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https://fr.mathworks.com/matlabcentral/cody/problems/528-find-the-largest-value-in-the-3d-matrix/solutions/1794697 | 1,575,799,318,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540508599.52/warc/CC-MAIN-20191208095535-20191208123535-00123.warc.gz | 379,128,151 | 15,602 | Cody
# Problem 528. Find the largest value in the 3D matrix
Solution 1794697
Submitted on 21 Apr 2019 by Fabio Squizzato
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
A = 1:9; A=reshape(A,[3 1 3]); y_correct = 9; assert(isequal(islargest(A),y_correct))
2 Pass
A = 9:17; A=reshape(A,[3 1 3]); y_correct = 17; assert(isequal(islargest(A),y_correct))
3 Pass
A = []; A(:,:,1) = magic(5); A(:,:,1) = eye(5); A(:,:,1) = 40*ones(5); y_correct = 40; assert(isequal(islargest(A),y_correct)) | 209 | 612 | {"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-2019-51 | latest | en | 0.634507 |
https://www.aqua-calc.com/one-to-one/density/kilogram-per-cubic-yard/long-ton-per-us-quart/1 | 1,571,742,793,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987817685.87/warc/CC-MAIN-20191022104415-20191022131915-00162.warc.gz | 802,030,338 | 8,840 | # 1 kilogram per cubic yard [kg/yd³] in long tons per US quart
## kg/yd³ to long tn/US qt unit converter of density
1 kilogram per cubic yard [kg/yd³] = 1 × 10-6 long ton per US quart [long tn/qt]
### kilograms per cubic yard to long tons per US quart density conversion cards
• 1
through
25
kilograms per cubic yard
• 1 kg/yd³ to long tn/qt = 1 × 10-6 long tn/qt
• 2 kg/yd³ to long tn/qt = 2 × 10-6 long tn/qt
• 3 kg/yd³ to long tn/qt = 4 × 10-6 long tn/qt
• 4 kg/yd³ to long tn/qt = 5 × 10-6 long tn/qt
• 5 kg/yd³ to long tn/qt = 6 × 10-6 long tn/qt
• 6 kg/yd³ to long tn/qt = 7 × 10-6 long tn/qt
• 7 kg/yd³ to long tn/qt = 9 × 10-6 long tn/qt
• 8 kg/yd³ to long tn/qt = 1 × 10-5 long tn/qt
• 9 kg/yd³ to long tn/qt = 1.1 × 10-5 long tn/qt
• 10 kg/yd³ to long tn/qt = 1.2 × 10-5 long tn/qt
• 11 kg/yd³ to long tn/qt = 1.3 × 10-5 long tn/qt
• 12 kg/yd³ to long tn/qt = 1.5 × 10-5 long tn/qt
• 13 kg/yd³ to long tn/qt = 1.6 × 10-5 long tn/qt
• 14 kg/yd³ to long tn/qt = 1.7 × 10-5 long tn/qt
• 15 kg/yd³ to long tn/qt = 1.8 × 10-5 long tn/qt
• 16 kg/yd³ to long tn/qt = 1.9 × 10-5 long tn/qt
• 17 kg/yd³ to long tn/qt = 2.1 × 10-5 long tn/qt
• 18 kg/yd³ to long tn/qt = 2.2 × 10-5 long tn/qt
• 19 kg/yd³ to long tn/qt = 2.3 × 10-5 long tn/qt
• 20 kg/yd³ to long tn/qt = 2.4 × 10-5 long tn/qt
• 21 kg/yd³ to long tn/qt = 2.6 × 10-5 long tn/qt
• 22 kg/yd³ to long tn/qt = 2.7 × 10-5 long tn/qt
• 23 kg/yd³ to long tn/qt = 2.8 × 10-5 long tn/qt
• 24 kg/yd³ to long tn/qt = 2.9 × 10-5 long tn/qt
• 25 kg/yd³ to long tn/qt = 3 × 10-5 long tn/qt
• 26
through
50
kilograms per cubic yard
• 26 kg/yd³ to long tn/qt = 3.2 × 10-5 long tn/qt
• 27 kg/yd³ to long tn/qt = 3.3 × 10-5 long tn/qt
• 28 kg/yd³ to long tn/qt = 3.4 × 10-5 long tn/qt
• 29 kg/yd³ to long tn/qt = 3.5 × 10-5 long tn/qt
• 30 kg/yd³ to long tn/qt = 3.7 × 10-5 long tn/qt
• 31 kg/yd³ to long tn/qt = 3.8 × 10-5 long tn/qt
• 32 kg/yd³ to long tn/qt = 3.9 × 10-5 long tn/qt
• 33 kg/yd³ to long tn/qt = 4 × 10-5 long tn/qt
• 34 kg/yd³ to long tn/qt = 4.1 × 10-5 long tn/qt
• 35 kg/yd³ to long tn/qt = 4.3 × 10-5 long tn/qt
• 36 kg/yd³ to long tn/qt = 4.4 × 10-5 long tn/qt
• 37 kg/yd³ to long tn/qt = 4.5 × 10-5 long tn/qt
• 38 kg/yd³ to long tn/qt = 4.6 × 10-5 long tn/qt
• 39 kg/yd³ to long tn/qt = 4.8 × 10-5 long tn/qt
• 40 kg/yd³ to long tn/qt = 4.9 × 10-5 long tn/qt
• 41 kg/yd³ to long tn/qt = 5 × 10-5 long tn/qt
• 42 kg/yd³ to long tn/qt = 5.1 × 10-5 long tn/qt
• 43 kg/yd³ to long tn/qt = 5.2 × 10-5 long tn/qt
• 44 kg/yd³ to long tn/qt = 5.4 × 10-5 long tn/qt
• 45 kg/yd³ to long tn/qt = 5.5 × 10-5 long tn/qt
• 46 kg/yd³ to long tn/qt = 5.6 × 10-5 long tn/qt
• 47 kg/yd³ to long tn/qt = 5.7 × 10-5 long tn/qt
• 48 kg/yd³ to long tn/qt = 5.8 × 10-5 long tn/qt
• 49 kg/yd³ to long tn/qt = 6 × 10-5 long tn/qt
• 50 kg/yd³ to long tn/qt = 6.1 × 10-5 long tn/qt
• 51
through
75
kilograms per cubic yard
• 51 kg/yd³ to long tn/qt = 6.2 × 10-5 long tn/qt
• 52 kg/yd³ to long tn/qt = 6.3 × 10-5 long tn/qt
• 53 kg/yd³ to long tn/qt = 6.5 × 10-5 long tn/qt
• 54 kg/yd³ to long tn/qt = 6.6 × 10-5 long tn/qt
• 55 kg/yd³ to long tn/qt = 6.7 × 10-5 long tn/qt
• 56 kg/yd³ to long tn/qt = 6.8 × 10-5 long tn/qt
• 57 kg/yd³ to long tn/qt = 6.9 × 10-5 long tn/qt
• 58 kg/yd³ to long tn/qt = 7.1 × 10-5 long tn/qt
• 59 kg/yd³ to long tn/qt = 7.2 × 10-5 long tn/qt
• 60 kg/yd³ to long tn/qt = 7.3 × 10-5 long tn/qt
• 61 kg/yd³ to long tn/qt = 7.4 × 10-5 long tn/qt
• 62 kg/yd³ to long tn/qt = 7.6 × 10-5 long tn/qt
• 63 kg/yd³ to long tn/qt = 7.7 × 10-5 long tn/qt
• 64 kg/yd³ to long tn/qt = 7.8 × 10-5 long tn/qt
• 65 kg/yd³ to long tn/qt = 7.9 × 10-5 long tn/qt
• 66 kg/yd³ to long tn/qt = 8 × 10-5 long tn/qt
• 67 kg/yd³ to long tn/qt = 8.2 × 10-5 long tn/qt
• 68 kg/yd³ to long tn/qt = 8.3 × 10-5 long tn/qt
• 69 kg/yd³ to long tn/qt = 8.4 × 10-5 long tn/qt
• 70 kg/yd³ to long tn/qt = 8.5 × 10-5 long tn/qt
• 71 kg/yd³ to long tn/qt = 8.6 × 10-5 long tn/qt
• 72 kg/yd³ to long tn/qt = 8.8 × 10-5 long tn/qt
• 73 kg/yd³ to long tn/qt = 8.9 × 10-5 long tn/qt
• 74 kg/yd³ to long tn/qt = 9 × 10-5 long tn/qt
• 75 kg/yd³ to long tn/qt = 9.1 × 10-5 long tn/qt
• 76
through
100
kilograms per cubic yard
• 76 kg/yd³ to long tn/qt = 9.3 × 10-5 long tn/qt
• 77 kg/yd³ to long tn/qt = 9.4 × 10-5 long tn/qt
• 78 kg/yd³ to long tn/qt = 9.5 × 10-5 long tn/qt
• 79 kg/yd³ to long tn/qt = 9.6 × 10-5 long tn/qt
• 80 kg/yd³ to long tn/qt = 9.7 × 10-5 long tn/qt
• 81 kg/yd³ to long tn/qt = 9.9 × 10-5 long tn/qt
• 82 kg/yd³ to long tn/qt = 0.0001 long tn/qt
• 83 kg/yd³ to long tn/qt = 0.000101 long tn/qt
• 84 kg/yd³ to long tn/qt = 0.000102 long tn/qt
• 85 kg/yd³ to long tn/qt = 0.000104 long tn/qt
• 86 kg/yd³ to long tn/qt = 0.000105 long tn/qt
• 87 kg/yd³ to long tn/qt = 0.000106 long tn/qt
• 88 kg/yd³ to long tn/qt = 0.000107 long tn/qt
• 89 kg/yd³ to long tn/qt = 0.000108 long tn/qt
• 90 kg/yd³ to long tn/qt = 0.00011 long tn/qt
• 91 kg/yd³ to long tn/qt = 0.000111 long tn/qt
• 92 kg/yd³ to long tn/qt = 0.000112 long tn/qt
• 93 kg/yd³ to long tn/qt = 0.000113 long tn/qt
• 94 kg/yd³ to long tn/qt = 0.000115 long tn/qt
• 95 kg/yd³ to long tn/qt = 0.000116 long tn/qt
• 96 kg/yd³ to long tn/qt = 0.000117 long tn/qt
• 97 kg/yd³ to long tn/qt = 0.000118 long tn/qt
• 98 kg/yd³ to long tn/qt = 0.000119 long tn/qt
• 99 kg/yd³ to long tn/qt = 0.000121 long tn/qt
• 100 kg/yd³ to long tn/qt = 0.000122 long tn/qt
• long tn/qt stands for long tn/US qt
#### Foods, Nutrients and Calories
Lamb, Australian, imported, fresh, leg, sirloin chops, boneless, separable lean only, trimmed to 1/8 inch fat, raw contain(s) 132 calories per 100 grams or ≈3.527 ounces [ price ]
PHILIPPINE BRAND, CALAMANSI JUICE DRINK, UPC: 716221051591 contain(s) 52 calories per 100 grams or ≈3.527 ounces [ price ]
#### Gravels, Substances and Oils
CaribSea, Freshwater, Instant Aquarium, Crystal River weighs 1 521.75 kg/m³ (94.99975 lb/ft³) with specific gravity of 1.52175 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Gallium(III) chloride [GaCl3] weighs 2 470 kg/m³ (154.19706 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | density ]
Volume to weightweight to volume and cost conversions for Peanut oil with temperature in the range of 10°C (50°F) to 140°C (284°F)
#### Weights and Measurements
The grain per cubic millimeter density measurement unit is used to measure volume in cubic millimeters in order to estimate weight or mass in grains
The time measurement was introduced to sequence or order events, and to answer questions like these: "How long did the event take?" or "When did the event occur?"
btu/h to µJ/s conversion table, btu/h to µJ/s unit converter or convert between all units of power measurement.
#### Calculators
Volume to Weight conversions for sands, gravels and substrates | 3,026 | 6,967 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2019-43 | latest | en | 0.161421 |
http://forums.xkcd.com/viewtopic.php?p=4186757 | 1,508,240,289,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187821088.8/warc/CC-MAIN-20171017110249-20171017130249-00805.warc.gz | 127,337,971 | 14,265 | For the discussion of the sciences. Physics problems, chemistry equations, biology weirdness, it all goes here.
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webgrunt
Posts: 123
Joined: Thu Apr 21, 2011 4:04 pm UTC
Greetings,
I don't understand something. According to what I've read, there's a law that states things progress from an ordered state to a disordered state. But according to some shows I've seen, early after or during the Big Bang, there was just a sort of chaotic energy soup before atoms formed and began to clump together forming stars which in turn formed more complex elements and eventually led to the complexity of the human brain, which in spite of all appearances to the contrary in the YouTube comments section, seems to be pretty damned ordered. Like, incredibly ordered.
So how does entropy hold up when we've gone from chaos to a bunch of brains that can work together to discover the secrets of the universe? What am I missing?
speising
Posts: 2042
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Location: wien
### Re: Lay question about entropy
That's life for you.
Entropy increases for the whole system, but it can always decrease locally, at the cost of even more entropy somewhere else (aka expenditure of energy).
Life could be defined as such an ordering process. In fact, everything life does is directed towards gathering energy and expending it to create order.
Tub
Posts: 299
Joined: Wed Jul 27, 2011 3:13 pm UTC
### Re: Lay question about entropy
Entropy is not order.
I know that entropy is often explained as being a measure of order, and that works out for a few simple examples.
But is it really more ordered to have all the particles of the universe randomly strewn about on planets and stars? I know that when my socks are randomly dispersed across the room, I consider that less ordered than a homogenous stack of socks. So shall we consider the homogenous particle soup ordered or unordered?
tl;dr: You can't use an intuition of order to determine entropy. You need to count microstates, put them into the formula and then you have it. And then you'll figure out that the total entropy has continually increased since the big bang. Local decreases are not a problem, only the total entropy of a system counts.
doogly
Dr. The Juggernaut of Touching Himself
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### Re: Lay question about entropy
We don't need the word life at all though, we just need the universe to be expanding and cooling. The entropy of the whole system is still increasing during this process. It is totally a very exciting chapter in Mukhanov's book. I recommend it if you are not in fact a lay person, but a grad student testing us.
https://www.amazon.com/Physical-Foundat ... DG7W1AF3B9
If you are actually a lay person, Carroll's book is p aight
https://www.amazon.com/Eternity-Here-Qu ... an+carroll
The only thing unsatisfying with this story is that maybe you also don't want the initial state to be very low entropy. You might like to think that a good initial state would be one that is "generic", and since low entropy states are "special", we are doing a shell game with our explanations. But we really do like entropy increasing, so thinking it was initially v v low is not a weird thought. Just, yeah, the cosmological implications do have some prickles.
LE4dGOLEM: What's a Doug?
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Tub
Posts: 299
Joined: Wed Jul 27, 2011 3:13 pm UTC
### Re: Lay question about entropy
doogly wrote:If you are actually a lay person, Carroll's book is p aight
Can confirm that his book "The Big Picture" is a good read for lay persons. His older book "From Eternity to Here" probably has more details on entropy, but I haven't read it. Can also recommend searching talks from Sean Carroll on youtube. His talks (i.e. not his lectures) are pretty laymen friendly. Can also recommend his blog.
Eebster the Great
Posts: 2689
Joined: Mon Nov 10, 2008 12:58 am UTC
### Re: Lay question about entropy
doogly wrote:The only thing unsatisfying with this story is that maybe you also don't want the initial state to be very low entropy. You might like to think that a good initial state would be one that is "generic", and since low entropy states are "special", we are doing a shell game with our explanations. But we really do like entropy increasing, so thinking it was initially v v low is not a weird thought. Just, yeah, the cosmological implications do have some prickles.
More specifically, the same statistical argument that shows that entropy is likely to be higher in the future also shows that, if we knew nothing about the past, we would expect it to be higher in the past as well. There ought to be some good reason why it was lower in the past (indeed, why it was extremely low following the Big Bang).
doogly
Dr. The Juggernaut of Touching Himself
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### Re: Lay question about entropy
I think you'd only expect that if the past were also infinite.
LE4dGOLEM: What's a Doug?
Noc: A larval Doogly. They grow the tail and stinger upon reaching adulthood.
Keep waggling your butt brows Brothers.
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Eebster the Great
Posts: 2689
Joined: Mon Nov 10, 2008 12:58 am UTC
### Re: Lay question about entropy
doogly wrote:I think you'd only expect that if the past were also infinite.
How do you figure?
doogly
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### Re: Lay question about entropy
Normally, forward in time is increase in entropy. You'd only say forward and backwards in time are going to look the same if you are in some eternal steady state, and you think it's just as likely that you're on the upswing or the downswing of a fluctuation.
LE4dGOLEM: What's a Doug?
Noc: A larval Doogly. They grow the tail and stinger upon reaching adulthood.
Keep waggling your butt brows Brothers.
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somitomi
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### Re: Lay question about entropy
Despite having passed Thermodynamics I. at university, my understanding of entropy comes mainly from this video The gist of it is that things can be disordered and complex at the same time (see also: my room).
An Imaginary Report on Hungarian Popular Music
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doogly
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### Re: Lay question about entropy
"Complexity" is just very difficult to define in any physically meaningful way.
LE4dGOLEM: What's a Doug?
Noc: A larval Doogly. They grow the tail and stinger upon reaching adulthood.
Keep waggling your butt brows Brothers.
Or; Is that your eye butthairs?
Eebster the Great
Posts: 2689
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### Re: Lay question about entropy
doogly wrote:Normally, forward in time is increase in entropy.
Well empirically, yes, but there is nothing in statistical mechanics that demonstrates that this ought to be the case in general, classical physics being time-symmetrical and all.
doogly
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### Re: Lay question about entropy
Passing to the statistical case breaks the reversability of the microphysics though
LE4dGOLEM: What's a Doug?
Noc: A larval Doogly. They grow the tail and stinger upon reaching adulthood.
Keep waggling your butt brows Brothers.
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Eebster the Great
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### Re: Lay question about entropy
doogly wrote:Passing to the statistical case breaks the reversability of the microphysics though
What do you mean? Microscopic physics is still reversible, unless you think weak CP violation explains the arrow of time.
doogly
Dr. The Juggernaut of Touching Himself
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### Re: Lay question about entropy
Right, but entropy isn't a thing in the microscopic picture, it's only defined in a statistical sense, at which point you break the reversability.
LE4dGOLEM: What's a Doug?
Noc: A larval Doogly. They grow the tail and stinger upon reaching adulthood.
Keep waggling your butt brows Brothers.
Or; Is that your eye butthairs?
Posts: 149
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### Re: Lay question about entropy
I think they meant something like (and I'm going to say this very poorly)
that even if the "time progressing" function is a bijection, that if some macrostates have more possible microstates than others, that the output of the function will be a microstate of a macrostate with more possible microstates more often than of one with fewer possible microstates.
I think?
whoops, they explained themself while I was writing this
I found my old forum signature to be awkward, so I'm changing it to this until I pick a better one.
morriswalters
Posts: 6885
Joined: Thu Jun 03, 2010 12:21 am UTC
### Re: Lay question about entropy
Well that got technical really quickly. Ginsberg's Theorem is an amusing take on it. In a practical sense it's why, no matter how long you wait, your coffee will never get any warmer, unless you add heat. The most common use of of the principle, at least that most people will recognize, is in their AC. The concept is enshrined in the bureaucracy, it's used by the patent office to weed out perpetual motion machines. Our brain had a chance to evolve because because we had a heat source.
Eebster the Great
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### Re: Lay question about entropy
doogly wrote:Right, but entropy isn't a thing in the microscopic picture, it's only defined in a statistical sense, at which point you break the reversability.
But the statistics of reversible physics can only be irreversible in peculiar circumstances. In the overwhelming majority of circumstances, they will be symmetrical. So that does not answer the question of why our particular circumstance is so extremely peculiar. It is predictable only in reference to another even more peculiar circumstance at another point in time (which we call the past). The fact that entropy is lower in the past and higher in the future cannot be predicted by statistics alone.
To put it another way, if we take a tenseless view of the universe, and we pick some point in time at random, we can equally well predict that the entropy at any other nearby point in time will be equal or greater. There is no law saying that it must be higher moving in one direction and lower moving in the other.
Tub
Posts: 299
Joined: Wed Jul 27, 2011 3:13 pm UTC
### Re: Lay question about entropy
Eebster the Great wrote:To put it another way, if we take a tenseless view of the universe, and we pick some point in time at random, we can equally well predict that the entropy at any other nearby point in time will be equal or greater. There is no law saying that it must be higher moving in one direction and lower moving in the other.
Careful with "picking at random". If we live in a universe where the overwhelming parts are in thermal equilibrium, then we predict the opposite - entropy in either direction from a random point should be equal or less.
Eebster the Great
Posts: 2689
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### Re: Lay question about entropy
Tub wrote:
Eebster the Great wrote:To put it another way, if we take a tenseless view of the universe, and we pick some point in time at random, we can equally well predict that the entropy at any other nearby point in time will be equal or greater. There is no law saying that it must be higher moving in one direction and lower moving in the other.
Careful with "picking at random". If we live in a universe where the overwhelming parts are in thermal equilibrium, then we predict the opposite - entropy in either direction from a random point should be equal or less.
In such a universe, what I said is still technically true. Entropy cannot decrease any more often than it increases in such a case, and regardless, most of the time it will remain constant.
Zamfir
I built a novelty castle, the irony was lost on some.
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### Re: Lay question about entropy
But the statistics of reversible physics can only be irreversible in peculiar circumstances. In the overwhelming majority of circumstances, they will be symmetrical. So that does not answer the question of why our particular circumstance is so extremely peculiar.
Is 'majority of circumstances' really a meaningful concept, here? For all we know, every theoretically possible universe has to start out like this. 'Start' is perhaps not the right word, just that every universe has to have a big-bang state somewhere, and the region of time near that event will have an observable arrow of time.
doogly
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### Re: Lay question about entropy
Zamfir wrote:Is 'majority of circumstances' really a meaningful concept, here?
Is it time for the measure problem? Is it time?
LE4dGOLEM: What's a Doug?
Noc: A larval Doogly. They grow the tail and stinger upon reaching adulthood.
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jewish_scientist
Posts: 606
Joined: Fri Feb 07, 2014 3:15 pm UTC
### Re: Lay question about entropy
Just wondering, does empty space* count as high entropy or low entropy?
*To anyone about to bring up quantum physics: shut up, you know what I mean and this is hard enough as it is without you bringing up a technicality from a different branch of physics.
doogly
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### Re: Lay question about entropy
The entropy of classical empty space is 0.
A classical vacuum is super boring, and you cannot avoid quantum mechanics for very long here.
LE4dGOLEM: What's a Doug?
Noc: A larval Doogly. They grow the tail and stinger upon reaching adulthood.
Keep waggling your butt brows Brothers.
Or; Is that your eye butthairs?
Eebster the Great
Posts: 2689
Joined: Mon Nov 10, 2008 12:58 am UTC
### Re: Lay question about entropy
Zamfir wrote:
But the statistics of reversible physics can only be irreversible in peculiar circumstances. In the overwhelming majority of circumstances, they will be symmetrical. So that does not answer the question of why our particular circumstance is so extremely peculiar.
Is 'majority of circumstances' really a meaningful concept, here? For all we know, every theoretically possible universe has to start out like this. 'Start' is perhaps not the right word, just that every universe has to have a big-bang state somewhere, and the region of time near that event will have an observable arrow of time.
That might be true, but it's not something you can derive from statistical mechanics, which is my point. Statistical mechanics do not explain the arrow of time as is often claimed. The low entropy of the Big Bang does.
WibblyWobbly
Can't Get No
Posts: 506
Joined: Fri Apr 05, 2013 1:03 pm UTC
### Re: Lay question about entropy
doogly wrote:The entropy of classical empty space is 0.
A classical vacuum is super boring, and you cannot avoid quantum mechanics for very long here.
So, quantum mechanics in a vacuum is like Liam Neeson looking for his daughter?
doogly
Dr. The Juggernaut of Touching Himself
Posts: 5190
Joined: Mon Oct 23, 2006 2:31 am UTC
Location: Somerville, MA
Contact:
### Re: Lay question about entropy
It is probably not the worst analogy someone has tried to use for quantum mechanics.
LE4dGOLEM: What's a Doug?
Noc: A larval Doogly. They grow the tail and stinger upon reaching adulthood.
Keep waggling your butt brows Brothers.
Or; Is that your eye butthairs?
WibblyWobbly
Can't Get No
Posts: 506
Joined: Fri Apr 05, 2013 1:03 pm UTC
### Re: Lay question about entropy
doogly wrote:It is probably not the worst analogy someone has tried to use for quantum mechanics.
Oh, I've certainly heard worse. Mostly from mainstream reporting on quantum mechanics. But they generally try to succeed and fail, where as I was trying to fail and succeeded. We're all winners.
Soupspoon
You have done something you shouldn't. Or are about to.
Posts: 2360
Joined: Thu Jan 28, 2016 7:00 pm UTC
Location: 53-1
### Re: Lay question about entropy
WibblyWobbly wrote:
doogly wrote:The entropy of classical empty space is 0.
A classical vacuum is super boring, and you cannot avoid quantum mechanics for very long here.
So, quantum mechanics in a vacuum is like Liam Neeson looking for his daughter?
"I don't know who you are. I don't know what you want. At least not at the same time. If you are looking for discrete values, I can tell you I don't have precision. But what I do have are a very particular set of interference patterns, interference patterns I have acquired over a very large light-cone. Interference patterns that make me a probability amplitude for superpositions like you. If you let my wave/particle duality go now, that'll be the end of it. I will not collapse your waveform for you, I will not entangle you. But if you don't, I will observe you, I will resolve your momentum, and I will cohere you."
WibblyWobbly
Can't Get No
Posts: 506
Joined: Fri Apr 05, 2013 1:03 pm UTC
### Re: Lay question about entropy
Soupspoon wrote:
WibblyWobbly wrote:
doogly wrote:The entropy of classical empty space is 0.
A classical vacuum is super boring, and you cannot avoid quantum mechanics for very long here.
So, quantum mechanics in a vacuum is like Liam Neeson looking for his daughter?
"I don't know who you are. I don't know what you want. At least not at the same time. If you are looking for discrete values, I can tell you I don't have precision. But what I do have are a very particular set of interference patterns, interference patterns I have acquired over a very large light-cone. Interference patterns that make me a probability amplitude for superpositions like you. If you let my wave/particle duality go now, that'll be the end of it. I will not collapse your waveform for you, I will not entangle you. But if you don't, I will observe you, I will resolve your momentum, and I will cohere you."
Good luck. | 4,645 | 18,835 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2017-43 | longest | en | 0.954408 |
http://temite.org/error-function/error-function-example.html | 1,519,388,406,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814700.55/warc/CC-MAIN-20180223115053-20180223135053-00471.warc.gz | 351,907,549 | 4,286 | # temite.org
Home > Error Function > Error Function Example
# Error Function Example
## Contents
So, if we are getting 404 geladen... Check This Out value.
## Error Function Calculator
Wird Press, William H.; Teukolsky, Saul A.; Vetterling, administrator is webmaster. The pairs of functions {erff(),erfcf()} and {erfl(),erfcl()} take and the complex z-plane in figures 2 and 3. programming for everyone.
Ausleihen des Videos verfügbar. Compute the error function for x = geladen... Error Function, Complimentary Error Function and Properties Complementary Error Function Calculator Dover, pp.299-300, 1972. − t 2 {\displaystyle e^{-t^ − 2}} is an even function.
The lower bound The lower bound Derivative Of Error Function Isn't that more expensive Wird https://www.mathworks.com/help/symbolic/erf.html erfc for real and complex arguments. Cody's algorithm.[20] Maxima provides both erf 1990, p.341) define without the leading factor of .
Please try Inverse Error Function desolate can I make a habitable world? Applied Mathematics Series. 55 (Ninth reprint with additional corrections Trademarks Privacy Policy Preventing Piracy © 1994-2016 The MathWorks, Inc. If you need to, you can adjust fastest approximation suitable for a given application. Sprache: Deutsch Herkunft der Inhalte: Deutschland p.297.
• For integer , (16) (17) (18) (19) (Abramowitz and Stegun 1972, p.299), where and erfc(x) for real arguments.
• Bitte versuche for real and complex arguments.
• return values of type float and long double respectively.
• Haskell: An erf package[18] exists that provides a typeclass for the real and complex arguments, which are also available in Wolfram Alpha.
• Watson, G.N.
## Derivative Of Error Function
Veröffentlicht am 18.08.2013Complete set of Video http://mathworld.wolfram.com/Erfc.html werden geladen... Copy file to current directory How Copy file to current directory How Error Function Calculator Anmelden Transkript 12.131 Aufrufe Complementary Error Function Table the first computational knowledge engine.
The relationship between the error function erf and normcdf is normcdf(x)=12(1−erf(−x2)).For expressions his comment is here G. Wikipedia® is a registered trademark of and erfc for real and complex arguments. How To Calculate Error Function In Casio Calculator do I have?
After division by n!, all the En for odd 0, x = ∞, and x = -∞. All generalised error functions for n>0 look similar this contact form damit dein Feedback gezählt wird.
To use these approximations for negative x, use the Error Function Matlab Math. to get translated content where available and see local events and offers. ISBN978-1-4020-6948-2. ^ Winitzki, Sergei (6 February 2008). "A handy of all odd numbers up to (2n–1).
## for modernizing math education.
It doesn't other feedback? Strategies for creating 3D text Please explain Inverse Error Function Calculator this preference below. IEEE Transactions on Wireless Communications, 4(2), 840–845, doi=10.1109/TWC.2003.814350. ^ Chang, Seok-Ho; Cosman, Pamela
MathWorld—A Wolfram Web Resource./ E. recommend that you select: . http://temite.org/error-function/error-if-function.html Not the answer the heat equation when boundary conditions are given by the Heaviside step function. | 801 | 3,259 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-09 | latest | en | 0.533961 |
https://www.manea.cambs.sch.uk/8th-june-2020-1/ | 1,716,175,928,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058147.77/warc/CC-MAIN-20240520015105-20240520045105-00571.warc.gz | 800,556,646 | 38,881 | # 8th June 2020
A message from Miss Bridges and Mrs Revell
As time goes by, we know it is getting harder for some of you to stay positive. We want you to know that we are very proud of all of the work that you have been doing at home, all of the kind gestures you have been doing for your family and how well you are coping. Remember as long as you are doing the best you can and aiming to be the best that you can be then you should be exceptionally proud of yourself.
Let's try and keep thinking positively, let's look for the good in everything and think hard about everything that is going well. You are all fabulous.
What was the highlight of your week last week?
Miss Bridges' favourite moment was watching one of the baby ducklings hatch from an egg- it was truely a magical moment!
Don't forget to look at the "egg-citing delivery" page for more duckling updates and challenges this week.
## A duckling hatching
This week's challenges:
Maths:
Challenge one: Let's start by thinking about our numbers. Can you build numbers using equipment around your house or in the garden? Can you build numbers over 20?
Can you compare two of your numbers using the symbols < and >? (remember the crocodile eats the largest number).
Can you say the number 1 more and 1 less than the number you have made? How about ten more and ten less?
Challenge two:
Lets look at subtraction.
You could do this by counting backwards on a numberline. Practise subtracting a one digit number from a 2 digit number e.g. 13-7 15-9 14-8. Try numbers over 20 if you find this easy.
## Subtract by counting back on a number line
Visit KeyStage2Maths.com for more.
Literacy
Challenge one: Let's practise writing sentences. Use the key vocabulary on the sheet to write sentences about the pictures. Can you add in describing words?
Challenge two: Have a go at reading one of the play scripts below with the help of somebody in your house. What do you think a play script is for?/ Remember to read the name of the character in your head.
Top | 466 | 2,037 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2024-22 | latest | en | 0.948773 |
https://www.jiskha.com/questions/1731254/diane-asked-her-dad-to-buy-18-bottles-of-soft-drink-each-bottle-contained-5-4-litres-the | 1,610,941,562,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514121.8/warc/CC-MAIN-20210118030549-20210118060549-00248.warc.gz | 844,623,025 | 5,155 | # Math
Diane asked her dad to buy 18 bottles of soft drink. Each bottle contained 5/4 litres. The glasses they had for the party could hold 1/5 of a litre. How many glasses could be filled
from the 18 bottles of soft drink?
1. 👍
2. 👎
3. 👁
1. 18 * (5/4) = 90/4 = 22.5 liters
22.5 / 0.2 = ________ glasses
1. 👍
2. 👎
👤
Ms. Sue
2. Note that Ms. Sue's answer is 112.5 glasses but that isn't the number that can be FILLED.
1. 👍
2. 👎
## Similar Questions
1. ### Mathematics
Perfume is sold in two different sizes of bottle The 80ml bottle is priced at £55 The 50ml bottle is usually priced at £45, but is on special offer If you buy one 50ml bottle of perfume, you get a second for half price By
2. ### Math!!
Raymond buys bottles of water at \$2.10 each and a large pizza at \$12.99. The total cost was \$21.39. How many bottles of water b did he buy? A. 2.10b+12.99=21.39; 4 bottles B. 2.10+12.99b=21.39; 4 bottles C. 2.10c= 21.39+12.99; 6
3. ### science
What volume of helium would be in a balloon the size of a soft drink bottle?
4. ### MATH
A market analyst creates a taste test to be conducted in a local shopping mall. The experiment asks 500 participants to select which of 2 soft drinks they prefer. The drinks are the same color and in similar cups. Participants
1. ### Chemistry
The price of a popular soft drink is \$0.980 for 24.0 oz (ounces) or \$0.780 for 0.500 L. 1 qt = 32.0 oz. What is the price per liter of the 24.0 oz bottle? What is the price per liter of the 0.500 L bottle?
2. ### physics
A soft-drink bottle resonates as air is blown across its top. what happens to the resonant frequency as the level of fluid in the bottle decreases?
3. ### Math
Kit bought a soft drink and a sandwich for \$9.00. What was the price of each if the sandwich cost 3.5 times as much as the soft drink? 3.5 x2 =\$7.00 sandwich 3.5 /7.00=2.50 drink check this for me
4. ### Chemistry
The price of a popular soft drink is \$0.97 for 24.0 oz or \$0.77 for 0.500L What is the price per liter of the 24.0oz bottle What is the price per liter of the 0.500 L bottle
1. ### Chemistry
I'm having trouble getting the right answer for a problem, I'm hoping that by posting my work someone can help me figure out where I've gone wrong. A two-liter soft drink bottle can withstand a pressure of 5 atm. Half a cup
2. ### Chemistry-Urgent
The partial pressure of CO2 inside a bottle of soft drink is 4.0 atm at 25°C. The solubility of CO2 is 0.12 mol/L. When the bottle is opened, the partial pressure drops to 3.0 ✕ 10-4 atm. What is the solubility of CO2 in the
3. ### Statistics
A machine is used to fill soda bottles in a factory. The bottles are labeled as containing 2.0 liters, but extra room at the top of the bottle allows for a maximum of 2.25 liters of soda before the bottle overflows. The standard
4. ### One more for Damon!
Sorry just this last one! 1. Raymond buys bottles of water at \$2.10 each and a large pizza at \$12.99. The total cost was \$21.39. How many bottles of water, b, did he buy? a. 2.10b + 12.99 = 21.39; 4 bottles b. 2.10 + 12.99b = | 948 | 3,080 | {"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} | 3.59375 | 4 | CC-MAIN-2021-04 | latest | en | 0.93316 |
http://roofgenius.com/tag/812-roof-pitch/ | 1,500,821,259,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424564.72/warc/CC-MAIN-20170723142634-20170723162634-00639.warc.gz | 280,822,929 | 9,984 | Sunday , July 23 2017
Home / Tag Archives: 8/12 roof pitch
# Tag Archives: 8/12 roof pitch
## The damage that wind causes to Roof Shingles
You could believe that wind hasn’t done damage to your roof if you shingles blowing off don’t come into your view. Consider once more. Wind could be the cause of numerous damages to the roof of yours and they are going to make its life lesser and leave it at …
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## Know about the aspect of roofing proposals
Every year, a great many building proprietors contract a roofer without getting to such an extent as a moment offer. While getting rid of offering evades the dreariness and worry of the offering procedure, it is a hazardous move financially. Quality, not cost, ought to be the main concern while …
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## 2 ways to determine or calculate roof pitch or roof slope using a carpenters level
How to determine or calculate roof pitch or roof slope Roof pitch is one of the most used terms in the roofing industry. (Also referred to as roof slope or roof slant) So what does the 7/12 in the example to the right mean? The 7 means that the roof rises …
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## Roof pitch determined 2 ways
How to determine or calculate roof pitch or roof slope Roof pitch is determined by the number of inches it rises vertically for every 12 inches it extends horizontally. (Also referred to as roof slope or roof slant.) So what does the 7/12 in the example to the right …
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## 8/12 Roof Pitch
8/12 Roof Pitch Information 8/12 Roof Pitch 33¾ Degrees Also referred to as 8/12 roof slope, 8 on 12, 8 to 12 and 8/12 roof angle
Read More »
## Roof Pitch Calculator
Roof Pitch to Angle Calculator (enter pitch in the first box – calculation is automatic) Roof Span: ft. in. Roof Rise: ft. in. Roof Pitch = / 12¨ Roof Pitch Calculator Why ? (Also referred to as roof slope or roof angle) Adding for roof pitch is needed if you …
Read More » | 462 | 1,918 | {"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-2017-30 | latest | en | 0.903719 |
https://www.double-entry-bookkeeping.com/efficiency-ratios/fixed-asset-turnover-ratio-calculator/ | 1,532,130,679,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592001.81/warc/CC-MAIN-20180720232914-20180721012914-00384.warc.gz | 859,897,556 | 12,691 | # Fixed Asset Turnover Ratio Calculator
## What does it do?
The fixed asset turnover ratio measures the amount of sales activity a business is deriving from its investment in fixed assets. This fixed asset turnover ratio calculator works out the value of the ratio based on input values for revenue and opening and closing fixed assets.
The value of the fixed asset turnover ratio will depend on the industry the business operates in. The higher the ratio, the higher the sales activity for a given level of fixed assets, and the higher the earnings are likely to be. Managements job is to ensure that the fixed assets are managed efficiently, and the trend of the ratio is upwards. Further details on this ratio can be found in our fixed asset turnover ratio tutorial.
## Formula
The fixed asset turnover calculation is carried out using the fixed asset turnover formula by dividing the revenue by the average fixed assets for the period.
`Fixed asset turnover ratio = Revenue / Average fixed assets`
Revenue sometimes referred to as sales or turnover, is found on the income statement. Fixed assets are found on the balance sheet of the business. The average of the opening and closing fixed asset levels is used to avoid distorting the result.
## Instructions
The Excel fixed asset turnover ratio calculator, available for download below, is used to compute the fixed asset turnover ratio by entering details relating to the revenue and the opening and closing fixed asset levels. The calculator is used as follows:
### Step 1
The revenue is entered. Revenue is found on the income statement of the business. It is sometimes referred to as sales, net sales or turnover.
### Step 2
The opening and closing fixed asset levels are entered. The opening and closing fixed asset levels are found on the balance sheets of the business. The figure used for fixed assets should be net of accumulated depreciation.
### Step 3
The fixed asset turnover ratio calculator works out the average value of the fixed assets during the accounting period, and the fixed asset turnover ratio. | 396 | 2,089 | {"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-2018-30 | longest | en | 0.925794 |
https://blogs.sw.siemens.com/embedded-software/2019/10/24/reunited-with-a-lost-wallet-a-story-of-genius/ | 1,708,997,647,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474669.36/warc/CC-MAIN-20240226225941-20240227015941-00745.warc.gz | 132,996,678 | 19,167 | # Reunited with a lost wallet – a story of genius
By Colin Walls
We live in a world that seems to be ever more complex. Much I love so much of the technology around me, I am often bewildered by how complicated it is [perhaps underneath the hood, resulting in a good user experience] and I have great respect for the smart people who create this magic. However, at heart, I am a simple man. I like simple things. I particularly enjoy an elegantly simple solution to a problem …
One of my favorite “simple solution” stories is about the guy who came up with a way for the toothpaste manufacturers to increase their sales. I wrote about that quite a while ago. I have no idea whether the tale is apocryphal or true; maybe it does not matter.
In the world of software, we are typically dealing with incredibly complex systems – so complex that it is amazing [to me] that they work at all! However, once in a while I come across something that has elegant simplicity. For example, imagine I have a number [which I will imaginatively call N] and I would like to print it out using a number base that I will call B – this can be in the range 2 to 10. Here is an algorithm to do it:
```PrintNumber(N, B)
divide N by B and save the remainder, R
if N <> 0
PrintNumber(N, B)
display(R)
end```
That is not a real programming language, but I hope that it demonstrates the logic. [Any coding geeks out there: yes, I know recursion is a Bad Thing, but it is cool, right?]
Finding simple solutions to things in everyday life is very rewarding. Last week, I saw a news story about a guy who solved someone’s problem is a very elegant way:
Tim was riding his bicycle in London and his wallet dropped out of his pocket. He was dismayed, not because there was much money in it, but there were several credit, debit and ID cards and it would be a very annoying job to cancel and replace them all. He back-tracked on his route, but did not find his wallet. Simon was cycling along the same road and spotted Tim’s wallet, which had clearly been lost. His first thought was to take it to a Police station, but he thought that he would try to find the owner himself first. There was nothing in the wallet with Tim’s address or phone number. He tried Facebook, but Tim’s name was too common. Then he had a brilliant idea. He sent Tim £0.04 using the account information on one of the cards in the wallet.
How did sending money help? In the UK [and I am sure other countries], a popular and efficient way to send money to someone else is by means of a “bank transfer”. All you need is their account number and “sort code” [another number] and you can send money almost instantly using online banking or your bank’s smartphone app. I do this all the time. When you make a transfer, there is the opportunity to add a “reference” – just a few characters of text – to identify the payment to the payee. Simon used this text to send Tim four messages, which put together, gave Tim the information he would need to recover his wallet. And it worked like a charm.
Tim rewarded Simon with a bottle of wine, but apparently has not reimbursed him the £0.04 …
IMHO, Simon is a genius and, clearly, a really nice guy.
Topics
This article first appeared on the Siemens Digital Industries Software blog at https://blogs.sw.siemens.com/embedded-software/2019/10/24/reunited-with-a-lost-wallet-a-story-of-genius/ | 779 | 3,381 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-10 | latest | en | 0.982928 |
https://www.studypool.com/services/90274/physics-questions-5 | 1,544,773,769,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825495.60/warc/CC-MAIN-20181214070839-20181214092339-00389.warc.gz | 1,077,558,723 | 21,782 | # physics questions
Jun 17th, 2015
Studypool Tutor
Price: \$10 USD
Tutor description
Q.1: Define SI Unit of current. (1) Q.2: Draw a graph of V vs I for a material that doesn’t obey Ohm’s Law. Name the material. (1) Q.3: Explain the significance of the expression: = . (1) Q.4: What focal length should the reading spectacles have for a person for whom the least distance of distinct vision is 50 cm?
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Showing Page: 1/8
Module 1 : A Crash Course in Vectors Lecture 1 : Scalar And Vector FieldsObjectives In this lecture you will learn the followingLearn about the concept of field Know the difference between a scalar field and a vector field. Review your knowledge of vector algebraLearn how an area can be looked upon as a vector Define position vector and study its transformation properties under rotationSCALAR AND VECTOR FIELDS This introductory chapter is a review of mathematical concepts required for the course. It is assumed that the reader is already familiar with elementary vector analysis. Physical quantities that we deal with in electromagnetism can be scalars or vectors. A scalar is an entity which only has a magnitude. Examples of scalars are mass, time, distance, electric charge, electric potential, energy, temperature etc. A vector is characterized by both magnitude and direction. Examples of vectors in physics are displacement, velocity, acceleration, force, electric field, magnetic field etc. A field is a quantity which can be specified everywhere in space as a function of position. The quantity that is specified may be a scalar or a vector. For instance, we can specify the temperature at every point in a room. The room may, therefore, be said to be a region of ``temperature field" which is a scalar field because the temperature is a scalar function of the position. An example of a scalar field in electromagnetism is the electric potential.
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2325 Tutors | 613 | 2,669 | {"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} | 3.828125 | 4 | CC-MAIN-2018-51 | latest | en | 0.875103 |
http://www.largeformatphotography.info/forum/showthread.php?144168-question-on-LAMOT-for-Durst-184&goto=nextoldest | 1,518,945,077,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891811795.10/warc/CC-MAIN-20180218081112-20180218101112-00092.warc.gz | 498,927,167 | 13,534 | # Thread: Enlarger: Neg to Lens to Print cheat sheet: 8X10, 5X7, 4X5
1. ## Enlarger: Neg to Lens to Print cheat sheet: 8X10, 5X7, 4X5
I had one, but I moved. I think it was DeVere and it gave guidelines for different size negs, enlarger lens combos, and final print size.
It listed distances from neg to lens to the baseboard.
Steve, you may have supplied it to me last time.
Thanks
2. ## Re: Enlarger: Neg to Lens to Print cheat sheet: 8X10, 5X7, 4X5
Randy,
Send me a PM with your e-mail address, and I can send you an Excel spreadsheet with the calculations for different lenses, negative size / diagonal measurement, negative stage to baseboard distance, etc.
Len
Done
Thanks Len
4. ## Re: Enlarger: Neg to Lens to Print cheat sheet: 8X10, 5X7, 4X5
You know in printing for some 40 years I've never had a need for one. If I need a larger print I raise the head or use a shorter lens. But to each his own.
5. ## Re: Enlarger: Neg to Lens to Print cheat sheet: 8X10, 5X7, 4X5
Luis,
In my case, I am constrained by the ceiling height where the enlarger is located, and purchased a 360mm lens, and found it greatly restricted the size of prints I could make, then purchased a 300mm lens to allow me to make up to 16x20 on a vertical projection.
So had I done the calculations prior, could have saved my purchase of a lens I did not use.
Hope the context helps in understanding the requirements,
Len
6. ## Re: Enlarger: Neg to Lens to Print cheat sheet: 8X10, 5X7, 4X5
Well, I've been printing for about 50 years...and would still really appreciate such a list! Thanks!
7. ## Re: Enlarger: Neg to Lens to Print cheat sheet: 8X10, 5X7, 4X5
Here is a table.
I record the baseboard and head height (my scale is in millimeters!) for each print size. I don't like to keep the \$250 lamp on more than needed when framing the print.
8. ## Re: Enlarger: Neg to Lens to Print cheat sheet: 8X10, 5X7, 4X5
and I did want horizontal #s for my Elwood!
Thanks, I think that is the one you gave me some years ago!
9. ## Re: Enlarger: Neg to Lens to Print cheat sheet: 8X10, 5X7, 4X5
Originally Posted by Len Middleton
Luis,
In my case, I am constrained by the ceiling height where the enlarger is located, and purchased a 360mm lens, and found it greatly restricted the size of prints I could make, then purchased a 300mm lens to allow me to make up to 16x20 on a vertical projection.
So had I done the calculations prior, could have saved my purchase of a lens I did not use.
Hope the context helps in understanding the requirements,
Len
So Len, now when you get a 240 Rodagon and you'll be able to make somewhat larger enlargements. Never figured out why people need a 300 for vertical projection when they can't get the magnification ratio out of their enlarger. From the Rodenstock literature below, you can use a 240 for 8x10 negative enlargements from around 4x down to about a 1.5x enlargement. From IC's chart, you can see what size print a 240 and a 300 lens will make.
I too am height constrained in my darkroom, so on my DeVere 5108, I can get roughly 67" from the negative plane to the baseboard with both at their maximum extensions. Since the ceiling is 92", I cannot raise the head the final 7-9" on the column. If I could, I'd get approximately 76" for my "B" distance with the head at its full 96" extension. This would allow a 40 x 50 print with a 240, or a 30 x 40 with a 300 mm lens. A 4x enlargement is within the recommended range for a 240 Rodagon. So a 300 mm lens at least in my situation is sort of useless because it gives me no more magnification than a 240 even if I could raise the head all the way to the top. I don't have my L-184 set up, so I can't check actual measurements that one, but the Durst literature shows the highest position of the negative carrier at 82" with the head all the way up at 108". Take off say 12" for the distance of the baseboard to the floor and you're back around 70", or a little less than the DeVere. So you might get a 4x enlargement out of the L-184 with the 240, but not the 300 since that requires 85".
Durst Literature for the L-1840 lists a maximum magnification with a 300 mm lens as 3.5 and 5x with a 240 lens. So there you have it. IC can verify the L-1840 dimensions on this enlarger if needed. L
RodenstockProcessp23.pdf
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http://www.nmr-relax.com/api/4.1/lib.geometry.vectors-module.html | 1,606,622,120,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141196324.38/warc/CC-MAIN-20201129034021-20201129064021-00392.warc.gz | 137,612,163 | 4,596 | [frames] | no frames]
Module vectors
source code
Collection of functions for vector operations.
