url
stringlengths 6
1.61k
| fetch_time
int64 1,368,856,904B
1,726,893,854B
| content_mime_type
stringclasses 3
values | warc_filename
stringlengths 108
138
| warc_record_offset
int32 9.6k
1.74B
| warc_record_length
int32 664
793k
| text
stringlengths 45
1.04M
| token_count
int32 22
711k
| char_count
int32 45
1.04M
| metadata
stringlengths 439
443
| score
float64 2.52
5.09
| int_score
int64 3
5
| crawl
stringclasses 93
values | snapshot_type
stringclasses 2
values | language
stringclasses 1
value | language_score
float64 0.06
1
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://hoursfinder.com/0-9-hours/12-beers-in-4-hours.html
| 1,653,172,107,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-21/segments/1652662541747.38/warc/CC-MAIN-20220521205757-20220521235757-00189.warc.gz
| 360,155,844
| 5,489
|
We collected information about 12 Beers In 4 Hours for you. Follow the liks to find out everything about 12 Beers In 4 Hours.
How Many Beers Can You Drink in an Hour and Still Drive ...
https://www.farrin.com/blog/how-many-beers-can-you-drink-in-an-hour-and-still-drive/
A 180-lb man may be able to drink 3.5 regular 12 ounce beers in one hour and keep his Blood Alcohol Concentration (BAC) under the legal limit of …
Standard Drinks Calculator - RUPissed.com
https://www.rupissed.com/standard-drinks-calculator.html
You have been drinking for 3 hours and have had 7 375ml cans of light beer in that time. 3 (hours) + 1 = 4 standard drinks "allowed" during that time 7 (can) x 0.80 (standard drink equivalents) = 5.6 standard drinks Whoops. Too many!. 2. You have been drinking for 2 hours and have had 3 bourbons in that time.
Blood Alcohol Content (BAC) Calculator (Beer, Wine & Liquor)
https://www.alcohol.org/bac-calculator/
Blood Alcohol Content (BAC) Calculator. Blood Alcohol Content, or BAC, refers to the percentage of alcohol in a person's bloodstream, and can be measured within 30-70 minutes after drinking. Contrary to popular belief, nothing can lower BAC except time; coffee, cold showers, and chugging glasses of water will not help you sober up any faster.
BAC Calculator
https://www.thecalculator.co/lifestyle/BAC-Calculator-158.html
- One beer is around 12 ounces (oz.) - One wine glass is around 5 oz. - One cocktail is usually around 4 oz. - One shot is around 1.5 oz. Regarding the alcohol percentage in beverage for 3 common drinks: - One beer has between 4-4.5% alcohol; - One wine glass has between 15-20% alcohol; - One shot has between 30-50% alcohol.
none
How Many Beers Does It Take to Get Drunk?
https://howmonk.com/how-many-beers-to-get-drunk/
For the average guy of 190 pounds (86kg) it takes 4 to 5 beers in 1 hour to get drunk, while for the average woman of 160lbs or 73kg, it’s 3 to 4 beers. The term “to get drunk” here means above 0.08% of blood alcohol content (BAC), and in US that means legally intoxicated (or legally drunk). Keep in mind that this is only an average ...
How many beers can you personally drink in two hours ...
https://www.quora.com/How-many-beers-can-you-personally-drink-in-two-hours-without-getting-drunk
Answer (1 of 7): This depends on a lot of factors such as whether you are male or female, how much you weigh, have you eaten before drinking, whether you’ve taken medication, and how your body personally metabolizes alcohol. For an example, let’s …
How To Calculate BAC - Blood Alcohol Content
https://www.drunkdrivingdefense.com/resources/how-to-calculate-bac/
It will show you your BAC after one beer. Here is how to estimate your blood alcohol content using our drunk calculator: 1. Count your number of drinks consumed. One drink = 1 ounce of 100-proof liquor, one 5 ounce glass of table wine, or one 12 ounce bottle of regular beer. 2.
Searching for 12 Beers In 4 Hours?
You can just click the links above. The info is collected for you.
| 789
| 3,027
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.890625
| 3
|
CC-MAIN-2022-21
|
latest
|
en
| 0.904283
|
https://en.m.wikibooks.org/wiki/Statics/Measurement_and_Units
| 1,713,601,625,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296817491.77/warc/CC-MAIN-20240420060257-20240420090257-00748.warc.gz
| 211,043,926
| 18,863
|
Statics/Measurement and Units
In statics and mechanics, units can be expressed in terms of three basic dimensions: length, mass, and time. All other units are created from combinations of these three basic units.
Force can be considered a fourth basic unit. Known as a derived measurement, it comes from Newton's 2nd Law:
${\displaystyle \ \ \!\mathrm {F} =\mathrm {m} \ \mathrm {a} \,\!}$
Here, force is defined as the amount of mass multiplied by the acceleration (length per second squared) that the mass achieves.
International System of Units (SI Units)
In the SI system of units, the three specified base units are the units of length, mass and time. A fourth unit, that of force, is derived from the base units.
• The unit of length is the meter (m).
• The unit of mass is the kilogram (kg).
• The unit of time is the second (s).
• The unit of force is the newton (N), where:
${\displaystyle \ \!\mathrm {N} =1\ \!\mathrm {kg} \ {\frac {\mathrm {m} }{\mathrm {s} ^{2}}}\,\!}$
When working with units that are either large multiples or small fractions of these units, prefixes are often used in order to keep the numbers manageable. For example,
${\displaystyle 1000\ \!\mathrm {m} =1\ \!\mathrm {km} }$
The following table gives a more detailed description of prefixes.
Prefix Abbrev. Factor
peta- P 1015
tera- T 1012
giga- G 109
mega- M 106
kilo- k 103
hecto- h 102
deca- da 101
deci- d 10-1
centi- c 10-2
milli- m 10-3
micro- μ 10-6
nano- n 10-9
pico- p 10-12
femto- f 10-15
atto- a 10-18
All other measurements are derived using variations of these four basic units and the listed prefixes.
Common SI units are listed in the following table.
Value Units (long) Units (abbrev.)
Acceleration meter per second squared
${\displaystyle {\frac {\mathrm {m} }{\mathrm {s} ^{2}}}}$
${\displaystyle \!\mathrm {rad} }$
Angular Acceleration radian per second squared
${\displaystyle {\frac {\mathrm {rad} }{\mathrm {s} ^{2}}}}$
${\displaystyle {\frac {\mathrm {rad} }{\mathrm {s} }}}$
Area square meter
${\displaystyle \!\mathrm {m} ^{2}}$
Density kilogram per cubic meter
${\displaystyle {\frac {\mathrm {kg} }{\mathrm {m} ^{3}}}}$
Energy joule
${\displaystyle \!\mathrm {J} }$ or ${\displaystyle \!\mathrm {Nm} }$
Force newton
${\displaystyle \!\mathrm {N} }$
Frequency hertz
${\displaystyle \!\mathrm {Hz} }$ or ${\displaystyle {\frac {1}{\mathrm {s} }}}$
Impulse newton-second
${\displaystyle \!\mathrm {Ns} }$ or ${\displaystyle \!\mathrm {kg} {\frac {\mathrm {m} }{\mathrm {s} }}}$
Length meter
${\displaystyle \!\mathrm {m} }$
Mass kilogram
${\displaystyle \!\mathrm {kg} }$
Force Moment newton-meter
${\displaystyle \!\mathrm {Nm} }$
Power watt
${\displaystyle \!\mathrm {W} }$ or ${\displaystyle {\frac {\mathrm {J} }{\mathrm {s} }}}$
Pressure pascal
${\displaystyle \!\mathrm {Pa} }$ or ${\displaystyle {\frac {\mathrm {N} }{\mathrm {m} ^{2}}}}$
Stress pascal
${\displaystyle \!\mathrm {Pa} }$ or ${\displaystyle {\frac {\mathrm {N} }{\mathrm {m} ^{2}}}}$
Time second
${\displaystyle \!\mathrm {s} }$
Velocity meter per second
${\displaystyle {\frac {\mathrm {m} }{\mathrm {s} }}}$
Volume (solids) cubic meter
${\displaystyle \!\mathrm {m} ^{3}}$
Volume (liquids) litre
${\displaystyle \!\mathrm {L} }$ or ${\displaystyle \!\mathrm {dm} ^{3}}$
Work joule
${\displaystyle \!\mathrm {J} }$ or ${\displaystyle \!\mathrm {Nm} }$
British and American Customary Units
While the International System of units is in common use throughout much of the world, engineers may still encounter British or American units. Therefore, it is a good idea to have some familiarity with them.
While the basic units in International System of units are length, mass, and time--with the unit of force defined in terms of these--in the British and American units, the base units are length, force and time, with mass being defined in terms of these.
• The unit of length is the foot (ft).
• The unit of force is the pound (lb), which is occasionally called pound-force (lbf).
• The unit of time is the second (s).
The unit of mass in British and American units is the slug. It is defined as the amount of mass accelerated at a rate of 1 ft/s^2 when 1 pound of force is applied.
${\displaystyle 1\ \!\mathrm {slug} =1\ \!\mathrm {lb} _{f}\ {\frac {\mathrm {ft} }{\mathrm {s} ^{2}}}}$
Occasionally, mass is described as a pound-mass. It is equal to the mass required to move one lb of weight when acted upon by the standard acceleration of gravity. On Earth, the standard acceleration of gravity is about 32.2 ft/s^2, this means that one slug is 32.2 lb(mass).
${\displaystyle 1\ \!\mathrm {slug} =32.2\ \!\mathrm {lb_{m}} }$
Conversion from one System of Units to Another
While we can do all our calculations in one set of units or the other, as long as we are consistent, there are times we will want to convert from one system to the other.
• Unit of Length 1 ft = 0.3048 m
• Unit of Force 1 lb = 4.448 N
• Unit of Mass 1 slug = 1 lb-s^2/ft = 14.59 kg
As mentioned earlier, the second is the same in both systems of units and so no conversion is required.
Common British and American Customary units and their SI equivalents are listed in the table below.
Value Conversion
Acceleration
${\displaystyle 1\ {\frac {\!\mathrm {ft} }{\!\mathrm {s} ^{2}}}=0.3048\ {\frac {\!\mathrm {m} }{\!\mathrm {s} ^{2}}}}$
Area
${\displaystyle 1\ \!\mathrm {ft} ^{2}=0.0929\ \!\mathrm {m} ^{2}}$
Energy
${\displaystyle 1\ \!\mathrm {ft} \ \!\mathrm {lb} =1.356\ \!\mathrm {J} }$
Force
${\displaystyle 1\ \!\mathrm {lb} =4.448\ \!\mathrm {N} }$
Impulse
${\displaystyle 1\ \!\mathrm {lb} \ \!\mathrm {s} =4.448\ \!\mathrm {N} \ \!\mathrm {s} }$
Length
${\displaystyle 1\ \!\mathrm {ft} =0.3048\ \!\mathrm {m} }$
${\displaystyle 1\ \!\mathrm {mile} =1.609\ \!\mathrm {km} }$
Mass
${\displaystyle 1\ \!\mathrm {lb} \ \!\mathrm {mass} =0.4536\ \!\mathrm {kg} }$
${\displaystyle 1\ \!\mathrm {slug} =14.59\ \!\mathrm {kg} }$
${\displaystyle 1\ \!\mathrm {ton} =907.2\ \!\mathrm {kg} }$
Moment
${\displaystyle 1\ \!\mathrm {lb} \ \!\mathrm {ft} =1.356\ \!\mathrm {N} \ \!\mathrm {m} }$
Power
${\displaystyle 1\ \!\mathrm {ft} \ {\frac {\!\mathrm {lb} }{\!\mathrm {s} }}=1.356\ \!\mathrm {W} }$
${\displaystyle 1\ \!\mathrm {hp} =745.7\ \!\mathrm {W} }$
Pressure
${\displaystyle 1\ \!\mathrm {psi} =6.895\ \!\mathrm {kPa} }$
Stress
${\displaystyle 1\ \!\mathrm {psi} =6.895\ \!\mathrm {kPa} }$
Velocity
${\displaystyle 1\ {\frac {\!\mathrm {ft} }{\!\mathrm {s} }}=0.3048\ {\frac {\!\mathrm {m} }{\!\mathrm {s} }}}$
${\displaystyle 1\ \!\mathrm {mph} =1.609\ {\frac {\!\mathrm {km} }{\!\mathrm {hr} }}}$
Volume (solids)
${\displaystyle 1\ \!\mathrm {ft} ^{3}=0.02832\ \!\mathrm {m} ^{3}}$
Volume (liquids)
${\displaystyle 1\ \!\mathrm {gal} =3.785\ \!\mathrm {L} }$
Work
${\displaystyle 1\ \!\mathrm {ftlb} =1.356\ \!\mathrm {J} }$
Example
According to the official National Hockey League rulebook, "The official size of the (hockey) rink shall be two hundred feet (200') long and eighty-five feet (85') wide." What are the dimensions in SI units?
Solution
From the above table:
${\displaystyle 1\ \mathrm {ft} =0.3048\ \mathrm {m} }$
Using dimensional analysis we find the length and width in meters.
${\displaystyle l\ =\ 200\ \mathrm {ft} \ \cdot \ {\frac {0.3048\ \mathrm {m} }{1\ \mathrm {ft} }}=\ 60.96\ \mathrm {m} }$
${\displaystyle w\ =\ 85\ \mathrm {ft} \ \cdot \ {\frac {0.3048\ \mathrm {m} }{1\ \mathrm {ft} }}=\ 25.96\ \mathrm {m} }$
Significant Digits
When we talk about measurements and calculations, we need to understand the degree of accuracy involved.
The accuracy of our calculations cannot be more precise than the accuracy of our measurements.
Suppose we are provided with a distance to an accuracy of one decimal place, say 9.8 m. We are told an object travels this distance in 0.81 seconds. It does not make sense to say the object is traveling at a velocity of 12.11111111 m/s, that is, to eight decimal places.
This is because neither the distance nor the time taken to travel this distance is specified to this degree of precision. In fact, they are both specified to an accuracy of only two significant digits.
For reasons we will discuss shortly, we can say the object is traveling at a velocity of 12.1 m/s.
For many calculations in statics, we work to at most three significant digits.
Rules for Finding the Correct Number of Significant Digits
In general, when making a calculation, the answer can not have more significant digits than any of the numbers used in calculating it. The number of significant digits in an answer is equal to the minimum number of significant digits used in the calculation.
Here are rules that will help outline whether or not a digit is significant or not.
1. Non-zero numbers are always significant.
2. Zeros placed in between two other digits are significant.
3. Zeros placed at the end of a number, after a decimal, are significant.
References
1 - Both the principal SI units used in mechanics and the US Customary units and their SI equivalents are taken from Beer, Ferdinand P. and El Russell Johnston, Jr. "Vector Mechanics For Engineers, Statics" 3rd edition, McGraw Hill c 1977. It should be possible to find similar tables in other texts on this subject.
2 - Rules for taking significant digits are taken from www.physics.uoguelph.ca
To do:I would like to add angle as a fundamental dimension. Define it as the ratio of the arc length to radius, in radians. We need this for rotational kinetics.
| 2,829
| 9,482
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 57, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.03125
| 4
|
CC-MAIN-2024-18
|
latest
|
en
| 0.808507
|
https://www.rcgroups.com/forums/showthread.php?3746657-Introducing-The-New-Xicoy-X90-By-Gaspar/page4
| 1,638,070,213,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-49/segments/1637964358443.87/warc/CC-MAIN-20211128013650-20211128043650-00271.warc.gz
| 1,095,082,392
| 22,522
|
Jan 22, 2021, 11:25 AM Registered User flow through I would think is more important than size because, if the UAT cannot pull fuel from the tanks faster than its using then your UAT will eventually go below the pick up level and you will flame out. Cavitation is similar in this regard, pressure difference and what not.
Jan 22, 2021, 11:26 AM
Registered User
Quote:
Originally Posted by T-CAT With this X90, would a standard (not the High Flow) 4oz MAP Air Trap be sufficient? I’m just planning the next stage of orders for supplies.
Yes it would
Jan 22, 2021, 12:49 PM Instagram: @tcat.rc.creations Thanks guys. I appreciate it . I was looking around reading various articles last night to see what additional information I could find and came upon a good one from Ultimate Jets. The article is linked below: https://ultimate-jets.eu/blogs/fuel-...ing-cavitation A quote from part of his article was: “So as good as the older system are, they only offer a few seconds of maximum thrust fuel flow buffer. However, for hard core aerobatics flyers and people who like to fly at full thrust for a large portion of the flight and keep high thrust level in the dive, the amount of air that is sucked by the clunk can be up to 15 seconds of full thrust fuel flow! For this reason I strongly recommend choosing and air trap that offers 30 seconds of full thrust fuel flow buffer capacity.” So, if I’m doing the math correctly with the engine fuel flow specs listed on Xicoy’s website for the X90, X45, and X120, then: 1. X90 fuel flow @max throttle = ~270g per minute 270g = ~9.1oz per min 9.1oz/2 = 4.55oz Air Trap to obtain the 30 second recommendation 2. X45 fuel flow @max throttle = ~145g per minute 145g = ~4.9oz per min 4.9oz/2 = 2.45oz Air Trap to obtain the 30 second recommendation 3. X120 fuel flow @max throttle = ~350g per minute 350g= ~11.8oz per min 11.8oz/2 = 5.9oz Air Trap to obtain the 30 second recommendation I’m comfortable rounding the numbers, and using a 4oz MAP Air Trap for the X90 and 2oz MAP Air Trap for the X45 . That should provide roughly 25 seconds of full thrust fuel flow buffer capacity if I calculated correctly . Last edited by T-CAT; Jan 22, 2021 at 05:19 PM.
Jan 22, 2021, 06:08 PM Registered User Like I said the above is all about fuel flow and its ability to flow through under various settings, so naturally it would make sense to go with a margin for error at fuel throttle. Cheers
Jan 22, 2021, 06:09 PM
Instagram: @tcat.rc.creations
Quote:
Originally Posted by fenderbean Like I said the above is all about fuel flow and its ability to flow through under various settings, so naturally it would make sense to go with a margin for error at fuel throttle. Cheers
Makes sense for sure.
Feb 14, 2021, 09:21 PM Instagram: @tcat.rc.creations Keith, Does the X90 have the exact same pump, manual shut off valve, and filter that the X45 has?
Feb 14, 2021, 11:20 PM ah dinnae ken The pump is different. It has more flow. Latest blog entry: Gliders_it graecalis
Feb 14, 2021, 11:27 PM
Instagram: @tcat.rc.creations
Quote:
Originally Posted by jimbo12 The pump is different. It has more flow.
I had a feeling that was the case, but wasn’t sure until confirmed. Thanks Jim .
Feb 15, 2021, 09:22 AM
Registered User
Quote:
Originally Posted by T-CAT Keith, Does the X90 have the exact same pump, manual shut off valve, and filter that the X45 has?
No its substantially larger than the X45 pump. If I was a betting man the same will be used on the larger Xicoy motors to come as well.
Same for the X90 & X120
### Images
Feb 15, 2021, 10:03 AM
Instagram: @tcat.rc.creations
Quote:
Originally Posted by fenderbean No its substantially larger than the X45 pump. If I was a betting man the same will be used on the larger Xicoy motors to come as well. Same for the X90 & X120
Excellent, thanks Keith.
Feb 26, 2021, 03:56 AM Suspended Account Gaspar!! ��������������������
Mar 24, 2021, 06:14 PM
Registered User
Xicoy X90 whats in the box (11 min 30 sec)
Mar 24, 2021, 06:15 PM
Registered User
Xicoy X90 ECU walkthtough (15 min 15 sec)
Xicoy X90 first start (5 min 24 sec)
Oct 17, 2021, 06:15 AM Registered User Any one already flying the X90? Real time feedback would be nice. Specifically if someone flying these in hot temperatures. Regards Chatty
Oct 19, 2021, 04:09 PM
Peter S.
Quote:
Originally Posted by c_makhija Any one already flying the X90? Real time feedback would be nice. Specifically if someone flying these in hot temperatures.
Hi Chatty,
until now I had 30 starts with my Viper / X90. Partially over 30° celsius in summer. The Xicoy X90 works perfect. I can absolutely recommend!
Regards Peter
C&C Model "Viper Jet" powered by Xicoy "X90" (4 min 58 sec)
| 1,306
| 4,713
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.875
| 3
|
CC-MAIN-2021-49
|
latest
|
en
| 0.927062
|
https://www.fractii.ro/reduced-fractions-simplified_200.php
| 1,680,383,691,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-14/segments/1679296950247.65/warc/CC-MAIN-20230401191131-20230401221131-00771.warc.gz
| 859,163,255
| 8,947
|
# The dataset no. 2: Common ordinary fractions reduced (simplified) to the lowest terms: all the operations run by the users, with explanations - more recent operations
• ## How to reduce (simplify) the fraction 497/ - 223 = ?
Last time calculated today, Apr 01 21:14 UTC (GMT)
• ## How to reduce (simplify) the fraction 763/153 = ?
Last time calculated today, Apr 01 21:14 UTC (GMT)
• ## How to reduce (simplify) the fraction 457/4,800 = ?
Last time calculated today, Apr 01 21:14 UTC (GMT)
• ## How to reduce (simplify) the fraction 332/609 = ?
Last time calculated today, Apr 01 21:14 UTC (GMT)
• ## How to reduce (simplify) the fraction 513/12,532 = ?
Last time calculated today, Apr 01 21:13 UTC (GMT)
• ## How to reduce (simplify) the fraction 503/2,926 = ?
Last time calculated today, Apr 01 21:13 UTC (GMT)
• ## How to reduce (simplify) the fraction 875/8 = ?
Last time calculated today, Apr 01 21:13 UTC (GMT)
• ## How to reduce (simplify) the fraction 509/1,769 = ?
Last time calculated today, Apr 01 21:13 UTC (GMT)
• ## How to reduce (simplify) the fraction 517/111 = ?
Last time calculated today, Apr 01 21:13 UTC (GMT)
• ## How to reduce (simplify) the fraction 697/987 = ?
Last time calculated today, Apr 01 21:13 UTC (GMT)
• ## How to reduce (simplify) the fraction 503/3,727 = ?
Last time calculated today, Apr 01 21:13 UTC (GMT)
• ## How to reduce (simplify) the fraction 505/918 = ?
Last time calculated today, Apr 01 21:13 UTC (GMT)
• ## How to reduce (simplify) the fraction 1,644/1,355 = ?
Last time calculated today, Apr 01 21:12 UTC (GMT)
• ## How to reduce (simplify) the fraction 523/822 = ?
Last time calculated today, Apr 01 21:12 UTC (GMT)
• ## How to reduce (simplify) the fraction 1,934/119 = ?
Last time calculated today, Apr 01 21:12 UTC (GMT)
• ## How to reduce (simplify) the fraction 565/1,009 = ?
Last time calculated today, Apr 01 21:12 UTC (GMT)
• ## How to reduce (simplify) the fraction 302/2,979 = ?
Last time calculated today, Apr 01 21:12 UTC (GMT)
• ## How to reduce (simplify) the fraction 1,372/51 = ?
Last time calculated today, Apr 01 21:12 UTC (GMT)
• ## How to reduce (simplify) the fraction 604/373 = ?
Last time calculated today, Apr 01 21:11 UTC (GMT)
• ## How to reduce (simplify) the fraction 614/785 = ?
Last time calculated today, Apr 01 21:11 UTC (GMT)
• ## How to reduce (simplify) the fraction 723/251 = ?
Last time calculated today, Apr 01 21:10 UTC (GMT)
• ## How to reduce (simplify) the fraction 621/1,093 = ?
Last time calculated today, Apr 01 21:10 UTC (GMT)
• ## How to reduce (simplify) the fraction 745/3,102 = ?
Last time calculated today, Apr 01 21:10 UTC (GMT)
• ## How to reduce (simplify) the fraction 602/3,467 = ?
Last time calculated today, Apr 01 21:10 UTC (GMT)
• ## How to reduce (simplify) the fraction 636/2,539 = ?
Last time calculated today, Apr 01 21:10 UTC (GMT)
• ## How to reduce (simplify) the fraction 637/3,586 = ?
Last time calculated today, Apr 01 21:10 UTC (GMT)
• ## How to reduce (simplify) the fraction 725/ - 283 = ?
Last time calculated today, Apr 01 21:10 UTC (GMT)
• ## How to reduce (simplify) the fraction 647/24,024 = ?
Last time calculated today, Apr 01 21:09 UTC (GMT)
• ## How to reduce (simplify) the fraction 678/3,467 = ?
Last time calculated today, Apr 01 21:09 UTC (GMT)
• ## How to reduce (simplify) the fraction 731/4,468 = ?
Last time calculated today, Apr 01 21:09 UTC (GMT)
• ## How to reduce (simplify) the fraction 699/742 = ?
Last time calculated today, Apr 01 21:09 UTC (GMT)
• ## How to reduce (simplify) the fraction 685/786 = ?
Last time calculated today, Apr 01 21:09 UTC (GMT)
• ## How to reduce (simplify) the fraction 707/333 = ?
Last time calculated today, Apr 01 21:09 UTC (GMT)
• ## How to reduce (simplify) the fraction 1,697/1,720 = ?
Last time calculated today, Apr 01 21:09 UTC (GMT)
• ## How to reduce (simplify) the fraction 900/1,241 = ?
Last time calculated today, Apr 01 21:08 UTC (GMT)
• ## How to reduce (simplify) the fraction 841/2,066 = ?
Last time calculated today, Apr 01 21:08 UTC (GMT)
• ## How to reduce (simplify) the fraction 716/235 = ?
Last time calculated today, Apr 01 21:08 UTC (GMT)
• ## How to reduce (simplify) the fraction 751/4,197 = ?
Last time calculated today, Apr 01 21:08 UTC (GMT)
• ## How to reduce (simplify) the fraction 887/5,527 = ?
Last time calculated today, Apr 01 21:08 UTC (GMT)
• ## How to reduce (simplify) the fraction 910/10,019 = ?
Last time calculated today, Apr 01 21:07 UTC (GMT)
• ## How to reduce (simplify) the fraction 480/7 = ?
Last time calculated today, Apr 01 21:07 UTC (GMT)
• ## How to reduce (simplify) the fraction 593/414 = ?
Last time calculated today, Apr 01 21:07 UTC (GMT)
• ## How to reduce (simplify) the fraction 801/4,214 = ?
Last time calculated today, Apr 01 21:07 UTC (GMT)
• ## How to reduce (simplify) the fraction 799/3,843 = ?
Last time calculated today, Apr 01 21:07 UTC (GMT)
• ## How to reduce (simplify) the fraction 803/10,073 = ?
Last time calculated today, Apr 01 21:07 UTC (GMT)
• ## How to reduce (simplify) the fraction 974/ - 19 = ?
Last time calculated today, Apr 01 21:07 UTC (GMT)
• ## How to reduce (simplify) the fraction 837/443 = ?
Last time calculated today, Apr 01 21:07 UTC (GMT)
• ## How to reduce (simplify) the fraction 851/972 = ?
Last time calculated today, Apr 01 21:06 UTC (GMT)
• ## How to reduce (simplify) the fraction 866/1,457 = ?
Last time calculated today, Apr 01 21:06 UTC (GMT)
• ## How to reduce (simplify) the fraction 869/20 = ?
Last time calculated today, Apr 01 21:06 UTC (GMT)
| 1,753
| 5,585
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14
|
latest
|
en
| 0.863861
|
https://www.hpmuseum.org/forum/showthread.php?tid=14476&pid=127885&mode=threaded
| 1,723,043,118,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722640694594.35/warc/CC-MAIN-20240807143134-20240807173134-00186.warc.gz
| 641,325,886
| 5,732
|
Wallis' product exploration
02-11-2020, 04:11 PM (This post was last modified: 06-27-2020 04:29 PM by Gerson W. Barbosa.)
Post: #17
Gerson W. Barbosa Senior Member Posts: 1,604 Joined: Dec 2013
RE: Wallis' product exploration
Try the following for even n and see what you get.
$$\frac{\pi }{2}\approx \left ( \frac{4}{3} \cdot \frac{16}{15}\cdot \frac{36}{35}\cdot\frac{64}{63} \cdots \frac{ 4n ^{2}}{ 4n ^{2}-1}\right )\left ( 1+\frac{1}{4n+\frac{3}{2-\frac{1}{4n+\frac{5}{2-\frac{3}{4n+\frac{7}{2-\frac{5}{4n+\frac{9}{2-\frac{7}{\dots \frac{ \ddots }{2-\frac{n-3}{4n+\frac{n+1}{2}}}}}}}}}}}} \right )$$
---
P.S.:
This will handle both parities:
$$\frac{\pi }{2}\approx \left ( \frac{4}{3} \times \frac{16}{15}\times \frac{36}{35}\times\frac{64}{63} \times \cdots \times \frac{ 4n ^{2}}{ 4n ^{2}-1}\right )\left ( 1+\frac{1}{4n+\frac{3}{2-\frac{1}{4n+\frac{5}{2-\frac{3}{4n+\frac{7}{2-\frac{5}{4n+\frac{9}{2-\frac{7}{\dots \frac{ \ddots }{4n \left ( n \bmod 2\right) + 2\left ( n+1 \bmod 2 \right )+\frac{1-n+\left ( 2n+1 \right )\left ( n \bmod 2 \right )}{4n \left (n+1 \bmod 2\right) + 2\left ( n \bmod 2 \right )}}}}}}}}}}} \right )$$
This approximation gives $$\frac{4}{3}n$$ correct significant digits.
« Next Oldest | Next Newest »
Messages In This Thread Wallis' product exploration - pinkman - 02-07-2020, 10:55 PM RE: Wallis' product exploration - Albert Chan - 02-08-2020, 02:13 AM RE: Wallis' product exploration - pinkman - 02-08-2020, 02:42 PM RE: Wallis' product exploration - Allen - 02-08-2020, 07:13 PM RE: Wallis' product exploration - Albert Chan - 02-08-2020, 07:53 PM RE: Wallis' product exploration - Allen - 02-08-2020, 08:58 PM RE: Wallis' product exploration - pinkman - 02-09-2020, 05:44 AM RE: Wallis' product exploration - Allen - 02-08-2020, 08:13 PM RE: Wallis' product exploration - Allen - 02-09-2020, 01:08 PM RE: Wallis' product exploration - Allen - 02-09-2020, 01:57 PM RE: Wallis' product exploration - Allen - 02-09-2020, 03:08 PM RE: Wallis' product exploration - pinkman - 02-09-2020, 02:14 PM RE: Wallis' product exploration - Gerson W. Barbosa - 02-09-2020, 10:58 PM RE: Wallis' product exploration - Gerson W. Barbosa - 02-11-2020, 12:53 AM RE: Wallis' product exploration - EdS2 - 02-10-2020, 10:35 AM RE: Wallis' product exploration - pinkman - 02-11-2020, 10:02 AM RE: Wallis' product exploration - Gerson W. Barbosa - 02-11-2020 04:11 PM RE: Wallis' product exploration - Gerson W. Barbosa - 02-11-2020, 10:48 PM RE: Wallis' product exploration - pinkman - 02-12-2020, 10:01 PM RE: Wallis' product exploration - Gerson W. Barbosa - 02-13-2020, 12:17 AM
User(s) browsing this thread: 2 Guest(s)
| 1,050
| 2,653
|
{"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}
| 3.90625
| 4
|
CC-MAIN-2024-33
|
latest
|
en
| 0.533429
|
https://www.vedantu.com/question-answer/atoms-are-there-in-20-mole-of-si-class-11-chemistry-cbse-60118dec63ac962acb6df9e4
| 1,721,455,707,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763515020.57/warc/CC-MAIN-20240720052626-20240720082626-00310.warc.gz
| 887,889,794
| 28,492
|
Courses
Courses for Kids
Free study material
Offline Centres
More
Store
# How many atoms are there in $2.0$ mole of $Si$?
Last updated date: 19th Jul 2024
Total views: 381.9k
Views today: 4.81k
Verified
381.9k+ views
Hint: We realize that the measure of substance having the very same number of atoms as are available in $12{\text{ }}g$ of ${C^{12}}$is known as mole. Moles is the proportion of the mass of substance in $g$ to the molar mass of the substance in $g/mol$. $1{\text{ }}mol$ of any substance contains $6.022 \times {10^{23}}$ atoms. $6.022 \times {10^{23}}$ is Avogadro's number.
Complete step by step answer:
Mole is determined as the amount of any substance comprising the same number of key units as the indistinguishable number of key units in unadulterated example of $^{12}C$ estimating precisely$12{\text{ }}g$.
One mole determines Avogadro's number of particles. i.e., ${N_A} = 6.022 \times {10^{23}} \cdot mo{l^{ - 1}}$. Avogadro's number of $12C$ particles, has a MASS of $12.0 \cdot g$ roughly.
Avogadro's number is along these lines the connection between miniature world particles, and atoms, which we can't see however whose presence we can deduce, with the full-scale universe of grams, and liters, and kilograms, that which we can quantify by certain methods in a lab.
We realize that $1{\text{ }}mol$ of any substance contains $6.022 \times {10^{23}}$particles. We are given $2.00$ moles of silicon. Accordingly,
Number of silicon atoms $= 2\;mol \times \dfrac{{6.022 \times {{10}^{23\;}}atoms}}{{1\;mol}}$
Number of silicon atoms $= 12.044 \times 1023{\text{ }}atoms$
Subsequently, $2.00$ moles of silicon contain $12.044 \times {10^{23}}$ atoms of silicon.
The Avogadro's number is named after the researcher Amedeo Avogadro who found it. The Avogadro's number is otherwise called Avogadro's constant. It is indicated by ${N_A}$. The Avogadro's number is the proportionality factor that relates the quantity of constituent particles for example the quantity of particles or atoms or particles in a single mole of a substance. The Avogadro's constant is a dimensionless number.
Note: The quantity of atoms of a compound is Avogadro's number for $\;1$ mole of compound. The number $6.022 \times {10^{23}}$ is known as Avogadro's number. The Avogadro's number gives the quantity of atoms, particles or atoms present in one mole of any substance.
| 664
| 2,378
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.84375
| 4
|
CC-MAIN-2024-30
|
latest
|
en
| 0.873177
|
https://discuss.codechef.com/t/talazy-error-with-previous-answer-dont-know-why/57677
| 1,686,213,933,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224654606.93/warc/CC-MAIN-20230608071820-20230608101820-00786.warc.gz
| 237,355,405
| 3,957
|
TALAZY Error with previous answer. Don't know why
I can’t find the reason my previous answer didn’t work on this question. My code is:
``````t = int(input())
for _ in range(t):
n, b, m = map(int, input().split())
faltam = n
tempo = 0
while faltam:
if n % 2 == 1:
n += 1
done = n / 2
faltam -= done
n = faltam
tempo += b + (m * done)
m *= 2
print(int(tempo - b))
``````
I got AC when I did:
``````t = int(input())
for _ in range(t):
n, b, m = map(int, input().split())
faltam = n
tempo = 0
while faltam:
done = (n + 1) // 2
faltam -= done
n = faltam
tempo += b + (m * done)
m *= 2
print(int(tempo - b))
``````
| 237
| 620
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.625
| 3
|
CC-MAIN-2023-23
|
latest
|
en
| 0.529339
|
http://math.stackexchange.com/questions/282062/cancellation-conditions-on-finitely-generated-projectives-over-a-commutative-rin
| 1,469,485,895,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-30/segments/1469257824395.52/warc/CC-MAIN-20160723071024-00012-ip-10-185-27-174.ec2.internal.warc.gz
| 148,053,367
| 17,963
|
# Cancellation conditions on finitely generated projectives over a commutative ring
A class $\mathcal{C}$ of $R$-modules is called
-separative if $A \oplus A \simeq A \oplus B \simeq B \oplus B$ implies $A \simeq B$ for each $A,B \in \mathcal{C}$
-cancellative if $A \oplus C \simeq B \oplus C$ implies $A \simeq B$ for all $A,B,C \in \mathcal{C}$.
According to literature, if $R$ is commutative then the class of finitely generated projectives over $R$ is separative iff it is cancelative. Even though I keep finding it as 'easy to see' in literature I seem unable to prove separative => cancelative. I would be grateful for any hint.
-
I'm not sure this is the intended "easy" argument, but here's what I could come up with. Fix a commutative ring $R$; the variables $A$, $B$, $C$, etc. will denote finitely generated projective $R$-modules. First, note that if $A$ locally has rank $>0$ on all of $\operatorname{Spec}(R)$, then for some $N\in\mathbb{N}$ there is a surjection $A^N\to R$. Indeed, the evaluation map (aka trace) $\operatorname{End}(A)=A\otimes \operatorname{Hom}(A,R)\to R$ is surjective, since locally it looks like the trace map $M_n(R)\to R$ for $n>0$. Choosing finitely many generators of $\operatorname{Hom}(A,R)$, we then get a surjection $A^N\to R$ for some $N$.
Now suppose $A\oplus C\simeq B\oplus C$. Note that $C$ is a direct summand of $R^n$ for some $n$; adding $R^n/C$ to both sides, we may assume $C=R^n$. Now let $n$ be minimal such that $A\oplus R^n\simeq B\oplus R^n$. Replacing $A$ and $B$ by $A\oplus R^{n-1}$ and $B\oplus R^{n-1}$, we may assume $n=1$, so we have $A\oplus R\simeq B\oplus R$ and wish to conclude $A\simeq B$. Note that the rank of $A$ is locally constant function on $\operatorname{Spec}(R)$, as is that of $B$, and these two functions must be equal. We lose no generality by restricting everything to the clopen subset of $\operatorname{Spec}(R)$ on which $A$ and $B$ have rank $>0$, since $A$ and $B$ are trivially isomorphic on the clopen set where they have rank $0$.
We thus may assume that $A$ and $B$ have rank $>0$ everywhere; let $N$ be such that there are surjections $A^N\to R$ and $B^N\to R$. These surjections split, so from the isomorphism $A\oplus R\simeq B\oplus R$ we get $A\oplus A^N\simeq B\oplus A^N$ and $A\oplus B^N\simeq B\oplus B^N$. We thus get that $$A^{2N}= A^N\oplus A^N\simeq A^{N-1}\oplus B\oplus A^N\simeq A^{N-2}\oplus B^2\oplus A^N\simeq\dots\simeq B^N\oplus A^N,$$ where at each step we combine one of the copies of $A$ in the first term with the $A^N$ in the last term and use $A\oplus A^N\simeq B\oplus A^N$. Swapping the roles of $A$ and $B$, we also have $B^N\oplus A^N\simeq B^{2N}$, so $A^{2N}\simeq B^{2N}$. By a similar argument, we have $A^n\simeq B^n$ for any $n\geq 2N$.
Now assume the finitely generated projective modules are separative, and suppose that $A\not\simeq B$. Let $k$ be the largest integer such that $A^k\not\simeq B^k$ (by the previous paragraph, such a largest integer exists and is $<2N$). Then by maximality of $k$, $A^{2k}\simeq B^{2k}$ and $A^{3k}\simeq B^{3k}$, so $$A^{2k}\oplus A^{2k}\simeq A^{2k}\oplus B^{2k}=(A^k\oplus B^k)\oplus (A^k\oplus B^k),$$ and also $$A^{2k}\oplus A^{2k}\simeq B^{2k}\oplus B^{2k}=B^{3k}\oplus B^k\simeq A^{3k}\oplus B^k=A^{2k}\oplus (A^k\oplus B^k).$$ Applying separativity to $A^{2k}$ and $A^k\oplus B^k$, we get $A^{2k}\simeq A^k\oplus B^k$. We can now apply separativity to $A^k$ and $B^k$ to get $A^k\simeq B^k$. This is a contradiction. Thus if the finitely generated projective modules are separative, we must have $A\simeq B$.
| 1,317
| 3,593
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.09375
| 3
|
CC-MAIN-2016-30
|
latest
|
en
| 0.857049
|
https://www.studystack.com/flashcard-357155
| 1,696,426,586,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233511369.62/warc/CC-MAIN-20231004120203-20231004150203-00755.warc.gz
| 1,080,117,233
| 20,264
|
Save
or
or
taken
Make sure to remember your password. If you forget it there is no way for StudyStack to send you a reset link. You would need to create a new account.
focusNode
Didn't know it?
click below
Knew it?
click below
Don't Know
Remaining cards (0)
Know
0:00
Embed Code - If you would like this activity on your web page, copy the script below and paste it into your web page.
Normal Size Small Size show me how
# kohl chapter 29words
What determines whether an object is a diffuse or polished reflector? wavelengths of the waves it reflects
Reverberations, caused by very reflective walls, can do what to sound? garble it
The law of reflection states that the angle of incidence must be equal to what? The angle of reflection
What is the line called that is perpendicular to the point where to incident ray strikes the barrier? Normal
Which type of mirror is best used for security? Convex
A light wave that strikes a boundary. incident wave
Name the two angles that occur when light bounces off a mirror and what is the name of the line that is drawn where the first ray strikes the barrier? Angle of Incidence, Angle of Reflection, and the Normal.
What is the study of the reflective properties of surfaces? Acoustics
Name the three types of mirrors and describe each one. plane mirrors are flat. then, there is a convex mirror that is curved outwards. Last, there is concave mirror that curves inward.
When waves interact with matter,what happens? They can be reflected, transmitted, or a combination of both. Waves that are transmitted can be refracted.
What is an image? a likeness or reproduction of an object.
What is refraction? the bending of a wave as it crosses the boundary between two media at an angle.
What is the relationship between the angle of incidence and the angle of reflection? the angle of incidence is equal to the angle of reflection
Where do divergent rays seem to originate? From behind the mirror.
What is an echo? Reflected sound
Created by: mquintero
Voices
Use these flashcards to help memorize information. Look at the large card and try to recall what is on the other side. Then click the card to flip it. If you knew the answer, click the green Know box. Otherwise, click the red Don't know box.
When you've placed seven or more cards in the Don't know box, click "retry" to try those cards again.
If you've accidentally put the card in the wrong box, just click on the card to take it out of the box.
You can also use your keyboard to move the cards as follows:
• SPACEBAR - flip the current card
• LEFT ARROW - move card to the Don't know pile
• RIGHT ARROW - move card to Know pile
• BACKSPACE - undo the previous action
If you are logged in to your account, this website will remember which cards you know and don't know so that they are in the same box the next time you log in.
When you need a break, try one of the other activities listed below the flashcards like Matching, Snowman, or Hungry Bug. Although it may feel like you're playing a game, your brain is still making more connections with the information to help you out.
To see how well you know the information, try the Quiz or Test activity.
Pass complete!
"Know" box contains: Time elapsed: Retries:
restart all cards
| 739
| 3,255
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40
|
latest
|
en
| 0.920136
|
https://questioncove.com/updates/4ea73265e4b0a7d5141cf981
| 1,695,740,822,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233510214.81/warc/CC-MAIN-20230926143354-20230926173354-00818.warc.gz
| 521,951,908
| 5,472
|
Mathematics 72 Online
OpenStudy (lukecrayonz):
In a dog show, 20% of the dogs are eliminated each day. If the competition starts with 2,000 dogs, to the nearest integer, how many dogs will be present at the start of the 5th day? 259 480 819 1229 In a dog show, 20% of the dogs are eliminated each day. If the competition starts with 2,000 dogs, to the nearest integer, how many dogs will be present at the start of the 5th day? 259 480 819 1229 @Mathematics
OpenStudy (anonymous):
2000(.8^5)
OpenStudy (anonymous):
655.36 so 655
OpenStudy (lukecrayonz):
Thats not a choice...
OpenStudy (anonymous):
2000(.8^4)
OpenStudy (anonymous):
819
OpenStudy (lukecrayonz):
@kp a little late! Haha
OpenStudy (lilymq):
yes she is correct it's 819
Latest Questions
| 228
| 766
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.15625
| 3
|
CC-MAIN-2023-40
|
longest
|
en
| 0.897445
|
https://www.codespeedy.com/convert-a-given-number-into-words-in-cpp/
| 1,675,583,730,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-06/segments/1674764500250.51/warc/CC-MAIN-20230205063441-20230205093441-00539.warc.gz
| 714,634,425
| 16,104
|
# How to convert a given number into words in C++
In this tutorial, we are going to learn to Convert a given number into words in C++. Example (123 to One Hundred Twenty Three).
Introduction
To perform this program we are going to broke the entered number into pieces. The number is broken into Ones and Tens.
## C++ Program to convert number into words
Code:
```#include<iostream>
using namespace std;
void convert(int value)
{
char *first[20] = {"ZERO", "ONE", "TWO", "THREE","FOUR","FIVE","SIX","SEVEN","EIGHT","NINE","TEN",
"ELEVEN","TWELVE","THIRTEEN","FOURTEEN","FIFTEEN","SIXTEEN","SEVENTEEN","EIGHTTEEN","NINETEEN"};
char *second[10] = {"", "TEN", "TWENTY", "THIRTY","FORTY","FIFTY","SIXTY","SEVENTY","EIGHTY","NINETY"};
if(value<0)
{
cout<<" ";
convert(-value);
}
else if(value>=1000)
{
convert(value/1000);
cout<<" THOUSAND";
if(value % 1000)
{
if(value % 1000 < 100)
{
cout << " AND";
}
cout << " " ;
convert(value % 1000);
}
}
else if(value >= 100)
{
convert(value / 100);
cout<<" HUNDRED";
if(value % 100)
{
cout << " AND ";
convert (value % 100);
}
}
else if(value >= 20)
{
cout << second[value / 10];
if(value % 10)
{
cout << " ";
convert(value % 10);
}
}
else
{
cout<<first[value];
}
return;
}
int main()
{
int number;
cout<<"Enter a number : ";
cin>>number;
convert(number);
return 0;
}```
Output:
```Enter a number : 786
SEVEN HUNDRED AND EIGHTY SIX```
Explanation:
Firstly, we create two arrays of string. The first one named “first” contains spelling of numbers from zero to ninety excluding “second” values like ten, twenty, etc.
Secondly, we need to separate the numbers from the value entered. We do this using “%”(Modulus Operator) and “/”(Division Operator).
Example:
If we want to extract digits from number “786”.
The value of (786%10) is 6
Then we divide 786 by 10 it gives 78.
Again to get the last digit 78%10 is 8.
To get the last digit again divide it by 10 78/10 it gives 7.
To get the final digit we perform the same step 7%10 gives 7.
So, the number 786 is now divided into a single digit.
In the code, we first classify the number entered into its range between (value < 0), (value >= 1000 ), (value >= 100), (value >= 20) and else the ones.
Every condition have is used to print the number and after that there postfix like hundred, thousand etc. After the print of the first digit, then condition calls the program itself. This continues till the last digit of the number is printed.
The function checks the single digit with the matching array and prints its correct conversion to words.
#### Conclusion
This is how we can convert any number into words. The same method can be put in use in many places any number can be given a special value and can be used to create a crypto program.
Also Checkout:
Reverse words in a string in C++ without using any built-in function in CPP
### One response to “How to convert a given number into words in C++”
1. Zailani says:
Thank you so much!
It came in handy to me.
| 823
| 2,981
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.8125
| 3
|
CC-MAIN-2023-06
|
latest
|
en
| 0.625427
|
https://placewit.medium.com/min-steps-to-1-using-dp-coding-question-57477413c860?source=post_internal_links---------5----------------------------
| 1,695,475,032,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233506481.17/warc/CC-MAIN-20230923130827-20230923160827-00049.warc.gz
| 518,717,086
| 34,381
|
--
# Problem Statement:
Given a positive integer ’n’, find and return the minimum number of steps that ’n’ has to take to get reduced to 1. You can perform any one of the following 3 steps:
1.) Subtract 1 from it. (n = n — 1) ,
2.) If n is divisible by 2, divide by 2.( if n % 2 == 0, then n = n / 2 ) ,
3.) If n is divisible by 3, divide by 3. (if n % 3 == 0, then n = n / 3 ).
# Input format :
The first and the only line of input contains an integer value, ‘n’.
# Output format :
Print the minimum number of steps.
Sample Input 1 :
`4`
Sample Output 1 :
`2`
Explanation of Sample Output 1 :
`For n = 4Step 1 : n = 4 / 2 = 2Step 2 : n = 2 / 2 = 1`
Sample Input 2 :
`7`
Sample Output 2 :
`3`
Explanation of Sample Output 2 :
`For n = 7Step 1 : n = 7 - 1 = 6Step 2 : n = 6 / 3 = 2 Step 3 : n = 2 / 2 = 1`
| 298
| 833
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.171875
| 3
|
CC-MAIN-2023-40
|
latest
|
en
| 0.719733
|
http://www.nabla.hr/FU-ImpIntDerIntSer4.htm
| 1,718,552,805,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198861665.97/warc/CC-MAIN-20240616141113-20240616171113-00843.warc.gz
| 44,834,631
| 5,200
|
Integral calculus
Differentiation and integration of infinite series
Differentiation of power series
Differentiation and integration of infinite series example
Differentiation and integration of infinite series
If f (x) is represented by the sum of a power series
with radius of convergence r > 0 and - r < x < r, then the function has the derivative
and the function has the integral
Thus, a power series can be differentiated and integrated term by term while the radius of convergence remains the same, with only (possible) exception at the endpoints of the interval of convergence.
Differentiation of power series
Recall that the exponential function f (x) = ex represented by the power series
is absolutely convergent for all real x since
by the ratio test
the limit L < 1 for any value of x.
Applying the power rule
thus, for all real x the function f (x) = ex is equal to its own derivative f ' (x).
Example: Find by representing the integrand function as the power series.
Solution: By substituting - x2 for x in the above power series expansion of ex we get
Let write down the initial sequence of nth order polynomials that describe the function for all real x,
Since every polynomial above is missing the preceding odd degree term, their coefficient an-1 = 0
thus, the coordinates of translations
Therefore, the polynomials that describe the function all are source polynomials of even degree translated in
the direction of the y axis by y0 = 1, as is shown in the picture below.
Note that the roots of odd indexed polynomials in the series correspond to the abscissas of successive even indexed polynomials, as shows the above graph.
On the graph of the bell-shaped curve, representing the probability density function of a normal distribution, at x = ± Ö2/2 denoted are the points of inflections.
Therefore, the power series representing the normal curve converges for all real x.
Hence, by integrating the series term by term obtained is
Functions contents G
| 435
| 1,986
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.125
| 4
|
CC-MAIN-2024-26
|
latest
|
en
| 0.904964
|
https://www.quizover.com/physics-ap/section/problems-exercises-conservative-forces-and-potential-by-openstax
| 1,539,876,662,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-43/segments/1539583511889.43/warc/CC-MAIN-20181018152212-20181018173712-00459.warc.gz
| 1,046,883,382
| 23,344
|
# 7.4 Conservative forces and potential energy (Page 3/8)
Page 3 / 8
## Using conservation of mechanical energy to calculate the speed of a toy car
A 0.100-kg toy car is propelled by a compressed spring, as shown in [link] . The car follows a track that rises 0.180 m above the starting point. The spring is compressed 4.00 cm and has a force constant of 250.0 N/m. Assuming work done by friction to be negligible, find (a) how fast the car is going before it starts up the slope and (b) how fast it is going at the top of the slope.
Strategy
The spring force and the gravitational force are conservative forces, so conservation of mechanical energy can be used. Thus,
${\text{KE}}_{\text{i}}+{\text{PE}}_{\text{i}}={\text{KE}}_{\text{f}}+{\text{PE}}_{\text{f}}$
or
$\frac{1}{2}{{\text{mv}}_{i}}^{2}+{\text{mgh}}_{i}+\frac{1}{2}{{\text{kx}}_{i}}^{2}=\frac{1}{2}{{\text{mv}}_{f}}^{2}+{\text{mgh}}_{f}+\frac{1}{2}{{\text{kx}}_{f}}^{2},$
where $h$ is the height (vertical position) and $x$ is the compression of the spring. This general statement looks complex but becomes much simpler when we start considering specific situations. First, we must identify the initial and final conditions in a problem; then, we enter them into the last equation to solve for an unknown.
Solution for (a)
This part of the problem is limited to conditions just before the car is released and just after it leaves the spring. Take the initial height to be zero, so that both ${h}_{\text{i}}$ and ${h}_{\text{f}}$ are zero. Furthermore, the initial speed ${v}_{\text{i}}$ is zero and the final compression of the spring ${x}_{\text{f}}$ is zero, and so several terms in the conservation of mechanical energy equation are zero and it simplifies to
$\frac{1}{2}{{\text{kx}}_{i}}^{2}=\frac{1}{2}{{\text{mv}}_{f}}^{2}\text{.}$
In other words, the initial potential energy in the spring is converted completely to kinetic energy in the absence of friction. Solving for the final speed and entering known values yields
$\begin{array}{lll}{v}_{f}& =& \sqrt{\frac{k}{m}}{x}_{i}\\ & =& \sqrt{\frac{\text{250}\text{.0 N/m}}{\text{0.100 kg}}}\left(\text{0.0400 m}\right)\\ & =& \text{2.00 m/s.}\end{array}$
Solution for (b)
One method of finding the speed at the top of the slope is to consider conditions just before the car is released and just after it reaches the top of the slope, completely ignoring everything in between. Doing the same type of analysis to find which terms are zero, the conservation of mechanical energy becomes
This form of the equation means that the spring’s initial potential energy is converted partly to gravitational potential energy and partly to kinetic energy. The final speed at the top of the slope will be less than at the bottom. Solving for ${v}_{\text{f}}$ and substituting known values gives
$\begin{array}{lll}{v}_{f}& =& \sqrt{\frac{{{\text{kx}}_{i}}^{2}}{m}-2{\text{gh}}_{f}}\\ & =& \sqrt{\left(\frac{\text{250.0 N/m}}{\text{0.100 kg}}\right)\left(\text{0.0400 m}{\right)}^{2}-2\left(\text{9.80}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}\right)\left(\text{0.180 m}\right)}\\ & =& \text{0.687 m/s.}\end{array}$
Discussion
Another way to solve this problem is to realize that the car’s kinetic energy before it goes up the slope is converted partly to potential energy—that is, to take the final conditions in part (a) to be the initial conditions in part (b).
Note that, for conservative forces, we do not directly calculate the work they do; rather, we consider their effects through their corresponding potential energies, just as we did in [link] . Note also that we do not consider details of the path taken—only the starting and ending points are important (as long as the path is not impossible). This assumption is usually a tremendous simplification, because the path may be complicated and forces may vary along the way.
## Phet explorations: energy skate park
Learn about conservation of energy with a skater dude! Build tracks, ramps and jumps for the skater and view the kinetic energy, potential energy and friction as he moves. You can also take the skater to different planets or even space!
## Section summary
• A conservative force is one for which work depends only on the starting and ending points of a motion, not on the path taken.
• We can define potential energy $\left(\text{PE}\right)$ for any conservative force, just as we defined ${\text{PE}}_{g}$ for the gravitational force.
• The potential energy of a spring is ${\text{PE}}_{s}=\frac{1}{2}{\text{kx}}^{2}$ , where $k$ is the spring’s force constant and $x$ is the displacement from its undeformed position.
• Mechanical energy is defined to be $\text{KE}+\text{PE}$ for a conservative force.
• When only conservative forces act on and within a system, the total mechanical energy is constant. In equation form,
where i and f denote initial and final values. This is known as the conservation of mechanical energy.
## Conceptual questions
What is a conservative force?
The force exerted by a diving board is conservative, provided the internal friction is negligible. Assuming friction is negligible, describe changes in the potential energy of a diving board as a swimmer dives from it, starting just before the swimmer steps on the board until just after his feet leave it.
Define mechanical energy. What is the relationship of mechanical energy to nonconservative forces? What happens to mechanical energy if only conservative forces act?
What is the relationship of potential energy to conservative force?
## Problems&Exercises
A $5\text{.}\text{00}×{\text{10}}^{5}\text{-kg}$ subway train is brought to a stop from a speed of 0.500 m/s in 0.400 m by a large spring bumper at the end of its track. What is the force constant $k$ of the spring?
$\text{7.81}×{\text{10}}^{5}\phantom{\rule{0.20em}{0ex}}\text{N/m}$
A pogo stick has a spring with a force constant of $2\text{.}\text{50}×{\text{10}}^{4}\phantom{\rule{0.20em}{0ex}}\text{N/m}$ , which can be compressed 12.0 cm. To what maximum height can a child jump on the stick using only the energy in the spring, if the child and stick have a total mass of 40.0 kg? Explicitly show how you follow the steps in the Problem-Solving Strategies for Energy .
#### Questions & Answers
derivation of ohms law
Kazeem Reply
derivation of resistance
Kazeem
R=v/I where R=resistor, v=voltage, I=current
Kazeem
Electric current is the flow of electrons
Kelly Reply
is there really flow of electrons exist?
babar
Yes It exists
Cffrrcvccgg
explain plz how electrons flow
babar
if electron flows from where first come and end the first one
babar
an electron will flow accross a conductor because or when it posseses kinectic energy
Cffrrcvccgg
electron can not flow jist trasmit electrical energy
ghulam
free electrons of conductor
ankita
electric means the flow heat current.
Serah Reply
electric means the flow of heat current in a circuit.
