url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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http://www.haskell.org/pipermail/haskell/2001-February/006576.html | 1,411,195,165,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657132883.65/warc/CC-MAIN-20140914011212-00131-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 554,751,187 | 2,216 | # realToFrac vs. fromRealFrac
Marcin 'Qrczak' Kowalczyk qrczak@knm.org.pl
3 Feb 2001 13:14:12 GMT
```Sat, 03 Feb 2001 23:34:37 +1100, Manuel M. T. Chakravarty <chak@cse.unsw.edu.au> pisze:
> fromIntegral :: (Integral a, Num b) => a -> b
> fromRealFrac :: (RealFrac a, Fractional b) => a -> b
>
> (`fromRealFrac' is also defined in Figure 7.)
>
> However, in Appendix A ``Standard Prelude'', we have
>
> realToFrac :: (Real a, Fractional b) => a -> b
> realToFrac = fromRational . toRational
>
> instead of a definition for `fromRealFrac'.
>
> Both GHC and Hugs go by Appendix A. What was the original
> intention?
I don't know. I always assumed that it's realToFrac. It yields a
quite consistent picture.
Classes provide conversions to and from two "universal" types:
toInteger (from Integral), fromInteger (to Num),
toRational (from Real), fromRational (to Fractional).
There are also conversions with both ends overloaded, defined as
compositions of appropriate conversions going through an universal
type. One of them converts from Real to Fractional and is thus called
realToFrac. The other converts from Integral to Num, and since Num
in the context of numeric types does not mean anything (all numeric
types are Num), it is omitted from the name fromIntegral.
Classes for these four conversions are chosen in a natural way.
toInteger and toRational require that appropriate types contains
*at most* integral and real values. fromInteger and fromRational
require that the given type contains *at least* integral and fractional
values (and all numeric types contain at least integral values).
It follows that for most injections (except between one complex type
to another) either of fromIntegral or realToFrac works. If both work,
they give the same result, although fromIntegral should be slightly
more efficient. Narrowing the range or precision is not considered
a failure to be an injection.
fromRealFrac taking RealFrac as an argument does not make much sense:
it is not necessary to have fractional numbers to be convertible to
Rational, because integral types will do too.
--
__("< Marcin Kowalczyk * qrczak@knm.org.pl http://qrczak.ids.net.pl/
\__/
^^ SYGNATURA ZASTĘPCZA
QRCZAK
``` | 596 | 2,252 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-41 | latest | en | 0.902549 |
https://xmphysics.com/2023/01/11/15-2-3-flux-change-vs-flux-cut/ | 1,686,168,199,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654012.67/warc/CC-MAIN-20230607175304-20230607205304-00065.warc.gz | 1,137,152,595 | 23,742 | # 15.2.3 Flux Change vs Flux Cut
Let’s recap our study of Faraday’s Laws so far.
Stationary Coil
First, there is the transformer emf: when a stationary coil of area A and N turns is facing a changing magnetic field B, the magnitude of the induced emf is proportional to the rate of change of magnetic flux linkage (of the coil).
$\displaystyle \varepsilon =\frac{{d\Phi }}{{dt}}=\frac{{d(NBA)}}{{dt}}=NA\frac{{dB}}{{dt}}$
Notice that B has to vary at a constant rate to maintain a constant ε.
Moving Wire
Next, there is the motional emf: when a conductor of length L is moving at velocity v perpendicularly across a constant magnetic field B, the magnitude of the induced emf is proportional to the rate of flux cutting (by the moving conductor).
$\displaystyle \varepsilon =BLv$
Notice that the conductor only has to maintain a constant v to maintain a constant ε.
Moving/Rotating/Deforming Coil
When a coil moves, or rotates, or deforms (e.g. shrinks or expands) in a constant magnetic field, emf may also be induced in the coil. This is fundamentally motional emf: a coil is made up of wire segments. Those wire segments are cutting magnetic flux as the coil moves/rotates/deforms. The magnitude of the induced emf in the coil can be obtained by summing up the motional emfs induced in all the segments.
$\displaystyle \varepsilon =\sum{{BLv}}$
Alternatively, we can also see that as the coil moves/rotates/deforms, the magnetic flux in the coil is changing. Hence, the magnitude of the induced emf can also be deduced from
$\displaystyle \varepsilon =\frac{{d\Phi }}{{dt}}=\frac{{d(BA)}}{{dt}}=B\frac{{dA}}{{dt}}$.
Both approaches are good, as long as you apply them correctly.
Example
A rectangular coil (30 cm by 20 cm) is moving at a constant speed of 5.0 cm s-1, entering and then leaving a region of uniform magnetic field of 0.80 T. The total resistance of the coil is 6.0 W. Determine the direction and magnitude of the induced current in the coil when it is
a) entering,
b) moving within, and
c) leaving the field.
Approach 1: Flux Cutting
a)
As the coil enters the field, only the PQ segment of the coil is cutting the flux (perpendicularly).
Using the FRHR, P and Q are the positive and negative terminals of the “battery” respectively. So the current is going to flow in an anticlockwise direction in the coil.
The emf induced between P and Q is therefore
$\displaystyle \varepsilon =BLv=(0.80)(0.30)(0.050)=0.012\text{ V}$
The induced current is therefore
$\displaystyle I=\frac{V}{R}=\frac{{0.012}}{{6.0}}=2.0\text{ mA}$
b)
When the entire coil is in the field, both the PQ and RS segments of the coil are cutting the flux. But note that PQ is trying to push current in the anticlockwise direction, whereas RS is trying to push current in the clockwise direction. The two emfs “cancel” each other. So the resultant emf and current is zero.
c)
As the coil leaves the field, only the RS segment of the coil is cutting the flux (perpendicularly).
Using the FRHR, R and S are the positive and negative terminals of the “battery” respectively. So the current is going to flow in a clockwise direction in the coil.
The magnitude of the emf and current is the same as part a).
Approach 2: Flux Changing
a) As the coil enters the field, a larger and larger area is exposed to the magnetic field. So the coil experiences an increasing magnetic flux (going into page). To oppose this change, the induced emf ought to be producing magnetic flux in the opposite direction (coming out of the page) (Lenz’s law). Using the RHGR, we can conclude that the induced current is going to be anticlockwise.
Since it takes $\displaystyle 20\div 5.0=4.0\text{ s}$ for the coil to complete the increase in flux,
$\displaystyle \varepsilon =\frac{{d\phi }}{{dt}}=\frac{{d(BA)}}{{dt}}=\frac{{(0.80)(0.30\times 0.20)}}{{4.0}}=0.012\text{ V}$
The induced current is therefore
$\displaystyle I=\frac{V}{R}=\frac{{0.012}}{{6.0}}=2.0\text{ mA}$
b) When the entire coil is in the field, the magnetic flux of the coil is constant. Immediately, we can conclude that there is no induced emf and current around the coil.
c)
As the coil leaves the field, a smaller and smaller area is exposed to the magnetic field. So it experiences a decreasing magnetic flux (going into the page). To oppose this change, the induced emf ought to be producing magnetic flux in the same direction (going into the page) (Lenz’s law). Using the RHGR, we can conclude that the induced current is going to be clockwise.
The magnitude of the emf is same as part a).
Concept Test
3025 | 1,193 | 4,585 | {"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": 9, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2023-23 | latest | en | 0.906332 |
https://www.cybrary.it/catalog/practice-labs-module/defining-simple-functions-0428ca06-7bf2-4702-bc8e-dd25e0dbf9df/?coursePermalink=/catalog/practice_labs/introduction-to-programming-using-java-script/ | 1,680,259,311,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949598.87/warc/CC-MAIN-20230331082653-20230331112653-00308.warc.gz | 800,928,094 | 35,313 | # Defining Simple Functions
Practice Labs Module
Time
18 minutes
Difficulty
Intermediate
Welcome to the Defining Functions Practice Lab. In this module, you will be provided with the instructions and devices needed to develop your hands-on skills.
Join over 3 million cybersecurity professionals advancing their career
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Overview
### Introduction
Welcome to the Defining Functions Practice Lab. In this module, you will be provided with the instructions and devices needed to develop your hands-on skills.
### Learning Outcomes
In this module, you will complete the following exercise:
• Exercise 1 - Implementing a Calculator Using Functions
After completing this lab, you will be able to:
• Structure Python programs by defining functions
### Exam Objectives
The following exam objectives are covered in this lab:
• 4.1 Construct and analyze code segments that include function definitions
### Lab Duration
It will take approximately 15 minutes to complete this lab.
### Exercise 1 - Implementing a Calculator Using Functions
A function is a block of program statements, which can be defined once and called repetitively in a program. You have been using built-in functions like print() all this while. This function helps you print any output in no time. However, not all your needs may be met through built-in functions. In these cases, Python allows you to define your own functions.
In Python, a user-defined function is declared with the keyword def and followed by the function name. For example:
def myfunc():
A colon follows the closing parenthesis to suggest that the following block of code will contain the function definition.
Any arguments are specified within the opening and closing parentheses just after the function name, as shown here:
def myfunc(arg1, arg2):
After defining the function name and arguments(s), a block of program statement(s) start at the next line and these statement(s) must be indented. For example:
def myfunc(arg1, arg2):
print (arg1)
print (arg2)
The above function named myfunc() takes two arguments and prints their values.
Once defined, you can call this function from within the Python program using the function name, parenthesis (opening and closing) and parameter(s). For example, to call the myfunc() function, you may use the following statement:
myfunc(3, 4)
The above statement will call the myfunc() function with 3 and 4 as arguments and the function will print 3 and 4 as the output.
You can call the same function to print the string “hello world” too, as follows:
myfunc(“hello”, “world!”)
A user-defined function can also return a value instead of directly printing the result. In Python, the return statement is used to return a value from a function. The return statement without an expression argument returns none.
If you modify the myfunc() function to return the parameters after adding number 1 to each parameter, the function will look as follows:
def myfunc(arg1, arg2):
newarg1 = arg1 + 1
newarg2 = arg2 + 1
return(newarg1, newarg2)
Now in this case, the myfunc() function will not print the result on its own. It will just return two values newarg1 and newarg2. The above function can be called and results can be printed using the following combined statement:
print(myfunc(3, 4))
The above statement will call the myfunc() function with 3 and 4 as arguments. This print function will print the values returned by the myfunc() function.
A user-defined function can be called multiple times. Your own user-defined functions can also be a third-party library/module for other users. In addition to fostering reusability and saving development time, user-defined functions help make the code well organized, easy to maintain, and developer-friendly.
In this exercise, you will perform three tasks. In the first task, you will download the Python program file from the Intranet and open it in the IDLE environment. In the second task, you will write specific functions to perform different mathematical operations on the numbers accepted at runtime and print the results. In the third task, you will modify the program to have the user-defined functions return values instead of directly printing them.
### Learning Outcomes
After completing this exercise, you will be able to:
• Structure Python programs by defining functions
Learning Partner
Comprehensive Learning
See the full benefits of our immersive learning experience with interactive courses and guided career paths. | 929 | 4,494 | {"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} | 3.234375 | 3 | CC-MAIN-2023-14 | latest | en | 0.838155 |
https://documen.tv/question/15-points-a-home-meal-delivery-business-delivers-34-5-pounds-of-food-each-day-the-meal-delivery-21490548-36/ | 1,632,224,954,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057202.68/warc/CC-MAIN-20210921101319-20210921131319-00632.warc.gz | 268,851,718 | 15,580 | ## 15 points A home meal delivery business delivers 34.5 pounds of food each day. The meal delivery company delivers meals that eac
Question
15 points
A home meal delivery business delivers 34.5 pounds of food each day. The meal delivery
company delivers meals that each weigh 24 ounces. How many meals will he deliver in 4
days? (1 lb. = 16 oz.) Fill in the blank. They will deliver
meals in 4 days. *
in progress 0
2 weeks 2021-08-29T03:19:47+00:00 1 Answers 0 views 0 | 136 | 473 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2021-39 | latest | en | 0.929711 |
http://sinclairzxworld.com/viewtopic.php?f=5&t=2236&sid=92b0a04fd13b3e9510569fc7f00e062f | 1,516,623,952,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891316.80/warc/CC-MAIN-20180122113633-20180122133633-00096.warc.gz | 304,574,240 | 8,074 | Random 10-digit generator
Anything Sinclair ZX Basic related; history, development, tips - differences between BASIC on the ZX80 and ZX81
Shaun_B
Posts: 379
Joined: Wed Apr 22, 2009 9:22 am
Random 10-digit generator
Attached is a listing that generates a random 10-digit ticket with each number from 0 to 9 inclusive in the sequence only once per run.
I wrote it to answer a code-golf question on Stack Exchange, with the ZX81 being an interesting beast as it's not very easy to golf on it (i.e., minimise and obfuscate the answers). I'm sure one of you guys could make it more golf-like. In fact, looking at it I could remove the GOSUB to increase the program speed a little; I could also change for FOR loop to could from 0 to 9.
Enjoy,
Shaun.
Attachments
zx81-ticket-generator.png (12.72 KiB) Viewed 515 times
XavSnap
Posts: 472
Joined: Sat May 10, 2008 3:23 pm
Re: Random 10-digit generator
May be faster! (don't use the recursive RND function! Only 10 loops)
1 LET A\$="0123456789"
2 LET B\$=""
3 LET R=INT(RND*LEN A\$)+1
5 LET B\$=B\$+A\$(R)
6 LET A\$=A\$( TO R-1)+A\$(R+1 TO)
7 IF A\$<>"" THEN GOTO 3
8 PRINT B\$
9 STOP
Shaun_B
Posts: 379
Joined: Wed Apr 22, 2009 9:22 am
Re: Random 10-digit generator
Thanks for the tip, my amended answer did credit you here.
The listing is as follows (forgive the GOTO abuse... actually, don't):
Code: Select all
``````1 LET A\$="0987654321"
2 LET R=INT (RND* LEN A\$)+1
3 PRINT A\$(R);
4 LET A\$=A\$( TO R-1)+A\$(R+1 TO )
5 GOTO 2+((A\$="")*4)
``````
Regards,
Shaun.
XavSnap
Posts: 472
Joined: Sat May 10, 2008 3:23 pm
Re: Random 10-digit generator
Cool !
dr beep
Posts: 1057
Joined: Thu Jun 16, 2011 7:35 am
Location: Boxmeer
Re: Random 10-digit generator
IMG_4887.PNG (94.58 KiB) Viewed 401 times
Shaun_B
Posts: 379
Joined: Wed Apr 22, 2009 9:22 am
Re: Random 10-digit generator
The problem with Code Golf on the Stack Exchange is that it expects answers to be least typing, rather than actual fewest bytes because modern programming simply counts the number of characters in your script as how much memory it's taking.
Regards,
Shaun.
1024MAK
Posts: 1822
Joined: Mon Sep 26, 2011 9:56 am
Location: Looking forward to summer in Somerset, UK...
Re: Random 10-digit generator
So to the age old question: is PRINT one character (as stored internally), or five...
Mark
Who is online
Users browsing this forum: No registered users and 2 guests | 750 | 2,417 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.847029 |
http://www.ck12.org/book/CK-12-Middle-School-Math-Concepts-Grade-7/section/4.7/ | 1,493,404,124,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123046.75/warc/CC-MAIN-20170423031203-00017-ip-10-145-167-34.ec2.internal.warc.gz | 480,863,318 | 37,373 | # 4.7: Differences of Integers Using a Number Line
Difficulty Level: At Grade Created by: CK-12
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Practice Differences of Integers Using a Number Line
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Over the past few months Mark has borrowed a lot of money from his sister. At this point he is $25 in debt to his sister! His sister makes him a deal and says that she will take away$8 of the debt if Mark cleans her room for her. How could Mark use a number line to find the difference of in order to determine how much he will still be in debt to his sister after cleaning his sister's room?
In this concept, you will learn how to subtract integers with the help of a number line.
### Subtracting Integers Using a Number Line
Integers are the set of whole numbers and their opposites.
There are many different strategies for subtracting integers. One strategy for subtracting integers is to use a number line. To subtract two integers using a number line:
1. First, draw a number line.
2. Then, find the location of the first integer on the number line.
3. Next, if the second integer is positive, move that many units to the left from the location of the first integer. If the second integer is negative, move the absolute value of that many units to the right from the location of the first integer. *Note that these are the opposite movements from when you ADD integers using a number line! This is because addition and subtraction are opposite operations.
Let's look at an example.
Use a number line to find the difference of .
Then, find the location of 4 (the first integer in your difference) on the number line.
Next, notice that the second integer, 3, is positive. This means you will be moving to the left. Starting at 4, move to the left 3 units.
You end up on 1. The answer is 1.
So .
Let's look at another example.
Use a number line to find the difference of .
Then, find the location of 4 (the first integer in your difference) on the number line.
Next, notice that the second integer, -3, is negative. This means you will be moving to the right. Starting at 4, move to the right 3 units.
You end up on 7. The answer is 7.
So .
Let's look at one more example.
Use a number line to find the difference of .
Then, find the location of -4 (the first integer in your difference) on the number line.
Next, notice that the second integer, -3, is negative. This means you will be moving to the right. Starting at -4, move to the right 3 units.
You end up on -1. The answer is -1.
So .
Remember that whether you move to the right or to the left from your starting point only depends on the sign of the second integer. The sign of the first integer just helps you to find the correct starting position on the number line.
### Examples
#### Example 1
Earlier, you were given a problem about Mark, who was in debt to his sister.
Mark is $25 in debt, but has the opportunity of getting rid of$8 of that debt if he cleans his sister's room. Mark wants to subtract in order to determine what his new debt would be if he cleaned his sister's room.
Then, find the location of -25 on the number line.
Next, notice that the second integer, -8, is negative. This means you will be moving to the right. Starting at -25, move to the right 8 units.
You end up on -17. The answer is -17.
So .
If Mark cleans his sister's room, he will be left with -$17 (he will be$17 in debt).
#### Example 2
Use a number line to find the difference of .
Then, find the location of -4 (the first integer in your difference) on the number line.
Next, notice that the second integer, 3, is positive. This means you will be moving to the left. Starting at -4, move to the left 3 units.
You end up on -7. The answer is -7. .
#### Example 3
Use a number line to find the difference of .
Then, find the location of -5 on the number line.
Next, notice that the second integer, 2, is positive. This means you will be moving to the left. Starting at -5, move to the left 2 units.
You end up on -7. The answer is -7. .
#### Example 4
Use a number line to find the difference of .
Then, find the location of 6 on the number line.
Next, notice that the second integer, -2, is negative. This means you will be moving to the right. Starting at 6, move to the right 2 units.
You end up on 8. The answer is 8. .
#### Example 5
Use a number line to find the difference of .
Then, find the location of -7 on the number line.
Next, notice that the second integer, -5, is negative. This means you will be moving to the right. Starting at -7, move to the right 5 units.
You end up on -2. The answer is -2. .
### Review
Subtract the following integers using a number line.
1.
1.
1.
1.
1.
1.
1.
1.
1.
1.
Subtract these integers without a number line.
1.
2.
3.
4.
5.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Integer The integers consist of all natural numbers, their opposites, and zero. Integers are numbers in the list ..., -3, -2, -1, 0, 1, 2, 3...
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Subjects: | 1,321 | 5,339 | {"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": 1, "texerror": 0} | 5.03125 | 5 | CC-MAIN-2017-17 | latest | en | 0.921547 |
https://readingicon.com/2018/12/30/physical-states-of-matter/ | 1,563,274,963,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195524522.18/warc/CC-MAIN-20190716095720-20190716121720-00388.warc.gz | 517,227,439 | 46,126 | # PHYSICAL STATES OF MATTER
CHAPTER-5
## Gaseous State
Q:what is difference between diffusion and effusion, explain with examples ?
1. Diffusion
Def: The movement of molecules from a higher concentration to a lower concentration is called diffusion
The characteristic smell of rotten egg spreads throughout the room. This is because of the concentration of hydrogen sulphide molecules at a particular place is higher, they start moving towards a place where their concentration is lower. This movement of molecules from a higher concentration to a lower concentration is called diffusion
• Effusion
Def: The escape of gas molecules through the hole one after the other without collision is called effusion
When gas is allowed to pass through a hole of the size of a molecule, only one molecule at a time can pass through the hole
#### Q: What is pressure & its units?
Def: Force exerted by a gas on unit area of a container is called its pressure
A gas exerts pressure on the walls of its container, as you can see when you blow a balloon.. All gases exert pressure. At any point a gas exerts an equal pressure in all directions.
At sea level at 0OC the atmospheric pressure is 760mm of Hg or 760 torr. This pressure is referred as one atmosphere.
So, 1atm = 760mm Hg = 760 torr
The SI unit of pressure is the pascal (Pa) which is very small unit, so that kilopascal (kPa) is used for atmospheric pressure under ordinary conditions.
## One pascal equals force of one Newton exerted on an area of one square meter.
1atm= 101.325kPa = 1.01325 x 105Pa.
Q: How gas effect volume affected by change in pressure & temperature?
Effect on the volume of a gas by a change in pressure and temperature
Effect of Temperature on Volume:
According to the kinetic molecular theory, gas molecules are in constant random motion. They move in straight line until they collide with another molecule or the walls of the container. The pressure a gas exerts in a container is due to the force exerted on the walls of the container. The average kinetic energy of gas molecules is directly proportional to the Kelvin temperature. So the average kinetic energy of a collision when a gas molecule collides with the wall of a container will not change when the pressure decreases at constant temperature. There are large empty spaces between the molecules. On increasing pressure on the gas, the distance between molecules decreases. So the volume of the gas decreases.
Effect of Pressure on VolumeOn theother hand when pressure of the gas is kept constant and temperature isincreased, the average kinetic energy of gas molecules increases. So, themolecules of the gas hit the wall of container more frequently andenergetically. This increases internal pressure. As a result, volume of the gasincreases to restore constant pressure
Do You Know? Earth’s atmosphere consists of gases that provide the oxygen and pressure necessary to support life as we know it. Atmospheric gases moderate temperature extremes and one of them (ozone) shields living things on the surface from harmful ultraviolet radiations. In contrast the moon has no atmosphere.
Q: what are variables used in gas laws?
A sample of a gas can be characterised by four variables:
1. Pressure (P)
2. Volume (V)
3. Temperature (T)
4. The number of moles of gas (n)
Q: What is boyle’s law?
Boyle’s law
Boyle’s law states that the volume of a fixed amount of a gas at a given temperature is inversely proportional to the applied pressure
So,
V x P = constant
P1V1 = P2V2
This means product of volume and pressure must be constant if inverse relationship exists between them.
Relationship between volume and pressure at constant temperature
Experiment No. Pressure (atm) Volume (dm3) V x P (dm3.atm) 1. 0.500 4.00 2.00 2. 1.00 2.00 2.00 3. 2.00 1.00 2.00 4. 4.00 0.500 2.00
Example 5.1: Accounting for pressure-volume changes in a gas using Boyle’s Law
Ethene is used as anaesthetic gas. The pressure on 2.5dm3 of ethene changes from 1.05 to 2.10 atm. The volume of ethene becomes 1.25dm3 if the temperature remains constant. Explain this change using Boyle’s law.
Solution:
P1 x V1 before change = 1.05 x atm x 2.5 dm3
= 2.625 atm. dm3
P2 x V2 after change = 2.1 atm x 1.25 dm3
= 2.625 atm. dm3
P1V1 = P2V2
Note: Relationship P1V1= P2V2 can be used to determine any variable knowing theother three
## SELF ASSESSMENT
1. A student obtained following data in an experiment at 20oC.
## solution:
S# Pressure : P (atm) Volume : V(dm3) Constant = P x V 1 0.350 0.707 0.350 x 0.707= 0.247 2 0.551 0.450 0.551 x 0.450 = 0.247 3 0.762 0.325 0.762 x 0.325 = 0.247 4 0.951 0.261 0.951 x 0.261 = 0.247 5 1.210 0.205 1.210 x 0.205 = 0.248
Ammonia gas is used as refrigerant 0.474 atm. pressure is required to change 2000cm3 sample of ammonia initially at 1.0 atm. to 4.22dm3 at constant temperature. Show that this data satisfies Boyle’s law.
Solution:
volume of ammonia initially = V1= 2000 cm3 = 2000/1000 = 2 dm3
pressure of ammonia initially= P1= 1 atm
Volume of ammonia after change = V2= 4.22 dm3
Pressure of ammonia after change = P2 = 0.474 atm
According to Boyle’s Law
P1V1 = P2V2
1 atm x 2 dm3 = 0.474 atm x 4.22 dm3
2.00 atm. dm3 = 2.00 atm dm3
## q. what is Charles’s gas law?
Charles’s Law
The law states that the volume of given mass of a gas varies directly with absolute temperature at constant pressure
Mathematically
V T
V= constant x T
= constant
Q: What is relationship between absolute temperature and charles’s law?
Ans: You can convert centigrade temperature to absolute or Kelvin temperature by adding 273 to it. Absolute temperature is represented by T.
T = toC + 27
Dependence of volume on absolute temperature
Straight line between the temperature and volume suggests that there is a direct relationship between volume and temperature measured at Kelvin scale at constant pressure.
This relationship is known as the Charles’s Law.
The law states that the volume of given mass of a gas varies directly with absolute temperature at constant pressure. Since the ratio of remains constant at constant pressure.
Example 5.2: Accounting for temperature-volume changes in a gas using Charles’s Law
Table 5.5 shows data of volume of a gas and its temperature for the given mass of a gas at 900 mm Hg.
Table 5.5: Temperature-volume data for a gas at 900mmHg
Temperature(oC) Volume(cm3) 0 107.9 5 109.7 10 111.7 15 113.6 20 115.5
Explain volume-temperature relationship using Charles’s Law.
Solution:
Table 5.6 Temperature-volume relationship
Temperature (oC) Volume (cm3) Temperature (K) 0 107.9 273 5 109.7 278 10 111.7 283 15 113.6 288 20 115.5 293
The ratio is fairly constant. Thus volume of the gas varies directly with the absolute temperature as stated by the Charles’s law.
SELF assessment EXERCISE 5.2
1. A chemist obtained data shown in table 5.7 in an experiment at 1 atm.
Table 5.7 Temperature-volume data of a gas at 1 atm
Temperature(oC) Volume(cm3) 25 117.5 30 119.4 35 121.3 40 123.2
Solution:
Temperature (oC) Volume (cm3) Temperature (K) 25 117.5 25+273=298 117.5/298=0.3942 30 119.4 30 + 273= 303 119.4/303=0.3940 35 121.3 35 + 273= 308 121.3/308=0.3938 40 123.2 40 + 273 = 313 123.2/313=0.3936
• A bacterial culture isolated from sewage produces 36.4 cm3 of methane (CH4) gas at 27oC and 760mm Hg. This gas occupies 33.124 cm3 at 0oC and same pressure. Explain volume-temperature relationship from this data.
SOLUTION:
Volume of methane initially: V1= 36.4 cm3
Temperature of methane initially = T1= 27 + 273= 300 K
Volume of methane after = V2= 33.124 cm3
Temperature of methane after = T2= 0+ 273= 273 K
According to Charles law
V1/T1 = V2/T2
36.4/300 = 33.124/ 273
0.121 cm3. K-1 = 0.121 cm3. K-1
• A perfect elastic balloon filled with helium gas has a volume of 1.25×103dm3 at 1.00atm and 25oC on ascending to a certain altitude where temperature is 15oC the volume of balloon becomes 1.208×103dm3. Show that this data satisfies the Charles’s law.
## Liquid State
It is a state of matter in which the constituent particles are loosely bound by intermolecular forces. Liquid change their shape with a fixed volume.
Typical Properties of Liquids
Some typical properties of liquids are given below:
1) Evaporation
2) Boiling Point
3) Vapour Pressure
4) Surface Tension
5) Freezing Point
6) Density etc.
7) Viscosity
8) Diffusion
9) Mobility
Define and Explain Evaporation Experimentally.
Evaporation
It refers to the conversion of a liquid to a gas or vapour at all temperatures but less than the Boiling point of the liquid is called Evaporation or vaporization.
Experimental Verification
In evaporation, we have to examine the movement of molecules in liquid. The molecules of a liquid are not motionless. The molecules which have low kinetic Energy move slowly, while others with high K.E move faster. This is clearly observed in an open container of Ether or Acetone. The volume of the liquid gradually decreases and finally no more liquid is left behind. This is because of Evaporation. Thus this spontaneous change of liquid into itsvapours is called Evaporation and it continues at all temperatures.
1. Evaporation is a Cooling Process?
2. Liquids Evaporate faster when heated?
i. Evaporation Causes Cooling
Evaporation Causes Cooling reason is that when high energy molecules leave the liquid and low energy molecules are left behind, the temperature of the liquid falls and heat moves from surrounding to liquid as a result temperature of surrounding also decreases. Hence- Evaporation is a cooling process and lower the temperature process of liquid.
ii) Liquids Evaporate faster
Higher the temperature, faster is the rate of evaporation and vice versa. This is because added heat increases the Kinetic Energy of the molecules. Hence liquids evaporate faster when heated.
Vapour Pressure
The pressure exerted by the vapours of a liquid in equilibriumwith its liquid at a given temperature is called vapourPressure.
Explanation
Let us consider a liquid enclosed in a container. Air is evacuated and the vessel is sealed. The liquid then starts evaporating and its molecules collide with each other. Some of the molecules are recaptured by the liquid phase. This process is called Condensation. When two opposing process proceed exactly the same rate then the system is said to be in a dynamic state of equilibrium and the number of molecules in the vapour phase exerts a definite pressure which is called the vapour pressure of the liquid as shown in the figures.
Effect of Temperature on Vapour Pressure
Vapour pressures of some liquid at various temperatures.
Table: Vapour pressure of some liquids at various temperatures
Vapour Pressure (kPa) of Several Substances at Various temperatures
0 C 20 C 40 C 60 C 80 C 100 C Water Ethanol Diethyl ether
We conclude from the table that the V.P of liquids changes with temperature. This is because an increase in temperature of a liquid increases the K.E of the molecules and as a result vapour pressure increases.
Boiling Point
The temperature at which vapour Pressure of a liquid becomes equal to the external or atmospheric Pressure i.e 760mm Hg or 101.325 KPa at sea level is called Boiling Point.
The B.Point of some of the liquids are given below
Liquid Water Chloroform Ether Acetone Ethanol Ethanoic acid B. Point (C) 100 61 35 56 78 119 B.Point (K) 373 334 308 329 351 392
Graphical Explanation
When a liquid is heated in an open container, average K.E of the molecules, increases and atmosphere exerts pressure on the liquid surface. Therefore, the temperature of the liquid gradually increases. This heating process overcome the attractive forces and increases the V.P of the liquids. At a certain temp, the V.P of the liquid becomes equal to the external or atmospheric Pressure. At this stage, bubbles of vapours from throughout the liquid, rise to surface and escape into the air and liquid starts boiling. This is called B.P of the liquids. The Graph clearly shows the variation in vapour Pressures at 101.325 KPa of four liquids with Boiling Points.
Variation in vapour pressures with temperature of four liquids.
Effect of External Pressure on Boiling Point
Boiling Points depend upon the External Pressure over the surface of the liquid. At Sea level, External Pressure is 1 atm = 760mm Hg = 101.325 KPa. As we go at higher altitude i.e a top of Mount Everest, the external pressure decreases to 34 KPa. So at sea Level, the normal B.P of water is 100°C at 101.325KPa while at higher altitude at Mount Everest, water boils at 70°C with 34 KPa external Pressure. Hence B.P is high at high External Pressure and vice versa which is clearly shown in the figures below.
Solid State
In Solid State atoms, molecules or ions are very closely packed having definite shape and volume. They are in compressible and do not flow but vibrate about their fixed positions as shown in the figure.
The typical properties of solids are:
1. Melting point/melting
2. Freezing Point/Freezing
3. Sublimation etc.
Melting and Freezing
During heating, the particles of a solid start vibrating with greater Kinetic Energy. This makes the molecules a little bit apart from each other. This process increases the volume of solids. On future heating a stage reaches when the particles leave their fixed positions and start moving in the form of a liquid state as shown in the figure. This is called melting.
The temperature at which a solid turns into a liquid is called melting point.
On cooling, the liquid freezes. Thus freezing of a liquid is the reverse of melting. So the temperature at which a liquid changes into the solid is called Freezing Point.
Themelting and freezing of substances occurs at the same temperature. At this
temperature, the liquid and solid substances are in equilibrium each other.
Solid liquid
The Melting Point of a solid depends on the strength of attractive forces that hold particles together in the fixed positions. Stronger the forces higher will be the melting point. Ionic solids comparatively having high M.P than covalent solids.
Sublimation
The vaporization of a solid substance into the vapour phase without passing through the liquid phase is called sublimation.
Example:
Iodine, Benzoic Acid, Ammonium, Chloride naphthalene etc.
Explanation
When impure sample of Iodine molecule is heated on sand bath. After some time dark violet black crystals of Iodine deposit the underside of the glass water placed on the top of a beaker containing Solid Iodine. The iodine vapour Sublimates from Iodine crystals in the bottom of the beaker and Condenses to form crystals on the glass watch as shown in the figure.
Sublimation of violet-black crystals of iodine
Types of Solids
On the basis of structure the solids have been divided into two classes based on their macroscopic appearance.
a) Amorphous Solids
b) Crystalline Solids
Amorphous is derived from Greek and “Amorphos” which means without shape. Hence Amorphous Solids are those in which atoms, ions or molecules are not arranged in a definite pattern, e.g. glass, Plastics, rubber etc..
Amorphous solids do not melt at a definite temperature but gradually soften when heated.
The solids in which atoms, molecules or ions are arranged in a regular repeating three- dimensional well- ordered pattern are called as crystalline solids. Crystals have regularshape. Particles forming the crystals are packed in a very exact and ordered pattern as shown in figure below.
Cubic crystalline shape ofNaCl
Allotropy
The phenomenon of the existence of an element in different forms which have different physical properties but same chemical properties is known as ALLOTROPY and different forms will be called ALLOTROPES or ALLOTROPIC FORMS.
Allotropes of Carbon
Carbon has many Allotropic forms which are divided into two classes.
1. Crystalline Allotropic forms: Diamond Graphite and Bucky Ball
2. Non-crystalline Allotropic forms e.g: coal, coke, charcoal,Lamp black etc
Crystalline Allotropes of Carbon
Carbon has three crystalline allotropes. The arrangements and properties are given below:
1. Diamond
It is the hardest known Substance in diamond; each carbon atom is covalently bonded to four other C-atoms which are arranged in the form of tetrahedron. Diamond is used for cutting glass and polishing hard surface because of rigid compact array as shown in the figure.
• Graphite
In Graphite, Carbon atoms are arranged in layers of hexagonal arrays. In these layers, each C-atoms is joined covalently by weak each Vender Waal’s forces. These layers slip over each other and makes Graphite Soft.
Graphite is used as electrodes, lubricant in machines, and black pigment.
3. Bucky Ball
It is new allotropic form which consists of forty to hundred carbon atoms. These atoms are arranged in a hollow cage like structure. Simplest molecule of Bucky Ball is made up of sixty C-atoms. Carbon atoms are arranged in Pentagons (Five member ring) and Hexagons (six member ring) just like a soccer Balls shown in the structure with the formula C60.
Allotropes of Phosphorus
Phosphours can exist in at least six different solid allotropic forms, three are common i.e white, Red and black phosphorus Phosphorus is a non-metal but here two are given.
1. White Phosphorus
It is a very reactive, poisonous, volatile, waxy allotrope which is soluble in benzene and carbon disulphide. It exist in the form of tetra atomic molecules (P4) and form a tetrahedral structure as shown below:
• Red Phosphorus
It is much less reactive and poisonous than white Phosphorus. The tetra atomic molecules of red Phosphorus combine to form macromolecules in long chains.
Allotropes of Sulphur
Sulphur consists of molecules that contain eight atoms, S8covalently bonded with each other. Sulphur exist two crystalline allotropes i,e. Rhombic and monoclinic sulphurs while one non crystalline (Amorphous) Sulphuri,e Plastic sulphur.
Compare the three Physical States of Matter.
Distinction between the three states of matter on the basis of particle model
Property Solids Liquids Gases 1. Shape and volume Solids posses definite shape and definite volume Liquid do not have definite shape but definite volume. Gases have neither a definite shape nor a definite volume. 2. Compress-ibility Solids are least compressible. It is due to the presence of small empty space between their molecules. It is slightly more that in solids. Gases are highly compressible. 3. Energy Molecules possess least energy. Molecules possess moderate values of energies. These molecules are highly energetic and have lot of energy in them. 4. Molecular Motion The molecules possess vibratory motion. Molecules have linear as well as vibratory motion and are restricted to the boundary of the liquid Molecules have complete freedom of motion. They move in all directions. 5. Intermolecular forces Mutual forces of attraction between the molecules are maximum. Forces of attraction between the molecules are small. Forces of attraction between molecules are almost negligible. 6. Packing of particles The molecules are very closely packed The molecules loosely held. The molecules are at a very large distance apart. 7. Diffusion No tendency to diffuse Slow tendency to diffuse High tendency to diffuse 8. Density D = m/v High Less than that of solids Very low densities
SOCIETY, TECHNOLOGY AND SCIENCE
1. Scientists use the power of reasoning to explain their observation. For instance when a balloon is filled with air, it expands. A scientist would explain it by saying that air molecules are free to move inside their container. There is no attractive or repulsive force between the molecules. As a result gas expands until it takes the shape of its container. Therefore, air expands to fill the interior of the balloon evenly.
2. In early 1600s, Galileo argued that suction pumps were able to draw water from a well because of the force of vacuum” inside the pump. After Galileo’s death, the Italian mathematician and physicist E. Torricelli proposed another explanation. He suggested that the air in the atmosphere has weight. The force of atmosphere pushing down on the surface of water drives the water into the suction pump when it is evacuated. In 1946 Torricelli invented a device called barometer. He measured atmospheric pressure is 760mm Hg. Torricelli’s work soon caught the attention of British scientist Robert Boyle. He modified barometric tube into a J-shaped tube. By adding mercury to the open end of the tube, he trapped a small volume of air in the sealed end. He studied what happened to the volume of the air as he added more mercury to the open end. Boyle’s from the studies discovered the pressure-volume relationship. J-tube was further modified and another device known as manometer was developed that can measure the pressure of any gas. This means instrumentation improves as science progresses.
3. Freeze-dried foods are light-weight and conveniently re-constituted by adding water. When salt is applied on meat it draws out considerable amount of water from the meat. After this meat is frozen and placed in a chamber attached to a vacuum pump. By lowering the pressure below the vapour pressure of ice, the ice crystals sublimate and the meat is dried without the loss of its flavour. Dried meat needs no refrigerator because bacteria such as salmonella which cause food poisoning cannot grow on salt and in the absence of moisture. Thus curing with the salt helps preserve meat.
EXERCISE MCQ’S
No. Ans. No. Ans. No. Ans. No. Ans. i) a ii) b iii) d iv) d v) a vi) c vii) b yiii) b ix) a x) d
1. Explain why volume of a gas decreases on increasing pressure on it at constant temperature?
Answer According to Boyles Law volume and pressure are inversely proportional to each other at constant temp. High pressure on gas molecules increases the intermolecular forces as a result of which volume of gas molecules decreases.
• How does temperature effect vapour pressure of a liquid?
• Water boils at 120°C in a pressure cooker, why?
Pressure Cooker is a closed container in which on heating B.P of water increases which help to increase the external pressure. A liquid can be made to boil at any temperature by changing the external pressure. So water boils at 120°C in a pressure Cooker because inside the pressure Cooker, external pressure increases.
• Is evaporation a cooling process?
Yes. evaporation is a cooling process. For more detail see question 14 (i)
• Can you make water boil at 70°C?
Yes. A liquid can be made to boil at any temperature by changing the external pressure. Lower the external pressure, lower, will be the boiling point of water. Water boils at 1000C with 532 mm Hg Pressure.
• Express the pressure 400mm Hg in kPa?
760mmHg = 101.325 Kpa
400 mm Hg of presuure =
= 53.33 KPa
Hence 400 mmHg = 53.33Kpa
Q12. The air in a perfectly elastic balloon occupies 855cm3, during the fall when the temperature is 20°C. During the winter, the temperature on a .particular day is -10°C, the balloon occupies 794.39cm3. If the pressure remains constant. Show that the given data proves the volume temperature relation according to the Charles Law.
V1 = 885cm3
T1= 20°C + 273 = 293K
V2 = 794.39cm3
T2 =-10°C + 273 = 263K
According to Charles Law = =
=
3.02 cm3K -1= 3.02 cm3K -1
Hence it proves Charles law
Q13. In the past, gas volume was used as a way to measure temperature using devices called gas thermometers. An experimenter obtains following data from gas thermometer.
Volume (dm3) Temperature (°C) 2.7 0°C 3.7 100°C 5.7 300°C
Solution
According to Charles Law = = =
V1= 2.7dm3 T1 = 0°C + 273 = 273 K
V2 = 3.7dm3 T2= 100°C + 273 =373K
V3 = 5.7dm3 T3 = 300°C + 273 = 573K
by putting the values
= =
0.00989 dm3 K-1 = 0.00992dm3 K-1 = 0.00994 dm3 K-1
9.89 cm3 K-1 = 9.92cm3 K-1= 9.94 cm3 K-1
Q.14 In automobile engine the gaseous fuel-air mixture enters the cylinder and is compressed by a moving piston before it is ignited. If the initial cylinder volume is 990cm3. After the piston moves up the volume is 90cm3. The fuel- air mixture initially has a pressure of 1. 0 atm and final pressure 11.0 atm. Do you think this change occurs according to the Boyles law.
V1 = 990cm3 V2 = 90cm3
P1 = 1 .0atm P2 = 11 .0 atm
According to Boyles Law
P1 V1 = P2V2
Putting the values
1. atm x 990cm3 =11.0 atm x 90cm3
990atm cm3 = 990 atm.cm3
Q15. A sample of neon that is used in a neon sign has a volume of 1500cm3 at a pressure of 636 torr. The volume of the gas after it is pumped into the glass tube of the sign is 1213.74cm3 when it shows a pressure of 786 torr. Show that this data obeys Boyle’s law?
V1= 1500cm3 P1= 636 torr
V2 = 1213.74cm3 P2= 786 torr
According to Boyel’s Law
P1V1 = P2V2
636 torr x 1500cm3 = 786 torr x 1213.74 cm3
954000 torr cm3 = 954000 torr cm3
Q16. Instrumentation changes as science progresses, comment on it?
The search of man for collecting knowledge and integrating it is called Science. So true knowledge and its integration can be progressed more with the help of instruments. Some examples will prove the truth.
• Exact heat changes during chemical reaction can be measured with the help of digital thermometer.
• pH of solution, urine and blood can be measured exactly by pH-meter not by pHpapers
• Stethoscope uses doctor to listen heart beats and breathing
• Barometer used for measuring air pressure easily
• Mass spectrometer used to find out the atomic mass of an atom
• Geologist easily detect the presence of oil and gas under earth with the help of instruments.
• Micro organism easily seem by micro scope for better analysis
• 1st date moon easily seen by powerful telescope
• In medicine, use of instruments plays a vital role in the micro analysis of blood, urine, cancer, breaking bones etc. In short, Instrumentation changes as science progresses.
ELT ASSESSMENT EXERCISE 5.1
Related To Gaseous State
1. A student obtained following data in an experiment at 20 C° Explain pressure- volume relationship using this data according Jo Boyle’s Law.
Solution
According to Boyles Law
Pressure x Volume = Constant
0.350 atm x 0.707dm = 0.248 atm. dm3
0.551 atm x 0.450dm3 = 0.248 atm. dm3
762 atm x 0.325dm3 = 0.248 atm. dm3
0.951 atm x 0.261 dm3 =0.248 atm. dm3
1.210 atm x 0.205 dm3= 0.248 atm. dm3
Thus the calculated results agrees with the Pressure-Volume relationship according to the Boyles Law.
• Ammonia gas is used as refrigerant 0.474 atm (P1) pressure is required to change 2000 cm3 (V1=2 dm3) sample of ammonia initially at 1.0 atm (P2) to 4.22 dm3 (V2) at constant temperature. Show that this data satisfies Boyle’s Law?
Solution
According to Boyles Law
P1V1=P2V2
V1= 2000cm3 = 2dm3
P1 = 0.474 atm
P2 = 1.0 atm
V2 = 4.22dm3
Putting all the values in above equation
P1V1 = P2V2
0.474 atm x 2 dm3 =1.0 atm x 4.22 dm3
0.948 atm. dm3≠ 4.22 atm dm3
The given data does not satisfied the Boyles Law
ELT ASSESSMENT EXERCISE 5.2
1. At Mount Everest, the atmospheric pressure is 34 KPa which is about 8850 m above the sea level. While at Murree atmospheric pressure is 99.298 KPa which is comparative near to sea level. So at high altitude, the atoms pressure decreases, obviously B.P of water also decreases. Hence B.P of water at Mount Everest is 70°C while at Murree98°C.
2. An egg in boiling water cooked longer while camping at an elevation of 0.5 Km in the mountain because at high altitude as compare 10*sea level, external pressure decreases which will decrease the Boiling point of the water. So it will take longer time than it does at home.
SELF ASSESSMENT EXERCISE 5.3
Give reason:
1. When you put nail polish remover on you palm, you feel a sensation of coldness.
2. Wet clothes dry quickly in summer than in winter.
Ans. 1
Nail Polish remover contains an organic solvent “Acetone”. It contains weak intermolecular forces and having high V.P. These high energy molecules escape from your palm and one feels sense of cooling. The other reason is that heat of body is used up to evaporate nail polish remover.
Ans.2
Greater the temperature of surrounding, greater the K.E possessed by the molecules and greater the escaping tendency from the surface of the liquids and Vice Versa. In summer, surrounding temperature is higher as compared to winter. So, wet clothes dry quickly in summer than in winter.
SELF ASSESSMENT EXERCISE 5.6
Sodium chloride, an ionic compound, has a high melting point of 801°C. Whereas molecular solid such as ice has relatively low melting point of 0°C. Explain this difference.
Ans.
In Sodium chloride, an ionic crystalline solid has strongelectrostatic forces of attraction are present between positively charged ionswith many negatively charged ions and Vice Versa while in molecular solids i.eice, weaker intermolecular forces are present. So M.P of sodium chloride(801°C) is higher than molecular solids (0°C) that is ice.
Sublimation
The vaporization of a solid substance into the vapour phase without passing through the liquid phase is called sublimation.
Example:
Iodine, Benzoic Acid, Ammonium, Chloride naphthalene etc.
Explanation
When impure sample of Iodine molecule is heated on sand bath. After some time dark violet black crystals of Iodine deposit the underside of the glass water placed on the top of a beaker containing Solid Iodine. The iodine vapour Sublimates from Iodine crystals in the bottom of the beaker and Condenses to form crystals on the glass.
On the basis of structure the solids have been divided into two classes based on their macroscopic appearance.
a) Amorphous Solids
b) Crystalline Solids
Amorphous is derived from Greek and “Amorphos” which means without shape. Hence Amorphous Solids are those in which atoms, ions or molecules are not arranged in a definite pattern, e.g. glass, Plastics, rubber etc..
Amorphous solids do not melt at a definite temperature but gradually soften when heated.
• Crystalline Solids
The solids in which atoms, molecules or ions are arranged in a regular repeating three- dimensional well- ordered pattern are called as crystalline solids. Crystals have regular shape. Particles forming the crystals are packed in a very exact and ordered pattern as shown in figure below.
Allotropy
The phenomenon of the existence of an element in different forms which have different physical properties but same chemical properties is known as ALLOTROPY and different forms will be called ALLOTROPES or ALLOTROPIC FORMS.
Allotropes of Carbon
Carbon has many Allotropic forms which are divided into two classes.
1. Crystalline Allotropic forms: Diamond Graphite and Bucky Ball
2. Non-crystalline Allotropic forms e.g: coal, coke, charcoal,Lamp black etc
Crystalline Allotropes of Carbon
Carbon has three crystalline allotropes. The arrangements and properties are given below:
1. Diamond
It is the hardest known Substance in diamond; each carbon atom is covalently bonded to four other C-atoms which are arranged in the form of tetrahedron. Diamond is used for cutting glass and polishing hard surface because of rigid compact array .
Graphite
In Graphite, Carbon atoms are arranged in layers of hexagonal arrays. In these layers, each C-atoms is joined covalently by weak each Vender Waal’s forces. These layers slip over each other and makes Graphite Soft. Graphite is used as electrodes, lubricant in machines, and black pigment.
3. Bucky Ball
It is new allotropic form which consists of forty to hundred carbon atoms. These atoms are arranged in a hollow cage like structure. Simplest molecule of Bucky Ball is made up of sixty C-atoms. Carbon atoms are arranged in Pentagons (Five member ring) and Hexagons (six member ring) just like a soccer Balls shown in the structure with the formula C60.
Allotropes of Phosphorus
Phosphours can exist in at least six different solid allotropic forms, three are common i.e white, Red and black phosphorus Phosphorus is a non-metal but here two are given.
1. White Phosphorus
It is a very reactive, poisonous, volatile, waxy allotrope which is soluble in benzene and carbon disulphide. It exist in the form of tetra atomic molecules (P4) and form a tetrahedral structure as shown below:
• Red Phosphorus
It is much less reactive and poisonous than white Phosphorus. The tetra atomic molecules of red Phosphorus combine to form macromolecules in long chains.
Allotropes of Sulphur
Sulphur consists of molecules thatcontain eight atoms, S8covalently bonded with each other. Sulphurexist two crystalline allotropes i,e. Rhombic and monoclinic sulphurs while onenon crystalline (Amorphous) Sulphuri,e Plastic s
Compare the three Physical States of Matter.
Or Distinction between the three states of matter on the basis of particle model
SOLID LIQUID GAS Has a definite shape and volume Has indefinite shape but definite volume Has indefinite shape and volume Particles are closely packed together bt have strong intermolecular forces Particles are not closely packed together bt have strong intermolecular forces Particles are free to move and have negligible intermolecular forces Particles can vibrate only along their mean position but cannot move. Particles move about freely and randomly in all directions with in surface of liquid Particles move about freely and randomly in all directions
SOCIETY, TECHNOLOGY AND SCIENCE
1. Scientists use the power of reasoning to explain their observation. For instance when a balloon is filled with air, it expands. A scientist would explain it by saying that air molecules are free to move inside their container. There is no attractive or repulsive force between the molecules. As a result gas expands until it takes the shape of its container. Therefore, air expands to fill the interior of the balloon evenly.
2. In early 1600s, Galileo argued that suction pumps were able to draw water from a well because of the force of vacuum” inside the pump. In 1946 Torricelli invented a device called barometer. He measured atmospheric pressure is 760mm Hg..
3. Freeze-dried foods are light-weight and conveniently re-constituted by adding water. When salt is applied on meat it draws out considerable amount of water from the meat. After this meat is frozen and placed in a chamber attached to a vacuum pump. By lowering the pressure below the vapour pressure of ice, the ice crystals sublimate and the meat is dried without the loss of its flavour. Dried meat needs no refrigerator because bacteria such as salmonella which cause food poisoning cannot grow on salt and in the absence of moisture. Thus curing with the salt helps preserve meat.
EXERCISE
1. Explain why volume of a gas decreases on increasing pressure on it at constant temperature?
Answer According to Boyles Law volume and pressure are inversely proportional to each other at constant temp. High pressure on gas molecules increases the intermolecular forces as a result of which volume of gas molecules decreases.
• How does temperature effect vapour pressure of a liquid? [SEE LEC # 3]
• Water boils at 120°C in a pressure cooker, why?
Pressure Cooker is a closed container in which on heating B.P of water increases which help to increase the external pressure. A liquid can be made to boil at any temperature by changing the external pressure. So water boils at 120°C in a pressure Cooker because inside the pressure Cooker, external pressure increases.
• Is evaporation a cooling process?
Yes. evaporation is a cooling process. [SEE LEC# 3]
• Can you make water boil at 70°C?
Yes. A liquid can be made to boil at any temperature by changing the external pressure. Lower the external pressure, lower, will be the boiling point of water. Water boils at 1000C with 532 mm Hg Pressure.
• Express the pressure 400mm Hg in kPa?
760mmHg is = 101.325 Kpa
1mmHg is = 101.325/760
400 mm Hg of pressure = = 53.33 KPa
Hence 400 mmHg = 53.33Kpa
Q.7: Differentiate between amorphous and crystalline solidsQ12. The air in a perfectly elastic balloon occupies 855cm3, during the fall when the temperature is 20°C. During the winter, the temperature on a .particular day is -10°C, the balloon occupies 794.39cm3. If the pressure remains constant. Show that the given data proves the volume temperature relation according to the Charles Law.
V1 = 885cm3
T1= 20°C + 273 = 293K
V2 = 794.39cm3
T2 =-10°C + 273 = 263K
According to Charles Law = =
=
3.02 cm3K -1= 3.02 cm3K -1
Hence it proves Charles law
Q13. In the past, gas volume was used as a way to measure temperature using devices called gas thermometers. An experimenter obtains following data from gas thermometer.
Volume (dm3) Temperature (°C) 2.7 0°C 3.7 100°C 5.7 300°C
Solution
According to Charles Law = = =
V1= 2.7dm3 T1 = 0°C + 273 = 273 K
V2 = 3.7dm3 T2= 100°C + 273 =373K
V3 = 5.7dm3 T3 = 300°C + 273 = 573K
by putting the values
= =
0.00989 dm3 K-1 = 0.00992dm3 K-1 = 0.00994 dm3 K-1
9.89 cm3 K-1 = 9.92cm3 K-1= 9.94 cm3 K-1
Q.14 In automobile engine the gaseous fuel-air mixture enters the cylinder and is compressed by a moving piston before it is ignited. If the initial cylinder volume is 990cm3. After the piston moves up the volume is 90cm3. The fuel- air mixture initially has a pressure of 1. 0 atm and final pressure 11.0 atm. Do you think this change occurs according to the Boyles law.
V1 = 990cm3 V2 = 90cm3
P1 = 1 .0atm P2 = 11 .0 atm
According to Boyles Law
P1 V1 = P2V2
Putting the values
1. atm x 990cm3 =11.0 atm x 90cm3
990atm cm3 = 990 atm.cm3
Q15. A sample of neon that is used in a neon sign has a volume of 1500cm3 at a pressure of 636 torr. The volume of the gas after it is pumped into the glass tube of the sign is 1213.74cm3 when it shows a pressure of 786 torr. Show that this data obeys Boyle’s law?
V1= 1500cm3 P1= 636 torr
V2 = 1213.74cm3 P2= 786 torr
According to Boyel’s Law
P1V1 = P2V2
636 torr x 1500cm3 = 786 torr x 1213.74 cm3
954000 torr cm3 = 954000 torr cm3
Q16. Instrumentation changes as science progresses, comment on it?
The search of man for collecting knowledge and integrating it is called Science. So true knowledge and its integration can be progressed more with the help of instruments. Some examples will prove the truth.
• Exact heat changes during chemical reaction can be measured with the help of digital thermometer.
• pH of solution, urine and blood can be measured exactly by pH-meter not by pH papers
• Stethoscope uses doctor to listen heart beats and breathing
• Barometer used for measuring air pressure easily
• Mass spectrometer used to find out the atomic mass of an atom
• Geologist easily detect the presence of oil and gas under earth with the help of instruments.
• Micro organism easily seem by micro scope for better analysis
• 1st date moon easily seen by powerful telescope
• In medicine, use of instruments plays a vital role in the micro analysis of blood, urine, cancer, breaking bones etc. In short, Instrumentation changes as science progresses.
ELT ASSESSMENT EXERCISE 5.2
1. At Mount Everest, the atmospheric pressure is 34 KPa which is about 8850 m above the sea level. While at Murree atmospheric pressure is 99.298 KPa which is comparative near to sea level. So at high altitude, the atoms pressure decreases, obviously B.P of water also decreases. Hence B.P of water at Mount Everest is 70°C while at Murree98°C.
2. An egg in boiling water cooked longer while camping at an elevation of 0.5 Km in the mountain because at high altitude as compare 10*sea level, external pressure decreases which will decrease the Boiling point of the water. So it will take longer time than it does at home.
SELF ASSESSMENT EXERCISE 5.3
Give reason:
1. When you put nail polish remover on you palm, you feel a sensation of coldness.
2. Wet clothes dry quickly in summer than in winter.
Ans. 1
Nail Polish remover contains an organic solvent “Acetone”. It contains weak intermolecular forces and having high V.P. These high energy molecules escape from your palm and one feels sense of cooling. The other reason is that heat of body is used up to evaporate nail polish remover.
Ans.2
Greater the temperature of surrounding, greater the K.E possessed by the molecules and greater the escaping tendency from the surface of the liquids and Vice Versa. In summer, surrounding temperature is higher as compared to winter. So, wet clothes dry quickly in summer than in winter.
SELF ASSESSMENT EXERCISE 5.6
Sodium chloride, an ionic compound, has a high melting point of 801°C. Whereas molecular solid such as ice has relatively low melting point of 0°C. Explain this difference.
Ans.
In Sodium chloride, an ionic crystalline solid has strong electrostatic forces of attraction are present between positively charged ions with many negatively charged ions and Vice Versa while in molecular solids i.e ice, weaker intermolecular forces are present. So M.P of sodium chloride (801°C) is higher than molecular solids (0°C) that is ice. | 10,330 | 41,762 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2019-30 | latest | en | 0.923935 |
http://nzmaths.co.nz/resource/castle | 1,371,678,335,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368709224828/warc/CC-MAIN-20130516130024-00060-ip-10-60-113-184.ec2.internal.warc.gz | 186,681,495 | 10,955 | Te Kete Ipurangi
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Level Four > Geometry and Measurement
# The Castle
Specific Learning Outcomes:
Construct 3-D shapes from 2-D drawings
Describe the symmetries of 3-D shapes
Draw 3-D shapes on isometric paper
Devise and use problem solving strategies to explore situations mathematically (guess and check, use equipment)
Description of mathematics:
This problem explores the relationship between 2-D and 3-D shapes and helps to visualise 3-D shapes from 2-D representations. It shows that different 3-D objects can be represented by the same set of 2-D objects. The problem also investigates the symmetry of objects.
Required Resource Materials:
small cubes
isometric dot paper
Copymaster of the problem (English)
Copymaster of the problem (Māori)
Activity:
### Problem
Ravi has designed plans for a castle that show the front, top, and side views.
Build a castle to Ravi’s design using the cubes.
What is the largest number of cubes that you can use in the construction of a castle from Ravi’s plans?
What is the smallest number of cubes that you can use in the construction of a castle from Ravi’s plans?
Can you build a symmetrical castle that is different from any of the ones you have built so far? How many symmetrical castles can you build to the Ravi’s specifications?
### Teaching sequence
1. Introduce the problem – create a scenario that the students are builders who have been given the task of constructing a castle from a set of plans. Make sure that the students understand how to use the 2-D views and the isometric paper.
2. Give blocks and isometric paper to pairs of students.
3. As the students work ask questions that focus their thinking on the models they are building.
What does the view from the top of your model look like? Side?
Is there any way to use the blocks to get the same view?
How do you know if you have found all the possibilities?
4. Share solutions – both models and isometric drawings.
5. If you are using this problem to reinforce symmetries require that the students record the symmetries on the drawing or describe them.
#### Extension
1. Is it possible to design plans for a castle so that it can only be built in one way (uniquely)?
2. On a base of three cubes by three cubes, design your own construction using top, side and front views only. Ask the questions above about your design. Stick to a three by three base. What design will give you the biggest difference between the largest number of cubes you can use and the smallest number?
#### Solution to the problem
Here we show one possible castle that could be drawn from Ravi’s design.
This castle looks exactly the same from each side. It has 22 blocks.
Actually this is the most blocks that can be used for Ravi’s design. If any other block is added, there will either be three consecutive blocks across a ‘face’ of the castle or there will be an extra block right in the middle of the top layer. Either of these situations can only occur if either the front or side views look like B. So no more blocks can be added.
There may be some dispute as to what constitutes the smallest castle. Here we assume that there have to be 9 blocks on the base. Given that, we have shown the smallest number of cubes in the castle below. It is assumed that the castle looks the same from the corner block opposite the nearest block.
By having ‘towers’ in opposite corners the castle has the correct front and side views. This castle has 14 blocks. Is this the only smallest castle?
It is possible to build other symmetrical castles. Just add the same tower to the two opposite corners that only have one block in the picture above. This tower can have one block in it or two. The version with two blocks has more symmetry than the other one. There are other possibilities too. How many can you find?
To make unique plans we need to have another view of the castle. Giving the view from the back might help in some cases. However, the back view of the two castles we have shown above, is the same! It would seem that four views are not good enough. So we can’t always represent 3-D objects accurately with 2-D snapshots of the kind we have used here, where we have tried a plan view and several elevation views. However, in this type of ‘cube’ building, we can determine the construction precisely if we use horizontal or vertical cross-sectional views. How many 2-D snapshots do we need, to precisely determine any Castle that is three cubes high?
AttachmentSize
Castle.pdf38.02 KB
CastleMaori.pdf41.19 KB
## Castles On The Ground
Construct 3D shapes from 2D drawings
Describe the symmetries of 3D shapes
Draw 3D shapes on isometric paper
Interpret information and results in context; make conjectures in a mathematical context;
Devise and use problem solving strategies to explore situations mathematically (guess and check, use equipment).
