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.chegg.com/homework-help/questions-and-answers/question-engineering-electromagnetics-william-h-hayat-jr-john--buck-7th-edd33-given-electr-q1170516 | 1,386,812,610,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164333763/warc/CC-MAIN-20131204133853-00047-ip-10-33-133-15.ec2.internal.warc.gz | 280,903,134 | 7,000 | # plz help me in Engineering Electromagnetics (charge and flux density)
0 pts ended
This question is from Engineering Electromagnetics by William H. Hayat & JR. John A. Buck 7th ed.
D3.3. Given the electric flux density, D = 0.3 r2 ar nC/m2 in free space:
(a) find E at point P(r = 2, θ = 25o, φ = 90o);
(b) find the total charge whithin the the sphere r = 3;
(C) find the total electric flux leaving the sphere r = 4.
Note: the answers sholud be: (a) 135.5ar V/m; (b) 305 nC; (c) 965 nC | 168 | 494 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2013-48 | latest | en | 0.856654 |
http://stackoverflow.com/questions/7707440/c-linked-list-implementation-crashing | 1,469,779,608,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257829972.49/warc/CC-MAIN-20160723071029-00240-ip-10-185-27-174.ec2.internal.warc.gz | 231,928,469 | 20,663 | Dismiss
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# C++ Linked list implementation crashing
I am trying to implement a linked list for a data structures class and I am having some difficulty with the searching portion of the algorithm.
Below is the offending code, which I have tried to implement following the pseudo-code in the MIT introduction to algorithms text:
``````//
// Method searches and retrieves a specified node from the list
//
Node* List::getNode(unsigned position)
{
for(unsigned i = m_listSize-1; (current != 0) && (i != position); --i)
current = current->next;
return current;
}
``````
The head at this point in the program is the 4th node, which contains the value of int 5. the problem appears to be in the body of the for-loop, where the pointer to the node object is assigned to the next node. But this is going beyond the head of the node, so it is essentially pointing at some random location in memory (this makes sense).
Shouldn't the algorithm be moving to the previous Node instead of the next Node in this case? Below is the pseudo-code:
``````LIST-SEARCH(L, k)
while x != NIL and key != k
do x <- next[x]
return x
``````
Also, here is the header file for my Linked list implementation. I haven't tried to implement it in Template form yet just to keep things simple:
``````#ifndef linkList_H
//
// Create an object to represent a Node in the linked list object
// (For now, the objects to be put in the list will be integers)
//
struct Node
{
// nodes of list will be integers
int number;
// pointer to the next node in the linked list
Node* next;
};
//
// Create an object to keep track of all parts in the list
//
class List
{
public:
// Contstructor intializes all member data
// methods to return size of list and list head
unsigned getListSize() const { return m_listSize; }
// retrieving and deleting a specified node in the list
Node* getNode(unsigned position);
private:
// member data consists of an unsigned integer representing
// the list size and a pointer to a Node object representing head
unsigned m_listSize;
};
#endif
``````
``````//
//
{
Node* theNode = new Node;
theNode = newNode;
theNode->next;
++m_listSize;
}
``````
-
Is the list doubly-linked? What does is look like in memory? – Dave Oct 10 '11 at 0:11
I think the problem may be in the addNode member function. Please supply that code so we can ensure that the list is constructed correctly. – Ed Heal Oct 10 '11 at 0:16
Code supplied above. – Dylan Oct 10 '11 at 0:20
Yes, I believe this should be doubly-linked. I haven't added the previous node field to the element. Not sure how to do that. – Dylan Oct 10 '11 at 0:23
Try this to construct the list:
``````void List::addNode(int number)
{
newNode = new Node;
newNode -> number = number;
newNode -> next = m_listHead ;
++m_listSize;
}
``````
It will add nodes to the head. Perhaps you may wish to store the pointer to the tail and insert the nodes there.
-
Unfortunately your code doesn't resemble the pseudo code you supply.
The pseudo-code is for searching a linked-list for a key, not a position.
``````Assign head to (node) x.
while x isn't null and the key inside the current node (x) doesn't match k
assign x->next to x
return x
``````
The returned value is either a pointer to the node that contains k or null
If you're trying to find the node at a given position your loop would be (note this is assuming you're going to use a zero-based index for accessing the list):
``````Assign head to (node) x
assign 0 to (int) pos
while x isn't null and pos not equal to given position
assign x->next to x
increment pos
return x
``````
The result will either be a pointer to the node at the given position or null (if you hit the end of the list first)
Edit: Your code is very close to the latter if that's what you're trying to do ... can you see the difference?
Edit because I like homework where the OP asks the right questions :)
``````Node* List::getNodeContaining(int searchValue)
{
while (current != 0 && current->number != searchValue)
{
current = current->next;
}
return current;
}
Node* List::getNodeAtPos(int position)
{
int pos = 0;
while (current != 0 && pos != position)
{
current = current->next;
pos++;
}
return current;
}
``````
-
Hmm. I understood the key to be the index of the loop and the k to the position supplied by the client code. Kind of in over my head here it seems... – Dylan Oct 10 '11 at 0:18
Ah, I see. The key field of the Node should be whatever the element is in the given node? (eg. "int number" for a list of numbers) – Dylan Oct 10 '11 at 0:22
Right :) If your method is supposed to be searching for what's inside the node, you need to be comparing the value passed in to each node's contents (`Node->number`) – Brian Roach Oct 10 '11 at 0:29
Well, it seems you are incrementing position. I am still slightly confused though; is "the key inside the current node" to mean what is pointed to by x (the current node)? I am trying to basically have the client code say "give me this bit of data (eg integer 4)" and the list will find that element. I guess if that is the case, I should probably be comparing position against current->key... – Dylan Oct 10 '11 at 0:34
See my latest edit. If you're trying to search for something in the list, you need to compare to the value contained inside the node. If you're looking for something at position x in the list, then you need to keep track of position. EDIT: (Gah - it's right now, made a mess of that edit) – Brian Roach Oct 10 '11 at 0:37
You list is very different from what a normal list ADT looks like. Rather than returning nodes, which would require the client know about the list implementation, you return and accept the type you're making a list of.
In this case you're making a list of integers, sou you'd want
``````public:
void add(int num); //prepends an Item to the list
int get(int pos);
``````
The implementations of both are simple. Add makes a new node, and links it in;
``````void List::add(int num)
{
Node *newNode = new Node;
newNode->number = num;
m_listSize++;
}
``````
Then get is easy too:
``````int List::get(int pos)
{
if(pos>m_listSize)
;//throw an error somehow | 1,591 | 6,334 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-30 | latest | en | 0.841736 |
http://stackoverflow.com/questions/5481375/collision-detection-not-working-right | 1,455,267,379,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701163512.72/warc/CC-MAIN-20160205193923-00114-ip-10-236-182-209.ec2.internal.warc.gz | 212,893,832 | 17,490 | # collision detection not working right
Okay, so my paddle collision is working fine:
``````if(velo.y > 0){
float ballHitX = position.x + velo.x * t;
if(t <= 1.0){
velo.y = -velo.y;
}
}
}
``````
But my wall collision isn't. (the ball goes up when under the paddle, and down when not)
``````if(velo.y < 0){
float t = ((position.y - radius) - (wall[2].y + wall[2].height))/ velo.y;
if(t <= 1.0){
velo.y = -velo.y;
}
}
``````
How do I stop this error and make it so the ball bounces off the wall?
-
My guess is that you're flipping it twice.
``````if(wall) {
velo = -velo;
}
velo = -velo;
}
``````
``````am i hitting the wall? nope
am i hitting the paddle? yep! okay flip velocity
``````
But when you do your wall, it goes like this:
``````am i hitting the wall? yep! okay flip velocity
am i hitting the paddle? yep! okay flip velocity
``````
So because yu're hitting both conditions, it flips twice.
You need to determine whether or not you've flipped already to prevent double flippage.
-
That worked. Thank you! ^_^ – CyanPrime Mar 30 '11 at 3:25 | 336 | 1,063 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-07 | longest | en | 0.741323 |
https://www.largeformatphotography.info/forum/showthread.php?142129-How-To-Understand-Densitometer-Readings&s=79b36afed753659c48f7e019fa0261be | 1,596,729,162,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439736972.79/warc/CC-MAIN-20200806151047-20200806181047-00276.warc.gz | 642,906,750 | 14,125 | 1. ## How To Understand Densitometer Readings
I have been lent a densitometer to experiment with.
Has anyone got any good online resources which I can read to better understand how to use them.
For example, With the emulsion side up. I have done measured a deep shadow on the negative and its reading 0.13 but at the moment I am not quite sure what that is telling me
2. ## Re: How To Understand Densitometer Readings
Ian - you cannot do much better than this:
It is an excellent, easy to follow introduction to sensitometry/densitometry, which is what you're doing.
To make a long story short, for transmission densitometry (reading densities of negatives), the densitometer reading is a measure of the opacity of the film. The higher the number, the more dense the film, and the more light it blocks. Low numbers would be relatively clear (shadow areas) and high numbers would be relatively dense (highlight areas).
Whether the negative should be read emulsion up or down depends on the densitometer type.
The densitometer may or may not need to be calibrated.
0.13 would be a very thin deep shadow. Assuming the densitometer is calibrated, you should first take a reading of an unexposed region of the film. This will give you the lowest density (film base + fog). Then you subtract that number from all your subsequent readings to give you the effective "net" density. The net density can be thought of as image density. For example suppose you read an unexposed part of the negative and the densitometer indicates 0.05, that would be your base+fog density. Then you would subtract 0.05 from all your other readings. 0.13-0.05=0.08, meaning the area you read has a net (image) density of 0.08. In Zone System terms that would be a very deep shadow on the low side of Zone I.
3. ## Re: How To Understand Densitometer Readings
Thanks Michael, I shall investigate that link you sent.
At this stage, I guess I am more interested to make sure I have enough density in the shadow areas to reveal some texture. I have read that Zone 1 is 0.1 above film base and fog so I am guessing that Zone 3 should be about 0.9 above film base and fog but I could and are probably wrong there
4. ## Re: How To Understand Densitometer Readings
Actually Zone III would be much lower than 0.9, closer to 0.4 depending on EI and contrast index.
Zone I is typically targeted at ~0.1 above film base + fog.
Keep in mind a densitometer reading can only be a guide, since "detail" is a function of contrast, not density. Most films will have good local contrast at Zone III densities if you are calibrating to a Zone I density of 0.1. Ultimately you have to judge shadow detail by looking at your prints (or digital output), not densitometer readings.
5. ## Re: How To Understand Densitometer Readings
Originally Posted by Michael R
Ian - you cannot do much better than this:
This has to be one of the best documents I have read to date
6. ## Re: How To Understand Densitometer Readings
Originally Posted by Michael R
The densitometer may or may not need to be calibrated.
0.13 would be a very thin deep shadow. Assuming the densitometer is calibrated, you should first take a reading of an unexposed region of the film. This will give you the lowest density (film base + fog). Then you subtract that number from all your subsequent readings to give you the effective "net" density. The net density can be thought of as image density. For example suppose you read an unexposed part of the negative and the densitometer indicates 0.05, that would be your base+fog density. Then you would subtract 0.05 from all your other readings. 0.13-0.05=0.08, meaning the area you read has a net (image) density of 0.08. In Zone System terms that would be a very deep shadow on the low side of Zone I.
Are there any charts that map densities to zones ?
7. ## Re: How To Understand Densitometer Readings
Originally Posted by Michael R
Whether the negative should be read emulsion up or down depends on the densitometer type.
If a transmission densitometer gives different readings for emulsion-up v. emulsion-down, the meter is broken.
Or else you have extremely bright ambient light sneaking into the measurement aperture.
This would be analogous to producing different enlargements depending on emulsion orientation.
It doesn't happen (except for the image being reversed).
- Leigh
8. ## Re: How To Understand Densitometer Readings
Density is one way of expressing the amount of ambient light transmitted through the emulsion.
This is of significance during printing (or viewing of a transparency).