Serah
What is electric
Manasseh Reply
electric means?
ghulam
electric means the flow of heat current in a circuit.
Serah
a boy cycles continuously through a distance of 1.0km in 5minutes. calculate his average speed in ms-1(meter per second). how do I solve this
Jenny Reply
speed = distance/time be sure to convert the km to m and minutes to seconds check my utube video "mathwithmrv speed"
PhysicswithMrV
d=1.0km÷1000=0.001 t=5×60=300s s=d\t s=0.001/300=0.0000033m\s
Serah
why we cannot use DC instead of AC in a transformer
kusshaf Reply
becuse the d .c cannot travel for long distance trnsmission
ghulam
what is physics
Chiwetalu Reply
branch of science which deals with matter energy and their relationship between them
ghulam
Life science
the
what is heat and temperature
Kazeem Reply
how does sound affect temperature
Clement Reply
sound is directly proportional to the temperature.
juny
how to solve wave question
Wisdom Reply
I would like to know how I am not at all smart when it comes to math. please explain so I can understand. sincerly
Emma
Just know d relationship btw 1)wave length 2)frequency and velocity
Talhatu
First of all, you are smart and you will get it👍🏽... v = f × wavelength see my youtube channel: "mathwithmrv" if you want to know how to rearrange equations using the balance method
PhysicswithMrV
nice self promotion though xD
Beatrax
thanks dear
Chuks
hi pls help me with this question A ball is projected vertically upwards from the top of a tower 60m high with a velocity of 30ms1.what is the maximum height above the ground level?how long does it take to reach the ground level?
mahmoud
please guys help, what is the difference between concave lens and convex lens
Vincent Reply
convex lens brings rays of light to a focus while concave diverges rays of light
Christian
for mmHg to kPa yes
Matthew
it depends on the size
Matthew Reply
please what is concave lens
Vincent
a lens which diverge the ray of light
rinzuala
concave diverges light
Matthew
thank you guys
Vincent
A diverging lens
Yusuf
What is isotope
Yusuf
each of two or more forms of the same element that contain equal numbers of protons but different numbers of neutrons in their nuclei, and hence differ in relative atomic mass but not in chemical properties; in particular, a radioactive form of an element. "some elements have only one stable isotope
Karthi
what is wire wound resistors?
Naveedkhan Reply
What are the best colleges to go to for physics
Matthew Reply
I would like to know this too
Trevor
How do I calculate uncertainty in a frequency?
Rebecca Reply
Calculate . ..
Olufunsho
### Read also:
#### Get the best College physics course in your pocket!
Source: OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
Google Play and the Google Play logo are trademarks of Google Inc.
Notification Switch
Would you like to follow the 'College physics' conversation and receive update notifications?
By By By Anindyo Mukhopadhyay By
| 2,621
| 9,927
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 22, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.75
| 5
|
CC-MAIN-2018-43
|
longest
|
en
| 0.856817
|
https://studysoup.com/tsg/270953/introductory-intermediate-algebra-for-college-students-4-edition-chapter-6-3-problem-6-1-358
| 1,585,447,357,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-16/segments/1585370493684.2/warc/CC-MAIN-20200329015008-20200329045008-00229.warc.gz
| 747,463,135
| 10,807
|
×
×
# Solved: In Exercises 101104, determine whether each
ISBN: 9780321758941 177
## Solution for problem 6.1.358 Chapter 6.3
Introductory & Intermediate Algebra for College Students | 4th Edition
• Textbook Solutions
• 2901 Step-by-step solutions solved by professors and subject experts
• Get 24/7 help from StudySoup virtual teaching assistants
Introductory & Intermediate Algebra for College Students | 4th Edition
4 5 0 286 Reviews
20
0
Problem 6.1.358
In Exercises 101104, determine whether each statement makes sense or does not make sense and explain your reasoning. Im working with a polynomial that has a GCF other than 1, but then it doesnt factor further, so the polynomial that Im working with is prime.
Step-by-Step Solution:
Step 1 of 3
Lab #1 The Penny Problem: Review The Scientific Method Scientists use the scientific method to answer questions and provide explanations about natural phenomena. It is a logical process based on careful observation and experimentation. 1. Observation: leads to a question and discussion about the observed phenomena. 2. Hypothesis: based on the observation scientists...
Step 2 of 3
Step 3 of 3
#### Related chapters
Unlock Textbook Solution
| 284
| 1,207
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.875
| 3
|
CC-MAIN-2020-16
|
latest
|
en
| 0.850734
|
http://www.cfd-online.com/Forums/cfx/127132-particle-tracking-print.html
| 1,444,010,023,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2015-40/segments/1443736676547.12/warc/CC-MAIN-20151001215756-00128-ip-10-137-6-227.ec2.internal.warc.gz
| 484,119,935
| 6,061
|
CFD Online Discussion Forums (http://www.cfd-online.com/Forums/)
- CFX (http://www.cfd-online.com/Forums/cfx/)
- - particle tracking (http://www.cfd-online.com/Forums/cfx/127132-particle-tracking.html)
sakurabogoda December 4, 2013 06:13
particle tracking
1 Attachment(s)
Hi,
I have a problem regarding particle tracking calculations of a cyclone. I am working with RSM, and using backup files, I can see particles have entered to cyclone cone region, but after that no more extension,and only can see increment of the velocity. But output file shows particle tracking.
Can anybody please describe me why this is happening like this? it is almost 5s passes and total time is 10s.
thanks...
ghorrocks December 4, 2013 07:24
Are you sure you have the boundary conditions correct? The correct inlet flow, and pressures on all exit ports?
sakurabogoda December 4, 2013 08:50
hi,
yes.. I have used velocity at inlet and pressure boundary condition at outlet..
One more thing is with the vector diagram, I can see flow is going out.
ghorrocks December 4, 2013 17:36
Yes, but have you set the correct pressures and flow rates? If the conditions are not set correctly then the device will not function as intended, and strange things can happen - like what you are seeing.
sakurabogoda December 4, 2013 22:49
hi Glenn,
Thanks a lot. But truely I cannot find a mistake with my model. I am attaching ccl file with this.
Could you pls have a look?
LIBRARY:
CEL:
EXPRESSIONS:
MFR = 2.828E-8[kg s^-1]
PR = 1E7*step(0.0001-t/1[s]) [s^-1]
END
END
MATERIAL: Air at 25 C
Material Description = Air at 25 C and 1 atm (dry)
Material Group = Air Data, Constant Property Gases
Option = Pure Substance
Thermodynamic State = Gas
PROPERTIES:
Option = General Material
EQUATION OF STATE:
Density = 1.185 [kg m^-3]
Molar Mass = 28.96 [kg kmol^-1]
Option = Value
END
SPECIFIC HEAT CAPACITY:
Option = Value
Specific Heat Capacity = 1.0044E+03 [J kg^-1 K^-1]
Specific Heat Type = Constant Pressure
END
REFERENCE STATE:
Option = Specified Point
Reference Pressure = 1 [atm]
Reference Specific Enthalpy = 0. [J/kg]
Reference Specific Entropy = 0. [J/kg/K]
Reference Temperature = 25 [C]
END
DYNAMIC VISCOSITY:
Dynamic Viscosity = 1.831E-05 [kg m^-1 s^-1]
Option = Value
END
THERMAL CONDUCTIVITY:
Option = Value
Thermal Conductivity = 2.61E-02 [W m^-1 K^-1]
END
ABSORPTION COEFFICIENT:
Absorption Coefficient = 0.01 [m^-1]
Option = Value
END
SCATTERING COEFFICIENT:
Option = Value
Scattering Coefficient = 0.0 [m^-1]
END
REFRACTIVE INDEX:
Option = Value
Refractive Index = 1.0 [m m^-1]
END
THERMAL EXPANSIVITY:
Option = Value
Thermal Expansivity = 0.003356 [K^-1]
END
END
END
MATERIAL: Particles
Material Group = Particle Solids
Option = Pure Substance
PROPERTIES:
Option = General Material
EQUATION OF STATE:
Density = 500 [kg m^-3]
Molar Mass = 1.0 [kg kmol^-1]
Option = Value
END
END
END
END
FLOW: Flow Analysis 1
SOLUTION UNITS:
Length Units = [m]
Mass Units = [kg]
Solid Angle Units = [sr]
Temperature Units = [K]
Time Units = [s]
END
ANALYSIS TYPE:
Option = Transient
EXTERNAL SOLVER COUPLING:
Option = None
END
INITIAL TIME:
Option = Automatic with Value
Time = 0 [s]
END
TIME DURATION:
Option = Total Time
Total Time = 10 [s]
END
TIME STEPS:
First Update Time = 0.0 [s]
Initial Timestep = 0.00001 [s]
Timestep Update Frequency = 1
Maximum Timestep = 0.005 [s]
Minimum Timestep = 0.00001 [s]
Option = Number of Coefficient Loops
Target Maximum Coefficient Loops = 4
Target Minimum Coefficient Loops = 2
Timestep Decrease Factor = 0.8
Timestep Increase Factor = 1.06
END
END
END
DOMAIN: cyclone
Coord Frame = Coord 0
Domain Type = Fluid
Location = CREATED_MATERIAL_17,CREATED_MATERIAL_16
BOUNDARY: Box walls
Boundary Type = WALL
Location = BOX
BOUNDARY CONDITIONS:
MASS AND MOMENTUM:
Option = No Slip Wall
END
WALL ROUGHNESS:
Option = Smooth Wall
END
END
FLUID: particles
BOUNDARY CONDITIONS:
PARTICLE WALL INTERACTION:
Option = Equation Dependent
END
VELOCITY:
Option = Restitution Coefficient
Parallel Coefficient of Restitution = 1.0
Perpendicular Coefficient of Restitution = 1.0
END
END
END
END
BOUNDARY: cyclone walls
Boundary Type = WALL
Location = \
VORTEX_FINDER_1,VORTEX_FINDER_2,TOP_CAP,OULLET_TUB E,INLET_TUBE,HOPPER\
_BASE,HOPPER,BODY
BOUNDARY CONDITIONS:
MASS AND MOMENTUM:
Option = No Slip Wall
END
WALL ROUGHNESS:
Option = Smooth Wall
END
END
FLUID: particles
BOUNDARY CONDITIONS:
PARTICLE WALL INTERACTION:
Option = Equation Dependent
END
VELOCITY:
Option = Restitution Coefficient
Parallel Coefficient of Restitution = 1.0
Perpendicular Coefficient of Restitution = 1.0
END
END
END
END
BOUNDARY: inlet
Boundary Type = INLET
Location = INLET
BOUNDARY CONDITIONS:
FLOW REGIME:
Option = Subsonic
END
MASS AND MOMENTUM:
Normal Speed = 10 [m s^-1]
Option = Normal Speed
END
TURBULENCE:
Option = Medium Intensity and Eddy Viscosity Ratio
END
END
FLUID: particles
BOUNDARY CONDITIONS:
MASS AND MOMENTUM:
Normal Speed = 10 [m s^-1]
Option = Normal Speed
END
PARTICLE MASS FLOW RATE:
Mass Flow Rate = MFR
END
PARTICLE POSITION:
Option = Uniform Injection
NUMBER OF POSITIONS:
Number per Unit Time = PR
Option = Direct Specification
END
END
END
END
END
BOUNDARY: oultet
Boundary Type = OPENING
Location = OUTLET
BOUNDARY CONDITIONS:
FLOW DIRECTION:
Option = Normal to Boundary Condition
END
FLOW REGIME:
Option = Subsonic
END
MASS AND MOMENTUM:
Option = Opening Pressure and Direction
Relative Pressure = 0 [Pa]
END
TURBULENCE:
Option = Medium Intensity and Eddy Viscosity Ratio
END
END
END
DOMAIN MODELS:
BUOYANCY MODEL:
Buoyancy Reference Density = 1.2 [kg m^-3]
Gravity X Component = 0 [m s^-2]
Gravity Y Component = -9.81 [m s^-2]
Gravity Z Component = 0 [m s^-2]
Option = Buoyant
BUOYANCY REFERENCE LOCATION:
Option = Automatic
END
END
DOMAIN MOTION:
Option = Stationary
END
MESH DEFORMATION:
Option = None
END
REFERENCE PRESSURE:
Reference Pressure = 101325 [Pa]
END
END
FLUID DEFINITION: Air
Material = Air at 25 C
Option = Material Library
MORPHOLOGY:
Option = Continuous Fluid
END
END
FLUID DEFINITION: particles
Material = Particles
Option = Material Library
MORPHOLOGY:
Option = Dispersed Particle Transport Solid
PARTICLE DIAMETER DISTRIBUTION:
Maximum Diameter = 18 [micron]
Mean Diameter = 5.4519154808 [micron]
Minimum Diameter = 0.742 [micron]
Option = Normal in Diameter by Number
Standard Deviation in Diameter = 4.7494103370 [micron]
END
END
END
FLUID MODELS:
COMBUSTION MODEL:
Option = None
END
FLUID: Air
FLUID BUOYANCY MODEL:
Option = Density Difference
END
END
FLUID: particles
EROSION MODEL:
Option = None
END
FLUID BUOYANCY MODEL:
Option = Density Difference
END
PARTICLE ROUGH WALL MODEL:
Option = None
END
END
HEAT TRANSFER MODEL:
Option = None
END
Option = None
END
TURBULENCE MODEL:
Option = SSG Reynolds Stress
BUOYANCY TURBULENCE:
Option = None
END
END
TURBULENT WALL FUNCTIONS:
Option = Scalable
END
END
FLUID PAIR: Air | particles
Particle Coupling = One-way Coupling
MOMENTUM TRANSFER:
DRAG FORCE:
Option = Schiller Naumann
END
Option = None
END
TURBULENT DISPERSION FORCE:
Option = None
END
VIRTUAL MASS FORCE:
Option = None
END
END
END
END
OUTPUT CONTROL:
MONITOR OBJECTS:
MONITOR BALANCES:
Option = Full
END
MONITOR FORCES:
Option = Full
END
MONITOR PARTICLES:
Option = Full
END
MONITOR POINT: point1
Cartesian Coordinates = -0.1305 [m], 0.21944 [m], 0 [m]
Option = Cartesian Coordinates
Output Variables List = Velocity,Velocity u,Velocity v,Velocity w
END
MONITOR POINT: point2
Cartesian Coordinates = 0.0[m],0.0[m],0.0[m]
Option = Cartesian Coordinates
Output Variables List = Velocity,Velocity u,Velocity v,Velocity w
END
MONITOR RESIDUALS:
Option = Full
END
MONITOR TOTALS:
Option = Full
END
END
RESULTS:
File Compression Level = Default
Option = Standard
END
TRANSIENT RESULTS: Transient Results 1
File Compression Level = Default
Option = Standard
OUTPUT FREQUENCY:
Option = Time Interval
Time Interval = 1 [s]
END
END
TRANSIENT STATISTICS: velocity max
Option = Maximum
Output Variables List = Velocity
END
TRANSIENT STATISTICS: velocity min
Option = Maximum
Output Variables List = Velocity
END
TRANSIENT STATISTICS: velocity rms
Option = Root Mean Square
Output Variables List = Velocity
END
TRANSIENT STATISTICS: velocity std
Option = Standard Deviation
Output Variables List = Velocity
END
END
SOLVER CONTROL:
Turbulence Numerics = First Order
Option = High Resolution
END
CONVERGENCE CONTROL:
Maximum Number of Coefficient Loops = 5
Minimum Number of Coefficient Loops = 1
Timescale Control = Coefficient Loops
END
CONVERGENCE CRITERIA:
Residual Target = 0.00001
Residual Type = RMS
END
PARTICLE CONTROL:
PARTICLE INTEGRATION:
Option = Forward Euler
END
PARTICLE TERMINATION CONTROL:
Maximum Number of Integration Steps = 1000000
Maximum Tracking Distance = 10000 [m]
Maximum Tracking Time = 10 [s]
END
END
TRANSIENT SCHEME:
Option = Second Order Backward Euler
TIMESTEP INITIALISATION:
Lower Courant Number = 0.00001
Option = Automatic
Upper Courant Number = 1
END
END
END
END
COMMAND FILE:
Version = 14.0
Results Version = 14.0
END
SIMULATION CONTROL:
EXECUTION CONTROL:
EXECUTABLE SELECTION:
Double Precision = Yes
END
PARALLEL HOST LIBRARY:
HOST DEFINITION: node01.local
Installation Root = /ansys_inc/v%v/CFX
Host Architecture String = linux-amd64
END
HOST DEFINITION: node02.local
Installation Root = /ansys_inc/v%v/CFX
Host Architecture String = linux-amd64
END
HOST DEFINITION: node03.local
Host Architecture String = linux-amd64
Installation Root = /ansys_inc/v%v/CFX
END
END
PARTITIONER STEP CONTROL:
MEMORY CONTROL:
Memory Allocation Factor = 1.2
END
PARTITIONING TYPE:
Option = MeTiS
MeTiS Type = k-way
Partition Size Rule = Automatic
Partition Weight Factors = 0.03030, 0.03030, 0.03030, 0.03030, \
0.03030, 0.03030, 0.03030, 0.03030, 0.03030, 0.03030, 0.03030, \
0.03030, 0.03030, 0.03030, 0.03030, 0.03030, 0.03030, 0.03030, \
0.03030, 0.03030, 0.03030, 0.03030, 0.03030, 0.03030, 0.03030, \
0.03030, 0.03030, 0.03030, 0.03030, 0.03030, 0.03030, 0.03030, \
0.03030
END
END
RUN DEFINITION:
Solver Input File = /home/sganegama2/mul10/mul10_2/M10.def
Run Mode = Full
INITIAL VALUES SPECIFICATION:
INITIAL VALUES CONTROL:
Use Mesh From = Solver Input File
Continue History From = Initial Values 1
END
INITIAL VALUES: Initial Values 1
Option = Results File
File Name = /home/sganegama2/cy10ms/10msS_001.res
END
END
END
SOLVER STEP CONTROL:
Runtime Priority = Standard
MEMORY CONTROL:
Memory Allocation Factor = 5
END
PARALLEL ENVIRONMENT:
Start Method = MPICH Distributed Parallel
Parallel Host List = node03.local*11,node02.local*11,node01.local*11
END
END
END
END
ghorrocks December 4, 2013 23:50
No, I do not think you understand my point. A cyclone will only work properly if the inlets and outlets are set to the correct pressures and flows. In this case you do not appear to either two outlets, you only have one. You need a clean exit at the top and a dirty exit at the bottom.
http://en.wikipedia.org/wiki/Cyclonic_separation
sakurabogoda December 5, 2013 00:09
Really?? I thought, according to practical, there is a dust collection hopper at base and I have used same way..
Is it a must to use a outlet/opening at the bottom? which BC is better?
many thanks..
ghorrocks December 5, 2013 00:12
Rather than guessing, how about you do some research on cyclone design (the link I posted is a good starting point) so you know how to design a good cyclone for your situation? That would seem to be the way forwards to me.
All times are GMT -4. The time now is 21:53.
| 3,462
| 11,449
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.609375
| 3
|
CC-MAIN-2015-40
|
longest
|
en
| 0.881861
|
https://www.esaral.com/q/evaluate-the-following-integrals-73309
| 1,675,595,253,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-06/segments/1674764500251.38/warc/CC-MAIN-20230205094841-20230205124841-00679.warc.gz
| 758,288,302
| 37,891
|
# Evaluate the following integrals:
Question:
Evaluate the following integrals:
$\int \frac{e^{3 x}}{4 e^{6 x}-9} d x$
Solution:
let $I=\int \frac{e^{2 x}}{4 e^{6 x_{-9}}} d x$
Let $\mathrm{e}^{3 \mathrm{x}}=\mathrm{t} \ldots \ldots(\mathrm{i})$
$\Rightarrow 3 e^{3 x} d x=d t$
$I=\frac{1}{3} \int \frac{1}{4 t^{2}-9} d t$
$=\frac{1}{12} \int \frac{1}{t^{2}-\frac{9}{4}} d t$
$I=\frac{1}{12} \int \frac{1}{t^{2}-\left(\frac{3}{2}\right)^{2}} d t$
$I=\frac{1}{36} \log \left|\frac{t-\frac{3}{2}}{t+\frac{3}{2}}\right|+c$
[since, $\left.\int \frac{1}{x^{2}-(a)^{2}} d x=\frac{1}{2 \times a} \log \left|\frac{x-a}{x+a}\right|+c\right]$
$I=\log \left|\frac{2 t-3}{2 t+3}\right|+c$
$I=\log \left|\frac{2 e^{2 x}-3}{2 e^{2 x}+3}\right|+c$ [using (i)]
| 348
| 758
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.34375
| 4
|
CC-MAIN-2023-06
|
longest
|
en
| 0.201831
|
http://slideplayer.com/slide/3935521/
| 1,656,362,959,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-27/segments/1656103341778.23/warc/CC-MAIN-20220627195131-20220627225131-00412.warc.gz
| 50,216,462
| 21,813
|
# 1 Slides based on those of Kenneth H. Rosen Slides by Sylvia Sorkin, Community College of Baltimore County - Essex Campus Graphs.
## Presentation on theme: "1 Slides based on those of Kenneth H. Rosen Slides by Sylvia Sorkin, Community College of Baltimore County - Essex Campus Graphs."— Presentation transcript:
1 Slides based on those of Kenneth H. Rosen Slides by Sylvia Sorkin, Community College of Baltimore County - Essex Campus Graphs
2 Definition 1. A simple graph G = (V, E) consists of V, a nonempty set of vertices, and E, a set of unordered pairs of distinct elements of V called edges. Simple Graph
3 A simple graph San Francisco Denver Los Angeles New York Chicago Washington Detroit How many vertices? How many edges?
4 A simple graph V = { Chicago, Denver, Detroit, Los Angeles, New York, San Francisco, Washington } SET OF VERTICES E = { {San Francisco, Los Angeles}, {San Francisco, Denver}, {Los Angeles, Denver}, {Denver, Chicago}, {Chicago, Detroit}, {Detroit, New York}, {New York, Washington}, {Chicago, Washington}, {Chicago, New York} } SET OF EDGES
5 A simple graph San Francisco Denver Los Angeles New York Chicago Washington Detroit The network is made up of computers and telephone lines between computers. There is at most 1 telephone line between 2 computers in the network. Each line operates in both directions. No computer has a telephone line to itself. These are undirected edges, each of which connects two distinct vertices, and no two edges connect the same pair of vertices.
6 Definition 2. In a multigraph G = (V, E) two or more edges may connect the same pair of vertices. A Non-Simple Graph
7 A Multigraph San Francisco Denver Los Angeles New York Chicago Washington Detroit THERE CAN BE MULTIPLE TELEPHONE LINES BETWEEN TWO COMPUTERS IN THE NETWORK.
8 Multiple Edges San Francisco Denver Los Angeles New York Chicago Washington Detroit Two edges are called multiple or parallel edges if they connect the same two distinct vertices.
9 Definition 3. In a pseudograph G = (V, E) two or more edges may connect the same pair of vertices, and in addition, an edge may connect a vertex to itself. Another Non-Simple Graph
10 A Pseudograph San Francisco Denver Los Angeles New York Chicago Washington Detroit THERE CAN BE TELEPHONE LINES IN THE NETWORK FROM A COMPUTER TO ITSELF (for diagnostic use).
11 Loops San Francisco Denver Los Angeles New York Chicago Washington Detroit An edge is called a loop if it connects a vertex to itself.
12 Undirected Graphs pseudographs simple graphs multigraphs
13 Definition 4. In a directed graph G = (V, E) the edges are ordered pairs of (not necessarily distinct) vertices. A Directed Graph
14 A Directed Graph San Francisco Denver Los Angeles New York Chicago Washington Detroit SOME TELEPHONE LINES IN THE NETWORK MAY OPERATE IN ONLY ONE DIRECTION. Those that operate in two directions are represented by pairs of edges in opposite directions.
15 Definition 5. In a directed multigraph G = (V, E) the edges are ordered pairs of (not necessarily distinct) vertices, and in addition there may be multiple edges. A Directed Multigraph
16 A Directed Multigraph San Francisco Denver Los Angeles New York Chicago Washington Detroit THERE MAY BE SEVERAL ONE-WAY LINES IN THE SAME DIRECTION FROM ONE COMPUTER TO ANOTHER IN THE NETWORK.
17 Types of Graphs TYPE EDGES MULTIPLE EDGES LOOPS ALLOWED? ALLOWED? Simple graph Undirected NO NO Multigraph Undirected YES NO Pseudograph Undirected YES YES Directed graph Directed NO YES Directed multigraphDirected YES YES
18 Definition 1. Two vertices, u and v in an undirected graph G are called adjacent (or neighbors) in G, if {u, v} is an edge of G. An edge e connecting u and v is called incident with vertices u and v, or is said to connect u and v. The vertices u and v are called endpoints of edge {u, v}. Adjacent Vertices (Neighbors)
19 a Degree of a vertex Definition 1. The degree of a vertex in an undirected graph is the number of edges incident with it, except that a loop at a vertex contributes twice to the degree of that vertex. b g f e c d deg( d ) = 1
20 a Degree of a vertex Definition 1. The degree of a vertex in an undirected graph is the number of edges incident with it, except that a loop at a vertex contributes twice to the degree of that vertex. b g f e c d deg( e ) = 0
21 a deg( b ) = 6 Degree of a vertex Definition 1. The degree of a vertex in an undirected graph is the number of edges incident with it, except that a loop at a vertex contributes twice to the degree of that vertex. b g f e c d
22 a deg( b ) = 6 Degree of a vertex Find the degree of all the other vertices. deg( a )deg( c )deg( f )deg( g ) b g f e c d deg( d ) = 1 deg( e ) = 0
23 a deg( b ) = 6 Degree of a vertex Find the degree of all the other vertices. deg( a ) = 2deg( c ) = 4deg( f ) = 3deg( g ) = 4 b g f e c d deg( d ) = 1 deg( e ) = 0
24 a deg( b ) = 6 Degree of a vertex Find the degree of all the other vertices. deg( a ) = 2deg( c ) = 4deg( f ) = 3deg( g ) = 4 TOTAL of degrees = 2 + 4 + 3 + 4 + 6 + 1 + 0 = 20 b g f e c d deg( d ) = 1 deg( e ) = 0
25 a deg( b ) = 6 Degree of a vertex Find the degree of all the other vertices. deg( a ) = 2deg( c ) = 4deg( f ) = 3deg( g ) = 4 TOTAL of degrees = 2 + 4 + 3 + 4 + 6 + 1 + 0 = 20 TOTAL NUMBER OF EDGES = 10 b g f e c d deg( d ) = 1 deg( e ) = 0
26 Theorem 1. Let G = (V, E) be an undirected graph G with e edges. Then deg( v ) = 2 e v V “The sum of the degrees over all the vertices equals twice the number of edges.” NOTE: This applies even if multiple edges and loops are present. Handshaking Theorem
27 Definition 6. A subgraph of a graph G = (V, E) is a graph H = (W, F) where W V and F E. Subgraph
28 C 5 is a subgraph of K 5 C 5 K 5
29 Definition 7. The union of 2 simple graphs G 1 = ( V 1, E 1 ) and G 2 = ( V 2, E 2 ) is the simple graph with vertex set V = V 1 V 2 and edge set E = E 1 E 2. The union is denoted by G 1 G 2. Union
30 W 5 is the union of S 5 and C 5 C 5 W 5 S 5 a b c e d a c e a b c e d a b c e d f f
31 A simple graph G = (V, E) with n vertices can be represented by its adjacency matrix, A, where entry a ij in row i and column j is 1 if { v i, v j } is an edge in G, a ij = 0 otherwise. Adjacency Matrix
32 Finding the adjacency matrix This graph has 6 vertices a, b, c, d, e, f. We can arrange them in that order. d W 5 a b c e a c e f
33 Finding the adjacency matrix a b c d e f d a 0 1 0 0 1 1 b c d e f FROM TO There are edges from a to b, from a to e, and from a to f W 5 a b c e a c e f
34 Finding the adjacency matrix a b c d e f d a 0 1 0 0 1 1 b 1 0 1 0 0 1 c d e f FROM TO There are edges from b to a, from b to c, and from b to f W 5 a b c e a c e f
35 Finding the adjacency matrix a b c d e f d a 0 1 0 0 1 1 b 1 0 1 0 0 1 c 0 1 0 1 0 1 d e f FROM TO There are edges from c to b, from c to d, and from c to f W 5 a b c e a c e f
36 Finding the adjacency matrix a b c d e f a 0 1 0 0 1 1 b 1 0 1 0 0 1 c 0 1 0 1 0 1 d e f FROM TO COMPLETE THE ADJACENCY MATRIX... d W 5 a b c e a c e f
37 Finding the adjacency matrix a b c d e f d a 0 1 0 0 1 1 b 1 0 1 0 0 1 c 0 1 0 1 0 1 d 0 0 1 0 1 1 e 1 0 0 1 0 1 f 1 1 1 1 1 0 FROM TO Notice that this matrix is symmetric. That is a ij = a ji Why? W 5 a b c e a c e f
38 Definition 1. A path of length n from u to v in an undirected graph is a sequence of edges e 1, e 2,..., e n of the graph such that edge e 1 has endpoints x o and x 1, edge e 2 has endpoints x 1 and x 2,... and edge e n has endpoints x n-1 and x n, where x 0 = u and x n = v. Path of Length n
39 One path from a to e This path passes through vertices f and d in that order. d W 5 a b c e a c f e
40 One path from a to a This path passes through vertices f, d, e, in that order. It has length 4. It is a circuit because it begins and ends at the same vertex. It is called simple because it does not contain the same edge more than once. d W 5 a b c e a c f e
41 Definition 3. An undirected graph is called connected if there is a path between every pair of distinct vertices of the graph. IS THIS GRAPH CONNECTED? Path of Length n W 5 a b c e a c e f
42 Theorem 1. There is a simple path between every pair of distinct vertices of a connected undirected graph. Theorem 1
43 Theorem 2. Let G be a graph with adjacency matrix A with respect to the ordering v 1, v 2,..., v n. The number of different paths of length r from v i to v j, where r is a postive integer, equals the entry in row i and column j of A r. NOTE: This applies with directed or undirected edges, with multiple edges and loops allowed. Paths of Length r between Vertices
Download ppt "1 Slides based on those of Kenneth H. Rosen Slides by Sylvia Sorkin, Community College of Baltimore County - Essex Campus Graphs."
Similar presentations
| 2,583
| 8,791
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.4375
| 3
|
CC-MAIN-2022-27
|
latest
|
en
| 0.842837
|
https://gmatclub.com/forum/because-of-the-erratic-pattern-of-sales-increases-this-year-retailers-294487-40.html
| 1,685,412,890,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224644915.48/warc/CC-MAIN-20230530000715-20230530030715-00380.warc.gz
| 324,660,143
| 67,572
|
It is currently 29 May 2023, 19:14
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Because of the erratic pattern of sales increases this year retailers
Intern
Joined: 13 Jan 2017
Posts: 32
GMAT Club Verbal Expert
Joined: 13 Aug 2009
Status: GMAT/GRE/LSAT tutors
Posts: 6428
Location: United States (CO)
GMAT 1: 780 Q51 V46
GMAT 2: 800 Q51 V51
GRE 1: Q170 V170
GRE 2: Q170 V170
GMAT Club Verbal Expert
Joined: 13 Aug 2009
Status: GMAT/GRE/LSAT tutors
Posts: 6428
Location: United States (CO)
GMAT 1: 780 Q51 V46
GMAT 2: 800 Q51 V51
GRE 1: Q170 V170
GRE 2: Q170 V170
Manager
Joined: 03 Jan 2019
Posts: 177
Intern
Joined: 05 Oct 2020
Posts: 2
GMAT Club Verbal Expert
Joined: 13 Aug 2009
Status: GMAT/GRE/LSAT tutors
Posts: 6428
Location: United States (CO)
GMAT 1: 780 Q51 V46
GMAT 2: 800 Q51 V51
GRE 1: Q170 V170
GRE 2: Q170 V170
Intern
Joined: 23 Apr 2019
Posts: 21
GMAT Club Legend
Joined: 15 Jul 2015
Posts: 4594
Location: India
GMAT 1: 780 Q50 V51
GRE 1: Q170 V169
Experts' Global Representative
Joined: 10 Jul 2017
Posts: 5095
Location: India
GMAT Date: 11-01-2019
GMAT Expert
Joined: 16 Oct 2010
Posts: 13773
Location: Pune, India
Manager
Joined: 28 Sep 2021
Posts: 160
Experts' Global Representative
Joined: 10 Jul 2017
Posts: 5095
Location: India
GMAT Date: 11-01-2019
GMAT Club Legend
Joined: 15 Jul 2015
Posts: 4594
Location: India
GMAT 1: 780 Q50 V51
GRE 1: Q170 V169
Intern
Joined: 30 Mar 2022
Posts: 20
Location: India
GMAT 1: 720 Q51 V35 (Online)
GPA: 4
Manhattan Prep Instructor
Joined: 30 Apr 2021
Posts: 503
GMAT 1: 760 Q49 V47
Manager
Joined: 08 Jun 2021
Status:In learning mode...
Posts: 160
Location: India
GMAT 1: 600 Q46 V27
Manhattan Prep Instructor
Joined: 30 Apr 2021
Posts: 503
GMAT 1: 760 Q49 V47
GMAT Club Legend
Joined: 15 Jul 2015
Posts: 4594
Location: India
GMAT 1: 780 Q50 V51
GRE 1: Q170 V169
Intern
Joined: 17 May 2021
Posts: 27
GMAT Club Verbal Expert
Joined: 13 Aug 2009
Status: GMAT/GRE/LSAT tutors
Posts: 6428
Location: United States (CO)
GMAT 1: 780 Q51 V46
GMAT 2: 800 Q51 V51
GRE 1: Q170 V170
GRE 2: Q170 V170
Moderators:
GMAT Club Verbal Expert
6428 posts
GMAT Club Verbal Expert
238 posts
| 947
| 2,574
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.625
| 3
|
CC-MAIN-2023-23
|
latest
|
en
| 0.796038
|
http://www.dummies.com/how-to/content/how-power-supplies-turn-ac-into-dc-in-electronic-c.navId-810951.html
| 1,469,930,004,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-30/segments/1469258948335.92/warc/CC-MAIN-20160723072908-00110-ip-10-185-27-174.ec2.internal.warc.gz
| 406,004,673
| 16,105
|
The task of turning alternating current into direct current is called rectification, and the electronic circuit that does the job is called a rectifier. The most common way to convert alternating current into direct current is to use one or more diodes, those handy electronic components that allow current to pass in one direction but not the other.
Although a rectifier converts alternating current to direct current, the resulting direct current isn't a steady voltage. It would be more accurate to refer to it as “pulsating DC.” Although the pulsating DC current always moves in the same direction, the voltage level has a distinct ripple to it, rising and falling a bit in sync with the waveform of the AC voltage that’s fed into the rectifier.
For many DC circuits, a significant amount of ripple in the power supply can cause the circuit to malfunction. Therefore, additional filtering is required to “flatten” the pulsating DC that comes from a rectifier to eliminate the ripple.
There are three distinct types of rectifier circuits you can build: half-wave, full-wave, and bridge. The following describes each of these three rectifier types.
Half-wave rectifier
The simplest type of rectifier is made from a single diode. This type of rectifier is called a half-wave rectifier because it passes just half of the AC input voltage to the output.
When the AC voltage is positive on the cathode side of the diode, the diode allows the current to pass through to the output. But when the AC current reverses direction and becomes negative on the cathode side of the diode, the diode blocks the current so that no voltage appears at the output.
Half-wave rectifiers are simple enough to build but aren't very efficient. That’s because the entire negative cycle of the AC input is blocked by a half-wave rectifier. As a result, output voltage is zero half of the time. This causes the average voltage at the output to be half of the input voltage.
Note the resistor marked RL. This resistor isn’t actually a part of the rectifier circuit. Instead, it represents the resistance imposed by the load that will ultimately be placed on the circuit when the power supply is put to use.
Full-wave rectifier
A full-wave rectifier uses two diodes, which enables it to pass both the positive and the negative side of the alternating current input. The diodes are connected to the transformer.
Notice that the full-wave rectifier requires that you use a center-tapped transformer. The diodes are connected to the two outer taps, and the center tap is used as a common ground for the rectified DC voltage. The full-wave rectifier converts both halves of the AC sine wave to positive-voltage direct current.
The result is DC voltage that pulses at twice the frequency of the input AC voltage. In other words, assuming the input is 60 Hz household current, the output will be DC pulsing at 120 Hz.
Bridge rectifier
The problem with a full-wave rectifier is that it requires a center-tapped transformer, so it produces DC that’s just half of the total output voltage of the transformer.
A bridge rectifier overcomes this limitation by using four diodes instead of two. The diodes are arranged in a diamond pattern so that, on each half phase of the AC sine wave, two of the diodes pass the current to the positive and negative sides of the output, and the other two diodes block current. A bridge rectifier doesn't require a center-tapped transformer.
The output from a bridge rectifier is pulsed DC, just like the output from a full-wave rectifier. However, the full voltage of the transformer’s secondary coil is used.
You can construct a bridge rectifier using four diodes, or you can use a bridge rectifier IC that contains the four diodes in the correct arrangement. A bridge rectifier IC has four pins: two for the AC input and two for the DC output.
| 805
| 3,860
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.625
| 3
|
CC-MAIN-2016-30
|
latest
|
en
| 0.923397
|
http://de.metamath.org/mpeuni/p1evtxdp1.html
| 1,716,557,900,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971058719.70/warc/CC-MAIN-20240524121828-20240524151828-00475.warc.gz
| 7,830,486
| 6,659
|
Mathbox for Alexander van der Vekens < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > p1evtxdp1 Structured version Visualization version GIF version
Theorem p1evtxdp1 40730
Description: If an edge 𝐸 (not being a loop) which contains vertex 𝑈 is added to a graph 𝐺 (yielding a graph 𝐹), the degree of 𝑈 is increased by 1. (Contributed by AV, 3-Mar-2021.)
Hypotheses
Ref Expression
p1evtxdeq.v 𝑉 = (Vtx‘𝐺)
p1evtxdeq.i 𝐼 = (iEdg‘𝐺)
p1evtxdeq.f (𝜑 → Fun 𝐼)
p1evtxdeq.fv (𝜑 → (Vtx‘𝐹) = 𝑉)
p1evtxdeq.fi (𝜑 → (iEdg‘𝐹) = (𝐼 ∪ {⟨𝐾, 𝐸⟩}))
p1evtxdeq.k (𝜑𝐾𝑋)
p1evtxdeq.d (𝜑𝐾 ∉ dom 𝐼)
p1evtxdeq.u (𝜑𝑈𝑉)
p1evtxdp1.e (𝜑𝐸 ∈ 𝒫 𝑉)
p1evtxdp1.n (𝜑𝑈𝐸)
p1evtxdp1.l (𝜑 → 2 ≤ (#‘𝐸))
Assertion
Ref Expression
p1evtxdp1 (𝜑 → ((VtxDeg‘𝐹)‘𝑈) = (((VtxDeg‘𝐺)‘𝑈) +𝑒 1))
Proof of Theorem p1evtxdp1
StepHypRef Expression
1 p1evtxdeq.v . . 3 𝑉 = (Vtx‘𝐺)
2 p1evtxdeq.i . . 3 𝐼 = (iEdg‘𝐺)
3 p1evtxdeq.f . . 3 (𝜑 → Fun 𝐼)
4 p1evtxdeq.fv . . 3 (𝜑 → (Vtx‘𝐹) = 𝑉)
5 p1evtxdeq.fi . . 3 (𝜑 → (iEdg‘𝐹) = (𝐼 ∪ {⟨𝐾, 𝐸⟩}))
6 p1evtxdeq.k . . 3 (𝜑𝐾𝑋)
7 p1evtxdeq.d . . 3 (𝜑𝐾 ∉ dom 𝐼)
8 p1evtxdeq.u . . 3 (𝜑𝑈𝑉)
9 p1evtxdp1.e . . 3 (𝜑𝐸 ∈ 𝒫 𝑉)
101, 2, 3, 4, 5, 6, 7, 8, 9p1evtxdeqlem 40728 . 2 (𝜑 → ((VtxDeg‘𝐹)‘𝑈) = (((VtxDeg‘𝐺)‘𝑈) +𝑒 ((VtxDeg‘⟨𝑉, {⟨𝐾, 𝐸⟩}⟩)‘𝑈)))
11 fvex 6113 . . . . . . 7 (Vtx‘𝐺) ∈ V
121, 11eqeltri 2684 . . . . . 6 𝑉 ∈ V
13 snex 4835 . . . . . 6 {⟨𝐾, 𝐸⟩} ∈ V
1412, 13pm3.2i 470 . . . . 5 (𝑉 ∈ V ∧ {⟨𝐾, 𝐸⟩} ∈ V)
15 opiedgfv 25684 . . . . 5 ((𝑉 ∈ V ∧ {⟨𝐾, 𝐸⟩} ∈ V) → (iEdg‘⟨𝑉, {⟨𝐾, 𝐸⟩}⟩) = {⟨𝐾, 𝐸⟩})
1614, 15mp1i 13 . . . 4 (𝜑 → (iEdg‘⟨𝑉, {⟨𝐾, 𝐸⟩}⟩) = {⟨𝐾, 𝐸⟩})
17 opvtxfv 25681 . . . . 5 ((𝑉 ∈ V ∧ {⟨𝐾, 𝐸⟩} ∈ V) → (Vtx‘⟨𝑉, {⟨𝐾, 𝐸⟩}⟩) = 𝑉)
1814, 17mp1i 13 . . . 4 (𝜑 → (Vtx‘⟨𝑉, {⟨𝐾, 𝐸⟩}⟩) = 𝑉)
19 p1evtxdp1.n . . . 4 (𝜑𝑈𝐸)
20 p1evtxdp1.l . . . 4 (𝜑 → 2 ≤ (#‘𝐸))
2116, 18, 6, 8, 9, 19, 201hevtxdg1 40721 . . 3 (𝜑 → ((VtxDeg‘⟨𝑉, {⟨𝐾, 𝐸⟩}⟩)‘𝑈) = 1)
2221oveq2d 6565 . 2 (𝜑 → (((VtxDeg‘𝐺)‘𝑈) +𝑒 ((VtxDeg‘⟨𝑉, {⟨𝐾, 𝐸⟩}⟩)‘𝑈)) = (((VtxDeg‘𝐺)‘𝑈) +𝑒 1))
2310, 22eqtrd 2644 1 (𝜑 → ((VtxDeg‘𝐹)‘𝑈) = (((VtxDeg‘𝐺)‘𝑈) +𝑒 1))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 383 = wceq 1475 ∈ wcel 1977 ∉ wnel 2781 Vcvv 3173 ∪ cun 3538 𝒫 cpw 4108 {csn 4125 ⟨cop 4131 class class class wbr 4583 dom cdm 5038 Fun wfun 5798 ‘cfv 5804 (class class class)co 6549 1c1 9816 ≤ cle 9954 2c2 10947 +𝑒 cxad 11820 #chash 12979 Vtxcvtx 25673 iEdgciedg 25674 VtxDegcvtxdg 40681 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-8 1979 ax-9 1986 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 ax-rep 4699 ax-sep 4709 ax-nul 4717 ax-pow 4769 ax-pr 4833 ax-un 6847 ax-cnex 9871 ax-resscn 9872 ax-1cn 9873 ax-icn 9874 ax-addcl 9875 ax-addrcl 9876 ax-mulcl 9877 ax-mulrcl 9878 ax-mulcom 9879 ax-addass 9880 ax-mulass 9881 ax-distr 9882 ax-i2m1 9883 ax-1ne0 9884 ax-1rid 9885 ax-rnegex 9886 ax-rrecex 9887 ax-cnre 9888 ax-pre-lttri 9889 ax-pre-lttrn 9890 ax-pre-ltadd 9891 ax-pre-mulgt0 9892 This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-3or 1032 df-3an 1033 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 df-eu 2462 df-mo 2463 df-clab 2597 df-cleq 2603 df-clel 2606 df-nfc 2740 df-ne 2782 df-nel 2783 df-ral 2901 df-rex 2902 df-reu 2903 df-rmo 2904 df-rab 2905 df-v 3175 df-sbc 3403 df-csb 3500 df-dif 3543 df-un 3545 df-in 3547 df-ss 3554 df-pss 3556 df-nul 3875 df-if 4037 df-pw 4110 df-sn 4126 df-pr 4128 df-tp 4130 df-op 4132 df-uni 4373 df-int 4411 df-iun 4457 df-br 4584 df-opab 4644 df-mpt 4645 df-tr 4681 df-eprel 4949 df-id 4953 df-po 4959 df-so 4960 df-fr 4997 df-we 4999 df-xp 5044 df-rel 5045 df-cnv 5046 df-co 5047 df-dm 5048 df-rn 5049 df-res 5050 df-ima 5051 df-pred 5597 df-ord 5643 df-on 5644 df-lim 5645 df-suc 5646 df-iota 5768 df-fun 5806 df-fn 5807 df-f 5808 df-f1 5809 df-fo 5810 df-f1o 5811 df-fv 5812 df-riota 6511 df-ov 6552 df-oprab 6553 df-mpt2 6554 df-om 6958 df-1st 7059 df-2nd 7060 df-wrecs 7294 df-recs 7355 df-rdg 7393 df-1o 7447 df-oadd 7451 df-er 7629 df-en 7842 df-dom 7843 df-sdom 7844 df-fin 7845 df-card 8648 df-cda 8873 df-pnf 9955 df-mnf 9956 df-xr 9957 df-ltxr 9958 df-le 9959 df-sub 10147 df-neg 10148 df-nn 10898 df-2 10956 df-n0 11170 df-xnn0 11241 df-z 11255 df-uz 11564 df-xadd 11823 df-fz 12198 df-hash 12980 df-vtx 25675 df-iedg 25676 df-vtxdg 40682 This theorem is referenced by: vdegp1bi-av 40753
Copyright terms: Public domain W3C validator
| 2,897
| 4,584
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.078125
| 3
|
CC-MAIN-2024-22
|
latest
|
en
| 0.203776
|
alphaentek.com
| 1,726,635,844,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651836.76/warc/CC-MAIN-20240918032902-20240918062902-00245.warc.gz
| 78,056,066
| 13,354
|
©
Music technology Units & symbols Equations & formulae Definitions & origins Sine waves Geometry deciBels & measurement
Useful formulae, equations, definitions & derivations.
For electrical engineering, electronics & music technology.
Electrical units & common symbols.
E = electromotive force (EMF) in Volts (V), also called Potential Difference (PD), or Voltage.
V is a less used alternative symbol for EMF in the SI or MKS system.
I = current in Amperes (A),
R = resistance in Ohms (Ω),
G = conductance in Siemens (S),
P = power in Watts (W).
C = capacitance in Farads (F),
L = inductance in Henrys (H).
X = reactance in Ohms (Ω),
B = susceptance in Siemens (S),
Z = impedance in Ohms (Ω),
Y = admittance in Siemens (S),
f = frequency in cycles per second (c/s), or Hertz (Hz)
Q (Quality), has no units, it is circuit magnification, the ratio of reactance to resistance. With a tuned circuit, refers to selectivity.
Formulae & equations.
Ohm's Law: Current = EMF (Voltage) divided by resistance.
The three forms of the equation are: I = E/R, R = E/I or E = I X R
Power: Watts = Voltage multiplied by current.
W = E X I = E2/R = I2 X I
One Horsepower (HP) = 746 Watts to the nearest integer, in Britain & America. To be be more exact it is 745.699872 Watts, the origin is 33,000 foot-pounds per minute or 550 foot-pounds per second. This was calculated by James Watt after measuring the work that pit ponies could do.
At one time the term 'brake horse power' or B.H.P. was used. This was because the standard measurement method involved absorbing the power in a weighted belt on drum brake. A means was needed to distinguish it from the archaic RAC automobile horsepower rating, which was essentially meaningless for modern engines.
A human athlete can produce about half a horsepower for extended periods. Some can achieve around 1.25 to 1.75 horsepower for brief sprints.
They specify 750 Watts in Europe, a big place, they need slightly bigger horses.
Resistors in series: Total resistance R is the sum of the individual resistances.
R = R1 + R2 + R3 etc.
Resistors in parallel: Easiest to work out using conductance. Total conductance G is the sum of the individual conductances.
G = 1/R1 + 1/R2 + 1/R3 etc. Then: R = 1/G
Another way sometimes quoted for 2 resistors is the product divided by the sum, (R1 X R2)/(R1 + R2). I personally cannot see the point in this 'short cut', it is more difficult.
Capacitors in parallel: Total capacitance C is the sum of the individual capacitances.
C = C1 + C2 + C3 etc.
Capacitors in series: Total capacitance C is the reciprocal of the sum of the individual reciprocals.
1/C = 1/C1 + 1/C2 + 1/C3 etc. Alternatively: C = 1/(1/C1 + 1/C2 + 1/C3 etc.)
Another way for 2 capacitors is the product divided by the sum, (C1 X C2)/(C1 + C2)
Inductors in series: Total inductance L is the sum of the individual inductances.
L = L1 + L2 + L3 etc.
Inductors in parallel: Total inductance L is the reciprocal of the sum of the individual reciprocals.
1/L = 1/L1 + 1/L2 + 1/L3 etc. Alternatively: L = 1/(1/L1 + 1/L2 + 1/L3 etc.)
Another way for 2 inductors is the product divided by the sum, (L1 X L2)/(L1 + L2)
These equations work equally for multiples or sub-multiples of the base unit. Microfarads, millihenrys, Kilohms etc. Provided the same multiple etc. is used for each component.
Reactance: Exhibited by coils (inductors) or condensers (capacitors). Controls current flow as does resistance. Called reactance because, unlike resistors, energy is not absorbed & converted into heat or light. Instead it is stored, being returned to the source when EMF is removed (an equal reaction).
Rather as a spring is compressed by a weight, returning stored energy by helping lift the weight.
Measured in Ohms (Ω). Series or parallel reactances, of one type, can be calculated using the same formulae as resistances.
Capacitive reactance: XC = 1/2πfC
Inductive reactance: XL = 2πfL. π (pi) = 3.14159 to 5 decimal places.
Susceptance, B, is the degree of ability of a simple reactance to conduct current. It is thus the reciprocal of reactance & is measured in Siemens (S).
Impedance Z is the complex vector combination of resistance & reactance in a practical circuit. Real circuits always have resistance, capacitance & inductance. Measured in Ohms (Ω).
Admittance, Y, the ability of a complex circuit to conduct current, is the reciprocal of impedance. Useful for calculating parallel impedances, it is measured in Siemens (S).
Resonance:
A coil & capacitor can be wired in parallel. Forming a tuned circuit, with a resonant frequency. At resonance the reactance of both components is equal. According to classical theory, current through a capacitor leads the Voltage by 90 degrees, that through an inductor lags by 90 degrees. Thus the two currents are 180 degrees out of phase.
Since they are equal at resonance, they cancel exactly, no net current flows.
So at resonance the combination exhibits infinite reactance. In parallel with this are circuit losses, essentially resistive, forming a finite impedance, nothing is perfect. In a practical circuit the ratio of frequency at resonance, to bandwidth is termed Q (quality). For RF this is typically around 50 to 100.
At low audio frequencies it is often less, due to practical component size & cost limitations.
The same components can be connected in series, also forming a tuned circuit. At resonance, since only one current must flow, the two Voltages are 180 degrees out of phase, cancelling exactly. So this arrangement has zero reactance, an effective short circuit, at this one frequency. In series with this are circuit losses. Q is again the ratio of resonant frequency to bandwidth.
Bandwidth is the difference between the frequencies, either side of the resonant peak, that are 3 dB down or half the power.
Another definition of Q for some purposes is circuit magnification. The ratio of impedance or gain at resonance to that at frequencies outside the range of interest. This figure is important in determining the degree of rejection of unwanted frequencies. Thus it may be called 'rejection ratio'. It can be improved, as in analogue radio use, by cascading a number of tuned circuits.
The relevant equations are:
At resonance: XL = XC, f = 1/(2π X √(LC)), Q = f/bandwidth
Definitions & descriptions.
Resistance (R) is the tendency of materials to oppose the flow of current through them.
Materials that present little opposition are termed conductors. Metals generally fall into this category, as does carbon. Resistors are usually carefully formulated & proportioned conductors.
Materials exhibiting almost complete opposition are called insulators. Most thermoplastics, ceramics, rubber & glass are examples.
Materials with intermediate properties are termed, logically, semiconductors. Examples are germanium, gallium arsenide, and suitably alloyed silicon. These have other properties, making them useful in electronics, particularly when controlled small proportions of suitable impurities are added. This subject requires at least its own page.
Many liquids, particularly aqueous solutions of salts, conduct electricity. The mechanism of conduction is different, so they are also treated separately.
The unit of resistance is the Ohm (Ω), named after Georg Simon Ohm, who formulated his famous law in the early Victorian era. Like many other pioneers he received little recognition in his lifetime. One Ohm allows one Ampere of current to flow when a potential difference of one Volt is applied, ten Ohms will allow one tenth of one Ampere when similiarly provoked.
Conductance (G) is a measure of the ability of materials to conduct current. It is the reciprocal or inverse of resistance. The unit of conductance is the Siemens, symbol S, named after Ernst Werner von Siemens. One Siemens conducts one Amp when one Volt is applied, ten Siemens will similiarly conduct ten Amps. It can therefore be thought of as Amperes per Volt.
A former name for the unit of conductance is the Mho, Ohm spelt backwards. In other respects it is the same
Parallel resistances are most easily calculated by addding their conductances (1/R). The reciprocal of the sum is the total resistance.
ElectroMotive Force (EMF), also called Voltage or Potential Difference. The usual symbol is E, V is used in the MKS system. It is the electrical force between two points that causes, or tries to cause, electrical current to flow. It is measured in Volts (V), named after Count Alessandro Giuspeppe Antonio Anastasio Volta, from Italy. One Volt causes a current of one Ampere to flow through a resistance of one Ohm.
A more complete definition of a Volt is: The unit of electrical potential equal to one Joule per Coulomb.
Current (I), is measured in Amperes, named after Andre Marie Ampere. 1 Ampere = one Coulomb (6.24 x 1018 electrons) per second, passing one point on a conductor. It is also defined as the current through a one Ohm resistor when one Volt is applied.
Electrical current is always a flow of electrons, from negative to positive. In a conductor they do not actually flow directly through, an electron arrives at an atom & joins its outer valency ring. In doing so one already there is displaced. Attracted towards the positive terminal it encounters another atom, displacing a further electron & so on. In a vacuum it is a true electron flow.
In the early days of electrical experimentation it was assumed that current flowed from positive to negative. In other words, that the moving particles were positive charges. This mistake persisted until Dr. A.J. Fleming's invention of the thermionic valve in 1904. His original diode proved beyond doubt that the travelling electrons are negative & flow from negative to positive.
Despite this many physicists, even today, insist that true current flows from positive to negative. Pigheadedly perpetuating a 200 year old mistake, despite a 1 hundred year old proof to the contrary. Books are still published showing 'conventional' current flowing from positive to negative whilst electrons flow the other way.
What a nonsense! When I was at school, in a 'science' class we were shown a so called educational film. This featured, in cartoon form, current flowing one way, whilst electrons went the other. I remember thinking at the time, 'this is rubbish, the opposite charges will collide & cancel out'. This is something they are still trying to push.
Millions of pounds are spent every year on research into positrons (anti-electrons) and anti-protons. Atoms are bombarded in cyclotrons & other mega-expensive machines. Whatever the potential uses for these exotic particles they do not contribute to current flow. They can only be isolated in high energy environments.
A lot of time & money is spent studying the theory of 'anti-matter'. This has atoms with a negative nucleus & orbiting positive charges. It will be handy to provide the astronomical energy requirements for interstellar space travel, along with warp drive & matter transporters.
It certainly can't exist in the same place as normal matter, otherwise the two would cancel. Unleashing a force sufficient to make an atom bomb seem like a firework. So we are not likely to be able to mine it on any planet where we can survive.
Electrical charge is measured in Coulombs, essentially a very small bucketful of electrons, named after Charles Augustin de Coulomb. One Coulomb = 6.24 x 1018 electrons or 1 Ampere for one second.
Energy, power used or work done is measured in Joules (J). Named after James Prescott Joule, one Joule is 6.24 X 1018 Electron Volts. Alternative specifications are: 1 Watt-second, 107 ergs, 1 Newton-Metre, 0.2390 calories or approximately 0.738 foot-pounds. The kinetic energy of a 2 kilogram mass moving at 1 metre per second. There are many other equivalents.
Capacitance (C) is measured in Farads, (F) named after Michael Faraday. One Farad is charged to a potential of one Volt by one Coulomb or one Ampere for one second. The Farad is inconveniently large for most purposes. Microfarads (μF), (F X 10-6), nanofarads (F X 10-9) or picafarads (F X 10-12) are in common use.
Inductance (L) is measured in Henrys (H), named after American physicist, Joseph Henry. The induced EMF across a one Henry inductor is one Volt, if the current through it changes by one Ampere per second, 1 H = 1 Vs/A. Convenient sub-multiples are millihenrys (mH) & microhenrys (μH). These are H X 10-3 & H X 10-6 respectively.
