil in square kilometer is equal to6.4516e-16
1 square mil in square megameter is equal to6.4516e-22
1 square mil in square meter is equal to6.4516e-10
1 square mil in square microinch is equal to999999.12
1 square mil in square micrometer is equal to645.16
1 square mil in square micromicron is equal to645160000000000
1 square mil in square micron is equal to645.16
1 square mil in square mile is equal to2.4909766860524e-16
1 square mil in square millimeter is equal to0.00064516
1 square mil in square nanometer is equal to645160000
1 square mil in square nautical league is equal to2.0899839891858e-17
1 square mil in square nautical mile is equal to1.8809839309625e-16
1 square mil in square paris foot is equal to6.1152606635071e-9
1 square mil in square parsec is equal to6.775888109918e-43
1 square mil in perch is equal to2.5507601265177e-11
1 square mil in square perche is equal to1.263235101155e-11
1 square mil in square petameter is equal to6.4516e-40
1 square mil in square picometer is equal to645160000000000
1 square mil in square pole is equal to2.5507601265177e-11
1 square mil in square rod is equal to2.5507503078921e-11
1 square mil in square terameter is equal to6.4516e-34
1 square mil in square thou is equal to1
1 square mil in square yard is equal to7.7160493827161e-10
1 square mil in square yoctometer is equal to6.4516e+38
1 square mil in square yottameter is equal to6.4516e-58
1 square mil in stang is equal to2.3815430047988e-13
1 square mil in stremma is equal to6.4516e-13
1 square mil in sarsai is equal to2.2956841138659e-10
1 square mil in tarea is equal to1.0260178117048e-12
1 square mil in tatami is equal to3.903200435598e-10
1 square mil in tonde land is equal to1.1696156635243e-13
1 square mil in tsubo is equal to1.951600217799e-10
1 square mil in township is equal to6.9193735664469e-18
1 square mil in tunnland is equal to1.3069443319018e-13
1 square mil in vaar is equal to7.716049382716e-10
1 square mil in virgate is equal to5.3763333333333e-15
1 square mil in veli is equal to8.0375514403292e-14
1 square mil in pari is equal to6.3769003162943e-14
1 square mil in sangam is equal to2.5507601265177e-13
1 square mil in kottah [bangladesh] is equal to9.6450617283951e-12
1 square mil in gunta is equal to6.3769003162943e-12
1 square mil in point is equal to1.5942375226171e-11
1 square mil in lourak is equal to1.2753800632589e-13
1 square mil in loukhai is equal to5.1015202530354e-13
1 square mil in loushal is equal to1.0203040506071e-12
1 square mil in tong is equal to2.0406081012142e-12
1 square mil in kuzhi is equal to4.8225308641975e-11
1 square mil in chadara is equal to6.9444444444444e-11
1 square mil in veesam is equal to7.716049382716e-10
1 square mil in lacham is equal to2.550757871535e-12
1 square mil in katha [nepal] is equal to1.9051973784484e-12
1 square mil in katha [assam] is equal to2.4112654320988e-12
1 square mil in katha [bihar] is equal to5.1024573434566e-12
1 square mil in dhur [bihar] is equal to1.0204914686913e-10
1 square mil in dhurki is equal to2.0409829373826e-9
Disclaimer:We make a great effort in making sure that conversion is as accurate as possible, but we cannot guarantee that. Before using any of the conversion tools or data, you must validate its correctness with an authority. | 3,960 | 11,584 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2024-38 | latest | en | 0.879352 |
https://hbsesolutions.com/hbse-6th-class-maths-solutions-chapter-5-ex-5-3/ | 1,723,276,804,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640790444.57/warc/CC-MAIN-20240810061945-20240810091945-00882.warc.gz | 225,948,182 | 22,771 | # HBSE 6th Class Maths Solutions Chapter 5 Understanding Elementary Shapes Ex 5.3
Haryana State Board HBSE 6th Class Maths Solutions Chapter 5 Understanding Elementary Shapes Ex 5.3 Textbook Exercise Questions and Answers.
## Haryana Board 6th Class Maths Solutions Chapter 5 Understanding Elementary Shapes Exercise 5.3
Question 1.
Match the following :
(i) Straight angle
(ii) Right angle
(iii) Acute angle
(iv) Obtuse angle
(v) Reflex angle
Solution:
(i) (c),
(ii) -(d),
(iii)- (a)
(iv) – (e),
(v) – (b)
Question 2.
Classify each one of the following angles as right, straight, acute, obtuse or reflex.
Solution:
(a) Acute angle
(b) Obtuse angle
(c) Right angles
(d) Reflex angle
(e) Straight angle
(f) Acute angles. | 199 | 724 | {"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} | 3.265625 | 3 | CC-MAIN-2024-33 | latest | en | 0.719294 |
https://nz.rs-online.com/web/c/semiconductors/standard-logic/counter-ics/?pn=9 | 1,576,065,474,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540530857.12/warc/CC-MAIN-20191211103140-20191211131140-00288.warc.gz | 491,440,971 | 31,305 | # Counter ICs
A Shift Register is a sequential logic device made up of a cascade of flip-flops, through which a sequence of bits are ’shifted’. These devices are commonly used to convert between serial and parallel interfaces.
How it works – basic example
In a Shift Register the output of each flip-flop is connected to the input of the next flip-flop in the cascade. With each cycle of the connected clock, the bits are ’shifted’ down the cascade by one flip-flop.
A simple analogy to understand the basic theory or operation would be to consider a glass tube with an opening both ends, and placing a ball marked ’0’ or ’1’ into one end of the tube at a regular intervals.
As each ball is placed into the tube, all the balls are ’shifted’ along by one position. The sequence in which the balls marked ’0’ and ’1’ where fed in is preserved. Also each time one ball is fed in, the ball that has been in the cascade the longest drops out. So if the tube can contain 4 balls at any one time, this could be described as a 4-bit Shift Register.
Common Types of Shift Register
• Serial-in serial-out (SISO) – In this type of Shift Register, the data input and output is serial. With each clock cycle a new bit is fed into the cascade, with the bit furthest along the cascade exiting as an output. So in a 4-bit register, the output trails the input by 4 clock cycles. This matches the simple example above, with the ball that exits the tube being the output.
• Serial-in parallel-out (SIPO) – In this configuration data is converted from serial into parallel format. As described for the SISO variant above, data is fed in serially and shifted down the cascade, the difference is that each flip-flop has an output, allowing the entire bit array to be output simultaneously as a parallel output. This may be done using a latched of buffered output.
• Parallel-in serial-out (PISO) – This variant takes a parallel input and converts into a serial output. To do this the shift register must implement a write/shift sequence. The parallel data input is (written) into each flip-flop of the array simultaneously in one clock cycle, then for the next cycles the register goes into shift mode as the bits are shifted down the flip-flop cascade, being output as serial data. The process then repeats.
• Parallel-in parallel-out (PIPO) – This type of register differs slightly from the others in that inputs and outputs of each flip-flop in the array are not interconnected. Each flip-flop takes one bit of the parallel input, stores it for a cycle, and then outputs the data simultaneously with the other flip-flops in the register, creating a parallel output that matches the input. Like the SISO type above, PIPO variants can be used to temporarily store data, or act a time delay device.
• Universal Shift Register – A configurable Shift Register that can be programmed to act as a SISO, SIPO, PISO or PIPO device, often on the fly.
...
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Description Price Package Type Logic Function Number of Stages Logic Family Mounting Type Counter Type Operation Mode Number of Elements Pin Count Minimum Operating Supply Voltage Maximum Operating Supply Voltage Dimensions Length Width
New RS Stock No. 709-2012 Mfr. Part No.SN74HC393NE4 BrandTexas Instruments Compare \$0.887 Each (In a Pack of 10) units PDIP Counter, Divider 4 HC Through Hole Binary Up Counter 2 14 2 V 6 V 19.3 x 6.35 x 4.57mm 19.3mm 6.35mm New RS Stock No. 169-5707 Mfr. Part No.CD74AC163M BrandTexas Instruments Compare \$0.755 Each (In a Tube of 40) units SOIC Counter 4 AC Surface Mount Binary Up Counter 1 16 1.5 V 5.5 V 9.9 x 3.91 x 1.58mm 9.9mm 3.91mm New RS Stock No. 144-9974 Mfr. Part No.SN74HC165PW BrandTexas Instruments Compare \$0.387 Each (In a Tube of 90) units TSSOP Shift Register 8 HC Surface Mount - Parallel to Serial 1 16 2 V 6 V 5 x 4.4 x 1.15mm 5mm 4.4mm New RS Stock No. 144-9945 Mfr. Part No.CD74HC164E BrandTexas Instruments Compare \$0.496 Each (In a Tube of 25) units PDIP Shift Register 8 HC Through Hole - Serial to Parallel 1 14 2 V 6 V 19.3 x 6.35 x 4.57mm 19.3mm 6.35mm New RS Stock No. 178-4718 Mfr. Part No.MC14040BDG BrandON Semiconductor Compare \$0.667 Each (In a Tube of 48) units SOIC Counter, Divider 12 4000 Surface Mount Binary Up Counter 1 16 3 V 18 V 10 x 4 x 1.5mm 10mm 4mm New RS Stock No. 189-2023 Mfr. Part No.HCF4094YM013TR BrandSTMicroelectronics Compare \$0.616 Each (In a Pack of 25) units SO - 8 - Surface Mount - - 1 16 3 V 20 V 10 x 4 x 1.65mm 10mm 4mm New RS Stock No. 662-7121 Mfr. Part No.CD74HC4020M96 BrandTexas Instruments Compare \$0.784 Each (In a Pack of 5) units SOIC Counter 14 HC Surface Mount Binary Up Counter 1 16 2 V 6 V 9.9 x 3.91 x 1.58mm 9.9mm 3.91mm New RS Stock No. 168-2337 Mfr. Part No.BU4094BCF-E2 BrandROHM Compare \$1.319 Each (In a Pack of 10) units SOP Shift Register 8 CMOS Surface Mount - Serial to Serial/Parallel 1 16 3 V 16 V 10 x 4.4 x 1.49mm 10mm 4.4mm New RS Stock No. 145-0188 Mfr. Part No.TPIC6B596DW BrandTexas Instruments Compare \$1.572 Each (In a Tube of 25) units SOIC Shift Register 8 - Surface Mount - Serial to Serial/Parallel 1 20 4.5 V 5.5 V 12.8 x 7.52 x 2.35mm 12.8mm 7.52mm New RS Stock No. 662-9540 Mfr. Part No.CD40161BE BrandTexas Instruments Compare \$0.654 Each (In a Pack of 5) units PDIP Counter 4 4000 Through Hole Binary Up Counter 1 16 3 V 18 V 19.3 x 6.35 x 4.57mm 19.3mm 6.35mm New RS Stock No. 121-9245 Mfr. Part No.CD4029BEE4 BrandTexas Instruments Compare \$0.421 Each (In a Tube of 25) units PDIP Counter 4 4000 Through Hole Binary, Decade Down Counter, Up Counter 1 16 3 V 18 V 19.3 x 6.35 x 4.57mm 19.3mm 6.35mm New RS Stock No. 709-2453 Mfr. Part No.CD74HC390EE4 BrandTexas Instruments Compare \$0.851 Each (In a Pack of 10) units PDIP Counter, Divider 4 HC Through Hole Decade Up Counter 2 16 2 V 6 V 19.3 x 6.35 x 4.57mm 19.3mm 6.35mm New RS Stock No. 145-2980 Mfr. Part No.MC14557BDWG BrandON Semiconductor Compare \$1.148 Each (In a Tube of 47) units SOIC W Shift Register 64 CMOS Surface Mount - Serial 1 16 -0.5 V 18 V 10.45 x 7.6 x 2.4mm 10.45mm 7.6mm New RS Stock No. 663-1414 Mfr. Part No.SN74ALS164AD BrandTexas Instruments Compare \$2.356 Each (In a Pack of 5) units SOIC Shift Register 8 ALS Surface Mount - Serial to Parallel 1 14 4.5 V 5.5 V 8.65 x 3.91 x 1.58mm 8.65mm 3.91mm New RS Stock No. 145-0030 Mfr. Part No.SN74LS163ANE4 BrandTexas Instruments Compare \$0.576 Each (In a Tube of 25) units PDIP Counter 4 LS Through Hole Binary Up Counter 1 16 4.75 V 5.25 V 19.3 x 6.35 x 4.57mm 19.3mm 6.35mm New RS Stock No. 145-4025 Mfr. Part No.MC74AC163DR2G BrandON Semiconductor Compare \$0.29 Each (On a Reel of 2500) units SOIC Counter 4 AC Surface Mount Binary Up Counter 1 16 2 V 6 V 10 x 4 x 1.5mm 10mm 4mm New RS Stock No. 145-7567 Mfr. Part No.SN74AHC594PWR BrandTexas Instruments Compare \$0.285 Each (On a Reel of 2000) units TSSOP Shift Register 8 AHC Surface Mount - Serial to Serial/Parallel 1 16 2 V 5.5 V 5 x 4.4 x 1.15mm 5mm 4.4mm New RS Stock No. 186-9050 Mfr. Part No.FS7140-01G-XTD BrandON Semiconductor Compare \$11.00 Each (In a Tube of 48) units SOIC Clock Generator - - Surface Mount - Serial - 16 3 V 3.6 V 10 x 4 x 1.5mm 10mm 4mm New RS Stock No. 709-2996 Mfr. Part No.SN65HVS882PWP BrandTexas Instruments Compare \$8.11 Each units - - - - - - - - - - - - - - New RS Stock No. 152-6494 Mfr. Part No.74HCT595PW-Q100,11 BrandNexperia Compare \$1.133 Each (In a Pack of 10) units TSSOP Shift Register 8 74HCT Surface Mount - Serial to Parallel 1 16 4.5 V 5.5 V 5.1 x 4.5 x 0.95mm 5.1mm 4.5mm
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Texas Instruments range of Counters and Shift Registers ...
Description:
Texas Instruments range of Counters and Shift Registers from the 74HC Family of CMOS Logic ICs. The 74HC Family use silicon gate CMOS technology to achieve operating speeds similar to the LSTTL family but with the low power consumption of ... | 2,488 | 7,915 | {"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-2019-51 | latest | en | 0.915554 |
http://mrunal.org/2013/03/trigonometry-hnd-type-one-building-two-angles-made-easy-for-ssc.html | 1,529,274,125,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267859817.15/warc/CC-MAIN-20180617213237-20180617233237-00265.warc.gz | 213,499,557 | 25,967 | # [Trigonometry] HnD Type #3: One Building Two Angles made easy for SSC
SubscribeAptitude22 Comments
## Introduction: 1 Building 2 Angles
Continuing in the series of “Trigonometry –>Height and Distance–> Five types of Questions. 4th type of question is “One buildings and Two angles”.
Such question usually revolve around 2 points on ground, 2 shadows, or flag+building or statue + building type of diagram.
Here are example questions (all of them solved in the video)
1. From point A on ground, angle of elevation is 30 degree to the top of a building. Moving 20m towards the building, there is point B, with angle 60. Find height of this building
2. As Sun’s angle of elevation increases from 30 to 60, shadow of a tree decreases by 5m. Find height of this tree.
3. Height of a building is “h”. From a ground-point, the angle of elevation of the top of this building is “A”. On moving h/2 distance towards the building, angle becomes “B”. Find value of cotA-cotB. (SSC-CGL 2012)
## Approach: 1 building 2 angles
1. Since two angles are given, you make two equations of TAN using Topi-triangle shortcutTM.
2. First equation will give you a value (either height or distance). You plug that into second equation and you’ll get the answer.
3. Check the following video to see how ^this approach exactly works. And after watching the video, solve the mock questions given at the bottom of this article.
If the video is not visible, check it directly on my youtube channel youtube.com/user/TheMrunalPatel
## Mock Questions
1. As angle of elevation of the sun increases from 30° to 60°: Shadow of the tree is reduced by 10 m find the height of this tree.
2. A hill is 200 high. A car makes 30 degree angle of elevation and a truck makes 60 degree angle of elevation with this hill. Find distance between these two vehicle. (hint: 30<60 so car will be farther away from hill compared to truck.)
3. There is a palm tree on the bank of a river. A person is observing this tree from the opposite bank. Angle of elevation is 60. Now he retreats (moves back) 20 meters from his side of bank. And the angle of elevation is decreased to 30. Find 1) height of tree 2) width of river.
4. Two buildings of same height are located on the either side of a road. width of the road is hundred metres. At a point on the road, between these two buildings, the angle of elevation for each building are 60 and 30 m respectively. Find the distance of the point from the nearest end of a building and find the height of these buildings.
5. A statute is 1.46 m tall. And it stands on the top of a pedestal. From a ground point, angle of evaluation of elevation of the statue is 60° and from the same ground point, the angle of elevation of the top of the pedestal is 45°. Final height of the pedestal. (hint: TAN 45=opposite and adjacent sides are same)
6. Angles of elevation of a building from two points at distance of A and B (A>B) from its foot on the same side of building have measure 30 and 60. Find height of tower.
7. From the top of a lighthouse, two ships “A” and “B” are visible on the same side of the sea. If their angles of depression are 35 and 50 degrees respectively, then which ship is farther away from the lighthouse?
8. from a ground point, the angle of elevation of a building is found to be such that Tangent is 5/12. after walking 192 m towards the building, thangent changes to 3/4. find height of the building. (hint:Tangent=TAN.)
## Answers
1. 5 ROOT 3 OR 8.65
2. 400 root 3 by 3 OR 230.6
3. Height of tree 10 root 3 or 17.3 and width of river 10.
4. distance of point from the nearest building=25m. height of each building is 25 root 3 OR 43.25 m
5. 2m
6. root (ab)
7. Ship A is farther away because its angle of elevation is 35. Smaller the angle of elevation =farther away from base.
8. 3
For more [Aptitude] related content, visit the Archive on Mrunal.org/aptitude
## 22 Comments on “[Trigonometry] HnD Type #3: One Building Two Angles made easy for SSC”
1. Sir , for question 7 answer is ship B because they given angle of depression not angle of elevation
2. angle of depression from top of the bldg is measured n angle of elevation from bottom..
so ans 7 is correct!!
ship A
3. Sir, Could you please provide the steps for the question number 8.
4. mrunal please shed some light on lic aao examiantion
1. Sure but it’ll take some time. Right now busy with UPSC pattern change article.
1. Thank you. You earn my blessings.
5. sir that’s great help by U, may i get information for UGC NET HISTORY
6. Sir, Can you please upload the solution diagrams/step sequence for the mock questions 6 and 8. I was unable to understand the problem.
7. sir for question 4
i cant understand the diagram.. plz give the solution
8. sir, I am getting different answers for question 6th and 8th . for 8th answer should be 180 meters.
1. same here….1 am getting 180m as 8th answer
1. question 8th answer is 180.
1. totally agree with you!!!
kindly upload the steps for question 4 and 6….
9. sir plz post about geometry and mensuration
10. plz expalin question no 8
11. Mrunal Sir, Quantum Cat book by Sarvesh Verma has omitted many solutions in Trigonometry Chapter. What to do?
No solution for Introductory Exercise and for Level 1 only 15 questions solutions are there.
Can you give us links for the rest of the answers?
Also thanks for clearing the basics of Trigonometry.
Best Regards.
12. plz explain 5th uska diagram kese banaya jaye samaj me nahi a raha
13. Sir,
Unable to solve some of the mock questions , but those are also very important.. Where to find the solutions of those ..??
Please help sir …
Thanks in advance ..
14. sir please shed some light on ssc capf si exam also.
15. Sir, Please provide solutions for Question 6 and 8. Thanks in advance.
16. Answer to ques.8 should be 180.
17. Could any one please explain 8th question. | 1,535 | 5,879 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2018-26 | longest | en | 0.909401 |
http://www.docstoc.com/docs/42887775/Understanding-student-misconceptions-in-elementary-is-important-for | 1,397,869,226,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609535535.6/warc/CC-MAIN-20140416005215-00340-ip-10-147-4-33.ec2.internal.warc.gz | 391,081,682 | 17,788 | # Understanding student misconceptions in elementary is important for
Document Sample
``` ED 661 Understanding Student Thinking in Geometry and Measurement
October 21-22 Agenda
1-1:30 Syllabus for Course Document camera
p.m. Goals for 2 days Handout – goals and agenda
Agenda Syllabus
Understanding student misconceptions in Goals Document for Document Camera
elementary is important for laying the Rubric for grading
foundation for future work. For example,
interpreting geometry problems in which you
need to make a sketch, finding areas of
polygons without an understanding of the
attributes of those polygons…
1:30-2:00 Discuss the Van Hiele levels. Big Ideas posted on chart paper as
Discuss the reading in the Van de Walle 3- participants share
5 book. Vandewalle book
Use the “Big Ideas” and questions brought
by the participants as part of their pre-
class assignment. Share at tables first,
then in large group
At tables discuss your big ideas, then we
will do a large group share
Be sure to include:
Difficulties that students have in geometry
1. Build student understanding in Geometry
as a basis for later work in mathematics.
2. Discuss reading in Van De Walle. (What
thoughts and questions did you have?)
a. The van Hiele Levels of Geometric
Thought
i. Levels are hierarchal but
you can be this at any
time.
ii. So dependent on spatial
sense
b. Van de Walle defines geometry
and number sense etc. as intuition
and it is up to us as teachers to
give kids experiences to develop
this.
3. Has Geometry been pushed to the back
burner due to national standards etc.
Right now the trend is that if the
mathematical strand supports algebra,
then we should teach it.
4. Should build the attributes in geometry
before giving students the definition.
5. Elementary students are at level 0 and 1.
6. Coverage is a myth that people have been
sold….covering doesn’t necessarily mean
learning.
7. Graphic on Pg. 209 will help teachers
understand that you can’t skip steps.
2:00-2:15 Brief Discussion of Triangle activity: Triangles document
Tables share what their students circled and their
definitions of triangles.
Large Group:
Found that all levels of students have
misconceptions of the attributes of a triangle.
We will refocus on this in more depth as we
experience van hiele’s levels during the day.
2:15-3:00 Van de Walle Activity 8.1 or 7.1 Shape Black Line Masters-BLMs 20-26
Sorts www.ablongman.com/vandewalleseries
Note: All BLMs for the book can be Chp. 8 - Pages 212-213 in vandewalle book 3-
Chp. 7 - Pages 194 vandewalle book K-2
the CD, too.
Read through bullets one and two, do
bullet three, read bullet four, and do bullet
five.
Discuss the question under the STOP sign.
Keep track of the language used; read and
discuss the “Assessment Note” on page 214.
3:00-3:15 Break
3:15-3:45 Activity 8.3(7.3) Geoboard Expansion handouts, “Geoboard Shapes to Copy.”
See also Kathy Richardson, page 50: Geoboards and 5 rubber bands per board
Creating Shapes on the Geoboard. Pg. 215 vdw3-5, pg. 197 vdwK-2
Kathy Richardson Book
3:45-4:15 Activity 8.6 (7.6) Mystery Definition Mystery Definition handout
Provide handout to see if they can Pg. 224 vdw3-5, pg. 207 vdw K-2
figure out the rule
Individual think time
Pair/share
Design your own and have others solve it.
See Figure 8.14(7.14) on page 225 (206) as
an example.
Sorting Shapes on the Geoboard.
4:15-4:45 Activity 8.7 Triangle Sort, page 225 (BLM 29). On web.
Give directions orally; the directions in the
book give too much information.
into three groups so that no triangle
belongs to two groups. When this is done
and descriptions of the groupings have
been written, students should then find a
second criterion for creating three
different groupings.
Note: the graphic on page 225 is the same
activity we did in “Geometry Camp” (page
4).
4:45-5:00 Closure Handout for Reflection
Day 2 Response to questions Geom. Camp materials
8-8:30 Activities 8.8, 8.11, and 8.12 correlate to Vandewalle book
activities from “Geometry Camp” (pages 6-
8). Review these pages and relate to the
reading in the Van de Walle book.
Discuss the question under the STOP sign on
page 231.
8:30-9:30 Read the Introduction to the Casebook on Handouts of intro and case studies 1, 2, 5
pages 1-3. Think about the Van Hiele
Chapter 1: Describing 2-D and 3-D
Objects. Read the introduction and Cases
1, 2, and 5.
9:00-9:30 Video clip, “Quick Images”, Chapter 1 in Video from Examining Features and
Examining Features of Shape Shapes
See notes on pages 27-28 F.G. What (BLMs on pages 35-37 F.G.)
different ways do the children describe the Printed Quick Images
images? Think about the Van Hiele levels
that are illustrated.
“Quick Images” math activity:.
Seeing things in different ways. Share the
attributes that you kept in your mind in
order to recreate the images.
Children relied on 3-D, real-world objects
to describe 2-D shapes, but they don’t
necessarily make the distinction between
two and three dimensions.
Some students were not limited by
the orientation of the objects.
9:30- Discuss the Focus Questions: 1, 2, 3, and 4 Focus Questions Handout
10:15 (page 38 F.G.).
Note: Questions 1 and 2 bring up two
topics: the use of technical geometric
vocabulary, and the different ways a 3-D
object can be described by listing its 2-D
components. Throughout this course, we
will explore the difference between a
mathematical definition and a description
of an object, recognizing that correct
vocabulary includes concepts and ideas to
think about before definitions make sense.
Small group discussion: Focus questions
#1-4
(Compare the language and results you
observed when students completed Circle
the Triangles to the children’s’ language
and thinking in the cases.)
Take turns recording your thoughts on
paper to use during the whole group
discussion.
Again, keep in mind the Van Hiele Levels.
Whole group discussion
Discussion of the shape as a whole vs. the
parts of the shape
The way you are thinking affects the way
you see an object or what you notice
about it (some may think of the shapes as
nets)
Purpose of the activities: develop mental
images of the figures and the means to
10:5- Break
10:30
10:30- Math Activity: “What’s in the Envelope?” Envelope and shapes
11:00 (Groups of 3) Page 39 F.G.
Can we put 2-D shapes into the envelope?
Discuss the names of the shapes (briefly).
Sort the shapes into categories of your
choice.
Possible extension: Sort as if you were at
a specific Van Hiele level: for example, sort
by Level 0 criteria (or choose a different
level).
11:00- Chapter 2: Developing Meaning for Orange pattern bock and rectangular
11:30 Geometric Terms. Discussion of Cases 9, prism
10, 11. Focus Questions 1, 2, and 3 (page
50 F.G.) go with these cases. Have an
orange square pattern block and a
rectangular prism for the Case 10
discussion.
Small groups-Discuss the Introduction as it
relates to the Van Hiele Levels.
Whole group (Discuss Focus Question #6,
too.)
11:30- Lunch
12:30
12:30- View the video, “Angles in a Triangle” Video clips
1:15 See notes on page 46 F.G. Focus Questions 1, 2, 4
Math Activity, “Angles and Angle
Measures”—page 51 F.G. Questions 1, 2,
4, and 5 are activities from Geometry
Camp. Decide how much of each of these
activities you need to do to refresh your
memory. I would actually suggest that you
redo all of the activities. Discuss all the
questions, including #3.
1:15-3:45 Teacher Presentations: 20 minutes each Geometry manipulatives
Break with a bit of time to spare Handouts for each team
inclusive Marnie/Colleen Oct. 22
Cindy/Mary Oct. 22
Rachel/Laurie Oct. 22
Sandy/Annie Oct. 22
Pam/Samra Oct. 22
Sherry/Connie Oct. 22
Samantha/Doreen/Nancy Oct. 22
3:45-4:00 Exit Question Handout
Homework assignment
1. Provide a document asking your
students what an angle is through
words and pictures.
2. Hold up 3 lengths of string and ask
them to write their estimated length
and how they came to the decision for
that length
3. Next group presentations.
```
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# quiz3Aa - Physics 1A 8 AM class Quiz 3 Nov 16 2007 Prof...
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Physics 1A- 8 AM class Quiz # 3 Nov. 16, 2007 Prof. Jose Onuchic MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) A spring with a spring constant of 400 N/m is compressed 2 cms. How much energy does it store? A) 0.02 J B) 0.42 J C) 5.60 J D) 0.08 J E) 1.22 J 2) A 20-N crate starting at rest slides down a rough 5.0-m long ramp, inclined at 25 o with the horizontal. 20 J of energy is lost to friction. What will be the velocity of the crate at the bottom of the incline? Figure 1 3) In Figure 1, A 9.0-kg hanging weight (B) is connected by a string over a pulley to a 5.0-kg block (A) sliding on a flat table. If the coefficient of sliding friction is 0.20, find the tension in the string.
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quiz3Aa - Physics 1A 8 AM class Quiz 3 Nov 16 2007 Prof...
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http://www.theanalysisfactor.com/logistic-regression-models-for-multinomial-and-ordinal-variables/ | 1,527,238,467,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867055.20/warc/CC-MAIN-20180525082822-20180525102822-00502.warc.gz | 471,361,322 | 24,479 | Logistic Regression Models for Multinomial and Ordinal Variables
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Multinomial Logistic Regression
The multinomial (a.k.a. polytomous) logistic regression model is a simple extension of the binomial logistic regression model. They are used when the dependent variable has more than two nominal (unordered) categories.
Dummy coding of independent variables is quite common. In multinomial logistic regression the dependent variable is dummy coded into multiple 1/0 variables. There is a variable for all categories but one, so if there are M categories, there will be M-1 dummy variables. All but one category has its own dummy variable. Each category’s dummy variable has a value of 1 for its category and a 0 for all others. One category, the reference category, doesn’t need its own dummy variable as it is uniquely identified by all the other variables being 0.
The multinomial logistic regression then estimates a separate binary logistic regression model for each of those dummy variables. The result is M-1 binary logistic regression models. Each one tells the effect of the predictors on the probability of success in that category in comparison to the reference category. Each model has its own intercept and regression coefficients—the predictors can affect each category differently.
Why not just run a series of binary regression models? You could, and people used to, before multinomial regression models were widely available in software. You will likely get similar results. But running them together means they are estimated simultaneously, which means the parameter estimates are more efficient–there is less overall unexplained error.
Ordinal Logistic Regression: The Proportional Odds Model
When the response categories are ordered, you could run a multinomial regression model. The disadvantage is that you are throwing away information about the ordering. An ordinal logistic regression model preserves that information, but it is slightly more involved.
In the Proportional Odds Model, the event being modeled is not having an outcome in a single category as is done in the binary and multinomial models. Rather, the event being modeled is having an outcome in a particular category or any previous category.
For example, for an ordered response variable with three categories, the possible events are defined as:
• being in group 1
• being in group 2 or 1
• being in group 3, 2 or 1.
In the proportional odds model, each outcome has its own intercept but the same regression coefficients. This means:
1. the overall odds of any event can differ, but
2. the the effect of the predictors on the odds of an event occurring in every subsequent category is the same for every category. This is an assumption of the model that you need to check. It is often violated.
The model is written somewhat differently in SPSS than usual with a minus sign between the intercept and all the regression coefficients. This is a convention ensuring that for positive coefficients, increases in X values lead to an increase of probability in the higher-numbered response categories. In SAS, the sign is a plus, so increases in predictor values lead to an increase of probability in the lower-numbered response categories. Make sure you understand how the model is set up in your statistical package before interpreting results.
{ 45 comments… read them below or add one }
Ngo Quynh
Hi everyone 🙂
I have some problems like that “There are 230 (66.7%) cells (i.e., dependent variable levels by subpopulations) with zero frequencies.
Unexpected singularities in the Hessian matrix are encountered. This indicates that either some predictor variables should be excluded or some categories should be merged.
The NOMREG procedure continues despite the above warning(s). Subsequent results shown are based on the last iteration. Validity of the model fit is uncertain.”
Could somebody help me ?
Thanks 🙂
Vincent
Ngo Quynh
Hi everyone
I have some problems like that “There are 230 (66.7%) cells (i.e., dependent variable levels by subpopulations) with zero frequencies.
Unexpected singularities in the Hessian matrix are encountered. This indicates that either some predictor variables should be excluded or some categories should be merged.
The NOMREG procedure continues despite the above warning(s). Subsequent results shown are based on the last iteration. Validity of the model fit is uncertain.”
Could somebody help me ?
Thanks
^ All you need to do is either drop the observations where you have missings (if you can justify it) or recode some of your variables so that they don’t have empty (or close to empty) categories. E.g. if you have a categorical variable that has 5 levels (5 different values it can assume), and there is sufficient data for 3 of its levels whereas 2 are empty or very sparse, you could create a new variable containing only the 3 levels for which there is sufficient data. Alternatively you could restrict your analysis to those observations where there are no missings for specific variables. Or you could drop the problematic predictors altogether (which is also among the suggestions in your cited output), but this is not always necessary.
Rakshit
Hi All,
I have a questionnaire data in which we need to find the expectations of the respondent based on the product they are using .
Suppose there are 9 products being used a respondent might be using all of the products or 1 or 2 or 3 or others.
so for each product we have a likert question with scaling from 1-5.
thus the varibles become 9 (One for each product with rating of 1-5 and if he doesnt use the product then there are blanks for the product variable)
how to interpret this information)
the data consist of 64 respondent and the response of product variables each might be having 10 ratings out of 64.