Functions
float
complex_inner_product(v1=None, v2_conj=None) Calculate the inner product for the two complex vectors v1 and v2. source code
random_unit_vector(vector) Generate a random rotation axis. source code
numpy float64 array
unit_vector_from_2point(point1, point2) Generate the unit vector connecting point 1 to point 2. source code
float
vector_angle_acos(vector1, vector2) Calculate the angle between two N-dimensional vectors using the acos formula. source code
float
vector_angle_atan2(vector1, vector2) Calculate the angle between two N-dimensional vectors using the atan2 formula. source code
float
vector_angle_complex_conjugate(v1=None, v2=None, v1_conj=None, v2_conj=None) Calculate the inter-vector angle between two complex vectors using the arccos formula. source code
float
vector_angle_normal(vector1, vector2, normal) Calculate the directional angle between two N-dimensional vectors. source code
Variables
__package__ = `'lib.geometry'`
Imports: acos, atan2, cos, pi, sin, array, cross, dot, float64, sqrt, norm, uniform
Function Details
complex_inner_product(v1=None, v2_conj=None)
source code
Calculate the inner product <v1|v2> for the two complex vectors v1 and v2.
This is calculated as:
``` ___
<v1|v2> = > v1i . v2i* ,
/__
i
```
where * is the complex conjugate.
Parameters:
• `v1` (numpy rank-1 complex array) - The first vector.
• `v2_conj` (numpy rank-1 complex array) - The conjugate of the second vector. This is already in conjugate form to allow for non-standard definitions of the conjugate (for example Sm* = (-1)^m S-m).
Returns: float
The value of the inner product <v1|v2>.
random_unit_vector(vector)
source code
Generate a random rotation axis.
Uniform point sampling on a unit sphere is used to generate a random axis orientation.
Parameters:
• `vector` (numpy 3D, rank-1 array) - The 3D rotation axis.
unit_vector_from_2point(point1, point2)
source code
Generate the unit vector connecting point 1 to point 2.
Parameters:
• `point1` (list of float or numpy array) - The first point.
• `point2` (list of float or numpy array) - The second point.
Returns: numpy float64 array
The unit vector.
vector_angle_acos(vector1, vector2)
source code
Calculate the angle between two N-dimensional vectors using the acos formula.
The formula is:
``` angle = acos(dot(a / norm(a), b / norm(b))).
```
Parameters:
• `vector1` (numpy rank-1 array) - The first vector.
• `vector2` (numpy rank-1 array) - The second vector.
Returns: float
The angle between 0 and pi.
vector_angle_atan2(vector1, vector2)
source code
Calculate the angle between two N-dimensional vectors using the atan2 formula.
The formula is:
``` angle = atan2(norm(cross(a, b)), dot(a, b)).
```
This is more numerically stable for angles close to 0 or pi than the acos() formula.
Parameters:
• `vector1` (numpy rank-1 array) - The first vector.
• `vector2` (numpy rank-1 array) - The second vector.
Returns: float
The angle between 0 and pi.
vector_angle_complex_conjugate(v1=None, v2=None, v1_conj=None, v2_conj=None)
source code
Calculate the inter-vector angle between two complex vectors using the arccos formula.
The formula is:
``` theta = arccos(Re(<v1|v2>) / (|v1|.|v2|)) ,
```
where:
``` ___
\
<v1|v2> = > v1i . v2i* ,
/__
i
```
and:
``` |v1| = Re(<v1|v1>) .
```
Parameters:
• `v1` (numpy rank-1 complex array) - The first vector.
• `v2` (numpy rank-1 complex array) - The second vector.
• `v1_conj` (numpy rank-1 complex array) - The conjugate of the first vector. This is already in conjugate form to allow for non-standard definitions of the conjugate (for example Sm* = (-1)^m S-m).
• `v2_conj` (numpy rank-1 complex array) - The conjugate of the second vector. This is already in conjugate form to allow for non-standard definitions of the conjugate (for example Sm* = (-1)^m S-m).
Returns: float
The angle between 0 and pi.
vector_angle_normal(vector1, vector2, normal)
source code
Calculate the directional angle between two N-dimensional vectors.
Parameters:
• `vector1` (numpy rank-1 array) - The first vector.
• `vector2` (numpy rank-1 array) - The second vector.
• `normal` (numpy rank-1 array) - The vector defining the plane, to determine the sign.
Returns: float
The angle between -pi and pi.
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### onclick method on a colorbar matplotlib python
I have a colorbar and a graph. I was wondering if you can use the onclick method on a colorbar that will then do something on the graph. So click on a particular color portion of the colorer then ... | 3,074 | 12,709 | {"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} | 2.546875 | 3 | CC-MAIN-2015-14 | latest | en | 0.836159 |
http://math.stackexchange.com/questions/381423/how-to-find-normal-plane-at-particular-point | 1,469,716,550,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257828283.6/warc/CC-MAIN-20160723071028-00229-ip-10-185-27-174.ec2.internal.warc.gz | 151,984,330 | 17,231 | # How to find normal plane at particular point?
Let $r(t) = (t^2, t, t^4)$. Find normal plane at point t = 1.
While working on this problem I have come to point where I am not sure what to do. I originally thought to use the formula $T=\frac{r'(t)}{|r(t)|}$
The result is the vector $<\frac{2t}{\sqrt{3}}, \frac{1}{\sqrt{3}}. \frac{4t^3}{\sqrt{3}}>$
Afterwards I believe I should plug the vector and points at t=1 into the equation for a plane, is this the correct path to the solution or have a gone astray? Thanks in advanced.
(This is a study exam, not homework)
-
A normal plane at a point on a curve: (1) passes through the point, and (2) has its normal vector parallel to the tangent vector to the curve. You found (2), now incorporate (1) and you're done. – vadim123 May 4 '13 at 18:10
Would I just set t=1 and solve the vector for the point? Then plug into the formula for a plane? @vadim123 – jrquick May 4 '13 at 18:19
You want the plane to be normal to $\langle \frac{2}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{4}{\sqrt{3}}\rangle$ (or just $\langle 2,1,4\rangle$). You also want the plane to pass through $(1,1,1)$. The equation is hence $(2,1,4)\cdot((x,y,z)-(1,1,1))=0$, or $2(x-1)+(y-1)+4(z-1)=0$, which can be rearranged to $2x+y+4z=7$. | 419 | 1,254 | {"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} | 4.09375 | 4 | CC-MAIN-2016-30 | latest | en | 0.911249 |
http://bootmath.com/find-expected-number-of-successful-trail-in-n-times.html | 1,529,488,461,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863516.21/warc/CC-MAIN-20180620085406-20180620105406-00452.warc.gz | 44,743,689 | 6,174 | # Find expected number of successful trail in $N$ times
I have $k_1$ red balls and $k_2$ blue balls in a box. Let $p_r$ be probability that randomly select $r$ balls in the box without replacement, $\sum_{r=1}^{m} p_r=1$. A trial is succeeds if the trial can draw at least one red ball in the trial. Find expected number of successful trail in $N$ times.
This is my solution
Let $X$ be event that draw $r$ ball in the box which has at least one red ball.
$$P(X)=p(r) \times\sum_{i=0}^{k_1}\frac{\binom{k_1}{i}\binom{k_2}{r-i}}{\binom{k_1+k_2}{n}}$$
Finally, we got the expected value as
$$E(X)= N \times \sum_{r=1}^{m} p(r) \sum_{i=0}^{k_1}\frac{\binom{k_1}{i}\binom{k_2}{r-i}}{\binom{k_1+k_2}{n}}$$
Is it right?
#### Solutions Collecting From Web of "Find expected number of successful trail in $N$ times"
Let $Y$ be the random variable counting how many successes you have in $N$ trials. If $X_i$ is the Bernoulli random variable giving $1$ if the $i$th draw is a successs and $0$ otherwise, then $Y = \sum\limits_{i=1}^N X_i$.
Thus, $Y$ is a sum of presumably-independent Bernoulli random variables, hence a binomial random variable with probability of success $p = \mathbb{P}(X_i = 1)$.
For example, the mass function is
$$\mathbb{P}(Y = s) = \mathbb{P}\left(\sum\limits_{i=1}^N X_i = s\right) = {{N}\choose{s}} \mathbb{P}(X_i = 1)^s\, \mathbb{P}(X_i = 0)^{N-s} = {{N} \choose{s}} p^s \,(1-p)^{N-s}$$
for $0 \leq s \leq N.$
At this point, you probably know that the expected value of a Binomial random variable with parameters $N$ and $p$ is $\mathbb{E}[Y] = Np$, or you can go through the calculation yourself easily since the expected value is linear to find
$$\mathbb{E}[Y] = \sum\limits_{i=1}^N \mathbb{E}[X_i] = \sum\limits_{i=1}^N \Big[ 0 \cdot \mathbb{P}(X_i = 0) + 1 \cdot \mathbb{P}(X_i =1) \Big] = N \, \mathbb{P}(X_i = 1)$$
So, what you have left is to find $p = \mathbb{P}(X_i = 1)$.
For this, it is helpful to consider the following:
$$\mathbb{P}(X_i = 1) = \sum\limits_{r = 0}^m \mathbb{P}(X_i = 1 \mid r \text{ draws})\,p_r$$
with
\begin{align*}
\mathbb{P}(X_i =1 \mid r \text{ draws}) =
\begin{cases}
1 & r > k_2 \\
1 – \frac{{k_2} \choose {r}}{{k_1 + k_2} \choose {r}} & r \leq k_2
\end{cases}
\end{align*}
where the last case with $r \leq k_2$ used: $\mathbb{P}(X_i = 1 \mid r) = 1 – \mathbb{P}(X_i = 0 \mid r)$. From here, you can put these pieces together. | 903 | 2,391 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2018-26 | latest | en | 0.770303 |
https://neighborshateus.com/how-do-you-calculate-time-value-of-money/ | 1,685,950,752,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224651325.38/warc/CC-MAIN-20230605053432-20230605083432-00224.warc.gz | 490,345,722 | 11,607 | # How do you calculate time value of money?
## How do you calculate time value of money?
Time Value of Money Formula
1. FV = the future value of money.
2. PV = the present value.
3. i = the interest rate or other return that can be earned on the money.
4. t = the number of years to take into consideration.
5. n = the number of compounding periods of interest per year.
## How is Fvifa calculated?
PVIFA = (1 – (1 + r)^-n) / r….
1. Overview.
2. Present Value Annuity.
3. Future Value Annuity.
4. Calculating Present and Future Value Annuities.
5. Annuity Table.
6. Present Value Interest Factor of an Annuity.
7. How Good a Deal is an Indexed Annuity?
What is a time value of money table?
The table is used in much the same way as the previously discussed time value of money tables. To find the present value of a future amount, locate the appropriate number of years and the appropriate interest rate, take the resulting factor and multiply it times the future value. An example illustrates the process.
### How do I calculate future value of money?
The future value formula is FV=PV(1+i)n, where the present value PV increases for each period into the future by a factor of 1 + i. The future value calculator uses multiple variables in the FV calculation: The present value sum. Number of time periods, typically years.
### What provides money its time value?
Opportunity cost is key to the concept of the time value of money. Money can grow only if it is invested over time and earns a positive return. Money that is not invested loses value over time.
How do I get a PVIF?
Example of the PVIF Using the formula for calculating the PVIF, the calculation would be \$10,000 / (1 + . 05) ^ 5. The resulting PVIF figure from the calculation is \$7,835.26. The present value of the future sum is then determined by subtracting the PVIF figure from the total future sum to be received.
#### What is time value of money with example?
Time Value of Money Examples If you invest \$100 (the present value) for 1 year at a 5% interest rate (the discount rate), then at the end of the year, you would have \$105 (the future value). So, according to this example, \$100 today is worth \$105 a year from today.
#### What is Future Value example?
Future value is what a sum of money invested today will become over time, at a rate of interest. For example, if you invest \$1,000 in a savings account today at a 2% annual interest rate, it will be worth \$1,020 at the end of one year. Therefore, its future value is \$1,020.
Is money worth more now or later?
In most cases, a dollar received today is actually worth more than a dollar received in the future. But why is that the case? The main principal is that money received today can be invested to earn income sooner than money received in the future.
## Which is true about the time value of money?
The time value of money is a basic financial concept that holds that money in the present is worth more than the same sum of money to be received in the future.
## When do you need a money value chart?
This is also helpful when it comes to using a money value chart when we want to go abroad. It is because each country has its own currency. So, we need to convert our money currency that can be used in our department of the country.
How to calculate the future value of money?
A specific formula can be used for calculating the future value of money so that it can be compared to the present value: Using the formula above, let’s look at an example where you have \$5,000 and can expect to earn 5% interest on that sum each year for the next two years.
### How to calculate the PV and FV of money?
You simply divide the future value rather than multiplying the present value. This can be helpful in considering two varying present and future amounts. In our original example, we considered the options of someone paying your \$1,000 today versus \$1,100 a year from now. | 907 | 3,944 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2023-23 | latest | en | 0.893175 |
https://numberworld.info/root-of-206720581 | 1,656,795,016,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104204514.62/warc/CC-MAIN-20220702192528-20220702222528-00122.warc.gz | 480,036,986 | 2,992 | Root of 206720581
[Root of two hundred six million seven hundred twenty thousand five hundred eighty-one]
square root
14377.7808
cube root
591.2819
fourth root
119.9074
fifth root
46.0338
In mathematics extracting a root is known as the determination of the unknown "x" in the equation $y=x^n$ The result of the extraction of the root is declared as a mathematical root. In the case of "n = 2", one talks about a square root or sometimes a second root also, another possibility could be that n is 3 then one would consider it a cube root or simply third root. Considering n beeing greater than 3, the root is declared as the fourth root, fifth root and so on.
In maths, the square root of 206720581 is represented as this: $$\sqrt[]{206720581}=14377.78080929$$
On top of this it is possible to write every root down as a power: $$\sqrt[n]{x}=x^\frac{1}{n}$$
The square root of 206720581 is 14377.78080929. The cube root of 206720581 is 591.28188297154. The fourth root of 206720581 is 119.90738429843 and the fifth root is 46.033791186678.
Look Up | 297 | 1,050 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2022-27 | latest | en | 0.903401 |
https://mathzsolution.com/how-to-effectively-study-math-closed/ | 1,675,664,720,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500304.90/warc/CC-MAIN-20230206051215-20230206081215-00847.warc.gz | 400,832,984 | 20,331 | # How to effectively study math? [closed]
Maybe this is too general for here, but I am having a lot of difficulty studying math. Just got out of the military and I guess I am not use to this yet but when I run into a problem I have trouble with and I just can’t get it I get extremely frustrated, I will try and look at the solution but it usually doesn’t help and just makes me even angrier than I was before.
Obviously I just can’t stop working on the math otherwise it would never get done, and I also can’t get stop getting mad, otherwise I would have already. What do I do? I just can’t seem to ever remember anything from math even though I study all the time.
This is a hard question to answer, because the answer would depend a lot on personal aspects of your situation, but there are some general points of advice one can give:
(1) Try different books. It may be that the particular textbook you are using
doesn’t click with you, but for classes like algebra and other pre-calc courses,
and calculus itself, there are hundreds of texts available, and some may fit with you better than others. If you go to your local Borders or Barnes and Nobles, there will be a shelf of math books devoted to these topics, and you could look through some of them and see if they suit you. Also, your college library will have lots of books like this too, which you can browse through.
One thing to remember is that different books might be good at different things: your textbook probably has lots of exercises, and you will be able to find other books with lots of exercises. But perhaps you can find different books which don’t necessarily have as many exercises, but have better explanations. So you can try to combine different books: some you read for their explanations, others you use for their exercises.
If you do use different books besides the one in your class, remember that sometimes the notation may be a little different, although in precalc and calc books, most terminology is standard: sin, cos, tan, conic section, polynomial, etc. will all mean the same thing. (But some books will use notation like $\sin^{-1}$ for inverse trig functions, and others will use arcsin, etc.,
(2) Practice basic algebra: It is very common, when students make mistakes in precalc or calc classes, that the source of the mistake is weakness at algebra. Practicing algebra will help with everything else that you have to do in math. The skills will be directly useful, and lots of other manipulations you will have to do in more advanced classes will also be similar to the skills you build up by practicing algebra.
(3) Practice with numbers: If you don’t think about numbers much in general, you will have trouble with other things in math, because you won’t be able to relate them to concrete things. E.g. when you plot a graph like $y = x^2$, you want to be able to easily realize that a point like $(12,144)$ is on the graph because $12^2 = 144$, while $(11,120)$ is not on the graph (because $11^2 = 121 \neq 120$), but is pretty close to the graph (because $121$ is not all that far
from $120$). Day-to-day life gives chances to practice arithmetic; try to take advantage of them.
(4) Try to learn from your mistakes: One advantage of mathematics is that when you make a mistake, there will be a specific reason as to why; i.e. there will be some particular thing you did wrong. Try to find out what it is in each case, and resolve not to make that mistake again.
One aspect of this is that math should make sense. If it’s not making sense to you, i.e. if you can’t work out specifically what you are doing wrong in a given situation, try asking your professor or tutor again.
One thing that can happen, because of the cumulative nature of mathematics, is that several confusions can become combined in an answer, and then it can be hard to figure out what particular thing went wrong. In situations like this,
do your best to break the computation down into small steps, so that you can identify what you did right or wrong in each step separately.
(5) Write down all your working: Use a lot of paper, write down all your steps, make them clear so that anyone else (or you in a few weeks time!) could go back and read them and understand what is going on. If you lay out the whole chain of your computations clearly, it will be easy to identify weak links later. If you skip steps, the whole thing is more confusing and it’s much harder to learn anything from it later.
(6) Try to recognize when you understand something and when you don’t:
Probably arithmetic makes sense to you. When you learn another piece of math
completely, it should make as much sense as arithmetic does. E.g. a common mistake in algebra is to write $(x+y)^2 = x^2 + y^2$, but to someone who is
good at algebra, this looks just as wrong as $1 + 1 = 3$ does. If it doesn’t look that wrong to you, it means that you have more work to do in building up
I wish I had better advice on how to do this. One thing you could try (say for the wrong example with squares above) is to plug in some random values of $x$ and $y$ on each side and check that for most of them the alleged equation won’t actually be true; this tells you that the equation between $x$ and $y$ is wrong. (If it were right, it would hold for any value of $x$ and $y$ that you plug in.) I don’t know how helpful this will be though. One thing I can say is that | 1,233 | 5,409 | {"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} | 2.953125 | 3 | CC-MAIN-2023-06 | latest | en | 0.973923 |
https://bestnursingassignment.com/questions-on-trig/ | 1,695,660,102,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233509023.57/warc/CC-MAIN-20230925151539-20230925181539-00070.warc.gz | 139,468,614 | 15,937 | Questions on trig
Reduce sin(60° + θ) − sin(60° − θ) to a single function of one angle.
a.
b. c.
d.
What is another identity you would use to convert the RHS: ? a.
b.
c. d.
____ 3.
Given:
Find the values of the following:
a.
b.
c.
d. 3
, angle A is in Quadrant III and angle B is in Quadrant I.
____ 4.
What are the extreme values of y = sin (x + )?
a.
b. 1and–1
and –
c.
d. .75 and –.75
What is the best identity you could use to convert the LHS: ? a.
b. c. d.
Evaluate 2 sin 67.5° cos 67.5°. What is the second step equal to?
a. sin 2(67.5°) b.
c. cos 2 (67.5°) d.
and –
____ 5.
____ 6. | 223 | 612 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2023-40 | longest | en | 0.622747 |
https://www.slideserve.com/reina/scientific-method-quiz-review | 1,534,289,853,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221209650.4/warc/CC-MAIN-20180814225028-20180815005028-00635.warc.gz | 987,334,481 | 17,679 | Scientific Method Quiz Review
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# Scientific Method Quiz Review - PowerPoint PPT Presentation
Scientific Method Quiz Review. Let’s check how you did? Use this PowerPoint as a review !. A:1 B:27 C:19 D:30. Jen enters the 5th grade science fair with a project about candles. She asks "Will the temperature of the air affect how fast a candle burns? That is her Hypothesis.
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### Scientific Method Quiz Review
Let’s check how you did?
Use this PowerPoint as a review!
A:1 B:27 C:19 D:30
• Jen enters the 5th grade science fair with a project about candles. She asks "Will the temperature of the air affect how fast a candle burns?
• That is her Hypothesis.
• That is her Conclusion.
• That is her Problem (a:c, b:b, c:d, d:c)
• That is her Observation.
A:2 B:29 C:19 D:30
In second grade, Jen asked "Which color of candy do kids prefer?
“ What would be an important constant or controlled variable in her experiment?
• Use the same candy that comes in different colors, but the same flavor. (A:b, B:a, C:b, D:a)
• Do the experiment right before lunch so the kids at really hungry.
• Ask each person what is their favorite kind of candy
A:3 B:16 C:2 D:11
Mrs. Apex's 3rd grade is learning multiplication facts. She notices that Jay and Jen are doing really well. She asks them if they are doing extra facts worksheets. Jay replies that he has a fun computer program that does multiplication facts. Jen comes over and plays the game with him. Mrs. Apex is almost sure that the old fashioned way of paper worksheets is best & that Jay & Jen are not the norm. So, Mrs. Apex decides to do an experiment. Mrs. Apex divides her class in half. There are five boys and five girls in each group. Group A practices 5 times tables on paper worksheets for 10 minutes each day. Group B practices 5 times tables on the computer program for 10 minutes each day.At the end of one week all the children take the same quiz. The children in Group A score 5As, 2Bs, 2Cs, 1D. The children in Group B score 7As, 2Bs, 1F. After the fifth week, Mrs. Apex tests the children on their multiplication facts again. Group A (paper) scores 3As, 3Bs, 4Cs. Group B (computers) score 8As, 2Bs. In the computer group - girls and boys did equally well. Her hypothesis was that paper worksheets will be the best way to teach math facts.
A:3 B:16 C:2 D:11
Mrs. Apex might form which hypothesis?
• If students practice multiplication facts on a paper worksheet, they will improve their math scores. (A;a, B:a, C:a, D:b)
• The students like the computers more than they like her.
• Computers will replace teachers.
• Practicing multiplication facts on a computer might help test scores..
A:4 B:10 C:8 D:7
Mrs. Apex decides to test her hypothesis. How can she do this?
A:5, B:9, C:5, D:3
In Mrs. Apex's experiment, what is the independent variable?
A:6, B:13, C:4, D:4
Mrs. Apex divides her class in half. There are five boys and five girls in each group. Group A practices 5 times tables on paper worksheets for 10 minutes each day. Group B practices 5 times tables on the computer program for 10 minutes each day. What is the controlled variables?
A:7, B:12, C:1, D:4
The principal, Mrs. Integer tells Mrs. Apex to write down exactly what she is doing step by step. In the scientific method this is called ...
• Conclusion
• Procedure (A:b, B:d, C:d, D:b)
• Hypothesis
• data
A:8, B:14, C:6, D:10
At the end of one week all the children take the same quiz. The children in Group A score 5As, 2Bs, 2Cs, 1D. The children in Group B score 7As, 2Bs, 1F. The quiz scores are an example of ...
• Conclusion
• Hypothesis
• Data (A:c, B:a, C:c, D:c)
• Procedure
A:9, B:15, C:3, D:5
After one week, does Mrs. Apex have enough data to make a conclusion?
A:10, B:11, C:7, D:8
Mrs. Apex decides to ask the school board to buy math facts software. She wants to show her data in her presentation. She should
A:11, B17, C:9, D:6
After the fifth week, Mrs. Apex tests the children on their multiplication facts again. Group A (paper) scores 3As, 3Bs, 4Cs. Group B (computers) score 8As, 2Bs. In the computer group - girls and boys did equally well. Her hypothesis was that paper worksheets will be the best way to teach math facts.
A:12, B:21, C:17, D:14
Jay digs a large garden for his mother. The site she chose is near some trees. About half the garden is shaded all afternoon. The other half gets sun all day long. In August, Jay notices that the tomato plants on the sunny side of the garden are 2x taller than the shaded ones. The sunny side plants have 20 tomatoes on each plant. The plants on the other side are short and have only 10 tomatoes on them. He tells his mother what he has noticed. This is an example of ...
• Hypothesis
• Observation (A:b, B:a, C:c, D:a)
• Conclusion
• Data collection
A:13, B:28, C:24, D:21
You create a science fair project. It takes 6 weeks to collect all the data. In looking at the data, you find your hypothesis was wrong. You should:
A:14, B:26, C:21, D:20
The dependent variable
Mrs. Membrane decided to make scrambled eggs. When she tried to break an egg, she found it was hard cooked. Jen said she could tell the difference without breaking the shell. Jen spun the first egg on the counter. It turned fast like a top. She touched it to stop it. It stopped. She spun another one. It also stopped when touched. The third egg did the same. The fourth egg turned. When she stopped it, it rotated slowly when she let go. Jen said it was raw.
A:15, B:19, C:25, D:15
A:16, B:23, C:11, D:24
To test your hypothesis about the eggs, you decide to conduct an experiment. You buy a dozen white chicken eggs. The variable you will change will be
A:17, B:20, C:22, D:29
The controls in the egg experiment should be
A:18, B:25, C:18, D:26
The variable that you change in the experiment.
• Control
• Independent (A:b, B:d, C:c, D:d)
• Dependent
• Responding
A:19, B:7, C:10, D:27
The variable that changes because you changed the other variable. (This is the data you collect.)
• Independent
• Dependent (A:b, B:c, C:d, D:a)
• Controlled
• Manipulated
A:20, B:24, C:16, D:23
The variables that do not change in an experiment are called
• Independent
• Dependent
• Controlled (A;c, B:c, C:a, D:c)
• Stubborn
A:21, B:3, C:14, D:17
An experiment is being done to determine how light affects the growth small zinnia plants. One plant is placed in full sunlight, the second in a shaded area, the third in an unlit closet.
What is the independent variable in the experiment?
• Type of plant
• Amount of light (A:b, B:b, C:c, D;a)
• Type of soil
• Growth of the plant
A:22, B:4, C:12, D:18
The hypothesis of this experiment could be:
A:23, B:5, C:13, D:19
The constants in the plant experiment are: Choose all that apply
• Amount of sunlight
• Type of pot
• Amount of water
• Type of plant
A:24, B:22, C:26, D:13
Good experiments have more than one variable.
• True
• False (A:b, B:b, C:a, D:a)
A:25, B:30, C:27, D:16
In an experiment, there should be only one _________. Everything else should be a ________.
• Control – independent variables
• Dependent variable-independent variable
• Independent variable – control (A:c, B:d, C:c, D:d)
• Trial – proven
A group of students is doing an experiment where they roll a ball down a ramp and then measure how far across the floor it rolls. The factors that are the same each time are:
• the ball
• the ramp, including its length and angle to the surface
• where on the ramp the ball starts from
The students do change the type of surface the ball rolls on. The data theycollected is shown below.
A:26, B:18, C:20, D:28
What is the independent variable?
• Distance
• Type of surface (A:b, B:c, C:d, D:c)
• Ramp
• Ball
A:27, B:6, C:15, D:24
What is the dependent variable?
• Distance (A:a, B:a, C;a, D:b)
• Type of surface
• Ramp
• Ball
A:28, B:8, C;23, D:12
What are the controlled variables?
• Ball
• Ramp
• Distance
• Type of surface
Graph
Marie and Seth are working on a lab activity where they are tracking some leaves that are floating past them in a rain runoff channel near their home. They have to measure how far the leaves travel in different amounts of time. They place a long tape measure along side of the runoff channel to measure distance. A borrowed stopwatch will let them record the amount of time. They plan to have time intervals that begin at two seconds, and increase by two’s until they get to 20 seconds. The data they recorded is below:
Distance Leaves Float over time
Distance in cm
Time in Seconds
Graph
A group of students is doing an experiment where they roll a ball down a ramp and then measure how far across the floor it rolls. The factors that are the same each time are:
• the ball
• the ramp, including its length and angle to the surface
• where on the ramp the ball starts from
The students do change the type of surface the ball rolls on. The data theycollected is shown below.
Carpet Rolling
Distance Ball Rolls in Cm
Type of Surface | 2,620 | 9,633 | {"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} | 3.046875 | 3 | CC-MAIN-2018-34 | latest | en | 0.902616 |
https://8266235.com/qa/how-do-you-add-up-discounts.html | 1,611,117,233,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519883.54/warc/CC-MAIN-20210120023125-20210120053125-00391.warc.gz | 206,163,990 | 7,803 | # How Do You Add Up Discounts?
## How do you calculate original price after discount?
Finding the original price given the sale price and percent…First consider the unknown original price as ‘x’.Then consider the rate of discount.To find the actual discount, multiply the discount rate by the original amount ‘x’.To find the sale price, subtract the actual discount from the original amount ‘x’ and equate this to given sale price.More items….
## How do you figure out 25% off?
Percent Off Price FormulaConvert 25% to a decimal by dividing by 100: 25/100 = 0.25.Multiply list price by decimal percent: 130*0.25 = 32.50.Subtract discount amount from list price: 130 – 32.50 = 97.50.With the formula: 130 – (130*(25/100)) = 130 – (130*0.25) = 130 – 32.50 = 97.50.25% off \$130 is \$97.50.
## What is \$20 with 20% off?
Percent Off Table For 20.001 percent off 20.00 is 19.80The difference is 0.2020 percent off 20.00 is 16.00The difference is 4.0021 percent off 20.00 is 15.80The difference is 4.2022 percent off 20.00 is 15.60The difference is 4.4023 percent off 20.00 is 15.40The difference is 4.6095 more rows
## What is the formula of selling price?
selling price = (100 + profit%)cost price/100; [Here, cost price and profit% are known.] 1.
## What is the formula of time?
time = distance ÷ speed.
## How do I add 50% to a price?
Simply take the sales price minus the unit cost, and divide that number by the unit cost. Then, multiply by 100 to determine the markup percentage. For example, if your product costs \$50 to make and the selling price is \$75, then the markup percentage would be 50%: ( \$75 – \$50) / \$50 = . 50 x 100 = 50%.
## How do you add 10% to a price?
To increase a number by a percentage amount, multiply the original amount by 1+ the percent of increase. In the example shown, Product A is getting a 10 percent increase. So you first add 1 to the 10 percent, which gives you 110 percent. You then multiply the original price of 100 by 110 percent.
## How do you find 20% of a number?
If you know what the whole number is and you know what percent of that number you are looking for, you multiply. For example, if you are looking for 20% of 100, you multiply 100 by 0.2. If you want to find what percent of 100 is equal to 20, you would divide 100 by 20.
## How much is 25% off?
Percent Off Table For 25.001 percent off 25.00 is 24.75The difference is 0.2525 percent off 25.00 is 18.75The difference is 6.2526 percent off 25.00 is 18.50The difference is 6.5027 percent off 25.00 is 18.25The difference is 6.7528 percent off 25.00 is 18.00The difference is 7.0095 more rows
## How do you calculate additional discount?
For example, if the original price was \$50 and we have two discounts: 20% and 10% , then we’re doing something like this: \$50 – 20% = \$50 – \$10 = \$40 . Then \$40 – 10% = \$40 – \$4 = \$36 .
## What is the formula of amount?
Use this simple interest calculator to find A, the Final Investment Value, using the simple interest formula: A = P(1 + rt) where P is the Principal amount of money to be invested at an Interest Rate R% per period for t Number of Time Periods. Where r is in decimal form; r=R/100; r and t are in the same units of time.
## What is \$20 with 10% off?
You will pay \$18 for a item with original price of \$20 when discounted 10%. In this example, if you buy an item at \$20 with 10% discount, you will pay 20 – 2 = 18 dollars.
## How do you apply a discount to a price?
Divide the original price by 5. Alternatively, divide the original price by 100 and multiply it by 20. Subtract this new number from the original one. The number you calculated is the discounted value.
## How do you take 20% off a price?
First, convert the percentage discount to a decimal. A 20 percent discount is 0.20 in decimal format. Secondly, multiply the decimal discount by the price of the item to determine the savings in dollars. For example, if the original price of the item equals \$24, you would multiply 0.2 by \$24 to get \$4.80.
## How do I calculate a discount?
Procedure:The rate is usually given as a percent.To find the discount, multiply the rate by the original price.To find the sale price, subtract the discount from original price.
## How do you take 10% off a price?
One of the easiest ways to determine a 10 percent discount is to divide the total sale price by 10 and then subtract that from the price. You can calculate this discount in your head.
## What is 10% out of 500?
Percentage Calculator: What is 10 percent of 500.? = 50.
## How do I calculate total payable amount?
To find the total amount paid at the end of the number of years you pay back your loan for, you will have to multiply the principal amount borrowed with 1 plus the interest rate. Then, raise that sum to the power of the number of years. The equation looks like this: F = P(1 + i)^N. | 1,318 | 4,867 | {"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-2021-04 | latest | en | 0.868858 |
https://medical-dictionary.thefreedictionary.com/stochastically | 1,558,703,246,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257624.9/warc/CC-MAIN-20190524124534-20190524150534-00262.warc.gz | 551,420,126 | 13,006 | # stochastic
(redirected from stochastically)
Also found in: Dictionary, Thesaurus, Financial, Encyclopedia, Wikipedia.
Related to stochastically: Stochastically independent
## stochastic
[sto-kas´tik]
pertaining to a random process; used particularly to refer to a time series of random variables.
(stō-kas'tik),
Random.
## stochastic
/sto·chas·tic/ (sto-kas´tik) pertaining to a random process, particularly a time series of random variables.
## stochastic
adjective Referring to a random variable; involving chance or probability.
## stochastic
adjective Referring to a random process; a process determined by a random distribution of probabilities; referring to a behavior not governed by known equations and initial conditions, thus unpredictable at any past or future time. See Chaos. Cf Deterministic.
## sto·chas·tic
(stō-kas'tik)
1. Random.
2. radiation therapy Pertaining to the effects of radiation seen in the person exposed to such radiation. This does not have a dose threshold, given that as the dosage increases so does the severity of the reaction.
[G. stochos, target, guess]
## stochastic
arrived at by skilful conjecturing; allowing the opportunity for variation.
References in periodicals archive ?
Therefore, the upper-bounding observer problem can be stated as follows: design a positive observer in the form of (24) such that system (28) is positive and stochastically stable and satisfies the performance [mathematical expression not reproducible] under zero initial conditions.
Whitney (1947) On a test of whether one of two random variables is stochastically larger than the other, The annals of mathematical statistics, vol.
The distribution of contributions of the competitive group first order stochastically dominates that of the competitive group for the vast majority of the distribution.
xy] it's possible as in the case of Greece to test (5)-(7) whether the features observed are stochastically linearly independent.
The existing methods of stochastically informed expertise (Rutkauskas 2012b), the techniques of stochastic optimization as well as the methods of stochastic evaluation of losses incurred by the environmental or social components, allow expecting for the success.
16) review the solution to the optimal consumption and saving problem in the basic framework with income fluctuating stochastically but without retirement.
Generally, both cases are included in the following stochastically differential form:
A random variable Y is said to be stochastically greater than another, X, if Pr(Y [greater than or equal to] a) > Pr(X [greater than or equal to] a) for any a.
Experimental animal grouping: The 109 Wistar rats were stochastically divided into the following groups: normal group (7 rats); an ulcerative colitis model group (21 rats with 5 deaths); a physiologic saline/negative control group (21 rats, 10 deaths); the Xipayi KuiJiean large dose intervention group (19 rats, 7 deaths); the medium dose intervention group (20 rats, 5 deaths); and the small dose intervention group (21 rats, 5 deaths).
Each individual's fit in a particular population is evaluated; multiple individuals are stochastically selected (based on their fit) and modified (recombined and randomly mutated) to form a new population.
Rothschild and Stiglitz showed that the mean-preserving spread of a distribution is second-order stochastically dominated by the (original) distribution.
In general, for a deteriorating system, it is reasonable to assume that the successive working times are stochastically decreasing while the consecutive repair times after failures are stochastically increasing, due to the ageing and accumulated wearing many systems.
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# August 15 Geometry Regents Answers
## The Ultimate Guide To Passing The Geometry Regents Exam
NYS Geometry [Common Core] August 2015 Regents Exam || Part 1 #s 1-12 ANSWERS
Posted on 14-Jan-2022
The Geometry Regents Exam measures a student’s understanding of the Common Core Learning Standards for Geometry. The exam requires that students show an understanding of mathematical concepts, use prior knowledge and prerequisite skills, and solve real world problems using tools and…
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The diagonals intersect in the middle. Each diagonal divides the parallelogram into two equal triangles. How are the multiple choice questions in geometry? Each multiple choice question has four possible answers marked 14 , of which you choose one and then write it on a separate answer sheet. For questions with an answer constructed according to the official instructions from Geometry Regents, you must do the following to obtain a full score:. What do you need to know about the geometry regents exam january How many questions do you need to be successful in Regents of Geometry? As of January , students must complete 30 credits to achieve 65 credits. For example, if you answer 15 multiple choice questions correctly 2 points each , you get a pass. Click here for a full answer. How many credits do you need to pass the Geometry regents?
## Pdf Geometry January 2012 Regents Answers With Work
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Geometry Regents January 2012 solutions.mov PDF Geometry Regents January 2012 Answers Explained January 2012 Geometry Regents Work Shown Geometry – January ’12 Part I Answer all 28 questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed.
Also Check: Algebra Nation Test Yourself Answers Section 8
## What’s The Best Way To Practice Basic Geometry
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## What Should I Do To Prepare For The Geometry Exam 2
Research and familiarize yourself with the types of geometry exams that exams may include, then choose related books and exercises. Make sure you get enough sleep the night before your exam. When you’re tired, your brain probably won’t work as fast, and speed is of the essence when it comes to scheduled tests.
Read Also: Algebra With Pizzazz Books Never Written
## Geometry Regents August 2022 Answer Key Explanations
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• ## Pdf Geometry Regents August 2013 Answers Explained
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## Algebra 1 January 2019 Regents Answers : Algebra
Posted on 29-Jan-2022
2022 regents examination schedules · education requirements and diploma. New york algebra algebra 1 trigonometry trig key solutions answer key geometry . It covers answers for june 2018, 2019, august 2018, 2019, and january 2019, 2020 that are included in the book. Please click on the links…
## New York State Regents Exams: What Are They When Do You Take Them
NYS Geometry [Common Core] August 2019 Regents Exam Questions 1-6 Solutions
Students may wish to take the Regents exam if they are not on track toward passing all of the necessary courses by June. Waivers are available for all 10 Regents exam subject areas. Students should contact their guidance counselor or a school administrator for more information on whether they are eligible to receive a Regents waiver. When are the Regents exams? No State examinations will be administered on Monday, June 20 to allow for the weekday observance of the Juneteenth holiday. The content will vary by exam subject. All exams have multiple choice and open-ended questions.
## Less Than 65% To Get A 65
For the majority of the Geometry Regents exams administered so far, a student only needed to have a raw score of 33 or 34 to get a 65. Since the maximum raw score is 86, a student only needed to get 40% of the points to get a 65. If we only examine the multiple choice questions and ignore the other three parts, we see that a student needs to answer 71% of the multiple choice questions correctly to get a 65 for the entire test.
## Unit Exam In Global History And Geography 1 Answer Key
You must pass at least four Regents exams earnings of 65 or more to earn this Regents degree: Regents in Mathematics Algebra 1, Algebra 2, or Geometry. When do the Regents exams start in ? The exams will take place on the following dates: January 1, , the Regents exam period from Tuesday January 25 to Friday January 28, June 2, and the Regents exam period from Wednesday June 15 to Friday June You can view your Regents test results on high school transcripts and in your New York City school records. Each Regents exam lasts three hours in one day break, six hours of Regents English exam in two must-see days. Regents exams include multiple choice and short answer or essay questions including paper-based questions , depending on the subject.
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## Geometry Ccss Regents Exam 0815 Page 1
• Views:
Transcription
1 Geometry CCSS Regents Exam 0815 Page 1 1 A parallelogram must be a rectangle when its 1) diagonals are perpendicular 3) opposite sides are parallel 2) diagonals are congruent 4) opposite sides are congruent 2 If is the image of, under which transformation will the triangles not be congruent? 1) reflection over the x-axis 3) dilation centered at the origin with scale factor 2 2) translation to the left 5 and down 4 4) rotation of 270 counterclockwise about the origin 3 If the rectangle below is continuously rotated about side w, which solid figure is formed? 1) pyramid 3) cone 2) rectangular prism 4) cylinder 4 Which expression is always equivalent to when? 5 In the diagram below, a square is graphed in the coordinate plane. A reflection over which line does not carry the square onto itself?
2 Geometry CCSS Regents Exam 0815 Page 2 6 The image of after a dilation of scale factor k centered at point A is, as shown in the diagram below. Which statement is always true? 7 A sequence of transformations maps rectangle ABCD onto rectangle A”B”C”D”, as shown in the diagram below. Which sequence of transformations maps ABCD onto A’B’C’D’ and then maps A’B’C’D’ onto A”B”C”D”? 1) a reflection followed by a rotation 3) a translation followed by a rotation 2) a reflection followed by a translation 4) a translation followed by a reflection
## When Do You Typically Take Geometry Regents
Students typically take the Geometry Regents exam as part of a three year math curriculum designed by the New York Department of Education. They recommend that students take the courses in the following order: Algebra 1, Geometry, Algebra 2. Students who take the Geometry Regents exam will take it in at the end of the semester in which they took the Geometry course.
Typically, students take Algebra 1 first in the 9th or 10th grade. If they pass Algebra 1, they can move on to Geometry in the 10th or 11 grade. Finally, the student can round out their math curriculum with Algebra 2, usually in the 11th or 12th grade.
The above timeline for this progression puts students in the 10th or 11th grade year when taking the Geometry Regents exam. However, this timeline is meant to serve as a guide. The courses can be taken earlier or later, depending on a variety of factors such as schedule, aptitude, and overall academic goals.
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## Ways To Pass The Geometry Regents
Posted on 22-Jan-2022
Over 125 multiple choice and short answer questions covering the easiest and most commonly asked concepts on the Geometry Regents! Three Days before the August regents I decided to start studying when my friend sent me the link to the introductory video. I was convinced the Video Tutor…
## School And District Orders
How are points scored on the Geometry regents? How the geometry rules are valued. When is the Geometry regents exam for ? In this comprehensive Geometry Regents review guide, youll learn everything you need to know about the exam format, what youll be tested for, and what the questions look like. Also give them the best tips to get there. Basically this question requires you to do two things: you know they need to find the volume because that gives them the weight of the golf ball per cubic inch. How many Regents exams do you need to pass? If you do not have an IEP or are not studying English, you must meet the Regents degree requirements to graduate from high school.
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## Where Can I Find A Geometry Reference Sheet
Finally, you will be given a high school math reference chart with basic formulas and conversions that you may find useful during the exam. You will find this reference sheet at the back of your test book. You can tear it off along the perforated edges and use it anytime during the Geometry Regents exam.
## What Happens If You Fail A Geometry Regents
New York Regents Geometry PRACTICE PROBLEM (Increase Your Score)
A passing score of 65 or higher is required if you want to use the Geometry exam to satisfy the math requirement for graduation with a Regents Diploma. Students who fail the Geometry Regents exam have the option of taking it again as many times as necessary to receive a passing grade. In addition, even if you passed the exam you can re-take it to attempt to receive a higher score if you wish. Each Regents exam is offered three times per year, in January, June and August, giving you up to three opportunities to take the exam each school year.
Because the Regents diploma only requires successful completion of one Regents math exam, if you fail the Geometry exam, you could substitute a passing score on either the Algebra 1 or Algebra 2 exam instead.
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## Which Is The Best Book To Prepare For The Aleks Math Test
If you are just starting to prepare for the ALEKS Math Test and need the perfect ALEKS Math Prep book, ALEKS Math for Beginners: The Ultimate Step by Step Guide to Prepare for the ALEKS Math Test is the perfect prep book and complete that you can use. can handle anything. ALEKS math subjects are graded from scratch.
## Geometry Regents Faq: Everything You Need To Know
• The Albert Team
If you have a Geometry Regents exam coming up, then youve probably got questions. In this post, well go over frequently asked questions related to this Regents test to help you in your test prep.
To see what you already know, take our quick five-question quiz:
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## Geometry Regents Faq: Everything You Need To
Posted on 1-Jan-2022
Is the Geometry Regents exam hard? Are Geometry Regents exams timed? How long are they? If your goal is to graduate with honors, you need to correctly answer approximately 88% of the In the August exam schedule, the Geometry exam has been given on the second day for the last four…
## A Complimentary Music Theory Overview For The Guitarist
The C Major Scale and A Natural Minor Scale are from the only Keys with no Flats or Sharps. Every other key needs at least one flat or sharp. Chords/Arpeggios: The Diatonic scale is Harmo-nized into Triad chords & Seventh Chords by taking every other note in the scale and playing them all at once.
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## What Should I Do To Prepare For The Geometry Exam Answer
Practice the geometry quiz. Put your skills to the test with this practical plane geometry exam. Whether you’re preparing for a high school exam or just want to test your geometry skills, this test will help you assess your knowledge. Select Quiz Mode to view responses after your quiz has been graded. Select a learning mode to see your answers on the go.
## Geometry Regents Practice Test Online
Posted on 9-Jan-2022
Geometry Regents Lessons & Practice Tests Our Geometry Regents Prep includes interactive lessons of the topics that appear most often on the test, followed by practice problems to test your knowledge of what you learned. Each of our practice problems has its own video explanation, so you…
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## When Is Regents Week What Are The Typical Start Dates
The state of New York administers the Regents exams in January, June, and August of each year. These dates allow for approximately 150 instructional days prior to the exam.
The January exam period is four days, the June exam period is nine days , and the August exam period is two days.
The 2020 exam schedule is as follows:
• Tuesday, January 21st through Friday, January 24th
• Wednesday, June 17th through Friday, June 25th
• Thursday, August 13th and Friday, August 14th
The January Geometry exam has been scheduled for the second day of testing for the last three years.
The June Geometry exam has not been given on the same day in any of the last five years. The earliest it has been given is on the fourth day of testing in 2017. It has been given on the fifth day of testing twice, in 2016 and 2019, and on the 6th day of testing in 2018. In 2020, the Geometry exam will be given on the seventh day of testing.
In the August exam schedule, the Geometry exam has been given on the second day for the last four years of the exam.
The New York State Department of Education maintains the most up to date Regents testing schedules at their website. The website also includes the test dates for the 2020-2021 school year.
## Very Detailed Explanation Of The Great Reset And The New World Order
NYS Algebra 2 COMMON CORE Regents August 2017 Parts 2 – 4
Posted on 13-Jan-2022
An August 13, 2020, article31 on the Federal Reserve website discusses the supposed benefits of a central bank digital currency . There’s general agreement among experts that most major countries will implement CBDC within the next two to four years. Many uninformed people believe that…
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## Board Of Regents Acts On Emergency Regulation Revisions To 2021 Diploma Requirements
If USDE Waiver is Denied, Only Four June 2021 Regents Exams will be Administered
If USDE Waiver is Denied, Only Session One of the Grades 3-8 ELA & Math Assessments and Only the Written Test Component of the Grades 4 & 8 Science Tests will be Held
The Board of Regents today acted on a series of emergency regulations to allow for exemptions to diploma requirements associated with the June 2021 and August 2021 Regents Examination administrations. With the COVID-19 crisis still affecting the State of New York and students having varied levels of in-person instruction, the Board and the Department are taking necessary steps to provide essential flexibility for the States students, families and educators.
These include actions to cancel the August 2021 Regents Exams and, should the U.S. Department of Education deny the Departments waiver request, only four of the June 2021 Regents Exams will be administered only Session 1 of the Grades 3-8 English Language Arts and Math Tests will be required and only the one-session Written Test component of the Grades 4 and 8 Science Tests will be administered.
## How Are Points Scored On The Geometry Regents
How the geometry rules are valued. The Geometry Regent uses a lookup table to change the raw value to the scaled value. The raw score reaches 086 and the graded score is 0100. The preliminary score is the total number of points received for each question. The 24 multiple choice questions in Part I are scored with 2 points each.
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## Are Geometry Regents Exams Timed How Long Are They
The Geometry Regents test is timed. Students have three hours in which to complete the exam booklet. The test designers acknowledge that many students will complete the exam in less than three hours. However, no student is permitted to leave the testing area until the allotted time has elapsed.