## Odd One Out
list a number of properties that distinguish squares from circles from cubes from pentagons.
devise and use problem solving strategies to explore situations mathematically (guess and check, make a drawing, use equipment).
## Peter's Third String
Determine the maximum area of a rectangle with a given perimeter
Consider how the area of a quadrilateral changes as its shape changes
Interpret a relationship from a graph
## QwikQure Algebra
Use the graph to describe the relationship
Devise and use problem solving strategies (draw a picture, make a table)
## The Topsy-Turvy Twins
Use a graph to describe a rule for continuing a number sequence
Use a graph to compare two number sequences
Devise and use problem solving strategies to explore situations mathematically (guess and check, be systematic, make a drawing). | 1,256 | 5,764 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | longest | en | 0.936916 |
https://socratic.org/questions/an-object-with-a-mass-of-7-kg-is-revolving-around-a-point-at-a-distance-of-8-m-i-3#610680 | 1,653,792,644,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663035797.93/warc/CC-MAIN-20220529011010-20220529041010-00149.warc.gz | 585,147,417 | 5,895 | # An object with a mass of 7 kg is revolving around a point at a distance of 8 m. If the object is making revolutions at a frequency of 1 Hz, what is the centripetal force acting on the object?
May 10, 2018
$\text{2208.5 N}$
#### Explanation:
Centripetal Force of a point object is given by the formula
${\text{F = m r ω}}^{2}$
${\text{F = m r (2 π f)}}^{2}$
$\text{F = 7 kg × 8 m × (2 × 3.14 × 1 Hz)"^2 = "2208.5 N}$ | 151 | 424 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 4, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2022-21 | latest | en | 0.856769 |
https://www.jiskha.com/display.cgi?id=1330654278 | 1,503,317,556,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886108268.39/warc/CC-MAIN-20170821114342-20170821134342-00452.warc.gz | 907,327,174 | 3,945 | # chemistry
posted by .
Calculate the standard reduction potential for the reaction of Cu(III) to Cu(II)
Use the following data:
1) Cu^+3 + 2e^- -> Cu^+, E°1 = 1.28 V
2) Cu^+2 + e^- -> Cu^+, E°2 = 0.15 V
3) Cu^+2 + 2e^- -> Cu(s), E°3 = 0.34 V
4) Cu^+ + e^- -> Cu(s), E°4 = 0.52 V
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Using the following data 1. CU3+ + 2e- ---> Cu+ E1= 1.28V 2. CU2+ + e- ---> Cu+ E2= 0.15V 3. Cu2+ + 2e- ----> Cu(s) E3= 0.34V 3. Cu+ + e- ----> Cu (s) E4= 0.52V calculate the standard reduction potential for the reaction …
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More Similar Questions | 981 | 2,861 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2017-34 | latest | en | 0.755386 |
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• Related Books
0 answersPost by Kumar Sandrasegaran on June 6, 2011Example IV should say "\$ 8.50 per hour for the first 40 hours ..."
Rates
• Rates and Unit Rates are types of ratios. A rate is a ratio that compares quantities measured in different units, such as miles per gallon. A unit rate is the rate for one unit of a quantity.
• To find a unit rate, divide the denominator by itself, and divide the numerator by the same number.
• Comparing unit rates can help you find the better buy when comparing different quantities of the same amount.
Rates
A car can go 70 miles on 2 gallons of gas. Find the unit rate.
• [(70 miles)/(2 gallons)] = [(m miles)/(1 gallon)]
• [(70 ÷2)/(2 ÷2)] = [35/1]
35 miles per 1 gallon
Unicorns can produce 12 rainbows every 4 hours. Find the unit rate.
• [(12 rainbows)/(4 hours)] = [(r rainbows)/(1 hour)]
• [(12 ÷4)/(4 ÷4)] = [3/1]
3 rainbows per 1 hour
A runner can run 21 miles in 3 hours. Find the unit rate.
• [(21 miles)/(3 hours)] = [(m miles)/(1 hour)]
• [(21 ÷3)/(3 ÷3)] = [7/1]
7 miles per 1 hour
A train travels 2 miles per chunk of coal. How many chunks of coal does a train need to travel 36 miles?
• [(2 miles)/(1 chunk of coal)] = [(36 miles)/(c chunks of coal)]
• [(2 ×18)/(1 ×18)] = [36/c]
c = 18 chunks of coal
Building a fence takes 36 nails. How many fences can be built using 144 nails?
• [(36 nails)/(1 fence)] = [(144 nails)/(f fences)]
• [(36 ×4)/(1 ×4)] = [144/f]
f = 4 fences
Every 33 ounces of frozen yogurt costs \$ 1. How much frozen yogurt can be bought with \$ 0.50?
• [(33 ounces)/\$ 1] = [(y ounces)/\$ 0.50]
• [(33 ÷2)/(\$ 1 ÷2)] = [(y ounces)/\$ 0.50]
y = [33/2] ounces = 16[1/2] ounces
Lisa can type 90 words per minute. How long would it take Lisa to type 450 words?
• [(90 words)/(1 minute)] = [(450 words)/(m minutes)]
• [(90 ×5)/(1 ×5)] = [450/m]
m = 5 minutes
One box of 32 - ounce cereal costs \$ 3.60. Another box of 45 - ounces cereal costs \$ 5.45. Which box of cereal is the better buy?
• [\$ 3.60/(32 ounces)] ≈ [\$ 0.11/(1 ounce)]
• [\$ 5.45/(45 ounces)] ≈ [\$ 0.12/(1 ounces)]
The 32-ounce box of cereal is the better buy.
Joe went to go buy a tub of ice cream. The tub of ice cream was 7 servings, and each serving was 350 calories. Joe ate the whole tub of ice cream. How many calories did Joe consume?
• [(1 serving)/(350 calories)] = [(7 servings)/(x calories)]
• [(1 ×7)/(350 ×7)] = [7/x]
x = 2450 calories
Melissa needs 4 packs of glue for her science project. Should she buy two 2 - packs of glue that cost \$ 2.50 each, or should she buy one 4 - pack of glue that costs \$ 4.95?
• [\$ 2.50/(2 glue sticks)] = [\$ 1.25/(1 glue stick)]
• [\$ 4.95/(4 glue sticks)] ≈ [\$ 1.24/(1 glue stick)]
4-pack
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Rates
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
• Intro 0:00
• What You'll Learn and Why 0:04
• Topics Overview
• Vocabulary 0:18
• Rate
• Unit Rate
• Finding a Unit Rate 1:00
• Example: Delivery Rate
• Using a Unit Rate 1:46
• Example: Miles and Gallon of Gas
• Comparing Unit Rates 2:52
• Example: Which is the Better Buy
• Extra Example 1: Calories 6:30
• Extra Example 2: Typing Speed 7:22
• Extra Example 3: Which is the Better Buy 8:23
• Extra Example 4: Wages 10:48 | 1,145 | 3,595 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2017-43 | latest | en | 0.821156 |
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# What is 18 over 94 in simplest form?
Updated: 8/21/2019
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6y ago
9/47
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Related questions
4/94/9
### What is the simplest form of 94 over 1000?
The simplest form of 94 over 1000 is 47/500
36/94 = 18/47
### What is 94 over 99 in simplest form?
94/99 is in its simplest form.
94
88/94 = 44/47
26/47
23/47
1/2
It is 47/50
### What is the simplest form of the fraction 47 over 94?
47 is half of 94, so 1 over 2
188/198 = 94/99 | 253 | 725 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.853437 |
https://pubs.aip.org/aapt/ajp/article/87/12/994/236672/Estimating-the-size-of-Earth-s-umbral-shadow-using | 1,716,465,938,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058625.16/warc/CC-MAIN-20240523111540-20240523141540-00229.warc.gz | 411,714,624 | 33,805 | We present a simple method to estimate the size of Earth's umbral shadow in a classroom setting. The method uses the published sky brightness curves obtained during a total lunar eclipse and requires only a conceptual understanding of lunar eclipses and simple geometric considerations. It is suitable for use in introductory and upper level astronomy courses.
Aristarchus of Samos was the first astronomer to estimate the relative sizes of the Earth and Moon and the distance to the Moon.1 His method involved observing the time between the Moon's entrance to and exit from the Earth's umbral shadow during a lunar eclipse. By comparing this transit time with the time for the Moon to traverse a distance equal to its own diameter, Aristarchus determined the size of Earth's shadow in terms of the lunar diameter. From a single lunar eclipse observation, he found that the umbral shadow at the distance to the Moon was 2 times the lunar diameter. Once Aristarchus determined the distance to the Moon, he used simple geometry to determine the size of Earth relative to the Moon using the diameter of the umbral shadow and the Earth-Moon distance.1,2 Today, we know that for a central lunar eclipse, in which the Moon passes through the center of Earth's shadow, the diameter of the umbral shadow averages 2.65 lunar diameters. The diameter of the umbral shadow can be as small as 2.578 lunar diameters if an eclipse occurs at lunar perigee or as large 2.735 lunar diameters if the eclipse occurs at apogee.
Cowley describes a classroom activity using lunar eclipse photographs to determine the Earth-Moon distance.3 His method relies on the measured curvature of Earth's shadow, as first done by Hipparchus, using photographs found in textbooks. Bruning used Cowley's method on a single photograph containing several phases of the July 5–6, 1987 lunar eclipse.4 The size of Earth's shadow can also be determined using observations of craters during a lunar eclipse: this technique uses the time it takes for the umbral shadow to traverse a crater of known diameter.5 We propose a new eclipse classroom activity, using published light curves of night sky brightness taken during lunar eclipses to estimate the size of Earth's umbral shadow at the location of the Moon, employing the method of Aristarchus. This method is accessible to non-science majors and provides them with the experience of using real data to reproduce an important scientific experiment.
Figure 1 shows a schematic representation of the Moon's motion through the penumbral and umbral shadows during a total, central eclipse. Figure 2 shows a plot of night sky brightness during the umbral phase of a total lunar eclipse, measured at zenith on the night of January 20–21, 2019 in Morehead, KY. The specific details of the instrumentation and observing conditions are available in another paper and are not particularly important for the activity.6 On the night of a lunar eclipse, the Full Moon will rise and the sky will begin to brighten. As the Moon enters Earth's umbral shadow (U1), the sky brightness will decrease. Once the entire Moon has entered the umbral shadow (U2), the sky will generally be about as bright as a New Moon night and will remain so until the Moon begins to exit the umbral shadow (U3).
Fig. 1.
A schematic of the progression of the Moon through the penumbral and umbral shadows during a central lunar eclipse. In this diagram, the diameter of the umbral shadow is 2.65 lunar diameters.
Fig. 1.
A schematic of the progression of the Moon through the penumbral and umbral shadows during a central lunar eclipse. In this diagram, the diameter of the umbral shadow is 2.65 lunar diameters.
Close modal
Fig. 2.
Sky brightness at zenith during the January 20–21, 2019 total lunar eclipse as a function of local time. The units of sky brightness are in magnitudes per square arc second: recall that the magnitude is an inverse scale so that larger numbers indicate a darker sky. These data were obtained by the authors using a Unihedron Sky Quality Light Meter (SQM) located in Morehead, KY. This plot was created with the Matplotlib library in PYTHON (Ref. 9).
Fig. 2.
Sky brightness at zenith during the January 20–21, 2019 total lunar eclipse as a function of local time. The units of sky brightness are in magnitudes per square arc second: recall that the magnitude is an inverse scale so that larger numbers indicate a darker sky. These data were obtained by the authors using a Unihedron Sky Quality Light Meter (SQM) located in Morehead, KY. This plot was created with the Matplotlib library in PYTHON (Ref. 9).
Close modal
The difference in the time between U2 and U1 and, likewise, between U4 and U3 represents the time it takes for the Moon to move one full diameter in the sky, assuming a central eclipse. The difference between U3 and U1 and, likewise, between U4 and U2 represents the total time for the Moon to travel through the full umbral diameter. To determine the size of the umbral shadow, you divide the average time for the Moon to traverse the umbral shadow by the average time it takes for the Moon to move one diameter,
$Dumbra=(tU3−tU1)+(tU4−tU2)(tU2−tU1)+(tU4−tU3)DMoon.$
(1)
Using the data from Fig. 2, we find an umbral shadow diameter of 1.9 ± 0.3 lunar diameters. Why is this value less than the average (and even the minimum)? The most likely interpretation is that the January 20–21, 2019 eclipse was not a central eclipse! Students can verify this by examining the eclipse data sheets available on NASA's Eclipse website.7
Our method assumes that the time elapsed between U1 and U2 and U3 and U4 actually represents the time for the Moon to move a distance equal to one full lunar diameter. However, this is only true for a central eclipse. It turns out that for non-central eclipses, this is an over-estimate of the time as the Moon will have to move further in the sky to be completely in the umbral shadow, as shown in Fig. 3. We can compare the average of these two times with the actual rate of motion in the sky. Recall that the synodic period of the Moon is 29.5 days. In one hour, the Moon moves
$t=360°29.5 days×1 day24 h=0.508°/h.$
(2)
So in 60 min, the Moon moves a distance equal to one full lunar diameter. If the average of the two time differences, tU2tU1 and tU4tU3, is more than 60 min, can we argue that the eclipse is not a central eclipse? (This is understood by comparing positions U1 and U2 (or U3 and U4) in Figs. 1 and 3. In Fig. 1, the Moon must travel exactly one diameter to fully enter or exit the umbra, while in Fig. 3, it must travel more than one diameter. Since the Moon travels one diameter in every 60 min, it will take longer than this to enter or exit the umbra in a non-central eclipse.) If the time in the denominator is over-estimated, then the quotient in Eq. (1) is underestimated, and hence, our method will tend to underestimate the size of the umbral shadow in a non-central eclipse. If the eclipse is central, the night sky brightness curve should be symmetric. Note the asymmetry of the light curve of our non-central eclipse: would asymmetry be a characteristic of a non-central eclipse light curve? (If students examine the “Eclipse Wise”8 data sheets for these three eclipses, they will find that the 1982 eclipse is very nearly a central eclipse, while those of 2004 and 2019 are definitely non-central. This eliminates non-centrality as a possible explanation for asymmetry in light curves. We posit that the observed asymmetry in the 2019 eclipse light curve is the result of changes in the lunar altitude, relative to the zenith observation point, during the eclipse. Without further data, we cannot test this hypothesis. This is a good lesson for students: more observations are often required to refute an explanation.) These questions should be accessible to general astronomy students in both a service course and physics majors. A question that you might ask physics majors: why do we not have to take into account Earth's motion around the Sun during the lunar eclipse? (The answer lies in Eq. (1): the diameter of the umbra is a ratio of two times, and the Earth's motion around the Sun affects both the numerator and the denominator in the same way.) This should stimulate discussions regarding the motion of the center of mass of the Earth-Moon system. Other ways to challenge upper level physics students include having them calculate the diameter of the penumbral shadow during the eclipse based on the known lunar distance10 and some simple geometry.11,12 More advanced physics students can consider whether stating a measurement uncertainty makes sense given that this method will tend to underestimate umbral diameters. You can really make this activity challenging for upper level physics majors: provide minimal guidance on the analysis technique and have them read the three articles with sky brightness data, applying this method to all three sets of data.
Fig. 3.
In a non-central eclipse, the distance that the Moon must traverse between the U1 and U2 stages of the eclipse is greater than that in the central eclipse in Fig. 1. Thus, the time between U1 and U2 stages in a non-central eclipse overestimates the time to move one full lunar diameter in the sky.
Fig. 3.
In a non-central eclipse, the distance that the Moon must traverse between the U1 and U2 stages of the eclipse is greater than that in the central eclipse in Fig. 1. Thus, the time between U1 and U2 stages in a non-central eclipse overestimates the time to move one full lunar diameter in the sky.
Close modal
Sky brightness curves for total lunar eclipses are rare. We have found only two other papers with night sky brightness light curves: one obtained during the July 6, 1982 total lunar eclipse, see Fig. 4, and another from the October 27–28, 2004 lunar eclipse, see Fig. 5.13,14 Although these light curves were obtained using very different instrumentation, the umbral portions of the light curves share the same characteristics as our data. Neither of these authors use their data to estimate the size of the umbral shadow, and so students can use these data to estimate the size of the umbral shadow under different lunar eclipse conditions using our method. Our method is unique in that it employs the same basic ideas as Aristarchus' original method but uses light curves.
Fig. 4.
Sky brightness curve from the July 6, 1982 total lunar eclipse. These data were obtained by Morton (Ref. 13) using the 31-in. refractor at Lowell Observatory. The telescope tracked a patch of sky 20° north of the Moon, tracking at lunar speed. This graph is reproduced using the WebPlotDigitizer (Ref. 15) online software using the automatic extraction feature.
Fig. 4.
Sky brightness curve from the July 6, 1982 total lunar eclipse. These data were obtained by Morton (Ref. 13) using the 31-in. refractor at Lowell Observatory. The telescope tracked a patch of sky 20° north of the Moon, tracking at lunar speed. This graph is reproduced using the WebPlotDigitizer (Ref. 15) online software using the automatic extraction feature.
Close modal
Fig. 5.
Sky brightness on the night of the October 28, 2004 total lunar eclipse. These data were obtained by Shawn Dvorak using a telescope equipped with a CCD while observing the variable star QQ Cas (Ref. 14).
Fig. 5.
Sky brightness on the night of the October 28, 2004 total lunar eclipse. These data were obtained by Shawn Dvorak using a telescope equipped with a CCD while observing the variable star QQ Cas (Ref. 14).
Close modal
Given that our method tends to underestimate the size of the umbral shadow for non-central eclipses, why would you use this method? Well, nearly sixty percent of all lunar eclipses between 1999 and 3000 A.D. are central, total eclipses.16 Our observation method,6 using a Unihedron Sky Quality Meter pointed at zenith,6 is simple and relatively inexpensive (about \$250 US) and requires only a meter, a mount, and a clear eclipse night. If there is a total lunar eclipse in your future, a class or a single student could take similar data and use this method.
Also, note that the night sky brightness curve during a total lunar eclipse bears a striking resemblance to exoplanet transit curves. Exoplanet transits can be central or non-central, and this affects the length of the flat portion of their light curve similar to a lunar eclipse. The radius of an exoplanet affects the depth of the light curve. On the other hand, the depth of the sky brightness light curve during an eclipse should only depend on the lunar altitude and local night sky brightness.17 The duration of exoplanet ingress and egress times coupled with the exoplanet transit time can be used to determine the radius of the exoplanet.18 These parallels make our method a useful educational precursor to the astrophysical study of exoplanets!
The authors thank the two anonymous referees: their comments and suggestions greatly improved the clarity of this paper.
1.
Alan H.
Batten
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Aristarchus of Samos
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Astronomy 101 Specials: Aristarchus and the Size of the Moon
,” <https://www.eg.bucknell.edu/physics/astronomy/astr101/specials/aristarchus.html>; accessed July 25, 2019.
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Global Robotic Telescopes Intelligent Array for e-Science (GLORIA)
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Measuring the size of the shadow of the Earth (Total Lunar Eclipse 2014)
,” <https://cordis.europa.eu/docs/projects/cnect/3/283783/080/deliverables/001-D29lunareclipseactivity1v02A.pdf>; accessed August 2, 2019.
6.
Jennifer J.
Birriel
and
J.
, “
Sky brightness at zenith during the January 2019 total lunar eclipse
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JAAVSO
47
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1
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), <https://www.aavso.org/apps/jaavso/article/3471/>.
7.
Robert M.
Candey
, “
Lunar Eclipse Page
,” <https://eclipse.gsfc.nasa.gov/lunar.html>; accessed August 2, 2019.
8.
Eclipse Wise
” Website, <http://eclipsewise.com/eclipse.html>; accessed September 17, 2019.
9.
J. D.
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The moon tonight
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11.
Eric
Rassmussen
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An American in Spain
,” <https://erikras.com/2011/12/16/how-big-is-the-earths-shadow-on-the-moon/>; accessed August 2, 2019.
12.
Pierre
Causeret
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How to Calculation the Duration of a Lunar Eclipse
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James C.
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Sky brightness and colour changes during the 1982 July lunar eclipse
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Shawn
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Serendipitous photometric observations of the 2004 lunar eclipse
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Ankit
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: Index to Five Millenium Catalog of Lunar Eclipses, <https://eclipse.gsfc.nasa.gov/LEcat5/LEcatalog.html>; accessed August 2, 2019.
17.
The Transit Method
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, https://www.iop.org/education/teacher/resources/exoplanet_physics/file_65609.pdf; accessed August 2, 2019.
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Astronomy with SalsaJ, “
Discover An Exoplanet: The Transit Method
,” < | 3,720 | 15,383 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 4, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-22 | latest | en | 0.897099 |
https://www.doubtnut.com/qna/38705552 | 1,723,664,772,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641121834.91/warc/CC-MAIN-20240814190250-20240814220250-00686.warc.gz | 569,799,743 | 44,994 | # If →a,→band→c are three non-coplanar non-zero vectors, then prove that (→a.→a)→b×→c+(→a.→b)→c×→a+(→a.→c)→a×→b=[→b→c→a]→a
Video Solution
Text Solution
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## As →a,→band→c are non- coplanar, →b×→a,→c×→aand→a×→b are also non-coplanar, So, any vector con be expressend as a linear combination of these vectors. →a=λ→b×→c+μ→c×→a+v→a×→b →a.→a=λ[→b→c→a],→a.→b=μ[→c→a→b],→a.→c=v[→a→b→c] →a=(→a.→a)→b×→c[→b→c→a]+(→a.→b)→c×→a[→c→a→b]+(→a.→c)→a×→b[→a→b→c]
|
Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One
## If →a,→b,→c are three non coplanar, non zero vectors then (→a.→a)(→b×→c)+(→a.→b)(→c×→a)+(→a.→c)(→a×→b) is equal to
A[abc]c
B[bca]a
C[cab]b
Dnone of these
• Question 2 - Select One
## If →a,→band→c1 are three non-coplanar vectors, then (→a+→b+→c).[(→a+→b)×(→a+→c)] equals
A0
B[abc]
C2[abc]
D[abc]
• Question 3 - Select One
## If →a,→band→c1 are three non-coplanar vectors, then (→a+→b+→c).[(→a+→b)×(→a+→c)] equals
A0
B[abc]
C2[abc]
D[abc]
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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 | 763 | 1,974 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-33 | latest | en | 0.772854 |
https://www.unitconverters.net/flow/cubic-meter-hour-to-liter-second.htm | 1,718,243,900,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861319.37/warc/CC-MAIN-20240612234213-20240613024213-00639.warc.gz | 950,868,435 | 3,316 | Home / Flow Conversion / Convert Cubic Meter/hour to Liter/second
# Convert Cubic Meter/hour to Liter/second
Please provide values below to convert cubic meter/hour [m^3/h] to liter/second [L/s], or vice versa.
From: cubic meter/hour To: liter/second
### Cubic Meter/hour to Liter/second Conversion Table
Cubic Meter/hour [m^3/h]Liter/second [L/s]
0.01 m^3/h0.0027777778 L/s
0.1 m^3/h0.0277777778 L/s
1 m^3/h0.2777777778 L/s
2 m^3/h0.5555555556 L/s
3 m^3/h0.8333333333 L/s
5 m^3/h1.3888888889 L/s
10 m^3/h2.7777777778 L/s
20 m^3/h5.5555555556 L/s
50 m^3/h13.8888888889 L/s
100 m^3/h27.7777777778 L/s
1000 m^3/h277.7777777778 L/s
### How to Convert Cubic Meter/hour to Liter/second
1 m^3/h = 0.2777777778 L/s
1 L/s = 3.6 m^3/h
Example: convert 15 m^3/h to L/s:
15 m^3/h = 15 × 0.2777777778 L/s = 4.1666666667 L/s | 343 | 820 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.333419 |
http://deerfieldbeachsdachurch.org/pdf/complex-numbers-in-n-dimensions-north-holland-mathematics-studies | 1,611,402,515,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703537796.45/warc/CC-MAIN-20210123094754-20210123124754-00107.warc.gz | 26,376,598 | 9,904 | # Complex Numbers in n Dimensions (North-Holland Mathematics by S. Olariu PDF
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By S. Olariu
specified structures of hypercomplex numbers in n dimensions are brought during this publication, for which the multiplication is associative and commutative, and that are wealthy adequate in homes such that exponential and trigonometric varieties exist and the innovations of analytic n-complex functionality, contour integration and residue could be defined.
The first form of hypercomplex numbers, known as polar hypercomplex numbers, is characterised via the presence in a fair variety of dimensions better or equivalent to four of 2 polar axes, and via the presence in a wierd variety of dimensions of 1 polar axis. the opposite kind of hypercomplex numbers exists as a different entity in basic terms while the variety of dimensions n of the distance is even, and because the placement of some extent is distinct because of n/2-1 planar angles, those numbers were referred to as planar hypercomplex numbers.
The improvement of the idea that of analytic services of hypercomplex variables was once rendered attainable by way of the lifestyles of an exponential kind of the n-complex numbers. Azimuthal angles, that are cyclic variables, look in those varieties on the exponent, and bring about the idea that of n-dimensional hypercomplex residue. Expressions are given for the effortless features of n-complex variable. particularly, the exponential functionality of an n-complex quantity is multiplied by way of features referred to as during this booklet n-dimensional cosexponential functions
of the polar and respectively planar variety, that are generalizations to n dimensions of the sine, cosine and exponential functions.
In the case of polar advanced numbers, a polynomial should be written as a made from linear or quadratic components, even though it is fascinating that numerous factorizations are as a rule attainable. in relation to planar hypercomplex numbers, a polynomial can consistently be written as a fabricated from linear elements, even though, back, a number of factorizations are mostly possible.
The ebook provides an in depth research of the hypercomplex numbers in 2, three and four dimensions, then offers the homes of hypercomplex numbers in five and six dimensions, and it keeps with an in depth research of polar and planar hypercomplex numbers in n dimensions. The essence of this booklet is the interaction among the algebraic, the geometric and the analytic points of the relations.
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Additional info for Complex Numbers in n Dimensions (North-Holland Mathematics Studies)
Example text | 864 | 4,489 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-04 | latest | en | 0.943216 |
https://www.mathworks.com/matlabcentral/answers/526138-how-to-print-symbolic-representation-of-4-1-1-4-2-1-4-3-1-rather-than-the-actual-numerical-output?s_tid=prof_contriblnk | 1,660,163,498,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571210.98/warc/CC-MAIN-20220810191850-20220810221850-00789.warc.gz | 783,702,403 | 26,697 | How to print symbolic representation of [4^1+1 4^2+1 4^3+1] rather than the actual numerical output?
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hmhuang on 17 May 2020
Commented: Ameer Hamza on 18 May 2020
I would like to add a line of texts like "points = 4^1+1 4^2+1 4^3+1" instead of "points = 5 17 65" to chart.
The current content of my code to print "points = 5 17 65" is as following:
for i = 1:3
n = 4^i + 1;
%{
do something with numeric values of n and i (i.e., i = 1, 2, 3)
%}
% Storing "number of points" in vector 'point'
point(i) = n;
end
disp(['Grid points = ' sprintf('%d ', point)])
Note that I do need the numerical values of n and i (i.e., i = 1, 2, 3 for n = 4^i + 1) to do some computations.
I found
syms i
may be appropriate for my intended use, but cannot figure out how to exactly apply it into my for-loop.
Thanks!
Ameer Hamza on 17 May 2020
Edited: Ameer Hamza on 17 May 2020
point = 1:3;
disp(['Grid points = ' sprintf('4^%d ', point)])
Result
Grid points = 4^1 4^2 4^3
Ameer Hamza on 18 May 2020
The syntax is not correct, according to MATLAB. But I didn't understand what you are trying to do here. Can you explain with an image what where you want to write the text()?
R2019b
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Start Hunting! | 432 | 1,309 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2022-33 | latest | en | 0.887542 |
http://physgre.s3-website-us-east-1.amazonaws.com/1992%20html/1992%20problem%2093.html | 1,531,813,168,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589618.52/warc/CC-MAIN-20180717070721-20180717090721-00292.warc.gz | 285,074,799 | 2,660 | ## Solution to 1992 Problem 93
Let $l$ be the length of the string. Then \begin{align*}x &= - l \cos \theta \\y &= - l \sin \theta\end{align*} implies that\begin{align*}\dot{x} &= l \dot{\theta} \sin \theta \\\dot{y} &= - l \dot{\theta} \cos \theta\end{align*} which implies that\begin{align*}\ddot{x} &= l \dot{\theta}^2 \cos \theta + l \ddot{\theta} \sin \theta \\\ddot{y} &= l \dot{\theta}^... Thus,\begin{align}a^2 &= \ddot{x}^2 + \ddot{y}^2 \label{eqn:3} \\&= l^2 \dot{\theta}^4 + l^2 \ddot{\theta}^2 \label{eqn:4}...So, we need to find $\dot{\theta}$ and $\ddot{\theta}$ as functions of $\theta$. For $\dot{\theta}$, this can be done by equating the kinetic energy that the mass has when its angle is $\theta$ to the change in the mass's potential energy between its initial position and its position when its angle is $\theta$:\begin{align*}m g l \sin \theta = \frac{1}{2} m l^2 \dot{\theta}^2\end{align*}Thus,\setcounter{equation}{2}\begin{align}\dot{\theta}^2 = \frac{2g}{l} \sin \theta \label{eqn:1}\end{align}In order to find $\ddot{\theta}$, we use Lagrange's equation\begin{align*}\frac{\partial L}{\partial \theta} = \frac{d}{dt} \frac{\partial L}{\partial \dot{\theta}}\end{align*}The kinetic energy of the particle is\begin{align*}T = \frac{1}{2}m l^2 \dot{\theta}^2\end{align*} The potential energy of the particle is\begin{align*}V = - l \sin \theta m g\end{align*}Therefore the Lagrangian of the particle is\begin{align*}L = \frac{1}{2}m l^2 \dot{\theta}^2 + l \sin \theta m g\end{align*}Using Langrange's equation, we find that\setcounter{equation}{3}\begin{align}\ddot{\theta} = \frac{g}{l} \cos \theta \label{eqn:2}\end{align}Plugging equations (3) and (4) into equation (1)-(2) gives\begin{align*}a^2 &= g^2 \left(4 \sin \theta^2 + \cos^2 \theta \right) \\&= g^2 \left(3 \sin \theta^2 +1 \right) \end...or\begin{align*}\boxed{a = g \sqrt{3 \sin \theta^2 +1 }} \\\end{align*}Therefore, answer (E) is correct. | 722 | 1,932 | {"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": 21, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2018-30 | latest | en | 0.573349 |
https://en.m.wikipedia.org/wiki/Non-perturbative | 1,542,618,328,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039745522.86/warc/CC-MAIN-20181119084944-20181119110944-00318.warc.gz | 611,285,210 | 8,520 | # Non-perturbative
In mathematics and physics, a non-perturbative function or process is one that cannot be accurately described by perturbation theory. An example is the function
${\displaystyle f(x)=e^{-1/x^{2}}}$.
The function e−1/x². The Taylor series is identically zero, but the function is not.
The Taylor series at x = 0 for this function is exactly zero to all orders in perturbation theory, but the function is non-zero if x ≠ 0.
The implication of this for physics is that there are some phenomena which are impossible to understand by perturbation theory, regardless of how many orders of perturbation theory we use. Instantons are an example.
Therefore, in theoretical physics, a non-perturbative solution or theory is one that does not require perturbation theory to explicate, or does not simply describe the dynamics of perturbations around some fixed background. For this reason, non-perturbative solutions and theories yield insights into areas and subjects perturbative methods cannot reveal. | 221 | 1,017 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.921846 |
https://www.wyzant.com/resources/answers/225705/complex_roots | 1,632,653,726,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057857.27/warc/CC-MAIN-20210926083818-20210926113818-00132.warc.gz | 1,066,508,851 | 15,795 | Samagra G.
asked • 06/27/16
# complex roots
if Z1 , Z2 , Z3, Z4 are roots of equation z4 + z3 + z2 + z + 1 = 0
then what is the sum of all the fourth and fifth power of roots ?
## 3 Answers By Expert Tutors
By:
Degonimia H. answered • 07/01/16
Tutor
New to Wyzant
Efficient work at reduced prices
Mark M. answered • 06/27/16
Tutor
5.0 (243)
Mathematics Teacher - NCLB Highly Qualified
Sanhita M. answered • 06/27/16
Tutor
4.7 (11)
Mathematics and Geology
Samagra G.
but z=1 is not satisfying the equation
Report
06/27/16
Sanhita M.
I'll get back to you as soon as I can resolve.
Report
06/27/16
Sanhita M.
As I tested the given polynomial w.r.t imaginary roots of 1 , the last term of the given polynomial having possible real factors +1, -1, and possible imaginary factors +√-1,-√-1, by synthetic division the results are having remainders equal to -1. Therefore, none of the factors of 1 is actually a root of the polynomial, hence the equation does not hold for any of the factors of 1. Moreover, it seems that either the expression is incomplete or wrong. If not, then please provide the solution when you get one.
Report
06/28/16
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Choose an expert and meet online. No packages or subscriptions, pay only for the time you need. | 400 | 1,400 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-39 | latest | en | 0.903565 |
https://www.business2community.com/gambling/blackjack-split | 1,716,840,743,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059045.25/warc/CC-MAIN-20240527175049-20240527205049-00223.warc.gz | 584,636,374 | 44,794 | A blackjack split allows players to split their original hand into two separate bets if the two cards are of the same value. Using a split is a useful tool in a bettor’s arsenal, as it can swing the percentages in your favor and give you an edge over the dealer’s hand.
Read on to discover exactly when to split in a game of blackjack and when it’s better not to.
## The Thrill of Splitting in Blackjack
The first question you need to know the answer to is when can you split in a game of blackjack? And the answer is you can do so when you have two cards of the same value in your blackjack games.
When you split, you are essentially separating your hand into two separate hands and you will receive an additional card for each. This also means that you will need to double your original stake.
While the core of blackjack strategy split rules and how to split during a game are the same at every in-person and online casino, certain rules can differ, so make sure to check the rules at the casino before you start playing.
• Splitting 10’s – Although, probability dictates that to split tens is never a good idea, should you want to, you need to know if a casino allows splitting of all cards with a value of ten, i.e. you can split a jack and a king, or if they only allow splits of exact pairs i.e. you are able to split two jacks but not a jack and a king.
• Doubling down after the split – Some casinos do not allow you to double down in blackjack after splitting.
• Splitting a split – Some casinos have rules about the number of splits you can make after the initial split.
• Splitting aces – Some casinos do not consider it a blackjack if you have split aces and received a 10, but rather just a score of 21. This means you will not get the blackjack payout of 3:2 or in many casinos now 6:5.
• Hitting after splitting aces – Should you split aces in blackjack most casinos will only allow you to hit only once.
## Understanding the Blackjack Split
Now you have an understanding of what a split is, you may want to know when to split and why. A split should only be done when you can maximize your profits by taking advantage of a dealer’s position, you have a probability of improving your additional hand or in some cases, to limit your losses.
Below we go over scenarios when it’s in a player’s favor to split in a game and when it’s better not to.
### When to Split In Blackjack Games
#### Aces and Eights
Probability dictates you should always split aces in blackjack. Cards with a value of 10 are the most commonly drawn in blackjack so splitting your aces gives you a good chance of hitting blackjack and should you draw a 9, 8,7, or even a 6 you will still have a solid hand with a good chance of winning or at least pushing.
If you stick with your pair of aces your hand has a value of twelve. Should you then hit and draw a 10, you are right back to have twelve as now both your aces have a value of one. This leaves you in a precarious position as a ten means you go bust in blackjack games.
Two eights is also another hand that most pros, and the probability, suggests you should split.
This is because a total of sixteen means you can only win if the dealer busts, and hitting on sixteen is likely to make you do so. Splitting the eights gives you a higher percentage chance of winning at least one of your hands, meaning you will not lose money.
However, there are a few experts who think that splitting pairs isn’t necessarily the best option and you should just cut your losses or hope the dealer busts, but the majority of blackjack pros will advise you to always split eights.
#### Dependent on the Dealer’s Upcard
Unlike with Aces, there are certain blackjack hands where splitting is a good idea depending on the dealer’s upcard.
• Twos, threes, and sevens – These three hands are seen as pretty awful and that’s because the probability of going bust in one or two hits is very high. Splitting these pairs is a good idea if the dealer’s upcard is between a four and seven. Should the dealer’s upcard be a two or a three you should also split, but only if the casino you are playing at lets you double down after a split. If not then your best bet is to hit
• Nines – While eighteen is a solid hand, it’s still very beatable. If a dealer is showing either two- six or an eight or a nine then it’s best to split your nines.
• Sixes – If the dealer’s upcard is two-six then splitting is your best option.
### When Splitting Pairs Isn’t Necessary
#### Tens
A score of twenty in blackjack results in a win for a player 70.2% of the time, 17.5% a push, and only a 12.2% chance of losing. This is why you should never split 10s as it just turns a very strong hand into potentially two weak hands.
#### Fours
A pair of fours is a fairly decent starting hand in blackjack as you can not go bust if you hit and should you get a ten or an Ace you’ll have a very solid hand. Splitting the hand gives you a fairly high probability of ending up with two weak hands and in the process doubling your stake.
However, should the dealer’s upcard be a 5 or a 6 and the casino lets you double down after splitting then you should take this opportunity as there is a 42 percent chance the dealer could go bust.
#### Fives
A pair of fives means you have a value of ten which is the perfect number to hit on as you can not go bust and anything from an Eight to Ace is going to leave you with a good to great hand. If you split pairs like this, you are going to be left with a lower-value hand or one that leaves you much more likely to bust.
## Try A Split in a Game of Blackjack
After reading this article you should now have a good understanding of the answer to what does it mean to split in blackjack? In addition, you should also have a clear idea of when it’s a good idea to split and when not to. Your next step is to find a great blackjack casino and try out what you have learned. | 1,346 | 5,923 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-22 | latest | en | 0.94122 |
https://jp.mathworks.com/matlabcentral/cody/problems/15-find-the-longest-sequence-of-1-s-in-a-binary-sequence/solutions/1896108 | 1,604,184,561,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107922463.87/warc/CC-MAIN-20201031211812-20201101001812-00514.warc.gz | 353,791,605 | 17,280 | Cody
# Problem 15. Find the longest sequence of 1's in a binary sequence.
Solution 1896108
Submitted on 9 Aug 2019 by Todd Snow
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = '0'; y_correct = 0; assert(isequal(lengthOnes(x),y_correct))
2 Pass
x = '1'; y_correct = 1; assert(isequal(lengthOnes(x),y_correct))
3 Pass
x = '01'; y_correct = 1; assert(isequal(lengthOnes(x),y_correct))
4 Pass
x = '10'; y_correct = 1; assert(isequal(lengthOnes(x),y_correct))
5 Pass
x = '00'; y_correct = 0; assert(isequal(lengthOnes(x),y_correct))
6 Pass
x = '11'; y_correct = 2; assert(isequal(lengthOnes(x),y_correct))
7 Pass
x = '1111111111'; y_correct = 10; assert(isequal(lengthOnes(x),y_correct))
8 Pass
x = '100101011111010011111'; y_correct = 5; assert(isequal(lengthOnes(x),y_correct))
9 Pass
x = '01010101010101010101010101'; y_correct = 1; assert(isequal(lengthOnes(x),y_correct))
10 Pass
x = '0101010111000101110001011100010100001110110100000000110001001000001110001000111010101001101100001111'; y_correct = 4; assert(isequal(lengthOnes(x),y_correct))
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https://www.zedzone.space/post/2012/08/maths-to-rescue.html | 1,702,131,618,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100912.91/warc/CC-MAIN-20231209134916-20231209164916-00475.warc.gz | 1,170,326,102 | 3,965 | Michael Zucchi
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# Maths to the rescue
I had a bit of a think about the expression I am evaluating and I noticed I could use logarithms to perform the calculation instead of multiplies. And actually, as I don't use the value itself other than as a threshold, the summation result can be used directly anyway - it just wont give as sharp peaks.
First, changed the inner loop to perform a summation of two possible values, in 16 bits. This removed the need for 32-bit masks, and the floating point multiplies (not that they were holding anything up). This took 2 classifiers down from 26 instructions to 18, and a test-case from 100 to 70ms. It also freed 3 quad registers but I haven't yet investigated whether this will let me do more (actually it will let me take the loop down to 16 instructions as I have 2 immediate loads in there).
However, this morning I had a closer look at the summation I now have.
` r = sum m { testbit ? p : n } (1)`
The bit testing code can't be further improved, but perhaps the select and sum can?
First, subtract n from the expression and move it outside of the select, and then separate the summation.
``` r = sum m { (testbit ? (p - n) : 0) - n }
= sum m { testbit ? (p - n) : 0 } - sum m { n }
= sum m { testbit ? (p - n) : 0 } - mn (2)```
Which leaves with a slightly simpler internal summation and a constant. This expression can be implemented without a select instruction but since we have select in NEON it doesn't gain any instructions on it's own.
But this is where a neat trick of binary arithmetic comes in to play. Unlike C logic which uses 0 and 1 for false and true, the result of a bit test in SIMD code (and opencl c on vectors) is 0 for false, and ~0 for true. But in two's compliment signed binary, ~0 is also equal to -1.
So ... if the expression can be converted into a sum of -1's, then all one needs to do is directly add up the result of the bit tests, and avoid the select altogether. It is rather trivial to convert (2) into this form (if I have my summation identities on target anyway).
``` r = sum m { (n-p) (testbit ? -1 : 0) } - mn
= (n-p) sum m { testbit ? -1 : 0 } - mn
= A sum m { testbit ? -1 : 0 } + B (3)```
Where A and B are constants.
Actually there is another benefit here, as the summand is either 0 or -1 the range of the working sum is far more limited. Because of the size of the template I couldn't get away with 8 bit arithmetic without the danger of overflow - if I was using normal 8 bit two's complement arithmetic. But NEON to the rescue again, it has saturating instructions which will not overflow. It may still lead to a slightly incorrect answer - but i'm not interested in the absolute result but the relative one, and i think it should suffice.
I have yet to code this up, but if it works this will get the two-classifier case down to only 12 instructions - for 8 pixels. Less than one instruction per classifier per template, or about one when including the data loading as well.
And as this all now runs in double registers instead of quad registers, and I don't need n or p in the inner loop, it freeds up a whole swag of registers. These can be used to unroll more of the loop or share more of the data, for example by calculating more than 8 results per pass, more than one classifier at once, and so on. This might squeeze out a tad more juice out of it, but i'm not expecting much more here.
Update: So I coded it, it works. I'm calculating 32 locations at once, and there's still a good few registers free. Test case down from 70ms to 50ms.
Even better, I noticed the 'limited dual issue' capability of the NEON processor could be exploited in a couple of cases. Interleave the vtbl with some shifts: down to 43ms. Interleave the vqadd's with some vext's: down to under 41ms. Noice.
For comparison, an integer version in C (gcc, -O2) takes about 700ms.
Time for beer.
Tagged beagle, hacking. | 1,303 | 4,684 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2023-50 | latest | en | 0.836338 |
http://www.gregthatcher.com/Stocks/StockFourierAnalysisDetails.aspx?ticker=UBNT | 1,524,497,443,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946077.4/warc/CC-MAIN-20180423144933-20180423164933-00583.warc.gz | 426,322,371 | 49,688 | Back to list of Stocks See Also: Seasonal Analysis of UBNTGenetic Algorithms Stock Portfolio Generator, and Best Months to Buy/Sell Stocks
# Fourier Analysis of UBNT (Ubiquiti Networks)
UBNT (Ubiquiti Networks) appears to have interesting cyclic behaviour every 31 weeks (2.9823*sine), 26 weeks (2.1884*sine), and 34 weeks (2.0692*sine).
UBNT (Ubiquiti Networks) has an average price of 36.59 (topmost row, frequency = 0).
Click on the checkboxes shown on the right to see how the various frequencies contribute to the graph. Look for large magnitude coefficients (sine or cosine), as these are associated with frequencies which contribute most to the associated stock plot. If you find a large magnitude coefficient which dramatically changes the graph, look at the associated "Period" in weeks, as you may have found a significant recurring cycle for the stock of interest.
## Fourier Analysis
Using data from 10/14/2011 to 4/16/2018 for UBNT (Ubiquiti Networks), this program was able to calculate the following Fourier Series:
Sequence #Cosine Coefficients Sine Coefficients FrequenciesPeriod
036.58914 0
12.99264 -12.84287 (1*2π)/341341 weeks
22.84788 -15.1037 (2*2π)/341171 weeks
34.4239 -1.36729 (3*2π)/341114 weeks
4.06782 -1.1625 (4*2π)/34185 weeks
51.08213 -5.15416 (5*2π)/34168 weeks
6-1.70029 -3.84031 (6*2π)/34157 weeks
7-1.69385 -1.12538 (7*2π)/34149 weeks
8-2.0058 -2.54393 (8*2π)/34143 weeks
9.0249 -2.5144 (9*2π)/34138 weeks
10-.41484 -2.06921 (10*2π)/34134 weeks
11.15582 -2.9823 (11*2π)/34131 weeks
12-.3633 -1.47546 (12*2π)/34128 weeks
13-.2041 -2.18842 (13*2π)/34126 weeks
14.351 -.0157 (14*2π)/34124 weeks
15-.63544 -.18027 (15*2π)/34123 weeks
16.72385 -.72057 (16*2π)/34121 weeks
17-.13862 -.16731 (17*2π)/34120 weeks
18-.58332 -.35134 (18*2π)/34119 weeks
19.18069 -.43486 (19*2π)/34118 weeks
20.18614 -.83213 (20*2π)/34117 weeks
21-.05299 -1.3733 (21*2π)/34116 weeks
22-.36019 -.75702 (22*2π)/34116 weeks
23-.7779 -.59567 (23*2π)/34115 weeks
24.33405 -.94358 (24*2π)/34114 weeks
25.5081 -.09936 (25*2π)/34114 weeks
26-.21882 -.49108 (26*2π)/34113 weeks
271.00753 -.31151 (27*2π)/34113 weeks
28-.04182 -1.01312 (28*2π)/34112 weeks
29.06744 -.67812 (29*2π)/34112 weeks
30-.11984 -.19652 (30*2π)/34111 weeks
31-.51751 -1.33893 (31*2π)/34111 weeks
32-.32363 -1.0847 (32*2π)/34111 weeks
33-.74876 -.89102 (33*2π)/34110 weeks
34-.68767 -.75899 (34*2π)/34110 weeks
35-.3214 -.35733 (35*2π)/34110 weeks
36-.00908 -.28175 (36*2π)/3419 weeks
37-.19515 -.31493 (37*2π)/3419 weeks
38.1985 -.93538 (38*2π)/3419 weeks
39-.07604 -.12409 (39*2π)/3419 weeks
40-.77792 -.74321 (40*2π)/3419 weeks
41.25128 -.32319 (41*2π)/3418 weeks
42-.44207 -.22925 (42*2π)/3418 weeks
43-.3197 -.50305 (43*2π)/3418 weeks
44-.41818 -.08918 (44*2π)/3418 weeks
45-.2622 .04916 (45*2π)/3418 weeks
46-.02693 -.01543 (46*2π)/3417 weeks
47-.04908 -.05778 (47*2π)/3417 weeks
48-.18121 -.54473 (48*2π)/3417 weeks
49.21903 -.47006 (49*2π)/3417 weeks
50-.1359 -.27095 (50*2π)/3417 weeks
51-.32634 -.20973 (51*2π)/3417 weeks
52-.04114 -.2051 (52*2π)/3417 weeks
53.11509 -.1585 (53*2π)/3416 weeks
54.17714 -.09245 (54*2π)/3416 weeks
55-.47333 -.51258 (55*2π)/3416 weeks
56.11883 -.24566 (56*2π)/3416 weeks
57-.14415 -.16587 (57*2π)/3416 weeks
58-.1115 -.45158 (58*2π)/3416 weeks
59.23623 -.51269 (59*2π)/3416 weeks
60-.26199 -.33682 (60*2π)/3416 weeks
61-.37219 -.24142 (61*2π)/3416 weeks
62.01434 -.26808 (62*2π)/3416 weeks
63-.03866 -.096 (63*2π)/3415 weeks
64.05308 -.50668 (64*2π)/3415 weeks
65.07508 -.45724 (65*2π)/3415 weeks
66-.31979 -.35417 (66*2π)/3415 weeks
67-.17631 -.44689 (67*2π)/3415 weeks
68-.51361 -.33797 (68*2π)/3415 weeks
69-.23497 -.455 (69*2π)/3415 weeks
70-.28411 -.31585 (70*2π)/3415 weeks
71-.45875 -.30015 (71*2π)/3415 weeks
72-.11989 -.3369 (72*2π)/3415 weeks
73-.11338 -.13473 (73*2π)/3415 weeks
74-.21269 -.2735 (74*2π)/3415 weeks
75.01981 -.11573 (75*2π)/3415 weeks
76-.33437 .05747 (76*2π)/3414 weeks
77-.415 -.19159 (77*2π)/3414 weeks
78-.03445 .04681 (78*2π)/3414 weeks
79-.32279 -.0258 (79*2π)/3414 weeks
80-.08577 -.39678 (80*2π)/3414 weeks
81-.28653 -.1553 (81*2π)/3414 weeks
82-.13703 -.12806 (82*2π)/3414 weeks
83-.18171 -.28962 (83*2π)/3414 weeks
84-.03996 -.05016 (84*2π)/3414 weeks
85-.22864 -.2363 (85*2π)/3414 weeks
86-.21958 -.004 (86*2π)/3414 weeks
87-.19205 -.02062 (87*2π)/3414 weeks
88-.26692 -.18167 (88*2π)/3414 weeks
89-.26826 .12772 (89*2π)/3414 weeks
90-.06863 -.01812 (90*2π)/3414 weeks
91-.12653 -.02524 (91*2π)/3414 weeks
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93-.23397 -.22148 (93*2π)/3414 weeks
94-.31401 -.04723 (94*2π)/3414 weeks
95-.16838 .13051 (95*2π)/3414 weeks
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99-.14243 -.16494 (99*2π)/3413 weeks
100-.12319 -.05771 (100*2π)/3413 weeks
101-.06376 -.07705 (101*2π)/3413 weeks
102-.24382 -.30879 (102*2π)/3413 weeks
103-.08703 -.19327 (103*2π)/3413 weeks
104-.14695 -.14948 (104*2π)/3413 weeks
105.05777 -.11101 (105*2π)/3413 weeks
106.01383 -.1108 (106*2π)/3413 weeks
107-.07313 -.35704 (107*2π)/3413 weeks
108-.07033 -.08296 (108*2π)/3413 weeks
109-.3311 -.04304 (109*2π)/3413 weeks
110-.02779 -.24618 (110*2π)/3413 weeks
111-.21775 -.14505 (111*2π)/3413 weeks
112-.31498 -.3014 (112*2π)/3413 weeks
113-.21296 -.28486 (113*2π)/3413 weeks
114-.34847 -.06868 (114*2π)/3413 weeks
115-.35077 -.10016 (115*2π)/3413 weeks
116-.20267 -.01934 (116*2π)/3413 weeks
117-.25601 -.08409 (117*2π)/3413 weeks
118-.15451 -.24409 (118*2π)/3413 weeks
119-.33019 -.15883 (119*2π)/3413 weeks
120-.33138 -.19729 (120*2π)/3413 weeks
121-.40372 .06862 (121*2π)/3413 weeks
122-.22358 -.01782 (122*2π)/3413 weeks
123-.18537 -.0098 (123*2π)/3413 weeks
124-.1804 .01014 (124*2π)/3413 weeks
125-.19077 -.04406 (125*2π)/3413 weeks
126-.13178 -.00808 (126*2π)/3413 weeks
127-.13662 -.05473 (127*2π)/3413 weeks
128-.12485 -.10893 (128*2π)/3413 weeks
129-.15657 -.03103 (129*2π)/3413 weeks
130-.04164 -.00678 (130*2π)/3413 weeks
131-.2208 -.01895 (131*2π)/3413 weeks
132-.01727 .03463 (132*2π)/3413 weeks
133.11643 .08444 (133*2π)/3413 weeks
134-.06954 -.05996 (134*2π)/3413 weeks
135-.01931 -.15835 (135*2π)/3413 weeks
136-.04049 -.09232 (136*2π)/3413 weeks
137-.0412 -.01198 (137*2π)/3412 weeks
138-.12462 -.18726 (138*2π)/3412 weeks
139-.09516 -.27438 (139*2π)/3412 weeks
140-.25444 -.11341 (140*2π)/3412 weeks
141-.27983 -.14395 (141*2π)/3412 weeks
142-.22037 .03474 (142*2π)/3412 weeks
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146-.05221 -.16947 (146*2π)/3412 weeks
147-.2634 -.03273 (147*2π)/3412 weeks
148-.11771 -.19269 (148*2π)/3412 weeks
149-.36606 -.05629 (149*2π)/3412 weeks
150-.28775 .06671 (150*2π)/3412 weeks
151-.27048 .00026 (151*2π)/3412 weeks
152-.34942 .01952 (152*2π)/3412 weeks
153-.22414 -.16933 (153*2π)/3412 weeks
154-.23382 .07622 (154*2π)/3412 weeks
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160-.25064 .19882 (160*2π)/3412 weeks
161-.06769 .07521 (161*2π)/3412 weeks
162-.15623 .25913 (162*2π)/3412 weeks
163-.17523 .07972 (163*2π)/3412 weeks
164-.09792 .10675 (164*2π)/3412 weeks
165-.2301 .02729 (165*2π)/3412 weeks
166-.20012 .07374 (166*2π)/3412 weeks
167-.09062 .10082 (167*2π)/3412 weeks
168-.25754 .09186 (168*2π)/3412 weeks
169.03501 -.07361 (169*2π)/3412 weeks
170-.06275 .17676 (170*2π)/3412 weeks
171-.06275 -.17676 (171*2π)/3412 weeks
172.03501 .07361 (172*2π)/3412 weeks
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177-.09792 -.10675 (177*2π)/3412 weeks
178-.17523 -.07972 (178*2π)/3412 weeks
179-.15623 -.25913 (179*2π)/3412 weeks
180-.06769 -.07521 (180*2π)/3412 weeks
181-.25064 -.19882 (181*2π)/3412 weeks
182-.11007 -.16108 (182*2π)/3412 weeks
183-.22435 -.06566 (183*2π)/3412 weeks
184-.25959 -.11393 (184*2π)/3412 weeks
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189-.34942 -.01952 (189*2π)/3412 weeks
190-.27048 -.00026 (190*2π)/3412 weeks
191-.28775 -.06671 (191*2π)/3412 weeks
192-.36606 .05629 (192*2π)/3412 weeks
193-.11771 .19269 (193*2π)/3412 weeks
194-.2634 .03273 (194*2π)/3412 weeks
195-.05221 .16947 (195*2π)/3412 weeks
196-.2607 .11501 (196*2π)/3412 weeks
197-.0861 .10162 (197*2π)/3412 weeks
198-.11382 .22709 (198*2π)/3412 weeks
199-.22037 -.03474 (199*2π)/3412 weeks
200-.27983 .14395 (200*2π)/3412 weeks
201-.25444 .11341 (201*2π)/3412 weeks
202-.09516 .27438 (202*2π)/3412 weeks
203-.12462 .18726 (203*2π)/3412 weeks
204-.0412 .01198 (204*2π)/3412 weeks
205-.04049 .09232 (205*2π)/3412 weeks
206-.01931 .15835 (206*2π)/3412 weeks
207-.06954 .05996 (207*2π)/3412 weeks
208.11643 -.08444 (208*2π)/3412 weeks
209-.01727 -.03463 (209*2π)/3412 weeks
210-.2208 .01895 (210*2π)/3412 weeks
211-.04164 .00678 (211*2π)/3412 weeks
212-.15657 .03103 (212*2π)/3412 weeks
213-.12485 .10893 (213*2π)/3412 weeks
214-.13662 .05473 (214*2π)/3412 weeks
215-.13178 .00808 (215*2π)/3412 weeks
216-.19077 .04406 (216*2π)/3412 weeks
217-.1804 -.01014 (217*2π)/3412 weeks
218-.18537 .0098 (218*2π)/3412 weeks
219-.22358 .01782 (219*2π)/3412 weeks
220-.40372 -.06862 (220*2π)/3412 weeks
221-.33138 .19729 (221*2π)/3412 weeks
222-.33019 .15883 (222*2π)/3412 weeks
223-.15451 .24409 (223*2π)/3412 weeks
224-.25601 .08409 (224*2π)/3412 weeks
225-.20267 .01934 (225*2π)/3412 weeks
226-.35077 .10016 (226*2π)/3412 weeks
227-.34847 .06868 (227*2π)/3412 weeks
228-.21296 .28486 (228*2π)/3411 weeks
229-.31498 .3014 (229*2π)/3411 weeks
230-.21775 .14505 (230*2π)/3411 weeks
231-.02779 .24618 (231*2π)/3411 weeks
232-.3311 .04304 (232*2π)/3411 weeks
233-.07033 .08296 (233*2π)/3411 weeks
234-.07313 .35704 (234*2π)/3411 weeks
235.01383 .1108 (235*2π)/3411 weeks
236.05777 .11101 (236*2π)/3411 weeks
237-.14695 .14948 (237*2π)/3411 weeks
238-.08703 .19327 (238*2π)/3411 weeks
239-.24382 .30879 (239*2π)/3411 weeks
240-.06376 .07705 (240*2π)/3411 weeks
241-.12319 .05771 (241*2π)/3411 weeks
242-.14243 .16494 (242*2π)/3411 weeks
243-.27831 .07981 (243*2π)/3411 weeks
244-.14386 .06037 (244*2π)/3411 weeks
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250-.12653 .02524 (250*2π)/3411 weeks
251-.06863 .01812 (251*2π)/3411 weeks
252-.26826 -.12772 (252*2π)/3411 weeks
253-.26692 .18167 (253*2π)/3411 weeks
254-.19205 .02062 (254*2π)/3411 weeks
255-.21958 .004 (255*2π)/3411 weeks
256-.22864 .2363 (256*2π)/3411 weeks
257-.03996 .05016 (257*2π)/3411 weeks
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260-.28653 .1553 (260*2π)/3411 weeks
261-.08577 .39678 (261*2π)/3411 weeks
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264-.415 .19159 (264*2π)/3411 weeks
265-.33437 -.05747 (265*2π)/3411 weeks
266.01981 .11573 (266*2π)/3411 weeks
267-.21269 .2735 (267*2π)/3411 weeks
268-.11338 .13473 (268*2π)/3411 weeks
269-.11989 .3369 (269*2π)/3411 weeks
270-.45875 .30015 (270*2π)/3411 weeks
271-.28411 .31585 (271*2π)/3411 weeks
272-.23497 .455 (272*2π)/3411 weeks
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274-.17631 .44689 (274*2π)/3411 weeks
275-.31979 .35417 (275*2π)/3411 weeks
276.07508 .45724 (276*2π)/3411 weeks
277.05308 .50668 (277*2π)/3411 weeks
278-.03866 .096 (278*2π)/3411 weeks
279.01434 .26808 (279*2π)/3411 weeks
280-.37219 .24142 (280*2π)/3411 weeks
281-.26199 .33682 (281*2π)/3411 weeks
282.23623 .51269 (282*2π)/3411 weeks
283-.1115 .45158 (283*2π)/3411 weeks
284-.14415 .16587 (284*2π)/3411 weeks
285.11883 .24566 (285*2π)/3411 weeks
286-.47333 .51258 (286*2π)/3411 weeks
287.17714 .09245 (287*2π)/3411 weeks
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292.21903 .47006 (292*2π)/3411 weeks
293-.18121 .54473 (293*2π)/3411 weeks
294-.04908 .05778 (294*2π)/3411 weeks
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311-.11984 .19652 (311*2π)/3411 weeks
312.06744 .67812 (312*2π)/3411 weeks
313-.04182 1.01312 (313*2π)/3411 weeks
3141.00753 .31151 (314*2π)/3411 weeks
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316.5081 .09936 (316*2π)/3411 weeks
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331-.41484 2.06921 (331*2π)/3411 weeks
332.0249 2.5144 (332*2π)/3411 weeks
333-2.0058 2.54393 (333*2π)/3411 weeks
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335-1.70029 3.84031 (335*2π)/3411 weeks
3361.08213 5.15416 (336*2π)/3411 weeks
337.06782 1.1625 (337*2π)/3411 weeks
3384.4239 1.36729 (338*2π)/3411 weeks
3392.84788 15.1037 (339*2π)/3411 weeks | 6,105 | 13,887 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2018-17 | latest | en | 0.763245 |
https://www.yumpu.com/en/document/view/28047318/slides-in-pdf-stanford-university | 1,566,583,454,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027318952.90/warc/CC-MAIN-20190823172507-20190823194507-00327.warc.gz | 1,034,129,171 | 18,877 | slides in pdf - Stanford University
slides in pdf - Stanford University
Techniques for
Fast Packet Buffers
Sundar Iyer, Nick McKeown
(sundaes,nickm)@stanford.edu
Departments of Electrical Engineering &
Computer Science, Stanford University
http://klamath.stanford.edu
Packet Buffer Architecture
Goal
Determine and analyze techniques
for building high speed (>40Gb/s)
electronic packet buffers.