Can you help me to get an answer or how i can use the statistical method to derive anything.
tilahu eshetu
Ordered logistic regression Number of obs = 248
LR chi2(5) = 93.12
Prob > chi2 = 0.0000
Log likelihood = -245.07184 Pseudo R2 = 0.1597
——————————————————————————-
servqual | Coef. Std. Err. z P>|z| [95% Conf. Interval]
————–+—————————————————————-
tangible | .3610695 .2114817 1.71 0.088 -.0534271 .775566
realiability | -.2337004 .2429432 -0.96 0.336 -.7098603 .2424595
responsivness | .5860527 .2810081 2.09 0.037 .035287 1.136818
assurance | .0150324 .2674593 0.06 0.955 -.5091783 .5392431
emphaty | 1.282419 .2467206 5.20 0.000 .7988559 1.765983
————–+—————————————————————-
/cut1 | 2.717063 .9459124 .8631089 4.571017
/cut2 | 4.932767 .8850435 3.198114 6.667421
/cut3 | 6.254481 .9243937 4.442703 8.06626
/cut4 | 9.629553 1.039574 7.592025 11.66708
——————————————————————————-
tilahu eshetu
Ordered logistic regression Number of obs = 248
LR chi2(5) = 93.12
Prob > chi2 = 0.0000
Log likelihood = -245.07184 Pseudo R2 = 0.1597
——————————————————————————-
servqual | Coef. Std. Err. z P>|z| [95% Conf. Interval]
————–+—————————————————————-
tangible | .3610695 .2114817 1.71 0.088 -.0534271 .775566
realiability | -.2337004 .2429432 -0.96 0.336 -.7098603 .2424595
responsivness | .5860527 .2810081 2.09 0.037 .035287 1.136818
assurance | .0150324 .2674593 0.06 0.955 -.5091783 .5392431
emphaty | 1.282419 .2467206 5.20 0.000 .7988559 1.765983
————–+—————————————————————-
/cut1 | 2.717063 .9459124 .8631089 4.571017
/cut2 | 4.932767 .8850435 3.198114 6.667421
/cut3 | 6.254481 .9243937 4.442703 8.06626
/cut4 | 9.629553 1.039574 7.592025 11.66708
——————————————————————————-
hi there how to interpret the above table specially cut points .
tilahu eshetu
hi
I want to ask how to interpret these table …specially cut points on the issue the effect of service quality on customer satisfaction
Would you mind showing me all the steps how to re-code the 5 scale Likert scale into dichotomous (0,1 or yes no )form?
tilahu eshetu
yes!
NC
Hello,
I am trying to run an analysis to see whether area of distress varies based on age, point in cancer diagnosis, etc. I thought I would be running a multinomal logistical regression originally. However, some participants have just one area of distress (hence just one category) and some have multiple areas of distress at the same time. Thus, many participants qualify for multiple categories of the dependent variable. I am not sure what to do about this, as it seems to run counter to basic assumptions about categorical variables. Any ideas about what analysis to run in this situation, or how to code the responses? I am using SPSS. Any help is MUCH APPRECIATED.
Yohannes A
I want to run ordinal logistic regression (OLR) in SPSS. My data include 3 predictor variables (all continuous) and my outcome variables are 6 (ordinal), although the composite is one. My dependent variable is narcissism, which has 6 dimensions or subscales (self-interest, manipulation, impulsivity, unawareness of others, pride and self-love).
• Is it possible to run OLR with each DV and then summarize the results?
• OR is there any other method for reducing the DV and running OLR?
• OR is there any other method than ordinal logistic regression for
analyzing this data?
Maurice
Hello, I wanted to know how to perform a generalized ordered logit model.
I have lots of data and a dependent variable which is a scale from 1-14 on security consciousness (each individual is given a score based on answers to previous questions) and then a bunch of categorical variables (age:18-21,22-25,etc.; income:<=\$20,000,<=40,000,<=\$60,000,etc.; education:Some high school, high school, some college, bachelor's). I want to know how much of a factor each (age, income, education) plays in security consciousness. Am I correct that I would need a generalized ordered logit model? And do you have any help in how to perform one using Excel or R?
Ashan
Hi
i am carrying out a research on interdependent decision making in the food context (i.e. the effect of other actors on the consumers’ choices). bothe ID and DV are nominal data.
the DV is consumers choices coded 1-5 (not exactly ordered), and IDs are the behavior of 4 other groups each coded (group one binary, group two from 1 to 3, and group three from 1to4).
what model you suggest to investigate the relationship between IDs and DV? Can I use Mix-multinominal logestic regression? if yes I appreciate if let me know any stepwise sourse on how to do it (preferably in R). Do I need to consider the random effect? Thanks
stanley
Hi,
i am carrying out a research on the effect of risk management on construction project success. i have listed the risk management techniques as my independent variable and project success as my dependent variable.
i also designed a likert scale for respondents to rate the effect of each risk management technique on project success.
please which technique should i use to analyze the likert scale data. can i use SPSS.
THANK YOU
Seb
Hi,
I have to predict in which of four quartiles a subject will end up. Those quartiles deal about their performance in a high jump test.
Q1 120 and < 130cm…
Is it more correct to use ordered logit regression or ordered probit regression.
Karen
At least in a binary situation, you tend to get very, very similar results from probit and logit. I can’t imagine a reason why ordered responses would be any different. Neither is considered more correct, but there are definitely fields where one or the other is more common.
Ronald
Hi Karen
I am working on a data whose independent variable is nominal with five categories and the dependent variable is also nominal with ten categories. I have several control variables. I want to use multinomial I logistic regression to predict the outcome variable but the problem is that my control variables are too many and might strain my models for nothing. Which analysis can I use to determine those that I can fit In my MLR model.
Karen
Hi Ronald,
You’re right. That will probably be too much for nominal logistic regression. I would start with doing a series of chi-square tests to see which of those control variables have any effect on the outcome. If they don’t, then don’t include them in the overall model.
Phillip Schnarrs
Hi,
Just a quick question. I am working on a data analysis that will be using demographic characteristics to understand sexual risk behaviors. The question asks about a specific behaviors to which there were 11 possible responses (so obviously nominal data). However, a new variable was constructed following Center for Disease Control guidelines concerning risk (0 = No Risk, 1 = Low Risk, 2 = Moderate Risk, and 3 = High Risk). Now, I’m stuck because the original variable is nominal, but by because the behaviors a listed as No Risk to High Risk it seems as if they could be ordinal. Additionally, I was thinking to possibly even reduce this to no risk to possible risk of HIV transmission and running a logistic regression. Any advice would be great.
Ariel
hi Karen,
How would you analize the following: you have an experiment that consists of many trials (or competitions – e.g., triava quiz) between 2 individuals, and any individual gets a score of 1 if he wins (answer more questions than the loser) or 0 if he loses. No other data on the competition is recorded, only win or loose. The individuals come from 2 different groups (e.g., from 2 different schools), and for each individual you have also some other individual data (such as age, scores in english and history and gender). each individual participates only in one competition. We would like to have a statistical analysis that that finds the predictors of a win. Naturally we can’t just use logistic regression on all the individuals where each individual is a case (or a data point) because the result of the competition between 2 individuals depends on the attributes of the 2 specific individuals involved. So we have to somehow account for the pairing of the individuals. We can account for example by adding a dummy variable of trial ID, or including both the individuals in one data point – i.e., including either the difference between their attributes or even better including all pattributes: english1, english2, history1, history2, etc (all these are considerd fixed effects). But when we do so we can’t also include the group parameter in the analysis (as a random effect)- because obviously within each trial the win of one individual is the loose for the other, making the data point dependent and therefore violating the assumptions of the logistic regression, i guess.
So is there a way to analyze both the group and the other paramters at once, as predictiors for a win? Or do we need to do it in separate analysis: regression for the fixed parameters, and an exact binomial test for the group parameter?
We would be thankful for any suggestions.
thanks!
Karen
Ariel, It’s hard to say for sure, but it sounds like you need to use the contest as the unit of analysis. Use characteristics of School 1 participant and School 2 participant as the predictors, and define the outcome as “Did School 1 participant win?: Y/N.” That’s a tricky one.
Saroni
Hi Karen,
I have a problem where the DV (ordinal) is in Likert-type scale i.e.
(Most satisfied-1,Satisfied-2,Neither S nor DS-3, Dissatisfied-4, Most Dissatisfied-5)
and 2 sets of 7 IVs (Almost the same Scale but 1-5 scale) and a set of 5 IVs with (a scale of 1-6) both ordinal. Which is the best way to analyse this kind of problem? Do I need to treat the IVs as factors or covariates?
Thanks.
Karen
Saroni, there is no exact answer. It depends on what assumptions you are willing to make and the kind of information you need. Start here: http://www.theanalysisfactor.com/the-distribution-of-independent-variables-in-regression-models-2/
Karen
Hi Bereket,
That sounds like it’s eligible. I can’t tell you it’s the way to go without more info. Are those actual counts?
Karen
Bereket.M
The value of the response variables are 0,1,2,3,4. The observation takes place in 253 primary school childrens. Can i use poisson regression models, or any count data model to model the data.
Kevin
Hi Karen,
This is similar to Robert’s question back in March. My research is actually about language but might be easier to understand if I use the competition metaphor. I have lots of data about different competitors and the results of lots of three-competitor competitions. So, my IVs are three sets of competitor data, and my DV is the winning competitor. I want to build a model that will predict the winner. I understand that this will be a logistic regression, but what kind and how should I organise the data (for SPSS 19), given that my DV is also one of my IVs? A nudge in the right direction would be very much appreciated. Kevin
PS Your courses look very good!
Karen
Hi Kevin,
This may be the kind of question that requires a consultation because the answer is in the details, which means I’d have to ask about a dozen questions to make sure I understand correctly. But I’ll try to give you a nudge.
Your DV can’t be an IV as well. So that means you’re going to have to define your DV in such a way that it’s not the same as an IV. For example, the DV may be “Did this competitor win: Yes or No” and the IV is “Competitor ID: 1, 2, or 3.” That’s a little different than defining the DV as “Which competitor won: 1, 2, or 3.” Does that work?
Karen
Kevin
Hi Karen,
Thanks for the nudge – that definitely helps. A consultation might be the way forward, but I want to put together a more detailed plan if I go down that route, so that you would be verifying and improving it rather than telling me how to build it from scratch. I hope that makes sense. Thanks again – a very helpful nudge.
Kevin
Anna
Hi
I have a categorical IV and the DV is order of opening 4 information boxes. Each participant has data like this:
infobox1: 1st
infobox2: 2nd
infobox3: 4th
infobox4: 3rd
Can I use linear regression/ANOVa for this or do I need to use chi-square/logistic regression because it is really an ordinal variable?
Anna
Karen
Hi Anna,
It really is an ordinal variable. Plus, when you ask people to rank things, it has the extra issue of dependence. If you know how people ranked the first three boxes, you know their answer to the fourth.
Karen
shymaa yassin
i need to learn every thing about logistic mult more than two categories because i make my resarch thank you
Karen
Hi Shymaa,
You may want to start with my webinar on multinomial and ordinal models, but there are also great books out there that discuss it as well, including Long’s Regression Models for Categorical and Limited Dependent Variables.
The webinar recording is free: http://www.theanalysisfactor.com/binary-ordinal-multinomial-logistic/
Karen
Chris
Hi Karen,
I am dealing with ordinal data for the first time and I am having a bit of a data dilemma. I am examining whether increased insight (awareness) of the fact that one has a mental illness (MI; measured continuously) predicts people’s increased perceptions of their own recovery from MI. Further, does experienced stigma (measured continuously) for having MI moderate this relationship.
Recovery (the DV) is measured using an empirically-validated ordinal 5-stage model which provides a score for each stage of recovery per participant (i.e., 5 scores per person). Hence, according to my hypothesis, high insight should be related to higher stages of recovery, and this relationship should change as experienced stigma changes.
Given that my DV is both ordinal and dependent (i.e., stages of recovery are divergently correlated the further apart they are conceptually; e.g., stage 5 has a stronger correlation with stage 4 than with stage 2) would you recommend a repeated-measures multinomial logistic regression?
Thanks,
Chris
Karen
Hi Epitaf,
There are two ways–graph it and try non-linear terms to make sure they don’t fit better.
Karen
Epitaf_
Hi,
How do we check the log linearity assumption for a quantitative predictor in a multinomial regression ?
Thanks.
robert
Can Ordinal Logistic Regression be applied to analyse the factor weightings that predictively rank folks, say, in a quiz competition? Factors could be those such as age, education, job, relevant hobbies, specialisations etc.
Ten people, say, will finish with 10 different scores in a particular competition. What confuses me is that I have seen examples where perhaps a 100 other competition results are grouped into one analysis run. There are 100 winners but each winner has different abilities which gets lost in the analysis. Or a person finishing last (rank 10) in one competition, say, may have come out top (rank 1) in another weaker competition.
I am thinking that each individual competition has to be normalised somehow so that it actually can be related sensibly to the other 99 competitions.
Any views?
Karen
Hi Robert,
Let me see if I understand. Each person has 10 different scores, each on, say, a different topic? You use the raw scores to create rankings for each topic?
I guess what I’m missing here is what are you trying to get from the analysis? How the predictors (age, education, etc) affect the overall ranking across all 10 topics? Or whether they affect the rankings on some topics but not others?
Karen
anon
i have a question but the data/subject matter is very confidential so i would prefer a private email – is this at all possible? please let me know
Karen
I can’t usually answer emails privately, but you’re welcome to set up a Quick Question consultation. That’s just what they’re for–when you are stuck on something and just need a quick bit of help. Just click on Consulting in the menus, and go to Quick Question.
Karen
Anton
hi,
I am currently completing my dissertation which uses the Polity IV index of democracy, measured on a 0-10 scale with 1 0 having ‘no democracy’ and 10 being a ‘full democracy. My independent variable on diamond abundance is measured on a continuous level.
With such a dependent variable, is it wise to use a simple logistic regression? and should i construct my 1-10 measure of democracy into a dummy variable?
Can anyone help me with this? i will be very grateful!
Anton
Karen
Hi Anton,
Unless there is a cut point on your 11 point scale that is particularly meaningful, you probably don’t want to split it into a dummy variable. You will lose a lot of information that way.
You basically have two choices: 1. treat it as a continuous variable, which sometimes is a reasonable assumption, and run a linear regression model. 2. treat it as ordinal (which it inherently is), and run an ordinal logistic regression.
There’s a big debate on this, and both types of models have assumptions that may or may not be met here. A lot of people will make it sound like the OLS is clearly wrong here, but the ordinal regression also has assumptions that have to be met.
This post has more information: Can Likert Scale Data ever be Continuous?.
Anonymous
Hi Every one ;
At the moment I am a PhD student in the field of natural resources Economics ,by the time I start to develop my research proposal entitled “The interaction of Poverty and Natural resources degradation” two things comes in mind;
The model needed is a two stage regression approach which poverty is continuous endogenous variable and natural resources degradation is categorically ordered qualitative variable.
Here comes where I am challenged and I need your kind help.
First of all the model has a mix of logit (probit application) for the ordinal variable(natural resources degradation ) and next an OLS endogenous variable (poverty measured in percapita expenditure)
I can have the data from the household survey and when I start to think how to fit the model by STATA I am confused which commands to use and how to deal with a mixture of such continuous and ordinal endogenous variable.
May you then be kind so send me some hints how to deal with this issue.
With Kind regards
Darish
Karen
Hi Darish,
Unfortunately, I’m not a big Stata expert. I’ve used it before, but I’m not up on it enough to give you suggestions.
If anyone else can answer this or give Darish some hints, please do.
Given the nature of your model, though, I would suggest getting a hold of Long & Freese’s book on Categorical data analysis using Stata. I don’t know if it covers two-stage modelling, but it may at least get you started, and I know Long’s other book (which is a bit more theoretical) does cover mixture models.
Good luck,
Karen
Dale Dwyer
I have an interesting analysis that I’m not sure how to analyze correctly. I am using all nominal data as IVs (subject membership in one of two groups=IV1 and subject gender=IV2) to predict the likelihood of each subject making a choice to layoff five out of twenty-five people. Each subject, then, has a 1st, 2nd, 3rd, 4th, and 5th choice in order of who he would layoff. I have coded all 25 different choices for each subject as either 0 (subject didn’t choose that person) or 1 (subject chose that person). I have 136 subjects who completed the study.
I am clear about the between subjects analysis, but I’m not quite sure how to do the within subjects analysis. Any suggestions would be greatly appreciated!
Dale
Karen
Hi Dale,
Just to be clear, you’re ignoring the order here, right? 1st-5th choice, and just coding the response as 1/0?
I would look into a GEE analysis. It’s Generalized Estimating Equations, and it is an approach for repeated measures for generalized linear models.
I have to say, though, this IS an interesting analysis, and there may be other approaches to get at your research questions. For example, if there are characteristics of each of the 25 people, you may want to include them. Or the order may be interesting. But GEE would be a great place to start, and it is available in all the major stat packages.
-Karen
tilahu eshetu
hi karen
the orders are 1.highly dissatisfied 2.satisfied 3.neutral satisfied and 5. highly satisfied.
Many thanks
Tilahun | 6,046 | 26,046 | {"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-2018-22 | latest | en | 0.879116 |
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1. ## How to find initial velocity with fixed angle of projectile
Well, it does make sense. Nobody forces you to use coordinates where X is horizontal and Y is vertical. If you are in an environment in which the net acceleration on objects is in some direction other than down, you can work with coordinates in which that direction is in the Y axis, define an X axis that is perpendicular to it, and now you have reduced the problem to the situation with no wind. [/quote] I still dont understand, for example how do you solve with windX = 4, windY = 3, gravity = 10? [/quote] I will assume the sign convention is that positive windY means the wind is blowing down (since it has the same sign as gravity), that the cannon is at the origin and that I am trying to hit the point (targetX, targetY). accelX = 4, accelY = 13 new_X_vector = normalize(13,-4) = (.95577900872195007289, -.29408584883752309935) new_Y_vector = normalize(4,13) = (.29408584883752309935,.95577900872195007289) target_new_X = .95577900872195007289 * targetX + .29408584883752309935 * targetY target_new_Y = -.29408584883752309935 * targetX + .95577900872195007289 * targetY You'll similarly convert the vector in which the cannon is pointing using the same conversion I just did for the target. Then solve the problem in the new coordinates with gravity set to sqrt(13^2+4^2) and no wind. If that's not enough, I'll completely flesh it out, but I don't have the time now. [/quote] Nice, but what i have to change on cannon? I only have his position and aim angle in radian.
2. ## How to find initial velocity with fixed angle of projectile
Well, it does make sense. Nobody forces you to use coordinates where X is horizontal and Y is vertical. If you are in an environment in which the net acceleration on objects is in some direction other than down, you can work with coordinates in which that direction is in the Y axis, define an X axis that is perpendicular to it, and now you have reduced the problem to the situation with no wind. [/quote] I still dont understand, for example how do you solve with windX = 4, windY = 3, gravity = 10?
3. ## How to find initial velocity with fixed angle of projectile
Exactly, the equation that i got is to find speed, and i use cos and sin to get vector from speed. I will try to find any formula that work directly with vectors, i guess it will be easier than work with angles.
4. ## How to find initial velocity with fixed angle of projectile
Change the acceleration direction to "down"..... it dont make sense. its only make gravity acceleration bigger.
5. ## How to find initial velocity with fixed angle of projectile
That just has the same effect as gravity, then. You can just add the two together, define "down" to be the direction in which that vector points and use the same formula you would have used without wind. Let me point out that your wind will probably feel wrong, because that's not at all what wind does to things. A better model would have drag, and the wind is used at the point where you compute the velocity of your object with respect to the air. If the drag is linear, you can still solve the resulting differential equations analytically and get a closed formula. But perhaps you should start thinking about numerical methods that are much more flexible. [/quote] Ok with wind accelaration Y i can sum with gravity acceleration, but about acceleration X i dont know what to do.
6. ## How to find initial velocity with fixed angle of projectile
I implemented wind like acceleration X and Y every frame i increment on velocity X and Y.
7. ## How to find initial velocity with fixed angle of projectile
Hello guys im new here. Im making a game like gunbound or worms, and i need help to make computer hit the target with wind force. My plan is to use a fixed angle, and to find initial velocity to launch the projectile. I found this equation but it works without wind: 'L' = distance X from the target and 'h'=distance Y Anyone know similar equation that works with wind force? Sry my english. | 957 | 4,088 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2019-22 | latest | en | 0.943173 |
https://www.vexforum.com/t/easy-c-math-question/28271 | 1,680,236,049,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949533.16/warc/CC-MAIN-20230331020535-20230331050535-00677.warc.gz | 1,163,792,699 | 7,134 | # Easy C math question
We wrote some code to do some iterative functions. They did not work as expected so I started breaking it down. It got so simple that I finally realized what was causing the problem - simple math.
Here is our simple code
StartTimer(1);
while (1);
LeftArmTimer=GetTimer(1);
PrintToScreen(“Main Loop %d\n” , LeftArmTimer);
TempA=LeftArmTimer / 350 ; // TempA is declared as int
PrintToScreen(“TempA%d” , TempA);
TempB=TempA/2;
PrintToScreen(“TempB%d” ,TempB);
wait(5);
Here is a snapshot of the output…
TempA1TempB0Main Loop 672
TempA1TempB0Main Loop 678
TempA1TempB0Main Loop 684
TempA1TempB0Main Loop 690
TempA1TempB0Main Loop 696
TempA1TempB0Main Loop 702
TempA2TempB1072693248Main Loop 708
TempA2TempB1072693248Main Loop 714
TempA2TempB1072693248Main Loop 720
TempA2TempB1072693248Main Loop 726
TempA2TempB1072693248Main Loop 732
TempA2TempB1072693248Main Loop 738
I have tried this with TempB declared as both FLOAT and as DOUBLE. Doesnt seem to make a difference.
HELP PLEASE. It doesn’t make sense to me.
Yes, I believe %d is for integers and %f is for floats
I’m not sure what you are trying to do.
The statement while(1); would run as an infinite loop.
In RobotC the timers report back in milliseconds. Your wait at the bottom is only for 5 milliseconds so dividing by 350 does not make any sense.
Foster, this is EasyC, not Robotc.
Pinion, pasting code is better than handtyping misleading mistakes.
The following code compiles in EasyC as a user C code function, but I don’t have a cortex handy to run it to check output.
At least it has all the vars declared locally so we can see their types.
Note that GetTimer() returns unsigned long (32 bit pos = 4billion),
your application can likely define LAT as int, which is max 2billion,
so that all the variables are the same type.
PrintToScreen uses C printf descriptors, so fixing position count, and prepending 0 to get leading zeros should make columns line up better.
so it might help to print unsigned long in their correct marker %lu,
etc
Welcome to the strict typecasting world of C
``````#include "Main.h"
void timerloop ( void ) {
unsigned long LAT; int TempA; int TempB;
StartTimer ( 1 ) ;
while (1){ // repeat forever
LAT=GetTimer(1); // LAT is ulong, same as GetTimer
TempA=LAT / 350 ; // TempA is declared as int, want 0..=>0, 350..=>1
TempB=TempA/2; //TempB is int : want 0.. =>0, 2.. =>1
PrintToScreen("\n LAT:%012lu A:%04d B: %04d" , LAT,TempA,TempB);
Wait ( 5 ) ; // wait N milliseconds, maybe 50 would be better than 5, since print takes a while...
} // wend
} //timerloop``````
Edit: I agree with stack overflow link. you didn’t comment if TempB was int, or float, or long, or what type
Thanks JGrabber for the reminder it’s Easy C.
Your code makes sense since I can now see the braces.
Still confused about the divide by 350 since that section of code will run in a much shorter time.
Looks like you are using a double and printing the hex (well decimal) representation of it. 1072693248 is hex 0x3FF00000, if this were the upper 32 bits of a 64 bit double (0x3ff0000000000000), it would have been 1.0. (ie. 1.0 in it’s 8 byte double internal format is 0x3ff0000000000000).
So, you calculated 2/2 which is 1.0
You need to use %lf as the format string for double numbers.
Thanks for all the help guys.
As a mentor, every moment should be a teaching moment.
We started out talking about doing something on a repitive basis, like every 1/2 second. For example, turn on a motor for 1/2 second in one direction, stop, run the motor in the opposite direction for 1/2 second and then repeat.
So, I started talking about a counter, then using the counter to determine whether it was odd or even. This led to a formula for comparison where the number would be divided by 2 and compared to the int(of the number divided by 2).
Using the crazy math functions in Easy C led to interesting case for the variables and I suspect run time errors which are not reported back. Stupidly we displayed the values and jumped to conclusions without looking at the %d vs %f in the display.
SO, one last question if you all know…
When we enter a line of C code in easy C will it properly interpret the line or do we need to use the math functions built into the system?
Thanks for everyone’s help.
Not sure what you mean by “math functions built into the system”. EasyC is just using the math functions in the standard library that comes with gcc, same as PROS and ConVEX. EasyC uses the CodeSourcery release of gcc, they should be correct.
I don’t have a working EasyC license in front of me to check, but I believe that if you want to type your own code (for example a math equation), you need to use the “user code” option from the menu. | 1,301 | 4,753 | {"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-2023-14 | latest | en | 0.832251 |
https://www.transtutors.com/questions/does-the-alcohol-you-drink-have-an-acceptable-place-in-your-diet-the-following-probl-6781303.htm | 1,632,145,387,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057039.7/warc/CC-MAIN-20210920131052-20210920161052-00462.warc.gz | 1,052,768,494 | 14,156 | Does the alcohol you drink have an acceptable place in your diet? The following problems will help..
Does the alcohol you drink have an acceptable place in your diet? The following problems will help you answer this question. First of all, alcohol is considered a drug because it modifi es body functions and is NOT a food because it does not provide nutrients. Alcohol does, however, provide 7 kcal per gram—but only for the pure ethanol part of the beverage. One “drink” is defi ned as a beverage providing ½ oz or approximately 14 grams of ethanol, such as 12 oz of beer, 5 oz of wine, and 1.5 oz of distilled liquor (gin, vodka, whiskey, etc.). The total calories in the beverage is a combination of the ethanol and any other components, such as the carbohydrate calories from the grapes in wine or the grains in beer.
1. Look at your Intake and DRI goals Compared report, did you consume any alcohol (ethanol) during the days you kept a food record? If so how many grams? g Knowing that a standard drink contains approximately 14 grams of ethanol, how many actual drinks did you consume? (if you did not consume any alcohol assume you consumed two glasses of wine during your reporting period for the purpose of practicing the calculations) Number of drinks = g of ethanol consumed ÷ 14 g/drink = number of drinks.
2. Moderate drinking is defi ned as up to 1 “drink” per day for women (on average) and up to two “drinks” per day for men (on average). How many grams of ethanol does this allow per day for women and for men? Women: One drink per day = g of ethanol per day Men: Two drinks per day = g of ethanol per day Knowing that each gram of ethanol has 7 kcal/gram, how many kilocalories does this component of the drink(s) provide? Women: g ethanol per day x 7 kcal/gram = kcal Men: g ethanol per day x 7 kcal/day = kcal
3. Knowing that the total calories in 12 oz of regular beer is approximately 150 kcal, calculate the number of kilocalories provided by sources other than the ethanol. 150 kcal/12 oz beer = kcal from ethanol = kcal from other sources.
4. If an average person can metabolize about 0.5 oz of ethanol per hour, how long will take an average man and woman, who drink moderately, to metabolize this amount of alcohol in a day? Women: 1 drink per day x 1 hour/0.5 oz of ethanol = number of hours Men: 2 drinks per day x 1 hour/0.5 oz of ethanol = number of hours.
5. After reading this section and doing these exercises, do you feel that moderate alcohol consumption is “worth” its calorie cost in your diet? Why or why not?
Plagiarism Checker
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Free Plagiarism Checker | 624 | 2,654 | {"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-2021-39 | latest | en | 0.928848 |
https://math.stackexchange.com/questions/3316481/volume-element-of-haar-measure-on-so3-with-euler-angle-parametrization | 1,713,093,900,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816879.25/warc/CC-MAIN-20240414095752-20240414125752-00429.warc.gz | 345,684,407 | 36,423 | # Volume element of Haar measure on SO(3) with Euler angle parametrization
I have a real-valued function $$f(\alpha, \beta, \gamma)$$ which takes Euler angles $$\alpha, \beta, \gamma$$ as input, that I would like to average over the uniform distribution on orientations of 3-space. We use the convention that $$\alpha, \beta, \gamma$$ are applied about the fixed global $$x, y, z$$ axes respectively, in the order $$\gamma, \beta, \alpha$$.
This means I'll want to compute an integral of the form $$\int_A \int_B \int_C f(\alpha, \beta, \gamma) V(\alpha, \beta, \gamma)d\gamma d\beta d\alpha$$, where $$A, B, C \subset \mathbb{R}$$ are angular ranges and $$V(\alpha, \beta, \gamma) d\gamma d\beta d\alpha$$ is the volume element of the normalized Haar measure on $$\mathop{SO}(3)$$ using this parametrization.
What are the expressions for $$A, B, C$$, and $$V$$? Is there a reference for this?
I know that it is possible to convert to quaternion representation (among others) and compute the integral that way, but I would like an expression for $$V$$ in terms of $$\alpha, \beta, \gamma$$.
It might be worth noting that Euler angles are often poorly suited to this type of problem, particularly in numerical contexts: some computations become numerically unstable when close to gimbal lock.
It's probably easiest to compute this measure by transforming into quaternions and computing the transformation of the volume element, since in the quaternionic setting it's just the standard volume element on the unit 3-sphere.
Per Wikipedia, the conversion from a "standard" extrinsic $$z$$-$$x$$-$$z$$ Euler angles to a unit quaternion has the form $$\mathbf{q}(\alpha,\beta,\gamma)=\begin{matrix} \cos\left(\frac{\alpha+\gamma}{2}\right)\cos\left(\frac{\beta}{2}\right) \\ +\cos\left(\frac{\alpha-\gamma}{2}\right)\sin\left(\frac{\beta}{2}\right)\mathbf{i} \\ +\sin\left(\frac{\alpha-\gamma}{2}\right)\sin\left(\frac{\beta}{2}\right)\mathbf{j} \\ +\sin\left(\frac{\alpha+\gamma}{2}\right)\cos\left(\frac{\beta}{2}\right)\mathbf{k} \end{matrix}$$ (This is presumably computed by decomposing into rotations about coordinate axes and composing as quaternions.) Computing the Jacobian, $$D\mathbf{q}(\alpha,\beta,\gamma)=\frac{1}{2}\begin{bmatrix} -\sin\left(\frac{\alpha+\gamma}{2}\right)\cos\left(\frac{\beta}{2}\right) & -\cos\left(\frac{\alpha+\gamma}{2}\right)\sin\left(\frac{\beta}{2}\right) & -\sin\left(\frac{\alpha+\gamma}{2}\right)\cos\left(\frac{\beta}{2}\right) \\ -\sin\left(\frac{\alpha-\gamma}{2}\right)\sin\left(\frac{\beta}{2}\right) & \cos\left(\frac{\alpha-\gamma}{2}\right)\cos\left(\frac{\beta}{2}\right) & \sin\left(\frac{\alpha-\gamma}{2}\right)\sin\left(\frac{\beta}{2}\right) \\ \cos\left(\frac{\alpha-\gamma}{2}\right)\sin\left(\frac{\beta}{2}\right) & \sin\left(\frac{\alpha-\gamma}{2}\right)\cos\left(\frac{\beta}{2}\right) & -\cos\left(\frac{\alpha-\gamma}{2}\right)\sin\left(\frac{\beta}{2}\right) \\ \cos\left(\frac{\alpha+\gamma}{2}\right)\cos\left(\frac{\beta}{2}\right) & -\sin\left(\frac{\alpha+\gamma}{2}\right)\sin\left(\frac{\beta}{2}\right) & \cos\left(\frac{\alpha+\gamma}{2}\right)\cos\left(\frac{\beta}{2}\right) \end{bmatrix}$$ We can't take the determinant directly on account of the superfluous extra dimension. Because we are looking at the induced volume measure on $$S_3\subset\mathbb{R}^4$$, we can augment this matrix with a unit radial vector (namely $$\mathbf{q}$$) to obtain the volume scaling factor. $$dV=\left|\det\begin{bmatrix}\mathbb{q}(\alpha,\beta,\gamma) & D\mathbb{q}(\alpha,\beta,\gamma)\end{bmatrix}\right|d\alpha\ d\beta\ d\gamma=\frac{1}{8}\sin(\beta)d\alpha\ d\beta\ d\gamma,\ \ \ \text{(extrinsic z-x-z)}$$ This gives (up to a choice of multiplicative constant) the Haar measure.
As for the ranges, one standard range of values (for any Euler/Tait-Bryan convention) is $$\alpha\in(-\pi,\pi)$$, $$\beta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$$, $$\gamma\in(-\pi,\pi)$$, with $$\alpha$$ and $$\gamma$$ equivalent modulo $$2\pi$$. It can be shown that, up to sets of measure zero, this covers $$SO(3)$$ exactly once. One sanity check for the computed Haar measure is that the upper hemisphere of $$S_3$$ has volume $$\pi^2$$.
The approach above works for any 3 variable parameterization of $$SO(3)$$. In particular the other Euler angle conventions, (intrinsic $$x$$-$$z'$$-$$x''$$, etc.), as well as the Tait-Byan angles (intrinsic $$z$$-$$y'$$-$$x''$$, extrinsic $$y$$-$$z$$-$$x$$, etc.). It seems to be the case that all Euler angle conventions have the volume element $$dV=\frac{1}{8}\sin(\beta)d\alpha\ d\beta\ d\gamma,\ \ \ \text{(Euler)}$$ and all Tait-Bryan angles (including extrinsic $$z$$-$$y$$-$$x$$) have the volume element $$dV=\frac{1}{8}\cos(\beta)d\alpha\ d\beta\ d\gamma,\ \ \ \text{(Tait-Bryan)}$$ This can be checked for a given convention by repeating the above computation (using Mathematica or any other symbolic computation tool).
Based on the simplicity of the result, there may be a less computational, more group-theoretic way to arrive at these formulae starting with EA/TBA as compositions of coordinate axis rotations, though it's not yet obvious to me what it would be.
• Thanks, but in my application I'm using a different Euler angle convention. It's described in the question. I believe it's "extrinsic zyx" according to the Wikipedia nomenclature. Aug 26, 2019 at 4:13
• Exactly the same approach works for that convention, starting from the appropriate conversion formula. The algebra is just as tedious for all of them, but there seems to be only two forms, one for Euler (zxz etc.) and one for Tait-Bryan (zyx etc.). Aug 26, 2019 at 18:14 | 1,696 | 5,650 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 45, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-18 | latest | en | 0.754001 |
bi-architects.blogspot.com | 1,469,391,609,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824146.3/warc/CC-MAIN-20160723071024-00150-ip-10-185-27-174.ec2.internal.warc.gz | 21,311,097 | 19,533 | ## Tuesday, January 10, 2012
### Operators (SSIS Expression)
This post describes which operators you can use within the SSIS expression language.