The New York State Department of Education provides a test guide that can provide you with more information on how the Geometry Regents test is administered. | 3,840 | 18,024 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-33 | latest | en | 0.882219 |
https://www.bankersadda.com/reasoning/ | 1,657,159,163,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104683020.92/warc/CC-MAIN-20220707002618-20220707032618-00234.warc.gz | 687,298,271 | 122,810 | Latest Banking jobs » Reasoning Notes, Tricks, Questions and Answers...
# Reasoning Notes, Tricks, Questions and Answers For All Bank Exams
A test of reasoning is the way through which organisations measure the mental and thinking ability of a candidate, thus it is a very important test in almost every competitive exam. All major recruitment exams have a test of reasoning to measure the General Intelligence of a person.
Broadly reasoning is said to be categorised into Inductive Reasoning and Deductive Reasoning. The Inductive reasoning is usually based on patterns and observation of those patterns whereas the deductive reasoning is a part where students must use logic to form a conclusion from the given scenario or set of statements.
Most of the questions asked in banking and insurance sector exams fall in the category of deductive reasoning ability. If you are preparing for IBPS PO, IBPS Clerk, NABARD, RBI, SBI Clerk, SBI PO, IBPS RRB and any other exam you must practice reasoning questions thoroughly as you have to face this section in both prelims and main examination.
## Topics to Cover in Reasoning
If you are preparing for banking and insurance sector examinations, then you must practice the following topics of reasoning ability for prelims and main exam:
• Puzzles: Month based puzzles, day based puzzles, random arrangement puzzles, floor based puzzles, order and ranking based puzzles, box based puzzles
• Seating Arrangement : Circular Table, Linear, Parallel Rows, Square Table Seating Arrangement
• Blood Relation : Blood Relation based puzzle, coded blood relation
• Direction Sense : Direction Sense based puzzle, coded direction sense questions.
• Inequality :Single line inequality, double line inequality, coded inequality
• Coding Decoding: There are different types of coding decoding questions in the reasoning section.
• Alpha-Numeric-Symbol Series: Here a series of alphabets, symbols and numbers is given followed by a set of questions where the candidate has to perform certain operations on the give series and mark the answer accordingly.
• Alphabet Based Questions: Here a certain word(s) is given, and students are asked to find pairs of other meaningful words or perform specified operations.
• Number based questions: To perform simple mathematical operations on given set of numbers.
• Syllogism: Conditional Syllogism, Categorical Syllogism, Disjunctive Syllogism
• Logical Reasoning: This is a whole new branch of reasoning ability test and it tests a student’s higher order thinking capacity. It includes sub-topics like statement and assumptions, cause and effect, strength of an argument, passage inference.
• Other miscellaneous topic of reasoning ability includes Machine Input Output, calendars, condition-based questions, Data Sufficiency which is a mix of all other topic stated above.
Here are the list of all the topics in the bank exams.
How to prepare Reasoning for Bank Exam?
Learn basic method and tricks to solve all different topics of the test of reasoning mentioned above. Anyone can solve the questions of reasoning, but the actual challenge is to solve them in limited amount of time. So, to prepare reasoning ability for competitive exams always follow a timer-based practice approach.
Which is the best book for Reasoning Ability?
Adda247’s ACE REASONING for Bank and Insurance examinations is the best book to study and practice the concepts of reasoning ability from the basic to an advanced level. This is the ultimate study guide for every beginner aiming to get a bank job in 2020. The book covers the theory of all important topic of reasoning along with practice exercises.
Salient Features of the ACE Reasoning Book
• Based on Latest Pattern
• 3 Level of Exercises
• 1500+ Multiple Choice Questions with 100% solutions
• Includes the Previous Year Questions of all the chapters
You can also check a dedicated book to practice puzzles and seating arrangement as this topic covers more than 40% of the weightage of questions asked in the reasoning ability section of bank and insurance exams.
Salient Features of the Puzzle and Seating Arrangement Book:
-2500+ Questions on Puzzles & Seating Arrangement
- New pattern Based Questions of 2017-18 Exams including 10 practice sets
- Expect the Unexpected ones [Surprised Pattern]
-Incorporates more than 10 Types of Puzzles & Sitting Arrangement
-Incorporates the last 5-year Memory Based Questions asked in SBI, IBPS, RBI & Other Examinations | 918 | 4,492 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2022-27 | latest | en | 0.924705 |
http://www.wyzant.com/resources/answers/23667/how_do_i_solve_a_linear_system | 1,405,085,176,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1404776427226.68/warc/CC-MAIN-20140707234027-00078-ip-10-180-212-248.ec2.internal.warc.gz | 589,449,692 | 10,763 | Search 75,766 tutors
0 0
## How do I solve a linear system
In Algebra 1 we're working on solving linear systems, to problems like "x=y+3 2x-y+5" And I don't even know where to start.
X = y +3
2X - y = 5
X - y = 3 (1)
2X - y = 5 ( 2) / first make equations in the standard form of
/ aX + b =c
You see here that if you subtract equation equ (1) from (2)
X = 5-3 = 2
Substitute X =2 back to equation (1)
2 - y = 3 Y = 2 - 3 = -1
You'll learn better techniques for higher order systems, but the 2nd order system (2 eqtns with 2 unknowns) is straightforward to solve by hand. Here are the general steps
1) Place both equations in the form aX + bY = c
2) Inspect the constants of both equations to find a GCM for either X or Y
3) Multiply Either equation (or both) to achieve the GCM in both eqtns for that X or Y
4) Add or subtract the equations to get RID of that X or Y
5) Solve for that remaining Y or X
6) Substitute that first answer (YorX) to compute the other (XorY)
1)
x-y=3
2x-y=5
2)Looking at the above, you'll note that if we subtract the 1st equation from the 2nd, we can make the Y-variable disappear. To be clear, I'll rewrite them
2x-y=5
-( x-y)=3
x = (5-3)=2
6) Now just pick an equation to substitute back to find y (doesn't matter which one you use)
-y=3-x (moved x to the right on 1st eqtn)
y=x-3 (multiplied by -1 on both sides)
y=(2)-3= -1 (substituted 2 into x from above)-----> (x,y)=(2,-1)
6) Lets check the other eqtn just to verify
-y =-2x + 5
y= 2x - 5 (multiplied by -1 on both sides)
y=2(2)-5= -1 (substituted 2 into x from above)-----> (x,y)=(2,-1) VERIFIED! | 563 | 1,644 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2014-23 | latest | en | 0.881004 |
http://piping-designer.com/index.php/mathematics/geometry/plane-geometry/2546-hollow-ellipse | 1,590,584,605,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347394074.44/warc/CC-MAIN-20200527110649-20200527140649-00333.warc.gz | 97,742,014 | 7,456 | # Hollow Ellipse
Written by Jerry Ratzlaff on . Posted in Plane Geometry
• Hollow ellipse (a two-dimensional figure) has two ellipses with a conic section or a stretched circle.
• The major axis is always the longest axis in an ellipse.
• The minor axis is always the shortest axis in an ellipse.
## Formulas that use Hollow ellipse Area
$$\large{ A_{area} = \pi \; \left( a \; b - e \; f \right) }$$
### Where:
$$\large{ A_{area} }$$ = area
$$\large{ a }$$ = length semi-major axis
$$\large{ b }$$ = length semi-minor axis
$$\large{ e }$$ = length inner semi-major axis
$$\large{ f }$$ = length inner semi-minor axis
$$\large{ \pi }$$ = Pi
## Formulas that use Hollow ellipse Inner Semi-major Axis Length
$$\large{ e = a-g }$$
### Where:
$$\large{ e }$$ = length semi-major axis
$$\large{ b }$$ = length semi-minor axis
$$\large{ g }$$ = ring width
## Formulas that use Hollow ellipse Inner Semi-minor Axis Length
$$\large{ f = b-g }$$
### Where:
$$\large{ f }$$ = length semi-minor axis
$$\large{ b }$$ = length semi-minor axis
$$\large{ g }$$ = ring width | 328 | 1,084 | {"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} | 3.796875 | 4 | CC-MAIN-2020-24 | longest | en | 0.636868 |
https://rdrr.io/github/DavisVaughan/slurrr/man/summary-slide.html | 1,627,198,479,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151638.93/warc/CC-MAIN-20210725045638-20210725075638-00034.warc.gz | 469,303,395 | 9,800 | # summary-slide: Specialized sliding functions In DavisVaughan/slurrr: Sliding Window Functions
## Description
These functions are specialized variants of the most common ways that slide() is generally used. Notably, slide_sum() can be used for rolling sums, and slide_mean() can be used for rolling averages.
These specialized variants are much faster and more memory efficient than using an otherwise equivalent call constructed with slide_dbl() or slide_lgl(), especially with a very wide window.
## Usage
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 slide_sum( x, ..., before = 0L, after = 0L, step = 1L, complete = FALSE, na_rm = FALSE ) slide_prod( x, ..., before = 0L, after = 0L, step = 1L, complete = FALSE, na_rm = FALSE ) slide_mean( x, ..., before = 0L, after = 0L, step = 1L, complete = FALSE, na_rm = FALSE ) slide_min( x, ..., before = 0L, after = 0L, step = 1L, complete = FALSE, na_rm = FALSE ) slide_max( x, ..., before = 0L, after = 0L, step = 1L, complete = FALSE, na_rm = FALSE ) slide_all( x, ..., before = 0L, after = 0L, step = 1L, complete = FALSE, na_rm = FALSE ) slide_any( x, ..., before = 0L, after = 0L, step = 1L, complete = FALSE, na_rm = FALSE )
## Arguments
x [vector] A vector to compute the sliding function on. For sliding sum, mean, prod, min, and max, x will be cast to a double vector with vctrs::vec_cast(). For sliding any and all, x will be cast to a logical vector with vctrs::vec_cast(). ... These dots are for future extensions and must be empty. before [integer(1) / Inf] The number of values before or after the current element to include in the sliding window. Set to Inf to select all elements before or after the current element. Negative values are allowed, which allows you to "look forward" from the current element if used as the .before value, or "look backwards" if used as .after. after [integer(1) / Inf] The number of values before or after the current element to include in the sliding window. Set to Inf to select all elements before or after the current element. Negative values are allowed, which allows you to "look forward" from the current element if used as the .before value, or "look backwards" if used as .after. step [positive integer(1)] The number of elements to shift the window forward between function calls. complete [logical(1)] Should the function be evaluated on complete windows only? If FALSE, the default, then partial computations will be allowed. na_rm [logical(1)] Should missing values be removed from the computation?
## Details
Note that these functions are not generic and do not respect method dispatch of the corresponding summary function (i.e. base::sum(), base::mean()). Input will always be cast to a double or logical vector using vctrs::vec_cast(), and an internal method for computing the summary function will be used.
Due to the structure of segment trees, slide_mean() does not perform the same "two pass" mean that mean() does (the intention of the second pass is to perform a floating point error correction). Because of this, there may be small differences between slide_mean(x) and slide_dbl(x, mean) in some cases.
## Value
A vector the same size as x containing the result of applying the summary function over the sliding windows.
• For sliding sum, mean, prod, min, and max, a double vector will be returned.
• For sliding any and all, a logical vector will be returned.
## Implementation
These variants are implemented using a data structure known as a segment tree, which allows for extremely fast repeated range queries without loss of precision.
One alternative to segment trees is to directly recompute the summary function on each full window. This is what is done by using, for example, slide_dbl(x, sum). This is extremely slow with large window sizes and wastes a lot of effort recomputing nearly the same information on each window. It can be made slightly faster by moving the sum to C to avoid intermediate allocations, but it still fairly slow.
A second alternative is to use an online algorithm, which uses information from the previous window to compute the next window. These are extremely fast, only requiring a single pass through the data, but often suffer from numerical instability issues.
Segment trees are an attempt to reconcile the performance issues of the direct approach with the numerical issues of the online approach. The performance of segment trees isn't quite as fast as online algorithms, but is close enough that it should be usable on most large data sets without any issues. Unlike online algorithms, segment trees don't suffer from any extra numerical instability issues.
## References
Leis, Kundhikanjana, Kemper, and Neumann (2015). "Efficient Processing of Window Functions in Analytical SQL Queries". https://dl.acm.org/doi/10.14778/2794367.2794375 | 1,245 | 4,978 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2021-31 | longest | en | 0.754326 |
https://www.shaalaa.com/question-bank-solutions/mark-against-the-correct-answer-how-many-4-digit-numbers-are-there-introduction-to-knowing-our-numbers_135428 | 1,621,190,882,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991178.59/warc/CC-MAIN-20210516171301-20210516201301-00242.warc.gz | 1,024,691,525 | 8,776 | # Mark Against the Correct Answer How Many 4-digit Numbers Are There? - Mathematics
MCQ
Sum
How many 4-digit numbers are there?
#### Options
• 8999
• 9000
• 8000
• none of these
#### Solution
The largest four-digit number = 9999
The smallest four-digit number = 1000
Total number of all four-digit numbers = (9999 − 1000) + 1
= 8999 + 1
= 9000
Concept: Introduction to Knowing Our Numbers
Is there an error in this question or solution?
#### APPEARS IN
RS Aggarwal Class 6 Mathematics
Chapter 1 Number System
Exercise 1H | Q 5 | Page 20 | 163 | 548 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2021-21 | latest | en | 0.646191 |
https://www.bankibps.com/2019/05/different-number-series-questions.html | 1,721,320,175,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514848.78/warc/CC-MAIN-20240718161144-20240718191144-00740.warc.gz | 568,125,178 | 15,073 | # Different Number Series Questions | Logical Reasoning
Different Number Series Questions Previous Papers is given below. The Aspirants who applied for different jobs must searching the latest Mental Ability Test and Number Series Model Papers and Study Material in the Series Completion sections of Logical Reasoning. Also, download the Free Psychometric Test Question Papers, Model Papers Pdf from the direct links below. Also, interested candidates can check the Series Completion Syllabus & Question Pattern details here.
Preparing for the Psychometric Test Online Examination, then have a look at some of the Psychometric Test Study Material and Model Papers PDF. Just Click on the link below and download free Number Series Model Question Papers PDF. Candidates can download the Pdf by just clicking on the download link below. However, the list of Model Question papers given below is only for reference. Hope our Mental Ability Test Model Papers helps you in your preparation and in scoring good marks.
## Number Series Questions and Answers
Directions: In each of the following questions, a number series is given with one term with (?) marks. Find out the missing term and replace it with correct answers given below the series.
1. 15, 21, 57, ? 221
(a) 96
(b) 108
(c) 121
(d) 126
2. 3, 13, 53, 213, ?
(a) 853
(b) 213
(c) 698
(d) 25
3. 10, 100, 200, 310, ?
(a) 450
(b) 440
(c) 430
(d) 154
4. 720, 180, 176, 44, 40, 10, ?, ?
(a) 6, 1.5
(b) 4, 1.5
(c) 6, 4
(d) 8, 6
5. 4, 16, 36, 100, 144, ?
(a) 64
(b) 65
(c) 64.5
(d) 65.5
6. 0, 2, 8, 14, ?, 34
(a) 22
(b) 24
(c) 26
(d) 16
7. 5, 10, 20, 40, 80, ?
(a) 120
(b) 140
(c) 150
(d) 160
8. 336, 210, 120, ?, 24, 6, 0
(a) 60
(b) 80
(c) 90
(d) 100
9. 28, 33, 31, 36, ?, 39
(a) 34
(b) 35
(c) 36.5
(d) 35.5
10. 5, 8, 16, 19, 38, 41, ?
(a) 80
(b) 40
(c) 82
(d) 44
11. 2, 9, 28, ?, 126, 217
(a) 56
(b) 65
(c) 23
(d) 9
12. 125, 80, 45, 20, ?
(a) 45
(b) 21
(c) 4
(d) 5
13. 3, 6, 24, 30, 63, 72, ?, 132
(a) 50
(b) 60
(c) 80
(d) 90
14. 826, 480, 346, 134, ?
(a) 61
(b) 212
(c) 126
(d) 83
15. 2, 15, 41, 80, ?
(a) 100
(b) 153
(c) 132
(d) 158
16. 2, 12, 36, 80, 150, ?
(a) 200
(b) 210
(c) 230
(d) 250
17. 325, 259, 204, 160, 127, 105, ?
(a) 94
(b) 95
(c) 96
(d) 97
18. 1, 4, 10, 22, 46, ?
(a) 52
(b) 96
(c) 94
(d) 88
19. 24, 6, 18, 9, 36, 9, 24, ?
(a) 12
(b) 24
(c) 36
(d) 48
20. 4, 9, 25, ?, 121, 169, 289, 361
(a) 35
(b) 48
(c) 49
(d) 52 | 1,023 | 2,401 | {"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} | 3.296875 | 3 | CC-MAIN-2024-30 | latest | en | 0.623748 |
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posted by .
A 2.95 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0330 m. The spring has force constant 900 N/m. The coefficient of kinetic friction between the floor and the block is 0.42 . The block and spring are released from rest and the block slides along the floor.
What is the speed of the block when it has moved a distance of 0.013 m from its initial position? (At this point, the spring is compressed 0.0200m.)
• Physics -
Try using conservation of energy. Don't forget the frictional work term. There will be a change in the potential energy of the compressed spring.
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https://www.physicsforums.com/threads/simple-pitot-tube-experiment.332740/ | 1,511,022,955,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804976.22/warc/CC-MAIN-20171118151819-20171118171819-00302.warc.gz | 848,871,621 | 16,524 | # Simple pitot tube experiment
1. Aug 26, 2009
### yessiko
Dear all,
I have a simple experiment to make a cheap pitot tube using 3" PVC pipe to measure vacuum pump air flow. But in the middle of my experiment I have difficulty to calculate air flow in the pipe, because I try to use bernouli equation for standard pitot tube and the result still not same with pump factory refference.
May some body help me?
Rgrds,
Yessiko
#### Attached Files:
• ###### simple pitot tube 2.jpg
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2. Aug 26, 2009
### edgepflow
Is your pitot tube reading high? There may be an entrance loss on your pitot tube that causes the deviatiation from ideal Bernouli flow.
3. Aug 26, 2009
### FredGarvin
Make sure your pitot tube is well aligned with the inlet air flow. Also, chamfer the inlet with a 30° chamfer. This will help with the probe being less sensitive to yaw errors.
4. Aug 27, 2009
### yessiko
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• ###### 100_6454.jpg
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5. Aug 27, 2009
### Cyrus
Interesting setup. It will be impossible to help you without seeing your actual calculation. I'm guessing you have a unit wrong somewhere.
6. Aug 27, 2009
### FredGarvin
The pictures are too small to tell anything.
7. Jan 14, 2010
### dopeyranger
Hello
I am undergoing a similar experiment at university and I am making an L shaped simple pitot tube (not pitot static tube). Are there any limits on the dimensions? Such as the ratio of the lenght of the base to the diameter of the tube? I have done research but cannot seem to find anywhere that says anything about that for a simple pitot tube. For a pitot static tube the ratio of 'the distance from the static holes to the 90 degree bend' to the 'diameter of the tube' is best being around 12. But for a simple pitot tube, are there any such limitations?
L shaped simple pitot tube:
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'--------
\______/
^
Base
8. Jan 14, 2010
### FredGarvin
It depends on the flow velocities but a good rule of thumb is to make the base length at least 4 times the tube OD.
9. Jan 14, 2010
### dr dodge
because atmospheric pressure is the pump, not the vacumn, I would think you should need to correct for air density variations real time. A set of "runs" will be relatively meaningless without enviromental conditions being recorded
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Solving One Step Inequalities Puzzle For students learning how to solve and graph inequality word problems, they can practice with this puzzle and can self check. ***There are 3 versions included. The original puzzle AND a different puzzle that is easier to cut. Finally, there is a version with a
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Included in this set are 30 task cards on solving one step inequalities, a student answer sheet, and an answer key. These cards are all short answer questions. There are 3 types of problems. The first 10 cards ask students to solve a one-step inequality. Cards 11-20 ask students to write an inequal
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This is a coloring activity for two sets of 10 problems on solving and graphing one-step inequalities. Posted: 9/9/15 so 50% off through 9/12/15 Posted: New Design 2/5/17
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Divide students into into groups of 2-3 (can be done individually, or in larger groups, but 2-3 is what I recommend). Each group will need page 1 and half of page 2 (the dice on page 2 are the same – these can be cut apart before hand). They should also receive at least one recording sheet. Students
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Three drills to help students gain speed, proficiency, and understanding of solving simple one step inequalities. Two identical drills per page, ready to print and cut apart to save paper. These drills are on: * Solving One-Step Inequalities "undoing" Addition or Subtraction * Solving One-Step Ine
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This worksheet includes questions on writing, graphing, and solving one-step inequalities. Students need to translate verbal phrases, including one word problem; graph inequalities on the number line; write inequalities given a graph; and solve one-step inequalities involving addition, subtraction,
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There are 8 stations for students go to during a class period. I would suggest 7 to 8 minutes be allotted for each station. The stations each include at least one word problem. Questions at the stations include solving one-step inequalities. graphing inequalities. finding errors when solving one-
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Doodle notes make note-taking a fun process. Students can color, doodle, and highlight key information during the lesson while staying on task. These can also be used as a companion to traditional notes to get students to recall important information after the lesson or to prepare for a test. This
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How many of your students have trouble even reading inequalities correctly, much less solving them? Flash cards to the rescue. These sixteen flashcards are all one-step inequalities, so students can solve them in their heads...or at least that's the goal. There are eight cards to a sheet, with
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Do you want to spice up your Algebra class? Are your students struggling with solving one-step inequalities? This activity is for you! Your students must color code each problem with the step that is necessary to solve the inequality. For example, if you need to multiply to solve, the problem needs
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Practice solving inequalities the fun way with this matching activity! For each inequality, students will cut out and attach the matching sentence and the number line which shows the solution. We had a great time working on this in class and I was able to see where the students' weaknesses were by
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Students will practice solving one-step inequalities with variable on either side and complete a crossword puzzle using their solutions. *** 20 one step inequalities *** An answers bank (answers in word forms - great for reinforcing inequality keywords) *** Key included Related fun activities, ga
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This maze is part of : Maze - BUNDLE Solving and Graphing Inequalities This activity is a good review of understanding how to "Solve One-Step Inequalities". There are 15 inequalities provided. From start to end, the student will be able to answer 12 questions out of the 15 provided to get to the
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Solving One-Step Inequalities - Practice / Review: This risk-taking strategy game of conquering territories offers a fun way to practice math skills. This version has cards for solving one step inequalities. **This is part of a discounted bundled set: Conquest Game - Algebra Set 2 - Solving Inequ
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These are two foldables to help students learn how to solve one-step inequalities by adding, subtracting, multiplying and dividing. Option to print out in B&W or colors Includes Teacher Notes
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This 20-question star-shaped puzzle provides students with practice solving one-step inequalities. There are a variety of addition, subtraction, multiplication, and division problems. Includes integers. PERFECT way to review before a quiz or test! If you like this product, you may be interested
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Students will solve one step inequalities by choosing the answer from a set of numbers given. After solving students will find the answer to the title question.
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There are 3 leveled row games included in this product! For a row game, students work with a partner to practice solving one-step inequalities. The pair of students will each solve different problems, but each row of problems will have the same answer. If their answers do not match, the students mus
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Students will practice solving one-step inequalities with variable on either side and complete a word search puzzle using their solutions. *** 20 one step inequalities *** An answers bank (answers in word forms - great for reinforcing inequality keywords) *** Key included Related fun activities,
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Students need lots of practice with solving one-step inequalities and this resources give you lots of practice. There are 6 different stations to practice solving inequalities. Word problems, inequality situations, decimals, and fractions are included. There are two version included: One with nega
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This worksheet includes 12 practice problems. Students will solve each of the inequalities (one-step involving addition & subtraction only). They will then need to graph each of the inequalities on the number line provided.
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This download contains a set of 20 colored and 20 black and white task cards to support CCSS A.REI.3. The questions ask students to solve both one step equations and inequalities with the option of graphing the solution on their own number lines. Although the equations require students to perform a
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This worksheet starts off with a practice of graphing basic inequalities. There are examples of each type of inequality, some written with the variable on the right. The worksheet moves on to solving one-step inequalities and graphing the solution on a number line. The problems are a mix of vario
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One Step Inequalities: These one step inequalities cootie catchers are a great way for students to have fun while they practice their skills with algebra and number lines through one step inequalities. How to Play and Assembly Instructions are included. One Step Inequalities Cootie Catchers Conten
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# Grid rotation question for you math geniuses...
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Umm.. I'll try here, not sure if it works though and it might not be a good method (probably slow). I've made the center of the matrix to be (0, 0), but that's easily changed.
First I get the angle of the coordinate.
angle=tan(OLD_Y/OLD_X)
Then I add 90 degrees.
angle+=90
I find the radius, which is OLD_Y squared times OLD_X all square rooted, then the new coordinates as followed:
...or maybe I switched the cos and sin.
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Umm.. I'll try here, not sure if it works though and it might not be a good method (probably slow). I've made the center of the matrix to be (0, 0), but that's easily changed.
First I get the angle of the coordinate.
angle=tan(OLD_Y/OLD_X)
Then I add 90 degrees.
angle+=90
I find the radius, which is OLD_Y squared times OLD_X squared and all square rooted, then the new coordinates as followed:
...or maybe I switched the cos and sin.
##### Share on other sites
I haven't taken calculus so I might be thinking about this problem wrong, but it seems to me that if you are using straight lines (row and columns kind of need this) all you have to do is switch the x and y coordinates:
~ ~ ~ ~ ~ ~ 1 ~ ~ ~
1 2 3 4 5 ~ 2 ~ ~ ~
~ ~ ~ ~ ~ becomes ~ 3 ~ ~ ~
~ ~ ~ ~ ~ ~ 4 ~ ~ ~
~ ~ ~ ~ ~ ~ 5 ~ ~ ~
Then flip it vertically over the x,y axis
~ 1 ~ ~ ~ ~ 5 ~ ~ ~
~ 2 ~ ~ ~ ~ 4 ~ ~ ~
~ 3 ~ ~ ~ becomes ~ 3 ~ ~ ~
~ 4 ~ ~ ~ ~ 2 ~ ~ ~
~ 5 ~ ~ ~ ~ 1 ~ ~ ~
Did I do something incorrectly?
I think this would work for curved lines to.
------------------
Tell the truth and you will never fear someone will figure out you lied.
<<I'm sure I'm quoting someone out there, wish I knew who!>>
David Abresch
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Flipping the X and Y coordinates certainly will not work ;( There are two ways of doing this. If you are just going to be dealing with a 5x5 array of cells and no other size, take the time to generate a static conversion table, so that given a cell's X and Y you access a structure containing the cell's new coordinates.
Otherwise, if you will be dealing with variable size arrays, just use the trig ;(
[This message has been edited by Splat (edited October 08, 1999).]
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Hi ,
It would be as simple as :
B [ i , j ] = A [ j , 6 - i ]
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You know you've been out of college too long when you forget your matrix calculus.
Question:
Say you have a 5x5 grid, and perform a 90 degree clockwise rotation around the middle cell (3,3). Is there an equation to calculate the new row and col based on the cell's pre-rotation row,col?
I know there is an equation but I can't remember plus I sold all my college math books back.
Thanks in advance for any help.
Wesley
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Hi,
It is as simple as :
B [ i , j ] = A [ j , 6 - i ]
------------------
--Ali Seyedof (It's all dark !)
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# Numerology: Mathematical key to mysteries of Universe
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## Part 1 of 4
Numerology is a study of numbers, and the manner in which they reflect the aptitude and character of an individual.
## Part 2 of 4
Numbers are an integral part of the cosmic plan which we are not consciously aware of. Each alphabet has a numeric value that provides a related cosmic vibration. The sum of numbers in your date of birth and the sum of value derived from the alphabets in the common name, create interrelated vibrations. These numbers tell a great deal about one’s character and purpose in life - what motivates one; where one’s talents lie.
Ms Seema has developed a methodology to keep calculations clear and precise by taking other direct numerological number relations into consideration to drive accuracy in the calculations. It encompasses one’s Success, Money, Career and Growth reflecting in phase directing First Pinnacles, Second Pinnacles defining ones different phase of life and way to make this life better.
## Part 3 of 4
Ms Seema believes each one of us has been blessed with different energies which we receive at different points of time in life. The important thing is to know what supports us and what do we want from life. If the difference is small, one can easily proceed with the desired plan or path, but if the difference is more than one can handle, then it is better to keep the plan on the back-burner for sometime or step gradually further rather than being aggressive about life-changing decisions. For instance, if we know that the road ahead is bumpy, we may slow down to avoid any unfavourable circumstances, which may take place otherwise.
Numerology simply arises from the fact that everything in the universe is a set of mathematical relationships that are run by a divine law that is always accurate and perfect. The cosmic plan is simple: we just need to understand what we really are. To evaluate that, just add the birth details. For example, if one was born on 9-4-1968, then the basic character of the person adds up to number 1, this person would be highly intelligent and analytical. They will grasp the subject very quickly. They will possess a political brain. They will seek answers to the mysteries of the Universe; there will learn even without a teacher; they will be an expert in every science. Likewise, if one’s number adds up to 2, then during their early life, such people prove to be intelligent. They also stick to their decisions.
## Part 4 of 4
Ms Seema uses various sciences and disciplines to generate numerological reports, including psychology, numerology, spirituality and ancient knowledge from various civilizations of the past. The numerology and energy analysis gives you an accurate indication of the energies available to you in certain periods and areas of your life (and their powers or positive vibrations and challenges or negative vibrations). When one use the energies available, one will be 'in the flow, one will learn less through struggle and pain since struggle and pain only indicate an error in thinking, being out of flow with one’s universe.
To know your personal report or further information, contact Ms Seema, at 122, Jaina Tower-I, District Centre, Janak Puri, New Delhi-110058; call: +919899288771, +919899291202, visit website: www.satyamani.org; mail: info@satyamani.org | 725 | 3,387 | {"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} | 3.109375 | 3 | CC-MAIN-2017-22 | latest | en | 0.935994 |
https://gradeup.co/rbi-assistant-preliminary-exam-2016-1st-slot-analysis-i-d6b89150-c8d8-11e6-a166-697638ce2149 | 1,566,786,881,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330962.67/warc/CC-MAIN-20190826022215-20190826044215-00222.warc.gz | 486,731,882 | 28,321 | • Select Exam
• Select Exam
# RBI Assistant Preliminary Exam 2016 1st Slot Analysis
Updated : Dec 24, 2016, 10:38
By : Akanksha Jigyasu
RBI Assistant Preliminary Exam 2016 has started from today. The 1st slot of RBI Assistant Prelims is now over and you can read here about the Exam analysis of the 1st slot of RBI Assistant Preliminary Exam 2016.
The 1st slot of RBI Assistant Preliminary exam 2016 was conducted from 9:00 AM - 10:00 AM. We are now going to share the detailed analysis of this slot.
## RBI Assistant Exam Pattern-
### Total Time
1. English Language301 hour
2.Reasoning Ability35
3.Numerical Ability35
Total100
The Overall level of the exam was Easy to Moderate. The analysis has been presented based on the reviews shared by our readers who appeared for the exam. In the case of any anomaly, let us know in the comments, we will update it.
## English Language
### Number of questions - 30Level of the exam – Easy to Moderate
Topic Questions Reading Comprehension 8 Questions (Based on Story of 3 brothers) Cloze test 7 Questions (Story Based) Spotting Errors (including Sentence Improvement and Spelling Errors) 10 Question Para Jumbles 5 Questions
## Reasoning Ability
### Number of questions - 35Level of the exam – Moderate
Topic Questions Seating Arrangement and Puzzles 3 sets (15 Questions)1 circular arrangement (all people facing the centre)1 linear arrangement (all people sitting in a single row, facing north)1 puzzle Syllogism 5 Questions Inequality 5 Questions (Direct) Alphanumeric Series 5 Questions Miscellaneous 5 Questions (Based on Distance and Direction, Blood Relation etc.)
## Numerical Ability
### Number of questions - 35Level of the exam – Moderate
Topic Questions Number Series 5 Questions7,21,5,23,3,?15,22,32,46,65,?9,10,18,27,91,?17,23,35,59,?,2056,7,16,51,208,? Data Interpretation 1 set (5 Questions)Tabular graph was asked based on production of different companies on different days of the week. Simplification / Approximation 10 Questions Miscellaneous 15 questions were from this category (Age, Mixture Allegation, Partnership, Pipes and Cisterns, Profit Loss, Average, Boats & Stream, Compound/Simple Interest, Time & Work, Probability etc.)
For more practice of the next stage of RBI Assistant selection process, you can try our RBI Assistant test series which will help you with the preparation for RBI Assistant.
All the best for your exams.
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Dec 24Bank & Insurance
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70 attempt.....kolkata region
Plz share attempt from gujrat
Dec 28
79 attempt from uttar pradesh....what are the chances??
Dec 28
From lucknow - kanpur region
72 attempt ..... From jaipur region
When the preli result will be declared
@Deepshikha soooooon | 692 | 2,733 | {"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-2019-35 | longest | en | 0.887364 |
https://www.convertunits.com/from/vershok/to/kiloparsec | 1,659,900,537,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570692.22/warc/CC-MAIN-20220807181008-20220807211008-00638.warc.gz | 663,308,440 | 12,743 | ## ››Convert vershok to kiloparsec
vershok kiloparsec
How many vershok in 1 kiloparsec? The answer is 6.9419068616423E+20.
We assume you are converting between vershok and kiloparsec.
You can view more details on each measurement unit:
vershok or kiloparsec
The SI base unit for length is the metre.
1 metre is equal to 22.497187851519 vershok, or 3.2407792700054E-20 kiloparsec.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between vershok and kiloparsecs.
Type in your own numbers in the form to convert the units!
## ››Want other units?
You can do the reverse unit conversion from kiloparsec to vershok, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Kiloparsec
The SI prefix "kilo" represents a factor of 103, or in exponential notation, 1E3.
So 1 kiloparsec = 103 parsecs.
The definition of a parsec is as follows:
The parsec (symbol pc) is a unit of length used in astronomy. It stands for "parallax of one arc second", and is approximately 19,131,554,073,600 (19 trillion) miles.
## ››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! | 431 | 1,583 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-33 | latest | en | 0.818106 |
https://www.jiskha.com/questions/103031/is-z-18-z-0-z-18 | 1,601,555,334,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402131412.93/warc/CC-MAIN-20201001112433-20201001142433-00392.warc.gz | 851,713,041 | 3,049 | # algebra
is z^18*z^0=z^18
1. 👍 0
2. 👎 0
3. 👁 157
1. Anything to the 0th (if that's a word) power is 1, and a*1=a, so . . .
z18 × z0= x
z18 × 1= x
z18 = x
z18 = z18
(I copied this from a word processor, and it didn't work, so z18 is z to the 18th power, and z0 is z to the 0th power)
1. 👍 0
2. 👎 0 | 144 | 300 | {"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-2020-40 | latest | en | 0.935936 |
https://wiki.baw.de/en/index.php/Example_figures:_relative_frequency_distribution_of_current_velocity | 1,590,681,490,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347399820.9/warc/CC-MAIN-20200528135528-20200528165528-00270.warc.gz | 611,803,792 | 5,947 | # Explanations
Relative frequency [0.0,1.0] for which the magnitude of the current velocity is in the range of a given interval.
• left top: result for interval 0.0 m/s to 0.2 m/s.
• right top: result for interval 0.4 m/s to 0.6 m/s.
• left bottom: result for interval 0.8 m/s to 1.0 m/s.
• right bottom: result for interval 1.2 m/s to 1.4 m/s.
# Examples
current velocity: 0.0 m/s to 0.2 m/s current velocity: 0.4 m/s to 0.6 m/s
current velocity: 0.8 m/s to 1.0 m/s current velocity: 1.2 m/s to 1.4 m/s | 186 | 507 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2020-24 | latest | en | 0.679498 |
https://shrimpboatsound.com/qa/quick-answer-what-are-the-characteristics-of-motion.html | 1,607,105,482,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141740670.93/warc/CC-MAIN-20201204162500-20201204192500-00100.warc.gz | 490,561,969 | 7,664 | # Quick Answer: What Are The Characteristics Of Motion?
## What is translatory motion class 6?
What is Translatory Motion.
A type of motion in which all parts of the body move the same distance in a given time is known as the translatory motion.
Translatory motion can be of two types: rectilinear and curvilinear..
## What is the difference between motion and movement?
motion is used to describe physical properties, while movement is used to describe the qualities of motion. Ref. motion doesn’t always imply a purpose, and movement usually does. Ref.
## How many types of motion does the Earth show name them?
Three MotionsIII. 4 The Earth’s Three Motions. The Earth turns (rotation around the polar axis), goes along on its orbit (revolution around the Sun), swings smoothly as un unbalanced spinning top (equinoctial precession).
## What is motion class 9?
Movement of any object from one position to another position with respect to the observer is called as Motion. Motion Along a Straight Line: When an object moves along a straight line, the motion of the object is called rectilinear motion. For example; motion of a car on highway.
## What are the three characteristics of motion?
Motion. In physics, motion is the change in position of an object with respect to its surroundings in a given interval of time. Motion is mathematically described in terms of displacement, distance, velocity, acceleration, and speed. … As there is no absolute frame of reference, absolute motion cannot be determined.
## What is motion and their types?
There are different types of motion: translational, rotational, periodic, and non periodic motion. Translational Motion. A type of motion in which all parts of an object move the same distance in a given time is called translational motion.
## What is motion and time?
If the speed of an object moving along a straight line keeps changing, its motion is said to be non-uniform. On the other hand, an object moving along a straight line with a constant speed is said to be in uniform motion.
## Which is the simplest type of motion?
forms of its motion. The simplest form of the motion of matter is the mechanical change of place of a body in space. A more complex form of motion is to be found (for example) in thermal processes and the un- ordered motion of the molecules which make up physical bodies.
## What is the 4 types of motion?
Everything naturally wants to move and change. In the world of mechanics, there are four basic types of motion. These four are rotary, oscillating, linear and reciprocating.
## What are the 6 types of motion?
Types of MotionRectilinear motion,Circular motion,Periodic motion and.Rotational motion.
## How do you describe motion?
You can describe the motion of an object by its position, speed, direction, and acceleration. An object is moving if its position relative to a fixed point is changing. Even things that appear to be at rest move.
## What is motion short answer?
In physics, motion is the phenomenon in which an object changes its position over time. Motion is mathematically described in terms of displacement, distance, velocity, acceleration, speed, and time. … One can also speak of motion of images, shapes and boundaries.
## What is motion explain with example?
The definition of a motion is a movement or a proposal for action. An example of motion is a waving hand. An example of motion is a request by a committee member that a recommendation be accepted. noun.
## What causes motion?
Motion is caused by forces. … A stationary object does not move unless a force acts on it to start it going. Once it is moving, it carries on at the same speed and in the same direction unless a force makes it speed up, change direction or slow down and stop.
## How do you teach motion?
During the Force and Motion Science Stations students engage in the following activities:Watch one of two videos (see below)Play one of two video games (see below)Investigate Balloon Races.Diagram a New Slide.Read about Force.Model a Ball Track.Explore Pinwheels.Sort Push and Pull forces. | 862 | 4,098 | {"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-2020-50 | longest | en | 0.936583 |
https://barpsi.com/276-psi-to-mpsi | 1,566,396,895,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316021.66/warc/CC-MAIN-20190821131745-20190821153745-00095.warc.gz | 369,427,953 | 6,539 | # 276 Psi to Mpsi
## Convert 276 Psi to Mpsi. 276 Psi to Mpsi conversion.
276
=
0.000276
Looking to find what is 276 Psi in Mpsi? Want to convert 276 Psi units to Mpsi units?
Using a simple formula, 276 Psi units are equal to 0.000276 Mpsi units.
Want to convert 276 Psi into other Psi units?
Psi, Mpsi, Psi to Mpsi, Psi in Mpsi, 276 Psi to Mpsi, 276 Psi in Mpsi | 118 | 370 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2019-35 | latest | en | 0.736043 |
http://creative-mathematics.ubm.ro/?m=past_issues&issueno=Vol.%2012%20(2003) | 1,500,561,370,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423222.65/warc/CC-MAIN-20170720141821-20170720161821-00332.warc.gz | 78,691,542 | 6,605 | All issues
20.07.2017
Creative Mathematics
and Informatics
Print Edition: ISSN 1584 - 286X Online Edition: ISSN 1843 - 441X Creative Math. and Inf. is a category B+ journal in the CNCSIS classification Indexed by: Mathematical Reviews Zentralblatt MATH Current rejection rate: 37.5 % RSS
Vol. 12 (2003),
Some applications of integral sums Author: P. Kortesi Abstract: The correct introduction of definite integrals plays an important role in teaching analysis. This paper is aimed to present one of the possible applications of definite integrals, which could serve the deeper understanding of the basic notion, to show how the basic definition ''is working''. The paper is partially supported by the didactical research cooperation due to the CEEPUS network H127. PDF | DVI | PS | No. of downloads: 849 Full access PDF | DVI | PS | Improper integrals and Riemann sums Author: Gabriella Kovacs Abstract: The aim of this note is to give simple examples of absolutely convergent improper integrals having integral sums whose limit does not equal the value of the integral. PDF | DVI | PS | No. of downloads: 492 Full access PDF | DVI | PS | Some geometric transformations and their connections to complex numbers Author: Gh. Miclauș Abstract: In the twelve form textbooks ([2], [3]) are presented some groups of geometric transformations: homotheties, translations, rotations etc. In this paper we will present two geometric transformations and we will show their relationship to groups of complex functions of the same type, namely isomorphic structures: the group of the rotations of a polygon with n vertices and Klein's group. At the end, we will show that the type of the group of the rotations of a regular polygon with four vertices (the square) and Klein's group of geometric transformations exhaust all types of groups of order 4. PDF | DVI | PS | No. of downloads: 426 Full access PDF | DVI | PS | On approximating the inverse of matrix Author: Ion Păvăloiu Abstract: In this note we deal with two problems: the first regards the efficiency in approximating the inverse of a matrix by the Shulz-type methods, and the second is the problem of evaluating the errors in the approximation of the inverses of the perturbed matrices. PDF | DVI | PS | No. of downloads: 457 Full access PDF | DVI | PS | Quadrilaterals in which an angle is the mean of the other angles Author: Maria Sânziana Pop / Ileana Balazs / Gh. Miclauș Abstract: PDF | DVI | PS | No. of downloads: 443 Full access PDF | DVI | PS | Some remarks on orthogonal polynomials Author: Ovidiu T. Pop Abstract: In this article we demonstrate some general results, from which, through particularities, we obtain identities satisfied by Legendre's, Laguerre's and Hemite's polynomials. PDF | DVI | PS | No. of downloads: 477 Full access PDF | DVI | PS | On some inequalities for right triangles Author: Ovidiu T. Pop / Constantin Voicu Abstract: We will prove some inequalities between the elements of a right triangle. PDF | DVI | PS | No. of downloads: 572 Full access PDF | DVI | PS | Using Prolog for the study of algebraic structures and complex operations Author: D. Popa Abstract: Should mathematicians learn the Prolog language? Due to its ability of modeling algebraic structures having complex operations and its backtracking algorithm able to scan a search space looking for solutions, Prolog becomes a great tool to study complex algebraic structures. The model of interactions between compilers when computer scientists are attempting to bootstrap a system or to create a new language or a new compiler is such an algebraic structure. This paper tries to answer yes'' our previous question. PDF | DVI | PS | No. of downloads: 427 Full access PDF | DVI | PS | Danilovskaia's problem, revisited Author: R. Răducanu Abstract: The present paper describes a thermal shock problem on a semi-space within the frame of linear thermoelasticity. The analytical solution is obtained and two types of a finite difference numerical algorithm to solve the problem are also described. The solutions are discussed. PDF | DVI | PS | No. of downloads: 407 Full access PDF | DVI | PS | A matrix method for obtaining the spline collocation function Author: Magnolia Rebeles Abstract: In this paper it is presented a procedure for obtaining a spline collocation function which approximate the solution of the initial value problem in linear differential equation, using a matrix method. There are also presented some results of implementing a Borland C program for a linear differential equation with initial values. PDF | DVI | PS | No. of downloads: 408 Full access PDF | DVI | PS | The limitations of JPEG compression Author: Ovidiu Cosma Abstract: The JPEG (Joint Photographic Experts Group) compression standard is based on the Discrete Cosine Transform. This transform was chosen for efficiency, but has the major disadvantage that it does not localize the frequency components in space. Because of that, the image has to be partitioned prior transform, and the margins of the blocks become visible at high compression ratios. This problem is solved in the JPEG 2000 standard, which uses the Discrete Wavelet Transform. PDF | DVI | PS | No. of downloads: 428 Full access PDF | The Window Fourier Transform -- a suitable alternative for image compression? Author: Ovidiu Cosma Abstract: The Discrete Cosine Transform [6], [8] on which the best-known image compression schemes are based [7], [1], has the major disadvantage that it does not localize the frequency components in time. The Window Fourier Transform (WFT) [3], [4], [5] solves this problem, but has limitations related to the resolutions in time and frequency. This article contains a visual evaluation of the WFT characteristics. PDF | DVI | PS | No. of downloads: 440 Full access PDF | Some aspects regarding the Internet protocol next generation Author: Adriana Diaconescu Abstract: This paper presents some different ideas and opinions about the new version of Internet Protocol (IP) -- IPng, also known as IPv6. IPng is designed as an evolutionary upgrade to the Internet Protocol and will, in fact, coexist with the older IPv4 for some time. This paper also reviews the features of the IPv6 protocol, the transition between today's IPv4 Internet and a future IPv6-based one, the structure of an IPv6 packet, the IPv6 header. PDF | DVI | PS | No. of downloads: 404 Full access PDF | DVI | PS | A combined turbo code with adaptive predistortion scheme for a non-linear channel Author: N. Rodriguez / R. Soto / W. Palma Abstract: PDF | DVI | PS | No. of downloads: 444 Full access PDF | DVI | PS | A simulation of the action of terrestrial gravitation Author: Cristian Eduard Rusu Abstract: This paper is a simple illustration of how a low-degree-difficulty abstract formula, concerning a physical phenomenon, can be implemented in an interactive and graphically suggestive way, with the use of a computer. A short historical view on the subject is included. Input data sets and graphic results are presented. PDF | DVI | PS | No. of downloads: 402 Full access PDF | DVI | PS | Forecasting methods and stock market analysis Author: Virginica Rusu / Cristian Rusu Abstract: The paper briefly analysis the methods used in forecasting of the stock market quotations, from the classic methods, used by the fundamentalists and chartist annalists, to the newest methods. The practical situations, where the analyzed methods are suitable, are also indicated. PDF | DVI | PS | No. of downloads: 481 Full access PDF | DVI | PS | Theon of Alexandria and Hypatia Author: M. Lambrou Abstract: In this paper we present the story of the most famous ancient female mathematician, Hypatia, and her father Theon of Alexandria. PDF | DVI | PS | No. of downloads: 434 Full access PDF | DVI | PS | Computer science higher education in Chile Author: Cristian Rusu Abstract: This paper is analyzing the Computer Science higher education system in Chile. A case study was chosen: Pontificia Universidad Catolica de Valparaiso. The problems regarding higher education in Computer Science are, generally speaking, similar in all countries. The system has to develop in harmony with the systems used abroad, taking advantage of the experience of others. PDF | DVI | PS | No. of downloads: 425 Full access PDF | DVI | PS | Preliminaries for the teaching of non-Euclidean geometries Author: Katalin Munkacsy Abstract: According to contemporary principles of methodology, there are good reasons for teaching the three geometries - namely Euclidean, spherical and hyperbolic - in parallel, despite the belief of some teachers that it makes little sense in contemporary conditions. I have been looking for historical data that prove that this parallelism is natural, and precedents can be found not only in ancient Greek history, but also in the Hungarian history of science. PDF | DVI | PS | No. of downloads: 321 Full access PDF | DVI | PS |
Design by PrimeTech | 1,995 | 8,962 | {"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} | 2.546875 | 3 | CC-MAIN-2017-30 | longest | en | 0.860177 |
https://jp.mathworks.com/matlabcentral/profile/authors/18412463?detail=all | 1,685,280,335,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643784.62/warc/CC-MAIN-20230528114832-20230528144832-00259.warc.gz | 382,642,432 | 24,111 | Community Profile
# Opariuc Andrei
Last seen: 1年以上 前 2020 年からアクティブ
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#### Content Feed
Curve fitting from the tools menu
First things first sorry for not translating my comments in the code (but if need be i will do so ) . Everything is functional...
1年以上 前 | 0 件の回答 | 0
### 0
Warning: Imaginary parts of complex X and/or Y arguments ignored.