Frequency (f) is measured in cycles per second (c/s). The European term for this value, Hertz (Hz) is now commonly used. Named after German Phyicist, Heinrich Hertz, the first to publish data on experiments with wireless transmission. Sine waves are cyclic, rotating in phase through 360 degrees continuously, hence cycles per second.
Kilocycles/s, Kc/s (f X 103), second Megacycles/s, Mc/s (f X 106) & Gigacycles/s, Gc/s (f X 109) are convenient multiples for practical use. Again Hz, instead of the logical c/s, is considered "politically correct" by the 'Establishment'.
There are a number of different methods for measuring AC Voltage & current. The image below represents a sine wave. Clicking on it will link to the Sine Wave page, with more information on derivation etc. The browser's 'Back' button will link back to here.
One cycle of a sine wave, 2π radians or 360 degrees. Plotted against angular displacement.
Moving the mouse cursor to the image will plot the wave. Horizontal scale factor is c/π.
The vertical scale is plus & minus 1 Volt for the purpose of the following discussion.
The obvious way is to measure the peak, or the peak to peak value. 1 Volt or 2 Volts respectively for this image. This is a useful means of indicating the electrical 'headroom' required to accomodate the sigmal.
It does not however directly give a value for the useful EMF, current or power. The secret here lies in the area enclosed by the curve, bounded by the reference or zero line. This gives the effective or heating power of an electrical waveform.
The accepted method of establishing this effective area is called RMS, a standard abbreviation for 'Root Mean Square'. The instantaneous amplitude is measured at a number of equally spaced points along one or more full cycles. Each of these values is squared & the results averaged. The square root of the answer is the RMS value.
Some waveforms are fully symmetrical, such as a triangle, square or sine wave. For these only one quadrant needs to be measured to obtain high accuracy. All others need the full cycle, reducing any errors requires more measured points.
For a pure sine wave Voltage there is a 'short cut', just one point, at 45 degrees, is precise. Multiplying the peak by the square root of one half (0.7071) gives this figure. Crest factor is the ratio of peak to RMS, 1.4142/1 (√2/1) in this case. This ratio is squared for power into a resistive linear load, it is thus 2/1.
Most analogue & digital Voltmeters average the waveform. Scaling is adjusted to the ratio of 1.1118/1 to give an RMS readout, this is generally only accurate for sine waves. Some more expensive meters have the extra circuitry to calculate a 'true RMS' reading.
For a triangle wave, the average & RMS readings are the same. With a true square wave, average, RMS & peak are the same. To obtain a true square wave an infinite number of odd harmonics are needed. The transition from positive to negative, or vice versa must occur in zero time. This requires infinite instantaneous power in a real circuit.
There is another way to obtain an RMS Voltage reading, regardless of a waveforms harmonic structure. Apply the Voltage across a perfect resistor & measure the temperature rise. This is slow & can suffer from ambient temperature or other losses. A useful technique for difficult waveforms, such as noise & narrow or asymmetric pulses.
Ratios denoted by the most common deciBel values. dB Voltage Power dB Voltage Power 0.5 1.06 1.12 25 17.8 316.23 1 1.12 1.26 30 31.623 1000 2 1.26 1.58 35 56.29 3,162.3 3 1.414 2.00 40 100 10,000 4 1.58 2.51 50 316.23 105 5 1.78 3.162 60 1000 106 6 1.99 3.98 70 3162.3 107 7 2.24 5.01 80 104 108 8 2.51 6.31 90 31,623 109 9 2.82 7.94 100 105 1010 10 3.162 10 110 316.233 1011 12 3.98 15.8 120 106 1012 15 5.62 31.62 130 316.234 1013 20 10 100 140 107 1014
The deciBel is one of the commonly misundertood aspects of sound & electrical measurement.
Not a unit of measurement as such, but a convenient means of specifying the relationship between 2 values.
The Bel, named after Alexander Graham Bell, is the common logarithm of a power or sound intensity ratio.
Fractional exponents are commonly encountered. To avoid this, the deciBel (dB), 1 tenth of a Bel, is used.
This table covers the most useful range for music & audio purposes.
Given a specified impedance, Voltage or current varies as the square root of power. The Voltage ratio columns also apply to current in this case. So 20dB (2 Bels) is a power ratio of 10 to the power of 2 (100). The Voltage & current ratio is the square root, or just 10 (101).
For values not in the table the deciBels can be added & the ratios multiplied. For example, to find a power ratio of 5,000 to 1: 1,000 is 30dB & 5 is roughly 7dB. So the answer is 37dB, within the limitations of this table. Conversely 28dB is 8dB + 20 dB, the Voltage ratio is therefore 2.51 X 10 or 25.1 to 1.
For greater accuracy with small ratios another table will be included at the end of this section. Feel free to download these tables if they are of use. We hope at some point to include the complete set of Henry Briggs logarithm tables.
The table clearly shows, with multiples of 10dB, that taking off the last zero gives the exponent for power ratios. The origin of the system lies in the realm of telephony. Here matching impedances for maximum power transfer is all important. This is why it relates directly to power rather than Voltage or current.
Originally called the Transmission Unit (T.U.), the name was changed in the 1920s. Early valve (vacuum tube) equipment has similiar requirements so the deciBel was adopted quickly throughout the sound engineering field in Britain & America.
Continental Europe had a similiar system called the Neper, from the original spelling of John Napier's (1550 to 1617) surname. He gave the world logarithms in 1614 & his version, to base e, is the basis of the Neper. It is thus a little smaller than the Bel.
It is interesting to note that common logarithms, to base 10 with log 1 = 0, arose out of discussions with Henry Briggs, in 1615 & 1616. Mr. Briggs published his first common log tables in 1617, giving credit to John Napier.
Within a few years a complete set were published, for all natural numbers (positive integers) from 1 to 100,000. Thes tables are to 14 places of decimals, better than most calculators can manage today. John Napier was also the man who gave us decimal notation.
Decimals were later adopted in America, for their currency after independence & by France after their revolution. Despite the system's advantages it only took over here in the second half of the 20th Century. The advent of cheap calculators finally ousted fractions from the academic mainstream. Some early 'mathematical' calculators had fractional input ability.
Audio measurement using deciBels
When the second term of the ratio is replaced by a constant, the deciBel becomes a unit of measurement. A number of such systems are in common use.
dBm, a measurement system relative to 1 milliWatt into 600Ω, the m stands for 'milliWatt'. The Voltage corresponding to this is 0.7746 Volts RMS. Stage instruments & instruments usually work at -10dBm. Studios & broadcasting use +4dBm as a normal level.
dBu, a similiar standard where 0dBu is 0.7746 Volts into any impedance, the u stands for 'unterminated'. Useful with more modern transistor equipment where impedance is less important.
dBv, an earlier name for dBu, the small v stands for Volts. No longer used as many writers etc. confused it with dBV (capital V).
dBV, similiar to dBu, but the standard reference is one Volt, the capital V stands for Volt. Our Bruel & Kjaer test equipment uses this standard.
VU, Volume Unit, a system for use with averaging sound level meters. Calibration is similiar to dBm, but the correlation only holds for steady sine waves. Normally set so that 0VU is +4dBm. These meters are of little use with digital recording.
dBW, a power measurement system where the reference standard is 1 Watt. Used for higher power systems & is less common.
dBK, in this standard 0dBK is one KiloWatt (1000 Watts). It is hard to see where this fits into most music use.
dBFS, a standard for digital recording, FS stands for Full Scale. 0dBFS is the level at which all bits of the digital code are set to 1, The signal can not get any greater than this, digital systems have zero headroom. For reproduction without clipping, the peak signal must be less than or equal to 0dBFS. Thus all working levels are negative figures.
Acoustic measurement
dBi, my proposed name for the unit of sound intensity, the i stands for intensity. The dB name is often quoted for 'unweighted' measurement. As this is the established name for a dimensionless ratio, it is confusion by & for the uneducated. A 20dB increase can change from loud to deafening, a 20dB level is very quiet. Where is the sense in that?
0dBi (or 0dB), the reference level, is 1 picoWatt (1 Watt x 10-12) per square metre. This miniscule figure is the measured threshold of hearing for a healthy young human @ 1 Kc/s. This is the acoustic equivalent of electrical power.
The level at which hearing damage can occur is 120dBi, one acoustic Watt. This is 1012, 1 Imperial billion (American trillion) times greater. So the range of human hearing is quite wide, I don't know how many people volunteered to go deaf to get this figure. One of the reasons for using a logarithmic scale is this wide range of hearing perception.
dBA, a similiar measure, but with the reference set to a curve approximating the human ear's frequency response. two other curves, dBB & dBC are rarely used.
dBSPL, sound pressure level. 0dBSPL is set at 20 microPascals (2 X 10-5 Newtons per square metre), the threshold of hearing. This measure is the equivalent of electrical Volts. 120dBSPL is again the level of potential hearing damage.
Acoustic displacement, or air movement is the equivalent of electrical current. Difficult to measure directly at the frequencies & amplitudes involved. It is approximately calculated from dBSPL readings at two close points, allowing for the specific mass & density of air.
A means of directly measuring minute air displacements at audio frequencies has now been devised. The technique is potentially highly accurate, allthough difficult to calibrate, since no precise standard exists. We may have the nearest thing to a perfect microphone, with essentially similiar mass to air.
Wrong Usage & Confusion.
There is an active campaign to get rid of the good old deciBel that has served us well. It would be replaced by measurements from the SI system of units. The reason is the ignorance & confusion that has plagued the dB for many years. It will be a shame if such a useful system for engineers becomes obsolete.
Just because those who don't know what they are talking about, insist on talking about it. The 'dumbing down' of much of the world's population has found its way into the technical, or pseudo technical writing 'profession'.
As an example take the specification of a typical stage performance loudspeaker driver. The 'sensitivity' figure is of interest, for an efficient model this may be 100dB, by this the writer usually means 100dBA. In some cases the reference power is included, less often the measurement distance, the frequency is seldom mentioned.
If the typical power of 1 Watt is included, it can be inferred that at 100 Watts the threshold of hearing damage (120dBA), will be reached. Exceeded by 3dB at 200 Watts. This is a common misconception, the sensitivity specification is an 'output', not a listener's auditory 'input'. If the distance is given, it is typically 1 Metre for a driver of this power.
No-one will be listening to a 100 or 200 Watt amplifier flat out at 1 metre. In an open space, sound follows the inverse square law, at 2 metres the intensity is one quarter, 6dB dowm. At 10 metres it is one hundredth, 20dB down. In a room full of people &/or soft furnishings sound will be further attenuated. It is reinforced by significant reverberation.
Formulae © Ron Lebar, Author. Updated: 14-5-2005. Loaded:
We Wish ALL the World Peace, Justice, Equality, Prosperity & an End to Fanaticism.
| 5,955
| 24,773
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.453125
| 3
|
CC-MAIN-2024-38
|
latest
|
en
| 0.937052
|
http://razhayesheitanparastan.com/46hffe/
| 1,571,164,282,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-43/segments/1570986660231.30/warc/CC-MAIN-20191015182235-20191015205735-00203.warc.gz
| 159,060,268
| 12,049
|
# Planet Earth Pole To Pole Worksheet
Monday, July 29th 2019. | Math Worksheets
Children Comprehend how to use technologies Surprisingly nicely. The start to suppose that doesn't have importance in their lifestyles. If a kid is in a unfamiliar, math homework help might be the absolute best bet to receive them together with the help they . If your young child is left to explore in their own freewill, you won't how much they absorb. 2nd grade, children practice addition a deal.
Teachers Cancan Begin by posing issues along with modeling , And ask students to utilize each other to find solutions. Because of this, it's essential for teachers and parents to participate children with mathematics pursuits and provide them easy and enjoyable z worksheets to repair. Teachers and parents the worksheets to direct the students. Students to how to add and subtract polynomials involving a of variables. They address the subtraction on the sheet to get the . Regrettably, almost all their up coming mathematics instruction depends on being in a position to fractional arithmetic. It's possible resources all the such as worksheets.
### Chemistry Lab Equipment Worksheet
#### Adding Subtracting Multiplying And Dividing Fractions Worksheet
##### Future Tense Spanish Worksheet
###### Stock Market Worksheets
Learning Internet Sites may be Quite effective because they feature games in to the educational procedure. There are sites provide a huge collection of worksheets. there are numerous websites you're ready to refer to of method. There are of provide much info about mathematics worksheets.
The student will Be Able to compose an Whole number in Fraction form. Students find extra details about the others' and answer some questions regarding their within this entertaining worksheet for children. The student may be able to detect that the reciprocal of a comprehensive number and a fraction. to write fractions in simplest form. He will have the ability to accurately divide fractions. If you would rather, you may the student to .
| 386
| 2,043
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.515625
| 3
|
CC-MAIN-2019-43
|
latest
|
en
| 0.940355
|
https://www.geeksforgeeks.org/conversion-of-regular-expression-to-finite-automata/?ref=rp
| 1,686,130,757,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224653631.71/warc/CC-MAIN-20230607074914-20230607104914-00430.warc.gz
| 863,818,132
| 34,723
|
Get the best out of our app
GeeksforGeeks App
Open App
Browser
Continue
# Conversion of Regular Expression to Finite Automata
As the regular expressions can be constructed from Finite Automata using the State Elimination Method, the reverse method, state decomposition method can be used to construct Finite Automata from the given regular expressions.
Note: This method will construct NFA (with or without ε-transitions, depending on the expression) for the given regular expression, which can be further converted to DFA using NFA to DFA conversion.
### State Decomposition Method
Theorem: Every language defined by a regular expression is also defined by a Finite Automata.
Proof: Let’s assume L = L(R) for a regular expression R. We prove that L = L(M) for some ε-NFA M with:
1) Exactly one accepting state.
2) No incoming edges at the initial state.
3) No outgoing edges at the accepting state.
The proof is done by structural induction on R by following the steps below:
Step 1: Create a starting state, say q1, and a final state, say q2. Label the transition q1 to q2 as the given regular expression, R, as in Fig 1. But, if R is (Q)*, Kleene’s closure of another regular expression Q, then create a single initial state, which will also be the final state, as in Fig 2.
Fig 1
Fig 2
Step 2: Repeat the following rules (state decomposition method) by considering the least precedency regular expression operator first until no operator is left in the expression. Precedence of operators in regular expressions is defined as Union < Concatenation < Kleene’s Closure.
Union operator (+) can be eliminated by introducing parallel edges between the two states as follows.
Fig 3: Removal of Union Operator
The concatenation operator (‘.’ or no operator at all) can be eliminated by introducing a new state between the states as follows.
Fig 4: Removal of Concatenation Operator
Kleene’s Closure (*) can be eliminated by introducing self-loops on states based on the following conditions:
1. If there is only one outgoing edge at the left-most state, i.e., A in transition A -> B, then introduce self-loop on state A and label edge A to B as an ε-transition, as shown in Fig 5.
Fig 5
2. Else if there is only one incoming edge at the right-most state, i.e., B in transition A -> B, then introduce self-loop on state B and label edge A to B as an ε-transition, as shown in Fig 6.
Fig 6
3. Else introduce a new state between two states having self-loop labeled as the expression. The new state will have ε-transitions with the previous states as follows, as shown in Fig 7.
Fig 7
### Example:
Construct Finite Automata for the regular expression, R = (ab + ba)*
Solution:
Step 1: As the given expression, R, is of the form (Q)*, so we will create a single initial state that will also be the final state, having self-loop labeled (ab + ba), as shown in Fig 8. (Refer Fig 2 above)
Fig 8
Step 2:
A. As the least precedency operator in the expression is a union(+). So we will introduce parallel edges (parallel self-loops here) for ‘ab’ and ‘ba’, as shown in Fig 9.
Fig 9
B. Now we have two labels with concatenation operators (no operator mentioned between two variables is concatenation), so we remove them one by one by introducing new states, q1 and q2 as shown in Fig 10 and Fig 11. (Refer Fig 4 above)
Fig 10
Fig 11
Step 3: As no operators are left, we can say that Fig 11 is the required finite automata (NFA).
My Personal Notes arrow_drop_up
| 834
| 3,484
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.890625
| 4
|
CC-MAIN-2023-23
|
latest
|
en
| 0.908264
|
http://www.geeksforgeeks.org/integer-literal-in-c-cpp-prefixes-suffixes/
| 1,513,225,709,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-51/segments/1512948539745.30/warc/CC-MAIN-20171214035620-20171214055620-00614.warc.gz
| 380,139,719
| 17,523
|
Integer literal in C/C++ (Prefixes and Suffixes)
Integer literal is a type of literal for an integer whose value is directly represented in source code. For example, in the assignment statement x = 1, the string 1 is an integer literal indicating the value 1, while in the statement x = 0x10 the string 0x10 is an integer literal indicating the value 16(in decimal), which is represented by 10 in hexadecimal (indicated by the 0x prefix).
Further, in x = “1” the “1” is a string literal(not a character or an integer literal), because it is in quotes. The value of the string is 1, which happens to be an integer string.
Integer literals are expressed in two types i.e.,
1. Prefixes which indicates the base. For example, 0x10 indicates the value 16 in hexadecimal having prefix 0x.
2. Suffixes which indicates the type. For example, 12345678901234LL indicates the value 12345678901234 as an long long integer having suffix LL.
Syntax
• Prefixes: They are basically represent in four types.
1. Decimal-literal(base 10):- a non-zero decimal digit followed by zero or more decimal digits(0, 1, 2, 3, 4, 5, 6, 7, 8, 9). For example, 56, 78.
2. Octal-literal(base 8):- a zero followed by zero or more octal digits(0, 1, 2, 3, 4, 5, 6, 7). For example, 045, 076, 06210.
3. Hex-literal(base 16):- 0x or 0X followed by one or more hexadecimal digits(0, 1, 2, 3, 4, 5, 6, 7, 8, 9, a, A, b, B, c, C, d, D, e, E, f, F). For example, 0x23A, 0Xb4C, 0xFEA.
4. Binary-literal(base 2):- 0b or 0B followed by one or more binary digits(0, 1). For example, 0b101, 0B111.
• Suffixes: They are represented in many ways according to their data types.
1. int:- No suffix are required because integer constant are by default assigned as int data type.
2. unsigned int: character u or U at the end of integer constant.
3. long int: character l or L at the end of integer constant.
4. unsigned long int: character ul or UL at the end of integer constant.
5. long long int: character ll or LL at the end of integer constant.
6. unsigned long long int: character ull or ULL at the end of integer constant.
```// C++ program to demonstrate the use of
// integer literal
#include <iostream>
using namespace std;
int main()
{
// PREFIXES
cout << 213 << '\n' // decimal integer literal
<< 0213 << '\n' // Octal integer literal
<< 0x213A << '\n' // hexadecimal integer literal
<< 0b101 << '\n' // binary integer literal
// SUFFIXES
// long long literal
<< 1234567890123456789LL << '\n'
// unsigned long long literal
<< 12345678901234567890ull << '\n'
// automatic conversion of unsigned long long even
// without long long prefix
<< 12345678901234567890u;
return 0;
}
```
```Output:
213
139
8506
5
1234567890123456789
12345678901234567890
12345678901234567890
1221300
```
Digit separator: In C++, integer literals may contain digit separators to allow digit grouping into more readable forms. This is particularly useful for bit fields, and makes it easier to see the size of large numbers (such as a million) at a glance by subitizing rather than counting digits. It is also useful for numbers that are typically grouped, such as credit card number or social security numbers.[a] Very long numbers can be further grouped by doubling up separators.
Typically decimal numbers (base-10) are grouped in three digit groups (representing one of 1000 possible values), binary numbers (base-2) in four digit groups (one nibble, representing one of 16 possible values), and hexadecimal numbers (base-16) in two digit groups (each digit is one nibble, so two digits are one byte, representing one of 256 possible values). Numbers from other systems (such as id numbers) are grouped following whatever convention is in use.
```// C++ program to demonstrate digit separator
#include <iostream>
using namespace std;
int main()
{
cout << 12345678901245LL <<'\n'
// long long int literal digit separator
<< 12'345'678'901'245LL <<'\n'
// binary literal digit separator
<< 0b1000'111'0 <<'\n'
<< 0X12A'2b4;
return 0;
}
```
```Output:
12345678901245
12345678901245
142
1221300
```
Reference:- https://en.wikipedia.org/wiki/Integer_literal
This article is contributed by Shubham Bansal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
GATE CS Corner Company Wise Coding Practice
Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.
1 Average Difficulty : 1/5.0
Based on 1 vote(s)
| 1,284
| 4,609
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.984375
| 3
|
CC-MAIN-2017-51
|
latest
|
en
| 0.763434
|
https://math.answers.com/questions/How_big_is_7_inch_in_cm
| 1,675,867,749,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-06/segments/1674764500813.58/warc/CC-MAIN-20230208123621-20230208153621-00543.warc.gz
| 400,295,119
| 50,519
|
0
# How big is 7 inch in cm?
Wiki User
2013-03-28 04:58:23
The answer is 17.78 cm (approx.). Inches and centimeters are both units of linear measurement. Inches are used in the imperial system whereas centimeters are used in the metric system. To convert from inches to cm, multiply the inch unit by 2.54.
Wiki User
2014-12-21 18:37:59
Study guides
20 cards
## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials
➡️
See all cards
3.81
2052 Reviews
Wiki User
2013-03-28 04:58:23
1 in = 2.54 cm so 7 in = 17.78 cm
Algebraic Steps / Dimensional Analysis Formula
7
in
*
2.54 cm
1 in
=
17.78
cm
| 215
| 665
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06
|
latest
|
en
| 0.816741
|
https://outschool.com/classes/pre-algebra-unit-2-algebraic-expressions-vWzTK6GO
| 1,716,902,438,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971059139.85/warc/CC-MAIN-20240528123152-20240528153152-00413.warc.gz
| 372,234,343
| 74,082
|
English
### Pre-Algebra Unit 2: Algebraic Expressions
Tina Dewey, M.Ed, State Certified Math Educator
Average rating:4.9Number of reviews:(462)
In this semester course, students will learn the concepts typically taught in the first semester of a Pre-Algebra course.
#### Class experience
###### US Grade 6 - 9
`This class is taught in English.`
`Please see the class description for the content covered in this class.`
```I have successfully taught students Pre-Algebra through Algebra 2 for thirty years in a variety of school settings. I've worked with students who have learning challenges and high-achieving students, and many students in between. I bring these experiences to the classroom. I know which content can be challenging, and have developed a variety of ways to help learners succeed.
Check out my reviews on Outschool! And, I would be happy to meet with you online to discuss your learners needs.```
`.css-1il00e6{display:-webkit-box;display:-webkit-flex;display:-ms-flexbox;display:flex;gap:1em;-webkit-flex-direction:column;-ms-flex-direction:column;flex-direction:column;}.css-k008qs{display:-webkit-box;display:-webkit-flex;display:-ms-flexbox;display:flex;}.css-1do1xce{-webkit-user-select:none;-moz-user-select:none;-ms-user-select:none;user-select:none;width:1em;height:1em;display:inline-block;fill:currentColor;-webkit-flex-shrink:0;-ms-flex-negative:0;flex-shrink:0;-webkit-transition:fill 200ms cubic-bezier(0.4, 0, 0.2, 1) 0ms;transition:fill 200ms cubic-bezier(0.4, 0, 0.2, 1) 0ms;font-size:inherit;vertical-align:-0.125em;text-align:center;width:1.25em;color:#368139;margin-right:0.5em;margin-top:0.2em;}.css-4j7l6r{margin:0;font-family:Ginto Normal,sans-serif;font-size:1.6rem;line-height:1.3;font-weight:500;letter-spacing:0.01rem;}Homework Offered.css-l2z0vi{margin-top:0.5em;}Weekly practice assignments will be assigned. Students learn math by DOING math! The focus will be on effort and learning from mistakes. I will be happy to correct assignments which are turned no later than 24 hours before the next scheduled class..css-1smi110{display:-webkit-box;display:-webkit-flex;display:-ms-flexbox;display:flex;margin-top:0.5em;margin-left:2px;}.css-1jafuj7{-webkit-user-select:none;-moz-user-select:none;-ms-user-select:none;user-select:none;width:1em;height:1em;display:inline-block;fill:currentColor;-webkit-flex-shrink:0;-ms-flex-negative:0;flex-shrink:0;-webkit-transition:fill 200ms cubic-bezier(0.4, 0, 0.2, 1) 0ms;transition:fill 200ms cubic-bezier(0.4, 0, 0.2, 1) 0ms;font-size:inherit;vertical-align:-0.125em;margin-right:0.5em;margin-top:0.2em;color:#5C5C5C;}.css-1757vrz{margin:0;font-family:Ginto Normal,sans-serif;font-size:1.6rem;line-height:1.3;font-weight:500;letter-spacing:0.01rem;color:#5C5C5C;}2 - 4 hours per week outside of classAssessments OfferedStudent progress is assessed through weekly practice assignments and participation in class. Optional test are also available upon request. .css-mt1z2p{-webkit-user-select:none;-moz-user-select:none;-ms-user-select:none;user-select:none;width:1em;height:1em;display:inline-block;fill:currentColor;-webkit-flex-shrink:0;-ms-flex-negative:0;flex-shrink:0;-webkit-transition:fill 200ms cubic-bezier(0.4, 0, 0.2, 1) 0ms;transition:fill 200ms cubic-bezier(0.4, 0, 0.2, 1) 0ms;font-size:inherit;vertical-align:-0.125em;text-align:center;width:1.25em;color:#A3A3A3;margin-right:0.5em;margin-top:0.2em;}Grades Offered`
`You will need access to a printer to print guided notes and homework assignments for each class.`
`Learners will not need to use any apps or websites beyond the standard Outschool tools.`
Average rating:4.9Number of reviews:(462)
Profile
`I began teaching in 1978 by forcing my little brothers to "play school" during summer breaks. I began my paid teaching career in 1991 by teaching at a public high school in Michigan. In my over 30 years of experience teaching, I have also taught... .css-gw2jo8{position:relative;display:inline-block;font-family:'Ginto Normal',sans-serif;font-style:normal;font-weight:500;font-size:1.6rem;text-align:center;text-transform:none;height:auto;max-width:100%;white-space:nowrap;cursor:pointer;-webkit-user-select:none;-moz-user-select:none;-ms-user-select:none;user-select:none;outline:none;border:none;background:none;padding:0;-webkit-transition:all ease-in-out 0.05s,outline 0s,;transition:all ease-in-out 0.05s,outline 0s,;line-height:1;color:#380596;}.css-gw2jo8:hover:not(:disabled),.css-gw2jo8:focus:not(:disabled){color:#380596;-webkit-text-decoration:underline;text-decoration:underline;}.css-gw2jo8:active:not(:disabled){color:#380596;}.css-gw2jo8:disabled{color:#C2C2C2;cursor:default;}.css-gw2jo8:focus-visible{outline-width:2px;outline-style:solid;outline-color:#4B01D4;outline-offset:2px;}`
#### \$95
for 6 classes
2x per week, 3 weeks
55 min
Completed by 20 learners
Live video meetings
Ages: 11-16
8-14 learners per class
| 1,405
| 4,899
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22
|
latest
|
en
| 0.685183
|
http://mathhelpforum.com/trigonometry/129554-determine-exact-value-trig-function-given-print.html
| 1,527,256,389,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-22/segments/1526794867092.48/warc/CC-MAIN-20180525121739-20180525141739-00362.warc.gz
| 192,312,041
| 2,562
|
Determine the exact value of (trig function) given...
• Feb 18th 2010, 06:37 PM
Incursion
Determine the exact value of (trig function) given...
A coordinate plane, and a coordinate. An example is: Coordinate (-3, -5) in quadrant 3?
• Feb 18th 2010, 06:46 PM
Prove It
Quote:
Originally Posted by Incursion
A coordinate plane, and a coordinate. An example is: Coordinate (-3, -5) in quadrant 3?
$\displaystyle x = r\cos{\theta}$ and $\displaystyle y = r\sin{\theta}$
where $\displaystyle r = \sqrt{x^2 + y^2}$.
| 161
| 512
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.21875
| 3
|
CC-MAIN-2018-22
|
latest
|
en
| 0.610866
|
https://ericlippert.com/2017/02/02/excessive-explanation-part-fifteen/?replytocom=62717
| 1,669,759,592,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-49/segments/1669446710711.7/warc/CC-MAIN-20221129200438-20221129230438-00275.warc.gz
| 263,242,009
| 25,170
|
# Excessive explanation, part fifteen
All right, we are getting through section four of this paper. Today: what is an “environment”?
```Now let Env = Id → V be the domain of environments η.
```
Just to be clear, this is not a function in the Exp language. This is a formal description of part of the runtime engine of the Exp language. Basically we are saying here that an “environment” is simply the part of the runtime of an implementation of the language that takes identifiers and maps them onto their values. And that just as we’ll use σ for type schemes and α for type variables, we’ll use η for environments.
You’re so accustomed to using environments that they are like the air you breathe; you don’t even think about it. When you’re writing a program in some language and you say `x = y + z;`, you’re saying “obtain the values associated with identifiers y and z from the environment, add them, and modify the environment so that next time we ask, x has that value”. We can characterize the accesses to y and z as the environment being a function that takes “y” or “z” and returns their values. We can characterize the assignment to x as being the construction of a new environment based on the previous environment.
```The semantic function ε : Exp → Env → V is given in [5].
```
Again this is not an function in the Exp language; this is again a formalization of the runtime engine.
Remember that in our notation, a function of two arguments is notated as a function of one argument that returns a function of one argument. So what we’re saying here is that the “semantic function” is a function that takes an expression and an environment and produces a value.
Again, this is making formal the basic operation of a runtime. The runtime has a bunch of expressions and the values of all the variables, and it evaluates the expressions in the context of the environment.
```Using it, we wish to attach meaning to assertions of the form
A ⊨ e:σ
where e ∈ Exp and A is a set of assumptions of the form x:σ, x∈Id.
```
This right here is the mission statement of this paper.
The double-turnstile means “the thing on the right is a logical consequence of the thing on the left”.
What we’re saying here is: we have (on the left) a set of judgments (called “assumptions” here) where all the expressions are identifiers. Say:
```x:int
square:int → int
identity:∀β(β → β)
```
We have on the right an arbitrary expression with a typing judgement, say
```λy.identity (square x) : ∀β(β → int)
```
I hope you agree that if we assume that x is an int, square is a function from int to int, and identity is an identity function, that the lambda given is a function that takes any type and returns an int. The assumptions on the left are sufficient to deduce that the judgment on the right is correct (if the assumptions are correct, of course).
The goal of this paper is to provide an algorithm which, given the facts on the left and the expression on the right, deduces the type schema on the right in a manner that allows us to confidently say that yes, we have sufficient evidence on the left to make this judgment on the right.
(Aside: is it possible to have a function with type scheme ∀β(β → β) that is not an identity function? Ponder this!)
```
If the assertion is closed, i.e. if A and σ contain no free type
variables, then the sentence is said to hold iff, for every
environment η, whenever η ⟦x⟧:σ', for each member x:σ' of A,
it follows that ε ⟦e⟧ η:σ.
```
OK, we’ve got some jargon and some notation here used without introduction.
First of all, “the assertion” and “the sentence” are being used interchangeably here. We are “asserting” that the assumptions about typed identifiers on the left are sufficient evidence to deduce the type of the expression on the right.
Second, remember when we abused the Exp function notation to say that an environment was a function `Id → V`? You would think then that we would continue to abuse that notation and say that given an environment η and an identifier x we would say that the value of x was η x. And is what we do, but to call out that x is being used here as an identifier rather than standing in for a value, we “quote” it with funny brackets. So the value of identifier x in environment η is notated η ⟦x⟧.
Similarly for the semantic function ε. It takes two things: an expression e and an environment η, and to emphasize that the expression is an expression of the Exp language, we bracket it as well. So ε ⟦e⟧ η is “the value of expression e when evaluated in environment η”.
So what the heck is this jargon-filled paragraph actually saying? We have a bunch of identifiers with types on one side, and the statement that this implies that expression on the other side has a particular type. The paragraph is saying that this means: choose any possible environment that has those identifiers having some values of the given types. No matter what environment you chose, evaluating the expression in that environment gives a value that matches the expression’s type.
That might be too abstract still. In our example:
```x:int, square:int → int, identity:∀β(β → β) ⊨
λy.identity (square x) : ∀β(β → int)
```
We’re saying that no matter what integer we choose for x, no matter what function from int to int “square” really is, and no matter what function matching its type scheme “identity” really is, we know that evaluating the expression on the right will produce a function from something to int. There is never any combination of values for the environment that would cause that expression to evaluate to something else.
```Further, an assertion holds iff all its closed instances hold.
```
That’s maybe not so clear but fortunately there is an example in the paper:
```
Thus, to verify the assertion
x:α, f: ∀β(β → β) ⊨ (f x): α
it is enough to verify it for every monotype μ in place of α.
This example illustrates that free type variables in an assertion
are implicitly quantified over the whole assertion, while explicit
quantification in a type scheme has restricted scope.
```
What we’re getting at here is that this is not the same as saying
``` x:∀α α, f: ∀β(β → β) ⊨ (f x): α
```
Because that scopes the alpha to just the type of x, but we want that alpha to be still “in scope” on the right hand side of the double turnstile. Basically it is saying that the assertion is logically:
``` ∀α ( x:α, f: ∀β(β → β) ⊨ (f x): α )
```
All right, that gets us through section four! The last paragraph in section four lays out what is going to happen in the rest of the paper:
```
The remainder of this paper proceeds as follows. First we present
an inference system for inferring valid assertions. Next we present
an algorithm W for computing a type scheme for any expression, under
assumptions A. We then show that W is sound, in the sense that any
type-scheme it derives is derivable in the inference system. Finally
we show that W is complete, in the sense that any derivable
type-scheme is an instance of that computed by W.
```
Next time: A notation for making chains of logical inferences, and another kind of turnstile.
## 14 thoughts on “Excessive explanation, part fifteen”
1. > (Aside: is it possible to have a function with type scheme ∀β(β → β) that is not an identity function? Ponder this!)
It is, if the function in question is not required to be total. For example, the function that diverges for every possible argument matches this type scheme.
In the real world, a function can diverge by entering an infinite loop, or terminating the process, or (in some frameworks) by throwing an exception, or by invoking another function that diverges.
• Am I missing something? Aren’t there lots of functions ∀β(β → β) that are not identities? Presumably the mentioned “square” function is not an identity, for example. Unless the problem is that Exp is so basic that it’s not really possible to define anything else, because there are no operators *, +, etc?
• square is a function from int to int; the requirement is that it be a function from any type to that type, and so square does not fit the bill.
• I don’t think I can edit or delete my above post. I think I see that it’s because it’s generic. It has to be something applicable for all types, not just some arithmetic type or whatever. But we don’t really have anything defined at the most base level (something of type “object” in a normal language I guess), so we can’t really do anything with a parameter of generic type β.
• Even if you could do something with “object”, how would that help you produce a value of type β?
In C#, there is at least one other function that fits: one that returns default(β). But there is no default operator in Exp.
• By “do something with”, I meant either modify or combine or duplicate, etc. But we can’t add βs, we can’t increment them, we can’t make new ones, and as you said we can’t even get a ‘default’ or ‘blank’ one.
2. > (Aside: is it possible to have a function with type scheme ∀β(β → β) that is not an identity function? Ponder this!)
I propose a function which returns 3 if β is an int and the function parameter otherwise.
• How are you going to write “if β is an int” in ML such that the type inferencer does not deduce that β can only be an int?
• Should be simple, just write … Oh, you can’t! Thanks, learned something new today.
3. > (Aside: is it possible to have a function with type scheme ∀β(β → β) that is not an identity function? Ponder this!)
let f a = f a
Which I thought would be a non-identity function from ∀β(β → β), however ghci tells me the type for this function is:
f :: t1 -> t
So it seems my assumption is off. Can you explain this and/or give the solution to your question?
Thanks!
• Your supposition is that your f is ∀β(β → β) but not an identity.
Suppose we call your function with a randomly chosen integer, say: (f 123)
Is the value that is returned always an integer?
If the answer is “yes it is always an integer” then we must demonstrate that it could be an integer other than 123 to show that f is a function ∀β(β → β) that is not an identity. Can you demonstrate that? Which integer is it?
If you cannot demonstrate that then we have no reason to suppose that it is not an identity.
If the answer is “no”, it is not always an integer, then it is not a function ∀β(β → β).
Of course, that’s a trick question because it is not really a “yes or no” question. The real answer is “this function never returns, so asking about the properties of the returned value is nonsensical”. You’ll note in a comment above that a commenter calls functions like this “diverging” functions.
That gives us our answer: since (f 123) does not return an integer I see no reason to suppose that f is ∀β(β → β).
This makes me think of a related question for you to ponder. Is there any function that gets typed as ∀α∀β(α → β) that does not diverge?
• There are lots of fun problems like this. Suppose we extend our type system to include generic lists, using the standard definition of lists: a list-of-α is either empty, or is an α followed by a list-of-α. Let’s define the function len as the function list-of-α → int that gives the length of the list: zero for an empty list, and so on, you know how this goes.
Suppose we have a function f whose type schema is ∀α(list-of-α → list-of-α). Can you make an argument that for any list-of-α x, that (len (f x)) is equal to (len x) ?
• Thanks for the explanation, that makes a lot of sense! As for your second question, I think it is clearer in this case that there exists no function which does not diverge such that ∀α∀β(α → β). You don’t know what α is, so there seems no way to reason about what you should return for β. It seems as though you would have to pick a random value of a random type to return, but I doubt that concept would fit within this functional framework.
| 2,823
| 11,918
|
{"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}
| 2.890625
| 3
|
CC-MAIN-2022-49
|
latest
|
en
| 0.916087
|
https://kalkicode.com/cuban-prime-numbers
| 1,652,796,474,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-21/segments/1652662517485.8/warc/CC-MAIN-20220517130706-20220517160706-00321.warc.gz
| 415,229,658
| 15,514
|
# Cuban prime numbers
Here given code implementation process.
``````// Java program for
// Cuban prime numbers
public class CubanPrimes
{
public int limit;
public boolean[] primes;
public CubanPrimes()
{
// This is prime number limit
this.limit = 1000000;
this.primes = new boolean[this.limit];
this.calculatePrime();
}
// Pre calculate prime number under limit
public void calculatePrime()
{
// Set initial all element is prime
for (int i = 2; i < this.limit; i++)
{
this.primes[i] = true;
}
for (int i = 2; i < this.limit; ++i)
{
if (this.primes[i] == true)
{
for (int j = 2 *i; j < this.limit; j += i)
{
this.primes[j] = false;
}
}
}
}
public boolean isCubanPrime(long num)
{
if (num <= 0)
{
return false;
}
if (num < limit)
{
// When number is under exist limit
return primes[(int) num];
}
// In case number is more than limit.
// Get square root of number
int max = (int) Math.sqrt(num);
// Check that number is cuban primes or not
for (int i = 3; i <= max; ++i)
{
if (primes[i] && (num % i) == 0)
{
return false;
}
}
// When number is cuban prime
return true;
}
public void nthCubanPrime(int n)
{
int count = 0;
long result = 0;
long num = 0;
for (long i = 0; count < n; ++i)
{
num = 1l + 3 *i *(i + 1);
if (this.isCubanPrime(num))
{
result = num;
count++;
}
}
System.out.print("\n " + n + "-th cuban prime is " + result);
}
public void printCubanPrime(int n)
{
if (n <= 0)
{
return;
}
System.out.print("\n Cuban prime numbers from 1 to " + n + " is \n");
int count = 0;
for (long i = 0; count < n; i++)
{
long num = 1l + 3 *i *(i + 1);
if (this.isCubanPrime(num))
{
count++;
System.out.print("\t" + num);
if (count % 5 == 0)
{
System.out.print("\n");
}
}
}
}
public static void main(String[] args)
{
// Test A
// Print initial 100 cuban prime number
// Test B
}
}``````
#### Output
`````` Cuban prime numbers from 1 to 100 is
7 19 37 61 127
271 331 397 547 631
919 1657 1801 1951 2269
2437 2791 3169 3571 4219
4447 5167 5419 6211 7057
7351 8269 9241 10267 11719
12097 13267 13669 16651 19441
19927 22447 23497 24571 25117
26227 27361 33391 35317 42841
45757 47251 49537 50311 55897
59221 60919 65269 70687 73477
74419 75367 81181 82171 87211
88237 89269 92401 96661 102121
103231 104347 110017 112327 114661
115837 126691 129169 131671 135469
140617 144541 145861 151201 155269
163567 169219 170647 176419 180811
189757 200467 202021 213067 231019
234361 241117 246247 251431 260191
263737 267307 276337 279991 283669
67-th cuban prime is 104347
154-th cuban prime is 812761``````
``````// Include header file
#include <iostream>
#include <math.h>
using namespace std;
// C++ program for
// Cuban prime numbers
class CubanPrimes
{
public: int limit;
bool *primes;
CubanPrimes()
{
this->limit = 1000000;
this->primes = new bool[this->limit];
this->calculatePrime();
}
// Pre calculate prime number under limit
void calculatePrime()
{
// Set initial all element is prime
for (int i = 2; i < this->limit; i++)
{
this->primes[i] = true;
}
for (int i = 2; i < this->limit; ++i)
{
if (this->primes[i] == true)
{
for (int j = 2 *i; j < this->limit; j += i)
{
this->primes[j] = false;
}
}
}
}
bool isCubanPrime(long num)
{
if (num <= 0)
{
return false;
}
if (num < this->limit)
{
// When number is under exist limit
return this->primes[(int) num];
}
// In case number is more than limit.
// Get square root of number
int max = (int) sqrt(num);
// Check that number is cuban primes or not
for (int i = 3; i <= max; ++i)
{
if (this->primes[i] && (num % i) == 0)
{
return false;
}
}
// When number is cuban prime
return true;
}
void nthCubanPrime(int n)
{
int count = 0;
long result = 0;
long num = 0;
for (long i = 0; count < n; ++i)
{
num = 1L + 3 *i *(i + 1);
if (this->isCubanPrime(num))
{
result = num;
count++;
}
}
cout << "\n " << n << "-th cuban prime is " << result;
}
void printCubanPrime(int n)
{
if (n <= 0)
{
return;
}
cout << "\n Cuban prime numbers from 1 to " << n << " is \n";
int count = 0;
for (long i = 0; count < n; i++)
{
long num = 1L + 3 *i *(i + 1);
if (this->isCubanPrime(num))
{
count++;
cout << "\t" << num;
if (count % 5 == 0)
{
cout << "\n";
}
}
}
}
};
int main()
{
// Test A
// Print initial 100 cuban prime number
// Test B
return 0;
}``````
#### Output
`````` Cuban prime numbers from 1 to 100 is
7 19 37 61 127
271 331 397 547 631
919 1657 1801 1951 2269
2437 2791 3169 3571 4219
4447 5167 5419 6211 7057
7351 8269 9241 10267 11719
12097 13267 13669 16651 19441
19927 22447 23497 24571 25117
26227 27361 33391 35317 42841
45757 47251 49537 50311 55897
59221 60919 65269 70687 73477
74419 75367 81181 82171 87211
88237 89269 92401 96661 102121
103231 104347 110017 112327 114661
115837 126691 129169 131671 135469
140617 144541 145861 151201 155269
163567 169219 170647 176419 180811
189757 200467 202021 213067 231019
234361 241117 246247 251431 260191
263737 267307 276337 279991 283669
67-th cuban prime is 104347
154-th cuban prime is 812761``````
``````// Include namespace system
using System;
// Csharp program for
// Cuban prime numbers
public class CubanPrimes
{
public int limit;
public Boolean[] primes;
public CubanPrimes()
{
// This is prime number limit
this.limit = 1000000;
this.primes = new Boolean[this.limit];
this.calculatePrime();
}
// Pre calculate prime number under limit
public void calculatePrime()
{
// Set initial all element is prime
for (int i = 2; i < this.limit; i++)
{
this.primes[i] = true;
}
for (int i = 2; i < this.limit; ++i)
{
if (this.primes[i] == true)
{
for (int j = 2 * i; j < this.limit; j += i)
{
this.primes[j] = false;
}
}
}
}
public Boolean isCubanPrime(long num)
{
if (num <= 0)
{
return false;
}
if (num < this.limit)
{
// When number is under exist limit
return this.primes[(int) num];
}
// In case number is more than limit.
// Get square root of number
int max = (int) Math.Sqrt(num);
// Check that number is cuban primes or not
for (int i = 3; i <= max; ++i)
{
if (this.primes[i] && (num % i) == 0)
{
return false;
}
}
// When number is cuban prime
return true;
}
public void nthCubanPrime(int n)
{
int count = 0;
long result = 0;
long num = 0;
for (long i = 0; count < n; ++i)
{
num = 1L + 3 * i * (i + 1);
if (this.isCubanPrime(num))
{
result = num;
count++;
}
}
Console.Write("\n " + n + "-th cuban prime is " + result);
}
public void printCubanPrime(int n)
{
if (n <= 0)
{
return;
}
Console.Write("\n Cuban prime numbers from 1 to " + n + " is \n");
int count = 0;
for (long i = 0; count < n; i++)
{
long num = 1L + 3 * i * (i + 1);
if (this.isCubanPrime(num))
{
count++;
Console.Write("\t" + num);
if (count % 5 == 0)
{
Console.Write("\n");
}
}
}
}
public static void Main(String[] args)
{
// Test A
// Print initial 100 cuban prime number
// Test B
}
}``````
#### Output
`````` Cuban prime numbers from 1 to 100 is
7 19 37 61 127
271 331 397 547 631
919 1657 1801 1951 2269
2437 2791 3169 3571 4219
4447 5167 5419 6211 7057
7351 8269 9241 10267 11719
12097 13267 13669 16651 19441
19927 22447 23497 24571 25117
26227 27361 33391 35317 42841
45757 47251 49537 50311 55897
59221 60919 65269 70687 73477
74419 75367 81181 82171 87211
88237 89269 92401 96661 102121
103231 104347 110017 112327 114661
115837 126691 129169 131671 135469
140617 144541 145861 151201 155269
163567 169219 170647 176419 180811
189757 200467 202021 213067 231019
234361 241117 246247 251431 260191
263737 267307 276337 279991 283669
67-th cuban prime is 104347
154-th cuban prime is 812761``````
``````package main
import "math"
import "fmt"
// Go program for
// Cuban prime numbers
type CubanPrimes struct {
limit int
primes []bool
}
func getCubanPrimes() * CubanPrimes {
var me *CubanPrimes = &CubanPrimes {}
// This is prime number limit
me.limit = 1000000
me.primes = make([] bool, me.limit)
me.calculatePrime()
return me
}
// Pre calculate prime number under limit
func(this CubanPrimes) calculatePrime() {
// Set initial all element is prime
for i := 2 ; i < this.limit ; i++ {
this.primes[i] = true
}
for i := 2 ; i < this.limit ; i++ {
if this.primes[i] == true {
for j := 2 * i ; j < this.limit ; j += i {
this.primes[j] = false
}
}
}
}
func(this CubanPrimes) isCubanPrime(num int) bool {
if num <= 0 {
return false
}
if num < this.limit {
// When number is under exist limit
return this.primes[num]
}
// In case number is more than limit.
// Get square root of number
var max int = int(math.Sqrt(float64(num)))
// Check that number is cuban primes or not
for i := 3 ; i <= max ; i++ {
if this.primes[i] && (num % i) == 0 {
return false
}
}
// When number is cuban prime
return true
}
func(this CubanPrimes) nthCubanPrime(n int) {
var count int = 0
var result int = 0
var num int = 0
for i := 0 ; count < n ; i++ {
num = 1 + 3 * i * (i + 1)
if this.isCubanPrime(num) {
result = num
count++
}
}
fmt.Print("\n ", n, "-th cuban prime is ", result)
}
func(this CubanPrimes) printCubanPrime(n int) {
if n <= 0 {
return
}
fmt.Print("\n Cuban prime numbers from 1 to ", n, " is \n")
var count int = 0
for i := 0 ; count < n ; i++ {
var num int = 1 + 3 * i * (i + 1)
if this.isCubanPrime(num) {
count++
fmt.Print("\t", num)
if count % 5 == 0 {
fmt.Print("\n")
}
}
}
}
func main() {
var task * CubanPrimes = getCubanPrimes()
// Test A
// Print initial 100 cuban prime number
// Test B
}``````
#### Output
`````` Cuban prime numbers from 1 to 100 is
7 19 37 61 127
271 331 397 547 631
919 1657 1801 1951 2269
2437 2791 3169 3571 4219
4447 5167 5419 6211 7057
7351 8269 9241 10267 11719
12097 13267 13669 16651 19441
19927 22447 23497 24571 25117
26227 27361 33391 35317 42841
45757 47251 49537 50311 55897
59221 60919 65269 70687 73477
74419 75367 81181 82171 87211
88237 89269 92401 96661 102121
103231 104347 110017 112327 114661
115837 126691 129169 131671 135469
140617 144541 145861 151201 155269
163567 169219 170647 176419 180811
189757 200467 202021 213067 231019
234361 241117 246247 251431 260191
263737 267307 276337 279991 283669
67-th cuban prime is 104347
154-th cuban prime is 812761``````
``````<?php
// Php program for
// Cuban prime numbers
class CubanPrimes
{
public \$limit;
public \$primes;
public function __construct()
{
\$this->limit = 100000;
\$this->primes = array_fill(0, \$this->limit, false);
\$this->calculatePrime();
}
// Pre calculate prime number under limit
public function calculatePrime()
{
// Set initial all element is prime
for (\$i = 2; \$i < \$this->limit; \$i++)
{
\$this->primes[\$i] = true;
}
for (\$i = 2; \$i < \$this->limit; ++\$i)
{
if (\$this->primes[\$i] == true)
{
for (\$j = 2 * \$i; \$j < \$this->limit; \$j += \$i)
{
\$this->primes[\$j] = false;
}
}
}
}
public function isCubanPrime(\$num)
{
if (\$num <= 0)
{
return false;
}
if (\$num < \$this->limit)
{
// When number is under exist limit
return \$this->primes[(int) \$num];
}
// In case number is more than limit.
// Get square root of number
\$max = (int) sqrt(\$num);
// Check that number is cuban primes or not
for (\$i = 3; \$i <= \$max; ++\$i)
{
if (\$this->primes[\$i] && (\$num % \$i) == 0)
{
return false;
}
}
// When number is cuban prime
return true;
}
public function nthCubanPrime(\$n)
{
\$count = 0;
\$result = 0;
\$num = 0;
for (\$i = 0; \$count < \$n; ++\$i)
{
\$num = 1 + 3 * \$i * (\$i + 1);
if (\$this->isCubanPrime(\$num))
{
\$result = \$num;
\$count++;
}
}
echo("\n ".\$n.
"-th cuban prime is ".\$result);
}
public function printCubanPrime(\$n)
{
if (\$n <= 0)
{
return;
}
echo("\n Cuban prime numbers from 1 to ".\$n.
" is \n");
\$count = 0;
for (\$i = 0; \$count < \$n; \$i++)
{
\$num = 1 + 3 * \$i * (\$i + 1);
if (\$this->isCubanPrime(\$num))
{
\$count++;
echo("\t".\$num);
if (\$count % 5 == 0)
{
echo("\n");
}
}
}
}
}
function main()
{
// Test A
// Print initial 100 cuban prime number
// Test B
}
main();``````
#### Output
`````` Cuban prime numbers from 1 to 100 is
7 19 37 61 127
271 331 397 547 631
919 1657 1801 1951 2269
2437 2791 3169 3571 4219
4447 5167 5419 6211 7057
7351 8269 9241 10267 11719
12097 13267 13669 16651 19441
19927 22447 23497 24571 25117
26227 27361 33391 35317 42841
45757 47251 49537 50311 55897
59221 60919 65269 70687 73477
74419 75367 81181 82171 87211
88237 89269 92401 96661 102121
103231 104347 110017 112327 114661
115837 126691 129169 131671 135469
140617 144541 145861 151201 155269
163567 169219 170647 176419 180811
189757 200467 202021 213067 231019
234361 241117 246247 251431 260191
263737 267307 276337 279991 283669
67-th cuban prime is 104347
154-th cuban prime is 812761``````
``````// Node JS program for
// Cuban prime numbers
class CubanPrimes
{
constructor()
{
this.limit = 1000000;
this.primes = Array(this.limit).fill(false);
this.calculatePrime();
}
// Pre calculate prime number under limit
calculatePrime()
{
// Set initial all element is prime
for (var i = 2; i < this.limit; i++)
{
this.primes[i] = true;
}
for (var i = 2; i < this.limit; ++i)
{
if (this.primes[i] == true)
{
for (var j = 2 * i; j < this.limit; j += i)
{
this.primes[j] = false;
}
}
}
}
isCubanPrime(num)
{
if (num <= 0)
{
return false;
}
if (num < this.limit)
{
// When number is under exist limit
return this.primes[num];
}
// In case number is more than limit.
// Get square root of number
var max = math.sqrt(num);
// Check that number is cuban primes or not
for (var i = 3; i <= max; ++i)
{
if (this.primes[i] && (num % i) == 0)
{
return false;
}
}
// When number is cuban prime
return true;
}
nthCubanPrime(n)
{
var count = 0;
var result = 0;
var num = 0;
for (var i = 0; count < n; ++i)
{
num = 1 + 3 * i * (i + 1);
if (this.isCubanPrime(num))
{
result = num;
count++;
}
}
process.stdout.write("\n " + n + "-th cuban prime is " + result);
}
printCubanPrime(n)
{
if (n <= 0)
{
return;
}
process.stdout.write("\n Cuban prime numbers from 1 to " + n + " is \n");
var count = 0;
for (var i = 0; count < n; i++)
{
var num = 1 + 3 * i * (i + 1);
if (this.isCubanPrime(num))
{
count++;
process.stdout.write("\t" + num);
if (count % 5 == 0)
{
process.stdout.write("\n");
}
}
}
}
}
function main()
{
// Test A
// Print initial 100 cuban prime number
// Test B
}
main();``````
#### Output
`````` Cuban prime numbers from 1 to 100 is
7 19 37 61 127
271 331 397 547 631
919 1657 1801 1951 2269
2437 2791 3169 3571 4219
4447 5167 5419 6211 7057
7351 8269 9241 10267 11719
12097 13267 13669 16651 19441
19927 22447 23497 24571 25117
26227 27361 33391 35317 42841
45757 47251 49537 50311 55897
59221 60919 65269 70687 73477
74419 75367 81181 82171 87211
88237 89269 92401 96661 102121
103231 104347 110017 112327 114661
115837 126691 129169 131671 135469
140617 144541 145861 151201 155269
163567 169219 170647 176419 180811
189757 200467 202021 213067 231019
234361 241117 246247 251431 260191
263737 267307 276337 279991 283669
67-th cuban prime is 104347
154-th cuban prime is 812761``````
``````import math
# Python 3 program for
# Cuban prime numbers
class CubanPrimes :
def __init__(self) :
self.limit = 1000000
self.primes = [False] * (self.limit)
self.calculatePrime()
# Pre calculate prime number under limit
def calculatePrime(self) :
i = 2
# Set initial all element is prime
while (i < self.limit) :
self.primes[i] = True
i += 1
i = 2
while (i < self.limit) :
if (self.primes[i] == True) :
j = 2 * i
while (j < self.limit) :
self.primes[j] = False
j += i
i += 1
def isCubanPrime(self, num) :
if (num <= 0) :
return False
if (num < self.limit) :
# When number is under exist limit
return self.primes[ num]
# In case number is more than limit.
# Get square root of number
max = math.sqrt(num)
i = 3
# Check that number is cuban primes or not
while (i <= max) :
if (self.primes[i] and(num % i) == 0) :
return False
i += 1
# When number is cuban prime
return True
def nthCubanPrime(self, n) :
count = 0
result = 0
num = 0
i = 0
while (count < n) :
num = 1 + 3 * i * (i + 1)
if (self.isCubanPrime(num)) :
result = num
count += 1
i += 1
print("\n ", n ,"-th cuban prime is ", result, end = "")
def printCubanPrime(self, n) :
if (n <= 0) :
return
print("\n Cuban prime numbers from 1 to ", n ," is ")
count = 0
i = 0
while (count < n) :
num = 1 + 3 * i * (i + 1)
if (self.isCubanPrime(num)) :
count += 1
print("\t", num, end = "")
if (count % 5 == 0) :
print(end = "\n")
i += 1
def main() :
# Test A
# Print initial 100 cuban prime number
# Test B
if __name__ == "__main__": main()``````
#### Output
`````` Cuban prime numbers from 1 to 100 is
7 19 37 61 127
271 331 397 547 631
919 1657 1801 1951 2269
2437 2791 3169 3571 4219
4447 5167 5419 6211 7057
7351 8269 9241 10267 11719
12097 13267 13669 16651 19441
19927 22447 23497 24571 25117
26227 27361 33391 35317 42841
45757 47251 49537 50311 55897
59221 60919 65269 70687 73477
74419 75367 81181 82171 87211
88237 89269 92401 96661 102121
103231 104347 110017 112327 114661
115837 126691 129169 131671 135469
140617 144541 145861 151201 155269
163567 169219 170647 176419 180811
189757 200467 202021 213067 231019
234361 241117 246247 251431 260191
263737 267307 276337 279991 283669
67 -th cuban prime is 104347
154 -th cuban prime is 812761``````
``````# Ruby program for
# Cuban prime numbers
class CubanPrimes
# Define the accessor and reader of class CubanPrimes
attr_accessor :limit, :primes
def initialize()
self.limit = 1000000
self.primes = Array.new(self.limit) {false}
self.calculatePrime()
end
# Pre calculate prime number under limit
def calculatePrime()
i = 2
# Set initial all element is prime
while (i < self.limit)
self.primes[i] = true
i += 1
end
i = 2
while (i < self.limit)
if (self.primes[i] == true)
j = 2 * i
while (j < self.limit)
self.primes[j] = false
j += i
end
end
i += 1
end
end
def isCubanPrime(num)
if (num <= 0)
return false
end
if (num < self.limit)
# When number is under exist limit
return self.primes[num]
end
# In case number is more than limit.
# Get square root of number
max = Math.sqrt(num).to_i
i = 3
# Check that number is cuban primes or not
while (i <= max)
if (self.primes[i] && (num % i) == 0)
return false
end
i += 1
end
# When number is cuban prime
return true
end
def nthCubanPrime(n)
count = 0
result = 0
num = 0
i = 0
while (count < n)
num = 1 + 3 * i * (i + 1)
if (self.isCubanPrime(num))
result = num
count += 1
end
i += 1
end
print("\n ", n ,"-th cuban prime is ", result)
end
def printCubanPrime(n)
if (n <= 0)
return
end
print("\n Cuban prime numbers from 1 to ", n ," is \n")
count = 0
i = 0
while (count < n)
num = 1 + 3 * i * (i + 1)
if (self.isCubanPrime(num))
count += 1
print("\t", num)
if (count % 5 == 0)
print("\n")
end
end
i += 1
end
end
end
def main()
# Test A
# Print initial 100 cuban prime number
# Test B
end
main()``````
#### Output
`````` Cuban prime numbers from 1 to 100 is
7 19 37 61 127
271 331 397 547 631
919 1657 1801 1951 2269
2437 2791 3169 3571 4219
4447 5167 5419 6211 7057
7351 8269 9241 10267 11719
12097 13267 13669 16651 19441
19927 22447 23497 24571 25117
26227 27361 33391 35317 42841
45757 47251 49537 50311 55897
59221 60919 65269 70687 73477
74419 75367 81181 82171 87211
88237 89269 92401 96661 102121
103231 104347 110017 112327 114661
115837 126691 129169 131671 135469
140617 144541 145861 151201 155269
163567 169219 170647 176419 180811
189757 200467 202021 213067 231019
234361 241117 246247 251431 260191
263737 267307 276337 279991 283669
67-th cuban prime is 104347
154-th cuban prime is 812761``````
``````// Scala program for
// Cuban prime numbers
class CubanPrimes(var limit: Int,
var primes: Array[Boolean])
{
def this()
{
this(1000000,Array.fill[Boolean](1000000)(false));
this.calculatePrime();
}
// Pre calculate prime number under limit
def calculatePrime(): Unit = {
var i: Int = 2;
// Set initial all element is prime
while (i < this.limit)
{
this.primes(i) = true;
i += 1;
}
i = 2;
while (i < this.limit)
{
if (this.primes(i) == true)
{
var j: Int = 2 * i;
while (j < this.limit)
{
this.primes(j) = false;
j += i;
}
}
i += 1;
}
}
def isCubanPrime(num: Long): Boolean = {
if (num <= 0)
{
return false;
}
if (num < limit)
{
// When number is under exist limit
return primes(num.toInt);
}
// In case number is more than limit.