The units is logarithmic. A factor of 2 equates to a value of 0.3, 4 = 0.6, 8 = 0.9, 10 = 1.0.
Each step of 0.3 equates to one aperture stop.
A density of 0.6 would equate to closing an aperture by 2 stops, or speeding the shutter by 4x.
- Leigh
9. ## Re: How To Understand Densitometer Readings
A helpful book from the heyday of densitometers is Sensitometry for Photographers by Jack Eggleston.
10. ## Re: How To Understand Densitometer Readings
Highly recommend the following book:
Photographic Sensitometry: The Study of Tone Reproduction
by Hollis N. Todd, Richard D. Zakia | Hardcover
You can easily find a copy up for auction for \$5-10.00. When I was at RIT, became good friends with Hollis Todd. The book at first seems overwhelming at first but once you skim its contents, it becomes easy reading.
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• | 1,336 | 5,557 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-34 | latest | en | 0.927328 |
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MHT CET 2021 21th September Evening Shift
+2
-0
The number of solutions of cos2$$\theta$$ = sin$$\theta$$ in (0, 2$$\pi$$) are
A
3
B
2
C
4
D
1
2
MHT CET 2021 21th September Evening Shift
+2
-0
If $$\theta+\phi=\alpha$$ and $$\tan \theta=k \tan \phi($$ where $$K>1)$$, then the value of $$\sin (\theta-\phi)$$ is
A
$$\mathrm{k} \tan \phi$$
B
$$\sin \alpha$$
C
$$\left(\frac{\mathrm{k}-1}{\mathrm{k}+1}\right) \sin \alpha$$
D
$$\mathrm{k} \cos \phi$$
3
MHT CET 2021 21th September Morning Shift
+2
-0
$$2 \sin \left(\theta+\frac{\pi}{3}\right)=\cos \left(\theta-\frac{\pi}{6}\right)$$, then $$\tan \theta=$$
A
$$\frac{-1}{\sqrt{3}}$$
B
$$-\sqrt{3}$$
C
$$\sqrt{3}$$
D
$$\frac{1}{\sqrt{3}}$$
4
MHT CET 2021 20th September Evening Shift
+2
-0
The principal solutions of $$\sqrt{3} \sec x+2=0$$ are
A
$$\frac{\pi}{6}, \frac{5 \pi}{6}$$
B
$$\frac{5 \pi}{6}, \frac{7 \pi}{6}$$
C
$$\frac{\pi}{3}, \frac{2 \pi}{3}$$
D
$$\frac{2 \pi}{3}, \frac{4 \pi}{3}$$
MHT CET Subjects
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EXAM MAP
Joint Entrance Examination | 498 | 1,193 | {"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} | 3.34375 | 3 | CC-MAIN-2023-50 | latest | en | 0.492605 |
https://schoollearningcommons.info/question/two-angles-are-complementary-one-of-them-is-40-degree-the-measure-of-the-other-one-is-24863258-86/ | 1,631,961,026,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056392.79/warc/CC-MAIN-20210918093220-20210918123220-00015.warc.gz | 564,415,515 | 13,365 | two angles are complementary one of them is 40 degree the measure of the other one is _________??
Question
two angles are complementary one of them is 40 degree the measure of the other one is _________??
in progress 0
5 days 2021-09-13T14:59:32+00:00 2 Answers 0 views 0
Ur ans is 50 degree
Step-by-step explanation:
As we know that complementary angles is equal to 90 degree
So, if one of them is 40 degree
then other is 90-40 = 50 degree
Hope this will help u | 135 | 475 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.932608 |
https://www.kidsacademy.mobi/printable-worksheets/online/age-9/learning-skills/addition-practice/ | 1,721,755,669,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518059.67/warc/CC-MAIN-20240723163815-20240723193815-00125.warc.gz | 722,600,685 | 56,177 | # Addition Practice Worksheets for 9-Year-Olds
Favorites
Interactive
• 9
• Interactive
In the desert, scorching heat and little water mean animals must adapt to survive. Show kids pictures of these animals and teach their names. Then, read the word problems in the worksheet, and have them check the box next to the pictures that portray each story. 80 words
Worksheet
Let your kids have fun with this baseball-themed addition worksheet! Read the three simple word sentences with them, then use the pictures to help them add with their fingers. Ask them to copy the fingers held up in the pictures, then check the box with the right answer. Kids who love baseball will enjoy this activity!
Worksheet
## Under the Sea Addition Worksheet
Ask your kids to identify undersea animals in a picture, then solve the word problems at the bottom with it! If they're into the nature channel or marine life, they'll love this worksheet. Check the box next to the correct answer for each one. 80 words
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## Sums and Differences Within 1 - Assessment 1 Worksheet
Help your child understand school topics better with your assistance. For math, go over their worksheet together and discuss the instructions and text. Solve the questions and check the right equation. This will take their learning to the next level.
Sums and Differences Within 1 - Assessment 1 Worksheet
Worksheet
## Reggie's Reuse Word Problems: Addition Worksheet
Help Reggie and your child save the planet by using this free worksheet! It reinforces addition word problems using one-to-one picture representation, and helps them pick out the correct number sentence and possible solutions. It's an easy way to help the planet and practice math!
Reggie's Reuse Word Problems: Addition Worksheet
Worksheet
## Number Line Fun Worksheet
Give your child a hand in math with this worksheet. It contains three number lines with points to which they should assign correct fractions. Support your child with this exercise and they'll get better at mathematics.
Number Line Fun Worksheet
Worksheet
## Regrouping with Base 10 Blocks Worksheet
Practice and master 3-digit subtraction with this handy worksheet! It uses base ten blocks to illustrate what's being taken away, helping children understand this tricky skill. Viewing the visual aid will help them get the hang of larger numbers quickly.
Regrouping with Base 10 Blocks Worksheet
Worksheet
Take your kids to Fairytale Land! They'll meet witches, dragons, fairies, elves, knights, and princesses. This free worksheet adds up the fun, letting kids use three addends to solve addition equations and find the right answers. With friends like these, math won't even seem like math!
Worksheet
## Little Witch Problem Solving Worksheet
Help this sweet little witch! Download this worksheet and use traceable lines to connect the problems with the correct answers. Practise basic addition involving three addends and have fun doing it!
Little Witch Problem Solving Worksheet
Worksheet
## Match and Solve: Cupcakes Problem Worksheet
This math worksheet will excite your kid! Read the word problem and look at the cupcakes; the slashes mean some are being taken away. Select the picture that matches the text, then complete the equation to find the solution!
Match and Solve: Cupcakes Problem Worksheet
Worksheet
## Valentines Day Word Problem Worksheet
This Valentine's Day worksheet will help your child practice using a number line to solve an addition problem. It's a great way to boost confidence and make solving word problems a breeze.
Valentines Day Word Problem Worksheet
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## Circus Math Printable
Let's take the kids to the circus and combine literacy and numeracy skills! Our circus math printable worksheet will encourage creativity and higher-order thinking by connecting addition to real-life examples. Kids will learn to love math while they explore the fun, real-life applications of math. Ignite their passion today!
Circus Math Printable
Worksheet
## Counting the Coins Money Worksheet
Count coins quickly and work out sums in cents. Remember the value of each coin!
Counting the Coins Money Worksheet
Worksheet
Learning Skills | 836 | 4,179 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-30 | latest | en | 0.898277 |
https://thetwomeatmeal.wordpress.com/tag/categorical/ | 1,580,202,008,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251776516.99/warc/CC-MAIN-20200128060946-20200128090946-00396.warc.gz | 682,488,630 | 13,373 | # Gracious Living
Quotients of Topological Spaces
November 28, 2010, 20:06
Filed under: Math, Topology | Tags: , ,
The last way to induce a topological space is by taking a quotient. This can be compared to taking a quotient of groups in the same way that the product topology corresponds to a direct product of groups, the subspace topology corresponds to a subgroup, and so on. The quotient topology is a bit less intuitive than the other constructions we’ve done, but once you get the hang of it, it turns out to be very useful.
More Products of Groups
November 27, 2010, 22:40
Filed under: Algebra, Math | Tags: , , , ,
Before looking at solvability and group classification, I want to mention a couple more ways of “building” groups. We’ve already seen how to find subgroups, and how to take the quotient by a normal subgroup, and how to find the direct product of a family of groups. Dual to the direct product is the free product, which generalizes the idea of a free group. The amalgamated free product is just a free product that we neutralize on the image of some map. Also, though the only really good example is the group $E(n)$ of Euclidean isometries, the semidirect product is worth a more formal look. Finally, though it’s mostly terminology, I define the direct sum, which is useful for studying abelian groups.
Products of Groups
November 19, 2010, 22:49
Filed under: Algebra, Math | Tags: , , , ,
Ugh, so, I’ve been really busy today and haven’t had the time to do a Banach-Tarski post. Since I really do want to see MaBloWriMo to the end, I’m going to take a break from the main exposition and quickly introduce something useful. There are a couple major ways of combining two groups into one. The most important one, called the direct product, is analogous to the product of topological spaces. I know this is sort of a wussy post — sorry.
Product and Disjoint Union Topologies
November 8, 2010, 15:55
Filed under: Math, Topology | Tags: , , , ,
As the subspace topology is the “best way” to topologize a subset, the product and disjoint union topologies are the “natural ways” to topologize Cartesian products and disjoint unions of topological spaces. Usually the disjoint union topology is only glossed over, while more time is spent on the product topology; I’m introducing them together in order to show you some of the similarities between them addressed by category theory. The relationship is one of duality, something like the intersection and union of sets. There’s this goofy mathematician way of putting “co” in front of dual constructions, so we could also call the disjoint union the “coproduct”. | 653 | 2,653 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2020-05 | latest | en | 0.915227 |
https://community.intel.com/t5/Intel-oneAPI-Math-Kernel-Library/OpenMP-Threading-for-FFT-Computation/td-p/1131632 | 1,701,973,295,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100677.45/warc/CC-MAIN-20231207153748-20231207183748-00071.warc.gz | 211,587,197 | 47,849 | Intel® oneAPI Math Kernel Library
Ask questions and share information with other developers who use Intel® Math Kernel Library.
6880 Discussions
## OpenMP* Threading for FFT Computation
Beginner
555 Views
As I understood from an example on https://software.intel.com/en-us/mkl-developer-reference-c-examples-of-using-openmp-threading-for-fft-computation, you calculate with 4 threads 4 different 2D-FFT (50x100 points). My question is - is it possible to calculate NxN FFT where I divide data between 2 processors where each will get NxM FFT and NXK FFT where M+K = N? I worry about the result, is it will be correct by following your example? Also, how do you do it inside and how I can apply it to fftw?
Thanks,
Sem
12 Replies
New Contributor I
555 Views
But if you use the fftw interface the function you are looking for is probably
`fftw_plan_with_nthreads(int nthreads);`
Beginner
555 Views
Hi Mathieu,
Thanks for your reply, but it is not exactly what I needed. Actually, I want to calculate, for instance, one part of 2D FFT on one socket and another part 2D FFT on another socket, after this to combine them in one that is will be correct. Like - FFT = FFT[0, N/2-1] + FFT[N/2, N].
Thanks,
Sem
Employee
555 Views
Hi Sem,
Yes, the example on https://software.intel.com/en-us/mkl-developer-reference-c-examples-of-u..., you calculate with 4 threads 4 different 2D-FFT (50x100 points).
is it possible to calculate NxN FFT where I divide data between 2 processors where each will get NxM FFT and NXK FFT where M+K = N?
From MKL FFT point of views, we take the problem as below you expected to do NxN FFT on 2 processors. Actually, both MKL FFT and FFTW support internal multi-threads, which means , MKL FFT or FFTW will use the existing processors (2 here) and distribute the compute tasks of NXN FFT to mult-threads to compute and then return the correct result. then developer don't need take care of how to seperate the task.