Example
OC768c = 40Gb/s; RTT*BW = 10Gb; 64 byte packets
Write Rate, R
1 packet
every 12.8 ns
Buffer
Memory
1 packet
every 12.8 ns
Use SRAM?
+ fast enough random access time, but
- too expensive, and
- too low density to store 10Gb of data.
Use DRAM?
+ high density means we can store data, but
- too slow (typically 50ns random access time)
Stanford University 2
Memory Hierarchy
Large DRAM memory holds the middle of FIFOs
54 53 52 51 50
10 9 8 7 6 5
95 94 93 92 91 90 89 88 87 86
15 14 13 12 11 10 9 8 7 6
1
2
86
85
84 83
82
11
10
9 8
7
Q
Writing
b cells
b cells
Arriving
Packets
60 59 58 57 56 55
97 96
1
2
5
4
4
3
3
2
2
1
1
1
2
Departing
Packets
R
R
91 90 89 88 87
Q
Small ingress SRAM
cache of FIFO tails
6
5
4
3
2
1
Q
Small ingress SRAM
cache of FIFO heads
Arbiter or
Scheduler
Requests
Stanford University 3
Questions
This talk…
How large does the SRAM cache need to be:
1. To guarantee that a packet is immediately
available in SRAM when requested, or
2. To guarantee that a packet is available within a
maximum bounded time?
What Memory Management Algorithm (MMA)
should we use?
Stanford University 4
Earliest Critical Queue First
(ECQF-MMA)
Cells
1. In SRAM: A Dynamic
Buffer of size Q(b-1)
2. Lookahead: Arbiter requests
for future Q(b-1) + 1 slots are
known
3. Compute: Find out the queue
which will runs into “trouble”
earliest
4. Replenish: “b” cells for the
“troubled” queue
Requests in lookahead buffer
Q(b-1) + 1
Q
b - 1
green!
Cells
Q
Stanford University 5
b - 1
Cells
Example of ECQF-MMA: Q=4, b=4
Cells
Cells
Requests in lookahead buffer
Requests in lookahead buffer
Requests in lookahead buffer
t = 0; Green Critical
t = 1
t = 2
Cells
Cells
Cells
Requests in lookahead buffer
Requests in lookahead buffer
Requests in lookahead buffer
t = 3
t = 4; Blue Critical
t = 5
Cells
Cells
Cells
Requests in lookahead buffer
Requests in lookahead buffer
Requests in lookahead buffer
t = 6
t = 7
t = 8; Red Critical
Stanford University 6
Results
1. Patient Arbiter: ECQF-MMA (earliest
critical queue first), minimizes the size
of SRAM buffer to Q(b-1) cells; and
guarantees that requested cells are
dispatched within Q(b-1)+1 cell slots.
2. Impatient Arbiter: MDQF-MMA
(maximum deficit queue first), with a
SRAM buffer of size Qb[2 +ln Q]
guarantees zero latency.
Stanford University 7
Implementation Numbers
(64byte cells, b = 8, DRAM T = 50ns)
1. VOQ Switch - 32 ports
– Brute Force : Egress. SRAM = 10 Gb, no DRAM
– Patient Arbiter : Egress. SRAM = 115kb, Lat. = 2.9 us, DRAM = 10Gb
– Impatient Arbiter : Egress. SRAM = 787 kb, DRAM = 10Gb
– Patient Arbiter(MA) : No SRAM, Lat. = 3.2us, DRAM = 10Gb
2. VOQ Switch - 32 ports, 16 Diffserv classes
– Brute Force : Egress. SRAM = 10Gb, no DRAM
– Patient Arbiter : Egress. SRAM = 1.85Mb, Lat. = 45.9us, DRAM 10Gb
– Impatient Arbiter : Egress. SRAM = 18.9Mb, DRAM = 10Gb
– Patient Arbiter(MA) : No SRAM, Lat. = 51.2us, DRAM = 10Gb
Stanford University 8
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http://zims-en.kiwix.campusafrica.gos.orange.com/wikipedia_en_all_nopic/A/Operation_(mathematics) | 1,611,477,416,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703547475.44/warc/CC-MAIN-20210124075754-20210124105754-00509.warc.gz | 186,436,096 | 6,203 | # Operation (mathematics)
In mathematics, an operation is a function which takes zero or more input values (called operands) to a well-defined output value.[1] The number of operands is the arity of the operation. The most commonly studied operations are binary operations (i.e., operations of arity 2), such as addition and multiplication, and unary operations (i.e., operations of arity 1), such as additive inverse and multiplicative inverse. An operation of arity zero, or nullary operation, is a constant.[2][3] The mixed product is an example of an operation of arity 3, also called ternary operation. Generally, the arity is taken to be finite. However, infinitary operations are sometimes considered,[2] in which context the "usual" operations of finite arity are called finitary operations.
## Types of operation
There are two common types of operations: unary and binary.[1] Unary operations involve only one value, such as negation and trigonometric functions. Binary operations, on the other hand, take two values, and include addition, subtraction, multiplication, division, and exponentiation.
Operations can involve mathematical objects other than numbers. The logical values true and false can be combined using logic operations, such as and, or, and not. Vectors can be added and subtracted. Rotations can be combined using the function composition operation, performing the first rotation and then the second. Operations on sets include the binary operations union and intersection and the unary operation of complementation. Operations on functions include composition and convolution.
Operations may not be defined for every possible value. For example, in the real numbers one cannot divide by zero or take square roots of negative numbers. Much like a function, the values for which an operation is defined form a set called its domain. The set which contains the values produced is called the codomain, but the set of actual values attained by the operation is its range. For example, in the real numbers, the squaring operation only produces non-negative numbers; the codomain is the set of real numbers, but the range is the non-negative numbers.
Operations can involve dissimilar objects: a vector can be multiplied by a scalar to form another vector (an operation known as scalar multiplication), and the inner product operation on two vectors produces a quantity that is scalar. An operation may or may not have certain properties, for example it may be associative, commutative, anticommutative, idempotent, and so on.[1]
The values combined are called operands, arguments, or inputs, and the value produced is called the value, result, or output. Operations can have fewer or more than two inputs (including the case of zero input and infinitely many inputs[2]).
An operation is like an operator, but the point of view is different. For instance, one often speaks of "the operation of addition" or "the addition operation", when focusing on the operands and result, but one switches to "addition operator" (rarely "operator of addition"), when focusing on the process, or from the more abstract viewpoint, the function + : S × SS.
## Definition
An operation ω is a function of the form ω : VY, where VX1 × ... × Xk. The sets Xk are called the domains of the operation, the set Y is called the codomain of the operation, and the fixed non-negative integer k (the number of arguments) is called the type or arity of the operation. Thus a unary operation has arity one, and a binary operation has arity two.[1] An operation of arity zero, called a nullary operation, is simply an element of the codomain Y. An operation of arity k is called a k-ary operation.[2] Thus a k-ary operation is a (k+1)-ary relation that is functional on its first k domains.
The above describes what is usually called a finitary operation, referring to the finite number of arguments (the value k). There are obvious extensions where the arity is taken to be an infinite ordinal or cardinal,[2] or even an arbitrary set indexing the arguments.
Often, the use of the term operation implies that the domain of the function is a power of the codomain (i.e. the Cartesian product of one or more copies of the codomain),[4] although this is by no means universal, as in the case of scalar multiplication, where a vector is multiplied by a scalar. | 913 | 4,358 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2021-04 | latest | en | 0.940084 |
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10.30.2006
OMG...
...something horrible has happened to me. I read these comments from the inside looking out.
10.27.2006
Bounded rationality
I'm discovering I'm a bad mathematician. That's a lie. I knew I was a bad mathematician. Nonetheless, I love theory. This is a problem.
Choices that decision makers make are boundedly rational if a) he would make the same choices when he has fewer options to choose from than when he has more options and b) he doesn't choose something between two options that he wouldn't have chosen when those two options are available plus more. The first one says that if the decision maker chooses chocolate when he gets to choose between broccoli, chocolate and yogurt then he'll choose chocolate when he only has chocolate and yogurt to choose from. The second one says that he won't choose broccoli over chocolate and he won't choose broccoli over yogurt if he doesn't choose broccoli over all three.
He's bounded rational because his choices may not be transitive. He may like chocolate over yogurt and yogurt over broccoli, but he's never compared chocolate to broccoli so its not given that he prefers the former to the later.
10.26.2006
Rhetoric of tyranny
As per usual, I put my students in small groups of 4 or 5 today in class. I find this method helps the more shy students participate, while allowing the know-it-alls to be big fish in smaller ponds. In any case, I have the students introduce themselves and each week I have them tell something about themselves. In the past couple of weeks the something else has been trivial like their favorite color or their favorite movie. This week I upped the ante and asked the students to tell their group mates what problem they think is the biggest the world faces.
I was sitting with one group where a student told the group the world biggest problem was that there was enough selection in foods. He wanted more variety. When it was my turn to share I said something like 'abject poverty like in Africa'. The 'not enough food' guy told me that my answer was trite. I think it was an 'I don't think that word means what you think it means' moment.
Anyway, another group had a student that said the President was the biggest problem the world faced. To prove her point, she said she had an article she'd send me after class. Here it is:
http://www.progressive.org/mag_mc100406?cheney
Absolute power corrupts absolutely. This can't be happening in the US.
But it is.
We are our own biggest enemy. We cannot allow this to continue.
Personal freedoms are more important that "security". Especially when
the security is false. To really understand what's going on you must go
is all about. If our soldiers are dying in Iraq to try to build a
democracy out of a land run by a tyrant that: 1) would arrest people
just because he didn't like what they stood for, 2) jail them without
reason or explanation, 3) not allow fair trail by jury, 4) torture them
until they died. That's why our soldiers are dying. And what will they
come home to? A country where people can be declared an "enemy
combatant" just because the President says so. People can be jailed
without reason or explanation. Not allowed a trial where they can be
presented with evidence against them. And be tortured until near death.
In fact, I could be arrested for this email because it might be viewed as
fomenting terrorism against the US. I could be tortured and locked away.
And as of Tuesday the 17th, it would all be legal. I'm sick that our
leaders are so corrupted by power that they even tried to pull this off.
I'm sick that the Congress rolled over an played dead and sacrificed my
personal freedoms for a false threat. I'm sick that the news wasn't
reported widely (perhaps out of fear of reprisals). Just google habeas
corpus and see how few stories pop up. I'm appalled. Al Queda's power is
an insignificant speck compared to the disastrous leadership of this
The comments are from the person who sent me this.
Hi xxx-
Thanks for sending this along. Let me respond first by saying that I'm very sympathetic to the issues raised by your friend. Personal liberty is a standard I often bear. I'm very suspect of government because I realize it is made up of people. People for better (e.g. market outcomes for the most part) or worse (e.g. greed) are self interested. Government provides the few people that run it with a significant amount of power and the means to subjugate others to their will.
I'll also say I'm a card carrying member of the ACLU and I donate to Amnesty International.
With that qualification, I'll respond to your friends points, point by point.
"1) would arrest people just because he didn't like what they stood for, 2) jail them without reason or explanation"
I assume she is referring to the detainees in places like Guantanamo and Abu Garab. The government tells us these people were detained not because of what they stand for but for their actions. These people were taken not because we didn't like them and there ideas, but because they were actively involved in aggression towards the U.S. Granted you have to believe the government claims to believe the people detained are justly detained.
"3) not allow fair trail by jury"
However, given recent Supreme Court decisions we won't have to take the governments word for it. Those rulings basically say that all the detainees must be given due process. Now, you can be uneased about what that process looks like (their cases won't be reviewed as per usual American legal processes, for example), but its very hard to say they won't be given fair trails.
"4) torture them until they died."
This is a red herring. We have no idea the extent of problems like Abu Graib. Is it an isolated incident? Was torture wide spread and sanctioned? Nobody knows. All we can do is denounce it and hold our leaders accountable (by voting against incumbents for example) for any exposed malfeasance. Torture is unacceptable. End of story. That said, Congress recently passed a law to specifically outlaw torture and to give it better definition. The law, by the way, was authored by John McCain a man who endured torture while a captive of the North Vietnamese.
"A country where people can be declared an "enemy combatant" just because the President says so"
The President has this right because it was given to him by an act of Congress. Congress, in essence, declared a state of war after 9/11 (this law is called AUMF). An "enemy combatant" is an old term to refer to people that are fighting against your country in a War. The President issued an order that explicitly defined them as "an individual who was part of or supporting the Taliban or al Qaida forces, or associated forces that are engaged in hostilities against the United States or its coalition partners. This includes any person who committed a belligerent act or has directly supported hostilities in aid of enemy armed forces." (See this wikipedia article).
The point of citing all of these laws is just my attempt to show that we're still being ruled by laws. The Congress can change the law (i.e. say we're no longer at war) and because the definition is pretty clear, the President can be shown to be in violation of the law. For example, its clear that your friend, by sending this email, would NOT fall under the definition of enemy combatant. Furthermore, all these laws can, and some of them have been, reviewed by the Supreme Court. In more than one instance, the President's interpretation of AUMF (i.e. the status of Guantanamo detainees) has been overruled by the Supreme Courts constitutional interpretation. There's a case in the lower courts right now that is contesting the President's wire tapping program.
Finally, your friend calls al Qaida a "... false threat."
I hope its obvious that this is just an incorrect statement. Al Qaida has shown itself to be a dire threat. The 3000 dead on 9/11 can attest to that. Its not unthinkable to think they're looking to get a nuclear weapon, for example. You don't have to stretch your imagination far to see that the impact of their successful use of one such weapon would be catastrophic in terms of lives and to our system of government. Al Qauida is a threat whether or not Bush is President.
In any case, while your friend seems to be in poor spirits about our system of government in this post-9/11 era, all that has transpired has actually lifted my confidence in our system. If you can believe it! Yes, the President did try to overreach his authority, but the President doesn't act in isolation. Our Courts, Congress and non-governmental institutions like the Press, Amnesty International and the ACLU have been there to check his power. In some sense, we've had to endure the pain of a Bush presidency so that we can get this right. We must create a system that can endure terrorist threats while balancing individual freedom.
I think you'd agree its important for us to get this right.
Thanks again for sending this too me.
Have hope,
Will
UPDATE: The guest on the Keith Olbermann video linked to above implies the McCain torture law, alluded to above, would allow the President to detain Americans. This seems false according to this wiki article. The stated purpose of the law is to apply to alien combatants. There is a troubling section in the definition of the combatant, though: "a person who, before, on, or after the date of the enactment of the Military Commissions Act of 2006, has been determined to be an unlawful enemy combatant by a Combatant Status Review Tribunal or another competent tribunal established under the authority of the President or the Secretary of Defense." Do these tribunals have jurisdiction over American citizens?
Gun control
127 kids died of a firearm accident in 2003. 910 died in drowning accidents.
Preventing accidental gun deaths is NOT a reason for gun control. (h/t Darwin Catholic)
10.21.2006
Data mining
I can't remember who I was talking to recently, but we were discussing data mining. Data mining is bad. Real bad. Everyone knows this, but why is it so bad?
Researchers in any discipline are suppose to check their biases at the lab door. They are to be objective observers of reality. The facts are the facts and the researcher's job is to uncover them.
In the physical sciences, I imagine this part of the job is a little bit easier than in social sciences. Its easy to separate your personal opinions from empirical observation when the thing being observed is inanimate. There are vested interests, of course. The scientist may feel very strongly about which way the data should come out. He would find it strange to have a result that refuted a law of thermodynamics or violated the speed of light. Not that physicists and chemists can't be passionate about their work, it just seems more likely for them to be able to separate their feelings from their observations. The picture gets a little more clouded if the scientist realizes if his experiments come out the 'wrong' way he may lose his funding. Someone who spent his whole career experimenting on ether, may be less than excited with experiments that show the atomic theory to be correct.
In softer sciences the object of study isn't inanimate. Worse, the lab rats for economics, sociology and political science are living breathing people with feelings and such. People, even the most autistic-like math-nerd theoretical economist types, tend to have strong feelings about how people act and should act as individuals and how they act and should act collectively.
Worse still, the motivation for most social scientists to enter their field is most likely tied up with their biases. The public economics researcher went into that field because he felt strongly about the role of the government in improving peoples lives. Certainly, the research he decides to do, the decision to study the impact of education on life outcomes or the effects of immigration on the working poor, stemmed from his particular opinions and experiences, his biases.
Checking your biases at the door becomes a lot harder when your so emotionally invested in the outcome. This is why social scientists have to be especially cautious about researcher bias.
Unlike in the physical sciences, most of the data in social science isn't acquired experimentally. In experiments, researchers control in the environment to very high degree. When a variable is tweaked, the experiment is set up such that it is known exactly the impact of the tweak. Thus, experimental data is context free and so repeatable. Additionally, experiments are generally engineered to reduce the noise to signal ratio. Very precise instruments measure to a very precise degree.
Most data in the social sciences is historical, entangled and dirty. They are path dependent, mired in context and contain a lot of noise. On the plus side, there's a lot more data essentially because everything is data. The government collects data. Business' collect data. Everyone collects data. And even where there was no data before (wages in 15th century Cairo), the clever researcher can discover it (http://gpih.ucdavis.edu/Datafilelist.htm#Africa ). The social scientist's job is to find the data, clean them, and disentangle them to find patterns and relationships. How does he do this?
First, before even looking for data or looking at it after found, he should have a theory of the data. What relationships should be found? What patterns are expected? How might one find such patterns? Next, he should delineate a research strategy, identify the data he will need and outline the methods that will be used to test the theory. Then, and only then, he should go to the data.
Data mining reverses this method. One starts with the data set and precedes to find relationships and patterns. This sounds innocuous at first. If the patterns are there, then why should this matter? Well, because of the sheer volume of data, the researcher will no doubt 'find' the relationships that correspond to his biases. Significant results (those that we have a 95% confidence in) are more likely to be insignificant if one is looking through piles of data to find the pattern he wants to find. If 1 out of 20 times, perceived patterns are just the result of randomness, then the more one 'mines' for patterns the more likely they'll be fooled. If you find 10 patterns that have 5% chance of being caused by chance, you're almost 40% likely to find at least one of them is random. If you dig for more and more patterns, you're more and more likely to find what you're looking for whether or not its is caused by randomness.
So, don't data mine is the mantra. Just don't do it.
However, on the off chance that a significant pattern is hiding in the data, it seems odd to dismiss data mining out of hand. The problem with data mining isn't data mining itself, its the bias the data miner brings to the job.
Its a wonder why data mining rates as such a sin in economics as to be on par with a certain Victorian era sin that caused hair to grow on the palm of the hands. Both sins are much committed but never discussed except to say they're bad. Is this sin always bad? Are there healthy ways to commit this sin?
This paper asks us to consider the difference between classical statistics (the kind of statistics that is more concerned with summarizing data) and econometrics (the kind of statistics that is more concerned with cleaning and disentangling data). Admitting those differences means that we must admit the researcher brings bias to his or her work. As such, we should have standard tests/corrections for researcher bias.
Here are some points that stuck out at me:
Point # 1: Researchers will always respond to incentives and will be more skeptical than standard statistical techniques suggest.
The most well-studied example of researcher initiative bias is data-mining (Leamer, 1978, Lovell, 1983). In this case, consider a researcher presenting a univariate regression explaining an outcome, which might be workers' wages or national income growth. That researcher has a data set with k additional variables other than the dependent variable. The researcher is selecting an independent variable to maximize the correlation coefficient (r) or r-squared or the t-statistic of the independent variable all of which are identical objective functions... If the true correlation between every one of the k independent variables and the dependent variable is zero, and if all of the independent variables are independent, then the probability that the researcher will able to produce a variable which is significant at the ninety-five percent level is 1－ .95k . If the researcher has ten potential independent variables, then the probability that he can come up with a false positive is 40 percent... Lovell (1983) provides a rule of thumb "when a search has been conducted for the best k
out of c candidate explanatory variables, a regression coefficient that appears to be significant at the level alpha_hat should be regarded as significant at only the level alpha = 1－ (1－alpha_hat)^(c/k)."
Point # 2: The optimal amount of data mining is not zero.
[E]conomic models have an uncountable number of potential parameters, non-linearities, temporal relationship and probability distributions. It would utterly impossible to test everything. Selective testing, and selective presentation of testing, is a natural response to vast number of possible tests.
Point # 5: Increasing methodological complexity will generally increase researcher initiative bias.
[New methodologies] clearly increase the degrees of freedom available to the researcher... A second reason for increased bias is introduced by the existence of particularly complex empirical methodologies. Methodological complexity increases the costs of competitors refuting or confirming results.
Point # 8: The search for causal inference may have increased the scope for researcher initiative bias.
The most worrisome aspect of causal inference is the prevalence of instrumental variables drawn from observational data selected, or even collected by the researcher. In that case, the scope for researcher effort is quite large, although some of this freedom is restricted when the researcher focuses on a clean public policy change that is directly targeted at a particular outcome. The problem will be most severe when the variables are general characteristics that may have only a weak correlation with the endogenous regressor.
10.06.2006
Somebody's trying to tell me something
While my department listserv is a debating the merits of the graduate student union (loss of liberty vs. well the other side hasn't really articulated their POV yet except to say that unions must be good), the following two articles, in succession, show up in my news reader
New Scientist - "Sense of justice discovered in the brain"
A brain region that curbs our natural self interest has been identified. The studies could explain how we control fairness in our society, researchers say... [U]sing a tool called the “ultimatum game”, researchers have identified the part of the brain responsible for punishing unfairness...
"Self interest is one important motive in every human," says Fehr, "but there are also fairness concerns in most people."
"In other words, this is the part of the brain dealing with morality," says Herb Gintis, an economist at the University of Massachusetts in Amherst, US. "[It] is involved in comparing the costs and benefits of the material in terms of its fairness. It represses the basic instincts."
and Tim Hartford - "The Undercover Economist: The great giveaway"
Selfishness is one of those issues where economists seem to see the world differently. It is not that economists are incapable of imagining - or even modelling - altruism. They can, but they usually don’t. And there is a good reason for that: people aren’t selfless... In fact, the closer you look at charitable giving, the less charitable it appears to be. A recent experiment by John List, an economist at the University of Chicago, and a team of colleagues, showed that donations are less than charitable after all. Using controlled trials to compare different methods of door-to-door fundraising, Professor List’s team discovered that it was much more effective to raise funds by selling lottery tickets than by asking for money. This hardly suggests a world populated by altruists seeking to do the maximum good with their charitable cash... Robert Frank, from Cornell, wryly observes that those organising fundraising drives for the vast US charity United Way tend to be disproportionately estate agents, insurance brokers, car dealers and other people with something to sell.
All I know is that *someone* is trying to tell me that justice matters (or maybe it doesn't).
I'd have no idea who to go for a good argument for justice. I find myself moved by arguments for liberty while arguments for justice often come off as whining. For all I know, this is only because those making the arguments for liberty are just better at making arguments than those that argue for justice.
Who is the Hayek or Friedman for justice?
It's all been said before
Some funny economics related quotes. My favorites:
• "Stiglitz's outlook is that markets are imperfect, but he is not. Where Marx offered dictatorship of the proletariat, Stiglitz would give us dictatorship of the Nobel Laureate. Between the two, we might be safer with Marx." - Arnold Kling
• "See, when the Government spends money, it creates jobs; whereas when the money is left in the hands of Taxpayers, God only knows what they do with it. Bake it into pies, probably. Anything to avoid creating jobs." - Dave Barry
• "There are two methods, or means, and only two, whereby man's needs and desires can be satisfied. One is the production and exchange of wealth; this is the economic means. The other is the uncompensated appropriation of wealth produced by others; this is the political means." - Albert Jay Nock
• "The best minds are not in government. If any were, business would steal them away." - Ronald Reagan
• "It's a very sobering feeling to be up in space and realize that one's safety factor was determined by the lowest bidder on a government contract."- Alan Shepherd
10.04.2006
Paper: Bar-Hillel, "Cycles of Intransitive Choice"
Transitivity is a property of preferences where if A is preferred to B (denoted A > B) and B preferred to C (B > C) then A must be preferred to C (A > C). The paper discussions various examples where this property doesn't hold.
The most famous "intransitivity" is in group decision making by majority rule. If person one has preferences A > B > C, person two has preferences B > C > A and person 3 has preferences C > A > B. A majority of people prefer A to B and B to C, but a majority also prefers C to A. Thus, for the group A > B > C > A > B and so on.
Group decision making is analogous to multi-dimensional decision making. If you're buying a new computer, you may think CPU speed, amount of memory and hard drive space are the important attributes. Let's say model A is fastest, B second fastest and C is the slowest. Likewise, B has the most memory, C the second most and A the least and C has the biggest hard drive, A the second biggest and B the smallest. You can see that this situation is very similar to the above and the decision rule "pick the model that has most preferred categories" gets you no where.
The paper also talks about "inherently comparative choice". The idea is that in the case where the satisfaction (or dissatisfaction) derived from a choose is dependent on the thing chosen and the things not chosen (e.g. regret), transitivity can be violated. A good example compares the difference in regard to stock market returns in the case where you bought a share of stock yesterday that doubled today versus the case where you've owned 2 shares of the same stock for a long while and it doubled. You may find yourself more pleased with the first situation because you where able to "time the market" even though the outcomes in the second case is better in a money-in-your-pocket sense. Comparing these situations to the situation where you had 4 shares of the stock yesterday but then sold one share, you may find yourself regretting that choice today when the stock price doubled. Even though you have higher return in this case than in the second case, the feeling of regret may make you to prefer the second case to the third. However, despite the regret in the third case, the money gain is so high that you still prefer the third to the first case. Thus, we have a violation of transitivity... timing the market with a small return is preferred to buy and hold with a medium return is preferred to regretting a sale with a large return is preferred timing the market with a small return, etc.
The paper concludes by remarking that it is implicit in revealed preference that choice is driven by preferences, but that perhaps preferences are "constructed" by choice and thus "choice procedures are geared towards justification and conflict resolution rather than towards optimization."
10.02.2006
Chapter: Kreps, "Notes on the theory of choice," chapter 2
Quick review of binary relations generally and preference relations specifically. The minimum requirement of binary relations to make them preference relations is asymmetry and negative transitivity. These two things imply the 'normal' qualities of preference relations (irreflexivity, transitivity and acyclicity). Phew, I was worried they might not.
It would be tedious to display the whole set of options to agents and have them state their preferences, e.g. "I prefer x to y, y to z and z to x... oops, I mean x to z." Another tact is choice theory or revealed preference theory. A choice function is a function that takes a subset of the total number of items to choose from and picks a subset, i.e. the agents choice(s). The interesting question is when are there correspondences between preference relations and choice functions. The weak axiom is explored via Houthakker's axiom and Sen's alpha and beta properties (pg 13-15).
Paper: Samuelson, "Exact Consumption-Loan Model of Interest"
The model is a one of multiple, identical generations (OLG), with each generation living 3 periods, working in the first 2 and being retired in the last. Total lifetime savings has to equal zero (i.e. grandparents don't go to the grave with their money and they don't leave estates).
In the stationary case, the case where interest rates don't change over time and population doesn't grow, the interest rate must zero. This is the case because everyone is assumed to be the same. There are no "savers", those that don't mind consuming in the future rather than today, who would get positive interest rates from "spenders", those that want to consume today. "Saver" and "spender" are relative, you're only a "saver" in relation to all the "spenders", terms and because everyone is the same there can't be relative differences.
If you let the population grow (exponentially) but keep interest rates constant over time, you find the interest rate must be equal to the population growth rate. Of course, this means that if the population is declining, you'd see a negative interest rate... Hmmmm.
Further, with equal interest rates over time, interest rate equal to the population growth rate is socially optimal.
But will each generation trade with the others. Its easy to see in the 2 period case, generations will not trade. The old won't buy bonds they'll be dead before they mature. Thus, the young consume all of their product and the old, well, consume their product (poor them). Recall, that we had interest rates on bonds equal to the population growth. So, in this case with no trading, why would there be any interest rate (not to mention a sensible one) on bonds?
Samuelson goes on to show that the optimal amount of savings won't be achieved in most settings and comments that this is where there is a role for government. He ends the essay by declaring that money is an asset that can be traded across generations, "as long as the new current generations of workers do not repudiate the old money, this gives workers of one epoch a claim on workers of a later epoch." Money, beyond being a medium of exchange, is the "social compact" that brings the economy into optimality.
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# It is possible that Native Americans originally have
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It is possible that Native Americans originally have [#permalink]
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22 Aug 2006, 13:52
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It is possible that Native Americans originally have migrated to the Western Hemisphere over a bridge of land that once existed between Siberia and Alaska.
(A) have migrated to the Western Hemisphere over a bridge of land that once existed
(B) were migrating to the Western Hemisphere over a bridge of land that existed once
(C) migrated over a bridge of land to the Western Hemisphere that once existed
(D) migrated to the Western Hemisphere over a bridge of land that once existed
(E) were migrating to the Western Hemisphere over a bridge of land existing once
[Reveal] Spoiler: OA
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Re: It is possible that Native Americans originally have [#permalink]
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22 Aug 2006, 17:01
D it is ..
migrated to some place... over a bridge..
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Re: It is possible that Native Americans originally have [#permalink]
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22 Aug 2006, 17:13
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D for me 100%
(A) this indicates that N.A. migrated before the bridge existed due to past perfect.
(B) were migrating is incorrect... it stretches the time frame to continious, thus changes the meaning
(C) this says that Western Hemisphere once existed where did it go?
(D) migrated to the Western Hemisphere over a bridge of land that once existed
(E) same as B
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Re: It is possible that Native Americans originally have [#permalink]
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22 Aug 2006, 21:07
(D) is 9 seconds. Reads like an old school history text. This is the Beiring Straight migration theory. That`s why so many native Americans look asian (eastern mongolian/siberian.)
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Re: It is possible that Native Americans originally have [#permalink]
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24 Aug 2006, 13:36
Yeah! OA D.
It wasn't a difficult one but I got it wrong by selecting B...
(B) were migrating to the Western Hemisphere over a bridge of land that existed once
I thought that the tense wasnt false, since they cd have migrated over a long time-> past continuous. But what decided me to take this rather than D, was that the "once" position changed. I believed that the adverb was better placed after the verb than before...
(D) migrated to the Western Hemisphere over a bridge of land that once existed
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Re: It is possible that Native Americans originally have [#permalink]
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08 Dec 2011, 05:06
I think, had migrated would be perfect.....
Ok with D, though!
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Re: It is possible that Native Americans originally have [#permalink]
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12 Dec 2011, 05:59
present perfect A is wrong
D wins
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Re: It is possible that Native Americans originally have [#permalink]
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19 Jan 2014, 23:28
karlfurt wrote:
It is possible that Native Americans originally have migrated to the Western Hemisphere over a bridge of land that once existed between Siberia and Alaska.
(A) have migrated to the Western Hemisphere over a bridge of land that once existed
(B) were migrating to the Western Hemisphere over a bridge of land that existed once
(C) migrated over a bridge of land to the Western Hemisphere that once existed
(D) migrated to the Western Hemisphere over a bridge of land that once existed
(E) were migrating to the Western Hemisphere over a bridge of land existing once
Present Perfect Tense we used where some work is started in past and just finished some time back. And Past Indefinite Tense we use where some work is completed in some particular time in the past.
In this context Migration happened. So D is the correct choice.
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Re: It is possible that Native Americans originally have [#permalink]
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17 Jun 2016, 20:17
Split 1 : Tense issues eliminate A, D and E
Split 2 : C says bridge existed to Western Hemisphere. Putting it in sentence shows it is clearly incorrect.
D is the winner.
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Re: It is possible that Native Americans originally have [#permalink]
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28 Jun 2016, 18:04
Lets draw the timeline.once existed is a past event. Anything that happen with respect to a past event comes in past tense.
past continuous B and E are just changing the meaning any making no sense at all.A is present tense.
between C and D, you migrate to some place , idiom problem.
(A) have migrated to the Western Hemisphere over a bridge of land that once existed - present tense.
(B) were migrating to the Western Hemisphere over a bridge of land that existed once - past continuous
(C) migrated over a bridge of land to the Western Hemisphere that once existed - you migrate to some place , idiom problem.
(D) migrated to the Western Hemisphere over a bridge of land that once existed - correct
(E) were migrating to the Western Hemisphere over a bridge of land existing once - past continuous
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Re: It is possible that Native Americans originally have [#permalink] 28 Jun 2016, 18:04
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# It is possible that Native Americans originally have
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne 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®. | 1,889 | 7,230 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43 | latest | en | 0.90485 |
http://open.math.uwaterloo.ca/modules/practiceTest.QuestionSheet?currPage=0&testId=76&actionID=next | 1,506,301,370,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690268.19/warc/CC-MAIN-20170925002843-20170925022843-00411.warc.gz | 263,613,063 | 6,723 | # Lesson: Orthogonal Diagonalization
1 point
## Transcript — Introduction
In the last lecture, we saw that every square matrix with all real eigenvalues is orthogonally triangularizable. In this lecture, we will determine which n-by-n matrices have the very special property that they are orthogonally similar to a diagonal matrix. That is, they have an orthonormal basis for Rn of eigenvectors.
We begin with a definition. Definition: A matrix A is said to be orthogonally diagonalizable if it is orthogonally similar to a diagonal matrix.
As mentioned, our goal is to figure out which matrices are orthogonally diagonalizable. But how do we even start looking for such matrices? A good trick in mathematics to try to find objects which satisfy a property is to work in reverse—that is, to assume that you have the property, and see what conditions this puts on the objects. So, we assume that we have an orthogonally diagonalizable matrix A. Then, by definition, this means that there exists an orthogonal matrix P such that (P-transpose)AP = D is diagonal. We are trying to find a condition on the matrix A, so let’s solve this equation for A to get A = PD(P-transpose). Since A and D are similar, they are very, um, similar. So, we need to think about what special properties D has since it is diagonal, and check to see if A has the same property.
Since D is diagonal, one nice property that it has is that D-transpose = D. Observe that this implies that A-transpose = (PD(P-transpose))-all-transposed. Using properties of transposes, this is ((P-transpose)-transpose)(D-transpose)(P-transpose), which equals PD(P-transpose), which is just A. So the fact that D = D-transpose, and A and D are similar, gives us that A also satisfies A = A-transpose.
We have proven the following theorem. Theorem 10.2.1: If A is orthogonally diagonalizable, then A-transpose = A.
This theorem says that a necessary condition for A to be orthogonally diagonalizable is that A-transpose = A. We want to know if this is, in fact, sufficient. That is, we want to determine if, in fact, every matrix A such that A-transpose = A is orthogonally diagonalizable. This result is, of course, true, and in fact, extremely important.
## The Principal Axis Theorem
The Principal Axis Theorem: If A is a matrix such that A-transpose = A, then A is orthogonally diagonalizable.
How are we going to prove this? We don’t seem to have enough information to even to try to diagonalize the matrix, let alone to find an orthonormal basis of eigenvectors for the matrix. The key here is to think of the Triangularization Theorem. However, to use the Triangularization Theorem, we first need to check the hypothesis that the matrix has all real eigenvalues.
Lemma 10.2.2: If A is a matrix such that A-transpose = A, then A has all real eigenvalues. The proof of this lemma requires properties of complex eigenvalues, and so we will delay the proof until Module 11.
We can now very easily prove the Principal Axis Theorem. Proof: By Lemma 10.2.2, A has all real eigenvalues, and so by the Triangularization Theorem, we have that there exists an orthogonal matrix P such that (P-transpose)AP = T is upper triangular. As we did in the proof of Theorem 10.2.1, we show that T must have the same properties as A since they are similar. We get that T-transpose = ((P-transpose)AP)-all-transposed, which, by properties of transposes, is (P-transpose)(A-transpose)P, which equals (P-transpose)AP since A-transpose = A. Hence, we have that T-transpose = T. But since T is upper triangular, taking the transpose gives us a lower triangular matrix, and so T is both upper and lower triangular, and so T is, in fact, diagonal.
Since the condition A-transpose = A is obviously very important, we give it a name. Definition: A matrix A such that A-transpose = A is called symmetric.
We have now proven that a matrix A is orthogonally diagonalizable if and only if it is symmetric. Although our proof of the Principal Axis Theorem is very quick and easy thanks to the Triangularization Theorem, there is one problem with it. It does not give us a good method for actually orthogonally diagonalizing a symmetric matrix. To figure out a good method for orthogonally diagonalizing a symmetric matrix, we prove a couple more theorems.
Theorem 10.2.4: A matrix A is symmetric if and only if (x dot product with Ay) is equal to (Ax dot y) for all vectors x and y in Rn. Proof: Suppose that A is symmetric. Then for any vectors x and y in Rn, we have (x dot Ay) is equal to x-transpose matrix-matrix multiplied by Ay, which is (x-transpose)(A-transpose)y since A is symmetric, which, by using properties of transposes, gives us ((Ax)-transpose)y, which is (Ax dot product y) as required.
On the other hand, if (x dot Ay) = (Ax dot y), then we have (x-transpose)Ay = ((Ax)-transpose)y, which equals (x-transpose)(A-transpose)y. Since this is valid for all vectors y in Rn, we can apply Theorem 3.1.4 to get (x-transpose)A = (x-transpose)(A-transpose). Now taking the transpose of both sides, we get (A-transpose)x = Ax. Now, since this is also valid for all x in Rn, we have that A-transpose = A by Theorem 3.1.4. Poof.
Take a minute to read over the proof, and make sure that you understand all the steps. Review Theorem 3.1.4 if necessary, and make sure that you really understand how it is being applied, and in particular the necessity of taking the transpose at the end.
Theorem 10.2.5: If A is a symmetric matrix with eigenvectors v1 and v2 corresponding to distinct eigenvalues lambda1 and lambda2, then v1 and v2 are orthogonal. Proof: We need to prove that v1 dot v2 = 0. We have (lambda1)(v1 dot v2) = ((lambda1)v1) dot product v2. We have assumed that v1 is an eigenvector corresponding to lambda1, so (lambda1)v1 = Av1, and hence, we have (Av1) dot v2. By Theorem 10.2.4, since A is symmetric, this equals v1 dot product (Av2), which gives v1 dot ((lambda2)v2) since v2 is an eigenvector of A with corresponding eigenvalue lambda2. And finally, we get that this equals (lambda2)(v1 dot v2). Since lambda1 is not equal to lambda2, this is only possible if v1 dot v2 = 0 as required.
This theorem is extremely helpful. It says that eigenvectors corresponding to different eigenvalues of a symmetric matrix A are necessarily orthogonal. Thus, finding an orthonormal basis for eigenvectors of a symmetric matrix should be relatively easy, as the eigenvectors corresponding to different eigenvalues are naturally orthogonal.
## Examples
We will now do two examples of orthogonally diagonalizing a symmetric matrix. Example: Orthogonally diagonalize the symmetric matrix A = [4, 0, 0; 0, 1, -2; 0, -2, 1]. Solution: We begin by diagonalizing the matrix like normal, so first, we find and factor the characteristic polynomial, and then find the eigenvalues and corresponding eigenvectors. We have the characteristic polynomial is the determinant of (A – (lambda)I), which equals –(lambda – 4)(lambda – 3)(lambda + 1). Thus, the eigenvalues are lambda1 = 4, lambda2 = 3, and lambda3 = -1, each with algebraic multiplicity 1.
For lambda1 = 4, we get (A – (lambda1)I) = [0, 0, 0; 0, -3, -2; 0, -2, -3], which row reduces to [0, 1, 0; 0, 0, 1; 0, 0, 0]. Thus, a basis for the eigenspace of lambda1 is {[1; 0; 0]}. For lambda2 = 3, we get (A – (lambda2)I) = [1, 0, 0; 0, -2, -2; 0, -2, -2], which clearly row reduces to [1, 0, 0; 0, 1, 1; 0, 0, 0]. And thus, a basis for the eigenspace of lambda2 is {[0; -1; 1]}. Finally, for lambda3 = -1, we get (A – (lambda3)I) = [5, 0, 0; 0, 2, -2; 0, -2, 2]}, which row reduces to [1, 0, 0; 0, 1, -1; 0, 0, 0]. Thus, a basis for the eigenspace of lambda3 is {[0; 1; 1]}.
Observe that, as predicted by Theorem 10.2.5, the eigenvectors [1; 0; 0], [0; -1; 1], and [0; 1; 1] do form an orthogonal set. Hence, if we normalize them—and we must normalize them because we need to get an orthonormal basis of eigenvectors of A so that we can form an orthogonal matrix P which diagonalizes A—then we get that A is diagonalized by the orthogonal matrix P = [1, 0, 0; 0, -1/(root 2), 1/(root 2); 0, 1/(root 2), 1/(root 2)] to D = [4, 0, 0; 0, 3, 0; 0, 0, -1].
It is important for you to be good at diagonalization. You should be able to easily find all eigenvalues of a given matrix, and you should be able to find a basis for the eigenspace of an eigenvalue lambda by just looking at the reduced row echelon form of (A – lambda)I).
Example: Orthogonally diagonalize the symmetric matrix A = [5, -4, -2; -4, 5, -2; -2, -2, 8]. Solution: We have the characteristic polynomial is equal to the determinant of (A – (lambda)I), which is the determinant of |5 – lambda, -4, -2; -4, 5 – lambda, -2; -2, -2, 8 – lambda|. Using elementary row and column operations to simplify the determinant, and then finding the determinant, we get the characteristic polynomial is equal to –lambda((lambda – 9)-squared). Thus, the eigenvalues are lambda1 = 9, with algebraic multiplicity 2, and lambda2 = 0, with algebraic multiplicity 1.
For lambda1 = 9, we get (A – (lambda1)I) = [-4, -4, -2; -4, -4, -2; -2, -2, -1], which row reduces to [1, 1, 1/2; 0, 0, 0; 0, 0, 0]. Thus, a basis for the eigenspace of lambda1 is {w1, w2} = {[-1; 1; 0], [-1; 0; 2]}. Wait—we have a problem. These eigenvectors are not orthogonal to each other. Notice that Theorem 10.2.5 only guarantees that eigenvectors corresponding to different eigenvalues are necessarily orthogonal. But we shouldn’t panic. We know that the eigenspace of lambda1 is a subspace of R3. What we really need is an orthogonal basis for this eigenspace. How could we find that? Yes, that’s right—we use the Gram-Schmidt procedure. By applying the Gram-Schmidt procedure to the basis {w1, w2} for the eigenspace of lambda1, we get an orthogonal basis for the eigenspace of lambda1 is {[-1; 1; 0] and [-1; -1; 4]}.
For lambda2 = 0, we get (A – (lambda2)I) is [5, -4, -2; -4, 5, -2; -2, -2, 8], which row reduces to give [1, 0, -2; 0, 1, -2; 0, 0, 0]. Thus, a basis for the eigenspace of lambda2 is {[2; 2; 1]}. Notice, it is easy to check that this vector is, in fact, orthogonal to the orthogonal basis for the eigenspace of lambda1, as predicted by Theorem 10.2.5. Hence, we now do have an orthogonal basis for R3 of eigenvectors of A. Normalizing these vectors, we take P = [2/3, -1/(root 2), -1/(root 18); 2/3, 1/(root 2), -1/(root 18); 1/3, 0, 4/(root 18)], and we get that (P-transpose)AP = D, which is [0, 0, 0; 0, 9, 0; 0, 0, 9].
These examples demonstrate that the procedure for orthogonally diagonalizing a symmetric matrix is exactly the same as normal diagonalization, except for a couple of key points. First, if we have a symmetric matrix A, then we automatically know by the Principal Axis Theorem that A is diagonalizable. By understanding the theory of diagonalization, this can lead to some shortcuts in finding eigenvalues, and it can help you in finding some computational errors. Second, Theorem 10.2.5 only guarantees that eigenvalues corresponding to different eigenvectors are orthogonal. Whenever we get an eigenvalue of geometric multiplicity greater than 1, remember to, if necessary, apply the Gram-Schmidt procedure to the basis for that eigenspace to get an orthogonal basis for the eigenspace. Third, when making the diagonalizing matrix P, make sure that you do have an orthonormal basis of eigenvectors for Rn of A. In particular, make sure that you remember to normalize the eigenvectors.
With enough practice, these should become very easy. Note that much of the rest of the course is about orthogonal diagonalization and its uses, so it is very important that you do practice enough.
This ends this lecture. | 3,318 | 11,674 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2017-39 | latest | en | 0.947551 |
https://gitlab.esrf.fr/tomotools/nabu/-/commit/9eb7a31c2a2990fb6653be3440cea1e5f7b1dcc2 | 1,675,230,817,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499911.86/warc/CC-MAIN-20230201045500-20230201075500-00484.warc.gz | 282,272,905 | 39,938 | Commit 9eb7a31c by Nicola Vigano
### Alignment: added band-pass filter computation and perform filtering in correlation
```
Signed-off-by: Nicola VIGANÒ <nicola.vigano@esrf.fr>```
parent 3353b97e
... ... @@ -7,14 +7,16 @@ import numpy as np try: import scipy.special as spspe __have_scipy__ = True except ImportError: import math __have_scipy__ = False def get_lowpass_filter(img_shape, cutoff_par): """Computes a low pass filter with the erfc. def get_lowpass_filter(img_shape, cutoff_par=None): """Computes a low pass filter using the erfc function. Parameters ---------- ... ... @@ -38,7 +40,9 @@ def get_lowpass_filter(img_shape, cutoff_par): numpy.array_like The computed filter """ if isinstance(cutoff_par, (int, float)): if cutoff_par is None: return 1 elif isinstance(cutoff_par, (int, float)): cutoff_pix = cutoff_par cutoff_trans_fact = None else: ... ... @@ -69,13 +73,13 @@ def get_lowpass_filter(img_shape, cutoff_par): else: res = np.array(list(map(math.erfc, k_pos_rescaled))) / 2 else: res = np.exp(- (np.pi ** 2) * (r ** 2) * (cutoff_pix ** 2) * 2) res = np.exp(-(np.pi ** 2) * (r ** 2) * (cutoff_pix ** 2) * 2) return res def get_highpass_filter(img_shape, cutoff_par): """Computes a high pass filter with the erfc. def get_highpass_filter(img_shape, cutoff_par=None): """Computes a high pass filter using the erfc function. Parameters ---------- ... ... @@ -99,4 +103,41 @@ def get_highpass_filter(img_shape, cutoff_par): numpy.array_like The computed filter """ return 1 - get_lowpass_filter(img_shape, cutoff_par) if cutoff_par is None: return 1 else: return 1 - get_lowpass_filter(img_shape, cutoff_par) def get_bandpass_filter(img_shape, cutoff_lowpass=None, cutoff_highpass=None): """Computes a band pass filter using the erfc function. The cutoff structures should be formed as follows: - tuple of two floats: the first indicates the cutoff frequency, the second \ determines the width of the transition region, as fraction of the cutoff frequency. - one float -> it represents the sigma of a gaussian which acts as a filter or anti-filter (1 - filter). Parameters ---------- img_shape: tuple Shape of the image cutoff_lowpass: float or sequence of two floats Cutoff parameters for the low-pass filter cutoff_highpass: float or sequence of two floats Cutoff parameters for the high-pass filter Raises ------ ValueError In case of malformed cutoff_par Returns ------- numpy.array_like The computed filter """ return get_lowpass_filter(img_shape, cutoff_par=cutoff_lowpass) * get_highpass_filter( img_shape, cutoff_par=cutoff_highpass ) | 645 | 2,564 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-06 | latest | en | 0.63988 |
http://extraconversion.com/volume/cubic-micrometers | 1,632,829,118,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060677.55/warc/CC-MAIN-20210928092646-20210928122646-00717.warc.gz | 23,155,001 | 21,955 | # Cubic micrometer
Cubic micrometer is a unit of volume. It is equal to 0.000000000000001 liter. Plural name is cubic micrometers.
## Sample Volume Conversion
10 kiloliter to liters, the result is 10000 liters 10 megaliter to liters, the result is 10000000 liters 10 US dry quart to cubic feet, the result is 0.3889 cubic feet 10 microliter to milliliters, the result is 0.01 milliliters 10 microliter to liters, the result is 1.0E-5 liters 10 cubic micrometer to cubic centimeters, the result is 1.0E-11 cubic centimeters 10 milliliter to dekaliters, the result is 0.001 dekaliters 10 kiloliter to hectoliters, the result is 100 hectoliters 10 centiliter to dekaliters, the result is 0.01 dekaliters 10 kiloliter to deciliters, the result is 100000 deciliters
## Table Conversion and Table Chart
### Cubic micrometer to Other Units
Cubic micrometer can convert to other units, such as acre foot, Canada cup, centiliter, cubic centimeter, cubic decimeter, cubic dekameter, cubic foot, cubic inch, cubic kilometer, cubic meter, cubic mile, cubic millimeter, cubic yard, deciliter, dekaliter, hectoliter, kiloliter, liter, megaliter, metric cup, metric tablespoon, metric teaspoon, microliter, milliliter, UK barrel, UK gallon, UK liquid ounce, UK pint, UK quart, UK tablespoon, UK teaspoon, UK wine barrel, US cup, US dry barrel, US dry gallon, US dry pint, US dry quart, US federal barrel, US gallons, US liquid barrel, US liquid gallon, US liquid ounce, US liquid pint, US liquid quart, US oil barrel, US tablespoon, US teaspoon and more. You can find the reference table below:
Unit to Other Unit Volume Conversion Volume Table
1 µm³ = 8.1071319217801E-22 ac ftcubic micrometers to acre feetcubic micrometers to acre feet table
1 µm³ = 4.3993849659818E-15 c (CA)cubic micrometers to Canada cupscubic micrometers to Canada cups table
1 µm³ = 1.0E-13 cLcubic micrometers to centiliterscubic micrometers to centiliters table
1 µm³ = 1.0E-12 cm³cubic micrometers to cubic centimeterscubic micrometers to cubic centimeters table
1 µm³ = 1.0E-15 dm³cubic micrometers to cubic decimeterscubic micrometers to cubic decimeters table
1 µm³ = 1.0E-21 dam³cubic micrometers to cubic dekameterscubic micrometers to cubic dekameters table
1 µm³ = 3.5314666574328E-17 ft³cubic micrometers to cubic feetcubic micrometers to cubic feet table
1 µm³ = 6.10237438368E-14 in³cubic micrometers to cubic inchescubic micrometers to cubic inches table
1 µm³ = 1.0E-27 km³cubic micrometers to cubic kilometerscubic micrometers to cubic kilometers table
1 µm³ = 1.0E-18 cubic micrometers to cubic meterscubic micrometers to cubic meters table
1 µm³ = 2.3991275756488E-28 mi³cubic micrometers to cubic milescubic micrometers to cubic miles table
1 µm³ = 1.0E-9 mm³cubic micrometers to cubic millimeterscubic micrometers to cubic millimeters table
1 µm³ = 1.307950613786E-18 yd³cubic micrometers to cubic yardscubic micrometers to cubic yards table
1 µm³ = 1.0E-14 dLcubic micrometers to deciliterscubic micrometers to deciliters table
1 µm³ = 1.0E-16 daLcubic micrometers to dekaliterscubic micrometers to dekaliters table
1 µm³ = 1.0E-17 hLcubic micrometers to hectoliterscubic micrometers to hectoliters table
1 µm³ = 1.0E-18 kLcubic micrometers to kiloliterscubic micrometers to kiloliters table
1 µm³ = 1.0E-15 Lcubic micrometers to literscubic micrometers to liters table
1 µm³ = 1.0E-21 MLcubic micrometers to megaliterscubic micrometers to megaliters table
1 µm³ = 4.0E-15 ccubic micrometers to metric cupscubic micrometers to metric cups table
1 µm³ = 6.6666666666667E-14 tbspcubic micrometers to metric tablespoonscubic micrometers to metric tablespoons table
1 µm³ = 2.0E-13 tspcubic micrometers to metric teaspoonscubic micrometers to metric teaspoons table
1 µm³ = 1.0E-9 µLcubic micrometers to microliterscubic micrometers to microliters table
1 µm³ = 1.0E-12 mLcubic micrometers to milliliterscubic micrometers to milliliters table
1 µm³ = 6.1102568971969E-18 bl (UK)cubic micrometers to UK barrelscubic micrometers to UK barrels table
1 µm³ = 2.1996924829909E-16 gal (UK)cubic micrometers to UK gallonscubic micrometers to UK gallons table
1 µm³ = 3.5195079727854E-14 oz (UK liquid)cubic micrometers to UK liquid ouncescubic micrometers to UK liquid ounces table
1 µm³ = 1.7597539863927E-15 pt (UK)cubic micrometers to UK pintscubic micrometers to UK pints table
1 µm³ = 8.7987699319635E-16 qt (UK)cubic micrometers to UK quartscubic micrometers to UK quarts table
1 µm³ = 7.0390159455708E-14 tbsp (UK)cubic micrometers to UK tablespoonscubic micrometers to UK tablespoons table
1 µm³ = 2.8156062295851E-13 tsp (UK)cubic micrometers to UK teaspoonscubic micrometers to UK teaspoons table
1 µm³ = 6.9831507396536E-18 bl (UK wine)cubic micrometers to UK wine barrelscubic micrometers to UK wine barrels table
1 µm³ = 4.2267528109322E-15 c (US)cubic micrometers to US cupscubic micrometers to US cups table
1 µm³ = 8.6484897727891E-18 bl (US dry)cubic micrometers to US dry barrelscubic micrometers to US dry barrels table
1 µm³ = 2.2702074456538E-16 gal (US dry)cubic micrometers to US dry gallonscubic micrometers to US dry gallons table
1 µm³ = 1.8161659565231E-15 pt (US dry)cubic micrometers to US dry pintscubic micrometers to US dry pints table
1 µm³ = 9.0808297826154E-16 qt (US dry)cubic micrometers to US dry quartscubic micrometers to US dry quarts table
1 µm³ = 8.5216790723083E-18 bl (US federal)cubic micrometers to US federal barrelscubic micrometers to US federal barrels table
1 µm³ = 2.641720526373E-16 gal[US]cubic micrometers to US gallonscubic micrometers to US gallons table
1 µm³ = 8.3864143251288E-18 bl (US liquid)cubic micrometers to US liquid barrelscubic micrometers to US liquid barrels table
1 µm³ = 2.6417205124156E-16 gal (US liquid)cubic micrometers to US liquid gallonscubic micrometers to US liquid gallons table
1 µm³ = 3.3814022558919E-14 oz (US liquid)cubic micrometers to US liquid ouncescubic micrometers to US liquid ounces table
1 µm³ = 2.1133764099325E-15 pt (US liquid)cubic micrometers to US liquid pintscubic micrometers to US liquid pints table
1 µm³ = 1.0566882049662E-15 qt (US liquid)cubic micrometers to US liquid quartscubic micrometers to US liquid quarts table
1 µm³ = 6.2898107438466E-18 bblcubic micrometers to US oil barrelscubic micrometers to US oil barrels table
1 µm³ = 6.7628045117839E-14 tbsp (US)cubic micrometers to US tablespoonscubic micrometers to US tablespoons table
1 µm³ = 2.0288411948901E-13 tsp (US)cubic micrometers to US teaspoonscubic micrometers to US teaspoons table
### Other Units to Cubic micrometer
Cubic micrometer also canvert from other unit such as acre foot, Canada cup, centiliter, cubic centimeter, cubic decimeter, cubic dekameter, cubic foot, cubic inch, cubic kilometer, cubic meter, cubic mile, cubic millimeter, cubic yard, deciliter, dekaliter, hectoliter, kiloliter, liter, megaliter, metric cup, metric tablespoon, metric teaspoon, microliter, milliliter, UK barrel, UK gallon, UK liquid ounce, UK pint, UK quart, UK tablespoon, UK teaspoon, UK wine barrel, US cup, US dry barrel, US dry gallon, US dry pint, US dry quart, US federal barrel, US gallons, US liquid barrel, US liquid gallon, US liquid ounce, US liquid pint, US liquid quart, US oil barrel, US tablespoon, US teaspoon and more. You can find the reference table below:
Unit to Other Unit Volume Conversion Volume Table
1 ac ft = 1.23348184E+21 µm³acre feet to cubic micrometersacre feet to cubic micrometers table
1 c (CA) = 2.273045E+14 µm³Canada cups to cubic micrometersCanada cups to cubic micrometers table
1 cL = 10000000000000 µm³centiliters to cubic micrometerscentiliters to cubic micrometers table
1 cm³ = 1000000000000 µm³cubic centimeters to cubic micrometerscubic centimeters to cubic micrometers table
1 dm³ = 1.0E+15 µm³cubic decimeters to cubic micrometerscubic decimeters to cubic micrometers table
1 dam³ = 1.0E+21 µm³cubic dekameters to cubic micrometerscubic dekameters to cubic micrometers table
1 ft³ = 2.831684671E+16 µm³cubic feet to cubic micrometerscubic feet to cubic micrometers table
1 in³ = 16387064069264 µm³cubic inches to cubic micrometerscubic inches to cubic micrometers table
1 km³ = 1.0E+27 µm³cubic kilometers to cubic micrometerscubic kilometers to cubic micrometers table
1 = 1.0E+18 µm³cubic meters to cubic micrometerscubic meters to cubic micrometers table
1 mi³ = 4.1681818430584E+27 µm³cubic miles to cubic micrometerscubic miles to cubic micrometers table
1 mm³ = 1000000000 µm³cubic millimeters to cubic micrometerscubic millimeters to cubic micrometers table
1 yd³ = 7.6455486121558E+17 µm³cubic yards to cubic micrometerscubic yards to cubic micrometers table
1 dL = 1.0E+14 µm³deciliters to cubic micrometersdeciliters to cubic micrometers table
1 daL = 1.0E+16 µm³dekaliters to cubic micrometersdekaliters to cubic micrometers table
1 hL = 1.0E+17 µm³hectoliters to cubic micrometershectoliters to cubic micrometers table
1 kL = 1.0E+18 µm³kiloliters to cubic micrometerskiloliters to cubic micrometers table
1 L = 1.0E+15 µm³liters to cubic micrometersliters to cubic micrometers table
1 ML = 1.0E+21 µm³megaliters to cubic micrometersmegaliters to cubic micrometers table
1 c = 2.5E+14 µm³metric cups to cubic micrometersmetric cups to cubic micrometers table
1 tbsp = 15000000000000 µm³metric tablespoons to cubic micrometersmetric tablespoons to cubic micrometers table
1 tsp = 5000000000000 µm³metric teaspoons to cubic micrometersmetric teaspoons to cubic micrometers table
1 µL = 1000000000 µm³microliters to cubic micrometersmicroliters to cubic micrometers table
1 mL = 1000000000000 µm³milliliters to cubic micrometersmilliliters to cubic micrometers table
1 bl (UK) = 1.6365924E+17 µm³UK barrels to cubic micrometersUK barrels to cubic micrometers table
1 gal (UK) = 4.54609E+15 µm³UK gallons to cubic micrometersUK gallons to cubic micrometers table
1 oz (UK liquid) = 28413062500000 µm³UK liquid ounces to cubic micrometersUK liquid ounces to cubic micrometers table
1 pt (UK) = 5.6826125E+14 µm³UK pints to cubic micrometersUK pints to cubic micrometers table
1 qt (UK) = 1.1365225E+15 µm³UK quarts to cubic micrometersUK quarts to cubic micrometers table
1 tbsp (UK) = 14206531250000 µm³UK tablespoons to cubic micrometersUK tablespoons to cubic micrometers table
1 tsp (UK) = 3551633000000 µm³UK teaspoons to cubic micrometersUK teaspoons to cubic micrometers table
1 bl (UK wine) = 1.43201835E+17 µm³UK wine barrels to cubic micrometersUK wine barrels to cubic micrometers table
1 c (US) = 2.36588238E+14 µm³US cups to cubic micrometersUS cups to cubic micrometers table
1 bl (US dry) = 1.1562712407273E+17 µm³US dry barrels to cubic micrometersUS dry barrels to cubic micrometers table
1 gal (US dry) = 4.4048838E+15 µm³US dry gallons to cubic micrometersUS dry gallons to cubic micrometers table
1 pt (US dry) = 5.50610475E+14 µm³US dry pints to cubic micrometersUS dry pints to cubic micrometers table
1 qt (US dry) = 1.10122095E+15 µm³US dry quarts to cubic micrometersUS dry quarts to cubic micrometers table
1 bl (US federal) = 1.173477658E+17 µm³US federal barrels to cubic micrometersUS federal barrels to cubic micrometers table
1 gal[US] = 3.78541178E+15 µm³US gallons to cubic micrometersUS gallons to cubic micrometers table
1 bl (US liquid) = 1.192404717E+17 µm³US liquid barrels to cubic micrometersUS liquid barrels to cubic micrometers table
1 gal (US liquid) = 3.7854118E+15 µm³US liquid gallons to cubic micrometersUS liquid gallons to cubic micrometers table
1 oz (US liquid) = 29573529687500 µm³US liquid ounces to cubic micrometersUS liquid ounces to cubic micrometers table
1 pt (US liquid) = 4.73176475E+14 µm³US liquid pints to cubic micrometersUS liquid pints to cubic micrometers table
1 qt (US liquid) = 9.4635295E+14 µm³US liquid quarts to cubic micrometersUS liquid quarts to cubic micrometers table
1 bbl = 1.589872956E+17 µm³US oil barrels to cubic micrometersUS oil barrels to cubic micrometers table
1 tbsp (US) = 14786764843750 µm³US tablespoons to cubic micrometersUS tablespoons to cubic micrometers table
1 tsp (US) = 4928922000000 µm³US teaspoons to cubic micrometersUS teaspoons to cubic micrometers table
## Acre foot [ ac ft ]
Acre-foot is commonly used in reference to large-scale water resources including canals and river flows.