// Get square root of number
var max: Int = scala.math.sqrt(num).toInt;
var i: Int = 3;
// Check that number is cuban primes or not
while (i <= max)
{
if (primes(i) && (num % i) == 0)
{
return false;
}
i += 1;
}
// When number is cuban prime
return true;
}
def nthCubanPrime(n: Int): Unit = {
var count: Int = 0;
var result: Long = 0;
var num: Long = 0;
var i: Long = 0;
while (count < n)
{
num = 1L + 3 * i * (i + 1);
if (this.isCubanPrime(num))
{
result = num;
count += 1;
}
i += 1;
}
print("\n " + n + "-th cuban prime is " + result);
}
def printCubanPrime(n: Int): Unit = {
if (n <= 0)
{
return;
}
print("\n Cuban prime numbers from 1 to " + n + " is \n");
var count: Int = 0;
var i: Long = 0;
while (count < n)
{
var num: Long = 1L + 3 * i * (i + 1);
if (this.isCubanPrime(num))
{
count += 1;
print("\t" + num);
if (count % 5 == 0)
{
print("\n");
}
}
i += 1;
}
}
}
object Main
{
def main(args: Array[String]): Unit = {
var task: CubanPrimes = new CubanPrimes();
// Test A
// Print initial 100 cuban prime number
// Test B
}
}``````
#### Output
`````` Cuban prime numbers from 1 to 100 is
7 19 37 61 127
271 331 397 547 631
919 1657 1801 1951 2269
2437 2791 3169 3571 4219
4447 5167 5419 6211 7057
7351 8269 9241 10267 11719
12097 13267 13669 16651 19441
19927 22447 23497 24571 25117
26227 27361 33391 35317 42841
45757 47251 49537 50311 55897
59221 60919 65269 70687 73477
74419 75367 81181 82171 87211
88237 89269 92401 96661 102121
103231 104347 110017 112327 114661
115837 126691 129169 131671 135469
140617 144541 145861 151201 155269
163567 169219 170647 176419 180811
189757 200467 202021 213067 231019
234361 241117 246247 251431 260191
263737 267307 276337 279991 283669
67-th cuban prime is 104347
154-th cuban prime is 812761``````
``````import Foundation;
// Swift 4 program for
// Cuban prime numbers
class CubanPrimes
{
var limit: Int;
var primes: [Bool];
init()
{
self.limit = 1000000;
self.primes = Array(repeating: false, count: self.limit);
self.calculatePrime();
}
// Pre calculate prime number under limit
func calculatePrime()
{
var i: Int = 2;
// Set initial all element is prime
while (i < self.limit)
{
self.primes[i] = true;
i += 1;
}
i = 2;
while (i < self.limit)
{
if (self.primes[i] == true)
{
var j: Int = 2 * i;
while (j < self.limit)
{
self.primes[j] = false;
j += i;
}
}
i += 1;
}
}
func isCubanPrime(_ num: Int) -> Bool
{
if (num <= 0)
{
return false;
}
if (num < self.limit)
{
// When number is under exist limit
return self.primes[num];
}
// In case number is more than limit.
// Get square root of number
let max: Int = Int(sqrt(Double(num)));
var i: Int = 3;
// Check that number is cuban primes or not
while (i <= max)
{
if (self.primes[i] && (num % i) == 0)
{
return false;
}
i += 1;
}
// When number is cuban prime
return true;
}
func nthCubanPrime(_ n: Int)
{
var count: Int = 0;
var result: Int = 0;
var num: Int ;
var i: Int = 0;
while (count < n)
{
num = 1 + 3 * i * (i + 1);
if (self.isCubanPrime(num))
{
result = num;
count += 1;
}
i += 1;
}
print("\n ", n ,"-th cuban prime is ", result, terminator: "");
}
func printCubanPrime(_ n: Int)
{
if (n <= 0)
{
return;
}
print("\n Cuban prime numbers from 1 to ", n ," is ");
var count: Int = 0;
var i: Int = 0;
while (count < n)
{
let num: Int = 1 + 3 * i * (i + 1);
if (self.isCubanPrime(num))
{
count += 1;
print("\t", num, terminator: "");
if (count % 5 == 0)
{
print(terminator: "\n");
}
}
i += 1;
}
}
}
func main()
{
// Test A
// Print initial 100 cuban prime number
// Test B
}
main();``````
#### Output
`````` Cuban prime numbers from 1 to 100 is
7 19 37 61 127
271 331 397 547 631
919 1657 1801 1951 2269
2437 2791 3169 3571 4219
4447 5167 5419 6211 7057
7351 8269 9241 10267 11719
12097 13267 13669 16651 19441
19927 22447 23497 24571 25117
26227 27361 33391 35317 42841
45757 47251 49537 50311 55897
59221 60919 65269 70687 73477
74419 75367 81181 82171 87211
88237 89269 92401 96661 102121
103231 104347 110017 112327 114661
115837 126691 129169 131671 135469
140617 144541 145861 151201 155269
163567 169219 170647 176419 180811
189757 200467 202021 213067 231019
234361 241117 246247 251431 260191
263737 267307 276337 279991 283669
67 -th cuban prime is 104347
154 -th cuban prime is 812761``````
``````// Kotlin program for
// Cuban prime numbers
class CubanPrimes
{
var limit: Int;
var primes: Array < Boolean > ;
constructor()
{
this.limit = 1000000;
this.primes = Array(this.limit)
{
false
};
this.calculatePrime();
}
// Pre calculate prime number under limit
fun calculatePrime(): Unit
{
var i: Int = 2;
// Set initial all element is prime
while (i < this.limit)
{
this.primes[i] = true;
i += 1;
}
i = 2;
while (i < this.limit)
{
if (this.primes[i] == true)
{
var j: Int = 2 * i;
while (j < this.limit)
{
this.primes[j] = false;
j += i;
}
}
i += 1;
}
}
fun isCubanPrime(num: Long): Boolean
{
if (num <= 0)
{
return false;
}
if (num < this.limit)
{
// When number is under exist limit
return this.primes[num.toInt()];
}
// In case number is more than limit.
// Get square root of number
val max: Int = Math.sqrt(num.toDouble()).toInt();
var i: Int = 3;
// Check that number is cuban primes or not
while (i <= max)
{
if (this.primes[i] && ((num % i)==0L) )
{
return false;
}
i += 1;
}
// When number is cuban prime
return true;
}
fun nthCubanPrime(n: Int): Unit
{
var count: Int = 0;
var result: Long = 0;
var num: Long ;
var i: Long = 0;
while (count < n)
{
num = 1L + 3 * i * (i + 1);
if (this.isCubanPrime(num))
{
result = num;
count += 1;
}
i += 1;
}
print("\n " + n + "-th cuban prime is " + result);
}
fun printCubanPrime(n: Int): Unit
{
if (n <= 0)
{
return;
}
print("\n Cuban prime numbers from 1 to " + n + " is \n");
var count: Int = 0;
var i: Long = 0;
while (count < n)
{
val num: Long = 1L + 3 * i * (i + 1);
if (this.isCubanPrime(num))
{
count += 1;
print("\t" + num);
if (count % 5 == 0)
{
print("\n");
}
}
i += 1;
}
}
}
fun main(args: Array < String > ): Unit
{
// Test A
// Print initial 100 cuban prime number
// Test B
}``````
#### Output
`````` Cuban prime numbers from 1 to 100 is
7 19 37 61 127
271 331 397 547 631
919 1657 1801 1951 2269
2437 2791 3169 3571 4219
4447 5167 5419 6211 7057
7351 8269 9241 10267 11719
12097 13267 13669 16651 19441
19927 22447 23497 24571 25117
26227 27361 33391 35317 42841
45757 47251 49537 50311 55897
59221 60919 65269 70687 73477
74419 75367 81181 82171 87211
88237 89269 92401 96661 102121
103231 104347 110017 112327 114661
115837 126691 129169 131671 135469
140617 144541 145861 151201 155269
163567 169219 170647 176419 180811
189757 200467 202021 213067 231019
234361 241117 246247 251431 260191
263737 267307 276337 279991 283669
67-th cuban prime is 104347
154-th cuban prime is 812761``````
## Comment
Please share your knowledge to improve code and content standard. Also submit your doubts, and test case. We improve by your feedback. We will try to resolve your query as soon as possible.
| 10,242
| 26,997
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.40625
| 3
|
CC-MAIN-2022-21
|
longest
|
en
| 0.301391
|
http://rational-equations.com/in-rational-equations/multiplying-fractions/teacher-addition-online-for.html
| 1,519,402,386,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-09/segments/1518891814801.45/warc/CC-MAIN-20180223154626-20180223174626-00723.warc.gz
| 293,629,891
| 11,732
|
Algebra Tutorials!
Home Rational Expressions Graphs of Rational Functions Solve Two-Step Equations Multiply, Dividing; Exponents; Square Roots; and Solving Equations LinearEquations Solving a Quadratic Equation Systems of Linear Equations Introduction Equations and Inequalities Solving 2nd Degree Equations Review Solving Quadratic Equations System of Equations Solving Equations & Inequalities Linear Equations Functions Zeros, and Applications Rational Expressions and Functions Linear equations in two variables Lesson Plan for Comparing and Ordering Rational Numbers LinearEquations Solving Equations Radicals and Rational Exponents Solving Linear Equations Systems of Linear Equations Solving Exponential and Logarithmic Equations Solving Systems of Linear Equations DISTANCE,CIRCLES,AND QUADRATIC EQUATIONS Solving Quadratic Equations Quadratic and Rational Inequalit Applications of Systems of Linear Equations in Two Variables Systems of Linear Equations Test Description for RATIONAL EX Exponential and Logarithmic Equations Systems of Linear Equations: Cramer's Rule Introduction to Systems of Linear Equations Literal Equations & Formula Equations and Inequalities with Absolute Value Rational Expressions SOLVING LINEAR AND QUADRATIC EQUATIONS Steepest Descent for Solving Linear Equations The Quadratic Equation Linear equations in two variables
Try the Free Math Solver or Scroll down to Resources!
Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
Ineq. #3:
Ineq. #4:
Ineq. #5:
Ineq. #6:
Ineq. #7:
Ineq. #8:
Ineq. #9:
Solve for:
Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg:
teacher addition online for course 3 illinois prentice hall mathematics
Related topics:
slope intercept formula | teaching how to multiply integers | How To Figure A 5th Grade Problem Of Ratio | teaching 5th grade expanded form with exponents | Solve Differential Equation In Excel | software algebra | examples of rational and irrational numbers | math trivia 6 | formulas of maths class 9th pak book | linear programming worded problems | +algebra calculations
Author Message
svalnase
Registered: 14.12.2004
From: Fort Wayne, Indiana
Posted: Friday 29th of Dec 09:18 I am in urgent need of help in completing a homework in teacher addition online for course 3 illinois prentice hall mathematics. I need to submit it by next week and am having a difficult time trying to figure out a few tricky problems. I tried some of the internet help sites but have not gotten what I want so far. I would be really glad if anyone can guide me.
ameich
Registered: 21.03.2005
From: Prague, Czech Republic
Posted: Sunday 31st of Dec 07:03 Hi! I guess I can give you ideas on how to solve your homework. But for that I need more specifics. Can you give details about what exactly is the teacher addition online for course 3 illinois prentice hall mathematics homework that you have to work out. I am quite good at working out these kind of things. Plus I have this great software Algebrator that I got from a friend which is soooo good at solving algebra assignment. Give me the details and perhaps we can work something out...
CHS`
Registered: 04.07.2001
From: Victoria City, Hong Kong Island, Hong Kong
Posted: Monday 01st of Jan 10:31 Algebrator indeed is a very good tool to help you learn math, without having to go to school. You won’t just get the answer to the question but the entire solution as well, that’s how you can build a strong mathematical foundation. And to score well in math, it’s important to have strong concepts. I would highly recommend using this software if you want to finish your project on time.
Dxi_Sysdech
Registered: 05.07.2001
From: Right here, can't you see me?
Posted: Wednesday 03rd of Jan 08:21 I remember having often faced problems with graphing function, simplifying expressions and perpendicular lines. A really great piece of algebra program is Algebrator software. By simply typing in a problem from workbook a step by step solution would appear by a click on Solve. I have used it through many math classes – College Algebra, Remedial Algebra and College Algebra. I greatly recommend the program.
Bnech
Registered: 20.08.2003
From: Slovenia
Posted: Thursday 04th of Jan 13:13 Is the software really that helpful? I’m worried because the software might not really help because it only solves the problem per ?e. I like to learn how a problem is answered and not only find out the answer. Nevertheless, could you give me a link for this software?
Jot
Registered: 07.09.2001
From: Ubik
Posted: Friday 05th of Jan 12:28 You should check out http://www.rational-equations.com/systems-of-linear-equations-1.html. Your algebra will get better in a short time, you shall see! Good luck !
| 1,188
| 5,032
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.875
| 4
|
CC-MAIN-2018-09
|
latest
|
en
| 0.828913
|
https://www.roadlesstraveledstore.com/what-is-rise-time-and-fall-time-of-a-cmos-inverter/
| 1,716,810,755,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971059040.32/warc/CC-MAIN-20240527113621-20240527143621-00729.warc.gz
| 849,811,376
| 15,670
|
# What is rise time and fall time of a CMOS inverter?
## What is rise time and fall time of a CMOS inverter?
Rise time (tr) is the time, during transition, when output switches from 10% to 90% of the maximum value. Fall time (tf) is the time, during transition, when output switches from 90% to 10% of the maximum value. Many designs could also prefer 30% to 70% for rise time and 70% to 30% for fall time.
What are the five regions of operation on CMOS inverter explain in detail?
CMOS Inverter – Circuit, Operation and Description
Region Vin nMOS
A < VTO, n Cut – off
B VIL Saturation
C Vth Saturation
D VIH Linear
What is VDD in CMOS inverter?
= VDD. • Output Low Voltage, V. OL. – minimum output voltage. • occurs when input is high (Vin = VDD)
### What is propagation delay in CMOS inverter?
The propagation delay tp of a gate defines how quickly it responds to a change at its. inputs, it expresses the delay experienced by a signal when passing through a gate. It is. measured between the 50% transition points of the input and output waveforms as. shown in the figure 16.1 for an inverting gate.
Does CMOS have high switching speed?
CMOS offers relatively high speed, low power dissipation, high noise margins in both states, and will operate over a wide range of source and input voltages (provided the source voltage is fixed).
What does a CMOS inverter do?
A CMOS inverter is a field-effect transistor that is composed of a metal gate that lies on top of an insulating layer of oxygen, which lies on top of a semiconductor. CMOS inverters are found in most electronic devices and are responsible for producing data within small circuits.
## What happens when CMOS inverter output is high?
You can easily see that the CMOS circuit functions as an inverter by noting that when VIN is five volts, VOUT is zero, and vice versa. Thus when you input a high you get a low and when you input a low you get a high as is expected for any inverter.
Why capacitor is used in CMOS inverter?
The effect of load capacitance is that it causes a transient current demand on the inverter output, which causes a number of secondary effects, two of which are: The output has a limited current capability, so this limits the maximum rate of change of the signal, slowing down the edges.
Which is faster CMOS or BJT?
Once the level of integration and power levels are taken into account, modern twists outperform bjt and bipolar-cmos (bicmos) in almost all cases. For current signal mode processing, bjt will be faster because current amplification does not involve big changes in connection potential.
### Is CMOS faster than TTL?
TTL chips are generally faster than CMOS gates (but see ACT series), however there are two logic technologies faster than TTL-Emitter-coupled logic (ECL) and gallium arsenide (GaAs). These chips come at considerable cost in power consumption and ease of interface to other logic families.
What is rise and fall in CMOS inverter?
Rise time (t r) is the time, during transition, when output switches from 10% to 90% of the maximum value. Fall time (t f) is the time, during transition, when output switches from 90% to 10% of the maximum value.
How is the propagation delay of an inverter calculated?
The delay is usually calculated at 50% point of input-output switching, as shown in above figure. Now, in order to find the propagation delay, we need a model that matches the delay of inverter.
## What does HL stand for in CMOS inverters?
In the plot of output voltage in figure 2, there are two time intervals marked by and . Here, the “p” in the subscript stands for propagation delay. The “hl” stands for high-to-low, and “lh” stands for low-to-high. The inverters in the circuit are operating between two voltages.
How is the input signal to a CMOS inverter driven?
The input signals to our CMOS inverter in the previous discussions was taken as an exact step function. But, for practical scenarios the inverter will also be driven by the output signal of some other logic gate. This means that the input signal to the inverter we are studying will be more of a “ramp-signal” rather than a step signal.
13/10/2020
| 978
| 4,168
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.515625
| 3
|
CC-MAIN-2024-22
|
latest
|
en
| 0.915102
|
https://biz.libretexts.org/Courses/Lumen_Learning/Book%3A_Business_Communication_Skills_for_Managers_(Lumen)/04%3A_Research/4.23%3A_Introduction_to_Internal_Data
| 1,696,290,341,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233511023.76/warc/CC-MAIN-20231002232712-20231003022712-00726.warc.gz
| 147,907,463
| 30,042
|
# 4.23: Introduction to Internal Data
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
## What you’ll learn to do: Process information from internal sources
Barron’s defines internal data as:[1]
Information, facts and data available from within a company’s information systems. Internal data is normally not accessible by outside parties without the company’s express permission.
While this definition seems straightforward, the complexity of gathering and analyzing these types of data can be more complicated than one might think. Processing data from internal sources requires patience, diligence and care. There are two primary concerns:
1. The act of gathering information about your own organization is not without political and ethical considerations. When a person from one part of the organization seeks data about another–or about the enterprise as a whole–it can have implications for the well-being and security of people’s circumstances. Think of a salesperson who is having their volume studied by an HR analyst or an accountant being questioned about how quickly their division processes invoices—inquiry into organizational phenomena can be a charged event.
2. Data can be dynamic and often hard to find. Some data are relatively straightforward and can be captured in time, or the gathering and publishing of the data is mandated by an outside element. For example, studying Apple from Apple’s 10-K is straightforward; however, learning more about their culture from people who currently work there is far more complicated, as illustrated by this Business Insider UK article. The reasons an organization would be so close-hold about their culture is usually justified by the competitive landscape in which it operates.
### Think about It
How would you address gathering data on your organization if you worked in a competitive and closely-guarded industry?
For our purposes here, the right approach to processing your data is to ask yourself early in your study, “is there any part of this project that could be problematic for any stakeholders involved?”
So, what is a stakeholder? A stakeholder is someone who has a particular interest in the organization; people who work in the organization are clear stakeholders as are customers, shareholders, suppliers and vendors. In short, conducting a (brief, at minimum) Stakeholder Analysis before your research is wise. While these types of analysis can become their own in-depth reports, your goal here is to address political, turf, and strategic sensitivities with regard to the gathering of your data; your stakeholder analysis does not need to be perfect or exhaustive, but consider it an important part of processing your data.
A YouTube element has been excluded from this version of the text. You can view it online here: http://pb.libretexts.org/bcsfm/?p=176
1. All Business, "Internal data." Barrons Dictionary. Web. 12 June 2018.
## Contributors and Attributions
CC licensed content, Original
• Introduction to Internal Data. Authored by: Freedom Learning Group. Provided by: Lumen Learning. License: CC BY: Attribution
| 977
| 4,082
|
{"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}
| 3.53125
| 4
|
CC-MAIN-2023-40
|
latest
|
en
| 0.623444
|
https://math.stackexchange.com/questions/141238/sliding-blocks-puzzle
| 1,708,958,420,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-10/segments/1707947474660.32/warc/CC-MAIN-20240226130305-20240226160305-00425.warc.gz
| 389,716,632
| 35,170
|
# Sliding blocks puzzle
Consider a 'game' played on a subset $S$ of an $n^2$ square grid as follows. There are 3 types of pieces, each occupying a square, 1 green, some red and the rest are blue, a move consists of shuffling the green piece with any of its 4 adjacent pieces (if they are within $S$). $S$ consists of squares, squares not in $S$ are static, $S$ can be any subset of the $n^2$ square.
If two board configurations are reachable from eachother, is it possible to obtain an upper bound on the number of moves needed, given only the board size $n$, is it polynomial in $n$?
Hint: consider the configurations with one red square and the rest (except the green) blue. Can you prove that any one can be converted to any other within a polynomial number of moves? Then you are there-if the number of red and blue squares differ, you can't fix that. If they don't, number the red and blue squares in the two configurations. Fix the upper left corner, then the square next to it, and so on. There are less than $n^2$ squares to fix. You still need to prove you won't disturb the ones you have already solved, but that shouldn't be hard except for the last square in a row.
| 290
| 1,180
|
{"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}
| 3.609375
| 4
|
CC-MAIN-2024-10
|
latest
|
en
| 0.951933
|
https://www.jiskha.com/display.cgi?id=1378160539
| 1,516,767,367,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-05/segments/1516084893300.96/warc/CC-MAIN-20180124030651-20180124050651-00266.warc.gz
| 950,423,993
| 3,226
|
# math
posted by .
what is the portion if the base is 900 and the rate is 12 3/4%
• math -
900 * 0.1275 = 114.75
## Similar Questions
1. ### Math
What is the portion if the base is 900 and the rate is 12¾ %?
2. ### math
What is the portion if the base is 900 and the rate is 12 3/4?
3. ### math
What is the portion if the base is 900 and the rate is 12¾ %
4. ### biz math
what is the portion if the base is 900 and the rate is 12 3/4?
5. ### math
What is the portion if the base is 900 and the rate is 12 3/4
6. ### math
What is the portion if the base is 900 and the rate is 12¾ %?
7. ### AICE math
A 30 foot ladder is leaning against a house(see figure). if the base of the ladder is pulled away from the house at a rate of 2 feet per second, the top will move down the wall at a a rate of r= 2x/(sqrt(900-x^2) ft/sec, where s ist …
8. ### Math
What is the portion if the base is 900 and the rate is 12¾ %?
| 297
| 922
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.328125
| 3
|
CC-MAIN-2018-05
|
latest
|
en
| 0.898864
|
http://www.jiskha.com/display.cgi?id=1265090750
| 1,498,357,148,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-26/segments/1498128320386.71/warc/CC-MAIN-20170625013851-20170625033851-00499.warc.gz
| 544,141,246
| 3,703
|
# algebra
posted by .
Carlos Martin received a statement from his bank showing a balance of \$56.75 as of March 15th. His checkbook shows a balance of \$87.37 as of March 20. The bank returned all the cancelled checks but two. One check was for \$5.00 and the other was for \$13.25. How much did Carlos deposit in his account between the March 15 and the March 20?
| 94
| 366
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26
|
latest
|
en
| 0.994661
|
https://activegaliano.org/how-tall-is-32-inches-in-feet-new-update/
| 1,695,979,816,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233510501.83/warc/CC-MAIN-20230929090526-20230929120526-00867.warc.gz
| 97,487,283
| 37,291
|
Skip to content
Home » How Tall Is 32 Inches In Feet? New Update
# How Tall Is 32 Inches In Feet? New Update
Let’s discuss the question: how tall is 32 inches in feet. We summarize all relevant answers in section Q&A of website Activegaliano.org in category: Blog Marketing. See more related questions in the comments below.
Table of Contents
## How many inches is 5 4 feet?
If you are 5 feet 4 inches tall, it means you are 64 inches tall, and here’s why: There are 12 inches in each foot, and if there are 5 feet, that’s 12 x 5, which is 60 inches. Plus the remaining 4 inches gives you a total of 64 inches.
See also How To Use Cv1 Coffee Maker? New
## What is the inch of 5/3 height?
Answer: 63 inches is 5’3.
Inch and feet are both imperial units of measuring length.
### How To Convert Feet to Inches and Inches to Feet
How To Convert Feet to Inches and Inches to Feet
How To Convert Feet to Inches and Inches to Feet
## What is the inch of 5 2 height?
5 foot 2 inches to centimeters.
More Heights.
Feet and Inches CM
5’2.99 159.9946 cm
## What is the inches of 5 5 height?
Human Height Conversion Table
ft in inches centimeters
5’3” 63in 160.02cm
5’4” 64in 162.56cm
5’5” 65in 165.10cm
5’6” 66in 167.74cm
Feb 5, 2010
## What height is a tall girl?
If you’re still unsure, try measuring yourself, since if you’re taller than 5 feet 6 inches, you can be considered tall.
## What is 3ft by 5ft in inches?
Feet to inches conversion table
Feet (ft) Inches (“)
2 ft 24 ″
3 ft 36 ″
4 ft 48 ″
5 ft 60 ″
## What is the average height for a 12 year old?
While the average height of 12-year-olds is around 58 inches for boys and 59 inches for girls, it could be normal for your child to be quite a bit shorter or taller than average. Puberty, family history and medical considerations all play into your child’s height and growth rate.
## What is the average height for a 5 3 12 year old?
How Tall Should a 12 Year Old Be? We can only speak to national average heights here in North America, whereby, a 12 year old girl would be between 137 cm to 162 cm tall (4-1/2 to 5-1/3 feet). A 12 year old boy should be between 137 cm to 160 cm tall (4-1/2 to 5-1/4 feet). Can I still grow after puberty?
## How do you write 5 feet?
There are two popular ways of writing feet: the abbreviation “ft.” or the single apostrophe (′). Inches can be represented using “in.” or with a double apostrophe (″). As an example, five feet, ten inches could be written as 5 ft. 10 in or 5′10″.
See also How Old Is Isaac From J House Vlogs? New
## Is 5.2 short for a girl?
Originally Answered: Is 5’2 really short for a female? Yes, it is.
## Is being 5’2 short for a girl?
About 68% of the female population is between the height of 5ft 2 inches and 5ft 9 inches. So, any woman shorter than 4ft 10 inches is typically considered short, and taller than 6 ft is assumed to be tall. In a nutshell, if a woman is less than 5ft 3 inches tall, she is considered short.
## How many cm is 5/6 feet?
5 feet 6 inches in cm = [(5×12)+6] x 2.54 = 66 x 2.54 = 167.64 cm.
### Measurement (Feet and Inches)
Measurement (Feet and Inches)
Measurement (Feet and Inches)
## Is 160 cm short for a girl?
No,this is not a short height for a woman specially who live in East it consider average height range for woman if u live in West it consider slightly short but not at all u having a good height as woman so, don’t worry be happy and if u do workout then u will definitely boom ur personality acc to ur height.
## What’s 5ft 5ins in Centimetres?
5’5 = 165.1 cm
Convert 5 ft 5 to centimeters.
## Is 5 5 feet short for a girl?
5’5” is just above average for females in United States, 5’4” is average mostly. In Asia you would be considered tall in most countries, In Scandinavia & Netherlands you’re below average but not short. 5’5” is a perfect height for females I think 🙂 any height is good!
## What is considered tall for a 13 year old?
Average Height For 13 Years Old Boys
Age (years) 50th percentile height for teen boys (inches and centimeters)
12 58.7 in. (149 cm)
13 61.4 in. (156 cm)
14 64.6 in. (164 cm)
15 66.9 in. (170 cm)
Dec 8, 2021
## How do you hug a taller girl?
Simply lean down to her level with your arms over her shoulders, hands touching her back, and lay your head on her shoulder or against the side of her head. This may feel like you’ve created an awkward angle, but it will help you feel better because you will be taller in this type of hug.
See also How To Cook Soup Dumplings Without A Steamer? Update
## What is considered tall for a 14 year old?
In a nutshell, the average height for teen girls and boys at their 14 years old is 62.5 inches (158.7 cm) and 64.5 inches (163.8 cm), respectively.
## How many inches is 4×6?
Conversion Tool
Standard Photo Size
Inches mm
4 x 6 102 x 152
5 x 7 127 x 178
6 x 8 152 x 203
## How many inches tall is a 6 foot tall man?
Answer: There are 72 inches in 6 feet.
## How CM is an inch?
The value of 1 inch is approximately equal to 2.54 centimeters. To convert inches to the centimeter values, multiply the given inch value by 2.54 cm. 1 cm = 0.393701 inches.
## Can you stop growing at 12?
They tend to grow most quickly between ages 12 and 15. The growth spurt of boys is, on average, about 2 years later than that of girls. By age 16, most boys have stopped growing, but their muscles will continue to develop.
### Convert Between Inches and Feet | Inches to Feet and Feet to Inches
Convert Between Inches and Feet | Inches to Feet and Feet to Inches
Convert Between Inches and Feet | Inches to Feet and Feet to Inches
## How tall is a 5 year old?
At the age of 5, a typical child is about 43 inches tall and weighs about 43 pounds, according to the CDC. However, children at this age can vary by as much as 5 inches in height. A standard height is around 39 to 48 inches for a 5-year-old boy or girl, and a normal weight is between 34 and 50 pounds.
## What is tall for an 11 year old?
So, in conclusion, the average height for 11 year olds boys is 56.4 inches (143.5 centimeters). On the other hand, 11-year-old teen girls are 3 inches higher than boys, which is 56.7 inches (or 144 centimeters). The 50th percentile height chart from the CDC was used to calculate this statistic.
Related searches
• how tall is 32′
• 40 inches in feet
• 30 inches in feet
• how tall is 32 feet
• how long is 32 inches compared to an object
• how tall is 58.25 inches in height
• how tall is 216 inches
• how tall is 33 inches in feet
• what is 32 feet tall
• 32 inches to feet and inches
• 32 inches in cm
• 64 inches in feet
• how tall is 3 feet
• how tall is 32 inches in height
• 48 inches in feet
• how high is 32 inches
## Information related to the topic how tall is 32 inches in feet
Here are the search results of the thread how tall is 32 inches in feet from Bing. You can read more if you want.
You have just come across an article on the topic how tall is 32 inches in feet. If you found this article useful, please share it. Thank you very much.
| 1,985
| 7,038
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40
|
latest
|
en
| 0.867012
|
https://infinitylearn.com/questions/chemistry/gaseous-mixture-contains-grams-oxygen-grams-nitrogen-grams
| 1,709,359,817,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-10/segments/1707947475757.50/warc/CC-MAIN-20240302052634-20240302082634-00394.warc.gz
| 312,413,492
| 12,385
|
Dalton's law
Question
# A gaseous mixture contains 8 grams of oxygen, 14 grams of Nitrogen, 9 grams of water vapour. Partial pressure of water vapour in the mixture is (Assume water vapour behaves like ideal gas)
Easy
Solution
## ${\mathrm{n}}_{{\mathrm{N}}_{2}}=\frac{14}{28}=\frac{1}{2};{\mathrm{n}}_{{\mathrm{O}}_{2}}=\frac{8}{32}=\frac{1}{4};{\mathrm{n}}_{{\mathrm{H}}_{2}\mathrm{O}}=\frac{9}{18}=\frac{1}{2}$${\mathrm{X}}_{{\mathrm{H}}_{2}\mathrm{O}}=\frac{{\mathrm{n}}_{{\mathrm{H}}_{2}\mathrm{O}}}{{\mathrm{n}}_{{\mathrm{N}}_{2}}+{\mathrm{n}}_{{\mathrm{O}}_{2}}+{\mathrm{n}}_{{\mathrm{H}}_{2}\mathrm{O}}}=\frac{\frac{1}{2}}{\frac{1}{2}+\frac{1}{4}+\frac{1}{2}}=\frac{1}{2}×\frac{4}{5}=\frac{4}{10}$
Get Instant Solutions
| 297
| 731
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.875
| 4
|
CC-MAIN-2024-10
|
latest
|
en
| 0.310349
|
https://discourse.pymc.io/t/difference-in-posterior-between-metropolis-and-nuts-sampler/14048
| 1,713,778,863,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296818105.48/warc/CC-MAIN-20240422082202-20240422112202-00055.warc.gz
| 187,414,989
| 7,507
|
# Difference in posterior between Metropolis and NUTS Sampler
Hello,
I try to sample different polymer-chains with length N in spherical-coordinates, and than calculate the center of mass in Cartesian coordinates. Theta and phi (angles between monomers) have a uniform prior distribution.
When running this with NUTS, I get a Gausian distribution for center of mass, with mean at 0. When i run this with Metropolis i get a Gausian, but for the x axis the mean is shifted. Seen in the plots below. I use 10 chains and 8000 draws. For me it looks like it converged(I am quite new to this). How can this happen, do I have a mistake in my parametrization?
(Using pymc 5.0.2)
My Model:
``````N=40
processors=10
with pm.Model() as chain:
theta = pm.Uniform("theta", lower=-np.pi/2, upper=np.pi/2, size=N)
phi = pm.Uniform("phi", lower=0, upper=2*np.pi, size=N)
ct = theta.cos()
st = theta.sin()
cp = phi.cos()
sp = phi.sin()
_x, _ = pt.scan(lambda _ct, _st, _cp, _sp, prior: prior + [_cp*_ct, _sp*_ct, _st],
sequences = [ct, st, cp, sp],
outputs_info = pt.tensor.zeros(3))
x = _x[:,0]
y = _x[:,1]
z = _x[:,2]
pm.Deterministic("z_mean", z.mean())
pm.Deterministic("x_mean", x.mean())
pm.Deterministic("y_mean", y.mean())
#This lets the energy-less distribution be uniform on the sphere
pm.Potential("sphere", ct.log().sum())
#step=[pm.step_methods.Metropolis([theta, phi])]
step=[pm.NUTS([theta, phi], target_accept=0.9)]
#datat = pm.smc.sample_smc(draws=500,cores=processors)
datat = pm.sample(draws = 2000, cores = processors, step= step)
``````
1 Like
Welcome!
It is not entirely clear what is going on here. If you have a runnable example (e.g., something simpler than what you are currently using) someone might be able to investigate further.
A couple of suggestions. First, I would install the most recent version of PyMC if you can mange to do so. Second, I would suggest avoiding explicit specification of the step methods and allow `pm.sample()` to automatically select step methods (including compound methods). There is not particular reason to use Metropolis except in relatively specific scenarios. If NUTS works for you model, you should probably use it. Third, 80,000 samples is likely excessive unless your model is sampling very very poorly (at which point you should just fix your model).
1 Like
Thanks for the quick response.
Updating to 5.6.1 solved the problem.
I guess the problem was fixed some time ago.
Regarding why i use Metropolis instead of NUTS;
I want to use SMC(for tempering), where i had the same problem(updating also solved this problem).
1 Like
Maybe i spoke to soon.
Updated to 5.11.0 it solved the problem for the Metropolis sampler.
Now i tried to use the SMC sample, resulting in the same problem. As seen in the pictures below, x_mean goes to 0 the more draws I use.
I don’t know if I should open a new thread for this?
Questions I have:
Why is this happening for x and not for y?
The different chains should be uncorrelated, so why are they converging to the same posterior which then converges to mean=0?(maybe I misunderstand how the SMC alg. works)
Also to note updating to 5.11.0 solved the problem for the “normal” Metropolis sampler but not for SMC.
This time I also added a exported jupyter notebook with my model. I couldn’t think of something more simpler. I excluded the z-axes.
For_forum_2d.py (1.7 KB)
| 891
| 3,384
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.8125
| 3
|
CC-MAIN-2024-18
|
latest
|
en
| 0.84831
|
https://www.unitconverters.net/pressure/kilogram-force-sq-cm-to-kilopascal.htm
| 1,660,142,888,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-33/segments/1659882571190.0/warc/CC-MAIN-20220810131127-20220810161127-00783.warc.gz
| 934,742,701
| 3,414
|
Home / Pressure Conversion / Convert Kilogram-force/sq. Cm to Kilopascal
# Convert Kilogram-force/sq. Cm to Kilopascal
Please provide values below to convert kilogram-force/sq. cm to kilopascal [kPa], or vice versa.
From: kilogram-force/sq. cm To: kilopascal
### Kilogram-force/sq. Cm to Kilopascal Conversion Table
Kilogram-force/sq. CmKilopascal [kPa]
0.01 kilogram-force/sq. cm0.980665 kPa
0.1 kilogram-force/sq. cm9.80665 kPa
1 kilogram-force/sq. cm98.0665 kPa
2 kilogram-force/sq. cm196.133 kPa
3 kilogram-force/sq. cm294.1995 kPa
5 kilogram-force/sq. cm490.3325 kPa
10 kilogram-force/sq. cm980.665 kPa
20 kilogram-force/sq. cm1961.33 kPa
50 kilogram-force/sq. cm4903.325 kPa
100 kilogram-force/sq. cm9806.65 kPa
1000 kilogram-force/sq. cm98066.5 kPa
### How to Convert Kilogram-force/sq. Cm to Kilopascal
1 kilogram-force/sq. cm = 98.0665 kPa
1 kPa = 0.0101971621 kilogram-force/sq. cm
Example: convert 15 kilogram-force/sq. cm to kPa:
15 kilogram-force/sq. cm = 15 × 98.0665 kPa = 1470.9975 kPa
| 373
| 1,010
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.578125
| 3
|
CC-MAIN-2022-33
|
latest
|
en
| 0.349484
|
https://plainmath.net/10102/profits-from-plumbing-increased-this-year-what-profit-company-earn-last
| 1,632,253,753,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-39/segments/1631780057227.73/warc/CC-MAIN-20210921191451-20210921221451-00530.warc.gz
| 506,613,694
| 8,424
|
Question
# The profits from Ed’s Plumbing Co. increased by $14,460, or 30%, this year. What profit did the company earn last year? Equations and inequalities ANSWERED asked 2021-02-09 The profits from Ed’s Plumbing Co. increased by$14,460, or 30%, this year. What profit did the company earn last year?
2021-02-10
Let x be the profit the company earned last year so:
30% of profit last year=increase this year
0.3x=14460
$$\displaystyle\frac{{3}}{{10}}\cdot{x}={14460}$$
$$\displaystyle{x}=\frac{{{14460}{\left({10}\right)}}}{{3}}$$
x=48200
So, the company earned a profit of \$48,200 last year.
| 185
| 597
|
{"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}
| 4.125
| 4
|
CC-MAIN-2021-39
|
latest
|
en
| 0.877739
|
https://www.coursehero.com/file/7392729/PrinCH27AISETB/
| 1,444,771,440,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2015-40/segments/1443738017788.94/warc/CC-MAIN-20151001222017-00065-ip-10-137-6-227.ec2.internal.warc.gz
| 1,193,511,666
| 112,924
|
# Prin_CH27.AISE_TB - Chapter 27 The Basic Tools of Finance...
Prin_CH27.AISE_TB
Showing pages : 1 - 2 of 40
This preview has blurred sections. Sign up to view the full version!
Chapter 27 The Basic Tools of Finance Multiple Choice 1. The field of finance primarily studies a. how society manages its scarce resources. b. the implications of time and risk for allocating resources over time. c. firms decisions concerning how much to produce and what price to charge. d. how society can reduce market risk. ANS: B PTS: 1 DIF: 1 REF: 19-1 TOP: Finance MSC: Definitional 2. Which of the following is the correct way to figure the future value of \$ X that earns r percent for N years? a. \$ X (1 + rN ) N b. \$ X (1 + r ) N c. \$ X (1 + rN ) d. \$ X (1 + r / N ) N ANS: B PTS: 1 DIF: 1 REF: 19-1 TOP: Future value MSC: Definitional 3. Which of the following is the correct way to figure the future value of \$1 put in an account that earns 5 percent for 20 years? a. \$1(1 + .05) 20 b. \$1(1 + .05 × 20) 20 c. \$1(1 + .05 × 20) d. \$1(1 + 20/.05) 20 ANS: A PTS: 1 DIF: 1 REF: 19-1 TOP: Future value MSC: Applicative 4. Which of the following is the correct way to figure the future value of \$100 put in an account that earns 4 percent for 10 years? a. \$100(1 + .04 10 ) b. \$100(1 + .04 × 10) c. \$100 x 10 x (1 + .04) d. \$100(1 + .04) 10 ANS: D PTS: 1 DIF: 1 REF: 19-1 TOP: Future value MSC: Applicative 5. The future value of a deposit in a savings account will be larger a. the longer a person waits to withdraw the funds. b. the higher the interest rate is. c. the larger the initial deposit is. d. All of the above are correct. ANS: D PTS: 1 DIF: 2 REF: 19-1 TOP: Future value MSC: Analytical 6. The future value of a deposit in a savings account will be smaller a. the longer a person waits to withdraw the funds. b. the lower the interest rate is. c. the larger the initial deposit is. d. All of the above are correct. ANS: B PTS: 1 DIF: 2 REF: 19-1 TOP: Future value MSC: Analytical 142 This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied, or distributed without the prior consent of the publisher.
143 Chapter 27/The Basic Tools of Finance 7. What is the future value of \$500 one year from today if the interest rate is 6 percent? a. \$503 b. \$515 c. \$530 d. None of the above is correct. ANS: C PTS: 1 DIF: 1 REF: 19-1 TOP: Future value MSC: Applicative 8. What is the future value of \$750 one year from today if the interest rate is 3 percent? a. 772.73 b. 772.50 c. 773.33 d. None of the above are correct to the nearest penny. ANS: B PTS: 1 DIF: 1 REF: 19-1 TOP: Future value MSC: Applicative 9. What is the future value of \$800 one year from today if the interest rate is 7 percent? a. \$747.66 b. \$756.00 c. \$856.00 d. None of the above are correct to the nearest penny. ANS: C PTS: 1 DIF: 1 REF: 19-1 TOP: Future value MSC: Applicative 10. What is the future value of \$333 at an interest rate of 3 percent one year from today?
This is the end of the preview. Sign up to access the rest of the document.
| 968
| 3,124
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-40
|
longest
|
en
| 0.801384
|
http://cgm.cs.mcgill.ca/~godfried/teaching/mir-web.html
| 1,542,163,017,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-47/segments/1542039741578.24/warc/CC-MAIN-20181114020650-20181114042650-00298.warc.gz
| 69,144,955
| 10,847
|
Music Information Retrieval
"Without music life would be a mistake." - Friedrich Nietzsche
# 1. Introduction to Pattern Recognition and Image Processing
1. Notes on Methods of Proof
2. Image Processing Basic Operators
3. Optical character recognition (brief introduction)
4. Tessellation Resources
5. Tessellation Tutorials
6. Grids:
7. M.I.T. reading machine for the blind
8. What is hysteresis?
9. Zacharia Nkgau's tutorial on hysteresis smoothing of monotonic polygons (with interactive Java applet)
10. Image Segmentation:
## 2. Smoothing, Approximation, Data-Compression and Fitting
1. Minkowski addition and subtraction (dilation and erosion)
2. Gaussian smoothing
3. More about Carl Friedrich Gauss
4. Polygonal Approximation:
1. Midpoint smoothing
2. Ramer-Douglas-Peucker algorithm (Iterative End-Points Fit):
3. Relative Convex Hull Smoothing:
5. Tutorial on polygonal approximation (Iri-Imai algorithm, Melkman-O'Rourke algorithm)
## 3. Differentiation, Sharpening, Enhancement, Caricatures and Shape Morphing
1. Differentiation and Edge Detection:
2. Enhancement and Lateral Inhibition:
3. The Laplacian:
4. Fundamentals of Visual Perception:
5. Shape Morphing:
## 5. Distance, Medial Axis Transforms and Skeletons
1. What is Distance?
1. Manhattan Metric (Taxicab Geometry)
2. Pascal Tesson's tutorial on taxicab geometry (with Java applet)
2. Minkowski metrics
3. Distance between sets:
4. Skeletons
1. Hilditch's algorithm
2. Skeletonization software
5. Medial Axis of Polygonal Sets (prairie-fire transformation)
6. Medial axis in 3D with applications
7. Medial axis software
8. Medial Axis of Pont Sets
## 6. Shape Decomposition, Geometric and Topological Features
1. Polygon Decomposition:
2. Convex hulls, concavities and enclosures:
3. Geometric Feature Extraction Methods
## 7. Processing Line Drawings
1. Basics of Freeman Chain Coding (PostScript)
2. Square, circular, and grid-intersect quantization
3. Probability of obtaining diagonal elements
4. Geometric Probability
6. More on Bertrand's paradox (with Java applet simulations)
8. Difference encoding & chain correlation functions
9. Minkowski metric quantization
## 8. Detection of Structure in Noisy Pictures and Dot Patterns
1. What is a line?
2. Point-to-curve transformations (Hough transform)
3. Point-Line duality
4. Hough Transforms:
5. GraphTheory:
6. Proximity graphs:
1. A Survey of Proximity Graphs
2. Minimal spanning tree (MST) of a dot pattern
3. MST interactive Java applet
4. Delaunay Triangulations and Voronoi diagrams
7. The shape of a set of points:
1. The relative neighbourhood graph of a finite planar set
2. Sphere-of-influence graphs and applet
3. Alpha shapes
1. François Bélair's Tutorial on Alpha Shapes (with interactive Java applet and a super-duper automated guided-tour demo)
2. Introduction to alpha shapes
3. Gallery of alpha shapes
4. Code for computing alpha-shapes (and convex hulls)
4. Beta skeletons:
1. Xiaoming Zhong's Tutorial on Beta Skeletons (with interactive Java applet)
5. Voronoi Diagram Based Methods:
## 9. Simple Classifiers, Neural Networks, and Machine Learning
1. Simple Classifiers
1. Template matching
2. Minimum-distance classifiers
3. Inner products
4. Linear discriminant functions
5. Decision boundaries
2. Mahalanobis Distance Classifiers
3. Learning from Examples
4. Neural Networks:
1. A Brief Tour of the Brain
2. Introduction to Neural Networks
3. Another Introduction to Neural Networks
4. Dr. Gurney's course on neural networks
5. A brief history of Neural Networks
6. Neural Network Basics (FAQ's)
7. Formal neurons, linear machines & perceptrons
8. Separability:
1. Linear separability
2. Separating points with circles
9. Pierre Lang's Neural Network for Character Recognition (with interactive Java applet that recognizes the characters you draw on the screen!)
# 10. Bayesian Decision Theory
1. Bayesian Decision Theory with Gaussian Distributions - A tutorial by Erin Mcleish
2. Introductory Statistics Course
3. Another Introduction to Probability and Statistics
4. Basics of Statistical Pattern Recognition (by Richard O. Duda)
6. Minimum risk classification
7. Minimum error classification
8. Discriminant functions (linear, quadratic, polynomial)
9. The bivariate Gaussian probability density function
10. Multivariate statistics
11. Lecture Notes on Statistical Pattern Recognition
12. Occam's Razor:
## 11. Feature Selection: Independence of Measurements, Redundancy, and Synergism
1. Independent and conditionally independent events
2. Class-conditional and unconditional independence assumptions in pattern recognition (Tutorial by Simon-Pierre Desrosiers)
3. Independence, uncorrelation and Gaussian distributions (PostScript notes by Julio Peixoto)
4. Information theory:
1. A primer on information theory (PostScipt)
2. Basic properties of Shannon's entropy and mutual information
3. Relative entropy and mutual information
4. From Euclid to entropy (PostScript)
5. Shannon's equivocation and the Fano bound
8. Calculating Information and Complexity
5. Feature Selection:
1. Independence, Redundancy and Synergism: A Tutorial by Irina Kezele
2. Feature Selection: Evaluation, Application, and Small Sample Performance (PostScipt)
3. Toward Optimal Feature Selection (PostScipt)
4. Dimensiobality Reduction: Francois Labelle's tutorial (with interactive Java applets)
5. Simon Plain's tutorial on feature selection (with interactive Java applets)
6. Feature evaluation criteria:
1. Kullback-Liebler information
2. The Divergence
3. The Affinity
4. The Mutual-Information criterion (PDF file)
5. Discrimination information and Kolmogorov variational distance (PDF file)
6. The Fisher Information
7. Feature selection methods (Richard Duda's course notes)
8. A survey of feature selection methods
9. The best K independent measurements are not the K best (PDF file)
10. Models of spatial dependence between features
## 12. Non-parametric Machine Learning
1. General Learning Resources
2. Perceptrons:
3. Non-parametric training of linear machines (Nilsson's book - Chapter 4)
4. Error-correction procedures
1. Rosenblatt's Perceptron Learning Algorithm (an interactive Java applet)
5. The fundamental learning theorem
6. Multi-layer networks
7. Competitive Learning:
## 13. Estimation of Density Functions, Parameters and Classifier Performance
1. Estimation of Parameters:
1. Robust estimators of location (Tutorial by Greg Aloupis)
2. Bias and variance of estimators
3. Maximum likelihood estimation
2. Density Estimation:
3. Estimators and Bias (Wolfram Research)
4. Dimensionality and sample size
5. Estimation of the probability of misclassification
6. Estimation of misclassification before 1974
1. Resubstitution
2. Holdout
3. Data Shuffling
4. Leave-One-Out
5. Bootstrap Methods
7. Ensembles of Classifiers
8. Boosting
9. Comparing the performance of several classifiers (the pitfalls)
## 14. Nearest Neighbor Decision Rules and Instance-Based Learning
1. Nearest Neighbor Decision Rules:
1. The nearest neighbor rule: a tutorial
2. The nearest neighbor rule with a reject option
3. The k-nearest neighbor rule applet
4. The Cover-Hart bounds and Jensen's inequality:
2. Efficient search methods for nearest neighbors:
3. Editing nearest neighbor rules to reduce storage:
4. Nearest neighbor computation software
5. Bibliography on Nearest Neighbor Methods
## 15. Using Contextual Information in Pattern Recognition
1. Using Context in Visual Perception
2. Infinite Monkey Theorem
3. Introduction to Markov Processes
5. Forward dynamic programming and the Viterbi algorithm:
1. A tutorial on the Viterbi algorithm
2. Viterbi algorithm demo for sentence recognition
6. Combined bottom-up and top-down algorithms
## 16. Unsupervised Machine Learning and Cluster Analysis
1. Unsupervised Learning:
2. Graph-theoretic methods:
1. Minimal spanning tree methods
2. Tutorial and Java applet by Mike Soss and Chrislain Razafimahefa
3. Hierarchical clustering:
| 1,896
| 7,951
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.609375
| 3
|
CC-MAIN-2018-47
|
latest
|
en
| 0.709881
|
https://www.physicsforums.com/threads/momentum-problem-driving-me-crazy.22644/
| 1,513,369,448,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-51/segments/1512948579564.61/warc/CC-MAIN-20171215192327-20171215214327-00626.warc.gz
| 822,508,730
| 15,019
|
# Momentum problem driving me crazy.
1. Apr 25, 2004
### Demondo
I asked this on IRC but with only 7 people there, including myself, I'm not too confident I'll get an answer anyway...
It seems like such an odd question because you aren't given the bullet's speed. Anyone have any idea what would happen? It is possible that the bullet would push the block, but that just doesn't seem likely at all because of it's speed. Or it could puncture just as deeply, but this time the block moves. But I'm really uncertain about what happens. Any help would be appraicted.
2. Apr 25, 2004
### Ebolamonk3y
Impulse... I think the bullet still goes in regardless... Probably not as deep or will not penetrate it... It will be a perfect inelastic collision where the 2 masses become one...
Say... If someone where to shoot you with a paintball gun in the face and you had your mask on... You would get paint all over your face. If you were to stand on ice or some frictionless surface and the same situation occured... You still get paint over your face and a fun slide in the direction of the paintball...
Thats what I think at least...
3. Apr 26, 2004
### Staff: Mentor
If you make a few simplifying assumptions, the problem is straightforward.
With the wood in the vise, all the kinetic energy of the bullet goes into penetrating the wood and thus is transformed into thermal energy. We'll make the simplifying assumption that the force between the bullet and wood is uniform: thus the given KE of the bullet leads to the given penetration, FxD = ΔKE.
With the wood free to move, assume that the bullet will again penetrate the wood making the collision perfectly inelastic. (Not unreasonable.) Apply conservation of momentum to get the speed of the block plus bullet post collision, then use that speed to calculate its KE. Now find the ΔKE during the collision--that's the energy used to penetrate the wood. In this case the penetration is less because some of the bullet's energy is used to accelerate the wood.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook
| 472
| 2,117
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.53125
| 4
|
CC-MAIN-2017-51
|
longest
|
en
| 0.959798
|
https://api.betterlesson.com/mtp/lesson/461878/print
| 1,519,282,518,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-09/segments/1518891814036.49/warc/CC-MAIN-20180222061730-20180222081730-00265.warc.gz
| 592,592,208
| 16,042
|
Andrea Palmer PROSPECT HILL ACADEMY CHARTER SCHOOL, SOMERVILLE, MA
6th Grade Math : Unit #2 - The College Project - Working with Decimals : Lesson #6
# Dividing with Decimals
Objective: SWBAT: • Make estimates of division problems • Divide whole numbers and decimals • Interpret a remainder
Standards: 6.NS.B.2 6.NS.B.3 MP1 MP6
Subject(s): Math
60 minutes
1 Do Now - 10 minutes
Notes:
• Before the lesson you will need to copy, cut, and organize the Division Matching cards. I like to label cards in sets (write #1 on all of first set, etc.) so that students know which cards are theirs.
• I use the data about division from the pre-test in Unit 1 to Create Homogeneous GroupsFor the do now I have students work in partners or groups of 3.
See my Do Now in my Strategy folder that explains my beginning of class routines.
Often, I create do nows that have problems that connect to the task that students will be working on that day. Here, I want students to connect what is being described in the word problem with a picture. Students must think about what is going on and what the action will look like.