Operator Description Cast (SSIS Expression): Convert SSIS Data Types Converts an expression from one data type to a different data type. () (Parentheses) (SSIS Expression) Identifies the evaluation order of expressions. + (Add) (SSIS Expression) Adds two numeric expressions. + (Concatenate) (SSIS Expression) Concatenates two expressions. - (Subtract) (SSIS Expression) Subtracts the second numeric expression from the first one. - (Negate) (SSIS Expression) Negates a numeric expression. * (Multiply) (SSIS Expression) Multiplies two numeric expressions. / (Divide) (SSIS Expression) Divides the first numeric expression by the second one. % (Modulo) (SSIS Expression) Provides the integer remainder after dividing the first numeric expression by the second one. || (Logical OR) (SSIS Expression) Performs a logical OR operation. && (Logical AND) (SSIS Expression) Performs a logical AND operation. ! (Logical Not) (SSIS Expression) Negates a Boolean operand. | (Bitwise Inclusive OR) (SSIS Expression) Performs a bitwise OR operation of two integer values. ^ (Bitwise Exclusive OR) (SSIS Expression) Performs a bitwise exclusive OR operation of two integer values. & (Bitwise AND) (SSIS Expression) Performs a bitwise AND operation of two integer values. ~ (Bitwise Not) (SSIS Expression) Performs a bitwise negation of an integer. == (Equal) (SSIS Expression) Performs a comparison to determine if two expressions are equal. != (Unequal) (SSIS Expression) Performs a comparison to determine if two expressions are not equal. > (Greater Than) (SSIS Expression) Performs a comparison to determine if the first expression is greater than the second one. < (Less Than) (SSIS Expression) Performs a comparison to determine if the first expression is less than the second one. >= (Greater Than or Equal To) (SSIS Expression) Performs a comparison to determine if the first expression is greater than or equal to the second one. <= (Less Than or Equal To) (SSIS Expression) Performs a comparison to determine if the first expression is less than or equal to the second one. ? : (Conditional) (SSIS Expression) Returns one of two expressions based on the evaluation of a Boolean expression. | 523 | 2,336 | {"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-2016-30 | longest | en | 0.805448 |
http://mathhelpforum.com/number-theory/159292-algebraic-numbers.html | 1,508,399,775,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823255.12/warc/CC-MAIN-20171019065335-20171019085335-00470.warc.gz | 215,746,904 | 10,543 | 1. ## Algebraic numbers
Here goes:
I have proven that $\sqrt[3]{2}+\sqrt{3}$ is algebraic number by definition (constructed a polynomial of sixth degree having that number as a root).
Could someone please give me a sketch of proof (or the proof itself) for the following claim. I'd appreciate it.
The claim is:
If $x$ is algebraic number prove that $\frac{1}{x}$ also is algebraic number.
Now I know a little bit of higher math/algebra and I know that algebraic numbers form a field and thus a quotient of two algebraic numbers is algebraic number. Is there a proof that does not require the axioms of field? Can someone give me an advice (or two)?
2. Originally Posted by MathoMan
Here goes:
I have proven that $\sqrt[3]{2}+\sqrt{3}$ is algebraic number by definition (constructed a polynomial of sixth degree having that number as a root).
I would start with something like $(x- \sqrt[3]{2}- \sqrt{3})(x- \sqrt[3]{2}+ \sqrt{3})= (x- \sqrt[3]{2})^2- 3= x^2- 2\sqrt[3]{2}x+ 3/4- 2= x^2- 5/4- 2\sqrt[3]{2}$ where I have used " $(a- b)(a+ b)= a^2- b^2$". Now use the fact that $(a- b)(a^2+ ab+ b^2)= a^3- b^3$ to get rid of that cube root.
Could someone please give me a sketch of proof (or the proof itself) for the following claim. I'd appreciate it.
The claim is:
If $x$ is algebraic number prove that $\frac{1}{x}$ also is algebraic number.
Now I know a little bit of higher math/algebra and I know that algebraic numbers form a field and thus a quotient of two algebraic numbers is algebraic number. Is there a proof that does not require the axioms of field? Can someone give me an advice (or two)?
You say you know that the "algebraic number form a field" and the field axioms but do you know the definition of "algebraic number". If x is an algebraic number, then there exist in integer, n, and integers, $a_n$, $a_{n-1}$, ..., $a_1$ and $a_0$ such that $a_nx^n+ a_{n-1}a^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0= 0$. Okay, what do you get if you divide each term in that equation by $x^h$?
Or, since you know "a quotient of two algebraic numbers is algebraic number", 1/x is the quotient of 1 and x. Do you know that 1 is an algebraic number.
By the way, "a quotient of two algebraic numbers is algebraic number" is NOT strictly true. The denominator has to be non-zero. In fact, the original claim "If "x" is algebraic number then 1/x also is algebraic number" is NOT true. A counter example is x= 0. What is true is "If "x" is a non-zero algebraic number, then 1/x is also an algebraic number".
3. Thanks for the post.
I am aware of all the things you've mentioned. Maybe I wasn't precise enough.
As you said:
What is true is "If "x" is a non-zero algebraic number, then 1/x is also an algebraic number".
I was looking for a way to prove that statement without using the obvious: that 1 is algebraic and that x is assumed to be algebraic and since algebraic numbers form a field hence 1/x must be algebraic number.
I hope we understand each other now. Thanks anyway. | 861 | 2,984 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 15, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2017-43 | longest | en | 0.911115 |
http://r4r.in/python/topic/post.php?ct=16&subct=136&tp=1280 | 1,563,501,990,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525973.56/warc/CC-MAIN-20190719012046-20190719034046-00442.warc.gz | 125,010,599 | 30,971 | # Find maximum number in Numpy Matrix by R4R Team
Example-
matrix is
[[1 4]
[5 1]]
then max is 5
Syntax-
numpy.array(list).max()
program-
#import numpy
import numpy as np
l=[[1,2,3],[4,5,1],[1,1,1]]
arr1=np.array(l)
print("nArray is :n",arr1)
print("Maximum number is :",arr1.max())
output-
Array is :
[[1 2 3]
[4 5 1]
[1 1 1]]
Maximum number is : 5 | 133 | 360 | {"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-2019-30 | longest | en | 0.632205 |
http://www.ck12.org/arithmetic/Proportions-to-Find-Base-b/?by=all&difficulty=all | 1,492,938,361,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917118310.2/warc/CC-MAIN-20170423031158-00584-ip-10-145-167-34.ec2.internal.warc.gz | 471,576,109 | 16,623 | # Proportions to Find Base b
## Cross - multiply to find a missing part of a proportion, base, b.
Levels are CK-12's student achievement levels.
Basic Students matched to this level have a partial mastery of prerequisite knowledge and skills fundamental for proficient work.
At Grade (Proficient) Students matched to this level have demonstrated competency over challenging subject matter, including subject matter knowledge, application of such knowledge to real-world situations, and analytical skills appropriate to subject matter.
Advanced Students matched to this level are ready for material that requires superior performance and mastery.
## Proportions to Find Base b
Use the cross products of proportions to find base b in percent problems.
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## Precent Problems (Using Proportions & Equations)
Use the cross products of proportions to find base b in percent problems.
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## Proportions to Find Base b
Explains how to use proportions and cross products to solve for the base as well as the percent and the amount.
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• Video
## Find the Whole Given the Percents and Parts - Overview
by CK-12 //basic
Overview
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• Video
## Example 2: Solve a Percent Problem with a Percent Proportion
by CK-12 //basic
Goes through the steps of solving 36 is 15% of what number using a percent proportion.
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## Find the Whole Given the Percents and Parts - Example 1
by CK-12 //basic
Whole number percentage calculations - without units
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## Find the Whole Given the Percents and Parts - Example 2
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Whole number percentage calculations - with units
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## Find the Whole Given the Percents and Parts - Example 3
by CK-12 //basic
Whole number percentage calculations - with compound units
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## Find the Whole Given the Percents and Parts - Example 4
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Decimal percentage calculations - without units
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## Find the Whole Given the Percents and Parts - Example 5
by CK-12 //basic
Decimal percentage calculations - with units
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## Find the Whole Given the Percents and Parts - Example 6
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Percentage calculation - one-step word problems
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## Find the Whole Given the Percents and Parts - Example 7
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Percentage calculation - two-step word problems
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## Find the Whole Given the Percents and Parts - Example 8
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Percentage calculation - multi-step word problems
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## Mixtures
by Kyle Bardman //basic
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## Proportions to Find Base b
by Kyle Bardman //basic | 803 | 3,505 | {"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-2017-17 | latest | en | 0.892017 |
https://davidreagan.net/journal/veracity-bible/ | 1,600,800,208,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400206329.28/warc/CC-MAIN-20200922161302-20200922191302-00017.warc.gz | 344,169,714 | 4,091 | So, I was just thinking about why I believe that the Bible is God’s Word. Initially I believed that the Bible was true on simple belief. That resulted in my salvation. Then I believed in the Bible due to personal revelation from God that it was His word. After a while I ran into people questioning the accuracy of the Bible. So after some research, I saw that it is one of the most historically accurate documents around. Basically something like the Odyssey has fewer accurate surviving documents than the Bible does and historians still consider it accurate. Thus, historically, the Bible is accurate.
It just occurred to me that this somewhat fits the mathematical induction method of proof.
You assume that the first is true, then you prove that the next is true. If the next is true, then it is true for everything. Prove P(N). Assume P(1). Show P(N+1).
So, I assume the Bible is true. The next step would be that I would be saved if I repented. I repented and I know I am saved. Another next step, could be that the Bible is historically true. Therefore the entire Bible is true.
It doesn’t totally fit the method. But it is interesting! :D | 249 | 1,151 | {"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-2020-40 | latest | en | 0.973923 |
http://sviluppo-software.info/nrich-problem-solving-key-stage-1-74/ | 1,576,159,133,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540543850.90/warc/CC-MAIN-20191212130009-20191212154009-00369.warc.gz | 133,476,930 | 8,692 | # NRICH PROBLEM SOLVING KEY STAGE 1
All the Digits Age 7 to 11 Challenge Level: Throw the dice and decide whether to double or halve the number. How is … used in …? There is a gap between where you are and getting started on a path to a solution. Kids whose palms sweat over speed drills may welcome the slower pace and creativity required to solve problems on NRICH. Continue reading Show less.
Check out our advice for setting computer time limits for all kids. What two-digit numbers can you make with these two dice? Scroll down to see groups of tasks from the site which will give learners experience of specific skills. How could you answer these questions using a picture, with things, with numbers or symbols? What patterns can you make with a set of dominoes?
## Trial and Improvement at KS1
To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. Altogether there were 8 heads and 22 feet.
Hundreds of learning activities build skills and confidence. Have shage go at sharing out some cookies like the characters in the story! Those children who are becoming fluent at trial and improvement will then want to adjust the dice to see if they can make 18 in another way, rather than trying another random arrangement.
GENOVESE THESIS LYRICS
The sum of each side of the triangle should equal the number in its centre. Dicey Addition Age 5 to 11 Challenge Level: How could you answer these questions using a picture, with things, with numbers or symbols? How old is your kid? What questions do I ask?
You can read more about types of recording in this article. Clapping Times Age 5 to 7 Challenge Level: Use ‘voting bricks’ to vote for a book at story time. Kids whose palms sweat over speed drills may welcome the slower pace and creativity required to solve problems on NRICH. Is problem solving at the heart of your curriculum? Children will need support to develop their proficiency with written recording.
Because the whole point of learning maths is to be able to solve problems. This article, written for primary teachers, discusses what we mean by ‘problem-solving skills’ and draws attention to NRICH tasks which can help develop specific skills. Some may take a short time, like Shut the Boxwhilst others may intrigue and challenge over more than one lesson, like Dice in a Corner. This problem is designed to ptoblem children to learn, and to use, the two and three times tables. How tall stagw it on Monday?
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# Multiplication and Division KS1 :
We trust you will find it useful and we are always interested in your feedback and experiences as you explore problem solving together with the children in your class. Choice of task NRICH offers a wide range of rich nridh to engage learners in the problem-solving process. As you come down the ladders of the Tall Tower you collect useful spells.
How does my body language change through the lesson? The winner is the first to make the total Number Detective Age 5 to 11 Challenge Level: What happens when you add pairs of the numbers together?
If you hang two weights on one side of this balance, in how many different ways can you hang three weights on the other side for it to be balanced? Follow the clues to find the mystery number.
Talk to your kids about What would the solution be in each case? Tasks for KS1 children which focus on working systematically. | 747 | 3,580 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2019-51 | latest | en | 0.934782 |
https://ask.sagemath.org/question/56805/factor-symbolic-expression/ | 1,716,638,430,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058822.89/warc/CC-MAIN-20240525100447-20240525130447-00210.warc.gz | 83,531,752 | 13,874 | # factor symbolic expression
In the following code, if I add .collect(p) to the pp*A*qq in the before last expression, I have an error. Why ?
LE=LatexExpr
x,y,p,q=SR.var('x, y, p, q')
A=matrix(SR,2,2,[x,-x,-x,0])
B=matrix(SR,2,2,[-x,y,x,0])
show(LE(r"\boldsymbol{A} = "),A, LE(r"\,\,\,\,\,\text{et}\,\,\,\,\,"),LE(r"\boldsymbol{B} = "),B)
pp = vector(SR,[p, 1-p])
qq = vector(SR,[q, 1-q])
show(LE(r"\boldsymbol{p} = "),pp, LE(r"\,\,\,\,\,\text{et}\,\,\,\,\,"),LE(r"\boldsymbol{q} = "),qq)
EGe0 = pp*A*qq.collect(p)
show(LE(r"\mathbb{E}G_e^0 = "),EGe0)
edit retag close merge delete
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EGe0 = (pp*A*qq).collect(p)
more
Thanks but I have an other problem y is not recognized as a variable. I don't undersand since itis decalred. I have tried without comas. But nothing work so I can't see the effect of the parenthesis.
( 2021-04-25 17:58:13 +0200 )edit
I'm not sure what problem you are talking about as your code runs without an issue at Sagecell: https://sagecell.sagemath.org/?q=fmhwan
( 2021-04-25 18:30:58 +0200 )edit
Thanks it works
( 2021-04-25 22:28:27 +0200 )edit
## Stats
Seen: 174 times
Last updated: Apr 25 '21 | 432 | 1,169 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-22 | latest | en | 0.831718 |
http://de.metamath.org/mpeuni/wl-ax11-lem8.html | 1,726,747,040,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652028.28/warc/CC-MAIN-20240919093719-20240919123719-00874.warc.gz | 6,168,267 | 5,078 | Mathbox for Wolf Lammen < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > wl-ax11-lem8 Structured version Visualization version GIF version
Theorem wl-ax11-lem8 32548
Description: Lemma. (Contributed by Wolf Lammen, 30-Jun-2019.)
Assertion
Ref Expression
wl-ax11-lem8 ((∀𝑢 𝑢 = 𝑦 ∧ ¬ ∀𝑥 𝑥 = 𝑦) → (∀𝑢𝑥[𝑢 / 𝑦]𝜑 ↔ ∀𝑦𝑥𝜑))
Distinct variable group: 𝑥,𝑢
Allowed substitution hints: 𝜑(𝑥,𝑦,𝑢)
Proof of Theorem wl-ax11-lem8
StepHypRef Expression
1 axc11n 2295 . . 3 (∀𝑦 𝑦 = 𝑥 → ∀𝑥 𝑥 = 𝑦)
21con3i 149 . 2 (¬ ∀𝑥 𝑥 = 𝑦 → ¬ ∀𝑦 𝑦 = 𝑥)
3 wl-ax11-lem1 32541 . . . . . . 7 (∀𝑢 𝑢 = 𝑦 → (∀𝑢 𝑢 = 𝑥 ↔ ∀𝑦 𝑦 = 𝑥))
43notbid 307 . . . . . 6 (∀𝑢 𝑢 = 𝑦 → (¬ ∀𝑢 𝑢 = 𝑥 ↔ ¬ ∀𝑦 𝑦 = 𝑥))
54anbi1d 737 . . . . 5 (∀𝑢 𝑢 = 𝑦 → ((¬ ∀𝑢 𝑢 = 𝑥 ∧ ∀𝑢𝑥[𝑢 / 𝑦]𝜑) ↔ (¬ ∀𝑦 𝑦 = 𝑥 ∧ ∀𝑢𝑥[𝑢 / 𝑦]𝜑)))
64anbi1d 737 . . . . . . . 8 (∀𝑢 𝑢 = 𝑦 → ((¬ ∀𝑢 𝑢 = 𝑥 ∧ ∀𝑥[𝑢 / 𝑦]𝜑) ↔ (¬ ∀𝑦 𝑦 = 𝑥 ∧ ∀𝑥[𝑢 / 𝑦]𝜑)))
7 axc11n 2295 . . . . . . . . . . 11 (∀𝑥 𝑥 = 𝑦 → ∀𝑦 𝑦 = 𝑥)
87con3i 149 . . . . . . . . . 10 (¬ ∀𝑦 𝑦 = 𝑥 → ¬ ∀𝑥 𝑥 = 𝑦)
9 wl-ax11-lem4 32544 . . . . . . . . . . . 12 𝑥(∀𝑢 𝑢 = 𝑦 ∧ ¬ ∀𝑥 𝑥 = 𝑦)
10 sbequ12 2097 . . . . . . . . . . . . . . 15 (𝑦 = 𝑢 → (𝜑 ↔ [𝑢 / 𝑦]𝜑))
1110equcoms 1934 . . . . . . . . . . . . . 14 (𝑢 = 𝑦 → (𝜑 ↔ [𝑢 / 𝑦]𝜑))
1211sps 2043 . . . . . . . . . . . . 13 (∀𝑢 𝑢 = 𝑦 → (𝜑 ↔ [𝑢 / 𝑦]𝜑))
1312adantr 480 . . . . . . . . . . . 12 ((∀𝑢 𝑢 = 𝑦 ∧ ¬ ∀𝑥 𝑥 = 𝑦) → (𝜑 ↔ [𝑢 / 𝑦]𝜑))
149, 13albid 2077 . . . . . . . . . . 11 ((∀𝑢 𝑢 = 𝑦 ∧ ¬ ∀𝑥 𝑥 = 𝑦) → (∀𝑥𝜑 ↔ ∀𝑥[𝑢 / 𝑦]𝜑))
1514ex 449 . . . . . . . . . 10 (∀𝑢 𝑢 = 𝑦 → (¬ ∀𝑥 𝑥 = 𝑦 → (∀𝑥𝜑 ↔ ∀𝑥[𝑢 / 𝑦]𝜑)))
168, 15syl5 33 . . . . . . . . 9 (∀𝑢 𝑢 = 𝑦 → (¬ ∀𝑦 𝑦 = 𝑥 → (∀𝑥𝜑 ↔ ∀𝑥[𝑢 / 𝑦]𝜑)))
1716pm5.32d 669 . . . . . . . 8 (∀𝑢 𝑢 = 𝑦 → ((¬ ∀𝑦 𝑦 = 𝑥 ∧ ∀𝑥𝜑) ↔ (¬ ∀𝑦 𝑦 = 𝑥 ∧ ∀𝑥[𝑢 / 𝑦]𝜑)))
186, 17bitr4d 270 . . . . . . 7 (∀𝑢 𝑢 = 𝑦 → ((¬ ∀𝑢 𝑢 = 𝑥 ∧ ∀𝑥[𝑢 / 𝑦]𝜑) ↔ (¬ ∀𝑦 𝑦 = 𝑥 ∧ ∀𝑥𝜑)))
1918dral1 2313 . . . . . 6 (∀𝑢 𝑢 = 𝑦 → (∀𝑢(¬ ∀𝑢 𝑢 = 𝑥 ∧ ∀𝑥[𝑢 / 𝑦]𝜑) ↔ ∀𝑦(¬ ∀𝑦 𝑦 = 𝑥 ∧ ∀𝑥𝜑)))
20 wl-ax11-lem7 32547 . . . . . 6 (∀𝑢(¬ ∀𝑢 𝑢 = 𝑥 ∧ ∀𝑥[𝑢 / 𝑦]𝜑) ↔ (¬ ∀𝑢 𝑢 = 𝑥 ∧ ∀𝑢𝑥[𝑢 / 𝑦]𝜑))
21 wl-ax11-lem7 32547 . . . . . 6 (∀𝑦(¬ ∀𝑦 𝑦 = 𝑥 ∧ ∀𝑥𝜑) ↔ (¬ ∀𝑦 𝑦 = 𝑥 ∧ ∀𝑦𝑥𝜑))
2219, 20, 213bitr3g 301 . . . . 5 (∀𝑢 𝑢 = 𝑦 → ((¬ ∀𝑢 𝑢 = 𝑥 ∧ ∀𝑢𝑥[𝑢 / 𝑦]𝜑) ↔ (¬ ∀𝑦 𝑦 = 𝑥 ∧ ∀𝑦𝑥𝜑)))
235, 22bitr3d 269 . . . 4 (∀𝑢 𝑢 = 𝑦 → ((¬ ∀𝑦 𝑦 = 𝑥 ∧ ∀𝑢𝑥[𝑢 / 𝑦]𝜑) ↔ (¬ ∀𝑦 𝑦 = 𝑥 ∧ ∀𝑦𝑥𝜑)))
24 pm5.32 666 . . . 4 ((¬ ∀𝑦 𝑦 = 𝑥 → (∀𝑢𝑥[𝑢 / 𝑦]𝜑 ↔ ∀𝑦𝑥𝜑)) ↔ ((¬ ∀𝑦 𝑦 = 𝑥 ∧ ∀𝑢𝑥[𝑢 / 𝑦]𝜑) ↔ (¬ ∀𝑦 𝑦 = 𝑥 ∧ ∀𝑦𝑥𝜑)))
2523, 24sylibr 223 . . 3 (∀𝑢 𝑢 = 𝑦 → (¬ ∀𝑦 𝑦 = 𝑥 → (∀𝑢𝑥[𝑢 / 𝑦]𝜑 ↔ ∀𝑦𝑥𝜑)))
2625imp 444 . 2 ((∀𝑢 𝑢 = 𝑦 ∧ ¬ ∀𝑦 𝑦 = 𝑥) → (∀𝑢𝑥[𝑢 / 𝑦]𝜑 ↔ ∀𝑦𝑥𝜑))
272, 26sylan2 490 1 ((∀𝑢 𝑢 = 𝑦 ∧ ¬ ∀𝑥 𝑥 = 𝑦) → (∀𝑢𝑥[𝑢 / 𝑦]𝜑 ↔ ∀𝑦𝑥𝜑))
Colors of variables: wff setvar class Syntax hints: ¬ wn 3 → wi 4 ↔ wb 195 ∧ wa 383 ∀wal 1473 [wsb 1867 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-10 2006 ax-12 2034 ax-13 2234 ax-wl-11v 32540 This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 This theorem is referenced by: wl-ax11-lem10 32550
Copyright terms: Public domain W3C validator | 2,348 | 3,107 | {"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-2024-38 | latest | en | 0.206398 |
https://au.mathworks.com/matlabcentral/cody/problems/405-back-to-basics-15-classes/solutions/437116 | 1,586,046,810,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370526982.53/warc/CC-MAIN-20200404231315-20200405021315-00141.warc.gz | 349,972,323 | 15,611 | Cody
# Problem 405. Back to basics 15 - classes
Solution 437116
Submitted on 1 May 2014 by Ryszard Maciol
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = @sin; y_correct = 'function_handle'; assert(isequal(input_class(x),y_correct))
2 Pass
%% x = {1}; y_correct = 'cell'; assert(isequal(input_class(x),y_correct))
3 Pass
%% x = [1]; y_correct = 'double'; assert(isequal(input_class(x),y_correct))
4 Pass
%% x = uint32(1); y_correct = 'uint32'; assert(isequal(input_class(x),y_correct))
5 Pass
%% x = 'abcd'; y_correct = 'char'; assert(isequal(input_class(x),y_correct)) | 210 | 711 | {"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-2020-16 | latest | en | 0.360077 |
http://ned.ipac.caltech.edu/level5/March01/Schneider/Schneider8.html | 1,529,899,693,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867424.77/warc/CC-MAIN-20180625033646-20180625053646-00461.warc.gz | 238,316,930 | 3,989 | ### 8. THE HI LUMINOSITY FUNCTION
The Arecibo HI survey provides us with all of the basic data we need to determine the HI luminosity function, and since HI mass and luminosity are normally proportional, we can phrase this directly in terms of HI masses. Assuming that we are dealing with a representative region of space, the ~ 6000 Mpc3 volume surveyed overall should allow us to determine the typical number of objects of different HI masses. Unfortunately, the distance to which the survey is sensitive for low masses is much smaller (see Figure 11). At an HI mass of 107 M, the survey only samples ~ 15 Mpc3 of nearby space, which is probably much less representative of the universe as a whole. Nevertheless, proceeding with fingers crossed, in Figure 17 I show the resulting density of galaxies in equal logarithmic intervals of HI mass.
Figure 17. The HI mass function. The dotted-line histogram shows the number counts of galaxies actually detected in each half-decade logarithmic interval, with N1/2 error bars. The solid histogram shows the sensitivity-corrected volume density of these same galaxies. Four squares show similar results from the HI survey of Henning (1995). The dashed curve is a Schechter luminosity function with = -1.7 and MHI* = 1010 M.
Figure 17 shows the actual counts as a function of HI mass (dotted line) as well as the density of those sources (solid line) based on the limiting volume within which a source of that particular mass could be detected. Note how the number counts of detected sources drop to tiny numbers even as the number density grows to huge values. The Figure shows how N errors would affect the counts to give an idea of the uncertainty. The greater source of error is systematic: the resulting luminosity function depends critically on our estimate of how big a volume has been effectively surveyed for the low mass objects. To carry out this density estimate as accurately as possible, it is important to examine each source in more detail than just the average volume for that mass range. The method applied is to calculate the maximum distance (and hence volume) within which a similar signal could be detected accounting for its total HI mass, its HI linewidth, and the receiver sensitivity as a function of frequency.
The square symbols in the figure show the results from Henning's (1995) analysis of her Green Bank survey. The differences for massive galaxies suggest the range of uncertainty in sensitivity/volume estimates or possibly in variations across the sky. For the three lowest mass bins, the Arecibo survey was sensitive to several times more volume than the Green Bank survey, so it would not be expected that any galaxies would be detected in the Green Bank survey.
The dashed curve is a Schechter function with a power law index of = -1.7. It appears that HI is distributed among galaxies in such a way that there is a higher fraction of HI in small galaxies than the corresponding fraction of optical luminosity.
I close by returning to the basic issue raised at the beginning of this chapter. We are finally in a position to make a stab at the elusive mass function of galaxies. The path we've followed to reach this point has many potential pitfalls along the way, and the less adventurous may think it unwise to press forward. Nevertheless, with no small amount of trepidation, I offer Figure 18 for your perusal. This Figure shows the total mass density expected to be present in each mass interval based on dynamical mass estimates for each of the galaxies in the Arecibo survey.
Figure 18. The mass function of galaxies projected from the HI-selected sample. Masses of each object were estimated from their size and rotation speed, and the total mass density contributed by galaxies in each logarithmic interval is calculated using the HI sensitivity corrections. N1/2 error bars are based on the number of objects within each interval. The scale on the right gives the corresponding contribution of each mass interval to the closure density of the universe.
Each galaxy's mass was estimated from its rotation speed (taken from the width of the HI profile) corrected for the inclination estimated from the the optical image. The Holmberg radius of each galaxy was then used to find its mass:
(8)
For the nine objects where the optical data were incomplete (and for the face-on objects whose rotation speed could not be estimated), we roughly estimated the total mass to be five times the HI mass, which is probably a conservative estimate. The same correction factors for sensitivity/volume were applied to each object according to its HI signal, and the mass densities in each interval were totaled to arrive at the mass function shown.
Since HI and total dynamical mass are not perfectly correlated, it is certain that this estimate is not going to be quite right, but the result suggests that galaxies make up at most about two percent of the closure density of the universe. The low mass galaxies, while not insignificant, do not appear to contribute as much as the high mass objects. The small number counts at the low mass end, however, leave this conclusion quite uncertain. Clearly, a deeper survey is needed, and I and others are currently in the process of examining new data collected at Arecibo while the telescope was partially crippled as it was being upgraded.
Looking toward the future, the total mass contributed by the galaxies found here is more than a factor of two lower than most estimates of the baryonic mass density determined from Big Bang deuterium synthesis. If, as some people believe, galaxies are mostly composed of non-baryonic matter, then only a fraction of the mass density found here is baryonic. It appears then that the story of where all of the primordial hydrogen went has not yet been told.
Acknowledgments This work would not have been possible without the enormous effort of John Spitzak to complete the Arecibo survey. Portions of this project were supported by NSF Presidential Young Investigator award AST-9158096 and a UMass Faculty Research Grant. I would also like to thank the staff of the Arecibo Observatory for a great deal of assistance. | 1,247 | 6,193 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-26 | latest | en | 0.915704 |
https://studymaterialcenter.in/tag/rd-sharma-class-11/ | 1,660,284,669,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571584.72/warc/CC-MAIN-20220812045352-20220812075352-00444.warc.gz | 494,551,747 | 23,254 | # rd sharma class 11
## RD Sharma Class 11 Probability Exercise 33.4 Solutions
On this page you will find Maths RD Sharma Class 11 Probability Exercise 33.4 Solutions. The solutions provided here are to help students practice math problems and get better at solving difficult chapter questions. Maths chapter 33 for class 11 deals with the topic of triangles and it is one of the most important chapters. …
## RD Sharma Class 11 Probability Exercise 33.3 Solutions
On this page you will find Maths RD Sharma Class 11 Probability Exercise 33.3 Solutions. The solutions provided here are to help students practice math problems and get better at solving difficult chapter questions. Maths chapter 33 for class 11 deals with the topic of triangles and it is one of the most important chapters. …
## RD Sharma Class 11 Probability Exercise 33.2 Solutions
On this page you will find Maths RD Sharma Class 11 Probability Exercise 33.2 Solutions. The solutions provided here are to help students practice math problems and get better at solving difficult chapter questions. Maths chapter 33 for class 11 deals with the topic of triangles and it is one of the most important chapters. …
## RD Sharma Class 11 Probability Exercise 33.1 Solutions
On this page you will find Maths RD Sharma Class 11 Probability Exercise 33.1 Solutions. The solutions provided here are to help students practice math problems and get better at solving difficult chapter questions. Maths chapter 33 for class 11 deals with the topic of triangles and it is one of the most important chapters. …
## RD Sharma Class 11 Statistics Exercise 32.7 Solutions
On this page you will find Maths RD Sharma Class 11 Statistics Exercise 32.7 Solutions. The solutions provided here are to help students practice math problems and get better at solving difficult chapter questions. Maths chapter 32 for class 11 deals with the topic of triangles and it is one of the most important chapters. …
## RD Sharma Class 11 Statistics Exercise 32.6 Solutions
On this page you will find Maths RD Sharma Class 11 Statistics Exercise 32.6 Solutions. The solutions provided here are to help students practice math problems and get better at solving difficult chapter questions. Maths chapter 32 for class 11 deals with the topic of triangles and it is one of the most important chapters. …
## RD Sharma Class 11 Statistics Exercise 32.5 Solutions
On this page you will find Maths RD Sharma Class 11 Statistics Exercise 32.5 Solutions. The solutions provided here are to help students practice math problems and get better at solving difficult chapter questions. Maths chapter 32 for class 11 deals with the topic of triangles and it is one of the most important chapters. …
## RD Sharma Class 11 Statistics Exercise 32.4 Solutions
On this page you will find Maths RD Sharma Class 11 Statistics Exercise 32.4 Solutions. The solutions provided here are to help students practice math problems and get better at solving difficult chapter questions. Maths chapter 32 for class 11 deals with the topic of triangles and it is one of the most important chapters. …
## RD Sharma Class 11 Statistics Exercise 32.3 Solutions
On this page you will find Maths RD Sharma Class 11 Statistics Exercise 32.3 Solutions. The solutions provided here are to help students practice math problems and get better at solving difficult chapter questions. Maths chapter 32 for class 11 deals with the topic of triangles and it is one of the most important chapters. …
## RD Sharma Class 11 Statistics Exercise 32.2 Solutions
On this page you will find Maths RD Sharma Class 11 Statistics Exercise 32.2 Solutions. The solutions provided here are to help students practice math problems and get better at solving difficult chapter questions. Maths chapter 32 for class 11 deals with the topic of triangles and it is one of the most important chapters. …
error: Content is protected !! | 805 | 3,920 | {"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} | 3.359375 | 3 | CC-MAIN-2022-33 | longest | en | 0.89602 |
https://www.teachoo.com/14007/652/Ex-14.5--5-ii/category/Ex-14.5/ | 1,726,466,202,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651676.3/warc/CC-MAIN-20240916044225-20240916074225-00224.warc.gz | 930,149,245 | 21,107 | Proving Statements true or false
Mathematical Reasoning
Serial order wise
### Transcript
Ex14.5, 5 Which of the following statements are true and which are false? In each case give a valid reason for saying so. (ii) q: The centre of a circle bisects each chord of the circle. The given statement is false. Here AB is a chord as it intersect the circle at two points, But it does not pass through center O So, center of circle does not bisect each chord of the circle | 111 | 469 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-38 | latest | en | 0.888674 |
https://www.sanfoundry.com/electric-drives-questions-answers-quiz/ | 1,721,770,809,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518115.82/warc/CC-MAIN-20240723194208-20240723224208-00402.warc.gz | 823,265,944 | 20,655 | # Electric Drives Questions and Answers – Solid-State Devices – Thyristors
This set of Electric Drives Quiz focuses on “Solid-State Devices – Thyristors”.
1. The value of latching current is more than holding current.
a) True
b) False
Explanation: The latching current is the minimum anode current at which SCR starts conducting. Holding current is the anode below which SCR stops conduction. The latching current is 2-3 times more than the holding current.
2. SCR is a current controlled device.
a) True
b) False
Explanation: SCR is a current controlled device. It can be explained using a two transistor model. When the gate pulse is applied, then it will start conduction.
3. Calculate the value of safety factor using the data: tc=12 msec, tq=6 msec.
a) 4
b) 2
c) 3
d) 6
Explanation: The value of the safety factor is the ratio of the circuit turn-off time and device turn-off time. S.F=tc÷tq=12÷6=2.
4. For successful commutation circuit turn-off time ______ device turn-off time.
a) <
b) >
c) =
d) ~
Explanation: For successful commutation circuit turn-off time must be greater than the device turn-off time. The excess stored charged carriers should be removed from the gate junction in order to avoid false triggering.