## Canada cup [ c (CA) ]
In Canada, a cup is equal to 8 Imperial fluid ounces (or 0.2273045 liter).
## Centiliter [ cL ]
The centiliter is a common metric unit of volume. It is equal to 10 cubic centimeters. In the kitchen, a centiliter is roughly equal to 2 U.S. teaspoons (or 0.704 British tablespoonful).
## Cubic centimeter [ cm³ ]
Cubic centimeter is a CGS unit of volume. It is equal to 10E-6 cubic meter, 1 milliliter, or about 0.0610237 cubic inch.
## Cubic decimeter [ dm³ ]
Cubic decimeter is a unit of volume. It is equal to 1 liter.
## Cubic dekameter [ dam³ ]
Cubic dekameter is a unit of volume. It is equal to 1000000 liters.
## Cubic foot [ ft³ ]
Cubic foot is a traditional unit of volume in English speaking countries. It is equal to 1728 cubic inces, or 28.316 85 liters.
## Cubic inch [ in³ ]
Cubic inch is a traditional unit of volume in English speaking countries. It is equal to 0.0163870640692641 cubic centimeter or 0.0163870640692641 milliliters.
## Cubic kilometer [ km³ ]
Cubic kilometer is a unit of volume. It is equal to 1000000000000 liters.
## Cubic meter [ m³ ]
Cubic meter is a SI unit of volume. It is equal to 1000 liters. A cubic meter is about 264.17 U.S. liquid gallons or 219.99 British Imperial gallons.
## Cubic micrometer [ µm³ ]
Cubic micrometer is a unit of volume. It is equal to 0.000000000000001 liter.
## Cubic mile [ mi³ ]
Cubic mile is a unit of volume. It is equal to 4168181843058.45 liters.
## Cubic millimeter [ mm³ ]
Cubic millimeter is a SI unit of volume. It is equal to 0.000001 liter.
## Cubic yard [ yd³ ]
Cubic yard is a traditional unit of volume in English speaking countries. It is equal to 27 cubic feet or 46656 cubic inches or 0.764555 cubic meters, or 764.554861215584 liters.
## Deciliter [ dL ]
Deciliter is a common metric unit of volume. It is equal to 0.1 liter or 100 cubic centimeters.
## Dekaliter [ daL ]
Dekaliter is a metric unit of volume. It is equal to 10 liters.
## Hectoliter [ hL ]
Hectoliter is a common metric unit of volume. It is equal to 100 liters, 0.1 cubic meter, 26.417 U.S. liquid gallons, 21.999 British imperial gallons, or 3.5315 cubic feet.
## Kiloliter [ kL ]
Kiloliter is a metric unit of volume. Kiloliter is identical to the cubic meter. It is equal to 35.3147 cubic feet, or 1000 liters.
## Liter [ L ]
Liter or litre is the common metric unit of volume. The unit is spelled liter in the U.S. and litre in Britain.
## Megaliter [ ML ]
Megaliter is a metric unit of volume. It is equal to 1000 cubic meters or 1000000 liters.
## Metric cup [ c ]
Metric cup is an informal metric unit of volume. It is equal to 0.25 liter, commonly used in recipes in Australia.
## Metric tablespoon [ tbsp ]
Tablespoon is a unit of volume used in food recipes. It is equal to 15 milliliters in Britain, Canada, and New Zealand, 20 milliliters in Australia.
## Metric teaspoon [ tsp ]
Teaspoon is a unit of volume used in food recipes. In metric kitchens in Britain, Canada, Australia, and New Zealand, a teaspoonful is exactly 5 milliliters.
## Microliter [ µL ]
Microliter is a metric unit of volume used in chemistry and medicine to measure very small quantities of liquid. It is equal to 0.001 milliliter or 1 cubic millimeter. It is also called lambda.
## Milliliter [ mL ]
Milliliter is a very common metric unit of volume. It is equal to 0.001 liter, exactly one cubic centimeter.
## UK barrel [ bl (UK) ]
Barrel is a unit of volume. In UK, it is equal to 163.65924 liters.
## UK gallon [ gal (UK) ]
The imperial gallon is designed to contain exactly 10 pounds of distilled water under precisely defined conditions. It is the same as 4.54609 liters.
## UK liquid ounce [ oz (UK liquid) ]
UK liquid ounce is a traditional unit of liquid volume. It is also called the fluid ounce to avoid confusion with the weight ounce. It is equal to 0.0284130625 liter.
## UK pint [ pt (UK) ]
UK pint is a unit of volume. It is equal to 0.56826125 liter.
## UK quart [ qt (UK) ]
UK quart is a unit of volume. It is equal to 1.1365225 liters.
## UK tablespoon [ tbsp (UK) ]
Tablespoon is a unit of volume used in food recipes. In Britain, it is equal to 14.20653125 milliliters.
## UK teaspoon [ tsp (UK) ]
Teaspoon is a unit of volume used in food recipes. In Britain, a traditional teaspoonful in the kitchen was equal to 1/8 Imperial fluid ounce or approximately 3.55 milliliters, but the medical teaspoonful was usually 5 milliliters.
## UK wine barrel [ bl (UK wine) ]
Barrel is a unit of volume. In UK, a barrel of wine is equal to 143.201835 liters of wine.
## US cup [ c (US) ]
US cup is a traditional unit of volume used in recipes in the United States. It is equal to 1/2 liquid pint, or 8 fluid ounces or 0.2365882375 liter.
## US dry barrel [ bl (US dry) ]
Barrel is a unit of volume. In USA, a dry barrel is equal to 115.627124072727 liters.
## US dry gallon [ gal (US dry) ]
US dry gallon is a historic British unit of dry volume still used implicitly in the U.S. US dry gallon is equal to 4.4048838 liters.
## US dry pint [ pt (US dry) ]
US dry pint is a unit of volume. It is equal to 0.550610475 liter.
## US dry quart [ qt (US dry) ]
US dry quart is a unit of volume. It is equal to 1.10122095 liters.
## US federal barrel [ bl (US federal) ]
Barrel is a unit of volume. In USA, a federal barrel is equal to 117.3477658 liters.
## US gallons [ gal[US] ]
The US gallon is a unit of volume use in US with the abbreviation gal (US).
## US liquid barrel [ bl (US liquid) ]
Barrel is a unit of volume. In USA, a liquid barrel is equal to 119.2404717 liters.
## US liquid gallon [ gal (US liquid) ]
US liquid gallon is a traditional unit of liquid volume. It is equal to 4 liquid quarts or exactly 3.785411784 liters.
## US liquid ounce [ oz (US liquid) ]
US liquid ounce is a traditional unit of liquid volume. It is also called the fluid ounce to avoid confusion with the weight ounce. It is equal to 0.0295735296875 liter.
## US liquid pint [ pt (US liquid) ]
US liquid pint is a unit of volume. It is equal to 0.473176475 liter.
## US liquid quart [ qt (US liquid) ]
US liquid quart is a unit of volume. It is equal to 0.94635295 liter.
## US oil barrel [ bbl ]
Barrel is a unit of volume. In USA, a barrel of oil or petroleum is equal to 158.9872956 liters.
## US tablespoon [ tbsp (US) ]
In the U.S., the tablespoon is equal to 1/2 fluid ounce. It is about 14.8 milliliters.
## US teaspoon [ tsp (US) ]
Teaspoon is a unit of volume used in food recipes. The U.S. teaspoon is equal to 1/3 tablespoon or 1/48 cup. It is about 4.9 milliliters.
## Acre-pie
Acre-pie es común en referencia a los recursos de agua a gran escala, incluidos los canales y los caudales de los ríos.
En Canadá, una taza equivale a 8 onzas de líquido Imperial (o 0.2273045 litros).
## Centilitro
El centilitro es una unidad de medida común de volumen. Es igual a 10 centímetros cúbicos. En la cocina, un centilitro es aproximadamente igual a 2 cucharaditas de EE.UU. (o 0.704 cucharada británica).
## Centímetro cúbico
Centímetro cúbico es una unidad CGS de volumen. Es igual a 10E-6 metro cúbico, un mililitro, o alrededor de 0.0610237 pulgadas cúbicas.
## Decímetro cúbico
Decímetro cúbico es una unidad de volumen. Es igual a 1 litro.
## Decámetro cúbico
Decámetro cúbico es una unidad de volumen. Es igual a 1000000 litros.
## Pie cúbico
Pie cúbico es una unidad tradicional de volumen en países de habla Inglés. Es igual a 1728 vincias cúbicos, litros o 28.316 85.
Pulgada cúbica es una unidad tradicional de volumen en países de habla Inglés. Es igual a 0.0163870640692641 centímetros cúbicos o mililitros 0.0163870640692641.
## Kilómetro cúbico
Kilómetro cúbico es una unidad de volumen. Es igual a 1000000000000 litros.
## Metro cúbico
Metro cúbico es una unidad SI de volumen. Es igual a 1000 litros. Un metro cúbico es de 264.17 galones EE.UU. líquido o 219.99 galones imperiales británicos.
## Micrómetro cúbico
Micrómetro cúbico es una unidad de volumen. Es igual a un litro 0.000000000000001.
## Milla cúbica
Milla cúbica es una unidad de volumen. Es igual a 4168181843058.45 litros.
## Milímetro cúbico
Milímetro cúbico es una unidad SI de volumen. Es igual a 0.000001 litros.
## Yarda cúbica
Yarda cúbica es una unidad tradicional de volumen en países de habla Inglés. Es igual a 27 pies cúbicos o 46656 centímetros cúbicos o 0.764555 metros cúbicos o litros 764.554861215584.
## Decilitro
Decilitro es una unidad de medida común de volumen. Es igual a 0.1 litros o de 100 centímetros cúbicos.
## Decalitro
Dekaliter es una unidad métrica de volumen. Es igual a 10 litros.
## Hectolitro
Hectolitro es una unidad de medida común de volumen. Es igual a 100 litros, 0.1 metros cúbicos, 26.417 galones de EE.UU. líquido, 21.999 galones imperiales británicos, o 3.5315 pies cúbicos.
## Kilolitro
Kilolitro es una unidad métrica de volumen. Kilolitro es idéntico al del metro cúbico. Es igual a 35.3147 pies cúbicos o litros de 1000.
## Litro
Litro es la unidad de medida común de volumen. La unidad se escribe liter en los EE.UU. y un litre en Gran Bretaña.
## Megalitro
Megaliter es una unidad métrica de volumen. Es igual a 1000 metros cúbicos o litros 1000000.
## Métricas taza
Taza métrica es una unidad métrica de volumen informales. Es igual a 0.25 litros, de uso común en las recetas en Australia.
Cucharada es una unidad de volumen utilizado en las recetas de alimentos. Es igual a 15 ml en Gran Bretaña, Canadá y Nueva Zelanda, 20 mililitros en Australia.
Cucharadita es una unidad de volumen utilizado en las recetas de alimentos. En las cocinas métricas en Gran Bretaña, Canadá, Australia y Nueva Zelanda, una cucharadita es exactamente 5 mililitros.
## Microlitro
Microlitro es una unidad métrica de volumen utilizada en la química y la medicina para medir cantidades muy pequeñas de líquido. Es igual a 0.001 ml o 1 milímetro cúbico.
## Mililitro
Mililitro es una unidad métrica muy común de volumen. Es igual a un litro 0.001, exactamente un centímetro cúbico.
## UK barril
Barril es una unidad de volumen. En el Reino Unido, que es igual a 163.65924 litros.
## UK galón
El galón imperial está diseñado para contener exactamente 10 libras de agua destilada en condiciones definidas con precisión. Es lo mismo que 4.54609 litros.
## UK onza líquido
Onza Reino Unido líquido es una unidad tradicional de volumen de líquido. También se conoce como la onza de líquido para evitar la confusión con la onza de peso. Es igual a un litro 0,0284130625.
## UK pinta
Pinta Reino Unido es una unidad de volumen. Es igual a un litro 0.56826125.
## UK cuarto
Cuarto del Reino Unido es una unidad de volumen. Es igual a 1.1365225 litros.
Cucharada es una unidad de volumen utilizado en las recetas de alimentos. En Gran Bretaña, es igual a 14.20653125 mililitros.
Cucharadita es una unidad de volumen utilizado en las recetas de alimentos. En Gran Bretaña, una cucharadita tradicionales en la cocina era igual a 1 / 8 oz Imperial líquido o mililitros aproximadamente 3.55, pero el médico fue generalmente cucharadita de 5 mililitros.
## UK barril de vino
Barril es una unidad de volumen. En el Reino Unido, un barril de vino es igual a 143.201835 litros de vino.
## US taza
Taza de EE.UU. es una unidad tradicional de volumen utilizado en las recetas en los Estados Unidos. Es igual a 1 / 2 litro de líquido, o de 8 onzas o litro 0.2365882375.
## US barril seco
Barril es una unidad de volumen. En EE.UU., el barril seco es igual a 115.627124072727 litros.
## US galón seco
Galones EE.UU. seca es una unidad histórica británica de volumen seco todavía se utiliza implícitamente en el galón de EE.UU. EE.UU. seco es igual a 4.4048838 litros.
## US pinta seco
Pinta EE.UU. seca es una unidad de volumen. Es igual a 0.550610475 litros.
## US cuarto seco
Cuarto EE.UU. seca es una unidad de volumen. Es igual a 1.10122095 litros.
## US barril federales
Barril es una unidad de volumen. En EE.UU., el barril federal es igual a 117.3477658 litros.
## Galón
El galón es una unidad de volumen que se emplea en los países anglófonos (especialmente Estados Unidos) o con influencia de estos (como Puerto Rico y Panamá), para medir volúmenes de líquidos.
## US barril líquido
Barril es una unidad de volumen. En EE.UU., un barril de líquido es igual a 119.2404717 litros.
## US galón líquido
Galones EE.UU. líquido es una unidad tradicional de volumen de líquido. Es igual a 4 cuartos de líquido o exactamente 3.785411784 litros.
## US onza líquido
EE.UU. onza líquido es una unidad tradicional de volumen de líquido. También se conoce como la onza de líquido para evitar la confusión con la onza de peso. Es igual a un litro 0.0295735296875.
## US pinta líquido
Pinta EE.UU. líquido es una unidad de volumen. Es igual a 0.473176475 litros.
## US cuarto líquido
Cuarto EE.UU. líquido es una unidad de volumen. Es igual a un litro 0.94635295.
## US barril de petróleo
Barril es una unidad de volumen. En EE.UU., el barril de petróleo o petróleo es igual a 158.9872956 litros.
En los EE.UU., la cucharada es igual a la onza de líquido 1 / 2. Se trata de 14.8 mililitros.
Cucharadita es una unidad de volumen utilizado en las recetas de alimentos. La cucharadita de EE.UU. es igual a 1 / 3 cucharada o una taza de 1 / 48. Se trata de alrededor de 4.9 ml.
## Acre pied
Acre-pied est généralement utilisé en référence aux ressources en eau à grande échelle, y compris les canaux et les débits des rivières.
Au Canada, une tasse est égale à 8 onces de liquide impériale (ou 0.2273045 litres).
## Centilitre
Le centilitre est une unité commune métriques de volume. Elle est égale à 10 centimètres cubes. Dans la cuisine, un centilitre est à peu près égal à 2 cuillères à café des États-Unis.
## Centimètre cube
Centimètre cube est une unité CGS de volume. Il est égal à 10E-6 mètre cube, 1 millilitre, soit environ 0.0610237 pouce cube.
## Décimètre cube
Décimètre cube est une unité de volume. Il est égal à 1 litre.
## Décamètre cube
Décamètre cube est une unité de volume. Il est égal à 1000000 litres.
## Pied cube
Pied cube est une unité traditionnelle de volume dans les pays anglophones. Il est égal à 1728 vinces cubes, ou 28.316 85 litres.
## Pouce cube
Pouce cube est une unité traditionnelle de volume dans les pays anglophones. Elle est égale à 0.0163870640692641 centimètre cube ou 0.0163870640692641 millilitres.
## Kilomètre cube
Kilomètre cube est une unité de volume. Elle est égale à 1000000000000 litres.
## Mètre cube
Mètre cube est une unité de mesure de volume. Elle est égale à 1000 litres. Un mètre cube est d'environ 264.17 gallons US liquide ou 219.99 gallons impériaux britanniques.
## Micromètre cube
Micromètre cube est une unité de volume. Elle est égale à 0.000000000000001 litre.
## Mille cube
Mille cube est une unité de volume. Elle est égale à 4168181843058.45 litres.
## Millimètre cube
Millimètre cube est une unité de mesure de volume. Elle est égale à 0.000001 litre.
## Yard cube
Yard cube est une unité traditionnelle de volume dans les pays anglophones. Il est égal à 27 pieds cubes ou 46656 pouces cubes ou 0.764555 mètres cubes, ou 764.554861215584 litres.
## Décilitre
Décilitre est une unité commune métriques de volume. Il est égal à 0.1 litre ou 100 centimètres cubes.
## Décalitre
Décalitre est une unité métrique de volume. Elle est égale à 10 litres.
## Hectolitre
Hectolitre est une unité commune métriques de volume. Elle est égale à 100 litres, 0.1 mètre cube, 26.417 gallons US liquide, 21.999 gallons impériaux britanniques, ou 3.5315 pieds cubes.
## Kilolitre
Kilolitre est une unité métrique de volume. Kilolitre est identique au mètre cube. Il est égal à 35.3147 pieds cubes, ou 1000 litres.
## Litre
Litre est l'unité de mesure commune de volume. L'unité est orthographié liter aux États-Unis et en Grande-Bretagne litre.
## Mégalitre
Megaliter est une unité métrique de volume. Elle est égale à 1000 mètres cubes ou 1000000 litres.
## Tasse métrique
Tasse métrique est une unité informelle métriques de volume. Elle est égale à 0.25 litre, couramment utilisés dans les recettes en Australie.
## Cuillère à soupe métrique
Cuillère à soupe est une unité de volume utilisé dans des recettes de cuisine. Elle est égale à 15 millilitres en Grande-Bretagne, le Canada et la Nouvelle-Zélande, 20 ml en Australie.
## Cuillère à café de métrique
Cuillère à café est une unité de volume utilisé dans des recettes de cuisine. Dans les cuisines métrique en Grande-Bretagne, le Canada, l'Australie et la Nouvelle-Zélande, une cuillerée à café est exactement 5 millilitres.
## Microlitre
Microlitre est une unité métrique de volume utilisé en chimie et en médecine pour mesurer de très petites quantités de liquide. Elle est égale à 0.001 ml ou 1 millimètre cube.
## Millilitre
Millilitre est une unité très commune métriques de volume. Elle est égale à 0.001 litre, soit exactement un centimètre cube.
## UK baril
Baril est une unité de volume. Au Royaume-Uni, elle est égale à 163.65924 litres.
## UK gallon
Le gallon impérial est conçu pour contenir exactement 10 livres d'eau distillée dans des conditions définies avec précision. Il est le même que 4.54609 litres.
## UK once liquide
Royaume-Uni once liquide est une unité traditionnelle de volume de liquide. Il est aussi appelé l'once liquide pour éviter la confusion avec l'once de poids. Elle est égale à 0.0284130625 litre.
## UK pinte
Pinte au Royaume-Uni est une unité de volume. Elle est égale à 0.56826125 litre.
## UK quart
Quart au Royaume-Uni est une unité de volume. Elle est égale à 1.1365225 litres.
## UK cuillerée à soupe
Cuillère à soupe est une unité de volume utilisé dans des recettes de cuisine. En Grande-Bretagne, elle est égale à 14.20653125 millilitres.
## UK cuillère à café
Cuillère à café est une unité de volume utilisé dans des recettes de cuisine. En Grande-Bretagne, une cuillerée à café dans la cuisine traditionnelle est égal à 1 / 8 once liquide impériale, soit environ 3.55 ml, mais la cuillère à café médical était habituellement de 5 millilitres.
## UK baril de vin
Baril est une unité de volume. Au Royaume-Uni, un tonneau de vin est égal à 143.201835 litres de vin.
## US tasse
Tasse des États-Unis est une unité traditionnelle de volume utilisé dans des recettes aux États-Unis. Elle est égale à 1 / 2 litre de liquide, ou 8 onces liquides ou 0.2365882375 litre.
## US baril sec
Baril est une unité de volume. Aux Etats-Unis, un baril à sec est égal à 115.627124072727 litres.
## US gallon sec
Gallon US à sec est une unité historique britannique de volume sec encore utilisé implicitement dans le gallon américain US sèche est égale à 4.4048838 litres.
## US pinte sec
US pinte sèche est une unité de volume. Il est égal à 0.550610475 litres.
## US quart sec
US quart sec est une unité de volume. Elle est égale à 1.10122095 litres.
## US baril fédéral
Baril est une unité de volume. Aux Etats-Unis, le baril fédéral est égal à 117.3477658 litres.
## Gallon
Le gallon (symbole : gal) est une unité de volume anglo-saxonne, utilisée pour mesurer les liquides, et ne faisant pas partie du système international d'unités.
## US baril liquide
Baril est une unité de volume. Aux Etats-Unis, le baril de liquide est égale à 119.2404717 litres.
## US gallon liquide
Gallon US liquide est une unité traditionnelle de volume de liquide. Il est égal à 4 quarts liquide ou exactement 3.785411784 litres.
## US once liquide
US once liquide est une unité traditionnelle de volume de liquide. Il est aussi appelé l'once liquide pour éviter la confusion avec l'once de poids. Elle est égale à 0.0295735296875 litre.
## US pinte liquide
US pinte de liquide est une unité de volume. Il est égal à 0.473176475 litres.
## US quart liquide
US quart liquide est une unité de volume. Elle est égale à 0.94635295 litre.
## US baril pétrole
Baril est une unité de volume. Aux Etats-Unis, un baril de pétrole ou de produits pétroliers est égal au 158.9872956 litres.
## US cuillerée à soupe
Aux États-Unis, la cuillère à soupe est égale à 1 / 2 once liquide. Il est environ 14.8 ml.
## US cuillère à café
Cuillère à café est une unité de volume utilisé dans des recettes de cuisine. La cuillère à café US est égal à 1 / 3 cuillère à soupe ou 1 / 48 tasse. Il est d'environ 4.9 millilitres.
## Acre-pé
Acre-pé é comumente usado em referência aos recursos hídricos de larga escala, incluindo os canais e fluxos dos rios.
No Canadá, um copo é igual a 8 onças de fluido Imperial (ou 0.2273045 litros).
## Centilitro
O centilitro é uma unidade de medida comum de volume. É igual a 10 centímetros cúbicos. Na cozinha, um centilitro é aproximadamente igual a 2 colheres de chá dos EUA.
## Centímetro cúbico
Centímetro cúbico é uma unidade CGS de volume. É igual a 10E-6 metros cúbicos, 1 mililitro, ou cerca de 0.0610237 centímetro cúbico.
## Decímetro cúbico
Decímetro cúbico é uma unidade de volume. É igual a 1 litro.
## Decâmetro cúbico
Decâmetro cúbico é uma unidade de volume. É igual a 1000000 de litros.
## Pé cúbico
Pé cúbico é uma unidade tradicional de volume, em países que falam Inglês. É igual a 1728 inces cúbicos, ou 28.316 85 litros.
Centímetro cúbico é uma unidade tradicional de volume, em países que falam Inglês. É igual a 0.0163870640692641 centímetros cúbicos ou 0.0163870640692641 mililitros.
## Quilômetro cúbico
Quilômetro cúbico é uma unidade de volume. É igual a 1000000000000 litros.
## Metro cúbico
Metro cúbico é uma unidade SI de volume. É igual a 1000 litros. Um metro cúbico é de cerca de 264.17 litros EUA líquidos ou 219.99 litros britânica Imperial.
## Micrômetro cúbico
Micrômetro cúbico é uma unidade de volume. É igual a 0.000000000000001 litros.
## Milha cúbica
Milha cúbica é uma unidade de volume. É igual a 4168181843058.45 litros.
## Milímetro cúbico
Milímetro cúbico é uma unidade SI de volume. É igual a 0.000001 litro.
## Jarda cúbica
Jarda cúbico é uma unidade tradicional de volume, em países que falam Inglês. É igual a 27 pés cúbicos ou 46.656 polegadas cúbicas ou 0.764555 metros cúbicos, ou 764.554861215584 litros.
## Decilitro
Decilitro é uma unidade de medida comum de volume. É igual a 0.1 litro, ou 100 centímetros cúbicos.
## Decalitro
Decalitro é uma unidade métrica de volume. É igual a 10 litros.
## Hectolitro
Hectolitro é uma unidade de medida comum de volume. É igual a 100 litros, 0.1 metros cúbicos, 26.417 galões líquidos EUA, 21.999 galões imperiais britânicos, ou 3.5315 metros cúbicos.
## Quilolitro
Quilolitro é uma unidade métrica de volume. Quilolitro é idêntico ao metro cúbico. É igual a 35.3147 metros cúbicos, ou 1000 litros.
## Litro
Litro é a unidade comum métricas de volume. A unidade está escrito liter em litre e os EUA na Grã-Bretanha.
## Megalitro
Megalitro é uma unidade métrica de volume. É igual a 1000 metros cúbicos ou 1000000 litros.
## Copo métrica
Copo Metric é uma unidade informal métricas de volume. É igual a 0.25 litros, comumente utilizados em receitas, na Austrália.
## Colher de sopa métrica
Colher de sopa é uma unidade de volume usado em receitas de comida. É igual a 15 ml na Grã-Bretanha, Canadá e Nova Zelândia, 20 ml, na Austrália.
## Colher de chá métrica
Colher de chá é uma unidade de volume usado em receitas de comida. Nas cozinhas métrica na Grã-Bretanha, Canadá, Austrália e Nova Zelândia, uma colher de chá é exactamente 5 ml.
## Microlitro
Microlitro é uma unidade métrica de volume utilizada em química e medicina para medir pequenas quantidades de líquido. É igual a 0.001 ml ou 1 milímetro cúbico.
## Mililitro
Mililitro é uma unidade muito comum métricas de volume. É igual a 0.001 litros, exatamente um centímetro cúbico.
## UK barril
Barril é uma unidade de volume. No Reino Unido, é igual a 163.65924 litros.
## UK galão
O galão imperial é projetado para conter exatamente 10 quilos de água destilada, sob condições definidas com precisão. É o mesmo que 4.54609 litros.
## UK onça líquido
Reino Unido onça líquida é uma unidade tradicional de volume líquido. É também chamada de onça fluida para evitar confusão com a onça de peso. É igual a 0.0284130625 litros.
## UK pinta
Pinta Reino Unido é uma unidade de volume. É igual a 0.56826125 litro.
## UK quarto
Quarto Reino Unido é uma unidade de volume. É igual a 1.1365225 litros.
## UK colher de sopa
Colher de sopa é uma unidade de volume usado em receitas de comida. Na Grã-Bretanha, é igual a 14.20653125 mililitros.
## UK colher de chá
Colher de chá é uma unidade de volume usado em receitas de comida. Na Grã-Bretanha, uma colher de chá tradicional da cozinha era igual a 1 / 8 onças Imperial fluido ou cerca de 3.55 ml, mas a médica foi teaspoonful geralmente 5 mililitros.
## UK barril de vinho
Barril é uma unidade de volume. No Reino Unido, um barril de vinho é igual a 143.201835 de litros de vinho.
## US copo
Copo dos EUA é uma unidade tradicional de volume usado em receitas nos Estados Unidos. É igual a 1 / 2 litro de líquido, ou 8 onças fluidas ou 0.2365882375 litros.
## US barril seco
Barril é uma unidade de volume. Nos EUA, o barril seco é igual a 115.627124072727 litros.
## US galão seco
EUA galão seco é uma unidade histórica britânica de volume seco ainda usado implicitamente no galão de EUA EUA seco é igual a 4.4048838 litros.
## US pinta seco
EUA pinta seca é uma unidade de volume. É igual a 0.550610475 litros.
## US quarto seco
EUA quarto seca é uma unidade de volume. É igual a 1.10122095 litros.
## US barril federal
Barril é uma unidade de volume. Nos EUA, o barril federal é igual a 117.3477658 litros.
## Galão
O galão (abreviação: gal) é uma unidade de medida de volume de líquidos, utilizada na comunidade anglo-saxónica.
## US barril líquido
Barril é uma unidade de volume. Nos EUA, o barril de líquido é igual a 119.2404717 litros.
## US galão líquido
EUA galão de líquido é uma unidade tradicional de volume líquido. É igual a 4 quarts líquido ou exatamente 3.785411784 litros.
## US onça líquido
EUA onça líquida é uma unidade tradicional de volume líquido. É também chamada de onça fluida para evitar confusão com a onça de peso. É igual a 0.0295735296875 litros.
## US pinta líquido
EUA pinta de líquido é uma unidade de volume. É igual a 0.473176475 litros.
## US quarto líquido
EUA quarto líquido é uma unidade de volume. É igual a 0.94635295 litro.
## US barril petróleo
Barril é uma unidade de volume. Nos EUA, o barril de petróleo ou derivados do petróleo é igual a 158.9872956 litros.
## US colher de sopa
Em os EUA, a colher de sopa é igual a 1 / 2 onça fluida. Trata-se de 14.8 mililitros.
## US colher de chá
Colher de chá é uma unidade de volume usado em receitas de comida. A colher de chá EUA é igual a 1 / 3 colher de sopa ou 1 / 48 xícara. É cerca de 4.9 mililitros.
## Acre fuß
Acre-fuß wird häufig in Bezug auf groß angelegte Wasserressourcen einschließlich Kanälen und Flussläufen verwendet.
In Kanada ist eine Tasse gleich 8 Imperial flüssige Unzen (oder 0.2273045 Liter).
## Zentiliter
Die zentiliter ist eine gemeinsame metrische Einheit des Volumens. Es ist gleich 10 Kubikzentimetern. In der Küche ist ein Centiliter etwa gleich 2 US Teelöffel.
## Kubikzentimeter
Kubikzentimeter ist ein CGS-Einheit des Volumens. Es ist gleich 10E-6 Kubikmeter, 1 Milliliter oder etwa 0.0610237 Kubikzoll.
## Kubikdezimeter
Kubikdezimeter ist eine Einheit der Lautstärke. Es ist gleich 1 Liter.
## Kubikdekameter
Kubikdekameter ist eine Einheit der Lautstärke. Es ist gleich zu 1000000 Liter.
## Kubikfuß
Kubikfuß ist eine traditionelle Einheit des Volumens in den englischsprachigen Ländern. Es ist gleich bis 1728 Kubikmeter vinzen oder 28.316 85 Liter.
## Kubikzoll
Kubikzoll ist eine traditionelle Einheit des Volumens in den englischsprachigen Ländern. Es ist gleich 0.0163870640692641 Kubikzentimeter oder Milliliter 0.0163870640692641.
## Kubikkilometer
Kubikkilometer ist eine Einheit der Lautstärke. Es ist gleich 1000000000000 Liter.
## Kubikmeter
Kubikmeter ist ein SI-Einheit des Volumens. Es ist gleich 1000 Liter. Ein Kubikmeter beträgt ca. 264.17 US-Gallonen oder 219.99 Flüssigkeit britischen Imperial Gallonen.
## Kubikmikrometer
Kubikmikrometer ist eine Einheit der Lautstärke. Es ist gleich 0.000000000000001 Liter.
## Kubikmeile
Kubikmeile ist eine Einheit der Lautstärke. Es ist gleich 4168181843058.45 Liter.
## Kubikmillimeter
Kubikmillimeter ist eine SI-Einheit des Volumens. Es ist gleich 0.000001 Liter.
## Kubikyard
Kubikyard ist eine traditionelle Einheit des Volumens in den englischsprachigen Ländern. Es ist gleich 27 Kubikfuß oder 46656 cubic inches oder 0.764555 Kubikmeter oder 764.554861215584 Liter.
## Deziliter
Deziliter ist eine gemeinsame metrische Einheit des Volumens. Es ist gleich 0.1 Liter bzw. 100 Kubikzentimeter.
## Dekaliter
Dekaliter ist eine metrische Einheit des Volumens. Es ist gleich 10 Liter.
## Hektoliter
Hektoliter ist eine gemeinsame metrische Einheit des Volumens. Es ist gleich 100 Liter, 0.1 Kubikmeter, 26.417 US Flüssigkeit Gallonen, 21.999 britische Imperial Gallonen oder 3.5315 Kubikfuß.
## Kiloliter
Kiloliter ist eine metrische Einheit des Volumens. Kiloliter ist identisch mit der Kubikmeter. Es ist gleich 35.3147 Kubikfuß oder 1000 Liter.
## Liter
Liter ist der gemeinsame metrische Einheit des Volumens. Das Gerät ist Dinkel Liter in den USA und in Großbritannien Liter.
## Megaliter
Megaliter ist eine metrische Einheit des Volumens. Es ist gleich 1000 Kubikmeter oder 1000000 Liter.
## Metrische tasse
Metric Tasse ist ein informelles metrische Einheit des Volumens. Es ist gleich 0.25 Liter, die gemeinhin in Rezepte in Australien verwendet.
## Metrischen esslöffel
Esslöffel ist eine Einheit des Volumens in Essen Rezepte verwendet. Es ist gleich 15 Milliliter in Großbritannien, Kanada und Neuseeland, 20 Milliliter in Australien.
## Metrischen teelöffel
Teelöffel ist eine Einheit des Volumens in Essen Rezepte verwendet. In metrischen Küchen in Großbritannien, Kanada, Australien und Neuseeland, ist ein Teelöffel genau 5 Milliliter.
## Mikroliter
Mikroliter ist eine metrische Einheit des Volumens in der Chemie und der Medizin eingesetzt, um sehr kleine Mengen Flüssigkeit zu messen. Es ist gleich 0.001 Milliliter oder 1 Kubikmillimeter.
## Milliliter
Milliliter ist eine sehr häufige metrische Einheit des Volumens. Es ist gleich 0.001 Liter, genau ein Kubikzentimeter.
## UK barrel
Barrel ist eine Einheit der Lautstärke. In Großbritannien ist es gleich 163.65924 Liter.
## UK gallone
Die kaiserliche Gallone entworfen, um genau 10 Pfund destilliertem Wasser unter genau festgelegten Bedingungen enthalten. Es ist die gleiche wie 4.54609 Liter.
## UK flüssigkeit unze
UK Flüssigkeit Unze ist eine traditionelle Einheit von Flüssigkeitsvolumen. Es ist auch die Flüssigunzen zu Verwechslungen mit dem Gewicht Unzen vermeiden genannt. Es ist gleich 0.0284130625 Liter.
## UK pinte
UK Pint ist eine Einheit der Lautstärke. Es ist gleich 0.56826125 Liter.
## UK quart
UK Quart ist eine Einheit der Lautstärke. Es ist gleich 1.1365225 Liter.
## UK esslöffel
Esslöffel ist eine Einheit des Volumens in Essen Rezepte verwendet. In Großbritannien ist es gleich 14.20653125 Milliliter.
## UK teelöffel
Teelöffel ist eine Einheit des Volumens in Essen Rezepte verwendet. In Großbritannien wurde ein traditionelles Teelöffel in der Küche gleich 1 / 8 Imperial Flüssigunze oder etwa 3.55 ml, aber die medizinische Teelöffel war in der Regel 5 Milliliter.
## UK weinfass
Barrel ist eine Einheit der Lautstärke. In Großbritannien ist ein Faß Wein gleich 143.201835 Liter Wein.
## US tasse
US-Tasse ist eine traditionelle Einheit des Volumens in den Rezepten in den Vereinigten Staaten verwendet. Es ist gleich 1 / 2 Pint Flüssigkeit oder 8 Flüssigunzen oder 0.2365882375 Liter.
## US trocken barrel
Barrel ist eine Einheit der Lautstärke. In den USA ist ein trockener Lauf gleich 115.627124072727 Liter.
## US trocken gallone
US trocken Gallone eine historische britische Einheit des trockenen Volumen noch implizit in den USA US-Gallonen trocken eingesetzt ist gleich 4.4048838 Liter.
## US trocken pinte
US trocken Pint ist eine Einheit der Lautstärke. Es ist gleich 0.550610475 Liter.
## US trocken quart
US trocken Quart ist eine Einheit der Lautstärke. Es ist gleich 1.10122095 Liter.
## US bundes barrel
Barrel ist eine Einheit der Lautstärke. In den USA ist ein föderaler Barrel gleich 117.3477658 Liter.
## Gallone
Die Gallone (englisch gallon) ist eine Raumeinheit (Trocken-, Flüssigkeitsmaß, siehe Raummaß). Es gibt unterschiedliche Definitionen der Gallone in den Maßsystemen verschiedener Länder.
## US flüssigkeit barrel
Barrel ist eine Einheit der Lautstärke. In den USA ist ein flüssiger Lauf gleich 119.2404717 Liter.
## US flüssigkeit gallone
US-Gallone Flüssigkeit eine traditionelle Einheit von Flüssigkeitsvolumen. Es ist gleich 4 Liter Flüssigkeit oder genau 3.785411784 Liter.
## US flüssigkeit unze
US Flüssigkeit Unze ist eine traditionelle Einheit von Flüssigkeitsvolumen. Es ist auch die Flüssigunzen zu Verwechslungen mit dem Gewicht Unzen vermeiden genannt. Es ist gleich 0.0295735296875 Liter.
## US flüssigkeit pinte
US Flüssigkeit Pint ist eine Einheit der Lautstärke. Es ist gleich 0.473176475 Liter.
## US flüssigkeit quart
US Flüssigkeit Quart ist eine Einheit der Lautstärke. Es ist gleich 0.94635295 Liter.
## US barrel erdöl
Barrel ist eine Einheit der Lautstärke. In den USA ist ein Barrel Öl oder Petroleum gleich 158.9872956 Liter.
## US esslöffel
In den USA ist der Esslöffel gleich 1 / 2 Flüssigunzen. Es ist etwa 14.8 Milliliter.
## US teelöffel
Teelöffel ist eine Einheit des Volumens in Essen Rezepte verwendet. Die US Teelöffel ist gleich 1 / 3 Esslöffel oder 1 / 48 Tasse. Es ist etwa 4.9 Milliliter. | 14,746 | 46,283 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2021-39 | latest | en | 0.607367 |
https://www.physicsforums.com/threads/first-order-non-linear-ode-problem.402539/ | 1,544,710,734,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824822.41/warc/CC-MAIN-20181213123823-20181213145323-00514.warc.gz | 993,406,961 | 16,687 | # Homework Help: First order, non linear ode problem
1. May 11, 2010
### trix312
1. The problem statement, all variables and given/known data
Finding the general solution of y'= (2y^2)/e^x
and the particular given that y(0) = 0
3. The attempt at a solution
I seperate variables etc. to get y = -1/(2c-2e^(-x)) where c is the constant. The problem then is obtaining the particular solution. If I substitute y and x for 0 i am left with 0= -1/(2c-2) which is unsolvable. what am i doing wrong ?
2. May 11, 2010
### Dick
Your equation is singular at y=0. There's another obvious solution you are missing.
3. May 11, 2010
### trix312
Ok.. I also get this as a solution y = -e^x/(ce^x-2) but this is also singular at y = 0.. i cant see the other obvious solution.
4. May 11, 2010
### Dick
That's the same as your first solution. If an ode is singular, it may have more than one solution. So there may be another one besides the one you found. How about y is identically equal to zero?
5. May 11, 2010
### trix312
yeah, i dont understand what you mean when you say that its singular. Does that mean the particular solution does not exist?
6. May 11, 2010
### Je m'appelle
The way I see it, you're confusing first-order ODEs with second-order ODEs.
In a first-order ODE, there is no particular solution, but instead a general solution, which is just a family of functions for different "c"s, that is, constants.
Once you're given an IVP (Initial Value Problem), then you'll find a value for the constant "c", which leads to a unique solution to your first-order ODE.
Don't use the term "particular solution" for first-order ODEs, use "unique solution" instead, that's because in second-order ODEs, you generally have two solutions called "complementary solution" and "particular solution", so be careful with the nomenclatures, as that can make a big confusion.
For example, your first-order ODE is
$$y' = \frac{2y^2}{e^x}$$
You said that you separated the variables, so that leads us to
$$\frac{dy}{y^2} = \frac{2}{e^x}dx$$
Which can also be rewritten as
$$\frac{dy}{y^2} = 2e^{-x} dx$$
So what do you do now? You integrate both sides
$$\int \frac{dy}{y^2} = \int 2e^{-x} dx$$
$$\frac{-1}{y} = -2e^{-x} + c$$
ATTENTION: Always remember your constants, after the integration process.
So I believe you can take it from here. Also, don't forget that the problem asks for a specific solution, that is, a unique solution, and not the general solution, so that means you'll have to solve for "C", and that can be done by using your initial value problem, which was given.
Last edited: May 11, 2010
7. May 11, 2010
### trix312
thanks for the detailed explanation. So I need to find the unique solution, and you have taken me to the part were i fail to find it. This is my blockade. The IVP is when y(0) = 0 and when i try to solve for "c" it is not possible. Uniqueness fails. I think this is what Dick was trying to tell me with y being identically equal to zero. Although I dont see exactly why that is so.
8. May 11, 2010
### l'Hôpital
When you separate variables, you are assuming y is not zero. Otherwise, you couldn't divide by y.
9. May 11, 2010
### Dick
I think I'm using 'singular' in the wrong way, but the solution y(x)=0 for all x solves your equation. In that case you can't divide both sides of your equation by y^2 and use separation of variables. Because y=0!
10. May 11, 2010
### Je m'appelle
Hmm, that's true, it can't be done by separation of variables then.
But is there a solution then?
11. May 11, 2010
### trix312
Ok. I think I sort of got the hang of what is happening in this problem. Yeah, there is no other possibility except y(x)= 0 . It is the only solution that would work for the given IVP. Thanks,
12. May 11, 2010
### Dick
y(x)=0. Isn't that a solution?
13. May 11, 2010
### Je m'appelle
trick problem
14. May 11, 2010
### Staff: Mentor
And I think you are confusing homogeneous differential equations with their nonhomogeneous counterparts.
Let's look at a simple example.
Homogeneous: y' - 2y = 0
Nonhomogeneous: y' - 2y = x
For the homogeneous equation, it's easy to see that the general solution is y = Ae2x. If this had been an initial value problem we could use the initial condition to solve for A.
For the nonhomogeneous equation, a particular solution is yp = -x. As before, the complementary solution is yc = Ae2x, so the general solution is y = yc + yp = Ae2x - x. Given an initial condition we could solve for A, arriving at a unique solution to the nonhomogeneous problem.
15. May 11, 2010
### Je m'appelle
You're absolutely correct Mark. What I meant was in fact homogeneous and nonhomogeneous equations just as you said, and somehow mislabelled it as first-orders and second-orders. Thanks for correcting me up.
16. May 11, 2010
### vela
Staff Emeritus
Actually, what particular solution means depends on the context. Some texts use the term to mean the solution to the IVP, as opposed to the general solution, and it's in this sense that the OP has used it. I don't see it as any sign of confusion on his or her part. | 1,426 | 5,121 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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} | 3.75 | 4 | CC-MAIN-2018-51 | latest | en | 0.916994 |
https://people.richland.edu/james/fall08/m116/dvds.html | 1,513,067,001,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948515311.25/warc/CC-MAIN-20171212075935-20171212095935-00354.warc.gz | 618,079,715 | 2,470 | # Math 116 - DVDs
There are DVDs to accompany College Algebra: A Graphing Approach, 4th edition, on reserve in the Kitty Lindsay Learning Resources Center (LRC). You must watch these videos in the LRC, you may not check them out. We are currently using the 5th edition of the textbook, but very little has changed.
The table below provides a guideline to let you know which DVD corresponds to each section and how long (in minutes) the lesson is.
Section DVD
#
Length
(Min)
Section Title
P.1 1 31 Real Numbers
P.2 1 46 Exponents and Radicals
P.3 1 35 Polynomials and Factoring
P.4 1 43 Rational Expressions
P.5 1 29 The Cartesian Plane
P.6 1 19 Exploring Data: Representing Data Graphically
1.1 2 30 Graphs of Equations
1.2 2 38 Lines in the Plane
1.3 2 32 Functions
1.4 2 33 Graphs of Functions
1.5 2 30 Shifting, Reflecting, and Stretching Graphs
1.6 2 9 Combinations of Functions
1.7 2 27 Inverse Functions
2.1 3 72 Modeling with Linear Equations
2.2 3 19 Solving Equations Graphically
2.3 3 23 Complex Numbers
2.4-2.5 3 66 Solving Equations Algebraically (2.4 on video)
2.6 4 41 Solving Inequalities Algebraically and Graphically (2.5 on video)
2.7 4 32 Exploring Data: Linear Models and Scatter Plots (2.6 on video)
3.2 4 31 Polynomial Functions of Higher Degree
3.3 4 37 Real Zeros of Polynomial Functions
3.4 4 19 The Fundamental Theorem of Algebra
3.5 4 11 Rational Functions and Asymptotes
3.6 4 24 Graphs of Rational Functions
3.7 4 10 Exploring Data: Quadratic Models
4.1 5 28 Exponential Functions and Their Graphs
4.2 5 35 Logarithmic Functions and Their Graphs
4.3 5 26 Properties of Logarithms
4.4 5 33 Solving Exponential and Logarithmic Equations
4.5 5 48 Exponential and Logarithmic Models
4.6 5 20 Exploring Data: Nonlinear Models
5.1 6 25 Solving Systems of Equations
5.2 6 13 Systems of Linear Equations in Two Variables
5.3 6 49 Multivariable Linear Systems
5.4 6 52 Matrices and Systems of Equations
5.5 6 17 Operations with Matrices
5.6 6 32 The Inverse of a Square Matrix
5.7 7 24 The Determinant of a Square Matrix
5.8 7 30 Applications of Determinants and Matrices
6.1 7 23 Sequences and Series
6.2 7 18 Arithmetic Sequences and Partial Sums
6.3 7 30 Geometric Sequences and Series
6.4 7 24 Mathematical Induction
6.5 7 28 The Binomial Theorem
6.6 7 24 Counting Principles
6.7 8 28 Probability
7.1-7.3 8 71 Conics (7.1 on video is untranslated conics)
7.1-7.3 8 51 Translations of Conics (7.2 on video is translated conics)
7.4 8 31 Parametric Equations (7.3 on video) | 832 | 2,499 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2017-51 | latest | en | 0.76119 |
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Question
# The shaded region consists of a chord and the arcs of the circle. Which of the following option is correct?
A
Yellow – Major segment, Purple – Minor segment
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Purple – Major segment, Yellow – Minor segment
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C
Yellow – Major arc, Purple – Minor arc
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Yellow – Minor arc, Purple – Major arc
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Solution
## The correct option is A Yellow – Major segment, Purple – Minor segment The shaded region consists of a chord and the arcs of the circle. Hence they are segments. The region in the yellow has the major arc so it is the major segment and the purple region consists of the minor arc so it is the minor segment. Also, the region in yellow has the centre of the circle.
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Special Issues
Volume 9, Issue 1, February 2020, Page: 16-25
Strong Fuzzy Chromatic Polynomial (SFCP) of Fuzzy Graphs and Some Fuzzy Graph Structures with Applications
Mamo Abebe Ashebo, Department of Mathematics, Wollega University, Nekemte, Ethiopia
Venkata Naga Srinivasa Rao Repalle, Department of Mathematics, Wollega University, Nekemte, Ethiopia
Received: Dec. 6, 2019; Accepted: Dec. 24, 2019; Published: Jan. 23, 2020
Abstract
In fuzzy graph theory, strong arcs have separate importance. Assign different colors to the end nodes of strong arcs in the fuzzy graph is strong coloring. Strong coloring plays an important role in solving real-life problems that involve networks. In this work, we introduce the new concept, called strong fuzzy chromatic polynomial (SFCP) of a fuzzy graph based on strong coloring. The SFCP of a fuzzy graph counts the number of k-strong colorings of a fuzzy graph with k colors. The existing methods for determining the chromatic polynomial of the crisp graph are used to obtain SFCP of a fuzzy graph. We establish the necessary and sufficient condition for SFCP of a fuzzy graph to be the chromatic polynomial of its underlying crisp graph. Further, we study SFCP of some fuzzy graph structures, namely strong fuzzy graphs, complete fuzzy graphs, fuzzy cycles, and fuzzy trees. Besides, we obtain relations between SFCP and fuzzy chromatic polynomial of strong fuzzy graphs, complete fuzzy graphs, and fuzzy cycles. Finally, we present dual applications of the proposed work in the traffic flow problem. Once SFCP of a fuzzy graph is obtained, the proposed approach is simple enough and shortcut technique to solve strong coloring problems without using coloring algorithms.
Keywords
Fuzzy Graph, Strong Coloring, Strong Fuzzy Chromatic Polynomial, Strong Fuzzy Graph, Complete Fuzzy Graph, Fuzzy Cycle, Fuzzy Tree, Traffic Flow Problems
Mamo Abebe Ashebo, Venkata Naga Srinivasa Rao Repalle, Strong Fuzzy Chromatic Polynomial (SFCP) of Fuzzy Graphs and Some Fuzzy Graph Structures with Applications, Pure and Applied Mathematics Journal. Vol. 9, No. 1, 2020, pp. 16-25. doi: 10.11648/j.pamj.20200901.13
Reference
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Binu, M., Mathew, S., and Mordeson, J. N., Wiener index of a fuzzy graph and application to illegal immigration networks. Fuzzy Sets Syst. In Press. 2019.
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[39]
Nagoorgani, A., Isomorphism properties on strong fuzzy graphs. Int. J. Algorithm Comput. Math. vol. 2, No. 1, pp. 39-47, 2009. | 2,522 | 7,419 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-50 | latest | en | 0.885289 |
https://lexique.netmath.ca/en/measurement-conversion/ | 1,723,145,305,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640740684.46/warc/CC-MAIN-20240808190926-20240808220926-00127.warc.gz | 299,219,311 | 13,956 | # Measurement Conversion
## Measurement Conversion
The set of operations or tables used to determine the equivalence between various unit of measurement systems or between various multiples or submultiples of the same unit.
### Examples
• To convert measurements in centimetres into metres, divide the number of centimetres by 100: 342 cm = (342 ÷ 100) m = 3.42 m.
• To convert degrees Celsius (°C) into degrees Fahrenheit (F), use the formula $$F=\frac{9}{5}\times C+32$$ | 117 | 476 | {"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} | 3.453125 | 3 | CC-MAIN-2024-33 | latest | en | 0.655949 |
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# Software Metrics using Constructive Cost Model. Describing Cocomo Model
Projektarbeit 2019 9 Seiten
## TABLE OF CONTENTS
ABSTRACT
INTRODUTION
CONSTRUCTIVE COST MODEL
COCOMO MODELS
Basic COCOMO
INTERMEDIATE COCOMO
ADVANCED COCOMO
LIMITATIONS OF COCOMO
COST ESTIMATION ACCURACY
COCOMO II 5
COCOMO II CALCULATION
COCOMO II USES
CONCLUSION
REFERENCES
## ABSTRACT:
Software products are said to be feasible if they are developed within the budget constraints. The cost estimation models are used to predict the effort and cost required to develop a project .These models give a base to predict the cost for developing a software project. The cost estimation can be used to develop a product utilizing optimum resources. In this paper we will be discussing Constructive Cost Model as the cost estimation model.
## INTODUCTION:
Software cost estimation model is an indirect measure, which is used by software personnel to predict the cost of a project. The development of software product varies depending upon the environment in which it is being developed. For projects with familiar environment it is easy to predict the cost of the project. The estimation model is useful for trade off between the developer and customer. Organization can realize of what is achievable and deliverable to the customer.[1]
For the organization to develop a cost estimation model the following things are required.
- List important or critical cost drivers.
- Prepare a scaling model for each cost driver.
- Find projects with similar environments.
- Compare the project with previous familiar projects.
- Evaluate the project whether it is feasible within the budget constraints.
- Incorporate the critical features in an iterative manner.[1]
Cost drivers are those critical features which have an impact on the project. The cost drivers may vary the cost of building a project. The most important cost driver is size of the project. Size of the project is measured in Kilo lines of code (KLOC). Function points are the empirical measurement to measure size of the project. Function points may vary the size of the project due to the variation in
- Number of inputs
- Number of outputs
- Number of inquires
- Number of files
- Number of interfaces[1]
## CONSTRUCTIVE COST MODEL (COCOMO):
Constructive Cost model was developed by Barry W Boehm in 1981[3]. It is an algorithmic cost model. Algorithmic cost model is developed based on relating the current project to previous projects. It is based on historical information. [2]
Cocomo is based on size of the project. The size of the project may vary depending upon the function points. [1]
## COCOMO MODELS:
### - BASIC COCOMO
- It is used for relatively small project.
- Only a few cost drivers are associated.
- Cost drivers depend upon project size mainly.
- Useful when the team size is small, i.e. small staff.
The effort (E) and schedule (S) of the project are calculated as follows
- Effort E = a * (KDSI) b * EAF Where KDSI is number of thousands of delivered source instructions a and b are constants, may vary depending on size of the project .[2]
- Schedule S= c * (E) d where E is the Effort and c, d are constants. [1]
- EAF is called Effort Adjustment Factor which is 1 for basic cocomo , this value may vary from 1 to 15.[3]
Abbildung in dieser Leseprobe nicht enthalten
Table 1 [3]
The above table 1 shows typical constant values that have been calculated from various projects.
The basic cocomo gives the magnitude of cost of the project. It varies depending upon size of the project. The various classes of software projects are
- Organic mode projects :
- Used for relatively smaller teams.
- Project is developed in familiar environment.
- There is a proper interaction among the team members and they coordinate their work.
- Bohem observed E=2.4(KDSI)[1].[05] E in person-months.
- And S=2.5(E)[0].[38]. [2]
- Semi-detached mode projects :
- It lies between organic mode and embedded mode in terms of team size.
- It consists of experienced and inexperienced staff.
- Team members are unfamiliar with the system under development.
- Bohem observed E=3(KDSI)[1].[12] E in person-months.
- And S=2.5(E)[0].[35].[2]
- Embedded mode projects :
- The project environment is complex.
- Team members are highly skilled.
- Team members are familiar with the system under development.
- Bohem observed E=3.6(KDSI)[1].[20] E in person-months.
- And S=2.5(E)[0].[32].[2]
### - Intermediate COCOMO
- It is used for medium sized projects.
- The cost drivers are intermediate to basic and advanced cocomo.
- Cost drivers depend upon product reliability, database size, execution and storage.[2]
- Team size is medium.
### - Advanced COCOMO
- It is used for large sized projects.
- The cost drivers depend upon requirements, analysis, design, testing and maintenance.
- Team size is large.
## LIMITATIONS OF COCOMO
- COCOMO is used to estimate the cost and schedule of the project, starting from the design phase and till the end of integration phase. For the remaining phases a separate estimation model should be used.
- COCOMO is not a perfect realistic model. Assumptions made at the beginning may vary as time progresses in developing the project.
- When need arises to revise the cost of the project. A new estimate may show over budget or under budget for the project. This may lead to a partial development of the system, excluding certain requirements.
- COCOMO assumes that the requirements are constant throughout the development of the project; any changes in the requirements are not accommodated for calculation of cost of the project.
- There is not much difference between basic and intermediate COCOMO, except during the maintenance and development of the software project.[1]
- COCOMO is not suitable for non-sequential ,rapid development, reengineering ,reuse cases models.[3]
## COST ESTIMATION ACCURACY
The cost estimation may vary due to changes in the requirements, staff size, and environment in which the software is being developed.
The calculation for cost estimation accuracy is given as follows
Absolute error= (Epred - Eactual)
Percentage error= (Epred - Epred)/Epred
Relative error= 1/n ∑ (Epred - Epred)/Epred
The above results give a more accurate estimation of costs for future projects.[3] The cost estimation model now becomes more realistic .
## COCOMO II
- COCOMOII was developed in 1995
- It could overcome the limitations of calculating the costs for non-sequential, rapid development, reengineering and reuse models of software.
- It has 3 modules
- Application composition: - good for projects with GUI interface for rapid development of project.
- Early design: - Prepare a rough picture of what is to be designed. Done before the architecture is designed.
- Post architecture: - Prepared after the architecture has been designed. [3]
## COCOMO II calculation
In COCOMO II the constant value b is replaced by 5 scale factors. [3]
- Effort (E) is calculated as follows
E = a * (KDSI) sf * π (EM)
Where a is constant, sf is scaling factor, EM is Effort Multiplier (7 for Early design, 17 for Post architecture). [4]
## COCOMO II USES
- Helps in making decisions based on business and financial calculations of the project.
- Establishes the cost and schedule of the project under development, this provides a plan for the project.
- Provides a more reliable cost and schedule, hence the risk mitigation is easy to accomplish.
- It overcomes the problem of reengineering and reuse of software modules.
- Develops a process at each level . Hence takes care of the capability maturity model.[5]
## CONCLUSION:
Constructive Cost Model developed by Barry W Boehm, is the most common and widely used cost estimation models for most software projects. The effort and schedule calculated by the model is based on two things, historical information and experience. Thus the reliability on cocomo has been increased. The website provided by NASA on cocomo, provides a cocomo calculator with cost drivers for a complex project. Cost drivers directly have an impact on the development of the project.
## References:
[1] Farshad Faghih,” Software Effort and Schedule Estimation”, http://www2.enel.ucalgary.ca/People/Smith/619.94/prev689/1997.94/reports/farshad.htm
[2] IAN SOMMERVILLE,” Software Engineering”, published by addision wesley, pg 514- 521.
[3]Seth Bowen, Samuel Lee, Lance Titchkosky,”Software cost estimation”,
http://www.computing.dcu.ie/~renaat/ca421/BLTCostEst.ppt
[4]Barry Boehm, “Cost Estimation With COCOMO II”, http://sunset.usc.edu/classes/cs577a_2002/lectures/19/ec19.pdf
[5]Center for Systems and Software engineering , “ COCOMO II ”, http://sunset.usc.edu/csse/research/COCOMOII/cocomo_main.html.
## Details
Seiten
9
Jahr
2019
Sprache
Englisch
Katalognummer
v457410
Note
10
Schlagworte
cocomo model advantages of cocomo model limitations of cocomo model | 2,084 | 8,933 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2019-43 | latest | en | 0.871923 |
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HW3_14 - Homework due 3/14 Calculate the pH of 0.20 L of an...
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Homework due 3/14 Calculate the pH of 0.20 L of an NH 3 -NH 4 NO 3 buffer that is 0.40 M in NH 4 NO 3 and 0.50 M in NH 3, a) initially, b) after 0.050 mole of HNO 3 have been added, c) after 0.040 moles KOH have been added to a fresh 0.20 L of the buffer. (K b for NH 3 = 1.8x10 -5 ) a) initially The first problem is to simply calculate the pH of the buffer as given. This can be done by substituting into the Henderson-Hasselbalch equation, where the acid form is NH 4 1+ (from the NH 4 NO 3 ), and the base form is NH 3 . NH 4 NO 3 NH 4 1+ + NO 3 1- {NO 3 1- is a spectator ion since it is from the strong acid HNO 3 } NH 3 + H 2 O NH 4 1+ + OH 1- or NH 4 1+ NH 3 + H 1+ pOH = pK b + log ([NH 4 1+ ]/[NH 3 ]) = -log 1.8x10 -5 + log (0.40/0.50) = pOH = 4.74 + (-0.10) = 4.64 pH = 14 - pOH = 14 - 4.64 = 9.36 Alternatively, pH = pK a + log [NH 3 ]/[NH 4 1+ ] pK a = 14 - pK b = 14 - 4.74 = 4.26 pH = 9.26 + log (0.50/0.40) = 4.26 + 0.10 = 9.36 b) After the addition of 0.050 mole HNO 3 . The H
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Ask a homework question - tutors are online | 601 | 1,605 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-09 | latest | en | 0.84637 |
https://countrymusicstop.com/how-to-simplify-2-2-update-new/ | 1,660,369,599,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571909.51/warc/CC-MAIN-20220813051311-20220813081311-00787.warc.gz | 187,045,364 | 21,335 | Home » How To Simplify 2 2? Update New
# How To Simplify 2 2? Update New
Let’s discuss the question: how to simplify 2 2. We summarize all relevant answers in section Q&A of website Countrymusicstop.com in category: MMO. See more related questions in the comments below.
## How do you simplify fractions step by step?
Here are the steps to follow:
1. Write down the factors for the numerator and the denominator.
2. Determine the largest factor that is common between the two.
3. Divide the numerator and denominator by the greatest common factor.
4. Write down the reduced fraction.
## What is 2 2 as a whole number?
Yes! We can divide the numerator (2) by the denominator (2) without any remainder. Yes, 2/2 is equal to the whole number 1!