What is your N size? Please see mkl use guide to check if the 2D FFT are multi-threaded in https://software.intel.com/en-us/mkl/documentation
Best Regards,
Ying
Beginner
555 Views
Hi Ying,
Thanks. My N can be 1024 and so on till 48540. I do not get how you combine the result in the end. I have checked and my result is not exactly correct when I calculated following by your method. I devide N*N/2, so one socket with 18 threads has N*N/2 data and another has also N*N/2. When I checked the result, it was incorrect. So, if you are saying that it is correct way to do one FFT by deviding it on different sockets, then I think I did some mistake, I need to check it again.
Thanks,
Sem
Employee
555 Views
Hi Sem,
Sorry for unclear reply. I mean, you can call N (1024, 48540) FFT directly. You don't have to divide the data yourself. MKL will distribute the compute tasks in two socket automatically.
Best Regards,
Ying
Beginner
555 Views
Hi, Ying,
Yes, I know and thanks again but I want to be able to manage data between threads and sockets by myself. This is why I have such kind of question.
Thanks,
Sem
Employee
555 Views
Hi Sem,
Get it. then you may want to implement one mult-thread FFT by parallel method actually. then you may refer to the source code of FFTW multhreads or on-line source like https://www.codeproject.com/Questions/706613/How-To-Write-an-FFT-Algorithm-With-Thread.
Best Regards,
Ying
Beginner
555 Views
Hi Ying,
Thank you. I thought I can just use your function. Another way to do it I think with MPI, do you have some example mkl fftw with MPI? Or you don't use them together?
Thanks,
Sem
New Contributor I
555 Views
I am not sure to understand well, but you can still write you 2D FFT with two set of 1D FFTs, basically 2D FFT is a FFT along one axis and after the second one. (If you do R2C the first one is R2C and after you have C2C along teh second axis ).
And if you like you can still write, you set of 1D FFT as 2 subset (n/2 first one lines and n/2 last one lines) and the same after with columns. I use this decomposition when I know that a line is only with zeros I don't compute it, that gives me the opportunity to win some small % of computation.
If you do it over two sockets, the (manuel) memory management can be problematic (at least complex), if you look at the algorithm that is a divide and conquer, you will see that at the last step you need memory of both sides.
Employee
555 Views
Hi Sem,
MKL provide Cluster FFT, which support MPI. It allows you to manage data between MPI threads and sockets by yourself. You can refer to MKL user guide and https://software.intel.com/en-us/mkl-linux-developer-guide-linking-with-intel-mkl-cluster-software ; and MKL provide the sample code under <install Dir /examples>
Agree with Mathieu, there are some ways to parallel the FFT. But the manual memory management can be problematic if your purpose here is get higher performance N FFT result, but not how to write a FFT multithread code.
Best Regards,
Ying
Beginner
555 Views
Hi Mathieu,
Thanks, yes, actually I am doing this way now. I thought there is a way to do it without transpose but just to sum results somehow from two parts of 2D FFT.
Hi Ying,
Thanks too, I have checked but did not find any example of implementation of mpi. Ok, I can do MPI + Multithreading in fftw, but would be good if the same I can do in mklfft.
Thanks,
Sem
New Contributor I
555 Views
Transpose ? What are you transposing ? Can you not just use the stride option ?
have maybe a look one : http://www.fftw.org/fftw3_doc/Advanced-Complex-DFTs.html
Basically, it looks like:
1) first compute the lines, that I suppose is ok
2) compute column-wise
fftw_plan_many_dft(1, numberOfLines, numberOfColumn,
in, numberOfLines,
numberOfColumn, 1,
out, numberOfLines,
numberOfColumn, 1,
FFTW_FORWARD, FFTW_PLAN_OPTION);
if you do C2C it's ok, if you do R2C you need to transform numberOfColumn in numberOfColumn/2+1, or something like that. I let you find out check that.
3) nothing else.
PS: I hope what I write is right ;)
PPS: in = out is fine too ! | 1,571 | 6,079 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.863709 |
https://www.texasgateway.org/resource/117-lab-1-chi-square-goodness-fit?book=297701 | 1,550,468,681,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247484689.3/warc/CC-MAIN-20190218053920-20190218075920-00304.warc.gz | 977,797,352 | 13,432 | ## Introduction
### Introduction
Stats Lab 11.1
#### Lab 1: Chi-Square Goodness-of-Fit
Class Time:
Names:
Student Learning Outcome
• The student will evaluate data collected to determine if they fit either the uniform or exponential distributions.
Collect the Data Go to your local supermarket. Ask 30 people as they leave for the total amount on their grocery receipts. Or, ask 3 cashiers for the last 10 amounts. Be sure to include the express lane, if it is open.
#### Note
You may need to combine two categories so that each cell has an expected value of at least five.
1. Record the values.
__________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________
Table 11.23
2. Construct a histogram of the data. Make five to six intervals. Sketch the graph using a ruler and pencil. Scale the axes.
Figure 11.9
3. Calculate the following:
1. $x¯$ =________
2. s = ________
3. s2 = ________
Uniform Distribution Test to see if grocery receipts follow the uniform distribution.
1. Using your lowest and highest values, X ~ U (_______, _______).
2. Divide the distribution into fifths.
3. Calculate the following:
1. lowest value = _________
2. 20th percentile = _________
3. 40th percentile = _________
4. 60th percentile = _________
5. 80th percentile = _________
6. highest value = _________
4. For each fifth, count the observed number of receipts and record it. Then determine the expected number of receipts and record that.
Fifth Observed Expected
1st
2nd
3rd
4th
5th
Table 11.24
5. H0: ________
6. Ha: ________
7. What distribution should you use for a hypothesis test?
8. Why did you choose this distribution?
9. Calculate the test statistic.
10. Find the p-value.
11. Sketch a graph of the situation. Label and scale the x-axis. Shade the area corresponding to the p-value.
Figure 11.10
13. State your conclusion in a complete sentence.
Exponential Distribution Test to see if grocery receipts follow the exponential distribution with decay parameter $1x¯1x$.
1. Using $1x¯1x¯$ as the decay parameter, X ~ Exp(_________).
2. Calculate the following:
1. lowest value = ________
2. first quartile = ________
3. 37th percentile = ________
4. median = ________
5. 63rd percentile = ________
6. 3rd quartile = ________
7. highest value = ________
3. For each cell, count the observed number of receipts and record it. Then determine the expected number of receipts and record that.
Cell Observed Expected
1st
2nd
3rd
4th
5th
6th
Table 11.25
4. H0: ________
5. Ha: ________
6. What distribution should you use for a hypothesis test?
7. Why did you choose this distribution?
8. Calculate the test statistic.
9. Find the p-value.
10. Sketch a graph of the situation. Label and scale the x-axis. Shade the area corresponding to the p-value.
Figure 11.11 | 755 | 3,029 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-09 | latest | en | 0.76582 |
https://justpaste.it/how-to-beat-online-casino | 1,601,409,446,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402088830.87/warc/CC-MAIN-20200929190110-20200929220110-00480.warc.gz | 455,714,362 | 6,400 | # How to Beat Online Casino
As practice shows, most newcomers to the online casino are interested in three questions: "Are there any honest online casinos?", "Can I win at an online casino?"
As practice shows, most newcomers to the online casino are interested in three questions: "Are there any honest online casinos?", "Can I win at an online casino?", "Are there any game systems guaranteeing winnings in a casino?". Some formulate the question a little differently: "How much did you win at an online casino?". Immediately answer the last question, I won amounts from several tens to thousands of dollars (but I had to lose, and a maximum of about \$ 200), my total income from playing casino exceeded \$ 10,000 (this is for several years of playing in the evenings after work), although I know Professional players who earn such amounts in a few months. On the first question about honest casinos, I tried to answer on the site "Honest online casino", I recommend you read it if you plan to play online at casinobonuspromotions.co.uk. Here I will try to answer the two remaining questions - can I still beat the online casino?
If to answer briefly, then "Yes, you can win", "No, there are no guaranteed strategies". Why no one can guarantee a win in the casino, I'll try to explain in more details on the page "Win-Win" systems. "The question of how to beat a casino is more complicated. As you know, casino games are built for chance - the desired card may come, or it may not come , your roulette number may drop out, or it may be quite different. It seems that everything depends only on luck: lucky you are in chocolate, no - in another substance. And at a short distance in several games this is really so. in an honest casino, depending on the luck It is important to remember that the rules of the game are such that the casino does not have a very big but mathematical advantage over the player.All the games are carefully calculated from the point of view of the probability theory and the payout in them is usually less than 100%. That is, on average, he gets \$ 90- \$ 99 on the average of \$ 100- \$ 99. I repeat, this is an average, at a short distance in one session it's not so noticeable, the chances to win are quite high, but the more you play, the more your result will be to Wed during the day - the expectation, which means your loss, alas. More details about the theory of probability in the application to the casino games I will try on the page "Mathematics games. | 549 | 2,496 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.961941 |
http://www.metodolog.ru/triz-journal/archives/1998/11/d/index.htm | 1,553,449,100,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203464.67/warc/CC-MAIN-20190324165854-20190324191854-00351.warc.gz | 331,662,911 | 4,760 | # The 39 Features of Altshuller’s Contradiction Matrix
Ellen Domb
+1(909)949-0857 FAX +1(909)949-2986
with Joe Miller, Ellen MacGran, and Michael Slocum
The contradiction matrix, one of the first tools of TRIZ, remains one of the most popular. (Ref.1) The matrix itself, and the 40 principles of problem solving to which it refers can be downloaded from the July, 1997, issue of The TRIZ Journal. The same issue has a tutorial article on how to use the matrix and the 40 principles. (Ref. 2)
One barrier to using the matrix has been the very brief statements of the features that are the lists of improving and worsening features. The following expanded list was prepared by comparing several different translations (ref. 3,4,5) and has proven useful in several test classes.