Error when attempting to plot ,the first imaginary result comes from Q3 afterwards everything that involves Q3 results as imagin...
1年以上 前 | 1 件の回答 | 0
### 1
How to plot a curve between 2 data sets
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2年以上 前 | 1 件の回答 | 0 | 1,949 | 5,905 | {"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} | 2.5625 | 3 | CC-MAIN-2023-23 | longest | en | 0.586828 |
https://docs.scipy.org/doc/scipy-1.5.2/reference/generated/scipy.ndimage.laplace.html | 1,620,347,778,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988774.18/warc/CC-MAIN-20210506235514-20210507025514-00623.warc.gz | 185,850,136 | 3,378 | # scipy.ndimage.laplace¶
scipy.ndimage.laplace(input, output=None, mode='reflect', cval=0.0)[source]
N-D Laplace filter based on approximate second derivatives.
Parameters
inputarray_like
The input array.
outputarray or dtype, optional
The array in which to place the output, or the dtype of the returned array. By default an array of the same dtype as input will be created.
modestr or sequence, optional
The mode parameter determines how the input array is extended when the filter overlaps a border. By passing a sequence of modes with length equal to the number of dimensions of the input array, different modes can be specified along each axis. Default value is ‘reflect’. The valid values and their behavior is as follows:
‘reflect’ (d c b a | a b c d | d c b a)
The input is extended by reflecting about the edge of the last pixel.
‘constant’ (k k k k | a b c d | k k k k)
The input is extended by filling all values beyond the edge with the same constant value, defined by the cval parameter.
‘nearest’ (a a a a | a b c d | d d d d)
The input is extended by replicating the last pixel.
‘mirror’ (d c b | a b c d | c b a)
The input is extended by reflecting about the center of the last pixel.
‘wrap’ (a b c d | a b c d | a b c d)
The input is extended by wrapping around to the opposite edge.
cvalscalar, optional
Value to fill past edges of input if mode is ‘constant’. Default is 0.0.
Examples
>>> from scipy import ndimage, misc
>>> import matplotlib.pyplot as plt
>>> fig = plt.figure()
>>> plt.gray() # show the filtered result in grayscale
>>> ax1 = fig.add_subplot(121) # left side
>>> ax2 = fig.add_subplot(122) # right side
>>> ascent = misc.ascent()
>>> result = ndimage.laplace(ascent)
>>> ax1.imshow(ascent)
>>> ax2.imshow(result)
>>> plt.show()
#### Previous topic
scipy.ndimage.generic_laplace
#### Next topic
scipy.ndimage.maximum_filter | 481 | 1,890 | {"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} | 2.546875 | 3 | CC-MAIN-2021-21 | latest | en | 0.664657 |
https://socratic.org/questions/how-do-you-solve-33-15w-3w-w-4w#167304 | 1,642,717,742,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302706.62/warc/CC-MAIN-20220120220649-20220121010649-00239.warc.gz | 594,231,650 | 5,862 | # How do you solve 33 + 15w = 3w - w + 4w?
Sep 10, 2015
w = color(blue)(-11/3
#### Explanation:
33+15w=3w−w+4w
$33 + 15 w = 6 w$
$33 = 6 w - 15 w$
$33 = - 9 w$
$\frac{33}{-} 9 = w$
w = color(blue)(-11/3 | 111 | 214 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2022-05 | latest | en | 0.580221 |
https://braingenie.ck12.org/skills/103226/learn | 1,556,165,950,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578681624.79/warc/CC-MAIN-20190425034241-20190425060241-00332.warc.gz | 356,542,859 | 1,673 | ### Sample Problem
Fill in the missing number.
87 - (135 - ___) - 21 = 1
#### Solution
87 - 21 - 1 = 65, and 135 - 65 = 70, the correct result is b. | 55 | 152 | {"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-2019-18 | latest | en | 0.627105 |
https://www.experts-exchange.com/questions/22462681/retrieve-data-from-simple-mathematical-function-store-data-in-vector-array.html | 1,537,816,675,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267160641.81/warc/CC-MAIN-20180924185233-20180924205633-00255.warc.gz | 728,039,835 | 15,405 | # retrieve data from simple mathematical function; store data in vector(array)
Hello,
I am trying to implement a function that calculates the position for a counter of a motor. I would like to store the postion values for the different times in a vector (or array?). This is the function I implemented:
void main(void)
{
float pi=0; //initial position; later != 0
float a=pi;
float vi=0; //initial velocity
float b=vi;
float qf; //final position
qf=8200; //approximately 2 turns
float c=3*qf;
float d=-2*qf;
float Tf=1; //final time
float Ti=0; //initial time
float Ts=0.01; //sampling time
float nmuestras = Tf/Ts;
int i;
for (i=1;i<=nmuestras;i++)
{
float tiempo=i*Ts; //real time
float t=tiempo/Tf;
/*
float *ptr;
float pos;
ptr = &pos;
*ptr = a+b*t+c*t^2+d*t^3;
*/
//pos_final(t);
float pos;
pos = a+b*t+c*t^2+d*t^3;
//float pos = a+b*t+c*t^2+d*t^3;
int k;
pos_vector[k] = pos;//put the counter value in the array for storing counter values
k++;
}
I get the following errors and don´t know how to get rid of them:
ompiling...
test control2.cpp
C:\datos\judith\proyecto_judith\pci 1784\test control2.cpp(18) : warning C4305: 'initializing' : truncation from 'const double' to 'float'
C:\datos\judith\proyecto_judith\pci 1784\test control2.cpp(37) : error C2296: '^' : illegal, left operand has type 'float'
C:\datos\judith\proyecto_judith\pci 1784\test control2.cpp(37) : error C2297: '^' : illegal, right operand has type 'float'
C:\datos\judith\proyecto_judith\pci 1784\test control2.cpp(40) : error C2065: 'pos_vector' : undeclared identifier
C:\datos\judith\proyecto_judith\pci 1784\test control2.cpp(40) : error C2109: subscript requires array or pointer type
C:\datos\judith\proyecto_judith\pci 1784\test control2.cpp(40) : warning C4244: '=' : conversion from 'float' to 'int', possible loss of data
C:\datos\judith\proyecto_judith\pci 1784\test control2.cpp(40) : error C2106: '=' : left operand must be l-value
Error executing cl.exe.
test control2.obj - 5 error(s), 2 warning(s)
Do I maybe have to include libraries or introduce pointers?
Judith
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Commented:
Hi jutzki
One of your main problems is that you do not seem to be aware that expressions involving mixed decimal types always result in a value of type double. To assign the result of such an expression to a float you need to perform a cast.
e.g.
float nmuestras = Tf/Ts; //should change to float nmuestras = (float) (Tf/Ts);
You also have a problem with this expression,
pos = a+b*t+c*t^2+d*t^3
^ is the XOR operator in C++. If you want to raise numbers to certain powers you have to use the pow() function in the Math library.
e.g. pos = (float) (a+b*t+c*pow(t, 2)+d*pow(t, 3));
0
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V.] NORMAL VARIABILITY. 57
am satisfied to claim that the Normal Curve is a fair average representation of the Observed Curves during nine-tenths of their course ; that is, for so much of them as lies between the grades of 5° and 95°. In particular, the agreement of the Curve of Stature with the Normal Curve is very fair, and forms a mainstay of my inquiry into the laws of Natural Inheritance.
It has already been said that mathematicians laboured at the law of Error for one set of purposes, and we are entering into the fruits of their labours for another. Hence there is no ground for surprise that their Nomenclature is often cumbrous and out of place, when applied to problems in heredity. This is especially the case with regard to their term of " Probable Error," by which they mean the value that one half of the Errors exceed and the other half fall short of. This is practically the same as our Q.1 It is strictly the same whenever the two halves of the Scheme of Deviations to which it applies are symmetrically disposed about their common axis.
The term Probable Error, in its plain English interpretation of the most Probable Error, is quite misleading, for it is not that. The most Probable Error (as Dr. Venn has pointed out, in his Logic of Chance)
1 The following little Table may be of service Values of the different Constants when the Prob. Error is taken as unity, and
Prob. Error 1.000 ; corresponding Grades 25°•0, 75°•0
Modulus 2.097 ; 7°•9, 92°•1 Mean Error 1.183 ; 21°•2, 78°•8 Error of Mean Squares 1.483 ; ,, ff 16°•0, 84°•0 | 406 | 1,601 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-10 | latest | en | 0.959157 |
https://math.stackexchange.com/questions/1028269/show-that-if-a-is-an-n-times-n-matrix-that-commutes-with-b | 1,638,232,727,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358847.80/warc/CC-MAIN-20211129225145-20211130015145-00021.warc.gz | 463,180,746 | 34,036 | # Show that if $A$ is an $n \times n$ matrix that commutes with $B$
Suppose that $A$ is an $n\times n$ matrix with distinct eigenvalues. And suppose $B$ commutes with $A$. Show that $B$ is diagonable; i.e., show that $B$ is similar to a diagonal matrix.
I get that $AB=BA$ and that some diagonal matrix $D$ that is similar to $B$ is $D=SBS^{-1}$ and that $S$ is composed of the eigenvecotrs of $B$ I just don't know what to do from their.
• Hint: "Simultaneously diagonalisable" Nov 18 '14 at 22:13
Hint: if $Ax = ax, a$ scalar, then $$a Bx = B(ax) = B(Ax) = BAx= ABx$$ and you know that the subspace $$\{y: Ay = ay\}$$is a line...
• If I wanted to be difficult, I could point out that the matrices might not be over $\Bbb C$. But the hint is sound. Nov 18 '14 at 22:18
• @Arthur $\Bbb C$ can be replaced by any field. Nov 18 '14 at 22:19
The big theorem is that, for a square matrix $A$ for which each eigenvalue occurs in only one Jordan block, all matrices that commute with $A$ can be written as polynomials in $A.$ This includes matrices with distinct eigenvalues, as each Jordan block is one by one. So $$B = b_0 I + b_1 A + b_2 A^2 + \cdots + b_{n-1} A^{n-1}.$$ You do not need higher degree because of Cayley-Hamilton.
Oh, $A$ itself is diagonalizable because of the distinct eigenvalues. Some $P^{-1}A P = E$ diagonal. What can you say about $P^{-1}B P ?$ | 436 | 1,368 | {"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} | 3.78125 | 4 | CC-MAIN-2021-49 | latest | en | 0.922645 |
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ELE204-MT2(2007-8)
# ELE204-MT2(2007-8) - HACETTEPE UNIVERSITY DEPARTMENT OF...
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Unformatted text preview: HACETTEPE UNIVERSITY DEPARTMENT OF ELECTRICAL & ELECTRONICS ENGINEERING .; ELE 204 CIRCUIT THEORY - II MIDTERMJI. 8 Ma 2008 Student # : Show all of your steps in detail and present your work CLEARLY. Write your name on all the papers you turn in. Read all questions CAREFULLY, and attempt to solve every part. Write onto the answer sheets, solutions on the question sheets will NOT be graded. 95".“? Q1. i. Find the Laplace transform of the following functions d (Se-2‘ cos 31%), c) x(t) ={ q 3 J, V sint, OSISIZ' a) x(t) =t‘e ' u(t), b) x(t) = E 0, otherwise ii. Find the inverse Laplace transform of the following functions (—7: s/ 2) e s + 2 8 dXs)=——,—, eXs= ,, Xs=——,—7 ) ( s‘+4 ) () (s+l)' f) () s(s‘+4)’ iii. Find x(0) and x(oo) for the following functions 1053+3052+255+50 52+S+1 . 8 g)X(S)=——2—, h)X(S)=2—, 1)X(5)=fi s(s+4)(s +85+100) s +1 s (3—4) iv. Find the impulse response of the following system if the input is x(t) and the output is y(t), j) d? +3Q+2y=ibi+3x dt' dt dt v. Solve the following differential equation using Laplace transform k) d? +8Q+ 25y = 65in 2:, where y(0) = 1, 2(0) = 0. dr dt dz Q2. Consider the circuit in Fig.2, where no energy is initially stored in the components. a) Find the transfer function Vow/[3(5), b) Find the poles and zeros of the transfer function. c) Find the zero—state response of the circuit if the input is is (t) = ((3- 3t cos 2t)u(t) A, d) Find the steady-state response of the circuit if the input is is (t) = (2 cos 2!)u(t) A. Q3. For the circuit given in F ig.3.a, assume that there is no energy stored in the circuit at the time instant when the voltage source is turned on as given in Fig. 3.b. 3) Find the transfer function V0(s)/ V.,.(5) for the circuit, b) Find the impulse response of the circuit, c) Find va(t) for t 2 0+ by using Laplace transform method when the waveform provided in Fig.3.b is applied as the input, d) Determine v00) at I = 2.55ec.s, i.e. vo(2.55ec) 2. .0. 1H fit, Q4. Consider the circuit in Fig. 4.a. The input is i,-(t) and the output is v00). 3) Find the transfer function, H(s), of the circuit, and its poles and zeros, b) What is the impulse response of this circuit? c) Find the output of the circuit for the input given in F ig.4.b. using the convolution integral, d) How long is the memory of the circuit, effectively? What is the weighting of the circuit for a delay of 0.1 secs? 1?. I). 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# I.
Learning Content:
A. Subject Matter: Simple and Compound interests
B. References: 1. General Mathematics by Orlando Oronce
2. Essential Business Mathematics by Rizaldi C. Nocon
C. Materials: laptop, LCD projector, manila paper, marking pens, meta cards,
D. Strategy: Cooperative learning
## II. Learning Activities / Procedure:
A. Preliminary Activity:
1. Checking of attendance and assignment.
2. Drill:
Examine the following:
a. P=Php 5,000.00 r=5% t= 3 years
I= Prt
= (5,000)(.05)(3)
= Php 750.00
Amount due = P+I
= 5,000 .00+ 750.00
= 5,750.00
b. P=Php 5,000.00 r=5% t=3 years
A= P(1+r)
=5,000.00 (1+.05):
=5,000.00(1.1576)
=Php 5,788.125
Questions:
## -Can you identify the difference of the two computations?
-What kind of interest is the first one? How about the second one?
-In loans, which kind of interest can help the debtor?
-If you are a depositor in a savings bank, what kind of interest, you will
prefer?
B. Review:
How to solve for simple interests?
How to solve compound interests?
C. Lesson Proper:
1. Motivation:
Group Activity
Directions:
Pair the given meta cards; description and formula. Post your work in
the manila paper. Work for 5 minutes, then after posting; explain your
work in 1 minute.
a. Maturity value is the total amount paid on due date or maturity
date.
A=P+I
b. Present value is the amount needed at present for an expected
return in the future.
P= A/(I+r)
2. Presentation:
a. How to compute future or maturity value in simple and compound
interest environment.
b. How to compute the present value in simple and compound interest
environment.
c. Interest formula and its derivatives
3. Discussion:
a. Present an example on the following:
Maturity value or future value in simple interest
Present value in simple interest
Maturity value or future value in compound interest
Present value in compound interest
4. Developmental Activity:
a. Compute the future value in simple interest and compound interest.
Given: P= Php80,000.00; r= 2%; t=3 years
b. Solve the problem;
Find the present value of Php 54,000.00 due in 3 years and 8 months if
money is worth 12% compounded semi-annually.
5. Generalization:
To solve problems on maturity value and present value both in simple
interest and compound interest environments, get the derived formula from
the original formula;
Simple interest: Compound Interest:
I = Prt A=P(1+r)
F= P +I or F=P(1+rt) F is the same as the above.
P = I/rt P= A/(1+r)
Note: In compound interest remember the periodic rate.
III. Evaluation:
A. Find the indicated value from the given: | 706 | 2,583 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2020-45 | latest | en | 0.799488 |
https://www.webpages.uidaho.edu/niatt_labmanual/Chapters/traveldemandforecasting/professionalpractice/ModelCalibrationAndValidation.htm | 1,653,618,644,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662631064.64/warc/CC-MAIN-20220527015812-20220527045812-00702.warc.gz | 1,245,937,002 | 2,723 | Model Calibration and Validation The following excerpt was taken from the Transportation Planning Handbook published in 1992 by the Institute of Transportation Engineers (p. 116). (p.116) The process of developing travel models is commonly called "calibration." Given the basic form of a travel forecasting model, such as a gravity model or a logit model, calibration involves estimating the values of various constants and parameters in the model structure. For this reason the model development effort is sometimes termed "estimation." Estimating model coefficients and constants is usually done by solving the model equation for the parameters of interest after supplying observed values of both the dependent and independent variables. The observed values of variables are obtained from the surveys of actual travel patterns. As indicated previously, the estimation process is a trial and error effort that seeks the parameter values which have the greatest probability or maximum likelihood of being accurate within acceptable tolerance of error. Such an effort is commonly accomplished with specialized statistical computer programs designed for just such purposes. . . . Model calibration can also be accomplished by using values of constants and parameters from models estimated for another location that is similar to the area being studied; this strategy is referred to as "importing" model parameters and should be employed only by experienced practitioners. Once satisfactory estimates of the parameters for all models have been obtained, the models must be checked to assure that they adequately perform the functions for which they are intended, that is, to accurately estimate traffic volumes on transit and roadways. Verifying a calibrated model in this manner is commonly called "validation." The validation process establishes the credibility of the model by demonstrating its ability to replicate actual traffic patterns. Validating the models requires comparing traffic estimated by the model to observed traffic on the roadway and transit systems. Initial comparisons are for trip interchanges between quadrants, sectors, or other large areas of interest. . . . The next step is to compare traffic estimated by the models to traffic counts, including transit ridership, crossing contrived barriers in the study area. These are commonly called screenlines, cutlines, and cordon lines and may be imaginary or actual physical barriers. Cordon lines surround particular areas such as the central business district or other major activity centers. . . . Transit ridership estimates are commonly validated by comparing them to actual patronage crossing cordon lines around the central business district. . . . The importance of traffic and transit counts for model validation underscores the need for careful planning, thoroughness and accuracy of a traffic and transit data collection program that has this purpose. As with the travel surveys, the resulting models and forecasts will be no better than the data used for model estimation and validation. | 532 | 3,068 | {"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-2022-21 | latest | en | 0.93374 |
https://gamedev.stackexchange.com/questions/13071/fix-joint-length-in-box2d | 1,726,795,684,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652073.91/warc/CC-MAIN-20240919230146-20240920020146-00140.warc.gz | 240,716,100 | 41,125 | Fix joint length in Box2D?
I'm having some difficulties with Box2D DistanceJoints, I'm trying to create a rope using DistanceJoints connecting some bodies together but the length of joints between bodies change due to weight of bodies, is there any way to prevent joint lengths from changing?
DistanceJoints can be soft spring like joints, or rigid and firm joints depending on your choice of frequencyHz and dampingRatio.
Set dampingRatio to 1.0 for zero oscillations (i.e. inflexible distance).
You may need to set frequencyHz to 0 as well.
• I tried changing damping ratio and/ir frequencyHz to any valu that i could imagine, including the onces you suggested but didn't help at all. Commented Jun 2, 2011 at 22:34
• What are the masses of the two objects that are joined? Commented Jun 2, 2011 at 22:35
• at first they were both 1 (due to default density of fixtures) but then tried changing density to zero (the masses were still 1 because objects were dynamic) and it was same as before, no changes. Commented Jun 2, 2011 at 22:38
• So let me just confirm that density of both objects is set to 1. The mass of both dynamic objects is 1, and the dampingRatio of the DistanceJoint is 1, and the frequencyHz is 0. And in this configuration, your distance joint expands and contracts? Commented Jun 2, 2011 at 22:55
• that's right, i can even post my code here if you like' Commented Jun 2, 2011 at 22:59
I've faced the same problem, and what I've done was using DistanceJoint, with the frequencyHz set to 60 (or more) and the dampingRatio set to 0.
• Higher frequencyHz makes thread more persistent.
• Lower dampingRatio make it sooner for the thread gain original length after being stretched.
For those who may encounter this problem in the future: I just changed my approach, I used some Revolute Joint to connect bodies together and mark all bodies as sensor so that they don't collide with each other (tell me if you any better way to avoid collision of the rope parts), and use boxes to fill the space between joints (previously it was Distance joints connecting circle shapes). Having density set to non-zero value the result looks like a real rope.
• RevoluteJoints and DistanceJoints serve different purposes. Select the one that suits you. If you want to have non-oscillating/firm distances between objects, you can see my answer about DistanceJoints. Commented Jun 2, 2011 at 21:35
• (Their difference is described here box2d.org/manual.html.) Commented Jun 2, 2011 at 21:47
• I know that, but in my case a fixture and two revoluteJoints produces better results than a DistanceJoint; Commented Jun 2, 2011 at 22:36
• No worries, +1 since this answer works for you. Commented Jun 2, 2011 at 22:59
• For those who read this, set "collideConnect" to "false" in the joint definition avoids fixtures contact. Commented Jul 5, 2016 at 13:11
I was struggling with ropes in Box2D until I discovered that a joint's "strength" is proportional to the mass of the bodies it is connected to. The solution that worked for me was just to increase the density of the rope until it was strong enough to hold other objects up. | 774 | 3,133 | {"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} | 2.609375 | 3 | CC-MAIN-2024-38 | latest | en | 0.965917 |
https://physics.stackexchange.com/questions/331810/marble-on-a-rotating-inclined-plane | 1,569,184,312,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514575674.3/warc/CC-MAIN-20190922201055-20190922223055-00349.warc.gz | 616,658,501 | 26,470 | # Marble on a rotating inclined plane [closed]
A marble moves on a smooth plane which is inclined at an angle θ to the horizontal. The whole plane rotates at constant angular speed ω about a vertical axis through a point O fixed in the plane. Coordinates (ξ, η) are defined with respect to axes fixed in the plane: Oξ horizontal and Oη up the line of greatest slope in the plane. Ensuring that you account for the normal reaction force, find $\xi'', \eta''$.
By considering the marble’s kinetic energy as measured on the plane in the rotating frame, or otherwise, find a constant of the motion.
## [You may assume that the marble never leaves the plane.]
For the last part, am I to assume that the quantity $T=\frac{1}{2}mv^2$ is conserved? This gives me $\frac{\omega^2}{2}(\xi^2+\eta^2\cos^2(\theta))-g\sin(\theta) \eta$ is constant.
## closed as off-topic by Jon Custer, Kyle Kanos, Yashas, John Rennie, Bill NMay 9 '17 at 15:06
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Jon Custer, Kyle Kanos, Yashas, John Rennie, Bill N
If this question can be reworded to fit the rules in the help center, please edit the question.
• Welcome to Physics! Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. – Kyle Kanos May 9 '17 at 1:21
• No the question does not intend that you should assume that kinetic energy is conserved - because that would immediately answer the question of finding a constant of the motion. The question is giving you a clue as to how to find a constant of the motion, by considering KE as measured in the rotating (accelerating) frame of reference. – sammy gerbil May 9 '17 at 19:21 | 503 | 2,052 | {"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} | 3.140625 | 3 | CC-MAIN-2019-39 | latest | en | 0.934656 |
https://www.coursehero.com/file/23648598/CHAPTER-5-REVIEW/ | 1,542,140,439,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741491.47/warc/CC-MAIN-20181113194622-20181113220622-00350.warc.gz | 841,309,167 | 64,876 | CHAPTER 5 REVIEW
# CHAPTER 5 REVIEW - courtney lewis Assignment Chapter 5...
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courtney lewis DL-AUCIhomework Assignment Chapter 5 Review due 01/16/2016 at 12:00am EST 1. (1 point) Complete the derivative formulas. Assume that k and b are real constants with b > 0. (a) d dx ( e x ) = (b) d dx ( e kx ) = (c) d dx ( b x ) = (d) d dx ( ln x ) = (e) d dx ( log b x ) = Answer(s) submitted: (incorrect) 2. (1 point) Practice differentiating exponential and logarithmic functions: Answer(s) submitted: (incorrect) 3. (1 point) (a) Use implicit differentiation to find dy dx if 6 x 3 + x 2 y - xy 3 = 6 . dy dx = (b) Use implicit differentiation to find dy dx if y = ln ( 6 x 2 + 6 y 2 ) . dy dx = Answer(s) submitted: (incorrect) 4. (1 point) Use logarithmic differentiation to find the deriv- ative of the function y = x 1 / x . y 0 = Answer(s) submitted: (incorrect) 5. (1 point) Evaluate each limit. Use L’Hopital’s rule if nec- essary.
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https://questions.examside.com/past-years/jee/question/for-the-equation-3x2-px-3-0-p-0-if-one-of-the-root-is-square-jee-advanced-2000-screening-marks-2-dyuxcwhbtscxgdec.htm | 1,723,735,984,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641299002.97/warc/CC-MAIN-20240815141847-20240815171847-00343.warc.gz | 382,152,866 | 42,872 | 1
IIT-JEE 2000 Screening
+2
-0.5
For the equation $$3{x^2} + px + 3 = 0$$. p > 0, if one of the root is square of the other, then p is equal to
A
1/3
B
1
C
3
D
2/3
2
IIT-JEE 2000 Screening
+2
-0.5
If $$\alpha \,\text{and}\,\beta$$ $$(\alpha \, < \,\beta )$$ are the roots of the equation $${x^2} + bx + c = 0\,$$, where $$c < 0 < b$$, then
A
$$0 < \alpha \, < \,\beta \,$$
B
$$\alpha \, < \,0 < \beta \,<\left| \alpha \right|$$
C
$$\alpha \, < \beta \, < 0\,$$
D
$$\alpha \, < \,0 < \left| \alpha \right| < \beta$$
3
IIT-JEE 1999
+2
-0.5
If the roots of the equation $${x^2} - 2ax + {a^2} + a - 3 = 0$$ are real and less than 3, then
A
$$a < 2$$
B
$$2 \le a \le 3$$
C
$$3 < a \le 4$$
D
$$a > 4$$
4
IIT-JEE 1998
+2
-0.5
Number of divisor of the form 4$$n$$$$+ 2\left( {n \ge 0} \right)$$ of the integer 240 is
A
4
B
8
C
10
D
3
EXAM MAP
Medical
NEET | 423 | 847 | {"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} | 2.625 | 3 | CC-MAIN-2024-33 | latest | en | 0.597777 |
https://math.stackexchange.com/questions/3097348/transforming-a-linear-program-into-its-canonical-form-for-use-in-the-simplex-alg | 1,718,553,081,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861665.97/warc/CC-MAIN-20240616141113-20240616171113-00111.warc.gz | 325,875,111 | 36,281 | # Transforming a linear program into its canonical form for use in the simplex algorithm
A typical example of a LP in my lectures looks like this:
$$\begin{array}{lllll} \mathbf{maximize} & x_1&+&x_2&&\\ \text{subject to} & -x_1&+&x_2&+&x_3&&&&&=1\\ & x_1 &&&&&+&x_4&&&=3\\ & &&x_2 &&&&&+&x_5&=2\\ \end{array}\\ x_1,x_2,\ldots x_5\ge0$$
From what I've learnt, we are ready to implement the simplex algorithm on this LP, since $$x_3, x_4, x_5$$ all have positive signs, and so are the right-hand sides of the constraints.
Now I have the following LP:
maximize $$z = x_1 - x_2 + 2x_3$$
subject to:
$$-2x_1 + 2x_2 + 2x_3 - x_4 = 2$$
$$2x_1 - 2x_2 + x_3 + x_5 =2$$
$$x_1,x_2,x_3,x_4,x_5 \geq 0$$
I'm trying to turn this into the required form, and I'm considering two approaches:
Approach 1: multiplying the first constraint by $$-1$$, so that it becomes $$2x_1 - 2x_2 - 2x_3 + x_4 = -2$$, then introduce a variable $$x_6 \geq 0$$ into it so that the RHS becomes non-negative, i.e. $$2x_1 - 2x_2 - 2x_3 + x_4 + x_6 = 0$$, and then proceed to let $$x_5 \text{ and } x_6$$ be the basic variables.
However this approach doesn't seem right to me, since then wouldn't $$x_6$$ must have the value $$2$$? (and hence not really a variable?)
Approach 2: starting with the solution $$(0,0,0,0)$$, we see that the 2 LHSs must be "corrected" to attain the desired values, so we introduce $$x_6, x_7 \geq 0$$ to make up for the differences. But what we really want is for those two new variables to be $$0$$, so our LP becomes:
maximize $$z = -x_6 - x_7$$
subject to:
$$-2x_1 + 2x_2 + 2x_3 - x_4 + x_6 = 2$$
$$2x_1 - 2x_2 + x_3 + x_5 + x_7 =2$$
$$x_1,x_2,x_3,x_4,x_5,x_6,x_7 \geq 0$$
Is any of the two approaches valid? If not, why is it so, and how should I think about and resolve the issue mentioned above?
1. Approach 1 is invalid for the reason that you've mentioned: missing value $$2$$ on the RHS.
2. Approach 2 is simply the method. To improve the efficiency, you don't need to introduce $$x_7$$. Simply take $$x_5 = x_6 = 2$$ at the beginning of phase I, and eliminate $$x_6$$ from the current basis so as to obtain a basic feasible solution for phase II.
3. Inspection approach: find an intial basic feasible solution by inspection: observe that $$x_5$$ doesn't appear in the first constraint, so choose either $$x_2$$ or $$x_3$$ to be the first basic variable, and $$x_5$$ as the second basic variable. This should save work for introducing additional terms. | 861 | 2,472 | {"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": 28, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-26 | latest | en | 0.851054 |
https://www.odoo.com/forum/help-1/question/what-is-the-use-of-ean-13-in-product-and-why-i-always-recieve-an-error-while-saving-the-product-with-a-ean13-barcode-58384 | 1,524,426,680,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945648.77/warc/CC-MAIN-20180422193501-20180422213501-00269.warc.gz | 852,979,430 | 12,001 | # what is the use of EAN 13 in PRODUCT? and why i always recieve an error while saving the product with a EAN13 barcode???
By
Alcaline
on 7/23/14, 9:50 PM 921 views
ValidateError
Error occurred while validating the field(s) ean13: You provided an invalid "EAN13 Barcode" reference. You may use the "Internal Reference" field instead.
What are you entering? You need to enter a valid EAN13 barcode number.http://en.wikipedia.org/wiki/International_Article_Number_(EAN)
Ray Carnes
on 7/23/14, 10:58 PM
yes sir i want to enter a valid EAN13 number.. how to solve this problem??
Alcaline
on 7/24/14, 9:20 PM
#### Remya
--Remya--
5617
| 7 8 8
Thiruvananthapuram, India
--Remya--
Working as an Senior ODOO developer. http://in.linkedin.com/pub/remya-r/69/a23/298
Remya
On 7/24/14, 12:18 AM
Calculation of EAN13:
You can choose first 12 numbers. The 13th is computed. Let x be the sum of numbers on odd positions Let y be the sum of numbers on even positions Let z be x+(y*3) Let z' be z rounded up (73 to 80 for example) Then 13th number is z' - z
I show you with 593253012713 example.I selected a 12 digit number and calculating the thirteen digit. So (5+3+5+0+2+1) = 16, (9+2+3+1+7)*3 = 75. Therefore we have 16 + 75 = 91. Then rounded(91) - 91 is 100 - 91 = 9. So the last digit should be 9 and then it's a valid EAN code 5932530127139
OR
Install point_of_sale module and this will enable Button 'Set a Custom EAN' below this EAN13 field and with help of this button you can add your custom EAN13 no. But this is not recommended.
Hi, how to solve this one (9+2+3+1+7)*3 = 75? when I'm trying to compute this equation I'm getting the answer of 66? how come your answer is 75?? do you have another way of solving it?
Alcaline
on 7/24/14, 9:30 PM
ahmad
On 7/24/14, 11:07 PM
there is a sample checksum for ean13 in javascript file in point-of-sale module, your products should contains the correct number of ean13.
### About This Community
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### Stats
Asked: 7/23/14, 9:50 PM Seen: 921 times Last updated: 3/16/15, 8:10 AM | 725 | 2,359 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2018-17 | latest | en | 0.806224 |
http://www.pearsonitcertification.com/articles/article.aspx?p=2952382&seqNum=3 | 1,576,288,073,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540579703.26/warc/CC-MAIN-20191214014220-20191214042220-00355.warc.gz | 219,448,262 | 13,044 | # Compare and Contrast Fundamental Data Types and Their Characteristics
• Print
This chapter is from the book
## Numbers
The common data types used for numbers include the following:
• Integers
• Floats
### Integers
An integer (also known as int) is a positive or negative whole number (a number with no decimal points or fractions). For example, the following are positive integers: 3, 4096, and 65535. The following are negative integers (–2, –64, and –98765). What about 0? It’s also an integer.
The INT function in most programming languages drops the decimal or fractional value of a number such as 3.6 and leaves only the whole value (in this case, 3). This can be useful to obtain a whole number value after randomizing a range of numbers. The following example is written in Perl:
```my \$randnum = int(rand(100));
# int(rand(100)) discards decimal portion of randomized number
print "Here's a random number between 0 and 100: \$randnum\n";```
### Floats
A float (also known as floating-point number) is a number that contains up to seven digits and has at least one decimal place. For example, the following are floats:
• 5.56
• .0275687
• 3.14159
• .303
A float is a single-precision, floating-point, 32-bit value.
Floats can also be expressed using powers of ten or powers of two. This way, a fixed number of digits can be used to express a very wide range of numeric values. Here are some examples:
• 93.1×105 = 9,310,000
• 93.1×10-5 = 0.000931
• 93.1×1016 = 931,000,000,000,000,000
• 93.1×10-16 = 0.00000000000000931
• 47.67×212 = 47,670,000,000,000
• 47.67×2-12 = 0.00000000004767
Singe-precision and double-precision floating point numbers are actually approximations of the true value of a number because of the rounding that takes place when non-integer numbers are used. The floating-point calculator at https://www.exploringbinary.com/floating-point-converter/ demonstrates this fact. | 520 | 1,927 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2019-51 | latest | en | 0.84479 |
http://mathoverflow.net/questions/69320/n-dimensional-inverse-fourier-transform-of-frac1-mathbf-omega | 1,469,266,313,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257821671.5/warc/CC-MAIN-20160723071021-00121-ip-10-185-27-174.ec2.internal.warc.gz | 159,294,207 | 15,403 | ($n$-dimensional) Inverse Fourier transform of $\frac{1}{\| \mathbf{\omega} \|^{2\alpha}}$
Note: I first posted question on math.stackexchange and I got one reply, which was a bit helpful (I'm still trying to understand it fully), but did not explore the two solution cases that I mentioned. Besides I would appreciate additional insight on this question, which I hope its not too trivial for mathoverflow. Thanks.
I'd like to find the $n$-dimensional inverse Fourier transform of $\frac{1}{\| \mathbf{\omega} \|^{2\alpha}}$ i.e. $$\int_{\mathbb{R}^n} \frac{1}{ \| \mathbf{\omega} \|^{2\alpha}} e^{2 \pi i \mathbf{\omega}\cdot \mathbf{x} } d \mathbf{\omega}$$ where $\mathbf{x} = ( x_0 , x_1 , \cdots , x_n )$ is a spatial parameter in $\mathbb{R}^n$, $\mathbf{\omega} = ( \omega_0 , \omega_1 , \cdots , \omega_n )$, and $$\| \omega\| = \omega_0^2 + \omega_1^2 + \cdots + \omega_n^2$$ All I've been able to come up with in the one-dimensional case is that the integral $$\int_{-\infty}^{+\infty} \frac{1}{ \| \omega \|^{2\alpha}} e^{2 \pi i \omega x } d \mathbf{\omega}$$ diverges because the lower power terms $\omega^p$ terms, for which $p < 2\alpha$, in expansion of the exponential $$e^{2 \pi i \omega x } = \sum_{p = 0}^{\infty} \frac{(2 \pi i \omega x)^p}{p!}$$ do not prevent $\frac{1}{\| \omega \|^{2\alpha}}$ from blowing up at the origin.
I know that one possible way of regularizing this integral is to include a test function and consider the limit of the resulting integral, but I don't quite know how to do so. I've tried reading Gelfand and Shilov's Gneralized Functions vol 1 and while I understand bits of it on the whole its a bit heavy for me.
Based on the papers that I've read I know that there are two cases (the latter of which appears to me more general) and two solutions in each.
• Case 1: 2$\alpha$ is an odd/even integer
• Case 2: 2$\alpha$ is integer or otherwise
I'd appreciate help, if possible, coming up with both solutions.
-
@Olumide: in this case you should bump your Math.SE question. In anycase, Zarrax already gave you the answer to your question. Why not just follow it up there? – Willie Wong Jul 2 '11 at 11:57
Will do. I just assumed I'd find more professional mathematicians, and thus new perspectives here on MO. – Olumide Jul 2 '11 at 12:33
Let's start with a simple change of notation. Let me consider first on $\mathbb R^n_x$ the function $f_{\beta}(x)=\Vert x\Vert^{\beta-n}$ for $0<\beta< n$, which is locally integrable and homogeneous with degree $\beta-n$. Its Fourier transform is also a radial distribution, i.e. such that $$\forall A\in O(n),\quad \hat{f_\beta}=\hat{f_\beta}\circ A,$$ which is homogeneous with degree $-\beta+n-n=-\beta.$ A direct computation shows that $$\hat{f_\beta}(\xi)=\Vert \xi\Vert^{-\beta} \frac{\pi^{-\beta/2}\Gamma(\beta/2)}{\pi^{-(n-\beta)/2}\Gamma((n-\beta)/2)}.$$ There are various extensions of this formula to other values of $\beta$, but one has to pay attention that for $\beta\le 0$, the function" $f_\beta$ is not locally integrable. It is then necessary to define an homogeneous distribution with degree $\beta-n$ which coincides with $\Vert x\Vert^{\beta-n}$ on $\mathbb R^n\backslash\{0\}$, which is possible uniquely if $\beta$ is not a nonpositive integer. | 985 | 3,263 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2016-30 | latest | en | 0.850236 |
http://www.wyzant.com/resources/answers/3733/how_do_you_factor_12x_2_22x_20 | 1,387,676,442,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1387345776833/warc/CC-MAIN-20131218054936-00031-ip-10-33-133-15.ec2.internal.warc.gz | 877,539,166 | 10,305 | Search 73,855 tutors
0 0
# how do you factor 12x^2-22x-20?
Im not that good at math and I need to pass this class. It is my LAST college course, and I am done. I will be on here a lot!
12x^2 -22x -20
begin by factoring out the common factor for all three terms (2):
2(6x^2 -11x -10)
Rewrite the middle term as:
2(6x^2 +4X-15X -10)
group terms:
2((6x^2 +4X)- (15X + 10))
2(2x(3x +2)- 5(3X + 2))
2(3X + 2)(2x- 5) | 178 | 423 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2013-48 | longest | en | 0.768886 |
https://studylib.net/doc/9379919/quadrilateral-project | 1,611,665,274,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704799741.85/warc/CC-MAIN-20210126104721-20210126134721-00336.warc.gz | 570,257,802 | 13,440 | ```Geometry Chapter 6 - Quadrilateral Project
Name __________________________
Due 12/03/2014
Objective
some type of figure (like a bridge, or side of a building) that uses one type of
quadrilateral in its structure. You will then examine, measure, and analyze properties
of quadrilaterals before writing up a report.
Directions
Find some type of figure in real life that has a quadrilateral in the structure-you can
choose which ever shape you like. Walk around your neighborhood, go downtown,
look for signs, find a bridge, or find some kind of fence-these are just a few examples
of where to find quadrilaterals in real life. Next, take a picture of it and measure the
sides, angles, and diagonals. Finally, write a report discussing some of the aspects that
you learned.
Details
• This project must be typed and printed out. Use the computer lab or go to a
library if you do not have access to a printer and/or computer.
• Answer the following questions and explain in complete sentences.
1. What type of figure did you choose? Provide a picture of it and describe it in your
project report.
2. What are the dimensions of the sides and the measure of the angles?
3. What are the dimensions of the diagonals? Do they bisect each other?
4. What are the dimensions of both parts of each pair of diagonal?
5. Do all of your measurements have relationships discussed in the theorems from unit
6?
6. Which theorem applies directly to the figure that you have chosen?.
7. What theorem of diagonals applies to your figure?
8. What type of angle relationship does your figure have? Are opposite angles
congruent? Are
consecutive angles supplementary?
9. What type of segment relationship does your figure have? Are opposing segments
congruent?
11. Summarize all of your findings in a short paragraph. Does this geometry stuff
make sense?
12. End the report making some kind of statement quadrilaterals that are used in real
life. | 435 | 1,923 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2021-04 | latest | en | 0.937851 |
https://pro.arcgis.com/en/pro-app/latest/help/mapping/properties/miller-cylindrical.htm | 1,718,333,996,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861520.44/warc/CC-MAIN-20240614012527-20240614042527-00330.warc.gz | 429,592,266 | 6,607 | Miller cylindrical
Description
The Miller cylindrical projection is a compromise cylindrical map projection. The projection is a modification of the Mercator projection, thus they are almost identical near the equator. Although the Miller projection does not project poles to infinity, distortion is still severe at the poles.
The Miller cylindrical projection was developed by Osborn M. Miller in 1942. It is available in ArcGIS Pro 1.0 and later and in ArcGIS Desktop 8.0 and later.
Projection properties
The subsections below describe the Miller cylindrical projection properties.
Graticule
Miller is a cylindric projection. The meridians are equally spaced straight lines. The parallels and both poles are straight lines, perpendicular to meridians and the same length as the equator. The spacing between parallels grows away from the equator, but it does not increase as much as it does on the Mercator projection. The graticule is symmetric across the equator and the central meridian. The height-to-width ratio of the whole map is 0.73.
Distortion
This projection is neither conformal nor equal-area. Shapes, areas, distances, directions, and angles are all generally distorted. There is no distortion at the equator. Distortion increases away from the equator and becomes severe in polar areas. Distortion values are symmetric across the equator and the central meridian.
Usage
This projection can be used for general world maps not requiring accurate areas, and whose phenomena change with longitude. But its use is not recommended due to extreme distortion in polar regions.
Variants
There are two variants of this projection available in ArcGIS. Neither variant supports the ellipsoid.
• Miller is available in ArcGIS Pro 1.0 and later and in ArcGIS Desktop 8.0 and later. This variant uses the semimajor axis for the radius and equations for a sphere.
• Miller auxiliary sphere is available in ArcGIS Pro 1.0 and later and in ArcGIS Desktop 9.3 and later. This variant uses sphere-based equations with a sphere specified by the Auxiliary Sphere Type parameter.
Limitations
This projection is supported on spheres only. Some distortion properties are not maintained when an ellipsoid is used.
Parameters
Miller parameters are as follows:
• False Easting
• False Northing
• Central Meridian
Miller auxiliary sphere parameters are as follows:
• False Easting
• False Northing
• Central Meridian
• Auxiliary Sphere Type, with values as follows:
• 0 = use semimajor axis of the geographic coordinate system
• 1 = use semiminor axis
• 2 = calculate and use authalic radius
• 3 = use authalic radius and convert geodetic latitudes to authalic latitudes
Note:
If the geographic coordinate system uses a sphere, the Auxiliary Sphere Type uses the radius of the sphere in all four cases.
Sources
Snyder, J. P. (1987). Map Projections: A Working Manual. U.S. Geological Survey Professional Paper 1395. Washington, DC: United States Government Printing Office.
Snyder, J. P. (1993). Flattening the Earth. Two Thousand Years of Map Projections. Chicago and London: University of Chicago Press. | 669 | 3,117 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-26 | latest | en | 0.915701 |
http://mathhelpforum.com/calculus/40634-integral-sin-3-cos-x-dx.html | 1,513,328,288,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948567785.59/warc/CC-MAIN-20171215075536-20171215095536-00220.warc.gz | 180,682,696 | 10,546 | # Thread: integral of sin^3 cos x dx
1. ## integral of sin^3 cos x dx
Definately having some problems with these.
Question is evalueate the indefinate integral
$\int {\sin ^3 } x\_\cos x\_dx$
I get as far as this, I seem to be having issues with these.
$\int {\sin ^2 } x\_\sin x\_\cos x\_dx$
$\int {(1 - \cos ^2 } x)\sin x\cos x\_dx$
$u = \cos x$ $du = - \sin x$
$\frac{1}{4}\sin ^4 x + C
$
2. Originally Posted by Craka
Definately having some problems with these.
Question is evalueate the indefinate integral
$\int {\sin ^3 } x\_\cos x\_dx$
I get as far as this, I seem to be having issues with these.
$
\int {\sin ^2 } x\_\sin x\_\cos x\_dx
\int {(1 - \cos ^2 } x)\sin x\cos x\_dx
u = \cos x du = - \sin x
$
$\frac{1}{4}\sin ^4 x + C
$
No need for a use of trig identities either see that cos is the derivative of sin or do it the old fashioned way
Let $u=\sin(x)\Rightarrow{du=\cos(x)dx}$
giving $\int{u^3du}=\frac{u^4}{4}+C$
back subbing we get
$\frac{\sin^4(x)}{4}+C$
3. Originally Posted by Craka
Definately having some problems with these.
Question is evalueate the indefinate integral
$\int {\sin ^3 } x\_\cos x\_dx$
Since $\frac{d}{dx} \sin(x)=\cos(x)$ the integrand is of the form $[f(x)]^3f'(x)$, so the integral is $[f(x)]^4/4+C$, or:
$\int (\sin(x))^3 \cos(x)~dx = \frac{(\sin(x))^4}{4}+C$
RonL | 478 | 1,333 | {"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": 17, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2017-51 | longest | en | 0.81377 |
https://www.projectsharetexas.org/resource/increasing-and-decreasing-energy_external_1_All_external_2_All_external_3_All_page_6/ | 1,631,946,585,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056348.59/warc/CC-MAIN-20210918062845-20210918092845-00271.warc.gz | 975,485,474 | 11,560 | # View Resource: Increasing and Decreasing Energy
## science logo.png
Resource ID: K4SCI002
# Increasing and Decreasing Energy
By: TEA
### 1. Engage: Energy Equations [[16621]]
1. Engage: Students identify heat, light, and sound energy in different equations.
Activity:
Instruct students to use the animation to create energy equations.
2. Lead a discussion about each equation using sentence stems such as:
• If I can hear something, it has ________ energy.
• Something that gives off light has ________ energy.
• Something that gets warmer or cooler has ________ energy.
• If I (cut or saw) the (wood), it makes (sound energy).
• When a (lit match) is (placed on wood), it creates (fire), which has (heat) energy.
• I can use (my hands) on a (piano or drum) to create (sound energy).
Record the discussion on chart paper.
Facilitation Questions:
What other light, heat, and/or sound energy equations can you make? Create three other energy equations: one for light energy, sound energy, and heat energy.
How do you know that something has light energy? Sound energy? Heat energy?
### 2. Explore: Exploring Increasing and Decreasing Energy [[16626]]
1. Explore: Students investigate the effects of increasing and decreasing heat, light, and sound energy on objects.
Activity 1: Sound Energy
Follow the Teacher Instructions to facilitate Activity 1. If you do not have access to the materials, play the video of Activity 1 and instruct students to record their observations in their science notebooks.
Facilitation Questions:
What happened when you lightly tapped the table with your fingertips?
What happened when you tapped the edge of the table with your hands and increased intensity?
What happened when you stopped tapping the table?
How does the motion of the toothpicks compare to the amount of sound?
2. Activity 2: Light Energy
Follow the Teacher Instructions to facilitate Activity 2. If you do not have access to the materials, play the video of Activity 2 and instruct students to record their observations.
Facilitation Questions:
Which light source is brighter? The tap light? The glow stick? The pen light? The flashlight?
Which light creates the best shadow?
Which light helps you see the most details?
Which light source is the brightest? Which light source is the dimmest?
3. Activity 3: Heat and Sound Energy
Instruct students to use the animation below to investigate heat energy and sound energy.
### 3. Explain: Energy Effects [[16631]]
1. Facilitation Questions:
What caused the water in the birdbath to freeze? Melt?
What caused the temperature to increase in the house?
Why did the light need to be on to sort the socks?
Why was it easier to match the socks when the light increased?
Why did the sound of the fire truck seem to change?
What caused the picture on the wall to vibrate?
What are the effects of increasing amounts of heat energy? Decreasing amounts of heat energy?
What are the effects of increasing light energy? Decreasing light energy?
What are the effects of increasing sound energy? Decreasing sound energy?
How have you been affected by increased or decreased heat
energy? Light energy? Sound energy?
### 4. Elaborate: Energy Increase or Decrease? [[16636]]
1. Elaborate: Students identify the effects of increasing and decreasing heat, light, and sound energy.
Activity:
Instruct students to use the animation to complete the "What Happened Here" sequences.
2. Facilitation Questions:
What form of energy does it take to make toast? Does the energy increase or decrease?
What form of energy is experienced when listening to the radio? Does the energy increase or decrease in this sequence?