### Simplifying Fractions
Simplifying Fractions
Simplifying Fractions
## How do you simplify in maths?
To reduce a fraction to its lowest terms by canceling to the lowest common factor for both numerator and denominator or to condense an algebraic expression by grouping and combining similar terms. Simplifying makes a algebric expression easily understandable and solvable.
## What is simplified?
to make less complex or complicated; make plainer or easier: to simplify a problem.
## Why do we simplify fractions?
We simplify fractions because it is always to work or calculate when the fractions are in the simplest form.
### Simplify 2m^2*2m^3
Simplify 2m^2*2m^3
Simplify 2m^2*2m^3
## What does simplify mean in fractions?
The word simplify means to make something easier to do or understand. So, reducing or simplifying fractions means we make the fraction as simple as possible. We do this by dividing the numerator and the denominator by the largest number that can divide into both numbers exactly.
## How do you simplify odd fractions?
To simplify an improper fraction, start by turning it into a mixed number by dividing the numerator by the denominator. Then, turn the remainder into a fraction by placing it over the denominator of the original fraction. If necessary, simplify the final fraction to get your answer.
## How do you change a mixed number to a fraction?
Calculator Use
Follow these 3 steps to convert mixed numbers to improper fractions: Multiply the whole number by the denominator. Add the answer from Step 1 to the numerator. Write answer from Step 2 over the denominator.
## What is 2 as a fraction?
Answer: 2% can be represented as 1/50 in fractional form.
### How to simplify (x-2)^2
How to simplify (x-2)^2
How to simplify (x-2)^2
## How do you write and simplify?
To simplify any algebraic expression, the following are the basic rules and steps:
1. Remove any grouping symbol such as brackets and parentheses by multiplying factors.
2. Use the exponent rule to remove grouping if the terms are containing exponents.
3. Combine the like terms by addition or subtraction.
4. Combine the constants.
## What is an example of simplify?
Simplify is to make something less complicated or less cluttered. An example of simplify is when you explain a tough math concept in really easy terms for a child to understand. An example of simplify is when you cut out a lot of the activities that were making you busy and stressed.
See also How To Install Map On Toyota Rav4 2021? Update New
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You have just come across an article on the topic how to simplify 2 2. If you found this article useful, please share it. Thank you very much. | 1,011 | 4,095 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.90625 | 5 | CC-MAIN-2022-33 | latest | en | 0.903889 |
https://forum.exercism.org/t/scrabble-dig-deeper-approaches/7231 | 1,702,156,193,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100972.58/warc/CC-MAIN-20231209202131-20231209232131-00651.warc.gz | 287,296,112 | 8,884 | # Scrabble Dig Deeper approaches
The dig deeper section on the exercises are a great way checking and understanding different approaches to the problems.
On the Scrabble exercise, the dig deeper solutions go over two possible ways to solve it, using a map, or using switch case, with the switch case being the faster one.
I was wondering if there was any place for the array of ints solution, that should be even faster, and still quite ‘vanilla’, at least on C/CPP. Is it not considered idiomatic or good practice in Go, and therefore not recommended?
Here is a quick example of what I mean. The tmpLetter is superfluous on the code below, and we can add to the score directly, making it even a bit faster.
``````package scrabble
// pointsTable stores the points for each letter. The first position is the points for the letter a,
// the second for letter b, and so forth.
var pointsTable = [26]int{
// A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z
1, 3, 3, 2, 1, 4, 2, 4, 1, 8, 5, 1, 3, 1, 1, 3, 10, 1, 1, 1, 1, 4, 4, 8, 4, 10}
// Score returns the score of a given word in scrabble.
func Score(word string) int {
var score int
var tmpLetter byte
for i := 0; i < len(word); i++ {
if word[i] > 'Z' {
tmpLetter = word[i] - 'a'
} else {
tmpLetter = word[i] - 'A'
}
score += pointsTable[tmpLetter]
}
return score
}
``````
Have you compared the performance of array vs map lookups? This is essentially a map lookup with a custom hashing function.
Hmm, interesting.
With this one, I get around 90ns/op:
`BenchmarkScore-8 13100623 87.07 ns/op 0 B/op 0 allocs/op`
``````package scrabble
// pointsTable stores the points for each letter. The first position is the points for the letter a,
// the second for letter b, and so forth.
var pointsTable = [26]int{
// A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z
1, 3, 3, 2, 1, 4, 2, 4, 1, 8, 5, 1, 3, 1, 1, 3, 10, 1, 1, 1, 1, 4, 4, 8, 4, 10}
// Score returns the score of a given word in scrabble.
func Score(word string) int {
var score int
for i := 0; i < len(word); i++ {
if word[i] > 'Z' {
score += pointsTable[word[i]-'a']
} else {
score += pointsTable[word[i]-'A']
}
}
return score
}
``````
With the proposed hashmap, around 950ns/op:
`BenchmarkScore-8 1251448 950.3 ns/op 0 B/op 0 allocs/op`
``````// Package scrabble scores a scrabble word.
package scrabble
import "unicode"
var lookup = map[rune]int{
'a': 1, 'e': 1, 'i': 1, 'o': 1, 'u': 1, 'l': 1, 'n': 1, 'r': 1, 's': 1, 't': 1,
'd': 2, 'g': 2,
'b': 3, 'c': 3, 'm': 3, 'p': 3,
'f': 4, 'h': 4, 'v': 4, 'w': 4, 'y': 4,
'k': 5,
'j': 8, 'x': 8,
'q': 10, 'z': 10,
}
// Score takes a word and returns its scrabble score.
func Score(word string) (score int) {
for _, ltr := range word {
score += lookup[unicode.ToLower(ltr)]
}
return score
}
``````
Since they are using different loops, and one uses runes instead of bytes, converts, etc, the comparison is not fair. Here is my attempt to make it a bit fairer.
I’m getting around 750ns/op.
`BenchmarkScore-8 1573298 747.1 ns/op 0 B/op 0 allocs/op`
``````// Package scrabble scores a scrabble word.
package scrabble
var lookup = map[rune]int{
'a': 1, 'e': 1, 'i': 1, 'o': 1, 'u': 1, 'l': 1, 'n': 1, 'r': 1, 's': 1, 't': 1,
'A': 1, 'E': 1, 'I': 1, 'O': 1, 'U': 1, 'L': 1, 'N': 1, 'R': 1, 'S': 1, 'T': 1,
'd': 2, 'g': 2,
'D': 2, 'G': 2,
'b': 3, 'c': 3, 'm': 3, 'p': 3,
'B': 3, 'C': 3, 'M': 3, 'P': 3,
'f': 4, 'h': 4, 'v': 4, 'w': 4, 'y': 4,
'F': 4, 'H': 4, 'V': 4, 'W': 4, 'Y': 4,
'k': 5, 'K': 5,
'j': 8, 'x': 8,
'J': 8, 'X': 8,
'q': 10, 'z': 10,
'Q': 10, 'Z': 10,
}
// Score takes a word and returns its scrabble score.
func Score(word string) (score int) {
for _, ltr := range word {
score += lookup[ltr]
}
return score
}
``````
Curiously, changing the lookup to map[byte]int, and looping over it using a ‘traditional’ for loop, it ballons over to around 1400ns/op. I’m curious as to why, as I thought it should be faster than the iterator loop, since it does not make any copies.
`BenchmarkScore-8 832183 1433 ns/op 0 B/op 0 allocs/op`
``````func Score(word string) (score int) {
for i := 0; i < len(word); i++ {
score += lookup[word[i]]
}
return score
}
``````
But the array solution seems to be much faster.
If speed is your goal, why not eliminate all the comparisons using a bigger lookup table? Make it handle the full ascii-range with scoring for both, uppercase and lowercase letters. Now, you can just iterate characters of the string making a table lookup and addition.
Below is my 8th version as I don’t use Go, but you probably get the idea:
``````[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 3, 3, 2,
1, 4, 2, 4, 1, 8, 5, 1, 3, 1, 1, 3, 10, 1, 0, 1, 1, 1, 4, 4, 8, 4, 10,
0, 0, 0, 0, 0, 1, 3, 3, 2, 1, 4, 2, 4, 1, 8, 5, 1, 3, 1, 1, 3, 10, 1,
1, 1, 1, 4, 4, 8, 4, 10 ] ( 0 >r swap ( a:@ n:r+ ) s:each! drop r> ) curry: score \ s -- n
``````
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Does that have any measurements speed improvement over the simpler array solution approve?
It’s the same method but using a bigger lookup table. I have not measured but it should be faster as the use of bigger lookup table eliminates all the comparisons and requires just a table lookup and addition per character.
It would be good to actually benchmark it to see if the difference is measurable It’s easy to assert that dropping a conditional will make a difference … but that’s not the same as actually demonstrating that to be the case.
1 Like
It also eliminates the need to subtract character values before table lookup and handles both uppercase and lowercase characters, so it saves more than just a conditional.
Yup. Integer subtraction is very low cost, though. The fact that it handles both upper and lowercase in the same lookup means a larger lookup, though, which might have a negative impact on the performance if you take the table creation into account
1 Like
But the smaller array might bei cache-friendlier.
On modern platforms with many processes, multiple CPUs, multiple levels of caches, with prefetching, instruction reordering, and speculative execution it’s hard to predict the effects of these kinds of micro-optimizations.
You have to benchmark and even doing that right ist not easy.
1 Like
It’s still a very small lookup table, so cache issues are not probable. It’s also precalculated and gets compiled directly into a word with my 8th version, so table creation cost should not be a big problem.
I used the same sized lookup table to calculate prime-hashes for words when finding anagrams to eliminate the need to sort and with large dictionary speed difference was easily noticeable.
That’s still a whole lot of conjectures and not much data. That larger lookup table may be more efficient. It may be less efficient. It may be very close. The best way to tell if it makes any difference - in any direction - would be some benchmarks.
1 Like
Okay, I made a real life test using a wordlist and it shows that using a bigger lookup table with this task makes program about two times faster.
``````var total-score
[ 1, 3, 3, 2, 1, 4, 2, 4, 1, 8, 5, 1, 3, 1, 1, 3, 10, 1, 1, 1, 1, 4, 4, 8, 4, 10 ]
( 0 >r swap ( dup 'Z n:> if 'a else 'A then n:- a:@ 0 ?: n:r+ ) s:each! drop r> ) curry: score \ s -- n
: app:main
"kotus_sanat.txt" f:slurp ( score total-score n:+! ) s:eachline
total-score @ . cr ;
``````
``````ohjaus@raspberrypi:~ \$ time /opt/8th/bin/rpi64/8th scrabble2.8th
1633402
real 0m0.523s
user 0m0.498s
sys 0m0.026s
ohjaus@raspberrypi:~ \$
``````
``````var total-score
[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 3, 3, 2,
1, 4, 2, 4, 1, 8, 5, 1, 3, 1, 1, 3, 10, 1, 0, 1, 1, 1, 4, 4, 8, 4, 10,
0, 0, 0, 0, 0, 1, 3, 3, 2, 1, 4, 2, 4, 1, 8, 5, 1, 3, 1, 1, 3, 10, 1,
1, 1, 1, 4, 4, 8, 4, 10 ] ( 0 >r swap ( a:@ 0 ?: n:r+ ) s:each! drop r> ) curry: score \ s -- n
: app:main
"kotus_sanat.txt" f:slurp ( score total-score n:+! ) s:eachline
total-score @ . cr ;
``````
``````ohjaus@raspberrypi:~ \$ time /opt/8th/bin/rpi64/8th scrabble.8th
1633402
real 0m0.264s
user 0m0.253s
sys 0m0.009s
ohjaus@raspberrypi:~ \$
``````
Did you run multiple iterations?
Also note, it might be worth benchmarking smaller inputs, too! It’s easy to optimize for a single test case, but forget that it might negatively impact other inputs
It’s easy to claim X is faster. “Faster” can be complex and messy and need to account for many different things. If we’re going to assert that it is faster in the Digging Deeper docs, we should actually test the hypothesis.
If an approach is faster for some inputs and not others, that’s also worth being explicit about it. Otherwise it introduces churn where someone else runs tests with different inputs and shows that the fastest solution is actually slower (in a different situation)!
Yes, the version using bigger lookup table was about 0.2 seconds faster on average. I used Finnish language dictionary as a test input and counted the total Scrabble score for every word inside the dictionary. With small test data both versions are fast enough to make measuring hard.
I’m getting roughly a 20% speed improvements with the benchmarks that the Go exercise ships with when using the bigger lookup table.
``````Small table:
BenchmarkScore-4 4893446 241.4 ns/op 0 B/op 0 allocs/op
Big table:
BenchmarkScore-4 6060932 198.1 ns/op 0 B/op 0 allocs/op
``````
`````` // Small table
if word[i] > 'Z' {
score += smallTable[word[i]-'a']
} else {
score += smallTable[word[i]-'A']
}
// vs big table
score += bigTable[word[i]]
``````
The `map` vs small table is a 15x speedup for me. | 3,641 | 9,949 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2023-50 | latest | en | 0.771034 |
https://thekidsworksheet.com/waves-worksheet-for-kids/ | 1,719,074,246,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862404.32/warc/CC-MAIN-20240622144011-20240622174011-00776.warc.gz | 496,089,571 | 22,180 | # Waves Worksheet For Kids
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Pin By Bradford Betz On Ojoj Worksheets Distance Time Graphs Worksheets Worksheet Template | 936 | 5,072 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.850986 |
https://anteupmagazine.com/2018/02/27/the-correct-answer-can-take-some-time/ | 1,620,623,174,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989030.87/warc/CC-MAIN-20210510033850-20210510063850-00330.warc.gz | 128,636,562 | 32,557 | # The correct answer can take some time
0
174
Take this short and quick quiz:
1. A bat and a ball cost \$1.10 in total. The bat costs
\$1 more than the ball. How much does the ball cost?
2. If it takes five machines five minutes to make five widgets, how long would it take 100 machines to make 100 widgets?
3. In a lake, there’s a patch of lily pads. Every day, the patch doubles in size. If it takes 48 days for the patch to cover the lake, how long would it take for the patch to cover half of the lake?
How do you think you did on the quiz? The answers seem obvious, right?
1. The ball costs 10 cents.
2. It would take 100 minutes for 100 machines to make 100 widgets.
3. It would take 24 days for the patch of lily pads to cover half the lake if it took 48 to cover the whole thing.
Well, this quiz produces the wrong answers for many people. We think too quickly; we don’t patiently assess the situation; we take thin slices of information and don’t bother to analyze the situations and our choices or decisions seem obvious. With time, deceleration and concentration, decision-making changes.
The correct answers are below. If you worked out the problem patiently and got the correct answers, that’s great; if not, you need some work on patience. Patience is a combination of attending, concentration, weighing options and weighing consequences.
In poker, the level of patience depends on the structure of the game. A satellite with the top 10 moving on requires more patience than a sit-and-go. Short-stacked in a deep tournament, I folded my way from eighth and min-cash to third to a much nicer payday.
But many poker players are more impulsive, saying and thinking, “I didn’t come here to fold; I came to play,” or “I am card dead and bored, I am going to start playing more hands.”
So much of how we play depends on what we expect and want from the game.
If patience fits your desired outcome and style, and yet you find you need work, then practice.
In poker, you have to have discipline, be patient, take the right opportunities, be aggressive, but not reckless, know when to take a chance, know when to bluff and know when you’re beat.
These are not only good poker skills, but good life skills, too. As always, keep your head in the game.
As for the quiz, the ball costs 5 cents (the bat is a dollar more at \$1.05, so they total \$1.10; if five machines could churn out five widgets in five minutes, then 100 machines could churn out 100 widgets in five minutes, and finally, a patch of lily pads that doubles in size every day and covers the whole lake in 48 days would have covered half of it in 47 days, doubling to cover the rest on the 48th day.
— Dr. Stephen Bloomfield is a licensed psychologist and avid poker player. Email him at editor@anteupmagazine.com. | 658 | 2,795 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-21 | longest | en | 0.953916 |
https://oeis.org/A175637 | 1,701,591,145,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100489.16/warc/CC-MAIN-20231203062445-20231203092445-00731.warc.gz | 494,994,243 | 4,455 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A175637 Numbers n such that the decimal digits of n are not present in k*n, k=2..9. 0
1, 3, 7, 9, 11, 77, 111, 707, 777, 909, 1111, 7777, 11111, 70707, 77777, 90909, 111111, 777777, 1111111, 7070707, 7777777, 9090909, 11111111, 77777777, 111111111, 707070707, 777777777, 909090909, 1111111111, 7777777777, 11111111111 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS There are four infinite patterns 1..1, 7..7, 7070..7 & 9090..9. LINKS Table of n, a(n) for n=1..31. EXAMPLE 7777*(1..9)= 7777,15554,23331,31108,38885,46662,54439,62216,69993. MATHEMATICA fQ[n_] := Intersection[ IntegerDigits@n, Union@ Flatten[ IntegerDigits[n* Range[2, 9]]]] == {}; k = 1; lst = {}; While[k < 10^9, If[ fQ@k, AppendTo[lst, k]]; k++ ]; lst CROSSREFS A140467 Sequence in context: A263926 A114788 A085074 * A110404 A366529 A190366 Adjacent sequences: A175634 A175635 A175636 * A175638 A175639 A175640 KEYWORD base,nonn AUTHOR Zak Seidov, Aug 01 2010 EXTENSIONS Corrected and extended the sequence, changed the comment line and added the Mathematica coding Robert G. Wilson v, Aug 03 2010 STATUS approved
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Last modified December 3 03:10 EST 2023. Contains 367531 sequences. (Running on oeis4.) | 621 | 1,888 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.693648 |
http://www.physicsforums.com/showthread.php?p=3985997 | 1,369,443,519,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705305291/warc/CC-MAIN-20130516115505-00028-ip-10-60-113-184.ec2.internal.warc.gz | 577,585,287 | 9,653 | ## Functional structure of Surface Heat
I am facing the following interesting question.
A closed room\hall contains several identical machines in it, they are fed by an electrical cable.
The machines can be turned on or off. When a machine is turned on, it consumes electrical energy and as a by product generates heat. The heat is radiated through the room walls to the outside air.
At present there are 2 machines in the room, but thed esigners consider adding mmore machines. The concern is that the outside surface area of the structure will become too hot.
What is the function that describes the room surface temperature in steady state as a function of the number of machines that are turned on? Is it power of 2 or of 3 of β (the number of machines that are turned on)?
It can be shown that the functional structure of the function that describes surface heat as a function of machines turned on is not dependent on the geometry of the room (the coefficients do). In order to simplify the analysis and find the functional structure as a function of β, assume that the room is a ball.
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Since you haven't had a reply i'll have a go... If we are talking about a real world building then for a first approximation I believe you can consider it a straightforward linear relationship. If one machine of power P raises the surface temperature by say 5C then two machines would raise it 10C . For example a typical power loss calculation for a room would be Power = (Troom-Tair)/Thermal Resistance Where Troom = Room air temperature Tair = Outside air temperature Power = Power loss through the walls. Under equilibrium conditions this is equal to the power going into the room. Thermal resistance = a constant. Typically the Thermal Tesistance would be made up several components in series. For example the thermal resistance of a wall would be the sum of the thermal resistances of it's component parts from the paint on the inside to the cladding or render on the outside. The above applies to normal building materials. If we are talking exotic materials such as thin layers of polished aluminium foil and temperatures high enough for radaition to dominate conduction then then you need a better more detailed answer that I can't provide. One word of caution...The outside surface temperature of some buildings is dominated by how sunny it is rather than what's inside generating heat.
Thank you CW, Thank you for your reply. The outside skeen is not conventional matrial but it is heat conductieve, thin and radiating. I thoght on a linear relationship too, but than measured 1 machine versus 2 mcahines on, waited long enough to reach steady state (same outside temperature), but the relationship is clearly non linear (I had also no machines on as another referecne point). My own interpretation is that heat radiation has an effect. Heat radiation is not a linear function. It is inaccurate to base the functional relationships on two observations that are too close one to the other. The main objective is to move 20 to 50 machines into the structure.
## Functional structure of Surface Heat
A black body emits thermal radiation based on the fourth power of temperature according to the stephan-Boltzman equation. Other bodies have an emissivity factor to account for a non black body surface and this factor can range from 0 to 1.
then there is absorbed energy, reflected energy and transmitted energy of the surface to factor in the model.
If you consider the surface to be a greybody then absorbed radiation equals emitted radiation of the surface.
In your model, these and a few more parameters that need to be taken into account. Is the structure radiating into space on into another sphere greybody or balackbody. Are the machines to be considered as radiating spheres within the enclosur sphere and/or is conduction and convection to be considered.
for example, a sphere enclosed by another sphere and both exchanging energy by radiation is basic textbook. Adding more spheres should just be an extension of the problem.
It can be shown that the functional structure of the function that describes surface heat as a function of machines turned on is not dependent on the geometry of the room (the coefficients do). In order to simplify the analysis and find the functional structure as a function of β, assume that the room is a ball.
For walls radiating to one another a shape factor from charts are available. If you choose to simpliy the problem by using spheres that would be an approximation of course, as the walls would not receive equal amounts of radiation.
Your model needs some more work done to it I think to figure out what more assumptions are necessary. Are you sure this isn't a homework question?
Thank you for your input, I appreciate your help. I assure you that it is not a homework, it is a very serious and fundamental problem. Yes, it can be assumed that the geometry is of a sphere, and the radiation is to the outside air. The boundery (the sphere surface) is very thin and the only way to get rid of the heat generated inside the sphere is through its surface. I don't know the inside temperature, but I know the heat energy per unit of time generated by each machine when it is on. All machines are identical, for simplicity you can assume that they consume negligible volume/space and all of them are in the center of the sphere (so we don't have to deal with their specific locations. Another way to approach and simplify is to assume that the machines are uniformely distributed in the volume of the sphere.
Tags heat analysis | 1,172 | 5,814 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.93981 |
https://ibacnet.org/3-digit-by-1-digit-division-worksheet/ | 1,568,625,027,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572516.46/warc/CC-MAIN-20190916080044-20190916102044-00240.warc.gz | 525,917,571 | 30,987 | # 3 Digit By 1 Digit Division Worksheet
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Top | 1,488 | 7,304 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2019-39 | latest | en | 0.886751 |
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Find the common difference of the arithmetic sequence 4, 7, 10, 13, ... Select the correct answer. PowerPoint PPT Presentation
Find the common difference of the arithmetic sequence 4, 7, 10, 13, ... Select the correct answer. 3 4 7 1. Find the n -th term of the arithmetic sequence. 4, 4 + s , 4 + 2 s , 4 + 3 s , ... Select the correct answer. 4 + s ( n - 1) 4 + sn s + 4( n - 1) s + 4 n.
Find the common difference of the arithmetic sequence 4, 7, 10, 13, ... Select the correct answer.
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- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
• 3
• 4
• 7
• 1
• 4 + s(n - 1)
• 4 + sn
• s + 4(n - 1)
• s + 4n
The 50th term of an arithmetic sequence is 67, and the common difference is 3. Find the first three terms. Select the correct answer.
• -80, -77, -74
• -80, -74, -68
• -83, -77, -71
• -83, -80, -77
A partial sum of an arithmetic sequence is given. Find the sum. 5 + 8 + 11 + ... + 62. Choose the answer from the following:
• 670
• 1,340
• 700
• 86
A man gets a job with a salary of \$30,000 a year. He is promised a \$2,100 raise each subsequent year. Find his total earnings for a 15 year period. Choose the answer from the following:
• \$670,500
• \$1,341,000
• \$686,250
• \$61,500 | 542 | 1,681 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2017-13 | longest | en | 0.854383 |
https://money.stackexchange.com/questions/152967/how-to-calculate-cost-basis-for-list-of-incongruent-batches-of-buy-and-sell-tran/153009 | 1,718,395,623,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861568.20/warc/CC-MAIN-20240614173313-20240614203313-00894.warc.gz | 362,798,828 | 41,881 | # How to calculate cost basis for list of incongruent batches of buy and sell transactions (forex trading)
How can I calculate the cost basis for my forex portfolio at any given time?
I'm trying to build a spreadsheet to track forex trading of cryptocurrency pairs. Specifically I'm looking at XMR-BTC
I'm not using any derivitives or limit orders. The idea is farily simple: Buy Monero (XMR) with Bitcoin (BTC) when the price is low. Sell Monero (XMR) for Bitcoin (BTC) when the price is high.
I want my spreadsheet to tell me something fairly simple: what's the average price (in BTC) that I've spent on Monero so far? This is very simple when you have only made a few purchases. It's also fairly simple when you've made sells of exactly the same quantity of your buys. But when you have a bunch of different buys and sells of different quantities, this is non-trivial.
As best I can tell, there is no precise way to measure this. There's many, including:
1. FIFO
2. LIFO
3. High Cost
4. Low Cost
5. Average Cost
6. Specific Identifier
7. Others??
So my question is, how can I calculate the cost basis of my portfolio at any point in the following set of transactions.
Note I'm not just interested in knowing how to determine the cost basis at the end of this set of transactions. I want to understand how to calculate the cost basis after every step of the following transactions.
I'm also not interested in tracking the cost basis in fiat. I only care about calculating the cost basis of my base currency (XMR) in the quote currency's units (BTC).
• `Tx 1` - `buy 3 XMR at 0.003929 BTC/XMR`
• `Tx 2` - `buy 1 XMR at 0.003847 BTC/XMR`
• `Tx 3` - `buy 3 XMR at 0.003741 BTC/XMR`
• `Tx 4` - `sell 3 XMR at 0.007678 BTC/XMR`
• `Tx 5` - `sell 2 XMR at 0.008008 BTC/XMR`
• `Tx 6` - `sell 2 XMR at 0.007912 BTC/XMR`
• `Tx 7` - `buy 1 XMR at 0.007084 BTC/XMR`
This data can also be portrayed in a table:
Tx Num Buy Qty Buy Cur Sell Qty Sell Cur Price (BTC/XMR) Date
1 3 XMR 0.011787 BTC 0.003929 2021-10-19
2 1 XMR 0.003847 BTC 0.003847 2021-12-15
3 3 XMR 0.011223 BTC 0.003741 2022-01-31
4 0.023034 BTC 3 XMR 0.007678 2022-09-01
5 0.016016 BTC 2 XMR 0.008008 2022-09-05
6 0.015824 BTC 2 XMR 0.007912 2022-09-08
7 1 XMR 0.007084 BTC 0.007084 2022-09-02
My understanding is that, following either FIFO or LIFO, the cost basis after `Day 4` would be calculated using the 3 XMR purchased either on `Day 1` or `Day 3`. What's especially interesting is: how do you calculate the cost basis after `Day 5` (which would be a combination of two distinct `buy` transactions)??
How can I calculate the cost basis for my above-described portfolio after each of the 7 days shown?
• Commented Oct 3, 2022 at 5:12
• As you've identified, there are many different answers to this question depending on how you are matching up buys and sells. You'd first need to determine that answer. Which would depend on things like whether you are trying to match the calculations of some other entity (i.e. your broker's site) or whether you are trying to match the calculations you are doing for tax purposes. Commented Oct 3, 2022 at 5:42
• This is to help determine the best price to sell above and buy under based on current buy-in (I have X Monero and I've paid an average of Y Bitcoin for it at this point in time). This is not for tax purposes. Commented Oct 3, 2022 at 16:37
• Commented Oct 5, 2022 at 15:33
• So you want to calculate 6 different answers on each day for the 6 different methods you've listed? Or do you want to calculate one answer on each day? The latter would be far more sensible in the vast majority of cases but then you'd need to tell us what algorithm you want to use. Commented Oct 5, 2022 at 20:50
If XMR is the base currency then use XMR numbers in place of dollar numbers. The cost basis of a transaction is just the number of XMR.
If BTC is the base currency then use BTC numbers in place of dollar numbers. The cost basis of a transaction is just the number of BTC.
Do a supplemental overall year-ending conversion to dollar result if wanted.
In the U.S., forex is mark-to-market accounting but not to be included on a Form 6781, for 60/40 long-term/short-term designation, unless the forex is also available on regulated futures exchanges.
However, crypto is not mark-to-market accounting and so it might be necessary to apply for mark-to-market accounting.
But with mark-to-market accounting then the overall year-ending conversion to dollar result is all that is required. Or itemized transactions might also be required and then just use your own method along with a letter of explanation.
Well, forex goes on either "Other income" of a Form 1040 Schedule 1, or goes on a trader-status Form 4797, or goes on a Form 6781.
Crypto goes on a Form 8949 .
• Sorry, this doesn't make any sense. I'm asking for how to calculate things, and I didn't see anything calculated. This question is also not related to taxes, so I don't know why that was mentioned. Commented Oct 5, 2022 at 22:53
• If XMR is the base currency then use XMR numbers in place of dollar numbers. The cost basis of a position is the number of XMR. You can't have two different types of cost basis and then say that nothing is calculated. Commented Oct 5, 2022 at 22:56
• I specifically asked for the basis in BTC. That's my main currency that I'm trying to stack. The valid answer to this question should provide the cost basis after each transaction and show how that was calculated. The reason I'm asking this question in the first place is because the internet is full of answers like yours, which are useless to actually understand how the basis is calculated. Commented Oct 5, 2022 at 22:57
• Well, the cost basis of XMR can be in BTC with BTC used as the base currency. Commented Oct 5, 2022 at 22:59
• I don't understand what you mean. Can you please show what you mean by typing out the mathematical equations after each transaction ? Commented Oct 5, 2022 at 23:00
This answer will show how to calculate:
2. the average sell, and
3. the break-even price
These calculations were determined using cointracking.info
Unfortunately, it's still a mystery how one would calculate the cost basis using FIFO or LIFO. I welcome other answers that show how to do these calculations :)
## Transaction 1
• `Tx 1` - `buy 3 XMR at 0.003929 BTC/XMR`
After the first day, the calculations are simple.
You currently have a total of 3 XMR.
You spent a total of `(3 XMR) * (0.003929 BTC/XMR) = 0.011787 BTC`
For the "Average Buy," we want to know the average price that we paid (in BTC) for each 1 coin of XMR.
To calculate the "Average Buy," we take the sum of the all of our weighted buys (obtained by multiplying the quantity of XMR bought by the price of 1 XMR in BTC) and divide it by the sum of the quantity of XMR bought.
Note For the "Average Buy" do not include any sell transactions
The algorithm for this is
[ (tx1 buy qty) * (tx1 price) + (tx2 buy qty) * (tx2 price) + ... + (txn buy qty) * (txn price) ] / [ (tx1 buy qty) + (tx2 buy qty) + ... + (txn buy qty) ]
``````[ (3 XMR) * (0.003929 BTC/XMR) ] / [ (3 XMR) ] = 0.003929 BTC/XMR
``````
Therefore, your average buy at this point is `0.003929 BTC/XMR`
##### Average Sell
For the "Average Sell," we want to know the average price that we sold (in BTC) for each 1 coin of XMR.
To calculate the "Average Sell," we take the sum of the all of our weighted sells (obtained by multiplying the quantity of XMR sold by the price of 1 XMR in BTC) and divide it by the sum of the quantity of XMR sold.
Note For the "Average Sell" do not include any buy transactions
The algorithm for this is
[ (tx1 sell qty) * (tx1 price) + (tx2 sell qty) * (tx2 price) + ... + (txn sell qty) * (txn price) ] / [ (tx1 sell qty) + (tx2 sell qty) + ... + (txn sell qty) ]
At this point, you haven't sold anything. There is no average sell price.
##### Break-Even Price
For the "Break-Even Price," we want to know at what price you would need to sell all of your XMR in order to break even (profit = 0 BTC).
To calculate the "Break-Even Price," we take the sum of the all of our weighted buys and sells (obtained by multiplying the quantity of XMR bought/sold by the price of 1 XMR in BTC) and divide it by the sum of the quantity of XMR bought/sold.
Note The Break-Even Price is very similar to the "Average Buy" and "Average Sell," except that it includes both buy and sell transactions.
The algorithm for this is
[ (tx1 qty) * (tx1 price) + (tx2 qty) * (tx2 price) + ... + (txn qty) * (txn price) ] / [ (tx1 qty) + (tx2 qty) + ... + (txn qty) ]
``````[ (3 XMR) * (0.003929 BTC/XMR) ] / [ (3 XMR) ] = 0.003929 BTC/XMR
``````
Therefore, your break-even price at this point is `0.003929 BTC/XMR`
## Transaction 2
• `Tx 1` - `buy 3 XMR at 0.003929 BTC/XMR`
• `Tx 2` - `buy 1 XMR at 0.003847 BTC/XMR`
After the second transaction, you've completed two buy transactions and currently hold a total of `3 XMR + 1 XMR = 4 XMR`.
For the "Average Buy," we simply sum how much we spent on XMR purchases (in bitcoin) and divide it by the sum of XMR that we own.
``````[ (3 XMR) * (0.003929 BTC/XMR) + (1 XMR) * (0.003847 BTC/XMR) ] / [ (3 XMR) + (1 XMR) ] = 0.0039085 BTC/XMR
``````
Therefore, your average buy at this point is `0.0039085 BTC/XMR`
Note that the previous buy price was `0.003929 BTC/XMR`. It makes sense that this new average buy price is slightly lower because our second purchase was made at a slightly less expensive price.
##### Average Sell
At this point, you haven't sold anything. There is no average sell price.
##### Break-Even Price
For the "Break-Even," we simply sum how much we spent or received on all XMR purchases & sales (in bitcoin) and divide it by the sum of XMR that we've bought and sold.
``````[ (3 XMR) * (0.003929 BTC/XMR) + (1 XMR) * (0.003847 BTC/XMR) ] / [ (3 XMR) + (1 XMR) ] = 0.0039085 BTC/XMR
``````
Therefore, your break-even price at this point is `0.0039085 BTC/XMR`
## Transaction 3
• `Tx 1` - `buy 3 XMR at 0.003929 BTC/XMR`
• `Tx 2` - `buy 1 XMR at 0.003847 BTC/XMR`
• `Tx 3` - `buy 3 XMR at 0.003741 BTC/XMR`
After the third transaction, you've completed three buy transactions and currently hold a total of `3 XMR + 1 XMR + 3 XMR = 7 XMR`.
``````[ (3 XMR) * (0.003929 BTC/XMR) + (1 XMR) * (0.003847 BTC/XMR) + (3 XMR) * (0.003741 BTC/XMR) ] / [ (3 XMR) + (1 XMR) + (3 XMR) ] = 0.003836714 BTC/XMR
``````
Therefore, your average buy at this point is `0.003836714 BTC/XMR`
##### Average Sell
At this point, you haven't sold anything. There is no average sell price.
##### Break-Even Price
``````[ (3 XMR) * (0.003929 BTC/XMR) + (1 XMR) * (0.003847 BTC/XMR) + (3 XMR) * (0.003741 BTC/XMR) ] / [ (3 XMR) + (1 XMR) + (3 XMR) ] = 0.003836714 BTC/XMR
``````
Therefore, your break-even price at this point is `0.003836714 BTC/XMR`
## Transaction 4
• `Tx 1` - `buy 3 XMR at 0.003929 BTC/XMR`
• `Tx 2` - `buy 1 XMR at 0.003847 BTC/XMR`
• `Tx 3` - `buy 3 XMR at 0.003741 BTC/XMR`
• `Tx 4` - `sell 3 XMR at 0.007678 BTC/XMR`
After the fourth transaction, you've completed three buy transactions and one sell transaction. You currently hold a total of `3 XMR + 1 XMR + 3 XMR - 3 XMR = 4 XMR`.
``````[ (3 XMR) * (0.003929 BTC/XMR) + (1 XMR) * (0.003847 BTC/XMR) + (3 XMR) * (0.003741 BTC/XMR) ] / [ (3 XMR) + (1 XMR) + (3 XMR) ] = 0.003836714 BTC/XMR
``````
Therefore, your average buy at this point is `0.003836714 BTC/XMR`
Note Because this sell transaction is not computed in the "Average Buy," there is no change from the previous transaction's average buy of `0.003836714 BTC/XMR`
##### Average Sell
``````[ (3 XMR) * (0.007678 BTC/XMR) / [ (3 XMR) ] = 0.007678 BTC/XMR
``````
Therefore, your average sell at this point is `0.007678 BTC/XMR`
##### Break-Even Price
``````[ (3 XMR) * (0.003929 BTC/XMR) + (1 XMR) * (0.003847 BTC/XMR) + (3 XMR) * (0.003741 BTC/XMR) - (3 XMR) * (0.007678 BTC/XMR) ] / [ (3 XMR) + (1 XMR) + (3 XMR) - (3 XMR) ] = 0.00095575 BTC/XMR
``````
Note For the sell transactions, we subtract both the weighted transaction amount from the top of the denominator and the quantity from the bottom of the denominator
Therefore, your break-even price at this point is `0.00095575 BTC/XMR`
Note that because we've now had our first sell transaction, the "Average Buy" and "Break-Even" prices begin to diverge.
## Transaction 5
• `Tx 1` - `buy 3 XMR at 0.003929 BTC/XMR`
• `Tx 2` - `buy 1 XMR at 0.003847 BTC/XMR`
• `Tx 3` - `buy 3 XMR at 0.003741 BTC/XMR`
• `Tx 4` - `sell 3 XMR at 0.007678 BTC/XMR`
• `Tx 5` - `sell 2 XMR at 0.008008 BTC/XMR`
After the fifth transaction, you've completed three buy transactions and two sell transaction. You currently hold a total of `3 XMR + 1 XMR + 3 XMR - 3 XMR - 2 XMR = 2 XMR`.
``````[ (3 XMR) * (0.003929 BTC/XMR) + (1 XMR) * (0.003847 BTC/XMR) + (3 XMR) * (0.003741 BTC/XMR) ] / [ (3 XMR) + (1 XMR) + (3 XMR) ] = 0.003836714 BTC/XMR
``````
Therefore, your average buy at this point is `0.003836714 BTC/XMR`
##### Average Sell
``````[ (3 XMR) * (0.007678 BTC/XMR) + (2 XMR) * (0.008008 BTC/XMR) ] / [ (3 XMR) + (2 XMR) ] = 0.00781 BTC/XMR
``````
Therefore, your average sell at this point is `0.00781 BTC/XMR`
##### Break-Even Price
``````[ (3 XMR) * (0.003929 BTC/XMR) + (1 XMR) * (0.003847 BTC/XMR) + (3 XMR) * (0.003741 BTC/XMR) - (3 XMR) * (0.007678 BTC/XMR) - (2 XMR) * (0.008008 BTC/XMR) ] / [ (3 XMR) + (1 XMR) + (3 XMR) - (3 XMR) - (2 XMR) ] = -0.0060965 BTC/XMR
``````
Therefore, your break-even price at this point is `-0.0060965 BTC/XMR`
Note Your break-even price becomes negative here because you've been so profitable that, even if the value of XMR drops to zero, you'd still have net positive gains
## Transaction 6
• `Tx 1` - `buy 3 XMR at 0.003929 BTC/XMR`
• `Tx 2` - `buy 1 XMR at 0.003847 BTC/XMR`
• `Tx 3` - `buy 3 XMR at 0.003741 BTC/XMR`
• `Tx 4` - `sell 3 XMR at 0.007678 BTC/XMR`
• `Tx 5` - `sell 2 XMR at 0.008008 BTC/XMR`
• `Tx 6` - `sell 2 XMR at 0.007912 BTC/XMR`
After the sixth transaction, you've completed three buy transactions and three sell transaction of equal amounts. You currently hold a total of `3 XMR + 1 XMR + 3 XMR - 3 XMR - 2 XMR - 2 XMR = 0 XMR`.
``````[ (3 XMR) * (0.003929 BTC/XMR) + (1 XMR) * (0.003847 BTC/XMR) + (3 XMR) * (0.003741 BTC/XMR) ] / [ (3 XMR) + (1 XMR) + (3 XMR) ] = 0.003836714 BTC/XMR
``````
Therefore, your average buy at this point is `0.003836714 BTC/XMR`
##### Average Sell
``````[ (3 XMR) * (0.007678 BTC/XMR) + (2 XMR) * (0.008008 BTC/XMR) + (2 XMR) * (0.007912 BTC/XMR) ] / [ (3 XMR) + (2 XMR) + (2 XMR) ] = 0.00781 BTC/XMR
``````
Therefore, your average sell at this point is `0.00781 BTC/XMR`
##### Break-Even Price
``````[ (3 XMR) * (0.003929 BTC/XMR) + (1 XMR) * (0.003847 BTC/XMR) + (3 XMR) * (0.003741 BTC/XMR) - (3 XMR) * (0.007678 BTC/XMR) - (2 XMR) * (0.008008 BTC/XMR) - (2 XMR) * (0.007912 BTC/XMR) ] / [ (3 XMR) + (1 XMR) + (3 XMR) - (3 XMR) - (2 XMR) - (2 XMR) ] = infinity!! BTC/XMR
``````
Therefore, your break-even price at this point is `infinity!! BTC/XMR`
Note Your break-even price cannot be calculated here because you've sold all your XMR and you cannot divide by zero
## Transaction 7
• `Tx 1` - `buy 3 XMR at 0.003929 BTC/XMR`
• `Tx 2` - `buy 1 XMR at 0.003847 BTC/XMR`
• `Tx 3` - `buy 3 XMR at 0.003741 BTC/XMR`
• `Tx 4` - `sell 3 XMR at 0.007678 BTC/XMR`
• `Tx 5` - `sell 2 XMR at 0.008008 BTC/XMR`
• `Tx 6` - `sell 2 XMR at 0.007912 BTC/XMR`
• `Tx 7` - `buy 1 XMR at 0.007084 BTC/XMR`
After the seventh transaction, you've bought and sold all your XMR for a profit, then you re-bought at a relative low. You currently hold a total of `3 XMR + 1 XMR + 3 XMR - 3 XMR - 2 XMR - 2 XMR + 1 XMR = 1 XMR`.
``````[ (3 XMR) * (0.003929 BTC/XMR) + (1 XMR) * (0.003847 BTC/XMR) + (3 XMR) * (0.003741 BTC/XMR) + (1 XMR) * (0.007084 XMR/BTC) ] / [ (3 XMR) + (1 XMR) + (3 XMR) + (1 XMR) ] = 0.004242625 BTC/XMR
``````
Therefore, your average buy at this point is `0.004242625 BTC/XMR`
##### Average Sell
``````[ (3 XMR) * (0.007678 BTC/XMR) + (2 XMR) * (0.008008 BTC/XMR) + (2 XMR) * (0.007912 BTC/XMR) ] / [ (3 XMR) + (2 XMR) + (2 XMR) ] = 0.00781 BTC/XMR
``````
Therefore, your average sell at this point is `0.00781 BTC/XMR`
##### Break-Even Price
``````[ (3 XMR) * (0.003929 BTC/XMR) + (1 XMR) * (0.003847 BTC/XMR) + (3 XMR) * (0.003741 BTC/XMR) - (3 XMR) * (0.007678 BTC/XMR) - (2 XMR) * (0.008008 BTC/XMR) - (2 XMR) * (0.007912 BTC/XMR) + (1 XMR) * (0.007084 XMR/BTC) ] / [ (3 XMR) + (1 XMR) + (3 XMR) - (3 XMR) - (2 XMR) - (2 XMR) + (1 XMR) ] = -0.020933 BTC/XMR
``````
Therefore, your break-even price at this point is `-0.020933 BTC/XMR` | 5,568 | 16,640 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2024-26 | latest | en | 0.901923 |
https://fungrim.org/entry/8814ad/ | 1,713,538,646,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817438.43/warc/CC-MAIN-20240419141145-20240419171145-00753.warc.gz | 228,634,231 | 3,743 | $\sum_{n=-\infty}^{\infty} \operatorname{sinc}^{3}\!\left(n\right) = \frac{3 \pi}{4}$
TeX:
\sum_{n=-\infty}^{\infty} \operatorname{sinc}^{3}\!\left(n\right) = \frac{3 \pi}{4}
Definitions:
Fungrim symbol Notation Short description
Sum$\sum_{n} f(n)$ Sum
Pow${a}^{b}$ Power
Sinc$\operatorname{sinc}(z)$ Sinc function
Infinity$\infty$ Positive infinity
Pi$\pi$ The constant pi (3.14...)
Source code for this entry:
Entry(ID("8814ad"),
Formula(Equal(Sum(Pow(Sinc(n), 3), For(n, Neg(Infinity), Infinity)), Div(Mul(3, Pi), 4))))
## Topics using this entry
Copyright (C) Fredrik Johansson and contributors. Fungrim is provided under the MIT license. The source code is on GitHub.
2021-03-15 19:12:00.328586 UTC | 241 | 706 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 6, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-18 | latest | en | 0.39489 |
https://www.pw.live/chapter-solids-and-fluids/exercise-1/question/27783 | 1,675,350,091,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500028.12/warc/CC-MAIN-20230202133541-20230202163541-00857.warc.gz | 945,708,207 | 21,856 | Question of Exercise 1
# Question Two identical blocks of mass M are linked by a thread wrapped around a pulley-block with a fixed axis. A small block of mass m is placed on one of the blocks as shown. If pulley and strings are ideal, the force exerted by small block m on block M is,
Option 1
Option 2
Option 3
Option 4
Maximum value of static friction is called
A: Limiting friction
B: Rolling friction
C: Normal reaction
D: Coefficient of friction
Solution:
Explanation:
Limiting friction is the maximum value of the static friction. When a body overcomes the limiting friction, the body starts moving.
Hence, option A is correct.
A running man has half the kinetic energy of that of a boy of half of his mass
The man speeds up by 1m/s , so that he has the same kinetic energy as that of the boy. The original speed of the man is?
Solution:
Two particles are projected simultaneously in the same
vertical plane, from the same point, both with different speeds and at different angles horizontally. The path followed by one, as seen by the other, is
A: a vertical line
B: a parabola
C: a hyperbola
D: a straight line making a constant angle ( ≠ 90 ^∘ ) with horizontal
Solution:
Explanation:-
which is a constant. So the path followed by one, as seen by the other, is a straight line, making a constant angle with the horizontal.
Hence, the correct option is (D).
Give two examples of the following Random motion
Solution:
Explanation:-
If an object in motion has no particular direction and its motion shifts unexpectedly, then it is said to have random motion.
The two examples of random motion are:
Motion of a football player.
Movement of a mosquito.
During an adiabatic process the pressure of a gas
is proportional to the cube of its adiabatic temperature.The value of Cp/Cv for that gas is :
A:4/3
B:2
C:5/3
D:3/2
Solution:
Explanation:- | 454 | 1,891 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2023-06 | latest | en | 0.934966 |
http://kea-monad.blogspot.com/2007/06/m-theory-lesson-66.html | 1,386,952,090,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164960531/warc/CC-MAIN-20131204134920-00031-ip-10-33-133-15.ec2.internal.warc.gz | 97,995,171 | 12,210 | occasional meanderings in physics' brave new world
Name:
Location: New Zealand
Marni D. Sheppeard
## Monday, June 11, 2007
### M Theory Lesson 66
Recall that Joan Birman et al studied knots in the Lorenz template with two generating holes X and Y. So knots are expressed as words in X and Y. In Robert Ghrist's paper Branched two-manifolds supporting all links he shows that the template $\mathcal{V}_0$ on more letters contains an isotopic copy of every (tame) knot and link. More specifically, for a parameter range $\beta \in [6.5,10.5]$ every link appears as a periodic solution to the equation which is used to model an electric circuit. This is cool stuff. In M Theory we like ribbon diagrams which are twisted into loops like in the Lorenz template diagram. The universal template $\mathcal{V}_0$ can be embedded in an infinite sequence of more complicated templates, which in turn are embeddable in $\mathcal{V}_0$. Ghrist also considers flows arising from fibrations, such as the 1-punctured torus fibration for the figure 8 knot complement. This fibration flow is also an example of a universal flow.
I was quite intrigued when a mathematical biologist at a conference told me recently that no one really knew why DNA had four bases rather than two. Apparently it isn't clear why self-replicating molecules fail to adopt a binary code in X and Y. Somebody else muttered something about hydrogen bonds and then, inspired and ignorant, I started rambling on about knot generation in templates. After all, DNA molecules need to know how to knot themselves.
CarlBrannen said...
You should be getting ready to defend yourself! The equations you gave remind me of what you get when you toss together a diode and some resistors, and combine it with an opamp.
It turns out that the simple circuit of a D flip-flop with its output connected back to its input with an inverter (not gate) will produce a transition to chaos as the frequency is increased. One could do the same thing with an opamp, provided you have a time delay built in.
Back when I was interested in the transition to chaos, I wrote down a very simple set of differential equations and got chaos to simulate on Mathematica. I don't know if this is a different set of differential equations than what is well known, since I never really studied the subject.
June 11, 2007 1:42 PM
Doug said...
Hi Kea and CarlB,
I looked more closely at the Ghrist web page. In his CV, under referenced publication, he lists two papers with LaValle.
http://www.math.uiuc.edu/~ghrist/cv.pdf
One paper, ’Nonpositive curvature and pareto-optimal coordination motion planning', SIAM Journal of Control and Optimization, 45(5), 1697-1713, 2006
[I could not find this on the web]
http://www.math.uiuc.edu/~ghrist/preprints/pareto.pdf
But a similar [perhaps identical] paper ’Nonpositive curvature and pareto-optimal coordination of robots’, SIAM Journal of Control and Optimization, 2007
[Is on the web]
http://msl.cs.uiuc.edu/~lavalle/mulrob.html
“Pareto-optimal” is game theory terminology.
http://en.wikipedia.org/wiki/Pareto_efficiency
I am more familiar with LaValle, “Planning Algorithms”.
This lead to my reading the Basar and Olsder book that I have often referenced.
[LaValle book available on-line]
http://planning.cs.uiuc.edu/
I am not famililiar with Joan Birman but I am intrigued.
Could the Lorentz knots be 'stringing loops' or 'looping strings' playing games?
Did Yau derive the concept of flop transitions from EM flip-flops?
Check out the Rossler Attractor, especially the section 'Links to other topics',
"The banding evident in the Rössler attractor is similar to a Cantor set rotated about its midpoint. Additionally, the half-twist in the Rössler attractor makes it similar to a Möbius strip."
http://en.wikipedia.org/wiki/R%C3%B6ssler_map
June 11, 2007 1:59 PM
Doug said...
RE: "... why DNA had four bases rather than two ..."
U_Utah has a great web page on ‘Purine and Pyrimidine Metabolism’.
http://library.med.utah.edu/NetBiochem/pupyr/pp.htm
Note that precursors of Purines could be”
Hypoxanthine = 6-oxy purine
or
Xanthine = 2,6-dioxy purine
Precursor(s) of Pyrimidines could be Orotic acid = 2,4-dioxy-6-carboxy pyrimidine.
This is not my final answer, because who knows for certain?
The binary code may be nested:
RNA or DNA
then
Purine or Pyrimidine
If Purine
If Pyrimidine
then generally Cytosine or
Uracil if RNA or
Thymine if DNA.
Note:
Uracil is more specific to RNA than Adenine and
Thymine is more specific to RNA than Adenine.
Why - Oxy or deoxy presence?
June 11, 2007 2:15 PM
Anonymous said...
Kea, I am a bit surprised that such a simple thing like pairing of AT and GC is unknown on this blog. Surely genetic code is binary because of such a pairing and surely information is read off in binary code. In addition, please, take a look at gr-qc/040029 and take a look how REAL alloy structures are related to gravity and Yamabe functionals
June 11, 2007 2:32 PM
Anonymous said...
Addendum. Sorry, it is late here, in US. The correct reference is gr-qc/0410029. In addition to this, you may take a look at hep-th/0701084 where these ideas are developed further in the style of Grisha's Perelman work(s)
June 11, 2007 3:08 PM
Mitchell said...
Anonymous - DNA is quaternary, not binary. AT is different from TA.
June 12, 2007 2:44 PM
Matti Pitkanen said...
I have considered some number theory inspired models for genetic code. Quaternary code is one of them studied also by Khrennikov.
Also 5-adicity has been suggested and I constructed for half year ago a model in which codons correspond to 5-adic numbers with non-vanishing digits: this means that codons correspond to numbers in interval [31,126].
The basic observation is that there are 20 primes in this interval. They would correspond naturally to primes labelling aminoacids. The variational principle states that the p-adic negentropy (log(x) is replaced with log(|x|_p) in Shannon entropy) associated with thermodynamical state in 5-adic thermodynamics and assigned to the partitions of integer n labelling a given codon is maximized as a function of p. Hence correspondence n-->p(n) characterizing code results.
There also other constraints and only few solutions are satisfying the constraints are obtained. See this.
June 12, 2007 6:28 PM
CarlBrannen said...
Kea, I've posted a blog on why DNA uses 4 nucleotides . Great topic.
June 14, 2007 6:28 PM | 1,632 | 6,437 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2013-48 | longest | en | 0.943103 |
https://ch.mathworks.com/matlabcentral/answers/674048-how-to-more-efficiently-populate-a-sparse-matrix | 1,611,744,629,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704821381.83/warc/CC-MAIN-20210127090152-20210127120152-00657.warc.gz | 258,819,376 | 23,882 | MATLAB Answers
# How to more efficiently populate a sparse matrix
30 views (last 30 days)
Toysey on 2 Dec 2020
Commented: Toysey on 2 Dec 2020
I have a code that solves the laplace equation in axisymmetric coordinates in a cone-shaped domain using a finite difference method. The solution is then of the form
phi = A\b;
where A is my operator matrix and b are the source terms. My issue is concerned with creating the matrix A which is sparse and diagonal. My current approach is:
Nz = someConstant; % Number of intervals in z
Nr = someConstant; % Number of intervals in r
N = Nr*Nz;
A = sparse(N,N); % N rows, N columns
b = sparse(N,1); % N rows, 1 column
for ii = 2:Nr-1
for jj = 2:Nz-1
n = ii + (jj - 1)*Nr; % Chosen indexing convention
r = R(ii,jj); % Array containing radial coordinates
dr = DR(jj); % Vector containing dr terms
dz = someConstant;
A(n, n ) = -2*(1/dz^2+1/dr^2); % \phi_{i,j}
A(n, n - 1 ) = 1/dr^2 - 1/(2*dr*r); % \phi_(i-1,j)
A(n, n + 1 ) = 1/dr^2 + 1/(2*dr*r); % \phi_(i+1,j)
A(n, n - Nr) = 1/dz^2; % \phi_(i, j-1)
A(n, n + Nr) = 1/dz^2; % \phi_(i, j+1)
b(n, 1 ) = 0; % Source term
end
end
Matlab warns me that 'this sparse indexing expression is likely to be slow' when filling in the terms in A but I don't know how to improve this. The particular concern is that because my domain is a cone shape dr depends on the z index.
(This code snippit does not include the boundary conditions so b = zeros(N,1) which will not give a solution)
#### 0 Comments
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### Accepted Answer
Bjorn Gustavsson on 2 Dec 2020
The general advice is to calculate and save the non-zero elements, and their array-indices, in 3 arrays and then once they've been calculated create your sparse array. Something like this (not your example converted):
Avals = zeros(5*n^2,1);
idx1 = Avals;
idx2 = idx1;
for i1 = 2:n
for i2 = 2:n
Avals(i2+n*(i1-1)+0) = -4; % Centre-point of Laplacian differentiation-operator
idx1(i2+n*(i1-1)+0) = i1;
idx2(i2+n*(i1-1)+0) = i2;
Avals(i2+n*(i1-1)+1) = 1; % One of the nearest neighbors
idx1(i2+n*(i1-1)+0) = i1-1; % ...etc
idx2(i2+n*(i1-1)+0) = i2;
end
end
A = sparse(idx1,idx2,Avals,n,n);
That way you only create space for your sparse array once, and the index and magnitude-arrays are also pre-allocated.
(In my experience I should test these generation-snippets with small enough arrays to manually inspect them.)
HTH
#### 1 Comment
Toysey on 2 Dec 2020
Thank you very much - that is exactly what I needed to know
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http://slidegur.com/doc/147629/week-03.2 | 1,477,684,975,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988725470.56/warc/CC-MAIN-20161020183845-00106-ip-10-171-6-4.ec2.internal.warc.gz | 234,509,011 | 8,724 | ### Week-03.2
```Dale & Lewis Chapter 3
Data Representation
Representing Real numbers
• Need to represent numbers such as 0.543 or -17.45
• The decimal system uses the “decimal point”
• Remember the definition of a number system, but
extended to negative exponents
dn-1 x bn-1 + dn-2 x bn-2 + … + d2 x b2 + d1 x b1 + d0 x b0
+ d-1 x b-1 + d-2 x b-2 + …
• So 0.543 is 0 x 100 + 5 x 10-1 + 4 x 10-2 + 3 x 10-3
• The decimal point position delimits where the exponents
become negative
Conversion from binary to decimal
• Use the definition of a number in a positional number
system with base 2
• Evaluate the formula using decimal arithmetic
• Example
10.1011 = 1 x 21 + 0 x 20 + 1 x 2-1 + 0 x 2-2 + 1 x 2-3 +
1 x 2-4
= 2 + 0.5 + 0.125 + 0.0625 = 2.6875
• Generalizing, the “decimal point” is called radix point in
non-decimal number systems
Conversion from decimal to binary
• Integer part: convert separately, as done before
• Fractional part:
− Repeatedly multiply by 2
− Integer part is the next digit
− Fraction is developed left to right
• Example: convert 3.14579 to binary
• Integer part is 11
.14579 x 2 = 0.29158
.0
.29158 x 2 = 0.58316
.00
.58316 x 2 = 1.16632
.001
.16632 x 2 = 0.33264
.0010
etc...
3.14579 in binary is 11.0010…
Rounding errors
• Try representing .1 (decimal) into binary
• How would one represent fractions with base 5 or 3?
• Infinite sequences, a computer with fixed number of bits
will make an approximation in the representation
• Needs to be taken into account, or it will lead to rounding
errors Patriot missile system
Floating point numbers
• Real numbers are represented using a concept inherited from
scientific notation
• Scientific notation is a way of writing numbers that are too big
or too small to be conveniently written in decimal form
• The following all represent +1234.56
+123456.0
+12345.6
+1234.56
+123.456
+12.3456
+1.23456
+0.123456
+0.0123456
x10-2
x10-1
x100
x101
x102
x103
x104 standard representation
x105
Floating point numbers
• The representations differ in that the decimal point “floats”
to the left or right by adjusting the exponent
• Any real number x can be written as
x = ±f x be
where b > 1 is the base, e is an integer for the exponent,
and the fraction is 1/b ≤ f < 1 (normalized)
• We know how to express each part in binary, but
convention for the integer e uses Excess notation
Excess notation
• Another way to express integers (we’ve seen signmagnitude and two’s complement)
• This shifts the list of numbers represented by bits to
include negative numbers, with 10…0 representing 0
• Convert binary excess-n to decimal: (binary value)-n
− Example for excess-8: 0101 5-8 = -3
• Convert decimal to binary excess-n: (decimal value)+n
− Example for excess-8: 6 6+8 = 1110
Excess notation
Floating point numbers in binary
• An example for “single precision” binary floating point
number (32 bits)
− S is the sign: 0 positive, 1 negative
− E is the exponent (here in 8 bits, 256 values, excess-128
notation)
− M is the manitssa (fractional part) has an implicit radix point
at the beginning (here 23 bits, 8,388,608 values)
SEEEEEEEEMMMMMMMMMMMMMMMMMMMMMMM
10010100101010011101001010001101
is -.01010011101001010001101 x 2-87
Floating point number examples
• Using 8-bit floating point representation (SEEEMMMM)
− Convert 3.14579 to binary
3.14579 = 11.0010…
11.0010… x 20
.110010… x 2+2
01101101 (rounded)
− Convert 0.1 to binary
0.1 = 0.0001100110011…
.0001100110011… x 20
.1100110011… x 2-3
00011101 (rounded)
``` | 1,142 | 3,500 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2016-44 | longest | en | 0.764959 |
http://demonstrations.wolfram.com/HeatTransferBetweenABarAndAFluidReservoirACoupledPDEODEModel/ | 1,511,522,668,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934807650.44/warc/CC-MAIN-20171124104142-20171124124142-00109.warc.gz | 71,033,530 | 11,824 | 11454
# Heat Transfer between a Bar and a Fluid Reservoir: A Coupled PDE-ODE Model
Consider a thin bar of length with initial temperature . The right end and the sides of the bar are insulated. For times , the left end is connected to a well-mixed insulated reservoir at an initial temperature . This Demonstration determines the transient temperature of the bar and the reservoir.
We use the following dimensionless variables:
is the dimensionless temperature,
is the dimensionless space coordinate,
is dimensionless time.
Here are the dimensionless equations describing the system.
For the bar:
,
with
,
and
.
For the reservoir:
,
with
.
Here and are the temperatures of the bar and reservoir,
is a mass-heat capacity ratio,
and represent the mass and heat capacities of the reservoir and the bar respectively.
The coupled system of one partial and one ordinary differential equation is solved using the built-in Mathematica function NDSolve. The temperatures of the bar and the reservoir are shown for different values of the mass heat capacity ratio and dimensionless time .