Explanation of the 39 Features of the Contradiction Matrix
No. Title Explanation Moving objects Objects which can easily change position in space, either on their own, or as a result of external forces. Vehicles and objects designed to be portable are the basic members of this class. Stationary objects. Objects which do not change position in space, either on their own, or as a result of external forces. Consider the conditions under which the object is being used. 1 Weight of moving object The mass of the object, in a gravitational field. The force that the body exerts on its support or suspension. 2 Weight of stationary object The mass of the object, in a gravitational field. The force that the body exerts on its support or suspension, or on the surface on which it rests. 3 Length of moving object Any one linear dimension, not necessarily the longest, is considered a length. 4 Length of stationary object Same. 5 Area of moving object A geometrical characteristic described by the part of a plane enclosed by a line. The part of a surface occupied by the object. OR the square measure of the surface, either internal or external, of an object. 6 Area of stationary object Same 7 Volume of moving object The cubic measure of space occupied by the object. Length x width x height for a rectangular object, height x area for a cylinder, etc. 8 Volume of stationary object Same 9 Speed The velocity of an object; the rate of a process or action in time. 10 Force Force measures the interaction between systems. In Newtonian physics, force = mass X acceleration. In TRIZ, force is any interaction that is intended to change an object's condition. 11 Stress or pressure Force per unit area. Also, tension. 12 Shape The external contours, appearance of a system. 13 Stability of the object's composition The wholeness or integrity of the system; the relationship of the system's constituent elements. Wear, chemical decomposition, and disassembly are all decreases in stability. Increasing entropy is decreasing stability. 14 Strength The extent to which the object is able to resist changing in response to force. Resistance to breaking . 15 Duration of action by a moving object The time that the object can perform the action. Service life. Mean time between failure is a measure of the duration of action. Also, durability. 16 Duration of action by a stationary object Same. 17 Temperature The thermal condition of the object or system. Loosely includes other thermal parameters, such as heat capacity, that affect the rate of change of temperature. 18 Illumination intensity * (jargon) Light flux per unit area, also any other illumination characteristics of the system such as brightness, light quality, etc.. 19 Use of energy by moving object The measure of the object's capacity for doing work. In classical mechanics, Energy is the product of force times distance. This includes the use of energy provided by the super-system (such as electrical energy or heat.) Energy required to do a particular job. 20 Use of energy by stationary object same 21 Power * (jargon) The time rate at which work is performed. The rate of use of energy. 22 Loss of Energy Use of energy that does not contribute to the job being done. See 19. Reducing the loss of energy sometimes requires different techniques from improving the use of energy, which is why this is a separate category. 23 Loss of substance Partial or complete, permanent or temporary, loss of some of a system's materials, substances, parts, or subsystems. 24 Loss of Information Partial or complete, permanent or temporary, loss of data or access to data in or by a system. Frequently includes sensory data such as aroma, texture, etc. 25 Loss of Time Time is the duration of an activity. Improving the loss of time means reducing the time taken for the activity. "Cycle time reduction" is a common term. 26 Quantity of substance/the matter The number or amount of a system's materials, substances, parts or subsystems which might be changed fully or partially, permanently or temporarily. 27 Reliability A system's ability to perform its intended functions in predictable ways and conditions. 28 Measurement accuracy The closeness of the measured value to the actual value of a property of a system. Reducing the error in a measurement increases the accuracy of the measurement. 29 Manufacturing precision The extent to which the actual characteristics of the system or object match the specified or required characteristics. 30 External harm affects the object Susceptibility of a system to externally generated (harmful) effects. 31 Object-generated harmful factors A harmful effect is one that reduces the efficiency or quality of the functioning of the object or system. These harmful effects are generated by the object or system, as part of its operation. 32 Ease of manufacture The degree of facility, comfort or effortlessness in manufacturing or fabricating the object/system. 33 Ease of operation Simplicity: The process is NOT easy if it requires a large number of people, large number of steps in the operation, needs special tools, etc. "Hard" processes have low yield and "easy" process have high yield; they are easy to do right. 34 Ease of repair Quality characteristics such as convenience, comfort, simplicity, and time to repair faults, failures, or defects in a system. 35 Adaptability or versatility The extent to which a system/object positively responds to external changes. Also, a system that can be used in multiple ways for under a variety of circumstances. 36 Device complexity The number and diversity of elements and element interrelationships within a system. The user may be an element of the system that increases the complexity. The difficulty of mastering the system is a measure of its complexity. 37 Difficulty of detecting and measuring Measuring or monitoring systems that are complex, costly, require much time and labor to set up and use, or that have complex relationships between components or components that interfere with each other all demonstrate "difficulty of detecting and measuring." Increasing cost of measuring to a saticfactory error is also a sign of increased difficulty of measuring. 38 Extent of automation The extent to which a system or object performs its functions without human interface. The lowest level of automation is the use of a manually operated tool. For intermediatel levels, humans program the tool, observe its operation, and interrupt or re-program as needed. For the highest level, the machine senses the operation needed, programs itself, and monitors its own operations. 39 Productivity * The number of functions or operations performed by a system per unit time. The time for a unit function or operation. The output per unit time, or the cost per unit output.
References:
1. G. Altshuller, Creativity as an Exact Science,Translated by Anthony Williams. Gordon & Breach, NY, 1988.
2. Ellen Domb, "Contradictions." The TRIZ Journal, http://www.triz-journal.com, July, 1997.
3. J. Terninko, A. Zusman, B. Zlotin, Step-by-Step TRIZ, Nottingham, NH, 1997.
4. The Invention Machine Laboratory ™ version 1.4, Help files
5. H. Altov (G. Altshuller pseudonym) And Suddenly the Inventor Appeared, Translated by Lev Shulyak, Technical Innovation Center, Worcester, MA, 1995. | 1,728 | 8,098 | {"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} | 3.78125 | 4 | CC-MAIN-2019-13 | latest | en | 0.931875 |
http://www.maplesoft.com/support/help/Maple/view.aspx?path=odeadvisor/erf | 1,502,962,379,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886103167.97/warc/CC-MAIN-20170817092444-20170817112444-00643.warc.gz | 619,054,030 | 22,601 | The Error Function (erf) ODE - Maple Programming Help
Home : Support : Online Help : Mathematics : Differential Equations : Classifying ODEs : Second Order : odeadvisor/erf
The Error Function (erf) ODE
Description
• The general form of the erf ODE is given by
> erf_ode := diff(y(x),x,x)+2*x*diff(y(x),x)-2*n*y(x) = 0;
${\mathrm{erf_ode}}{≔}\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}{}{y}{}\left({x}\right){+}{2}{}{x}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}{}{y}{}\left({x}\right)\right){-}{2}{}{n}{}{y}{}\left({x}\right){=}{0}$ (1)
where n is an integer. See Abramowitz and Stegun, "Handbook of Mathematical Functions", section 7.2.2. The solution of this type of ODE can be expressed in terms of the WhittakerM and WhittakerW functions.
Examples
> $\mathrm{with}\left(\mathrm{DEtools},\mathrm{odeadvisor}\right)$
$\left[{\mathrm{odeadvisor}}\right]$ (2)
> $\mathrm{odeadvisor}\left(\mathrm{erf_ode}\right)$
$\left[{\mathrm{_erf}}\right]$ (3)
> $\mathrm{dsolve}\left(\mathrm{erf_ode}\right)$
${y}{}\left({x}\right){=}{\mathrm{_C1}}{}{{ⅇ}}^{{-}{{x}}^{{2}}}{}{\mathrm{KummerM}}{}\left({1}{+}\frac{{1}}{{2}}{}{n}{,}\frac{{3}}{{2}}{,}{{x}}^{{2}}\right){}{x}{+}{\mathrm{_C2}}{}{{ⅇ}}^{{-}{{x}}^{{2}}}{}{\mathrm{KummerU}}{}\left({1}{+}\frac{{1}}{{2}}{}{n}{,}\frac{{3}}{{2}}{,}{{x}}^{{2}}\right){}{x}$ (4) | 510 | 1,292 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 7, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2017-34 | latest | en | 0.375108 |
https://math.answers.com/questions/What_number_would_you_get_if_you_added_9_tens_to_seven_hundred_twelve | 1,716,153,326,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057922.86/warc/CC-MAIN-20240519193629-20240519223629-00497.warc.gz | 343,343,952 | 48,567 | 0
# What number would you get if you added 9 tens to seven hundred twelve?
Updated: 9/26/2023
Wiki User
6y ago
(9 × 10) + 712 = 802
Wiki User
6y ago
Wiki User
6y ago
802
Earn +20 pts
Q: What number would you get if you added 9 tens to seven hundred twelve?
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### How do you say this number in words 4732412?
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1i seven
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### Twelve and nine thousand six hundred and forty-seven millionths?
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### What is the number to4793470112038760?
Four quadrillion, seven hundred ninety-three trillion, four hundred seventy billion, one hundred twelve million, thirty-eight thousand, seven hundred sixty. | 401 | 1,630 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-22 | latest | en | 0.820564 |
http://mathhelpforum.com/discrete-math/54752-how-many-bit-strings-length-10-a.html | 1,503,056,423,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104634.14/warc/CC-MAIN-20170818102246-20170818122246-00011.warc.gz | 269,104,070 | 10,641 | # Thread: how many bit strings of length 10 ...
1. ## how many bit strings of length 10 ...
how many bit strings of length 10 either start with 000 or end with 1111??????
Thank you very much
2. Hello,
Originally Posted by Narek
how many bit strings of length 10 either start with 000 or end with 1111??????
Thank you very much
Well, keep 000 at the beginning.
Then there are 7 remaining bits that are to be chosen.
Each with two possible values : 0 & 1.
So it's 2^7.
Now, keep 1111 in the end.
There are 6 remaining bits that are to be chosen.
Each with two possible values : 0 & 1.
So it's 2^6.
Total number : 2^7+2^6 (if you have to include the case where it both starts with 000 and ends with 1111)
3. ## Thank you
Thank you my friend,
you always explain so clearly and I really appreciate it.
4. Originally Posted by Moo
Total number : 2^7+2^6 (if you have to include the case where it both starts with 000 and ends with 1111)
But with that total, you've counted doubly those strings that begin with 000 and end with 1111. There are 2^3 such strings by the same reasoning, so you have to subtract those from the sum.
5. ## so the answer is ... ?
(2^7 + 2^6) - (2^3) = 184
,
,
,
,
,
### how many bit string length 10 starting with 11 contain 0
Click on a term to search for related topics. | 373 | 1,313 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | longest | en | 0.936473 |
http://acm.zju.edu.cn/onlinejudge/showContestProblem.do?problemId=2851 | 1,556,150,575,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578675477.84/warc/CC-MAIN-20190424234327-20190425020327-00285.warc.gz | 4,086,475 | 3,811 | Welcome to ZOJ
Contests Information Problems Runs Statistics Ranklist Clarification
73 - ZOJ Monthly, December 2008 - D
Team Forming
Time Limit: 5 Seconds Memory Limit: 32768 KB Special Judge
ZJU ICPC Summer Camp 2008 is going to end shortly, and 18 members will be selected to form the ZJU ICPC teams, which will participate in the regionals this year. However, team forming is usually a headache for the coaches, since in order to get the best achievement, the members should be selected carefully so the abilities of the teams are maximized. This is a diffcult task, as it is hard to define what the ability of a team is, and the coaches could only make a forming plan that is intuitively good in the past. This year they've found a new method to measure the abilities of the team members: radar charts. For example, if we just consider six aspects of a member's abilites: Greedy, DP, Math, Geometry, Graph and Others, and compute the relative value of each aspect with the performance in summer camp (once the problems are classilied, this is an easy job). Then we can get a chart like this picture:
To make things simpler, we believe that in each aspect, the ability of a team is the maximum ability of three members, and define the area of the resulting diagram to be the overall ability of this team. Then it is very easy to write a program to find the best team forming plan to achieve the maximum abilities for all teams. But as you may already know, the coaches are too lazy to write programs, so it's your turn to write a program to prove your skill now. But note that in order to get a high rank in the world finals, there should always be one or two teams whose abilites are as good as possible. So your program should first form one team with the best overall ability or two teams with the highest total abilites, then form the remaining teams to get the maximum total abilities.
Input
Each case begins with three positive integers A, N and F, where A is the number of aspects to be considered (3 <= A <= 20), N is the number of ICPC team members (N <= 18, and N is a multiple of 3) and F is either 1 or 2 (F <= N / 3), means the first F team(s) should be the best one(s). Then N lines followed, each describes a member with his/her name and relative abilities in the A aspects. The names of members are distinct and each contains up to 31 characters that is either a letter or an underscore ('_'). The values of relative abilites are between 0 and 100, inclusive, with no more than two digits after the decimal points.The radar charts are drawn in the same order as the input (the Ath aspect is next to the first one, of course).
There's a blank line between every two successive cases. Processing to the end of file.
Output
For each test case, print N / 3 lines each represents a team with the names of three members. Any solution that first maximize the total abilites of the first F team(s) and then maximize the total abilites of remaining teams will be accepted.
Sample Input
```6 6 1
sghao126 90.33 57.13 86.78 84.88 83.79 90.78
lyt 76.66 82.56 78.52 55.49 31.02 45.37
gy 71.71 81.2 79.84 82.65 69.16 74.89
liux0229 74.16 61.03 81.33 73.39 85.01 80.7
Charizard 94.32 86.96 87.51 76.02 77.25 77.38
hhanger 93.37 78.93 76.47 84.88 75.79 77.38
```
Sample Output
```sghao126 liux0229 Charizard
lyt gy hhanger
``` | 889 | 3,352 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2019-18 | longest | en | 0.951735 |
https://www.jiskha.com/display.cgi?id=1301959137 | 1,516,618,311,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891277.94/warc/CC-MAIN-20180122093724-20180122113724-00146.warc.gz | 935,538,516 | 3,991 | # ordering decimals
posted by .
Ms.Sue could you please show us another exmple i understood 0.6 0.65 0.7 how about with a double digit number like:
0.25_____0.15 or 0.425_____0.475
• ordering decimals -
The first pair could be read as 25 cents and 15 cents. What comes between them?
For the second pair, the second digit after the decimal point is different in each one. What comes between 2 and 7?
0.425, 0.426, 0.427 . . .
0.435, 0.445, 0.455, 0.465.