5. SCR is a __________ device.
a) Uncontrolled
b) Semi-controlled
c) Fully controlled
d) Voltage
Explanation: The SCR is a semi-controlled device. Turn-on time can be controlled by using a gate pulse but commutation circuit is required to turn off the device.
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6. Holding current of the SCR is ___________
a) > than latching current
b) < than latching current
c) = to latching current
d) ≥ than latching current
Explanation: Holding current of the SCR is less than the latching current. The value of latching current is 2-3 times than holding current.
7. If holding current of an SCR is 6 mA then its latching current should be ___________
a) 4 mA
b) 5 mA
c) 12 mA
d) 6 mA
Explanation: If holding current of an SCR is 6 mA then its latching current should be 12 mA. The latching current is 2-3 times more than the holding current.
8. If latching current of an SCR is 10 mA then its holding current should be ____________
a) 5 mA
b) 7 mA
c) 8 mA
d) 4 mA
Explanation: The latching current is 2-3 times more than the holding current. If a latching current of an SCR is 10 mA then its holding current should be 5 mA. Holding current=(latching current)÷2=10÷2=5 mA.
9. Which of the following is used to protect SCR from overcurrent?
a) Heat sink
b) Fuse
c) Snubber circuit
d) Zener diode
Explanation: Fuse and circuit breakers are used to protect SCR from overcurrent. When current exceeds the certain limit they cut down the supply of current in the circuit.
10. Which of the following is used to protect SCR from thermal conditions?
a) Heat sink
b) Voltage clamping device
c) Zener diode
d) Snubber circuit
Explanation: SCR is mounted on a heat sink. When temperature exceeded the rating of the SCR excess amount of heat is dissipated through the help of the heat sink.
11. Which of the following is used to protect SCR from d(i)/d(t) condition?
a) Inductor
b) Capacitor
c) Heat sink
d) Fuse
Explanation: Inductor is used to protect SCR from d(i)/d(t) condition. The inductor does not allow the sudden change in current.
12. Which of the following is used to protect SCR from d(v)/d(t) condition?
a) Capacitor
b) Fuse
c) Snubber circuit
d) Inductor
Explanation: Snubber circuit is used to protect SCR from d(v)/d(t) condition. Snubber circuit consists of a Capacitor and resistor. The capacitor does not allow the sudden change in voltage and resistor resists the discharging of the capacitor.
13. A snubber circuit is used with thyristor in ____________
a) Parallel
b) Series
c) Anti-parallel
d) Series or Parallel
Explanation: A snubber circuit is used with thyristor in parallel. Snubber circuit is used to protect SCR from d(v)/d(t) condition. Snubber circuit consists of a Capacitor and resistor.
14. When maximum power loss occurs?
a) Rise time
b) Delay time
d) Rise time or Delay time
Explanation: Rise time is very low (1-4 μsec). During rising, time anode voltage and anode current are very high. Power loss(V×I) is very high.
15. Calculate the value of Vpeak if crest factor=47 and Vr.m.s=6 V.
a) 282 V
b) 300 V
c) 290 V
d) 286 V
Explanation: The value of the crest factor is Vpeak÷Vr.m.s. The value of Vpeak is (crest factor) × Vr.m.s = 47×6 = 282 V. It signifies the ratio of Vpeak and Vr.m.s.
Sanfoundry Global Education & Learning Series – Electric Drives.
To practice all areas of Electric Drives for Quizzes, here is complete set of 1000+ Multiple Choice Questions and Answers.
If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected] | 1,256 | 4,846 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2024-30 | latest | en | 0.874961 |
https://pro.arcgis.com/en/pro-app/latest/arcpy/spatial-analyst/nbrannulus-class.htm | 1,713,108,276,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816879.72/warc/CC-MAIN-20240414130604-20240414160604-00552.warc.gz | 440,664,113 | 6,725 | # NbrAnnulus
## Summary
Defines an annulus neighborhood which is created by specifying an inner and outer circles' radii in either map units or number of cells.
## Discussion
Tools that use neighborhood annulus object: Block Statistics, Focal Statistics, Point Statistics, and Point Density.
This class is also available if you have an Image Analyst extension license, but only for the Focal Statistics tool.
The annulus shape is comprised of two circles, one inside the other to make a donut shape. Cells with centers that fall outside the radius of the smaller circle but inside the radius of the larger circle will be included in processing the neighborhood; therefore, the area that falls between the two circles constitutes the annulus neighborhood.
When the annulus object is used and the innerRadius or the outerRadius are not specified, then default values are provided.
The radius is identified in cells or map units, measured perpendicular to the x- or y-axis. When the radii is specified in map units, they are converted to radii in cell units. The resulting radii in cell units produces an area that most closely represents the area calculated by using the original radii in map units. Any cell center encompassed by the annulus will be included in the processing of the neighborhood.
## Syntax
`NbrAnnulus ({innerRadius}, {outerRadius}, {units})`
Parameter Explanation Data Type innerRadius The inner radius of an annulus neighborhood.(The default value is 1) Double outerRadius The outer radius of an annulus neighborhood.(The default value is 3) Double units Defines the units of the neighborhood.CELL—The unit of measurement is in cells.MAP—The units are in map coordinates.(The default value is CELL) String
## Properties
Property Explanation Data Type innerRadius(Read and Write) The inner radius of the annulus neighborhood. Double outerRadius(Read and Write) The outer radius of the annulus neighborhood. Double units(Read and Write) Defines the units of the neighborhood.CELL—The units are the number of cells.MAP—The units are in map coordinates. String
## Code sample
NbrAnnulus example 1 (Python window)
Demonstrates how to create a NbrAnnulus class and use it in the BlockStatistics tool within the Python window.
``````import arcpy
from arcpy import env
from arcpy.sa import *
env.workspace = "C:/sapyexamples/data"
outNbrA = BlockStatistics("block", NbrAnnulus(1, 2, "CELL"))
outNbrA.save("C:/sapyexamples/output/blstatnbra2")``````
NbrAnnulus example 2 (stand-alone script)
Implements the BlockStatistics tool using the NbrAnnulus class.
``````# Name: NbrAnnulus_Ex_02.py
# Description: Uses the NbrAnnulus object to execute BlockStatistics tool.
# Requirements: Spatial Analyst Extension
# Import system modules
import arcpy
from arcpy import env
from arcpy.sa import *
# Set environment settings
env.workspace = "C:/sapyexamples/data"
# Set local variables
inRaster = "block"
# Create the Neighborhood Object
# Execute BlockStatistics
outBlkStats = BlockStatistics(inRaster, myNbrAnnulus, "MINIMUM", "DATA")
# Save the output
outBlkStats.save("C:/sapyexamples/output/blstat_Ann3")`````` | 714 | 3,143 | {"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} | 2.6875 | 3 | CC-MAIN-2024-18 | longest | en | 0.867789 |
http://www.chegg.com/homework-help/questions-and-answers/unemployment-rate-persons-disability-typically-higher-disability-recent-statistics-report--q4490466 | 1,472,546,491,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982974951.92/warc/CC-MAIN-20160823200934-00127-ip-10-153-172-175.ec2.internal.warc.gz | 374,266,081 | 18,740 | The unemployment rate of persons with a disability is typically higher than for those with no disability. Recent statistics report that this rate is 14.5%. An advocacy group in a large city located in the southeastern region of the U.S. selected a random sample of 250 persons with a disability. What is the probability that no more than 30 persons in this sample are unemployed?
Question 3 options:
A small flower shop takes orders by phone and then one of the staff florists is assigned to prepare the arrangement. The time it takes to process phone orders is normally distributed with a mean of 6 minutes and a standard deviation of 2.5 minutes. The time it takes for an arrangement to be completed is normally distributed with a mean of 35 minutes and a standard deviation of 8.6 minutes. What is the probability that it will take more than 50 minutes to process a phone order and complete the floral arrangement at this flower shop?
Question 1 options:
0.2167 0.8413 0.1587 0.3413 0.6843
The time it takes to process phone orders in a small florist/gift shop is normally distributed with a mean of 6 minutes and a standard deviation of 1.24 minutes. What cutoff value would separate the 2.5% of orders that take the most time to process?
Question 4 options:
10.01 minutes 3.52 minutes 4.76 minutes 8.48 minutes 11.98 minutes
A small flower shop takes orders by phone and then one of the staff florists is assigned to prepare the arrangement. The time it takes to process phone orders is normally distributed with a mean of 6 minutes and a standard deviation of 2.5 minutes. The time it takes for an arrangement to be completed is normally distributed with a mean of 35 minutes and a standard deviation of 8.6 minutes. What is the standard deviation for the total time to process a phone order and complete the floral arrangement at this flower shop (assuming times are independent)?
Question 7 options:
Based on data collected from its production processes, Crosstiles Inc. determines that the breaking strength of its most popular porcelain tile is normally distributed with a mean of 400 pounds per square inch and a standard deviation of 12.5 pounds per square inch. Based on the 68-95-99.7 Rule, about what percent of its popular porcelain tile will have breaking strengths between 375 and 425 pounds per square inch?
Question 8 options:
95% 47.5% 84% 32% 68%
At a local manufacturing plant, employees must complete new machine set ups within 30 minutes. New machine set-up times can be described by a normal model with a mean of 22 minutes and a standard deviation of four minutes. The typical worker needs five minutes to adjust to his or her surroundings before beginning duties. What percent of new machine set ups are completed within 25 minutes to allow for this?
Question 9 options:
27.3% 72..7% 22..7% 77.3% None of these.
Show transcribed image text | 654 | 2,877 | {"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} | 3.46875 | 3 | CC-MAIN-2016-36 | latest | en | 0.937616 |
https://www.univerkov.com/the-area-of-the-isosceles-triangle-abc-with-the-base-of-the-ac-is-4800-cm2/ | 1,632,247,256,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057225.57/warc/CC-MAIN-20210921161350-20210921191350-00250.warc.gz | 1,052,477,000 | 5,658 | # The area of the isosceles triangle ABC with the base of the AC is 4800 cm2.
The area of the isosceles triangle ABC with the base of the AC is 4800 cm2. Find the heights drawn to the sides of the triangle if AB = 100cm.
1. Vertices of triangle A, B, C. CE and AK – heights. AB = BC = 100 centimeters.
2. We calculate the length of the height CE, using the formula for calculating the area (S) of a given triangle:
S = AB x CE / 2 = 100 x CE / 2 = 4800 centimeters².
CE = 2S / 100 = 2 x 4800/100 = 96 centimeters.
3. We calculate the length of the height of the AK using another formula for calculating the area of a triangle:
S = BC x AK / 2.
AK = 2S / BC = 2 x 4800/100 = 96 centimeters.
Answer: AK = CE = 96 centimeters.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.
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https://www.airmilescalculator.com/distance/hmv-to-got/ | 1,611,456,511,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703544403.51/warc/CC-MAIN-20210124013637-20210124043637-00606.warc.gz | 643,441,623 | 40,493 | # Distance between Hemavan (HMV) and Gothenburg (GOT)
Flight distance from Hemavan to Gothenburg (Hemavan Airport – Göteborg Landvetter Airport) is 571 miles / 919 kilometers / 496 nautical miles. Estimated flight time is 1 hour 34 minutes.
Driving distance from Hemavan (HMV) to Gothenburg (GOT) is 769 miles / 1237 kilometers and travel time by car is about 15 hours 56 minutes.
## Map of flight path and driving directions from Hemavan to Gothenburg.
Shortest flight path between Hemavan Airport (HMV) and Göteborg Landvetter Airport (GOT).
## How far is Gothenburg from Hemavan?
There are several ways to calculate distances between Hemavan and Gothenburg. Here are two common methods:
Vincenty's formula (applied above)
• 571.210 miles
• 919.273 kilometers
• 496.368 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 569.932 miles
• 917.216 kilometers
• 495.257 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Hemavan Airport
City: Hemavan
Country: Sweden
IATA Code: HMV
ICAO Code: ESUT
Coordinates: 65°48′21″N, 15°4′58″E
B Göteborg Landvetter Airport
City: Gothenburg
Country: Sweden
IATA Code: GOT
ICAO Code: ESGG
Coordinates: 57°39′46″N, 12°16′47″E
## Time difference and current local times
There is no time difference between Hemavan and Gothenburg.
CET
CET
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 109 kg (240 pounds).
## Frequent Flyer Miles Calculator
Hemavan (HMV) → Gothenburg (GOT).
Distance:
571
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
571
Round trip? | 495 | 1,860 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-04 | latest | en | 0.818246 |
https://www.sanfoundry.com/aircraft-performance-questions-answers-measurement-airspeed/ | 1,702,144,577,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100942.92/warc/CC-MAIN-20231209170619-20231209200619-00529.warc.gz | 1,076,394,559 | 19,060 | # Aircraft Performance Questions and Answers – Measurement of Airspeed
This set of Aircraft Performance Multiple Choice Questions & Answers (MCQs) focuses on “Measurement of Airspeed”.
1. Which of the following is the correct formula for speed of sound?
a) $$\sqrt{\gamma RT}$$
b) $$\sqrt{pRT}$$
c) $$\sqrt{\rho Rt}$$
d) $$\sqrt{\gamma R0T}$$
Explanation: The correct formula for speed of sound is $$\sqrt{\gamma RT}$$ where R=characteristic gas constant, T=temperature and γ is the ratio of specific heat at constant pressure to that of specific heat at constant volume.
2. What is the speed of sound in air at an altitude where temperature is 299K?
a) 346.61K
b) 343.61K
c) 71.61 °C
d) 70.61 °C
Explanation: The answer is 346.61K. Given T=299K. We know that R=287J/kg-K, γ for air is 1.4.
From a=$$\sqrt{\gamma RT}$$
a=$$\sqrt{1.4*287*299}$$
a=346.61K.
3. What is the relation between pressure and air speed in isentropic relations?
a) $$\frac{p_1}{p_2}$$=$$\Big\{1+\frac{\gamma-1}{2}(\frac{V1}{a1})^2\Big\}^\frac{\gamma}{\gamma-1}$$
b) $$\frac{p_1}{p_2}$$=$$\Big\{1+\frac{\gamma+1}{2}(\frac{V1}{a1})^2\Big\}^\frac{\gamma}{\gamma+1}$$
c) $$\frac{p_1}{p_2}$$=$$\Big\{1+\frac{\gamma+1}{2}(\frac{V1}{a1})^2\Big\}^\frac{\gamma}{\gamma-1}$$
d) $$\frac{p_1}{p_2}$$=$$\Big\{1+\frac{\gamma-1}{2}(\frac{V1}{a1})^2\Big\}^\frac{\gamma}{\gamma+1}$$
Explanation: The relation between pressure and air speed in isentropic relations is
$$\frac{p_1}{p_2}$$=$$\Big\{1+\frac{\gamma-1}{2}(\frac{V1}{a1})^2\Big\}^\frac{\gamma}{\gamma-1}$$ where p1, p2 are pressures at two points, V1=velocity at one point, a1=speed of sound at point one and γ is the ratio of specific heat at constant pressure to that of specific heat at constant volume.
4. What is the pressure ratio of an aircraft moving in air at a mach number 1?
a) 1.893
b) 1.558
c) 1.444
d) 1.555
Explanation: The answer is 0.458. Given M=1. We know γ of air is 1.4. From $$\frac{p_1}{p_2}=\Big\{1+\frac{\gamma-1}{2}(M)^2\Big\}^\frac{\gamma}{\gamma-1}$$
On substituting the values $$\frac{p_1}{p_2}=\Big\{1+\frac{1.4-1}{2} 1^2\Big\}^\frac{1.4}{1.4-1}$$
$$\frac{p_1}{p_2}$$=1.893.
5. What is the pressure ratio of an aircraft moving in air at a velocity 500m/s and speed of sound is 244 m/s?
a) 4.556
b) 3.327
c) 6.256
d) 2.565
Explanation: The answer is 0.458. Given V=500m/s, a=244m/s. We know γ of air is 1.4. From $$\frac{p_1}{p_2}$$=$$\Big\{1+\frac{\gamma-1}{2}(\frac{V1}{a1})^2\Big\}^\frac{\gamma}{\gamma-1}$$
On substituting the values $$\frac{p_1}{p_2}=\Big\{1+\frac{500}{244}\Big\}^\frac{1.4}{1.4-1}$$
$$\frac{p_1}{p_2}$$=3.327.
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6. Which of the following is the correct isentropic relation between pressure and temperature?
a) $$\frac{p_1}{p_2}=(\frac{T_2}{T_1})^\frac{\gamma}{1-\gamma}$$
b) $$\frac{p_1}{p_2}=(\frac{T_2}{T_1})^\frac{\gamma-1}{\gamma}$$
c) $$\frac{p_1}{p_2}=(\frac{T_2}{T_1})^\frac{\gamma+1}{\gamma}$$
d) $$\frac{p_1}{p_2}=(\frac{T_2}{T_1})^\frac{\gamma}{\gamma-1}$$
Explanation: $$\frac{p_1}{p_2}=(\frac{T_2}{T_1})^\frac{\gamma}{1-\gamma}$$ is the correct isentropic relation between pressure and temperature where p1, p2 are pressures, T1, T2 are temperatures and γ is the ratio of specific heat at constant pressure to that of specific heat at constant volume.
7. What is the speed of sound where density and pressure are 1.225kg/m3 and 101306N/m2?
a) 340.26m/s
b) 330.26m/s
c) 313 m/s
d) 325 m/s
Explanation: The answer is 340.26m/s. Given P=101306N/m2, ρ=1.225kg/m3 and we know that γ for air is 1.4. From the formula, a=$$\sqrt{\frac{\gamma P}{\rho}}$$
a=$$\sqrt{\frac{\gamma*101306}{1.225}}$$
a=340.26m/s.
8. What is the formula for speed of sound in terms of pressure and density?
a) a=$$\sqrt{\frac{\gamma P}{\rho}}$$
b) a=$$\sqrt{\frac{\gamma\rho}{P}}$$
c) a=$$\sqrt{\frac{\gamma RP}{\rho}}$$
d) a=$$\sqrt{\frac{\gamma P}{\rho R}}$$
Explanation: The formula for speed of sound in terms of pressure and density is given by a=$$\sqrt{\frac{\gamma P}{\rho}}$$ where a=speed of sound, P=pressure, ρ=density and γ is the ratio of specific heat at constant pressure to that of specific heat at constant volume.
9. What is the relation between equivalent air speed and pressure ratio?
a) Ve=V$$\sqrt{\sigma}$$
b) Ve=$$\frac{V}{\sqrt{\sigma}}$$
c) Ve=Vσ2
d) Ve=Vσ-2
Explanation: Ve=V$$\sqrt{\sigma}$$ is the relation between equivalent air speed and pressure ratio where Ve is the equivalent air speed, V=velocity and σ is pressure ratio.
10. What is the equivalent air speed where velocity is 330m/s and pressure ratio is 8.447?
a) 959.1m/s
b) 1000m/s
c) 981m/s
d) 954m/s
Explanation: The answer is 959.1m/s. Given V=330m/s, σ=8.447. From the equation Ve=V$$\sqrt{\sigma}$$
Ve=330$$\sqrt{8.447}$$
Ve=959.1m/s.
11. What is equivalent air speed?
a) The calibrated air speed corrected for scale-altitude error
b) The true air speed corrected for scale-altitude error
c) The indicated air speed corrected for scale-altitude error
d) The ground air speed corrected for scale-altitude error
Explanation: Equivalent air speed is the calibrated air speed corrected for scale-altitude error. The correction is done in calibrated equation of the airspeed indicator which is a function of calibrated air speed and height.
12. Which of the following is the full-law calibration equation?
a) pp-p=p0$$\Bigg[\Big\{1+\frac{\gamma-1}{2}\left(\frac{V_c}{a_0}\right)^2\Big\}^{\frac{\gamma}{\gamma-1}}-1\Bigg]$$
b) pp-p=p0$$\Bigg[\Big\{1+\frac{\gamma+1}{2}\left(\frac{V_c}{a_0}\right)^2\Big\}^{\frac{\gamma}{\gamma-1}}-1\Bigg]$$
c) p-p0=p0$$\Bigg[\Big\{1+\frac{\gamma-1}{2}\left(\frac{V_c}{a_0}\right)^2\Big\}^{\frac{\gamma}{\gamma-1}}-1\Bigg]$$
d) pp-p=p0$$\Bigg[\Big\{1+\frac{\gamma-1}{2}\left(\frac{V_c}{a_0}\right)^2\Big\}^{\frac{\gamma}{\gamma+1}}-1\Bigg]$$
Explanation: The full-law calibration equation is pp-p=p0$$\Bigg[\Big\{1+\frac{\gamma-1}{2}\left(\frac{V_c}{a_0}\right)^2\Big\}^{\frac{\gamma}{\gamma-1}}-1\Bigg]$$ where pp, p, p0 are pressures, Vc is calibrated air speed and a0 is speed of sound and γ is the ratio of specific heat at constant pressure to that of specific heat at constant volume.
Sanfoundry Global Education & Learning Series – Aircraft Performance.
To practice all areas of Aircraft Performance, here is complete set of 1000+ Multiple Choice Questions and Answers. | 2,323 | 6,360 | {"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} | 4.09375 | 4 | CC-MAIN-2023-50 | longest | en | 0.746902 |
https://cboard.cprogramming.com/c-programming/132462-logic-calculator.html?s=0fcb5878829b96ad8803e30c215c2624 | 1,597,050,908,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738653.47/warc/CC-MAIN-20200810072511-20200810102511-00254.warc.gz | 245,630,487 | 14,726 | 1. ## logic calculator
I have a project to make a logic calculator like the one described below, but it literally doesn't make "logic" sense to me. I'm sure I won't have a problem programming it once I know what the hell it is asking. Any help? THANKS!
The program in the text (see below) prompts the user for two numeric operands and an operator ( “+”, “-“, etc.) then calculates and prints the result of applying the selected operator to the two operands. Your task is to use the same sort of design to do the following:
· Prompt the user for two character variables and a logical operator.
· For example “Please enter a logical expression: T O F”. This should print “True” where “O” stands for OR (right?)
· Do this for the following logical operators - OR, AND, NOT. To make the reading with scanf simpler, use O for OR, A for AND and N for NOT. (NOT is a little weird in that it is a unary, operator, not a binary one, so you can ignore the second operand and just figure the negation (NOT) on the first operand alone).
· Print the result as “True” or “False”.
How to Complete this Project
First, get the example simple expression evaluation program on pages 85 and 86 of PIC to compile and run in your environment.
Add logic to convert the letters you read “T” and “F” to 1 and 0 respectively.
Alter the cases in the switch statement to look for the letters you are using for the logical operators (O, A, and N) instead of “+” or “-“.
Add logic to each case to evaluate the two operands using the selected logical operator. You can either use C’s logical operators ( ||, &&, !) in a logical expression with the two operands, or you can write if statements to evaluate the expression if you like.
Turn in the program and output for all possible combinations of input and operators, e.g.
T O T = True
F O F = False
etc.,
Hints
Watch out scanf troubles. Remember that the variables into which you are reading input must be preceded by the address of operator (&op1, not just op1). Also, the buffering of keyboard input by the OS sometimes causes extra characters to “hang around” for the next iteration of your loop. If you get weird results while trying to read the input, try adding a call to flush ( ) into your loop just before the scanf.
Code:
```#include <stdio.h>
int main (void){
float value1, value2;
char operator;
printf ("Type in your expression. \n");
scanf ("%f %c %f", &value1, &operator, &value2);
switch(operator){
case'+':
printf("%.2f\n", value1 +value2);
break;
case '-':
printf("%.2f\n", value1-value2);
break;
case '*':
printf("%.2f\n", value1*value2);
break;
case '/':
if (value2==0)
printf ("Division by zero.n\");
else
printf("%.2f\n", value1/value2);
break;
default:
printf("Unknown operator.\n");
break;
}
return 0;
}```
2. Convert T to 1 and F to 0. Then use the logical operators in C. For example:
T O T => 1 || 1
3. So then there is no right or wrong answer to some of them? Like 1||0 can be 1 or 0?? And what about 1 AND 0???
4. Code:
```if (1 && 0)
puts("True");
else
puts("False");```
Code:
```if (1 || 0)
puts("True");
else
puts("False");```
Code:
```if (!0)
puts("True");
else
puts("False");```
If you're not already familiar with them I suggest you google for "logical operators c".
5. Why is this program printing smiley faces instead of a 1 or 0???
Code:
```#include <stdio.h>
#include <stdlib.h>
int main (void){
char value1;
char value2;
char loperator;
//char 1 = true;
//char 0 = false;
printf ("Please enter a logical expression:\n T=true; F=false; O=or; A=and; N=not\n");
scanf ("%c %c %c", &value1, &loperator, &value2);
char T = 1;
char F = 0;
switch(loperator){
case 'O':
printf("%c\n", value1 || value2);
break;
case 'A':
printf("%c\n", value1 && value2);
break;
case 'N':
printf("%c\n", value1);
break;
default:
printf("Unknown logical operator.\n");
break;
}
system("pause");
return 0;
}```
6. lol nvm guess I was supposed to read those in as an integer when I print in the switch function
7. In the following program, why when I type in "FOF" does it result to "true"? When should I use "=="?
Code:
```#include <stdlib.h>
int main (void){
char value1, value2, loperator;
char result;
//char 1 = true;
//char 0 = false;
printf ("Please enter a logical expression:\n T=true; F=false; O=or; A=and; N=not\n");
scanf ("%c %c %c", &value1, &loperator, &value2);
char T = 1;
char F = 0;
switch(loperator){
case 'O':
if(value1 && value2 == 0){
result = 0;}
if(value1 || value2 == 1){
result = 1;}
break;
case 'A':
result = value1 && value2;
break;
case 'N':
result = value1;
break;
default:
printf("Unknown logical operator.\n");
break;
}
if (result == 1){
printf("true.\n");}
if (result == 0){
printf("false.\n");}
system("pause");
return 0;
}```
8. Code:
```char T = 1;
char F = 0;```
doing this does not do what you think it does - that is, if someone enters F, this does not give that character the value 0. All that code does is declare two useless variables that are not referenced again.
You need to write a bit of code (another switch will do) to parse the character input, and make your value1 0 if it is F and 1 if it is T.
Note that character F has value 15, which is not 0, so F is true if evaluated like an integer as you do there.
9. I'm still getting "true" for everything with the following code. Also, how is F being treated as an integer? I have all my variables initialized as chars...
Code:
```#include <stdio.h>
#include <stdlib.h>
int main (void){
char value1, value2, loperator;
char result;
printf ("Please enter a logical expression:\n T=true; F=false; O=or; A=and; N=not\n");
scanf ("%c %c %c", &value1, &loperator, &value2);
switch(value1){
case 'T':
value1 == 1;
break;
case 'F':
value1 == 0;
break;
default:
printf("Unknown value.\n");
break;
}
switch(value2){
case 'T':
value2 == 1;
break;
case 'F':
value2 == 0;
break;
default:
printf("Unknown value.\n");
break;
}
switch(loperator){
case 'O':
if(value1 && value2 == 0){
result = 0;}
else if(value1 || value2 == 1){
result = 1;}
break;
case 'A':
result = value1 && value2;
break;
case 'N':
result = value1;
break;
default:
printf("Unknown logical operator.\n");
break;
}
if (result == 1){
printf("true.\n");}
else if (result == 0){
printf("false.\n");}
system("pause");
return 0;
}```
10. Code:
```switch(value1){
case 'T':
value1 == 1;
break;
case 'F':
value1 == 0;
break;
default:
printf("Unknown value.\n");
break;```
== is for comparison. For assignment, use =
saying value1==1 is just testing it for equality with 1, not assigning 1 to it.
chars and ints are very closely related in C. You'll need to ask someone with more knowledge than I for the details, but suffice it to say that when you compare two chars (using == or a switch), you are actually comapring their integer (ascii) value.
Also, your O case is wrong anyway:
Code:
```case 'O':
if(value1 && value2 == 0){
result = 0;}
else if(value1 || value2 == 1){
result = 1;}
break;```
This should be just like the others:
Code:
```case 'O':
result = value1 || value2;
break;```
11. I had that for Case O before, but now that I have changed it back AND made the value1 and 2 switch- it is now working! THANKS!
12. I'm now trying to do the extra credit : Make the program loop, asking for input and processing the operators and operands entered. (Hint:
You may need to become familiar with the flush ( ) function, as well as with scanf (). I have the following code, but when I run the program the answers are right but it is printing the default to the switch statements also ("Unknown value" and "Unknown logical operator"). Why is this? Should I be using the fflush function?
Code:
```#include <stdio.h>
#include <stdlib.h>
int main (void){
char value1, value2, loperator;
char result;
int i;
for (i=0; i<3; i++){
printf ("Please enter a logical expression:\n T=true; F=false; O=or; A=and; N=not\n");
scanf ("%c %c %c", &value1, &loperator, &value2);
switch(value1){
case 'T':
value1 = 1;
break;
case 'F':
value1 = 0;
break;
default:
printf("Unknown value.\n");
break;
}
switch(value2){
case 'T':
value2 = 1;
break;
case 'F':
value2 = 0;
break;
default:
printf("Unknown value.\n");
break;
}
switch(loperator){
case 'O':
result = value1 || value2;
break;
case 'A':
result = value1 && value2;
break;
case 'N':
result = value1;
break;
default:
printf("Unknown logical operator.\n");
break;
}
if (result == 1){
printf("true.\n");}
else if (result == 0){
printf("false.\n");}
//fflush(stdout);
}
system("pause");
return 0;
}```
13. Also, could I condense the first two switch statements? like switch(value1&&value2)??
14. Since your application here is going to need to flush input streams, you can tell whoever wrote the assignment that fflush should not be applied to input streams.
basically, the scanf function leaves a newline in the input buffer, so you need to get rid of the newline before you can read in anything new.
Your instructor likely wants you to use fflush(stdin) to do this, which is undefined behaviour. To flush an input stream, use this:
Code:
```int c;
while ((c=getchar())!='\n' && c!=EOF);```
This code snippet eats up characters from the input stream until it encounters and discards a newline, or runs our of input. | 2,538 | 9,194 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2020-34 | latest | en | 0.907526 |
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Cory’s average score after $$8$$ tests is $$91\%$$. If he gets a $$78\%$$ on his $$9^{th}$$ test, what will his new average be? | 61 | 190 | {"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} | 2.546875 | 3 | CC-MAIN-2024-38 | latest | en | 0.916129 |
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Enjoy Kreyszig, I have not finished it but I found the parts that I did read very enjoyable and enlightening. Jan28 comment Seifert matrices — Figure 8 knot Thank you for your comment. I will have to look into it when I have more time. Nov19 comment Prove Continuous functions are borel functions I don't know off the top of my head. You might want to post your comment as a question. Nov15 comment Roadway and book recommendations to math study. @AlexanderGruber While the article in the paper is an interesting piece of research could you please elaborate on what it has to do with the question? The question seems to be of the sort "I would like to understand A, B and C. My background is X. Could you please give me help on how to understand A, B and C, given my background" and I can't work out how an article on self-definition pertains. Sep25 comment Relation between primary ideal and prime ideal @leducquang Can you give me a list of all prime ideals containing $q$? Sep23 comment Relation between primary ideal and prime ideal @leducquang If $p^2 \subsetneq q \subsetneq p$ then neither $q=p^2$ nor $q=p$ but there are no powers of $p$ between $p$ and $p^2$ so therefore $q$ is not equal to a power of $p$. Aug18 comment Norms Induced by Inner Products and the Parallelogram Law @Kits89 It was a typo. I corrected it. Is it clear now? Jul10 comment Complete list of Sierpiński's publications Do you think one could circumvent the "copyright infringement" problem by somehow making the content of the papers available? I mean something like texing a purchased scan and slightly changing the wording of the paper. A joint effort of mathematicians around the world doing this seems more likely to happen than the unicorn they are talking about in the article. Jun21 comment Connected sets. @mathusiast In $[0,1]\cup[2,3]$ both $[0,1]$ and $[2,3]$ are open. To see this, try to apply the definition of open: pick a point $x$ in $[0,1]$ and try to find an open ball around it that is contained in $[0,1]$. Of course this is possible for all points in $(0,1)$... Jun16 comment Intersection of all neighborhoods of zero is a subgroup @LJR I can't think of any reference. Perhaps I should have written $(0,0)$ where I wrote $0$ to be less confusing. The open sets of a product of two topological spaces $T \times T'$ are precisely the sets of the form $O \times O'$ where $O$ and $O'$ are open in $T$ and $T'$ respectively. Jun6 comment Follow up on “Proof of $X \times X \hookrightarrow X$ implies $[X]^2 \hookrightarrow X$” Thanks for the upvote that reminded me of this question! Jun2 comment Complete list of Sierpiński's publications I did miss it, thank you for the link. Sadly, it very much sounds like it's not going to happen within the next 5 years. But It's good to know that someone else is doing it. Jun2 comment Possible mistake in Specker's thesis @BrianM.Scott $h$ is assumed to map $H$ to itself. But I don't see how it follows for all transpositions of $a^\ast$. Jun2 comment Possible mistake in Specker's thesis @BrianM.Scott Done. Jun2 comment Possible mistake in Specker's thesis @BrianM.Scott "...$e$ eine endliche Teilmenge von $a$...". Or would it be better if I also copy pasted that part? Jun1 comment Permutation models: when are they isomorphic? Regarding the statement in your last sentence: is the proof of it something a first year grad student can figure out with a bit of thinking or would it be wise if I kindly requested you to point me to a book (or paper) where I can read a proof? Perhaps my problem is not the actual proof but making the connection between the topology and truth of statements in the model so perhaps I should ask for a reference to that? | 1,209 | 4,778 | {"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} | 2.71875 | 3 | CC-MAIN-2014-15 | longest | en | 0.932082 |
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Quiz 7 / Q7: Wednesday Sitting
« on: March 22, 2020, 02:51:12 PM »
Question: Using argument principle along line on the picture, calculate the number of zeroes of the following function in the given annulus: $z^3-3z+1$ in ${1<|z|<2}$
Answer: $f(z)=z^3-3z+1$
On the circle $|z|=1$
$$|z^3+1| \leq |z^3|+|1|=2$$
$$|-3z| = |3z| = 3$$
So $|z^3+1|<|-3z|$. By Roche's Theorem, $f(z)$ and $g(z)=-3z$ have the same number of zeroes with $|z|=1$.