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Note: To run this Demonstration you need Mathematica 7+ or the free Mathematica Player 7EX Download or upgrade to Mathematica Player 7EX I already have Mathematica Player or Mathematica 7+ | 507 | 2,472 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2017-47 | longest | en | 0.869047 |
https://institutoflash786.org/2019/11/29/computer-scientists-expand-the-frontier-of-verifiable-knowledge/ | 1,632,505,405,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057564.48/warc/CC-MAIN-20210924171348-20210924201348-00339.warc.gz | 374,684,189 | 32,347 | # Computer Scientists Expand the Frontier of Verifiable Knowledge
The universe of problems that a computer can check has grown. The researchers’ secret ingredient? Quantum entanglement.
By interrogating entangled quantum computers, a person can verify the answers to enormously complicated problems.
Imagine someone came along and told you that they had an oracle, and that this oracle could reveal the deep secrets of the universe. While you might be intrigued, you’d have a hard time trusting it. You’d want some way to verify that what the oracle told you was true.
This is the crux of one of the central problems in computer science. Some problems are too hard to solve in any reasonable amount of time. But their solutions are easy to check. Given that, computer scientists want to know: How complicated can a problem be while still having a solution that can be verified?
Turns out, the answer is: Almost unimaginably complicated.
In a paper released in April, two computer scientists dramatically increased the number of problems that fall into the hard-to-solve-but-easy-to-verify category. They describe a method that makes it possible to check answers to problems of almost incomprehensible complexity. “It seems insane,” said Thomas Vidick, a computer scientist at the California Institute of Technology who wasn’t involved in the new work.
The research applies to quantum computers — computers that perform calculations according to the nonintuitive rules of quantum mechanics. Quantum computers barely exist now but have the potential to revolutionize computing in the future.
The new work essentially gives us leverage over that powerful oracle. Even if the oracle promises to tell you answers to problems that are far beyond your own ability to solve, there’s still a way to ensure the oracle is telling the truth.
## Until the End of the Universe
When a problem is hard to solve but easy to verify, finding a solution takes a long time, but verifying that a given solution is correct does not.
For example, imagine someone hands you a graph — a collection of dots (vertices) connected by lines (edges). The person asks you if it’s possible to color the vertices of the graph using only three colors, such that no connected vertices have the same color.
This “three-color” problem is hard to solve. In general, the time it takes to find a three-coloring of a graph (or determine that none exists) increases exponentially as the size of the graph increases. If, say, finding a solution for a graph with 20 vertices takes 320nanoseconds — a few seconds total — a graph with 60 vertices would take on the order of 360 nanoseconds, or about 100 times the age of the universe.
But let’s say someone claims to have three-colored a graph. It wouldn’t take long to check whether their claim is true. You’d just go through the vertices one by one, examining their connections. As the graph gets bigger, the time it takes to do this increases slowly, in what’s called polynomial time. As a result, a computer doesn’t take much longer to check a three-coloring of a graph with 60 vertices than it does to check a graph with 20 vertices.
“It’s easy, given a proper three-coloring, to check that it works,” said John Wright, a physicist at the Massachusetts Institute of Technology who wrote the new paper along with Anand Natarajan of Caltech.
In the 1970s computer scientists defined a class of problems that are easy to verify, even if some are hard to solve. They called the class “NP,” for nondeterministic polynomial time. Since then, NP has been the most intensively studied class of problems in computer science. In particular, computer scientists would like to know how this class changes as you give the verifier new ways to check the truth of a solution.
## The Right Questions
Prior to Natarajan and Wright’s work, verification power had increased in two big leaps.
To understand the first leap, imagine that you’re colorblind. Someone places two blocks on the table in front of you and asks whether the blocks are the same or different colors. This is an impossible task for you. Moreover, you can’t verify someone else’s solution.
But you’re allowed to interrogate this person, whom we’ll call the prover. Let’s say the prover tells you that the two blocks are different colors. You designate one block as “Block A” and the other as “Block B.” Then you place the blocks behind your back and randomly switch which hand holds which block. Then you reveal the blocks and ask the prover to identify Block A.
If the blocks are different colors, this couldn’t be a simpler quiz. The prover will know that Block A is, say, the red block and will correctly identify it every single time.
But if the blocks are actually the same color — meaning the prover erred in saying that they were different colors — the prover can only guess which block is which. Because of this, it will only be possible for the prover to identify Block A 50 percent of the time. By repeatedly probing the prover about the solution, you will be able to verify whether it’s correct.
In 1985 a trio of computer scientists proved that such interactive proofs can be used to verify solutions to problems that are more complicated than the problems in NP. Their work created a new class of problems called IP, for “interactive polynomial” time. The same method used to verify the coloring of two blocks can be used to verify solutions to much more complicated questions.
The second major advance took place in the same decade. It follows the logic of a police investigation. If you have two suspects you believe committed a crime, you’re not going to question them together. Instead, you’ll interrogate them in separate rooms and check each person’s answers against the other’s. By questioning them separately, you’ll be able to reveal more of the truth than if you had only one suspect to interrogate.
“It’s impossible for [two suspects] to form some sort of distributed, consistent story because they simply don’t know what answers the other is giving,” Wright said.
In 1988 four computer scientists proved that if you ask two computers to separately solve the same problem — and you interrogate them separately about their answers — you can verify a class of problems that’s even larger than IP: a class called MIP, for multi-prover interactive proofs.
With a multi-prover interactive approach, for example, it’s possible to verify three-colorings for a sequence of graphs that increase in size much faster than the graphs in NP. In NP, graph sizes increase at a linear rate — the number of vertices might grow from 1 to 2 to 3 to 4 and so on — so that the size of a graph is never hugely disproportionate to the amount of time needed to verify its three-coloring. But in MIP, the number of vertices in a graph grows exponentially — from 21 to 22 to 23 to 24 and so on.
As a result, the graphs are too big even to fit in the verifying computer’s memory, so it can’t check three-colorings by running through the list of vertices. But it’s still possible to verify a three-coloring by asking the two provers separate but related questions.
In MIP, the verifier has enough memory to run a program that allows it to determine whether two vertices in the graph are connected by an edge. The verifier can then ask each prover to state the color of one of the two connected vertices — and it can cross-reference the provers’ answers to make sure the three-coloring works.
The expansion of hard-to-solve-but-easy-to-verify problems from NP to IP to MIP involved classical computers. Quantum computers work very differently. For decades it’s been unclear how they change the picture — do they make it harder or easier to verify solutions?
The new work by Natarajan and Wright provides the answer.
## Quantum Cheats
Quantum computers perform calculations by manipulating quantum bits, or “qubits.” These have the strange property that they can be entangled with one another. When two qubits — or even large systems of qubits — are entangled, it means that their physical properties play off each other in a certain way.
In their new work, Natarajan and Wright consider a scenario involving two separate quantum computers that share entangled qubits.
This kind of setup would seem to work against verification. The power of a multi-prover interactive proof comes precisely from the fact that you can question two provers separately and cross-check their answers. If the provers’ answers are consistent, then it’s likely they’re correct. But two provers sharing an entangled state would seem to have more power to consistently assert incorrect answers.
And indeed, when the scenario of two entangled quantum computers was first put forward in 2003, computer scientists assumed entanglement would reduce verification power. “The obvious reaction of everyone, including me, is that now you’re giving more power to the provers,” Vidick said. “They can use entanglement to correlate their answers.”
Despite that initial pessimism, Vidick spent several years trying to prove the opposite. In 2012, he and Tsuyoshi Ito proved that it’s still possible to verify all the problems in MIP with entangled quantum computers.
Natarajan and Wright have now proved that the situation is even better than that: A wider class of problems can be verified with entanglement than without it. It’s possible to turn the connections between entangled quantum computers to the verifier’s advantage.
To see how, remember the procedure in MIP for verifying three-colorings of graphs whose sizes grow exponentially. The verifier doesn’t have enough memory to store the whole graph, but it does have enough memory to identify two connected vertices, and to ask the provers the colors of those vertices.
With the class of problems Natarajan and Wright consider — called NEEXP for nondeterministic doubly exponential time — the graph sizes grow even faster than they do in MIP. Graphs in NEEXP grow at a “doubly exponential” rate. Instead of increasing at a rate of powers of 2 — 21, 22, 23, 24 and so on — the number of vertices in the graph increases at a rate of powers of powers of 2 — 221,222,223,224 and so on. As a result, the graphs quickly become so big that the verifier can’t even identify a single pair of connected vertices.
“To label a vertex would take 2n bits, which is exponentially more bits than the verifier has in its working memory,” Natarajan said.
But Natarajan and Wright prove that it’s possible to verify a three-coloring of a doubly-exponential-size graph even without being able to identify which vertices to ask the provers about. This is because you can make the provers come up with the questions themselves.
The idea of asking computers to interrogate their own solutions sounds, to computer scientists, as advisable as asking suspects in a crime to interrogate themselves — surely a foolish proposition. Except Natarajan and Wright prove that it’s not. The reason is entanglement.
“Entangled states are a shared resource,” Wright said. “Our entire protocol is figuring out how to use this shared resource to generate connected questions.”
If the quantum computers are entangled, then their choices of vertices will be correlated, producing just the right set of questions to verify a three-coloring.
At the same time, the verifier doesn’t want the two quantum computers to be so intertwined that their answers to those questions are correlated (which would be the equivalent of two suspects in a crime coordinating their false alibis). Another strange quantum feature handles this concern. In quantum mechanics, the uncertainty principle prevents us from knowing a particle’s position and momentum simultaneously — if you measure one property, you destroy information about the other. The uncertainty principle strictly limits what you can know about any two “complementary” properties of a quantum system.
Natarajan and Wright take advantage of this in their work. To compute the color of a vertex, they have the two quantum computers make complementary measurements. Each computer computes the color of its own vertex, and in doing so, it destroys any information about the other’s vertex. In other words, entanglement allows the computers to generate correlated questions, but the uncertainty principle prevents them from colluding when answering them.
“You have to force the provers to forget, and that’s the main thing [Natarajan and Wright] do in their paper,” Vidick said. “They force the prover to erase information by making a measurement.”
Their work has almost existential implications. Before this new paper, there was a much lower limit on the amount of knowledge we could possess with complete confidence. If we were presented with an answer to a problem in NEEXP, we’d have no choice but to take it on faith. But Natarajan and Wright have burst past that limit, making it possible to verify answers to a far more expansive universe of computational problems.
And now that they have, it’s unclear where the limit of verification power lies.
“It could go much further,” said Lance Fortnow, a computer scientist at the Georgia Institute of Technology. “They leave open the possibility that you could take another step.” | 2,729 | 13,254 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-39 | latest | en | 0.948051 |
http://www.enotes.com/homework-help/topic/science?pg=5&filters%25255B0%25255D=All | 1,444,743,908,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443738006925.85/warc/CC-MAIN-20151001222006-00182-ip-10-137-6-227.ec2.internal.warc.gz | 552,614,787 | 15,804 | # Science Homework Help
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Asked by xxenderbossxx on via web | 2,701 | 11,921 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2015-40 | latest | en | 0.932442 |
https://www.bankersadda.com/sbi-po-quantitative-aptitude-for-22/ | 1,597,518,727,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439741154.98/warc/CC-MAIN-20200815184756-20200815214756-00480.warc.gz | 582,269,682 | 55,043 | # SBI PO Quantitative Aptitude For Prelims: 22nd February
Dear Aspirants,
### Quantitative Aptitude Quiz For SBI PO
Numerical Ability or Quantitative Aptitude Section has given heebie-jeebies to the aspirants when they appear for a banking examination. As the level of every other section is only getting complex and convoluted, there is no doubt that this section, too, makes your blood run cold. The questions asked in this section are calculative and very time-consuming. But once dealt with proper strategy, speed, and accuracy, this section can get you the maximum marks in the examination. Following is the Quantitative Aptitude quiz to help you practice with the best of latest pattern questions.
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Q1. A milkman has 20 L of pure milk. He sells 5 L of it at profit of 20% and he mixes 5 L water in remaining quantity of milk. If he sells whole quantity of mixture at cost price then find his percentage profit?
25%
20%
24%
28%
30%
Solution:
Q2. The sum of the radius and height of a cylinder is 18m. The total surface area of the cylinder is 792 m², what is the volume of the cylinder ? (in m³)
1848
1694
1716
1724
1440
Solution:
Q3. In a 90 litre mixture of milk and water, percentage of water is only 30%. The milkman gave 18 litres of this mixture to a customer and then added 18 litres of water to the remaining mixture. What is the percentage of milk in the final mixture ?
64%
48%
52%
68%
56%
Solution:
Q4. If the area of a circle is 616 cm², what would be the total surface area of a hemisphere having the same radius as the circle ?
1848 cm²
1648 cm²
2218 cm²
1808 cm²
1765cm²
Solution:
Q5. There are two mixtures of honey and water, the quantity of honey in them being 40% and 75% of the mixture respectively. If 5 gallons of the first are mixed with 8 gallons of the second, what will be the ratio of honey to water in the new mixture ?
11 : 2
8 : 5
9 : 11
2 : 11
3 : 22
Solution:
Directions (6-10): There are total five departments in a company. There are total 90 employees in Finance department which is 25% of total employees in the company. 2/9 of the total employees of the company are working in HR department. Employees working in Sales department is 25% more than that in HR department. Ratio between employees working in Security and Housing department is 4 : 5.
Q6. Find number of employees working in HR department is what percent more than number of employees working in Security department?
250%
200%
150%
100%
50%
Solution:
Q7. Find the average number of employees working in Sales, Finance and Housing department?
60
70
80
90
100
Solution:
Q8. Number of employees in Housing department is how much more than number of employees in Security department?
10
20
30
40
50
Solution:
Q9. In Security department, 40% are female employees then find total male employees working in Security department?
16
40
32
8
24
Solution:
Total number of male employees working in Security department = 60/100Ă—40=24
Q10. Ratio between total number of male and female employees in HR department is 2 : 3. Find total number of female employees working in HR department?
32
48
64
40
56
Solution:
Total number of female employees working in HR department = 3/5 Ă— 80=48
Directions (11-15): In each of the following questions two equations are given. Solve the equations and give answer—
Q11.
If x < y
If x ≤ y
x = y or relationship between x and y cannot be established.
If x ≥ y
If x > y
Solution:
Q12.
If x < y
If x ≤ y
x = y or relationship between x and y cannot be established.
If x ≥ y
If x > y
Solution:
Q13.
If x < y
If x ≤ y
x = y or relationship between x and y cannot be established.
If x ≥ y
If x > y
Solution:
Q14.
If x < y
If x ≤ y
x = y or relationship between x and y cannot be established.
If x ≥ y
If x > y
Solution:
Q15.
If x < y
If x ≤ y
x = y or relationship between x and y cannot be established.
If x ≥ y
If x > y
Solution:
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http://msccc.net/core/?page_id=534 | 1,553,440,383,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203462.50/warc/CC-MAIN-20190324145706-20190324171706-00049.warc.gz | 137,015,087 | 19,461 | # Operations and Algebraic Thinking
Represent and solve problems involving addition and subtraction.
1. Use addition and subtraction within 20 to solve word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.
2. Solve word problems that call for addition of three whole numbers whose sum is less than or equal to 20, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.
Understand and apply properties of operations and the relationship between addition and subtraction.
3. Apply properties of operations as strategies to add and subtract. [Students need not use formal terms for these properties.] Examples: If 8 + 3 = 11 is known, then 3 + 8 = 11 is also known. (Commutative property of addition.) To add 2 + 6 + 4, the second two numbers can be added to make a ten, so 2 + 6 + 4 = 2 + 10 = 12. (Associative property of addition.)
4. Understand subtraction as an unknown-addend problem. For example, subtract 10 – 8 by finding the number that makes 10 when added to 8. Add and subtract within 20. | 294 | 1,258 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-13 | latest | en | 0.919896 |
https://brainly.in/question/307215 | 1,485,035,792,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281226.52/warc/CC-MAIN-20170116095121-00436-ip-10-171-10-70.ec2.internal.warc.gz | 800,963,395 | 9,581 | # A resistance of 5ohm is connected in the left gap of a meter bridge and 15ohm in the other gap.sketch the diagram and then find the position of the balancing point.
1
by divallin
2016-03-22T14:25:31+05:30
For meter bridge :-
X = R(100-l) / l
, where X & R are resistances & l is balancing length
Putting values :-
= 5 = 15(100-l) / l
= l = 300 - 3l
= 4l = 300
= l = 300/4 = 75cm | 141 | 389 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2017-04 | latest | en | 0.791236 |
http://tex.stackexchange.com/questions/46798/how-do-we-draw-a-bird-in-latex/46966 | 1,469,772,673,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257829972.19/warc/CC-MAIN-20160723071029-00263-ip-10-185-27-174.ec2.internal.warc.gz | 245,416,116 | 27,674 | # How do we Draw a Bird in LaTeX
Yes, it's that time of the month again. It's time for somebody to ask how we can draw X in LaTeX. This time I'm asking the question and X is a bird.
The reason why I'm interested in this question is that drawing "non-geometrical" shapes is difficult. Addign symmetry makes it even more difficult and I'd like to see the different solutions that are used.
I welcome any answers but I'm especially keen on birds with symmetry. All solutions have to be original.
The following is a possible example of what I'm looking for.
First the design, which is inspired by Escher, but based on an original idea:
Next the LaTeX output:
Following the suggestion of @Hooked, I provide a possible solution below.
-
And the question is ... what, exactly? Seriously, you appear to have already drawn a bird. So before I try to better it, I'd like to know a bit more: why are you doing this, and what's wrong with what you've already done? – Loop Space Mar 4 '12 at 21:50
Thanks for this ! – projetmbc Mar 4 '12 at 22:02
Andrew: While this really isn't a question and might not really belong here, I found it nice and don't mind. – Ben Mar 4 '12 at 23:01
To keep it a proper question - why not erase your solution and write it as an answer? That way the upvotes will help determine the "best" bird to answer the question. – Hooked Mar 9 '12 at 19:23
@MarcvanDongen presumably the site is structured to allow answering ones own question, there's even a badge for it! tex.stackexchange.com/badges/14/self-learner If you do it this way, it allows us to upvote the question for it's merits separately from the answer you may give (of which we may have a different criteria for). – Hooked Mar 11 '12 at 3:58
## 6 Answers
No symmetry no Escher's style but ... that look like birds. First you need to save the next code in a file names bird1.pgf. It's not exactly the code given by Inkscape. I export the code with LaTeX and PSTricks then I transform the code to get something lighter.
%%Creator: inkscape 0.48.2
%%Please note this file requires PSTricks extensions
\m 294.2539 630.192
\c 402.19387 856.17313262 487.970 847.918 488.203 848.384
\c 418.76941 793.96220262 351.3144 706.885 294.253 630.191
\o
\m 492.937 872.945
\c 561.627 1035.579 624.256 909.310 624.256 909.310
\c 624.256 909.310 553.546 979.011 492.9373 872.945
\o
\m 594.269 881.372
\c 594.269 875.794 589.746 871.271 584.167 871.2714
\c 578.588 871.271 574.066 875.794 574.066 881.3729
\c 574.066 886.951 578.588 891.474 584.167 891.4744
\c 589.746 891.474 594.269 886.951 594.269 881.3729
\o
\m 612.533 855.021
\c 615.077 835.925 686.994 862.510 675.314 862.510
\c 684.625 869.475 622.031 865.087 612.533 855.021
\o
\m 527.722 809.837
\c 535.020 790.648 629.957 829.080 622.287 821.021
\c 566.611 762.5112 513.586 794.030 527.722 809.837
\o
\m 459.507 512.230
\c 583.604 590.988 648.910 788.7036 647.107 803.656
\c 647.107 803.656 651.605 695.9141 619.59 643.544
\c 568.314 559.659 472.752 504.8105 459.507 512.230
\o
\m 277.499 613.928
\l 241.428 601.428
\c 241.428 601.428 181.071 587.857 116.78571 534.642
\c 52.500 481.428 53.5714 481.428 53.571432 481.428
\c 53.5714 481.428 113.207 513.232 157.14286 540.357
\c 196.708 564.784 182.094 558.488 277.49999 613.928
\o
\m 361.508 504.392
\c 361.508 504.392 352.922 525.605 280.69625 530.151
\c 208.470 534.697 52.928 455.760 53.411917 455.905
\c 129.959 478.800 152.115 489.060 251.40183 503.887
\c 346.187 518.043 361.508 504.392 361.50845 504.392
\o
\m 487.803 503.150
\c 487.803 503.150 463.304 478.906 479.994 446.076
\c 496.684 413.247 496.684 414.762 496.684 414.762
\c 487.619 437.921 482.945 448.602 487.803 503.150
\o
\m 441.374 498.150
\c 441.374 498.150 416.876 473.906 433.566 441.076
\c 450.255 408.247 450.255 409.762 450.255 409.762
\c 441.191 432.921 436.516 443.602 441.374 498.150
\o
\m 521.785 414.642
\c 573.214 417.857 572.5 380.000 572.5 380.000
\c 572.5 380.000 560 401.785 521.785 414.642
\o
\m 494.64285 393.214
\c 546.07142 396.428 545.357 358.571 545.357 358.571
\c 545.35714 358.571 532.857 380.357 494.642 393.214
\o
\m 468.928 374.285
\c 520.357 377.499 519.642 339.642 519.642 339.642
\c 519.642 339.642 507.142 361.428 468.928 374.285
\o
\m 458.427 389.0716
\c 395.702 393.445 396.573 341.929 396.573 341.929
\c 396.573 341.929 411.819 371.575 458.427 389.071
\o
\m 308.571 610.000
\c 385.714 548.571 530 741.42 530 741.428
\c 530 741.428 495.714 645.714 414.28 601.428
\c 339.269 560.630 310 602.857 308.57 610.000
\o\s
\endinput
It's the first time, I created a vector object with Inkscape. I take an example (.png) and with a bezier tool (pen) I draw the bird. If someone know how to transform a file.png in a file.eps I will be happy. I think it's possible with Inkscape to vectorize a bipmap but I don't know how to do.
For the first birds,I use TikZ so if you don't want to download tikzrput and pgfornament you can comment the last pictures. Then I try with \rput pgf version and the last I try with pgfornament.
\documentclass[11pt]{scrartcl}
\PassOptionsToPackage{dvipsnames,svgnames}{xcolor}
\usepackage{tikz,tikzrput} % altermundus.com/pages/tkz/tikzrput/
\usepackage{pgfornament} % altermundus.com/pages/tkz/ornament/
\makeatletter
\newcommand{\callornament}[1]{%
\begingroup
\def\i{\pgfusepath{clip}}%
\let\o\pgfpathclose
\let\s\pgfusepathqfillstroke
\def\p ##1##2{\pgfqpoint{##1bp}{##2bp}}%
\def\m ##1 ##2 {\pgfpathmoveto{\p{##1}{##2}}}%
\def\l ##1 ##2 {\pgfpathlineto{\p{##1}{##2}}}%
\def\r ##1 ##2 ##3 ##4 {\pgfpathrectangle{\p{##1}{##2}}{\p{##3}{##4}}}%
\def\c ##1 ##2 ##3 ##4 ##5 ##6 {%
\pgfpathcurveto{\p{##1}{##2}}{\p{##3}{##4}}{\p{##5}{##6}}}%
\@@input #1\relax
\endgroup}
\makeatother
\begin{document}
\begin{tikzpicture}[scale=.2,fill=MidnightBlue,draw=black]
\callornament{bird1.pgf}
\begin{scope}[fill=yellow,draw=black,cm={-1,0,0,1,(50,10)}]
\callornament{bird1.pgf}
\end{scope}
\end{tikzpicture}
\rput{-30}(7,-2){\tikz[scale=.2,fill=SpringGreen] \callornament{bird1.pgf} ; }
\gdef\OrnamentsFamily{bird}
\tikzset{pgfornamentstyle/.style={fill=Goldenrod}}%
\rput(0,-2){\pgfornament[scale=.3]{1}}
\end{document}
-
The process of turning a raster image into vector form is called tracing, I believe, for which there exists at least the programs AutoTrace and Potrace. But often I find, especially with relatively simple graphics, that just writing it out in SVG produces the most compact code. – morbusg Mar 6 '12 at 21:50
@morbusg Inkscape also has some auto tracing tools, see e.g. inkscape.org/doc/tracing/tutorial-tracing.html – Torbjørn T. Mar 19 '12 at 7:40
@TorbjørnT. Thanks for the information. I never used Inskape but this is a soft very interesting – Alain Matthes Mar 19 '12 at 7:47
I made a bird. It is butt ugly, but it flies. -and it tessellates the plane.
\documentclass{standalone}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}[scale=0.01]
\newcommand\bird[1]{%
\begin{scope}[#1]
\filldraw (0,0) --%
(10,-5) -- (19,8) -- (35,7) -- (38,-12) -- (60,0) --%
(75,23) -- (80,40) -- (52,37) -- (74,50) -- (52,50) -- (60,60)--%
(38,48) -- (35,67) -- (19,68) -- (10,55) -- (0,60) --%
(-8,50) -- (14,50) -- (-8,37) -- (20,40) -- (15,23) -- cycle;
\begin{scope}[draw=gray]
\draw (35,15) -- +(-25:25) (40,20) -- +(-20:25) (42,25) -- +(-15:25) (40,30) -- +(-10:30);
\draw (1,55) circle (1);
\draw[clip] (24,55) circle (4);
\filldraw[gray] (27,52) circle (4);
\end{scope}
\end{scope}
}
\bird{shift={(0,0)}, fill=white};
\bird{shift={(60,0)}, fill=black};
\bird{shift={(120,0)}, fill=white};
\bird{shift={(180,0)}, fill=black};
\bird{shift={(240,0)}, fill=white};
\bird{shift={(0,60)}, fill=black};
\bird{shift={(60,60)}, fill=white};
\bird{shift={(120,60)}, fill=black};
\bird{shift={(180,60)}, fill=white};
\bird{shift={(240,60)}, fill=black};
\end{tikzpicture}
\end{document}
I do not know if I passes the arguments in a smart way!?, and my short attempt to use a foreach loop failed.
-
Sweet! Here is rough for loop. \foreach[count=\xi] \x in {0,60,...,240}{ \pgfmathparse{int(mod(\xi,2))} \ifnum \pgfmathresult>0 \bird{shift={(\x,0)}, fill=white}; \bird{shift={(\x,60)}, fill=black}; \else \bird{shift={(\x,60)}, fill=white}; \bird{shift={(\x,0)}, fill=black}; \fi } – percusse Mar 18 '12 at 4:19
Utterly hideous! +1 – qubyte Mar 23 '12 at 1:47
For PSTricks fans:
\documentclass{minimal}
\usepackage{pst-fun}
\begin{document}
\pspicture[showgrid](6,5)
\rput(0,2){\psBird}
\rput{-30}(2,2){\psBird}
\endpspicture
\end{document}
Note: Compile it with xelatex or latex-dvips-ps2pdf sequence.
-
\newcounter{\Terrorist} :P – percusse Mar 6 '12 at 12:25
@percusse:\newcommand{\nickname}{Counter Terrorist}\renewcommand{\nickname}{Damien Walters}. – kiss my armpit Mar 6 '12 at 12:31
@YiannisLazarides: Pixels have been repeated and tiled here. :-) – kiss my armpit Mar 6 '12 at 12:34
@Yiannis: AFAIS, the question asks for drawing a bird, OP just himself drew a symmetric bird. So this answer is just fine in every respect (+1). – morbusg Mar 6 '12 at 12:43
@YiannisLazarides “I'm especially keen” != “requirement”;) – morbusg Mar 6 '12 at 13:17
Update
The model for the birds comes from here but I draw myself my new ugly birds with a simple drawing. It's possible to make a code shorter. Now the code with the foreach is ugly. I was not able to find something elegant.
\documentclass[border=6pt]{standalone}
\usepackage{tikz}
\begin{document}
\newcommand\myuglybird[2]{%http://www.tess-elation.co.uk/birds---an-introduction/birds-1-1
\begin{scope}[rotate=45,#1]
\draw[#2] (0,0) -- ++(0, 1) -- ++( .5,.5) -- ++( .5,-.5)
-- ++(0, 1) -- ++( .5,.5) -- ++( .5,-.5)
-- ++(0,-1) -- ++( .5,.5) -- ++( .5,-.5)
-- ++(0,-1) -- ++(-.5,.5) -- ++(-.5,-.5)
-- ++(0,-1) -- ++(-.5,.5) -- ++(-.5,-.5)
-- ++(0, 1) -- ++(-.5,.5) -- cycle;
\draw ( .2, .2) -- ( .2, .6)
( .4, .4) -- ( .4, .8)
( .6, .4) -- ( .6, .8)
( .8, .2) -- ( .8, .6)
(1.2,-.8) -- (1.2, .2)
(1.4,-.6) -- (1.4, .4)
(1.6,-.6) -- (1.6, .4)
(1.8,-.8) -- (1.8, .2)
(2.2, .2) -- (2.2, .6)
(2.4, .4) -- (2.4, .8)
(2.6, .4) -- (2.6, .8)
(2.8, .2) -- (2.8, .6)
(1.3,2.3) -- (1.5,2.1) -- (1.7,2.3)
(1.5,2.5) -- (1.5,2.1)
(1.7,2.0) circle (2pt)
(1,0) -- ++(.5,.5) --++(.5,-.5) ;
\end{scope}
}%
\begin{tikzpicture}
\draw[thick] (-3,5) rectangle (12,11);
\clip (-3,5) rectangle (12,11);
\foreach \y in {0,1,...,5}{%
\foreach \x in {0,1,...,9}{%
\pgfmathsetmacro{\z}{mod(\x+\y,4)}
\ifcase\z
\def\col{blue}
\or
\def\col{red}
\or
\def\col{green}
\or
\def\col{yellow}
\fi
\pgfmathsetmacro{\xx}{2*\x-mod(\y,2)-1}
\pgfmathsetmacro{\yy}{-\x+3*\y}
\ifnum 2=\y \pgfmathsetmacro{\yy}{-\x+3*\y-1} \fi
\ifnum 3=\y \pgfmathsetmacro{\yy}{-\x+3*\y-1} \fi
\ifnum 4=\y \pgfmathsetmacro{\yy}{-\x+3*\y-2} \fi
\ifnum 5=\y \pgfmathsetmacro{\yy}{-\x+3*\y-2} \fi
\ifnum 6=\y \pgfmathsetmacro{\yy}{-\x+3*\y-2} \fi
\myuglybird{shift={(\xx cm,\yy cm)}}{shade, top color=\col!30}
} }
\end{tikzpicture}
\end{document}
I keep my first ugly bird. An ugly bird with tesselation
\documentclass[border=6pt]{standalone}
\usepackage{tikz}
\begin{document}
\newcommand\myuglybird[1]{%
\draw[#1] (1.7,2.5) circle (2pt) (0,0) --++(0.25,-3)--++(0.75,0)--++(-0.25,0.5)--++(0,1.75)--++(1.75,0.75)
--++(-1,+2)--++(1,1)--++(-1.75,-0.75)--++(-0,-1.75)--++(+0.25,-0.5)--++(-0.75,0)--++(-0.25,+3)--++(-1,-1)--++(1,-2)--cycle;
}%
\begin{tikzpicture}
\draw[thick] (0,0) rectangle (12,15);
\clip (0,0) rectangle (12,15);
\foreach \y in {0,1,...,6}{%
\foreach \x in {0,1,...,5}{%
\pgfmathsetmacro{\z}{\x+\y}
\ifodd \z \def\col{lightgray} \else \def\col{gray}\fi
\myuglybird{fill=\col,xshift=2.5*\x cm,yshift=3*\y cm}
} }
\end{tikzpicture}
\end{document}
-
It's not ugly it's just a taek won do bird :) – percusse Mar 18 '12 at 18:40
Is the ugly bird based on "Escher's design 128" (melusine.eu.org/syracuse/var/syracuse/metapost/galeries/escher/…)? – Marc van Dongen Mar 19 '12 at 8:34
@MarcvanDongen Yes and no : I've some books with a lot of pictures from Escher and I took my bad inspiration in one of these books but you are right it's a really bad version of "Escher's design 128". Now I need to find a personal design but this will take time ! – Alain Matthes Mar 19 '12 at 8:46
@MarcvanDongen The best pictures for me are the pictures with birds and fishes at the same time. Really awesome ! The ugly bird is very similar to Bip Bip fr.wikipedia.org/wiki/Bip_Bip_et_Coyote – Alain Matthes Mar 19 '12 at 8:51
Thanks for the explanation. – Marc van Dongen Mar 19 '12 at 9:45
As promised in the question, I decided to separate the question and a possible solution.
Here is a possible solution.
\documentclass{article}
\usepackage{graphicx}
\usepackage{tikz}
\usetikzlibrary{calc}
\usetikzlibrary{intersections}
\newcommand*\BirdWidth[0]{1.7}
\newcommand*\BirdHeight[0]{1.0}
\newcommand\DrawBird[1]{%
\path (#1)
coordinate (b)
let \n{w}={\BirdWidth},
\n{h}={\BirdHeight},
\n{r}={1.5*\n{w}},
\n{a2}={atan2(0.5*\n{w},\n{h})},
\n{ang}={\n{a2}-90},
\n{xdiff}={0.25*\n{w}+\n{r}*cos(\n{ang})},
\n{ydiff}={0.5*\n{h}+\n{r}*sin(\n{ang})},
\n{rx}={sqrt(\n{ydiff}*\n{ydiff}+\n{xdiff}*\n{xdiff})} in
(b) + (+0.5*\n{w},\n{h}) coordinate (r)
+ (\n{xdiff},\n{ydiff}) coordinate (c b r)
($(b)!0.5!(r)$) coordinate (m b r)
($(c b r)!\n{rx}cm!(m b r)$) coordinate (m r)
(r)++ (-\n{w},0) coordinate (l)
+ (\n{xdiff},-\n{ydiff}) coordinate (c b l)
($(b)!0.5!(l)$) coordinate (m b l)
($(c b l)!\n{rx}cm!(m b l)$) coordinate (m l)
\foreach \label/\angle in {0/-10,1/-6,2/-3,3/12,4/8,5/6,6/0} {
($(c b r)!1!\angle:(m r)$) coordinate (r\label)
($(c b l)!1!-\angle:(m l)$) coordinate (l\label)
}
\foreach \label\angle/\rat in {0/-2.5/1.075,1/1/1.086,2/3.5/1.094,3/6.5/1.1} {
($(c b r)!\rat!\angle:(m r)$) coordinate (ri\label)
($(c b l)!\rat!-\angle:(m l)$) coordinate (li\label)
}
\foreach \label\angle\rat in {0/6.50/1.03,1/6.55/1.06,3/5.45/1.17,4/3.95/1.21,5/-2.4/1.035} {
($(c b r)!\rat!\angle:(m r)$) coordinate (ro\label)
($(c b l)!\rat!-\angle:(m l)$) coordinate (lo\label)
}
($(ro1)!1.6!(ri3)$) coordinate (ro2)
($(lo1)!1.6!(li3)$) coordinate (lo2)
(r) ++ (-0.40,+0.000) coordinate (t1)
+ (+0.13,+0.120) coordinate (t2)
+ (+0.30,+0.300) coordinate (t3)
+ (-0.15,+0.300) coordinate (t4)
+ (-0.20,+0.120) coordinate (t5)
++ (-0.25,+0.000) coordinate (t6)
+ (-0.35,-0.070) coordinate (t7)
++ (-0.67,+0.000) coordinate (t8)
+ (+0.04,+0.190) coordinate (t9)
++ (+0.13,+0.420) coordinate (t10)
+ (-0.40,+0.030) coordinate (t11)
++ (-0.79,-0.060) coordinate (t12)
+ (+0.08,-0.015) coordinate (t13)
++ (+0.17,-0.050) coordinate (t14)
+ (-0.07,+0.002) coordinate (t15)
++ (-0.14,-0.005) coordinate (t16)
+ (+0.12,-0.060) coordinate (t17)
+ (+0.20,-0.160) coordinate (t18)
;
\filldraw (b) .. controls (r3) and (r4) .. (r5)
.. controls (ro1) .. (ri3)
.. controls (ri2) and (ri1) .. (ri0)
.. controls (ro5) .. (r2)
.. controls (r1) and (r0) .. (r)
-- (t1) .. controls (t2) .. (t3)
-- (t4) .. controls (t5) .. (t6)
.. controls (t7) .. (t8)
.. controls (t9) .. (t10)
.. controls (t11) .. (t12)
.. controls (t13) .. (t14)
.. controls (t15) .. (t16)
.. controls (t17) and (t18) .. (l)
.. controls (l3) and (l4) .. (l5)
.. controls (lo1) .. (li3)
.. controls (li2) and (li1) .. (li0)
.. controls (lo5) .. (l2)
.. controls (l1) and (l0) .. (b)
(l5) .. controls (l6) .. (l2);
\draw (ri3) .. controls (ro2) and (ro3) .. (ro4);
\filldraw (l) ++ (+0.16,+0.33) circle (0.8pt);
}
\begin{document}
\begin{tikzpicture}[scale=4,fill=yellow!70!gray,draw=yellow!50!gray,line width=2pt]
\foreach \shift in {0,1} {
\DrawBird{\shift*\BirdWidth,0}
}
\begin{scope}[rotate=180,xscale=-1]
\DrawBird{0.5*\BirdWidth,-\BirdHeight}
\end{scope}
\end{tikzpicture}
\end{document}
-
Here is a proposition mixing Escher, Penrose and Picasso (do you see the birds?):
The code (derived form my answer to Penrose tiling in TikZ):
\documentclass[tikz]{standalone}
\usepackage{tikz}
\usetikzlibrary{calc}
\pgfmathsetmacro{\invphi}{2/(1+sqrt(5))}
% default styles
\tikzset{
% borders style
penrose line/.style={draw=black,line join=round},
% kites and darts styles
penrose kite/.style={penrose line},
penrose dart/.style={penrose line},
% the three paths (and the three corresponding reverse paths)
}
\newcommand\penrosedrawkite[3]{% ver, angle, len
\path let
\p1=(#1),
\p2=($(\p1) + (#2+36:#3)$),
\p3=($(\p1) + (#2:#3)$),
\p4=($(\p1) + (#2-36:#3)$)
in
[penrose kite] (\p1)
to[penrose path 1] (\p2)
to[penrose rev path 2] (\p3)
to[penrose path 2] (\p4)
to[penrose rev path 1] (\p1);
}
\newcommand\penrosekite[5]{% n, ver, angle, len, rot
\ifnum#1=0 % draw or recursive decomposition ?
\ifnum#5=1 % draw kite if current semikite is clockwise
\penrosedrawkite{#2}{#3}{#4}
\fi
\else
{
% decomposition (semikite => 2 semikites and 1 semidart)
\edef\dep{#1}
\edef\ver{#2}
\edef\angle{#3}
\edef\len{#4}
\edef\rot{#5}
\pgfmathtruncatemacro{\n}{\dep-1}
\edef\namex{\ver\n}
\pgfmathsetlengthmacro{\newlen}{\len*\invphi}
\ifnum#5=1 % anticlockwise or clockwise ?
\path (\ver) ++(\angle-36:\len) coordinate (\namex);
\pgfmathtruncatemacro{\newanglea}{mod(\angle+108,360)}
\penrosekite{\n}{\namex}{\newanglea}{\newlen}{1}
\penrosekite{\n}{\namex}{\newanglea}{\newlen}{0}
\penrosedart{\n}{\ver}{\angle}{\newlen}{1}
\else
\path (\ver) ++(\angle+36:\len) coordinate (\namex);
\pgfmathtruncatemacro{\newanglea}{mod(\angle-108,360)}
\penrosekite{\n}{\namex}{\newanglea}{\newlen}{0}
\penrosekite{\n}{\namex}{\newanglea}{\newlen}{1}
\penrosedart{\n}{\ver}{\angle}{\newlen}{0}
\fi
}
\fi
}
\newcommand\penrosedrawdart[3]{
\path let
\p1=(#1),
\p2=($(\p1) + (#2:#3)$),
\p3=($(\p1) + (#2-36:#3*\invphi)$),
\p4=($(\p1) + (#2-72:#3)$)
in [penrose dart] (\p1)
to[penrose rev path 1] (\p2)
to[penrose path 2] (\p3)
to[penrose rev path 2] (\p4)
to[penrose path 1] (\p1);
}
\newcommand\penrosedart[5]{% n, ver, angle, len, rot
\ifnum#1=0 % draw or recursive decomposition ?
\ifnum#5=1 % draw dart if current semidart is clockwise
\penrosedrawdart{#2}{#3}{#4}
\fi
\else
{
% decomposition (semidart => 1 semikite and 1 semidart)
\edef\dep{#1}
\edef\ver{#2}
\edef\angle{#3}
\edef\len{#4}
\edef\rot{#5}
\pgfmathtruncatemacro{\n}{\dep-1}
\edef\namex{\ver\n}
\pgfmathsetlengthmacro{\newlen}{\len*\invphi}
\path (\ver) ++(\angle:\len) coordinate (\namex);
\ifnum#5=1 % anticlockwise or clockwise
\pgfmathsetmacro{\newanglea}{mod(\angle-144,360)}
\pgfmathsetmacro{\newangleb}{mod(\angle-36,360)}
\penrosedart{\n}{\namex}{\newanglea}{\newlen}{1}
\penrosekite{\n}{\ver}{\newangleb}{\newlen}{0}
\else
\pgfmathtruncatemacro{\newanglea}{mod(\angle+144,360)}
\pgfmathtruncatemacro{\newangleb}{mod(\angle+36,360)}
\penrosedart{\n}{\namex}{\newanglea}{\newlen}{0}
\penrosekite{\n}{\ver}{\newangleb}{\newlen}{1}
\fi
}
\fi
}
\pgfmathsetlengthmacro{\len}{8cm}
\pgfmathsetmacro{\recurs}{int(3)}
\begin{document}
\begin{tikzpicture}
\tikzset{
penrose path 1/.style={to path={
-- ($(\tikztostart)!.4!(\tikztotarget)$)
-- ($(\tikztostart)!.4!30:(\tikztotarget)$)
-- ($(\tikztotarget)!.4!-30:(\tikztostart)$)
-- ($(\tikztotarget)!.4!(\tikztostart)$)
-- (\tikztotarget)
\pgfextra{
\pgfinterruptpath
\draw ($(\tikztotarget)!.5!-10:(\tikztostart)$) circle(2pt);
\fill ($(\tikztotarget)!.5!-8:(\tikztostart)$) circle(1pt);
\endpgfinterruptpath
}
}},
penrose rev path 1/.style={to path={
-- ($(\tikztostart)!.4!(\tikztotarget)$)
-- ($(\tikztostart)!.4!-30:(\tikztotarget)$)
-- ($(\tikztotarget)!.4!30:(\tikztostart)$)
-- ($(\tikztotarget)!.4!(\tikztostart)$)
-- (\tikztotarget)
\pgfextra{
\pgfinterruptpath
\draw ($(\tikztostart)!.5!-10:(\tikztotarget)$) circle(2pt);
\fill ($(\tikztostart)!.5!-8:(\tikztotarget)$) circle(1pt);
\endpgfinterruptpath
}
}},
penrose path 2/.style={to path={
-- ($(\tikztostart)!.4!(\tikztotarget)$)
-- ($(\tikztostart)!.4!30:(\tikztotarget)$)
-- ($(\tikztotarget)!.4!-30:(\tikztostart)$)
-- ($(\tikztotarget)!.4!(\tikztostart)$)
-- (\tikztotarget)
\pgfextra{
\pgfinterruptpath
\draw ($(\tikztostart)!.5!10:(\tikztotarget)$) circle(2pt);
\fill ($(\tikztostart)!.5!8:(\tikztotarget)$) circle(1pt);
\endpgfinterruptpath
}
}},
penrose rev path 2/.style={to path={
-- ($(\tikztostart)!.4!(\tikztotarget)$)
-- ($(\tikztostart)!.4!-30:(\tikztotarget)$)
-- ($(\tikztotarget)!.4!30:(\tikztostart)$)
-- ($(\tikztotarget)!.4!(\tikztostart)$)
-- (\tikztotarget)
\pgfextra{
\pgfinterruptpath
\draw ($(\tikztotarget)!.5!10:(\tikztostart)$) circle(2pt);
\fill ($(\tikztotarget)!.5!8:(\tikztostart)$) circle(1pt);
\endpgfinterruptpath
}
}},
}
\tikzset{
penrose line/.style={draw=black,line width=.2pt,line join=round,rounded corners=3pt},
}
\foreach \level in {0,...,4}{
\begin{scope}[rotate=\level*72]
\coordinate (a) at (0,0);
\penrosekite{\recurs}{a}{0}{\len}{0}
\penrosekite{\recurs}{a}{0}{\len}{1}
\end{scope}
}
\end{tikzpicture}
\end{document}
-
I'm not sure this looks like a bird or birds to me. But it is certainly very cool looking! – cfr yesterday
Like the TikZ to path solution. – Marc van Dongen yesterday
For me, this looks like the poor birds were squished with a big stone or something... =( – JBFWP286 yesterday | 8,657 | 20,990 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2016-30 | latest | en | 0.957469 |
www.appelhuis.nl | 1,632,606,242,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057775.50/warc/CC-MAIN-20210925202717-20210925232717-00277.warc.gz | 670,716,425 | 6,903 | [email protected]
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Both of these concrete calculators make an allowance for the fact that material losses volume after being mixed to make concrete Calculator are provided for a general mix and a paving mix the different ratio of materials are General mix 15 cementallin ballast or 121893189 cementsharp sandgravel and.
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PreMix Concrete Bag Calculator Calculated at 2160 kg per 1m179 Allow extra for waste Cubic Metres 20 kg Bags 25 kg Bags 30 kg Bags 40 kg Bags per bag. | 2,305 | 10,207 | {"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} | 2.921875 | 3 | CC-MAIN-2021-39 | latest | en | 0.759295 |
http://binarynode.info/index.php/ebooks/computer-simulation-methods-in-theoretical-physics | 1,620,260,383,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988724.75/warc/CC-MAIN-20210505234449-20210506024449-00057.warc.gz | 6,445,497 | 8,949 | # Computer Simulation Methods in Theoretical Physics by Dieter W. Heermann
By Dieter W. Heermann
Computational equipment relating many branches of technology, comparable to physics, actual chemistry and biology, are awarded. The textual content is essentially meant for third-year undergraduate or first-year graduate scholars. notwithstanding, energetic researchers eager to know about the recent strategies of computational technological know-how also needs to take advantage of examining the e-book. It treats all significant tools, together with the robust molecular dynamics process, Brownian dynamics and the Monte-Carlo procedure. All tools are taken care of both from a theroetical perspective. In each one case the underlying concept is gifted after which useful algorithms are displayed, giving the reader the chance to use those tools without delay. For this goal workouts are incorporated. The e-book additionally positive factors entire application listings prepared for program.
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The pioneering paintings of Edwin T. Jaynes within the box of statistical physics, quantum optics, and likelihood idea has had an important and lasting impact at the learn of many actual difficulties, starting from basic theoretical questions via to sensible purposes corresponding to optical snapshot recovery.
State-Selected and State-to-State Ion-Molecule Reaction Dynamics. P. 2. Theory
The purpose of this sequence is to assist the reader receive common information regarding a large choice of themes within the wide box of chemical physics. specialists current analyses of matters of curiosity to stimulate new study and inspire the expression of person issues of view.
Extra info for Computer Simulation Methods in Theoretical Physics
Sample text
Note that only the average temperature is fixed. We have already encountered a method to constrain the kinetic energy to a given value. 47,57,58]. After reaching the desired energy or temperature the system was left to itself. , do n = 1, max time step loop 1. Compute the forces. 2. Compute rn+l = 91 (rn , vn , F" ) • 3. Compute vn+ 1 = 92 (vn , F" , (F"+1 ) ) • 4. Compute the kinetic energy. 5. Scale the velocities vn+1 ..... vn+1 {J • end time step loop The functions gl and g2 denote the recursion relations.
Repeat from step I until the system has reached equilibrium. The success of the procedure depends on the initial positions and the distribution of the velocities. A common practice is to set up the system on a lattice and assign velocities according to a Boltzmann distribution. Sometimes, instead of the velocities being scaled, they are all set to zero. In any case, one has to check the velocity distribution after the equilibration phase has been reached to make sure that it has the equilibrium Maxwell-Boltzmann form.
11 A problem of more a current research effort is, can you think of ways to parallelize the molecular dynamics simulation? 12 Can you think of another way to introduce periodic boundary conditions? (Hint: 40, Quaterions). 50 4. Stochastic Methods This chapter is concerned with methods which use stochastic elements to compute quantities of interest. These methods are not diametrically opposed to the deterministic ones. Brownian dynamics provides an example where the two methods are combined to form a hybrid technique. | 681 | 3,428 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | longest | en | 0.905058 |
https://stacks.math.columbia.edu/tag/0FDU | 1,656,240,987,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103205617.12/warc/CC-MAIN-20220626101442-20220626131442-00686.warc.gz | 589,982,958 | 6,338 | Remark 42.33.8. There is a more general type of bivariant class that doesn't seem to be considered in the literature. Namely, suppose we are given a diagram
$X \longrightarrow Z \longleftarrow Y$
of schemes locally of finite type over $(S, \delta )$ as in Situation 42.7.1. Let $p \in \mathbf{Z}$. Then we can consider a rule $c$ which assigns to every $Z' \to Z$ locally of finite type maps
$c \cap - : \mathop{\mathrm{CH}}\nolimits _ k(Y') \longrightarrow \mathop{\mathrm{CH}}\nolimits _{k - p}(X')$
for all $k \in \mathbf{Z}$ where $X' = X \times _ Z Z'$ and $Y' = Z' \times _ Z Y$ compatible with
1. proper pushforward if given $Z'' \to Z'$ proper,
2. flat pullback if given $Z'' \to Z'$ flat of fixed relative dimension, and
3. gysin maps if given $D' \subset Z'$ as in Definition 42.29.1.
We omit the detailed formulations. Suppose we denote the collection of all such operations $A^ p(X \to Z \leftarrow Y)$. A simple example of the utility of this concept is when we have a proper morphism $f : X_2 \to X_1$. Then $f_*$ isn't a bivariant operation in the sense of Definition 42.33.1 but it is in the above generalized sense, namely, $f_* \in A^0(X_1 \to X_1 \leftarrow X_2)$.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 420 | 1,388 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 2, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2022-27 | latest | en | 0.728418 |
https://economicswriters.com/this-problem-i-need-a-detailed-explanation/ | 1,652,750,998,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662515466.5/warc/CC-MAIN-20220516235937-20220517025937-00007.warc.gz | 312,352,940 | 30,324 | # this problem i need a detailed explanation
An audio engineer finds that her mixes sound best when the reverb () is less than or equal to this expression: and the reverb is more than or equal to this expression: , where represents the number of instruments in the mix. These expressions can be written as inequalities:
Hereâ€s a graph** of these two inequalities:
a. Explain what steps you would take to find the points of intersection of the two curves on the graph
b. Determine which of the following points lie in the optimal range:
(8, 80), (16, 100), (22, 80 ), (20, 80), (16, 60), (19.5, 73.6)
c. How do you know which of these points is in the optimal range when you use the graph?
d. How would you prove which of these points are in the optimal range by using algebra?
**For your convenience, you can access a more interactive graph here. You can use the magnification tools on the graph to zoom (though the interactive graph does not have any shading for the inequalities).
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The post this problem i need a detailed explanation appeared first on Custom Nursing Help. | 333 | 1,423 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2022-21 | latest | en | 0.931444 |
https://www.talkstats.com/tags/data-analysis/page-3 | 1,659,924,886,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570741.21/warc/CC-MAIN-20220808001418-20220808031418-00075.warc.gz | 868,457,759 | 13,315 | # data analysis
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https://www.weegy.com/?ConversationId=TK2A2APJ | 1,631,828,239,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780053759.24/warc/CC-MAIN-20210916204111-20210916234111-00153.warc.gz | 1,106,958,983 | 9,986 | solve for x 7x 9=6x-5
Question
Updated 11/17/2014 5:39:53 AM
Confirmed by jeifunk [11/17/2014 5:39:53 AM]
s
Original conversation
User: solve for x 7x 9=6x-5
Weegy: 7x+9=6x-5
7x-6x=-5-9
x = -14
User: can you solve (x+2)(x-3)-(x^2-4)
Question
Updated 11/17/2014 5:39:53 AM
Confirmed by jeifunk [11/17/2014 5:39:53 AM]
Rating
8
(x+2)(x-3)-(x^2-4)
= x^2-3x+2x-6-x^2+4
= -x-2
Confirmed by jeifunk [11/17/2014 5:39:48 AM]
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https://www.teacherspayteachers.com/Browse/Search:multiplication%20and%20division%20equations | 1,527,028,538,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864968.11/warc/CC-MAIN-20180522205620-20180522225620-00274.warc.gz | 842,327,254 | 82,399 | showing 1-52 of 3,857 results
These Multiplication and Division Number Puzzles help students learn their multiplication facts. They are simple puzzles to determine an unknown number. There are three version of the number puzzles: • One version has four problems / answers in one puzzle. It’s simply a fun way to get four pr
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Get this as part of a bundle: Number Puzzles for Third Grade BUNDLE
\$4.29
60 Ratings
4.0
PDF (3.68 MB)
Are your students working on solving one step equations? This QR code scavenger hunt is just what you need to give them a little extra practice in a fun and engaging way. This fun scavenger hunt will get your students up and moving around the room using technology to practice solving one step
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\$3.75
11 Ratings
4.0
PDF (1.1 MB)
This 21- question, leaf- shaped puzzle provides students with practice solving one-step equations involving multiplication & division. There are TWO options (equations with whole numbers OR integers). You will receive: • 2 pages worth of puzzle pieces • student recording sheet • answer key • d
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\$3.00
11 Ratings
4.0
ZIP (623.61 KB)
This document consists of 6 pages (intended to be used as three two-sided tasks) aligned to the 3rd Grade Math Common Core Standard 3.OA.4 "Determine the unknown whole number in a multiplication or division equation relating three whole numbers. For example, determine the unknown number that makes t
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\$3.00
11 Ratings
3.9
PDF (135.08 KB)
Help students understand basic algebra for multiplication and division with this scaffolded packet. Students start by figuring out what number a pig clip art is covering. They can also create their own problems by cutting out and placing their own pigs. These add your own pig pages are available in
Types:
CCSS:
\$3.25
13 Ratings
4.0
PDF (13.89 MB)
These differentiated practice pages are to assess students ability to find a missing number in a multiplication or division equation. It contains problems for students to find a missing digit, compare multiplication and division equations where students must decide if the equation is equal or not, a
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CCSS:
\$2.00
11 Ratings
4.0
PDF (64.62 KB)
Whenever my students hear “algebra,” I see a classroom full of panicked faces! To ease their worried looks, I decided to make an inviting and engaging math center for my students! Students are asked to join the Construction Crew to fix up and build the algebraic equations wherever a variable is.
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\$3.00
9 Ratings
4.0
PDF (9.07 MB)
This is an excellent way to introduce solving one step equations with multiplication or division to the whole class. It would also be suitable to have a students work on independently if they missed the lesson or to put on your webpage as a review resource. Please be sure to rate & leave feed
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\$2.00
3 Ratings
4.0
PPT (195.5 KB)
#### Videos matching "multiplication and division equations" (BETA)
Dip your students toes into multiplication and division equations with this game "I Have Who Has". The game gets progressively more challenging, and is in alignment with 4.OA.2 standards including multiplication and division comparisons. *To be clear this is NOT to be used for a lesson with variab
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\$1.00
7 Ratings
4.0
PDF (141.2 KB)
This file contains 21 problems that are designed to give students extra practice working on Multiplication and Division Equations. I have created this file so that it can be copied double-sided and used as a worksheet, or you can copy the file, cut out the cards, and use them as Task Cards. In addi
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\$1.50
3 Ratings
4.0
PDF (191.68 KB)
These notes introduce applying multiplication and division properties to solve equations. The notes include a key and guided questions for students to work through with the teacher.
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\$1.00
2 Ratings
4.0
PDF (41.14 KB)
Solve addition and subtraction equations to solve the riddle! Answer key included. Aligns with Pearson digits 6 lesson 3-4.
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\$1.00
2 Ratings
4.0
PDF (417.84 KB)
Multiplication and Division Equations Interactive Notebook Foldable by Math Doodles Included: - Black and white student copy - Red ink KEY and colored KEY - PowerPoint to show how to fill in the blanks - Folding Movie and printing directions No Cutting Mess! Make 2 sided copy, students
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\$3.00
1 Rating
4.0
ZIP (11.6 MB)
Your students will have fun while solving one step and two step word problems! Just print the problems and tape around your classroom. Give each student or pair of students a record sheet and watch them solve multiplication and division problems and pick the correct equation. The problems are a co
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\$3.00
1 Rating
4.0
PDF (1.57 MB)
Let these cute little monsters get your kiddos excited about solving multiplication and division equations. Start out with a slide show where students will solve problems in a whole group setting. Then move on to a worksheet or two for practice. Continue on with a set of task cards for independ
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\$5.00
1 Rating
4.0
PDF (9.64 MB)
Differentiated by interest. Students will be given one and two digit numbers and asked to write a multiplication or division sentence using a variable. All the numbers are the same in each problem, but the wording is different for each page depending on the student's interest. The interests included
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\$2.00
1 Rating
4.0
DOCX (18.57 KB)
This is a handout that students can work in a group or together as a whole class. Each word problem has a tape diagram and a place to define the variable and write an equation. There are 5 examples and 5 practice problems. Shrink the document if you want to use it in an interactive notebook.
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\$1.00
1 Rating
3.5
DOCX (19.77 KB)
Worksheet aligns with common core 6.EE.B.7. Worksheet uses slightly larger numbers to encourage students to show solving using the inverse property.
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\$1.50
1 Rating
4.0
DOC (27.5 KB)
This bundle includes differentiated worksheets, hands-on sorting activities and answer keys. Great resource for homework, classwork, centers and enrichment activities! Cheat sheets that include steps and tiered worksheets to meet every student's individual needs! Answer keys included!
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\$3.00
1 Rating
4.0
PDF (7.19 MB)
Use this differentiated BUMP game to help your students practice solving one step multiplication & division equations! Simply print & go! Recording sheets & ANSWER KEYS are included! Black & white AND color versions of each game board are included. There are 2 BUMP games in this
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CCSS:
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PDF (1.61 MB)
Every steps to MATH success lessons have a guided notes to along with them. This guided note page goes along along with the lesson solving 1-step multiplication and division equations.
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DOC (27 KB)
This activity is a great way for students to review the rules for solving multiplication and division equations. It can be done as a whole group, in pairs or small groups, and the answer key is provided. Great for review to check for understanding or to use as a station activity. Students LOVE this
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PDF (887.84 KB)
Trying to teach your students how to create multiplication and division equations with one unknown variable? Having a tough time getting them to see how it works? Try this powerpoint presentation! It is completely animated, just hit the space bar and the next step shows up! Take the pain out of solv
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ZIP (15.26 MB)
All steps to MATH success lessons include a step-by-step explanation, many example problems, formative assessment opportunities throughout the lesson, and a review game. This lesson cotains 45 slides of examples and review for maximum student learning opportunities.
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NOTEBOOK (466.87 KB)
Doodle notes that are aligned with PA Common Core Standards for Chapter 6 Lesson 2 of the Glencoe Math Course 2 Volume 1 Book. These are the doodle notes that I have used in my classroom as a part of my interactive student notebooks (ISNs). It includes a Doodle note for your students and then one wi
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DOCX (53.72 KB)
This is a game that allows kiddos to practice inverse multiplication and division equations. The are 27 game cards.
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FREE
2 Ratings
4.0
PDF (8.9 MB)
This song teaches multiplication and division when calculating a mathematical problem.
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A set of 230Task Cards for your students to practice multiplication (and a few division) skills. This is a bundle of my 7 Multiplication Task Card products. **Answer Keys and Recording Sheets included for all!** Descriptions Below: Multiplication Word Problem Task Cards Common Core, One step an
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 4,437 | 18,265 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-22 | latest | en | 0.899776 |
https://brilliant.org/discussions/thread/a-nice-but-easy-proof/ | 1,586,165,881,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371620338.63/warc/CC-MAIN-20200406070848-20200406101348-00437.warc.gz | 341,566,367 | 20,968 | # A nice but easy proof!
Prove that the value of, $\sum_{n=1}^{n}\dfrac{1}{n^2}<2$.Feel free to post your innovative methods! I have posted mine below!
4 years, 5 months ago
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We basically have to prove that,$\sum_{n=2}^{n}\dfrac{1}{n^2}<1$ $\text{Motivation of proof}:$,the first idea that struck me was the use of telescoping series,if i could write,$\dfrac{1}{n^2}<\text{something}$ and write,$\dfrac{1}{(n-1)^2}<\text{another something}$,and when we add the terms of $L.H.S$ we would get the required expression and when we add the terms of the $R.H.S$ we would get a telescoping series.The next thing that came to my mind was that,$n^2>(n)(n-1)$(we have taken the greater than sign as then when we take the reciprocal the sign would get reversed),since $\dfrac{1}{(n)(n-1)}=\dfrac{1}{n-1}-\dfrac{1}{n}$,and we would get a telescoping series like this, $\dfrac{1}{2^2}<\dfrac{1}{1}-\dfrac{1}{2}\\ \dfrac{1}{3^2}<\dfrac{1}{2}-\dfrac{1}{3}\\ .\\ .\\ .\\ \dfrac{1}{n^2}<\dfrac{1}{n-1}-\dfrac{1}{n}\\ \text{adding,we get}:\\ \sum_{n=2}^{n}\dfrac{1}{n^2}<1-\dfrac{1}{n}<1$.Hence proved.And done!