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6. ### English
1. May I take your order? 2. Would you like to order now?
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Jason has 4 tiles. Each tile has a number printed on it. The numbers are 2,3,6, and 8. A decimal number is formed using the tiles and the clues. Be a math detective and find the number. Clues: digit in tens place is the greatest number. …
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I am a two digit no. If you subtract my second digit from my first digit the difference is 6. (i.e. 9-3= 6 or 8-2=6 or 7-1=6) The product of my digit is 0. What number am I?
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More Similar Questions | 673 | 2,419 | {"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} | 3.6875 | 4 | CC-MAIN-2018-05 | latest | en | 0.874391 |
https://www.coursehero.com/file/6353883/ConcCalcCMU032909/ | 1,513,344,483,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948569405.78/warc/CC-MAIN-20171215114446-20171215140446-00497.warc.gz | 701,172,574 | 221,078 | ConcCalcCMU032909
# ConcCalcCMU032909 - CONCEPT CALCULUS by Harvey M Friedman...
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CONCEPT CALCULUS by Harvey M. Friedman Ohio State University http://www.math.ohio-state.edu/%7Efriedman/ Pure and Applied Logic Colloquium Carnegie Mellon University March 26, 2009
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We have discovered an unexpectedly close connection between the logic of mathematical concepts and the logic of informal concepts from common sense thinking. Our results indicate that they are, in a certain precise sense, the same. This connection is new and there is the promise of establishing similar connections involving a very wide range of informal concepts. We call this development the Concept Calculus . We begin with some background concerning the crucial notion of interpretation between theories that is used to state results in Concept Calculus. We then give a survey of major results in Concept Calculus. In particular, we establish the mutual interpretability of formal systems for set theory and formal systems for a variety of informal concepts from common sense thinking .
INTERPRETATION POWER The notion of interpretation plays a crucial role in Concept Calculus. Interpretability between formal systems was first precisely defined by Alfred Tarski. We work in the usual framework of first order predicate calculus with equality. An interpretation of S in T consists of • A one place relation defined in T which is meant to carve out the domain of objects that S is referring to, from the point of view of T. • A definition of the constants, relations, and functions in the language of S by formulas in the language of T, whose free variables are restricted to the domain of objects that S is referring to (in the sense of the previous bullet). • It is required that every axiom of S, when translated into the language of T by means of i,ii, becomes a theorem of T. In ii, we usually allow that the equality relation in S need not be interpreted as equality – but rather as an equivalence relation.
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INTERPRETATION POWER CAUTION : Interpretations do not necessarily preserve truth. They only preserve provability. We give two illustrative examples. Let S consist of the axioms for strict linear order together with “there is a least element”. • ¬(x < x) • x < y y < z x < z. • x < y y < x x = y. • ( x)( y)(x < y x = y). Let T consist of the axioms for strict linear order together with “there is a greatest element”. I.e., • ¬(x < x) • x < y y < z x < z. • x < y y < x x = y. • ( x)( y)(y < x y = x).
INTERPRETATION POWER S • ¬(x < x) • x < y y < z x < z. • x < y y < x x = y. • ( x)( y)(x < y x = y). T • ¬(x < x) • x < y y < z x < z. • x < y
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ConcCalcCMU032909 - CONCEPT CALCULUS by Harvey M Friedman...
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Ask a homework question - tutors are online | 785 | 3,322 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2017-51 | latest | en | 0.923087 |
http://mathoverflow.net/questions/64573/how-to-verify-the-weak-convergence | 1,469,434,085,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824217.36/warc/CC-MAIN-20160723071024-00241-ip-10-185-27-174.ec2.internal.warc.gz | 151,120,267 | 15,076 | # How to verify the weak convergence?
Given a finite measure on a compact, take $f_n\in L^1$ with norms $\leq 1$ and suppose that $\int f_n g$ tends to a limit for all continuous $g$. Is it true that then $\int f_n g$ converge for any $g\in L^\infty$? How can one prove this?
-
mathoverflow.net/faq#whatnot Voting to close. – Yemon Choi May 11 '11 at 6:33
The answer, by the way, is "no". Could you say a little more about where you came across this problem, which special cases you have already tried, and so forth? – Yemon Choi May 11 '11 at 6:36
Yemon, I agree this is a homework problem (or if not, it should be), but (possibly) at graduate level, so I think you're being a little harsh; it is conceivable that a non-(functional analyst) research mathematician, or even a (non-functional) analyst, might find this a bit tricky. There are many interesting special cases of $f_n$ where the answer is "yes"! – Zen Harper May 11 '11 at 6:49
In the other direction, asking what extra assumptions on $(f_n)$ and the measure will guarantee the answer "yes" seems to me like an interesting question, possibly connected with Tauberian theories. This is exactly the kind of question which can make research interesting, in my opinion, since I doubt there will be any precise necessary/sufficient characterisation from general theory which isn't just tautological. – Zen Harper May 11 '11 at 7:29
The needed "something extra" is a condition that guarantees that the sequence has weakly compact closure in $L^1$, such as no subsequence of $f_n$ is equivalent to the unit vector basis for $\ell^1$. – Bill Johnson May 11 '11 at 21:16
If $(f_n)$ is a bounded sequence in $L^1(X,\mu)$, it is true that the set of $g\in L^\infty(X,\mu)$ such that $\int_X f_n g\, d\mu$ has a limit in $\mathbb{C}$ is a norm-closed linear subspace of $L^\infty(X,\mu)$. Thus, from a dense set of test functions $g$ you can infer the weak convergence of the sequence $(f_n)$ (to an element of ($L^\infty)^*$, of course, in general not in $L^1$). In conclusion, in general continuous functions are not enough (they are a closed subspace, usually proper, of $L^\infty$), but e.g. simple functions, which are uniformly dense, will do. | 602 | 2,203 | {"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} | 3.15625 | 3 | CC-MAIN-2016-30 | latest | en | 0.937117 |
http://www.mathisfunforum.com/viewtopic.php?pid=292014 | 1,398,403,392,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1398223210034.18/warc/CC-MAIN-20140423032010-00485-ip-10-147-4-33.ec2.internal.warc.gz | 709,563,885 | 4,264 | Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
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## #1 2013-11-25 23:40:37
ninjaman
Member
Offline
### trig problem
hello
i have a problem that im stuck on, the question is this
from a ship at sea, the angles of elevation of the top and bottom of a vertical lighthouse standing on the edge of a vertical cliff are 31degrees, and 26 degrees respectively. if the light house is 25m high, how high is the cliff?
so far i have h+25/tan31 = h/tan26
rearrange to, (h+25)tan26 = h tan31
im lost after this
any help would be appreciated.
thanks
simon:)
## #2 2013-11-26 00:18:35
bob bundy
Moderator
Offline
### Re: trig problem
hi Simon,
You've done the correct thing so far and I got this, too.
So multiply the bracket:
Move the h terms together:
Factorise the h:
And divide by the bracket ........
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #3 2013-11-26 09:26:55
ninjaman
Member
Offline
### Re: trig problem
hello bob
thanks for the help!
I divided by the brackets? (I think), meaning I did what was in the brackets first
(tan31-tan26) = 0.1131280305 * h = 25 tan 26
then, divided the first half by 0.1131280305 to move it over and under to get
h = 25 tan 26
0.1131280305 = 107.78332
is this correct, I have a feeling you meant something else by divide by the brackets and now im way off.
cheers
all the best
simon
## #4 2013-11-26 10:28:14
bob bundy
Moderator
Offline
### Re: trig problem
hello Simon,
That's spot on; well done!
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #5 2013-11-27 03:56:52
ninjaman
Member
Offline
### Re: trig problem
that's great!
thanks for your help bob!
all the best
simon:) | 584 | 1,872 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2014-15 | longest | en | 0.834797 |
https://www.effortlessmath.com/math-topics/shsat-math-formulas/ | 1,624,120,008,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487648373.45/warc/CC-MAIN-20210619142022-20210619172022-00519.warc.gz | 661,667,851 | 21,230 | # SHSAT Math Formulas
Are you preparing students for the SHSAT Math test? Looking for a comprehensive list of all math formulas students need to know before the test day? If so, then you are in the right place.
Below you will find a list of all Math formulas your students MUST have learned before test day, as well as some explanations for how to use them and what they mean. Keep this list around for a quick reminder when your students need it. Let your students review them all, then help them to begin applying them!
## Mathematics Formula Sheet
### Place Value
The value of the place, or position, of a digit in a number.
Example: In 456, the 5 is in “tens” position.
### Fractions
A number expressed in the form $$\frac{a}{b}$$
Adding and Subtracting with the same denominator:
$$\frac{a}{b}+\frac{c}{b}=\frac{a+c}{b}$$
$$\frac{a}{b}-\frac{c}{b}=\frac{a-c}{b}$$
Adding and Subtracting with the different denominator:
$$\frac{a}{b}+\frac{c}{d}=\frac{ad+cb}{bd}$$
$$\frac{a}{b}-\frac{c}{d}=\frac{ad-cb}{bd}$$
Multiplying and Dividing Fractions:
$$\frac{a}{b} × \frac{c}{d}=\frac{a×c}{b×d}$$
$$\frac{a}{b} ÷ \frac{c}{d}=\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{ad}{bc}$$
### Comparing Numbers Signs
Equal to $$=$$
Less than $$<$$
Greater than $$>$$
Greater than or equal $$≥$$
Less than or equal $$≤$$
### Rounding
Putting a number up or down to the nearest whole number or the nearest hundred, etc.
Example: 64 rounded to the nearest ten is 60 , because 64 is closer to 60 than to 70.
### Whole Number
The numbers $$\{0,1,2,3,…\}$$
### Estimates
Find a number close to the exact answer.
### Decimals
Is a fraction written in a special form. For example, instead of writing $$\frac{1}{2}$$ you can write $$0.5$$.
### Mixed Numbers
A number composed of a whole number and fraction. Example: $$2 \frac{2}{ 3}$$ Converting between improper fractions and mixed numbers: $$a \frac{c}{b}=a+\frac{c}{b}= \frac{ab+ c}{b}$$
### Factoring Numbers
Factor a number means to break it up into numbers that can be multiplied together to get the original number. Example:$$12=2×2×3$$
### Divisibility Rules
Divisibility means that you are able to divide a number evenly. Example: 24 is divisible by 6, because $$24÷6=4$$
### Greatest Common Factor
Multiply common prime factors
Example:$$200=2×2×2×5×5 60=2×2×3×5$$
GCF $$(200,60)=2×2×5=20$$
### Least Common Multiple
Check multiples of the largest number
Example: LCM (200, 60): 200 (no), 400 (no), 600 (yes!)
### Integers
$$\{…,-3,-2,-1,0,1,2,3,…\}$$
Includes: zero, counting numbers, and the negative of the counting numbers
### Real Numbers
All numbers that are on number line. Integers plus fractions, decimals, and irrationals etc.) ($$\sqrt{2},\sqrt{3},π$$, etc.)
### Order of Operations
PEMDAS
(parentheses / exponents / multiply / divide / add / subtract)
### Absolute Value
Refers to the distance of a number from , the distances are positive as absolute value of a number cannot be negative. $$|-22|=22$$
or $$|x| =\begin{cases}x \ for \ x≥0 \\x \ for \ x < 0\end{cases}$$
$$|x|<n⇒-n<x<n$$
$$|x|>n⇒x<-n or x>n$$
### Ratios
A ratio is a comparison of two numbers by division.
Example: $$3 : 5$$, or $$\frac{3}{5}$$
### Percentages
Use the following formula to find part, whole, or percent
part $$=\frac{percent}{100}×whole$$
### Proportional Ratios
A proportion means that two ratios are equal. It can be written in two ways:
$$\frac{a}{b}=\frac{c}{d}$$ , $$a: b = c: d$$
### Percent of Change
$$\frac{New \ Value \ – \ Old \ Value}{Old Value}×100\%$$
### Markup
Markup $$=$$ selling price $$–$$ cost
Markup rate $$=$$ markup divided by the cost
### Discount
Multiply the regular price by the rate of discount
Selling price $$=$$ original price $$–$$ discount
### Expressions and Variables
A variable is a letter that represents unspecified numbers. One may use a variable in the same manner as all other numbers: Addition: $$2+a$$ : $$2$$ plus a
Subtraction: $$y-3$$ : $$y$$ minus $$3$$
Division: $$\frac{4}{x}$$ : 4 divided by x
Multiplication: $$5a$$ : $$5$$ times a
### Tax
To find tax, multiply the tax rate to the taxable amount (income, property value, etc.)