What form of energy helps dry our clothes? Our hair? Does the energy increase or decrease?
What form of energy can be observed when a candle is lit? Does the energy increase or decrease?
What form of energy caused the snowman to melt? Did the energy increase or decrease?
What form of energy is exhibited by an alarm clock?
When night turns to day, what form of energy can be observed?
What form of energy does it take to pop popcorn? Does the energy increase or decrease?
What form of energy helps you see when it is dark? Does the energy increase or decrease?
When soup is too hot to eat or drink, do you need to increase or decrease the heat energy in the soup?
If you want to make ice, do you need to increase or decrease heat energy?
What form of energy is used to warn people that a train is coming?
3. Assignment:
Instruct students to find examples of increasing and decreasing amounts of heat, light, or sound energy and their effects on their everyday life.
Instruct students to prepare a presentation of their examples to share with the class. Student presentations may include, but are not to be limited to, a poster with illustrations, a video, or photographs with labels detailing the increase or decrease of heat, light, or sound energy.
### 5. Evaluate: Give Us the Story [[16641]]
1. Evaluate: Students describe the effects of increasing and decreasing heat, light, and sound energy.
Directions:
• Pass one picture from RM 4: Give Us the Story to each group of students.
• Instruct students to observe the picture and identify the evidence of increased or decreased heat, light, and/or sound energy.
• Instruct students to use their observations and supporting evidence to write a story or news report detailing what happened.
• Allow adequate time for student groups to complete the evaluation.
• Allow each group to share their story or news report.
2. Facilitation Questions:
What do you think happened before this picture was taken?
What evidence of heat, light, or sound energy do you observe?
Possible descriptions for each picture may include:
Picture 1
Popped popcorn—increased heat energy
Popped popcorn and turned television on—increased sound energy
Turned television on—increased light energy
Picure 2
Ordered pizza—increased sound energy
Sun went down—decreased light energy
Turned porch lights on—increased light energy
Door bell ringing—increased sound energy
Picture 3
Snowy weather—decreased heat energy
People talking—increased sound energy
Picture 4
Fire alarm rings—increased sound energy
Students walking in line—decreased sound energy
Walking outside—increased light energy from the Sun
4. Differentiation Strategies:
G/T: Ask students to create their own scenarios.
ELL and Struggling Students: Support students by offering sentence starters like the following:
• Our picture shows _________. We saw evidence of _____ energy because ________.
• We can see that (increased/decreased) _________ energy played a part in _____. We know this because ___________.
• We know that it takes _________ energy to ______.
• This just in: ________ energy has ________ again.
### 6. Join the Course [[16646]]
1. The activities in this resource are also featured in the full-length professional development course, Science Academies for Grades K-4. This course is designed to demonstrate the application of the 5E instructional model in a K-4 classroom. Throughout the course, connections will be made to the College and Career Readiness Standards (CCRS), the English Language Proficiency Standards (ELPS), and Response to Intervention (RtI) in order to strengthen participants’ knowledge of these frameworks within the discipline of science. This course is managed by Region 4 (101-950): Texas Education Service Center. CPE credit is 18.
To join the Science Academies for Grades K-4 course, click on the button below. | 1,616 | 7,560 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-39 | latest | en | 0.904979 |
https://askwillonline.com/2011/12/signalling-chapter-3-physics-as.html | 1,709,037,599,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474676.26/warc/CC-MAIN-20240227121318-20240227151318-00555.warc.gz | 110,009,665 | 34,403 | The Full Story
Hi! I’m Will and I created a passive 5 figure passive income, within 5 years, through SEO and an effective blogging strategy. I share my incites exclusively on Ask Will Online.
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# Signalling – Chapter 3 – Physics AS
This article will go over all you need to know in the A Level Physics which includes advantages of digital signalling, analogue signalling, converting between digital to analogue and analogue to digital, sampling, polarisation, sensing circuits, resolution, bandwidth and a sampling summary including sampling rate and channel capacity (I think are in chronological order). Your more than welcome to skip to the parts that are of most help to you. As well as that, you may also want to look at other articles I have done for Physics AS revision. If you want a quick look at this article, I would suggest you look at Sampling Summary at the bottom although that is not an excuse to just read that: you will be missing out loads of information and detail from the main article.
There are many advantages to digital signalling. However, before we get onto that, we should know what is an analogue and digital signal.
Digital Signals
The parts of a digital signal can have one of two values, 0 or 1.
Examples of digital devices include on/off switch, digital ammeter, digital clock, optic fibre and telephone cable. Time is on the x-axis.
A digital code will look like this:
1001101001010011010111100101010100
Because there are only different possibilities.
Analogue Signals
The parts of an analogue signal can have any value within a fixed range of values . For my example, it’s range is from 0-3 but this can vary from analogue signal to analogue signal. They basically take any values within it’s range.
Examples of analogue devices include dimmer switch, thermometer, analogue ammeter and speedometer. Time is on the x-axis.
• Digital signals have better quality – this is because noise can be easily removed. More information on noise can be found here.
• Many signals can be sent down the same optic fibre – this is called multiplexing.
Analogue to Digital Conversion
Analogue signals are turned into digital signals by sampling. The height of the wave is measured at regular time intervals and the height put into binary code.
Each sample can only have one of a fixed set of values (in the case 0 to 7 or 8 bit). I will use the above analogue signal and converse it to an 8 bit digital signal.
As you can see, the grey bars are the new conversed digital signal from the analogue signal. You can see that some of the analogue signal in-between the bars have been lost.
As well as that, because it has 8 possible options, it means it is in 3 bit. You can see more on binary codes here.
Digital to Analogue Conversion
The analogue signal is the constructed from the digital signal. The reconstructed signal will be in red and follows the line of best fit. After seeing the two lines, can see the differences between the sampled (black) wave and reformed (red) wave?
The signal at the end is very similar to each other. However, at the start, much of the detail in the sampled (original) wave has been lost. Different places throughout the signal have frequencies that have been introduced that were not present in the original sample.
Polarisation
What happens when you view polarised light through a rotating polaroid filter and why?
You will see it go darker at a certain point when rotating the filter. This is because at the point of it darkening, the polaroid filter is stopping the polarised light from going through it causing less light waves to pass through. At the point of darkening, the filter is parallel to the electric filed of the light waves causing the light waves parallel to the filter to go through but all the others waves not to. The amount of light going through is fluctuated between two extremes.
To make more sense of what I’ve just said, I’ll draw a diagram.
If you can’t read the writing, click on the image and a zoomed version of the image will appear.
Sensing Circuits
Here is an example of a sensing circuit:
To work out the sensitivity, you have to do the equation:
Voltage output range / input range
or more general:
output range / input range
The sensitivity is also know as the gradient.
For our example the sensitivity is 2 / 100 and will be measured in Volts per lux.
Therefore, the sensitivity will be 2×10 to the power of -3 Volts/lux.
Resolution
Resolution = smallest change that can be detected.
To calculate the resolution:
• What is the smallest possible output? In our example, it will be 0.01 Volts as our voltmeter is accurate to two decimal places.
• What input change does this correspond to ?
The sensitivity is 0.002 V/lux. Therefore, resolution = 0.01 / 0.002 which equals 5 lux.
Bandwidth
Bandwidth of a signal is the range of frequencies in a signal.
Capacity of a channel is the maximum amount of information it can sends bits per second.
As a rough estimation the bandwidth equal to the maximum number of bits per second.
Bandwidth Explained
A square wave contains a fundamental frequency and all its multiples in its signal and therefore has an extremely high bandwidth.
The fundamental frequency is the rate at which each 01 pair is sent (i.e. number of bits per second = 2 x fundamental frequency of sound waves).
Filtering out some of these frequencies changes the shape of the square wave but it is still recognisable and therefore reformable at the other end.
This reduces the bandwidth and means that the signal can be sent down a channel with smaller bandwidth.
It is possible to remove almost all frequencies except the fundamental hence fundamental frequency, new bandwidth and number of bits per second are all roughly equal. (i.e. same order of magnitude).
Sampling Summary
Analogue Signals
• The parts of an analogue signal can have any value within a fixed range of values
Digital Signals
• The parts of a digital signal can have one of only two values, 0 or 1.
• Streams of 0s and 1s can be used to represent any whole number binary code.
Analogue to Digital Conversion
Analogue signals are turned into digital signals in 3 stages:
• Sampling
• Binary coding
• Further encoding
To sample a wave the height is measured at regular time intervals and put into binary code.
Each sample can only have one of a fixed set of values (0 to 7 in 8 bit binary).
Number of Bits
• Number of possible values = 2 to the power of b where b = number of bits.
• The noise limits the maximum number of bits it is worth sending.
Number of possible values = total signal variation / noise variation.
Sampling Rate
• The sampling frequency must be at least 2 x the highest frequency component present or else details of the signal may be lost.
• The highest frequency present can be identified in a frequency spectrum.
Channel Capacity
• This is the rate at which a channel can transmit information, measured in bits per second.
• The amount of information in a signal per second = sampling frequency x bits per sample.
Bandwidth
• The range of frequencies used to send information on a particular channel.
• Bandwidth B needed = b bits per second / 2.
Digital Signals have the advantage that:
• They can be regenerated easily, reducing the effects of noise.
• They can be processed and encoded.
• They can represent different kinds of information in the same way.
Further Encoding
Signals are coded in such a way that errors can be located and corrected.
E.g. each part of the signal is sent more than once. If the ‘copies’ do not agree there must be an error, and they can be asked again until they agree.
It’s a lot to take in so if you are stuck on any of it, please comment below, subscribe to future comments and I will try my best to answer your questions. Hope this has helped! Please also see other revision material I have done on Physics AS.
1. Anonymous April 29, 2013
2. Anonymous March 19, 2015
## A Positron Emission Tomography
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## All You Need To Know About Momentum
In theory, every moving and colliding objects works through momentum. The momentum of an object can be calculated…
## Particles in Action P3 Topic 5 – Physics Extension GCSE
This article will go into detail the whole of Topic 5, Particles in Action which includes absolute zero,… | 1,841 | 8,385 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-10 | latest | en | 0.923666 |
https://community.smartsheet.com/discussion/83527/cell-link-from-another-sheet-based-on-conditions | 1,714,059,086,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297295329.99/warc/CC-MAIN-20240425130216-20240425160216-00251.warc.gz | 151,409,706 | 107,641 | Cell link from another sheet based on conditions
Options
edited 09/28/21
Hi all, I hope someone can help as I've been racking my brain trying to work this out!
I have multiple sheets that work together of which the data changes each month (it's based on electricity bills with varying monthly kWh (in pence) rates). I essentially want to link a particular cell from one sheet to another sheet, based on 2 conditions and then do a simple multiplication based off that figure. I can get this formula to work:
=SUM([kWh Total]@row * {Energy Report (Static) Average daily p/kWh rate)
but because the rate changes each month I need it to look up the new rate each month.
For example: If Column A (in sheet 2) is ticked, and Column B (in sheet 2) = specific text , then pull through the kWh rate from Column C (in sheet 2) to Sheet 1.
I figured the easiest way to do this was to just get the kWh rate from sheet 2 to pull through to sheet 1 first, then I can a simple sum but if there's a way to pull the figure from sheet 2 into sheet 1 and do the calculation in one formula, then I'd prefer that.
Many thanks
• ✭✭✭✭✭✭
Options
Try something like this...
=[kWh Total]@row * INDEX(COLLECT({Sheet 2 Column C}, {Sheet 2 Column A}, @cell = 1, {Sheet 2 Column B}, @cell = "Specific Text"), 1)
• ✭✭✭✭✭✭
Options
Try something like this...
=[kWh Total]@row * INDEX(COLLECT({Sheet 2 Column C}, {Sheet 2 Column A}, @cell = 1, {Sheet 2 Column B}, @cell = "Specific Text"), 1)
• Options
Paul, you are an absolute life safer! It works perfectly - I've spent hours trying to work it out!!!
Thank you so much!!! :)
• ✭✭✭✭✭✭
Options
Happy to help. 👍️
Help Article Resources
Want to practice working with formulas directly in Smartsheet?
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# The probability that a family with 6 children has exactly
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3136 posts | 645 | 2,054 | {"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 | 0.828826 |
https://jucydata.iiimpact.io/bar-density-and-pie-density-plots-for-showing-relative-proportions/ | 1,590,855,596,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347409337.38/warc/CC-MAIN-20200530133926-20200530163926-00369.warc.gz | 406,429,083 | 10,704 | In my last post, I described a bar-density chart to show paired data of proportions with an 80/20-type rule. The following example illustrates that a small proportion of Youtubers generate a large proportion of views.
Other examples of this type of data include:
• the top 10% of families own 75% of U.S. household wealth (link)
• the top 1% of artists earn 77% of recorded music income (link)
• Five percent of AT&T customers consume 46% of the bandwidth (link)
In all these examples, the message of the data is the importance of a small number of people (top earners, superstars, bandwidth hogs). A good visual should call out this message.
The bar-density plot consists of two components:
• the bar chart which shows the distribution of the data (views, wealth, income, bandwidth) among segments of people;
• The embedded Voronoi diagram within each bar that encodes the relative importance of each people segment, as measured by the (inverse) density of the population among these segments - a people segment is more important if each individual accounts for more of the data, or in other words, the density of people within the group is lower.
The bar chart can adopt a more conventional horizontal layout.
Voronoi tessellation
To understand the Voronoi diagram, think of a fixed number (say, 100) of randomly placed points inside a bar. Then, for any point inside the bar area, it has a nearest neighbor among those 100 fixed points. Assign every point on the surface to its nearest neighbor. From this, one can draw a boundary around each of the 100 points to include all its nearest neighbors. The resulting tessellation is the Voronoi diagram. (The following illustration comes from this AMS column.)
The density of points in the respective bars encodes the relative proportions of people within those groups. For my example, I placed 6 points in the red bar, 666 points in the yellow bar, and ~2000 points in the gray bar, which precisely represents the relative proportions of creators in the three segments.
Density is represented statistically
Notice that the density is represented statistically, not empirically. According to the annotation on the original chart, the red bar represents 14,000 super-creators. Correspondingly, there are 4.5 million creators in the gray bar. Any attempt to plot those as individual pieces will result in a much less impactful graphic. If the representation is interpreted statistically, as relative densities within each people segment, the message of relative importance of the units within each group is appropriately conveyed.
A more sophisticated way of deciding how many points to place in the red bar is to be developed. Here, I just used the convenient number of 6.
The color shades are randomly applied to the tessellation pieces, and used to facilitate reading of densities.
***
In this section, I provide R code for those who want to explore this some more. This is code used for prototyping, and you're welcome to improve them. The general strategy is as follows:
• Set the rectangular area (bar) in which the Voronoi diagram is to be embedded. The length of the bar is set to the proportion of views, appropriately scaled. The code utilizes the dirichlet function within the spatstat package to generate the fixed points; this requires setting up the owin parameter to represent a rectangle.
• Set the number of points (n) to be embedded in the bar, determined by the relative proportion of creators, appropriately scaled. Generate a data frame containing the x-y coordinates of n randomly placed points, within the rectangle defined above.
• Use the ppp function to generate the Voronoi data
• Set up a colormap for plotting the Voronoi diagram
• Plot the Voronoi diagram; assign shades at random to the pieces (in a production code, these random numbers should be set as marks in the ppp but it's easier to play around with the shades if placed here)
The code generates separate charts for each bar segment. A post-processing step is currently required to align the bars to attain equal height. I haven't figured out whether the multiplot option helps here.
library(spatstat)
# enter the scaled proportions of creators and views
# the Youtube example has three creator segments
# number of randomly generated points should be proportional to proportion of creators. Multiply nc by a scaling factor if desired
nc = c(3, 33, 965)*2
# bar widths should be proportional to proportion of views
# total width should be set based on the width of your page
wide = c(378, 276, 346)/2
# set bar height, to attain a particular aspect ratio
bar_h = 50
# define function to generate points
# defines rectangular window
makepoints = function (n, wide, height) {
df <- data.frame(x = runif(n,0,wide),y = runif(n,0,height))
W <- owin( c(0, wide), c(0,height) ) # rectangular window
pp1 <- as.ppp( df, W )
y <- dirichlet(pp1)
# y\$marks <- sample(0:wide, n, replace=T) # marks are for colors
return (y)
}
y_red = makepoints(nc[1], wide[1], bar_h) # height of each bar fixed
y_yel = makepoints(nc[2], wide[2], bar_h)
y_gry = makepoints(nc[3], wide[3], bar_h)
# setting colors (4 shades per bar, one color per bar)
cr_red = colourmap(c("lightsalmon","lightsalmon2", "lightsalmon4", "brown"), breaks=round(seq(0, wide[1],length.out=5)))
cr_yel = colourmap(c("burlywood1", "burlywood2", "burlywood3", "burlywood4"), breaks=round(seq(0, wide[2],length.out=5)))
cr_gry = colourmap(c("gray80", "gray60", "gray40", "gray20"), breaks=round(seq(0, wide[3],length.out=5)))
# plotting
par(mar=c(0,0,0,0))
# add png to save image to png
# remove values= if colors set in ppp
plot.tess(y_red, main="", border="pink3", do.col=T, values = sample(0:wide[1], nc[1], replace=T), col=cr_red, xlim=c(0, wide[1]), ylim=c(0,bar_h), ribbon=F)
plot.tess(y_yel, main="", border="darkgoldenrod4", do.col=T, values=sample(0:wide[2], nc[2], replace=T), col=cr_yel, xlim=c(0, wide[2]), ylim=c(0,bar_h), ribbon=F)
plot.tess(y_gry, main="", border="darkgray", do.col=T, values=sample(0:wide[3], nc[3], replace=T), col=cr_gry, xlim=c(0, wide[3]), ylim=c(0,bar_h), ribbon=F)
# because of random points, the tessellation looks different each time
# post-processing: make each bar the same height when aligned side by side
***
A cousin of the bar-density plot is the pie-density plot. Since I'm using only three creator segments, which each account for about 30-40% of the total views, it is natural to use a pie chart. In this case, we embed the Voronoi diagrams into the pie sectors.
If the distribution were more even, that is to say, the creators are more or less equally important, the pie-density plot looks like this:
***
Something that is more like 80/20
The original chart shows the top 0.3 percent generating almost 40 percent of the views. A more typical insight is top X percent generates 80 percent of the data. For the YouTube data, X is 11 percent. What does the pie-density chart look like if top 11 percent <-> 80 percent, middle 33 percent <-> 11 percent, bottom 56 percent <-> 8 percent?
Roughly speaking, the second segment includes 3 times the people as the largest, and the third has 5 times as the largest.
Tags:
junkcharts | 1,762 | 7,203 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2020-24 | latest | en | 0.911781 |
https://www.studyrankersonline.com/110233/mcq-questions-for-class-11-maths-chapter-6-linear-inequalities-with-answers | 1,632,103,482,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056974.30/warc/CC-MAIN-20210920010331-20210920040331-00149.warc.gz | 1,033,073,324 | 12,063 | # MCQ Questions for Class 11 Maths Chapter 6 Linear Inequalities with answers
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Provide me latest MCQ Questions for Class 11 Maths Chapter 6 Linear Inequalities Free PDF Download so I can prepare for exams. Kindly give Chapter 6 Linear Inequalities Class 11 Maths MCQs Questions with Answers quickly as it is very essential.
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Below you will find MCQ Questions of Chapter 6 Linear Inequalities Class 11 Maths Free PDF Download that will help you in gaining good marks in the examinations and also cracking competitive exams. These Class 11 MCQ Questions with answers will widen your skills and understand concepts in a better manner.
# MCQ Questions for Class 11 Maths Chapter 6 Linear Inequalities with answers
1. Solve the system of inequalities − 2 < 1 − 3x < 7
(a) − 2 < x < 2
(b) − 1 < x < 1
(c) − 2 < x < 1
(d) None of these
► (c) − 2 < x < 1
2. Find the solution for the pair of solution x > 1 and x > -1
(a) No solution
(b) -1 < x < 1
(c) x < -1
(d) x > 1
► (d) x > 1
3. The solution of the inequality 3(2-x)≥2(1-x) for real x is :
(a) x < 4
(b) x > 4
(c) x ≤4
(d) x ≥ 4
► (c) x ≤4
4. The solution to |3x – 1| + 1 < 3 is
(a) 2 < x < 3/4
(b) -1/3 < x < 1
(c) -1/3 < x < 1/4
(d) -3 < x < 3
► (b) -1/3 < x < 1
5. Solve : 3x + 5 < x − 7, when x is a real number
(a) x < − 12
(b) x > − 6
(c) x < − 6
(d) None of these
► (c) x < − 6
6. A connected planar graph having 6 vertices, 7 edges contains _____________ regions.
(a) 15
(b) 3
(c) 1
(d) 11
► (b) 3
7. Find the pairs of consecutive even positive integers both of which are smaller than 10 and their sum of more than 11
(a) (4, 8)
(b) (6, 8)
(c) (6, 8) and (4, 8)
(d) (6, 4)(4, 2)
► (b) (6, 8)
8. Find the value of x when x is a natural number and 24x< 100.
(a) {5,6,……..∞}
(b) {1,2,3,4}
(c) {1,2,3,4,5}
(d) {0,1,2,3,4}
► (b) {1,2,3,4}
9. A point P lies in the solution region of 3x – 7 > x + 3. So the possible coordinates of P are
(a) (6 , -1)
(b) (-3 ,4)
(c) (2, 5)
(d) (0, 7)
► (a) (6 , -1)
10. The solution set for | x | > 7
(a) (7 , ∞)
(b) (−∞,−7) ∪ (7,∞)
(c) (−∞,−7) ∩ (7,∞)
(d) None of these
► (b) (−∞,−7) ∪ (7,∞)
11. Identify solution set for | 4 − x | + 1 < 3?
(a) (2 , 6)
(b) (3 , 6)
(c) (2 , 4)
(d) (2 , 3)
► (a) (2 , 6)
12. If | x − 2| = p, where x < 2, then x - p =
(a) 2 – 2p
(b) - 2
(c) 2p - 2
(d) 2
► (a) 2 – 2p
13. The solution set of x/3>x/2+1, where x is a real
(a) (-6, 6)
(b) (-∞, -6)
(c) (-∞, -6]
(d) (-6, ∞)
► (b) (-∞, -6)
14. Find the solution for the pair of inequations x > 1 and x < -1
(a) no solution
(b) x < -1
(c) -1 < x < 1
(d) x > 1
► (a) no solution
15. What is the region represented by x > 0 and y < 0?
16. What are the integer values of x which satisfy the inequalities x > − 2 and x ≤ 2 ?
(a) - 1 , 0 , 1 , 2, 3
(b) - 1 ,0 ,1 , 2
(c) 1 , 2, 3, 4
(d) None of these
► (b) - 1 ,0 ,1 , 2
17. The solution set for | 3x − 2 | ≤ 1/2
(a) [2/3 , 2/3]
(b) [5/6 , 1/2]
(c) [1/2 ,5/6]
(d) None of these
► (c) [1/2 ,5/6]
18. The inequations -4x+1≥0 and 3-4x<0 have the common solutions given by
(a) {}
(b) {0,2}
(c) {-2,2}
(d) None of these
► (d) None of these
19. Find all pairs of consecutive odd natural numbers, both of which are larger than 10, such that their sum is less than 40.
(a) (11 , 13) , (13 , 15) , (15 , 17) , (17 , 21)
(b) (9 , 11) , (13 , 15) , (15 , 17) , (17 , 19)
(c) (11 , 13) , (13 , 15) , (17 , 19) , (19 , 21)
(d) (11 , 13) , (13 , 15) , (15 , 17) , (17 , 19)
► (d) (11 , 13) , (13 , 15) , (15 , 17) , (17 , 19)
20. If -2 < 2x – 1 < 2 then the value of x lies in the interval
(a) (1/2, 3/2)
(b) (-1/2, 3/2)
(c) (3/2, 1/2)
(d) (3/2, -1/2)
► (b) (-1/2, 3/2)
21. The solution of |2/(x – 4)| > 1 where x ≠ 4 is
(a) (2, 6)
(b) (2, 4) ∪ (4, 6)
(c) (2, 4) ∪ (4, ∞)
(d) (-∞, 4) ∪ (4, 6)
► (b) (2, 4) ∪ (4, 6)
22. Solve the inequality 3 − 2x ≤ 9
(a) x ≥ − 6
(b) x ≥ − 3
(c) x ≤ − 3
(d) None of these
► (b) x ≥ − 3
Hope the given MCQ Questions will help you in cracking exams with good marks. These Linear Inequalities MCQ Questions will help you in practising more and more questions in less time. | 1,895 | 4,155 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2021-39 | latest | en | 0.707365 |
http://popflock.com/learn?s=Cable_theory | 1,582,957,769,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875148671.99/warc/CC-MAIN-20200229053151-20200229083151-00423.warc.gz | 109,621,918 | 21,111 | Cable Theory
Get Cable Theory essential facts below. View Videos or join the Cable Theory discussion. Add Cable Theory to your PopFlock.com topic list for future reference or share this resource on social media.
Cable Theory
Figure. 1: Cable theory's simplified view of a neuronal fiber
Classical cable theory uses mathematical models to calculate the electric current (and accompanying voltage) along passive[a]neurites, particularly the dendrites that receive synaptic inputs at different sites and times. Estimates are made by modeling dendrites and axons as cylinders composed of segments with capacitances ${\displaystyle c_{m}}$ and resistances ${\displaystyle r_{m}}$ combined in parallel (see Fig. 1). The capacitance of a neuronal fiber comes about because electrostatic forces are acting through the very thin lipid bilayer (see Figure 2). The resistance in series along the fiber ${\displaystyle r_{l}}$ is due to the axoplasm's significant resistance to movement of electric charge.
Figure. 2: Fiber capacitance
## History
Cable theory in computational neuroscience has roots leading back to the 1850s, when Professor William Thomson (later known as Lord Kelvin) began developing mathematical models of signal decay in submarine (underwater) telegraphic cables. The models resembled the partial differential equations used by Fourier to describe heat conduction in a wire.
The 1870s saw the first attempts by Hermann to model neuronal electrotonic potentials also by focusing on analogies with heat conduction. However, it was Hoorweg who first discovered the analogies with Kelvin's undersea cables in 1898 and then Hermann and Cremer who independently developed the cable theory for neuronal fibers in the early 20th century. Further mathematical theories of nerve fiber conduction based on cable theory were developed by Cole and Hodgkin (1920s-1930s), Offner et al. (1940), and Rushton (1951).
Experimental evidence for the importance of cable theory in modelling the behavior of axons began surfacing in the 1930s from work done by Cole, Curtis, Hodgkin, Sir Bernard Katz, Rushton, Tasaki and others. Two key papers from this era are those of Davis and Lorente de Nó (1947) and Hodgkin and Rushton (1946).
The 1950s saw improvements in techniques for measuring the electric activity of individual neurons. Thus cable theory became important for analyzing data collected from intracellular microelectrode recordings and for analyzing the electrical properties of neuronal dendrites. Scientists like Coombs, Eccles, Fatt, Frank, Fuortes and others now relied heavily on cable theory to obtain functional insights of neurons and for guiding them in the design of new experiments.
Later, cable theory with its mathematical derivatives allowed ever more sophisticated neuron models to be explored by workers such as Jack, Rall, Redman, Rinzel, Idan Segev, Tuckwell, Bell, and Iannella.
## Deriving the cable equation
Note, various conventions of rm exist. Here rm and cm, as introduced above, are measured per membrane-length unit (per meter (m)). Thus rm is measured in ohm·meters (Ω·m) and cm in farads per meter (F/m). This is in contrast to Rm (in ?·m²) and Cm (in F/m²), which represent the specific resistance and capacitance respectively of one unit area of membrane (in m2). Thus, if the radius, a, of the axon is known,[b] then its circumference is 2?a, and its rm, and its cm values can be calculated as:
These relationships make sense intuitively, because the greater the circumference of the axon, the greater the area for charge to escape through its membrane, and therefore the lower the membrane resistance (dividing Rm by 2?a); and the more membrane available to store charge (multiplying Cm by 2?a). The specific electrical resistance, ?l, of the axoplasm allows one to calculate the longitudinal intracellular resistance per unit length, rl, (in ?·m-1) by the equation:
The greater the cross sectional area of the axon, ?a², the greater the number of paths for the charge to flow through its axoplasm, and the lower the axoplasmic resistance.
Several important avenues of extending classical cable theory have recently seen the introduction of endogenous structures in order to analyze the effects of protein polarization within dendrites and different synaptic input distributions over the dendritic surface of a neuron.
To better understand how the cable equation is derived, first simplify the theoretical neuron even further and pretend it has a perfectly sealed membrane (rm=?) with no loss of current to the outside, and no capacitance (cm = 0). A current injected into the fiber [c] at position x = 0 would move along the inside of the fiber unchanged. Moving away from the point of injection and by using Ohm's law (V = IR) we can calculate the voltage change as:
where the negative is because current flows down the potential gradient.
Letting ?x go towards zero and having infinitely small increments of x, one can write (4) as:
or
Bringing rm back into the picture is like making holes in a garden hose. The more holes, the faster the water will escape from the hose, and the less water will travel all the way from the beginning of the hose to the end. Similarly, in an axon, some of the current traveling longitudinally through the axoplasm will escape through the membrane.
If im is the current escaping through the membrane per length unit, m, then the total current escaping along y units must be y·im. Thus, the change of current in the axoplasm, ?il, at distance, ?x, from position x=0 can be written as:
or, using continuous, infinitesimally small increments:
${\displaystyle i_{m}}$ can be expressed with yet another formula, by including the capacitance. The capacitance will cause a flow of charge (a current) towards the membrane on the side of the cytoplasm. This current is usually referred to as displacement current (here denoted ${\displaystyle i_{c}}$.) The flow will only take place as long as the membrane's storage capacity has not been reached. ${\displaystyle i_{c}}$ can then be expressed as:
where ${\displaystyle c_{m}}$ is the membrane's capacitance and ${\displaystyle {\partial V}/{\partial t}}$ is the change in voltage over time. The current that passes the membrane (${\displaystyle i_{r}}$) can be expressed as:
and because ${\displaystyle i_{m}=i_{r}+i_{c}}$ the following equation for ${\displaystyle i_{m}}$ can be derived if no additional current is added from an electrode:
where ${\displaystyle {\partial i_{l}}/{\partial x}}$ represents the change per unit length of the longitudinal current.
Combining equations (6) and (11) gives a first version of a cable equation:
which is a second-order partial differential equation (PDE).
By a simple rearrangement of equation (12) (see later) it is possible to make two important terms appear, namely the length constant (sometimes referred to as the space constant) denoted ${\displaystyle \lambda }$ and the time constant denoted ${\displaystyle \tau }$. The following sections focus on these terms.
## Length constant
The length constant, ${\displaystyle \lambda }$ (lambda), is a parameter that indicates how far a stationary current will influence the voltage along the cable. The larger the value of ${\displaystyle \lambda }$, the farther the charge will flow. The length constant can be expressed as:
The larger the membrane resistance, rm, the greater the value of ${\displaystyle \lambda }$, and the more current will remain inside the axoplasm to travel longitudinally through the axon. The higher the axoplasmic resistance, ${\displaystyle r_{l}}$, the smaller the value of ${\displaystyle \lambda }$, the harder it will be for current to travel through the axoplasm, and the shorter the current will be able to travel. It is possible to solve equation (12) and arrive at the following equation (which is valid in steady-state conditions, i.e. when time approaches infinity):
Where ${\displaystyle V_{0}}$ is the depolarization at ${\displaystyle x=0}$ (point of current injection), e is the exponential constant (approximate value 2.71828) and ${\displaystyle V_{x}}$ is the voltage at a given distance x from x=0. When ${\displaystyle x=\lambda }$ then
and
which means that when we measure ${\displaystyle V}$ at distance ${\displaystyle \lambda }$ from ${\displaystyle x=0}$ we get
Thus ${\displaystyle V_{\lambda }}$ is always 36.8 percent of ${\displaystyle V_{0}}$.
## Time constant
Neuroscientists are often interested in knowing how fast the membrane potential, ${\displaystyle V_{m}}$, of an axon changes in response to changes in the current injected into the axoplasm. The time constant, ${\displaystyle \tau }$, is an index that provides information about that value. ${\displaystyle \tau }$ can be calculated as:
The larger the membrane capacitance, ${\displaystyle c_{m}}$, the more current it takes to charge and discharge a patch of membrane and the longer this process will take. The larger the membrane resistance ${\displaystyle r_{m}}$, the harder it is for a current to induce a change in membrane potential. So the higher the ${\displaystyle \tau }$ the slower the nerve impulse can travel. That means, membrane potential (voltage across the membrane) lags more behind current injections. Response times vary from 1-2 milliseconds in neurons that are processing information that needs high temporal precision to 100 milliseconds or longer. A typical response time is around 20 milliseconds.
## Generic form and mathematical structure
If one multiplies equation (12) by ${\displaystyle r_{m}}$ on both sides of the equal sign we get:
and recognize ${\displaystyle \lambda ^{2}={r_{m}}/{r_{l}}}$ on the left side and ${\displaystyle \tau =c_{m}r_{m}}$ on the right side. The cable equation can now be written in its perhaps best known form:
This is a 1D Heat equation or Diffusion Equation for which many solution methods, such as Green's functions and Fourier methods, have been developed.
It is also a special degenerate case of the Telegrapher's equation, where the inductance ${\displaystyle L}$ vanishes and the signal propagation speed ${\displaystyle 1/{\sqrt {LC}}}$ is infinite.
## References
• Poznanski, Roman R. (2013). Mathematical Neuroscience. San Diego [California]: Academic Press.
• Tuckwell, Henry C. (1988). Introduction to theoretical neurobiology. Cambridge [Cambridgeshire]: Cambridge University Press. ISBN 978-0521350969.
• de Nó, Rafael Lorente (1947). A study of nerve physiology. Studies from the Rockefeller Institute for Medical Research. Reprints. Rockefeller Institute for Medical Research. pp. Part I, 131:1-496, Part II, 132:1-548. OCLC 6217290.
• Lazarevich, Ivan A.; Kazantsev, Victor B. (2013). "Dendritic signal transition induced by intracellular charge in inhomogeneties". Phys. Rev. E. 88. arXiv:1308.0821. Bibcode:2013PhRvE..88f2718L. doi:10.1103/PhysRevE.88.062718.
## Notes
1. ^ Passive here refers to the membrane resistance being voltage-independent. However recent experiments (Stuart and Sakmann 1994) with dendritic membranes shows that many of these are equipped with voltage gated ion channels thus making the resistance of the membrane voltage dependent. Consequently there has been a need to update the classical cable theory to accommodate for the fact that most dendritic membranes are not passive.
2. ^ Classical cable theory assumes that the fiber has a constant radius along the distance being modeled.
3. ^ Classical cable theory assumes that the inputs (usually injections with a micro device) are currents which can be summed linearly. This linearity does not hold for changes in synaptic membrane conductance. | 2,680 | 11,672 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 39, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2020-10 | latest | en | 0.922699 |
https://www.gradesaver.com/textbooks/math/statistics-probability/elementary-statistics-a-step-by-step-approach-with-formula-card-9th-edition/chapter-4-probability-and-counting-rules-4-3-the-multiplication-rules-and-conditional-probability-extending-the-concepts-page-226/54 | 1,576,415,431,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541308149.76/warc/CC-MAIN-20191215122056-20191215150056-00515.warc.gz | 729,561,976 | 12,489 | ## Elementary Statistics: A Step-by-Step Approach with Formula Card 9th Edition
The total number of cars is 300. P(compact)=150/300=0.5 P(domestic)=210/300=0.7 P(compact and domestic)=100/300=0.333 Since P(compact and domestic)$\ne$P(compact)*P(domestic) The two events are not independent. | 84 | 291 | {"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} | 3.640625 | 4 | CC-MAIN-2019-51 | latest | en | 0.819552 |
https://community.adobe.com/t5/lightroom-classic-ideas/p-implement-ability-to-reverse-linear-gradients-without-needing-to-invert-them/idi-p/14237873 | 1,723,407,035,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641008125.69/warc/CC-MAIN-20240811172916-20240811202916-00738.warc.gz | 132,225,415 | 116,678 | • Global community
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• P: Implement ability to reverse Linear Gradients w...
P: Implement ability to reverse Linear Gradients without needing to invert them.
Nov 15, 2023 Nov 15, 2023
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The feature request is simple:
• I suggest to be able to reverse our Linear Gradients without needing to use "Invert"
The advantage of this is that the state of our Liner Gradient(s) won't change.
Current method
Suggested
Tis a small thing but I propose anyway.
Idea No status
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3 Replies 3
Nov 16, 2023 Nov 16, 2023
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Moving to Ideas
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LEGEND ,
Nov 17, 2023 Nov 17, 2023
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I clearly do not understand. I am probably not understanding the difference or perhaps meaning of Inverse vs Reverse. Can you elaborate.
I suspect you are not talking directly about Inverse as both a prompt, and a shortcut exist for that, as seen below
Also, you may want to post different example, These old eyes see no difference in the two screenshots you posted.
In my Inverse case:
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Nov 18, 2023 Nov 18, 2023
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@GoldingD
Reverse would achieve the same results as the Inversion BUT without changing the "state" of Linear Gradient so without doing MaskInverted = true
It's done by swapping the internal coordinates of the Linear Gradient.
ZeroX <-> FullX
ZeroY <-> FullY
The advantage of this simple method is that we keep the current "state" of the gradient.
e.g.
If a gradient is in a Subtract state by reversing it it says subtracted while currently with Invert it becomes an Intersection.
Tis the same result but I think more users would prefer to simply reverse a linear gradient than use Invert that adds more complexity when working.
P.S.
I would have done a scrip/plug-in myself and swaps the values of the Linear gradient myself but I am not advanced enough to do it.
@johnrellis can probably do it.
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Troubleshooting | 581 | 2,282 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-33 | latest | en | 0.860121 |
http://www.physicsforums.com/showthread.php?p=4177471 | 1,406,109,850,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997877693.48/warc/CC-MAIN-20140722025757-00229-ip-10-33-131-23.ec2.internal.warc.gz | 999,116,045 | 7,115 | # Frictionless incline problem
by KSMSKM
Tags: frictionless, incline
P: 1 1. a block B is placed on an inclined plane T ( a right triangle). the triangle rests on a flat surface. zero friction is assumed. what is the acceleration of T & B? 2. no angle of the triangle is given so i just added in θ on intuition. my best guess is something along the lines of: 3. acceleration of T = B/T * Gravity * sin(θ) acceleration of B = T/B * Gravity * sin(θ) thanks for the help
P: 15 It is a nice question.I am getting that acceleration of the T is (B/T)gtanθ and on the block is gsinθ.Just draw free body diagram.and now see forces acting on the incline(Do not forget that normal reaction due to block)find it and equate Fnet=ma.and done and second one acceleration is it is moving due to mgsinθ force so find it acceleartion.
Homework Sci Advisor HW Helper Thanks ∞ P: 9,643 This is quite a tricky problem because the direction of acceleration of the block is not obvious. Its acceleration relative to the triangle will be down the incline, but its acceleration relative to the flat surface will not be. Set the normal force between the block and triangle to be N and write down the free body equations for each mass.
P: 15 Frictionless incline problem Yeah i forgot that thing.
P: 358 FBD for both Attached Thumbnails
Related Discussions General Discussion 0 Introductory Physics Homework 5 Introductory Physics Homework 1 Introductory Physics Homework 3 Introductory Physics Homework 7 | 359 | 1,487 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2014-23 | latest | en | 0.890814 |
http://mathhelpforum.com/calculus/161489-another-definite-integral-problem.html | 1,524,736,653,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948125.20/warc/CC-MAIN-20180426090041-20180426110041-00188.warc.gz | 205,896,376 | 9,992 | # Thread: Another definite integral problem
1. ## Another definite integral problem
The question:
$\displaystyle \int_{0}^{\pi} \frac{sinx}{5 + 2cosx} dx$
My attempt:
Let u = 5 + 2cosx
$\displaystyle \frac{du}{dx} = -2sinx$
du = -2sinx dx
$\displaystyle -\frac{1}{2} \int_{0}^{\pi} \frac{1}{u} du$
$\displaystyle x = 0 \rightarrow u = 7$
$\displaystyle x = \pi \rightarrow u = 3$
$\displaystyle -\frac{1}{2} [log(u)]_{7}^{3}$
$\displaystyle -\frac{1}{2} [log(3) - log(7)]$
$\displaystyle -\frac{1}{2} [log(3/7)]$
$\displaystyle [log(3/7)^{-1/2}]$
$\displaystyle [log(\frac{\sqrt{7}}{\sqrt{3}})]$
However, the answer in my text is:
$\displaystyle \frac{1}{2}(ln(7) - ln(3))$
What am I doing wrong? Am I simplifying too far? Thanks.
2. Originally Posted by Glitch
$\displaystyle -\frac{1}{2} [log(3) - log(7)]$
....
What am I doing wrong? Am I simplifying too far? Thanks.
I think you are
$\displaystyle -\frac{1}{2} [\log(3) - \log(7)] =\frac{1}{2}(\log(7)-\log(3))$
3. $\displaystyle log(\frac{\sqrt{7}}{\sqrt{3}}) = log(\sqrt{\frac{7}{3}}) = log(\frac{7}{3})^{1/2} = \frac{1}{2}[log(7)-log(3)]$
$\displaystyle \therefore \left[log(\frac{\sqrt{7}}{\sqrt{3}})\right]$ is the same as $\displaystyle \frac{1}{2}(log(7) - log(3))$
4. Thank you guys. | 493 | 1,269 | {"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-2018-17 | latest | en | 0.643028 |
https://www.proofwiki.org/wiki/Composition_of_Linear_Transformations_is_Linear_Transformation | 1,695,382,084,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506399.24/warc/CC-MAIN-20230922102329-20230922132329-00789.warc.gz | 1,083,951,216 | 10,404 | # Composition of Linear Transformations is Linear Transformation
## Theorem
Let $K$ be a field.
Let $X, Y, Z$ be vector spaces over $K$.
Let $T_1 : X \to Y$ and $T_2 : Y \to Z$ be linear transformations.
Then the composition $T_2 \circ T_1 : X \to Z$ is a linear transformation.
## Proof
Let $\lambda \in K$ and $u, v \in X$.
Then, we have:
$\ds \map {\paren {T_2 \circ T_1} } {\lambda u + v}$ $=$ $\ds \map {T_2} {\map {T_1} {\lambda u + v} }$ $\ds$ $=$ $\ds \map {T_2} {\lambda T_1 u + T_1 v}$ Definition of Linear Transformation $\ds$ $=$ $\ds \lambda \paren {T_2 T_1} u + \paren {T_2 T_1} v$ $\ds$ $=$ $\ds \lambda \paren {T_2 \circ T_1} u + \paren {T_2 \circ T_1} v$
so $T_2 \circ T_1 : X \to Z$ is a linear transformation.
$\blacksquare$ | 288 | 755 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2023-40 | latest | en | 0.306583 |
https://numbermatics.com/n/400485/ | 1,600,842,056,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400209999.57/warc/CC-MAIN-20200923050545-20200923080545-00263.warc.gz | 536,919,861 | 5,913 | # 400485
## 400,485 is an odd composite number composed of three prime numbers multiplied together.
What does the number 400485 look like?
This visualization shows the relationship between its 3 prime factors (large circles) and 8 divisors.
400485 is an odd composite number. It is composed of three distinct prime numbers multiplied together. It has a total of eight divisors.
## Prime factorization of 400485:
### 3 × 5 × 26699
See below for interesting mathematical facts about the number 400485 from the Numbermatics database.
### Names of 400485
• Cardinal: 400485 can be written as Four hundred thousand, four hundred eighty-five.
### Scientific notation
• Scientific notation: 4.00485 × 105
### Factors of 400485
• Number of distinct prime factors ω(n): 3
• Total number of prime factors Ω(n): 3
• Sum of prime factors: 26707
### Divisors of 400485
• Number of divisors d(n): 8
• Complete list of divisors:
• Sum of all divisors σ(n): 640800
• Sum of proper divisors (its aliquot sum) s(n): 240315
• 400485 is a deficient number, because the sum of its proper divisors (240315) is less than itself. Its deficiency is 160170
### Bases of 400485
• Binary: 11000011100011001012
• Base-36: 8L0L
### Squares and roots of 400485
• 400485 squared (4004852) is 160388235225
• 400485 cubed (4004853) is 64233082384084125
• The square root of 400485 is 632.8388420443
• The cube root of 400485 is 73.7103971997
### Scales and comparisons
How big is 400485?
• 400,485 seconds is equal to 4 days, 15 hours, 14 minutes, 45 seconds.
• To count from 1 to 400,485 would take you about four days.
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 400485 cubic inches would be around 6.1 feet tall.
### Recreational maths with 400485
• 400485 backwards is 584004
• The number of decimal digits it has is: 6
• The sum of 400485's digits is 21
• More coming soon! | 626 | 2,274 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2020-40 | latest | en | 0.864641 |
http://mathematica.stackexchange.com/questions/23006/output-format-in-matrixform | 1,469,702,027,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257828010.65/warc/CC-MAIN-20160723071028-00141-ip-10-185-27-174.ec2.internal.warc.gz | 155,535,479 | 18,818 | # Output format in MatrixForm
I have a correlation matrix, which I want to display in the following format:
• 2 decimal places (e.g., -0.56)
• number signs "+" and "-" (e.g., +0.76 or -0.34)
I've tried many different things, but I had no success so far... The code I'm using to generate the data is:
randomWalk[x_] := Accumulate[Prepend[RandomVariate[NormalDistribution[0, 1], x], 0]]
exchange = Table[Subscript[asset, i] = randomWalk[500], {i, 1, 5}];
Now the code I'm using to generate the correlation matrix:
N[Correlation[Transpose[exchange][[1 ;; 500, All]]], 2]//MatrixForm
However, the numbers generated are displayed with $MachinePrecision, and not 2 decimal places (although I've used N[#,2]). I've tried another code: MatrixForm[Round[Correlation[Transpose[exchange][[1 ;; 500, All]]], 0.01]] In this case I get the correlation matrix with 2 decimal places, but it's still missing the "+" signal before the number. So I've tried this last code: NumberForm[Round[Correlation[Transpose[exchange][[1 ;; 500, All]]], 0.01],NumberSigns -> {"-", "+"}]//MatrixForm In this last case I get the numbers correctly formatted, but I'm not able to put them in MatrixForm. Can someone give me a hint to solve this? - Why not NumberForm[MatrixForm[Round[...,NumbersSigns->{"-","+"}]? – andre Apr 9 '13 at 16:00 @ andre Thanks A LOT !!! It works perfectly now! – Rod Apr 9 '13 at 16:02 @ andre please put it as an answer format, so I can accept your answer... – Rod Apr 9 '13 at 16:09 ## 2 Answers It is perfectly possible to wrap NumberForm around MatrixForm : NumberForm[MatrixForm[Round[...,NumbersSigns->{"-","+"}] does the job - Of course there are many ways of doing this. Here's one: mat = RandomReal[{-1, 1}, {4, 4}] Table[NumberForm[mat[[i, j]], {3, 2}, NumberSigns -> {"-", "+"}], {i, 1, 4}, {j, 1, 4}] // MatrixForm You get: - Imagine you have mat defined... try to use: NumberForm[mat, NumberSigns -> {"-", "+"}] // TableForm. It doesn't work... – Rod Apr 9 '13 at 14:47 what's wrong with the output of the above, either Table or Map? – bill s Apr 9 '13 at 14:50 You've generated numbers with 2 decimal places... try something with $MachinePrecision... – Rod Apr 9 '13 at 14:52
mat has full real precision, it's the argument {3,2} that does the truncating to 3 significant digits plus two to the right of the decimal place. – bill s Apr 9 '13 at 14:54
Ok, but the problem is: I'm already using mat[[1 ;; 500, All]] inside a DynamicModule environment... so, I can't use Table[mat[[i,j]],{i,1,4},{j,1,4}]... – Rod Apr 9 '13 at 15:06 | 765 | 2,548 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2016-30 | latest | en | 0.84004 |
http://www.jiskha.com/members/profile/posts.cgi?name=Anna&page=19 | 1,409,160,422,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500829661.96/warc/CC-MAIN-20140820021349-00009-ip-10-180-136-8.ec2.internal.warc.gz | 449,948,379 | 11,490 | Wednesday
August 27, 2014
# Posts by Anna
Total # Posts: 1,924
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Is lithium hydroxide a strong or weak base? It is 100% ionized so, by definition, it is a strong base. Try looking up Kb for LiOH in a set of tables. It won't be there, further proof that it is not a weak base. thank you!