- 4 years, 5 months ago
I like your motivations haha :P
- 4 years, 5 months ago
From where you got this motivation ?
- 4 years, 5 months ago
I am sorry but i don't understand the meaning of your comment.Could you please explain?
- 4 years, 5 months ago
Simplification: Source of the question = ?
- 4 years, 5 months ago
- 4 years, 5 months ago
Well, yours is the simplest way. For the sake of mentioning, one can also write a proof by induction for this.
- 4 years, 5 months ago
How do you do a proof by induction (that is fundamentally different from his approach)?
Staff - 4 years, 5 months ago
@Calvin Lin Oops, although Induction is tempting at first look, it isn't the best way to go about. I haven't found an inductive proof yet 😕.
- 4 years, 5 months ago
Right. The inductive proof that I know, is to show that for $n \geq 2$,
$\sum_{i=1}^n \frac{1}{i^2 } < 2 - \frac{1}{n}.$
This is similar to what Adarsh did.
Staff - 4 years, 5 months ago
from the basel problem we get $\sum_{n=1}^\infty (\frac{1}{n^2}) = \frac{\pi^2}{6}<2$. although this is probably not intended, but still works as a good proof. we prove the basel proble: Euler's original derivation of the value $\frac{π^2}{6}$ essentially extended observations about finite polynomials and assumed that these same properties hold true for infinite series.
recall the Taylor series expansion of the sine function
$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots.$ Dividing through by x, we have
$\frac{\sin(x)}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \cdots.$ Using the Weierstrass factorization theorem, it can also be shown that the left-hand side is the product of linear factors given by its roots, just as we do for finite polynomials
\begin{aligned} \frac{\sin(x)}{x} &= \left(1 - \frac{x}{\pi}\right)\left(1 + \frac{x}{\pi}\right)\left(1 - \frac{x}{2\pi}\right)\left(1 + \frac{x}{2\pi}\right)\left(1 - \frac{x}{3\pi}\right)\left(1 + \frac{x}{3\pi}\right) \cdots \\ &= \left(1 - \frac{x^2}{\pi^2}\right)\left(1 - \frac{x^2}{4\pi^2}\right)\left(1 - \frac{x^2}{9\pi^2}\right) \cdots. \end{aligned} If we formally multiply out this product and collect all the x2 terms (we are allowed to do so because of Newton's identities), we see that the x2 coefficient of sin(x)/x is
$-\left(\frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \cdots \right) = -\frac{1}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2}.$ But from the original infinite series expansion of sin(x)/x, the coefficient of x2 is −1/(3!) = −1/6. These two coefficients must be equal; thus,
$-\frac{1}{6} = -\frac{1}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2}.$ Multiplying through both sides of this equation by $-\pi^2$ gives the sum of the reciprocals of the positive square integers.
$\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}.$
- 4 years, 5 months ago
You are right,it wasn't intended but if you write this,you are expected to prove it too.
- 4 years, 5 months ago
fine
- 4 years, 5 months ago
Thanks a lot for adding the proof!
- 4 years, 5 months ago
Isn't this reimann zeta(2)? Then the value will be 1.64 only.
- 4 years, 4 months ago
Yes,it is but that isn't the intended proof,or else you will have to prove how you found the value of $\zeta{(2)}$.
- 4 years, 4 months ago
Oh so that's why proof! Btw my method was just like yours
- 4 years, 4 months ago | 2,021 | 6,136 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 32, "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} | 3.25 | 3 | CC-MAIN-2020-16 | longest | en | 0.85489 |
https://giasutamtaiduc.com/surface-area-of-a-square-pyramid-formula.html | 1,712,993,578,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816586.79/warc/CC-MAIN-20240413051941-20240413081941-00835.warc.gz | 252,816,243 | 23,669 | # Surface Area of a Square Pyramid Formula
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## Surface Area of a Square Pyramid Calculator
Are you in search of a surface area of a square pyramid calculator to estimate all types of areas of the Great pyramid?! If yes, you’re in the right place. You can calculate the total surface area, base area, lateral surface area, and face area of any square pyramid using our tool.
We also discuss how to find the surface area of a square pyramid using slant height and base length and how to calculate surface area using slant height and base perimeter. You can also find answers to some interesting numerical problems like the surface area of a pyramid of Giza and the amount of groundsheet required for any tent.
## How do I calculate the surface area of a square pyramid?
To find the surface area of a square pyramid:
1. Determine how many faces are there on a square pyramid: there are 4 triangular faces. Sum up their areas.
2. Find the area of a square base.
3. Add up 4 triangular faces area to 1 square base area to find the surface area of a square pyramid.
## What is the formula for the surface area of a square pyramid?
The formula to calculate the surface area of a square pyramid is:
where:
• SA is the surface area of the square pyramid;
• a is the length of the base edge; and
• h is the height of the square pyramid.
## How do I calculate the base area of a square pyramid?
The base of a square pyramid is a square. Hence, the base area of the square pyramid of base edge a is a2.
## How to find the lateral area of a square pyramid
First of all, let’s explain what the lateral area of a square pyramid is. The lateral surface area or lateral area of a square pyramid is the total area of its 4 triangular faces. We calculate the lateral area as:
## How do I find the face area of a square pyramid?
The face area of a square pyramid is the area of one of its four triangular faces. We calculate the area of a triangular face, the face area (FA), using the formula:
## How to find the surface area of a square pyramid using slant height
To find the surface area using the slant height, we use the formula:
• SA = a2 + 2×a×l
🔎 Proof: The surface area of a square pyramid is the sum of the areas of its square base and four triangular faces:
• SA = BA + (4 × FA)
The area of a triangle is half of the product of its base length (a) and height (l):
• FA = a×l/2
Therefore, the area of four triangular faces or the lateral surface area of the square pyramid is:
• 4 × FA = 2 × a × l
Thus, the lateral surface area (LSA) of the square pyramid of slant height l is
• LSA = 2 × a × l
and the total surface area is
• SA = a2 + 2 × a × l
## How to use the surface area of a square pyramid calculator
❓ The Pyramid of Giza has a height of 480 feet. If the length of each side of the base is approximately 756 feet, what is its total surface area?
Okay, let’s see how to use this total surface area of a square pyramid calculator to solve the problem given above:
1. Identify the measurements:
• Height (h) = 480 ft
• Base edge (a) = 756 ft
2. Switch the units from cm to ft and cm2 to ft2 (or the desired units for the area) using the drop-down list near each variable in the square-based pyramid surface area calculator.
3. Enter the values for height (480 ft) and base edge (756 ft).
4. Ta-da! You got the results in no time! Our tool uses the surface area square pyramid formula to find:
• Slant height (l) = 611 ft
• Total surface area (SA) = 1,495,322 ft2
• Base area (BA) = 571,536 ft2
• Lateral surface area (LSA) = 923,786 ft2
• Face area (FA) = 230,947 ft2
## FAQ
### How many faces are on a square pyramid?
A square pyramid has 5 faces: 4 equal triangles (side faces) and 1 square (base face).
### How do I calculate the surface area of a square pyramid using its base perimeter?
The surface area of a square pyramid is half of the product of its base perimeter and its slant height:
• SA = P × l / 2
where P is the base perimeter and l is the slant height.
### How many square meters of groundsheet is required for a tent of base 5m and height 1.8m?
You need (5 m)2 = 25 m2 of groundsheet for the tent of base length 5 m irrespective of its height. The amount of groundsheet in square meters is the base area of the tent. The base area of a tent is BA = a2, where a is the base length.
## Surface Area of a Square Pyramid
In this section, we will learn about the surface area of a square pyramid. A pyramid is a 3-D object whose all side faces are congruent triangles and whereas its base can be any polygon. One side of each of these triangles coincides with one side of the base polygon. A square pyramid is a pyramid whose base is a square. The pyramids are named according to the shape of their bases. Just like other three-dimensional shapes, a square pyramid also has two types of areas.
• Total Surface Area (TSA)
• Lateral Surface Area (LSA)
Let us learn about the surface area of a square pyramid along with the formula and a few solved examples here. You can find a few practice questions in the end.
## What is the Surface Area of a Square Pyramid?
The word “surface” means ” the exterior or outside part of an object or body”. So, the total surface area of a square pyramid is the sum of the areas of its lateral faces and its base. We know that a square pyramid has:
• a base which is a square.
• 4 side faces, each of which is a triangle.
All these triangles are isosceles and congruent, each of which has a side that coincides with a side of the base (square).
So, the surface area of a square pyramid is the sum of the areas of four of its triangular side faces and the base area which is square.
## Formula of Surface Area of a Square Pyramid
Let us consider a square pyramid whose base’s length (square’s side length) is ‘a’ and the height of each side face (triangle) is ‘l’ (this is also known as the slant height). i.e., the base and height of each of the 4 triangular faces are ‘a’ and ‘l’ respectively. So the base area of the pyramid which is a square is a × a = a2 and the area of each such triangular face is 1/2 × a × l. So the sum of areas of all 4 triangular faces is 4 ( ½ al) = 2 al. Let us now understand the formulas to calculate the lateral and total surface area of a square pyramid using height and slant height.
### Total Surface Area of Square Pyramid Using Slant Height
The total surface area of a square pyramid is the total area covered by the four triangular faces and a square base. The total surface area of a square pyramid using slant height can be given by the formula,
Surface area of a square pyramid = a2 + 2al
where,
• a = base length of square pyramid
• l = slant height or height of each side face
### Total Surface Area of a Square Pyramid Using Height
Let us assume that the height of the pyramid (altitude) be ‘h’. Then by applying Pythagoras theorem (you can refer to the below figure),
### Lateral Surface Area of a Square Pyramid
The lateral surface area of a square pyramid is the area covered by the four triangular faces. The lateral surface area of a square pyramid using slant height can be given by the formula,
Lateral surface area of a square pyramid = 2 al
or,
where,
• a = base length of square pyramid
• l = slant height or height of each side face
• h = height of square pyramid
### Lateral surface area of a square pyramid
The lateral surface area of a pyramid is the area occupied by its lateral surfaces or side faces. The formula for calculating the lateral surface area of a square pyramid using slant height is given as follows,
Lateral surface area (LSA) = ½ × perimeter × slant height
We know that,
The perimeter of a square = 4s
So, LSA = ½ × 4s × l = 2sl
Lateral surface area of a square pyramid (LSA) = 2sl square units
Where,
“s” is the base length of a square pyramid,
“l” is the slant height or height of each side face.
The slant height of the pyramid (l) = √(s2/4 + h2)
The formula for calculating the lateral surface area of a square pyramid using the height is given as follows,
Lateral surface area of a square pyramid = 2s√(s2/4 + h2) square units
where,
“s” is the base length of a square pyramid,
“h” is the height of the square pyramid.
### Total surface area of a square pyramid
The total surface area of a square pyramid is the sum of the areas of its lateral faces and its base area. The formula for calculating the total surface area of a square pyramid is given as follows,
Total surface area of a pyramid (TSA) = Lateral surface area of the pyramid (LSA) + Base area
The lateral surface area of the square pyramid (LSA)= 2sl square units
Base area = s2 square units
So, TSA = 2sl + s2
Total surface area of a square pyramid (TSA) = 2sl + s2 square units
where,
“s” is the base length of a square pyramid,
“l” is the slant height or height of each side face.
Slant height of the pyramid (l) = √(s2/4 + h2)
The formula for calculating the lateral surface area of a square pyramid using the height is given as follows,
Total surface area of a square pyramid (TSA) = 2s√(s2/4 + h2) + s2 square units
where,
“s” is the base length of a square pyramid,
“h” is the height of the square pyramid.
## How to Calculate Surface Area of Square Pyramid?
The surface area of a square pyramid can be calculated by representing the 3D figure into a 2D net. After expanding the 3D figure into a 2D net we will get one square and four triangles.
The following steps are used to calculate the surface area of a square pyramid :
• To find the area of the square base: a2, ‘a’ is the base length.
• To find the area of the four triangular faces: The area of the four triangular side faces can be given as: 2al, ‘l’ is the slant height. If slant height is not given, we can calculate it using height, ‘h’ and base
• Add all the areas together for the total surface area of a square pyramid, while the area of 4 triangular faces gives the lateral area of the square pyramid.
• Thus, the surface area of a square pyramid is a2 + 2al and lateral surface area as 2al in squared units.
Now, that we have seen the formula and method to calculate the surface area of a square pyramid, let us have a look at a few solved examples to understand it better.
## Examples on Surface Area of a Square Pyramid
Example 2: The height of a square pyramid is 25 units and the base area of a square pyramid is 256 square units. Find its surface area.
Solution
Let the side of the base (square) be ‘a’ units.
Then it is given that a2 = 256 ⇒ a = 16 units.
The height of the given square pyramid is h = 25 units.
Answer: The surface area of the given square pyramid = 1095.96 square units.
## FAQs on Surface Area of a Square Pyramid
### What Is the Surface Area of the Square Pyramid?
The surface area of a square pyramid is the sum of the areas of all its 4 triangular side faces with the base area of the square pyramid. If a, h, and l are the base length, the height of the pyramid, and slant height respectively, then the surface area of
### How Do You Find the Lateral Area of a Square Pyramid?
To find the lateral area of a square pyramid, find the area of one side face (triangle) and multiply it by 4. If a and l are the base length and the slant height of a square pyramid, then the lateral area of the square pyramid = 4 (½ × a × l) = 2al.
If h is the height of the pyramid, then the lateral area = 2a
### What Is the Area of One of the Triangular Faces of a Square Pyramid?
If a and l are the base length and the slant height of a square pyramid, then the area of one of the 4 triangular side faces is, ½ × a × l.
### How Do You Find the Lateral Area and Surface Area of a Square Pyramid?
The lateral area of a square pyramid is the sum of the areas of the side faces only, whereas the surface area is the lateral area + area of the base. The lateral area of a square pyramid = 2al (or) 2a
To get the total surface area, we need to add the area of the base (which is a2) to each of these formulas. The total surface area of a
where,
• a = Length of the base (square)
• l = Slant height
• h = Height of the pyramid
### How To Calculate Surface Area of a Square Pyramid Without Slant Height?
We know, slant height of a square pyramid is given in terms of
where,
• a = Length of the base (square)
• l = Slant height
• h = Height of the pyramid
### What Is the Base Area of a Square Pyramid?
The base of a square pyramid is square-shaped. Thus, the base area of square pyramid can be calculated using the formula, Base Area of Pyramid = a2, where, a is the length of the base of square pyramid.
### How Many Bases Does a Square Pyramid Have?
A square pyramid is a pyramid with a square-shaped base. A square pyramid thus has only one base.
### Which Two Shapes Make up a Square Pyramid?
The base of a square pyramid is a square and its side faces are triangles. So the two shapes that make up a square pyramid are square and triangle.
## Solved Examples based on Square Pyramid Formulas
Example 1: Determine the total surface area of a square pyramid if the base’s side length is 15 cm and the pyramid’s slant height is 21 cm.
Solution:
Given,
The side of the square base (s) = 15 cm
Slant height, (l) = 21 cm
The perimeter of the square base (P) = 4s = 4(15) = 60 cm
The lateral surface area of a square pyramid = (½) Pl
LSA = (½ ) × (60) × 21 = 630 sq. cm
Now, the total surface area = Area of the base + LSA
= s2 + LSA
= (15)2 + 630
= 225 + 630 = 855 sq. cm
Therefore, the total surface area of the given pyramid is 855 sq. cm.
Example 2: Determine the lateral surface area of a square pyramid if the side length of the base is 18 inches and the pyramid’s slant height is 22 inches.
Solution:
Given,
The side of the square base (s) = 18 inches
Slant height, (l) = 22 inches
The perimeter of the square base (P) = 4s = 4(18) = 72 inches
The lateral surface area of a square pyramid = (½) Pl
LSA = (½ ) × (72) × 22 = 792 sq. in.
Therefore, the lateral surface area of the given pyramid is 792 sq. in.
Example 3: What is the slant height of the square pyramid if its lateral surface area is 200 sq. in. and the side length of the base is 10 inches?
Solution:
Given data,
Length of the side of the base of a square pyramid (s) = 10 inches
The lateral surface area of a square pyramid = 200 sq. in
Slant height (l) = ?
We know that,
The lateral surface area of a square pyramid = (½) Pl
The perimeter of the square base (P) = 4s = 4(10) = 40 inches
⇒ 200 = ½ × 40 × l
⇒ l = 10 in
Hence, the slant height of the square pyramid is 10 inches.
Example 4: Calculate the side length of the base of the square pyramid if its lateral surface area is 480 sq. cm and the slant height is 24 cm.
Solution:
Given data,
The lateral surface area of a square pyramid = 480 sq. cm
Slant height (l) = 24 cm
Let the length of the side of the base of a square pyramid be “s”.
We know that,
The lateral surface area of a square pyramid = (½) Pl
The perimeter of the square base (P) = 4s
⇒ 480 = ½ × 4s × 24
⇒ s = 10 cm
Hence, the side length of the base of the square pyramid is 10 cm.
Example 5: Determine the total surface area of a square pyramid if the base’s side length is 14 cm and the pyramid’s height is 24 cm.
Solution:
Given,
The side of the square base (s) = 14 cm
The height of the square pyramid (h) = 24 cm
The slant height of the pyramid (l) = √[(s/2)2 + h2]
l = √[(14/2)2 + 242)] = √(49+576)
= √625 = 25 cm
The perimeter of the square base (P) = 4s = 4(14) = 56 cm
The lateral surface area of a square pyramid = (½) Pl
LSA = (½ ) × (56) × 25 = 700 sq. cm
Now, the total surface area = Area of the base + LSA
= s2 + LSA
= (14)2 + 700
= 196 + 700 = 896 sq. cm
Therefore, the total surface area of the given pyramid is 896 sq. cm.
Example 6: Determine the surface area of a square pyramid if the base’s side length is 10 cm and the pyramid’s slant height is 15 cm.
Solution:
Given,
The side of the square base (s) = 10 cm
Slant height (l) = 15 cm
We know that,
The total surface area of a square pyramid (TSA) = 2sl + s2 square units
= 2 × 10 × 15 + (10)2
= 300 + 100 = 400 sq. cm
Therefore, the surface area of the given pyramid is 400 sq. cm.
## FAQs on Square Pyramid Formula
Question 1: What is the area of the base of a Square Pyramid?
Base of a square pyramid is shaped as square. Thus, the area of the base of square pyramid can be calculated using the formula, for area of square.
Area of base Square Pyramid = a2,
where,
a is the length of the base of the square pyramid.
Question 2: How many bases does a Square Pyramid have?
A square pyramid is a pyramid with only one base in shape of a square. So, a square pyramid has only one base.
Question 3: Which two shapes make up a Square Pyramid?
The base of a square pyramid is a square and all its side faces are triangles. So there are two shapes that make up a square pyramid which are square and triangle.
Question 4: What is the area of one of the triangular faces of a Square Pyramid?
Suppose, a is the length of base and l is the slant height of a square pyramid, then the area of any one of the four triangular side faces is,
Area = ½ × a × l.
## Here is how to derive the surface area of a square pyramid.
Surface area of the square pyramid = area of the base + area of 4 triangles.
The area of the square base is s2
The area of one triangle is (s × l)/2
Since there are 4 triangles, the area is 4 × (s × l)/2 = 2 × s × l
Therefore, the surface area, call it SA is SA = s2 + 2 × s × l
## A couple of examples showing how to find the surface area of a square pyramid.
Example #1:
Find the surface area of a square pyramid with a base length of 5 cm, and a slant height of 10 cm.
SA = s2 + 2 × s × l
SA = 52 + 2 × 5 × 10
SA = 25 + 10 × 10
SA = 25 + 100
SA = 125 cm2
Example #2:
Find the surface area with a base length of 3 cm, and a slant height of 2 cm.
SA = s2 + 2 × s × l
SA = 32 + 2 × 3 × 2
SA = 9 + 6 × 2
SA = 9 + 12
SA = 21 cm2
Example #3:
Find the surface area with a base length of 1/2 cm, and a slant height of 1/4 cm.
SA = s2 + 2 × s × l
SA = (1/2)2 + 2 × 1/2 × 1/4
SA = (1/2)×(1/2) + 2 × 1/2 × 1/4
SA = 1/4 + 1 × 1/4
SA = 1/4 + 1/4
SA = 2/4
SA = 1/2 cm2
Question: What is the surface area of the square pyramid which has the base length of 8 cm and side 5 cm?
Solution:
Given,
Base length = 8 cm
Side length = 5 cm
Using the formula:
## General Formula for Surface Area of Pyramid
The general formula for the surface area of the pyramid is as follows:
The lateral surface area of the regular pyramid formula is given by:
Lateral Surface Area of Regular Pyramid = (½) Pl Square units
Similarly, the total surface area of the regular pyramid formula is given by:
Total Surface Area of Regular Pyramid = (½)Pl + B Square units
Where “l” is the slant height of a pyramid
“P” is the base perimeter of a pyramid
“B” represents the base area of a pyramid
## Surface Area of Pyramid Formulas
Generally, if we are asked to find the surface area of the pyramid without any specifications, it represents the total surface area. Now, let us discuss the surface area of pyramid formulas with different bases.
### Triangular Pyramid
In the triangular pyramid, the base of the pyramid is a triangle.
Surface Area of Triangular Pyramid = 1⁄2(a × b) + 3⁄2(b × s) square units
Where, b = side length, a = height, s = slant height.
### Square Pyramid
In the Square pyramid, the base of the pyramid is a square.
Surface Area of Square Pyramid = 2bs + b2 Square units
Where, b = side length, s = slant height.
### Pentagonal Pyramid
In the pentagonal pyramid, the base of the pyramid is a pentagon.
Surface Area of Pentagonal Pyramid = 5⁄2(a × b) + 5⁄2(b × s) Square units
Where, b = side length, s = slant height, a = apothem length.
### Hexagonal Pyramid
In the hexagonal pyramid, the base of the pyramid is a hexagon.
Surface Area of Hexagonal Pyramid =3(a × b) + 3(b × s) Square units
Where, b = side length , s = slant height, a= apothem length.
## Surface Area of Pyramid Examples
Go through the solved examples on the surface area of the pyramid:
Example 1:
Determine the surface area of the square pyramid, given that side length = 4 cm and slant height = 8 cm.
Solution:
Given: Side length, b = 4 cm
Slant height, s = 8 cm
We know that the surface area of square pyramid = 2bs + b2 Square units
Now, substitute the known values in the formula
S.A = 2(4)(8) + (4)2
S.A = 64+16 = 80
Hence, the surface area of the square pyramid is 80 cm2.
Example 2:
Compute the base area of the pentagonal pyramid, given that side length = 6.4 m and apothem = 16 m.
Solution:
Given: Side length, b = 6.4 m
Apothem = 16 m
We know that the base of a pentagonal pyramid is a pentagon. Hence, base area = 5⁄2(a × b) square units.
Base area = (5/2)(16×6.4)
Base area = 5(8×6.4) = 5(51.2) = 256
Thus, the base area of a pentagonal pyramid is 256 m2.
### Practice Questions
Solve the following problems:
1. Determine the surface area of the triangular pyramid, given that side length = 2 cm, height = 4 cm and slant height = 5 cm.
2. Compute the base area of the square pyramid whose side length is equal to 7 cm.
3. Find the surface area of the square pyramid given that side length = 5 cm and slant height = 7 cm.
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## Frequently Asked Questions on Surface Area of Pyramid
### What is the surface area of a pyramid?
The surface area of the pyramid is the total area occupied by the surface of the pyramid.
### What is the general formula for the lateral surface area of a regular pyramid?
If “P” is the base perimeter and “l” is the slant height, then the lateral surface area of the regular pyramid formula is given by (½) Pl Square units.
### What is the general formula for the total surface area of a regular pyramid?
If “P” is the base perimeter, “l” is the slant height, and “B” is the base area, then the total surface area of the regular pyramid formula is given by (½)Pl + B Square units.
### Surface Area Of A Pyramid
A pyramid is a solid with a polygonal base and several triangular lateral faces. The lateral faces meet at a common vertex. The number of lateral faces depends on the number of sides of the base. The height of the pyramid is the perpendicular distance from the base to the vertex.
A regular pyramid has a base that is a regular polygon and a vertex that is above the center of the polygon. A pyramid is named after the shape of its base. A rectangular pyramid has a rectangle base. A triangular pyramid has a triangle base.
We can find the surface area of any pyramid by adding up the areas of its lateral faces and its base.
Surface area of any pyramid = area of base + area of each of the lateral faces
If the pyramid is a regular pyramid, we can use the formula for the surface area of a regular pyramid.
Surface area of regular pyramid = area of base + 1/2 ps
where p is the perimeter of the base and s is the slant height.
If the pyramid is a square pyramid, we can use the formula for the surface area of a square pyramid.
Surface area of square pyramid = b2 + 2bs
where b is the length of the base and s is the slant height.
Example:
Calculate the surface area of the following pyramid.
Solution:
Sketch a net of the above pyramid to visualize the surfaces.
Since the given pyramid is a square pyramid, we can use any of the above formulas.
Using the formula for the surface area of any pyramid:
Area of base = 6 × 6 = 36 cm2
Area of the four triangles = 1/2 × 6 × 12 × 4 = 144 cm2
Total surface area = 36 + 144 = 180 cm2
Using the formula for a regular pyramid
Surface area of regular pyramid = area of base + 1/2 ps
= 6 × 6 + 1/2 × 6 × 4 × 12 = 180 cm2
Using the formula for a square pyramid
Surface area of square pyramid = b2 + 2bs
= 6 × 6 + 2 × 6 × 12 = 180 cm2
Related Topics
Completing the Square Formula
Chi Square Formula
Associative Property Formula
Commutative Property Formula ⭐️⭐️⭐️⭐️⭐
Square Root Property Formula ⭐️⭐️⭐️⭐️⭐️
Chi Square Formula
Root Mean Square Formula ⭐️⭐️⭐️⭐️⭐️
Diagonal Of A Square Formula ⭐️⭐️⭐️⭐️⭐
Perfect Square Trinomial Formula ⭐️⭐️⭐️⭐️⭐️
Perimeter of a Square Formula ⭐️⭐️⭐️⭐️⭐️
R Squared Formula ⭐️⭐️⭐️⭐️⭐️
Regression Sum of Squares Formula ⭐️⭐️⭐️⭐️⭐️
Regular Square Pyramid Formula ⭐️⭐️⭐️⭐️⭐️
Secant Square x Formula ⭐️⭐️⭐️⭐️⭐️
Sin squared x formula ⭐️⭐️⭐️⭐️⭐️
Square Formula ⭐️⭐️⭐️⭐️⭐️
Sum of Squares Formula
Area of a Square Formula
Formula for calculating the perimeter of a square
Surface Area of a Square Pyramid Formula
Volume of a Square Pyramid Formula | 6,643 | 24,810 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.839768 |
https://www.jove.com/de/science-education/12616/position-and-displacement | 1,722,737,329,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640388159.9/warc/CC-MAIN-20240804010149-20240804040149-00783.warc.gz | 669,526,552 | 37,497 | # Position and Displacement
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Position and Displacement
### Nächstes Video3.2: Average Velocity
To describe an object's motion, its position needs to be specified relative to a convenient frame of reference. An arbitrary set of axes that describes an object's position and motion is known as a frame of reference.
Generally, Earth or stationary objects on Earth are used as frames of reference. For example, an airplane's position during its takeoff could be described with respect to Earth as a whole or the airport's control tower. To describe an object's position along a one-dimensional motion, we often use the variable x.
Now, suppose an object moves relative to a frame of reference—for example, a car moves to the left relative to a building, then the car's position changes. This change in position is called displacement.
However, the numerical value of x along a straight line defines an object's position where it might be located. Any change in position along this line is indicated by displacement, where Δx is displacement and is calculated by subtracting the initial position from the final position. It is a vector quantity since it indicates direction, and, depending on the choice of positive direction, it can be either positive or negative.
## Position and Displacement
The position of an object defines its location relative to a convenient frame of reference at any particular time. A frame of reference is an arbitrary set of axes from which the position and motion of an object are described. Earth is often used as a frame of reference, and we often describe the position of an object as it relates to stationary objects on Earth. For example, a rocket launch could be described in terms of the position of the rocket with respect to Earth as a whole. On the other hand, a cyclist's position could be described in terms of where they are in relation to the buildings they pass. In other cases, we use reference frames that are not stationary but are in motion relative to Earth. To describe the position of a person in an airplane, for example, we use the airplane, not Earth, as the reference frame. To describe the position of an object undergoing one-dimensional motion, we often use the variable x.
If an object moves relative to a frame of reference, then the object's position changes. For example, if a professor moves to the right relative to a whiteboard, then the professor's position has changed. This change in position is called displacement. The word displacement implies that an object has moved or has been displaced. Although the position is the numerical value of x along a straight line where an object might be located, displacement gives the change in position along this line. Since displacement indicates direction, displacement is a vector quantity and can be either positive or negative, depending on which direction is chosen as positive.
Also, an analysis of motion can have many displacements embedded in it. For example, if the right is positive, and an object moves 2 m to the right, then 4 m to the left, the individual displacements are 2 m and −4 m, respectively. Displacement is denoted by Δx, which is the change in position. We always solve for displacement by subtracting the initial position from the final position.
This text is adapted from Openstax, University Physics Volume 1, Section 3.1: Position, Displacement, and Average Velocity. | 730 | 3,579 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.884837 |
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## Introduction to Abstract and Linear Algebra
Introduction to Abstract and Linear Algebra. Instructor: Dr. Sourav Mukhopadhyay, Department of Mathematics, IIT Kharagpur. Abstract and Linear Algebra are applicable to every discipline, be it engineering and technology, economics or social sciences. It is essential for the students to get acquainted with the subject of Abstract and Linear Algebra at an early stage. The present course has been designed to introduce the subject to undergraduate/postgraduate students in science and engineering. The course contains a good introduction to each topic and an advance treatment of theory at a fairly understandable level to the students at this stage. (from nptel.ac.in)
Lecture 19 - Rings
Go to the Course Home or watch other lectures:
Lecture 01 - Set Theory Lecture 02 - Set Operations Lecture 03 - Set Operations (cont.) Lecture 04 - Set of Sets Lecture 05 - Binary Relation Lecture 06 - Equivalence Relation Lecture 07 - Mapping Lecture 08 - Permutation Lecture 09 - Binary Composition Lecture 10 - Groupoid Lecture 11 - Group Lecture 12 - Order of an Element Lecture 13 - Subgroup Lecture 14 - Cyclic Group Lecture 15 - Subgroup Operations Lecture 16 - Left Cosets Lecture 17 - Right Cosets Lecture 18 - Normal Subgroup Lecture 19 - Rings Lecture 20 - Field Lecture 21 - Vector Spaces Lecture 22 - Subspaces Lecture 23 - Linear Span Lecture 24 - Basis of a Vector Space Lecture 25 - Dimension of a Vector Space Lecture 26 - Complement of Subspace Lecture 27 - Linear Transformation Lecture 28 - Linear Transformation (cont.) Lecture 29 - More on Linear Mapping Lecture 30 - Linear Space Lecture 31 - Rank of a Matrix Lecture 32 - Rank of a Matrix (cont.) Lecture 33 - System of Linear Equations Lecture 34 - Row Rank and Column Rank Lecture 35 - Eigenvalue of a Matrix Lecture 36 - Eigenvector Lecture 37 - Geometric Multiplicity Lecture 38 - More on Eigenvalue Lecture 39 - Similar Matrices Lecture 40 - Diagonalisable | 443 | 1,993 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.757746 |
http://www.instructables.com/id/Volume-of-a-Cylinder-2/ | 1,529,640,751,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864343.37/warc/CC-MAIN-20180622030142-20180622050142-00502.warc.gz | 432,893,360 | 12,052 | # Volume of a Cylinder
6,358
7
Published
## Introduction: Volume of a Cylinder
There is more than one way to calculate volume. This Instructable will show you how to calculate volume using water.
I selected this empty can because I wanted to know its internal volume.
First, I dried the can out with a towel and made sure that it was clean.
## Step 1: Place Empty Can on Scale.
Next , I placed the can on a scale and changed the units to grams.
This can happens to have a mass of 48 grams.
## Step 2: Tare Out the Empty Can.
My scale has a nifty feature that lets you tare out the mass of the scale.
I pressed the tare button and the scale resets back to zero with the can still resting on the scale.
## Step 3: Fill With Water.
Now, I filled the can with water up to the brim. I made sure that I did not spill any.
The mass of the water in the can is 454 grams. This is the true mass of the water because the mass of the can was already accounted for.
Ok, here is how I can figure out the volume of the can. The density of water is 0.9982 g/cc at 68 degrees Fahrenheit.
If we take the weight of the water, 454 grams and divide this by the density of the water at 68F we will be left with the volume of the water.
454 grams / 0.9982 g/cc = 454.8 cubic centimeters of water.
This turns out to be 15.38 ounces of volume for this particular can.
Using this same technique the volume of a sphere or any enclosed object can be calculated.
That is how you can find the volume of a cylinder using water. I hope you enjoyed this Instructable.
While your at it check out my other Instructables:
https://www.instructables.com/id/Digital-Measurment-Estimator/
https://www.instructables.com/id/Hanging-around/
https://www.instructables.com/id/476_better_than_a_Bank/
## Recommendations
• ### 3D CAM and CNC Class
606 Enrolled
• ### Make it Move Contest
We have a be nice policy.
## Questions
You should also realise that since your scales only read to the nearest gram, the volume you calculated is only accurate to 3 significant figures at best ( ie, 15.4 fl.oz., not 15.38). I note your comments that you were not looking to achieve high precision, but in that case you should report your results only to the degree of accuracy that your method allows. Likewise using density values for water of 5 significant figures is unnecessary when your scales are two orders of magnitude less accurate. Just something to be aware of when taking scientific measurements.
If you know the radius or the circumfrence and the hieght of the can you can calculate the volume by the formula- pi * r(radius)^2 * h
for eg if the height is 7m and radius is 2m then it goes like this
22/7 * 2 * 2 *7=88 Meter cube
Good stuff. The only issue is you measured the capacity volume of the can, not the can. The can's volume found using water is by using a graduated container larger than the can, filled with water. Set the water at a desired height and then submerse the can. The difference in volume is the volume of the can. Sorry. Just being nitpicky. By the way. 68 F is 20 C
Yes you are correct, I did not specify that I am measure the internal volume of the can. Thanks
68oF is about 25oC - was this the temperature you weighed the can at? (it's a bit warm for where I live)
Also, "ounces" are units of weight unless you mean fluid ounces.
L
(interesting user name)
2 replies
Thanks for looking at my Instructable. I appreciate your comments and I would like to respond to your concerns.
I made a couple of assumptions that maybe I shouldn't have. I thought since the subject was volume it was naturally assumed units of measure to be fluid ounces rather than ounces in weight.
My second assumption was regarding the temperature. The temperature was not recorded at the time of measure. I did test inside with the air temperature of about 72 degrees, however the water was straight out of the tap and was probably about 55 degrees or so.
If you are interested, below is a link showing some the density of water at various temperatures.
I used the temperature that best matched my ambient air/water temperature.
As you can see the density of water at 100c is 0.95836 g/cc as compared to 0c of 0.99984 g/cc Water is 4.14% less dense at boiling than at freezing. So yes there would be some error due to effects of temperature.
Another error that I failed to account for is the use of a kitchen scale. This is hardly what I would consider to be accurate. Since there would be error associated with this device as well the total volume of water in the can is approximately close to what the true volume of the can.
I suspect that because the temperature of the water was at approximately 55 degrees F and making the assumption that the scale is accurate the actual volume of the can would be slightly larger than what I calculated above.
The intended purpose of this Instructable was demonstrate the concept. I really was not trying to achieve a fine degree of accuracy including values of uncertainty.
Thanks, I hope this helps,
I read my thermometer wrong last night, 68oF is more like 20oC - I read across from 78 (the 70 isn't printed), so sorry about that.
The level of accuracy is not that important, as long as you know how accurate your method is: it still gives you a volume.
The kitchen scale may or may not be accurate, but it's important to recognise that the precision of the scale is a different thing. And for this volume +/- 1g is less than 0.5%
L | 1,284 | 5,479 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2018-26 | latest | en | 0.907836 |
https://mkyong.com/java/java-display-double-in-2-decimal-points/ | 1,582,063,229,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143815.23/warc/CC-MAIN-20200218210853-20200219000853-00193.warc.gz | 492,781,338 | 10,854 | # Java – Display double in 2 decimal places
In Java, there are few ways to display `double` in 2 decimal places.
Note
## 1. DecimalFormat
DecimalExample.java
``````
package com.mkyong;
import java.math.RoundingMode;
import java.text.DecimalFormat;
public class DecimalExample {
private static DecimalFormat df2 = new DecimalFormat("#.##");
public static void main(String[] args) {
double input = 3.14159265359;
System.out.println("double : " + input);
System.out.println("double : " + df2.format(input)); //3.14
// DecimalFormat, default is RoundingMode.HALF_EVEN
df2.setRoundingMode(RoundingMode.DOWN);
System.out.println("\ndouble : " + df2.format(input)); //3.14
df2.setRoundingMode(RoundingMode.UP);
System.out.println("double : " + df2.format(input)); //3.15
}
}
``````
Output
``````
double : 3.14159265359
double : 3.14
double : 3.14
double : 3.15
``````
## 2. String.format
StringFormatExample.java
``````
package com.mkyong;
public class StringFormatExample {
public static void main(String[] args) {
double input = 3.14159265359;
System.out.println("double : " + input);
System.out.println("double : " + String.format("%.2f", input));
System.out.format("double : %.2f", input);
}
}
``````
Output
``````
double : 3.14159265359
double : 3.14
double : 3.14
``````
## 3. BigDecimal
BigDecimalExample.java
``````
package com.mkyong;
import java.math.BigDecimal;
import java.math.RoundingMode;
public class BigDecimalExample {
public static void main(String[] args) {
double input = 3.14159265359;
System.out.println("double : " + input);
BigDecimal bd = new BigDecimal(input).setScale(2, RoundingMode.HALF_UP);
double newInput = bd.doubleValue();
System.out.println("double : " + newInput);
}
}
``````
Output
``````
double : 3.14159265359
double : 3.14
``````
## References
##### mkyong
Founder of Mkyong.com, love Java and open source stuff. Follow him on Twitter. If you like my tutorials, consider make a donation to these charities.
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Dmitry Fucintv
Hi, mkyong? Can I use code from your examples anyway I want? I mean create my own example based on yours and post it somewhere?
Guest
KuchiMonster
When using the String.format example do you know how it rounds? Is it HALF / UP / or DOWN.
I know it does not have those constants but the formula behind the .format method must round it one of those 3 ways right?
Guest
mostafa
God bless you man 😀
Guest
ajay katoch
Hello, Can you share Java code or script for finding P-value of large data sets:- eg:- Input File (.txt format/or any) Name X Y Z A_1 1.7085586 0.73179674 3.3962722 A_2 0.092749596 -0.10030079 -0.47453594 A_3 1.1727467 0.15784931 0.0572958 A_4 -0.91714764 -0.62808895 -0.6190882 A_5 0.34570503 0.10605621 0.30304766 This should be the Input format in form of text file or Excel sheet and the outPut should be:- Name p_Value A_1 0.129618297992839 A_2 0.436354399799269 A_3 0.323631285199105 A_4 0.0179166576112724 A_5 0.0763318283011515 I have done it in R programming, but I want to this by using Java, but I am a beginner in Java so …..… Read more »
Guest
aboudi
thank you…
Guest
chiefrocker86
package com.mkyong.loan;
public class Test{
public static void main(String[] args) {
double input = 32.123456;
System.out.println(“double : ” + input);
System.out.printf(“double : %.2fn”, input);
}
}
Guest
Nick
I realize that when comes to 7.00 and it become 7. But I want it as 7.00. Any Idea?
Guest
Nick
I fixed it, just simply change from “#.##” to “#.00”
Guest
Mustafa dev
cool
Guest
Vaishnavi Yerpude
Thanks .it worked | 1,085 | 3,642 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.365144 |
http://robotc.net/forums/viewtopic.php?p=25007 | 1,529,643,920,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864354.27/warc/CC-MAIN-20180622045658-20180622065658-00341.warc.gz | 277,561,510 | 7,970 | View unanswered posts | View active topics It is currently Fri Jun 22, 2018 1:05 am
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while loops and timers issue
Author Message
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Joined: Mon Jul 08, 2013 12:53 pm
Posts: 1
while loops and timers issue
I am teaching a class on RobotC basics and Vex and am new to robotc. Students and I are trying to write code for a simple program that has use a timer for 10 to 20 seconds to control the robot. While under the timer loop (while statement) the robot needs to go forward until it senses an object turn left or right then go forward turn left or right then go straight. After 10 seconds gives the user control of the robot through joystick controller. So far we can either get it to sense the object or go straight for 10 seconds then give up control but not both. Example code below. Thoughts?
Code:task main(){int normalSpeed;int slowSpeed;normalSpeed = 103;slowSpeed = 63; ClearTimer(T1); while(time1[T1] < 10000) { if(SensorValue[sonarSensor] > 3) { motor[leftMotor] = 103; motor[rightMotor]= 63; } else // see wall { motor[rightMotor]= 0; motor[leftMotor] = 0; } wait1Msec(500); SensorValue[leftEncoder] = 0; SensorValue[rightEncoder] = 0; while(SensorValue[rightEncoder] < 900) { motor[leftMotor] = 103; motor[rightMotor] = -103; } SensorValue[leftEncoder] = 0; SensorValue[rightEncoder] = 0; while(SensorValue[leftEncoder] < 900) { if(SensorValue[leftEncoder] > SensorValue[rightEncoder]) { motor[rightMotor] = slowSpeed; motor[leftMotor] = normalSpeed; } if(SensorValue[rightEncoder] > SensorValue[leftEncoder]) { motor[rightMotor] = slowSpeed; motor[leftMotor] = normalSpeed; } if(SensorValue[leftEncoder] == SensorValue[rightEncoder]) { motor[rightMotor] = normalSpeed; motor[leftMotor] = normalSpeed; } } SensorValue[leftEncoder] = 0; SensorValue[rightEncoder] = 0; while(SensorValue[leftEncoder] < 900) { motor[leftMotor] = 103; motor[rightMotor] = -103; } while(SensorValue[leftEncoder] < 1440) { if(SensorValue[leftEncoder] > SensorValue[rightEncoder]) { motor[rightMotor] = slowSpeed; motor[leftMotor] = normalSpeed; } if(SensorValue[rightEncoder] > SensorValue[leftEncoder]) { motor[rightMotor] = slowSpeed; motor[leftMotor] = normalSpeed; } if(SensorValue[leftEncoder] == SensorValue[rightEncoder]) { motor[rightMotor] = normalSpeed; motor[leftMotor] = normalSpeed; { } while(1 == 1) { //Driving Motor Control motor[leftMotor] = vexRT[Ch3] / 2; motor[rightMotor] = vexRT[Ch2] / 2; //Arm Control if(vexRT[Btn6U] == 1) { motor[armMotor] = 40; } else if(vexRT[Btn6D] == 1) { motor[armMotor] = -40; } else { motor[armMotor] = 0; } }}}}}
Attachments: final test program.c [3.48 KiB] Downloaded 374 times
Mon Jul 08, 2013 12:57 pm
Joined: Tue Oct 09, 2012 10:34 am
Posts: 192
Re: while loops and timers issue
Hi jgarrett,
Let me make a couple of suggestions, and if you still can't figure it out, let me know:
1. In ROBOTC, select all your code, then select Edit>>Code Formatting>>Format Selection from the menu. This will properly indent your code so that it's easier to see which bits of code are grouped together within while and if statements. I suspect your code isn't doing what you think it is or want it to.
2. Remember that a while loop will cause only its code to be run until it's condition is false. If you want your program to do two things at once, such as go straight and detect objects, they will both need to be inside the same while loop.
Best,
--Ryan
_________________
Ryan Cahoon
RVW Software Developer
Robot Potato Head; Virtual NXT
Wed Jul 10, 2013 7:27 pm
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# FASS509 Quantitative Research Methods 1 (online distance learning)
## Module description
• Uncertainty and probability.
• Random sampling and random variables. Different approaches to sampling. What is a "random" sample and why is it important for statistical inference?
• Variable types - discrete vs. continuous.
• Distributions - what they are and why they are important.
• Basic descriptive statistics - mean, median, mode, standard deviation, interquartile range.
• Credible intervals. The difference between probability and effect size. Credible intervals as limits of (un)certainty. Factors affecting the width of credible intervals.
• Simple hypothesis testing using credible intervals - the single mean, single proportion, difference between two means, and difference between two proportions.
## Aims and objectives
To provide students with:
• a fundamental understanding of probability
• a basic understanding of random sampling and why it is important.
• an introduction to basic descriptive statistics
• a simple introduction to inferential statistics for one- and two-sample problems, based around the notion of credible intervals.
• To teach students how to perform the relevant calculations using a spreadsheet (Microsoft Excel or Open Office Calc).
On successful completion of this module students will be able to:
• Understand what probability is.
• Distinguish between probability and effect size.
• Understand how and why researchers take random samples from larger populations.
• Have a general idea of what a distribution is.
• Understand what basic descriptive measures (means, medians, modes, standard deviations, and interquartile ranges) can tell us.
• Understand what a credible interval is and what it can tell us.
• Know how to use a credible interval to test simple one- and two-sample hypotheses.
• Carry out the above calculations using a spreadsheet (Microsoft Excel or Open Office Calc).
Everitt, B. (2008), Chance Rules: An Informal Guide to Probability, Risk and Statistics, 2nd Ed. New York, Springer.
Franklin, J. (2009), What Science Knows and How It Knows It, New York, Encounter Books.
Gigerenzer, G. (2002), Reckoning with Risk: Learning to Live with Uncertainty, London, Penguin.
Hubbard, D.W. (2010), How to Measure Anything: Finding the Value of Intangibles in Business, 2nd Ed. Hoboken, NJ, John Wiley.
Paulos, J.A. (1995), A Mathematician Reads the Newspaper: Making Sense of The Numbers in the Headlines, London, Penguin.
Savage, S.L. (2009), The Flaw of Averages: Why We Underestimate Risk in the Face of Uncertainty, Hoboken, NJ, John Wiley.
NB these are not required texts - this is a background reading list.
## Timing and Location
Term: Michaelmas Date(s): 12/10/15 - 20/11/15 Number of sessions: 6 x 1 hour sessions Runs weekly, starting in week 2 Timing and Location: online distance learning
Minimum quota: 6
Maximum quota: 30
Apart from delivery mode, the syllabuses for FASS508: Quantitative Research Methods Introduction (face-to-face) and FASS509: Quantitative Research Methods 1 (online distance learning) are basically identical, so you should only need to take one of these. Either will provide the background necessary to prepare you for FASS512: Quantitative Research Methods 2.
## RTP 2015-16
Information about the RTP for 2015-16 is being added to these pages as and when it becomes available. Registration is now open and must be done on the RTP registration form.
| Home | Who's who? | Research Training | News and Events | Resources |
Research Training Programme, Faculty of Arts and Social Sciences, Lancaster University, Lancaster LA1 4YD, UK
Tel: +44 (0) 1524 510852 E-mail: fass-rtp@lancaster.ac.uk | 847 | 3,793 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2016-07 | latest | en | 0.872214 |
http://www.lofoya.com/Solved/9/a-man-arranges-to-pay-off-a-debt-of-rs-3600-by-40-annual-installments | 1,516,390,661,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888113.39/warc/CC-MAIN-20180119184632-20180119204632-00281.warc.gz | 489,620,474 | 15,246 | # Easy Algebra Solved QuestionAptitude Discussion
Q. A man arranges to pay off a debt of Rs 3600 by 40 annual installments which are in A.P. When 30 of the installments are paid he dies leaving one-third of the debt unpaid. The value of the $8^{th}$ installment is:
✖ A. Rs 35 ✖ B. Rs 50 ✔ C. Rs 65 ✖ D. Rs 70
Solution:
Option(C) is correct
Let the first installment be '$a$' and the common difference between any two consecutive installments be '$d$'
Using the formula for the sum of an A.P.
$S=\dfrac{n}{2}[2a+(n−1)d]$
We have,
\begin{align*}
3600&= \dfrac{40}{2}[2a+(40−1)d]\\
\Rightarrow 180&=2a+39d \tag{1} \label{eq1}\\
\text{And }2400&= \dfrac{30}{2}[2a+(30−1)d] \\
\Rightarrow 160&=2a+29d \tag{2} \label{eq2}
\end{align*}
On solving both the equations we get:
$d=2$ and $a=51$
Note: To check out the calculations, check comment by Anita.
Value of $8^{th}$ installment;
$=51+(8−1)2$
$= \text{Rs. } 65$
## (5) Comment(s)
Priyanka
()
How $a = 51$ ,came???
Anita
()
It came by solving the two equations arrived at the solution,
$180=2a+39d$ -------- (1)
$160=2a+29d$ -------- (2)
Now, $(1)*29-(2)*39$
$180*29=2a*29+39d*29$
-$(160*39=2a*39+29d*39)$
$=>180*29- 160*39=2a*29-2a*39$
$=> -1020=-20a$
$=> a=51$
It's not that difficult if you give it a proper try.
Anita
()
Although it would be easier to find $d$ first and then find the value of $a$.
To find $d$, just do $(2)-(1)$,
$=>(180-160)=2a+39d -(2a+29d)$
$=> 20=10d$
$=>d=2$
Now, put the value of $d$ in $(1)$,
$180=2a+39*2$
$a=\dfrac{180-78}{2}$
$=\dfrac{102}{2}$
$=51$
Priyanka
()
2400 = 30/2 [2a + ( 30 - 1 )d ]
how 2400 came??
Anita
()
Since the man dies after paying 30 instalments and leaving one-third of the debt unpaid.
Total debt to be paid = 3600
Debt remaining= $\frac{1}{3}$ of 3600
$=1200$
So, debt paid in the 30 installments,
$=3600-1200$
$=2400$
Which takes to the below equation,
$2400= \dfrac{30}{2}[2a+(30−1)d]$
Hope this helps. | 752 | 1,953 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2018-05 | latest | en | 0.697417 |
https://whatisconvert.com/12-feet-second-in-kilometers-hour | 1,643,140,397,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304872.21/warc/CC-MAIN-20220125190255-20220125220255-00669.warc.gz | 635,081,525 | 7,577 | # What is 12 Feet/Second in Kilometers/Hour?
## Convert 12 Feet/Second to Kilometers/Hour
To calculate 12 Feet/Second to the corresponding value in Kilometers/Hour, multiply the quantity in Feet/Second by 1.0972799999991 (conversion factor). In this case we should multiply 12 Feet/Second by 1.0972799999991 to get the equivalent result in Kilometers/Hour:
12 Feet/Second x 1.0972799999991 = 13.167359999989 Kilometers/Hour
12 Feet/Second is equivalent to 13.167359999989 Kilometers/Hour.
## How to convert from Feet/Second to Kilometers/Hour
The conversion factor from Feet/Second to Kilometers/Hour is 1.0972799999991. To find out how many Feet/Second in Kilometers/Hour, multiply by the conversion factor or use the Velocity converter above. Twelve Feet/Second is equivalent to thirteen point one six seven Kilometers/Hour.
## Definition of Foot/Second
The foot per second (plural feet per second) is a unit of both speed (scalar) and velocity (vector quantity, which includes direction). It expresses the distance in feet (ft) traveled or displaced, divided by the time in seconds (s, or sec). The corresponding unit in the International System of Units (SI) is the metre per second. Abbreviations include ft/s, ft/sec and fps, and the rarely used scientific notation ft s−1.
## Definition of Kilometer/Hour
The kilometre per hour (American English: kilometer per hour) is a unit of speed, expressing the number of kilometres travelled in one hour. The unit symbol is km/h. Worldwide, it is the most commonly used unit of speed on road signs and car speedometers. Although the metre was formally defined in 1799, the term "kilometres per hour" did not come into immediate use – the myriametre (10,000 metres) and myriametre per hour were preferred to kilometres and kilometres per hour.
## Using the Feet/Second to Kilometers/Hour converter you can get answers to questions like the following:
• How many Kilometers/Hour are in 12 Feet/Second?
• 12 Feet/Second is equal to how many Kilometers/Hour?
• How to convert 12 Feet/Second to Kilometers/Hour?
• How many is 12 Feet/Second in Kilometers/Hour?
• What is 12 Feet/Second in Kilometers/Hour?
• How much is 12 Feet/Second in Kilometers/Hour?
• How many km/h are in 12 ft/s?
• 12 ft/s is equal to how many km/h?
• How to convert 12 ft/s to km/h?
• How many is 12 ft/s in km/h?
• What is 12 ft/s in km/h?
• How much is 12 ft/s in km/h? | 624 | 2,401 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2022-05 | latest | en | 0.879973 |
http://www.alanzucconi.com/2016/12/14/simplified-collision-geometry/ | 1,524,755,284,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948285.62/warc/CC-MAIN-20180426144615-20180426164615-00074.warc.gz | 350,316,322 | 18,129 | # Simplified Collision Geometry
This tutorial shows how to automatically generate simplified colliders for 3D models imported into Unity. The tutorial uses Google SketchUp as an example, but its knowledge and code is agnostic to whichever modelling tool you are using.
#### Introduction
Modern games often features very details 3D models. It is not uncommon for characters to render thousands of vertices and triangles. However, collisions with the player are often calculated on a simplified mesh. The standard player controllers Unity comes with, for instance, replies on capsule colliders. When it comes to collisions, there is often no need to be that accurate about the shape of the player. Collision between spheres, cubes, capsules and similar geometrical primitives can be performed very efficiently. Calculating and resolving collision between the individual faces of each 3D models is doable; Unity offers a special component called MeshCollider that is specifically designed for this. However, its dramatic cost rarely translates into an equally significant improvement in the overall gaming experience.
#### Step 1. The Problem
A standard approach in modern games relies on the idea that physical objects uses simplified collision meshes. This introduces an extra step in the development pipeline of a game. On top of creating 3D models, developers now also have to create collision meshes and to place them in the right position. Depending on the software you are using, this task could be massively simplified. If you are using 3D modelling tools such as Google SketchUp (Unity Manual: Importing Objects From SketchUp), unfortunately, this is not the case. SketchUp does not provide any way to distinguish between render and collision geometry. When a model is imported in Unity, you only have the option to generate a MeshCollider.
A possible approach to this problem relies on the idea that you can add, automatically, all the geometry you need. The first thing to understand is that once a model is imported from SketchUp, Unity will preserve the hierarchy of its components. SketchUp supports the concepts of groups and components; they represents containers for the actual geometrical entities (vertices and triangles) that make up your models. Both groups and components are translated into Unity as game objects. For components, the name that Unity chooses is the name used in SketchUp, plus an incremental number. If your Sketchup model contains two instances of the Table components, Unity will name them Table #1 and Table #2.
#### Step 2. The Solution
The idea behind this simple tutorial is to iterate over the imported game objects, and to add colliders only for certain specific components. Since an imported model consists of many nested game objects, we will need to iterate through all of them recursively. If you are unfamiliar with recursion, it works like this:
1. We start from the game object that represents the imported model;
2. We iterate over the game objects it contains;
1. If the current game object satisfies our criteria (it’s name starts with Table ), then we add a collider and terminates;
2. If it doesn’t, we iterate on its children repeating [2]
In terms of C# code:
Adding a BoxCollider with AddComponent has the same effect of adding a collider manually from the inspector. Unity will automatically resize it to fit the bounding box of the 3D mesh. This generates a box around the object, which works in most cases.
#### Step 3. The Custom Editor
To test CreateCollidersRecursively, we can invoke it from the Start method of a C# script.
The main issues with this approach is that the code is executed each time you start the game. There’s no reason why this couldn’t be done before starting the game, reducing its loading time. In order to do this, we need a way to execute the script in the editor, rather than in the game. This is possible by creating a custom editor script. If you are unfamiliar with this technique, Unity allows to extends the functionality and the look of its editor with custom scripts. We can use the attribute MenuItem to a function so that we can invoke it directly from the editor, before running the game.
This new setup allows us to generate colliders without damaging the final player’s experience. If you have more than one models, you can simply place them under the same game object and invoke the script on that one.
#### Step 4. Improvements
Deleting previous colliders. If you have run the CreateCollidersRecursively function multiple times, you might have noticed that colliders are added multiple times. This can cause many issues. For this reason, we should ensure that the script deletes all previous colliders before creating new ones.
Instantiating prefabs. We can use this very technique to automatically instantiate other objects that Sketckup does not natively support. Lights, audio sources, collectibles, enemies… they can all be created in the recursive step of the function. For instance, the following snipped allows to create a light from a prefab when the name of the group matches “Light”.
#### Conclusion
This tutorial briefly covered how to recursively loop over the components of 3D model in Unity. This was used to automatically add colliders and to instantiate prefabs such as lights, audio sources and collectibles. | 1,119 | 5,349 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-17 | latest | en | 0.914026 |
https://brainmass.com/statistics/regression-analysis/regression-analysis-of-test-scores-and-stress-154517 | 1,481,061,758,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542002.53/warc/CC-MAIN-20161202170902-00029-ip-10-31-129-80.ec2.internal.warc.gz | 770,011,858 | 19,286 | Share
Explore BrainMass
# Regression Analysis of Test Scores and Stress
See attached.
7. It is recommended that you use the computer EXCEL sheet attached to help you answer the following questions: Write the answers in the provided space. An answer saying "see attached page" is not an option. There are almost no calculations. Most of the answers are on the EXCEL spreadsheet.
X Y
Stress Test
Score Score
6.5 81
4.0 96
2.5 93
7.2 70
8.1 63
3.4 84
5.5 73
sum 37.2 560
Given the above data for a regression analysis answer the following questions.
A. What is the independent and what is the dependent variable?
B. Develop the simple linear regression equation.
C. Find (Calculate) the Correlation Coefficient.
D. What percent of the total variation is explained by the regression equation?
E. What is the value of Y6 ? i.e. what is the expected test score with a stress level of 6.0?
#### Solution Summary
Step by step method for regression analysis is discussed here. Regression coefficients, coefficient of determination, scatter diagram and significance of regression model are explained in the solution. This solution is presented in an attached Excel file, Word file and PDF copy.
\$2.19 | 285 | 1,202 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2016-50 | longest | en | 0.901697 |
https://physics.stackexchange.com/questions/279376/hydrostatic-force-on-a-gate | 1,606,192,196,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141171077.4/warc/CC-MAIN-20201124025131-20201124055131-00541.warc.gz | 435,710,664 | 34,841 | Hydrostatic force on a gate
here is a copy of the problem statement.
I have attempted to do the problem, but some concepts I really dont understand. My professor showed us the answer and his free body diagram which I drew at the top and is in the middle. Here are the issues/concepts I don't understand.
Here is my work:
Force Balance $$R_1+R_2+F_p=W$$
$$R_1=R_2$$
$$R_1+R_1+F_p=W \rightarrow R_1=\frac{W-F_p}{2}$$
I know that force due to pressure is $\int_A pdA ; p=\rho g z ; dA = Wdl ; dl=\frac{dz}{\sin\theta}$ so the force due to pressure is $dF_p=\rho g z W \frac{dz}{\sin\theta}$
I know that the torque due to the pressure is $d\tau_p = ldF_p$, but since l is changing, I must relate l to z using a triangle. Doing this I yield $l = \frac{z}{\sin\theta}$
Doing some plugging and substitution for the $dF_p$ and the relationship of $l = \frac{z}{\sin\theta}$ I get,
$d\tau_p=\rho g W \frac{z^2 dz}{\sin^2\theta}$
Integration this with respect from z=0 to z=H, $d\tau_p=\rho g W \frac{1}{\sin^2\theta}\int_0^H z^2dz$
$\tau_p=\frac{\rho g W L^2 H}{3}$
Then computing the Torque due to the reaction force I get
$\tau_R = R\cos\theta$
And lastly the toque due to the weight of the door is
$\tau_w=W\cos\theta \frac{L}{2}$
Here are my questions
• What is the reaction force? I think I understand that R2 is coming from the ground pushing up on the door, but I don't understand R1 or why it is vertical in the y direction. Is R1 due to the liquid, or something completely else? And why are R1 and R2 equal?
• Second, I don't understand why Fp, force due to pressure, is not normal to the surface. In his FBD it is completely in the y direction.
• Lastly, since this is a hydrostatic situation, would I be setting the $\tau_p+\tau_R-\tau_w=0$? I assume that Reaction 1 has no effect as it is right at the hinge..I think?