After most students have finished most of the matches, I call the class together. I ask students to share their number sentences and pictures for #4 and #8. There is a subtle difference in these problems, which I want students to notice. A common mistake is that students confuse the situations that represent 24/4= 6 and 24/6= 4
2 Estimation - 5 minutes
Here I want students to quickly review the vocabulary word quotient and then work on estimating. I give students a couple minutes to make and write explanations for their estimates. I stress to students that estimates need to be relatively quick.
I walk around and monitor student work. I’m looking to see what strategies students are using. Some students may round numbers to whole numbers and divide. Other students may make connections to multiplication. I am interested to see what students do with #3. If they struggle I ask, “Do you think the quotient is going to be bigger or smaller than 1.89? Why?”
I call one student to explain one of their estimates. These students are students I observed using a particular strategy as I was walking around. I am not giving exact answers at this time. We will work on finding exact quotients later in the lesson. It is important that students are able to use their number sense to make reasonable estimates.
Estimation Reflection
Grappling with Complexity
In this section, I called on a couple students for each problem. I continued to emphasize that students need to make the number easy to work with using mental math. Interestingly, a few students in each class said that problem 3 was “impossible”. I asked them why, and they explained that you can divide a number by something that is larger than that number. I asked them to round 1.89 to a number that is easier to work with, and they responded with 2. I told them that they had \$2 that they had to share equally between 4 people. How much money would each person get? Some students explained that each person would get fifty cents, because fifty cents times four equals \$2. I stated that the problem wasn’t impossible, but that when you divide a number by something larger than that number your quotient will be less than one. Making the connection to money helped students understand the problem.
3 Practice - 15 minutes
We do number 1 together. I ask students how they want to find the exact quotient. I also show them the algorithm. I stress that students need to use their number sense and the patterns they noticed earlier to decide if their final answer makes sense (MP1)
Students work independently on the rest of the problems. They can check in with their partner if they are stuck. I Post A Key so that students can check their work when they finish a page. I am looking that students are making reasonable estimates and that they are successfully dividing using a strategy of their choice.
If students successfully complete the practice problems, they can play the Leftovers From 100 game.
If students are struggling, I may intervene in one of the following ways:
• Ask them what their estimate is and how they got it.
• Let them use a multiplication chart.
• Give them a word problem situation that represents the problem.
• Have them use the grids to organize their problems.
• Pull a small group of students who are struggling to work together.
4 Sharing \$\$ - 10 minutes
I have students participate in a Think Write Pair Share. Working with money forces students to interpret the remainder. I have a few students share and explain their strategies. Some students will use the picture, others will use the algorithm, others will make connections to multiplication. Here students are using MP 6: Attend to precision.
We move on to the next page where I ask students to take a couple minutes to analyze the examples. What do they notice? What are the similarities? The differences? I give the students a couple minutes to jot down notes.
I want students to see that all of the answers are technically correct, although not as useful as others. Problem A solves the problem but the answer doesn’t help. I also want students to start moving away from the partial quotient method and towards the standard algorithm. Problem B turns the remainder into a fraction, but we then need to figure out how to do that.
A common struggle is that students don’t know where to put a decimal point. This can be easily remedied if students are able to make reasonable estimates. For example, offer \$67.50 as an answer. Hopefully students will quickly eliminate this as a possibility, since it is way too big!
I want students to notice the differences between C and D. In problem C the person ignored the decimal point until the end, and then used estimation to decide where it goes. Problem D kept the decimal point in the dividend and brought it up to the quotient. I ask students how could these students check their answers. I want students to get into the habit of using multiplication to check their answers.
Sharing \$\$ Reflection
Diverse Entry Points
This problem was a great way to get students to understand dividing with decimals. Some students who knew the traditional algorithm used it to find their answer of \$6.75. Other students used a traditional method or the partial quotient method to figure out that each person would get 6 ¾ dollars, which is the same as \$6.75. Other students drew pictures to show each person’s money. There were able to give each person a \$5 bill and a \$1 bill. To split up the remaining \$3, they traded it in for 12 quarters and each person received 3 quarters. I had a student who drew a picture present first, followed by students who used the partial product method and the traditional method. I told students that if someone presented a strategy that they did not use, they needed to draw it on their own paper. I wanted students to make connections between all of the methods.
5 More Practice - 10 minutes
Just like the previous Practice section, we do number 1 together. I ask students how they want to find the exact quotient. I stress that students need to use their number sense and estimates to decide if their final answer makes sense.
Students work independently on the rest of the problems. They can check in with their partner if they are stuck. I Post A Key so that students can check their work when they finish a page. I am looking that students are making reasonable estimates and that they are successfully dividing using a strategy of their choice.
If students successfully complete the practice problems, they can play the Leftovers From 100 game.
If students are struggling, I may intervene in one of the following ways:
• Ask them what their estimate is and how they got it.
• Let them use a multiplication chart.
• Give them a word problem situation that represents the problem.
• Have them use the grids to organize their problems.
• Pull a small group of students who are struggling to work together.
Transitioning From the Partial Quotients Method
Student Led Inquiry
I realized that a number of my students were using the partial quotient method for division, instead of the traditional method. In this section, I modeled how the traditional method and the partial quotients method are connected.
One problem I did was 812.6 divided by 5. Unit 2.5 Transitioning From the Partial Quotients Method.jpgI first had students help me create a reasonable estimate for the problem. Then I solved the problem under the document camera, step by step in both ways. I had students copy down both methods on their paper. I showed them how the partial quotient method starts to break down when there is a remainder. I emphasize to students that in sixth grade they need to learn the traditional method, which will help them divide with decimals. This transition takes time, so I will have to continue to model for students.
6 Closure and Ticket to Go - 10 minutes
I begin the Closure by asking students to look at questions 3 and 4. What was your estimate? Why? How did you find the exact quotient. I look for students who used different strategies and I have them show and explain their work. If there is a common mistake I see students making, I will present it and ask students to address it here. I am also interested to see what students estimated with #4. Even if students struggle with estimating, they should be able to see that they answer is going to be smaller than 2.34, since you are dividing it by a number that is larger.
With about five minutes left I pass out the Ticket to Go for students to complete independently. Then I pass out the HW Dividing with Decimals. I may also assign one of the “Facts about College” or “Working During College” pages for homework, depending on how much students finished.
| 2,150
| 9,955
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.5625
| 5
|
CC-MAIN-2018-09
|
latest
|
en
| 0.923757
|
https://socratic.org/questions/how-do-you-find-the-equation-of-a-circle-that-passes-through-the-points-1-1-1-5-
| 1,713,712,415,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296817780.88/warc/CC-MAIN-20240421132819-20240421162819-00126.warc.gz
| 482,160,090
| 6,185
|
# How do you find the equation of a circle that passes through the points (1, 1), (1, 5), and (5, 5)?
Jun 25, 2016
${\left(x - 3\right)}^{2} + {\left(y - 3\right)}^{2} = 8$
#### Explanation:
Note that the centre of the circle is equidistant from all three points, the distance being the radius of the circle.
Any point equidistant from two points must lie on the perpendicular bisector of the line segment joining those two points. That is, on the line through the midpoint of the line segment, perpendicular to the line segment.
The perpendicular bisector of the line segment joining the points $\left(1 , 1\right)$ and $\left(1 , 5\right)$ is $y = 3$
The perpendicular bisector of the line segment joining the points $\left(1 , 5\right)$ and $\left(5 , 5\right)$ is $x = 3$
These intersect at the centre of the circle $\left(3 , 3\right)$
The distance between $\left(1 , 1\right)$ and $\left(3 , 3\right)$ is:
$\sqrt{{\left(3 - 1\right)}^{2} + {\left(3 - 1\right)}^{2}} = \sqrt{{2}^{2} + {2}^{2}} = \sqrt{8}$
So the equation of the circle may be written:
${\left(x - 3\right)}^{2} + {\left(y - 3\right)}^{2} = 8$
graph{((x-3)^2+(y-3)^2-8)((x-3)^2+(y-3)^2-0.02)((x-1)^2+(y-1)^2-0.02)((x-1)^2+(y-5)^2-0.02)((x-5)^2+(y-5)^2-0.02)(y-3)(x-3+0.0001y)=0 [-7.625, 12.375, -2.72, 7.28]}
| 477
| 1,292
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 12, "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}
| 4.78125
| 5
|
CC-MAIN-2024-18
|
latest
|
en
| 0.728331
|
https://mixomics-users.discourse.group/t/simulating-multilevel-data/551
| 1,618,105,254,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-17/segments/1618038060603.10/warc/CC-MAIN-20210411000036-20210411030036-00278.warc.gz
| 514,390,746
| 4,614
|
# Simulating Multilevel Data
Hey, I have a question on how to simulate data from
“A novel approach for biomarker selection and the integration of repeated measures experiments from two assays”.
I want to compare using Xw to X for PLSDA filter methods. I want to simulate data similar to the block correlation used for the subject random effect and error term in the paper. However I am using only two treatment groups.
This is my code:
library(MASS)
library(rlist)
library(Matrix)
mujk=list(c(rep(2,40),rep(0,160)),
c(rep(1,40),rep(0,160))) #means separating certain genes for both treatments
rho1=0.8 #correlation for subject random effect
rho2=0.7 #correlation for error
varR=3 #variance of subject random effect
varE=0.5 #variance of error
sigmarandom=list((diag(20) + (rho1varR)(1-diag(20)))-diag(20)+(diag(20)varR),(diag(20) + (rho1varR)*(1-diag(20)))-diag(20)+(diag(20)*varR)) #homogenous covariance matrix rho =0.8 for subject random effect block diagonal
sigmaerror=list((diag(20) + (rho2varE)(1-diag(20)))-diag(20)+(diag(20)varE),(diag(20) + (rho2varE)*(1-diag(20)))-diag(20)+(diag(20)*varE)) #homogenous covariance matrix rho =0.7 for error random effect block diagonal
Rsig=as.matrix(bdiag(rep(sigmarandom,5)))
Esig=as.matrix(bdiag(rep(sigmaerror,5)))
perswitch=c(1,-1) #gives labels to treatments
lmmper=list(2) #treatments
lmmS=list(12) #total subjects to simulate
for(s in 1:12){ #12 is number of subjects
###random effect is changing
Randeff=mvrnorm(n=1,mu=rep(0,200),Sigma=Rsig)# Random effect for subject same for each #observation on subject
for(p in 1:2){ #for each treatment
lmmper[[p]]=c(s,perswitch[p],mujk[[p]]+Randeff+mvrnorm(n=1,mu=rep(0,200),Sigma=Esig))#(sample ID, Y, X)
}
lmmS[[s]]=lmmper
}
lmmX=lapply(1:length(lmmS),function(i){list.rbind(lmmS[[i]])})
lmmX=data.frame(list.rbind(lmmX))
colnames(lmmX)[1:2]<-c(“sample”,“response”)
lmmY=data.frame(lmmX[,1:2])
X=lmmX[,-c(1,2)]
Y=lmmY\$response
I am curious if I am following the logic correctly from the paper. I can simulate the data using a block diagonal right? If so does my approach look correct or am I misunderstanding how to conduct the simulation?I am comparing filter methods with PLSDA on this design and am not seeing any difference in feature selection. Makes me think I am doing something wrong.
I appreciate the help!
Dear @Neystale
Unfortunately I wont be able to (efficiently) help you here. The person who developed the simulated data is Prof Benoit Liquet, you could try to contact him, I dont have access to his code.
Potentially you could reuse the data from vac18.simulated if you are after a two factor design and remove some treatment, but this would be suboptimal. I confirm that the effects should be strong through and you should be able to extract the relevant variables!
Kim-Anh
| 821
| 2,809
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17
|
latest
|
en
| 0.618456
|
http://doyourmath.com/algebra-help/i-don-t-understand-why-you-don-t-take-out-a-1
| 1,539,857,667,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-43/segments/1539583511761.78/warc/CC-MAIN-20181018084742-20181018110242-00490.warc.gz
| 104,124,984
| 4,460
|
## The problem
Click to see the original problem
I don’t understand why you don’t take out a -1.
## Answer provided by our tutors
Make sure that you have actually entered an equation or inequality (it needs to contain =, <, >, >= or <= symbol).
For example, -5x^2+30x+1 is not an equation – it is an expression.
On the other hand, -5x^2+30x+1 = 0 is an equation which can be solved (for either one of the variables).
| 116
| 422
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.78125
| 3
|
CC-MAIN-2018-43
|
latest
|
en
| 0.897989
|
https://statquest.org/tag/statistics/
| 1,680,163,783,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-14/segments/1679296949107.48/warc/CC-MAIN-20230330070451-20230330100451-00700.warc.gz
| 627,502,017
| 13,575
|
# StatQuest: Linear Discriminant Analysis (LDA), clearly explained
Here it is, folks! By popular demand, a StatQuest on linear discriminant analysis (LDA)!
Also, because you asked for it, here’s some sample R code that shows you how to get LDA working in R.
If all went well, you should get a graph that looks like this:
# RNA-seq: The Pitfalls of Technical Replicates
Science, like most things, isn’t perfect. Because individual mice (or other biological samples, like humans) are different and because the way we measure things isn’t perfect, every time we do an experiment, we get slightly different results.
We cope with this variation by doing the same experiment a bunch of times. Once we’ve collected a lot of samples, we typically calculate the average and the standard deviation. If we’re worried about the biological variation (the variation from each individual mouse or human) we do lots of biological replicates, and if we’re worried about technical variation, we do technical replicates (remeasure the same individual multiple times). Often we’ll do both.
For simple and inexpensive experiments, it’s easy to do lots of replicates, both biological and technical. However, RNA-seq is time consuming and expensive, and we want to squeeze as much as we can out of the samples that we have (which can be very limited!) The good news is that because of the way technical variation affects RNA-seq results, we only need to do biological replicates with this type of experiment.
Here’s an example of biological variation alone (imagine there is no technical variation). In this example, we’re measuring read counts for an imaginary gene “X”. Each mouse has either more or less than the average for all mice. If we do enough samples, our estimate of the average of all mice will be pretty good because the measurements that are above the mean will be canceled out by the measurements that are below the mean.
This figure shows how biological variation gives us measurements that are always a little larger or smaller than the mean.
When we add technical variation to the mix, RNA-seq will result in random values that either add or subtract read counts from each mouse’s measurement. Here’s an example where I’ve colored the biological variation orange and the technical variation green.
This figure shows how technical variation can increase or decrease the read counts we get for each sample. It also shows the equation for the average read count.
When we calculate the average measurement from our samples, we end up with a term for biological variation and a term for technical variation. Just like before, the term for biological variation will go to zero as we add samples, because the positive values will be canceled out by the negative values. However, because the technical variation is similarly distributed, it will cancel itself out for the same reason. Thus, without doing a single technical replicate (and only doing biological replicates) we can reduce the affects of biological and technical variation and get a good estimate of the mean.
This last figure shows how the biological and technical variation can be separated into separate terms. Each of these terms goes to zero as we add more biological replicates.
One problem that is often overlooked is that technical replicates alone will only eliminate the term for technical variation. The term for biological variation will only be reinforced, and this can lead to misleading results. See the video for more details.
# PCA clearly explained
Update: A lot of people ask to see example code that shows how to actually do PCA. Here’s how to do it in R
[sourcecode language=”R”]
## In this example, the data is in a matrix called
## data.matrix
## columns are individual samples (i.e. cells)
## rows are measurements taken for all the samples (i.e. genes)
## Just for the sake of the example, here’s some made up data…
data.matrix <- matrix(nrow=100, ncol=10)
colnames(data.matrix) <- c(
paste("wt", 1:5, sep=""),
paste("ko", 1:5, sep=""))
rownames(data.matrix) <- paste("gene", 1:100, sep="")
for (i in 1:100) {
wt.values <- rpois(5, lambda=sample(x=10:1000, size=1))
ko.values <- rpois(5, lambda=sample(x=10:1000, size=1))
data.matrix[i,] <- c(wt.values, ko.values)
}
pca <- prcomp(t(data.matrix), scale=TRUE)
## plot pc1 and pc2
plot(pca\$x[,1], pca\$x[,2])
## get the name of the sample (cell) with the highest pc1 value
rownames(pca\$x)[order(pca\$x[ ,1], decreasing=TRUE)[1]]
## get the name of the top 10 measurements (genes) that contribute
## most to pc1.
gene_score_ranked <- sort(gene_scores, decreasing=TRUE)
top_10_genes <- names(gene_score_ranked[1:10])
top_10_genes ## show the names of the top 10 genes
## get the scree information
pca.var <- pca\$sdev^2
scree <- pca.var/sum(pca.var)
plot((scree[1:10]*100), main="Scree Plot", xlab="Principal Component", ylab="Percent Variation")
[/sourcecode]
RNA-seq results often contain a PCA (Principal Component Analysis) or MDS plot. Usually we use these graphs to verify that the control samples cluster together. However, there’s a lot more going on, and if you are willing to dive in, you can extract a lot more information from these plots. The good news is that PCA only sounds complicated. Conceptually, it’s actually quite simple.
In this video I clearly explain how PCA graphs are generated, how to interpret them, and how to determine if the plot is informative or not. I then show how you can extract extra information out of the data used to draw the graph.
# The Standard Error (and a bootstrapping bonus!!!)
The standard error isn’t just some error that we all make (as in “Dude, did you see Jimmy at the bar last night? He made the standard error…”), it’s a measure of how much we can expect the mean to change if we were to re-do an experiment.
Let’s start by thinking about error bars:
We’ve all seen error bars in plots before. The most common types of error bars are standard deviations, standard errors and confidence intervals. Standard deviations describe how the data that you collected varies from one measurement to the next. In contrast, standard errors describe how much the mean value might change if we did the whole experiment over again. Whoa!!! I know it sounds crazy, so let’s talk about.
First, let’s consider a simple (but expensive and time consuming) way to calculate the standard error. There’s a simple formula for calculating the standard error, but it won’t help us understand what’s really going on, so we’ll ignore it for now.
Assume we took a sample of 5 measurements. From these 5 measurements, we can easily calculate (and plot) the mean.
Now, we can easily imagine that if we took another sample of 5 measurements, the mean might be slightly different from the one we got from the first sample. So let’s take a bunch of samples.
After taking a bunch of samples, we can calculate (and plot) the mean for each one. In the figure we can see that the means are more tightly clustered together than the original measurements. This is because it’s much more common to get a single extreme value in a single sample, than get an entire sample worth of extreme values, which is what we would need to get an extreme value for a mean. Now that we have calculated the means, we can calculate their standard deviation.
The standard error is just the standard deviation of all those means – it describes how much the values for the mean differs in a bunch of samples.
Now, like I said before, there’s a simple formula for calculating the standard error, but there’s also a simple, time and cost effective way to do it without having to use that formula. This method is called “bootstrapping”. What’s cool about bootstrapping is that you can use it to calculate the standard error of anything, not just the mean, even if there isn’t a nice formula for it. For example, if you had calculated the median instead of the mean, you’d be in deep trouble if you needed to calculate its standard error and you didn’t have bootstrapping.
Bootstrapping is introduced in the video, and in the next StatQuest, we’ll use it to calculate confidence intervals, which are super cool since they let you compare samples visually.
If you wan to try out bootstrapping yourself, here’s some R code:
[sourcecode language=”R”]
## bootstrap demo…
# Step 1: get some data
# (in this case, we will just generate 20 random numbers)
n=20
data <- rnorm(n)
## Step 2: plot the data
stripchart(data, pch=3)
## Step 3: calculate the mean and put it on the plot
data.mean <- mean(data)
abline(v=data.mean, col="blue", lwd=2)
## Step 4: calculate the standard error of the mean using the
## standard formula (there is a standard formula for the standard
## error of the mean, but not for a lot of other things. We use
## the standard formula in this example to show you that the
## bootstrap method gets the about the same values
data.stderr <- sqrt(var(data))/sqrt(n)
data.stderr ## this just prints out the calculated standard err
## Step 5: Plot 2*data.stderr lines +/- the mean
## 2 times the standard error +/- the mean is a quick and dirty
## approximatino of a 95% confidence interval
abline(v=(data.mean)+(2*data.stderr), col="red")
abline(v=(data.mean)-(2*data.stderr), col="red")
## Step 6: Use "sample()" to bootstrap the data and calculate a lot of
## bootstrapped means.
num.loops <- 1000
boot.means <- vector(mode="numeric", length=num.loops)
for(i in 1:num.loops) {
boot.data <- sample(data, size=20, replace=TRUE)
boot.means[i] <- mean(boot.data)
}
## Step 7: Calculate the standard deviation of the bootstrapped means.
## This is the bootstrapped standard error of the mean
boot.stderr <- sqrt(var(boot.means))
boot.stderr
## Step 8: Plot 2*boot.stderr +/- the mean.
## Notice that the 2*data.stderr lines and the 2*boot.stderr lines
## nearly overlap.
abline(v=(data.mean)+(2*boot.stderr), col="green")
abline(v=(data.mean)-(2*boot.stderr), col="green")
[/sourcecode]
| 2,326
| 9,961
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.171875
| 3
|
CC-MAIN-2023-14
|
longest
|
en
| 0.911762
|
https://finnstats.com/index.php/2022/06/13/nonlinear-regression-analysis-in-r/
| 1,686,044,761,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224652494.25/warc/CC-MAIN-20230606082037-20230606112037-00531.warc.gz
| 278,384,572
| 51,679
|
# Nonlinear Regression Analysis in R
Nonlinear Regression Analysis in R, when at least one of the parameters in a regression appears to be nonlinear, it is called nonlinear regression.
It is often used to sort and analyze data from many businesses, such as retail and banking.
It also aids in drawing conclusions and forecasting future trends based on the user’s online behavior.
The process of creating a nonlinear function in R is known as nonlinear regression analysis. This approach uses independent factors to predict the result of a dependent variable using model parameters that are depending on the degree of association between variables.
When the variance in sample data is not constant or when errors are not normally distributed, generalized linear models (GLMs) calculate nonlinear regression.
When the following sorts of regressions are regularly used, a generalized linear model is used.
Proportions are used to express count data (e.g. logistic regressions).
Data is not expressed as proportions when it is counted (e.g. log-linear models of counts).
Binary response variables exist (for example, “yes/no,” “day/night,” “sleep/awake,” and “purchase/not buy”).
The data shows that the coefficient of variation is constant (e.g. time data with gamma errors).
## Nonlinear Regression Analysis in R
One of the most prevalent types of nonlinear regression in statistics is logistic regression. It’s a method for estimating the likelihood of an occurrence based on one or more independent factors.
Using probability theory, logistic regression determines the associations between the listed factors and independent variables.
Logistic Regression is a technique for predicting the outcome of In circumstances where the rate of growth does not remain constant throughout time, models are commonly used.
For example, when a new technology is first brought to the market, its demand rises quickly at first, but then progressively declines.
Logistic regression is defined using logit() function:
`f(x) = logit(x) = log(x/(1-x))`
Assume that p(x) indicates the probability of an event, such as diabetes, occurring based on an independent variable, such as a person’s age. The following will be the probability p(x).
`P(x)=exp(β0+ β1x1 )/(1+ exp(β0+ β1x1)))`
Here β is a regression coefficient.
We obtain the following when we take the logit of the previous equation.
`logit(P(x))=log(1/(1-P(x)))`
When we solve the equation above, we get.
`logit(P(x))=β0+ β1x1`
The Sigmoid Function is a logistic function that is represented by an S-shaped curve.
When a new technology is introduced to the market, demand typically rises quickly in the initial few months before progressively declining over time.
This is a case of logistic regression in action. In circumstances where the rate of growth does not remain constant over time, Logistic Regression Models are commonly used.
Multivariate logit() Function
In the case of several predictor variables, the logistic function is represented by the following equation.
`p = exp(β0+ β1x1+ β2x2+—– βnxn)/(1+exp(β0+ β1x1+ β2x2+…+βnxn))`
The predicted probability is p, the independent variables are x1,x2,x3,…,xn, and the regression coefficients are 0, 1, 2,…n.
Manually estimating coefficients is a time-consuming and error-prone technique that requires numerous difficult and lengthy calculations.
As a result, sophisticated statistical software is typically used to make such estimates. In statistical analysis, coefficients must be calculated.
A relationship between three or more variables that specify the simultaneous influence of two or more interacting variables on a dependent variable is known as interaction.
We can use interacting variables to calculate logistic regression with three or more variables in a relationship where two or more independent factors influence the dependent variable.
An enumerated variable in logistic regression can have an order but not a magnitude. Because arrays have both order and magnitude, they are inappropriate for storing enumerated variables.
Dummy or indicator variables are used to store enumerated variables. There are two possible values for these dummy or indicator variables: 0 and 1.
You must test the accuracy of a Logistic Regression Model’s predictions after it has been developed. The following are some examples of adequacy checking techniques:
Residual Deviance – A high residual variation indicates that the Logistic Regression Model is insufficient. The best value for the Logistic Regression Model’s residual variance is 0.
Parsimony – Logistic Regression Models with fewer explanatory variables are more reliable and can also be more valuable than models with a high number of explanatory variables.
The process of determining the threshold probability for a response variable based on explanatory variables is referred to as classification accuracy. It makes it simple to understand the results of a Logistic Regression Model.
Cross-validation is used to check the accuracy of predictions for the Logistic Regression Model, and it is repeated numerous times to increase the Logistic Regression Model’s prediction accuracy.
### Applications:
Loan Acceptance – Organizations that provide banks or loans can use logistic regression to evaluate if a consumer would accept a loan offered to them based on their previous behavior.
Customer age, experience, income, family size, CCAvg, Mortgage, and other explanatory variables are among the several explanatory variables.
Delayed Airplanes – A likely delay in flight timing can also be predicted using logistic regression analysis. Different arrival airports, varied departure airports, carriers, weather conditions, the day of the week, and a categorical variable for different departure hours are all explanatory variables.
MLE-based line estimation
The parameter values that appear in the regression model are used to construct regression lines for models. As a result, you must first estimate the regression model’s parameters. To improve the accuracy of linear and nonlinear statistical models, parameter estimation is used.
Maximum Likelihood Estimation is a method for estimating the parameters of a regression model (MLE).
Nonlinear regression in action: This example is based on the link between the length of a deer’s jaw bone and its age.
The dataset is read from the jaws.txt file; the file’s path is used as a parameter.
`deer<-read.table("E:\Website\Images\jaws.txt",header=T)`
Getting the model to fit — The nls() command accepts a nonlinear equation as an argument with beginning values for the a, b, and c parameters. The outcome is saved in the model object.
```bone<-deer\$bone
age<-deer\$age
model1<-nls(bone~a-b*exp(-c*age),start=list(a=120,b=110,c=0.064))```
The summary() command is used to display information about a model object. As demonstrated below, the model object is passed as a parameter to the summary() command:
`summary(model1)`
```Formula: bone ~ a - b * exp(-c * age)
Parameters:
Estimate Std. Error t value Pr(>|t|)
a 115.2528 2.9139 39.55 < 2e-16 ***
b 118.6875 7.8925 15.04 < 2e-16 ***
c 0.1235 0.0171 7.22 2.44e-09 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 13.21 on 51 degrees of freedom
Number of iterations to convergence: 5
Achieved convergence tolerance: 2.383e-06```
For the changed regression model, use the nls() command on the new model. The outcome is saved in the model2 object.
`model2<-nls(bone~a*(1-exp(-c*age)),start=list(a=120,c=0.064))`
When comparing the models, keep the following in mind: To compare the output objectsmodel1 and model2, use the anova() tool. These items are then passed to the anova() command as parameters.
`anova(model1,model2)`
Analysis of Variance Table
```Model 1: bone ~ a - b * exp(-c * age)
Model 2: bone ~ a * (1 - exp(-c * age))
Res.Df Res.Sum Sq Df Sum Sq F value Pr(>F)
1 51 8897.3
2 52 8929.1 -1 -31.843 0.1825 0.671```
Taking a look at the New Model2’s components
`summary(model2)`
```Formula: bone ~ a * (1 - exp(-c * age))
Parameters:
Estimate Std. Error t value Pr(>|t|)
a 115.58056 2.84365 40.645 < 2e-16 ***
c 0.11882 0.01233 9.635 3.69e-13 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 13.1 on 52 degrees of freedom
Number of iterations to convergence: 5
Achieved convergence tolerance: 1.369e-06```
## Summary
In R programming, we learned the entire notion of nonlinear regression analysis. We learned about R logistic regression and its applications, as well as MLE line estimation and R nonlinear regression models.
Note:
Models with Generalized Additive Functions (GAM)
We can see that the relationship between y and x is nonlinear at times, but we don’t have any theory or mechanical model to suggest a specific functional form (mathematical equation) to represent it.
Generalized Additive Models (GAMs) are particularly beneficial in these situations since they fit a nonparametric curve to the data without asking us to specify a mathematical model to characterise the nonlinearity.
GAMs are valuable because they allow you to figure out the relationship between y and x without having to choose a specific parametric form. The function gam() command in R implements generalised additive models.
Many of the features of both glm() and lm() are present in the gam() command, and the result can be modified using the update() command.
After a GAM has been fitted to data, you can use all of the standard methods including print, plot, summary, anova, forecast, and fitted. The mgcv library includes the gam function.
| 2,218
| 9,704
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.53125
| 4
|
CC-MAIN-2023-23
|
latest
|
en
| 0.918101
|
http://mathhelpforum.com/algebra/84299-can-equation-solved-x-impossible.html
| 1,481,413,094,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-50/segments/1480698543577.51/warc/CC-MAIN-20161202170903-00248-ip-10-31-129-80.ec2.internal.warc.gz
| 171,518,607
| 9,978
|
Thread: can this equation be solved for x or it impossible?
1. can this equation be solved for x or it impossible?
can this equation be solved for x or is it impossible?
x(y) + z + x(w) = t + t(y) + z + t(w)
trying to get to x = ...?
2. Hello, bdud!
can this equation be solved for x or is it impossible?
. . $xy + z + xw \:=\: t + ty + z + tw$
Well, we can subtract $z$ from both sides:. $xy + xw \:=\:t + ty + tw$
Now this situation arises quite often in mathmatics ... get used to it.
Factor: . $x(y + w) \:=\:t(1 + y + w)$
Divide by $(y+w)\!:\quad x \;=\;\frac{t(1+y+w)}{y+w}\quad\hdots$ . See?
| 205
| 609
|
{"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}
| 3.6875
| 4
|
CC-MAIN-2016-50
|
longest
|
en
| 0.837702
|
https://www.brighthubengineering.com/structural-engineering/60388-so-you-want-to-build-a-structure/
| 1,596,519,665,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-34/segments/1596439735860.28/warc/CC-MAIN-20200804043709-20200804073709-00096.warc.gz
| 561,613,474
| 12,580
|
Using Structural Engineering to Create Complex Structures from Simple Construction Elements
Page content
Basic Elements of Structural Construction
Never Push On a Rope
Creating a safe, durable, cost effective structural engineering design is usually not a trivial process. A thorough understanding of stresses, strains, material properties, environmental effects, computer modeling, physics, mathematics, construction, manufacturing, architecture, and related engineering concepts are typically required to successfully execute a design. And while many structures appear incredibly complex, often they are constructed using multiple quantities of only a few basic elements.
A useful collection of these structural engineering basics includes beams, columns, cables, and plates. Beams, columns, and cables are considered to be elements much longer in a single dimension, i.e., length vs. width and depth. Plates, including shells, have two dimensions greater than the third. The type of load each element carries helps to define them as well. Beams are typically compressively loaded to bend normal to the main axis, while columns carry axial tensile or compressive loads. Cables (including wires, ropes, and chains) carry only axial tensile loads and are considered flexible elements. Plates carry bending and shells carry compressive loads in two dimensions.
Useful Complexity from Simple Elements
Catenaries Don’t Have Feathers
Using these simple elements in complex geometries creates many useful structural forms. For example, a chain hanging freely but with ends anchored across two columns creates a catenary curve. Catenary suspension bridges, such as a simple foot bridge, place the load directly on this catenary element. Suspension bridges which support the deck with vertical chains or cables attached between the deck and the flexible suspension cable force the catenary shape into a parabolic form. In both cases the suspension cable experiences tensile stresses. However, inverting the catenary shape creates an ideal arch, a spanning structural form which can withstand purely compressive loads. Building on this form with rigid elements, such as stone blocks, has created some of the most graceful and enduring structures throughout history. Over 75 years ago Catalan architect Antoni Gaudí is said to have used this technique to design some of the arches and vaults in the Church of Colònia Güell. By creating catenaries using small suspended chains and viewing them in a mirror, accurate models could be viewed of the stonework required for constructing the building. Not a bad technique for the period, but modern design and analysis uses computational techniques is a more accurate and efficient method.
Structural Analysis
And Never Pull On Dirt
Consequently before a single cable or block is put in place, a fair amount of structural analysis takes place to ensure these basic elements are assembled correctly using properly selected and specified materials. Using concrete as an anchoring medium for the end of a suspension cable, for example, would be highly inappropriate as concrete has very poor tensile strength. Attaching the cable to a steel eyebolt embedded in concrete tangentially opposite of the tensile stress running in the cable would be more appropriate, as this would load the concrete in compression. And so forth. Future articles in this series will further examine the construction of complex elements using these basic structural elements.
| 642
| 3,493
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.890625
| 3
|
CC-MAIN-2020-34
|
latest
|
en
| 0.902272
|
http://gmatclub.com/forum/the-simple-facts-are-these-the-number-of-people-killed-each-11220.html?fl=similar
| 1,394,254,020,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2014-10/segments/1393999653106/warc/CC-MAIN-20140305060733-00024-ip-10-183-142-35.ec2.internal.warc.gz
| 77,532,379
| 34,280
|
Find all School-related info fast with the new School-Specific MBA Forum
It is currently 07 Mar 2014, 20:47
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# The simple facts are these: the number of people killed each
Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:
Senior Manager
Joined: 19 Feb 2004
Posts: 415
Location: Lungi
Followers: 1
Kudos [?]: 1 [0], given: 0
The simple facts are these: the number of people killed each [#permalink] 02 Nov 2004, 00:06
00:00
Difficulty:
5% (low)
Question Stats:
0% (00:00) correct 0% (00:00) wrong based on 0 sessions
The simple facts are these: the number of people killed each year by bears is about the same as the number of
people killed by lightning of golf courses. And the number of people killed by lightning on golf courses each year is about the same as the number of people electrocuted by electric blenders. All the horrible myths and gruesome stories aside, therefore a grizzly bear is in fact about as dangerous as an electric blender or a game of golf.
Which one of the following is an assumption that the author relies upon in the passage?
(A) Most incidents involving grizzly bears are fatal.
(B) Grizzly bears are no longer the danger they once were.
(C) The number of fatalities per year is an adequate indication of something¡¯s dangerousness.
(D) A golf course is a particularly dangerous place to be in a thunderstorm.
(E) Something is dangerous only if it results in death in the majority of cases.
Director
Joined: 16 Jun 2004
Posts: 893
Followers: 1
Kudos [?]: 7 [0], given: 0
[#permalink] 02 Nov 2004, 00:26
C for me.
Why? Author narrates like this 'number killed by bears = number killed lightning in a golf course = number killed by blenders. So bears are as dangerous as golf and blenders'
In effect author implies that number killed determines how dangerous that is. So, C.
E if negated sounds naive and is the 'extreme type' with the word 'only'
Senior Manager
Joined: 21 Jun 2004
Posts: 340
Followers: 1
Kudos [?]: 2 [0], given: 0
[#permalink] 02 Nov 2004, 15:23
I will vote for c.
The use of 'only' suggested me to strike out E.
_________________
510 on my first GMAT. 610 on second GMAT.!
The struggle continues.
[#permalink] 02 Nov 2004, 15:23
Similar topics Replies Last post
Similar
Topics:
The simple facts are these: the number of people killed each 8 25 Mar 2004, 07:50
As a part of a game, 4 people each choose one number from 1 7 16 Apr 2006, 18:07
The simple facts are these: the number of people killed each 2 18 Feb 2008, 05:16
1 A number of people each wrote down one of the first 30 posit 6 30 Nov 2008, 18:53
Roland: the alarming fact is that 90% of the people in this 1 16 Jun 2011, 09:35
Display posts from previous: Sort by
# The simple facts are these: the number of people killed each
Question banks Downloads My Bookmarks Reviews Important topics
Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.
| 947
| 3,641
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.703125
| 4
|
CC-MAIN-2014-10
|
longest
|
en
| 0.925066
|
http://mathhelpforum.com/calculus/128440-just-starting-calculus-computing-average-guidance-help-please.html
| 1,481,121,724,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-50/segments/1480698542213.61/warc/CC-MAIN-20161202170902-00092-ip-10-31-129-80.ec2.internal.warc.gz
| 174,354,975
| 13,347
|
1. ## Just starting calculus...computing the average...guidance/help please
A bonus question on one of my first calculus quizzes tomorrow for the semester:
V=Potential Energy
V=2g(1-costheta) ... compute the average Potential Energy over the interval [pi/4, pi/3]
Solve in terms of g
So I set it up and went through the motions...but am kind of stuck
v= 2g - 2gcos(pi/3) - 2g - 2gcos(pi/4) / pi/3 - pi/4
... then what?
2. You might want to factor out a g on the right side... I'm assuming g is unknown? If not, you could just sub in for g... but, otherwise, factoring will do wonders.
3. Originally Posted by seichan
You might want to factor out a g on the right side... I'm assuming g is unknown? If not, you could just sub in for g... but, otherwise, factoring will do wonders.
ok continuing...
v= 2g - 2gcos(pi/3) - 2g - 2gcos(pi/4) / pi/3 - pi/4
v= -2g(1/2+sqrt(2)/2) / pi/3 - pi 4
...since this question involves cos, can you change pi/3 and pi/4 to 1/2 and sqrt(2)/2, respectively?
if so v = -2g(1/2+sqrt(2)/2) / 1/2 - sqrt(2)/2
...any more guidance?
4. If you want to solve for g, you might be interested in isolating the g, such that you have a function of the form
g=(something)
5. Except the whole thing is already equal to v .... hm
6. I am assuming you do not know the value of V or of g? If so, then you are done. If you know the value of V, than solving such that g=(something) will give you an answer. If you know g, then you should just substitute the value for g in...
7. Sorry if I am being unclear....we are solving for V....in terms of G. We do not know either, G on earth is 9.81 m/s, but he said "We could be on another planet for all we know, so leave it in terms of g."
8. Alright then, you should be done you already solved for V in terms of g, and, if he wants it left in terms of g, you're done!
9. Originally Posted by Dreamshot
Sorry if I am being unclear....we are solving for V....in terms of G. We do not know either, G on earth is 9.81 m/s, but he said "We could be on another planet for all we know, so leave it in terms of g."
Please do not confuse "g" and "G". they have completely different meanings!
10. Unfortunately, the original method is wrong so pretty much everything everyone has said is off the point. The problem is NOT just simplifying that equation, the equation itself is wrong.
You do NOT find the average of a function, f(x) on an interval, [a, b], by $\frac{f(b)- f(a)}{b- a}$, any more than you find the average of a list of numbers by simply averaging the first and last values. you have to sum the numbers which, for functions, means "integrate".
The average of f(x) on the interval [a, b] is
$\frac{1}{b-a} \int_a^b f(x)dx$
The average you want is
$\frac{2g\int_{\pi/4}^{\pi/3} (1- cos(\theta))d\theta}{\frac{\pi}{3}- \frac{\pi}{4}}$
11. Originally Posted by HallsofIvy
The average of f(x) on the interval [a, b] is
$\frac{1}{b-a} \int_a^b f(x)dx$
provided $f$ is continuous on $[a,b]$.
12. Originally Posted by HallsofIvy
Unfortunately, the original method is wrong so pretty much everything everyone has said is off the point. The problem is NOT just simplifying that equation, the equation itself is wrong.
You do NOT find the average of a function, f(x) on an interval, [a, b], by $\frac{f(b)- f(a)}{b- a}$, any more than you find the average of a list of numbers by simply averaging the first and last values. you have to sum the numbers which, for functions, means "integrate".
The average of f(x) on the interval [a, b] is
$\frac{1}{b-a} \int_a^b f(x)dx$
The average you want is
$\frac{2g\int_{\pi/4}^{\pi/3} (1- cos(\theta))d\theta}{\frac{\pi}{3}- \frac{\pi}{4}}$
This helps a lot, but we haven't done derivatives yet or integration for that matter. Could you show me a process, if possible, of solving the problem without using those two techniques?
13. Halls of Ivy is correct but for the sake of argument suppose the question
is what is the average rate of change of the potential energy then
[V(pi/3) -V(pi/4)]/(pi/3-pi/4)
Be careful with parenteheses 2g[1-cos(pi/3)- (1-cos(pi/4)]/pi/12
= 2g[1/2-1+sqrt(2)/2]/pi/12
= 24g[-1/2 +sqrt(2)/2]/pi
=g/pi [ -12 +12sqrt(2)]
g is of course the acceleration due to gravity = -32ft/sec^2 or
- 9.8m/sec^2
G is the universal gravitation constant G = 6.673 x10^(-11)Nm^2/kg^2
14. Originally Posted by Calculus26
Halls of Ivy is correct but for the sake of argument suppose the question
is what is the average rate of change of the potential energy then
[V(pi/3) -V(pi/4)]/(pi/3-pi/4)
Be careful with parenteheses 2g[1-cos(pi/3)- (1-cos(pi/4)]/pi/12
= 2g[1/2-1+sqrt(2)/2]/pi/12
= 24g[-1/2 +sqrt(2)/2]/pi
=g/pi [ -12 +12sqrt(2)]
g is of course the acceleration due to gravity = -32ft/sec^2 or
- 9.8m/sec^2
G is the universal gravitation constant G = 6.673 x10^(-11)Nm^2/kg^2
On this step:
2g[1-cos(pi/3)- (1-cos(pi/4)]/pi/12
Wouldn't it be 2g[1-cos(pi/3)-2g(1-cos(pi/4)]/pi/12 ?
Since its an average rate of change, do you include the entire equation?
15. I factored out the 2g so the entire equation for V was included
| 1,609
| 5,074
|
{"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": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4
| 4
|
CC-MAIN-2016-50
|
longest
|
en
| 0.89919
|
https://maker.pro/riotous/tutorial/how-to-make-a-riotous-powered-power-logger
| 1,675,504,735,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-06/segments/1674764500095.4/warc/CC-MAIN-20230204075436-20230204105436-00666.warc.gz
| 392,303,587
| 145,703
|
RIOTOUS
# How to Make a RIOTOUS-Powered Power Logger
May 10, 2018 by Robin Mitchell
Share
This wireless power logger measures voltage, current, and power draw with the RIOTOUS V0.2 and a VB.net-powered program.
Power logging can be very useful in situations where the average power draw of a circuit needs to be determined. In this project, we will create a wireless power logger that can be used to measure voltage, current, and power draw, which streams its data to a VB.net-powered program using RIOTOUS V0.2.
## Schematic
### The Power Logger Circuit
The RIOTOUS-based circuit consists of a microcontroller (the PIC16F1825), an ESP-01 module for Wi-Fi connectivity, analog circuitry for recording voltage and current (U2), and basic power regulation (U1). To record how much power is consumed by a device, two variables need to be obtained: voltage drop across the device, and the current flowing through that device. Voltage draw is very easy to determine since the external device is being powered by the VCC rail, so an analog measurement of this can be taken directly by the microcontroller using an analog pin.
However, since VCC will be larger than the 3.3V supply that the PIC uses, the VCC voltage is first fed into a potential divider (RV2 and R8), which divides the voltage by 10. This divided voltage is then buffered by a unity gain amplifier (U2B), which is then fed into one of the PICs analog-digital inputs. By doing this, we can measure VCC values up to 33V (which is far beyond what the AMS1117 can handle anyway).
The second parameter we need to record is current, which is slightly trickier to measure. In our circuit, we use a sense resistor (R4), which is essentially a resistor that sits in series with the device under test. Because the resistor is in series, it will have a voltage drop across it, and the voltage drop will be proportional to the current flowing through it (thanks to V = IR). Rearranging this formula to make I the subject reveals...
Because the resistor and device are in series, the current flowing through the sense resistor must be the same as the current that flows through the device (thanks to Kirchhoff’s law). Therefore, we can determine the current flow through the device by dividing the voltage drop across the sense resistor by its resistance (which is known). Before the voltage drop is measured by the PIC, it is first multiplied by 10 with the use of U2A. This makes it easier to detect smaller currents, but at a cost of limiting the maximum current we can measure.
### The Circuit Code
The firmware on the PIC16F1825 uses the RIOTOUS framework to handle the ESP8266 as well as incoming and outgoing data. On start up, the logger tries to connect to the specified Wi-Fi network and then connect to the server (which is our power logger program written in VB.net). Once a connection has been established, the PIC waits until a “start logging†command is sent. When this command has been sent, the PIC starts taking periodic readings of both voltages (VCC and the voltage drop across the resistor). Upon taking readings, the supply value is multiplied by the sense resistor voltage divided by its resistance to give the power consumption. This value is then sent to the server program for logging on a chart using a VB.net form application.
Here's the code that reads the ADC peripheral to measure the current and voltage:
Next, we add in the calculation code that determines the power consumption:
Finally, we insert the code that converts the result to a string and sends it to the server:
### Server Code
The server side is the application that uses RIOTOUS to establish a RIOTOUS server that our logger can connect to. Then, as readings are streamed in the VB.net form, application plots the data onto a chart to show the rolling power consumption.
The logger having just started
The logger after a minute
### Construction
This circuit can be built using many different circuit board techniques including stripboard, breadboard, and even a PCB. My first attempt at this project was using a PCB, but due to a daft error that I had made (using the programming ports as my analog-read channels), I had to redesign the project. However, instead of making another PCB, I opted to use a breadboard, which provided a challenge!
The ESP-01 module uses 0.1†headers, which is helpful for hobbyists, but the headers have two rows, which means it cannot be used on a breadboard. Therefore, I had to use a piece of stripboard to create an adaptor that would allow it to work with a breadboard.
### Author
Robin Mitchell
Graduated from the University Of Warwick in Electronics with a BEng 2:1 and currently runs MitchElectronics.
| 1,033
| 4,715
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.796875
| 3
|
CC-MAIN-2023-06
|
longest
|
en
| 0.935284
|
https://www.coursehero.com/file/6661779/t02-bv03-cacl3-94160-20111021fri-f11/
| 1,527,001,431,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-22/segments/1526794864790.28/warc/CC-MAIN-20180522131652-20180522151652-00398.warc.gz
| 729,269,874
| 601,304
|
{[ promptMessage ]}
Bookmark it
{[ promptMessage ]}
t02-b(v03)_cacl3_94160_20111021fri_f11
# t02-b(v03)_cacl3_94160_20111021fri_f11 - TALLAHASSEE...
This preview shows page 1. Sign up to view the full content.
t02-b(v03)_cacl3_94160_ 20111021fri_f11.docx // page 1 of 1 TALLAHASSEE COMMUNITY COLLEGE MAC 2313-94160 / CALC3 / FALL2011 LAST _____________________ TEST #02B / §14.3, 14.4, 14.5, 14.6 , and 1 “carry - over” problem from Test #1. THOMAS’ CALC 11 -th ed ET. FIRST ____________________ D.Jones / DATE of TEST: FRI. 10/21/2011 INSTRUCTIONS : Do all work & put all answers on the work-sheets which I am providing. Box your final answer. Do no more than 2 problems per page. Draw a horizontal line between problems on the same page. Write on one side only. Draw axes w/ a straightedge. There are 6 problems on this part. Numbers 4 & 5 are worth 9 points each, and the other 4 problems are worth 8 points each for a total of 50 points. You may use a TI 83 or 84 calculator, or its equivalent, to check your work, but all answers must be exact (and 10 digit calculator answers are NOT exact). ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ 1. If 2 4 3 , , 5 4 f x y z x yz xy yz , then find , , and x y z f f f . ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ 2. If 2 0 3 3 0 and 1, 2 xy y x P , u se Implicit Differentiation and the “F -formula from Sec. 14.4 (Thm 8:p.986: x y dy F dx F
This is the end of the preview. Sign up to access the rest of the document.
{[ snackBarMessage ]}
### What students are saying
• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.
Kiran Temple University Fox School of Business ‘17, Course Hero Intern
• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.
Dana University of Pennsylvania ‘17, Course Hero Intern
• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.
Jill Tulane University ‘16, Course Hero Intern
| 686
| 2,504
|
{"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}
| 3.625
| 4
|
CC-MAIN-2018-22
|
latest
|
en
| 0.768597
|
https://brilliant.org/problems/iceberg-floating-in-the-ocean/
| 1,586,519,914,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-16/segments/1585371896913.98/warc/CC-MAIN-20200410110538-20200410141038-00401.warc.gz
| 360,377,900
| 8,864
|
# Iceberg floating in the ocean
An iceberg having a specific gravity of $0.92$ is floating on the salt water of specific gravity of $1.03$. If the volume of the iceberg above the water surface is $1000~m^3$, which of the following is the most approximate total volume of the iceberg in $m^3$?
Details:
unit weight of water = $9.81\frac{kN}{m^3}$
×
| 97
| 351
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 5, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.9375
| 3
|
CC-MAIN-2020-16
|
longest
|
en
| 0.903469
|
http://www.edaboard.com/entry1820.html
| 1,513,374,530,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-51/segments/1512948579567.73/warc/CC-MAIN-20171215211734-20171215233734-00626.warc.gz
| 340,680,683
| 14,806
|
# Tahmid
#### Modalities of Using the PIC (16F877A) CCP Module - Compare Section
by , 7th December 2012 at 18:02 (10911 Views)
Here I will show you how to use the compare section of the CCP (Compare, Capture, Pulse Width Modulation) module of the PIC16F877A. You can use this concept for any PIC with a CCP/ECCP module.
There are 2 modules associated with the compare section – the CCP module which we’ll use for the compare function, and the Timer 1 module. The function of the compare module is to compare (obviously, as the name suggests) the value of the CCPR1 register against the Timer 1 value. The CCPR1 register is actually composed of two 8-bit registers that together form a 16-bit register. The high 8-bits – the high byte – make up the CCPR1H register and the low 8-bits – the low byte – make up the CCPR1L register.
For example, if the value of CCPR1H is 30 and the value of CCPR1L is 47, what is the value of the 16-bit CCPR1 register?
Solution:
CCPR1H = 30 = 0x1E
CCPR1L = 47 = 0x2F
So, the high byte = 0x1E. The low byte = 0x2F.
CCPR1 = 0x1E2F = 7727
Like CCPR1, TMR1 is a 16-bit register composed of two 8-bit registers – TMR1H and TMR1L. If TMR1H = 30 and TMR1L = 47, TMR1 = 7727.
Back to the function of the compare mode. The 16-bit value of CCPR1 is compared against the 16-bit value of TMR1. RC2 pin is associated with the CCP module. Upon match of the CCPR1 register against TMR1 (when CCPR1 is equal to TMR1), one of the following can happen to RC2 (RC2 is the pin associated with the CCP module) depending on the setting of the CCP1CON register:
• RC2 is driven high
• RC2 is driven low
• RC2 is unaffected
If RC2 is to be used, TRISC2 must be cleared to make RC2 a digital output.
Upon match of CCPR1 and TMR1 (when CCPR1 is equal to TMR1), CCP1IF (bit 2 of PIR1) is set. If the CCP interrupt is enabled, a CCP interrupt is generated. To enable the CCP interrupt, CCP1IE (bit 2 of PIE1) has to be set.
Another interesting setting of the CCP module is that it can be set up so that upon compare match of CCPR1 and TMR1, the following happen:
• CCP1IF is set
• CCP interrupt is generated if CCP1IE is set
• RC2 is unaffected
• Timer 1 is reset (TMR1 = 0)
• If the ADC module is enabled, an A/D conversion is started
What happens when there is a compare match of CCPR1 and TMR1 depends on the setting of the CCP1CON register. Here is the CCP1CON register:
• When CCP1CON = 8 (0x08), CCP1 module is configured for compare mode and is set up so that upon a compare match of CCPR1 and TMR1, RC2 is driven high.
• When CCP1CON = 9 (0x09), CCP1 module is configured for compare mode and is set up so that upon a compare match of CCPR1 and TMR1, RC2 is driven low.
• When CCP1CON = 10 (0x0A), CCP1 module is configured for compare mode and is set up so that upon a compare match of CCPR1 and TMR1, RC2 is unaffected.
• When CCP1CON = 11 (0x0B), CCP1 module is configured for compare mode and is set up so that upon a compare match of CCPR1 and TMR1, RC2 is unaffected, TMR1 is reset and an A/D conversion is started if the ADC module is enabled.
Remember that in all compare modes, whenever there is a compare match of CCPR1 and TMR1, CCP1IF is set.
Now let’s take a look at a code example to make things clear. Let’s say we’re running a PIC16F877A with a crystal oscillator of frequency 20MHz. Let’s use the compare mode to generate a 50Hz output with 50% duty cycle.
Solution:
Tcy = (1/20000000)*4*106 us = 0.2us
Since one TMR1 increment takes 1 instruction cycle (with prescaler 1:1), one TMR1 overflow takes 65536 instruction cycles. That’s 65536 * 0.2us = 13107.2us = 13.1072ms. For 50Hz, time period = 20ms. So with 50% duty cycle, on time is 10ms.
The number of instruction cycles in 10ms is equal to (10000/0.2) = 50000.
When TMR1 = 50000 after counting up from 0, 10ms will have elapsed. So, we’ll assign 50000 to CCPR1 so that a compare match occurs every 10ms. We’ll use the CCP setting for trigger special event where TMR1 is reset upon compare match.
CCPR1 = 50000 = 0xC350
Therefore, CCPR1H = 0xC3 and CCPR1L = 0x50
For our required mode of operation, CCP1CON = 11. As CCP interrupt will be used, CCP1IE must be set. GIE and PEIE also have to be set.
We’ll use RC0 as our output pin. Every 10ms, the state of RC0 must be toggled. We’ll do the toggling in the interrupt service routine (ISR). We chose to use RC0. Since in our mode of operation, RC2 is unaffected, we can use any pin really and I arbitrarily chose to use RC0. If any of the pins that are multiplexed to the ADC module or the comparator module is to be used, remember to disable the analog circuitry and use that pin as digital output.
The code should be quite easy to understand. Here it is:
Code:
```//Programmer: Syed Tahmid Mahbub
//Target Microcontroller: PIC16F877A
//Compiler: mikroC PRO for PIC
//Using Compare Section of PIC CCP Module - using PIC16F877A
void interrupt(){
if (CCP1IF_bit == 1){
RC0_bit = ~ RC0_bit; //Toggle RC0
CCP1IF_bit = 0;
}
}
void main() {
PORTC = 0;
TRISC = 0;
CCP1CON = 11;
CCP1IE_bit = 1;
GIE_bit = 1;
PEIE_bit = 1;
CCPR1H = 0xC3;
CCPR1L = 0x50;
//CCPR1 = 0xC350 = 50000
T1CON = 0x01; //Prescaler 1:1, start Timer 1
while(1){
//Do whatever else is required
}
}```
Here is the output waveform:
Now let’s take a look at another code example. Let’s say we’re running a PIC16F877A with a crystal oscillator of frequency 20MHz. Let’s use the compare mode to generate a 50Hz output with 75% duty cycle.
Solution:
Tcy = 0.2us
In the previous example, as duty cycle was 50%, the simple thing to do was to toggle the output pin. Here we cannot do that as duty cycle is 75%.
This time, we’ll use RC2 as the output pin.
Time period = 20ms. With 75% duty cycle, on time is 15ms. Off time is 5ms.
Number of instruction cycles for 5ms = (5000/0.2) = 25000. Number of instruction cycles for 15ms = 75000. This is beyond the maximum for TMR1. So, we’ll utilize the TMR1 prescaler. We’ll use a prescaler 1:2.
So, for 5ms, number of TMR1 increments = 12500. For 15ms, number of TMR1 increments = 37500.
We’ll change the CCP1CON setting from 8 to 9 to 8 to set the output high to low to high upon consecutive compare matches. We’ll need to alternately change CCPR1 between 12500 and 37500 as required.
12500 = 0x30D4
37500 = 0x927C
In the 2 modes of operation we chose to use, TMR1 is not cleared/reset upon compare match. So, in the interrupt, we must clear TMR1 manually to start TMR1 counting from 0.
If you understand how this code works, you should be clear about the compare module by now.
Here is the code:
Code:
```//Programmer: Syed Tahmid Mahbub
//Target Microcontroller: PIC16F877A
//Compiler: mikroC PRO for PIC
//Using Compare Section of PIC CCP Module - using PIC16F877Avoid interrupt(){
if (CCP1IF_bit == 1){
TMR1H = 0;
TMR1L = 0;
//TMR1 must be cleared manually
if (CCP1CON == 8){
CCP1CON = 9;
CCPR1H = 0x92;
CCPR1L = 0x7C;
//CCPR1 = 37500 --> 15ms
}
else{
CCP1CON = 8;
CCPR1H = 0x30;
CCPR1L = 0xD4;
//CCPR1 = 12500 --> 5ms
}
CCP1IF_bit = 0;
}
}
void main() {
TRISC = 0;
PORTC = 4;
CCP1CON = 9; //Clear RC2 on match
CCPR1H = 0x92;
CCPR1L = 0x7C;
//CCPR1 = 37500 --> 15ms
CCP1IE_bit = 1;
GIE_bit = 1;
PEIE_bit = 1;
T1CON = 0x11; //Prescaler 1:2, start Timer 1
while(1){
//Do whatever else is required
}
}```
Here is the output waveform:
The compare section of the CCP module is very useful for timing-dependent applications. As shown above, you can use it for PWM at frequencies that are too low for the PWM section of the CCP module.