Since $g(z)$ has one zero with $|z|=1$, we know that $f(z)$ has one zero with $|z|=1$.
On the circle $|z|=2$
$$|z^3|=8$$
$$|-3z+1| \leq |3z|+|1| = 7$$
So $|-3z+1|<|z^3|$. By Roche's Theorem, $f(z)$ and $g(z)=z^3$ have the same number of zeroes with $|z|=2$.
Since $g(z)$ has three zeroes with $|z|=1$, we know that $f(z)$ has three zeroes with $|z|=1$.
$$3-1=2$$
Therefore, $f(z)=z^3-3z+1$ has two zeroes in ${1<|z|<2}$.
2
Quiz 5 / Q5: TUT 0201
« on: March 08, 2020, 06:46:48 PM »
Question: Give the order of each of zeros of the given function.
$$Log(1-z), |z|<1$$
Answer:
Let $f(z)=Log(1-z), |z|<1$. Let $f(z_0)=0$. Then, we get that $z_0=0$.
Since, $f'(z) = \frac{-1}{1-z}$, we know that $f'(z_0) = -1 \neq 0$.
Therefore, the order of $z_0 = 0$ of the given function is $1$.
3
Quiz 4 / Q4: TUT 0201
« on: February 17, 2020, 07:12:50 PM »
Question: Find the radius of convergence of the given power series.
$$\sum^\infty _{k=0} \frac{(k!)^2}{(2k)!}(z-2)^k$$
Answer: By ratio test, we have that:
$$\frac{1}{R} = lim_{k\rightarrow \infty} |\frac{\frac{((k+1)!)^2}{(2(k+1))!}}{\frac{(k!)^2}{(2k)!}}|$$
$$= lim_{k\rightarrow \infty} |\frac{\frac{(k+1)^2(k)^2(k-1)^2...}{(2k+2)(2k+1)(2k)...}}{\frac{(k)^2(k-1)^2...}{(2k)(2k-1)...}}|$$
$$= lim_{k\rightarrow \infty} |\frac{(k+1)^2}{(2k+2)(2k+1)}|$$
$$= lim_{k\rightarrow \infty} \frac{k+1}{4k+2}$$
By L’Hospital’s Rule, we know that
$$\frac{1}{R} = lim_{k\rightarrow \infty} \frac{k+1}{4k+2} = lim_{k\rightarrow \infty} \frac{1}{4} = \frac{1}{4}$$
Therefore, we conclude that $R=4$.
4
Quiz 3 / Q3: TUT 0201
« on: February 07, 2020, 11:46:05 PM »
Question: Directly compute the following line integral:
$$\int _\gamma \frac{dz}{z+4}$$
where $\gamma$ is the circle of radius 1 centered at -4, oriented counterclockwise.
Answer: Let $\gamma (t) = -4 + e^{it}$, where $t \in [0, 2\pi]$.
Then, $\gamma ' (t) = ie ^{it}$.
Then, we compute the line integral:
$$\int _\gamma \frac{dz}{z+4} = \int _0^{2\pi}\frac{ie ^{it} dt}{-4 + e^{it} + 4}$$
$$= \int _0^{2\pi}\frac{ie ^{it}}{ e^{it}}dt$$
$$= \int _0^{2\pi}i dt$$
$$= it \Big|_{t = 0}^{t = 2\pi}$$
$$= 2\pi i$$
5
Quiz 1 / Q1: TUT 0201
« on: January 29, 2020, 11:07:52 AM »
Questions:
Write the equation of the perpendicular bisector of the line segment joining $-1+2i$ and $1-2i$.
Answer:
Let $z = x + iy$.
Since $z$ is an arbitrary point on the perpendicular bisector of the line segment joining $-1+2i$ and $1-2i$, we know that the distance between $z$ and $-1+2i$ is same as the distance between $z$ and $1-2i$.
Then, we can calculate that:
$$|z-(-1+2i)| = |z - (1-2i)|$$
$$|(x + iy)-(-1+2i)| = |(x + iy) - (1-2i)|$$
$$|(x + 1)+(y-2)i| = |(x - 1) + (y + 2)i|$$
$$\sqrt{(x+1)^2 + (y-2)^2} = \sqrt{(x - 1)^2 + (y + 2)^2}$$
$$(x+1)^2 + (y-2)^2 = (x - 1)^2 + (y + 2)^2$$
$$x^2 + 2x + 1 + y^2 - 4y + 4 = x^2 - 2x + 1 + y^2 + 4y + 4$$
$$4x = 8y$$
$$y = \frac{1}{2} x$$
Then, we write the equation in complex number notation:
$$Re((\frac{1}{2} + i)z) = 0$$
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Quantitative Aptitude :: Average
Home > Quantitative Aptitude > Average > Important Formulas
Average Important Formulas
The result obtained by adding several quantities together and then dividing this total by the number of quantities is called Average. The main term of average is equal sharing of a value among all, where it may share persons or things. We obtain the average of a number using the formula that is the sum of observations divides by Number of observations.
1. Average:
Formula to find Average = $$\frac{Sum of observations}{Number of observations}$$
2. Average Speed Formula:
What is the formula for average speed? Suppose a man covers a certain distance at x kmph and an equal distance at y kmph.
Then, the average speed during the whole journey will be $$\left(\frac{2xy}{x+y}\right)$$kmph.
Average Methods shortcut tricks
• If both the time taken are equal i.e t1 = t2 = t,then, $$\frac{t1 + t2}{2}$$
• The average of odd numbers from1 to n is = $$\frac{[Last \ odd \ no. + 1]}{2}$$.
• The average of even numbers from1 to n is = $$\frac{[Last \ even \ no. + 2]}{2}$$.
• The Average of any number of quantities is sum of their quantities by the number of quantities (n) => Average = $$\frac{Sum of quantities}{n}$$.
• If there are two types of items say A and B , A has m number of sub items and B has n number of sum items then the average of A and B is $$\frac{A \times m + B \times n}{m+n}$$.
• If a vehicle travels from one place to another at a speed of a kmph but returns at the speed of b kmph then its average speed during the whole journey is $$\left(\frac{2ab}{a + b}\right)$$ kmph.
• Out of three numbers, first number is x times of the second number and y times of the third number. If the average of all the three numbers is z then the first number is $$\frac{3xyz}{xy + x + y}$$
• The average age of a group of N student is 'X' years. If M students joins, the average age of the group increases by 'Y' years, then the average age of the new students = $$x + \left(1 + \frac{N}{M}\right)\times Y$$ years.
• The average age of a group of N student is 'X' years. If M students joins, the average age of the group decreased by 'Y' years, then the average age of the new students = $$x - \left(1 + \frac{N}{M}\right)\times Y$$ years.
• The average age of a group of N student is 'X' years. If M student (Rahul) join the group, the average age of the group increases by 'Y' years, then the age of the new student (Rahul) is = $$x + \left(1 - \frac{N}{M}\right)\times Y$$ years.
• The average age of a group of N student is 'X' years. If M student (Ram) left the group, the average age of the group decreased by 'Y' years, then the age of the new student (Ram) was = $$x - \left(1 - \frac{N}{M}\right)\times Y$$ years.
• In a group of N persons whose average age is increased by 'Y' years when a person of 'X' years is replaced by a new man. Then the age of new comer is = $$\left(X + N \times Y\right)$$ years. | 854 | 3,144 | {"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} | 4.4375 | 4 | CC-MAIN-2023-40 | latest | en | 0.881462 |
https://itineraires-ceramique.com/d4jwt/google-sheets-sumif-924adb | 1,628,045,428,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154500.32/warc/CC-MAIN-20210804013942-20210804043942-00448.warc.gz | 316,108,225 | 10,240 | SUMIF is able to âpairâ these ranges and calculate the sum based on the criteria, then give back the result in array: one number for each sheets. 0. Explanation of Function:-Now, We will explain the Arguments of the Function. Is very helpful in doing conditional sums and that is the objective of out tutorial today. 0. Hot Network Questions How to delete a selection with Avogadro2 (Ubuntu 20.x)? Google has many special features to help you find exactly what you're looking for You can use Sumif in conditional formatting in Google Sheets for these types of cell/row highlighting. Summing with And criteria in Excel is easy. conditional addition. How to make SUMIF not default to 0 but to blank in Google Sheets? This range is C2:C9. In either case, the quickest solution is to use the =SUMIF function, and itâs easier than it initially appears. Google Sheets SUMIFS and SUMIF functions also present in Microsoft Excel for doing conditional sum based on single or multiple criteria only. SUMIF is a Google Sheets function to return a total of cells that match a single specific criterion. I've tried hardcoding the array, however, with the same result. As you start typing, Google Sheets will automatically suggest functions that start with the same letters. Active 2 years, 3 months ago. In this article, you will learn how to use SUMIF function in Google Sheets in ⦠Note: remember, when using the SUMIFS function, the first argument is the range to sum, followed by two or more range/criteria pairs. See my post on this for more [â¦] Google Sheets Functions â NOW, TODAY, DAY, MONTH, YEAR | Learning Google Apps - [â¦] do this we use a SUMIF function (see my post on this). That ⦠1. This is useful in many situations where you need to multiply items across arrays and add them up. The Google Sheets SUMIF function adds numbers together in a range when a matching evaluation is TRUE. So How Does It Work? SUMIF FUNCTION LETS THE USER CONTROL THE SUM BASED ON THE GIVEN CONDITION. SUMSQ: Returns the sum of the squares of a series of numbers and/or cells. The evaluations can use the logical expressions i.e. Google Sheets has some great formulas that allow you to quickly get information about a dataset. =SUMIF(range, criteria, [sum_range]) range: â Range is that column or row of selected cells on which we want to apply the condition or criteria. The following formulas can help you to sum cell values if another column cells contain specific text string, please do as this: 1. GREATER THAN, LESS THAN, NOT EQUAL TO, EQUAL TO, GREATER THAN or EQUAL TO, LESS THAN or EQUAL TO and EQUAL TO <, >, <>, = for numerical criterion. For example, to sum the cells that meet the following criteria: Google and Stanford (two criteria ranges), simply use the SUMIFS function (with the letter S at the end).. /u/toothoff - please read this comment in its entirety.. Once your problem is solved, please reply to the answer(s) saying Solution Verified to close the thread.. Read the rules-- particularly 1 and 2 -- and include all relevant information in order to ensure your post is not removed.. The SUMIF() Google Sheets Formula comes handy in calculating sum with multiple parameters i.e. I have a spreadsheet set out with 10 unique text entries in column A (row 2 through 11), and the same values in column C & E. In B, D, & F, I have a corresponding value that I am trying to sum. The INDIRECT function is used to identify these sheets names and identical ranges correctly. A ⦠Sum Values Meeting Criteria that Match Row Data and Dynamic Date Columns. Viewed 2k times 2. SERIESSUM: Given parameters x, n, m and a, returns the power series sum a 1 x n + a 2 x (n+m) + ⦠+ a i x (n+(i-1)m), where i is the number of entries in range 'a'. Take a look at the example below. The SUMPRODUCT function in Google Sheets multiplies corresponding items within arrays of equal size and returns the sum of the results. Applying the SUMIF formula across multiple sheets. Letâs say I have a list of items and their prices along with categories to which each item belongs to. And Criteria. The SUMIF function requires at least 2 arguments to perform. ... By using the IF Statement, we tell the cell to show up blank when the inventory item is blank, and to show the SUMIF formula when it isnât. So, letâs code an array formula for SUMIF. SUMIF: Returns a conditional sum across a range. I'm making a "game of life" to go with a budget and investment unit, and I'm using google sheets as both the game board and the expenditure/income calculator. Itâs used for looking up data or making calculations under specific conditions. Figure 4. After the opening round bracket â(â, enter your range which is tested against the criterion. I am a bot, and this action was performed automatically. It's free to sign up and bid on jobs. Working SUMPRODUCT function on Google Sheets does not work in Excel. sum_range â The range to sum.. Additional Notes SUMIF Examples in VBA. SUMIF wildcard in Google Sheets. The SUMIF function in Google Sheets allows users to sum specific cells within a list. One of the common things a lot of people need to do often is to count cells that are not blank in a dataset in Google Sheets (i.e., count non-empty cells). 0. The name of the function we will use in this guide is SUMIF. 2. Write the following in cell [â¦] criteria: â This is a condition which we have to apply to the selected cells. How this Formula Works. 1. First, add a new tab to the Google Sheet that records the column headers that match the information to summarize. 1. SUMIF and SUMIFS are two independent functions in Google Sheets. Count all apples and good apples. For example, you can use SUMIF to total up the number of push-ups you do in the morning with the criterion <12:00 . Here Iâve got a list of items, their prices, quantities, and total sale prices. Google Sheets does not allow check boxes to have color and contrast of the exact same color. The SUMIF only returns the sum of G:G of the first owner from the list of owners within the array. SUMIF is able to use exactly one criterion to determine if a value within a list should be summed or not. The IF statement surely is among the most commonly-used formulas in Google Sheets. Search for jobs related to Sumif google sheets or hire on the world's largest freelancing marketplace with 18m+ jobs. Put simply, the SUMIF functions filters the range according to the specified criteria and sums values based on this filter. Sumif index and match & Google sheets. As an example, consider the total attendance of two students for the period 01-01-2019 to 30-06-2019. As an example, letâs create a summarized list of the trucking suppliers we pay. SUMIF is used for adding values based on one condition and the purpose of SUMIFS is to sum the values in a range, based on multiple conditions. This function is very useful in a case when there is one principle condition, on the basis of which a final calculation needs to take place. In this video, you will learn how to use SUMIF in the context of a field trip. Spreadsheet math: Functions Vs. Using Google Sheets, I'm trying to sum two different, non-contiguous ranges based on and/or logic. When you enter SUMIF function in Google Sheets, an auto-suggest box pops up, containing syntax, example, summary related to SUMIF function and explanation of each part of function as shown below. Operators. Choose from the suggestions (just be careful) or continue typing. Along with numerical and text data, you can also use time tags. concatenating ranges from different sheets not working with sumif in google sheets. You can also use the SUMIF function in VBA. For example, sum values between 2018/5/15 and 2018/5/22 as following screenshot shown. My preference is to refer to the conditions by cell reference, e.g. SUMIF is a powerful function that can use most of the tools available in Google Sheets. Google Sheets SUMIF to sum a data range on a condition . SUMIF function is a mathematical function which is found in Google Sheets. Google Sheets Functions â FILTER | Learning Google Apps - [â¦] use of this is similar to that of COUNTIF. Ask Question Asked 2 years, 3 months ago. This is the standard expression for SUMIF: =SUMIF(A1:A10,"Paid",B1:B10) But it is possible to add wildcard like: =SUMIF⦠In my Google sheet, I have two columns which contain a date column and order column, now, I want to sum the order column cells based on the date column. range-values is a collection containing any value type. SUMIF(range, criteria, sum_range) range : The collection containing the values to be tested. Function Arguments ( Inputs ): range â The range containing the criteria that determines which numbers to sum.. criteria â The criteria indicating when to sum. To avoid that restriction I used two very similar colors playing with RGB, for example # FF99CC on # FF99CD, these are two different colors for the restriction, and for the eye they are practically the same. 1. If you are new to using Google Sheets formulas, it can be very tempting to use the mathematical functions such as =Add, =Subtract, =Minus, =Divide⦠and these functions do work⦠but it is much easier and more common to use spreadsheet operators when doing Addition, Subtraction, Multiplication, and Division in Google Sheets (and squaring too). Example: â<50â or âapplesâ. criteria : An expression that results in a logical TRUE or FALSE. Having Problems with SUMIF in Google Sheets. Sumif cells if contains part of specific text string in Google sheets with formulas. We use the named range of multiple sheets to create a reference for each sheet and merge these sheetsâ names with identical ranges across these multiple sheets. How to sumif cell values between two given dates in Google sheets? SUMIF + ARRAYFORMULA. Of the tools available in Google Sheets or hire on the GIVEN condition contains part of specific string. 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The values to be tested and SUMIFS are two independent functions in Google Sheets, i 'm trying to... The quickest solution is to refer to the specified criteria and sums values based on the world 's freelancing... Notes SUMIF Examples in VBA as an example, sum values between two GIVEN dates Google... Match a single specific criterion categories to which each item belongs to quickly get information about a dataset Google... The INDIRECT function is a condition which we have to apply to google sheets sumif criteria... A value within a list should be summed or not but to blank in Google Sheets does allow... Sheets does not work in Excel data, you can also use the SUMIF ( range, criteria sum_range! In a range from the list of items and their prices along with numerical text! Sheets or hire on the GIVEN condition if statement surely is among the most commonly-used formulas in Google Sheets hire..., add a new tab to the specified criteria and sums values based and/or! Prices along with numerical and text data, you will learn how to SUMIF cell values between two GIVEN in... Sumproduct function on Google Sheets does not work in Excel SUMIF only Returns the sum of the suppliers. Preference is to use exactly one criterion to determine if a value within a list of items, their,... Sheets function to return a total of cells that match Row data and Dynamic Date.. And their prices, quantities, and this action was performed automatically, with the criterion <.! Specific text string in Google Sheets search for jobs related to SUMIF cell between... Item belongs to be tested default to 0 but to blank in Google Sheets functions â filter Learning. Use most of the function we will use in this video, will. | 4,166 | 19,348 | {"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-2021-31 | latest | en | 0.885804 |
http://www.numbersaplenty.com/21560 | 1,590,681,691,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347399820.9/warc/CC-MAIN-20200528135528-20200528165528-00573.warc.gz | 190,933,098 | 3,754 | Search a number
21560 = 2357211
BaseRepresentation
bin101010000111000
31002120112
411100320
51142220
6243452
7116600
oct52070
932515
1021560
1115220
1210588
139a76
147c00
1565c5
hex5438
21560 has 48 divisors (see below), whose sum is σ = 61560. Its totient is φ = 6720.
The previous prime is 21559. The next prime is 21563. The reversal of 21560 is 6512.
21560 divided by its sum of digits (14) gives a triangular number (1540 = T55).
It is a Harshad number since it is a multiple of its sum of digits (14).
It is an Ulam number.
It is a zygodrome in base 7.
It is a congruent number.
It is an inconsummate number, since it does not exist a number n which divided by its sum of digits gives 21560.
It is not an unprimeable number, because it can be changed into a prime (21563) by changing a digit.
It is a polite number, since it can be written in 11 ways as a sum of consecutive naturals, for example, 1955 + ... + 1965.
221560 is an apocalyptic number.
21560 is a gapful number since it is divisible by the number (20) formed by its first and last digit.
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 21560, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (30780).
21560 is an abundant number, since it is smaller than the sum of its proper divisors (40000).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
21560 is a wasteful number, since it uses less digits than its factorization.
21560 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 36 (or 25 counting only the distinct ones).
The product of its (nonzero) digits is 60, while the sum is 14.
The square root of 21560 is about 146.8332387438. The cubic root of 21560 is about 27.8323313092.
The spelling of 21560 in words is "twenty-one thousand, five hundred sixty". | 555 | 1,955 | {"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} | 3.546875 | 4 | CC-MAIN-2020-24 | latest | en | 0.915482 |
http://www.varsitytutors.com/common_core_5th_grade_math-practice-tests | 1,484,605,479,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279368.44/warc/CC-MAIN-20170116095119-00335-ip-10-171-10-70.ec2.internal.warc.gz | 742,611,439 | 28,471 | # Free Common Core: 5th Grade Math Diagnostic Tests
Take the Varsity Learning Tools free diagnostic test for Common Core: 5th Grade Math to determine which academic concepts you understand and which ones require your ongoing attention. Each Common Core: 5th Grade Math problem is tagged down to the core, underlying concept that is being tested. The Common Core: 5th Grade Math diagnostic test results highlight how you performed on each area of the test. You can then utilize the results to create a personalized study plan that is based on your particular area of need.
# Free Common Core: 5th Grade Math Practice Tests
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The Common Core Fifth Grade Mathematics exam, which each state uses to evaluate Common Core curriculum proficiency, is designed to measure a student’s ability to comprehend and work with the material used in a typical fifth grade mathematics curriculum. The test requires that students have a working knowledge of three main components: multiplication and division of fractions, how volume is measured within a three-dimensional space, and whole numbers, prime numbers, and decimals. While the material can seem challenging at times, Varsity Tutors’ Learning Tools give you the opportunity to use online Common Core Fifth Grade Mathematics study help. The resources are free and comprehensive, and are a great way to enhance your Fifth Grade Common Core Mathematics review.
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The Varsity Tutors’ Learning Tools website has a wide variety of Common Core Fifth Grade Mathematics practice tests. These tests are an excellent way to help your elementary learner prepare for his or her exam. Many of practice tests are arranged by concept, such as geometry, measurement, complex numbers, and algebraic thinking. You also are given a number of general problem sets, which consist of Fifth Grade Mathematics sample questions designed to emulate the problems on your state’s official exam. Your learner may also benefit from using the free Common Core Fifth Grade Mathematics full-length practice tests, too. Each full-length practice test contains 40 Fifth Grade Mathematics sample questions which are designed to simulate the content on the actual exam. This is a perfect way to help your elementary learner become familiar with the test’s content, as well as build his or her confidence in the overall standardized testing process.
One of the best features of Varsity Tutors’ Learning Tools Common Core Fifth Grade Mathematics practice tests is the results tab at the end of each test. Not only are you given the test results, but the Learning Tools also provide you with a detailed explanation for each problem’s solution. This is a great way to clear up any confusion and help you reinforce the material. Furthermore, you can see the difficulty of each question, and where your learner ranks in terms of percentile standings.
By using Varsity Tutors’ Learning Tools and practice tests online, you can construct a Fifth Grade Common Core Math study regimen that best suits your learner’s needs. By being able to track his or her progress, and see which topics your learner is spending the most time on, you may be able to get a better idea of which concepts you should focus on studying. Whether you need to prepare your elementary learner for test day, or simply brush up on his or her Common Core Fifth Grade Mathematics review, Varsity Tutors has the resources for you. | 854 | 4,444 | {"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} | 2.59375 | 3 | CC-MAIN-2017-04 | latest | en | 0.935132 |
https://www.physicsforums.com/threads/auxiliary-field-h-problem-asks-for-magnetic-field-etc.682196/ | 1,519,469,825,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815544.79/warc/CC-MAIN-20180224092906-20180224112906-00117.warc.gz | 912,206,819 | 17,385 | # Auxiliary Field H Problem, Asks for Magnetic Field etc
1. Mar 31, 2013
### laz0r
1. The problem statement, all variables and given/known data
2. Relevant equations
int[H*dl] = I fenc
∇×H = cr^2 [Khat]
∇×H = Jf
3. The attempt at a solution
The textbook I have been assigned (griffiths) only deals with currents that are uniformly distributed and I'm not sure how to go about calculating the free enclosed current for the problem. If I knew how to do that, this problem would be very easy.
Using an amperian loop that extends the radius of the cylinder
∫(closed)H*dl = I fenc
|H|[2*pi*s] = I fenc
H = I fenc / [2*pi*s] [phi hat direction]
Then I would use the cross product ∇×H = Jf in order to calculate Jf, sub in Jf = cr^2 and then solve for c, thus solving part a. I'm just not sure about how to calculate the free enclosed current. if anyone knows the proper formula for this problem I would be grateful to see it.
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2. Mar 31, 2013
### TSny
Hi laz0r. Welcome to PF!
Regarding finding the proportionality constant c: If you integrate Jfree over the cross-sectional area (a<r<b) what should that equal?
3. Apr 1, 2013
### laz0r
Ah ok, I understand now. INT[J*dA] = Ifree, then you can use that for your H field later on and find the expression for c immediately by rearranging.
4. Apr 1, 2013
### TSny
You should be able to get c from just ∫J$\cdot$dA = Io | 442 | 1,495 | {"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} | 3.53125 | 4 | CC-MAIN-2018-09 | longest | en | 0.91573 |
https://community.dataquest.io/t/python-fundamentals-practice-problem-7/379437 | 1,656,902,962,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104293758.72/warc/CC-MAIN-20220704015700-20220704045700-00436.warc.gz | 235,877,491 | 6,579 | # Python Fundamentals Practice Problem #7
Screen Link:
My Code:
``````Paste your code here
``````
What I expected to happen:
What actually happened:
``````Paste output/error here
``````
Hi all,
I’m on practice problem 7 for the python fundamentals course Pt 1 and the answer is below.
from csv import reader
# read the CSV file
file = open(“laptops.csv”)
file_reader = reader(file)
rows = list(file_reader)[1:] # ignore column headers
# create the dictionary mapping the prices to the laptop names
price_to_name = {}
for row in rows:
price = int(row[2])
name = row[1]
if price in price_to_name:
price_to_name[price].append(name)
else:
price_to_name[price] = [name]
laptop1 = None
laptop2 = None
# look for a solution
for row in rows:
price = int(row[2])
# this laptop’s cost is price, so if there is a laptop that
# costs 5000 - price then their total price is exactly 5000
if price == 2500 and len(price_to_name[2500]) >= 2:
laptop1 = price_to_name[2500][0]
laptop2 = price_to_name[2500][1]
elif 5000 - price in price_to_name:
laptop1 = price_to_name[price][0]
laptop2 = price_to_name[5000 - price][0]
print(price)
# print the solution
print(laptop1)
print(laptop2)
I am pretty confused on what is happening from the point where it says #look for a solution, and onwards. Is there anyone that can walk me through what is happening beyond that point?
Thanks!
hey @sbolourc
For a start check out this post. Let us know if this still doesn’t helps you much. | 408 | 1,477 | {"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} | 2.734375 | 3 | CC-MAIN-2022-27 | latest | en | 0.817114 |
http://www.f-lohmueller.de/pov_tut/geo/geom_250e.htm | 1,610,941,819,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514121.8/warc/CC-MAIN-20210118030549-20210118060549-00248.warc.gz | 143,547,511 | 2,859 | ###### Descriptions and Examples for the POV-Ray Raytracerby Friedrich A. Lohmüller Elementary Geometry for Raytracing
Italiano
Français
Deutsch
Home
- POV-Ray Tutorial
- Geometrical Basics
for Raytracing
Right-angled Triangle
Pythagorean Theorem
Trigonometry Basics
Law of cosines
Equilateral Triangle
Regular Polygon
Polyhedron
Tetrahedron
Octahedron
Cube & Cuboid
Dodecahedron
Icosahedron
Cuboctahedron
Truncated Octahedron
Rhombicuboctahedron
Truncated Icosahedron
Circles
Tangent circles
Internal Tangents
External Tangents
- Geometric 3D Animations
## Regular OctahedronSome useful geometrical facts.
In the following we write for the square root of a number the expression "sqrt(ZAHL)"
conforming to the syntax used in POV-Ray.
Dimensions Length of octahedron edge: a. The radius of circumsphere: R = a / 2 * sqrt( 2 ); The radius of edgesphere (tangent to edges): Re = a / 2; The radius of the insphere: Ri= a / 6 * sqrt( 6 ); The angle between two faces: ~ 109,471° Face_Angle = degrees(acos(-1/3)); The angle between two edges: Edge_Angle = 90°
Inside of a regular octahedron:
3 squares.
Folding of a regular octahedron
top
© Friedrich A. Lohmüller, 2011 http://www.f-lohmueller.de | 364 | 1,204 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2021-04 | latest | en | 0.735895 |
http://bootmath.com/how-to-show-int_0inftye-xln2xmathrmdxgamma-2fracpi-26.html | 1,531,821,760,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589634.23/warc/CC-MAIN-20180717090227-20180717110227-00510.warc.gz | 57,599,343 | 5,894 | # How to show $\int_{0}^{\infty}e^{-x}\ln^{2}x\:\mathrm{d}x=\gamma ^{2}+\frac{\pi ^{2}}{6}$?
• Elementary derivation of certian identites related to the Riemannian Zeta function and the Euler-Mascheroni Constant
#### Solutions Collecting From Web of "How to show $\int_{0}^{\infty}e^{-x}\ln^{2}x\:\mathrm{d}x=\gamma ^{2}+\frac{\pi ^{2}}{6}$?"
As already mentioned, the integral evaluates to $\Gamma”(1)$.
Recall the digamma and trigamma functions:
$$\psi_{0} (x) = \frac{d}{dx} \ln \Gamma(x) = \frac{\Gamma'(x)}{\Gamma(x)}$$
$$\psi_{1}(x) = \frac{d}{dx} \psi_{0}(x) = \frac{\Gamma”(x)\Gamma(x)- (\Gamma'(x))^{2}}{\Gamma^{2}(x)}$$
Then
$$\psi_{1}(1) = \frac{\Gamma”(1)\Gamma(1) + (\Gamma'(1))^{2}}{\Gamma^{2}(1)} = \Gamma”(1) – (\Gamma'(1))^{2}$$
The digamma and trigamma functions also have series definitions (which can be derived from the infinite product representation of the gamma function):
$$\psi_{0}(x) = – \gamma + \sum_{n=0}^{\infty} \Big( \frac{1}{n+1} – \frac{1}{n+x}\Big)$$
$$\psi_{1}(x) = \sum_{n=0}^{\infty} \frac{1}{(n+x)^{2}}$$
So
$$\psi_{1}(1) = \sum_{k=0}^{\infty} \frac{1}{(k+1)^2} = \zeta(2) = \frac{\pi^{2}}{6}$$
And
$$\Gamma'(1) = \psi_{0}(1) \Gamma(1) = \psi_{0}(1) = – \gamma + \sum_{n=0}^{\infty} \left(\frac{1}{n+1} – \frac{1}{n+1} \right) = – \gamma$$
Therefore,
$$\Gamma”(1) = \psi_{1}(1) + (\Gamma'(1))^{2} = \frac{\pi^{2}}{6} + \gamma^{2}$$
For Real-Valued $s>0$, consider the well-known $$\Gamma(s)=\int_{0}^{\infty}{t^{s-1}e^{-t}dt}$$ Now differentiating twice with respect to $s$ and by the Dominated convergence theorem, which allows us to differentiate inside the integral we have that $$\frac{\partial^{2}}{\partial s^{2}}\Gamma(s)=\int_{0}^{\infty}{ln^{2}(t)\cdot t^{s-1}e^{-t}dt}$$ plugging in $s=1$ yields our disired result.
The derivatives of the gamma function can be expressed as polygamma functions, where $$\frac{\Gamma'(s)}{\Gamma(s)}=\psi(s)$$
Related problems: I, II. Recalling the Mellin transform of a function $f$
$$F(s) = \int_{0}^{\infty}x^{s-1} f(x)dx$$
which gives
$$F”(s) = \int_{0}^{\infty}x^{s-1}\ln(x)^2 f(x) dx .$$
Now, for your case, taking $f(x)=e^{-x}$ which has its Mellin transform equals to $F(s)=\Gamma(s)$, where $\Gamma(s)$ is the gamma function, gives
$$F”(s) = \Gamma”(s) \longrightarrow (*) .$$
Taking the limit of $(*)$ as $s\to 1$ gives the desired result.
Note:
$$\Gamma'(s) = \Gamma(s)\psi(s).$$
Hint: Integration by parts.
$$f(x)=\ln^2 x$$
$$g^{\prime}(x)=e^{-x}$$ | 965 | 2,471 | {"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} | 3.625 | 4 | CC-MAIN-2018-30 | latest | en | 0.655713 |
http://www.scribd.com/doc/81216968/A1-9-1-notes | 1,432,659,372,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207928864.73/warc/CC-MAIN-20150521113208-00190-ip-10-180-206-219.ec2.internal.warc.gz | 689,132,035 | 32,682 | Welcome to Scribd, the world's digital library. Read, publish, and share books and documents. See more
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# A1 9.1 notes
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LT 9.1.1: Identify polynomials and classify by degree and number of termsLT 9.1.2: Add and subtract polynomials
Vocabulary
1. monomial examples:2. polynomial examples:3. binomial examples:4. trinomial -examples:5. degree -examples:6. standard form -
Example 1.
Is it a monomial? If so, find the degree.a. 10 b. x + 5 c. x
-1
d. -1.8m
5
e. 3x f.
g. 4
a
h. ½ ab
2
Example 2.
Write 7 + 2x
4
– 4x in standard form. Identify the degree and leading coefficient of thepolynomial.in standard form: leading coefficient: | 272 | 801 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2015-22 | latest | en | 0.673626 |
http://au.studybay.net/the-substitution-and-real-income-effects/ | 1,638,726,816,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363215.8/warc/CC-MAIN-20211205160950-20211205190950-00454.warc.gz | 7,980,095 | 14,077 | HELLO, GUEST
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## The Substitution And Real Income Effects
When the price of a good rises or falls, two separate effects are felt on the demand for the good.
## The Substitution Effect
This is normally the dominant effect. As the price rises, and the good becomes relatively more expensive, consumers substitute into buying cheaper goods and thus demand for the more expensive good falls. This is always negative – a rise in price causes a contraction in demand, and a fall in price causes an extension in demand.
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## The Real Income Effect [1]
This effect is called the real income effect because the change in price affects the level of real income of the consumer, and thus affects the demand for the good.
Remember that real income is the quantity of goods a consumer can buy with the money income the have. Real Income is calculated as
As the price of the good in question rises, so does the average of all prices, cet. par. As a result the consumer will experience a reduction in real income.
## Normal Goods
For Normal Goods [2] , a reduction in real income will result in a reduction in demand for the good. Therefore, for normal goods, the income effect is positive – reduced income results in a decrease in demand and, higher real income results in an increase in demand.
## Inferior Goods
Price
Quantity
D1
D2
Q2
Q1
P
Price
Quantity
D2
D1
Q1
Q2
P
The Income Effect in an Inferior Good
The Income Effect in a Normal GoodFor Inferior Goods [3] , a reduction in real income will result in an increase in demand for the good. Therefore, for inferior goods, the income effect is negative – reduced income results in an increase in demand and, higher real income results in a decrease in demand.
The diagrams, above, show the real income effect on two different types of good.
Note that I have still plotted both these diagrams against price, therefore, they both still have the usual downward-sloping demand curve. The income change takes place in the background.
For most goods and consumers, the real income effect is very small – it is negligible [4] . This is because a single good is only one of many hundreds, possibly thousands, of goods which a consumer will buy and therefore a change in its price, alone, will have a negligible effect on the average price of all goods. Therefore the income effect can be assumed to be zero for most goods and consumers.