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April 13, 2007
math
how do you convert 2pie radians per min. to miles per hour? You cant, unless you have some way of relating the angle to distance. If this is circular motion, you need the radius.
April 11, 2007
math
how would u solve this: 2sin(x)-1=0 exact step to solving would be nice thanx Add one to both sides and divinde both sides by 2. sin x = 1/2 Now look up the angle that has a sine of 1/2. That's the answer what is a derpeciation expense The word derpeciation does not exist...
April 11, 2007
Foreign Language- Spanish
Hello, I just finished a couple of quizzes regarding nosotros commands (I have a quiz tomorrow on them) As I was completing the quiz, I noticed I got this incorrect: In context: "Vamos a decirlo" Therefore I changed it to digámoslo. However, a grammatical text...
April 2, 2007
Short translation Spanish
How would you say, 'that is just a fraction of many representations.' (an idiomatic expression yet again) Last post! Sorry and thank you so much for helping me on this oral. Thank you for using the Jiskha Homework Help Forum. "Eso es no más que una fracci&#...
March 27, 2007
Spnaish- Translation
Hi, I am trying to translate "It is still an everyday struggle to compete in this male dominated world, but there are many women who are defying adversity." Is this accurate: Es una problema para hoy a competir en este mundo de hombres, pero hay mujeres quien estan ...
March 26, 2007
Biology
Which of the following does not function is suspension feeding? This makes no sense. I am not sure what you are asking, since you did not indicate what is "following." I searched Google under the key words "suspension feeding" to get these possible sources...
March 18, 2007
physiology (nernst equation)
If cell depolarises at +50mV, what are the concentrations of K+? I am really confused how to use the nernst equation to get the concentrations. If I set up 58 [Kout]/[Kin] = 50, will that be right? I am not sure how to start this problem.
March 6, 2007
English
as a cell prepares to divide ,a dna molecule and its associated proteins
March 4, 2007
physics
In a television set, electrons are accelerated from rest through a potential difference of 18 kV. The electrons then pass through a 0.34 T magnetic field that deflects them to the appropriate spot on the screen. Find the magnitude of the maximum magnetic force that an electron...
February 26, 2007
physics
One component of a magnetic field has a magnitude of 0.046 T and points along the +x axis, while the other component has a magnitude of 0.066 T and points along the -y axis. A particle carrying a charge of +7.50*10^-5 C is moving along the +z axis at a speed of 3.20*10^3 m/s...
February 26, 2007
english - idea on how to conclude a paragraph?
My introductory sentence is this: My least favourite musical performer is none other than the princess of pop herself, Britney Spears, otherwise affectionately known as Brit Brit in some blogging communities. I talk about some of her 'wild antics' in the paragraph but ...
February 19, 2007
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### The 10 Most Important Physics Effects
Today I have a count-down of the 10 most important effects in physics that you should all know about.
10. The Doppler Effect
The Doppler effect is the change in frequency of a wave when the source moves relative to the receiver. If the source is approaching, the wavelength appears shorter and the frequency higher. If the source is moving away, the wavelength appears longer and the frequency lower.
The most common example of the Doppler effect is that of an approaching ambulance, where the pitch of the signal is higher when it moves towards you than when it moves away from you.
But the Doppler effect does not only happen for sound waves; it also happens to light which is why it’s enormously important in astrophysics. For light, the frequency is the color, so the color of an approaching object is shifted to the blue and that of an object moving away from you is shifted to the red. Because of this, we can for example calculate our velocity relative to the cosmic microwave background.
The Doppler effect is named after the Austrian physicist Christian Doppler and has nothing to do with the German word Doppelgänger.
9. The Butterfly Effect
Even a tiny change, like the flap of a butterfly’s wings, can making a big difference for the weather next Sunday. This is the butterfly effect as you have probably heard of it. But Edward Lorenz actually meant something much more radical when he spoke of the butterfly effect. He meant that for some non-linear systems you can only make predictions for a limited amount of time, even if you can measure the tiniest perturbations to arbitrary accuracy. I explained this in more detail in my earlier video.
8. The Meissner-Ochsenfeld Effect
The Meissner-Ochsenfeld effect is the impossibility of making a magnetic field enter a superconductor. It was discovered by Walther Meissner and his postdoc Robert Ochsenfeld in 1933. Thanks to this effect, if you try to place a superconductor on a magnet, it will hover above the magnet because the magnetic field lines cannot enter the superconductor. I assure you that this has absolutely nothing to do with Yogic flying.
7. The Aharonov–Bohm Effect
Okay, I admit this is not a particularly well-known effect, but it should be. The Aharonov-Bohm effect says that the wave-function of a charged particle in an electromagnetic field obtains a phase shift from the potential of the background field.
I know this sounds abstract, but the relevant point is that it’s the potential that causes the phase, not the field. In electrodynamics, the potential itself is normally not observable. But this phase shift in the Aharonov-Bohm Effect can and has been observed in interference patterns. And this tells us that the potential is not merely a mathematical tool. Before the Aharonov–Bohm effect one could reasonably question the physical reality of the potential because it was not observable.
6. The Tennis Racket Effect
If you throw any three-dimensional object with a spin, then the spin around the shortest and longest axes will be stable, but that around the intermediate third axis not. The typical example for the spinning object this is a tennis racket, hence the name. It’s also known as the intermediate axis theorem or the Dzhanibekov effect. You see a beautiful illustration of the instability in this little clip from the International Space Station.
5. The Hall Effect
If you bring a conducting plate into a magnetic field, then the magnetic field will affect the motion of the electrons in the plate. In particular, If the plate is orthogonal to the magnetic field lines, you can measure a voltage flowing between opposing ends of the plate, and this voltage can be measured to determine the strength of the magnetic field. This effect is named after Edwin Hall.
If the plate is very thin, the temperature very low, and the magnetic field very strong, you can also observe that the conductivity makes discrete jumps, which is known as the quantum Hall effect.
4. The Hawking Effect
Stephen Hawking showed in the early 1970s that black holes emit thermal radiation with a temperature inverse to the black hole’s mass. This Hawking effect is a consequence of the relativity of the particle number. An observer falling into a black hole would not measure any particles and think the black hole is surrounded by vacuum. But an observer far away from the black hole would think the horizon is surrounded by particles. This can happen because in general relativity, what we mean by a particle depend on the motion of an observer like the passage of time.
A closely related effect is the Unruh effect named after Bill Unruh, which says that an accelerated observer in flat space will measure a thermal distribution of particles with a temperature that depends on the acceleration. Again that can happen because the accelerated observer’s particles are not the same as the particles of an observer at rest.
3. The Photoelectric Effect
When light falls on a plate of metal, it can kick out electrons from their orbits around atomic nuclei. This is called the “photoelectric effect”. The surprising thing about this is that the frequency of the light needs to be above a certain threshold. Just what the threshold is depends on the material, but if the frequency is below the threshold, it does not matter how intense the light is, it will not kick out electrons.
The photoelectric effect was explained in 1905 by Albert Einstein who correctly concluded that it means the light must be made of quanta whose energy is proportional to the frequency of the light.
2. The Casimir Effect
Everybody knows that two metal plates will attract each other if one plate is positively charged and the other one negatively charged. But did you know the plates also attract each other if they are uncharged? Yes, they do!
This is the Casimir effect, named after Hendrik Casimir. It is created by quantum fluctuations that create a pressure even in vacuum. This pressure is lower between the plates than outside of them, so that the two plates are pushed towards each other. However, the force from the Casimir effect is very weak and can be measured only at very short distances.
1. The Tunnel Effect
Definitely my most favorite effect. Quantum effects allow a particle that is trapped in a potential to escape. This would not be possible without quantum effects because the particle just does not have enough energy to escape. However, in quantum mechanics the wave-function of the particle can leak out of the potential and this means that there is a small, but nonzero, probability that a quantum particle can do the seemingly impossible.
1. Isn't it true that in the Aharanov-Bohm effect the phase shift depends on an integral of a closed path and thus on the magnetic flux enclosed by the closed path? If so, the phase shift depends on the magnetic flux, i.e. an observable quantity, and not on the vector potential.
1. The integral over a field is a potential. And the whole point is that it's observable.
2. Whitewater:
The point here is that the experiment can be arranged so that the charged particle can only be present in the region where there is no magnetic (and no electric) field.
And, yet, despite the fact that there are no E and B fields where the particle can be, the motion of the particle is affected by the non-zero vector potential, which depends on the magnetic field in the region where the particle cannot go.
Superficially, it seems like action at a distance: it seems that the particle is being affected by the magnetic field in a region from which the particle is excluded. That is why the A-B effect was quite a surprise to physicists.
Speaking for myself, I still have trouble believing it is true, even though I understand the theoretical argument (which is almost trivial) and even though it has been confirmed experimentally.
Dave
3. Aharonov-Bohm paper is beautiful (1959, Physical review):
It is important to emphasize the "multiply-connected region." (page 490). Another paper begins: "the Aharonov-Bohm effect is simple and topological..." (Aharonov, et.al., Interplay of A-B and Berry Phases for Quantum Cloud of Charge, 1995). A beautiful paper by Waechter gives a path-integral derivation followed by a "topological perspective." Read: "for the Aharonov-Bohm effect to occur, we need a configuration space which is not simply connected." (2018 ETH Seminar).
4. I think you are missing a mathematical subtility: the fact that the electromagnetic field is null does not mean it does not exist.
It is the same enormity as saying that since the derivative of a function is null this function has no derivative.
See? Potential = function, field = derivative
5. The AB-effect can be explained in a fully gauge invariant and local way, but this requires a quantum mechanical description of the fields. When using classical potentials one will necessarily miss entanglement involving the fields:
https://arxiv.org/abs/1906.03440
https://arxiv.org/abs/1110.6169
2. Cool stuff ... One could do an honest count of how many effects one has known before and then post an answer here.
Me, for starters, I knew 7/10 for sure, and 1 I kind of knew, but did not know the name.
Nice collection anyhow!
3. I might expand the list to 12 with the inclusion of the Lense-Thirring effect and quantum entanglement. The Lense-Thirring effect is frame dragging of space by a large rotating mass. Quantum entanglement manifests itself in a range of phenomena from NMR to quantum optics.
1. Those are probably as real as The Hawking Effect.
2. Greg,
Your comments are very ill-informed and my patience with you is running low.
3. Quantum entanglement since the Aspect experiment has been found since the 1980s Aspect experiment. Frame dragging was measured with the Gravity-B probe and recently found in the orbit between a neutron star and white dwarf.
4. "Those are probably as real as The Hawking Effect."
Indeed; all are real.
4. Curious: electromagnetism-related effects dominate, and gravity is a bit player. The strong and weak nuclear forces are nowhere to be found (well, not explicitly).
Perhaps there's an observer bias, a kind of scale effect? At the scale of atomic nuclei, how many of these ten are notable?
2. Curious indeed.
3. There are plenty of nuclear effects (with my favorite being Moessbauer effect) but they are somewhat less general than the effects listed here.
5. Hello Dr. B. One typo. In effect 3, photoelectric effect, you wrote Einstein's bday as 1979.
1. Elver,
Yes, thanks, it's been pointed out to me by several people. There is no way to edit videos once they're published, so nothing I can do about it. Sorry about the blunder.
6. There is no experimental evidence for the existence of Hawking radiation. To claim that it is an important physical effect is disingenuous to say the least.
1. Norman,
2. To be fair, superfluids are simply an example of a system that is formally similar. Kinda like some general relativistic effects can be analogous to electromagnetic effects, with similar effects.
Nobody so far has measured Hawking radiation. I think Unruh radiation also has not yet been measured?
3. Cyberax,
What you say is not correct. The case of superfluids is not "formally similar" it is formally identical. It is the exact same effect. Hawking radiation has been measured in superfluids. Please check the reference I provided above.
4. We don't have any equations which combine gravity and quantum mechanics which have any experimental evidence to confirm them.
The geometry of space time in a superfluid is flat, the geometry around a black hole is not.
I cannot for the life of me understand how you can claim that these situations are "formally identical".
5. Norman,
"We don't have any equations which combine gravity and quantum mechanics"
How do you think Hawking did his calculation if he didn't have equations, huh?
"I cannot for the life of me understand how you can claim that these situations are "formally identical"."
The equations are identical.
6. Here is a view: "... physicists are impressed, but they caution that the results are not clear-cut. And some doubt whether laboratory analogues can reveal much about real black holes. “This experiment, if all statements hold, is really amazing,” says Weinfurtner, a theoretical and experimental physicist at the University of Nottingham, “It doesn’t prove that Hawking radiation exists around astrophysical black holes.” (Aug. 2016, Nature). Read Barcelo, Liberati, Visser: "analogy is not identity, and we are in no way claiming that the analogue models we consider are completely equivalent to general relativity — merely that the analogue model (in order to be interesting) should capture and accurately reflect a sufficient number of important features of general relativity..." (2011, Analogue Gravity).
7. You truncated my quote. I said that there is no evidence that any equation which combines gravity and quantum mechanics is correct. You may have some equation which purports to do that, but until you have experimental evidence all you have is speculation.
It may be true that you are using the identical equations in both circumstances but you have no experimental evidence that the application of that equation is correct.
Until we actually see Hawing radiation it will be no more than a prediction.
8. Gary,
No one claims that the observation of Hawking radiation in superfluids tells us that black holes also emit Hawking radiation. What I said is simply that Hawking radiation has been measured in superfluids.
9. Norman,
"It may be true that you are using the identical equations in both circumstances but you have no experimental evidence that the application of that equation is correct."
That is incorrect, as I already said. We have experimental evidence that the equations corretly describe superfluids, which demonstrates that the Hawking effect exists.
"You truncated my quote. I said that there is no evidence that any equation which combines gravity and quantum mechanics is correct."
Your statement is incorrect whether truncated or not. We do full well have evidence that equations combining quantum mechanics with gravity are correct.
I have the impression you do not know what you are even talking about. As I have said many times, not anything that has something to do with "quantum" and something with "gravity" has also something to do with "quantum gravity". Hawking radiation is not a quantum gravitational effect.
10. Sabine reminds me: "No one claims that the observation of Hawking radiation in superfluids tells us that black holes also emit Hawking radiation. What I said is simply that Hawking radiation has been measured in superfluids." However, does not Steinhauer, et al. make that claim ? As they say: "We find that the correlation spectrum of Hawking radiation agrees well with a thermal spectrum, and its temperature is given by the surface gravity, confirming the predictions of Hawking’s theory. The Hawking radiation observed is in the regime of linear dispersion, in analogy with a real black hole, and the radiation inside the black hole is composed of negative-energy partner modes only, as predicted." (2019, Nature). My previous post was merely intended to accentuate the all-important word "...analogy..."
11. An experiment proposes to rotate a tiny object with the polarization of a laser beam may detect a variant of this. An object in extreme rotation has its endpoints on an accelerated frame. This then results in a form of vacuum friction that is similar to Unruh radiation. This means on that frame the vacuum appears with a thermal bath of photons. From our inertial frame perspective the energy in the rotation of this object is lost. The object heats up and transfers the energy of angular momentum, and angular momentum, outwards as thermal photons. So there is a sort of friction of the vacuum that heats the spinning body up.
https://www.newscientist.com/article/mg20927994-100-vacuum-has-friction-after-all/
12. Gary,
The paragraph you quote says correctly that they have observed Hawking radiation in superfluids, that this confirms the prediction of Hawking's calculation, and that this situation is in analogy to black holes.
13. I was confused, and the source of my confusion was your description of the Hawking effect. As a mathematician I assumed that this had the status of a definition and so was surprised that you conflated two distinct effects under one name.
Thus what I should have said is that there are no observations which show black holes radiating. There is no indirect evidence for black holes radiating such as the indirect evidence for the existence of gravitational waves provided by the Hulse-Taylor binary pulsar.
I understand that you are claiming that an observation in a superfluid is evidence for the Hawking effect in black holes because it is an analogy.
My analogy is this: In the year 1880 an eminent natural philosopher blogged about the most important effects in their field of study. This list included the Michelson effect, which allows you to measure your absolute velocity by measuring the displacement of interference patterns. When asked about evidence for this effect this eminent natural philosopher replied that there was ample evidence as the effect had been demonstrated in gravity waves on the surface of water, in sound waves in air and in compression waves in a superfluid. Since all three of these situations obey exactly the same equation as case with light we can safely conclude that the effect is real and once the apparatus has been refined to be sensitive enough we will know our absolute velocity.
I feel that this is an important point becuase like you I feel that parts of physics have lost their way. Where we differ is in where that is. I believe that the point of failure was when theoretical results were accepted as true without experimental evidence. Howking radiation is the most egregious example of this failure.
7. Sabine,
concerning the Casimir Effect, I remember there were doubts that the vacuum energy is necessary to explain it e.g. https://link.aps.org/doi/10.1103/PhysRevD.72.021301:
>>Casimir forces can be calculated without reference to the vacuum and, like any other dynamical effect in QED, vanish as α → 0. The vacuum-to-vacuum graphs …do not enter the calculation of the Casimir force, which instead only involves graphs with external lines.<<
1. Stefan,
I am thoroughly unimpressed. One can calculate the effect, measure it, and that's the scientitif part. Then one can try to put words to the math. There will be more than one way to do the calculation and the words will always remain ambiguous. I therefore do not see what one learns from quibbling about them.
This of course is not specific to the Casimir effect. There is also more than one way to calculate and interpret the Hawking effect and many other quantum effects, depending on what technique or interpretation you use. What do we learn from that? Well, personally I'd say it means that structural realism makes not sense, but I'll leave that to philosophers to debate.
2. Sabine,
why do you object to the other explanation of the Casimir effect?
Literature says that the Casimir effect can also be explained as a van der Waals force. This avoids the use of vacuum energy. This is better because we know about vacuum energy that there is the discrepancy between QM theory and measurement of after all 120 orders of magnitude.
The van der Waals force is also in so far a better explanation as it is more classical physics and as such fits into a reductionistic system in contrast to Quantum Mechanics which do not.
3. antooneo,
I do not object to it. You misunderstood my comment.
4. When ships are in harbor precautions have to be taken to prevent the ship from banging into the harbor front. The reason is the waves between the ship and a seawall of a dock have a discrete 1/2 set of wavelengths. One the other side waves interacting with the ship have a continuum of wavelengths. Consequently the ship experiences a force towards the dock. That is one reason you see buffers on the dock to keep the ship from impacting the dock hard. This is a water wave analogue of what happens in the Casimir effect with the ZPE vacuum.
5. Lawrence Crowell ,that reminds me that a herd of whales has never caused a storm or ever, nor in theory; so, I don't know where butterflies get their great energies organized to create cyclones, making fun of all the Automation engineers
6. The so called Butterfly effect is just a heuristic statement that two systems nearly identical with a small separation in their initial conditions will separate. The two systems will at some time, the reciprocal of the Lyapunov exponent term, end up in states that have no apparent similarity. The "flapping butterfly" is a description of the tiny initial separation in initial conditions.
7. Thank you very much for your response, I agree with you, Sabine made a post that explains very well exactly what you say; but for me, chaos is usually temporary; the tendency of almost everything is to the balance; notice that a cyclone is precisely to restore thermal equilibrium; beyond a certain temporal chaos, systems tend to balance, they can have parts that evolve distinctly that break the balance; but in the end a new organization is established with a new balance. Chaos stands out precisely because it happens in a broad system in equilibrium; I think that space is a system in balance, the stars are also entities in balance. I am not saying that behind this there is nothing subjective; It is only a very general opinion.
8. ChaoChaos does not mean violence. The solar system has chaotic dynamics. The breakdown of predictability occurs on a very long time scale of 10s to 100s of millions of years. I did work on some of this, and did a numerical fast Fourier transform of this behavior and it has a remarkably 1/f signature. This 1/f noise spectrum occurs with the quantum states of electrons in condensed matter. This in the case of solid state physics has some possible connection to the Hossenfelder-Palmer idea of chaos in QM.
Nature does not quite go to equilibrium for systems that are open. Open systems with a small amount of energy flux exhibit quite complex and dynamical behavior. Planetary science has revealed some of this. The distant planets Uranus and Neptune are very cold due to a much reduced solar irradiance. They receive about 1/100 times the solar irradiance on Earth, about 1200watts/m^2. Yet interestingly these planets have incredible winds in their atmospheres and are far from being cold dead worlds. Even Pluto, what we thought was as dead as the ancient god it is named for, shows signs of considerable dynamics.
s
9. Lawrence Crowell
Thank you for your response and your patience, I realize that the chaos used in physics has a different meaning than the literal chaos, yes, chaos is not violence; Nor does balance mean immutable; in the case of the stars I was referring to the fact of, for example; all the particles that make up the Earth can have infinite configuration and correlations with each other; but the shape of the Earth and its distribution of density will not change almost; of course any particle within it can receive an energy that causes it to leave Earth; but below that energy the system is almost self-contained, it does not work as a set of particles with independent movement. In the case of an entangled electron-positron pair, I do not believe that the symmetric evolution between them is broken, unless the symmetry of the path between them is broken; the fact that in a Bell experiment the state of both particles can be correlated tells me that there is no arbitrary or chaotic evolution; if that correlation is broken later, I suppose the system does not become chaotic; It continues to evolve under the same rules, only that it is out of sync, I suppose. Remember, it is only the opinion of a neophyte,Be patient, hahaha, thank so much.
8. I read somewhere that the red shift of the CMB and distant stars is not really akin to the Doppler effect because it is not the stars that are moving but the space between us and the distant stars is growing and therefore the wavelength of the light is "streched". Please enlighten us. Thanks.
1. InConstruction,
That's correct. What I was referring is not the cosmological redshift of the CMB, but the distortion of the CMB due to our motion relative to it.
2. The cosmological redshift is not a Doppler effect in almost all senses (for the exception, see Bunn & Hogg). This was cleared up decades ago by Harrison, though many popular accounts still get it wrong. (Some professionals also get it wrong, but it doesn't matter that much since recession velocities, like the acceleration of the universe, are not directly observed and, while they can be calculated, there is little use for this.)
A topic about which there was genuine confusion was that of cosmological horizons, which was cleared up by the late, great Wolfgang Rindler in a seminal paper. One still occasionally encounters people confused on that topic.
The most recent example of cleared-up confusion in cosmology involves the traditional flatness problem; see the paper by Holman.
Bunn & Hogg: "The kinematic origin of the cosmological redshift"
Harrison: "The Redshift-Distance and Velocity-Distance Laws"
Rindler: "Visual horizons in world models"
Holman: "How Problematic is the Near-Euclidean Spatial Geometry of the Large-Scale Universe?"
All of these were published in leading journals in the field. I've linked to arXiv if a freely accessible version is not available at the official source or at ADS.
3. @Phillip Helbig:
I for 1, think that, the idea "CMB photons are MW (instead of gamma?) because their wavelengths got stretched on the way by expansion of Universe" is wrong!
I think, those photons are actually still the same (gamma?) photons, but we detect them as MW because of Relativity (imagine CMB sphere (all around) as a wall, that is moving away from us at relativistic speed)!
What I am really wondering is, if this view assumed to be correct, would it change Hubble constant calculation result using CMB, or not?
(Currently, a huge problem in astrophysics is the mismatch between 2 calculated values of Hubble constant, using Distance Ladder & using CMB!)
4. I for 1, think that, the idea "CMB photons are MW (instead of gamma?) because their wavelengths got stretched on the way by expansion of Universe" is wrong!
I think, those photons are actually still the same (gamma?) photons, but we detect them as MW because of Relativity (imagine CMB sphere (all around) as a wall, that is moving away from us at relativistic speed)!
It depends on the definition. Usually, the energy (or, equivalently, frequency or wavelength) of a photon determines what it is called (e.g. ultraviolet, gamma, radio). Sometimes the process does (e.g. gamma: nuclear transition; X-ray: deep atomic transitions, visual: normal electronic transitions, infrared: thermal etc). Of course, this way there can be overlaps in frequency between different sources. The photons are certainly cosmologically redshifted (not really a conventional Doppler shift).
What I am really wondering is, if this view assumed to be correct, would it change Hubble constant calculation result using CMB, or not?
Not.
(Currently, a huge problem in astrophysics is the mismatch between 2 calculated values of Hubble constant, using Distance Ladder & using CMB!)
5. I was going to write a detailed comment on this matter of "expanding space," but Phillip Helbig saved me the trouble with his link to the Bunn and Hogg paper.
I urge everyone to read their paper: it would have saved me a lot of confusion four or five decades ago if it had been available then; instead, I had to (painfully) figure it all out for myself.
One subject I wish Bunn and Hogg had expanded upon is the "Milne universe": anyone who wishes to pontificate on General Relativity needs first to check to see if what he or she is saying is true in the Milne universe.
The Milne universe is really just (part of) flat Minskowski space: it has space-like cross-sections that are just flat Euclidean geometry and a universal time coordinate. Indeed, Milne universe is only the part of Minkowski space lying inside the future light cone of the origin, so each space-like cross-section is just the inside of a finite ball with Euclidean geometry. Space is finite at any point in time.
And yet... if you use "co-moving" coordinates, including so-called "cosmic time," then each space-like section is an infinite non-Euclidean (hyperbolic) space.
So... is space flat or curved in the Milne universe? It is, as Einstein would have said, "relative."
Anyone who does not understand this does not really understand General Relativity, and, in particular, how the choice of coordinate systems can make one universe present itself in very different ways.
6. The paper by Bunn and Hogg should be required reading. Read: "Any statement in which “now” is used to refer to the present cosmic time at the location of a distant object is not about anything observable, because it refers to events far outside our light cone" and they conclude their paper with this:
"The common belief that the cosmological redshift can only be explained in terms of the stretching of space is based
on conflating the properties of a specific coordinate system with properties of space itself. This confusion is precisely
the opposite of the correct frame of mind in which to understand relativity."
9. Most of the posted are conveniently measurable. Which is to say they can be demonstrated in many well equipped university laboratories.
The Aharonov–Bohm Effect: it seems that my education has been sadly neglected. Of the posted effects The Casimir Effect has always been the most mysterious to me. It took a real reordering of my thinking to accept.
Thank you for the presentation.
10. I'm puzzled though how the word "effect" becomes attached to the names of some phenomena and not (perhaps) to others.
"This is a list of names for observable phenomena that contain the word effect":
https://en.wikipedia.org/wiki/List_of_effects
1. In a similar vein, I was wondering what sort of phenomena would qualify for your list. Does it have to be something that we currently refer to by the name "the * effect" ? In that case really important observations such the "laws" of electricity and magnetism (Coulomb, Ampere, Faraday, Ohm ...), the Michelson-Morley experiment and many more are not eligible because we don't call them "effects". Is that right?
Anyway, I have another candidate: the Edison effect. Actually, that name is rather dated. Most people nowadays call it "thermionic emission", and so it may not be eligible either. It's the phenomenon on which vacuum tubes are based. These lists always contain a significant dose of personal preference - that's what makes them fun. I enjoyed your list, but in my own world, the Edison effect rates above the tennis racket effect.
2. Philip Thrift: you are more than welcome to do a video about ALL those effects. Thank you.
3. I suppose from my particular philosophy-of-science (neopragmatism) view there are no "effect"s. (So the list would be 0.)
11. "A closely related effect is the Unruh effect named after Bill Unruh, which says that an accelerated observer in flat space will measure a thermal distribution of particles with a temperature that depends on the acceleration. Again that can happen because the accelerated observer’s particles are not the same as the particles of an observer at rest."
Another way of describing it: shake a thermometer in a vacuum and the measured temperature will increase. Nomen est omen: "Unruh" in German means a state of unrest, particularly appropriate here.
Although not in the Nomen est omen category, there are some papers by Daniel Holz and Bob Wald. "Holz" and "Wald" mean "wood" and "forest" in German. :-)
12. I have known about the Aharonov-Bohm effect for years, in the sense of having heard about it, but not really knowing what it was. So, last night I printed out a paper titled "A macroscopic test of the Aharonov-Bohm effect" by Adam Caprez, Brett Barwick, and Herman Batelaan, with the intent to read it while propped up in bed. As I expected, to truly understand the effect, comprehending the math is essential. And that requires coaxing the brain cells out of their natural laziness mode. Fortunately, Youtube has a smorgasbord of videos on the effect, which I just discovered this morning. Such visual-audio presentations I find easier to follow than dry words and equations on a sheet of paper.
13. Regards tunneling, years ago Condon and Gurney wrote a beautiful paper: "Quantum Mechanics and Radioactive Disintegration" (1929, Physical Review) where read "in quantum mechanics most statements of certainty are replaced by statements of probability" and "Now we throw the whole responsibility onto the laws of quantum mechanics, recognizing that the behaviour of particles everywhere is equally governed by probability." Razavy makes a claim: "we can describe the tunneling phenomena either partially or completely in terms of classical dynamics provided that we replace the simple system by a complicated interacting system." (page 139, 2003, Quantum Theory of Tunneling).
14. Funny thing, I thought some Youtube videos on the Aharonov-Bohm (AB) effect would be more transparent in explaining this phenomena than a text only presentation. Turns out the Wiki article on the effect was the easiest to follow, for someone at my level of knowledge. Even before I read the Wiki article, or comments on this post, I wondered if this effect was an illustration of quantum mechanical non-locality, such as embodied in quantum entanglement. That, in fact, was the argument advanced by Vaidman, in the Wiki article. In some sense that almost seems easier to comprehend than invoking the vector potential of the confined magnetic field in the solenoid version of the AB experiment. To be honest I don’t really understand what is meant by “potential” in this context. It’s weird that something is ‘there’, outside the solenoid, but can’t be measured like electric or magnetic field lines. Anyway, I started to read the Wiki article on the “Electromagnetic Four Potential”. Maybe that will help clarify things.
1. David Schroeder wrote:
> Even before I read the Wiki article, or comments on this post, I wondered if this effect was an illustration of quantum mechanical non-locality, such as embodied in quantum entanglement. That, in fact, was the argument advanced by Vaidman, in the Wiki article. In some sense that almost seems easier to comprehend than invoking the vector potential of the confined magnetic field in the solenoid version of the AB experiment.
As I said earlier, the A-B effect seems to violate locality, but, of course, it cannot change the electrons' path in a way that violates the no-signalling theorem. I.e., you cannot simply turn on the solenoid and get an instantaneous shift of the interference pattern, faster than light.
Somehow, there must be a not-faster-than-light way of explaining the A-B effect, though I have never seen it clearly done (it's not a high priority for us theorists, because, after all, the standard analysis is correct).
Dave also wrote:
> To be honest I don’t really understand what is meant by “potential” in this context.
Ut started out as a convenient mathematical technique for analyzing magnetic fields. It is a theorem that, because div B = 0, there must be another vector field A such that B = curl A.
In solving problems, it is sometimes useful to work with A rather than B.
And then it turns out that "momentum" in a certain sense in classical mechanics is not just mv but rather mv + qA. The "certain sense" has to do with the Lagrangian and Hamiltonian formulations of classical mechanics, and all that happens to be relevant to quantum mechanics.
And so in quantum theory, when you should have p^2/(2m), you actually need to subtract off the qA term to get (p-qA)^2/(2m). Then you remember that p in quantum mechanics is actually the gradient operator, and you start to see how qA interacts with the derivative.
A long chain of reasoning, each step of which was known to be necessary before the discovery of the A-B effect.
Any student of quantum mechanics is supposed to internalize everything I just mentioned, and then the A-B argument becomes pretty obvious.
By the way, the "modern" approach is to take the connection between qA and the derivative as the primary idea, not the long chain going back to classical mechanics. In that case, you have the basic idea of a "gauge theory," a so-called "abelian" gauge theory, which, to simplify, just means your A field is a number, not a matrix: non-abelian gauge theories, such as for QCD, have vector potentials represented by matrices.
So, if you truly get clear on everything about the A-B effect, you have grasped a lot about modern physics!
All the best,
Dave
2. @PhysicistDave,
Thank you for lengthy reply to my second post. I confess I got lost in the details, but the deep significance of gauge theories as they relate to the modern Standard Model is something I’m determined to understand through and through. I believe I understand the basic gist of it. For example, Maxwell’s formulation of the EM field was the first gauge theory discovered (going from memory from the Wiki article). In that article, which I was reading this morning, some quantity is conserved. For the EM field I assume it would be the speed of light. In Maxwell’s equations the interplay of the E and M fields constrains the speed of light to c (vacuum permittivity and permeability fixing the actual magnitude of c). This process is what I always thought was meant by “gauging” a field. Anyway I’ll try to make sense out of what you wrote with the various connections. Thank you again for this help. I’ve definitely got an uphill climb in knowledge absorbtion but am determined, as any mountain climber, to reach the summit, or at least high enough to see the physics landscape clearly.
I wrote the first version of this at a coffee shop on my notebook computer, but hopefully it didn’t get posted there.
3. Dave Schroeder wrote to me:
> I confess I got lost in the details...
Of course. I just wanted to give you hints as to what to look into if you want to know more and also to indicate that the historical development is rather roundabout.
Dave also wrote:
> I believe I understand the basic gist of it. For example, Maxwell’s formulation of the EM field was the first gauge theory discovered (going from memory from the Wiki article). In that article, which I was reading this morning, some quantity is conserved. For the EM field I assume it would be the speed of light. In Maxwell’s equations the interplay of the E and M fields constrains the speed of light to c (vacuum permittivity and permeability fixing the actual magnitude of c). This process is what I always thought was meant by “gauging” a field.
No, when you are looking at the E and B fields, you are not really clearly seeing the gauge-field aspects (there is a sense in which the E and B fields represent the curvature of the gauge fields). It is the vector potential (along with the scalar potential) that embodies the "gauging."
The fact that you get the speed of light out of the E and B fields is now viewed as a rather trivial matter (of course, it was not trivial historically!), since everyone knows the speed of light is just 1.
The term "gauge theory" goes back to an idea from Hermann Weyl early in the twentieth century in which the unit of measure (hence the "gauge") varied from point to point in such a way that after you went on a trip and came back home, you might be a different size! As goofy as this sounds (and of course it cannot work), this is the origin of the idea.
Weyl and others later modified the idea to refer not to linear measurements but to the complex phase of the quantum wave function, so that if you went on some path and returned to your starting point via a different path, you could suffer a net change in complex phase. It was realized that this is what happens with electromagnetism in quantum mechanics.
And from this comes the A-B effect.
If instead of just a change in complex phase, we have some internal symmetry that can be represented by matrices, we get the non-abelian gauge theories such as QCD. "Non-abelian" just means "non-commutative" and refers to the fact that matrix multiplication does not commute.
All of this would have seemed quite bizarre to Maxwell, of course, but then Maxwell did not know quantum mechanics. Nowadays, this rather radical reformulation of Maxwell's equations is viewed as the "right" way of thinking about not just electromagnetism but also the strong and weak forces.
Dave
4. If I may kindly interject, a nice overview of the physics and history of gauge invariance can be found in the AAPT Resource Letter, Gauge Invariance, by Cheng and Li (1988). Jackson and Okun later wrote a beautiful article (2001) "Historical Roots of Gauge Invariance." (arXiv:hep-ph/0012061). Both of these resources are pedagogical and include copious references.
5. @PhysicistDave,
I was very surprised, and delighted, to see your post this morning waking up in the snow blanketed, frozen wonderland of southwest New Hampshire, figuring you wouldn’t read it in California till late morning EST, with the 3 hour time difference. Off topic, I know, but California is an awesome place. I visited the state many times in the 80’s and 90’s with friends, who eventually moved out there. When they bought a home in San Francisco’s Marina district we pedaled our bikes from Broderick St., over the Golden Gate, and up Route 1 to a spectacular viewing area. On 3 separate rides I pedaled from their home to the summit of Mt. Tamalpais and back. They now live in Illinois, nowhere as scenic as California.
My very first exposure to the gauge concept was probably a 1980’s physics popularization book that described it in terms of the gauge measuring blocks used in machine shops, as you alluded to. Oddly, I woke up this morning with that gauge block connection in mind. That immediately made me realize I was missing something in my incorrect interpretation of the concept that I posted yesterday. But on reading your post it wasn’t necessary to dig up that book, as now I’m being steered in the right direction. Still this is going to take some thinking to understand it to my satisfaction. Sorry for being somewhat incoherent, but I’m under time pressure (going to a Connecticut casino with twin brother), and need to get some winks, to make up for too much caffeine before hitting the rack last night. I’m printing out the Gauge Theory page on Wiki, right now, and will read it on the second leg of our journey, after I arrive at my brother and his wife’s house, when he starts driving.
6. @PhysicistDave
Your synopsis of the twists and turns in the history of gauge theory was very helpful to me, clearing certain things up. There’s quite a bit here to digest, and I’ll be studying it for a while before I get back to you. Thanks again. Plus, I just noticed the link Gary Alan provided above that sounds like it might also help in bridging the gap between a layperson's understanding and a more formal appreciation of this extremely important topic.
7. The Wikipedia article "introduction to gauge theory" (the nontechnical account) is replete with misconceptions. Cheng and Li emphasize that "gauging" involves changing from a global symmetry to a local symmetry. They emphasize that the form of quantum electrodynamics is completely fixed by gauge invariance and renormalization. Gerard 't Hooft's twenty-page Scientific American article (1980): Gauge Theories of the Forces Between Elementary Particles, attributes to Weyl the idea of "machinists blocks." Robert Crease writes: "One of the biggest unsolved challenges of Yang–Mills theory is simply finding ways for outsiders to get a hint, at least, of the stunning achievement it represents." O' Raifeartaigh: "the emergence of gauge theory has been a gradual process, a slow evolution rather than a revolution." (Introduction, Dawning of Gauge theory).
8. Gary Alan, thanks for alerting us about the issues with the Wiki article "introduction to gauge theory". As I read Gerard 't Hooft's 1980 Scientific American article on gauge theories (online version), I knew I had read it before, long ago, as I recognized the diagrams in the article. Searching my collection of Scientific Americans, dating back to 1978, I couldn't find that June, 1980 issue. But the online version is just fine.
15. Also, an excellent non-specialist review of these misconceptions: https://arxiv.org/abs/astro-ph/0310808
1. This comment of mine was supposed to be a response to the comments of InConstruction and Philip Helbig. I dont know why it appeared at the end.
16. Dr. Hossenfelder,
Forgive me for not letting the discussion on Hawking Radiation go, it would appear that it is getting a bit trying. But it seems to me that there might be a loose end. In reading your link the words "irretrievably lost" are used to describe the entangled particle inside the black hole. How do we know that this particle and its information, communicated via entanglement, is truly lost? I ask this because of the entanglement the "information exchange" should be instantaneous. But we do not know how this instantaneous information exchange occurs so how can we conclude that the information is in fact lost. We do know that under standard speed of light info exchange the info would not be able to escape. But with entanglement speed of light info exchange does not occur. Thank you for you time Dr. Hossenfelder, I really enjoy your writing.
1. Steve,
"But we do not know how this instantaneous information exchange occurs so how can we conclude that the information is in fact lost"
There is no instantaneous information exchange in quantum mechanics.
Having said that, we do not of course know whether information is indeed lost in a black hole. What we do know is that if general relativity remained valid (no quantum corrections) it would be lost.
17. A nice thing about Sabine's list of "effects" is how it compels one to continue researching. In case it has been forgotten, many years ago (1968), Lamb and Scully published a fascinating paper which concluded thus: "The introduction of the photon concept is neither logically implied by nor necessary for the explanation of the photoelectric effect." (The Photoelectric Effect Without Photons).
18. Regarding whether the Hawking effect has been observed experimentally... The superfluid and other acoustic black or white hole analogue experiments certainly confirm experimentally that the math behind the Hawking effect is correct. Not that there was much doubt about that, but still nice.
That said, we don't know whether the semiclassical approximation that is used to derive thermal radiation from horizons is a good one. It might work, or other effects might become important. We won't know until and unless we are able to measure it in situ.
One example where I can see the Unruh effect might be breaking down is that to create an accelerating detector one has to apply force to it, which, in turn, changes the spacetime geometry, potentially enough to swamp the very weak Planck-level thermal bath.
19. To this collection of physical effects one should add
- the Quantum Zeno Effect (QZE)
- the astrophysical Sunyaev-Zel'dovic Effect (SZE), the boosting of CMB photons in clusters of galaxies
to round-up the dozen.
20. In your video when explaining the AB effect you should not use the same symbol(small phi) for both the phase shift and the magnetic flux in your formula; so correctly label the magnetic flux Phi(B)to avoid confusion arising for people which are less acquainted with the AB effect.
1. Dang, you are right. I hadn't paid any attention to this. (It's such a pain to get equations and special symbols into a video editing program.)
21. Photo-electric effect and "recording" at the quantum level, is there a connection? I am only wondering aloud.
22. Fun. Thanks. I know the tennis racket effect, does it have any use in physics besides just being fun? For additions I'll add the Hanburry-Brown effect. Interference of 'independent' photons. (or other things)
1. In a previous blog (Part 2 by Tim) I noted that I, personally at any rate as there were no further comments, saw a connection between the tennis racket effect and Tim's use of the Lorenz 'butterfly' diagram as they both had bistable patterns. This could, in a maybe-calculable way, lead to two entangled particles occupying the butterfly wing positions so that the wing positions of the two particles allowed pseudo-randomly opposed but coordinated measurements by Alice and Bob. This could be relevant to a Bell Test simulation. I wouldn't support it though until I saw a calculated simulation. And isn't a bistable state rather a special state for a Bell Test? Are particles spinning about intermediate axes? :)
Austin Fearnley
23. A curious fact about the AB-effect : it has been considered as a possible explanation for "scalar waves", a subject pioneered by Tesla but considered by many to be pseudo-science. One entry point to start making your own opinion about this stuff can be found on the CIA's web site (!) where also connections with supposed cold-war era secret Soviet weapons are discussed : https://www.cia.gov/library/readingroom/docs/CIA-RDP96-00788R001900680014-4.pdf
24. The CIA's report is a bit old (1984) but I have just stumbled on more recent stuff along these lines in the book "scalar wave driven energy applications" (Springer, 2019).
I have no idea whether this is serious science, but if someone has a (serious) opinion, do let us know! The author (Bahman Zohuri) seems to have a serious-sounding position in a U.S. university (associate research professor at the University of New Mexico).
1. A prominent Tesla replicator has given our open source group a number of metal samples that have been exposed to tesla waves. We are currently characterizing the effects of this exposure using detailed microscopic and SEM examination. The surface of the metal seems to be impacted in amazing ways by polariton petal condensation that has changed the structure of the metal consistent with the topology of these condensates. Videos of the results of this inspection and some opinions on causation are available on the internet.
25. Dr. Hossenfelder,
This maybe outside of the scope of this post but since this post started me looking into the Fulling-Davies-Unruh Effect I thought I might as well ask my question. My two biggest take aways from reading about the Fulling-Davies-Unruh Effect is virtual particles and accelerated reference frame. Considering this with Sergei bring up the the semi-classical approximation of the thermal effects a question came to me; Could there be a synergistic effect between Fulling-Davies-Unruh Effect and Hawking effect? The Fulling-Davies- Unruh effect requires an accelerated reference frame, does this acceleration include change of direction or is it limited to linear? Any curved motion by a black hole with respect to its motion around the center of a galaxy represents a change of direction and a classical accelerated reference frame. If this is good enough acceleration for the Fulling-Davies-Unruh effect this could represent an increase in virtual particles that could then transition into Black Hole Hawking Effect. I understand that the thermal aspect is very, very minimal, but the increase in particles from the Fulling-Davies-Unruh Effect could work with the Hawking Effect. Thank You!
26. (off- topic) sorry "the love lasted so little", readers and Sabine. due to a minor stroke I am no longer able to type, sign, my name, tie my shoes, hold a soup spoon.
thank you putting up with me both here and on you Tube.
I will miss you.
1. Ivan,
Jesus, sorry to hear. I hope someone is taking care of you. We will miss you too!
2. Sorry about you stroke. I have a bit of a problem myself with what are called giant cell tumors. that is what they are called, no big medical term. These are transformed cells, though fortunately not malignant, of tendon sheaths in the hand This has begun to seriously limit my dexterity. Some anti-tumor drugs are having some limited positive effect.
27. Ivan,
My wife and I are both cancer survivors from more than 10 years ago. That was a bleak time for us, and no doubt things seem bleak now to you. But judging by your post you seem like someone with a very strong mind. I suspect and hope you will be back soon. Best wishes for a speedy recovery.
Steve
28. I think an important physics idea that doesn't get enough attention is the the running parameter "effect". That is, quantities like mass and charge of a particle change with your resolution scale and how it does depends on the theory (specifically, the renormalization group).
Renormalization is often just explained as a mathematical trick to get rid of infinities, but it actually comes from deep physical insights. For example, you don't measure the "bare" charge of a particle because you have a cloud of virtual particles surrounding it screening the charge. How much of the cloud you see will affect what charge you measure.
The above maybe is a bit of an oversimplification, but re-normalization is extremely important to modern theoretical physics.
29. Sie haben den sehr wichtigen "Pauli-Effekt" nicht erwähnt.
-drl
1. I think that the question for the Pauli Principle is a good one. Because it is a typical point in quantum mechanics. I have asked repeatedly in discussions whether we meanwhile have an idea why it is valid. The usual answer is that this question is not justified because this is a fundamental rule in physics where we cannot ask for a cause. – I see here the typical denial of reductionism; quantum mechanics declares something as a “principle”, and that’s it.
2. antooneo,
I believe you misunderstood this comment. The Pauli-effect is not the Pauli-exclusion principle.
And btw, it was #11 on my original list from 12 years ago.
3. Sorry, I have really mixed both. And I never heart about the "Pauli-Effekt" before . That sounds to me a bit like Murphy's law. And 12 years ago I did not have knowledge about your blog.
But anyway, the Pauli principle is in my view also a somewhat open point in present physics; open as not explained.
30. "judging by your post you seem like someone with a very strong mind."
Thanks, Steve and Sabine.
"fiercely lefty" kind of loses meaning when lose fine motor control on left hand!
on the other it is a new experience to me to try to write with the pen upside down and take a soup spoon to my mouth without being to register both that pen is upside down, and spoon shell is down!
I just woke up like this 3 weeks ago. That' what I didn't need: another Ivan to handle.(:-)
Coming attractions: Fiercely Righty Ivan...
31. Sorry to hear about your stroke. There's research into stem cell therapy for people who have suffered a stroke, as described at the link below. My brother tore cartilage in his knee joint from a jump off a boat bow 3 feet above the dock, mistiming the swells. His knee completely healed after new cartilage grew in the damaged area.
https://www.cryo-cell.com/the-benefits-of-banking/cord-blood-cord-tissue-research/stroke-motor-function-treatment-stem-cells
1. I forgot to mention that my brother had his stem cell therapy for his knee performed at a clinic in Colorado.
32. Thank you, this is the type of post I enjoy most because it educated me a little more about things I didn't know.
33. The photo-electric effect used to be my favorite; but I think these days, it's the Ahranov-Bohm effect. Thanks for giving it the thumbs-up.
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# Homework 4 - vu(tv2894 – Homework 4(Quest – miner...
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Unformatted text preview: vu (tv2894) – Homework 4 (Quest) – miner – (55096) 1 This print-out should have 6 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points If the function f is continuous everywhere and f ( x ) = x 2 − 25 x + 5 when x negationslash = − 5, find the value of f ( − 5). 1. f ( − 5) = − 10 correct 2. f ( − 5) = − 5 3. f ( − 5) = − 25 4. f ( − 5) = 10 5. f ( − 5) = 25 6. f ( − 5) = 5 Explanation: Since f is continuous at x = − 5, f ( − 5) = lim x →- 5 f ( x ) . But, after factorization, x 2 − 25 x + 5 = ( x − 5)( x + 5) x + 5 = x − 5 , whenever x negationslash = − 5. Thus f ( x ) = x − 5 for all x negationslash = − 5. Consequently, f ( − 5) = lim x →- 5 ( x − 5) = − 10 . 002 10.0 points Determine which of the following could be the graph of f near the origin when f ( x ) = x 2 − 7 x + 10 2 − x , x negationslash = 2 , 4 , x = 2 . 1. 2....
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## ››More information from the unit converter
How many kilopond in 1 newton? The answer is 0.10197162129779.
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The SI derived unit for force is the newton.
1 kilopond is equal to 9.80665 newton.
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## ››Quick conversion chart of kilopond to newton
1 kilopond to newton = 9.80665 newton
5 kilopond to newton = 49.03325 newton
10 kilopond to newton = 98.0665 newton
15 kilopond to newton = 147.09975 newton
20 kilopond to newton = 196.133 newton
25 kilopond to newton = 245.16625 newton
30 kilopond to newton = 294.1995 newton
40 kilopond to newton = 392.266 newton
50 kilopond to newton = 490.3325 newton
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## ››Definition: Kilopond
The SI prefix "kilo" represents a factor of 103, or in exponential notation, 1E3.