### Distributive Property
$$a(b+c)=ab+ac$$
### Polynomial
$$P(x)=a_{0} x^n+ a_{1} x^{n-1}+$$⋯$$+a_{n-2} x^2+a_{n-1} x+an$$
### Systems of Equations
Two or more equations working together.
example: $$\begin{cases}-2x+2y=4\\-2x+y=3\end{cases}$$
### Equations
The values of two mathematical expressions are equal.
$$ax+b=c$$
### Functions
A function is a rule to go from one number (x) to another number (y), usually written $$y=f(x)$$.For any given value of x, there can only be one corresponding value y. If $$y=kx$$ for some number k (example: $$f(x)= 0.5 x$$), then y is said to be directly proportional to x. If y$$=\frac{k}{x }$$ (example: f(x $$=\frac{5}{x}$$), then y is said to be inversely proportional to x. The graph of $$y=f(x )+k$$ is the translation of the graph of $$y=f(x)$$ by $$(h,k)$$ units in the plane. For example, $$y=f(x+3)$$ shifts the graph of $$f(x)$$ by 3 units to the left.
### Inequalities
Says that two values are not equal
$$a≠b$$ a not equal to b
$$a<b$$ a less than b
$$a>b$$ a greater than b
$$a≥b$$ a greater than or equal b
$$a≤b$$ a less than or equal b
### Solving Systems of Equations by Elimination
Example: \cfrac{\begin{align} x+2y =6 \\ + \ \ -x+y=3 \end{align}}{}
\cfrac{ \begin{align} 3y=9 \\ y=3 \end{align} }{\begin{align} x+6=6 \\ ⇒ x=0 \end{align}}
### Lines (Linear Functions)
Consider the line that goes through points $$A(x_{1},y_{1})$$ and $$B(x_{2},y_{2})$$.
### Distance from A to B:
$$\sqrt{(x_{1}-x_{2})^2+(y_{1}-y_{2})^2 }$$
### Parallel and Perpendicular lines:
Have equal slopes. Perpendicular lines (i.e., those that make a $$90^°$$ angle where they intersect) have negative reciprocal slopes: $$m_{1}$$ .$$m_{2}=-1$$.
Parallel Lines (l $$\parallel$$ m)
### Mid-point of the segment AB:
M ($$\frac{x_{1}+x_{2}}{2} , \frac{y_{1}+y_{2}}{2}$$)
### Slope of the line:
$$\frac{y_{2}- y_{1}}{x_{2} – x_{1} }=\frac{rise}{run}$$
### Point-slope form:
Given the slope m and a point $$(x_{1},y_{1})$$ on the line, the equation of the line is
$$(y-y_{1})=m \ (x-x_{1})$$.
### Intersecting lines:
Opposite angles are equal. Also, each pair of angles along the same line add to $$180^°$$. In the figure above, $$a+b=180^°$$.
### Slope-intercept form:
given the slope m and the y-intercept b, then the equation of the line is:
$$y=mx+b$$.
### Transversal: Parallel lines:
Eight angles are formed when a line crosses two parallel lines. The four big angles (a) are equal, and the four small angles (b) are equal.
### Parabolas:
A parabola parallel to the y-axis is given by $$y=ax^2+bx+c$$.
If $$a>0$$, the parabola opens up.
If $$a<0$$, the parabola opens down. The y-intercept is c, and the x-coordinate of the vertex is: $$x=-\frac{b}{2a}$$.
### Factoring:
“FOIL”
$$(x+a)(x+b)$$
$$=x^2+(b+a)x +ab$$ “Difference of Squares”
$$a^2-b^2= (a+b)(a-b)$$
$$a^2+2ab+b^2=(a+b)(a+b)$$
$$a^2-2ab+b^2=(a-b)(a-b)$$ “Reverse FOIL”
$$x^2+(b+a)x+ab=$$ $$(x+a)(x+b)$$
You can use Reverse FOIL to factor a polynomial by thinking about two numbers a and b which add to the number in front of the x, and which multiply to give the constant. For example, to factor $$x^2+5x+6$$, the numbers add to 5 and multiply to 6, i.e.: $$a=2$$ and $$b=3$$, so that $$x^2+5x+6=(x+2)(x+3)$$. To solve a quadratic such as $$x^2+bx+c=0$$, first factor the left side to get $$(x+a)(x+b)=0$$, then set each part in parentheses equal to zero. For example, $$x^2+4x+3= (x+3)(x+1)=0$$ so that $$x=-3$$ or $$x=-1$$.
To solve two linear equations in x and y: use the first equation to substitute for a variable in the second. E.g., suppose $$x+y=3$$ and $$4x-y=2$$. The first equation gives y=3-x, so the second equation becomes $$4x-(3-x)=2 ⇒ 5x-3=2$$ $$⇒ x=1,y=2$$.
### Exponents:
Refers to the number of times a number is multiplied by itself.
$$8 = 2 × 2 × 2 = 2^3$$
### Scientific Notation:
It is a way of expressing numbers that are too big or too small to be conveniently written in decimal form.
In scientific notation all numbers are written in this form: $$m \times 10^n$$
Decimal notation:
5
$$-25,000$$
0.5
2,122.456
Scientific notation:
$$5×10^0$$
$$-2.5×10^4$$
$$5×10^{-1}$$
$$2,122456×10^3$$
### Square:
The number we get after multiplying an integer (not a fraction) by itself. Example: $$2×2=4,2^2=4$$
### Square Roots:
A square root of $$x$$ is a number r whose square is $$x : r^2=x$$
$$r$$ is a square root of $$x$$
### Pythagorean Theorem:
$$a^2+b^2=c^2$$
### Right triangles:
A good example of a right triangle is one with a=3 , b=4 and c=5, also called a $$3-4-5$$ right triangle. Note that multiples of these numbers are also right triangles. For example, if you multiply these numbers by 2, you get a=6, b=8 and
$$c=10(6-8-10)$$ , which is also a right triangle.
### All triangles:
Area $$=\frac{1}{2}$$ b . h
Angles on the inside of any triangle add up to $$180^\circ$$.
The length of one side of any triangle is always less than the sum and more than the difference of the lengths of the other two sides.
An exterior angle of any triangle is equal to the sum of the two remote interior angles. Other important triangles:
### Equilateral:
These triangles have three equal sides, and all three angles are $$60^\circ$$.
### Isosceles:
An isosceles triangle has two equal sides. The “base” angles (the ones opposite the two sides) are equal (see the $$45^\circ$$ triangle above).
### Similar:
Two or more triangles are similar if they have the same shape. The corresponding angles are equal, and the corresponding sides are in proportion. For example, the $$3-4-5$$ triangle and the $$6-8-10$$ triangle from before are similar since their sides are in a ratio of to .
### Circles
Area $$=πr^2$$
Circumference $$=2πr$$
Full circle $$=360^\circ$$
(Square if l=w)
Area=lw
### Parallelogram
(Rhombus if l=w)
Area=lh
Regular polygons are n-sided figures with all sides equal and all angles equal.
The sum of the inside angles of an n-sided regular polygon is
$$(n-2) .180^\circ$$.
### Area of a parallelogram:
$$A = bh$$
### Area of a trapezoid:
$$A =\frac{1}{2} h (b_{1}+b_{2})$$
### Surface Area and Volume of a rectangular/right prism:
$$SA=ph+2B$$
$$V=Bh$$
### Surface Area and Volume of a cylinder:
$$SA =2πrh+2πr^2$$
$$V =πr^2 h$$
### Surface Area and Volume of a Pyramid
$$SA=\frac{1}{2} \ ps+b$$
$$V=\frac{1}{3}\ bh$$
### Surface Area and Volume of a Cone
$$SA =πrs+πr^2$$
$$V=\frac{1}{3} \ πr^2 \ h$$
### Surface Area and Volume of a Sphere
$$SA =4πr^2$$
$$V =\frac{4}{3} \ πr^3$$
(p $$=$$ perimeter of base B; $$π ~ 3.14$$)
### Solids
Rectangular Solid
Volume =lwh
Area =2(lw+wh+lh)
Right Cylinder
Volume $$=πr^2 \ h$$
Area $$=2πr(r+h)$$
### Simple interest:
$$I=prt$$
(I = interest, p = principal, r = rate, t = time)
### mean:
mean: $$\frac{sum \ of \ the \ data}{of \ data \ entires}$$
### mode:
value in the list that appears most often
### range:
largest value $$-$$ smallest value
### Median
Middle value in the list (which must be sorted)
Example: median of
$$\{3,10,9,27,50\} = 10$$
Example: median of
$$\{3,9,10,27\}=\frac{(9+10)}{2}=9.5$$
### Sum
average $$×$$ (number of terms)
### Average
$$\frac{sum \ of \ terms}{number \ of \ terms}$$
### Average speed
$$\frac{total \ distance}{total \ time}$$
### Probability
$$\frac{number \ of \ desired \ outcomes}{number \ of \ total \ outcomes}$$
The probability of two different events A and B both happening is:
P(A and B)=p(A) .p(B)
as long as the events are independent (not mutually exclusive).
### Powers, Exponents, Roots
$$x^a .x^b=x^{a+b}$$
$$\frac{x^a}{x^b} = x^{a-b}$$
$$\frac{1}{x^b }= x^{-b}$$
$$(x^a)^b=x^{a.b}$$
$$(xy)^a= x^a .y^a$$
$$x^0=1$$
$$\sqrt{xy}=\sqrt{x} .\sqrt{y}$$
$$(-1)^n=-1$$, if n is odd.
$$(-1)^n=+1$$, if n is even.
If $$0<x<1$$, then
$$0<x^3<x^2<x<\sqrt{x}<\sqrt{3x}<1$$.
### Simple Interest
The charge for borrowing money or the return for lending it.
Interest = principal $$×$$ rate $$×$$ time
OR
$$I=prt$$
### Positive Exponents
An exponent is simply shorthand for multiplying that number of identical factors. So $$4^3$$ is the same as (4)(4)(4), three identical factors of 4. And $$x^3$$ is just three factors of x, $$(x)(x)(x)$$.
### Negative Exponents
A negative exponent means to divide by that number of factors instead of multiplying.
So $$4^{-3}$$ is the same as $$\frac{1}{4^3}$$ and
$$x^{-3}=\frac{1}{x^3}$$
### Factorials
Factorial- the product of a number and all counting numbers below it.
8 factorial $$=8!=$$
$$8×7×6×5×4×3×2×1=40,320$$
5 factorial $$=5!=$$
$$5×4×3×2×1=120$$
2 factorial $$=2!=2× 1=2$$
### Multiplying Two Powers of the SAME Base
When the bases are the same, you find the new power by just adding the exponents
$$x^a .x^b=x^{a+b }$$
### Powers of Powers
For power of a power: you multiply the exponents.
$$(x^a)^b=x^{(ab)}$$
### Dividing Powers
$$\frac{x^a}{x^b} =x^a x^{-b}= x^{a-b}$$
### The Zero Exponent
Anything to the 0 power is 1.
$$x^0= 1$$
### Permutation:
When different orderings of the same items are counted separately, we have a permutation problem:
$$_{n}p_{r}=\frac{n!}{(n-1)!}$$
### Combination:
The fundamental counting principle, as demonstrated above, is used any time the order of the outcomes is important. When selecting objects from a group where order is NOT important, we use the formula for COMBINATIONS:
The fundamental counting principle, as demonstrated above, is used any time the order of the outcomes is important. When selecting objects from a group where order is NOT important, we use the formula for COMBINATIONS:
$$_{n}C_{r}=\frac{n!}{r!(n-1)!}$$
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Help save the environment | 4,595 | 14,032 | {"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": 3, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2021-25 | longest | en | 0.801187 |
https://myyachtguardian.com/how-far-is-4000-feet-in-miles-update-new/ | 1,674,892,690,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499524.28/warc/CC-MAIN-20230128054815-20230128084815-00058.warc.gz | 423,177,979 | 36,420 | # How Far Is 4000 Feet In Miles? Update New
Let’s discuss the question: how far is 4000 feet in miles. We summarize all relevant answers in section Q&A of website Myyachtguardian.com in category: Blog MMO. See more related questions in the comments below.