• Not to sound rude, but I'm not sure what you are getting at. I have attached a picture of me attempting a solution to my question. I needed assistance understanding some concepts which I posted. If you could elaborate that would be great. – JuliusDariusBelosarius Sep 10 '16 at 17:02
• You just put a picture of a sheet full of formulas. How are we supposed to understand what your thoughts were? We have a nice system here that can help you to structure your question with equations. – Bernhard Sep 10 '16 at 17:12
• OK. I can convert it to Latex if that is what you mean. I posted the picture as I want. I apologize for that, but appreciate the feedback on proper question asking. I just figured it would be easier since I also have drawings that accompany the problem and I don't believe I can make those in latex. – JuliusDariusBelosarius Sep 10 '16 at 17:32
• I think @Bernhard is (somewhat less than politely) trying to point out that posting images of math is generally not a good idea for several reasons: 1) We can't edit it if we see a typo, 2) It's not searchable, 3) It's harder to read. Please go through and type up your work using mathjax, being very careful to use clear and consistent notation. You will still have to include the diagrams as images, and that's fine. Also, when asking for help solving specific problems, always identify a specific conceptual issue that has you stuck. – DanielSank Sep 10 '16 at 18:41
• @DanielSank I know understand! Thank you for that. Ill get right on that shortly. I just wanted to know why it is taboo to post images, and now I am aware. – JuliusDariusBelosarius Sep 10 '16 at 18:59
1 Answer
All you need to do is balance the torque due to the weight of the plate with the torque due to water pressure on the plate. Take moments about the hinge to eliminate any reaction there ($R_1$), and assume the reaction at the gasket is zero ($R_2=0$) for minimum weight of plate.
In answer to your questions :
1. Ignore reaction forces (see above).
2. Force due to pressure of water is normal to the plate. The FBD seems to consider only vertical components of force.
3. $\tau_R=0$ because we are assuming $R_2=0$; that leaves $\tau_p=\tau_W$. | 1,120 | 4,039 | {"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} | 4.09375 | 4 | CC-MAIN-2020-50 | latest | en | 0.933641 |
http://www.dicegamedepot.com/dice-n-games-blog/pig-rules | 1,508,640,921,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825057.91/warc/CC-MAIN-20171022022540-20171022042540-00123.warc.gz | 427,247,397 | 23,715 | ## Dice game rules: Pig
Posted by Dice Game Depot
Pig is a simple dice game which in its basic form is playable with just a single die. You win by being the first player to score 100 or more points.
To play you'll need 2 to 10 players, one 6-sided dice, and a pencil and some paper for keeping score.
Skull dice (where the 1 is replaced with a skull) work well for this game.
## So how do you play Pig?
Choose a player to go first. That player throws a die and scores as many points as the total shown on the die providing the die doesn’t roll a 1. The player may continue rolling and accumulating points (but risk rolling a 1) or end his turn.
If the player rolls a 1 his turn is over, he loses all points he accumulated that turn, and he passes the die to the next player.
Play passes from player to player until a winner is determined.
## How do you win?
The first player to accumulate 100 or more points wins the game.
## Are there any rule or scoring variations?
Two-Dice Pig
This variant is the same as Pig, except two standard dice are rolled. If neither shows a 1, their sum is added to the turn total. If a single 1 is rolled, the player scores nothing and the turn ends. If two 1s are rolled, the player’s entire score is lost and their turn ends.
Big Pig
This variant is the same as Two-Dice Pig, except rolling double 1s ends the player’s turn, scores 25 points, and eliminates any other points the player may have accumulated that turn. If any other doubles are rolled, the player adds twice the value of the dice to the turn total.
Hog
This variant is the same as Pig, except the player rolls only once per turn with an arbitrary number of dice. | 396 | 1,676 | {"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} | 2.6875 | 3 | CC-MAIN-2017-43 | latest | en | 0.965433 |
http://www.iliamnaair.com/topic/adjacent-angles-math-definition | 1,601,490,992,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402127397.84/warc/CC-MAIN-20200930172714-20200930202714-00711.warc.gz | 162,406,277 | 5,878 | Apex of an angle is the end point of the rays that form the sides of the angle. If we speak of a common node and a common side, we mean that the node and the side are divided by the two angles. Plural noun geometry. two angles with the same vertex and a common side between them.
## Name the sides of a right-angled triangle
When you have problems memorizing the definition, just recall SOH CAH TOA. When you have a TI-83 graphic computer, choose Degrees from the Modus drop-down list and change Float to 4 to get the answer correctly with 4 digits after the comma. Use the QUIT button to go back to the home page and clear the page if necessary.
When using a different pocket calculator, refer to the Owner's Manual or ask your instructor how to change the setting to "Degree Mode" and view the digits after the comma to 4 digits. Rate the following points with 4 digits after the comma::
## One way to recall
The two angles PQR and JKL in the above illustration are complimentary as they always sum to 90°. Often the two angles are adjacent, in this case they make a right one. The two smaller angles in the right quadrilateral are always complimentary. Why? An elbow is 90° and all three result in 180°.
The two smaller ones must therefore be added to 90 and are therefore by definition complementary). A similar approach is used for additional angles that are added to 180°. It is sometimes difficult to notice what is what between supplementing (adds to 180°) and supplementing (adds to 90°). A right corner is 90 degree and, yes, "compliment" and "supplement" are not the same term, but it is a way of remembering what it is.
Set a line to the right of the character 'c' in 'complementary' to form a '9'. Complimentary angles are added to 90. Throughout the years we have used advertisements to promote the site so that it can stay free for all. Admittedly, the ad revenues are declining and I have always disliked the advertisements.
If we achieve the objective, I will delete all advertisements from the website.
Definitions and examples of adjacent angles | Definitions of adjacent angles - Geometry
Contiguous" means "next" or "neighboring" angles are angles that are directly adjacent to each other. Adjoining angles divide a shared apex and a shared side, but do not overlay. When two adjacent angles make a right corner (180o), they are complementary. Shown in the illustration are a and a adjacent angles.
You have a shared node M and a shared page OA. Rename all adjacent angle couples in the graph. Adjoining angles are angles that lie directly next to each other. Adjoining angles divide a shared apex and a shared side, but do not overlay. Corresponding to the above picture, a and b, e and a are the couples of the neighboring angel in the figure above. | 624 | 2,783 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2020-40 | latest | en | 0.93538 |
https://puzzling.stackexchange.com/questions/28156/reductio-ad-absurdum/28157 | 1,627,438,098,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153515.0/warc/CC-MAIN-20210727233849-20210728023849-00540.warc.gz | 491,561,214 | 37,224 | This is a riddle with diminishing returns.
It's acrostic again. Can't kick that habit.
No special knowledge is needed.
When you find the answer, you will be in no doubt.
Faker! They cried and cut off his tail.
Call for the axe, put his head in a pail
Sharpen the axe and take off his top
Single him out, let him endlessly drop
Atop the trash heap. Take his last right away.
Nothing left. Now we're done. Goodbye and good day.
PHONEY $$~\to~$$ PHONE $$~\to~$$ HONE $$~\to~$$ ONE $$~\to~$$ ON $$~\to~$$ O
## Argument:
Faker! They cried and cut off his tail.
A faker is PHONEY; cutting of the tail Y yields PHONE
Call for the axe, put his head in a pail
You do calls with your PHONE; cutting of the head P yields HONE
Sharpen the axe and take off his top
To HONE is another word for to sharpen; taking of the top H yields ONE
Single him out, let him endlessly drop
The ONE is singled out; dropping the end E yields ON
Atop the trash heap. Take his last right away.
Atop is ON; taking the right N away leaves O
Nothing left. Now we're done. Goodbye and good day.
O or 0 or zero stands for nothing
• Congratulations! I fear it must have been too easy for you. I'll be vaguer next time. ;) – Hugh Meyers Mar 1 '16 at 17:25
• It was not too easy. The "faker" gave me SWINDLER or FRAUD or PHONY/PHONEY, and "sharpen" gave me WHET or HONE. – Gamow Mar 1 '16 at 17:29
• Acrostics can go by letter, syllable or word. In this case the first word of each line. They don't exactly form a message but I wasn't sure what else to call it. – Hugh Meyers Mar 1 '16 at 17:32
• I see. Beautiful! – Gamow Mar 1 '16 at 17:33 | 478 | 1,617 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2021-31 | latest | en | 0.952418 |
https://question.pandai.org/note/read/kssr-mt-y3-2022-03-03/KSSR-Y3-MT-03-03 | 1,720,801,194,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514450.42/warc/CC-MAIN-20240712161324-20240712191324-00336.warc.gz | 386,032,471 | 14,361 | ## Percentages
3.3 Percentages
A percent is a way of expressing a number as a fraction of 100 (percent means per hundredth). t is often marked with a percent sign, %.
Example 26 out of 100 is $\frac{26}{100}$. $\frac{26}{100}$ in percentage is written as 26%. We read it as twenty-six percent.
## Percentages
3.3 Percentages
A percent is a way of expressing a number as a fraction of 100 (percent means per hundredth). t is often marked with a percent sign, %.
Example 26 out of 100 is $\frac{26}{100}$. $\frac{26}{100}$ in percentage is written as 26%. We read it as twenty-six percent. | 175 | 598 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 4, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2024-30 | latest | en | 0.809516 |
http://dailybrainteaser.blogspot.com/2011/10/sherlock-holmes-cipher-puzzle.html | 1,484,762,947,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280310.48/warc/CC-MAIN-20170116095120-00211-ip-10-171-10-70.ec2.internal.warc.gz | 65,828,221 | 27,456 | ## Search This Blog
### Sherlock Holmes Cipher Puzzle
Sherlock Holmes Cipher Puzzle - 25 October
Sherlock, A detective who was mere days from cracking an international smuggling ring has suddenly gone missing. While inspecting his last-known location, you find a note:
710 57735 34 5508 51 7718
Currently there are 3 suspects: Bill, John, and Todd. Can you break the detective's code and find the criminal's name?
1. BILL is the answer and i believe the note says "BILL IS BOSS HE SELLS OIL". I am not sure though, i can be wrong as well!!!
1. BILL IS THE BOSS, HE SELLS OIL .
2. Ur genious .congrates correct answer
3. Look at the numbers upside down
2. BILL IS BOSS HE SELLS OIL
3. Bill. If you read the message upside down, you'll notice that the numbers resemble letters and that those letters form legible sentences. The message is 'Bill is boss. He sells oil.'
4. Bill is the criminal's name. I know it has been already answered. I just want to know that I got the correct answer too!! :))
5. I Think the code is indicating only name that is of JOHN because no groups of numbers are similar at the end of code and BILL and TODD both ends with double letter i.e. LL&DD
6. can i get the whole procedure to solvE this for BILL
1. revers the order of the numbers so it is now
8177 15 8055 43 54775 017
put bills name under 8177 then fill the rest in
8177 15 8055 43 54775 017
Bill i b ll il
test with different letters
8177 15 8055 43 54775 017
Bill is boss he sells oil.
2. Just read it upside down.
7. I know that all have given answer and i am one of them
8. my god .......... the logic for a 4 star question is just childish .........
9. Bill. reading upside down- BILL IS BOSS HE SELLS OIL just like mysterious letter puzzle in professor Layton :)
10. Bill is the boss who sells oil
11. People in there dreams cannot solve and get it....jus becoz answer is given...I don't know why people are writing it..
12. People in there dreams cannot solve and get it....jus becoz answer is given...I don't know why people are writing it..
13. Even breaking the code, I don't see how this identifies the criminal. A discarded unsigned note is hardly damning evidence. There's no proof Sherlock wrote it, or that the "Bill" in the note is the same "Bill" who is a suspect in smuggling. It's not like selling oil is even illegal.
Also, there's the possibility that Sherlock is using retro slang to comment about how cool Bill is for being an oil merchant.
14. This is from a Professor Layton game | 644 | 2,512 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | longest | en | 0.90675 |
https://www.ic.sunysb.edu/Class/phy141md/doku.php?id=phy131studiof15:lectures:chapter3&rev=1438028494&do=diff | 1,606,323,702,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141183514.25/warc/CC-MAIN-20201125154647-20201125184647-00067.warc.gz | 722,229,377 | 4,302 | # Differences
This shows you the differences between two versions of the page.
phy131studiof15:lectures:chapter3 [2015/07/27 16:21]mdawber [3.P.003] phy131studiof15:lectures:chapter3 [2015/08/31 09:05] (current)mdawber [Vectors - Components] Both sides previous revision Previous revision 2015/08/31 09:05 mdawber [Vectors - Components] 2015/08/31 08:47 mdawber [Vectors - Multiplication by a scalar] 2015/07/27 16:34 mdawber [Vectors - Multiplication by a scalar] 2015/07/27 16:33 mdawber [Vectors - Components] 2015/07/27 16:21 mdawber [3.P.003] 2015/07/27 16:20 mdawber [3.P.003] 2015/07/27 16:08 mdawber [Adding and subtracting vectors] 2015/07/20 15:20 mdawber [Vectors - Components] 2015/07/20 13:50 mdawber [Adding and subtracting vectors] 2015/07/20 13:48 mdawber created Next revision Previous revision 2015/08/31 09:05 mdawber [Vectors - Components] 2015/08/31 08:47 mdawber [Vectors - Multiplication by a scalar] 2015/07/27 16:34 mdawber [Vectors - Multiplication by a scalar] 2015/07/27 16:33 mdawber [Vectors - Components] 2015/07/27 16:21 mdawber [3.P.003] 2015/07/27 16:20 mdawber [3.P.003] 2015/07/27 16:08 mdawber [Adding and subtracting vectors] 2015/07/20 15:20 mdawber [Vectors - Components] 2015/07/20 13:50 mdawber [Adding and subtracting vectors] 2015/07/20 13:48 mdawber created Line 64: Line 64: {{vectorcomponentadd.png}} {{vectorcomponentadd.png}} + The angles $\theta_{1}$ and $\theta_{2}$ are defined with respect to the positive $x$ axis, ie. $\theta_{1}$ is negative and $\theta_{2}$ is positive. | $v_{1x}=v_{1}\cos\theta_{1}$ <html>    | $v_{2x}=v_{2}\cos\theta_{2}$ <html>    | | $v_{1x}=v_{1}\cos\theta_{1}$ <html>    | $v_{2x}=v_{2}\cos\theta_{2}$ <html>    | Line 71: Line 72: | $v_{Rx}=v_{1x}+v_{2x}$ <html>    | $\tan\theta_{R}=\frac{v_{Ry}}{v_{Rx}}$ <html>    | | $v_{Rx}=v_{1x}+v_{2x}$ <html>    | $\tan\theta_{R}=\frac{v_{Ry}}{v_{Rx}}$ <html>    | | $v_{Ry}=v_{1y}+v_{2y}$ <html>    | $v_{R}=\sqrt{v_{Rx}^2+v_{Ry}^2}$ <html>    | | $v_{Ry}=v_{1y}+v_{2y}$ <html>    | $v_{R}=\sqrt{v_{Rx}^2+v_{Ry}^2}$ <html>    | + + ===== 3.P.027 ===== + + ===== 3.P.042 ===== + + ===== 3.P.051 ===== + + ===== 3.P.062 ===== + + + ===== Vectors - Multiplication by a scalar ===== ===== Vectors - Multiplication by a scalar ===== + + Multiplication of a vector by a scalar can change the magnitude, but not the direction of the vector, ie. each component of the vector is multiplied by the scalar in the same way. + ===== 3.P.071 ===== | 1,073 | 2,702 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2020-50 | latest | en | 0.473397 |
https://groups.yahoo.com/neo/groups/boatdesign/conversations/topics/1135 | 1,438,403,516,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042988458.74/warc/CC-MAIN-20150728002308-00259-ip-10-236-191-2.ec2.internal.warc.gz | 862,997,066 | 17,377 | ## peas
Expand Messages
• Most of you will have seen this already. http://apci.net/~michalak/ If not, do check it out. In particular, if you ve heard me rambling about the seas of
Message 1 of 3 , Apr 1, 2000
• 0 Attachment
Most of you will have seen this already. http://apci.net/~michalak/
If not, do check it out.
In particular, if you've heard me rambling about the 'seas of peas' theory
in the past, and wondered what on Earth I was gibbering about, this is an
opportunity to find out more.
And any ideas why these boats should be so fast? Could it be that the old
speed/length rule just doesn't apply, and these boats plane or semi plane in
a very smooth way that's actually not at all obvious when it happens?
Cheers,
Gav
• This may be heresy..... I would like to see (or hear) the Bolger sharpie-shape theory discussed in its most complete form. Bolger is capable of both subtle
Message 2 of 3 , Apr 3, 2000
• 0 Attachment
This may be heresy.....
I would like to see (or hear) the Bolger sharpie-shape theory
discussed in its most complete form. Bolger is capable of both subtle
thought and pedantic simplification, so I am reluctant take issue
with him based on the short, simple explanations that he has put in
articles.
However...
The proposition that the curve of the sides (i.e. angle of entry of
the side) should match the curve of the bottom (i.e. angle of entry
of the bottom) BECAUSE then the bottom pushes water down equally as
the sides push water out seems to me to be flawed. Water under the
boat is constrained because water can not be compressed. Water on the
side of the boat can be forced upward. It seems to me that the Bolger
Box form leads to higher pressure under the boat than on the sides.
According to the Bolger Theory, this is desireable because it means
water passes around the chine only in the bottom-to-side direction
and not in the higher drag side-to-bottom direction.
I do believe that he has proved on the water that the Bolger Box
makes a superior boat in practice.
Now, the question of why Bolger Boxes can be capable of high speeds.
(Not that I believe in an 8kt Micro.) I think it is because of
Bolger's oft repeated observation that a shallow boat can not make a
deep wave. I will have to post my complete line of thought sometime,
but the jist is this:
The forward part of a boat depresses the water surface. As the boat
goes by, water pressure pushes the surface back to the equilibrium
level. (This pressure is set by the density of water, and is
independant of boat length and speed.) If the water is not pushed up
quickly enough to support the aft section of the boat, the stern
falls, the boat gets out of trim and resistance increases. The boat
has reached 'hull speed.' If the water is pushed up quickly, the boat
stays in trim, resistance does not rise, and the boat can accelerate.
It seems obvious that shorter the distance the water is displaced,
the quicker it can rise to support the stern of the boat, and the
faster the boat can go before it reaches hull speed. The Square Boats
displace the surface about the smallest amount of any shape of boat,
which may explain why they can generate unexpectedly high top speeds.
Peter
• This may be heresy..... I would like to see (or hear) the Bolger sharpie-shape theory discussed in its most complete form. Bolger is capable of both subtle
Message 3 of 3 , Apr 3, 2000
• 0 Attachment
This may be heresy.....
I would like to see (or hear) the Bolger sharpie-shape theory
discussed in its most complete form. Bolger is capable of both subtle
thought and pedantic simplification, so I am reluctant take issue
with him based on the short, simple explanations that he has put in
articles.
However...
The proposition that the curve of the sides (i.e. angle of entry of
the side) should match the curve of the bottom (i.e. angle of entry
of the bottom) BECAUSE then the bottom pushes water down equally as
the sides push water out seems to me to be flawed. Water under the
boat is constrained because water can not be compressed. Water on the
side of the boat can be forced upward. It seems to me that the Bolger
Box form leads to higher pressure under the boat than on the sides.
According to the Bolger Theory, this is desireable because it means
water passes around the chine only in the bottom-to-side direction
and not in the higher drag side-to-bottom direction.
I do believe that he has proved on the water that the Bolger Box
makes a superior boat in practice.
Now, the question of why Bolger Boxes can be capable of high speeds.
(Not that I believe in an 8kt Micro.) I think it is because of
Bolger's oft repeated observation that a shallow boat can not make a
deep wave. I will have to post my complete line of thought sometime,
but the jist is this:
The forward part of a boat depresses the water surface. As the boat
goes by, water pressure pushes the surface back to the equilibrium
level. (This pressure is set by the density of water, and is
independant of boat length and speed.) If the water is not pushed up
quickly enough to support the aft section of the boat, the stern
falls, the boat gets out of trim and resistance increases. The boat
has reached 'hull speed.' If the water is pushed up quickly, the boat
stays in trim, resistance does not rise, and the boat can accelerate.
It seems obvious that shorter the distance the water is displaced,
the quicker it can rise to support the stern of the boat, and the
faster the boat can go before it reaches hull speed. The Square Boats
displace the surface about the smallest amount of any shape of boat,
which may explain why they can generate unexpectedly high top speeds.
Peter
Your message has been successfully submitted and would be delivered to recipients shortly. | 1,376 | 5,759 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-32 | longest | en | 0.948719 |
https://socratic.org/questions/for-what-values-of-x-if-any-does-f-x-1-x-3-x-7-have-vertical-asymptotes | 1,686,083,402,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653071.58/warc/CC-MAIN-20230606182640-20230606212640-00374.warc.gz | 564,556,983 | 6,100 | # For what values of x, if any, does f(x) = 1/((x-3)(x-7)) have vertical asymptotes?
Jan 7, 2016
$x = 3$ vertical asymptote
$x = 7$ vertical asymptote
graph{1/((x-3)(x-7)) [-10, 10, -5, 5]}
#### Explanation:
Vertical asymptotes could be find in the points where the Exist Conditions of $f \left(x\right)$ are not satisfied
The Field of Existence of $f \left(x\right)$ is:
x in ]-oo,3[uu]3,7[uu]7,+oo[
because
$f \left(x\right) = \frac{N \left(x\right)}{D \left(x\right)}$
FE (Field of Existence):
$D \left(x\right) \ne 0$
$\left(x - 3\right) \left(x - 7\right) \ne 0$
${x}_{1} \ne 3$
${x}_{2} \ne 7$
Now these values of $x$ are vertical asymptotes if:
${\lim}_{x \rightarrow {x}_{1 , 2}^{\pm}} f \left(x\right) = \pm \infty$
${\lim}_{x \rightarrow {3}^{-}} f \left(x\right) = {\lim}_{x \rightarrow {3}^{-}} \frac{1}{\left(2.9 - 3\right) \left(2.9 - 7\right)} =$
${\lim}_{x \rightarrow {3}^{-}} \frac{1}{\left({0}^{-}\right) \left(- 4.1\right)} = {\lim}_{x \rightarrow {3}^{-}} \frac{1}{0} ^ + = + \infty$
${\lim}_{x \rightarrow {3}^{+}} f \left(x\right) = {\lim}_{x \rightarrow {3}^{+}} \frac{1}{\left(3.1 - 3\right) \left(3.1 - 7\right)} =$
${\lim}_{x \rightarrow {3}^{+}} \frac{1}{\left({0}^{+}\right) \left(- 3.9\right)} = {\lim}_{x \rightarrow {3}^{+}} \frac{1}{{0}^{-}} = - \infty$
${\lim}_{x \rightarrow {7}^{-}} f \left(x\right) = {\lim}_{x \rightarrow {7}^{-}} \frac{1}{\left(6.9 - 3\right) \left(6.9 - 7\right)} =$
${\lim}_{x \rightarrow {7}^{-}} \frac{1}{\left(3.9\right) \left({0}^{-}\right)} = {\lim}_{x \rightarrow {7}^{-}} \frac{1}{{0}^{-}} = - \infty$
${\lim}_{x \rightarrow {7}^{+}} f \left(x\right) = {\lim}_{x \rightarrow {7}^{+}} \frac{1}{\left(7.1 - 3\right) \left(7.1 - 7\right)} =$
${\lim}_{x \rightarrow {7}^{+}} \frac{1}{\left(4.1\right) \left({0}^{+}\right)} = {\lim}_{x \rightarrow {7}^{+}} \frac{1}{{0}^{+}} = + \infty$ | 802 | 1,866 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 20, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2023-23 | latest | en | 0.388204 |
http://www.justintools.com/unit-conversion/data-storage.php?k1=nanobytes&k2=terabits | 1,566,445,099,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316718.64/warc/CC-MAIN-20190822022401-20190822044401-00489.warc.gz | 264,460,942 | 18,252 | Please support this site by disabling or whitelisting the Adblock for "justintools.com". I've spent over 10 trillion microseconds (and counting), on this project. This site is my passion, and I regularly adding new tools/apps. Users experience is very important, that's why I use non-intrusive ads. Any feedback is appreciated. Thank you. Justin XoXo :)
# Convert [Nanobytes] to [Terabits], (nB to Tbit)
## DATA STORAGE
1 Nanobytes
= 8.0E-21 Terabits
*Select units, input value, then convert.
Embed to your site/blog Convert to scientific notation.
Category: data storage
Conversion: Nanobytes to Terabits
The base unit for data storage is bytes (Non-SI/Derived Unit)
[Nanobytes] symbol/abbrevation: (nB)
[Terabits] symbol/abbrevation: (Tbit)
How to convert Nanobytes to Terabits (nB to Tbit)?
1 nB = 8.0E-21 Tbit.
1 x 8.0E-21 Tbit = 8.0E-21 Terabits.
Always check the results; rounding errors may occur.
Definition:
In relation to the base unit of [data storage] => (bytes), 1 Nanobytes (nB) is equal to 1.0E-9 bytes, while 1 Terabits (Tbit) = 125000000000 bytes.
1 Nanobytes to common data-storage units
1 nB =1.0E-9 bytes (B)
1 nB =1.0E-12 kilobytes (KB)
1 nB =1.0E-15 megabytes (MB)
1 nB =1.0E-18 gigabytes (GB)
1 nB =1.0E-21 terabytes (TB)
1 nB =8.0E-9 bits (bit)
1 nB =8.0E-12 kilobits (kbit)
1 nB =8.0E-15 megabits (Mbit)
1 nB =8.0E-18 gigabits (Gbit)
1 nB =8.0E-21 terabits (Tbit)
Nanobytes to Terabits (table conversion)
1 nB =8.0E-21 Tbit
2 nB =1.6E-20 Tbit
3 nB =2.4E-20 Tbit
4 nB =3.2E-20 Tbit
5 nB =4.0E-20 Tbit
6 nB =4.8E-20 Tbit
7 nB =5.6E-20 Tbit
8 nB =6.4E-20 Tbit
9 nB =7.2E-20 Tbit
10 nB =8.0E-20 Tbit
20 nB =1.6E-19 Tbit
30 nB =2.4E-19 Tbit
40 nB =3.2E-19 Tbit
50 nB =4.0E-19 Tbit
60 nB =4.8E-19 Tbit
70 nB =5.6E-19 Tbit
80 nB =6.4E-19 Tbit
90 nB =7.2E-19 Tbit
100 nB =8.0E-19 Tbit
200 nB =1.6E-18 Tbit
300 nB =2.4E-18 Tbit
400 nB =3.2E-18 Tbit
500 nB =4.0E-18 Tbit
600 nB =4.8E-18 Tbit
700 nB =5.6E-18 Tbit
800 nB =6.4E-18 Tbit
900 nB =7.2E-18 Tbit
1000 nB =8.0E-18 Tbit
2000 nB =1.6E-17 Tbit
4000 nB =3.2E-17 Tbit
5000 nB =4.0E-17 Tbit
7500 nB =6.0E-17 Tbit
10000 nB =8.0E-17 Tbit
25000 nB =2.0E-16 Tbit
50000 nB =4.0E-16 Tbit
100000 nB =8.0E-16 Tbit
1000000 nB =8.0E-15 Tbit
1000000000 nB =8.0E-12 Tbit
(Nanobytes) to (Terabits) conversions | 1,024 | 2,269 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-35 | latest | en | 0.796775 |
https://www.oschina.net/question/861073_122501 | 1,558,931,617,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232260658.98/warc/CC-MAIN-20190527025527-20190527051527-00262.warc.gz | 875,355,607 | 12,853 | ## java提高精度问题（斐波那契数列）
```public class fibonacci2 {
public static void main(String args[])
{
double n = 1000;
double nums = fibonacci(n);
double nums_1 = fibonacci(n-1);
double nums_2 = nums_1/nums;
System.out.println(nums_2);
}
static double fibonacci(double n)
{
double a=1,b=1,c;
if(n==0) return 0;
if(n==1) return 1;
if(n==2) return 1;
else
{
for(int i=3;i<=n;i++)
{
c = a+b;
a = b;
b = c;
}
}
return (double) b;
}
} ```
0
```import java.math.BigDecimal;
import java.math.BigInteger;
import java.math.MathContext;
public class Fib {
public static void main(String[] args) {
int n = 1000;
BigDecimal d1 = new BigDecimal(fibonacci(n));
BigDecimal d2 = new BigDecimal(fibonacci(n - 1));
System.out.println(d2.divide(d1, new MathContext(100)));
}
static BigInteger fibonacci(int n) {
BigInteger a = BigInteger.ONE;
BigInteger b = BigInteger.ONE;
BigInteger c = BigInteger.ONE;
if (n < 3)
return BigInteger.ONE;
for (int i = 3; i <= n; i++) {
a = b;
b = c;
}
return c;
}
}```
Output: 0.6180339887498948482045868343656381177203091798057628621354486227
052604628189024497072072041893911375
0
0
BigDecimal nums_2 = nums_1.divide(nums,new MathContext(70));
0
0
BigDecimal
0
0 | 371 | 1,184 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2019-22 | latest | en | 0.145642 |
https://lessonotes.com/v2/nursery-2/lesson-notes-for-nursery-2-2nd-term-week-2-numeracy-topic-is-countingidentification-of-numbers-1---190.html | 1,702,156,586,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100972.58/warc/CC-MAIN-20231209202131-20231209232131-00883.warc.gz | 417,851,297 | 3,901 | # Lesson Notes By Weeks and Term - Nursery 2
Counting/identification of numbers 1 - 190
SUBJECT: Numeracy
TERM: 2nd Term
Week: 2
CLASS – Nursery 2
AGE – 5 years
DURATION – 5 periods of 35 minutes each
TOPIC- Counting/identification of numbers 1-190
CONTENT – Writing 1-150, Number endings. Fill in the missing number 100-140. Number spellings 1-30. Introduction to 2x table. Solid shapes
INSTRUCTIONAL OBJECTIVES – By the end of the lesson, the pupils should be able to –
1. Count and identify numbers 1-190
2. Write numbers 1-150
3. Spell and write numbers 1-30 in words
4. Identify solid shapes
5. Recite the 2x table
INSTRUCTIONAL MATERIALS – flash cards, charts, counters
INSTRUCTIONAL TECHNIQUES – Explanation, repetition, questions and answers
ENTRY BEHAVIOUR – The pupils can count numbers 1-170 and spell numbers 1-25
SET INDUCTION –She uses number rhyme to arouse their interest
INSTRUCTIONAL PROCEDURE
PERIOD I: Counting numbers 1-190 and writing 1-150
STEPS TEACHER’S ACTIVITY PUPILS’ ACTIVITY IIntroduction The teacher reviews the previous lesson introduces the new topic Pupils participate IICounting Using the number chart, she counts number 1-190 repeatedly. Pupils repeat after her IIIWriting numbers She counts and writes numbers 1-150 on the board, she also uses flashcards Pupils count and write in their book IVNumber ending She write the number endings on the board and also uses flashcards Pupils count Evaluation Count and write numbers 1-150 Conclusion She marks their books and corrects them where necessary
PERIODII& III: Number spelling and 2x table
STEPS TEACHER’S ACTIVITY PUPILS’ ACTIVITY IIntroduction The teacher reviews the previous lesson introduces the new topic Pupils participate IINumber spelling Using flashcards, she reviews spelling 1-25 and then introduces 26-30. She also writes them on the board Pupils spell the number IIIDrilling She drills them on the spelling.Twenty six = 26Twenty seven = 27Twenty eight = 28Twenty nine = 29Thirty = 30 IVFill in the missing gaps The teacher writes and fill in the missing gaps 100-160100 ______ 102 _____ 104 ____106 ______ 108 _____ 110 ____112 ______ 114 _____ 116 ____118 ______ 120 _____ 122 ____124 ______ 126 _____ 128 ____130 ______ 132 _____ 134 ____136 ______ 138 _____ 140 ____142 ______ 144 _____ 146 ____148 ______ 150 _____ 152 ____154 ______ 156 _____ 158 ____160 Evaluation Write numbers 26-30 in words26 =27 =28 =29 =30 =Fill in the missing gaps100 ______ 102 _____ 104 ____106 ______ 108 _____ 110 ____112 ______ 114 _____ 116 ____118 ______ 120 _____ 122 ____124 ______ 126 _____ 128 ____130 ______ 132 _____ 134 ____136 ______ 138 _____ 140 ____142 ______ 144 _____ 146 ____148 ______ 150 _____ 152 ____154 ______ 156 _____ 158 ____160 Conclusion She marks their books and corrects them where necessary
PERIOD V: Addition of numbers, 2x table
STEPS TEACHER’S ACTIVITY PUPILS’ ACTIVITY IIntroduction The teacher reviews the basic shapes and introduces the new topic Pupils respond IIAddition of numbers The teacher does simple addition of numbers 6 8 + 4 + 8 10 17 9+ 8 17 Pupils watch and listen III2x table She writes the 2x table on the board and recites it severally as the pupils recite after her2 x 1 = 22 x 2 = 42 x 3 = 62 x 4 = 82 x 5 = 102 x 6 = 122 x 7 = 142 x 8 = 162 x 9 = 182 x 10 = 202 x 11 = 222 x 12 = 24 Pupils participate Evaluation Add up1. 9 + 5 ___ 2. 9+ 6___ 3. 7+ 4___Complete the 2x table1. 2 x 5 =2. 2 x 6 =3. 2 x 7 =4. 2 x 8 = Conclusion She marks their books and corrects them where necessary | 1,027 | 3,645 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2023-50 | latest | en | 0.836023 |
https://byjus.com/question-answer/construct-a-triangle-abc-in-which-bc-6-cm-angle-b-45-circ-and-ab-28/ | 1,718,772,554,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861797.58/warc/CC-MAIN-20240619025415-20240619055415-00373.warc.gz | 133,657,196 | 26,755 | 1
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Question
# Construct a triangle ABC in which BC = 6 cm, ∠B=45∘ and AB - AC = 2.5 cm Steps of construction: Step 1: Draw BC = 6 cm Step 2: Construct ∠YBC=45∘ Step 3 will be -
A
Join AC, ΔABC is the required triangle.
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B
Draw perpendicular bisector of CD intersecting BY at A.
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C
From ray BY, cut-off line segement BD = 2.5 cm
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D
Join CD
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Solution
## The correct option is C From ray BY, cut-off line segement BD = 2.5 cm Step 3: From ray BY, cut-off line segement BD = 2.5 cm. Join CD Step 4: Draw perpendicular bisector of CD intersecting BY at A. Step 5: Join AC, ΔABC is the required triangle.
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Join BYJU'S Learning Program | 339 | 1,098 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2024-26 | latest | en | 0.82145 |
http://mathhelpforum.com/discrete-math/34080-proving-urgent.html | 1,480,894,013,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541426.52/warc/CC-MAIN-20161202170901-00214-ip-10-31-129-80.ec2.internal.warc.gz | 180,490,773 | 10,321 | 1. ## Proving (URGENT)
Prove that if n is an element of Z(the integers) and log2n is rational, then log2n is an integer.
2. Consider this:
$log_2 (n)^k = k log_2 (n)$
When n = 2, $log_2 (2) = 1$
If n is not a power of 2, then
$log_2 n = \frac{ln (n)}{ln (2)}$
is ln 2 rational?
3. Originally Posted by vincentngtf
Suppose that $\log_2 n = p/q$ where $p/q$ is a positive rational number. This means $2^{p/q} = n\implies \left( 2^{p/q} \right)^q = n^q\implies 2^p = n^q$. Now the fundamental theorem of arithmetic allows us to factorize $n$, the thing is that all factors much be $2$'s because the LHS is made out of two, thus, $n=2^m$. This means, $2^p = 2^{mq}$. Thus, $p=mq$ which means $q|p$ which tells us that $p/q$ is an integer. | 274 | 740 | {"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": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2016-50 | longest | en | 0.871509 |
https://topic.alibabacloud.com/zqpop/angle-divisor-how-to-use_12410.html | 1,709,269,691,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474948.91/warc/CC-MAIN-20240301030138-20240301060138-00786.warc.gz | 566,709,407 | 17,559 | # angle divisor how to use
Read about angle divisor how to use, The latest news, videos, and discussion topics about angle divisor how to use from alibabacloud.com
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### Java report tool FineReport common functions usage Summary (mathematical and trigonometric functions), javafinereport
Java report tool FineReport common functions usage Summary (mathematical and trigonometric functions), javafinereport ABS ABS (number): returns the absolute value of a specified number. Absolute Value refers to a value without positive or negative
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1. filter: Import flash. Filters. * package, If a filter is applied to a display objectcacheAsBitmapThe property value is settrue. If all filters are cleared, they are restored.cacheAsBitmapThe original value. If the width or height of the result
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JavaScript: Use atan2 to draw arrows and curves, and use javascriptatan2 Recently, I am engaged in Canvas plotting and know that JavaScript provides a trigonometric function such as atan2 (y, x. At first glance, I don't know. After all, the
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AbsABS (number): Returns the absolute value of the specified digit. An absolute value is a numeric value that has no sign.Number: Any real number that requires an absolute value.Example:ABS (-1.5) equals 1.5.ABS (0) equals 0.ABS (2.5) equals
### JavaScript: Use atan2 to draw arrows and curves
recently engaged Canvas drawing, knowing that a trigonometric functions such as atan2 (y,x) are available in JavaScript . At first glance, do not know, after all, in high school, learned trigonometric functions are:sin,cos,arcsin,arccos,tan,arctan ,
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ABS (x ):Basic syntax and crystal syntax.Parameter: X is the number or currency in which you want to return the absolute value.Return Value: Numeric ValueOperation: ABS (x) returns the absolute value of X.Example:ABS (1, 1.50)1.50 is returned. ABS (-
### Level 2015 C + + 4th week Project function
"Project 1-seeking greatest common divisor" reference solution(1) Enter two numbers and find out the greatest common divisor#include usingnamespacestd;//自定义函数的原型（即函数声明）int main(){ int a,b,g; cin>>a>>b; g=gcd(a,b); cout"最大公约数是:
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inherited1, the purpose of inheritance is to extend the function of the class 2, a subclass of Java can only inherit a parent class 3, Java does not allow multiple inheritance, allowing multiple layers of inheritance 4. The private method in the
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• Alibaba Cloud offers highly flexible support services tailored to meet your exact needs. | 979 | 4,312 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-10 | latest | en | 0.726566 |
https://www.education.com/resources/first-grade/money-math/CCSS/ | 1,643,200,810,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304947.93/warc/CC-MAIN-20220126101419-20220126131419-00159.warc.gz | 786,281,344 | 37,292 | # Search 1st Grade Common Core Money Math Educational Resources
26 filtered results
26 filtered results
Money Math
Common Core
Sort by
Money, Money, Money!
Workbook
Money, Money, Money!
This workbook is all about helping first graders ace their money math. Counting coins, word problems, and coin matching will help familiarize kids with the basics of money.
Math
Workbook
Counting Money: Chinese New Year
Worksheet
Counting Money: Chinese New Year
Start the lunar new year off right by giving your child's math skills a boost with this colorful coin-counting activity.
Math
Worksheet
Coin Toss: Frog
Worksheet
Coin Toss: Frog
Young learners cut out and assemble this simple template, then compete to see who can toss the most coins through the frog's open mouth.
Math
Worksheet
Money Match: Pennies
Worksheet
Money Match: Pennies
This worksheet lets your first grader play with money, while he practices counting and working with pennies.
Math
Worksheet
Learning to Count Coins
Worksheet
Learning to Count Coins
How many coins can you count? Help your little cashier learn to add up coin amounts to figure out the price of these items!
Math
Worksheet
Counting Money
Worksheet
Counting Money
Buy erasers, gum, and yo-yo's with this coin counting worksheet. Kids will use their math skills to find the coins needed to buy these items and more.
Math
Worksheet
Money Mania: Coins
Worksheet
Money Mania: Coins
Children practice recognizing the different coins and adding up their amounts.
Math
Worksheet
Money Word Problems #7
Worksheet
Money Word Problems #7
Encourage your young mathematician to help Stacy, Martin, and Gregory calculate how much money they have. Try this money math word problems worksheet.
Math
Worksheet
Money Word Problems #10
Worksheet
Money Word Problems #10
Help Nina, William, and Deana figure out how much money they have. Kids will solve money word problems to figure out how much cash each character has.
Math
Worksheet
Coin Math
Worksheet
Coin Math
Money math can be a fun way to practice addition. Your beginning coin counter can practice matching groups of coins with equal values!
Math
Worksheet
Worksheet
Students draw ten and one dollar bills for practice adding multiples of ten to two-digit numbers in this money math worksheet.
Math
Worksheet
Counting Coins: Dimes
Worksheet
Counting Coins: Dimes
To shop for fruit, kids read prices, count dimes, and match the prices to the equivalent amount of dimes on this first grade math worksheet.
Math
Worksheet
Comparing Money Amounts #4
Worksheet
Comparing Money Amounts #4
Your kid can gain some comfort with currency by using this money worksheet. Kids will compare two visual representations of money and decide which is bigger.
Math
Worksheet
Play Money
Worksheet
Play Money
Use these play bills to add excitement and real life context to your math lessons. Students will love using these one, five, ten, twenty, and hundred dollar bills to practice place value and money addition.
Math
Worksheet
How Much Money Do I Have? #9
Worksheet
How Much Money Do I Have? #9
Math
Worksheet
Coins: Discover the Dimes
Worksheet
Coins: Discover the Dimes
Kids completing this first grade math worksheet learn the name and appearance of dimes, find and identify dimes, and distinguish dimes from other coins.
Math
Worksheet
Dime Word Problems
Worksheet
Dime Word Problems
Students draw dimes and pennies to practice solving for 10 more and 10 less than a given two-digit number in this money math worksheet.
Math
Worksheet
Comparing Money Amounts #3
Worksheet
Comparing Money Amounts #3
Decimals can be daunting, but your kid can gain comfort with them without writing. This worksheet lets your kid work with money without writing decimal amounts.
Math
Worksheet
Coins: Catch the Quarters
Worksheet
Coins: Catch the Quarters
With this first grade math worksheet, kids learn the name and appearance of quarters, find and identify quarters, and distinguish quarters from other coins.
Math
Worksheet
Counting Coins: Present Math II
Worksheet
Counting Coins: Present Math II
Let your first grader flex his financial savvy with this fun worksheet. Your child must circle the coins that add up to the value of each present.
Math
Worksheet
Counting Coins: Present Math I
Worksheet
Counting Coins: Present Math I
In this first grade math worksheet, your child will circle the coins needed to purchase each item. This exercise teaches coin recognition and addition.
Math
Worksheet
Coin Amounts #6
Worksheet
Coin Amounts #6
These spilled coins need sorting! See if your child identify the amount of a coin at a glance.
Math
Worksheet
Coin Amounts #7
Worksheet
Coin Amounts #7
Learning to identify coins can be tricky at first, but with this matching activity your child will be an expert in no time!
Math
Worksheet
Counting Coins: Pennies and Nickels II
Worksheet
Counting Coins: Pennies and Nickels II
Teach your first grader about money with this simple counting worksheet. Your child will count each group of coins, then write their total value on the line. | 1,112 | 5,065 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2022-05 | longest | en | 0.856124 |
https://dcasler.com/9550/ | 1,722,713,407,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640377613.6/warc/CC-MAIN-20240803183820-20240803213820-00508.warc.gz | 158,531,188 | 12,318 | # Extra Class Worked Example, Question E5A16
Frank Dambach, KK6GPI, asked for a detailed, worked-through example of determining the resonant frequency of an RLC circuit. He chose Extra question E5A16. The question has to do with determining the resonant frequency of a parallel RLC circuit for R=33 ohms, L=50 microhenries, and C=10 picofarads.
The 33 ohm resistor in the question is extraneous information. It’s what in my engineering student days we called “the dimensions of the doorknob,” meaning information that has no bearing on the answer. The frequency of resonance is determined solely by the inductor and the capacitor.
However, if we were determining the Q for the circuit, the resistance enters into the equation. But that’s not the task at hand.
(My apologies for the focus issues early in the video. The camera hunts for something at the center of the scene to focus on.)
This entry was posted in Amateur Extra Class License Training, Ham Radio blog entries, Video and tagged . Bookmark the permalink.
### 2 Responses to Extra Class Worked Example, Question E5A16
1. Dave says:
You’re welcome!
2. Frank Dambach says:
Thanks sooo much Dave! I figured out how to do this on a scientific calculator but had forgotten completely how to treat the exponents when squaring and moving from denominator to numerator. This was a big help and I hope other followers will agree. Thanks again! | 316 | 1,405 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-33 | latest | en | 0.935526 |
https://virtuescience.com/13.html | 1,585,773,882,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506121.24/warc/CC-MAIN-20200401192839-20200401222839-00426.warc.gz | 754,205,692 | 8,827 | <12 Number Data-Base Random Number 14>
## The Number 13: Properties and Meanings
13 is a Prime Number.
13 can be Partitioned in 101 ways.
13 can be Partitioned 7 times with each term no larger than 2.
13 can be Partitioned 21 times with each term no larger than 3.
13 can be Partitioned 39 times with each term no larger than 4.
13 can be Partitioned 57 times with each term no larger than 5.
13 can be Partitioned 71 times with each term no larger than 6.
13 can be Partitioned 82 times with each term no larger than 7.
13 can be Partitioned 89 times with each term no larger than 8.
13 can be Partitioned 94 times with each term no larger than 9.
13 can be Partitioned 97 times with each term no larger than 10.
13 can be Partitioned 99 times with each term no larger than 11.
13 can be Partitioned 100 times with each term no larger than 12.
13 is the number of + signs needed to write the Partitions of 5
13 is a Fibonacci Number.
13 is a Centered Square Number.
13 is a Centered 12-gonal Number.
13 is a Delannoy Number.
The hypotenuse of the Pythagorean 5-12-13 triangle= 13
The 13 Archimedean Solids.
13 space groups are Monoclinic.
The Chemical Element Aluminum has an atomic number of 13.
The sacred cord of Druids has thirteen segments.
There are thirteen lunations in the year.
The Sumerians used a zodiac including 13 constellations and 26 main stars.
The thirteen gates of the Human body of the woman: 2 eyes, 2 ears, 2 nostrils, the mouth, 2 breasts, the navel, the anus, the urethra and the vagina.
The card deck includes 13 hearts, 13 spades, 13 squares, 13 clubs.
The Jewish sage Moses Maimonides established 13 principles of the Jewish faith.
The Chinese abacus consists of 13 columns of beads.
The 13 ancient Crystal Skulls.
Benjamin Franklin
Virtues:
1. TEMPERANCE. Eat not to dullness; drink not to elevation. 2. SILENCE. Speak not but what may benefit others or yourself; avoid trifling conversation. 3. ORDER. Let all your things have their places; let each part of your business have its time. 4. RESOLUTION. Resolve to perform what you ought; perform without fail what you resolve. 5. FRUGALITY. Make no expense but to do good to others or yourself; i.e., waste nothing. 6. INDUSTRY. Lose no time; be always employ'd in something useful; cut off all unnecessary actions. 7. SINCERITY. Use no hurtful deceit; think innocently and justly, and, if you speak, speak accordingly. 8. JUSTICE. Wrong none by doing injuries, or omitting the benefits that are your duty. 9. MODERATION. Avoid extreams; forbear resenting injuries so much as you think they deserve. 10.CLEANLINESS. Tolerate no uncleanliness in body, cloaths, or habitation. 11.TRANQUILLITY. Be not disturbed at trifles, or at accidents common or unavoidable. 12. CHASTITY. Rarely use venery but for health or offspring, never to dulness, weakness, or the injury of your own or another's peace or reputation. 13. HUMILITY. Imitate Jesus and Socrates.
Film: Thirteen.
Film: 13 Ghosts.
Film: 13 going on 30.
Film: The 13th Warrior.
In the year 13 AD Tiberius made his triumphant procession through Rome after siege of Germany.
The year 13 AD was the last year of shijianguo era of the Chinese Xin Dynasty.
<12 Number Data-Base Random Number 14>
Share any properties and meanings for particular Numbers...contact me directly, thanks.
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An exclusive community: HighTSQ.com
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Charging a small amount helps to keep out spammers and trolls plus it helps support me and my work. | 950 | 3,624 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2020-16 | longest | en | 0.886445 |
http://csharphelper.com/blog/category/puzzles/ | 1,632,060,148,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056890.28/warc/CC-MAIN-20210919125659-20210919155659-00276.warc.gz | 14,148,238 | 11,786 | # Category Archives: puzzles
## New Book: The Modern C# Challenge
My latest book, The Modern C# Challenge, is now available. It’s a collection of 100 programming challenges that let you test your ability in a wide variety of programming topics, many of which are not usually covered in traditional programming … Continue reading
## Solution: The 40 pound stone puzzle
This is the solution to The 40 pound stone puzzle. If you haven’t read the puzzle yet, click here before you read the solution. This puzzle similar to some of the puzzles in my book Interview Puzzles Dissected. The … Continue reading
Posted in algorithms, books, games, puzzles | | Comments Off on Solution: The 40 pound stone puzzle
## Puzzle: The 40 pound stone
This is a variation of a puzzle given on Car Talk. It’s similar to some of the puzzles in my book Interview Puzzles Dissected. A farmer uses a double-pan balance and a 40 pound stone to measure 40 pound weights: … Continue reading
Posted in algorithms, books, games, puzzles | | Comments Off on Puzzle: The 40 pound stone
## Make the basis for a card game in C#
The goal of this example is to provide some support for building a card game. It shows how to make a class to represent cards and to to tell what card is displayed in a PictureBox. It’s actually relatively straightforward. … Continue reading
Posted in games, graphics, image processing, puzzles | | 1 Comment
## Verify sizes of playing cards in C#
Recently I wanted to use some playing cards in a program so I looked online for some nice images. I could have included 55 or so separate images (including jokers and backs), but I decided it would be easier to … Continue reading
Posted in games, graphics, image processing, puzzles | | Comments Off on Verify sizes of playing cards in C#
## Five hats puzzle solution
If you haven’t read it yet, read the Five hats puzzle. Recall that the warden has three white hats and two black hats. If the first prisoner, Art, saw two black hats, then he would know that he had a … Continue reading
Posted in books, games, puzzles | | Comments Off on Five hats puzzle solution
## Five hats puzzle
This “hats puzzle” was described on December 26, 2016 on NPR’s Car Talk show. Sadly Tommy died in 2014 and the show was a rebroadcast so this puzzler has probably been used before. My book Interview Puzzles Dissected doesn’t include … Continue reading
Posted in books, games, puzzles | | Comments Off on Five hats puzzle
## Sample Puzzles for Interview Puzzles Dissected
Here are some sample puzzles (in PDF format) from my book Interview Puzzles Dissected. Three Birds – Calculate the total number of miles three birds fly while zigzagging between each other during a trip from New York to Philadelphia. We … Continue reading
## Book Errata Page: Interview Puzzles Dissected
This is the errata page for my book Interview Puzzles Dissected, Solving and Understanding Interview Puzzles. If you find mistakes, please post them here in the Leave a Reply box at the bottom of the page. If you have questions, … Continue reading
Posted in books, games, miscellany, puzzles | | 1 Comment
## Book Discussion Page: Interview Puzzles Dissected
This is a discussion page for my book Interview Puzzles Dissected, Solving and Understanding Interview Puzzles. Please post questions, thoughts, and suggestions here in the “Leave a Reply” box at the bottom of the page. I will moderate posts and … Continue reading
Posted in books, games, miscellany, puzzles | | 3 Comments | 799 | 3,513 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.921188 |
gravitypushing.org | 1,566,295,666,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315321.52/warc/CC-MAIN-20190820092326-20190820114326-00164.warc.gz | 85,993,169 | 6,136 | Derived Value for a Universal Gravitation Constant for Pushing Gravity
Assumptions
1. Gravity comes from outer space and pushes.
2. Gravity is streams of particles travelling in straight lines.
Taking the earth as a sphere and placing a sphere of 1kg mass on the earth's surface, the centre of the imaginary sphere is where the two spheres touch and gravity from one side of the imaginary sphere minus gravity from the other side = net gravitational force.
Let: Sg = force of gravity from outer space on unit mass ρ = mean earth density = 5514.736kg/m3 D = mean diameter of earth = 12742000m d = diameter in metres of 1kg kgf = force in Newtons of 1kg at the surface of the mean density earth and mean diameter earth at the pole and defined as 9.82192738N which is equal to G x 1kg x earth mass / (earth mean radius)2 G = Newtonian universal gravitational constant (6.6738 x 10-11N m2/kg2) Gp = effectivity on bulk material SR = shielding ratio
SR is the ratio of shielded space gravity force to unshielded space gravity force passing through the imaginary sphere. The imaginary sphere whose surface passes through the centre of the earth will have gravity streams coming from space through its surface, but a proportion is affected by the earth. This proportion divided by the space gravity through an unaffected sphere is the "shielding ratio". Any gravity stream though, that is not parallel to a line between the earth centre and the net gravitational force position will have a diminishing force in that plane as the angle between the gravity stream and the line from earth to net gravitational force position increases to 90 degrees.
It is well known that the average line length of lines through a circle from a point on its circumference is 2/3 of the diameter, and the average length to the perpendicular from the circumference is 2/3 of the diameter. The value for SR that applies in all cases can be found by moving the centre of the earth to a distance r from where the net gravitational force is to be calculated. If this is made very large in relation to the earth's diameter the streams of particles passing through the earth to where the net gravitational force is measured can be assumed to be parallel to each other.
Gravitational force shielding = cross sectional area of earth times 2/3 = (D2Π)/6
Unshielded gravitational force passing through the imaginary hemisphere is equal to the surface area of a hemisphere with an average effective r in a particular direction. This can be written as:
(4Πr2)/3
This is equivalent to the net force of gravity from the hemisphere surface to the net gravitational force position.
Therefore SR = [(D2GpΠ)/6]/[(4ΠGpr2)/3] = (D2/8r2Π)/6 (1)
This equation is also derived when the net gravitational force is at the surface of the earth, in this case SR is equal to 0.5 since r is D/2. It will be seen that this relationship applies to any planet or object diameter and its distance from the net gravitational force.
If a stream of gravity particles hit the surface of the earth and pass through it, the amount of force lost by the stream in passing through the earth's diameter is defined as:
Earths density x distance through earth x the effectivity
The force in any particular direction is the sum of all the forces with forces perpendicular to the particular direction having no effect.
net force = sum of all forces
With reference to Figure 1 gravity from one side of the imaginary sphere minus gravity from the other = net gravitational force.
With d on the surface of D: the force per kg = [Sg - (dρGp)SR] - [Sg - (DρGp)SR] (2) re-arranging = [Sg - (dρGp)SR - Sg + (DρGp)SR] (3) simplifying (3) = [(DρGp)SR - (dρGp)SR] (4) when force = 1kgf then d is negligible relative to D and therefore DρGpSR = 1kgf/kg (5)
From (1)
When r = D/2 ie. net gravitational force is on the surface of D then SR = 0.5 Putting SR = 0.5 into (5) and putting 1kgf = 9.8258816N Gp = 2x9.82192738/(Dρ) (6)
If gravity comes from an infinite distance to a point, the number of force lines in a sphere will be constant and with a constant effectivity for any 1kg mass then GpNm2/kg per kg is constant
Gp = 2.795531562x10-10Nm2/kg2 (7)
The net gravitational force per kg of an object r distance from the earth's centre is given by combining (1) and (5) and using Gp
D2GpρD/(8r2) = force per kg (8)
In general terms
Let: M1 = mass at the net gravitational force (kg) D0 = diameter of object creating a net gravitational force (m) ρ = density of object creating a net gravitational force (kg/m3) r0 = distance between center of M1 and centre of D0(m) Then (8) becomes Force = (ρD02M1GpD0)/8r02 (9)
Comparing Newton's formula for gravitational force with the one derived above gives a derived value for G.
From Newton Force F = M1M2G/r2 (10) where M1 = 1kg F = gravitational force of 1kg on the surface of a mean diameter and density of earth at the pole r = distance between M1 and M2 centres earth mass M2 = (D3Πρ)/6 (11) where D = mean diameter of earth and ρ = mean density of earth Also from (9) F = (ρD3M1Gp)/8r2 equating (9) and (10) and using (11) (M1ΠρD3G)/6r2 = (ρD3M1Gp)/8r2 rearranging G = 3Gp/4Π = 6.67384x10-11Nm2/kg2 Therefore the derived value for Newton's universal gravitation constant is: G = 6.67384x10-11Nm2/kg2
Newton's attractive (pulling) force between two spheres M1 and M2 separated by distance r is given by Force = M1M2G/r2
The force pushing the same two spheres together is given by:
Force = M1Gp(ρD3/8r2) where D is the diameter of M2 and D3 is D2 times the length of D which in the case of a sphere is D but in the case of a spiral galaxy is the combined length of the stars in a column.
If M1 and M2 are both spheres r distance apart then: M1Gp(ρD3/8r2) = M1M2G/r2
Spiral Galaxy
The force on a body at the extremity of a spiral galaxy is influenced by the diameter, density and length of all the bodies in a radial line passing through the imaginary sphere. The sphere can be centred on the extreme body location and the radius is the length to the centre of the galaxy and the length through the galaxy is the length through the column of stars (DL). This will give a force towards the galaxy centre which is more than calculated by Newtonian calculations. The orbital velocity of stars is likely to remain fairly constant since
centripetal force = Mass x velocity2/radius
and as the radius increases the mass of the line or column of stars increases.
From equation (9) Force = (ρD02M1GpD0)/8r02 and for a spiral galaxy it becomes Force = (ρD02M1GpDL)/8r02 where DL is the summed length of all the stars in the column.
Big Bang Mass
Considering the gravitational force on the earth caused by the sun, both equation (9) and (10) are in agreement except in one case gravity is pushing and in the other it's pulling. Assuming gravity is pushing and comparing (9) with the standard equation for centripetal force F = Mω2r, the mean centripetal force is 0.0007% less than the gravitational force.
If gravity comes from outer space it may be wholly or partly produced in stars. To make up the difference between the centripetal and gravitational force, the sun would need to produce a force of 1.7814 x 10-8N per kg of mass. Gravity produced in the sun will diminish with the square of the distance from the sun. The effect of the sun's gravity on a space probe sent from earth would reduce the probe's acceleration in the direction of the sun by 2.7834 x 10-12m/s2 at 80AU distance from the sun. A further consequence of this reasoning is that if the sun had its mass increased with constant density a point would be reached when the gravity coming out of the sun equalled the gravity coming from outer space. At this point when the sun's net surface gravity force would be zero, the sun would have a mass of 7.03241 x 1039kg. If production of gravity is also dependent on the temperature of the star, cooler bodies collecting under space gravitational force could have a greater mass than could be sustained after warming up by compression and nuclear reactions. This suggests singularities could occur on the galactic scale and also the universal scale.
Retardation of the Earths Orbit Velocity Around the Sun
At the atomic scale the passage of gravity through the nucleus of an atom may be very different from passage through bulk material.
From equation (6) Gp is constant at 2.795531562x10-10Nm2/kg2
On the atomic scale the constant may increase in magnitude from C to Ca. Calculations for electron velocity and energy, and the corresponding forces at the principle quantum number, are shown to be balanced by forces from equation (9) if C is changed to Ca and
Ca = 6.354783124x1029Nm2/kg2
If the gravity particle stream mass rate is equal to the electron orbital mass rate the gravity particle velocity is 6.59 x 1014 x c (where c = speed of light). Feynman found that over the past 200 years there had been several attempts to show that gravity pushed, but the retardation forces on the earth in its orbit around the sun associated with these proposals would have a far greater effect than that which is observed.
In the present case though, the constant associated with atomic forces is Ca and on bulk matter C. Using Ca, a minimum for the product of gravity mass (gm) and change in gravity velocity (gvel) can be found which will balance the atomic forces, and from that the minimum change in gvel can be found if it is assumed that the gravity mass rate is equal to the elementary particle mass rate in a given time. The centripetal force of an orbiting mass secured in orbit by a tie or arm is effectively secured by the interaction of the atoms in the tie and this is the same interaction as gravity with elementary particles.
If the gravity mass rate balances the atomic forces at any point in time, then half the earth mass will face forwards and half rearwards and the gravity mass rate will equal the earth mass rate.
The earth's velocity though space is negligible compared to the postulated speed of gravity and results in no reduction in earth's velocity.
This gives a time for the earth to be absorbed by the sun that is longer than the generally accepted length of time for this to happen due to other considerations. The above reasoning shows that all known gravitational forces can be calculated if gravity is a pushing force. If gravity is produced in stars, cold condensed matter may shed matter progressively in ripples as it warms up, a galactic mass could be sufficient for this process. This could also explain the big bang singularity. | 2,576 | 10,587 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2019-35 | latest | en | 0.921584 |
https://developer.arm.com/docs/100069/0605/advanced-simd-programming/polynomial-arithmetic-over-01 | 1,580,119,259,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251696046.73/warc/CC-MAIN-20200127081933-20200127111933-00446.warc.gz | 397,124,908 | 22,974 | You copied the Doc URL to your clipboard.
ARM Compiler armasm User Guide : Polynomial arithmetic over {0,1}
# Polynomial arithmetic over {0,1}
The coefficients 0 and 1 are manipulated using the rules of Boolean arithmetic.
The following rules apply:
• + 0 = 1 + 1 = 0.
• 0 + 1 = 1 + 0 = 1.
• 0 * 0 = 0 * 1 = 1 * 0 = 0.
• 1 * 1 = 1.
That is, adding two polynomials over {0,1} is the same as a bitwise exclusive OR, and multiplying two polynomials over {0,1} is the same as integer multiplication except that partial products are exclusive-ORed instead of being added. | 170 | 575 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2020-05 | latest | en | 0.8814 |
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