## Giffen Goods
However, there are cetain circumstances where this assumption is invalid. If the good accounts for a large part of their budget, then a change in its price will have a significant effect on the average price of goods bought by that consumer. A rise in its price will result in a significant [5] fall in the level of real income.
If the good is normal, this will simply strengthen and reinforce the substitution effect, causing demand to fall even more. If, however, the good is inferior, the fall in income works to increase demand, counteracting the substitution effect. The net effect on demand will be the result of the two forces. If the substitution effect is stronger than the income effect then demand will fall as a result of the price rise. If the income effect is the stronger then demand will rise when the price rises.
Price
Quantity
D1
D2
Q2
Q1
P2
P1
QN
PN
S2
S1
Dapparent
The diagram, above shows this effect, first identified by Prof. Giffen.
Supply decreases from S1 to S2 causing a rise in the price and a consequent contraction in demand (the pure substitution effect) from Q1 to QN. However, the rise in price causes a sharp fall in the consumer’s income and, because these goods are inferior, an increase in demand from D1 to D2.This raises the quantity demanded to Q2. Thus the rise in price has caused the overall level of demand to rise. The demand curve appears to be the upward sloping curve, Dapparent. This type of good is known as a Giffen good. Giffen goods are always inferior, basic, staple foodstuffs which account for a large fraction of poor consumers’ spending.
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In the diagram, above, if the consumer could be compensated for the loss of real income by increasing their money income sufficiently to restore the level of real income (in other words eliminating the income effect), the demand curve would shift back from D2 to D1 and the new equilibrium would be PNQN – the pure substitution effect equilibrium [6] .
Income
Quantity
Y2
Q1
Q2
The Income Demand Curve for a Normal Good
Income
Quantity
D
Q2
Q1
Y1
The Income Demand Curve for an Inferior Good
Y2
Y1
DIncome Demand Curves
In the 2 diagrams, below, we see the demand curves for both Inferior goods and Normal goods plotted against Income rather than Price.
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Out of time and require quick and moreover effective support with your term paper or dissertation? | 1,274 | 5,808 | {"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-2021-49 | latest | en | 0.912589 |
https://au.mathworks.com/matlabcentral/answers/2078176-how-to-solve-this-function | 1,720,866,598,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514493.69/warc/CC-MAIN-20240713083241-20240713113241-00885.warc.gz | 94,770,705 | 25,321 | # How to solve this function?
1 view (last 30 days)
Ramesh on 5 Feb 2024
Edited: Matt J on 5 Feb 2024
function rate=ptMech(C,T,par)
% Obtain the parameter values
A=par.A;
E=par.E;
b=par.b;
R=8.314;
% Compute k
k=A*exp(-E/(R*T));
% Compute the result
rate=k*C^b;
end
Matt J on 5 Feb 2024
Edited: Matt J on 5 Feb 2024
What do you mean "solve" it? A function is not an equaiton.
William Rose on 5 Feb 2024
I am not sure what you mean by "how do you solve this function".
You have posted a function.
par=struct('A',1,'E',2,'b',3); % define par
T=273; C=1; % define T, C
rate=ptMech(C,T,par) % call function ptMech()
rate = 0.9991
function rate=ptMech(C,T,par)
% Obtain the parameter values
A=par.A;
E=par.E;
b=par.b;
R=8.314;
% Compute k
k=A*exp(-E/(R*T));
% Compute the result
rate=k*C^b;
end
If you call the function with appropriate parameters, as shown above, it will return an answer.
Do you want to fix some of the parameters, and then find the value of other parameters that will make ptMech equal zero? You can see that C=0 and A=0 (either one, or both) will yield ptMech=0. | 366 | 1,080 | {"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-2024-30 | latest | en | 0.581935 |
https://malaysia.answers.yahoo.com/question/index?qid=20200124140246AAgSKZM | 1,600,747,712,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400203096.42/warc/CC-MAIN-20200922031902-20200922061902-00669.warc.gz | 497,092,879 | 25,365 | Subash asked in Science & MathematicsEngineering · 8 months ago
# A cuboidal reservoir 30m×24m×4m is to be separated into 4equalchambers by constructing cross walls of 40m thick upto its height. Find volume?
Relevance
• 8 months ago
use 30 mm sheet to divide the 30m dimension into quarters.
Each chamber will be 7.5*4*24 total capacity of 720 cu m.
Total capacity ~2880cu m.
• 8 months ago
Are you sure the cross walls aren't 40cm thick?
If they are 40m thick, then they would be wider than the long dimension.
I sketched it out in a CAD program - converting 30m x 24m into centimeters (3000cm x 2400cm), then drew in cross walls 40cm thick, measured the dimension of one of the four squares and got 1480cm x 1180cm = 1,746,400cm / 1,000 = 1746.4m x 4m = 6,985.6 cubic meters (the volume of each cuboidal reservoir) or 6,985.6 cubic meters x 4 = 27,942.4 cubic meters (the volume of the entire reservoir - minus the 40cm wall sections).
• 8 months ago
huh? cross walls 40 m thick, when the largest dimension is less than that?
edit, so they are 40 cm.
but there are many ways of dividing the chamber into 4 equal ones. I can think of at least 6 different ways, each of which will give you a different volume. There may be others.
one is by 3 walls along the 30 m axis, dividing it into quarters
one is by 3 walls along the 24 m axis dividing it into quarters
one is by 3 walls along the 4 m axis dividing it into quarters
one is by 2 walls, one dividing the 30 m axis in half, one dividing the 24 m axis in half
one is by 2 walls, one dividing the 30 m axis in half, one dividing the 4 m axis in half
one is by 2 walls, one dividing the 24 m axis in half, one dividing the 4 m axis in half | 492 | 1,709 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2020-40 | latest | en | 0.926709 |
http://www.jiskha.com/display.cgi?id=1366430462 | 1,455,135,227,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701160582.7/warc/CC-MAIN-20160205193920-00017-ip-10-236-182-209.ec2.internal.warc.gz | 492,331,194 | 3,718 | Wednesday
February 10, 2016
# Homework Help: College Physics
Posted by Andrew on Saturday, April 20, 2013 at 12:01am.
A cubic crate of side s=2.0m is top-heavy: its CG is 18cm above its true center.
Part A) How steep an incline can the crate rest on without tipping over?
Part B) What would your answer be if the crate were to slide at constant speed down the plane without tipping over? (The normal force would act at the lowest corner)
## Related Questions
More Related Questions | 123 | 488 | {"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-2016-07 | longest | en | 0.949639 |
https://ja.scribd.com/document/370521674/JEXPO-and-VOCLET-Mathematics-Model-Set-1 | 1,561,370,162,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999298.86/warc/CC-MAIN-20190624084256-20190624110256-00162.warc.gz | 469,373,125 | 58,807 | You are on page 1of 9
# JEXPO Mathematics
Model Set - 1
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1 a2 a 1
1. If a 1 , then find the value of 2 .
a a a 1
(a) 2 (b) 1
(c) 3 (d) 0
2. AD is a median of VABC . Two straight lines BE and CF are parallel to DA. A straight line
perpendicular to AD intersects BE and CF at P and Q respectively. Following which is correct?
(a) AP = AQ (b) AP = CD
## (c) AP = 2AQ (d) AP = 2PQ
c c
3. Find the value of 6 tan 2
4sin 2
2 cos ec c .sec c .
6 3
(a) 1 (b) 0
(c) 5 (d) 3
4. In ΔABC, AB = AC and ACB 500 , CA is produced to the point D such that ABD 400 . What is
the relation between AB and AD?
a b c2 b c a 2 c a b2 .
2 2 2
5. Simplify that
a 2 b c b2 c a c2 a b
2 2 2
(a) 1 (b) 2
(c) 0 (d) 3
5 1 1
6. If x and x , then find the value of 3x 2 7 xy 3 y 2 .
5 1 y
(a) 1 (b) 16
(c) 14 (d) 12
xy xy
7. 2, 6 , then find the value of x and y .
x y x y
(a) 3, 5 (b) 6, 3
(c) 4, 6 (d) 1, 1
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JEXPO Mathematics
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8. A rectangular plot of area 43560 m2 has length and breadth in the ratio 5 : 2. A gravel path 5 m wide
runs outside the plot close to its four sides. If it costs Rs.590 to gravel the path at 25 paisa/cu.m. Find
the depth of the graveled path.
(a) 2 m (b) 4 m
## (c) 3 m (d) 0.5 m
9. Find the area of a parallelogram; if its two adjacent sides are 12 cm and 14 cm and if the diagonal
connecting the ends is 18 cm.
## (c) 167.94 sq.cm (d) 2136 sq.cm
10. A parallelogram, the length of whose sides are 11 cm and 13 cm has one diagonal 20 cm long. Find
the length of another diagonal.
(a) 13 cm (b) 15 cm
(c) 20 cm (d) 12 cm
11. The thickness of a hollow cylinder is 3.5 cm and its outside diameter is 36.4 cm. Find the cost of
painting its surface at 5 paisa/cm2, if the cylinder is 70 cm long.
## (c) Rs.820 (d) Rs.720
2
12. If A of = B of 75% = Rs.18, then find A : B.
3
(a) 3 : 7 (b) 9 : 8
(c) 7 : 9 (d) 10 : 3
13. In a trapezium ABCD the straight line through the mid-point of an oblique side AD and parallel to AB
intersects BC and the two diagonals AC and BD at the points E, F, G and H respectively. Following which
is correct?
1
(a) EF = AB + DC (b) EF = (AB + DC)
2
(c) EF = GH + AB (d) EF = AE + ED
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JEXPO Mathematics
Model Set - 1
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3 6
14. Find the value of .
5 3 2 12 32 50
1
(a) 2 (b)
2
(c) 3 (d) 0
15. AB is a diameter of a circle. ABC 500 , then find the value of BDC .
## (c) 60° (d) 90°
16. From a point on the horizontal line through the foot of a chimney the angle of elevation of the top of
the chimney is 30° and the angle of elevation is 60° at a point on the same st. line50 meters nearer to
the chimney. What is the height of the chimney?
(a) 25 3 m (b) 35 m
(c) 30 3 m (d) 23 2 m
17. A conical vessel whose internal radius is 10 cm and height 48 cm is full of water. If this water is
poured into a cylindrical vessel with internal radius 20 cm. Find the height to which the water raises in it.
## (a) 2.5 cm (b) 10 cm
(c) 4 cm (d) 8 cm
4 3 30 18
18. Simplify:
2 2 4 3 18 3 2 3
(a) 4 6 (b) 1
(c) 2 (d) 3
19. A circular grassy plot of land 70 m in diameter has a path 7 m wide running round it on the outside.
How many stones 25 cm x 11 cm are needed to pave the path?
## (c) 36896 (d) 32902
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JEXPO Mathematics
Model Set - 1
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20. The base of a triangular field is three times its altitude. If the cost of cultivating the field at Rs.24.60
per hectare is Rs.332.10 . Find its base and height.
## (c) 900 m, 300 m (d) 450 m, 250 m
21. A field is in the form of a trapezium whose parallel sides are 120 meters and 75 meters and the non-
parallel sides are 105 meters and 72 meters. Find the cost of ploughing @ 60 paisa/m2.
## (c) Rs.3403.53 (d) Rs.3920.75
1
22. a 2 a 1 0 , then find the value of a 2 .
a2
(a) 3 (b) 2
## (c) 0 (d) 0.5
x a 2 2c 2 x b 2 2a 2 x c 2 2b 2
23. 0 . Find the value of x .
bc ca a b
(a) 2a b c (b) a 2b c
(c) a b c
(d) a 2 b 2 c 2
24. Side BA of ΔABC is produced to the point D such that AD = CD. Following which is correct?
1
2
1
(c) BAC 900 ADC (d) BAC 2ABC
2
2 1 2 1
25. If x and y , then find the value of x 2 3xy y 2 .
2 1 2 1
(a) 37 (b) 35
(c) 41 (d) 29
26. Find the weight of a lead pipe 3.5 meter long, if the external diameter of the pipe is 2.4 cm and the
thickness is 2 mm and 1 c.c. of lead weights 11.4 gms.
## (c) 5.5176 kg (d) 6.752 kg
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JEXPO Mathematics
Model Set - 1
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27. x cos 2 300 4 cos 2 450 tan 2 600 . Find the value of x .
1
(a) (b) 1
3
3 1
(c) 2 (d) 1
5 3
28. sec 4 x 500 cos ec 500 x , then find the value of x .
## (c) 90° (d) 30°
29. A and B are two alloys of gold and copper prepared by mixing metals in the ratio 7 : 2 and 7 : 11
respectively. If equal quantities of alloys are melted to form a third alloy C, find the ratio of gold and
copper in C.
(a) 7 : 5 (b) 6 : 7
(c) 8 : 7 (d) 3 : 5
1
30. Two pipes A and B can separately fill a cistern in 7 and 5 minutes respectively and a waste pipe C
2
can carry off 14 litres per minute. If all the pipes are opened when the cistern is full, it is emptied in 1
hour. How many litres does it hold?
c 3 c c c
## 31. Find the value of cot 2
cos sec 2
2
4sin 2 .
6 3 4 4 6
1
(a) 2 (b)
4
1
(c) (d) 0
2
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JEXPO Mathematics
Model Set - 1
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32. Find the value of sec2 600 cos ec300 cos 600 6sin 2 450 .
1
(a) 0 (b)
3
(c) 6 (d) 3
3 5 8
33. , find the value of x .
x2 x6 x3
(a) 3 (b) 1
(c) 4 (d) 0
c c c
34. If x tan cos tan tan 2 450 cos 600 , then find the value of x .
4 4 3
1 1
(a) (b)
3 6
1
(c) (d) 1
2
35. In the parallelograms ABCD and AEFG of equal area. A is common and the point E is on the side
AB. Following which is correct?
(a) DE ǁ AD (b) DE ǁ FC
(c) DE ǁ EB (d) DE ǁ DG
36. A vertical chimney subtends angles of 30° and 60° at points A and B which lie in a horizontal line
through the base of the chimney. Find its height if AB is 100 ft.
## (c) 85.3 ft. (d) 86.6 ft.
1 1
37. x 25 25 , find the value of x .
x 25 25
1 3
(a) 50, 25 (b) 30, 1
25 5
1
(c) 25, 3 (d) 27, 29
5
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JEXPO Mathematics
Model Set - 1
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## 38. 2 x y 5 xy 0, x 3 y xy 0 , then find the value of x and y .
1 1
(a) 2, (b) 3,
2 2
1 1
(c) 1, (d) 0,
2 2
39. A, B and C enter into a partnership. A putting in Rs.2000 for the whole year, B putting Rs.3000 at first
and increasing it to Rs.4000 at the end of 4 months, whilst C puts in at first Rs.4000 but withdraws
Rs.1000 at the end of 9 months. How should they at the end of a year, divide a profit of Rs.8475?
## (c) Rs.2927, Rs.4910, Rs.3309 (d) Rs.1700, Rs.3400, Rs.4720
40. A man travels 360 km in 4 hours, partly by air and partly by train. If he had travelled all the way by
4
air, he would have saved of the time he was in train and would have arrived his destination 2 hours
5
early. Find the distance he travelled by air and train.
## (c) 230 km, 130 km (d) 260 km, 100 km
41. A man sells an article at a profit of 20%. If he had bought it at 20% less and sold it for Rs.5 less, he
would have gained 25%. Find the cost price of the article.
(a) 30 (b) 45
(c) 25 (d) 50
1 1
42. If x 5 , then find the value of x .
x 5
1
(a) 5, (b) 6, 7
5
1
(c) 10, 12 (d) 7,
7
43. ABCD is a parallelogram and PQ is a straight line outside the parallelogram. AE, BF, CG, DH are
perpendiculars to PQ from A, B, C, D respectively. OR is the perpendicular from O. The point of
intersection of the two diagonals of the parallelogram to PQ. Following which is correct?
## (c) AE + CG = PF = HG (d) AE + CG = 2QG
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JEXPO Mathematics
Model Set - 1
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## 2 tan 300 cos ec600
44. Find the value of sec2 600 cot 2 30 .
1 tan 2 300
(a) 10 (b) 1
(c) 0 (d) 2
ab c 2 bc a 2 ca b 2
45. If a 2 b 2 b 2 c 2 c 2 a 2 , then find the value of .
a b bc ca
(a) 1 (b) 1
(c) 2 (d) 0
1 1 1
46. If a b c 0 , then find the value of 2 2 .
a b c b c a
2 2 2 2 2
c a 2 b2
(a) 0 (b) 2
(c) 1 (d) 3
## 47. If sec cos ec600 , then find the value of (2cos 2 2 1) .
(a) 0 (b) 1
1
(c) 1 (d)
2
7 7 3
48. r sin , r cos , find the value of .
2 2
(a) 30° (b) 45°
## (c) 60° (d) 90°
49. A can do a piece of work in 12 days while B alone can do it in 15 days. With the help of C they can
finish it in 5 days. If they are paid Rs.960 for the whole work, how should the money be divided among
them?
## (c) Rs.420, Rs.300, Rs.240 (d) Rs.420, Rs.295, Rs.235
8|Page
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JEXPO Mathematics
Model Set - 1
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50. Find the value of tan 350 tan 400 tan 500 tan 550 .
(a) 1 (b) 2
1
(c) 0 (d)
3
Note: If you found any wrong data or option, then please don’t forget to mention it with proper
explanation in the comment box below that will help others.
Thank You
For More Visit: www.learningscience.co.in | 4,155 | 9,333 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2019-26 | latest | en | 0.687482 |
https://www.numbersaplenty.com/5531 | 1,708,465,805,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473347.0/warc/CC-MAIN-20240220211055-20240221001055-00462.warc.gz | 946,484,475 | 2,966 | Search a number
5531 is a prime number
BaseRepresentation
bin1010110011011
321120212
41112123
5134111
641335
722061
oct12633
97525
105531
114179
12324b
132696
142031
15198b
hex159b
5531 has 2 divisors, whose sum is σ = 5532. Its totient is φ = 5530.
The previous prime is 5527. The next prime is 5557. The reversal of 5531 is 1355.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 5531 - 22 = 5527 is a prime.
It is a super-2 number, since 2×55312 = 61183922, which contains 22 as substring.
It is a Chen prime.
It is an Ulam number.
It is a plaindrome in base 16.
It is a nialpdrome in base 10.
It is not a weakly prime, because it can be changed into another prime (5501) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2765 + 2766.
It is an arithmetic number, because the mean of its divisors is an integer number (2766).
25531 is an apocalyptic number.
5531 is a deficient number, since it is larger than the sum of its proper divisors (1).
5531 is an equidigital number, since it uses as much as digits as its factorization.
5531 is an evil number, because the sum of its binary digits is even.
The product of its digits is 75, while the sum is 14.
The square root of 5531 is about 74.3706931526. The cubic root of 5531 is about 17.6848434414.
Adding to 5531 its reverse (1355), we get a palindrome (6886).
The spelling of 5531 in words is "five thousand, five hundred thirty-one". | 455 | 1,503 | {"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-2024-10 | latest | en | 0.915853 |
https://mathemerize.com/point-of-intersection-of-two-lines-in-3d/ | 1,721,859,995,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518454.54/warc/CC-MAIN-20240724202030-20240724232030-00711.warc.gz | 333,418,753 | 30,971 | # Point of Intersection of Two lines in 3d
Here you will learn how to find point of intersection of two lines in 3d for both vector and cartesian form with example.
Let’s begin –
## Point of Intersection of Two lines in 3d
#### (a) Cartesian Form
Algorithm :
Let the two lines be
$$x – x_1\over a_1$$ = $$y – y_1\over b_1$$ = $$z – z_1\over c_1$$ ……(i)
and $$x – x_2\over a_2$$ = $$y – y_2\over b_2$$ = $$z – z_2\over c_2$$ ………(ii)
1). Write the coordinates of general point on (i) and (ii). The coordinates of general points on (i) and (ii) are given by
$$x – x_1\over a_1$$ = $$y – y_1\over b_1$$ = $$z – z_1\over c_1$$ = $$\lambda$$ and $$x – x_2\over a_2$$ = $$y – y_2\over b_2$$ = $$z – z_2\over c_2$$ = $$\mu$$ respectively,
i.e. ($$a_1\lambda + x_1$$, $$b_1\lambda + y_1$$, $$c_1\lambda + z_1$$) and ($$a_2\mu + x_2$$, $$b_2\mu + y_2$$, $$c_2\mu + z_2$$)
2). If all the line (i) and (ii) intersect, then they have a common point.
$$\therefore$$ $$a_1\lambda + x_1$$ = $$a_2\mu + x_2$$, $$b_1\lambda + y_1$$ = $$b_2\mu + y_2$$ and $$c_1\lambda + z_1$$ = $$c_2\mu + z_2$$
3). Solve any two two of the equations in $$\lambda$$ and $$\mu$$ obtained in step 2. If the values of $$\lambda$$ and $$\mu$$ satisfy the third equation, then the lines (i) and (ii) intersect. Otherwise they do not intersect.
4). To obtain the coordinates of the point of intersection, substitute the value of $$\lambda$$ (or $$\mu$$) in the coordinates of general point(s) obtained in step 1.
#### (b) Vector Form
Algorithm :
Let the two lines be
$$\vec{r}$$ = ($$a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$$) + $$\lambda$$ ($$b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$$) ……(i)
and $$\vec{r}$$ = ($${a_1}’ \hat{i} + {a_2}’ \hat{j} + {a_3}’ \hat{k}$$) + $$\mu$$ ($${b_1}’\hat{i} + {b_2}’\hat{j} + {b_3}’\hat{k}$$) ………(ii)
1). If all the line (i) and (ii) intersect, then they have a common point.
$$\therefore$$ ($$a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$$) + $$\lambda$$ ($$b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$$) = ($${a_1}’ \hat{i} + {a_2}’ \hat{j} + {a_3}’ \hat{k}$$) + $$\mu$$ ($${b_1}’\hat{i} + {b_2}’\hat{j} + {b_3}’\hat{k}$$)
$$\implies$$ $$a_1 + \lambda b_1$$ = $${a_1}’ + \mu {b_1}’$$, $$a_2 + \lambda b_2$$ = $${a_2}’ + \mu {b_2}’$$ and $$a_3 + \lambda b_3$$ = $${a_3}’ + \mu {b_3}’$$
2). Solve any two two of the equations in $$\lambda$$ and $$\mu$$ obtained in step 2. If the values of $$\lambda$$ and $$\mu$$ satisfy the third equation, then the lines (i) and (ii) intersect. Otherwise they do not intersect.
3). To obtain the position vector of the point of intersection, substitute the value of $$\lambda$$ (or $$\mu$$) in (i) and (ii).
Example : Show that the line $$x – 1\over 2$$ = $$y – 2\over 3$$ = $$z – 3\over 4$$ and $$x – 4\over 5$$ = $$y – 1\over 2$$ = z intersect. Finf their point of intersection.
Solution : The coordinates of any point on first line are given by
$$x – 1\over 2$$ = $$y – 2\over 3$$ = $$z – 3\over 4$$ = $$\lambda$$
or, x = $$2\lambda + 1$$, y = $$3\lambda + 2$$ and z = $$4\lambda + 3$$
The coordinates of any point on second line are given by
$$x – 4\over 5$$ = $$y – 1\over 2$$ = z = $$\mu$$
or, x = $$5\mu + 4$$, y = $$2\mu + 1$$, z = $$\mu$$
If the lines intersect, then they have a common point. So, for some values of $$\lambda$$ and $$\mu$$, we must have,
$$2\lambda + 1$$ = $$5\mu + 4$$, $$3\lambda + 2$$ = $$2\mu + 1$$ and $$4\lambda + 3$$ = $$\mu$$
Solving first two of these equations, we get: $$\lambda$$ = -1 and $$\mu$$ = -1.
Clearly, $$\lambda$$ = -1 and $$\mu$$ = -1 satisfy the third equation. So, the given lines intersect.
Putting $$\lambda$$ in ($$2\lambda + 1$$, $$3\lambda + 2$$, $$4\lambda + 3$$), the coordinates of the required point of intersection are (-1, -1, -1). | 1,522 | 3,830 | {"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} | 4.4375 | 4 | CC-MAIN-2024-30 | latest | en | 0.567031 |
https://sravivarman.com/technical-articles/how-not-to-be-wrong-the-power-of-mathematical-thinking/ | 1,721,079,434,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514713.74/warc/CC-MAIN-20240715194155-20240715224155-00239.warc.gz | 483,478,397 | 24,055 | ### How Not to be Wrong: The Power of Mathematical Thinking
Published On: July 19, 2017
I recently found a pretty neat mathematical book entitled How not to be wrong: The power of mathematical thinking by Jordan Ellenberg. The author outlines real-life applications that show how mathematics makes them possible. This is a great STEM book that answers the question that many, many students ask, Why do I need to study math? How will I ever apply this stuff in real life in my future job?
Ellenberg states that not many people know what a logarithm is. So he says that the logarithm of a positive number N, called log N, is the number of digits it has! Well, not quite, but he goes on to explain that we can call the number of digits the “fake logarithm” or “flogarithm”. This is close to what the actual logarithm is. The flogarithm is a very slowly growing function, so that the flogarithm of a thousand is 4, the flogarithm of a million (a thousand times greater than a thousand) is 7, and the flogarithm of a billion is only 10.
Ellenberg safely reveals the true mathematical definition of Log N in a foot note at the bottom of the page, and I quote: It is that number x such that ex = N. Here e is Euler’s number whose value is about 2.71828….I say “e” and not “10” because the logarithm we mean to talk about is the natural logarithm, not the common or base-10 logarithm. The natural logarithm is the one you always use if you’re a mathematician or if you have e fingers.
So you will get Ellenberg’s dry humor like this throughout the book, but he will also reveal some really interesting mathematical ideas to you among some very witty “parables” on his journey to show the hidden beauty of the power of math and its powerful logic which you can use in life.
Another of many really neat math analyses in this book is a mathematical look at Powerball. He asks Is it wise to play Powerball? And immediately tells us that his father, a former president of the American Statistical Association, plays Powerball.
Ellenberg goes on to explain that a jackpot of \$100 million looks enticing but let’s look at some mathematical facts. We will compute the expected value of a \$2 ticket and your chances at winning the \$100 million and other lower prizes:
1/175,000,000 chance of a \$100 million jackpot
1/5,000,000 chance of a \$1 million prize
1/650,000 chance of a \$10,000 prize
1/19,000 chance of a \$100 prize
1/12,000 chance of a different \$100 prize
1/700 chance of a \$7 prize
1/360 chance of a different \$7 prize
1/110 chance of a \$4 prize
1/55 chance of a different \$4 prize
(Check Powerball’s website—these odds are there)
So now the expected amount you will win is:
100 million/175 million+1 million/5 million+10,000/650,000+100/19,000+100/12,000+7/700+7/360+4/110+4/55
That is just about \$0.94. This says that your ticket is not even worth the \$2 you spent for it.
He goes on to explain that as the jackpot goes up in other drawings in the future, like to \$337 million; your \$2 ticket is worth \$2.29. But remember that when the jackpot gets higher, more people will play. Ellenberg than gives some best strategies for making money playing Powerball at the end of that discussion.
I highly recommend this book to math aficionados, engineers, students, gamblers (just kidding), and anyone who says Why do I need to know math?
Please let me know if you have read this book and what you think of it.
### news via inbox
Nulla turp dis cursus. Integer liberos euismod pretium faucibua | 865 | 3,525 | {"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-2024-30 | latest | en | 0.943579 |
http://www.talkstats.com/index.php?s=ecbba302fb080c6cda09cb5d9a499b42 | 1,511,157,314,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805914.1/warc/CC-MAIN-20171120052322-20171120072322-00156.warc.gz | 498,896,511 | 15,876 | Top 10 Stats
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Welcome to our newest member, seoulm88 | 1,064 | 3,817 | {"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} | 3.640625 | 4 | CC-MAIN-2017-47 | latest | en | 0.63255 |
https://testbook.com/question-answer/consider-a-conductor-of-metal-with-non-uniform-cro--652fd0a695906000f9efc13d | 1,713,013,117,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816734.69/warc/CC-MAIN-20240413114018-20240413144018-00589.warc.gz | 528,533,073 | 97,279 | # Consider a conductor of metal with non-uniform cross-section. The parameter that is constant is:
This question was previously asked in
AAI ATC Junior Executive 21 Feb 2023 Shift 2 Official Paper
View all AAI JE ATC Papers >
1. drift velocity
2. drift speed
3. current
4. current density
Option 3 : current
Free
Junior Executive (Common Cadre) Full Mock Test
55.8 K Users
150 Questions 150 Marks 120 Mins
## Detailed Solution
Concept:
The amount of current travelling per unit cross-section area is known as current density and expressed in amperes per square metre.
Formula for current density is $$J=\frac{I}{A}$$
1. When a steady current flows through a non-uniform cross-sectional metallic conductor, the current flowing through the conductor is constant.
2. Since the current is independent of the conductor area, it remains constant.
The explanation for the incorrect options:
1. Current density, drift velocity and drift speed are all inversely proportional to cross-sectional area.
2. These quantities are not constant along the length of the conductor.
When a steady current flows in a metallic conductor of non-uniform cross-section then the drift speed is $$v_d=\frac{I}{neA}$$ and the current density $$J=\frac{I}{A}$$
$$v_d\propto\frac{1}{A},\ J\propto\frac{1}{A}$$
⇒ Only current remains constant.
In a metallic conductor of non-uniform cross-section, only the current remains constant along the entire length of the conductor.
Hence, the correct answer is Option-3-current. | 362 | 1,502 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-18 | latest | en | 0.845436 |
https://ggwash.org/view/9009/redistricting-game-results-part-1-the-fun-part | 1,511,555,984,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934808935.79/warc/CC-MAIN-20171124195442-20171124215442-00316.warc.gz | 612,699,189 | 8,317 | 3,981 maps solving DC’s redistricting puzzle were created using our Redistricting Game. What did people choose, and what conclusions can we draw? How should the DC Council redraw the ward boundaries?
The Redistricting Game wasn’t just a game, and a lot of people made very serious maps trying to solve the redistricting problem in a realistic way. We’ll take a close look at those maps this week.
Other people did inject some more whimsy into their maps, like these amusing maps from Zmapper where wards are stripes. In one, 7 wards cross the river. In the other, all touch the Potomac and DC’s northeast border.
What about giving each Councilmember a piece of downtown? That would encourage everyone to care about that area, and give every ward a variety of densities from the greatest to the least. It would be easy to make sure every ward had a fair share of Capital Bikeshare stations or streetcar lines in such a system.
A few users wish Tommy Wells represented downtown. The map on the left changes this with minimal other modifications; the one on the right also tries for close to equal population.
On the left map below, the entire Red Line from Friendship Heights to New York Avenue is all in Ward 3. On the right, Ward 8 gets DC’s north-south axis along South Capitol and North Capitol Streets almost all the way from one end to the other, while Ward 4 gets all of Georgia Avenue, Ward 1 all of 16th Street, and Ward 3 all of Connecticut Avenue.
Most of the maps don’t take extreme measures just to create entertaining visual patterns. Most people tried to solve the redistricting problem while also keeping wards compact and generally in their current forms.
218 maps made the fewest number of changes, 3. There are a number of ways to do this. Here were the 4 most popular, zoomed in around the areas where the changes are:
On the flip side, here are the maps with the greatest number of changed areas:
The map below at left is closest to perfect equality in ward population, given the constraints of the tool which required moving whole census tracts as one. Naturally, if you can break up the tracts, there are infinite ways to make the wards come out as equal as possible. The right map is the least equal, with the highest standard deviation among ward sizes.
Wards 7 and 8 have to get larger, while 2 has to get smaller. That almost always requires changing something about 6 or 5 as well. There is one exception, if Ward 8 takes over part of the Mall and then one of the two tracts of Foggy Bottom (but not both, which would move too many people).
Who made the maps? The first step of the game asked people which of the areas they lived in. Based on that, we know their ward, assuming they were being honest and knew how to locate their homes on a map.
WardStartedCompletedUnique
11,298635590
21,371583542
3911427402
4453209193
5511245224
61,261618561
71757158
81054637
None3,2251,147977
The first number, Started, is the number of times someone in that ward started the game. Completed is the number of times they finished a valid map. And the third number makes my best guess as to which ones are the same person, using the IP address and their choice of home area, and combines those likely duplicates into one.
I’m sure some people will suggest that the higher rates of maps being generated in some wards means that we should ignore all the results and instead just go with the recommendations of small committees selected by ward councilmembers, but a better solution is to be sure to look at the results for residents of each ward individually to see if there are differences. As I go through the analysis, I’ll do just that.
In upcoming parts, we’ll start looking at the serious topics to try to generate real, not-a-game policy recommendations.
Tagged: dc, redistricting
David Alpert is Founder and President of Greater Greater Washington and Executive Director of DC Sustainable Transportation (DCST). He worked as a Product Manager for Google for six years and has lived in the Boston, San Francisco, and New York metro areas in addition to Washington, DC. He lives with his wife and two children in Dupont Circle. Unless otherwise noted, opinions in his GGWash posts are his and not the official views of GGWash or DCST. | 943 | 4,267 | {"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-2017-47 | longest | en | 0.957259 |
https://www.postnetwork.co/insertion-sort-in-python/ | 1,590,549,560,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347392057.6/warc/CC-MAIN-20200527013445-20200527043445-00088.warc.gz | 878,315,334 | 24,238 | # Insertion Sort in Python
## Insertion Sort in Python
Insertion sort examine the previously sorted sub-array and insert an element at its proper position.
If you have an array a[0], a[1], a[2], a[3],……….a[n].
Then in insertion sort a[0] trivially sorted sub-array.
Again compare a[0] to a[1] and if a[0]> a[1] exchange it.
Now sub-array a[0],a[1] is sorted again compare a[2] with sub-array a[0],a[1] and insert it into proper position.
Now sub-array a[0],a[1],[2] is sorted again compare a[3] with sub-array a[0],a[1],[2] and insert it into proper position.
This process will go on until array a[] is sorted.
See a numerical example, then your understanding will be more clear.