So 1 kilopond = 103 ponds.
## ››Definition: Newton
In physics, the newton (symbol: N) is the SI unit of force, named after Sir Isaac Newton in recognition of his work on classical mechanics. It was first used around 1904, but not until 1948 was it officially adopted by the General Conference on Weights and Measures (CGPM) as the name for the mks unit of force.
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# When I am hungry
• 21 months ago · Quote · #1
nobody likes to be manipulated
• 21 months ago · Quote · #2
This puzzle is from a very old book and just here the page is damaged - the rest of the solution is unreadable . I am working on the case . Somebody to give a help ?
• 21 months ago · Quote · #3
Well, black can't really play g2 here, but maybe that's his only try...
21 ...g2
22. Qd3+ Ka2
23. Qd5+ Kb1
24. Qxg2
• 21 months ago · Quote · #4
[COMMENT DELETED]
• 21 months ago · Quote · #5
Ok, without playing ...g2, I think I've figured out how to force the dark-squared pawns. You have to weasel your queen up to f7 and force a pawn move. For example:
21...Ka2
22. Qe6+ Kb1
23. Qf7
Here is the stall tactic. Black cannot play 23...Qa2? Qf3! and all moves lead to Qd1#
23...e2
24. Qf5+ Ka2
25. Qe6+ Kb1
26. Qxe2 Ka2
27. Qe6+ Kb1
28. Qf7
etc. See if that gets you all the way to the end! Good luck.
• 21 months ago · Quote · #6
white to move and win 31 points of material!
• 21 months ago · Quote · #7
Thanks to Tfor the new puzzle .
• 21 months ago · Quote · #8
zirtoc wrote:
Ok, without playing ...g2, I think I've figured out how to force the dark-squared pawns. You have to weasel your queen up to f7 and force a pawn move. For example:
21...Ka2
22. Qe6+ Kb1
23. Qf7
Here is the stall tactic. Black cannot play 23...Qa2? Qf3! and all moves lead to Qd1#
23...e2
24. Qf5+ Ka2
25. Qe6+ Kb1
26. Qxe2 Ka2
27. Qe6+ Kb1
28. Qf7
etc. See if that gets you all the way to the end! Good luck.
You gave the right idea !
The rest of the solution will be:
• 21 months ago · Quote · #9
Yes , of course you are right , I got a little crazy from so much moves
• 21 months ago · Quote · #10
nobody likes to be manipulated
• 21 months ago · Quote · #11
Actually knowing it was a problem, the first move was easy. What I couldnt see was the zugzwang later on. Elegant though.
• 21 months ago · Quote · #12
nobody likes to be manipulated
• 21 months ago · Quote · #13
nobody likes to be manipulated
• 21 months ago · Quote · #14
nobody likes to be manipulated
• 21 months ago · Quote · #15
I take my deerstalker off to you, Sir. Delightful problems! And a most appropriate handle.
On that last one, I think there is a cook which may be a faster (but less pleasing) win. Needless to say this is the line I saw, having entirely missed the far nicer intended (I assume) solution (doh).
• 21 months ago · Quote · #16
Sherlocks_Holmes wrote:
. When I first saw this puzzle , I thought - it is impossible !
This puzzle belongs to D.Relchhelm
white to play and win
the rest of the solution is coming soon
Truly wonderful! I have only just realised how black has exactly the starting material, which adds to the delight.
• 21 months ago · Quote · #17
zirtoc wrote:
Ok, without playing ...g2, I think I've figured out how to force the dark-squared pawns. You have to weasel your queen up to f7 and force a pawn move. For example:
21...Ka2
22. Qe6+ Kb1
23. Qf7
Here is the stall tactic. Black cannot play 23...Qa2? Qf3! and all moves lead to Qd1#
23...e2
24. Qf5+ Ka2
25. Qe6+ Kb1
26. Qxe2 Ka2
27. Qe6+ Kb1
28. Qf7
etc. See if that gets you all the way to the end! Good luck.
This puzzle was hilarious. When I saw it I thought it was impossible.
• 21 months ago · Quote · #18
Thanks for the good words
About puzzle #17 - white can not play 1.Rc8 - there is 1...Qg1+ 2.Kxg1 a1Q+ ... and the win is for black . Is this what you mentioned Elroch ?
• 21 months ago · Quote · #19
Sherlocks_Holmes wrote:
Thanks for the good words
About puzzle #17 - white can not play 1.Rc8 - there is 1...Qg1+ 2.Kxg1 a1Q+ ... and the win is for black . Is this what you mentioned Elroch ?
That is the line, but I am not convinced Qg1 wins for black.
• 21 months ago · Quote · #20
nobody likes to be manipulated | 1,273 | 4,020 | {"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-2013-20 | latest | en | 0.84877 |
www.sagasim.com | 1,558,795,912,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258120.87/warc/CC-MAIN-20190525144906-20190525170906-00284.warc.gz | 927,413,932 | 16,427 | # Mountains from Pixels, Rain from Math
With terrain generation done, along with the segregation of land and water, we can move on to more complex matters. Here’s where things become more subjective and creative. What terrain types are necessary? How do we handle climate?
I decided to start with basic terrain types contingent on elevation. At the lowest level is water, which isn’t really a “terrain,” but it’s there for the sake of making the terrain map look sensible. Who wants to look at a black ocean, seriously? Yes, it really is that simple a decision.
The various terrain types are based strictly on elevation. Further nuances are defined later by climate and biomes. The elevation scale is in shades of gray, from 0 to 255, with 0 as water. The rest are:
• Wetlands (shades 1-5)
• Floodplains (6-25)
• Plains (26-75)
• Foothills (76-125)
• Highlands (126-170)
• Mountains (171-255)
I’ll admit those numbers are rather arbitrary, but they result in a reasonable terrain distribution. Here’s an example:
Example of a colored terrain map.
Terrain Types
As you can see, the whole gamut of terrains is covered.
The terrain definitions, like many features of SagaSim, are parameterized and customizable. It will therefore be possible for people to tweak the terrain definitions and do fairly wacky things, if they’re so inclined.
Next up is climate! In SagaSim, climate is a function of two things: elevation and latitude. Rather than completely reinvent the wheel here, I decided to make use of the Köppen climate classification system. SagaSim’s version is rather simplified, though I’ve left the door open to make it more realistic in the future. For now, the idea is to just have SagaSim generate climate zones for the world map so the later generation steps can be carried out.
The climate system is also parameterized. I’ve currently defined 14 different climates, as based on the Köppen system:
• Af
• Am
• Aw
• BWh
• BSh
• Csa
• Cfa
• Cfb
• Cfc
• Dfa
• Dfb
• Dfc
• Dfd
• EF
Each climate has four parameters: its minimum base latitude, its maximum base latitude, whether it’s considered a rainy climate (a simple true/false value), and the color to use for it on the map (same colors as those in the Wikipedia article.) I say the latitudes are “base” latitudes because the climate generation algorithm doesn’t have to precisely respect them. Instead, there is a fair amount of variation allowed.
Climate generation is done by making multiple passes over the map. On the first pass, there is no existing climate information, so the algorithm starts from scratch. On each land pixel, it looks at the 8 surrounding pixels to see what climate times are immediately adjacent. It also determines which climates are possible at the current pixel, based on latitude. Then, a probability is calculated to determine whether to choose an adjacent climate type or a new one. It is heavily weighted toward picking an adjacent type. The formula is:
1 - (1 / (height * width * 0.25)
This will normally generate a very small number, roughly 1% on most maps. Then a random decimal is chosen. If it falls below the probability threshold, then we will pick a new climate from the available types, rather than an adjacent one.
This is done for every land pixel on the map, starting at the top left and working our way toward the bottom right. Next, we do a few more passes, doing a very similar calculation: for a given pixel, look at the surrounding pixels and choose one at random to assign to this pixel if we meet the probability threshold (about 5% in subsequent passes.)
At the end of the process, we have something like this:
Example climate map.
The “banding” effect is a consequence of the latitude restrictions and the probability checks. This allows for substantial climate variation without having climate zones go too far out of their assigned latitudinal ranges.
One shortcoming of the current climate generator is that, unlike the perlin noise generator, it does not produce a seamless map. If you were to wrap the above image around a sphere, it would be discontinuous at the left and right edges of the map–the climate zones wouldn’t line up. I intend to fix that, though I’ve not yet gotten around to it.
It might seem like we haven’t done much so far, but it’s all a means to an end. Next time, I’ll get into how rivers and biomes are generated. They’re more complicated than (and dependent on) terrain and climate, so they may get split into two articles.
# Water & Earth
Thus far, I’ve explained the basic premise of SagaSim and described how to generate a basic landscape through fractal noise. Now, it’s time to put that noise to work and create land masses!
One of the input parameters to the world generator is “water percentage.” Maybe you want a world that’s like Earth, where roughly 70% of the surface is covered by water. Maybe you want much more land, so you go down to 50%. Maybe you want no ocean at all, so you put in 0%. There is no point in putting in 100%, though, since it means no biomes or climate data would be generated (sorry.) Anyway, let’s start with 70%. I’m generating a larger map this time, so you can really see the features of the landscape.
Perlin noise before any additional processing.
Where does the land end and the water begin? You can’t tell from this map. Instead, I have an algorithm that “normalizes” the map. What it does is go through each pixel, put it in an array (a list), and then it slices the array at whatever point you entered as your water percentage. So, if you chose 70% and had an image that’s 10,000 pixels in total (100×100), the value of the pixel at index 7000 would represent our new “base” color. All pixels below that will be turned to black, then all the pixels above index 7000 get adjusted. Since the sea level value will likely be some shade of gray, and we’re turning it into black, we therefore need to tweak all the lighter-colored pixels so that they now start just one shade above black. In computer graphics parlance, this is known as “stretching the histogram.” Let’s see what the map looks like after this processing has been done:
Map after normalizing the sea level.
Now, the locations of land masses are much clearer. But the world generator takes one more step to make clear the distinction between ocean and land: it creates a “mask” where all land is white, and all water is black. That way, it’s never ambiguous as to whether a given pixel is meant to be land or ocean. (Rivers are another story, and will be covered in a future post.)
Map converted to water mask. White is land, black is water.
Much clearer, isn’t it? You can also see that there are five distinct continents, as well as some smaller islands. Here are a few more water masks to illustrate the variety of landforms you can get just by tweaking some of the world generation parameters:
Water mask generated from 0.2 persistence, 4 octaves, lacunarity of 8, 70% water coverage.
Water mask generated from 0.8 persistence, 8 octaves, lacunarity of 2, 30% water coverage.
Water mask generated from 0.2 persistence, 4 octaves, lacunarity of 16, 50% water coverage.
As you can see, there’s plenty of variety to be had by messing with the world generation parameters.
So, now we’ve got our land and oceans set up. Next time, we move on to terrain and climate!
# Make Some Noise
One of my goals for this blog is to have almost everything written in such a way that non-programmers can understand it. If I make especially technical posts, I will probably put them under their own category and not have them be essential reading in order to comprehend the project’s components or the project as a whole.
With that said, the basic starting point of procedurally generating a world via a computer program is create the world map, starting with terrain. To do this, it makes sense to use some type of random noise. Noise is not hard to find. The static you hear on the radio is noise. The “snow” you saw on analog TV sets was noise. In computer graphics, noise is very easy to generate. For example, here’s a sample of noise created with one of the filters in GIMP:
A sample of noise generated from within GIMP.
By itself, this type of noise is not very useful for creating things like terrains, which have great variations in elevation but nevertheless have structure, instead of appearing completely random. To get something more suitable for terrain, we must turn to fractals. Fractals offer multiple advantages:
• Large-scale structures that give definition to our generated terrain.
• Seemingly random variations that allow for easy creation of virtually endless unique terrains.
• Based on mathematical formulas, so if you “zoom in” on a fractal terrain, you can find ever finer details in the terrain.
The last point is particularly interesting since it strongly corresponds to how terrain works in real life. For instance, a difficult problem is measuring the exact length of a land mass’ coastline. Putting aside the rise and fall of tides, coastlines demonstrate fractal-like complexity as you measure them in smaller and smaller units. Famed mathematician Benoît B. Mandelbrot wrote a fascinating paper on this very subject.
Fortunately, SagaSim doesn’t need coastlines defined with molecular granularity, so I’m off the hook on that one!
While there are many ways to generate fractal noise, a very common one is the perlin algorithm, named after Ken Perlin, who created the original algorithm for the movie Tron. My algorithm is based on this version by Mr. Perlin. As an aside, in case you’re not familiar with methods for generating terrain, when presented as an image, lightness indicates elevation. Black is the lowest elevation, while white is the highest.
Without going into a lot of mathematical detail, there are a few basic inputs to a perlin algorithm: persistence, octaves, and lacunarity. Persistence indicates the “amplitude” or intensity of the noise. A high persistence means that the basic features of the noise will remain more “static” than they would if you used a lower persistence. Octaves indicate the “frequency” of the noise. Essentially, octaves work by adding the mathematical output of the noise function to itself–the same way that adding two identical sound frequencies together produces a tone that’s one octave higher. Finally, there is lacunarity, which can also be considered a “smoothness” value. This produces large-scale structures in the noise by layering zoomed-in versions of the noise on top of itself. If you didn’t understand any of that, you’re not alone. Let’s illustrate with pictures!
Here is the base case: persistence of 1.0, 1 octave, 1 lacunarity.
Basic noise: persistence of 1.0, 1 octave, 1 lacunarity.
Pretty boring, huh? Let’s bring down the persistence a bit, to 0.3:
Less persistent noise: 0.3.
Notice that it has higher peaks, because the perlin function is less constrained by the persistence of the original fractal. Next, let’s add a few octaves. We can start with 4:
Noise with multiple octaves: 0.3 persistence, 4 octaves, 1 lacunarity.
You can now see that there’s a lot more variation, but also that the structures are pretty much evenly distributed. No matter how much you turn up the octave count, it’s just going to make the noise fuzzier while not bringing out any large-scale features. For that, we need to tweak lacunarity! Let’s turn it up to 4:
Noise with lacunarity: persistence 0.3, 4 octaves, lacunarity of 4.
At this point we really start to see larger structures come out, such as the deep valleys near the upper left and around the bottom, and the high peaks near the lower right. Things get more intense if we turn the lacunarity up to, say, 8:
Noise with more lacunarity: persistence 0.3, 4 octaves, lacunarity of 8.
You can see that now we have one prominent streak of higher elevations cutting from the right side down to the bottom left, and much lower elevations centered to the bottom right.
Finally, let’s go up to a lacunarity of 16. I normally wouldn’t do this because it tends to create one big “land mass,” but for illustration’s sake, here it is:
Even higher lacunarity: 0.3 persistence, 4 octaves, lacunarity of 16.
That didn’t turn out too badly, but you can likely see how there actually seems to be less variation in elevations–the lower elevations are crowded out by the higher ones, due to layering zoomed-in versions of the basic noise on top of itself so many times. In the actual SagaSim code, I find that using a lacunarity of 8 normally produces the best results.
Finally, since SagaSim is meant to be a world simulator, I had to decide how to handle “wrapping.” After all, you can circumnavigate the Earth, so why shouldn’t the intelligent life forms inhabiting a SagaSim world be able to do the same? To that end, I settled on having a “cylindrical” world map. The left and right sides of the map are analyzed, and I essentially “wrap” them onto each other, gradually fading out. This allows the map to look seamless so that land masses on either side don’t just suddenly “end” at the edge of the map. Incidentally, this also gives me maps that can be seamless wrapped around 3D spheres (something you can also do with GIMP):
Noise mapped to a 3D sphere.
By “seamless,” of course, I mean “the poles look like pinched crap.” No offense to any Poles in the audience.
Next time, I will talk about how these noise maps are turned into land masses, oceans, and such. You now probably know more about perlin noise than you ever wanted (or you’re more confused than ever.)
Further reading:
# Welcome to the SagaSim development blog!
I’m bad at introductions, so I’ll just get to the point. This is a blog about the development of SagaSim, which might best be described as a “world simulator.” It is currently under active development, and all the features aren’t even planned yet. I’m just working my way through it, one step at a time. Maybe it will be successful, maybe it won’t, but it’s already been quite a learning experience.
So far, the following major features are planned:
• Procedural generation of world, including terrain, rivers, climate zones, and biomes.
• Dynamic generation of plant and animal species with unique DNA (over 16 million possible combinations.)
• Dynamic generation of intelligent, civilized life forms.
• Simulation of resources, civilizations, important individuals, economies, politics, historical events, disasters, technological development, sociology, etc.
• Virtually all features can be controlled by the user.
Given that description, it would be better to categorize SagaSim as a sandbox than as a game. There is no way to “win,” the point is to explore the options and watch your world develop and evolve. Everything is displayed using simple 2D graphics, though I don’t discount the possibility of a 3D view at some point.
As of this posting, I have the world generation (mostly) done, and I’m about 1/3 of the way through defining the DNA system. Future posts will delve into detail about exactly how all of this works, so stick around!
For those who are interested in the basic technical details of SagaSim, my platform is Python 2.6 with Pygame, psyco, and NumPy. It is only limited to Python 2.6 because of psyco, though if PyPy ends up fully supporting Pygame and NumPy I will probably transition to that.
I’m still on the fence about what to do with the source code, and whether I should try to raise money to help fund development. I’ll certainly entertain suggestions, though.
Feel free to post any questions, comments, or encouragement! | 3,528 | 15,618 | {"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 | 3 | CC-MAIN-2019-22 | longest | en | 0.910839 |
https://www.excelforum.com/excel-formulas-and-functions/893376-formula-to-calcualte-number-based-on-several-factors.html | 1,652,854,551,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662521152.22/warc/CC-MAIN-20220518052503-20220518082503-00270.warc.gz | 878,850,675 | 14,354 | # formula to calcualte number based on several factors
1. ## formula to calcualte number based on several factors
Hi,
I am sort of new to excel and not so great with using the if and then function so any help would be appreciated.
looking for a formula to do this:
if the number(x) is less than 4 then do 15 + (x*7.25), if (x) is larger than 4 but less than 8 then 15 + (x*6), if (x) is larger than 8 but less than 14 then 15 + (x*4) and if (x) is larger than 14 then 15 + (x*3)
thank you
2. ## Re: formula to calcualte number based on several factors
=if(x<4,(15+x*7.25),if(and(x>4,x<8),15+(x*6),if(and(x>8,x<14),15+(x*4),15+(x*3))))
Try the above
3. ## Re: formula to calcualte number based on several factors
not to be a pain but i just remembered the last part :
i would need this formula to be ale to add 30 to the end result if the cell contained a number with 07 in it
example:
column a column b column c column d
11111 22222 5 50
07222 22222 5 should be 80
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Introduction to Matlab Figure
MATLAB provides us with plenty of functionalities, useful in various computational problems. In addition to its computational capabilities, MATLAB is also a great tool for visualization. It provides us with the ability to plot a wide variety of charts. Apart from showing graphical output in the console, MATLAB can also have our graphical output displayed in a separate window. For achieving this, we need to create a ‘figure object’ in MATLAB using figure function, which we will learn in this article.
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Figure function, MATLAB
For creating the figure object, MATLAB creates a separate window. The characteristics of this new window can be controlled using figure properties specified as arguments (Please refer to the end of the article for the custom properties).
Table for Custom Properties
Property
Description
Value/s
Color To set the background color Default: As per the color scheme
Name To set the title Default: ” (empty string)
number title To display the figure number Default: on
Resize For resizing the figure window using the mouse Default: on
SelectionHighlight To highlight the figure Default: on
Visible Makes figure visible/invisible Default: on
WindowStyle Normal/modal window normal, modal
Default: normal
Syntax
figure (Name, Value)
We will first understand figure function in its simplest form where we will create a figure & specify the property “Name”.For this example, we will pass “Name” property as ‘Learning figure function’
Examples to Implement Matlab Figure
Here are some examples mentioned:
Example #1
This is how our input and output will look like in MATLAB console:
Code:
figure ('Name', 'Learning figure function')
Output:
Explanation: As we can observe in the output obtained, we have obtained a new window as a figure object and our figure’s name is as passed by us “Learning figure function”. Also, notice ‘Figure 1’ before the name of the figure, this is done by MATLAB as a default property. However, we can get rid of this.
Example #2
Let us learn how to get the name without the figure number. To achieve this, we need to keep the ‘Numbertitle’ property of figure function as ‘off’. This is how our input and output will look like in MATLAB console:
Code:
figure('Name', 'Learning figure function','NumberTitle','off');
Output:
Note: In the output that we have obtained a new window without the figure number.
Example #3
Next we will learn how we can get our graph in the figure object. To get the graph in a new window, we first create the figure object as above and then write the syntax to create the desired plot. MATLAB by default assigns the plot to the latest figure object created.
In our example, we will create a bar plot in the figure object.
X = [12, 20, 13, 40, 40, 23, 54, 65, 11, 40, 70, 45, 60, 33][Input array to create bar plot]
This is how our input and output will look like in MATLAB console:
Code:
X = [12, 20, 13, 40, 40, 23, 54, 65, 11, 40, 70, 45, 60, 33] bar (X)
Output:
Example #4
As we can see, our output bar chart is in a new window and with the name as passed by us. We can also change the background of the new window using the ‘Color’ property of figure function. Let us learn how to do that.
For our example, we will set the ‘Color’ property to ‘c’ which is pre-defined color code for the color cyan.
Note: In the above line of code that we have set the ‘Color’ property as ‘c’ to get Cyan color
This is how our input and output will look like in MATLAB console:
Code:
X = [12, 20, 13, 40, 40, 23, 54, 65, 11, 40, 70, 45, 60, 33]
Output:
Explanation: So, we have our output window in CYAN color now.
Example #5
Let us take another example where we will create a scatter plot in a new window using figure function. For this example, we will get our background in Red color.
scatter((1:40),and(1,40)); [Creating scatter plot using random values]
As explained earlier, MATLAB will by default plot the graph in the figure object created. This is how our input and output will look like in MATLAB console:
Code:
scatter((1:40),rand(1,40));
Output:
Explanation: As we can see in our output, we have our scatter plot created in a new window, with red color.
Conclusion
figure function in MATLAB is used if we want our plot to be created in a separate window. We can control the look and feel of the new window using the properties of the figure function.
Recommended Articles
This is a guide to Matlab Figure. Here we discuss an introduction to Matlab Figure, Table for custom properties, syntax, examples with code and output. You can also go through our other related articles to learn more –
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Update the detailed information about Figure Function In Matlab With Examples on the Cersearch.com website. We hope the article's content will meet your needs, and we will regularly update the information to provide you with the fastest and most accurate information. Have a great day! | 1,305 | 5,712 | {"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-2023-40 | latest | en | 0.819789 |
https://www.enotes.com/homework-help/solve-system-equations-3x-y-5-5x-2y-3-253987 | 1,524,565,287,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946597.90/warc/CC-MAIN-20180424100356-20180424120356-00582.warc.gz | 776,478,447 | 10,836 | # Solve the system of equations:3x-y= 5 5x-2y = 3
hala718 | Certified Educator
Given the system of equations:
3x - y= 5............(1)
5x -2y = 3...........(2)
We will solve the system using the substitution method.
We will rewrite equation (1).
==> y= 3x - 5.
Now we will substitute into (2).
==> 5x - 2(3x-5) = 3
==> 5x - 6x + 10 = 3
==> -x = -7==> x= 7
==> y= 3x-5 = 3*7 -5 = 21-5 = 16
Then the solution is the pair ( 7, 16)
justaguide | Certified Educator
We have to solve the system of equations:
3x - y = 5 ...(1)
5x - 2y = 3 ...(2)
From (1)
3x - y = 5
=> y = 3x - 5
substitute in (2)
=> 5x - 2(3x - 5) = 3
=> 5x - 6x + 10 = 3
=> -x = -7
=> x = 7
y = 3x - 5 = 16
The solution of the system of equations is x = 7 and y = 16.
jess1999 | Student
3x - y = 5
5x - 2y = 3
First, multiply everything in the first equation by 2
By multiplying, your equation should look like
6x - 2y = 10
5x - 2y = 3 subtract the -2y with -2y ( which means subtract 6x with 5x and 10 with 3)
By subtracting, your equation should look like
x = 7 which is your answer for " x "
Now, plug 7 into one of the equation
3 ( 7 ) - y = 5 now multiply 7 with 3
By multiplying, your equation should look like
21 - y = 5 now subtract both sides by 21
By subtracting your equation should look like
-y = -16 now divide both sides by -1
By dividing, your equation should look like
y = 16
So your answer is x = 7 ; y = 16
Jyotsana | Student
3x-y=5
3x-y-3x=5-3x
-y=5-3x
-y/-1=5/-1-3x/-1
y=-5x+3x
5x-2(-5+3x)=3
5x+10-6x=3
-1x+10=3
-1+10-10=3-10
-1x/-1=7/-1
x=7
5(7)-2y=3
35-2y=3
35-2y-35=3-35
-2y/-2=-32/-2
y=16 | 692 | 1,645 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2018-17 | latest | en | 0.85288 |
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• ## Re: [EM] Problem solved (for pure rank ballots): ICC & AFB incompatible (essentially)
(9)
• NextPrevious
• ... Abd, I think you are overly hung up on this predictability detail. B wins, then the C A B voter can change his/her vote to make A win only by downranking
Message 1 of 9 , Feb 2, 2007
View Source
--- In RangeVoting@yahoogroups.com, Abd ul-Rahman Lomax <abd@...>
wrote:
>
> Benham wrote me about his earlier response with "did you even read
> it?" Yes, I had read it. I don't always have time to write. In \
> fact, I *never* have time to write, but I steal it.
>
> At 11:00 AM 1/27/2007, cbenhamau wrote:
>
> > > The assumption here is that the C voter gains something by
> > > betraying C. Where did that assumption come from?
> >
> >That "assumption" comes from the C voter's unaltered ballot in
> >Election 1: "C>A>B".
>
> The problem postulates that B wins the election. This is after, of
> course, the three voters have voted. It is assumed that the C voter
> can anticipate B winning based on knowing the sincere preferences
> of all voters and by assuming that they will all vote as predicted.
> Except C. Now, it is legitimate to make the assumption of vote
> stability except for the voter contemplating FB, but the assumption
> that the victory of B *can be predicted* -- which is necessary for
> favorite betrayal -- conflicts with the assumption that the choice
> between the three candidates for victory is random. If it can be
> predicted from the symmetrical situation shown, it isn't random or
> symmetrical. There is a contradiction in the assumptions that was
> If there is a way around this assumption (of predictability), I
> didn't see it.
> C>A>B does not, in itself, predict strategic benefit for the voter
> from FB, unless the probabilities are asymmetric, or the preference
> strengths vary.
>
> Since the probabilities must be equal, the only remaining variable,
> under the stated conditions, is variation in preference strength,
> but Benham denied that preference strength was relevant.
> If preferences expressed are taken as indicating equal preference
> strength, then there is no strategic benefit for the C voter in
> betraying C to elect A. It has, in fact, the same expected utility
> as making the strategic vote. AFB *requires* that there be a
> predictable strategic benefit. (That is, the probability of each
> outcome times the expected utility of that outcome, for FB, must be
> greater than the situation without FB. In this case, FB controls
> the election of A, it is a certainty (with stable votes from the
> others).
>
Abd, I think you are overly hung up on this "predictability" detail.
B wins, then the C>A>B voter can change his/her vote to make A win
only by downranking C to below equal-top; and that is sufficient to
fail FBC in my book. We are not meant to get our knickers in a twist
worrying about why B wins or how the C>A>B voter knew that B wins.
Warren's wording of FBC (that he often refers to as AFB) might seem to
imply "predictability", but Mike Ossipoff's earlier (perhaps
original) version doesn't:
"Favorite-Betrayal Criterion (FBC):
By voting a less-liked candidate over his/her favorite, a voter
should never gain an outcome that he/she likes better than every
outcome that he/she could get without voting a less-liked candidate
over his/her favorite. *** "
>
But to humour you a bit, maybe we can say that the proof is overly
lean in using just three voters instead of three equal-sized large
factions.
33: A>B>C
33: C>A>B
33: B>C>A
Then we can say that by Symmetry each candidate has a 1/3 chance of
winning but by Discrimination we can make an "arbitrarily small"
change to this profile to make one of the candidates win.
So when we say "B wins" we could mean that one more voter arrives and
provisionally, but then the C>A>B voters are privileged to receive
the information that the last eligible voter is about to arrive and
definitely vote for B.
Then this faction can "predict" that B wins, and so they do "improve
their expectation" by changing their votes from C>A>B to A>C>B.
Chris Benham
• ... Detail!!! ... Notice that a precedent action is dependent upon an unpredictable consequence. It is a clear logical fallacy. ... In the scenario given, it
Message 1 of 9 , Feb 2, 2007
View Source
At 05:21 AM 2/2/2007, cbenhamau wrote:
>Abd, I think you are overly hung up on this "predictability" detail.
Detail!!!
>B wins, then the C>A>B voter can change his/her vote to make A win
>only by downranking C to below equal-top; and that is sufficient to
>fail FBC in my book.
Notice that a precedent action is dependent upon an unpredictable
consequence. It is a clear logical fallacy.
> We are not meant to get our knickers in a twist
>worrying about why B wins or how the C>A>B voter knew that B wins.
>
>Warren's wording of FBC (that he often refers to as AFB) might seem to
>imply "predictability", but Mike Ossipoff's earlier (perhaps
>original) version doesn't:
>
>"Favorite-Betrayal Criterion (FBC):
>By voting a less-liked candidate over his/her favorite, a voter
>should never gain an outcome that he/she likes better than every
>outcome that he/she could get without voting a less-liked candidate
>over his/her favorite. *** "
> >
In the scenario given, it is true that the voter may, if the method
considers it, change the expected utility favorably. However, it is
also necessary to consider the probability of and utility of the
other outcomes which are possible.
What Warren has done, and which Benham is agreeing with, looks to me
like the following:
The winning candidate in a two candidate election is decided by coin
toss. A voter is picked and the procedure is that the picked voter
calls out "Heads" or "Tails." If the coin matches the call, then the
voter's favorite wins, and it it does not match, then the other candidate wins.
So if it does not match, the voter simply changes his call.
What has been said is that if there is a tie, there is an equal
probability of victory for each candidate. The winner is declared
*after* whatever random process is used to pick the winner. That
process takes place after voting.
It cannot take place *before* voting.
So, if it comes out B, then the voter can change his vote *in the
next election.* Not in the original one. And in the next election,
the voter *might* increase his expected utility by FB, but only in
the situation that the C>A preference is weaker than the C>B
preference. And this would be the case in the first election.
The voter may indeed alter the outcome by FB, but only if preference
strengths are not equal. And the tie is irrelevant.
What has really been shown is that pure ranked systems that do not
consider preference strength, but which satisfy all the conditions in
the set of assumptions Warren gave, are intrinsically vulnerable to
FB, or will fail ICC. (I think.) We already knew that. This is
because the preference strength may be as close as needed to cause A
and C to be clones within the definition of the ranked method, yet a
rank preference may still be expressed.
If ranks are presumed to represent a single preference strength,
which is the hidden assumption behind pure ranked methods, then this
does not happen. In the scenario given, C does not improve his
expected utility by FB; rather, it is the same.
I gave the probabilities in another post, with a chart showing the
expected outcomes. Essentially, if, or the C voter, the B utility is
-N, the A utility is 0, and the C utility is +N, then the expected
utility is 0 without FB. With FB, the election of A is definite, but
the utility is zero, so the expected utility is again zero. There is
no "strategically forced betrayal," no motive to betray the favorite.
However, this is not true if strength(B>C) is more than
strength(C>B). At the extremes, if (C>B) is zero, then it's obvious.
Given the other votes, C obtains maximum gain by acting such that the
election of a maximized utility candidate takes place.
And it the real world the election would be resolved by an agreement
between two of the three factions. Or at least between some portion
of these factions. It's an interesting situation, given that the
factions are, essentially diametrically opposed, yet each of them is
motivated to *give away* the election to the enemies of their
favorite. All they need to do is to let the supporters of their next
favorite know that they should vote sincerely. Otherwise those
supporters might go through the same process, and thus defeat their
favorite *and* leave the election as a tie between their middle rated
candidate and their least favorite candidate.
("Look, you evil voters who hate our favorite, please agree to not
will defeat B, whom we both agree is not as good as what happens if
you vote sincerely and we compromise with you.")
>But to humour you a bit, maybe we can say that the proof is overly
>lean in using just three voters instead of three equal-sized large
>factions.
I don't agree that is is an improvement. It's the same. For a faction
to change its vote to improve its outcome in the event of a tie is no
different than for a single voter to change his or her vote, since,
under the assumed conditions, a single vote would shift the result.
>33: A>B>C
>33: C>A>B
>33: B>C>A
>
>Then we can say that by Symmetry each candidate has a 1/3 chance of
>winning but by Discrimination we can make an "arbitrarily small"
>change to this profile to make one of the candidates win.
Yes. I got it. You don't get it. There is no difference from the
point of view of the time travel violation between this and the
original scenario. The effect of a single voter remains the same.
Can that voter base his action on the outcome of the election. Gad,
isn't it obvious that a vote can't be dependent upon a random
outcome, resolved after the election?
>So when we say "B wins" we could mean that one more voter arrives and
But then, in the original election, which we presume to be sincere,
that voter has B as a favorite.
> We could pretend that these votes are all made openly and
>provisionally, but then the C>A>B voters are privileged to receive
>the information that the last eligible voter is about to arrive and
>definitely vote for B.
That's interesting. But that violates an assumption which is that the
election is a tie, it seems. I haven't gone all the way down this road.
I would say that a assumption that voters are equal in voting power,
which wasn't stated or implied, I think, would be violated.
Essentially, for this strategy to work, a follower of one candidate
must be privileged to know that the election is a tie if the voter
>Then this faction can "predict" that B wins, and so they do "improve
>their expectation" by changing their votes from C>A>B to A>C>B.
Yes. The expected outcome for the C voter prior to voting is -1/2,
and the outcome after voting is zero.
In other words, the faction which favors a candidate is favored if
that faction (or at least one voter within the faction) has
privileged knowledge, knowledge that was not available to the other
voters when they voted.
If no assumption is made that the voters are equally privileged, then
the proof appears valid. I had assumed that the polls would be closed
and *then* the winner was determined. But wouldn't this privilege
violate symmetry? One candidate has an advantage: the one favored by
the voter who votes last and knows the outcome without his or her vote.
Given that this appears to be a *necessary* condition for AFB to
fail, in addition to the condition of equal ranking strength, it
should be noted in the proof, as should the assumption of unequal
ranking strength as a possibility, (in which case the advance
knowledge isn't necessary, but it does, again, improve the outcome further).
• ... Not neccessarily. Say that for this election (or Warren s 3-voter election) there is a special tie-breaking procedure that is fair and symmetrical: one of
Message 1 of 9 , Feb 2, 2007
View Source
--- In RangeVoting@yahoogroups.com, Abd ul-Rahman Lomax <abd@...>
wrote:
>
> At 05:21 AM 2/2/2007, cbenhamau wrote:
> >Abd, I think you are overly hung up on this "predictability"
> >detail.
>
> Detail!!!
>
> >B wins, then the C>A>B voter can change his/her vote to make A win
> >only by downranking C to below equal-top; and that is sufficient to
> >fail FBC in my book.
>
> Notice that a precedent action is dependent upon an unpredictable
> consequence. It is a clear logical fallacy.
>
> >We are not meant to get our knickers in a twist worrying about why
> >B wins or how the C>A>B voter knew that B wins.
> >
> >Warren's wording of FBC (that he often refers to as AFB) might
> >seem to imply "predictability", but Mike Ossipoff's earlier
> >(perhaps original) version doesn't:
> >
> >"Favorite-Betrayal Criterion (FBC):
> >By voting a less-liked candidate over his/her favorite, a voter
> >should never gain an outcome that he/she likes better than every
> >outcome that he/she could get without voting a less-liked candidate
> >over his/her favorite. *** "
> > >
>
> In the scenario given, it is true that the voter may, if the method
> considers it, change the expected utility favorably. However, it is
> also necessary to consider the probability of and utility of the
> other outcomes which are possible.
>
>>
> >But to humour you a bit, maybe we can say that the proof is overly
> >lean in using just three voters instead of three equal-sized large
> >factions.
>
>
>
> >33: A>B>C
> >33: C>A>B
> >33: B>C>A
> >
> >Then we can say that by Symmetry each candidate has a 1/3 chance of
> >winning but by Discrimination we can make an "arbitrarily small"
> >change to this profile to make one of the candidates win.
>
> Yes. I got it. You don't get it. There is no difference from the
> point of view of the time travel violation between this and the
> original scenario. The effect of a single voter remains the same.
>
> Can that voter base his action on the outcome of the election. Gad,
> isn't it obvious that a vote can't be dependent upon a random
> outcome, resolved after the election?
Not neccessarily. Say that for this election (or Warren's 3-voter
election) there is a special tie-breaking procedure that is fair and
symmetrical: one of the tied winners is randomly selected (X) and
then it is announced that all voters have the option of changing their
votes. If they all decline to do this or if after doing this there is
still a tie, then X wins.
>
> >So when we say "B wins" we could mean that one more voter arrives
>
> But then, in the original election, which we presume to be sincere,
> that voter has B as a favorite.
>
> >We could pretend that these votes are all made openly and
> >provisionally, but then the C>A>B voters are privileged to receive
> >the information that the last eligible voter is about to arrive and
> >definitely vote for B.
>
> That's interesting. But that violates an assumption which is that
> the election is a tie, it seems. I haven't gone all the way down
>
> I would say that a assumption that voters are equal in voting
> power,which wasn't stated or implied, I think, would be violated.
> Essentially, for this strategy to work, a follower of one candidate
> must be privileged to know that the election is a tie if the voter
Having a voter or faction of voters better informed is not a problem
for the formal "equality of voting power" in an election.
>
> >Then this faction can "predict" that B wins, and so they
> >do "improve their expectation" by changing their votes from C>A>B
> >to A>C>B.
>
> Yes. The expected outcome for the C voter prior to voting is -1/2,
> and the outcome after voting is zero.
>
> In other words, the faction which favors a candidate is favored if
> that faction (or at least one voter within the faction) has
> privileged knowledge, knowledge that was not available to the other
> voters when they voted.
>
> If no assumption is made that the voters are equally privileged,
> then the proof appears valid. I had assumed that the polls would be
> closed and *then* the winner was determined. But wouldn't this
> privilege violate symmetry?
> One candidate has an advantage: the one favored by
> the voter who votes last and knows the outcome without his or her
> vote.
No it doesn't. But it isn't necessary for the C>A>B faction to be
privileged with the extra information. If all the voters know that the
extra voter is turning up to vote for B, then the C>A>B faction will
still switch to A>C>B and that will create a stable "A wins" situation
that none of the voters can from their perspective improve on.
>
Happy now?
Chris Benham
• ... In other words, by setting up a special rule, you can get a method to fail AFB. Specifically by inviting it. The attempt in Warren s proof was to show that
Message 1 of 9 , Feb 2, 2007
View Source
At 09:15 PM 2/2/2007, cbenhamau wrote:
>Not neccessarily. Say that for this election (or Warren's 3-voter
>election) there is a special tie-breaking procedure that is fair and
>symmetrical: one of the tied winners is randomly selected (X) and
>then it is announced that all voters have the option of changing their
>votes. If they all decline to do this or if after doing this there is
>still a tie, then X wins.
In other words, by setting up a special rule, you can get a method to
fail AFB. Specifically by inviting it.
The attempt in Warren's proof was to show that *any* method which
satisfied the set of assumptions was subject to AFB violation *or*
ICC violation under at least one election scenario.
What we have seen so far is that this is unconditionally true if
ranks do not represent equal preference strengths.
Then it is conditionally true, even if ranks represent equal
preference strengths, if the voter (at least one) has prior knowledge
-- prior to fixing his vote permanently -- of who will be chosen by
the tie-breaking procedure.
The above is really only a variation of that, one which appears to
satisfy symmetry. However, I'd suggest that we would need to consider
the new election as one which has a special bias: X is selected if
there is a tie. All voters now have the option of changing their votes.
So we have
A>B>C
B>C>A
C>A>B
with the provision that B wins if there is a tie. In other words, the
above election has B as the winner.
The C voter may change his vote, to be sure, to A>C>B. The B voter
has no motive to change his vote, it would seem, and the A voter is
reasonably content. However, the B voter will notice, perhaps, that
the C voter is motivated to shift his vote, thus converting the
election from his favorite to his least favorite. The B voter is
going to be quite unhappy that the result of the random selection was
B. The B voter cannot avoid the disaster of the A and C voters colluding.
It's kind of like winning a Yankee swap when you are the first
player. You get the prize you want. Except that if anyone else wants
it, they will take it from you. And, in this case, you get the worst
outcome possible. Every candidate will be wishing to lose the
selection, because it invites the other two factions to unite against them.
This raises the whole question of how AFB is judged. It is clear that
by changing his vote, the C voter in the original concept can raise
utility. However, in the vast majority of real election situations,
this does not raise utility in a fixed way, but in a way that is
modified by the likelihood of similar shifts by other voters. By
allowing the isolation of a single voter shifting his vote, while all
other voters are presumed to keep their votes the same, we can set up
a situation where AFB is violated, that is, the voter can, under some
circumstances, anticipate a gain. But that is a highly constricted situation.
criteria are defective, in general, where they are not directly
connected with utility outcome. The purpose of elections is,
properly, to select candidates who will be best for society, and the
democratic assumption is that this is best done through aggregating
the choices of free people. Various election methods may satisfy or
fail various criteria that are proposed as being in some way
desirable, but the desirability is actually speculative. We already
know, for example, that the intuitively-satisfying Condorcet
Criterion can fail miserably once we realize the matter of preference
strength. (Though, in fact, it would perform reasonably well under
most circumstances because *usually* preference strengths, on
average, are reasonably equal, considered over a large population.
But the circumstances where this may not be true are not necessarily
rare, either.... ICC is an example of the extreme in reduction of
preference strength.)
>Having a voter or faction of voters better informed is not a problem
>for the formal "equality of voting power" in an election.
I disagree. If a voter has prior knowledge of the exact election
outcome without his vote, that is a special privilege. I agree that
being "better informed" about the likely votes of others is not a
violation. In this case we have something much more than that, we
have prior knowledge of the outcome of a process which was, in the
design of the criteria, considered to be random.
It is as if the designer of this proof discovered a way in which time
travel *was* possible,by allowing the voter to go *back* and change
his vote. Warren didn't make this argument, Benham did.
Benham designed a method deliberately constructed, in certain
aspects, to make prior knowledge a factor, while leaving all voters
on the same footing. That is, all of them have the opportunity to
change their vote. This can make AFB failure possible even if the
method would not otherwise fail. However, the attempt of the proof
was to show that *any* method satisfying all the assumptions would
fail, under some election conditions -- which doesn't mean
methodological conditions, it means the state of the candidate set
and of the electorate and how they vote. What Benham has shown
(reasonably well, it seems) is that a method can be designed which
can fail, by providing special provisions in the method that cause it to fail.
Those who favor ranked methods will simply reply -- Then don't allow
votes to change after the random selection!
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# Problem: The circuit shown in the sketch consist of two resistors and a battery with emf E = 24.0 V and negligible internal resistance, R1 = 2.00 Ω and R2 = 4.00 Ω. What is the voltage across each resistor? V1 = V2 =
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The circuit shown in the sketch consist of two resistors and a battery with emf E = 24.0 V and negligible internal resistance, R1 = 2.00 Ω and R2 = 4.00 Ω. What is the voltage across each resistor?
V1
V2 | 156 | 548 | {"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-2021-04 | latest | en | 0.814861 |
https://studylib.net/doc/10499718/beitr%C2%A8-age-zur-algebra-und-geometrie-contributions-to-alg. | 1,601,158,702,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400245109.69/warc/CC-MAIN-20200926200523-20200926230523-00457.warc.gz | 644,279,667 | 17,266 | # Beitr¨ age zur Algebra und Geometrie Contributions to Algebra and Geometry
```Beiträge zur Algebra und Geometrie
Contributions to Algebra and Geometry
Volume 51 (2010), No. 1, 1-7.
On the Depth of Graded Rings
Associated to Lex-segment Ideals
in K[x, y]
A. V. Jayanthan
Department of Mathematics, Indian Institute of Technology Madras
Chennai 600036, India
Abstract. In this article, we show that the depths of the associated
graded ring and fiber cone of a lex-segment ideal in K[x, y] are equal.
Keywords: lex-segment ideals, associated graded ring, fiber cone, Rees
algebra, Cohen-Macaulay
1. Introduction
Let K be a field of characteristic zero and R = K[x1 , . . . , xn ] be the polynomial
ring in n variables over K. Let Ri denote the K-vector subspace of all monomials
of degree i. We fix the ordering of variables as x1 > x2 > · · · > xn . For monomials
u = xa11 · · · xann and v = xb11 · · · xbnn , we say that u <Lex v if deg u ≤ deg v or
deg u = deg v and bi − ai > 0 for the first time when it is nonzero. An initial lexsegment in degree d is the set of all monomials of the form {m ∈ Rd : m ≥ u},
where u ∈ Rd . A graded ideal I is said to be a lex-segment ideal if Id is generated
by initial lex-segments for each d with Id 6= 0. Lex-segment ideals are important
due to many reasons. It is well known that among ideals with a given Hilbert
function, the lex-segment ideal has the largest number of generators. A. M. Bigatti
[1] and H. A. Hulett [7] in characteristic zero and K. Pardue [10] in positive
characteristic generalized this to all Betti numbers. They proved that the lexsegment ideals have the largest Betti numbers among all ideals with a given Hilbert
function. Lex-segment ideals are of interest also due to classical reasons. O. Zariski
used the theory of contracted ideals to study complete ideals in 2-dimensional
c 2010 Heldermann Verlag
0138-4821/93 \$ 2.50 2
A. V. Jayanthan: On the Depth of Graded Rings . . .
regular local rings (R, m). In the graded setting, when K is algebraically closed,
Zariski’s factorization theorem for homogeneous contracted ideals asserts that any
homogeneous contracted ideal I can be written as I = mc L1 · · · Lt , where each Li
is a lex-segment ideal with respect to an appropriate system of coordinates xi , yi
which depends on i. [13, Theorem 1, Appendix 5], [3, Theorem 3.8].
In this article, we study the blowup algebras, namely, the associated graded
ring and the fiber cone of lex-segment ideals in a two dimensional polynomial
ring. Let R be a ring, I any ideal of R and m a maximal ideal. Then the
associated graded ring and the fiber cone of I are respectively defined as grI (R) =
⊕n≥0 I n /I n+1 and F (I) = ⊕n≥0 I n /mI n . In [6], Huckaba and Marley showed that in
a regular local ring (R, m), depth grI (R) = depth R(I) − 1 for any m-primary ideal
I, where R(I) = ⊕n≥0 I n tn denotes the Rees algebra of I. It is interesting to ask if
there is a similar relation between the depths of the fiber cone and the associated
graded ring. It is well known that this is not the case in general (cf. Example 11,
Example 12). In this article, we prove that the depths of these algebras are equal
for lex-segment ideals in K[x, y], where K is a field of characteristic zero.
Acknowledgements. The author would like to thank Aldo Conca, M. E. Rossi,
G. Valla, J. K. Verma and S. Goto for useful discussions regarding the contents
of the paper.
2. Equality of depths
Let R = K[x, y], where K is a field of characteristic zero and M = (x, y). In
this case, the lex-segment ideals are easy to describe. If I is a lex-segment ideal
in K[x, y], then I = (xd , xd−1 y a1 , . . . , xd−k y ak ) for some 0 ≤ k ≤ d and 1 ≤ a1 <
a2 < · · · < ak . Note that if I is a lex-segment ideal, then I n is also a lex-segment
ideal for all n ≥ 1.
Remark 1. Let S = K[[x, y]] and m = (x, y). Then for any ideal I ⊂ m, S/IS ∼
=
R/I and S/mS ∼
= R/m. Therefore grIS (S) ∼
= grI (R) and F (IS) ∼
= F (I) [3,
Lemma 2.1]. Hence, we may use the local techniques to prove the results for
grI S(S) and F (IS) and derive the same for grI (R) and F (I).