If clearly understood, it is very easy to use and I hope I could make the compare mode of the CCP module clear. Your comments and feedback are welcome.
Tags: None
Categories
Uncategorized
Hi Tahmid,
I have a project trying to build a thermal sensor using TC74 and PIC18F4520 and having the temperature to display by an LCD, but i do not know how to start programming it. I am using MPLAB to write my program C language. I look forward to your reply on how to start on my project.
Thank You Very Much
Best Regards
InNeedOfHelp
In this project you are using Digital Oscilloscope can I know about it a bit more ?
InNeedOfHelp, you should start by going through example codes for I2C and TC74 interfacing. Then, you can adapt them to your own requirements.
debjit625, the digital oscilloscope is part of the simulation software Proteus (ISIS).
Originally Posted by Tahmid
debjit625, the digital oscilloscope is part of the simulation software Proteus (ISIS).
Thanks ,I thought its some PC oscilloscope
Hello Tahmid,can I have some help regarding the interface of condense to PIC16F877A for switch on and off of an appliance using compare mode of PIC16F877A?
Thanks
| 2,640
| 8,665
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.8125
| 3
|
CC-MAIN-2017-51
|
latest
|
en
| 0.858019
|
https://flatearthsoc.com/flat-earth-news-blog-8-4-greg-davies-nicknames.html
| 1,566,442,156,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-35/segments/1566027316718.64/warc/CC-MAIN-20190822022401-20190822044401-00051.warc.gz
| 467,199,024
| 9,076
|
It is a well-known fact that clouds are continually seen moving in all manner of directions - yes, and frequently, in different directions at the same time - from west to east being as frequent a direction as any other. . Now, if the Earth were a globe, revolving through space from west to east at the rate of nineteen miles in a second, the clouds appearing to us to move towards the east would have to move quicker than nineteen miles in a second to be thus seen; whilst those which appear to be moving in the opposite direction would have no necessity to be moving at all, since the motion of the Earth would be more than sufficient to cause the appearance. But it only takes a little common sense to show us that it is the clouds that move just as they appear to do, and that, therefore, the Earth is motionless. We have, then a proof that the Earth is not a globe.
155) Some people claim to have seen the curvature of the Earth out their airplane windows. The glass used in all commercial airplanes, however, is curved to remain flush with the fuselage. This creates a slight effect mixed with confirmation bias people mistake for being the alleged curvature of the Earth. In actuality, the fact that you can see the horizon at eye-level at 35,000 feet out both port/starboard windows proves the Earth is flat. If the Earth were a ball, no matter how big, the horizon would stay exactly where it was and you would have to look DOWN further and further to see the horizon at all. Looking straight out the window at 35,000 feet you should see nothing but "outer-space" from the port and starboard windows, as the Earth/horizon are supposed to be BELOW you. If they are visible at eye level outside both side windows, it’s because the Earth is flat!
130) From “Earth Not a Globe!” by Samuel Rowbotham, “Take two carefully-bored metallic tubes, not less than six feet in length, and place them one yard asunder, on the opposite sides of a wooden frame, or a solid block of wood or stone: so adjust them that their centres or axes of vision shall be perfectly parallel to each other. Now, direct them to the plane of some notable fixed star, a few seconds previous to its meridian time. Let an observer be stationed at each tube and the moment the star appears in the first tube let a loud knock or other signal be given, to be repeated by the observer at the second tube when he first sees the same star. A distinct period of time will elapse between the signals given. The signals will follow each other in very rapid succession, but still, the time between is sufficient to show that the same star is not visible at the same moment by two parallel lines of sight when only one yard asunder. A slight inclination of the second tube towards the first tube would be required for the star to be seen through both tubes at the same instant. Let the tubes remain in their position for six months; at the end of which time the same observation or experiment will produce the same results--the star will be visible at the same meridian time, without the slightest alteration being required in the direction of the tubes: from which it is concluded that if the earth had moved one single yard in an orbit through space, there would at least be observed the slight inclination of the tube which the difference in position of one yard had previously required. But as no such difference in the direction of the tube is required, the conclusion is unavoidable, that in six months a given meridian upon the earth's surface does not move a single yard, and therefore, that the earth has not the slightest degree of orbital motion."
80.) It is supposed,"in the regular course of the Newtonian theory, that the Earth is, in June, about 190 millions of miles (190,000,000) away from its position in December. Now, since we can, (in middle north latitudes), see the North Star, on looking out of a window that faces it – and out of the very same corner of the very same pane of glass in the very same window – all the year round, it is proof enough for any man in his senses that we have made no motion at all. It is a proof that the Earth is not a globe.
Imagine an ant walking along the surface of an orange, into your field of view. If you look at the orange “head on”, you will see the ant’s body slowly rising up from the “horizon” because of the curvature of the orange. If you would do that experiment with the ant approaching along a long road rather than a round object, the effect would change: The ant would slowly "materialize" into view (depending on how sharp your vision is).
```It is in evidence that, if a projectile be fired from a rapidly moving body in an opposite direction to that in which the body is going, it will fall short of the distance at which it would reach the ground if fired in the direction of motion. Now, since the Earth is said to move at the rate of nineteen miles in, a second of time, "from west to east," it would make all the difference imaginable if the gun were fired in an opposite direction. But, as, in practice, there is not the slightest difference, whichever way the thing may be done, we have a forcible overthrow of all fancies relative to the motion of the Earth, and a striking proof that the Earth is not a globe.
```
6 And God said, Let there be a firmament in the midst of the waters, and let it divide the waters from the waters. 7 And God made the firmament, and divided the waters which were under the firmament from the waters which were above the firmament: and it was so. 8 And God called the firmament Heaven. And the evening and the morning were the second day.
We read in the inspired book, or collection of books, called THE BIBLE, nothing at all about the Earth being a globe or a planet, from beginning to end, but hundreds of allusions there are in its pages which could not be made if the Earth were a globe, and which are, therefore, said by the astronomer to be absurd and contrary to what he knows to be true! This is the groundwork of modern infidelity. But, since every one of many, many allusions to the Earth and the heavenly bodies in the Scriptures can be demonstrated to be absolutely true to nature, and we read of the Earth being "stretched out" "above the waters," as "standing in the water and out of the water," of its being "established that it cannot be moved," we have a store from which to take all the proofs we need, but we will just put down one proof - the Scriptural proof - that Earth is not a globe.
48.) In Mr. Proctor's "Lessons in Astronomy," page 15, a ship is represented as sailing away from the observer, and it is given in five positions or distances away on its journey. Now, in its first position, its mast appears above the horizon, and, consequently, higher than the observer's line of vision. But, in its second and third positions, representing the ship as further and further away, it is drawn higher and still higher up above the line of the horizon! Now, it is utterly impossible for a ship to sail away from an observer, under the, conditions indicated, and to appear as given in the picture. Consequently, the picture is a misrepresentation, a fraud, and a disgrace. A ship starting to sail away from an observer with her masts above his line of sight would appear, indisputably, to go down and still lower down towards the horizon line, and could not possibly appear – to anyone with his vision undistorted – as going in any other direction, curved or straight. Since, then the design of the astronomer-artist is to show the Earth to be a globe, and the points in the picture, which would only prove the Earth to be cylindrical if true, are NOT true, it follows that the astronomer-artist fails to prove, pictorially, either that the Earth is a globe or a cylinder, and that we have, therefore, a reasonable proof that the Earth is not. a globe.
4) Rivers run down to sea-level finding the easiest course, North, South, East, West and all other intermediary directions over the Earth at the same time. If Earth were truly a spinning ball then many of these rivers would be impossibly flowing uphill, for example the Mississippi in its 3000 miles would have to ascend 11 miles before reaching the Gulf of Mexico.
In Figure 12, the ship now appears beyond the pier. Notice that the inferior mirage of the lettering on the hull is much more obvious now. In Figure 13, the lettering and its inferior mirage have not merged. In Figure 14, the lettering is difficult to see. This probably is because most of the lettering is below the horizon, and what appears to be the bottom of the hull is an inferior mirage of the top of the hull. This is clearly seen by the inferior mirage of the first layer of red containers below the turquoise. In Figure 15, the inferior mirage of the bottom layer of containers is more obvious, and the inferior mirage of the bottom of the bridge castle is beginning to show up. Clearly, at least half of the turquoise visible here is an inferior mirage. Most of the hull is below the curvature of the earth. Unfortunately, at this point the sun was about to set, so light levels were dropping quickly, forcing me to use longer exposures. At that point, I stopped taking photographs.
88.) If we could – after our minds had once been opened to the light of Truth – conceive of a globular body on the surface of which human beings could exist, the power – no matter by what name it be called – that would hold them on would, then, necessarily, have to be so constraining and cogent that they could not live; the waters of the oceans would have to be as a solid mass, for motion would be impossible. But we not only exist, but live and move; and the water of the ocean skips and dances like a thing of life and beauty! This is a proof that the Earth is not a globe.
5) The sun is much closer than we have been told. It is, in fact, in our atmosphere. You can clearly see that it is not 93 million miles away. Many times you can see the sun’s rays shooting out of a cloud forming a triangle. If you follow the rays to their source it will always lead to a place above the clouds. If the sun was truly millions of miles away, all the rays would come in at a straight angle. Also the sun can be seen directly above clouds in some balloon photos, creating a hot spot on the clouds below it and in other photos you can clearly see the clouds dispersing directly underneath the close small sun.
The higher up you climb, the farther you will see. Usually, we tend to relate this to Earthly obstacles—like the fact we have houses or other trees obstructing our vision on the ground, and climbing upwards we have a clear view—but that’s not the true reason. Even if you stood on a completely clear plateau with no obstacles between you and the horizon, you would see much farther from the greater height than you would on the ground.
On January 25th, 2016 Atlanta rapper B.o.B., who has self-identified as a member of the Flat Earth Society, tweeted a photograph of himself against a skyline, then tweeted a screenshot from Flat Earth Movement literature that proclaimed that Polaris (the North Star) can be seen 20° south of the Equator. Neil DeGrasse Tyson answered the rapper's question, writing "Polaris is gone by 1.5 deg S. Latitude. You’ve never been south of Earth’s Equator, or if so, you've never looked up."
If the Earth were a globe, there would, very likely, be (for nobody knows) six months day and six months night at the arctic and antarctic regions, as astronomers dare to assert there is: - for their theory demands it! But, as this fact - the six months day and six months night - is; nowhere found but in the arctic regions, it agrees perfectly with everything else that we know about the Earth as a plane, and, whilst it overthrows the "accepted theory," it furnishes a striking proof that Earth is not a globe.
75.) Considerably more than a million Earths would be required to make up a body like the Sun -the astronomers tell us: and more than 53,000 suns would be wanted to equal the cubic contents of the star Vega. And Vega is a "small star!" And there are countless millions of these stars! And it takes 30,000,000 years for the light of some of those stars to reach us at 12,000,000 miles in a minute! And, says Mr. Proctor, "I think a moderate estimate of the age of the Earth would be 500,000,000 years! "Its weight," says the same individual, "is 6,000,000,000,000,000,000,060 tons!" Now, since no human being is able to comprehend these things, the giving of them to the world is an insult – an outrage. And though they have all risen from the one assumption that Earth is a planet, instead of upholding the assumption, they drag it down by the weight of their own absurdity, and leave it lying in the dust – a proof that Earth is not a globe.
```118) Furthermore, the velocity and path of the Moon are uniform and should therefore exert a uniform influence on the Earth’s tides, when in actuality the Earth’s tides vary greatly and do not follow the Moon. Earth’s lakes, ponds, marshes and other inland bodies of water also inexplicably remain forever outside the Moon’s gravitational grasp! If “gravity” was truly drawing Earth’s oceans up to it, all lakes, ponds and other bodies of standing water should certainly have tides as well.
```
169) So-called “satellite” TV dishes are almost always positioned at a 45 degree angle towards the nearest ground-based repeater tower. If TV antennae were actually picking up signals from satellites 100+ miles in space, most TV dishes should be pointing more or less straight up to the sky. The fact that “satellite” dishes are never pointing straight up and almost always positioned at a 45 degree angle proves they are picking up ground-based tower signals and not “outer-space satellites.”
As previously mentioned, the reaction of bodies of water with sunlight is very different from that of land. Being largely transparent, light penetrates deeply into water, so that the sun’s light is absorbed throughout a thick layer from the surface to some depth rather than just on the surface, as with land. Additionally, water has a high specific heat, which means that its temperature increases very slowly as heat is added. Consequently, water exposed to sunlight does not change temperature appreciably throughout the day, so there is no heating of air in contact with the water. If anything, during summer afternoons, when land is rapidly heating, bodies of water frequently are cooler than air temperature. The cooler water chills the air in direct contact with it, so the air lying just above water often is cooler than air higher up. Since air temperature normally decreases with height, this temperature reversal from the norm is called a temperature inversion. Temperature inversions are common over bodies of water during late spring and into summer. Since this temperature structure is the reverse of what causes inferior mirages, inferior mirages are far less commonly noticed over water. This happens particularly during the summer, when inferior mirages are common over land.
| 3,291
| 15,082
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.890625
| 3
|
CC-MAIN-2019-35
|
latest
|
en
| 0.969344
|
http://www.cram.com/flashcards/b1-w5-m-buehler-2-389006
| 1,490,409,929,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-13/segments/1490218188773.10/warc/CC-MAIN-20170322212948-00476-ip-10-233-31-227.ec2.internal.warc.gz
| 493,762,747
| 24,425
|
• Shuffle
Toggle On
Toggle Off
• Alphabetize
Toggle On
Toggle Off
• Front First
Toggle On
Toggle Off
• Both Sides
Toggle On
Toggle Off
Toggle On
Toggle Off
Front
### How to study your flashcards.
Right/Left arrow keys: Navigate between flashcards.right arrow keyleft arrow key
Up/Down arrow keys: Flip the card between the front and back.down keyup key
H key: Show hint (3rd side).h key
A key: Read text to speech.a key
Play button
Play button
Progress
1/17
Click to flip
### 17 Cards in this Set
• Front
• Back
What is relationship between single channel current and actual membrane current? A single channel conductance x the open probability is equal to the Actual membrane current. If you multiply the actual current x the # of channels equal the macroscopic current. How many gates in the K+-delayed rectifier channel? 1 gate What is the Probability of the K+ delayed rectifier channel being opened? P= n^4 How many gates in the Na+ channel? 2 gates What is the Probability of the Na+ channel being opened? P= m^3xh What states does the K+ delayed rectifier channel undergo before it becomes inactivted? it goes from C1--> C2 --> Open--> Inactive C2=resting state C1= What is special about the A-type K channels? they have a ball and chain mechanism that inactivates channel When is there no conductance of Na+ after sequential stimulation? During the Absolute Refractory period i.e since their is no Na+ conductance then their is no AP When is there partial conductance of Na+ after sequential stimulus? During the Relative Refractory Period: Last about 10 ms i.e Na+ conductance slowly increases What controls the Space response (i.e Slow Depolarization) prior to reaching AP treshold? Transient outward K+ channels What causes pacemaker activity? Ca++ dependant K-channels produce a slow after hyperpolirization causing a long resting period in where intracellular calcium prevent Vm from reaching AP threshold eventually Ca++ is lowered by Ca++-pumps which slowly repolarize What Shapes Action Potentials? Ca++dependant K-channels What causes a FAST After-hyperpolarization? voltage gated delayed rectifier K channels What causes a SLOW Afterhyperpolarization? Ca dependant K channels Where does Ca++ come in through? voltage gated Ca channels What is the equation for single channel current i.e ohms law single channels currents? i(open)= signal channel conductance x(Vm-Nerst potential) What is the macroscopic membrane current equation? i(open) x Open Probability x number of open channels
| 581
| 2,514
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.625
| 3
|
CC-MAIN-2017-13
|
longest
|
en
| 0.859089
|
https://www.coursehero.com/file/6189712/Ch7-aumlfrac12%C5%93aumlcedil%C5%A1egravesectpoundccedilshy-2010/
| 1,496,024,466,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-22/segments/1495463612008.48/warc/CC-MAIN-20170529014619-20170529034619-00035.warc.gz
| 1,050,781,703
| 25,693
|
Ch7-作业解答-2010
# Ch7-作业解ç”-2010
This preview shows pages 1–13. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: Ch7À } ^ * ª È À 7-3 èª*ÐÙxE 4ô 27 7 x (n)^ È}Àª* 7-37 b7 1 7-3 7 -5 ¸ª E x Ù Ð * 7 -5 ;X À] }ª* 8 y ( n ) Hwª*8“D¸s Á y (−1) = 0 • n 3 x(n) = u (n) − u (n − 5) 2 7-5 b Š E x Ù Ð * ª Á ˜ y ( n) = x ( n) + 3 x(n) = u (n) − u (n − 5) 1 y (n − 1) 3 y ( n) X 7-9 N E x Ù Ð * ª ` 7-9ˆ¸ 3 7-9 À ª * } ˆ 7-11_ 2 ª Á ‡ ° 8 _} ♠ ℵ ° ♠ ℑ 8 α −2=0 α =2 y ( n) = C • 2 n 1 1 y (0) = ‡ 8 C= °Áª* 2 2 1 y (n) = • 2n = 2n −1 2 ˜ ° Á 7-14 ‡ª y (n) = −5 y (n − 1) + n⊥ ℵ } ª* y (−1) = 0 ª 7-16 À }_ y (n) + 2 y (n − 1) + y (n − 2) = 3n⊥ ℵ } *♠ y (−1) = 0 y (0) = 0 _ À } 7 -17ª y ( n) + y (n − 2) = sin n⊥ ª} À* y (−1) = 0 y (−2) = 0 7-27 4 7-27 7-30 û 1 + ¨ Œ ì 5 á h( n ) n g ( n² ) ‡ * Á ª n2 Á * ´‡ (ª h( n² ) ‡ * Á ª g (n) Áª E x Ù Ð * 7 -32‹0 n y ( n) n h( n) n x( n) n y ( n) n 1 x(n) n h(n) 3ÚxE n “ ”@ E x Û y (n) Ø}Àª* ^ 7-32 a y (n) ‡Áª* ´ 7-32 b y (n) ؇Áª* µ 7-32 c 5 7-32 7-33 7 E x Û 7-33 @ h1 (n) n h2 (n) n h1 (n) = δ (n) − δ (n − 3) h2 (n) = (0.8) n u (n) x ( n) = u ( n) 1 y ( n) y (n) = [ x( n) ∗ h1 (n)] ∗ h2 (n) 2 y ( n) y (n) = x(n) ∗ [h1 (n) ∗ h2 (n)] 5 á쌨+ 6 7-33 17 y (n) = [ x(n) ∗ h1 (n)] ∗ h2 (n) = 5(1 − 0.8n +1 )u ( n) − 5(1 − 0.8n − 2 )u (n − 3) 27 = 5(1 − 0.8n +1 )u ( n) − 5(1 − 0.8n − 2 )u (n − 3) ª D *“ ω ℑ 8♠ 17 * s ¸ 2Hℜ ...
View Full Document
## This note was uploaded on 03/28/2011 for the course ELECTRICAL 30230104 taught by Professor Yongren during the Fall '11 term at Tsinghua University.
### Page1 / 13
Ch7-作业解ç”-2010
This preview shows document pages 1 - 13. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online
| 1,213
| 2,494
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-22
|
longest
|
en
| 0.419734
|
https://www.physicsforums.com/threads/solving-initial-velocity-for-falling-object.797520/
| 1,508,329,487,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-43/segments/1508187822930.13/warc/CC-MAIN-20171018104813-20171018124813-00790.warc.gz
| 1,005,018,141
| 17,040
|
# Solving initial velocity for falling object
1. Feb 12, 2015
### heythere1010
1. The problem statement, all variables and given/known data
A stone is dropped from a bridge that is 65 m high. A second stone is thrown downwards 2 seconds later. Both the stones are heard hitting the water at the same time. The speed of sound is a constant 343 m/s. What is the initial velocity of the second stone.
2. Relevant equations
d=vi*t+1/2at^2
3. The attempt at a solution
65=0+1/2(-9.8)(t)^2
I know t-2 is used for the time of the second stone. I don't know which equations to use to make them equal.
2. Feb 12, 2015
### Staff: Mentor
What value did you get for the time of fall for the first stone? So what must be the time (numeric value) that the second stone has to make the complete journey?
3. Feb 12, 2015
### heythere1010
5.2. Do I just sub it in to the second equation with t-2 and solve for initial velocity?
4. Feb 12, 2015
### haruspex
Sure.
5. Feb 12, 2015
### heythere1010
So the speed of sound is irrelevant?
6. Feb 12, 2015
### haruspex
The sounds of hitting the water were heard at the same time. Did they hit the water at the same time?
| 339
| 1,167
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43
|
longest
|
en
| 0.928367
|
https://www.answerminer.com/blog/create-graph-pie-chart-guide
| 1,596,952,605,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-34/segments/1596439738425.43/warc/CC-MAIN-20200809043422-20200809073422-00015.warc.gz
| 601,821,096
| 62,175
|
Create a Graph: A Pie Chart Visualization Guide
After the bar chart visualization guide, we are continuing with one of the most common charts. Pie charts are widely used in everyday statistics, in the business world, and in education. In this article, we will give you some tips about pie chart visualization even though we recommend that you use other charts if it is possible. If there is no other way to visualize your data, this article will be a good guide for you.
What Is a Pie Chart?
According to Wikipiedia: “Pie chart (or a circle chart) is a circular statistical graphic, which is divided into slices to illustrate numerical proportion. In a pie chart, the arc length of each slice (and consequently its central angle and area), is proportional to the quantity it represents. While it is named for its resemblance to a pie which has been sliced, there are variations on the way it can be presented. The earliest known pie chart is generally credited to William Playfair’s Statistical Breviary of 1801:
“A table is nearly always better than a dumb pie chart; the only worse design than a pie chart is several of them…” – Tufte, The Visual Display of Quantitative Information
Do not forget the Five common mistakes
Pie Charts and Donuts
Pie charts are mostly used to compare part-to-whole relationships in the case of discrete or linear data. We recommend using with a low number of slices. In addition to pie charts, donut charts are also a commonly used circle visualization method.
The biggest disadvantage of pie charts is that human eyes can only recognize friendly rates like 25%,-50%,-75%, or 100%. To read more, check out our previous article about circle visualization.
Standard Pie Chart: Used for Describing Whole-to-Part Relationships
Donut Chart: Can Contain Additional Information or Style Elements
Pie Chart Guide
Below you will read four tips about how to create pie charts.
1. Don’t use pie charts for comparing. Instead of a pie chart, use a stacked bar chart.
1. Use only two or three categories on one pie chart and a maximum five.
1. The total value of the pie chart must be 100%.
1. Order slices clockwise starting with the largest part.
Source: http://www.slideshare.net/Visage/data-visualization-101-how-to-design-chartsandgraphs
• Popular and easy to understand if only using two or three categories
• Indicates a part-to-whole relationship between values
• Looks nice
• Angles are difficult to estimate
• Difficult to compare quantities
• Can’t visualize complex relationships
• Valid for one point in time only
In the end here is a short guide how to create perfect pie chart.
http://bit.ly/2dbz2ko
Connect your data and start the exploration
| 578
| 2,700
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.890625
| 4
|
CC-MAIN-2020-34
|
latest
|
en
| 0.891984
|
https://english.stackexchange.com/questions/526228/are-all-billionaires-also-millionaires
| 1,716,874,948,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971059067.62/warc/CC-MAIN-20240528030822-20240528060822-00572.warc.gz
| 200,184,487
| 37,104
|
# Are all billionaires also millionaires? [closed]
Are all billionaires also millionaires? In other words, should the definition of millionaire be taken to include billionaires, or exclude them?
• Please add dictionary definitions, 107, linked and attributed. // Would you say "He's a millionaire ... perhaps even a billionaire" is incorrect? Feb 26, 2020 at 17:52
• Would you say that an octogenarian is also a septuagenarian_ ? Feb 26, 2020 at 18:13
• @Nigel J Would most people say that a square is a rectangle? That a shrimp and a prawn are similar? When precising definitions may well be involved, one has to define terms, not assume a particular definition. Feb 26, 2020 at 18:47
• What does the dictionary say? The definition I looked at didn't place an upper limit on a millionaire's wealth. Feb 27, 2020 at 1:41
• It was not a good decision to close this question. Feb 27, 2020 at 11:32
No, billionaires are not called "millionaires", even though by definition they may actually be millionaires. The relevant conversational principle is the maxim of quantity:
Grice's Maxims. The maxim of quantity, where one tries to be as informative as one possibly can, and gives as much information as is needed, and no more. ... The maxim of manner, when one tries to be as clear, as brief, and as orderly as one can in what one says, and where one avoids obscurity and ambiguity.
(by googling "maxim of quantity")
• Billionaires wouldn't normally be called millionaires, but they'd probably be included in an accurate count of "total millionaires" unless that counting exercise involved producing a separate tally for "total billionaires". Feb 26, 2020 at 18:36
• @FF The maxim of adapting to the register being used. To most people, squares aren't rectangles. Feb 26, 2020 at 18:45
• I went to an LCC junior school in London from 1959, where I was taught the difference between a rectangle and an oblong. Feb 26, 2020 at 18:53
• @EdwinAshworth - Anyone with a basic highschool education should know that squares are rectangles mathematically, but in non-maths contexts I believe you're right that most people would assume that references to "rectangles" would not include squares. (So in other words the answer is "maybe"?) Feb 27, 2020 at 1:39
• @nnnnnn No; the answer is to either agree (defining if necessary) or fully understand the prevailing usage in the situation in which you're conversing, or (if you're lecturing, writing a book ...) explain your terminology. Then stick to it for the duration. The answer is not 'maybe', but 'yes' or 'no'. But we can't say which until the context is spelled out. Feb 27, 2020 at 12:37
It is a distinction that is made and should be made. To do otherwise is to deceive.
To say that Michael Bloomberg is a millionaire when he actually has a net worth over 50 billion dollars is such a misrepresentation that it would probably be considered a lie.
Also being a billionaire means you have passed a threshold.
https://www.harpersbazaar.com/uk/culture/culture-news/a25166069/jk-rowling-earnings-worth/
According to the 2018 Sunday Times Rich List, Rowling's net worth is estimated at around 700 million. She famously lost her billionaire status, due to the amount of money she's donated to charity.
| 794
| 3,248
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.84375
| 3
|
CC-MAIN-2024-22
|
latest
|
en
| 0.961736
|
https://www.understood.org/en/articles/10-multisensory-techniques-for-teaching-math?_sp=20501afa-d39c-4cc0-9dc4-0f0c361f63e3.1658343596857
| 1,725,893,889,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651103.13/warc/CC-MAIN-20240909134831-20240909164831-00385.warc.gz
| 973,631,465
| 49,636
|
# 10 multisensory techniques for teaching math
Multisensory instruction isn’t just for reading. It can also help kids who have trouble with math. The use of sight, touch, hearing, and movement can make it easier to understand what numbers and symbols represent. These 10 multisensory techniques can help kids learn math.
## Visualizing with beads or cereal
Using beads, dried beans, or cereal as is a great way to have kids represent math operations. For instance, kids might solve an addition sentence by adding two sets of beads together. Or they might find out how much is left after subtracting some beads. Kids can also group together different amounts of the items for multiplication and division.
By moving these items around and seeing how the quantities change, kids have a concrete way of understanding how these math operations work. Manipulatives can also help kids develop number sense and understand amounts.
## Building with colored cubes and tiles
Kids can use cubes or tiles to build shapes. This gives them a concrete idea of the measurement and properties of the figures they create.
Tiles and cubes also work great when teaching number patterns and operations. For instance, you can stack cubes in groups of 2, 4, 6, and 8. Then ask kids to build the next stacks in the pattern, adding two cubes each time (10, 12, and so on). After the pattern is complete, help kids make the connection between the stacks and the numbers they represent.
## Drawing math problems
Drawing math problems is a good next step after working with hands-on materials like beads or colored tiles. It’s a way for kids to show their thinking — and it takes them one step closer to writing number sentences with numerals and symbols.
For instance, ask kids to solve the multiplication problem 4 × 6 by drawing 6 groups of 4 apples. Or kids can color in 4 rows of 6 square units on graph paper. When they’re finished, they’ll see 4 groups of 6, or 24 square units colored in.
## Tapping out numbers
The act of tapping out numbers can help kids connect symbols to actual amounts, and “feel” the value. This is especially useful for working with multiples.
For instance, ask kids to list multiples of 4. They begin tapping sets of 4, counting as they go. Every fourth number gets a louder tap and is written down (“1, 2, 3, 4! 5, 6, 7, 8! 9, 10, 11, 12!”). In the end, kids have a list they can use to answer multiplication and division problems.
## Making musical connections
There are many ways to connect math and music. Kids can use songs to help memorize math rules, for instance.
Playing musical notes can help kids learn grouping or fractional parts. You can play one note on a keyboard and hold it for a count of four. This is the “whole note.” Next, you can ask, “How many quarter notes make a whole note?” After some discussion, kids can play four short notes that total the same length as the whole note.
## Putting movement into math
Working movement into math practice is an engaging way to help kids retain what they’ve learned. There are many ways to do this. For instance, kids can demonstrate angles by rotating their body while standing in a hula-hoop.
Here’s another example. Write numbers on the outside of a large ball. (These could be whole numbers, fractions, or decimals.) Pass the ball around. Each time someone catches it, they have to do a math operation with the two numbers their hands land on.
## Bundling sticks
One way to introduce kids to regrouping and place value is to have them bundle craft sticks together in groups of 10. (You can also use coffee stirrers.)
For instance, ask kids to solve 45 – 9 using sticks. By collecting 4 bundles of ten and 5 single (or “ones”) sticks, they can see how each place in the number 45 holds value. Then, to subtract 9, they need to break apart one of the bundles to make 15 individual sticks. After taking 9 out, this leaves 3 bundles and 6 sticks remaining, or 36.
## Building with base 10 blocks
These blocks come in different sizes that represent 1000s (a “cube”), 100s (a “flat”), 10s (a “long”), and 1s (a “unit”). Kids can form numbers with them to identify place value. (They can also use them to perform operations, show regrouping, and find patterns.)
For example, give kids the number 145 and ask them to “build” it using the blocks. Kids need to select one 100-block, four 10-blocks, and five 1-blocks. Then ask, “Which digit has the greatest value: 1, 4, or 5?”
## Creating a hundreds chart
A hundreds chart can help kids who struggle with math to see number relationships. For example, give kids a black-and-white 100s grid. (A grid is a large square broken into 100 smaller square units). Ask kids to shade in 1/4 of the whole grid. Then ask them to find the number of square units they colored in (25). The connection is that 1/4 is the same as 25 out of 100, or 25 percent.
## Using pizza slices
Cutting a pizza into slices is a great way to help teach fractions. You can make several pizzas out of construction paper, then cut them into slices of different sizes. This way, kids can “see” fractions like 1/8 or 1/4 by selecting slices of pizza. Using different colors for different size slices lets kids match equivalent fractions like 2/8 and 1/4. Kids can also combine slices to make a “whole” pizza pie!
| 1,240
| 5,316
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38
|
latest
|
en
| 0.918478
|
https://stats.stackexchange.com/questions/81861/how-to-learn-bayesian-network-structure-from-the-dataset
| 1,718,585,864,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198861674.39/warc/CC-MAIN-20240616233956-20240617023956-00690.warc.gz
| 497,290,991
| 39,823
|
# How to learn Bayesian Network Structure from the dataset?
I need to learn a Bayesian Network Structure from a dataset. I read the book titled "Learning Bayesian Networks" written Neapolitan and Richard but I have no clear idea.
According to the book from the data i can:
1) Create all the DAG Pattern, where a DAG Pattern is an equivalence class of DAG (in the respect of Markov Equivalence).
2) I can create all multinomial augmented bayesian newtowrk correlated to any of the equivalence class;
3) I use a score function to find the best multinomial augmented bayesian newtowrk;
Now i have not understood how to work this scoring function. In the literature, there is more than one? Can you help me understand precisely how to work the main scoring function?
I have also read that this research is hyper-exponential compared to the number of variables N, is that right? instead, there is some other method more efficient?
The score function measures whether the DAG structure that has been learnt is a good fit to the dataset. Of course, you can define the score function in several ways, depending on the dataset, and the ultimate objective of learning the DAG structure. One commonly used score function is the log-posterior.
Given dataset $D$ and a vector $\mathbf{X}$ of variables, the log posterior score function $S(D,G)$ is defined as $$S(D,G) := \log{p_{pr}(G)} + \log{p(D|G)}$$ where $p_{pr}$ is the prior over the DAGs. Let the set of parameters be $\theta \in \Theta$. $p(D|G)$ is the marginal likelihood $$p(D|G)= \int_{\Theta}{p(D|G, \theta) \cdot p_{pr}(\theta)d\theta}$$
The bnlearn R Package defines several score functions depending on the nature of the data (whether it is categorical, continuous or mixed).
### Categorical data (multinomial distribution):
• the multinomial log-likelihood;
• the Akaike Information Criterion (AIC);
• the Bayesian Information Criterion (BIC);
• a score equivalent Dirichlet posterior density (BDe);
• a sparse Dirichlet posterior density (BDs);
• a Dirichlet posterior density based on Jeffrey's prior (BDJ);
• a modified Bayesian Dirichlet for mixtures of interventional and observational data;
• the K2 score;
### Continuous data (multivariate normal distribution):
• the multivariate Gaussian log-likelihood;
• the corresponding Akaike Information Criterion (AIC);
• the corresponding Bayesian Information Criterion (BIC);
• a score equivalent Gaussian posterior density (BGe);
### Mixed data (conditional Gaussian distribution):
• the conditional Gaussian log-likelihood;
• the corresponding Akaike Information Criterion (AIC);
• the corresponding Bayesian Information Criterion (BIC).
For $n$ variables, the number of possible DAGs is super-exponential. Here is a link to the integer sequence. As you can see, the number grows very fast. https://oeis.org/A003024
There are a number of packages you can use in R.
One example that I am familiar with is bnlearn. A large number of the algorithms in this package use local search. This means that for the most part the procedure is:
1. Generate a random DAG structure
2. Score the structure using some methodology (in bnlearn it is by default AIC or BIC)
3. Score all neighbors of the randomly selected structure (meaning the same structure as the original structure but changed by one arch).
4. Select the neighbor with a better score as the next DAG structure to explore.
5. Stop when you have reached the optimal score.
This algorithm may converge to a local maximum.
| 785
| 3,502
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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}
| 3.125
| 3
|
CC-MAIN-2024-26
|
latest
|
en
| 0.782082
|
http://www.chess.com/forum/view/community/alarming-account?quote_id=3298540&page=2
| 1,386,561,838,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2013-48/segments/1386163870408/warc/CC-MAIN-20131204133110-00086-ip-10-33-133-15.ec2.internal.warc.gz
| 277,253,377
| 15,887
|
# Alarming account
• 4 years ago · Quote · #21
...almost. haha
• 4 years ago · Quote · #22
RetGuvvie98 wrote:
bigpoison wrote:
So does HECTARE.
Let's stick to the Queen's bloody English system, eh?
How long does it take a man and a mule to plow a hectare, anyway?
First, users would need to know how much land is in a hectare, so I submit the following from wikipedia:
Metric 10,000 square metres 0.01 square kilometre 1 square hectometre = 100 metres × 100 metres (a square with sides 100 metres long) 10 decares 100 ares 10,000 centiares
as to how long it would take to plow a square area 100 meters on each side, that would depend on whether the man or the mule is pulling the plow. (leave it to a chessnerd to think of that one)
If in the standard form, (mule pulling, man walking behind, holding the reins and guiding), then it would depend on age of the mule, size of the mule, whether the mule is fed/rested or has been is worn out from plowing lots of other land all month, as well as whether the land was ever previously plowed, disced and cultivated, or not, what had been planted there in the last season, as well as how deep you want it plowed. a few too many variables to give a good answer to you without some additional information.
You've forgotten one very important variable: composition of earth. Clay, loam, sand, rock?
It's funny that an acre is "supposedly" the amount of land a man and a mule can plow in a day. I've never bothered to consider all of these different variables.
• 4 years ago · Quote · #23
bigpoison wrote:
RetGuvvie98 wrote:
bigpoison wrote:
So does HECTARE.
Let's stick to the Queen's bloody English system, eh?
How long does it take a man and a mule to plow a hectare, anyway?
First, users would need to know how much land is in a hectare, so I submit the following from wikipedia:
Metric 10,000 square metres 0.01 square kilometre 1 square hectometre = 100 metres × 100 metres (a square with sides 100 metres long) 10 decares 100 ares 10,000 centiares
as to how long it would take to plow a square area 100 meters on each side, that would depend on whether the man or the mule is pulling the plow. (leave it to a chessnerd to think of that one)
If in the standard form, (mule pulling, man walking behind, holding the reins and guiding), then it would depend on age of the mule, size of the mule, whether the mule is fed/rested or has been is worn out from plowing lots of other land all month, as well as whether the land was ever previously plowed, disced and cultivated, or not, what had been planted there in the last season, as well as how deep you want it plowed. a few too many variables to give a good answer to you without some additional information.
You've forgotten one very important variable: composition of earth. Clay, loam, sand, rock?
It's funny that an acre is "supposedly" the amount of land a man and a mule can plow in a day. I've never bothered to consider all of these different variables.
can we get back to the point please
• 4 years ago · Quote · #24
Which is?
• 4 years ago · Quote · #25
TheGrobe wrote:
Which is?
that dark days are coming for chess.com as cheater_1 has returned!!!!!!
• 4 years ago · Quote · #26
I think maybe you're blowing it out of proportion. It's not cheater_1, it's a cheap facsimile, and even if it was what's one more troll...?
• 4 years ago · Quote · #27
I think discussing land cultivation is immensely more interesting than getting worried about whether or not somebody is going to start playing chess at chess.com in a month. 'Course that's just me.
• 4 years ago · Quote · #28
were a mule a sphere, then there would be no question as there is no place on a sphere to fasten a harness for a plow.
• 4 years ago · Quote · #29
RetGuvvie98 wrote:
were a mule a sphere, then there would be no question as there is no place on a sphere to fasten a harness for a plow.
Superglue!
• 4 years ago · Quote · #30
This has been attempted a couple times before, and yet all of them are lacking in several fundamental factors. Usually, it is the lack of a complex use of the English language, and of course, emphasising words at the most effective moments.
Yes, cheater_1 was a sad, misguided human being, but we have to admit he fulfilled his role of site imbecile rather well. And, it would appear, no-one is able to emulate him.
• 4 years ago · Quote · #31
Nytik wrote:
This has been attempted a couple times before, and yet all of them are lacking in several fundamental factors. Usually, it is the lack of a complex use of the English language, and of course, emphasising words at the most effective moments.
Yes, cheater_1 was a sad, misguided human being, but we have to admit he fulfilled his role of site imbecile rather well. And, it would appear, no-one is able to emulate him.
We need a new Cheater_1. The discussions were so fun :P (or someone else good at being a site imbecile)
• 4 years ago · Quote · #32
xbigboy wrote:
RetGuvvie98 wrote:
were a mule a sphere, then there would be no question as there is no place on a sphere to fasten a harness for a plow.
Superglue!
yes but then by the force of rolling the sphere would also tip over the plow and there for it will not move as the resistance is greater than force of the sphere and a harness allows the force through or other wise known as inertia there for the movement of the sphere is decelerated and there for needs more force to put it into a state of constant movement. However a better way to put the sphere in a force of motion then if you put it into space then if the sphere is big in enough then it will either attach a forward or backward motion around a planet or some other form of gravity to which it will be going very fast or if the sphere is is around the size of a very lage rock then is could bring it into geostationary orbit and then make the plow spin round either very slowy or very quickly and that is how you get make a sphere and a plow work at the same time.
although there is another problem in which you would need a way in which you can attach the plow up to the sphere. to do this you need a different form of harness to cause enough friction and inertia to move the plow around and the sphere as well so it needs to made out of a material that is cheap and easy to use also it needs to be able to manufacture quickly enough to work with the the sphere and the crops at the same time but it needs to be strong as the amount of force created will be great but the force of the sphere has to be greater. also it needs to be light weight and aerodynamic to cause maxium performance.
once all of these factors have been considered your can sucsessfully plow although you need to work out a new type of plowing Technic but i will explain that a bit later if you want
littlehotpot
• 4 years ago · Quote · #33
xbigboy wrote:
Nytik wrote:
This has been attempted a couple times before, and yet all of them are lacking in several fundamental factors. Usually, it is the lack of a complex use of the English language, and of course, emphasising words at the most effective moments.
Yes, cheater_1 was a sad, misguided human being, but we have to admit he fulfilled his role of site imbecile rather well. And, it would appear, no-one is able to emulate him.
We need a new Cheater_1. The discussions were so fun :P (or someone else good at being a site imbecile)
We've got one.
• 4 years ago · Quote · #34
bigpoison wrote:
xbigboy wrote:
Nytik wrote:
This has been attempted a couple times before, and yet all of them are lacking in several fundamental factors. Usually, it is the lack of a complex use of the English language, and of course, emphasising words at the most effective moments.
Yes, cheater_1 was a sad, misguided human being, but we have to admit he fulfilled his role of site imbecile rather well. And, it would appear, no-one is able to emulate him.
We need a new Cheater_1. The discussions were so fun :P (or someone else good at being a site imbecile)
We've got one.
Who?
• 4 years ago · Quote · #35
recathe_1
thecare_1
hearcet_1
theecar_1
1heat_rec
• 4 years ago · Quote · #36
ilikeflags wrote:
recathe_1
thecare_1
hearcet_1
theecar_1
1heat_rec
oh
by the way do you think my theory will work
• 4 years ago · Quote · #37
littlehotpot wrote:
bigpoison wrote:
xbigboy wrote:
Nytik wrote:
This has been attempted a couple times before, and yet all of them are lacking in several fundamental factors. Usually, it is the lack of a complex use of the English language, and of course, emphasising words at the most effective moments.
Yes, cheater_1 was a sad, misguided human being, but we have to admit he fulfilled his role of site imbecile rather well. And, it would appear, no-one is able to emulate him.
We need a new Cheater_1. The discussions were so fun :P (or someone else good at being a site imbecile)
We've got one.
Who?
Browse the forums for a while and you'll figure it out. Here's a hint: ilikeflags and this fella are good buddies.
• 4 years ago · Quote · #38
1re_cheat
c1eat_hec
eat_1cher
• 4 years ago · Quote · #39
ilikeflags wrote:
1re_cheat
c1eat_hec
eat_1cher
how many more people are there on chess.com where there member name is an anagram of cheater_1?
• 4 years ago · Quote · #40
Uh, guys... Maybe he IS for real...
Back to Top
| 2,444
| 9,359
|
{"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}
| 3.34375
| 3
|
CC-MAIN-2013-48
|
latest
|
en
| 0.953581
|
http://learnenglish.britishcouncil.org/fr/grammar/intermediate-to-upper-intermediate/conditionals-2?page=31
| 1,596,953,097,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-34/segments/1596439738425.43/warc/CC-MAIN-20200809043422-20200809073422-00227.warc.gz
| 61,246,107
| 19,180
|
# Conditionals 2
Do you know how to use third and mixed conditionals?
Look at these examples to see how third and mixed conditionals are used.
We would have walked to the top of the mountain if the weather hadn't been so bad.
If we'd moved to Scotland when I was a child, I would have a Scottish accent now.
If she was really my friend, she wouldn't have lied to me.
Try this exercise to test your grammar.
Grammar test 1
Conditionals 2: Grammar test 1
## Grammar explanation
Do you know how to use third and mixed conditionals?
### Third conditionals and mixed conditionals
Conditionals describe the result of a certain condition. The if clause tells you the condition (If I hadn't been ill) and the main clause tells you the result (I would have gone to the party). The order of the clauses does not change the meaning.
If I hadn't been ill, I would have gone to the party.
I would have gone to the party if I hadn't been ill.
Conditional sentences are often divided into different types.
### Third conditional
The third conditional is used to imagine a different past. We imagine a change in a past situation and the different result of that change.
If I had understood the instructions properly, I would have passed the exam.
We wouldn't have got lost if my phone hadn't run out of battery.
In third conditional sentences, the structure is usually: If + past perfect >> would have + past participle.
### Mixed conditionals
We can use mixed conditionals when we imagine a past change with a result in the present or a present change with a result in the past.
#### 1. Past/Present
Here's a sentence imagining how a change in a past situation would have a result in the present.
If I hadn't got the job in Tokyo, I wouldn't be with my current partner.
So the structure is: If + past perfect >> would + infinitive.
#### 2. Present/Past
Here's a sentence imagining how a different situation in the present would mean that the past was different as well.
It's really important. If it wasn't, I wouldn't have called you on your holiday.
And the structure is: If + past simple >> would have + past participle.
Do this exercise to test your grammar again.
Grammar test 2
Conditionals 2: Grammar test 2
### Language level
Intermediate: B1
Upper intermediate: B2
Thanks Kirk for your help! Much more clear now!
Regards!
Hello everybody
I have a serious doubt with mixed conditionals. Could anybody tell me if we can use "should" into a 3rd conditional phrase?
e.g: If she had taken the second choice, she should have told the manger
Hello gaullanos,
No, this sentence doesn't make sense. Please note that 'should have' is different from 'should' - 'should' refers to the present or future, whereas 'should have' refers to the past.
Depending on what you want to say, you might want to say 'If she took the second choice, she would have to tell the manager' or 'If she had taken the second choice, she would have had to tell the manager'. Does either of these make sense?
I hope this helps.
Best wishes,
Kirk
The LearnEnglish Team
Thank you very much for your answer and for your help.
If he didn’t have to work tomorrow he wouldn’t be so miserable today.
is it right to say " if he will not have to work tomorrow....." ?
is there a specific structure of the mixed conditional?
hi i have a qshn may u please help me ?
what is the answer and more important why ?
I preferred that she.....here
1=stayed 2=has stayed 3=staying 4=stay ?
dont we use s after she ? i mean it must be stays but there is no such a thing!
4 stay
Hello StrangeGr,
We don't answer exercise questions for users here - that is your job! However, I can tell you that 'prefer' can be followed by several structures:
prefer to do sth
prefer sb to do sth
prefer doing sth
prefer sb doing sth
prefer (that) sb do sth ['do' is the subjunctive form; it is also possible to use 'does' here in modern English]
prefer (that) sb did sth
Best wishes,
Peter
The LearnEnglish Team
hi i am new here , i want to improve my english ,i heard somewhere that by regular communication you can learn any language ,so that's why i want to talk to some one.
Hello....
I'm here too..
I want have a discussion with some one.
If you ready,it's ok.
| 1,002
| 4,240
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-34
|
latest
|
en
| 0.978515
|
http://mathhelpforum.com/pre-calculus/217009-log-simplification.html
| 1,508,402,232,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-43/segments/1508187823255.12/warc/CC-MAIN-20171019065335-20171019085335-00607.warc.gz
| 215,133,878
| 10,984
|
1. Log Simplification?
I am stuck on something that shouldn't be particularly complicated.
I have logx(xlogyy5)
Should I get 1+logx5? This doesn't look right to me.
Instructions much appreciated.
2. Re: Log Simplification?
I think your approach is right. In case you find a discrepancy you may paste the same.
3. Re: Log Simplification?
Originally Posted by Quixotic
I am stuck on something that shouldn't be particularly complicated.
I have logx(xlogyy5)
Should I get 1+logx5? This doesn't look right to me.
Instructions much appreciated.
It's correct.
-Dan
4. Re: Log Simplification?
Hello, Quixotic!
$\text{I have: }\:\log_x(x\log_yy^5)$
$\text{Should I get: }\:1+\log_x5\,?$ . Yes!
$\log_5(x\log_yy^5) \;=\;\log_5(x\cdot 5\log_yy) \qquad\quad \text{Note that: }\log_yy =1$
. . . . . . . . . . $=\;\log_5(x\cdot 5\cdot 1)$
. . . . . . . . . . $=\;\log_5(5x)$
. . . . . . . . . . $=\;\log_55 + \log_5x \qquad\quad\text{Note that: }\log_55 = 1$
. . . . . . . . . . $=\; 1 + \log_5x$
5. Re: Log Simplification?
Thanks everyone!
| 364
| 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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43
|
longest
|
en
| 0.803068
|
https://ourcodingclub.github.io/tutorials/brms/
| 1,702,041,164,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-50/segments/1700679100745.32/warc/CC-MAIN-20231208112926-20231208142926-00751.warc.gz
| 495,988,697
| 16,363
|
# Bayesian modelling using the brms package
## From research question to final report, unleashing the full potential of brms
Have you ever thought frequentist statistics were confusing? Have you ever felt like your mind was getting lost between p-values, random 0.05 thresholds, and confidence intervals that had little to do with confidence? But do you still need to use statistics for your degree and are looking for more straightforward methods? Then you are in the right place! These are questions and topics that I have also struggled with, and I found Bayesian models to be easier to deal with. Hopefully, you will too after completing this tutorial.
The following tutorial is an introduction to Bayesian modelling but it assumes a prior understanding of modelling and data distribution. If you are just getting started with R coding, you should check out this introduction tutorial from the Coding Club. To make sure everything is clear in your mind, you can also check out these tutorials as well beforehand.
• Meta-analysis for biologists using MCMCglmm (an introduction to the MCMCglmm package) available here
• Generalised linear models in Stan (using the Rstanarm and brms packages to run Stan models) available here
This tutorial should teach you how to create, assess, present and troubleshoot a brm model.
All the files you need to complete this tutorial can be downloaded from this repository. Click on `Code/Download ZIP` and unzip the folder, or clone the repository to your own GitHub account.
# Tutorial Structure:
1. All you need to know about Bayesian stats
2. Building a simple model
3. Extracting results and assessing the model
4. Building up the complexity
6. Potential issues and how to solve them
# All you need to know about Bayesian stats
Lets start with a little theoretical explanation of Bayesian statistics. This method was invented by Reverend Bayes in the 1770s and published after his death. The Bayes theorem was revolutionary in that it introduced the possibility to calculate conditional probabilities, basically the probability of an event happening while knowing that another event already happened. Say you are taking part in a raffle where they pick out numbers between 0 and 100. Let’s say you want to know the probability of getting a number below 50. Now, we could calculate that probability, but I don’t want to confuse you with math formulas (and it’s not necessary to understand this concept). But we would get a certain probability of this event happening (the event being picking out a number between 0 and 49). So, now imagine that same situation, but you also get the info that all numbers in the box are bigger than 80. And now your understanding has changed because you acquired prior information, and you can say that there is absolutely no chance of getting a number between 0 and 49. So this is the main idea behind Baye’s theorem: prior knowledge will influence the probability of an event taking place.
But when we look at Bayesian models, things take place on a bigger scale. The prior information in your model is going to be a distribution of probabilities rather than just one probability. This is because the event turned into multiple events. To make this a little more concrete, we can take an example: if you are looking at the abundance of a species, the “event”” will be that you find a specific value of abundance for that species on a specific day, so the probability of getting an abundance of 50 for example. And if you measure abundance every month for 5 years, you get 60 values of abundance for this species, which will represent 60 “events”. So if we plot all those measures of abundance, and the number of times you got those measures, we get a distribution of the probability of getting those measures, which could look like this:
Now this example brought us back to our first situation, where we know nothing about the system (the species and their abundance), apart from what our data tells us. But what if our data is skewed, what if we measured the wrong thing, what if we made mistakes? Just like with the raffle, prior understanding of the system can help us make a more informed guess, and hopefully overcome those weaknesses.
The main idea behind prior distributions, is that the data you have collected, the abundance of our species around the world, is biased by your sampling methods or any other issue you might have encountered. Maybe, you weren’t able to sample younger individuals, or you sampled a proxy for abundance instead of actual counts. Therefore, the data that you have can be considered incomplete compared to the reality. However, if somebody before you sampled this species in a different way, and found that the population has a specific distribution (for example a large birth rate but very few individuals that reach adulthood) you can add this information into your model, and fill the gap in your data. The model is going to take your data, take this additional information (called prior distribution), and create a new spread of data that should reflect the distribution of your species’ population in the real world (called posterior distribution).
So if we look at this in a graph, it would look like this:
The light green line represent the data you sampled, the blue line represents the prior distribution, and the dark green line represents the posterior distribution that your model created. If you are very keen, you can check out the math behind the theory explained here but it is not necessary to understand and carry out this tutorial.
The important bit, is that the model is going to create a posterior distribution for your whole abundance data, and then a specific distribution of all the possible values for each of your variables (time and abundance and any other one you want to look at). These posterior distributions all have a mean and a standard error. And the mean value of the distribution of your explanatory variable (time in our example) can be used to represent the slope of the effect of time on you response variable (abundance in our example).
So, compared to other statistical methods (like a linear model for example), where the model finds the best fit line between two variables using only your data, the Bayseian model is going to find the “real” distribution of your data, and give you the most probable value of the estimate of the effect of one variable on another.
Hopefully, you understand this bit of theory, but if you don’t that’s also fine! You can come back to this later and check out the references at the end to get a different explanation as well.
# Building a simple model
Let’s start our analysis!
Start by opening Rstudio and setting the working directory with the file path leading to the folder you just downloaded, and load a first package.
``````# Set the working directory
setwd("your_filepath")
library(tidyverse)
``````
We can now load the data. This dataset is a subset of the LPI dataset available in its entirety here. Today, we’ll be looking at red knot counts from a study carried out in France, on the Atlantic coast and the Channel coast.
``````# Load the data
``````
And we can start by checking out what our data look like .
``````head(France) # to get the first observations in each column
str(France) # what type of variables do we have
``````
As for every modelling exercise, we first need a research question to focus on. For this tutorial, we’ll look at a simple one: Has the red knot population in France increased over time?
In other words, we will be looking at the effect of time, our explanatory variable or fixed effect, on the abundance of red knots, our response variable. These will be the parameters we include in our model.
## Data distribution
Now that we know what variables to look at, another important information to include in the model is the type of distribution of the data. To find out which it is, we can plot our data as a histogram.
``````(hist_france <- ggplot(France, aes(x = pop)) +
geom_histogram(colour = "#8B5A00", fill = "#CD8500") +
theme_bw() +
ylab("Count\n") +
xlab("\nCalidris canutus abundance") + # latin name for red knot
theme(axis.text = element_text(size = 12),
axis.title = element_text(size = 14, face = "plain")))
``````
This histogram should look something like this:
The data we have is abundance data also known as count data. This means the numbers we will include in the models are restricted in some way, they can’t be negative and they are full numbers as well. We can see that our data follows a poisson distribution, and this is one of the things we’ll have to tell our model.
Now looking at our variables,
``````unique(France\$year)
``````
We can see here that our data starts in 1976, and ends in 2010. This is something we will have to include in the model as well. If we just include `year`, the model will start at 1976, but really we want the model to read this as Year 1. See the next part for the exact syntax to avoid this.
## First model
To answer our research question, we will create a Bayesian model, using the brms package. If you don’t have this package already installed, uncomment the first line.
``````# install.packages("brms")
library(brms)
``````
We can now write our model. The `brms` package sometimes gets hidden by the `stats` package, so it’s always better to include `brms::brm` to call the modelling function. The first argument in the brackets is the response variable (red knot abundance or `pop` for us) and the variables placed after the `~` sign are the fixed and random effects, our explanatory variables (time or `year` for us).
As explained earlier, we want to change the year variable to start at one. We can do that in the model by using: `I(year - 1975)` . “I” specifying integer, and “year-1975” to make the year variable start at 1. NB: Another way to do this is to make a new column with year the way you want it to be read in the model, instead of specifying it during the model. ```France <- France %>% mutate(year_2 = I(year - 1975))```
The family argument corresponds to the distribution of our data, and as we saw earlier, that should be `poisson`. You can look at the `brmsfamily` R Documentation page to find the other family options and their characteristics. N.B: Setting the family argument to poisson log-transforms our data (just something to keep in mind for later).
The `iter` argument defines how many times you want the model to run. The Bayesian model runs many times by picking random values and assessing how the distribution changes and fits the data, before deciding on a perfect fit of the posterior distribution (which should end up bring a mix your data and the prior distribution). The higher the number of iterations, the longer it will take for the model to run.
The `warmup` argument refers to the number of first iterations that the model should disregard (or chuck out) before creating the posterior distribution. This is done to make sure that the first random number that the model choses doesn’t influence the final convergence of the model. The `chains` argument defines the number of independent times the model will run the iterations. Again, this is done to insure thorough exploration of all the possible values for our posterior distribution.
Additionally, you should note that we haven’t added a prior distribution in this model. This doesn’t mean that the model doesn’t use any. The brm function has a default prior that is very uninformative (pretty much flat so it won’t change you data a lot). This is useful if you don’t have any prior information to give to the model, and this means that your posterior distribution will be very close to the distribution of your data. We will be using this default prior for now to understand how the model works.