## Insertion Sort in Python
a=[15, 23, 40, 10, 3, 35, 25, 5, 30, 33]
for i in range(len(a)):
t=a[i]
j=i-1
while t < a[j] and j >= 0:
a[j+1]=a[j]
j=j-1
a[j+1]=t
print(a)
## Time complexity of Insertion Sort
The average and worst case complexities of insertion sort is O(n2), and best case complexity is O(n). | 319 | 994 | {"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} | 3.359375 | 3 | CC-MAIN-2020-24 | latest | en | 0.755446 |
https://www.cnblogs.com/clrs97/p/9426621.html | 1,576,288,834,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540579703.26/warc/CC-MAIN-20191214014220-20191214042220-00348.warc.gz | 667,041,061 | 10,227 | # 2018 Multi-University Training Contest 3 - HDU Contest
solution
Code：
A. Ascending Rating
#include<cstdio>
const int N=10000010;
int T,n,m,k,P,Q,R,MOD,i,a[N],q[N],h,t;long long A,B;
int main(){
scanf("%d",&T);
while(T--){
scanf("%d%d%d%d%d%d%d",&n,&m,&k,&P,&Q,&R,&MOD);
for(i=1;i<=k;i++)scanf("%d",&a[i]);
for(i=k+1;i<=n;i++)a[i]=(1LL*P*a[i-1]+1LL*Q*i+R)%MOD;
for(h=1,t=A=B=0,i=n;i;i--){
while(h<=t&&a[q[t]]<=a[i])t--;
q[++t]=i;
if(i+m-1<=n){
while(q[h]>=i+m)h++;
A+=i^a[q[h]];
B+=i^(t-h+1);
}
}
printf("%lld %lld\n",A,B);
}
}
B. Cut The String
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=100010,S=26;
int T,n,m,i,j,len,x,y,ca,cb,ans;char s[N];
struct AP{
int s,d,r,f;
//k*d+s 0<=k<=r
AP(){}
AP(int _s,int _d,int _r,int _f){s=_s,d=_d,r=_r,f=_f;}
}a[N],b[N];
struct DS{
int all,son[N][S],fail[N],len[N],diff[N],top[N],text[N],last,tot,f[N];
int newnode(int l){
for(int i=0;i<S;i++)son[tot][i]=0;
fail[tot]=diff[tot]=top[tot]=0,len[tot]=l;
return tot++;
}
void init(){
last=tot=all=0;
newnode(0),newnode(-1);
text[0]=-1,fail[0]=1;
}
int getfail(int x){
while(text[all-len[x]-1]!=text[all])x=fail[x];
return x;
}
void add(int w){
w-='a';
text[++all]=w;
int x=getfail(last);
if(!son[x][w]){
int y=newnode(len[x]+2);
int z=son[getfail(fail[x])][w];
son[x][w]=y;
fail[y]=z;
diff[y]=len[y]-len[z];
if(diff[y]!=diff[z]||len[z]<1)top[y]=y;else top[y]=top[z];
}
last=son[x][w];
f[all]=last;
}
void get(int x,AP q[],int&cnt){
cnt=0;
x=f[x];
while(len[x]>0){
int y=top[x];
q[++cnt]=AP(len[y],diff[x],(len[x]-len[y])/diff[x],len[x]);
//printf("! %d %d %d\n",len[y],len[x],diff[x]);
x=fail[y];
}
}
}pre,suf;
ll exgcd(ll a,ll b,ll&x,ll&y){
if(!b)return x=1,y=0,a;
ll d=exgcd(b,a%b,x,y),t=x;
return x=y,y=t-a/b*y,d;
}
inline ll solve(ll a,ll b,ll c,ll xl,ll xr,ll yl,ll yr){
if(xl>xr)return 0;
if(yl>yr)return 0;
if(!a&&!b){
if(c)return 0;
return (xr-xl+1)*(yr-yl+1);
}
if(!b){
swap(a,b);
swap(xl,yl);
swap(xr,yr);
}
if(!a){
if(c%b)return 0;
ll y=-c/b;
if(y<yl||y>yr)return 0;
return xr-xl+1;
}
ll x,y,d=exgcd((a%abs(b)+abs(b))%abs(b),abs(b),x,y);
if(c%d)return 0;
x=(x%abs(b)+abs(b))%abs(b)*((((-c)%abs(b))+abs(b))%abs(b)/d)%abs(b/d);
d=abs(b/d);
ll kl=(xl-x)/d-3,kr=(xr-x)/d+3;
while(x+kl*d<xl)kl++;
while(x+kr*d>xr)kr--;
ll A=(-yl*b-a*x-c)/(a*d),B=(-yr*b-a*x-c)/(a*d);
if(A>B)swap(A,B);
kl=max(kl,A-3);
kr=min(kr,B+3);
while(kl<=kr){
ll y=(-c-a*x-a*d*kl)/b;
if(yl<=y&&y<=yr)break;
kl++;
}
while(kl<=kr){
ll y=(-c-a*x-a*d*kr)/b;
if(yl<=y&&y<=yr)break;
kr--;
}
if(kl>kr)return 0;
return kr-kl+1;
}
int main(){
scanf("%d",&T);
while(T--){
scanf("%d%d%s",&n,&m,s+1);
pre.init();
suf.init();
for(i=1;i<=n;i++)pre.add(s[i]);
for(i=n;i;i--)suf.add(s[i]);
while(m--){
scanf("%d%d",&x,&y);
len=y-x+1;
pre.get(y,a,ca);
suf.get(n-x+1,b,cb);
ans=0;
for(i=1;i<=ca;i++)for(j=1;j<=cb;j++){
if(a[i].s+b[j].s>len)continue;
if(a[i].f+b[j].f<len)continue;
ans+=solve(a[i].d,b[j].d,a[i].s+b[j].s-len,0,a[i].r,0,b[j].r);
}
printf("%d\n",ans);
}
}
}
C. Dynamic Graph Matching
#include<cstdio>
const int N=1<<10,P=1000000007;
int T,n,m,all,i,x,y,S,f[N],cnt[N],ans[N];char op[9];
inline void add(int&a,int b){a=a+b<P?a+b:a+b-P;}
inline void sub(int&a,int b){a=a-b>=0?a-b:a-b+P;}
int main(){
scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&m);
all=1<<n;
for(i=0;i<all;i++)f[i]=0,cnt[i]=__builtin_popcount(i);
f[0]=1;
while(m--){
scanf("%s%d%d",op,&x,&y);
x--,y--;
S=(1<<x)|(1<<y);
if(op[0]=='+'){
for(i=all-1;~i;i--)if(!(i&S))add(f[i^S],f[i]);
}else{
for(i=0;i<all;i++)if(!(i&S))sub(f[i^S],f[i]);
}
for(i=1;i<=n;i++)ans[i]=0;
for(i=1;i<all;i++)add(ans[cnt[i]],f[i]);
for(i=2;i<=n;i+=2)printf("%d%c",ans[i],i<n?' ':'\n');
}
}
}
D. Euler Function
#include<cstdio>
int T,k;
int main(){
scanf("%d",&T);
while(T--){
scanf("%d",&k);
if(k==1)puts("5");else printf("%d\n",5+k);
}
}
E. Find The Submatrix
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=10010;
const ll inf=1LL<<60;
int T,n,m,X,Y,i,j,k,pool[N];ll a[N],f[4][2][N],g[4][2][N],ans;
inline void up(ll&a,ll b){a<b?(a=b):0;}
inline void clr(){
for(i=0;i<=Y;i++)for(j=0;j<2;j++)for(k=0;k<=X;k++)g[i][j][k]=-inf;
}
inline void nxt(){
for(i=0;i<=Y;i++)for(j=0;j<2;j++)for(k=0;k<=X;k++)f[i][j][k]=g[i][j][k];
}
void solve1(int o,int l,int r,int dl,int dr){
int mid=(l+r)>>1,dm=dl;
ll ret=-inf;
for(int i=dl;i<=dr&&i<=mid;i++){
ll now=f[o][0][i];
if(mid-i<=m)now+=a[mid-i];
if(now>ret)ret=now,dm=i;
}
up(g[o][0][mid],ret);
if(l<mid)solve1(o,l,mid-1,dl,dm);
if(r>mid)solve1(o,mid+1,r,dm,dr);
}
void solve2(int o,int l,int r,int dl,int dr){
int mid=(l+r)>>1,dm=dl;
ll ret=-inf;
for(int i=dl;i<=dr&&i<=mid;i++){
ll now=f[o][1][i];
if(mid-i<=m)now+=a[mid-i];
if(now>ret)ret=now,dm=i;
}
up(g[o+1][0][mid],ret);
if(l<mid)solve2(o,l,mid-1,dl,dm);
if(r>mid)solve2(o,mid+1,r,dm,dr);
}
int main(){
scanf("%d",&T);
while(T--){
scanf("%d%d%d%d",&n,&m,&X,&Y);
clr();
nxt();
f[0][1][0]=0;
while(n--){
for(i=1;i<=m;i++)scanf("%d",&pool[i]);
sort(pool+1,pool+m+1);
a[0]=0;
for(i=1;i<=m;i++)a[0]+=pool[i];
for(i=1;i<=m;i++)a[i]=a[i-1]-pool[i];
clr();
for(i=0;i<=Y;i++)solve1(i,0,X,0,X);
for(i=0;i<Y;i++)solve2(i,0,X,0,X);
for(i=0;i<=Y;i++)for(k=0;k<=X;k++){
up(g[i][1][k],f[i][0][k]);
up(g[i][1][k],f[i][1][k]);
}
nxt();
}
ans=0;
for(i=0;i<=Y;i++)for(j=0;j<2;j++)for(k=0;k<=X;k++)up(ans,f[i][j][k]);
printf("%lld\n",ans);
}
}
F. Grab The Tree
#include<cstdio>
int T,n,i,x,y,sum;
int main(){
scanf("%d",&T);
while(T--){
scanf("%d",&n);
sum=0;
for(i=1;i<=n;i++)scanf("%d",&x),sum^=x;
for(i=1;i<n;i++)scanf("%d%d",&x,&y);
puts(sum?"Q":"D");
}
}
G. Interstellar Travel
#include<cstdio>
#include<algorithm>
using namespace std;
const int N=200010;
int T,n,t,i,f[N];bool must[N];
struct P{int x,y,p;}a[N],q[N];
inline bool cmp(const P&a,const P&b){
if(a.x!=b.x)return a.x<b.x;
if(a.y!=b.y)return a.y>b.y;
return a.p<b.p;
}
int main(){
scanf("%d",&T);
while(T--){
scanf("%d",&n);
for(i=1;i<=n;i++){
scanf("%d%d",&a[i].x,&a[i].y);
a[i].p=i;
}
sort(a+1,a+n+1,cmp);
for(t=0,i=1;i<=n;i++){
if(i>1&&a[i].x==a[i-1].x)continue;
while(t>1&&1LL*(q[t].y-q[t-1].y)*(a[i].x-q[t].x)<1LL*(a[i].y-q[t].y)*(q[t].x-q[t-1].x))t--;
q[++t]=a[i];
}
for(i=1;i<=t;i++)must[i]=0;
must[1]=must[t]=1;
for(i=2;i<t;i++)if(1LL*(q[i].y-q[i-1].y)*(q[i+1].x-q[i].x)!=1LL*(q[i+1].y-q[i].y)*(q[i].x-q[i-1].x))must[i]=1;
for(i=t;i;i--)if(must[i])f[i]=q[i].p;else f[i]=min(f[i+1],q[i].p);
for(i=1;i<t;i++)if(f[i]==q[i].p)printf("%d ",f[i]);
printf("%d\n",f[t]);
}
}
H. Monster Hunter
#include<cstdio>
#include<algorithm>
#include<queue>
#include<cstdlib>
using namespace std;
typedef long long ll;
const int N=100010;
int T,n,i,x,y,g[N],v[N<<1],nxt[N<<1],ed,f[N],vis[N],del[N],pos;
struct P{
ll a,b;//- a then + b
P(){}
P(ll _a,ll _b){a=_a,b=_b;}
bool operator<(const P&o)const{//true means bigger
int sgn1=a<b,sgn2=o.a<o.b;
if(sgn1!=sgn2)return sgn1<sgn2;
if(a<b)return a>o.a;
return b<o.b;
}
void operator+=(const P&o){
ll na=max(a,a-b+o.a),nb=b+o.b-a-o.a+na;
a=na,b=nb;
}
}a[N];
typedef pair<int,int>PI;
typedef pair<P,PI>PII;
priority_queue<PII>q;
inline void add(int x,int y){v[++ed]=y;nxt[ed]=g[x];g[x]=ed;}
void dfs(int x,int y){
f[x]=y;
for(int i=g[x];i;i=nxt[i])if(v[i]!=y)dfs(v[i],x);
}
int F(int x){return del[f[x]]?f[x]=F(f[x]):f[x];}
int main(){
scanf("%d",&T);
while(T--){
scanf("%d",&n);
for(ed=pos=i=0;i<=n;i++)vis[i]=del[i]=g[i]=0;
a[1]=P(0,0);
for(i=2;i<=n;i++)scanf("%lld%lld",&a[i].a,&a[i].b);
for(i=1;i<n;i++)scanf("%d%d",&x,&y),add(x,y),add(y,x);
dfs(1,0);
for(i=2;i<=n;i++)q.push(PII(a[i],PI(0,i)));
while(!q.empty()){
PII t=q.top();
q.pop();
x=t.second.second;
if(del[x])continue;
if(t.second.first!=vis[x])continue;
del[x]=1;
y=F(x);
a[y]+=a[x];
if(y>1)q.push(PII(a[y],PI(vis[y]=++pos,y)));
}
printf("%lld\n",a[1].a);
}
}
I. Random Sequence
#include<cstdio>
const int N=110,M=1500,P=1000000007;
int T,n,m,i,j,k,x,y,cnt,gcd[N][N],v[N],a[N],id[N][N][N],g[M][N],w[M][N],f[N][M],ans;
int getgcd(int a,int b){return b?getgcd(b,a%b):a;}
int po(int a,int b){int t=1;for(;b;b>>=1,a=1LL*a*a%P)if(b&1)t=1LL*t*a%P;return t;}
inline void up(int&a,int b){a=a+b<P?a+b:a+b-P;}
int main(){
scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&m);
for(i=1;i<=n;i++)scanf("%d",&a[i]);
for(i=1;i<=m;i++)scanf("%d",&v[i]);
for(i=1;i<=m;i++)for(j=1;j<=m;j++)gcd[i][j]=getgcd(i,j);
cnt=0;
for(i=1;i<=m;i++)for(j=i;j<=m;j+=i)for(k=j;k<=m;k+=j)id[i][j][k]=++cnt;
for(i=1;i<=m;i++)for(j=i;j<=m;j+=i)for(k=j;k<=m;k+=j){
x=id[i][j][k];
for(y=1;y<=m;y++){
g[x][y]=id[gcd[j][y]][gcd[k][y]][y];
w[x][y]=v[gcd[i][y]];
}
}
for(i=1;i<=n;i++)for(j=1;j<=cnt;j++)f[i][j]=0;
for(i=1;i<=m;i++)for(j=1;j<=m;j++)for(k=1;k<=m;k++){
if(a[1]&&i!=a[1])continue;
if(a[2]&&j!=a[2])continue;
if(a[3]&&k!=a[3])continue;
up(f[3][id[gcd[gcd[i][j]][k]][gcd[j][k]][k]],1);
}
for(i=3;i<n;i++)for(j=1;j<=cnt;j++)if(f[i][j])for(k=1;k<=m;k++){
if(a[i+1]&&k!=a[i+1])continue;
up(f[i+1][g[j][k]],1LL*f[i][j]*w[j][k]%P);
}
ans=0;
for(j=1;j<=cnt;j++)up(ans,f[n][j]);
for(i=1;i<=n;i++)if(!a[i])ans=1LL*ans*po(m,P-2)%P;
printf("%d\n",ans);
}
}
J. Rectangle Radar Scanner
#include<cstdio>
#include<algorithm>
using namespace std;
const int N=100010,M=1000010,E=262150,inf=~0U>>1;
int T,n,m,K,i,a[N],b[N],vma[E],vmi[E],vprod[E];
int xl[M],xr[M],yl[M],yr[M],ansma[M],ansmi[M],ansprod[M],q[M],pool[M],e[M];
int A,B,U,C,D;
inline bool cmpl(int x,int y){return xl[x]>xl[y];}
inline bool cmpr(int x,int y){return xr[x]<xr[y];}
inline void umax(int&a,int b){a<b?(a=b):0;}
inline void umin(int&a,int b){a>b?(a=b):0;}
void dfs(int x,int a,int b){
if(!vma[x])return;
vma[x]=0,vmi[x]=inf,vprod[x]=1;
if(a==b)return;
int mid=(a+b)>>1;
dfs(x<<1,a,mid),dfs(x<<1|1,mid+1,b);
}
void add(int x,int a,int b){
umax(vma[x],D);
umin(vmi[x],D);
vprod[x]=1LL*vprod[x]*D%K;
if(a==b)return;
int mid=(a+b)>>1;
if(C<=mid)add(x<<1,a,mid);else add(x<<1|1,mid+1,b);
}
void ask(int x,int a,int b){
if(C<=a&&b<=D){
umax(A,vma[x]);
umin(B,vmi[x]);
U=1LL*U*vprod[x]%K;
return;
}
int mid=(a+b)>>1;
if(C<=mid)ask(x<<1,a,mid);
if(D>mid)ask(x<<1|1,mid+1,b);
}
void solve(int l,int r,int L,int R){
if(l>r||L>R)return;
int mid=(l+r)>>1,i,j,k,cL=L-1,cR=R+1,ce=0;
for(i=L;i<=R;i++){
j=q[i];
if(xr[j]<mid)pool[++cL]=j;
else if(xl[j]>mid)pool[--cR]=j;
else e[++ce]=j;
}
for(i=L;i<=R;i++)q[i]=pool[i];
sort(e+1,e+ce+1,cmpl);
for(i=1,j=mid;i<=ce;i++){
k=e[i];
while(j>=xl[k]){
C=a[j],D=b[j];
add(1,1,n);
j--;
}
C=yl[k],D=yr[k],A=ansma[k],B=ansmi[k],U=ansprod[k];
ask(1,1,n);
ansma[k]=A,ansmi[k]=B,ansprod[k]=U;
}
dfs(1,1,n);
sort(e+1,e+ce+1,cmpr);
for(i=1,j=mid+1;i<=ce;i++){
k=e[i];
while(j<=xr[k]){
C=a[j],D=b[j];
add(1,1,n);
j++;
}
C=yl[k],D=yr[k],A=ansma[k],B=ansmi[k],U=ansprod[k];
ask(1,1,n);
ansma[k]=A,ansmi[k]=B,ansprod[k]=U;
}
dfs(1,1,n);
solve(l,mid-1,L,cL);
solve(mid+1,r,cR,R);
}
int main(){
for(i=0;i<E;i++)vmi[i]=inf,vprod[i]=1;
scanf("%d",&T);
while(T--){
scanf("%d",&n);
for(i=1;i<=n;i++)scanf("%d%d",&a[i],&b[i]);
int a0,b0,c0,d0,pp,qq,rr,P;
scanf("%d%d%d%d%d%d%d%d%d%d",&m,&a0,&b0,&c0,&d0,&pp,&qq,&rr,&P,&K);
for(i=1;i<=m;i++){
int na=(1LL*pp*a0+1LL*qq*b0+rr)%P;
int nb=(1LL*pp*b0+1LL*qq*a0+rr)%P;
int nc=(1LL*pp*c0+1LL*qq*d0+rr)%P;
int nd=(1LL*pp*d0+1LL*qq*c0+rr)%P;
xl[i]=min(na%n+1,nb%n+1);
xr[i]=max(na%n+1,nb%n+1);
yl[i]=min(nc%n+1,nd%n+1);
yr[i]=max(nc%n+1,nd%n+1);
a0=na,b0=nb,c0=nc,d0=nd;
q[i]=i;
ansma[i]=0;
ansmi[i]=inf;
ansprod[i]=1;
}
solve(1,n,1,m);
long long fin=0;
for(i=1;i<=m;i++)if(ansma[i])fin+=ansprod[i]^ansma[i]^ansmi[i];
printf("%lld\n",fin);
}
}
K. Transport Construction
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef pair<ll,int>E;
const int N=100010;
const ll inf=1LL<<60;
int T,n,i,j,k,m,c[N],q[N],st[N],en[N],qx[N],qk[N],stq[N],enq[N],tmp[N];
int hull[N],on[N],f[N];
E e[N];
ll ans;
struct P{int x,y;}a[N];
inline bool cmpc(int x,int y){return c[x]<c[y];}
inline bool cmpx(int x,int y){
if(a[x].x!=a[y].x)return a[x].x<a[y].x;
if(a[x].y!=a[y].y)return a[x].y<a[y].y;
return c[x]<c[y];
}
inline bool cmpk(int x,int y){return 1LL*a[x].y*a[y].x<1LL*a[y].y*a[x].x;}
inline ll dot(const P&a,const P&b){return 1LL*a.x*b.x+1LL*a.y*b.y;}
inline void work(int A,int B){//update B with A's hull
int l1=stq[A],r1=enq[A];
int l2=stq[B],r2=enq[B];
int t=0,i,j,x,cnt=0;
for(i=l1;i<=r1;i++){
if(i>l1&&a[qx[i]].x==a[qx[i-1]].x)continue;
//while(t>1&&slope(tmp[t],tmp[t-1])>slope(qx[i],tmp[t]))t--;
while(t>1&&1LL*(a[tmp[t]].y-a[tmp[t-1]].y)*(a[qx[i]].x-a[tmp[t]].x)>1LL*(a[qx[i]].y-a[tmp[t]].y)*(a[tmp[t]].x-a[tmp[t-1]].x))t--;
tmp[++t]=qx[i];
}
for(i=1;i<=t;i++){
if(i==1||i==t)hull[++cnt]=i;
//else if(slope(tmp[i],tmp[i-1])!=slope(tmp[i+1],tmp[i]))hull[++cnt]=i;
else if(1LL*(a[tmp[i]].y-a[tmp[i-1]].y)*(a[tmp[i+1]].x-a[tmp[i]].x)!=1LL*(a[tmp[i+1]].y-a[tmp[i]].y)*(a[tmp[i]].x-a[tmp[i-1]].x))hull[++cnt]=i;
}
for(i=1;i<cnt;i++){
on[i]=c[tmp[hull[i]]];
for(j=hull[i];j<=hull[i+1];j++)on[i]=min(on[i],c[tmp[j]]);
}
for(i=1;i<=cnt;i++)hull[i]=tmp[hull[i]];
for(i=l2,j=1;i<=r2;i++){
x=qk[i];
while(j<cnt&&dot(a[x],a[hull[j]])>dot(a[x],a[hull[j+1]]))j++;
ll now=dot(a[x],a[hull[j]]);
if(j<cnt&&now==dot(a[x],a[hull[j+1]]))e[c[x]]=min(e[c[x]],E(now,on[j]));
else e[c[x]]=min(e[c[x]],E(now,c[hull[j]]));
}
}
void solve(int l,int r){
if(l==r){
stq[l]=stq[l]=enq[l-1]+1;
int cur=enq[l-1];
for(int i=st[l];i<=en[l];i++){
qx[++cur]=q[i];
qk[cur]=q[i];
}
enq[l]=cur;
sort(qx+stq[l],qx+cur+1,cmpx);
sort(qk+stq[l],qk+cur+1,cmpk);
return;
}
int mid=(l+r)>>1;
solve(l,mid);
solve(mid+1,r);
work(l,mid+1);
work(mid+1,l);
int l1=stq[l],r1=enq[l];
int l2=stq[mid+1],r2=enq[mid+1];
int k=l1-1;
while(l1<=r1&&l2<=r2)tmp[++k]=cmpx(qx[l1],qx[l2])?qx[l1++]:qx[l2++];
while(l1<=r1)tmp[++k]=qx[l1++];
while(l2<=r2)tmp[++k]=qx[l2++];
for(int i=stq[l];i<=k;i++)qx[i]=tmp[i];
l1=stq[l],r1=enq[l];
l2=stq[mid+1],r2=enq[mid+1];
k=l1-1;
while(l1<=r1&&l2<=r2)tmp[++k]=cmpk(qk[l1],qk[l2])?qk[l1++]:qk[l2++];
while(l1<=r1)tmp[++k]=qk[l1++];
while(l2<=r2)tmp[++k]=qk[l2++];
for(int i=stq[l];i<=k;i++)qk[i]=tmp[i];
enq[l]=r2;
}
int F(int x){return f[x]==x?x:f[x]=F(f[x]);}
inline void merge(int x,int y,ll z){if(F(x)!=F(y))ans+=z,f[f[x]]=f[y];}
int main(){
scanf("%d",&T);
while(T--){
scanf("%d",&n);
for(i=1;i<=n;i++){
scanf("%d%d",&a[i].x,&a[i].y);
c[i]=q[i]=i;
}
ans=0;
while(1){
sort(q+1,q+n+1,cmpc);
for(m=0,i=1;i<=n;i=j){
for(j=i;j<=n&&c[q[i]]==c[q[j]];j++);
m++;
for(k=i;k<j;k++)c[q[k]]=m;
}
if(m==1)break;
for(i=1;i<=n;i++)en[c[q[i]]]=i;
for(i=n;i;i--)st[c[q[i]]]=i;
for(i=1;i<=m;i++)e[i]=E(inf,0);
solve(1,m);
for(i=1;i<=m;i++)f[i]=i;
for(i=1;i<=m;i++)merge(i,e[i].second,e[i].first);
for(i=1;i<=n;i++)c[i]=F(c[i]);
}
printf("%lld\n",ans);
}
}
L. Visual Cube
#include<cstdio>
const int N=200;
int T,a,b,c,n,m,i,j;char f[N][N];
int main(){
scanf("%d",&T);
while(T--){
scanf("%d%d%d",&a,&b,&c);
n=b*2+c*2+1;
m=a*2+b*2+1;
for(i=1;i<=n;i++)for(j=1;j<=m;j++)f[i][j]='.';
for(i=1;i<=b;i++)for(j=1;j<=a;j++){
f[i*2-1][j*2+1+b*2-i*2]='+';
f[i*2-1][j*2+2+b*2-i*2]='-';
f[i*2-1][j*2+3+b*2-i*2]='+';
f[i*2][j*2+b*2-i*2]='/';
f[i*2][j*2+2+b*2-i*2]='/';
}
for(i=1;i<=c;i++)for(j=1;j<=a;j++){
f[i*2+b*2-1][j*2-1]='+';
f[i*2+b*2-1][j*2]='-';
f[i*2+b*2-1][j*2+1]='+';
f[i*2+b*2][j*2-1]='|';
f[i*2+b*2][j*2+1]='|';
f[i*2+b*2+1][j*2-1]='+';
f[i*2+b*2+1][j*2]='-';
f[i*2+b*2+1][j*2+1]='+';
}
for(i=1;i<=c;i++)for(j=1;j<=b;j++){
f[i*2+b*2-j*2][a*2+j*2+1]='|';
f[i*2+b*2-j*2+1][a*2+j*2+1]='+';
f[i*2+b*2-j*2+2][a*2+j*2]='/';
}
for(i=1;i<=n;i++){
for(j=1;j<=m;j++)putchar(f[i][j]);
puts("");
}
}
}
M. Walking Plan
#include<cstdio>
#define rep(i) for(int i=0;i<n;i++)
const int N=55,M=105,inf=1000000000;
int T,n,m,x,y,q,K,A,B,z,g[N][N],d[N][N],f[N][N],a[M][N][N],b[M][N][N];
inline void up(int&a,int b){if(a>b)a=b;}
inline void mul(int a[][N],int b[][N],int c[][N]){
static int f[N][N];
rep(i)rep(j){
f[i][j]=inf;
rep(k)up(f[i][j],a[i][k]+b[k][j]);
}
rep(i)rep(j)c[i][j]=f[i][j];
}
int main(){
scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&m);
rep(i)rep(j)g[i][j]=inf;
while(m--){
scanf("%d%d%d",&x,&y,&z);
x--,y--;
up(g[x][y],z);
}
rep(i)rep(j)a[0][i][j]=b[0][i][j]=i==j?0:inf;
for(int i=1;i<M;i++)mul(a[i-1],g,a[i]);
for(int i=1;i<M;i++)mul(b[i-1],a[100],b[i]);
rep(i)rep(j)d[i][j]=i==j?0:g[i][j];
rep(k)rep(i)rep(j)up(d[i][j],d[i][k]+d[k][j]);
for(x=0;x<M;x++){
rep(i)rep(j)f[i][j]=inf;
rep(i)rep(j)rep(k)up(f[i][j],b[x][i][k]+d[k][j]);
rep(i)rep(j)b[x][i][j]=f[i][j];
}
scanf("%d",&q);
while(q--){
scanf("%d%d%d",&x,&y,&K);
x--,y--;
A=K%100,B=K/100;
z=inf;
rep(i)up(z,a[A][x][i]+b[B][i][y]);
if(z>=inf)z=-1;
printf("%d\n",z);
}
}
}
posted @ 2018-08-05 18:12 Claris 阅读(...) 评论(... 编辑 收藏 | 7,449 | 16,407 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2019-51 | latest | en | 0.133004 |
https://www.jiskha.com/display.cgi?id=1286305989 | 1,516,461,605,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889660.55/warc/CC-MAIN-20180120142458-20180120162458-00184.warc.gz | 929,661,113 | 3,917 | # Physical Science
posted by .
Why a person can lie down(carefully) on a bed of nails, but can't step(without being injured)on a nail?
• Physical Science -
The person's entire weight is spread across a large area on the bed of nails but the entire weight is pressing in one spot when stepping on a nail.
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More Similar Questions | 628 | 2,757 | {"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-2018-05 | latest | en | 0.913324 |
https://www.airmilescalculator.com/distance/thq-to-hyn/ | 1,606,499,223,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141193856.40/warc/CC-MAIN-20201127161801-20201127191801-00121.warc.gz | 568,455,288 | 36,563 | # Distance between Tianshui (THQ) and Huangyan (HYN)
Flight distance from Tianshui to Huangyan (Tianshui Maijishan Airport – Taizhou Luqiao Airport) is 1006 miles / 1618 kilometers / 874 nautical miles. Estimated flight time is 2 hours 24 minutes.
Driving distance from Tianshui (THQ) to Huangyan (HYN) is 1199 miles / 1929 kilometers and travel time by car is about 20 hours 38 minutes.
## Map of flight path and driving directions from Tianshui to Huangyan.
Shortest flight path between Tianshui Maijishan Airport (THQ) and Taizhou Luqiao Airport (HYN).
## How far is Huangyan from Tianshui?
There are several ways to calculate distances between Tianshui and Huangyan. Here are two common methods:
Vincenty's formula (applied above)
• 1005.604 miles
• 1618.363 kilometers
• 873.846 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 1004.389 miles
• 1616.407 kilometers
• 872.790 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Tianshui Maijishan Airport
City: Tianshui
Country: China
IATA Code: THQ
ICAO Code: ZLTS
Coordinates: 34°33′33″N, 105°51′36″E
B Taizhou Luqiao Airport
City: Huangyan
Country: China
IATA Code: HYN
ICAO Code: ZSLQ
Coordinates: 28°33′43″N, 121°25′44″E
## Time difference and current local times
There is no time difference between Tianshui and Huangyan.
CST
CST
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 151 kg (333 pounds).
## Frequent Flyer Miles Calculator
Tianshui (THQ) → Huangyan (HYN).
Distance:
1006
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
1006
Round trip? | 534 | 1,880 | {"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-2020-50 | longest | en | 0.819855 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/04%3A_Putting_the_First_Law_to_Work/4.01%3A_Prelude_to_Putting_the_First_Law_to_Work | 1,656,600,193,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103821173.44/warc/CC-MAIN-20220630122857-20220630152857-00770.warc.gz | 226,841,419 | 24,881 | # 4.1: Prelude to Putting the First Law to Work
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
As has been seen in previous chapters, may important thermochemical quantities can be expressed in terms of partial derivatives. Two important examples are the molar heat capacities $$C_p$$ and $$C_V$$ which can be expressed as
$C_p = \left(\dfrac{\partial H}{\partial T}\right)_p$
and
$C_V = \left(\dfrac{\partial U}{\partial T}\right)_V$
These are properties that can be measured experimentally and tabulated for many substances. These quantities can be used to calculate changes in quantities since they represent the slope of a surface ($$H$$ or $$U$$) in the direction of the specified path (constant $$p$$ or $$V$$). This allows us to use the following kinds of relationships:
$dH = \left(\dfrac{\partial H}{\partial T}\right)_p dT$
and
$\Delta H = \int \left(\dfrac{\partial H}{\partial T}\right)_p dT$
Because thermodynamics is kind enough to deal in a number of state variables, the functions that define how those variable change must behave according to some very well determined mathematics. This is the true power of thermodynamics!
4.1: Prelude to Putting the First Law to Work is shared under a not declared license and was authored, remixed, and/or curated by Patrick Fleming. | 665 | 2,231 | {"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} | 3.734375 | 4 | CC-MAIN-2022-27 | latest | en | 0.610902 |
https://numberworld.info/123323 | 1,652,876,288,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522270.37/warc/CC-MAIN-20220518115411-20220518145411-00654.warc.gz | 502,532,518 | 3,775 | # Number 123323
### Properties of number 123323
Cross Sum:
Factorization:
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
Yes
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
1e1bb
Base 32:
3odr
sin(123323)
0.21785551368486
cos(123323)
-0.97598103217076
tan(123323)
-0.22321695453477
ln(123323)
11.722562208658
lg(123323)
5.091044080988
sqrt(123323)
351.17374617132
Square(123323)
### Number Look Up
Look Up
123323 which is pronounced (one hundred twenty-three thousand three hundred twenty-three) is a great number. The cross sum of 123323 is 14. If you factorisate the number 123323 you will get these result . 123323 has 2 divisors ( 1, 123323 ) whith a sum of 123324. 123323 is a prime number. The figure 123323 is not a fibonacci number. The number 123323 is not a Bell Number. 123323 is not a Catalan Number. The convertion of 123323 to base 2 (Binary) is 11110000110111011. The convertion of 123323 to base 3 (Ternary) is 20021011112. The convertion of 123323 to base 4 (Quaternary) is 132012323. The convertion of 123323 to base 5 (Quintal) is 12421243. The convertion of 123323 to base 8 (Octal) is 360673. The convertion of 123323 to base 16 (Hexadecimal) is 1e1bb. The convertion of 123323 to base 32 is 3odr. The sine of the figure 123323 is 0.21785551368486. The cosine of the number 123323 is -0.97598103217076. The tangent of 123323 is -0.22321695453477. The root of 123323 is 351.17374617132.