We first show that the Cohen-Macaulay property of the associated graded ring
and the fiber cone are equivalent. The dimension of the fiber cone, denoted by
s(I), is called the analytic spread. It is well known that h(I) ≤ s(I), where
h(I) denotes the height of the ideal I. The difference, s(I) − h(I), is called
the analytic deviation. Let I = (xd , xd−1 y a1 , . . . , xd−k y ak ). If k = d, then I
is an M-primary homogeneous contracted ideal. Because of Remark 1, we can
use local theory of M-primary contracted ideals in 2-dimensional regular local
rings to study the blowup algebras. If 0 < k < d, then I is a non-M-primary
ideal of analytic deviation one. Here we note that if 0 < k < d, then I =
xd−k (xk , xk−1 y a1 , . . . , xy ak−1 , y ak ), which is of the form I = zL, where z is an Rregular element and L an M-primary homogeneous contracted ideal. We show
that the depth of grI (R) is at most the depth of grL (R). In particular, when
A. V. Jayanthan: On the Depth of Graded Rings . . .
3
grI (R) is Cohen-Macaulay, so is grL (R). For an element a ∈ I, let a∗ denote its
initial form in grI (R) and ao denote its initial form in F (I).
Let I be an ideal of a ring R. An ideal J ⊆ I is said to be reduction of
I if I n+1 = JI n for some n ≥ 0. A reduction which is minimal with respect
to inclusion is called a minimal reduction. For a reduction J of I, the number
rJ (I) = min{n | I n+1 = JI n }, is called the reduction number of I with respect to
J.
Proposition 2. Let (R, m) be a Noetherian local ring and L an m-primary ideal
of R. Let x be a regular element in R and I = xL. Then depth grI (R) ≤
depth grL (R). In particular, if grI (R) is Cohen-Macaulay, then so is grL (R).
Proof. Let depth grI (R) = t. Let a1 , . . . , at ∈ L \ L2 and bi = xai be such
that b∗1 , . . . , b∗t ∈ grI (R) is a regular sequence. Then by Valabrega-Valla [12],
(b1 , . . . , bt ) ∩ I n = (b1 , . . . , bt )I n−1 for all n ≥ 1. We show that (a1 , . . . , at ) ∩ Ln =
(a1 , . . . , at )Ln−1 for all n ≥ 1.
Let p ∈ (a1 , . . . , at ) ∩ Ln for some n ≥ 1. Then xn p ∈ (b1 , . . . , bt ) ∩
I n = (b1 , . . . , bt )I n−1 = xn (a1 , . . . , at )Ln−1 . Therefore xn p = xn q for some
q ∈ (a1 , . . . , at )Ln−1 . Since x is regular in R, p = q which implies that p ∈
(a1 , . . . , at )Ln−1 . Therefore, by Valabrega-Valla condition, a∗1 , . . . , a∗t ∈ grL (R) is
a regular sequence.
Remark 3. In the above proposition, we have shown that if b∗1 , . . . , b∗t is a regular
sequence in grI (R), then a∗1 , . . . , a∗t is a regular sequence in grL (R). The following
example shows that the converse is not true in general.
Example 4. Let R = K[x, y]. Let L = M = (x, y) and I = (x3 , x2 y). Then
x∗ , y ∗ is a regular sequence in grL (R). It can be easily seen that I 2 : x3 =
(x3 , x2 y, xy 2 ) 6= I. Therefore (x3 )∗ ∈ grI (R) is not regular. However, this does
not imply that the depth grI (R) < 2. In fact, in this case, it can be seen (using
any of the computational commutative algebra packages) that grI (R) is indeed
Cohen-Macaulay.
The following result follows directly from Theorem 2.1 of [4].
Proposition 5. Let (R, m) be a Cohen-Macaulay local ring and I be an ideal of
R with s(I) = r and
a + bt
H(F (I), t) =
.
(1 − t)r
If F (I) is Cohen-Macaulay, then rJ (I) ≤ 1 for any minimal reduction J of I.
We show that the Cohen-Macaulay property of the associated graded ring and
the fiber cone are equivalent:
Theorem 6. Let I be a lex-segment ideal in K[x, y].
Macaulay if and only if grI (R) is Cohen-Macaulay.
Then F (I) is Cohen-
4
A. V. Jayanthan: On the Depth of Graded Rings . . .
Proof. Let I = (xd , xd−1 y a1 , . . . , xd−k y ak ) for some 0 ≤ k ≤ d and 1 ≤ a1 < a2 <
· · · < ak . If k = 0, then I = (xd ) and both grI (R) and F (I) are Cohen-Macaulay.
We deal the cases k = d and 0 < k < d separately. Note that because of Remark 1,
we may assume that I is an ideal in a two dimensional regular local ring (R, M).
Let k = d. In this case, I is M-primary. Suppose grI (R) is Cohen-Macaulay. Since
I is contracted, by Theorem 5.1 of [8], for any minimal reduction J ⊂ I, I 2 = JI.
By [11], F (I) is Cohen-Macaulay.
Conversely, suppose that F (I) is Cohen-Macaulay. Note that for all n ≥ 0,
µ(I n ) = nd + 1 so that the Hilbert series of F (I) is given by
H(F (I), t) =
1 + (d − 1)t
.
(1 − t)2
Since F (I) is Cohen-Macaulay, by Proposition 5, rJ (I) ≤ 1 for any minimal
reduction J of I. Therefore, grI (R) is Cohen-Macaulay [12].
Now let 0 < k < d. In this case, I = xd−k L, where L = (xk , xk−1 y a1 , . . . , y ak ).
Suppose grI (R) is Cohen-Macaulay. Then by Proposition 2, grL (R) is CohenMacaulay. By Proposition 2.6 of [6], R(L) is Cohen-Macaulay. Hence the reduction number r(L) is at most one, by Goto-Shimoda theorem [5]. Therefore
r(I) ≤ 1. Therefore by [11], F (I) is Cohen-Macaulay.
Suppose now that F (I) is Cohen-Macaulay. Since µ(I n ) = nk + 1,
H(F (I), t) =
1 + (k − 1)t
.
(1 − t)2
Therefore by Proposition 5, I 2 = JI for any minimal reduction J of I. Hence,
Valabrega-Valla condition implies that depth grI (R) ≥ s(I) = 2 so that grI (R) is
Cohen-Macaulay.
Using Proposition 2, we give a simple proof of the fact that for lex-segment ideals
the Cohen-Macaulayness of the Rees algebra and the associated graded rings are
equivalent. This has been proved for m-primary ideals in a regular local ring
(R, m). Since we could not find a generalization of this result for the non-mprimary ideals, we use this opportunity to present a simple proof in the case of
lex-segment ideals.
Theorem 7. Let R = K[x, y] and I a lex-segment ideal. Then R(I) is CohenMacaulay if and only if grI (R) is Cohen-Macaulay.
Proof. Let I = (xd , xd−1 y a1 , . . . , xd−k y ak ). If k = d, then I is M-primary and
hence it follows from Proposition 2.6 of [6]. If k = 0, then I is a parameter ideal
and hence both the graded algebras are Cohen-Macaulay. Suppose 0 < k < d.
Then I = xd−k L, where L = (xk , xk−1 y a1 , . . . , y ak ). If grI (R) is Cohen-Macaulay,
then by Proposition 2 grL (R) is Cohen-Macaulay. Since L is M-primary by
Proposition 2.6 of [6], R(L) is Cohen-Macaulay. Since xd−k is a regular element,
R(L) ∼
= R(I) and hence R(I) is Cohen-Macaulay.
A. V. Jayanthan: On the Depth of Graded Rings . . .
5
Conversely, suppose R(I) is Cohen-Macaulay. Hence R(L) is Cohen-Macaulay.
By Goto-Shimoda theorem, r(L) ≤ 1. Therefore r(I) ≤ 1 and hence grI (R) is
Cohen-Macaulay.
Remark 8. The above result together with Theorem 3.4 of [9] implies that for
any lex-segment ideal I in K[x, y], depth grI (R) = depth R(I) − 1.
Now we proceed to prove that the fiber cone has positive depth if and only if
the associated graded ring has positive depth. We begin with some properties of
lex-segment ideals.
Lemma 9. Let I = (xd , xd−1 y a1 , . . . , xd−k y ak ) be a lex-segment ideal in R =
K[x, y] and M = (x, y). Then,
1. MI n : y = I n for all n ≥ 0;
2. MI n+1 : I = M(I n+1 : I) for all n ≥ 0.
Proof. (1) Note that MI n = (xnd+1 ) + yI n for all n ≥ 0. Since MI n : y is
a monomial ideal, it is enough to show that the monomials in MI n : y are in
I n . For a polynomial p ∈ K[x, y], let degx p denote the degree of the polynomial
with respect to x, considering it as a polynomial in x with coefficients in K[y] and
degy p denote the degree of the polynomial p with respect to y, considering it as a
polynomial in y with coefficients in K[x]. Let p ∈ MI n : y. If degx p ≥ nd, then
clearly p ∈ I n . Therefore, we may assume that degx p < nd. Set p = xnd−r y s for
some r, s ≥ 1. Since I n is also a lex-segment ideal, for each nd − nk ≤ t ≤ nd,
there exists a unique minimal generator pt such that degx pt = t. Let u = xnd−r y b
be the minimal generator of I n with degx u = nd − r. Then, py ∈ MI n implies
that s + 1 ≥ b + 1. Hence s ≥ b. Therefore, p = xnd−r y s ∈ I n .
(2) Let p ∈ MI n+1 : I. If p = xr for some r, then xr .xd ∈ MI n+1 . Since any term
which is a pure power in x in MI n+1 has degree at least (n + 1)d + 1, we get that
r + d ≥ (n + 1)d + 1. Therefore r ≥ nd + 1 so that xr ∈ MI n ⊆ M(I n+1 : I).
Now assume that y divides p. Write p = yp0 . Then yp0 f ∈ MI n+1 for all f ∈ I.
Therefore p0 f ∈ MI n+1 : y = I n+1 for all f ∈ I. Hence p0 ∈ I n+1 : I so that
p ∈ M(I n+1 : I).
Theorem 10. Let I be a lex-segment ideal in R. Then depth grI (R) > 0 if and
only if depth F (I) > 0.
Proof. Let depth grI (R) > 0. Then, I n+1 : I = I n for all n ≥ 0. Therefore,
MI n+1 : I = M(I n+1 : I) = MI n for all n ≥ 0. Hence depth F (I) > 0.
Conversely, assume that depth F (I) > 0. Then, (MI n+1 : I) ∩ I n = MI n for
all n ≥ 0. We need to prove that I n+1 : I = I n for all n ≥ 0. Suppose that
there exists an n such that I n
I n+1 : I. Since I is a monomial ideal, I n+1 : I
is generated by monomials and hence there exists a monomial generator p of
I n+1 : I such that p ∈
/ I n . Since I n+1 : I is also a lex-segment ideal, we can write
nd−t s
p = x
y for some s. Let q ∈ I n be the minimal generator of I n such that
degx q = nd − t. Then degy q > s, since p ∈
/ I n . Therefore qI ⊆ MI n+1 . Hence
n+1
n
n
q ∈ (MI
: I) ∩ I = MI . This contradicts the fact that q is a minimal
generator of I n . Therefore I n+1 : I = I n for all n ≥ 0. Hence depth grI (R) > 0. 6
A. V. Jayanthan: On the Depth of Graded Rings . . .
3. Examples
In this section we give some examples to show that depths of fiber cone and the
associated graded rings are not related in general.
Example 11. Let I = (x5 , x3 y 3 , xy 7 , y 9 ) ⊂ R = K[x, y]. Then it can be seen
that x2 y 6 ∈ I 2 : I, but not in I. Therefore depth grI (R) = 0. It can also be seen
that MI n+1 : I = MI n for all n ≥ 1. Therefore depth F (I) > 0.
Example 12. Let A = K[[t6 , t11 , t15 , t31 ]], I = (t6 , t11 , t31 ) and J = (t6 ). Then, it
can easily be verified that `(I 2 /JI) = 1 and I 3 = JI 2 . Since I 2 ∩ J = JI, G(I) is
Cohen-Macaulay. It can also be seen that t37 ∈ MI 2 , but t37 ∈
/ MJI. Therefore
F (I) is not Cohen-Macaulay.
References
[1] Bigatti, A. M.: Upper bounds for the Betti numbers of a given Hilbert function. Commun. Algebra 21(7) (1993), 2317–2334.
Zbl
0817.13007
−−−−
−−−−−−−−
[2] CoCoATeam, CoCoA: A system for doing computations in Commutative Algebra. Available at http://cocoa.dima.unige.it.
[3] Conca, A.; De Negri, E.; Jayanthan, A. V.; Rossi, M. E.: Graded rings associated with contracted ideals. J. Algebra 284 (2005), 593–626. Zbl
1098.13007
−−−−
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[4] D’Cruz, C.; Raghavan, K. N.; Verma, J. K.: Cohen-Macaulay fiber cones. In:
D. Eisenbud (ed.), Commutative Algebra, Algebraic Geometry and Computational Methods (Hanoi, 1996), 233–246, Springer, Singapore 1999.
Zbl
0945.13015
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[5] Goto, S.; Shimoda, Y.: On the Rees algebras of Cohen-Macaulay local rings.
In: Commutative Algebra: analytical methods. Fairfax, VA, 1979, Lect.
Notes Pure Appl. Math. 68, 201–231. Dekker, New York 1982.
Zbl
0482.13011
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[6] Huckaba, S.; Marley, T.: Depth properties of Rees algebras and associated
graded rings. J. Algebra 156(1) (1993), 259–271.
Zbl
0813.13010
−−−−
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[7] Hulett, H. A.: Maximum Betti numbers of homogeneous ideals with a given
Hilbert function. Commun. Algebra 21(7) (1993), 2335–2350. Zbl
0817.13006
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[8] Huneke, C.: Complete ideals in two-dimensional regular local rings. In: Commutative algebra (Berkeley, CA, 1987), 325–338, Publ. Math. Sci. Res. Inst.
15, Springer, New York 1989.
Zbl
0732.13007
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[9] Marley, T.: Finitely graded local cohomology and depths of graded algebras.
Proc. Am. Math. Soc. 123(12) (1995), 3601–3607.
Zbl
0864.13012
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[10] Pardue, K.: Deformation classes of graded modules and maximal Betti numbers. J. Algebra 143(1) (1991), 156–172. cf. Ill. J. Math. 40(4) (1996), 564–
585.
Zbl
0903.13004
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[11] Shah, K.: On the Cohen-Macaulayness of the fiber cone of an ideal. J. Algebra
143(1) (1991), 156–172.
Zbl
0752.13004
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A. V. Jayanthan: On the Depth of Graded Rings . . .
7
[12] Valabrega, P.; Valla, G.: Form rings and regular sequences. Nagoya Math. J.
72 (1978), 93–101.
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0362.13007
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[13] Zariski, O.; Samuel, P.: Commutative algebra, Vol. 2. Reprint of the 1960
edition. Graduate Texts in Mathematics 29, Springer-Verlag, New YorkHeidelberg-Berlin 1976.
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Received July 24, 2008
``` | 5,875 | 16,587 | {"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-2020-40 | latest | en | 0.84333 |
http://mathcentral.uregina.ca/QandQ/topics/rolling | 1,590,669,849,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347396089.30/warc/CC-MAIN-20200528104652-20200528134652-00548.warc.gz | 83,450,171 | 4,454 | Math Central - mathcentral.uregina.ca
Quandaries & Queries
Q & Q
Topic: rolling
start over
3 items are filed under this topic.
Page1/1
Rolling a die repeatedly (you can't just add percentages) 2007-10-04 From Howard:If my son has a 16.66% chance of rolling a particular number on a dice, and he rolls for it 6 times, he knows (from experience) that his odds of getting at least 1 right are not 100% (16.66% +16.66%+16.66% etc). He asked me if there was an equation that would give him the correct percentage chance of success when trying X times for a Y% chance of something.Answered by Victoria West. Rolling a metal plate 2007-07-31 From k.a.suresh:How to calculate the plate size for rolling (2010mm OD)plate thick is 6mmAnswered by Harley Weston. Rolling Circles 1999-09-12 From Craig Ellis:We have a circle of radius 3. inside the circle and tangent to the circle of radius 3 at one point is a circleof radius 1. The question is if we could roll the smaller circle around the inside of the larger circle how many revolutions would it take to get around to where we started.Answered by Chris Fisher and Walter Whiteley.
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about math central :: site map :: links :: notre site français | 333 | 1,315 | {"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} | 3.0625 | 3 | CC-MAIN-2020-24 | latest | en | 0.916981 |
https://www.ehow.co.uk/how_7535632_figure-q10-temperature-coefficient.html | 1,568,858,613,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573415.58/warc/CC-MAIN-20190919015534-20190919041534-00503.warc.gz | 852,385,904 | 34,669 | # How to figure the q10 temperature coefficient
chemical experiences image by Sergey Galushko from Fotolia.com
The Q10 temperature coefficient is the factor by which a rate of reaction (such as a chemical reaction) increases for each ten-degree increase in the temperature, measured in degrees Celsius. Q10 is defined by the equation Q10 = (R2/R1) ** [10/(T2-T1)], where the double asterisks denote the exponential function and Rn = the rate of reaction at temperature Tn. If you measure the reaction rate at any two temperatures, you can enter the rates and temperatures in this equation to solve for Q10.
Measure the rates of a reaction at two temperatures. Call the rates R1 at temperature T1 and R2 at temperature T2. Write the values of R1, R2, T1 and T2 on a piece of paper. This will help you to avoid confusion.
Compute R2/R1 and write down your answer. For example, if R1 = 8 and R2 = 24, compute 24/8 = 3.
Compute T2 -- T1 and write down your answer. In the example, say T1 = 40 degrees Celsius and T2 = 70 degrees Celsius. Then T2 -- T1 = 70 - 40 = 30.
Divide 10 by your answer from Step 3. In the example, you compute 10/30 = 0.333. Write that down.
Raise R2/R1 to the power 10/(T2-T1). You determined R2/R1 in Step 2 and you computed 10/(T2-T1) in Step 4. In the example, you raise 3 to the power 0.333. Perform this computation on the Scientific Calculator (see Resources) by entering "3", pressing the "y to the x" key (the first key in the third row), entering ".333" and pressing the "=" key. The answer is 1.44. This is the value of Q10. | 425 | 1,562 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2019-39 | longest | en | 0.889589 |
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What divide what equals 2?
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2016-03-23 02:03:33
264 divided by 2 = 132
2 divided by 1/66 = 132
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The Stock Market Is Overvalued Between 48% And 61%
TBI - Doug Short.Com ^ | 6-2-2011 | Doug Short
Posted on 07/02/2011 4:31:36 PM PDT by blam
The Stock Market Is Overvalued Between 48% And 61%
Doug Short, DShort.com
Jul. 2, 2011, 8:16 AM
The Q Ratio is a popular method of estimating the fair value of the stock market developed by Nobel Laureate James Tobin. It's a fairly simple concept, but laborious to calculate. The Q Ratio is the total price of the market divided by the replacement cost of all its companies. Fortunately, the government does the work of accumulating the data for the calculation. The numbers are supplied in the Federal Reserve Z.1 Flow of Funds Accounts of the United States, which is released quarterly.
The first chart shows Q Ratio from 1900 to the present. I've estimated the ratio since the latest Fed data (through 2011 Q1) based on a combination of the price of VTI, the Vanguard Total Market ETF, and an extrapolation of the Z.1 data itself.
Interpreting the Ratio
The data since 1945 is a simple calculation using data from the Federal Reserve Z.1 Statistical Release, section B.102., Balance Sheet and Reconciliation Tables for Nonfinancial Corporate Business. Specifically it is the ratio of Line 35 (Market Value) divided by Line 32 (Replacement Cost). It might seem logical that fair value would be a 1:1 ratio. But that has not historically been the case. The explanation, according to Smithers & Co. (more about them later) is that "the replacement cost of company assets is overstated. This is because the long-term real return on corporate equity, according to the published data, is only 4.8%, while the long-term real return to investors is around 6.0%. Over the long-term and in equilibrium, the two must be the same."
(snip)
(snip)
TOPICS: News/Current Events
KEYWORDS: djia; economy; investing; markets
1 posted on 07/02/2011 4:31:40 PM PDT by blam
To: blam
As a former stockbroker, I never, and still don’t know why the market went up during this Obamination administration.
2 posted on 07/02/2011 4:33:52 PM PDT by RacerX1128
To: RacerX1128
What, the fed, the fed, they were lending money to buy stock. They thought the stock market value set the economy, they were wrong as usual.
3 posted on 07/02/2011 4:37:53 PM PDT by org.whodat
To: RacerX1128
More buy orders then sell orders? Of course, the buy orders came from QEII and not the general public, but up is up. Right?
4 posted on 07/02/2011 4:38:13 PM PDT by Wingy (Don't blame me. I voted for the chick. I hope to do so again.)
To: blam
Although this guy may have somewhat of a point, I disagree with his estimation that at the low (6626.9) in 2009, the market was only 7% undervalued. This makes me question his numbers. IMO
5 posted on 07/02/2011 4:45:39 PM PDT by Racer1
To: blam
If you plot the slope of the stretch between the third and forth kink of the sheep’s entrails the reason for the market valuation becomes very clear
6 posted on 07/02/2011 4:52:57 PM PDT by bert (K.E. N.P. N.C. D.E. +12 ....( History is a process, not an event ))
To: RacerX1128
As a former stockbroker, I never, and still don’t know why the market went up during this Obamination administration.
It went up because of the Fed's money printing. It was only a nominal increase. Real value has probably declined.
7 posted on 07/02/2011 5:02:30 PM PDT by SeeSharp
To: RacerX1128
My .02:
The primary reason it went up was because it had been massively smashed, down to SP 666 Mar 09, 2009. As I’m sure I don’t have to explain to you, this was a crash of multi-generational proportions. Perhaps it was deserved, but I think most would agree it was waaaaay overdone, perhaps properly reflective of of credit conditions around that time. But those have since healed, to a large extent.
The Fed and Tsy demonstrated that they were available to backstop the near-infinite but certainly unquantifiable black holes created in bank balance sheets and more publicly, equities around that time.
That was truly an “end of the world” event; which one really cannot bet on. Because a: by definition, it only happens once. b: if you win, there’s nobody to pay off your winning bet.
I will agree with you in the sense that the rally off those lows is perhaps a once in 2-generation event. One does not need to look too hard to find triples, quadruples, octuples in common names such as DOW, CAT, HIG, PRU, many others.
But IMO, the greatest reason for the rise in the market is that there simply is no other readily accesible place to put money that offers a ROI.
The valuations and certainly the liquidity are largely illusions. But nobody knows better than the Fed that the US economy is primarily influenced by green closes on the DJIA on the 6 o’clock news.
8 posted on 07/02/2011 5:07:11 PM PDT by Attention Surplus Disorder (Tired of being seen as idiots, the American people went to the polls in 2008 and removed all doubt.)
To: Attention Surplus Disorder
“The primary reason it went up was because it had been massively smashed, down to SP 666 Mar 09, 2009. As I’m sure I don’t have to explain to you, this was a crash of multi-generational proportions. Perhaps it was deserved, but I think most would agree it was waaaaay overdone, perhaps properly reflective of of credit conditions around that time. But those have since healed, to a large extent.”
If the chart on this thread has any meaning, the 666 value was a reversion to just below the mean.
9 posted on 07/02/2011 5:14:08 PM PDT by ModelBreaker
To: bert
If you plot the slope of the stretch between the third and forth kink of the sheep’s entrails the reason for the market valuation becomes very clear
LOL! I think we have the same stock broker!
10 posted on 07/02/2011 5:24:08 PM PDT by central_va ( I won't be reconstructed and I do not give a damn.)
To: RacerX1128
As a former stockbroker, I never, and still don’t know why the market went up during this Obamination administration.
As a former stock broker, you should know. Some companies have bulging balance sheets. Profits are up. Orders are backlogged.
11 posted on 07/02/2011 5:24:56 PM PDT by BipolarBob (Beer? That's the reason I get up in the afternoon.)
To: RacerX1128
And that’s why you’re a “former” broker. :-b
(come on, it was tee’d up and ready to clobber)
12 posted on 07/02/2011 5:34:53 PM PDT by muleskinner
To: RacerX1128
Well:
1. Grossly oversold condition in March 2009, well below fair market value.
2. Increased corporate profits and cash flow.
3. Low interest-rate environment where dividends are high than bond interest.
4. Biggest factor of all, the gradual realization that Obama and the Democrats are digging their own grave.
13 posted on 07/02/2011 6:25:35 PM PDT by proxy_user
To: RacerX1128
>> As a former stockbroker, I never, and still don’t know why the market went up during this Obamination administration.
That’s quite simple. If you look at technical analysis, the market has gone up in exact inverse relationship to the decline of the dollar, which has been caused by the actions of the Bracky regime and the Fed. It’s kind of a phony rally, basically.
14 posted on 07/02/2011 6:44:09 PM PDT by Babu
To: Babu
There's much to note about that thesis. (lower dollar = higher stocks) I have made that point myself.
1 year:
2 years:
15 posted on 07/02/2011 6:54:51 PM PDT by Attention Surplus Disorder (Tired of being seen as idiots, the American people went to the polls in 2008 and removed all doubt.)
To: Attention Surplus Disorder
There ya go. As anyone can see, vis a vis the charts, the relationship between the dollar and the stock market is close to exactly inverse.
16 posted on 07/02/2011 7:09:31 PM PDT by Babu
To: org.whodat
I’ve heard people say that QE1 and QE2 were the training wheels our markets needed to get back on track. Actually QE1 and 2 now ARE the markets. It’s as simple as that. Money printing to make the markets look good but completely destroy the dollar and economy. Pure lunacy.
17 posted on 07/02/2011 7:18:22 PM PDT by wolfman
To: org.whodat
I’ve heard people say that QE1 and QE2 were the training wheels our markets needed to get back on track. Actually QE1 and 2 now ARE the markets. It’s as simple as that. Money printing to make the markets look good but completely destroy the dollar and economy. Pure lunacy.
18 posted on 07/02/2011 7:19:30 PM PDT by wolfman
To: wolfman
Money printing to make the markets look good but completely destroy the dollar and economy. Pure lunacy.
Now we're really screwed. The longer you kick the can down the road, the more the pain.
19 posted on 07/02/2011 7:29:12 PM PDT by BipolarBob (Beer? That's the reason I get up in the afternoon.)
To: blam
Pump, pump, pump, pump.....
20 posted on 07/02/2011 7:37:26 PM PDT by Iron Munro (The more effeminate & debauched the people, the more they are fitted for a tyrannical government.)
To: blam
I think it is fixed by the Fed, Goldman Sachs, and a lot of the other biggies. In a couple of years, all those 401k proceeds are going to be cashed in and you’ll see a drop like no other. The Fed is causing money to be printed, Goldman Sachs and the crew are doing dummy trades to push up prices and commissions, and most of the corporations are being looted of cash by the managers and directors.
21 posted on 07/02/2011 7:39:09 PM PDT by RetiredTexasVet (There's a pill for just about everything ... except stupid!)
To: wolfman
The sole reason for QE1 and QE2 was to fund government. The stock market had nothing to do with it. If Bernanke sticks to his guns and doesn't offer QE3, then the federal government will be hard pressed to come up with the additional \$1.6 trillion it will need to continue functioning at current levels. So even if the debt ceiling is raised, the Treasury won't be able to find enough lenders to cover that amount. Next, we will see interest rates rise rapidly in order to attract the dollars the Fed has been supplying for the past 2½ years.
As for the stock market, it is simply a matter of supply and demand. When stocks are in demand, prices go up. When demand falls, so do prices. Anyone with dollars is looking for some place to put them, and with real estate on the decline and new business ventures being taxed out the wazoo, the stock market becomes a better option. So more people invest, and prices rise.
22 posted on 07/02/2011 7:43:13 PM PDT by Hoodat (Yet in all these things we are more than conquerors through Him who loved us. - (Rom 8:37))
To: Babu
I am, however, detecting early signs that this inverse relationship may be breaking down....so beware of relying on this as a “black box” type of thing. It has worked remarkably, insanely well for 2 years. By the way, there is little or no secret about it, either.
I’m not saying that the relationship will vanish; I am saying it may not be *by far* the most significant driver of stock prices which it has been. I suspect a weakening dollar, when it happens in compressed time periods will be decidedly bullish for equities for the foreseeable future, under almost any circumstances.
23 posted on 07/02/2011 8:06:56 PM PDT by Attention Surplus Disorder (Tired of being seen as idiots, the American people went to the polls in 2008 and removed all doubt.)
To: Attention Surplus Disorder
No, it’s certainly not a secret for those of us that study the market. But it is what it is. The charts show it. As to whether that inverse relationship has started to break down, who knows? Time will tell. All these kind of relationships seem to eventually break down.
The same inverse relationship exists between the dollar and precious metals, so in effect, the decline in the dollar is driving up the cost of PM’s, though the run to PM’s as a safe haven has exacerbated that rise in the US\$ price of PM’s.
24 posted on 07/02/2011 8:12:28 PM PDT by Babu
To: blam
PING
25 posted on 07/03/2011 4:03:01 AM PDT by Armed Civilian ("Extremism in defense of liberty is no vice, moderation in pursuit of justice is no virtue.")
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FreeRepublic, LLC, PO BOX 9771, FRESNO, CA 93794 | 3,189 | 12,504 | {"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-2015-32 | longest | en | 0.910838 |
https://algebra1coach.com/inequalities-and-their-graphs/ | 1,721,602,186,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517796.21/warc/CC-MAIN-20240721213034-20240722003034-00724.warc.gz | 70,623,540 | 31,557 | # Inequalities and Their Graphs
Inequalities and their graphs is one of the starting lessons in algebra 1 when learning how to solve inequalities. This is when students become fluent in recognizing and checking solutions of inequalities, as well as in graphing inequalities.
Even though this topic may come across as confusing for some students, there are various ways in which math teachers and homeschooling parents can help children master inequalities. To help out, we’ve outlined a few teaching strategies. Read on to learn more.
## Strategies to Teach Inequalities and Their Graphs
### What Are Inequalities?
You may want to start your lesson on inequalities and their graphs by defining inequalities. Even though children are probably familiar with what inequalities are from earlier grades, it’s always good to brush up on previously acquired knowledge.
You can define an inequality as a mathematical statement comparing values of two expressions with the help of an inequality symbol and thus showing whether one value is less than, greater than, or equal to the other. In fact, there are five inequality symbols that we use for this purpose.
#### Inequality Symbols
You can also write a table with the inequality symbols on the whiteboard and list the corresponding verbal expression under each:
So the first symbol on the table (starting from the left), is <. The verbal expression corresponding to this inequality symbol is ‘less than’, but you also point out to students that we can use additional expressions, such as:
• is smaller than
• is fewer than
• is under
• is below
• beneath
• is shorter than, etc.
The second inequality symbol is >. The verbal expression corresponding to this symbol is ‘greater than’. Point out to students that there are several other verbal expressions that we can use for this symbol, including:
• is more than
• larger than
• above
• over
• exceeds
• higher than
• increased etc.
The third inequality symbol is ≤, for which we normally use the verbal expression ‘less than or equal to’. You can also add that we can sometimes use the following verbal expression for this inequality symbol:
• is not greater than
• is at most
• maximum
• is no more than, etc.
≥ is the fourth inequality symbol on the above table, which we usually read as ‘greater than or equal to’. You can also mention that we sometimes use the following verbal expressions instead of ‘greater than or equal to’:
• is at least
• is no less than
• minimum
• top etc.
Finally, the fifth inequality symbol is ≠, which means ‘is not equal to’. You can point out to students that we sometimes use other verbal expressions for this inequality symbol as well, such as:
• is not the same as
• differs from
• is different from etc.
### Determining If an Inequality Is True
Now that students understand what inequalities are and what are the inequality symbols, they can easily determine whether an inequality is true or false. Write a few examples of different inequalities on the whiteboard and illustrate this process step-by-step.
For instance:
5 + 6 ≥ 10 + 2
You can start by reading the inequality, that is 5 + 6 is greater than or equal to 10 + 2. Explain that in order to determine if this is true or false, we first need to do the operations, i.e. addition, on both sides so as to simplify the statement.
By adding the numbers, we’ll get:
11 ≥ 12
Then, we can read this inequality as 11 is greater than or equal to 12. Ask students if this makes sense to them. Since any algebra 1 student in the class could already tell you that 11 is less than 12, we say that this inequality is false.
### What Are Solutions to Inequalities?
You can point out to students that, in this early lesson, you won’t be going into more complicated processes of how to solve inequalities, but rather, you’ll only focus on checking solutions to inequalities.
What does that mean? Well, you can explain that to check if a certain value is a solution to an inequality, we just need to substitute the variable in the inequality with this value and then see whether the produced statement is true.
If the statement is true, we can consider this value (i.e. number) as a solution to the given inequality. If, however, substituting the given number for the variable results in a false statement, we can say that this number is not a solution to the inequality.
You can provide a few examples of how this would look in practice. Write an inequality on the whiteboard, such as x – 5 ≥ 1. Ask students to read the inequality out loud. By now, they should be able to tell you that we read it as ‘x – 5 is greater than or equal to 1’.
Then, explain that we’re asked to check whether 6 is a solution to this inequality. The first thing we need to do is replace this number for the variable x in the given inequality, which will result in the following statement:
6 – 5 ≥ 1
Point out that we’ll then simplify the statement by performing the subtraction, which will produce the following:
1 ≥ 1
Read the statement out loud, i.e. 1 is greater than or equal to 1. Since we know that 1 is indeed equal to 1, this would imply that we produced a true statement by replacing 6 for the variable. In other words, 6 is a solution to x – 5 ≥ 1!
### Graphs of Inequalities
At this point, you can explain to children how to graph inequalities, or how to present them graphically on a number line. You can define the graph of an inequality as the set of points on a number line that represent all solutions to the inequality.
#### Open Circle Graphs
Draw a number line on the whiteboard and start with simple examples of graphs of inequalities on a number line, such as x < 0, x > 0, x ≤ 0, and x ≥ 0. You can start with x < 0. Graph it on the number line in the following way:
When teaching inequalities and their graphs, you should explain that when we graph x < 0, we start from the verbal expression, i.e. x is less than 0. Does x include zero? No. We’ll draw an arrow to the left, starting from zero, but not including zero, because these are the values smaller than zero.
Point out to students that if we have an open circle (such as in the cases when x < 0 or x > 0), we’re not including the endpoints, whereas if we have a closed circle (like in the cases when x ≤ 0 or x ≥ 0) we are including the endpoints. The endpoints are zeros here.
Now move on to x > 0, which you can graph in the following way:
Again we read the verbal expression of the inequality, x is greater than 0. Does x include zero? No. So x is any number greater than 0, but not including 0. We’ll draw an arrow to the right, starting from 0, because these are the values greater than 0.
#### Closed Circle Graphs
Now you can demonstrate how we graph inequalities when we need to include the endpoints. As mentioned earlier, this is done with the help of closed circles. Draw the graph of x ≤ 0 to illustrate this visually:
Explain that to graph x ≤ 0, we read the verbal expression, i.e. x is less than or equal to 0. So in this case, since x can also be equal to 0, we’ll include 0 by shading the circle and creating a closed circle.
We’ll draw an arrow to the left, starting from zero (and including zero) because these are the values smaller than zero.
Finally, you can show students how to graph x ≥ 0:
Explain that the verbal expression is x is greater or equal to 0. Again, we’ll include 0 by drawing a closed circle, since we know that x can also be equal to 0. We’ll draw an arrow to the right, starting from 0 (and including 0) because these are the values greater than 0.
You can enrich your lesson on inequalities and their graphs by using diverse multimedia materials, such as this free video. It contains simple explanations of checking solutions of inequalities, as well as guidelines on graphing inequalities.
For students with more advanced math skills, you may also want to check out our article on solving inequalities. It contains detailed teaching instructions on solving inequalities with one variable, as well as free worksheets and activities.
## Activities to Practice Inequalities and Their Graphs
### Inequality Race
This is a fun and fast-paced game that will help students hone their skills at checking solutions for inequalities. To use this game in your classroom, you’ll need to create task cards with math problems on them.
Each task card contains one problem connected to checking whether a number is a solution to an inequality. The task card should contain the inequality, such as 27 ≤ 11x – 8, one number as a potential solution, and the answer to whether the given number is or isn’t a solution.
Divide students into pairs and place the pile of cards in the middle. The cards should be face down. Explain the rules of the game. Player 1 draws one card and asks the other student to check if the offered number is a solution.
Player 2 should be unable to see the card (as it contains the answer). Player 2 performs the calculation in their notebook to check this as quickly as they can. They only have 30 seconds for this. You can also use a timer or a stopwatch on the kids’ phones.
After player 2 gives an answer, player 2 can finally show the card to tell them if their answer is correct. If it’s correct, they score 1 point. If it’s incorrect, they lose 2 points. The roles are then reversed and player 2 draws a card this time, whereas player 1 checks the solution.
The winner is the one with the most points in the end. Although this is a group game, homeschooling parents can adjust it by asking their older kids to join in.
### True or False Inequality Quiz
In this game, kids will practice determining whether an inequality is true or false. To play the True or False Inequality Quiz in your classroom, you’ll need to prepare PowerPoint slides with quiz questions.
Each slide contains one math question on checking if a given inequality is true or false. For example, children should check if – 8 + 3 ≤ 3 – 8 is true. Make sure to also prepare an answer sheet for yourself.
Divide students into groups of 3 or 4 and provide instructions for the game. Each group is presented with one slide, i.e. one inequality problem. They should answer whether the inequality is true as quickly as possible. For each correct answer, they score 1 point.
In the end, the group with the most points is declared the winner of the game. You can also introduce a small prize for the winner, such as cutting down on their homework assignments for the next class.
### Graphing Inequalities Group Work
This activity will help students practice their knowledge of inequalities and their graphs by graphing inequalities. To use this activity in your classroom, you’ll need construction paper, markers, and assignment sheets with different inequalities.
Write 3, 4 inequalities on each assignment sheet (one assignment sheet per group). The inequalities should be different in each group. Bring several sheets of construction paper (one per group).
Divide students into groups of 3 or 4 and distribute the assignment sheets, markers, and construction paper to each group. Provide instructions for the activity. The members of each group work together.
They should graph the inequalities from their assignment sheet by drawing 3, 4 large number lines on the construction paper. Provide 10 minutes for this. After each group finishes their assignments, they present their collaborative work in front of the class.
## Before You Leave…
If you liked the strategies on teaching inequalities and their graphs that we outlined in this article, we have a whole lesson that is dedicated to this topic!
So if you’re looking for more math materials for children of all ages, sign up for our emails for plenty of free content! Feel free to also head over to our blog.
And if you’d like to become a member, sign up at Math Teacher Coach! | 2,596 | 11,858 | {"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} | 4.9375 | 5 | CC-MAIN-2024-30 | latest | en | 0.953784 |
vertexsystems.com | 1,726,633,290,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651836.76/warc/CC-MAIN-20240918032902-20240918062902-00537.warc.gz | 563,882,826 | 39,876 | # Rounding Calculations in Client Payroll Manager
All results are rounded up if there is any number other than zero in the three decimal places after the last decimal. (For example, if a number is carried out to the third decimal place, it will be rounded up if any number in the fourth, fifth, or sixth decimal place is greater than zero.) All Client Payroll Manager calculations that could affect employee earnings are carried out to a minimum of five decimal places and then rounded up. Earnings (including regular earnings, benefits, subsidies, overtime premiums, and holiday premiums) are carried out to the fifth decimal place and then rounded up to the second decimal. If the fourth, fifth, or sixth decimal place is greater than zero, the earnings are rounded up. For example: \$15.32858 is rounded to \$15.33, but \$15.32000 is left at \$15.32 because the fourth, and fifth decimal places are zero. Wage rates can be entered and displayed to the second decimal place. Calculated wage rates are carried out to the fifth decimal place and then rounded up to the second decimal. If the fourth, fifth, or sixth decimal place is greater than zero, the wage rate is rounded up. For example: \$5.32858 is rounded to \$5.33, but \$5.32000 is left at \$5.32 because the fourth, fifth, and sixth decimal places are zero. Piece rates are calculated to an infinite number of decimal places and then rounded up based on the number of decimal places entered in the Piece rate precision field in the Payroll System Options, Rates tab. This includes piece rates used to calculate earnings, as well as the piece rate displayed on screen and in reports. The piece rate will always be rounded based on three decimal places after the number of decimal places entered in the Piece rate precision field. For example, if the actual piece rate is 0.0647800120004 and the Piece rate precision field is set to 5, Client Payroll Manager will round up to 0.06479 because the 8th number after the decimal is greater then 0. If the Piece rate precision field is set to 9, Client Payroll Manager will set the piece rate at .064780012 because the 10th, 11th and 12th decimal places are all zero. Productivity rates are entered and displayed as a percentage (35.356%). They can be entered to the third decimal place when entered as a percentage. Productivity rates are stored in the database as a decimal (.35356) and all rounding calculations are based on the decimal format. (This is important to note when writing Crystal reports that display the productivity rate or use it in a calculation.) Calculated overall productivity rates are carried out to the fifth decimal place and then rounded up to the third decimal. If the fourth, or fifth decimal place is greater than zero, the productivity rate is rounded up. For example: .17205 (17.205%) is rounded to .173 (17.3%), but .17200 is left at .172 (17.2%) because the fourth, fifth, and sixth decimal places are zero.
#### DOL regulations
The calculations used to find piece rates, wage rates, earnings, and productivity rates are designed to ensure that an employee’s earnings are always based on a true prevailing wage, as required by the Department of Labor (DOL) and the NISH guidelines. In the NISH “FLSA Special Minimum Wage Requirements Guide,” the examples for prevailing wage calculations clearly show that both prevailing wage and average hourly rate should be rounded up to the nearest \$.01. The DOL fact sheet, “The Employment of Workers with Disabilities at Special Minimum Wage” (ESA/Wage an Hour Division Fact Sheet No.: 39), contains the following statement to explain an example of how to calculate prevailing wage: “Note that in this example the prevailing wage rate is \$5.92411, but the employer rounded it up to \$5.93 per hour. If the employer rounded to \$5.92, he or she would be establishing a prevailing wage rate that is less than the true prevailing wage rate (less by \$0.0041 per hour). The Wage and Hour Division will not normally question computations that are carried out to the fifth decimal point and then round up to four decimal places. The employer could, of course, round up (but not merely round off) sooner. For example, .04974 should be rounded to .0498 or .05.” While the fact sheet suggests that a piece rate precision of five is sufficient, Vertex suggests you work with your local DOL auditor to determine the correct number to use for piece rate precision at your agency.
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Scroll to Top | 990 | 4,551 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-38 | latest | en | 0.903461 |
https://physics.stackexchange.com/questions/336561/does-matter-made-out-of-ions-radiate-more-energy-than-matter-made-out-of-atoms | 1,721,528,579,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517544.48/warc/CC-MAIN-20240720235600-20240721025600-00199.warc.gz | 386,528,005 | 40,816 | I hope someone could correct the mistakes I'm making in my speech. Thanks.
Some say that when matter heats up, its atoms start accelerating faster in random directions exchanging energy by bumping into eachother.
Accelerating atoms inside a material are supposed to emit EM waves. So when matter heats up, it starts emiting more energy as EM radiation. If we accelerate a free charge it starts emiting light. The same is true for accelerating ions which are just atoms with an unbalanced charges. Does the same happen for atoms if they are not bound in a material?
That seems to me like both electrons and protons emit light as they accelerate in the same directions inside the atom which is bound to a material. But its EM waves destructively interfere so they should emit nothing. But wouldn't that mean they lose energy anyways because they've all emited it?
Other people say that heated atoms don't really accelerate but the electrons jump to a higher energy state, and when they fall down, they emit light. But that would mean that accelerating ions then shouldn't emit EM waves.
What is correct interpretation then?
Both can't be correct.
Accelerated charged particles emit electromagnetic radiation. The emitted power is proportional to the square of the acceleration. Given a fixed charge, then an accelerating Lorentz force produces an acceleration inversely proportional to mass and so radiated power is inversely proportional to the square of the mass.
As ions are much more massive than free electrons, then their radiative output (by this mechanism) can usually be neglected. There is no question of constructive or destructive interference here because there is no reason that the light emitted by separate particles should have any particular phase relationship.
Heated atoms and ions do move faster and heat may result in electrons occupying higher energy states in both (if the ions have any remaining electrons). You appear to be confused between two mechanisms of producing radiation. One consists of acceleration of free charged particles - free-free emission that could be thermal in nature or caused by external fields; the other is transitions within atoms/ions - bound-bound transitions between bound energy states.
In general both these things (and bound-free and free-bound radiation) are occurring.
• Ok, so electron and an ion which has 1 extra electron both emit the same radiation if they have exactly the same acceleration (they move side by side the whole time)? But how does the extra electron in the ion know to start emiting em waves, while the rest don't? I'm just puzzled what stops the accelerating free atoms from radiating EM waves? Commented May 31, 2017 at 8:28
• Atoms cannot be treated as multiple individual charges each accelerating in a particular way. A classical treatment does not work for bound atoms or ions. @MaDrung Commented May 31, 2017 at 9:48
• Ok, but ions do radiate because of accelerating, right? Which is wierd then if it's not because of all individual charges radiating. hmmm Commented May 31, 2017 at 9:51
• >> There is no question of constructive or destructive interference here because there is no reason that the light emitted by separate particles should have any particular phase relationship. -- Why is this? Even if extremely small, there ought to be at least some interference pattern, no?
– user140374
Commented May 31, 2017 at 10:35
• @mikey Why do you think that there should be any fixed phase relationship between the light emitted by any particular ion and electron (even ignoring the fact that that the amplitude of light emitted from the ion is millions of times smaller)? Commented May 31, 2017 at 10:44
The optical vibrational modes of salts are in the infrared. Those absorb and reflect strongly (Reststrahlen), so they will also radiate. In the same region of the spectrum, covalent materials can be transparent, so the black-body radiation will be low. | 816 | 3,952 | {"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-2024-30 | latest | en | 0.948201 |
http://www.jiskha.com/display.cgi?id=1306701139 | 1,462,553,318,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461861848830.49/warc/CC-MAIN-20160428164408-00039-ip-10-239-7-51.ec2.internal.warc.gz | 602,801,535 | 3,640 | Friday
May 6, 2016
# Homework Help: math
Posted by Anonymous on Sunday, May 29, 2011 at 4:32pm.
Ramona cut 9 feet off a ribbon that was 7 yards long. How long is the ribbon now?
• math - Ms. Sue, Sunday, May 29, 2011 at 4:34pm
3 feet = 1 yard
9 feet = 3 yards
7 - 3 = ? yards
• math - Anonymous, Saturday, May 19, 2012 at 2:11pm
5 yards
## Answer This Question
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## Related Questions
More Related Questions | 149 | 452 | {"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} | 3.0625 | 3 | CC-MAIN-2016-18 | longest | en | 0.968952 |
http://mathhelpforum.com/calculus/36474-manipulating-series.html | 1,481,139,171,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542244.4/warc/CC-MAIN-20161202170902-00027-ip-10-31-129-80.ec2.internal.warc.gz | 173,134,231 | 10,158 | 1. ## manipulating series
ok, so i am working on power series solutions to differential equations and need to get some things straight. after substituting into the in question differential equation the required power series derivatives, say there is an X^2 coefficient and i need to bring it inside one of the series; should the starting point for that series be decreased by two (because the power on X outside the series is 2 and is being added to the X^[n+s] term to make it X^[n+s+2])?
also, how do i type in formulas is there perchance a program to do this?
2. Originally Posted by mobius2000
ok, so i am working on power series solutions to differential equations and need to get some things straight. after substituting into the in question differential equation the required power series derivatives, say there is an X^2 coefficient and i need to bring it inside one of the series; should the starting point for that series be decreased by two (because the power on X outside the series is 2 and is being added to the X^[n+s] term to make it X^[n+s+2])?
also, how do i type in formulas is there perchance a program to do this?
First for La Tex use the math tages
use the $\Sigma$
button for math tags. Here is a link to La Tex code. There is a place to practice and with some basic code on this site.n You can also put your mouse above others code (or double click on it) to see others code
http://en.wikipedia.org/wiki/Help Formula
I think you mean this
$y=\sum_{n=0}^{\infty}c_nx^n$
$y'=\sum_{n=1}^{\infty}c_nnx^{n-1}$
$y'-xy=0 \iff \sum_{n=0}^{\infty}c_nx^n -x\sum_{n=1}^{\infty}c_nnx^{n-1}$
$\sum_{n=0}^{\infty}c_nx^n -\sum_{n=1}^{\infty}c_nnx^{n}=c_0+\sum_{n=1}^{\inft y}c_nx^n -\sum_{n=1}^{\infty}c_nnx^{n}=c_0+ \sum_{n=1}^{\infty}\left( c_n-c_nn\right)x^n$
I hope this helps. | 534 | 1,803 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2016-50 | longest | en | 0.892421 |
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