``````france1_mbrms <- brms::brm(pop ~ I(year - 1975),
data = France, family = poisson(), chains = 3,
iter = 3000, warmup = 1000)
# saveRDS(france1_mbrms, "france1_mbrms.RDS")
# you can save the model as an RDS (Rdata) that way you don't need to run the model again if you come back to this code
``````
BE AWARE, a brms model always takes a while to run, and these two messages will often appear before it starts sampling, “Compiling Stan program…recompiling to avoid crashing R session”, but everything is still okay!
# Extracting results and assessing your model
## Summary of the model
By running this line of code, we can extract the summary of the model, just like you would get the results of an ANOVA for example.
``````summary(france1_mbrms)
# fixef(france1_mbrms) # to get more detailed values for estimates
# coef(model_name) # if you have group-level effects (hierarchical data)
``````
The Console should look something like this. Now this may seem a little confusing for now, but just wait, you should be able to understand all of it in a bit.
Important note: Because of the stochastic nature of Bayesian statistics, every time you (re)run a model, your output will be slightly different, so even if you use the same effects in your model, it would always be slightly different to whatever was printed here. TLDR; Do not worry if your results do not exactly match the below image!
The top of the summary output is simply a recap of the model we ran (you can look at it if you don’t remember which model this was, but we are going to skip this).
The interesting part is what is written under Population-Level Effects. The model gives us an `Estimate` aka the mean of our posterior distribution for each variable. As explained earlier, these estimates can be used as the intercept and slope for the relationship between our two variables. `Est.Error` is the error associated with those means (the standard error).
The other important part of that summary is the 95% Credibility Interval (CI), which tells us the interval in which 95% of the values of our posterior distribution fall. The thing to look for is the interval between the values of `l-95% CI` and `u-95% CI`. If this interval is strictly positive or negative, we can assume that the effect is significant (and positive or negative respectively). However, if the interval encompasses 0, then we can’t be sure that the effect isn’t 0, aka non-significant. In addition, the narrower the interval, the more precise the estimate of the effect.
In our case, the slope 95% CI does not encompass 0 and it is strictly positive, so we can say that time has a significantly positive effect on red knot abundance.
## Assessing model fit
Now that we have our results, we should assess how our model converged and if it fits the data well.
If we just look at our summary from earlier, we already get a bit of information about this. The `Bulk_ESS` a,d `Tail_ESS` are the effective sample size measures for each parameter. These should be high (>1000) to be correct (which is the case for our model). Secondly, the `Rhat` values for each effect should be equal to 1 if the model converged well. For now, everything looks good in our model.
Another way to assess convergence is to use the `plot` function.
``````plot(france1_mbrms)
``````
This should show up like this. We call this the trace or caterpillar plots. If you focus on the right hand plots, you want to see a sort of fuzzy caterpillar, or a festive tinsel. If this is the case, it means your model explored all the possible values it could look at, so it converged well. On the x-axis of those trace plots, we have the iterations done after the warmup (so 3000-1000 = 2000 in our case). And on the y-axis are all the values of the mean of the posterior distribution that have been assessed by our model.
On the left side, the density plots shows all of those mean values again, plotted by the amount of times the model got this value (so the distribution of means basically). And if you look closely, the mean of this density plot is going to be the mean value that has been found by the model most often, so probably the most “correct” one. And that value should be very close to the actual estimate that the summary function gave us. In our case, the top plot is the intercept and that density plot seems to be centered around 8.70, which is the estimate value that we got in the summary!
The second plot you want to look at is the `pp_check` plot. The main use of this function is to check if you model predicts your data accurately (using the estimates). If it does, then you can use that model to generate new data and make accurate predictions.
``````pp_check(france1_mbrms) # posterior predictive checks
``````
The thin light blue lines on this plot represent 10 random draws or distributions created by the model (you can increase this by including `ndraws = 100` in the code). The dark blue line represent the posterior distribution (which is considered to fit our data well so it can be used to compare model predictions with reality). As you can see here, the two distributions look similar so everything is good.
# Building up the complexity
Now that we know how to make a very basic model, we can start adding complexity little by little.
We know that our red knot population grows over the years, but it could be that each year, the previous population level has an effect on the next year. This means that the population could be growing due to random variations every year, rather than throughout a whole time period. In the brms package, you can include random effects very easily by adding ` + (1| random variable)`. Here we can just use the variable “year” because random effects will automatically become factors.
``````france2_mbrms <- brms::brm(pop ~ I(year - 1975) + (1|year),
data = France, family = poisson(), chains = 3,
iter = 3000, warmup = 1000)
summary(france2_mbrms)
plot(france2_mbrms)
``````
As we can see in the plot, the model converged well. The summary doesn’t show an estimate of the effect for random variables, but it accounts for it during the sampling. We can still see that there is no effect of year as a random variable because our estimates have not changed compared to the first model.
If we look at the data further in detail…
``````unique(France\$Location.of.population) # observations come from 2 locations
``````
…we can see that the observations come from two different locations: the Atlantic coast and the Channel coast, this is something that we will have to account for in our model.
Whenever your data distribution is grouped or separated into categories, you should include that information in your model to check if the groups are significantly different. In our case, those two locations correspond to two different bodies of water, which may support different numbers of red knot individuals.
If we check this by plotting the data…
``````(boxplot_location <- ggplot(France, aes(Location.of.population, pop)) +
geom_boxplot() + # could be a significant effect between locations so should look at that
theme_bw() +
xlab("Location\n") +
ylab("\nCalidris canutus abundance") +
theme(axis.text = element_text(size = 12),
axis.title = element_text(size = 14, face = "plain")))
``````
Your boxplot should look something like this, and you can see that there is a difference between our two sampling sites. By including this categorical variable into our model, we can check if this difference is significant.
As a side note, we will be including location as a fixed effect because we only have 2 locations. If you want to include it as a random effect, your variable should have at least 5 “levels” or categories.
The code for the model would look like this.
``````france3_mbrms <- brms::brm(pop ~ I(year - 1975) + Location.of.population,
data = France, family = poisson(), chains = 3,
iter = 3000, warmup = 1000)
summary(france3_mbrms)
plot(france3_mbrms)
pp_check(france3_mbrms)
``````
Now if we look at our model plots we can see it converged well and it fits the data even more than the previous models.
The summary also tells us that the effect of location is significant! The estimate is -0.06 for `Location.of.populationChannelCoast`. This means that the Channel coast population has a significantly lower abundance than the Atlantic coast population.
## The LOO method to assess fit
Another assessment you can do for your model is to look at the leave-one-out cross validation (LOO) method. The LOO assesses the predictive ability of posterior distributions (a little like the `pp_check` function). It is a good way to assess the fit of your model. You should look at the `elpd` estimate for each model, the higher value the better the fit. By adding `compare = TRUE`, we get a comparison already done for us at the bottom of the summary. The value with an elpd of 0 should appear, that’s the model that shows the best fit to our data.
``````loo(france1_mbrms,france2_mbrms, france3_mbrms, compare = TRUE)
``````
Since the third model shows the best fit, this is the one we will focus on. And now, we can move on to presenting our results in a report!
Although the code for the model and the summary output are interesting on their own, they might be little hard to understand. A good graph and figure legend can present your findings in a much clearer way.
## Plotting the model
The main plot we would want to present in a report is the relationship between our two main variables (abundance and time), basically the line created with the intercept and slope values from our summary output. In addition to that line, we can also add the credibility interval, because that shows the confidence that we have in that estimate. And finally, adding the raw data points (abundance counts every year), we can show how well the model fits the original data.
In this long and seemingly complex piece of code, we are using our original data and adding the posterior distribution through a pipe. Once this is done, we can plot the raw data, add the regression line and the credibility interval. The rest is just making it pretty.
``````library(tidybayes)
(model_fit <- France %>%
ggplot(aes(x = year, y = pop)) +
stat_lineribbon(aes(y = .prediction), .width = c(.95, .80, .50), # regression line and CI
alpha = 0.5, colour = "black") +
geom_point(data = France, colour = "darkseagreen4", size = 3) + # raw data
scale_fill_brewer(palette = "Greys") +
ylab("Calidris canutus abundance\n") + # latin name for red knot
xlab("\nYear") +
theme_bw() +
theme(legend.title = element_blank(),
legend.position = c(0.15, 0.85)))
``````
Now that you have your plot, you can save it and add it in your report, with an informative figure caption (for example Fig 1: The abundance of red knot birds in France significantly increased between 1976 and 2012 (β=0.04, 95% CI=0.04-0.04)).
``````# ggsave(filename = "france3_fit.png", model_fit, device = "png")
``````
Another useful plot would be one showing the trendline in each location, since we saw there was a significant difference between the two.
`````` (location_fit <- France %>%
group_by(Location.of.population) %>%
ggplot(aes(x = year, y = pop, color = ordered(Location.of.population), fill = ordered(Location.of.population))) +
stat_lineribbon(aes(y = .prediction), .width = c(.95, .80, .50), alpha = 1/4) +
geom_point(data = France) +
scale_fill_brewer(palette = "Set2") +
scale_color_brewer(palette = "Dark2") +
theme_bw() +
ylab("Calidris canutus abundance\n") +
xlab("\nYear") +
theme_bw() +
theme(legend.title = element_blank()))
``````
## Reporting the results in a report
An important thing to remember, is that when you are trying to report the results by going back to the original units (reporting the abundance change in terms of number of bids for example), you might have to transform your estimate a little.
This is because the model transforms your data depending on the distribution of your values. If you used a gaussian (normal) distribution, you won’t need to worry about this. But if you used a poisson distribution like we did, the model will have log-transformed your data. This means you won’t be able to report it in the original units if you don’t back-transform it again.
This isn’t very hard to do. In our case, if we want to report the number of new birds every year, using the estimates of our first model, we just need two small steps:
• First, get the actual value of increase by adding the mean to the intercept estimate (this is necessary because we know our population doesn’t start at 0.04 but rather at 8.70) = 8.74
• Second, get the exponential of that value to undo the log-transformation that our data went through to get an estimate in number of birds = 6247.896
We can now say that there were on average 6247.896 new red knot birds in France every year between 1976 and 2010.
# Potential issues and how to solve them
A lot of small thing can cause big problems in the models we are using here. A common warning message about divergent transitions, for example: “There were 132 divergent transitions after warmup. Increasing adapt_delta above 0.8 may help. See http://mc-stan.org/misc/warnings.html#divergent-transitions-after-warmup” A few divergent transitions can be ignored, but the higher the number the more concerning. Going to the website R suggests will give you more information on what this error means. They suggest increasing `adapt-delta` about 0.9 and increasing `max_treedepth` beyond 12.
``````#france3_mbrms <- brms::brm(pop ~ I(year - 1975) + Location.of.population,
# data = France, family = poisson(), chains = 3,
# iter = 3000, warmup = 1000,
# control = list(max_treedepth = 15, adapt_delta = 0.99)
``````
However, there are other ways to adjust your model to avoid these issues.
### Solution 1: Scaling the variable
Another way of transforming your data to help the model deal with it, is to scale your variables. Scaling changes your data by centering it on 0 (mean = 0) and changing the values to have a standard deviation of 1.
``````France\$year.scaled <- scale(I(France\$year - 1975), center = T) # scaling time
France\$pop.scaled <- scale(France\$pop, center = T) # scaling abundance
``````
The other thing that changes here is the distribution of the data, from poisson to normal, which you will have to change in the model as well
``````hist(France\$pop.scaled) # you can see that the distribution changed
# so will have to change it in the model as well
# france4_mbrms <- brms::brm(pop.scaled ~ year.scaled + (1|location),
# data = France, family = gaussian(), chains = 3,
# iter = 3000, warmup = 1000)
``````
If you are interested in learning more about scaling data, check out our scaling tutorial.
### Solution 2: Changing the units of the variable
A lot of abundance data can have very large numbers. Say if the smallest value of abundance in your data is 50,000. This can cause a problem when the model runs iterations because it will be looking at a whole range of values between 0 and 50,000 even though nothing is interesting there. The model might fail to converge properly or take a very long time to do so.
A way to solve this is to change the units from single counts to thousands of counts for example.
As I explained earlier, we used non-informative, default priors in the previous models. However, increasing the information you give to your model will probably help it converge faster. The prior information you give depends highly on the understanding you have of the specific systme you are working on, but here is the way you would include a prior in your model.
First, you would need to define the prior, by including the parameters of a new distribution of your data. The `normal` or `cauchy` arguments describe the shape of that distribution, and the numbers in the brackets describe the width and height of that shape (in this order: (mean, standard deviation)). You can set a prior for each variable that you want to include in your model. And here is a random example of what that would look like for our model:
``````prior1 <- c(set_prior(prior = 'normal(0,6)', class='b', coef='year'),
# global slope belongs to a normal distribution centered around 0
set_prior(prior = 'normal(0,6)', class='Intercept', coef=''))
# global intercept
set_prior(prior = 'cauchy(0,2)', class='sd'))
# if we had group-level intercepts and slopes
# france5_mbrms <- brms::brm(pop ~ year + location, data = France,
# family = poisson(), chains = 3, prior = prior1,
# iter = 3000, warmup = 1000)
# The intercept here will be very different than your previous models, but that is because we are using the "year" variable and not the adjusted year variable, but you will see that the fixed effects look the same. You could change this by making a new column where the year variable to starts at 1 and using that to specify the priors and in the model.
``````
As you can see in the comments part above, the prior would be included in the model with a `prior = prior1` argument.
### Solution 4: Increasing iterations
Finally, increasing the number of iterations by a few thousands (and the warmup accordingly) might also help your model converge better by letting it run for longer.
Et voilà! This is the end of the tutorial, I hope you managed to understand everything and increase your knowledge about Bayesian modelling even a little. Take the time to go back to the theory part if you are interested and look at the other Coding Club tutorials on that topic as well to test your understanding.
In this tutorial you learned:
• How a Bayesian model works and what is the theory behind it
• How to create a simple model using the brms package and extract the results
• How to assess the convergence and fit of this model
• How to present your results in a report
• How to build a more complex model using the brms package
• Some solutions in case your model doesn’t converge well
| 6,826
| 29,966
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.453125
| 3
|
CC-MAIN-2023-50
|
latest
|
en
| 0.943018
|
https://myhomeworkgeeks.com/stats-case-study-projectanswers-1bids-1other-questions-10stats-case-study-projectanswers-1bids-1other-questions-10/
| 1,642,729,453,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-05/segments/1642320302715.38/warc/CC-MAIN-20220121010736-20220121040736-00086.warc.gz
| 483,561,443
| 13,229
|
# stats case study projectAnswers 1Bids 1Other questions 10,stats case study projectAnswers 1Bids 1Other questions 10
1. Develop 10-page research paper using the attached dataset (Prescott_Parcel.xslx)2. Assignment possible points: 1502. Paper should comprise of the following structures:- Introduction – Paper must define the problem scope and outline the objectives of the analysis in a clear, concise, and professional manner (1 page)- Data Presentation: Paper must present a summary of data in an accurate and professional manner, and provide accurate interpretation of the data. Make sure you must provide a table showing summary statistics (descriptive statistics) of selected variables in your regression model and provide a brief description of those data. (1 page)- Model Specification & Development: Paper must specify the appropriate linear regression model in an accurate and professional manner. Provide accurate equations for selected linear regression model(s) with the detail description of the notation of each variable. If you estimate more than 2 regression models, provide the detail justifications for doing it(2 pages)- Interpretation of Results: Paper must provide a clear, concise, consistent and accurate summary of results from linear regression model(s). Discuss the statistical significance of each variable, and justify the sign of each variable. Also, discuss the meaning of slope for each variable. The interpretations must be understandable to the general public. (2.5 pages) – Also keep in mind that 1E-10=0.0000000001- Policy Recommendations & Conclusions: Paper must address policy recommendations and/or implications in the business world in a professional manner. In other words, students must provide a clear recommendation and plan of action that government agency or property developers must take in their future decision making process, based on the results from linear regression model(s). (2.5 pages)- Integration of Biblical Principle: Paper must identify and integrate biblical principles into the decision making process in an appropriate and insightful manner (1 page)
Haven’t Found The Relevant Content? Hire a Subject Expert to Help You With
stats case study projectAnswers 1Bids 1Other questions 10,stats case study projectAnswers 1Bids 1Other questions 10
Why Hire a Professional Essay Writer from MyHomeworkGeeks
Quality Urgent Essays
Domyhomeworkk online platform provides the best essay writing service ever received elsewhere, thanks to our professional essay writers. Experience gained in the 10+ years of being in the assignment writing industry is also an added advantage.
A lot is done to maintain us as the best essay writing service provider. Besides owning a degree from a recognized university, a writer must pass the rigorous tests we take them through before they are considered eligible to offer urgent essay help on our website.
Affordable Urgent Assignment Help
Who said a reliable website should not be cheap also? While it is human nature to associate prices with quality, that does not always work. At domyhomeworkk.com, we guarantee you cheap and urgent essay help without compromising the quality.
24/7 Available Urgent Essay Writing Service
Whether it is a ‘write my essay for me cheap’ or ‘edit my essay’ order you want to request, always feel free to reach us at any time of the day. Our experts work in shifts to ensure that you access urgent essay writing services at any time of the day.
Plagiarism Free Essays Online
The punitive measures that come with plagiarized content are so harsh, and we understand that. We consequently ensure to write each client’s papers from scratch no matter how urgent the essay is. To acknowledge borrowed content, our quick writer references and cites the work.
Our support agents are available 24 hours a day 7 days a week and committed to providing you with the best customer experience. Get in touch whenever you need any assistance.
Try it now!
## Calculate the price of your order
We'll send you the first draft for approval by at
Total price:
\$0.00
How it works?
Fill in the order form and provide all details of your assignment.
Proceed with the payment
Choose the payment system that suits you most.
Our Services
So much stress and so little time? Take care of yourself: let us help you with your tasks. We offer all kinds of writing services.
## Custom Essay Writing Services
No matter what kind of academic paper you need and how urgent you need it, you are welcome to choose your academic level and the type of your paper at an affordable price. We take care of all your paper needs and give a 24/7 customer care support system.
| 951
| 4,676
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05
|
latest
|
en
| 0.820656
|
https://converths.com/how-many-days-is-a-million-seconds/
| 1,701,415,487,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-50/segments/1700679100276.12/warc/CC-MAIN-20231201053039-20231201083039-00605.warc.gz
| 224,844,421
| 47,401
|
How fast can you figure out the question – how many days in one million seconds?
## In one million seconds there are 11.5740740740741 days.
Easy Online Calculator
People nowadays use second, minute, hour, day, week, month and year to measure time, which is clear to everyone. So how to convert 1 million seconds into days? How many many days is 1 million seconds? -1 million seconds is equal to 1,000,000 seconds. In 1 million seconds how many days?
As we know, there are 60 seconds in a minute, 60 minutes in an hour, 24 hours in a day, 7 days in a week, 365 days in a year. When we say ‘a year’, which usually means a common year with exactly 365 days. How many days is 1000000 seconds? Let’s break it down.
So,
. 1 day = 24 hours
. 1 hour = 60 minutes
. 1 minute = 60 seconds
1 day = 24h x 60m x 60s = 86,400 seconds (PS: hour= h, minute= m, second= s)
Then,
. 1 million seconds = 1,000,000 seconds
1 million seconds = 1,000,000s ÷ 86,400s = 11.5740740740741 days
(PS: second= s, hour= h, days =d)
1,000,000 seconds = 11.5740740740741 days
## How Many Days is a Million Seconds – Video
Got a different answer? And what’s the advantage of the way you calculate? We’d love to hear your voice.
Leave your comment below, share with a friend and never stop wondering.❤️
| 378
| 1,278
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.78125
| 4
|
CC-MAIN-2023-50
|
latest
|
en
| 0.919574
|
http://wiki.zcubes.com/Manuals/calci/POINT
| 1,642,789,904,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-05/segments/1642320303512.46/warc/CC-MAIN-20220121162107-20220121192107-00104.warc.gz
| 74,508,514
| 6,084
|
# Manuals/calci/POINT
POINT(X,Y,Z)
• and are any real numbers.
## Description
• This function gives the point for the coordinates of and .
• Point is the exact location which don't have a size, only position.
• A set of points lying in a straight line, then they are called collinear.
• A set of points lying in a same plane, then they are called co-planar.
• This function will give the result as error when any one of the argument is non-numeric.
## Examples
1. =POINT(2,4,8) = 2 4 8
2. =POINT(-7,10,16) = -7 10 16
3. =POINT(17.4,29.02,41.65) = 17.4 29.02 41.65
POINTS
| 181
| 577
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05
|
latest
|
en
| 0.862229
|
https://hoven-in.appspot.com/Home/Aptitude/Volume-and-Surface-Areas/maths-aptitude-quiz-questions-and-mock-test-on-volume-and-surface-areas-009.html
| 1,713,032,612,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296816832.57/warc/CC-MAIN-20240413180040-20240413210040-00463.warc.gz
| 284,730,267
| 6,784
|
# Volume and Surface Areas Quiz Set 009
### Question 1
A room has a floor size of 17 × 31 sq. m. What is the height of the room if the sum of the areas of the base and roof is equal to the sum of the areas of the four walls?
A
\$10{47/48}\$ m.
B
\$12{11/47}\$ m.
C
\$9{29/50}\$ m.
D
\$13{21/50}\$ m.
Soln.
Ans: a
Let h be the height, and l and b be the length and breadth. We are given lb + lb = lh + hb + lh + hb. Combining the terms and cancelling 2, we get lb = h(l + b), which gives h = \${lb}/{l + b}\$ = \${17 × 31}/{17 + 31}\$ = \${527/48}\$, which is same as: \$10{47/48}\$ m.
### Question 2
How much water flows per hour through a pipe of radius 10 cm, if water flows at 10 km/h?
A
100 π cu. m.
B
102 π cu. m.
C
98 π cu. m.
D
104 π cu. m.
Soln.
Ans: a
In one hour, a water column of length 10 km is delivered through the cylindrical pipe. The equivalent volume is π × \${10 × 10 × 10 × 1000}/{100 × 100}\$, which can easily be cancelled to get 100π cu. m.
### Question 3
What is the volume of a cone generated by rotating a right angled triangle with sides 25, 7 and 24 cm? The rotation is done about the side of length 24 cm.
A
392 π sq. cm.
B
393 π sq. cm.
C
391 π sq. cm.
D
\$131{2/3}\$ π sq. cm.
Soln.
Ans: a
The radius of the base of the cone r = 7 cm, and height h = 24 cm. The volume is \$1/3\$π\$(r^2 × h)\$ = \$1/3\$π\$(7^2 × 24)\$ = 392π.
### Question 4
What is the length of a 10 mm wire that has been drawn out of a lump of 77 cu. cm. tin? Take π = 22/7.
A
2450 cm.
B
2452 cm.
C
2448 cm.
D
2454 cm.
Soln.
Ans: a
The wire is a cylinder with r = 1/10 cm, and length L. The volume of tin is equal to the volume of the wire. So π \${1/10} × {1/10} × L\$ = 77, which gives L = \${10 × 10 × 77 × 7}/22\$ = 2450 cm.
### Question 5
An iron tank 2 m × 3 m is filled with water upto a height of 2 m. What surface area of the tank is at a higher risk of corrosion?
A
26 cu. m.
B
28 cu. m.
C
24 cu. m.
D
30 cu. m.
Soln.
Ans: a
If L, B and H are the dimensions of the water column, then the wet area is 2(BH + LH) + LB = 2 × (3 × 2 + 2 × 2) + 2 × 3. The result is 26 cu. m.
| 780
| 2,141
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18
|
latest
|
en
| 0.751913
|
https://www.read-any-book.net/ebook/elementary-properties-of-clock-regulated-queues-160452
| 1,713,038,116,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296816832.57/warc/CC-MAIN-20240413180040-20240413210040-00360.warc.gz
| 897,767,893
| 45,433
|
# Elementary Properties of Clock Regulated Queues
Elementary Properties of Clock Regulated Queues
O E Percus
The book Elementary Properties of Clock Regulated Queues was written by author Here you can read free online of Elementary Properties of Clock Regulated Queues book, rate and share your impressions in comments. If you don't know what to write, just answer the question: Why is Elementary Properties of Clock Regulated Queues a good or bad book?
In our eReader you can find the full English version of the book. Read Elementary Properties of Clock Regulated Queues Online - link to read the book on full screen. Our eReader also allows you to upload and read Pdf, Txt, ePub and fb2 books. In the Mini eReder on the page below you can quickly view all pages of the book - Read Book Elementary Properties of Clock Regulated Queues
What reading level is Elementary Properties of Clock Regulated Queues book?
To quickly assess the difficulty of the text, read a short excerpt:
22) ^0) = iif^(l-f).
As for u'(j), either the queue has length j and the message is taken first, or the length is ; — 1 and it is taken second. Together, then, w{j) = (1 — |)p(j) + |p(j — 1): (2. 23) w(j) = ^-f i-dl-f=\ for J > 0. It readily follows from (2. 16) that (2. 24) „■(. , = f. , .. I^l_£LfI.
The average waiting time is thus ^ dx '^-^ 4(1 -p)' The fact that this is identical to the mean queue length (2. IS) divided by the mean arrival rate p is an immediate consequence of Little's form
...ula [4]. However. (2. 24) allows all of the moments to be found just as easily.
3. The Output Process of the Queue A queue transforms an input time process into an output time process. Since the i. I. D. Input assumption proliferates throughout the literature, it is of interest to find the characteristics of the associated output, which of course serves in part as input to the next stage. As a first step, let us examine the probability ^„ of returning to the origin for the first time at the nth step (i.
What to read after Elementary Properties of Clock Regulated Queues?
You can find similar books in the "Read Also" column, or choose other free books by O E Percus to read online
10
Tokens
Elementary Properties of Clock Regulated Queues
+Write review
| 559
| 2,245
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.21875
| 3
|
CC-MAIN-2024-18
|
latest
|
en
| 0.917062
|
http://mathhelpforum.com/calculus/127660-integration-parts-help.html
| 1,529,916,142,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267867579.80/warc/CC-MAIN-20180625072642-20180625092642-00591.warc.gz
| 202,554,244
| 9,723
|
# Thread: Integration by Parts Help
1. ## Integration by Parts Help
Hey everybody,
My class is learning integration by substitution and integration by parts right now, and I'm having a lot of trouble with them.
Right now, I need to find the antiderivative of $\displaystyle [(x^2)/(x^2 + 1)]dx$.
I've tried letting $\displaystyle u = x^2$ and $\displaystyle dv = [1/(x^2 + 1)]dx$, but I don't know if that's correct, because now I'm stuck at a weird place where I have to find the antiderivative of $\displaystyle [x arctan(x)]dx$, and I don't think we're supposed to have to do that much at this level.
2. $\displaystyle \int\frac{x^2}{x^2+1}~dx = \int 1-\frac{1}{x^2+1}~dx$
$\displaystyle \int x\left(\frac{x}{x^2+ 1}\right)dx$
and let u= x, $\displaystyle dv= \frac{x}{x^2+1}dx$.
| 251
| 788
|
{"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}
| 3.421875
| 3
|
CC-MAIN-2018-26
|
latest
|
en
| 0.929215
|
https://www.doubtnut.com/qna/644006329
| 1,726,386,080,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651616.56/warc/CC-MAIN-20240915052902-20240915082902-00451.warc.gz
| 671,882,060
| 41,276
|
# 10n+3(4n+2)+5 is divisible by (n∈N)
A
7
B
5
C
9
D
7
Video Solution
Text Solution
Generated By DoubtnutGPT
## To prove that 10n+3(4n+2)+5 is divisible by 9 for all n∈N, we will use the principle of mathematical induction.Step 1: Base CaseWe start by checking the base case, n=1.P(1)=101+3(41+2)+5Calculating this gives:P(1)=10+3(43)+5=10+3(64)+5=10+192+5=207Now we check if 207 is divisible by 9:207÷9=23(exact division)Thus, P(1) is true.Step 2: Inductive HypothesisAssume that the statement is true for some integer k, i.e.,P(k)=10k+3(4k+2)+5 is divisible by 9.This means there exists an integer λ such that:10k+3(4k+2)+5=9λ.Step 3: Inductive StepWe need to show that P(k+1) is also true:P(k+1)=10k+1+3(4(k+1)+2)+5.This can be rewritten as:P(k+1)=10⋅10k+3(4k+3)+5.Using the property of exponents, we can express 4k+3 as:4k+3=4k⋅43=64⋅4k.Thus,P(k+1)=10⋅10k+3(64⋅4k)+5.This simplifies to:P(k+1)=10⋅10k+192⋅4k+5.Step 4: Substitute the Inductive HypothesisFrom the inductive hypothesis, we know:10k+3(4k+2)+5=9λ.We can express 3(4k+2) as 3(4k⋅16):P(k)=10k+48⋅4k+5=9λ.Now, substituting this back into our equation for P(k+1):P(k+1)=10⋅10k+192⋅4k+5=10(9λ−48⋅4k−5)+192⋅4k+5.This leads to:P(k+1)=90λ+192⋅4k+5−10⋅5.=90λ+192⋅4k−50+5.=90λ+192⋅4k−45.Step 5: Check Divisibility by 9Now we need to check if P(k+1) is divisible by 9:P(k+1)=90λ+192⋅4k−45.Since 90λ is divisible by 9 and 192⋅4k can be expressed as 9m (where m is an integer), we can conclude that:P(k+1) is divisible by 9.ConclusionBy the principle of mathematical induction, since P(1) is true and P(k)⇒P(k+1) is true, we conclude that:10n+3(4n+2)+5 is divisible by 9 for all n∈N.
|
Updated on:7/8/2024
### Knowledge Check
• Question 1 - Select One
## If 10n+3.4n+2+k is divisible by 9,∀n∈N then the least positive integral value of k is
A(a) 1
B(b) 3
C(c) 5
D(d) 7
• Question 2 - Select One
## For all n∈N,2n+1+32n−1 is divisible by: (i) 5 (ii) 7 (iii) 14 (iv) 135
A5
B7
C14
D135
Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Rajasthan Board, MP Board, Telangana Board etc
NCERT solutions for CBSE and other state boards is a key requirement for students. Doubtnut helps with homework, doubts and solutions to all the questions. It has helped students get under AIR 100 in NEET & IIT JEE. Get PDF and video solutions of IIT-JEE Mains & Advanced previous year papers, NEET previous year papers, NCERT books for classes 6 to 12, CBSE, Pathfinder Publications, RD Sharma, RS Aggarwal, Manohar Ray, Cengage books for boards and competitive exams.
Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation
| 1,121
| 2,940
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.625
| 5
|
CC-MAIN-2024-38
|
latest
|
en
| 0.754464
|
https://www.unitconverters.net/power/deciwatt-to-newton-meter-second.htm
| 1,566,581,805,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-35/segments/1566027318952.90/warc/CC-MAIN-20190823172507-20190823194507-00195.warc.gz
| 1,010,062,054
| 3,978
|
Home / Power Conversion / Convert Deciwatt to Newton Meter/second
# Convert Deciwatt to Newton Meter/second
Please provide values below to convert deciwatt [dW] to newton meter/second, or vice versa.
From: deciwatt To: newton meter/second
### Deciwatt to Newton Meter/second Conversion Table
Deciwatt [dW]Newton Meter/second
0.01 dW0.001 newton meter/second
0.1 dW0.01 newton meter/second
1 dW0.1 newton meter/second
2 dW0.2 newton meter/second
3 dW0.3 newton meter/second
5 dW0.5 newton meter/second
10 dW1 newton meter/second
20 dW2 newton meter/second
50 dW5 newton meter/second
100 dW10 newton meter/second
1000 dW100 newton meter/second
### How to Convert Deciwatt to Newton Meter/second
1 dW = 0.1 newton meter/second
1 newton meter/second = 10 dW
Example: convert 15 dW to newton meter/second:
15 dW = 15 × 0.1 newton meter/second = 1.5 newton meter/second
| 294
| 872
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.65625
| 3
|
CC-MAIN-2019-35
|
longest
|
en
| 0.585439
|
https://brainly.com/question/34994
| 1,484,616,574,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-04/segments/1484560279379.41/warc/CC-MAIN-20170116095119-00233-ip-10-171-10-70.ec2.internal.warc.gz
| 802,635,736
| 8,996
|
2014-04-15T00:04:12-04:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
So we know what the variables are already.
Now, 3 x 2 = 6.
So our fraction is 6/6.
6/6 is also equal to 1.
• Brainly User
2014-04-15T03:01:40-04:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
| 195
| 792
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.546875
| 3
|
CC-MAIN-2017-04
|
latest
|
en
| 0.952581
|
http://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.201281.html
| 1,368,954,029,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2013-20/segments/1368697232084/warc/CC-MAIN-20130516094032-00032-ip-10-60-113-184.ec2.internal.warc.gz
| 309,168,209
| 4,460
|
# SOLUTION: R
Algebra -> Algebra -> Probability-and-statistics -> SOLUTION: R Log On
Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help! Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!
Algebra: Probability and statistics Solvers Lessons Answers archive Quiz In Depth
Click here to see ALL problems on Probability-and-statistics Question 201281: RAnswer by javeria13(7) (Show Source): You can put this solution on YOUR website!OKAY.. all you have to do is make a tree diagram. first on a paper write this down. A= 1,2,3,4,5,6 B=1,2,3,4,5,6 B=1,2,3,4,5,6 C=1,2,3,4,5,6 C=1,2,3,4,5,6 C=1,2,3,4,5,6 So there are a total of 36 outcomes beacuse there are 36 possible numbers that could occur. Out of those the probability of(B,4) occuring is 2/36 or 1/18 (reduced answer) as shown above because there are 2 B'S and each has one 4 in it.
| 288
| 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}
| 3.5
| 4
|
CC-MAIN-2013-20
|
latest
|
en
| 0.872753
|
http://jwilson.coe.uga.edu/EMT668/EMAT6680.2001/Bell/emat6700.hsb/sierpinski.hsb/sierpinhsb.html
| 1,539,917,543,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-43/segments/1539583512268.20/warc/CC-MAIN-20181019020142-20181019041642-00185.warc.gz
| 198,883,475
| 2,826
|
Program Sierpinski
Purpose: to generate Sierpinski's triangle with a seemingly chaotic algorithm. Uses drawing, random numbers, and a "For" loop on the TI-83.
This algorithm begins with an equilateral triangle. When performing the process on paper, a random point is chosen within the triangle as a starting point. Then a die is rolled (or some other method of randomization) choosing a vertex. The midpoint between the original random point and the random vertex is then plotted. Starting with the newly plotted point, the process is then repeated. We can plot this on paper several times, but after a while it becomes tedious. Writing a program to perform the procedure many times will save us from having to do it.
Because the program is used to graph a picture, we must set up the graph it will be plotted on. Turn functions off, clear any drawings, and turn the statistics plots off with the following commands: FnOff, ClrDraw, and PlotsOff. To make sure the X and Y axes do not interfere with our graph, also use the AxesOff command. All of these commands (along with any other) can be found in the Catalog by pressing the yellow 2nd followed by the zero key.
Now set the domain and range. The "VARS" key (3rd row, 4th column) contains menus of variables. The variables Xmax, Xmin, Ymax, Ymin are contained in the "Window..." menu.
Let Xmin=0 and Xmax=1.
Let Ymin=0 and Ymax=1.
Then a random number (between 0 and 1) is stored as the first x coordinate, likewise for the y.
Now comes the "For" loop. To obtain a good picture, command the program to loop (plot a new point) a large number of times--more than you would ever plot on paper. I choose 3000. Note: the more points you choose they longer the program will run. 3000 takes about 6 minutes to run.
How can we simulate the random selection of a vertex? Since the calculator function "rand" automatically chooses something between 0 and 1, divide those selections into thirds. That is, if the random number is less than 1/3, designate one vertex; if the random number is between 1/3 and 2/3, designate the second vertex; if the random number is greater than 2/3, designate the third vertex. How can we determine the half way point between the plotted point and the newly chosen vertex?
Think in terms of the range in your window. For the equilateral triangle to fit in this domain and range, what must the coordinates of the vertices be?
Obviously the vertices are the origin, (1,0) and (0.5,0.866). How can you determine the value of the y coordinate of point C?
"If" statements can be used to test the value of the random number N.
When random number N is less than or equal to 1/3, we will plot a new point half way between the orignal (X,Y) and vertex A (the origin). This is the easiest one to consider. The new point would be (0.5X, 0.5Y), so store those as new values of X and Y.
For random numbers greater than 1/3 and less than or equal to 2/3, choose vertex C. How can you calculate the midpoint of (X,Y) and vertex C?
Go ahead and write two more "If" statements for vertices B and C which store the midpoint as the new X and Y coordinates.
Now that our new X and Y values are stored, they can be plotted using the "Pt-On" command. It is found in the "Draw" menu under "Points." Access the "Draw" menu by pressing the yellow "2nd" button followed by the "PGRM" button, which we are all too familiar with.
The "For" loop can now "End." Store your new graph for good measure using the "Store Pic" command under the "DRAW" menu (right arrow over to "STO").
| 864
| 3,541
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.625
| 4
|
CC-MAIN-2018-43
|
latest
|
en
| 0.922441
|
https://jackschaedler.github.io/circles-sines-signals/dft_leakage.html
| 1,722,899,510,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722640455981.23/warc/CC-MAIN-20240805211050-20240806001050-00473.warc.gz
| 260,381,120
| 5,105
|
WHEN THE INPUT DOESN’T “FIT”
THE PHENOMENON OF LEAKAGE
Up until now, I’ve tried to contrive signals which will produce clean DFT output. Unfortunately, almost every “real-world” signal that you’ll process using the DFT will produce somewhat messy results. You probably noticed in the last section that signals which were composed of only one sinusoid were capable of producing spectra with many strange peaks and valleys when we started zero-padding them.
As it turns out, when any frequency component of the input signal does not perfectly match up with an available bin frequency, the energy related to that frequency component will “spill” out into all of the other bins. This is a phenomenon known as spectral leakage.
Figure 1 Allows you to see the effect of spectral leakage by altering the frequency of the input signal. The slider at the top of the figure allows you to change the input frequency smoothly from 1Hz to 3Hz. Notice that when the frequency of the input signal perfectly matches one of the available bin frequencies, there is no leakage. As soon as the frequency varies even slightly from a bin frequency, all of the other bins will assume a non-zero magnitude.
Figure 1. DFT Response to Different Input Frequencies
LEAKAGE
SAMPLING A CONTINUOUS SPECTRUM
To understand leakage, we must appreciate that the Discrete Fourier Transform actually produces a sampled version of a Continuous Fourier Transform. The continuous frequency spectrum of a sampled sine wave is approximated by the Sinc function shown in Figure 2.1
Figure 2. The Sinc Function. The DFT of a Sampled Sine Wave is Approximated by the Sinc Function.
The Discrete Fourier Transform actually gives us a sampled version of this continuous spectrum. In Figure 3, each sample of the DFT output is plotted as a blue square on this continuous curve. When the input signal has an integral number of periods, all of the DFT samples will lie at either zero magnitude or on the very tip of the continuous curve’s main “bump”. As soon as the frequency of the input signal drifts away from an available bin frequency, the other bins assume a non-zero magnitude due to the shape of the continuous curve.
Figure 3. Viewing Leakage as the Sampling of a Continuous Spectrum Frequency
Now that we understand the sampled nature of the DFT, the results we saw when zero-padding our signal in the previous section might make a bit more sense.
1. The continuous curve has this distinct shape because we’re actually sort of misusing the Fourier Transform. The Fourier Transform assumes that the input signal is periodic and infinitely long. When we feed a finite-length input signal into the Fourier Transform we are actually implicitly multiplying the input signal with a rectangular window (A rectangular window is just a signal with ones for each its samples) The spectrum of a rectangular window is approximated by the so-called sinc function which is defined as, sin(x) / x (shown in Figure 2). When we multiply our input signal with this rectangular window, the DFT output will be affected. Specifically, the output spectra of our signal will be convolved with the spectra of a rectangular window. When we convolve the two signals we are left with this characteristic sinc spectrum. To really understand this phenomenon, we need to introduce and understand the notion of convolution. If there's enough interest, I'll create a section on convolution and attempt to dig further into this topic. In the meantime, let's slowly back away and go read Wikipedia's page on the convolution theorem.
WINDOWING
REDUCING LEAKAGE
A first step towards understanding leakage is to think about the discontinuities that would exist at the endpoints of our input signal if we were to loop it. Figure 4 shows an imaginary “unrolling” of our input signal in the top row of the figure.2 Discontinuities (instantaneous jumps) are shown in red. Notice that whenever leakage occurs, there are discontinuities present. When the leakage is at its worst, there is no jump, but rather a very sharp V-shaped corner in the waveform.
We might assume that these discontinuities are the cause of the leakage. In order to remove the discontinuities, we can window our signal.3 Windowing is a fancy term for multiplying the input signal with a “window” signal. We can craft a window signal which tapers off to zero at its ends in order to remove the discontinuities in our looped signal. In Figure 4, the window is shown in grey. You can see that the windowed signal loops smoothly without discontinuities and ugly corners. The continuous spectrum of a this windowed sine wave is shown in green. You’ll notice that the effect of leakage would be less pronounced if we chose to sample the windowed curve.
Figure 4. Windowing The DFT's Input Signal Frequency
You should be aware that there are many types of windows, and each can be appropriate depending upon the circumstances and needs of your analysis task. I suggest that you refer to the Wikipedia article on windowing if you want to know more about the various window functions and their properties.
2. To accommodate the visualization of the looping, please note that I’ve reduced the frequency of the input signal to the range between 1 and 2 Hz.
3. Windowing isn’t a panacea. For example, windowing can be very destructive to low frequency signals.
| 1,122
| 5,365
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.09375
| 3
|
CC-MAIN-2024-33
|
latest
|
en
| 0.920186
|
https://www.newscientist.com/article/mg21929300.300-meet-the-nasa-scientist-devising-a-starship-warp-drive
| 1,477,497,487,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-44/segments/1476988720962.53/warc/CC-MAIN-20161020183840-00142-ip-10-171-6-4.ec2.internal.warc.gz
| 974,618,800
| 20,020
|
## SUBSCRIBE AND SAVE 50%
### Reinventing Energy Summit:
Meet the people shaping the future of energy - 25 November in London
Interview
14 August 2013
# Meet the NASA scientist devising a starship warp drive
To pave the way for rapid interstellar travel, NASA propulsion researcher Harold “Sonny” White plans to manipulate space-time in the lab
The idea that nothing can exceed the speed of light limits our interstellar ambitions. How do we get round this?
Within general relativity, there are two loopholes that allow you to go somewhere very quickly, overcoming the restriction of the speed of light. One is a wormhole and the other is a space warp.
What is a space warp and how can it help?
A space warp works on the principle that you can expand and contract space at any speed. Take a terrestrial analogy. In airports we have moving walkways that help you cover distance quicker than you would otherwise. You are walking along at 3 miles an hour, and then you step onto the walkway. You are still walking at 3 miles an hour, but you are covering the distance much more quickly relative to somebody who isn’t on the belt.
What would a starship with warp drive be like?
Imagine an American football, for simplicity, that has a toroidal ring around it attached with pylons. The football is where the crew and robotic systems would be, while the ring would contain exotic matter called negative vacuum energy, a consequence of quantum mechanics. The presence of this toroidal ring of negative vacuum energy is what’s required from the math and physics to be able to use the warp trick.
What would it be like to travel at warp speed?
You would have an initial velocity as you set off, and then when you turn on the ring of negative vacuum energy it augments your velocity. Space would contract in front of the spacecraft and expand behind it, sending you sliding through warped space-time and covering the distance at a much quicker rate. It would be like watching a film in fast forward.
Even if travelling at warp speed is theoretically possible, don’t the huge energy requirements make it unlikely?
When the idea was first proposed mathematically in 1994 it required a vast amount of negative vacuum energy which made the idea seem impossible. I did some work in 2011 and 2012 as part of the 100 Year Starship symposium and discovered ways to reduce the energy requirements by many orders of magnitude, so for a 10-metre diameter spacecraft with a velocity of 10 times light speed, I can reduce the negative energy needed.
How close are you to making this a reality?
We are very much in the science rather than the technology phase. We have got some very specific and controlled steps to take to create a proof of concept, to show we have properly understood and applied the math and physics. To that end we will try to generate a microscopic instance of a warp bubble in the lab and measure it.
If successful is the next stop Alpha Centauri?
We don’t just go from the lab to an interstellar mission. There will be intermediate steps, other things we would do with this long before we get to some of the romantic pictures of a captain on the bridge telling the helmsman to engage warp drive.
### Profile
Harold “Sonny” White is advanced propulsion theme lead at NASA’s Johnson Space Center in Houston, Texas. He is also a keynote speaker at this week’s Icarus Interstellar Starship Congress
More on these topics:
| 720
| 3,432
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-44
|
latest
|
en
| 0.947884
|
https://openstax.org/books/physics/pages/11-problems
| 1,722,899,658,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722640455981.23/warc/CC-MAIN-20240805211050-20240806001050-00614.warc.gz
| 355,698,251
| 81,374
|
Physics
# Problems
PhysicsProblems
## 11.1Temperature and Thermal Energy
14.
What is 35.0 °F in kelvins?
1. 1.67 K
2. 35.0 K
3. -271.5 K
4. 274.8 K
15.
Design a temperature scale where the freezing point of water is 0 degrees and its boiling point is 70 degrees. What would be the room temperature on this scale?
1. If room temperature is 25.0 °C, the temperature on the new scale will be 17.5 °.
2. If room temperature is 25.0 °C, the temperature on the new scale will be 25.0°.
3. If the room temperature is 25.0 °C, the temperature on the new scale will be 35.7°.
4. If the room temperature is 25.0 °C, the temperature on the new scale will be 50.0°.
## 11.2Heat, Specific Heat, and Heat Transfer
16.
A certain quantity of water is given 4.0 kJ of heat. This raises its temperature by 30.0 °F. What is the mass of the water in grams?
1. 5.7 g
2. 570 g
3. 5700 g
4. 57 g
17.
5290 J of heat is given to 0.500 kg water at 15.00 °C. What will its final temperature be?
1. 15.25° C
2. 12.47 ° C
3. 40.3° C
4. 17.53° C
## 11.3Phase Change and Latent Heat
18.
How much energy would it take to heat 1.00 kg of ice at 0 °C to water at 15.0 °C?
1. 271 kJ
2. 334 kJ
3. 62.8 kJ
4. 397 kJ
19.
Ice cubes are used to chill a soda with a mass msoda = 0.300 kg at 15.0 °C. The ice is at 0 °C, and the total mass of the ice cubes is 0.020 kg. Assume that the soda is kept in a foam container so that heat loss can be ignored, and that the soda has the same specific heat as water. Find the final temperature when all ice has melted.
1. 19.02 °C
2. 90.3 °C
3. 0.11 °C
4. 9.03 °C
Order a print copy
As an Amazon Associate we earn from qualifying purchases.
| 578
| 1,660
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.84375
| 4
|
CC-MAIN-2024-33
|
latest
|
en
| 0.84811
|
https://essaysontime.us/2020/09/20/assignment-testing-for-correlation-and-bivariate-regression-in-spss/
| 1,607,112,845,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-50/segments/1606141743438.76/warc/CC-MAIN-20201204193220-20201204223220-00409.warc.gz
| 284,422,222
| 8,642
|
Assignment: Testing for Correlation and Bivariate Regression in SPSS
You had the accident precedent in the week to accomplish an word sapidity on interrelation and ultimate direct retirement and conquer equal feedback. Hopefully you are bewildered encircling the undeveloped these criterions hold; equivalent grave is that you identify some of their weaknesses. Now, it is uninterruptedly frequently interval to put all of that good-tempered-tempered brainstorming to use and vindication a collective exploration investigation delay the interrelation and ultimate direct retirement. As you arise the Assignment, be strong and pay arrest consideration to the assumptions of the criterion. Specifically, establish strong that your variables are metric equalize variables that can amply be interpreted in these criterions. For this Assignment, you allure ponder interrelation and bivariate retirement criterioning. To arrange for this Assignment: Review this week’s Learning Resources and resources program akin to retirement and interrelation. Using the SPSS software, disclosed the High School Longitudinal Study factsset (whichever you select) Based on the factsset you chose, invent a exploration investigation that can be vindicationed delay a Pearson interrelation and bivariate retirement. For this Assignment: Write a 2- to 3-paragraph decomposition of your interrelation and bivariate retirement results for each exploration investigation. Do not obliviate to evaluate if the interrelation and bivariate retirement assumptions are met and relation the movables extent. In your decomposition, spread-out the facts for the output. Based on your results, furnish an interpretation of what the implications of collective vary dominion be. Use appertinent APA format, citations, and referencing for your decomposition, exploration investigation, and spread-out of output. By Day 7 Submit your Assignment: Testing for Interrelation and Bivariate Regression.
| 368
| 1,961
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50
|
latest
|
en
| 0.873749
|
https://math.stackexchange.com/tags/integration/info
| 1,721,814,766,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763518198.93/warc/CC-MAIN-20240724075911-20240724105911-00216.warc.gz
| 331,844,192
| 19,226
|
Tag Info
For questions about the properties of integrals. Use in conjunction with (indefinite-integral), (definite-integral), (improper-integrals) or another tag(s) that describe the type of integral being considered. This tag often goes along with the (calculus) tag.
Integration is a major part of .
There are two main kinds of integrals:
• definite integrals (e.g. proper and improper integrals), which often have numerical values
• indefinite integrals, which group families of functions with the same derivative.
Several techniques to solve integrals have been developed, including integration by parts, substitution, trigonometric substitution, and partial fractions.
Integration can be used to find the area under a graph and find the average of the function. Also, it can be used to compute the volume of certain solids and to find the displacement of a particle.
| 182
| 877
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.5625
| 3
|
CC-MAIN-2024-30
|
latest
|
en
| 0.882462
|
https://www.physicsforums.com/threads/poles-and-residues.798686/
| 1,527,143,746,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-22/segments/1526794865928.45/warc/CC-MAIN-20180524053902-20180524073902-00451.warc.gz
| 820,959,523
| 18,311
|
# Homework Help: Poles and residues
1. Feb 18, 2015
### nmsurobert
1. The problem statement, all variables and given/known data
Show that all the singular points of the following functions are poles.
Determine the order of each pole and the corresponding residue.
z/cosz
2. Relevant equations
maybe cosz = 1/2 ez+e-z is relevant but i dont use it here
3. The attempt at a solution
the pole i found was z = π/2 + 2nπ and -π/2 +2nπ
i attempted to find the residues using limits.
limz→π/2 + 2nπ (z-(π/2 + 2nπ)) z/cosz
doing some annoying algebra and just evaluating at π/2 + 2nπ i get 0/0. so i can use L'hopitals rule, right?
doing so i get my limit to equal π/2 + 6nπ
is that right?
i know i can use lorent series but i honestly dont know where to start with that mess haha
2. Feb 18, 2015
### Dick
That's pretty bad. cosz = (1/2)(eiz+e-iz) so your relevant equation is wrong and you do want to use it to show the only poles are on the real axis. Given that you do have the right poles on the real axis. And yes, you can use l'hopital. But show how you got that limit.
3. Feb 18, 2015
### nmsurobert
oops i forgot the i's. but doesnt just putting the poles into the original equation shows that theyre poles?
i didn't realize it was "pretty bad" LOL
4. Feb 18, 2015
### Dick
It shows they are poles, it doesn't show they are the only poles. That's not my only complaint. The limit to compute the residue is bad and I can't tell why because you didn't show your work.
5. Feb 18, 2015
### nmsurobert
lim z → (z-(π/2 + 2nπ)) z/cosz
z2-π/2z - 2nπz
then evaluate at z = π/2 + 2nπ
π2/4 + π/2(2nπ) +π/2(2nπ) + 4n2π22/4- π/2(2nπ) π/2(2nπ) - 4n2π2 = 0
cos(π/2 + 2nπ) = 0
0/0
l'hopital rule
2z - π/2 - 2nπ/ - sinz evaluate at π/2 + 2nπ
2(π/2) + 4nπ -π/2 -2nπ / -sin(π/2 + 6nπ)
= -π/2 - 2nπ
so my first calculation was different but its still the same method.
6. Feb 18, 2015
### Dick
"Different" is what I wanted since the first one was bad. That one looks ok (with some typos). Now what about the -π/2+2nπ series and what about showing that the ONLY poles are on the real axis?
7. Feb 18, 2015
### nmsurobert
im gonna have to do more reading because i dont know what youre talking about with "-π/2+2nπ series" i thought i take the limit and the limit is the residue.
i also need to do some more reading because i dont know how to find any other poles.
im happy ive made it this far though, i had no idea what the hell i was looking at 2 hours ago.
8. Feb 18, 2015
### nmsurobert
thank you though, even though youre kind of a dick. ;-)
9. Feb 19, 2015
### Dick
You said "the pole i found was z = π/2 + 2nπ and -π/2 +2nπ". So far you've only found the residue at π/2 + 2nπ. What about -π/2 +2nπ?
10. Feb 19, 2015
### nmsurobert
sorry, i went to sleep then to class lol.
the next one should be -π/2 +2nπ.
using the same messy method
11. Feb 19, 2015
### Dick
Right again. It's not really all that bad is it? Now can you show they are the only poles using your relevant equation? And what's the order of the poles? Yes, I'm kind of a dick :-;
12. Feb 19, 2015
### nmsurobert
i think theyre both order 1.
im not sure how to use the equation to show that. im guessing it has something to do with using the ln|z| + iarg though. not sure if thats right.
13. Feb 19, 2015
### Dick
You are correct about the order, can you say why? As to why they are the only poles, if $cos(z)=0$ then $e^{iz}=-e^{-iz}$, right? Put $z=a+ib$ and show $b$ must be 0.
| 1,174
| 3,488
|
{"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}
| 3.75
| 4
|
CC-MAIN-2018-22
|
latest
|
en
| 0.92874
|
https://brainly.com/question/118524
| 1,484,987,640,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-04/segments/1484560281001.53/warc/CC-MAIN-20170116095121-00051-ip-10-171-10-70.ec2.internal.warc.gz
| 799,972,928
| 9,126
|
2014-09-14T23:00:35-04:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
For cross multiplication, you divide by a number divisible by the number and the number angled across from it. 5 and 25 are both divisible by 5 so you divide both numbers by 5. 5=1 and 25=5. Which temporary makes it 1/8 and 12/5. Since both 8 and 12 are divisible by 4, divide both of them by 4., 8 by 2 times and 12 by 3 times(8=2 and 12=3) So the result is 1/2*3/5 which is 3/10.
2014-09-14T23:56:22-04:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
You don't cross-multiply when you're multiplying fractions.
Multiplying fractions is really simple:
-- The numerator of the product is the product of the two numerators.
The two numerators are 5 and 12, so the numerator of the product is 60 .
-- The denominator of the product is the product of the two denominators.
The two denominators are 8 and 25, so the new denominator is 200.
The new fraction is 60/200 .
You could simplify it, and then it would look like 3/10 .
| 403
| 1,535
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.859375
| 4
|
CC-MAIN-2017-04
|
latest
|
en
| 0.932926
|
https://fehrplay.com/domashniy-uyut/11103-kak-sdelat-gantelyu-v-domashnih-usloviyah-prakticheskie-sovety.html
| 1,586,431,989,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-16/segments/1585371833063.93/warc/CC-MAIN-20200409091317-20200409121817-00016.warc.gz
| 450,361,051
| 7,335
|
# How to make a dumbbell at home? Practical tips
Home cosiness
You want to have a beautiful body, but hiking ingym require too much time and money? There is an alternative - classes with weighting at home. Although it should be noted that the purchase of dumbbells and barbells, too, will cost a tidy sum. The following question arises: "How to make a dumbbell at home?". To obtain the necessary inventory, not much is needed: cement and sand, empty bottles, a piece of ordinary iron pipe, metal cans.
To begin with, determine what weight dumbbells are for youwill be needed. If you need an inventory of 0.2 kg to 1 kg, then to solve the problem of how to make a dumbbell at home, it will be enough to take small plastic bottles and fill them with water or sand. To get a little more weight, add a little water to the sand container and close the lid tightly. For classes, you can use elongated cans with any contents. Their weight is usually printed on the package.
And how to make a dumbbell at home, ifYou need a more solid weight, for example, 2-6, or even 8 kg? This will require a little more effort. Take 4 metal jars, for example, from paint, fill two of them with cement mortar, insert a piece of metal pipe into this solution. Make sure that the pipe is perpendicular. After the solution has solidified, repeat the operation for the other side of the pipe. With dumbbells received, you should handle them carefully, do not throw them sharply on the floor, because concrete - a substance that is heavy, but brittle, it can crack or crack. By the way, the rod is similarly made. Dumbbells need a short handle and a light weight, and for the rod select a longer tube and larger jars.