## How far away is 1000 feet in miles?
1000 feet is equal to 0.189 miles. Therefore, a person will walk 1000 feet in 3.78 minutes at a speed of 3 miles per hour. 1000 feet is less than 1/5 of a mile.
## How many feet is a whole mile?
How Many Feet in a Mile? Why a Mile Is 5,280 Feet | Reader’s Digest.
### How High could we Build?
How High could we Build?
How High could we Build?
## How far is a 4000 meter run?
The total distance run is 4000 meters, or nearly 2.5 miles. Aside from the 400 meter segment, which is a sprint, all legs are a middle distance run. Prior to going metric, the distance medley relay consisted of a 440-yard leg, an 880-yard leg, a 1320-yard leg and a one-mile leg.
## How many survey feet are in a mile?
Foot (US Survey) to Mile (US Survey) Conversion Table
Foot (US Survey) ft] Mile (US Survey) [mi]
1 ft 0.0001893939 mi
2 ft 0.0003787879 mi
3 ft 0.0005681818 mi
5 ft 0.0009469697 mi
## How long is a mile?
mile, any of various units of distance, such as the statute mile of 5,280 feet (1.609 km). It originated from the Roman mille passus, or “thousand paces,” which measured 5,000 Roman feet.
## How many floors is 1000 feet?
Feet to Other Units Conversion Chart
Feet [ft] Output
1000 Feet in Step is Equal to 400
1000 Feet in Story is Equal to 92.36
1000 Feet in Stride is Equal to 200
1000 Feet in Stride [Roman] is Equal to 205.95
## How long does it take to walk a mile?
Most people can expect to walk a mile in 15 to 22 minutes, according to data gathered in a 2019 study spanning five decades. The average walking pace is 2.5 to 4 mph, according to the Centers of Disease Control and Prevention.
## How many feet is a 10th of a mile?
Answer: There are 528 feet in 1/10 of a mile.
## How many feet is 3 miles answer without units?
And 3 miles have 5280×3=15840 feet.
## How many miles is the 5000 meters?
The 5000 metres or 5000-metre run is a common long-distance running event in track and field, approximately equivalent to 3 miles 188 yards or 16,404 feet 2 inches.
### ✅ How Many Feet In A Mile
✅ How Many Feet In A Mile
✅ How Many Feet In A Mile
## How far away is 3000 m?
A distance of 3,000 meters is approximately 1.86 miles or 3 kilometers. The 3,000-meter run is a middle-distance track event in track and field competitions.
## What is a 400 meter run in miles?
There are 0.24860 miles in a 400 meters.
## How many blocks is a mile?
The general rule is eight blocks equal a mile, provided you are counting the side of the block that is 660 feet long rather than the 330 foot side.
## Why does the UK use miles?
Answer has 7 votes. Historically the road network in England was established by the Romans who measured in miles. The metric system was first introduced to France by Napoleon at a time when they were at war with England. This is why the English were reluctant to adopt metrification.
## How long is 1 mile in a car?
Since the speed you drive is measured in miles per hour (mph), an easy way to estimate travel time is to compare your average speed to a single hour. For example, if you’re driving 60 mph, that means you’ll travel 60 miles in one hour, so it’ll take just one minute to travel one mile.
## How many feet is a story?
The height of each storey in a building is based on ceiling height, floor thickness, and building material — with a general average of about 14 feet.
## How many feet high is a 10 story building?
The building is 10 stories tall. This is usually around 14 feet 43 m.
## How many floors is the tallest building in the world?
World Records
At over 828 metres (2,716.5 feet) and more than 160 stories, Burj Khalifa holds the following records: Tallest building in the world.
## Is it better to walk slow or fast?
Researchers found that obese people who walk at a slower pace burn more calories than when they walk at their normal pace. In addition, walking at a slower, 2-mile-per-hour pace reduces the stress on their knee joints by up to 25% compared with walking at a brisk 3-mile-per-hour pace.
### World First – Skydiver Luke Aikins Jumps 25000 Feet Into Net With No Parachute
World First – Skydiver Luke Aikins Jumps 25000 Feet Into Net With No Parachute
World First – Skydiver Luke Aikins Jumps 25000 Feet Into Net With No Parachute
## Is walking 1 mile a day enough?
It depends on your goals. For older adults or sedentary people aiming to start a fitness plan, walking a mile a day might be enough. For other individuals, 150 minutes of moderate aerobic exercise per week is the recommended minimum, which is likely more than 1 mile a day.
## Is 2 miles walking a day good?
Walking 2 miles a day is a great weight loss strategy, and it does wonders for your mental health, heart health and general well being. The best part is that unlike other popular workouts, this exercise is incredibly inexpensive to start and stick with.
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Creating a 3D World
Team ROGUE:
Greg Brown, Olivia Ortiz, Ravenna Thielstrom
Abstract
2
Introduction
2
Background/Theory 2
Completed Project Design 5
Results
6
Discussion/Conclusion
7
2
Abstract:
Using MATLAB, we designed a three-dimensional cityscape. We developed three
functions to make rectangular prisms, cylinders, and spheres, as collections of two-dimensional
patches. We used these functions to rapidly populate a three-dimensional plot with the varied
scenery of a cityscape. Lighting, colors, and shading added to the realism of the city. A series of
for-loops that adapt the camera’s position, angle, and target move the observer along a preset,
real-time path to showcase a comprehensive fly-through of the city.
Introduction:
Our goal in completing this project was to design a three-dimensional landscape—
whether it be a city, forest, or barn yard—on MATLAB and use the camera-properties of
MATLAB to show it to others. While part of our motivation was to complete the project, our
group was drawn to the computer-related aspects of engineering. Coding MATLAB is very
computer-related, so this project became a way to further explore our interests. We found the
idea of making a three-dimensional landscape particularly interesting because we had only used
MATLAB in two-dimensions in class, and graphic design is an interesting subject. Likewise, we
had never worked with the different camera properties of MATLAB and were curious as to how
to use them.
Background/Theory:
Data in MATLAB may be plotted on the z-axis in addition to the x- and y-axes. In class,
we created patches on the x-and-y plane. By aggregating groups of patches that also contain zdata, different three-dimensional shapes can be created; the perspective of each face of these
3
shapes will change according to the camera properties of position and target to create an illusion
of a three-dimensional landscape.
The greater complexity of objects visible in the demo is emerges from the composition of
multiple basic three-dimensional objects, namely rectangular prisms, cylinders, and spheres. We
created functions that would assemble patches in a predefined way so that we could easily
develop a cityscape. Most of the properties that are applied to patches are meant to be applied to
every patch, so that functions can easily be called to set the shape properties that vary: location,
dimensions, and color.
Each function varies in complexity. The rectangular prism function, for instance, creates
six separate patches, one for each of the sides of a rectangular prism. Because the dimensions of
each shape could be specified, the rectangular prism function was versatile: it could be used to
make flat surfaces as well as rectangular structures like buildings. The cylinder and sphere
functions both internally used functions that would return matrices of x-, y-, and z-coordinates
for their respective shapes. It was necessary to create our own functions, nonetheless, because
this data had to be edited to accommodate location and plotted with the necessary properties; the
function that returned values for a cylinder also did return values for the bases. The cylinder
function creates a three-dimensional shape by using a for-loop that creates a given number of
“sides” that make up the cylinder’s round portion to create the illusion of a smoothly curving
surface and then adding two bases. The sphere function effectively creates a set of patches using
another function, the ‘surf’ function, which better accommodates the complexity of a sphere’s
many “faces,” which are not all rectangular.
When MATLAB displays a 3D graph to a user, the resulting image depends on how the
camera is positioned and where it is pointing. In order to alter the default camera properties, we
4
can call the campos, camtarget, and camva functions, which respectively control position, target,
and view angle. Position is defined as a point of x, y, and z coordinates somewhere on the graph
where the camera is located. Target is defined as a point of x, y, and z coordinates where the
camera is aimed at. To “move” the camera around, therefore, we changed the position and target
of the camera incrementally with a brief pause for every iteration of a for-loop. A smaller view
angle “zooms in” the camera by shrinking the window that the camera looks through—whatever
can be seen through this window is what fills the screen when the program is run. This helps to
reduce clipping, which occurs when the camera gets too close to the buildings, by basically
zooming in to a point that most of the clipping occurs out of the camera’s window of sight.
Patches have a number of properties that can be recalled and modified. Patch properties
that we modified at various points were ‘Clipping,’ ‘EdgeColor,’ and ‘BackFaceLighting.’ We
turned ‘Clipping’ to ‘off’ so that MATLAB will show portions of a patch outside the axes,
although additional modifications to camera view angle were still necessary. Turning
‘EdgeColor’ to ‘none’ removes the black outline to add realism, particularly in the cases of
cylinders and spheres which aggregate many patches to create the illusion of a smooth surface.
Turning ‘BackFaceLighting’ to ‘unlit’ makes faces less reflective by preventing light from
changing their coloration; this was applied just for the roads, which would unrealistically turn
bright white at certain angles. Turning ‘EdgeLighting’ to ‘phong’ changes the shading of patches
so that differences in shading as a result of lighting manifest as a gradient rather than along a
line. This issue was most apparent on the surfaces of spheres and tops of buildings; as such,
finding that applying it to every patch in the rectangular prism function decreased the processing
speed, we decided to only apply change this property for the top patch.
5
MATLAB plots have many properties; two that we changed were color and lighting. We
changed the background of the plot and the plot-window by changing their ‘Color’ properties
with ‘set(gcf,...)’ and ‘set(gca,...).’ We modified the lighting by recalling ‘light’ and changing its
‘Position’ and ‘Style’ properties, which change where the light comes from and whether it is a
localized source.
Completed Project Design:
The previously described functions were used to populate the city with buildings and
other structures. The base of the city is a grey rectangle. The rows and columns of roads were
iterated to create 36 blocks in which structures could be added. Many buildings are singular,
simple rectangles; others are composed of multiple shapes. The sphere function was made to
create buildings with “domes” since the bottom hemisphere is embedded in a rectangle or
cylinder below. A green with trees was added as well; the trees consist of a cylindrical trunk and
spherical bunch of leaves. The tree placement and trunk height is actually randomized, so each
time the script is run they will appear in a new orientation, and the number of trees can be easily
manipulated since they are created with an iterative for-loop. As the script is run, the camera
(with a constant view angle of 30 degrees) will follow several paths through the city, each path
defined in our code by a separate for-loop. Within the for-loop, the camera position and target
are changed as needed: keeping the position coordinates constant changing the target coordinates
will cause the camera to rotate in place, keeping the target and changing the position will result
in the camera flying along a path but always focused on one spot, regardless of its position to the
camera, and changing both at the same time by the same factor will cause the camera to look a
constant distance ahead of itself, so that the view remains the same. With these tactics, the
6
camera essentially circles around the city to look at it from all angles, taking wildly different
routes along the way, and specifically lingering on a shot of the city park to show off the trees.
Results:
We achieved our goal in completing the project by designing a three-dimensional
landscape and learning how to show it with different camera-functions. Considering we started
with virtually no knowledge, we managed a great deal. Because our initial goal was so general,
though, it was easy to accomplish. Ideas and potential goals that surfaced after the city began
construction, however, were less easily obtained or not obtained at all. The idea of making a
cityscape likely emerged because the rectangular prism function was developed first, and a city
is primarily rectangular. makeCylinder and makeSphere were completed later and showcased in
specific parts of the city to demonstrate that they worked as intended. makeSphere in particular
was difficult to complete because the patches needed to create one are more complex and
numerous; for instance, even the patches comprising a cylinder, excluding the bases, are
rectangular, but some values returned by the sphere function create rectangular patches while
others create triangular patches. This complexity required that the ‘surf’ function be used instead.