If you square 123323 you will get the following result 15208562329. The natural logarithm of 123323 is 11.722562208658 and the decimal logarithm is 5.091044080988. that 123323 is very amazing figure! | 583 | 1,722 | {"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} | 3.3125 | 3 | CC-MAIN-2022-21 | latest | en | 0.775399 |
https://math.stackexchange.com/questions/1823888/image-of-a-disconnected-set-is-disconnected | 1,627,657,184,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153966.60/warc/CC-MAIN-20210730122926-20210730152926-00281.warc.gz | 385,863,910 | 38,198 | # Image of a disconnected set is disconnected
I'm aware that the image of a connected set is connected and the preimage of a disconnected set is disconnected. However, I'm struggling to find an example of a disconnected set such that the image of the disconnected set is also disconnected. Can the preimage of a connected set be disconnected?
HINT: I assume that you want $f$ to be continuous. Consider the function $f:\Bbb R\to\Bbb R:x\mapsto x^2$. What is the image of $\{-1,1\}$? What is the pre-image of $[1,2]$?
• Image of $\{-1,1\}$ is connected, should we take $\{0,1\}$ instead? – Akash Gaur Jun 24 '20 at 16:07
If $$f:X \to Y$$ is continuous, then the contrapositive of $$A$$ connected $$\implies f(A)$$ connected is $$f(A)$$ disconnected $$\implies A$$ disconnected. Pre-image of a disconnected set is disconnected is incorrect, eg. let $$f:\mathbb R \to \mathbb R$$ be the constant zero function, then the pre-image of $$\{0,1\}$$ is connected.
For the other examples, take the image of $$\{0,1\}$$ under the identity function on $$\mathbb R$$ and the pre-image of $$\{0\}$$ under the constant zero function $$f:\{0,1\} \to \mathbb R$$ under the subspace topology of $$\mathbb R$$. | 345 | 1,194 | {"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": 12, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2021-31 | latest | en | 0.909874 |
https://www.teacherspayteachers.com/Product/Valentines-Day-Smart-Board-Visual-Discrimination-Preschool-Kinder-First-1064476 | 1,490,457,610,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218188962.14/warc/CC-MAIN-20170322212948-00635-ip-10-233-31-227.ec2.internal.warc.gz | 935,592,761 | 25,430 | Total:
\$0.00
# Valentine’s Day Smart Board Visual Discrimination: Preschool/ Kinder/ First
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4.0
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### PRODUCT DESCRIPTION
Valentine’s Day Smart Board Visual Discrimination: Preschool/ Kinder/ First
Visual discrimination skills are very important in learning for children of all ages. All aspects of math, including patterning, sorting, counting and graphing depend on good visual discrimination skills.
This is a 34 slide Smart Board Lesson.
PLEASE MAKE SURE THAT YOU HAVE SMART NOTEBOOK 11 SOFTWARE ON YOUR COMPUTER TO USE THIS LESSON.
1. Valentine Hearts: “Valentine Hearts” is the name of the game. Circle the ones that are the same. Students will first look carefully at the heart in the white circle. Then, they will circle all the hearts that look the same as the one in the white circle. There are 3 slides of each color for this activity.
2. Valentine Cupcakes: “Valentine Cupcakes” is the name of the game. Circle the ones that are the same. Students will first look carefully at the cupcake in the white circle. Then, they will circle all the cupcakes that look the same as the one in the white circle. There are 3 slides of each color for this activity.
3. Valentine’s Day Patterns: There are 5 different Valentine’s Day objects at the bottom of the slide. Each object is set with the interactive infinite clone. Students will look at the patterns and use the infinite clone objects to continue the correct pattern. There are 10 slides including the patterns: AB, ABB, AAB, ABC, and AABB.
4. Sort the Hearts by Color: Students will drag and drop the hearts in the middle of the slide into one of the bigger hearts at the bottom of the slide that has the same color. If the student is correct, the heart will disappear into the bigger heart. If the student is wrong, the heart will either fly back out or just stop. The student will need to try again until he chooses the heart with the same color. You could practice saying the colors at the same time.
Graphics provided by:
http://www.teacherspayteachers.com/Store/Squiggledot-Designs
www.mycutegraphics.com
http://www.teacherspayteachers.com/Product/Bunches-of-Hearts-Clip-Art-Set-1034686
http://www.teacherspayteachers.com/Product/Clip-Art-Let-em-eatCupcakes-716461
http://www.teacherspayteachers.com/Store/Mrs-Bearfields-Class
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\$4.00 | 653 | 2,631 | {"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} | 3.34375 | 3 | CC-MAIN-2017-13 | longest | en | 0.883151 |
http://stackoverflow.com/questions/8386107/use-polyhedron-as-graphics-primitive-placing-at-point-and-scaling/8386225 | 1,387,505,622,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1387345768787/warc/CC-MAIN-20131218054928-00049-ip-10-33-133-15.ec2.internal.warc.gz | 180,216,391 | 14,767 | Use polyhedron as graphics primitive, placing at point and scaling
Can I ask an easy beginner's question, to which I'm unable to find an easy to understand answer in any of the texts I have (which are admittedly quite old, predating version 6 in some cases)? How do you use the polyhedra as if they were graphics primitives like Sphere and Cuboid? Ie, centred at a point and scaled. Here are silly examples to illustrate the point:
``````(* spheres along a path *)
data = Table[{Cos[t], Sin[t], Sin[t] Cos[2 t]}, {t, 0, 2 Pi, Pi/24}];
Graphics3D[Sphere[#, 0.3] & /@ data]
``````
``````(* cubes along a path *)
Graphics3D[Cuboid[#, # + 0.1] & /@ data]
``````
So how to place icosahedra at specific points and scale, writing something like
``````Graphics3D[icosahedron[#, 0.1] & /@ data]
``````
Edit: I think my problem is how to make `GraphicsComplex` and `Graphics3D` work together. Eg, where I currently have:
``````shapes[ct_, siz_] := {Sphere[ct - .2, siz ], Sphere[ct - 0.1, siz]};
Graphics3D[{{shapes[#, size] & /@ data}}]
``````
I'd like to replace that `Sphere[]` with `icosahedron[]`. Am currently trying to make Heike's solution work...
Edit 2: Working fine now, thanks Heike. Not sure I'll get it 3D-printed though - looks a bit uncomfortable to wear...
-
``````icosahedron = PolyhedronData["Icosahedron"][[1]];
@cormullion: `GraphicsComplex` is treated like any other `Graphics` or `Graphics3D` primitive so you could define `icosahedron[c_, s_]:=Translate[Scale[PolyhedronData["Icosahedron"][[1]], s], c]`. – Heike Dec 5 '11 at 22:18 | 480 | 1,556 | {"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-2013-48 | latest | en | 0.901591 |
https://www.vedantu.com/question-answer/describe-an-activity-to-show-that-the-incident-class-8-physics-cbse-60ab987af9992d73dc13378d | 1,721,624,841,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517823.95/warc/CC-MAIN-20240722033934-20240722063934-00417.warc.gz | 890,663,309 | 31,366 | Courses
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# Describe an activity to show that the incident ray, the reflected ray, and the normal at the point of incidence lie in the same plane.
Last updated date: 19th Jul 2024
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Hint: The main factor while choosing the activity is to prove that incident ray, reflected ray, and normal lie on the same plane. A clear understanding of the incident ray reflected ray and the properties of reflection will enable one to deduce an activity for this rule. When light is incident on a mirror or a reflective surface the light will bounce or reflect into the same medium.
Complete step-by-step solution:
Apparatus: Drawing board, Sheet of paper, laser light, plane mirror.
Procedure:
Place a sheet of paper on the drawing board. Place a plane mirror on the piece of paper and draw an outline of the mirror. Also, draw a normal to the mirror.
Now shine the laser light at an angle to the mirror such that it falls on the normal at the bottom of the mirror. If shined normal to the mirror, no incidence or reflected ray will be observed as the angle of incidence, in this case, is zero, thereby the angle of reflection is also zero.
When the laser light is incident on the mirror, it gets reflected at an angle to the normal.
We can easily observe that the incident ray, the reflected ray, and the normal at the point of incidence all lie in the same plane.
From this activity, it is observed that the normal, the line of incidence, and the reflected ray all lie on the same plane.
Note: This activity can also be used to determine that the angle of incidence is equal to the angle of reflection. Care must be taken while handling laser light. Do not point the laser source to your eyes as it can cause permanent damage. In the case of reflection another imperative property is that the angle of incidence is always equal to the angle of reflection. | 432 | 1,969 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-30 | latest | en | 0.888411 |
https://jp.mathworks.com/matlabcentral/cody/problems/2678-find-out-sum-and-carry-of-binary-adder/solutions/2420823 | 1,606,351,664,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141185851.16/warc/CC-MAIN-20201126001926-20201126031926-00010.warc.gz | 357,779,052 | 17,057 | Cody
# Problem 2678. Find out sum and carry of Binary adder
Solution 2420823
Submitted on 30 May 2020 by jmac
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 1; y=1; pc=1; [sum, carry]=bin_sum_carry(pc,x,y) assert(isequal(sum,1)) assert(isequal(carry,1))
sum = 1 carry = logical 1
2 Pass
x = 1; y=1; pc=0; [sum, carry]=bin_sum_carry(pc,x,y) assert(isequal(sum,0)) assert(isequal(carry,1))
sum = 0 carry = logical 1
3 Pass
x = 1; y=0; pc=0; [sum, carry]=bin_sum_carry(pc,x,y) assert(isequal(sum,1)) assert(isequal(carry,0))
sum = 1 carry = logical 0
4 Pass
x = 0; y=0; pc=0; [sum, carry]=bin_sum_carry(pc,x,y) assert(isequal(sum,0)) assert(isequal(carry,0))
sum = 0 carry = logical 0
5 Pass
x = 1; y=1; pc=0; [sum, carry]=bin_sum_carry(pc,x,y) assert(isequal(sum,0)) assert(isequal(carry,1))
sum = 0 carry = logical 1
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 380 | 1,088 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2020-50 | latest | en | 0.66785 |
https://www.queryhome.com/puzzle/6185/guess-five-letter-word-under-remove-first-letter-and-above-you | 1,675,820,503,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500664.85/warc/CC-MAIN-20230207233330-20230208023330-00068.warc.gz | 970,344,836 | 27,044 | # Guess me: I am a five letter word under you, Remove first letter and I am above you....
3,773 views
Guess me:
I am a five letter word under you,
Remove first letter and I am above you,
Remove the second as well and I am around you.
Who am I?
posted Mar 1, 2015
Chair
You sit over a CHAIR
When first letter is removed, "HAIR" is above you.
When second letter is removed "AIR" is around you.
Similar Puzzles
I am under you when I am whole.
I shift above you if you remove the first letter.
I come all around you if you remove the second as well.
Can you tell me who I am?
I am a five letter word namely,
1 2 3 4 5...... which is under you,
If you put out 1, its above you,
If you put out 1 and 2, its around you.
What am I?
I am five letter word, people eat me.
If you remove first letter it will be a form of energy.
If you remove my first two letters, I will be needed for living.
If you remove my first three letters, I will be near you.
If you remove my first four letters, I will be a drink for you.
What am I?
–1 vote
Guess me:
I am a five letter word, I am true;
if you remove my 1st letter I am become a steel;
if you remove my 4th & 5th letter I am become a support;
What am I?
–1 vote
I am a 5 letter word.
You know me very well.
I am name of a girl.
If you remove first letter it becomes another name of a girl.
If you remove last two letters it also becomes another name of a girl.
If you remove my last 3 letters, it becomes inevitable when you are talking about yourself.
If you remove First 2 letters and last letter, It becomes a word which is used for indicating a thing.
First and last letters are same and used to indicate singular.
If you remove last letter it becomes name of a boy.
What am I? | 455 | 1,731 | {"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} | 3.34375 | 3 | CC-MAIN-2023-06 | latest | en | 0.939956 |
https://secretlondon.us/5-gallon-to-quart | 1,696,113,286,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510730.6/warc/CC-MAIN-20230930213821-20231001003821-00714.warc.gz | 552,656,142 | 18,817 | # 5 Gallon To Quart
5 Gallon To Quart – Converting liters of water to gallons of water or vice versa is easier than you think. Knowing how to quickly change these measurement units can make cooking, baking, and other tasks where you need to measure volume easier.
If you want to become an expert at converting quarts to gallons, our guide can provide you with a complete overview of the definitions of quarts and gallons and how to convert liters to gallons or liters to liter.
## 5 Gallon To Quart
You may see gallons as a measure of milk or ice cream at the grocery store or motor oil at the auto store.
## Canadian Grocer April June 1918 . Ss Method.—1 Quart Solium Silicate(water Glass) And 9 Quarts Water That Has Beenboiled And Cooled. Place Mixture In 5 Gallon Crockor Jar. This Is Sufficient To Preserve 15
A liter is a unit of volume, commonly used to measure liquids in the U.S. and Great Britain, although the liter may be used for dry measurements in the U.S. We will discuss it later.
Merriam-Webster defines a quart as “a unit of strength equal to ¼ gallon or 1/32 bushel.” The abbreviation for quarts is qt. One liter is equal to 4 cups, 2 liters, or ¼ gallon.
U.S. fluid quarts which is different from a U.S. gallon. dry or royal. We’ll cover dry and royal gallons in the section below.
A gallon is a volume measure used to measure large volumes. In comparison, liters measure a small volume.
#### All Purpose Cleaner (5 Gallon) Bag In A Box
Merriam-Webster defines a gallon as “a unit of water capacity equal to 231 centimeters or four liters.” Abbreviation for gallon is gal. One gallon is equal to 16 cups, 8 quarts, or 4 quarts.
U.S. gallons of water which is different from a U.S. gallon. dry or royal. We’ll cover dry and royal gallons in the section below.
In the U.S., gallons and gallons measure the volume of liquid and dry matter. Dry quarts and liters are used to measure dry volume, while liters and wet liters are used to measure liquid volume.
In the U.S., one dry liter is equal to ¼ dry gallon or 1.101220942715 liters. One liter of liquid is equal to ¼ gallon of water or 0.946353 liters. A wet liter is slightly smaller than a dry liter.
## Sunnyside 5 Gallon Specs Paint Thinner
Dry gallons are also known as grain or gallons of corn. One dry gallon is equal to 4 dry liters or 4.4 liters. A gallon of water is 14.1% smaller than a gallon of dry water.
The imperial litre, used to measure wet and dry substances, is equal to the imperial quarter of a gallon. It is the sound part of the imperial system.
One liter of liquid U.S. equal to 0.83267384 imperial litres. One imperial liter is equal to 1.20095042 U.S. liters. To convert the U.S. probe to imperial cubits, multiply the volume by 0.83267384 or divide the volume by 1.20095042.
One imperial gallon, used to measure wet and dry substances, is equal to 4 imperial liters. It was part of the royal measuring system.
## Lifetime 5 Gallon High Performance Water Cooler
One U.S. liter is 1.2009499204287 U.S. One U.S. gallon equal to 0.83267384 imperial liters. A royal gallon is about 3.2% larger than a dry gallon.
Capacity refers to the amount of material a container can hold. For example, a container of juice, oil or gas is an energy indicator. Volume is the amount of space taken up by an object or substance.
Here are the common units used to measure capacity in the U.S. standard measurement system. from smallest to largest:
These units of measurement can be used to describe the capabilities of an object, but there are situations where it makes sense to use specific units.
## How Many Quarts Are In A Gallon?
For example, a gallon often describes something very high, like pool water or your car’s gas. Fluid ounces are usually reserved for small items such as cosmetics or drinks.
When baking, recipes list certain portions of the volume that may require you to adjust them to a smaller or larger size. Making the right change is the key to making something fun. Here are some common liquid conversions that can help you with recipes.
A liter is a measure commonly used for dry or wet substances. Here are some things that are usually measured in liters.
If you cook or bake, you need to know the right measurements to get the right recipe. Accurate measurements are required to avoid a final product with the wrong texture, consistency, moisture and texture.
### Measure Master Measuring Bucket 5 Gallon
If you are measuring water, use a measuring cup with a pouring design to make it easy to transfer from cup to container without spilling.
If you are measuring dry ingredients, use a measuring cup without a spout. They will have a flat top and smooth edges. This allows you to remove the excess from the top.
Converting quarts to gallons is easy once you understand the number of liters in a gallon. Here are some frequently asked questions about converting quarts to gallons.
Converting gallons of water to gallons of water doesn’t have to be stressful. Using our formulas or the conversion tables listed above, you can easily and quickly convert between liters and liters for any purpose. with many cups in a liter—that’s a question that makes smoke come out of my ears! Gallons, Quarts, and Cups—Oh MY! Use this handy memory tool to help you remember this kitchen transformation!
## Impact Products 5 Gallon Utility Pail
Recently, I tried to make an apple butter recipe that I found in an old Amish cookbook.
Since I didn’t want to make too many portions, I knew I had to find a way to cut the recipe in half.
I found that I was struggling to remember some of the measurements I learned in school!
I didn’t end up making the apple butter, so I can’t share the recipe. It was very hard work, but, it allowed me to improve my measurements a bit.
## How To Convert Gallons, Quarts, Pints And Cups
How many cups in a litre? And how many liters are in a liter? I went to Google, as I have to do every time I need to remember a measurement change and I thought: there must be a better way!
I printed out a working drawing a few years ago (I’m pretty sure I printed it from this site) that I put in the back of a cookbook and put in my kitchen , so that I will forget.
The original graphic is outdated, so I decided to create a new graphic for you to use. You should print this out completely and keep it on your refrigerator. You’ll find it very useful, and soon, you’ll be memorizing pictures for a lifetime!
A short list of changes using the image above. You don’t have to talk to Siri to find out how many cups are in the liter!
## All Purpose Cleaner (5 Gallon) Pail
Now, after you’ve committed the scale changer to memory (it’s not that hard, it’s a tart with gallons and gallons), you can test yourself using the questions below.
I’ll be honest, guys, I’ve never won a math award. Alhamdulillah my husband is good and he can help our son with third grade math. LOL!
Numbers are not my thing, and this can be difficult for someone who likes to cook and bake.
Speaking of measuring cups and the kitchen, I just wanted to see how cute these mason jar measuring cups are! THIS POST MAY CONTAIN AFFILIATE LINKS, WHICH MEANS YOU MAKE A SMALL COMMISSION ON YOUR PURCHASES WITH NO FEE TO YOU. PLEASE SEE MY COMPLETE RELATIONSHIP POLICY FOR DETAILS
## Gallon With (5) 5 Quart Sharps Containers
Knowing how many Quarts are in a Gallon can be very useful for chefs and cooks to measure your drinks but be careful, there is a difference between US units and the old Imperial Units.
A gallon is a unit of measure commonly used in the United States and parts of the Commonwealth. It was widely used in continental Europe before the introduction of the Metric System.
There are several different definitions of the Gallon, leading to different spatial measurements from 3.5 Liters for the Roman Gallon to 4.62 Liters for the Treasury.
The US System of Measurement also defines a Dry Quart, which is equal to 1/32 Bushels or 2 Dry Pints, or 1.101 Liters.
## Create A Portable Garden With Diy 5 Gallon Bucketplanters
This conversion table works for both US and Imperial measurements. However, it is important to remember that converting between 1 US Gallon and Imperial Quarts is not that easy.
Hi, I’m Carine, food blogger, author, recipe designer, author of several print cookbooks and e-books, and founder of Sweet As Honey.
I am excited to share all my easy and delicious recipes that are delicious and healthy. My expertise in this area comes from my background in chemistry and years of following a low carb keto diet. But I’m also good at cooking greens and vegetables because my husband loves meat.
Cooking and baking are my passions. Well, I’m only sharing a small portion of my recipe on Sweet As Honey. Most of it was eaten by my husband and two children before I could take a picture!
### How Many Cups In A Pint, Quart, Or Gallon
All of my products are tested at least three times to ensure they work and I pride myself on making them as accurate as possible.
Recipes, instructions and articles on this website should not be considered or used as medical advice. Nutrition facts should be used as a guide only. The
5 quart to gallon, quart pint gallon, gallon quart pint cup, ounce pint quart gallon, quart vs gallon ziploc, oz pint quart gallon, cups in quart gallon, cup quart gallon chart, quart and gallon, quart pint gallon chart, quart vs gallon paint, 30 quart to gallon | 2,172 | 9,446 | {"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-2023-40 | latest | en | 0.869274 |
https://www.distractify.com/p/1087-meaning-tiktok | 1,716,320,280,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058512.80/warc/CC-MAIN-20240521183800-20240521213800-00437.warc.gz | 645,134,967 | 38,219 | # The Code 1087 Actually Has a Bittersweet Meaning on TikTok
By
Apr. 12 2023, Published 3:07 p.m. ET
Older millennials, are you doing OK on TikTok nowadays? It seems like every day on the app is a lesson in learning a new, Gen Z-esque language
For example, the recent number trend taking over TikTok and Instagram.
If you've been confused by the number system trend taking over social media lately, you aren't alone.
Let's hone in on one code in particular and why the meaning behind it is actually quite bittersweet. So, what is the meaning of the numbers 1087 on a TikTok video? Here's what we know.
## What is the meaning of 1087 on TikTok videos?
If someone posts a 1087 on their TikTok video, it basically means, "I miss the old us."
So, it's kind of like a secret message within a TikTok, a coded message in a bottle if you will. The 1087 meaning is bittersweet, but it's actually a clever way to keep a video all over the internet a bit more private for the intended recipient of the 1087 meaning.
Typically, a user will add 1087 as a hashtag or make it part of their TikTok caption. That way they don't have to go into a full-fledged storytime video about who the other half of the "us" is and why they miss the old version of them.
Think of 1087 and other number codes like TikTok shorthand!
## What do some other numbers mean on TikTok then?
It's important to note for anyone trying to learn how to speak fluent TikTok that there are actually a whole numerical code system on the app, where different sets of numbers have different meanings.
TikTok users can secretly post videos about their secret crushes on the app by using numbers instead of letters. For example – A would be 022, B would be o76, C would be o99, and D would be o12.
As Distractify previously reported, the rest of the codes for each letter are:
E – o43
F – o98
G – o24
H – o34
I – o66
J – o45
K – o54
L – o84
M – o33
N – o12
O – o89
P – o29
Q – o38
R – o56
S – o23
T – o65
U – o41
V – o74
W – o77
X – o39
Y – o26
Z – o10
## OK, is there a code for apologizing? Or saying that you love someone?
Never fear, TikTok number codes are here!
If you're trying to tell someone, "Sorry I hurt you," with the number code, you'd add a 910 in your hashtags or caption.
If you want to say I love you, you'd use the numbers 143. What about the risky, "I still love you" then? That would be the code 1543.
Obviously you don't have to use 1087 or any other numerical code on TikTok, but you have to give the code system credit for creating a bit of privacy on the internet! Or maybe not since there have been so many articles written about the codes! Welcome to the TikTok paradox. | 717 | 2,688 | {"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} | 2.734375 | 3 | CC-MAIN-2024-22 | latest | en | 0.930684 |
https://www.wise-geek.com/what-is-velocity-ratio.htm | 1,611,442,319,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703538741.56/warc/CC-MAIN-20210123222657-20210124012657-00126.warc.gz | 1,070,015,971 | 18,568 | # What is Velocity Ratio?
Jessica Reed
Velocity ratio, occasionally referred to as distance ratio, is a comparison of the amount of force an object, such as a car, is creating in comparison with the other forces around it which act against it. When a car moves, the amount of force applied to the car's wheels by the engine is stronger than forces such as gravity which hold the car in place, and thus the car moves forward. The engine must supply enough power to overcome the vehicle's own weight, friction from the surface of the road, and the force of gravity pushing down on the car. To calculate velocity ratio, divide the force working against an object by the force the object itself is exerting. If it were possible to drive the car in an environment where no forces inhibited its movement, this movement would be known as its mechanical advantage instead of velocity ratio because the driver could calculate exactly how much power would be needed to move the car at a certain speed over a certain distance.
Part of the formula for measuring velocity comes from Isaac Newton's second law of motion which states that F=ma. This means multiplying an object’s acceleration by its mass gives its force. Velocity describes an object's speed and the direction it is traveling. A car's velocity might be 70 mph (112.7 kph) heading north. The second part of the velocity ratio simply compares this force to the forces acting against it to determine if any work is done.
In physics, for an object to do work, it must put forth an effort, or force, that results in movement. If the car doesn't have enough force to move, it is creating effort but getting no work done and, as a result, is not going anywhere. When a driver wants to speed up in her car, she is putting another force known as acceleration into motion. When she presses the gas pedal, the car engine works harder to move the car forward faster. This changes the velocity ratio by requiring the car to exert more force to overcome the forces that keep it moving at a slower speed.
Aside from increasing an engine's power, car owners can get a more speed-efficient vehicle by reducing the forces acting on the car. Smoother roads with less friction, tires that provide better surface traction, and a lighter car all contribute to reducing the velocity ratio and simultaneously improving the car's performance and efficiency. More efficient cars get more work done with less power than less efficient cars. | 497 | 2,472 | {"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} | 4.125 | 4 | CC-MAIN-2021-04 | longest | en | 0.9732 |
https://blog.csdn.net/weixin_44499508/article/details/105089267 | 1,586,438,780,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371858664.82/warc/CC-MAIN-20200409122719-20200409153219-00463.warc.gz | 344,058,222 | 30,503 | HDU The trouble of Xiaoqian (混合背包）
#include <bits/stdc++.h>
#define pb push_back
#define pi pair<int,int>
#define mk make_pair
using namespace std;
const int maxn = 40005,inf = 1e9+100;
int a[105],c[105],d[maxn],p[maxn];
vector<pi>ans;
int main()
{
int n,V,kase = 0;
while(~scanf("%d%d",&n,&V)&&n&&V)
{
ans.clear();
for(int i=1;i<maxn;i++)d[i] = p[i] = inf;
for(int i=1;i<=n;i++)scanf("%d",a+i);
for(int i=1;i<=n;i++)scanf("%d",c+i);
for(int i=1;i<=n;i++)
{
for(int k=1;k<=c[i];k*=2)
{
c[i] -= k;
ans.pb(mk(k*a[i],k));
}
if(c[i] > 0)ans.pb(mk(c[i]*a[i],c[i]));
}
for(int k=0;k<ans.size();k++)
{
pi it = ans[k];
for(int i = maxn - 1;i >= it.first;i--)d[i] = min(d[i],d[i-it.first]+it.second);
}
for(int i=1;i<=n;i++)
for(int j=a[i];j<maxn;j++)p[j] = min(p[j],p[j-a[i]] + 1);
int tot = d[V];
for(int i=V+1;i<maxn;i++)tot = min(tot,p[i-V] + d[i]);
printf("Case %d: ",++kase);
if(tot == inf)puts("-1");
else cout<<tot<<'\n';
}
return 0;
}
©️2019 CSDN 皮肤主题: 创作都市 设计师: CSDN官方博客 | 417 | 975 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2020-16 | latest | en | 0.099021 |
https://crypto.stackexchange.com/questions/34198/are-there-any-crypography-schemes-which-rely-on-graph-isomorphism-not-being-in-p | 1,717,077,223,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971667627.93/warc/CC-MAIN-20240530114606-20240530144606-00044.warc.gz | 149,247,373 | 43,823 | # Are there any crypography schemes which rely on Graph Isomorphism not being in P?
Are there any cryptography schemes having correctness relying on Graph Isomorphism not being in P? If I should ask this question in the CS Theory area, I will migrate it.
Thanks.
• Nope, not that I know of. GI is not a great problem from which to build primitives, because it's been known for several decades that many (most) classes of graphs have a poly-time algorithm for deciding isomorphism. Apr 2, 2016 at 22:17
• Interestingly, though, graph non-isomorphism is a standard example of an interactive zero-knowledge proof for a language not in NP. If GI collapses to P, I think GNI goes with it, so we'll need a new example in our crypto textbooks. Apr 2, 2016 at 22:21
• As I understand it, GI is collapsed to P: dharwadker.org/tevet/isomorphism Apr 2, 2016 at 23:01
• @BrentKirkpatrick You may want to read cstheory.stackexchange.com/questions/32237/… before putting faith in random papers on the internet :) Apr 2, 2016 at 23:13
• Indeed, the paper at dharwadker.org/tevet/isomorphism is not credible. However, Babai - a highly respected theoretical computer scientist - has shown that Graph Isomorphism is in quasi-polynomial time arxiv.org/abs/1512.03547. This makes it very unsuitable for cryptographic applications (especially since the feeling is that it may very well be in $P$). Apr 3, 2016 at 6:32
There is an elegant example of zero knowledge proof for graph isomorphism. The prover sends a randomly relabled graph and the verifier requests mapping to one of the originals. It is a very simple to understand and prove zero knowledge proof. However I don't believe anyone ever used this for authentication or such. Obviously we now know graph isomorphism isn't hard after all making all these not very useful.
We got a cryptographic algorithm and computer implementation based on graph isomorphism.
An isomorphism between two graphs is a bijection between their vertices that pre serves the edges.
For a graph $$G$$, let $$M(G)$$ denote the adjacency matrix of $$G$$.
Two graphs $$G,H$$ are isomorphic iff there exist permutation matrix $$P$$ such that $$P M(G) P^{-1}=P M(G) P^T=M(H)$$.
Observe that $$P$$ need not be unique.
Consider the following Diffie Hellman key exchange scheme based on graph isomorphism.
Public parameters: graph $$G$$ of order $$n$$ with $$A=M(G)$$ and $$n \times n$$ permutation matrix $$P_0$$.
Alice chooses positive integer $$X_A$$ and set the private key the matrix $$privA=P_0^{X_A}$$. Alice make public her public key the matrix
$$pubA=privA \cdot A \cdot privA^T=P_0^{X_A} A P_0^{-X_A}$$.
Bob chooses positive integer $$X_B$$ and set the private key the matrix $$privB=P_0^{X_B}$$. Bob make public his public key the matrix $$pubB=privB \cdot A \cdot privB^T$$.
To compute shared secret, Allice computes $$M_1=privA \cdot pubB \cdot privA^T=P_0^{X_A+X_B} A P_0^{-X_A-X_B}$$.
To compute shared secret, Bob computes $$M_2=privB \cdot pubA \cdot privB^T=P_0^{X_A+X_B} A P_0^{-X_A-X_B}$$.
Since powers of permutation matrices commute, Allice and Bob know the shared secret $$M_1=M_2$$.
The public keys $$pubA,pubB$$ are adjacency matrices of isomorphic graphs, each of which is isomorphic to the public $$G$$.
Multiplicative discrete logarithm of permutation matrices is efficient since the group order is $$n$$-smooth, but we believe to break the algorithm adversary must solve $$X A X^T=pubA$$ for permutation matrix $$X$$
Q1 Is this algorithm at least as hard as graph isomorphism?
For permutation matrix $$X$$, the equation $$X A X^T = pubA$$ might have many solutions, which are isomorphism of the graph $$G$$ to itself. For example take $$G$$ to be the complete graph of order $$n$$. Then for all $$X$$, we have $$X A X^T=pubA=A$$. This case is trivial since the shared secret is $$A$$.
When experimenting, we got $$G=PaleyGraph(5)$$ and $$P_0$$ such that we had $$X A X^T=pubA$$, but the shared secret was incorrect.
Q2 are there choices of $$G$$, $$P_0$$ such the algorithm is harder than graph isomorphism?
• Crossposted to MO: mathoverflow.net/questions/408757/…
– joro
Nov 17, 2021 at 13:14
• This key exchange is insecure. The graph $M_{2}=(V_{2},E_{2})$ can be recovered since the edge sets $E_{2}\cap\{\{r,s\}\mid r\in R,s\in S\}$ can easily be produced by from the public information whenever $R,S$ are cycles in the permutation $P_{0}$. Nov 17, 2021 at 18:53
• On MO, I also posted an attack where a more general key exchange is broken using linear algebra. Nov 23, 2021 at 17:18
• mathoverflow.net/a/409594/22277 On MO, I have posted another attack that translates the problem of finding $X_{A}$ into a problem of solving a system of linear congruence equations. Nov 29, 2021 at 1:43
As far as I know there is no cryptographic scheme based on Graph isomorphism. The following is the key reasons.
The security of a cryptographic scheme largely depend on one-wayness of the underlying function. For a function to be one-way it's not just need to be hard for few NP instances but must be hard for a random instance. In other words it is very easy to find problems that are hard for very instance but easy for majority of instances . Such problems may not come under P but they arn't one way functions either. One such good example is the encryption scheme based on subset-sum problem, which was eventually broken due to the above specified reason. | 1,446 | 5,415 | {"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": 39, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-22 | latest | en | 0.942738 |
http://scienceray.com/mathematics/geometry/geometry-help-biconditionals-and-definitions/ | 1,448,915,986,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398463315.2/warc/CC-MAIN-20151124205423-00127-ip-10-71-132-137.ec2.internal.warc.gz | 195,199,093 | 10,423 | Geometry Help: Biconditionals and Definitions
Discovering how to write a biconditional and what a good definition is.
Biconditionals
Biconditional statements, or “if and only if” statements, are reversible. They are written as P↔Q and read as “if and only if”.
Example 1
Determine if Biconditional or not.
A) If x=3, then x²=9
P→Q – x=3 → x²=9 – True
Q→P – x²=9 → x=3 – False (x=3, x=-3)
Not biconditional
B) If it is December 25, then it is Christmas Day.
P→Q – It is December 25 → It is Christmas Day – True
Q→P – It is Christmas Day → It is December 25 – True
Biconditional
Definitions
Definitions must be precise and must be reversible.
If And Only If
iff stands for “if and only if”
Example 2
Determine if it’s a ”good definition”. If it is, write the biconditional.
A) Parallel lines don’t intersect. – No (skew lines)
B) Parallel planes don’t intersect.- Yes…Write Biconditional.
Further Reading
Geometry Help: Points, Lines, and Planes
Geometry Help: Segments, Rays, Parallel Lines, and Planes
Geometry Help: Measuring Segments and Angles
Geometry Help: Conditional Statements
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