What to replace dumbbells, if at hand there is nocement mortar and iron pipes? They will rescue plastic bottles with water or other fillers, for example, all with the same sand. Choose their weight depending on the exercise. Not always the shape of the bottles is convenient and suitable for grasping. You can choose such plastic containers, which already have a handle in their design. Another option for solving the problem of how to make a dumbbell at home is the use of bottles packed in a sturdy bag with strong handles. Such a bag can be used, performing dumbbell lifts to the chest, while performing various types of wiring. If you need weighting to perform bench presses from a prone position, it is better to use a wide, but not very high bag with a short handle.
If you have a familiar welder, you can dohe order for the manufacture of self-made dumbbells. To do this, he will need short trimming of pipes and sheet steel. The specialist will be able to cut metal circles of different diameters independently and make holes in them. Will only clean them. Do not forget to order special locks, which will keep pancakes on the handle.
We hope that our advice will help you make the necessary inventory, and then become the owner of a beautiful body!
| 658
| 2,955
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3
| 3
|
CC-MAIN-2020-16
|
latest
|
en
| 0.926608
|
http://bookshadow.com/weblog/2019/03/18/leetcode-numbers-with-repeated-digits/
| 1,627,309,400,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-31/segments/1627046152129.33/warc/CC-MAIN-20210726120442-20210726150442-00426.warc.gz
| 8,804,992
| 20,233
|
## 题目描述：
LeetCode 1015. Numbers With Repeated Digits
Given a positive integer `N`, return the number of positive integers less than or equal to `N` that have at least 1 repeated digit.
Example 1:
```Input: 20
Output: 1
Explanation: The only positive number (<= 20) with at least 1 repeated digit is 11.
```
Example 2:
```Input: 100
Output: 10
Explanation: The positive numbers (<= 100) with atleast 1 repeated digit are 11, 22, 33, 44, 55, 66, 77, 88, 99, and 100.
```
Example 3:
```Input: 1000
Output: 262
```
Note:
1. `1 <= N <= 10^9`
## Python代码：
``````class Solution(object):
def numDupDigitsAtMostN(self, N):
"""
:type N: int
:rtype: int
"""
self.S = str(N)
ans = self.solve(self.S, 0) + self.calc9(len(self.S) - 2)
return N - ans
def solve(self, S, D):
pcnt = collections.Counter(self.S[:D] or [-1])
if len(S) == 1:
return sum(1 for x in range(not D, int(S) + 1) if not pcnt[str(x)])
ans = 0
for x in range(not D, int(S[0])):
if pcnt[str(x)] == 0:
ans += self.arrange(9 - D, len(S) - 1)
if pcnt[S[0]] == 0:
ans += self.solve(S[1:], D + 1)
return ans
def calc9(self, M):
return 9 * sum(self.arrange(9, i) for i in range(M + 1))
def arrange(self, a, b):
if a < b or a < 0 or b < 0: return 0
return self.factorial(a) / self.factorial(a - b)
def factorial(self, n):
if not n: return 1
return reduce(operator.mul, range(1, n + 1))
``````
Pingbacks已关闭。
| 486
| 1,365
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.734375
| 4
|
CC-MAIN-2021-31
|
longest
|
en
| 0.462348
|
http://stats.stackexchange.com/questions/14636/how-to-choose-the-tolerance-parameter-for-abc
| 1,386,922,322,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2013-48/segments/1386164920374/warc/CC-MAIN-20131204134840-00049-ip-10-33-133-15.ec2.internal.warc.gz
| 165,986,158
| 13,743
|
# How to choose the tolerance parameter for ABC?
I have the following sorted data (sampling from parametric space [1,5]) with respect to their distances of parameter Theta. i.e., Let say N = 1000,
Theta : 1.1, 1.7, 1.9, 2.4, 2.8, . . . , 4.9
Distance : 0.2, 0.3, 0.5, 0.9, 1.1, . . . , 1.9
From literature I know that more than 3% (i.e., 0.03*N) data are not useful, but I don't know where to cutoff? Could you suggest any re-sampling methods? How can I treat this problem? Classification or Regression?
This data is basically output of rejection sampling (see http://en.wikipedia.org/wiki/Approximate_Bayesian_computation). Now I hope that would be clear.
-
What is theta, and what is the distance measured from? What is the goal? What do you mean that the data is not useful? – Aniko Aug 22 '11 at 18:59
In place of "from literature" you might cite a specific paper. – Karl Aug 22 '11 at 22:23
I edited the title and tags to better match your question (hopefully correctly). – Aniko Aug 23 '11 at 15:31
One approach to choosing the cutoff value $\epsilon$ for ABC rejection sampling is the following (similar to Aniko's answer). Simulate several test data sets from known parameter values which are vaguely similar to your observed data (e.g. by performing ABC with a relatively large $\epsilon$). From the ABC output for a test data set, some criterion of performance compared to the true parameters can be calculated, such as mean squared error. Calculate this for all test data sets at many $\epsilon$ values, and choose $\epsilon$ to optimise the mean criterion (as this is a Monte Carlo estimate of its expectation). This requires many repetitions of the ABC algorithm, but can be done efficiently by using the same $N$ data simulations in every ABC algorithm (although this introduces some dependency between simulations).
In general, there is not a lot of published work on the choice of $\epsilon$. I think the approach above has been used somewhere and I will edit if I remember the references. An alternative is in "Choosing the Summary Statistics and the Acceptance Rate in Approximate Bayesian Computation" by Michael Blum. Other methods that I'm aware of apply only to SMC or MCMC methods.
-
But how I know the true parameter value in advance? – love-stats Sep 28 '11 at 19:07
Based on your edit, it appears that you are looking for guidance in selecting the tolerance parameter $\epsilon$ for ABC sampling. I don't know much about the topic, but $\epsilon$ should be small. A simple possibility is to select several different values and see whether the resulting posterior distributions look similar (based on new sets of samples). The largest value that still gives the same posterior can be used.
| 658
| 2,727
|
{"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}
| 3.203125
| 3
|
CC-MAIN-2013-48
|
latest
|
en
| 0.893308
|
http://www.math.mcgill.ca/~darmon/courses/13-14/algebra/blog.html
| 1,511,274,852,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-47/segments/1510934806388.64/warc/CC-MAIN-20171121132158-20171121152158-00485.warc.gz
| 455,385,257
| 9,349
|
# 189-235A: Algebra 1
## Blog
Week 1 (Sept 4 and 6). This week I gave a brief motivational overview of Abstract Algebra. One of the historical origins of the subject can be traced to the algebraic solution of the cubic equation discovered by the mathematicians of the 16th Century Italian Renaissance Niccolo Tartaglia, Scipione del Ferro and Girolamo Cardano. For a delightful account of the dramatic story surrounding this discovery, see Chapter 6 of
Journey through genius: the great theorems of mathematics by William Dunham.
Cardano's solution to the cubic was a turning point because it went beyond what the ancients had been able to achieve, suggesting that there might be a lot more to mathematics than was contained in Archimedes and Euclid. It also created a compelling case for the introduction and use of complex numbers. Attempts by mathematicians and philosophers to come to terms with the ``imaginary quantities" in Cardano's formula for the (very real, both in a mathematical and ontological sense) solutions of the cubic were an important impetus for the birth of modern abstract algebra.
Friday's lecture concluded with my beginning to write down the definition of an abstract ring (I didn't get to the end of it, and will conclude at the start of Monday's lecture.) This is to give you a feeling for the kind of general abstract structure we are aiming to study. But starting from week 2, we will be backtracking somewhat, and about very special instances of this concept, starting with the ring of integers which are, in many ways, the prototypical concrete example on which the abstract notion of ring is based.
Week 2 (Sept 9 -13). This week we made the first steps in describing the arithmetic in Z, following loosely sections 8 and 9 of the on-line notes. But our treatment was a bit different from the one given in the on-line notes. For instance on Monday, we introduced the set of positive integers as a set equipped with a zero element and a successor function, satisfying the axiom of induction. We then gave an inductive definition of addition and multiplication, and proved that the resulting operations satisfy the familiar rules (associativity, commutatitvity, distributivity of multiplication over addition...) The resulting (slightly pedantic) proofs are a good illustration of the general strategy of proof by induction: for more on that, see also Section 2.3. of the on-line notes. On Wednesday and Friday we turned to the Euclidean division algorithm and the algorithm for the gcd, following closely sections 8 and 9 of the notes.
This development of the integers served partly as a pretext to cover some of the basic "mathematical vocabulary", concerning proof, sets, functions, complex numbers that will be used throughout the course. This is the material of Part I of the on-line notes, which I am asking you to cover as independent reading. For some of you this might be a leisurely review; on the other side of the spectrum, if you have never been exposed to this material before, you may have to spend a bit more time reading and absorbing the material in Part 1, preferably in advance of the lectures. The sooner you are comfortable with the language, point of view, and concept of proof, the easier it will be to follow the lectures and do the assignments.
Organisational remark. My colleagues in the mathematics department know that some of you have not yet been able to register for the course. With high probability, the problem will sort itself out by the drop-add period if, as is often the case, sufficiently many students decide to drop 235A in favor of other classes that they might be trying out at the same time. If the course is still overbooked by then, you should go see Angela White on the 10th floor of Burnside. I have been assured that everything will be done so that no one who needs the course for her or his program is turned away because of limits on registration.
Week 3 (Sept 16-20).
Organisational remarks: This week I will be out of town on Thursday and Friday, and for this reason I am rescheduling my office hours exceptionally to Wednesday from 2:00-4:30.
On Friday, Antonio Lei will be filling in for me.
The first assignment . Note that your first assignment is due on Monday of next week. You can either hand it in during class or drop it in the assignment slot on the 10th floor of Burnside Hall, before 2:00 PM. I aim to return these assignments to you, like clockwork, on the Monday of the following week. Because of the number of students enrolled in the class, an army of TA's and graders will have to coordinate their work with a military precision to rival the planning that went into the invasion of the beaches of Normandy on D-day. For this reason, among others, absolutely no late assignments will be accepted. If you are unable to complete your work by the 2:00 PM deadline, just hand in what you've written up, and continue to think about the questions you haven't had time to complete. You are also encouraged to discuss them, of course, with your classmates, with me, or with any of the TA's. But such exchanges can only be profitable if you've spent a healthy amount of time before-hand thinking about the questions on your own.
Please remember to clearly indicate your name and the name/number of the course for which you are returning the assignment.
This week we will follow rather closely the material in sections 8-10 in the notes, culminating in the proof of the fundamental theorem of arithmetic (Section 10 of the notes). This theorem, which asserts that every integers can be expressed uniquely as a product of primes (raised eventually to certain powers) is the most fundamental fact about the multiplicative structure of the integers. There is a good reason why it is called the fundamental theorem of arithmetic, and it is also, in some sense, one of the most fundamental of all mathematical truths. It is also at the origins of the most important practical application of number theory, to public key cryptography. This application is based on the facts that
(a) it is not so hard to produce large primes, of 100 digits or so, on a computer (or a cell phone, for that matter), and
(b) given a product pq of two such primes p and q, it is virtually impossible to figure out what the primes p and q are, in any reasonably amount of time, even with a battery of dedicated super computers working on the problem in parallel for months (or years, or centuries).
We will explain (a) and (b) more thoroughly in the coming weeks.
Time permitting, on Friday we will also embark on the first steps in our discussion of congruences (Section 12 and part of Section 13.)
Week 4 (Sept 23-27).
On Friday of last week, Antonio introduced you to the notion of congruences, which turns out to be a very powerful piece of notation, and will furnish us with our first interesting example, beyond the prototypical one of the integers, of a ring. This week will be devoted to various topics related to congruences: notably the fact that Z/nZ is a ring, and that it is even a field if and only if n is prime. We use this to prove Fermat's little theorem, following up with a general discussion of linear congruence equations. Applications of congruences to the RSA cryptosystem will be discussed, as well as some discussion of the problem of primality testing.
Have you completely absorbed the language of set theory? You probably have if you laugh at the following joke.
Week 5 (Sept 30-Oct 4).
This week, we will begin by wrapping up our discussion of congruences by discussing a few further topics, such as Wilson's theorem, and (more importantly) the Chinese remainder theorem and its applications to efficient parallel computing.
We will then move on to the theory of rings of polynomials with coefficients in a field, following more or less the development in Chapter 4 of Eyal Goren's notes.
Office hours: This week, I again have to be absent on Friday afternoon after the lecture, so my office hours will be rescheduled to Wednesday, from 12:30-3:00.
Week 6 (Oct 7-11).
The week's lectures are devoted to rings of polynomials with coefficients in a field F, following more or less the development in Part 4 of Eyal Goren's notes. Highlights include the proof of the unique factorisation theorem which asserts that every polynomial in F[x] factors uniquely into a product of irreducible polynomials. This has strong implications, notably, on the number of distinct roots that a polynomial in F[x] can have.
A theme which emerges from the unique factorisation theorem (and from its proof, which is basically the same as the one we gave for the integers) is the strong analogy that exists between the rings Z and the polyomial rings F[x]. Seeking to develop this analogy even furhter, we introduce the notion of congruence modulo a polynomial and show that it enjoys all the same formal properties of the congruences modulo an integer which were discussed in previous weeks. In the same way that the study of congruences modulo an integer n led us to a new ring, the ring Z/nZ of residue classes modulo n, we are led to construct the rings F[x]/(f(x)) of polynomials modulo a given polynomial f(x). The remainder of the week (Wednesday and Friday) is devoted to the study of these so-called ``quotient rings", which are used, notably, to construct a finite field with 8 elemens, and related algebraic structures.
Time permitting, we will also discuss the factorisation of polynomials over finite fields, a topic that will also be taken up partly in assignment 3.
Remark on the midterm. This is our last full week of lectures before the midterm, which will be on the Wednesday of the following week. The midterm will be based on all the material that we've covered so far, including the material of this week's lectures. (But not the material of Monday, which may be devoted to some review of previous material.)
Week 7 (Oct 14-18).
This week was a short one, with no lecture on Monday because of the Thanksgiving holiday, and the midterm exam on Wednesday. In Friday's lecture I gave corrections for the midterm exam, and we had time in the second half to prove a new, and important, theorem about the field F':=F[x]/f(x) where f(x) is an irreducible polynomial. Namely, we showed that this new field contains the field F as a subfield in a natural way, and that it also contains a root of the polynomial f(x), viewed as a polynomial in F'[x] via the natural inclusion of F into F'. This general procedure for creating new rings (or fields) from existing ones by a process of "adjunction of roots" is an important one in modern abstract algebra--for instance it demystifies the complex numbers and the "imaginary quantities", like the square root of -1, whose "existence" was the object of intense philosophical debate until well into the 17th century.
Week 8 (Oct 21-25).
Having seen the striking list of formal analogies between our study of the integers and the set of polynomials with coefficients in a field, we are now ready and motivated to move to the next level of abstraction and discuss general rings, of which the integers, and the ring F[x] of polynomials, are two prototypical examples.
This week was largely devoted to covering Part 5 of Eyal Goren's notes. After briefly recalling the general ring axioms (which have already been seen before, of course) and some of their basic consequences, we introduce the basic notion of an ideal in a ring, an all-important concept which is meant to make up for the fact that the general notion of the gcd is not as well-behaved in a general ring as it is in Z and F[x].
We proved the theorem that every ideal in the ring Z of integers or in the ring F[x] of polynomials over a field is always principal. We then saw examples of ideals that are not principal, for instance in the ring Z[x] of polynomials with integer coefficients, or in the ring F[x,y] of polynomials in two variables over a field F.
Then, on Friday, I discussed the notion of products of ideals in rings like Z[sqrt(-5)], and explained how the non-unique factorisation of the number 6 into irreducible elements turned into an ostensibly unique factorisation of the principal ideal generated by 6, into four ideals. Very little was proved in this lecture, but this calculation gives a glimpse of one of the historical motivations of the introduction of ideals, by Dedekind. The terminology "ideal" mirrors Dedekind's notion of an "ideal number"---these are extra elements that one has to add to the set of irreducible elements of the ring Z[sqrt(-5)] under multiplication, in order to salvage a statement like unique factorisation (which now becomes unique factorisation into ideals, and not irreducible elemments.)
Week 9 (Oct 28 - Nov 1).
On Monday we introduced the notion of the quotient of a ring by an ideal and worked out a few examples. The notion of quotient (of one algebraic structure by another) represents a pinnacle of abstraction, as far as what you will see in this course. The best way to become familiar and comfortable with the notion is to work out many examples. In fact, you have already worked with quotients before in this class (a bit like Monsieur Jourdain in Molière's "Le Bourgeois Gentilhomme", who discovered he had been speaking in prose without ever being aware of it) when you performed calculations with the congruence rings Z/nZ, and the rings F[x]/(f(x)).
On Wednesday we discussed homomorphisms of rings, and proved that the kernel of any homomorphism is an ideal in the domain ring. This set the stage for the fundamental isomorphism theorem for rings to be covered on Friday, which establishes a dictionary between (surjective) homomorphisms having a given ring R as its domain, and the set of ideals in R.
Week 10 (Nov 4 - Nov 8).
This week we embarked on a new topic, namely, the theory of groups. These new mathematical structures are defined by a much simpler set of axioms than rings: a group is just a set endowed with a single binary operation satisfying the associative law, as well as possessing a (unique) neutral element for this operation, relative to which every element posseses an inverse. Groups turn out to be the ideal structure with which to describe and model the abstract notion of symmetry, which is pervasive in mathematics as well as in physics and computer science.
The most important examples of groups which we discussed were
• the groups GLn(F) of invertible n times n matrices with entries in a field F, the group operation being the usual matrix multiplication, and the identity element being the usual identity matrix.
• the permutation group Sn consisting of all bijective functions from the set {1,...,n} to itself, the group operation being the composition of functions. We introduced a powerful, versatile notation to describe elements of Sn, the so-called cycle notation, and explained how elements in Sn can be composed with each other rather efficiently in terms of this notation. Other advantages of the cycle notation is that it becomes easy to determine at a glance certain basic invariants of a permutation, such as its order: it is simply the least common multiple of the lengths of the disjoint cycles describing the permutation.
• the dihedral group D8 is the group of rigid motions preserving a square. It consists of the four rotations 1, r, r2, and r3, and the four reflections V, H, D1 and D about the horizonal, vertical, an digaonal axes of symmetry of the square.
We then described the notion of a homomorphism from one group to another, and discussed the important case of a homomorphism from a finite group G to a permutation group. We proved Cayley's Theorem that every finite group admits an injective homomorphism to a group Sn, where n can be taken to be the cardinality of G. The idea of the proof was to associate to each element of G a bijection from G to itself, arising from left multiplication by that element.
Further examples of homomorphisms were given, for instance, various examples of homomorphisms from D8 obtained by considering the effect of a rigid motion of the square on various naturally occuring structures inside the square (its vertices, sides, axes of symmetry, etc.)
Week 11 (Nov 11 - Nov 15).
This week we discussed the notion of cosets of a group H in a group G. We proved that a group G can always be written as a disjoint union of its cosets, which all have the same cardinality as that of H. As a result, we deduced Lagrange's theorem that, if G is a finite group and H is a subgroup of G, the cardinality of the latter must divide that of the former.
We then discussed the structure of the set G/H of all cosets for H in G, and asked when the ``natural" rule aH . bH = (ab) H gives rise to a well-defined operation on the set G/H. When it is well-defined, this rule gives rise to a composition law on G/H satisfying all the axioms of a group, and one says that G/H ```inherits" a group structure from the group structure on G. We saw that the answer to the question "When is G/H a group?" is "Sometimes, but not always!". More precisely, it is necessary and sufficient that H be a normal subgroup of G, i.e., that it be stable under conjugation by arbitrary elements of G. This means that if h belongs to H, and g is an arbitrary element of G, then hg = g-1 h g must belong to H as well.
This led to a study of the all-important nottion of the conjugacy relation in a group, and to the notion of conjugacy class, which is key to classifying the possible normal subgroups of a group.
Week 12 (Nov 18 - Nov 22).
On Monday we continued briefly with our discussion of conjugacy classes, normal subgroups, and their associated quotients, by describing a few illustrative examples, namely,
• the dihedral group of order 8;
• the symmetric group S4 and its normal subgroup of order 4.
We stated and proved the first isomorphism theorem for groups, and explained how it could be use to identify the structure of G/H in a few specific examples.
On Wednesday and Friday, I was out of town, and Antonio Lei filled in for me. His lectures followed sections 30-31 of the online notes, and were devoted to the all-important notion of an action of a group on a set, which is just another way of talking about permutation representations (homomorphisms from an abstract group to a permutation group) that were introduced already in week 10. In that week we saw that a finite group G can always be realised as a subgroup of a permutation group (Cayley's theorem) but one can then ask ``In how many fundamentally distinct ways can this be done?" We gave what amounts to a complete classification of the sets which can be equipped with a finite group action, in terms of the notion of cosets.
Week 13 (Nov 25 - Nov 29).
This week will be devoted to covering a few topics in group theory that are left in the notes, including sections 32, 33.4, and 33.5.
Week 14 (Dec 2 - Dec 3).
This week will consist of two consecutive Mondays from the point of view of the McGill calendar, and hence we will have two hours of lecture, but on Monday and Tuesday rather than on Monday and Wednesday!
These two lectures will be devoted to review of the material. I strongly advise you to work out a lot of problems from the posted "practice finals" in the "assignments" web site, and to come to these two lectures armed with precise questions! You will get a lot more out of listening to solutions to problems you have always spent time struggling with yourself (even if with only partial success.)
In addition, there will be extra review sessions by the TA's that week, with the following schedule:
Monday, December 2, 3:30-5:00. Dylan Atwell-Duval, in BH 1018.
Tuesday, December 3, 12:30-2:30. Henri Darmon, BH1111 or 10th floor lounge of Burnside Hall.
Tuesday, December 3, 4:00-6:00. Bahare Mirza, in Burnside Hall lounge (10th floor.)
Wednesday, December 4, 3:30-5:30. Henri Darmon, in Burnside Hall lounge (10th floor).
| 4,424
| 19,976
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.234375
| 3
|
CC-MAIN-2017-47
|
latest
|
en
| 0.94932
|
https://goprep.co/ex-7.a-q9-find-the-area-of-an-isosceles-triangle-each-of-i-1njpuq
| 1,642,461,002,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-05/segments/1642320300624.10/warc/CC-MAIN-20220117212242-20220118002242-00499.warc.gz
| 360,669,068
| 29,999
|
Q. 94.4( 13 Votes )
# <span lang="EN-US
Let,
a = 13 cm
b = 13 cm
And,
C = 20 cm
Now,
s =
=
= = 23 cm
We know that,
Area of triangle =
=
=
= 10
= 10 × 8.306
= 83.06 cm2
Therefore,
Area of isosceles triangle = 83.06 cm2
Rate this question :
How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses
RELATED QUESTIONS :
A piece of land iRS Aggarwal & V Aggarwal - Mathematics
The question consRS Aggarwal & V Aggarwal - Mathematics
The dimensions ofNCERT Mathematics Exemplar
A hand fan is madRD Sharma - Mathematics
Find the area of NCERT Mathematics Exemplar
| 247
| 865
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05
|
latest
|
en
| 0.809067
|
https://cr4.globalspec.com/thread/76754/What-Makes-Turbine-Output-Difference-Between-50-Hz-and-60-Hz
| 1,675,727,311,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-06/segments/1674764500365.52/warc/CC-MAIN-20230206212647-20230207002647-00422.warc.gz
| 199,068,348
| 28,909
|
Previous in Forum: Help with Torque Please Next in Forum: How to Bend Pipe 28" for Flow Gas in Workshop, Not in Pipe Company
Associate
Join Date: Nov 2011
Posts: 42
# What Makes Turbine Output Difference Between 50 Hz and 60 Hz?
03/21/2012 9:30 PM
For example, in the same gas turbine model, I have seen that the ouput of 50 Hz gas turbine is higher than 60 Hz. Could someone explain what makes this difference? Is it like both turbine have the same turbine structure, but 50 Hz turbine is bigger than 60 Hz turbine in size?
Interested in this topic? By joining CR4 you can "subscribe" to
Check out these comments that don't yet have enough votes to be "official" good answers and, if you agree with them, rate them!
Guru
Join Date: May 2009
Location: Richland, WA, USA
Posts: 21011
#1
### Re: What makes turbine output difference between 50 Hz and 60 Hz?
03/21/2012 9:50 PM
P = kNT; the 60Hz one should produce more.
__________________
In vino veritas; in cervisia carmen; in aqua E. coli.
Guru
Join Date: Jan 2008
Posts: 1753
#6
### Re: What makes turbine output difference between 50 Hz and 60 Hz?
03/23/2012 12:00 AM
IF, the same general size, same magnetic flux, I agree.
On the other hand (and it is a not a quite technical answer), if the turbine maker buys off the self standard size generator, all bets are off. And forget about optimum output.
Guru
Join Date: Dec 2008
Location: Long Island NY
Posts: 15286
#2
### Re: What makes turbine output difference between 50 Hz and 60 Hz?
03/21/2012 9:52 PM
The gas turbine produces the same amount of mechanical power at the shaft output. It depends on the generator attached to the turbine what frequency and voltage gets produced by the generator.
__________________
"Don't disturb my circles." translation of Archimedes last words
Guru
Join Date: Jul 2005
Location: Stoke-on-Trent, UK
Posts: 4347
#4
### Re: What makes turbine output difference between 50 Hz and 60 Hz?
03/22/2012 5:13 PM
He didn't mention a generator, though reference to 50 and 60Hz suggests that's the application. 60 Hz of course means 20% higher turbine speed than 50Hz, and it's not unreasonable to expect it to produce more power at higher speed, as the torque probably doesn't change much.
A generator produces a given Hz only at certain speeds, depending on the number of poles, 3000, 1500, 1000 etc rpm for 50 Hz, and 3600, 1800, 1200 etc rpm for 60 Hz. The maximum generator output power obviously depends on the power of the turbine.
__________________
Give masochists a fair crack of the whip
4
Power-User
Join Date: Dec 2010
Posts: 377
#3
### Re: What makes turbine output difference between 50 Hz and 60 Hz?
03/22/2012 2:10 AM
This is simple electrical theory describe as below:
Generator are generally operated by gas turbines directly. For a given number of poles, Frequency of power generated by generator is given by
f=NP/120,
where N = RPM of machine, P = Number of poles in generator
Assume generator is with 2 pole,
for generating power at 60 Hz, RPM of the gas turbines will be
N1=fx120/P or 60x120/2 = 3600
but for generating power at 50 Hz, gas turbine will run at
N2 = 50x 120/2 or 3000 Rpm.
Now Power generated by Gas turbine
Q(kw) = (torque N.m) x 2 pi x (rotational Speed RPM)/60000
It is evident that for generating same power at 3000 RPM, gas turbine should have higher torque value as compared to Gas turbine operating at 3600 rpm.
Hence Gas turbine for 50 Hz generator will be bigger then 60Hz generator for same power.
Associate
Join Date: Jun 2011
Posts: 48
#5
### Re: What makes turbine output difference between 50 Hz and 60 Hz?
03/22/2012 11:38 PM
Thus, the higher the torque developed at lower frequency for same power generated, the sturdy the turbine should be the reason why low frequency rotating machine are bigger compared to high frequency machine.
i also agree with guru that the 60 hz turbine - generator should produce more power than the 50 Hz, the poster must have miss something in his observation..:)
Guru
Join Date: Jun 2010
Location: srilanka
Posts: 2725
#7
### Re: What Makes Turbine Output Difference Between 50 Hz and 60 Hz?
03/23/2012 3:18 AM
Why don't you post the name plate details. Think about the mechanical aspects of gas/steam turbine-power vs speed.
__________________
pnaban
Power-User
Join Date: Oct 2007
Location: S. California, USA
Posts: 279
#8
### Re: What Makes Turbine Output Difference Between 50 Hz and 60 Hz?
03/23/2012 11:18 AM
All magnetic devices (transformers, inductors, magnetic heads, rotary machines, etc.) get larger as the frequency goes down because more magnetic material is required to keep the core from saturating. It is all about the magnetic flux density, B and the magnetic field, H
Guru
Join Date: May 2007
Location: NYC metropolitan area.
Posts: 3230
#9
### Re: What Makes Turbine Output Difference Between 50 Hz and 60 Hz?
03/23/2012 5:48 PM
It is amazing to see how many assumptions there are in this thread. There are two types of gas turbines in use by utilities, Industrial Turbines in which the compressor and output shafts operate at a fixed speed, and those that utilize aircraft-style jet engines where the compressor speeds are much higher and the only coupling between the engine and the turbine driving the generator is the hot gas streaming from the engine.
OP is probably talking about an industrial turbine in which case the limiting factor is not the generator but the materials and design of the turbine blades. As the turbine speed increases by 20% the rotational forces on the rotor blades increase by 40%. If the machine is rated for 50/60 Hz operation then these additional forces have to be factored into the overall design and usually result in a derating of the output shaft power as the speed goes up.
On the other hand jet-engine types are usually constrained by the output capabilities of the generator because designing the final stage turbine for those additional stresses is much more economic than designing all the stages for them. The manufacturer simply designs for 60 Hz operation (the jet engine doesn't care what the final speed is) and allows for 50 Hz operation as long as the output is derated (if necessary) consistent with Volts/Hz and the MVA rating at the lower speed of the generator.
All of the above is predicated upon identical operating conditions (temperature, fuel, altitude, hunidity, etc.), economics (cost, heat rate, availability, etc.), and the fact that any critical frequencies are well below the 50 Hz operational mode.
__________________
“Tell me and I forget. Teach me and I remember. Involve me and I learn.” Ben Franklin.
Guru
Join Date: Jan 2008
Posts: 1753
#10
### Re: What Makes Turbine Output Difference Between 50 Hz and 60 Hz?
03/23/2012 8:20 PM
Thank you for the elaboration on the turbines. In short, the turbine is a given. The off-the-shelf generator comes in discrete sizes, a given too.
With that, there are roughly 4 areas of deratings:
Turbine power / torque
Local air pressure / temperature
Frequencies / rpm
Discrete generator sizes
The engineering answer comes out after applying these factors to the theoretical calculations.
Active Contributor
Join Date: Dec 2010
Location: Brasil
Posts: 21
#11
### Re: What Makes Turbine Output Difference Between 50 Hz and 60 Hz?
03/24/2012 9:06 AM
The same can happend with diesel engines. Some are not suitable for both frequencies, and other have a derating from 50 to 60 hz. See an example at http://www.cumminsgdrive.com/accelsite/media/1703/QSX15-G8.pdf.
I know it is due to the tunning between turbo and others performance components, but not the details.
Associate
Join Date: Nov 2011
Posts: 42
#12
### Re: What Makes Turbine Output Difference Between 50 Hz and 60 Hz?
03/26/2012 8:59 AM
Thanks all of your answers. I really appreciate it. The reference gas turbine was SGT6-8000H and SGT5-8000H manufactured by Simens. SGT6-8000H is 60 Hz and its output is 274 MW, but SGT5-8000H is 50 Hz and its output is 375 MW. SGT5-8000H's output is quite higher than SGT6-8000H, isn't it?
2
Guru
Join Date: May 2007
Location: NYC metropolitan area.
Posts: 3230
#13
### Re: What Makes Turbine Output Difference Between 50 Hz and 60 Hz?
03/27/2012 3:58 PM
Your observation is correct but just because the only difference between them is 5 vs 6 in the product name does not make them twins. They are totally different machines, a quick look at the specs on the Siemens websites
http://www.energy.siemens.com/us/en/power-generation/gas-turbines/sgt6-8000h.htm#content=Technical%20Data
http://www.energy.siemens.com/us/en/power-generation/gas-turbines/sgt5-8000h.htm#content=Technical%20Data
reveals that the 50 Hz version weighs 160 tons more than the 60 Hz version, it's also longer, taller and wider by about 20% on all dimensions. Since the linear dimensions are 20% larger one would expect that the area dimensions should be about 40% greater resulting in approximately 40% greater power related parameters. Let's compare...Output 375 vs 275 MW (37%), Mass Flow rate 820 vs 600 kg/s (37%); yep, they're different sized machines based upon the same overall design scaled up.
__________________
“Tell me and I forget. Teach me and I remember. Involve me and I learn.” Ben Franklin.
Associate
Join Date: Nov 2011
Posts: 42
#14
### Re: What Makes Turbine Output Difference Between 50 Hz and 60 Hz?
03/28/2012 1:27 AM
Thank you!
If 60 Hz model is only scaled down, will 60 Hz model work well if 50 Hz machine works without problem or 60 Hz has to be tested since what?
Guru
Join Date: May 2007
Location: NYC metropolitan area.
Posts: 3230
#15
### Re: What Makes Turbine Output Difference Between 50 Hz and 60 Hz?
03/29/2012 11:27 PM
Not necessarily, and probably no. Scaling has its limits, and one of those limits is the law of unintended consequences. The problem of resonance or critical frequencies is always present in the design and operation of a large rotating mass such as a turbine.
A 50 Hz machine may have no critical frequencies below its operating point but start vibrating uncontrollably at say 58 and never reach 60 Hz without throwing a blade and self-destructing. That's what overspeed protection is for. So if you're planning on copying a design for one speed and scaling to the next speed without adequate modeling and testing expertise I wish you luck.
__________________
“Tell me and I forget. Teach me and I remember. Involve me and I learn.” Ben Franklin.
Interested in this topic? By joining CR4 you can "subscribe" to
| 2,713
| 10,584
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.515625
| 3
|
CC-MAIN-2023-06
|
latest
|
en
| 0.887893
|
http://waiferx.blogspot.com/2011/07/physics-presentation-interactions.html
| 1,532,250,404,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-30/segments/1531676593142.83/warc/CC-MAIN-20180722080925-20180722100925-00191.warc.gz
| 389,893,799
| 23,736
|
## 20110720
### Physics presentation: interactions
Yes, beer is good. (Perhaps there's an extra letter in this sign that shouldn't be there...) But if you walked into a microbrewery with a wide selection of fine beers on tap, how do you know which beer is good? You don't have the time, money, or wherewithal to order and drink a pint of each different type of beer. So what do you do?
Some microbreweries offer what they call a "Brew-Ski," where a small sample of each of their beers on tap is placed, literally, on a "ski," in order to introduce you to their line-up. (These samplers are sometimes called "flights.")
Sampling individual beers separately is fine, but what would happen if all the different beers from a "Brew-Ski" were combined into the same glass? Is this so wrong?
Which leads us to our discussion of different types of mechanical interactions--forces. Since we will investigate the details of these forces later in this (algebra-based college physics) course to, let's settle for the "Brew-Ski" approach, where we'll look at a selection of important forces, briefly. And since many situations in the real world involve more than one type of force acting on the same object at the same time, we'll consider the result of combining different types of beers--that is, forces--into a net force.
Here's our "Brew-Ski" force line-up. Remember, we'll go into more detail on each of these as the course progresses, but for now, this will just be an introductory tasting.
Weight, or the force of gravity, is the easy-cheesy force. For our purposes its magnitude is always equal to the product of the object's mass m, and the gravitational strength constant g ("acceleration due to gravity"), regardless of the motion or location of the object. (Movie link: "Biggest Cliff Jump on Youtube (100+ Feet).")
The normal force is the supporting contact force exerted by a surface. Its magnitude can vary from zero (no or barely any contact) up to ∞, but practically speaking a surface can only exert up to a maximum value, depending on structural integrity of the underlying material, until it fails. In this case, the floor can exert a normal force to support the weight of the contents of the room, but only up to when the room gets filled with too much water. (Movie link: "Collapsing floor by filling room with water.")
The tension force is the force exerted along a string, rope, or cable. Its magnitude can vary from zero (slack or barely pulling) up to ∞, but practically speaking a rope will have a maximum value, depending on the strength of its material/construction, until it fails. Here, a towing "snatch" strap exerts a tension force to pull on a stuck vehicle, but only up to when it is pulled too much. (Movie link: "Broken snatch strap.")
The static and kinetic friction forces apply to unsticking or subsequently sliding an object across a surface, respectively. The magnitude of the static friction force can vary from zero (no or barely trying to unstick an object) up to a maximum value, at which point the object becomes unstuck, and just begins to move. The magnitude of the kinetic friction force magnitude is always a constant value, once the object is already unstuck and moving. Note that the maximum magnitude of "stiction" (static friction) is greater than the magnitude of "sliption" (kinetic friction). (Movie link: "Static vs. Sliding Friction.")
So much for the "Brew-Ski" overview of different types of forces. Let's move on to more complex situations, and the tools used to handle them.
Most everyday situations involve more than one type of interaction happening at the same time--consider the forces acting on the board: weight, normal force, tension, static friction force (assuming that the girl is not sliding off of the board), and kinetic friction force (assuming that the board is already sliding across the sand).
Each force is represented with a vector--the magnitude (length) represents the strength of the interaction, and the direction represents, well, the direction of the force exerted. Typically many forces, all acting on the same object can be depicted on a free body diagram.
Forces all acting on the same object all add together to result in a net force (Fnet), which is what the summation operator signifies. Since mathematically vectors can be broken up into x- and y- components, and then separately added together to find the resulting x- and y- components of the net force.
Looking ahead, how many Newton's laws are there? (Three.) But there are only two ways to classify motion--constant, or changing, corresponding to Newtons first law, or second law. Or only two ways to classify net force--zero, or non-zero, corresponding to Newton's first law, or second law.
So if there's only two types of motion, and two types of net forces, what's up with Newton's third law? As it turns out, Newton's third law has nothing to do with motion or net force, but something else entirely, something much more universal and encompassing than considering a particular type of motion or net force...
| 1,107
| 5,066
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.5
| 4
|
CC-MAIN-2018-30
|
latest
|
en
| 0.953655
|
https://www.mbatious.com/topic/960/quant-boosters-hemant-malhotra-set-11/26
| 1,632,684,687,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-39/segments/1631780057913.34/warc/CC-MAIN-20210926175051-20210926205051-00666.warc.gz
| 888,863,286
| 25,389
|
# Quant Boosters - Hemant Malhotra - Set 11
• Roll's theroem
it states that if two value a and b are such that f(a)>0 and
f(b) < 0
then there must be a real root in between a and b
Example -
if f(x)=f(x)=2^x-x^2+1
f(3)=0
f(3.2) < 0
f(4)=2^4-16+1>0
so one root will lie between 3 and 4
now f(-1)=1/2-1+1>0
f(-2)=1/4-4+1
so one root lie between -1 and -2
and f(3)=0 so one root 3
so 3 roots
every equation of an odd degree has at least one real root
x^3-6x^2+15x+3=0 has atleast one real root and max real roots=3
now question
f(x)=x^3-6x^2+15x+3
f'(x) ((differentiation)
3x^2-12x+15=3((( x-2)^2+1))
so f'(x)>0 means function is increasing so it will cut x axis only once so one real root only
so this equation has one real root either positive or negative
now f(0)=3
f(-1) < 0
so one root will lie between o and -1 so negative real root
• Q9) x^3 - ax^2 + bx - a = 0 has three real roots then which of the following is true
a) a = 11
b) a not equal to 1
c) b = 1
d) b not equal to 1
• x^3 - ax^2 + bx - a = 0
(x-p)(x-q)(x-r)=x^3-(p+q+r)x^2+(pq+qr+rx)x-pqr
now compare
b=pq+qr+rp
a=p+q+r=pqr
(p+q+r/pqr)=1
sk 1/pq+1/qr+1/rp=1
so pq+qr+rp > 3 so b does not equal to 1
• Q10) root(x+1)-root(x-1)=root(4x-1) find number of real values of x
a) 1
b) 0
c) 2
d) not
• When you solve, you will get x=5/4 but that value will not satisfy initial equation so zero roots
• Q11) Find integral value of a for which x^2 - 2(4a - 1)x + 15a^2 - 2a - 7 > 0 is valid for any x
a) 2
b) 3
c) 4
d) none of the above
• if f(x) > 0 for all values of x
then D < 0
4(4a-1)^2-4(15a^2-2k-7) < 0
a^2-6a+8 < 0
2 < a < 4
so integral value is 3
• Q12) If roots p,q,r in HP of equation x^3 - 3ax^2 + 3bx - c = 0 then
a) q = 1/p
b) q = b
c) q = c/b
d) q = b/c
• approach 1
p+q+r = 3a
pq + qr + rp = 3b
pqr = c
q = 2pr/p+r
1 = 2c/q / [3b-c/q]
3b = 3c/q
q=c/b
approach 2
roots are in HP
so 1/p,1/q,1/r are the roots of
(1/x)^3-3z(1/x)^2+3b(1/x)-c=0
so -cx^3+3bx^2-3ax+1=0 are in AP
so 1/p+1/q+1/r=3b./c
so 3/q=3b/c
so q=c/b
• Q13) For all real values of find min and max value of (x^2-3x+4)/(x^2+3x+4)
• x^2-3x+4/(x^2+3x+4) = y
now make a quadratic in x
x^2(y-1)+3x(y+1)+4(y-1) = 0
now real x so D > = 0
9(y+1)^2-16(y-1)^2 > = 0
so (7y-1)(y-7) < = 0
so 1/7 < = y < = 7
• Q14) Number of possible value of integer p for which x^2 + px + 16 = 0 has integral roots
a) 4
b) 6
c) 2
d) not
• For roots to be integral Discriminant should be perfect square
D=k^2
a^2-64=k^2
a^2-k^2=64
Now easy ? Answer is 6
• Q15) In how many ways we arrange 6 boys and 4 girls such that
a) all girls are together
b) all boys are together
c) all boys are together , all girls together
d) no girls are together
• a) all girls are together so make a group of them and take it as a one unit
so they could be arranged in 4! ways so now we have BBBBBB(G) so total 7 persons , number of ways to arrange those 7!
so total ways =7! * 4
b) all boys are together so make it as one entity so they could be arranged in 6! ways
so total number of ways = 6! * 5!
c) all boys are together and all girls are together
so BBBBBBGGGG
6!*4!
and GGGGBBBBBB
so 4 * 6!
so total ways = 2 * 4! * 6!
d) no two girls are together
so _ B _ B _ B _ B _ B _ B _
first arrange 6 boys so ways 6!
now 6 boys will leave 7 spaces
now select 4 spaces and arrange 4 girls there
so 7c4*4!
so total ways = 6! * 7c4 * 4!
• Q16) Five boys and five girls form a line with the boys and girls alternating . The number of ways of making the line
• GBGBGBGBGB
so 5!*5!
now BGBGBGBGBG
so 5!*5!
so total ways = 5! * 5! + 5! * 5! = 2 * (5!)^2
• Q17) Number of arrangements that can be made with the letters of the word " MATHEMATICS " in which all vowels come together
• there are 4 vowels A,E,A,I , number of ways to arrange them = 4!/2!
now considering four vowels as one letter ( bcz we want them together ) we have 8 letters
M,T,H,M,T,C,S and one letter combining the vowels
so number of ways to arrange them = 8!/(2!)*2! ( bcz of 2 M and 2 T )
so total ways =8!/2!*2! * 4!/2!
• Q18) Find the rank of the word " SACHIN" in dictionary
61
62
63
61
61
46
62
61
| 1,700
| 4,082
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.875
| 4
|
CC-MAIN-2021-39
|
latest
|
en
| 0.72672
|
https://www.physicsforums.com/threads/how-much-water-is-in-the-atmosphere.709786/
| 1,660,314,999,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-33/segments/1659882571719.48/warc/CC-MAIN-20220812140019-20220812170019-00344.warc.gz
| 830,136,136
| 15,312
|
# How much water is in the atmosphere?
Anna-Banana
At any given moment do scientists have an idea of how much water is in the atmosphere? Can we compare it to how much fresh water is stored on earth as a liquid?
Staff Emeritus
Homework Helper
the mean moisture content of the atmosphere is equivalent to approx. 25 mm of liquid water covering the entire surface of the globe. (See the section 'Water Vapor in Earth's Atmosphere')
Staff Emeritus
At any given moment do scientists have an idea of how much water is in the atmosphere?
Thanks to weather satellites, yes, they do. You've seen weather satellite images that show the Earth as ocean and land, partly covered by clouds. Those are visible images of the Earth as seen from space. Looking at the Earth in infrared gives a very different picture. Several wavelengths are particularly sensitive to water vapor. For example, here's a 6.7 micron channel image:
1 person
Gold Member
http://ga.water.usgs.gov/edu/earthhowmuch.html details all the forms of water on Earth
"About 3,100 mi3 (12,900 km3) of water, mostly in the form of water vapor, is in the atmosphere at any one time. If it all fell as precipitation at once, the Earth would be covered with only about 1 inch of water."
Edit: Dang --- did it again ---- old post
keepit
According to the USGS Water Science School the amount of water in the atmosphere is .04 of a percent of the total of earth's fresh water and .001 percent of the amount of earth's total water. The increase in the amount of water vapor the atmosphere can hold doubles with an increase of 10 degrees (centigrade i believe).
Last edited:
| 370
| 1,625
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33
|
latest
|
en
| 0.940312
|
http://www.usmint.gov/kids/teachers/lessonPlans/viewLP.cfm?id=358
| 1,472,555,101,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-36/segments/1471982976017.72/warc/CC-MAIN-20160823200936-00044-ip-10-153-172-175.ec2.internal.warc.gz
| 746,256,097
| 12,823
|
# Using a Coordinate Grid
### Summary
The student will manipulate ordered pairs in the first quadrant of a coordinate plane.
• Quarter
### Coin Program(s)
• 50 State Quarters
### Objectives
The student will manipulate ordered pairs in the first quadrant of a coordinate plane.
• Math
### Minor/supporting Subject Area Connections
• Art
• Language Arts
### Class Time
Sessions: Two
Session Length: 45-60 minutes
Total Length: 91-120 minutes
### Groupings
• Whole group
• Small groups
• Pairs
• Individual work
### Terms and Concepts
• Coordinate grid
• Coordinate plane
• Ordered pairs
• Horizontal
• Vertical
### Materials
• “Coordinate Grid” worksheet (page 20), one per student
• Paper
• Pencil
• “Enlarge a Quarter” worksheet (page 21), one per student
• One picture of a quarter, enlarged (see “Reproducible Coin Sheets,” pages 31 and 32)
• Markers or crayons
• Rulers
### Preparations
• Make copies of the following:
• “Coordinate Grid” worksheet (page 20)
• “Enlarge a Quarter” worksheet (page 21).
• Enlarge a picture of a quarter for display.
• Using a ruler, draw a grid over the quarter with a black marker.
### Worksheets and Files
Lesson plan, worksheet(s), and rubric (if any) at www.usmint.gov/kids/teachers/lessonPlans/pdf/354-359.pdf.
1. Put list of terms on the board and review definitions.
2. Demonstrate that finding an ordered pair on a grid (5,4) would be 5 over to the right (horizontally) and 4 up (vertically). Pass out the “Coordinate Grid” worksheet (page 20) and ask students to complete it.
3. Display the enlarged picture of the quarter with the grid. Demonstrate how to enlarge the picture on the board as follows: Measure the distance between grid lines, Multiply the distance between grid lines by three, and draw a larger grid on the board, transfer the image from the paper to the board by drawing the contents of each grid block in the appropriate box on the board.
4. Distribute the “Enlarge a Quarter” worksheet (page 21). Have students use a pencil and a ruler to draw in grid lines (to make a grid five boxes across, and five boxes down, gridlines should be three centimeters apart). NOTE: For younger students, you may wish to draw in gridlines before copying sheets.
5. Students will enlarge the picture, using the grid as a guide.
6. Students can erase the grid lines, and then color in their picture. These pictures can be displayed in the classroom.
### Enrichments/Extensions
• Students can create a picture (e.g., a tree, a star) or design made up of points plotted on a grid, and determine the ordered pair for each point. The student can then read the ordered pairs to a partner, who will plot them on a grid, connecting the dots to achieve the same picture.
• Students can create their own coordinate grid puzzle, like that on page 20. Students can write a question referring to some aspect of the lesson, and then “hide” the answer in a grid. By assigning a letter to various points in the grid, and then listing the ordered pairs in the correct order, students will make a puzzle for a partner to decipher.
Use the worksheets and class participation to assess whether the students have met the lesson objectives.
### Games
Discipline: Math
Domain: 5.G Geometry
Cluster: Graph points on the coordinate plane to solve real-world and mathematical problems
Standards:
• 5.G.1. Use a pair of perpendicular number lines, called axes, to define a coordinate system, with the intersection of the lines (the origin) arranged to coincide with the 0 on each line and a given point in the plane located by using an ordered pair of numbers, called its coordinates.
• Understand that the first number indicates how far to travel from the origin in the direction of one axis, and the second number indicates how far to travel in the direction of the second axis, with the convention that the names of the two axes and the coordinates correspond (e.g., x-axis and x-coordinate, y-axis and y-coordinate).
• 5.G.2. Represent real world and mathematical problems by graphing points in the first quadrant of the coordinate plane, and interpret coordinate values of points in the context of the situation.
Discipline: Mathematics
Domain: 3-5 Data Analysis and Probability
Cluster: Formulate questions that can be addressed with data and collect, organize, and display relevant data to answer them.
| 987
| 4,359
|
{"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}
| 4.03125
| 4
|
CC-MAIN-2016-36
|
latest
|
en
| 0.78095
|
https://www.redhotpawn.com/forum/help/how-does-the-rating-system-work.26024
| 1,537,578,301,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-39/segments/1537267158001.44/warc/CC-MAIN-20180922005340-20180922025740-00211.warc.gz
| 846,785,462
| 6,762
|
Please turn on javascript in your browser to play chess.
#### Help Forum
1. KJCavalier
Happier Now!!
08 Jul '05 09:57
I am just curious as to know how the rating system work. if anyone could explain I would be appreciated
2. Ragnorak
08 Jul '05 09:59
Originally posted by KJCavalier
I am just curious as to know how the rating system work. if anyone could explain I would be appreciated
The help section explains it better than most people on here could.
D
3. Bowmann
Non-Subscriber
08 Jul '05 15:53
Firstly, a completely random number between 0 and .9 is generated, multiplied by the day of the month and rounded off.
Then, this number (again, by random determination) is either added to or subtracted from your current rating.
4. Aiko
Nearing 200000...!
08 Jul '05 16:30
Originally posted by Bowmann
Firstly, a completely random number between 0 and .9 is generated, multiplied by the day of the month and rounded off.
Isn't it easier to just pick a random number from the range available days in the current month? It is not so random if 'lower ranked' numbers have a higher expectancy of getting picked...
Like Ragnorak said: http://www.redhotpawn.com/help/index.php?help=faq
5. 09 Jul '05 01:34
Erm, it doesn't. Ignore them little numbers unless you want to be completely surprised by someones ability or lack of it.
6. invigorate
Only 1 F in Uckfield
09 Jul '05 11:21
Originally posted by KJCavalier
I am just curious as to know how the rating system work. if anyone could explain I would be appreciated
Basically If you beat someone higher than you, you go up a lot, if you beat someone lower than you, you go up a bit.
If you draw the affect on your rating is minimal.
If you lose to someone lower than you, you drop heavily, if you lose to someone higher than you, you drop a bit.
The most you can go up or down in one game is 32 rating points.
Finally like all good statistical formulae it works. If you play honestly and regularly you will find your rating tracks your skill level accurately. You will no longer be a name but a number +/- 50
Best regards
Mr 1250!
7. KJCavalier
Happier Now!!
13 Jul '05 10:03
Originally posted by invigorate
Basically If you beat someone higher than you, you go up a lot, if you beat someone lower than you, you go up a bit.
If you draw the affect on your rating is minimal.
If you lose to someone lower than you, you drop heavily, if you lose to someone higher than you, you drop a bit.
The most you can go up or down in one game is 32 rating points.
F ...[text shortened]... ill level accurately. You will no longer be a name but a number +/- 50
Best regards
Mr 1250!
I thank everyone for their responses(even slightly humorous ones) I read the FAQ and saw how they figured it out. My algebra is a bit to rusty for that though. but enjoy playing the game so I will be around for a while
| 730
| 2,853
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-39
|
latest
|
en
| 0.943412
|
http://www.maa.org/external_archive/joma/Volume8/Kalman/index.html?device=mobile
| 1,454,785,737,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-07/segments/1454701147492.21/warc/CC-MAIN-20160205193907-00278-ip-10-236-182-209.ec2.internal.warc.gz
| 515,071,141
| 4,840
|
# The Journal of Online Mathematics and Its Applications
Volume 8. March 2008. Article ID 1663
# The Most Marvelous Theorem in Mathematics
## Introduction
My favorite theorem in all of mathematics is what I call Marden's theorem. It relates the roots of a polynomial p(x) and those of its derivative p′(x). This is a familiar idea in calculus, where Rolle's theorem tells us that a root of the derivative must occur between any pair of roots of the original function. That is one kind of relationship between the roots of p and the roots of its derivative.
Figure 1: Rolle's Theorem: Between roots of p(x) there is a root of p′(x).
However, Marden's theorem is set in the complex plane. In that context, the polynomial p(z) has the same algebraic form that we meet in calculus, for example, p(z) might be given by z3 + a2 z2 + a1 z + a0. But now, the coefficients aj are permitted to be fixed complex numbers, and the variable z likewise can take on any complex value. We can still discuss roots of p -- values of z where p(z) = 0 -- and we compute p′(z) in the familiar way. For the example, we have p′(z) = 3z2 + 2a2 z + a1. If you know where the roots of p are, what can you say about the roots of p? Does Rolle's theorem still hold?
Where the real numbers are envisioned as lying along a line, complex numbers occupy a plane. The roots of p and p are points in that plane. One might ask whether roots of p must be found between roots of p, as in Rolle's theorem. But note that there is some ambiguity about the meaning of between when dealing with points in a plane. One obvious idea would be to ask whether a root of p must lie on the line segment between roots of p. But this cannot be true. Consider, if p is a cubic (that is, of third degree), and if the roots are not all in a line, then these roots determine a triangle. It is not possible for there to be a root of the derivative on the line between each pair of roots of p, as p has only two roots. If some version of Rolle's Theorem is to hold, the two roots of p′ must somehow be shared among the three sides of the triangle.
To see how this might occur, let us examine an example. Suppose p(z) = (z2 + 1)(z − 1) = z3z2 + z − 1. The roots are at 1, i, and i. Meanwhile, p′(z) = 3z22z + 1, which has roots at . As expected, these roots of the derivative do not lie on the line segments joining the roots of p, as illustrated in Figure 2 below, where the roots of p appear as black dots and the roots of p as blue dots.
Figure 2: Roots of p (blue dots) do not lie on segments joining the roots of p (black dots).
On the other hand, notice how the roots of p' can be found near to each side of the triangle, and as a result, they are completely surrounded by the roots of p. It turns out that this is always the case, as asserted by Lucas' Theoremall of the roots of the derivative must lie in the convex hull of the roots of the original polynomial. In particular, if p(z) is a third degree polynomial with roots forming a triangle (as in the example) then the roots of the derivative p′(z) must lie inside or on this triangle. That is what Lucas' theorem says. But we can say more, and that is where Marden's theorem comes in.
Marden's theorem gives a wonderfully geometric recipe for finding the exact positions of the roots of p when p is a cubic polynomial with distinct noncollinear roots in the complex plane. These roots are the vertices of a triangle, and a unique ellipse can be inscribed in the triangle tangent to the sides at their midpoints. Like any ellipse, it has two special points called foci. The foci are the roots of p′(z). This situation is illustrated below in Figure 3. The roots of p are at the vertices of the triangle, the midpoints of the sides of the triangle are indicated in red, and the foci of the ellipse are shown in blue.
Figure 3: A Marvelous Theorem: p(z) = 0 at the vertices of the triangle; p′(z) = 0 at the foci of the inscribed ellipse.
When I first read of this result in my third year of college, I thought it was the most marvelous theorem in mathematics, a feeling that persists to this day. I call it Marden's theorem because I read it in a book by Marden, although Marden himself credits Siebeck with the original discovery. My purpose in this exposition is to celebrate Marden's theorem and to share it with a broad audience. I will present an interactive dramatization of Marden's theorem, as well as some of the history, and a proof. The proof is completely elementary and yet draws on background information from several diverse parts of elementary mathematics. To fully understand the proof, the reader must be familiar with the geometric properties of ellipses, fundamentals of complex analysis, linear transformations, and algebraic properties of polynomials. I have endeavored to provide the necessary background for all of these topics, so that anyone with a command of first year calculus can work through whatever details are unfamiliar or interesting. Most of this background is included as hyper−text, meaning that the reader can reach it through links, at his or her option. You may already have noticed this in connection with Rolle's Theorem above, as well as some of the terminology. The presentation of the background topics also includes some interactive dramatizations which illustrate fundamental ideas.
The overall organization comprises four independent sections, plus a list of references. You can choose which of these are of interest, and may read them in any order. One possibility is to systematically explore the mathematical background. Alternatively, as you read the proof, you can follow links to each background topic at the point where it is relevant.
Note that one of the motivations for presenting this material in a hyper−text format is to include much more detail than would be possible in a print presentation. You are encouraged to explore the details that you find most interesting, without feeling compelled to read everything. For a more traditional printed presentation, or an overview in general terms, see Kalman [2] and [3].
| 1,456
| 6,131
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.34375
| 4
|
CC-MAIN-2016-07
|
latest
|
en
| 0.926352
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.