Originally, we had hoped to use more diverse shapes in our city. It was originally
intended to add other shapes, such as triangular prisms; however, we have amply demonstrated
that a variety of functions can be created and applied. The specific variety that we have used
does not particularly matter. Another potential direction the project could have taken was to add
more realistic details like windows, streetlamps, and textures. This idea was not thoroughly
explored, but we did encounter some performance limitations in what we tried to do already, so
additional patches and textures could have limited our ability to display the city. Another area to
7
explore would be adding motion of objects to the city. Specifically, we attempted to add a car,
but it would be difficult to animate while the camera was moving through the city. Patches must
be deleted after each “frame” so that the car can move without leaving behind a “trail” of its
previous patches, but the functions we developed were not amenable to the deletion of patches
since we could not easily assign handles to them. Another option would be to clear the figure
between each frame, but recreating the whole city, which consists of perhaps thousands of
patches, is extremely intensive. The movement of the car, because it happens frame-by-frame
and simultaneously with the motion of the camera, would have to be added during the for-loops
regulating the camera’s motion. Adaptation to the functions used to make the city, perhaps, could
be made to circumvent these issues, but it would be another project entirely.
One detail that we completely succeeded with was the creation of trees. By making
spheres directly on top of cylinders, we created balloon-ish trees. To add more excitement, we
randomized their placement and their heights in the large park space. All of our trees are
deciduous most likely Norway Maples. If we had a triangular prism-making function, there
would have also been pine trees.
Discussion/Conclusion
For the most part, we found our efforts in MATLAB to be fruitful—the buildings were
easy to construct and place within the city blocks, and once we figured out the how to change the
camera properties, manipulating the camera into a fly-through was fairly easy. We were even
somewhat surprised at how realistic the final city ended up looking, which is due in a large part
to the extra measures we went to in lighting the city correctly. Smoothness of the buildings was a
last-minute but highly effective detail. Something that could have run more effectively, however,
8
was the speed of the fly-through, which varied drastically if we added or subtracted large overall
properties, and tended to increase during certain paths and not others. This made the final
animation look lot more jerky on a big screen than we had prepared it to look on a small
computer screen, and theoretically could not be easily fixed since it was the processing speed of
MATLAB itself that slowed the program, not the set time to pause between steps.
Another problem that never quite worked out was the issue of clipping. We did a lot of
investigating into how MATLAB operates in order to shed some light on this problem, but in the
end there was always still some clipping present in our run of the fly-through, no matter what we
did to repair it (such as adjusting the camera view angle and the “Clipping” property of patches).
An obvious extension of this would be to simply make the city bigger—add more streets
and buildings so that the “city” is not just a couple of blocks floating in a square in the sky.
Another would be to add moving cars, or details like windows, as discussed above.
9
``` | 2,652 | 12,761 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-30 | latest | en | 0.944307 |
https://www.sanfoundry.com/statistical-quality-control-written-test-questions-answers/ | 1,719,337,185,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866218.13/warc/CC-MAIN-20240625171218-20240625201218-00355.warc.gz | 846,073,588 | 19,556 | # Statistical Quality Control Questions and Answers – What is Experimental Design – 2
This set of Statistical Quality Control written test Questions & Answers focuses on “What is Experimental Design – 2”.
1. Which of these can be obtained by using the Experimental design?
a) Reduced process capability
b) Increased variability
c) Increased cost
d) Reduced distance from the nominal value
Explanation: With the use of the designed experiments, it is possible to increase the conformance of the product to the nominal value. So it can be used in designing a new product.
2. Which of these can be used in the DFSS activities?
a) Design of experiments
b) Histogram
c) Acceptance sampling
d) Stem and Leaf plot
Explanation: The DFSS activities are said to be the Design for Six-sigma activities. The process of design of experiments is very useful in these activities.
3. The DFSS stands for _______________
a) Deleting for smaller standards
c) Design for six-sigma
d) Development for six-sigma
Explanation: The DFSS stands for the Design for Six – Sigma activities. These activities are used to setup a process environment to get the six-sigma output.
4. SPC and Design of experiments are very closely interrelated.
a) True
b) False
Explanation: SPC and experimental design of experiments are two very interrelated tools for process improvement and optimization. For example, if a low Cp process is in-control; Designed experiments could be used to increase process capability.
5. SPC is a passive statistical method.
a) True
b) False
Explanation: SPC is a passive statistical method as, in this procedure; we have to wait for the process information to come, to decide a step to control the process. However, if the process is in-control, we can’t do very much to reduce the variability.
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!
6. Which of these can be used in the process of engineering design?
a) Design of experiment
b) Control charting
c) Acceptance sampling
d) Cusum charts
Explanation: The design of experiments can also play a major role in engineering design activities, where new products are developed and existing ones are improved. So this is also used in DFSS activities.
7. Which of these is not an application of experimental design in the field of the Engineering design?
a) Evaluation and comparison of basic design configurations
b) Evaluation of material alternatives
c) Evaluation of key product design parameters that impact performance
d) Evaluation of correct process to develop the product
Explanation: The Design of the experiment procedure helps in engineering design by helping in evaluation of material alternatives, key product design parameters, and basic design configurations.
8. Use of designed experiments does not give _________
a) Improved yield
b) Reduced variability
c) Closer conformance to the nominal
d) Increased development time
Explanation: The use of the design of the experiments procedure is quite useful in process development. This gives an improved yield, reduced variability and closer conformance to nominal, and reduced development time.
Sanfoundry Global Education & Learning Series – Statistical Quality Control. | 663 | 3,230 | {"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} | 3.28125 | 3 | CC-MAIN-2024-26 | latest | en | 0.907236 |
https://rockyj.in/2011/04/08/mapreduce-in-english.html | 1,670,626,771,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711552.8/warc/CC-MAIN-20221209213503-20221210003503-00683.warc.gz | 530,254,808 | 3,338 | ## MapReduce in English
08/04/2011
It took me a lot of time to understand MapReduce, I read the definition from Google, then I read the explanation in "CouchDB : The Definitive Guide" and finally I read the Wikipedia entry. I found the Wikipedia entry to be the most useful and simple to understand. So before I myself forget what I have understood, I want to blurt it out in this blog.
To understand what is MapReduce one must first think (like with most technology) why one needs MapReduce. The answer is simply to distribute work. Now work can be distributed to a cluster of workstations or even the cores of a single CPU but the idea behind the MapReduce algo is to "Divide and Rule".
The quintessential example given to explain MapReduce is how you would count the number of occurrence of each word in a document (or a set of documents). I will use the same example but twist it just a little bit, so I say, if I give you a book and "N" number of assistants, how would you count the occurrences of each word in the book?
Most likely answer is, that you would tear the book and hand over some pages to each of your assistants asking them to count the word occurrences in their respective pages.
In computing we will do this is two steps, first we will need a function (somethings which the assistants do) that takes in a document (be it 1 page or 10 pages or 1 paragraph in our example) and once a word is found it simply flags count to one.
Next, me the master would then take all the words and their flags (just like a Map) and sort them on the keys i.e. the words themselves. So we have something like -
``````a : 1
a : 1
a : 1
a : 1
Once : 1
Once : 1
the : 1
the : 1
the : 1
...
``````
So on and so forth.
The Map function is therefore designed in such a way that it is distributable and all the workers are doing the same thing. Now comes the Reduce function, again whatever it does has to be distributable and uniform. So in my example I would give each for my assistants a word (lets say "Once" in example above) or a set of words along with a list of its count ((1, 1) for "Once"). The Reduce function can then simply add the count for each word and give me back the result which is exactly what I need (the number of occurrences of each word).
If it is still confusing, let me revisit my example. I tear the book and give pages to each of my assistants who simply give me back a Map of each word they read with a count of 1. I then sort the words (simple if I am a computer) and then give the workers back a set of words with their associated occurrences (a list just saying (1, 1, 1, ..)). The workers then for each word sum the elements of the list and hand me back the result.
In essence, MapReduce enables us to break down very complex problems into smaller parts and helps us bring down the computation time dramatically. I my example above I have simplified things to a great extent to make it understandable but technically a lot more may be going on. For example, one of my assistants can become a sub-master and sub-divide the work to other assistants and sort the maps himself.
MapReduce is very powerful and with the advent of multi-core processors and cheap hardware it just might be a non-negotiable skill in every developer's arsenal. Google used it to completely regenerate Google's index of the World Wide Web. Need I say more :). | 785 | 3,372 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2022-49 | longest | en | 0.94432 |
https://balsammed.net/present-angles-pairs-worksheet/ | 1,620,349,696,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988774.18/warc/CC-MAIN-20210506235514-20210507025514-00570.warc.gz | 116,877,695 | 8,874 | Present angles pairs worksheet ideas
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Present Angles Pairs Worksheet. Search for angles that are adjacent as well as supplementary in the figures presented in this PDF worksheet and recognize them as linear pairs. Once you find your worksheet click on pop-out icon or print icon to worksheet. Subtract 9 from each side. Identifying an Angle in the Linear Pair The image featured in this 7th grade printable worksheet contains numerous linear pairs.
Angle Pairs Lesson Plans Worksheets Lesson Planet From lessonplanet.com
Some of the worksheets for this concept are 3 parallel lines and transversals Identify pairs of lines and angles Work section 3 2 angles and parallel lines Parallel lines and transversals special angle pairs with Angle pairs in two lines cut by a transversal Name answer key Parallel lines transversals work. This chance we present you several perky photos that weve gathered special for you this time we decide to be focus about Angle Pairs Worksheet. You might use the particular very same worksheet for a lot of of your. Showing top 8 worksheets in the category - Angle Pairs. Geometry angles worksheet 4th grade 7th grade geometry worksheets. 18012020 Great Geometry Section 15 Angle Pair Relationships Practice Worksheet Answers Free Printables Worksheet.
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Apply appropriate properties of adjacent complementary linear vertical corresponding alternate and same-side angles to find the measures of the indicated angles. Sep 2 2015 - Plug into our printable pairs of angles worksheets and get to the bottom of the angle pair relationships and the special properties they exhibit. Linear Pair And Vertical Angles. Search for angles that are adjacent as well as supplementary in the figures presented in this PDF worksheet and recognize them as linear pairs. Once you find your worksheet click on pop-out icon or print icon to worksheet.
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Showing top 8 worksheets in the category - Angle Pairs. Some of the worksheets for this concept are 3 parallel lines and transversals Identify pairs of lines and angles Work section 3 2 angles and parallel lines Parallel lines and transversals special angle pairs with Angle pairs in two lines cut by a transversal Name answer key Parallel lines transversals work. Showing top 8 worksheets in the category - Angle Pairs. The students are asked to set up and solve linear equations. These relationships are linear pair angles complementary angles and vertical angles.
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Pairs Of Angles Homework - Displaying top 8 worksheets found for this concept. Feb 19 2015 - Plug into our printable pairs of angles worksheets and get to the bottom of the angle pair relationships and the special properties they exhibit. Identifying an Angle in the Linear Pair The image featured in this 7th grade printable worksheet contains numerous linear pairs. Just before preaching about Pairs Of Angles Worksheet Answers you need to are aware that Training can be your critical for a greater down the road in addition to understanding wont just avoid as soon as the school bell rings. The students are asked to set up and solve linear equations.
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Source: byveera.blogspot.com
In all likelihood students of grade 6 grade 7 and grade 8 have learned that two angles are linear if they are adjacent angles formed by two intersecting lines. The students are asked to set up and solve linear equations. Some of the worksheets for this concept are 3 parallel lines and transversals Identify pairs of lines and angles Work section 3 2 angles and parallel lines Parallel lines and transversals special angle pairs with Angle pairs in two lines cut by a transversal Name answer key Parallel lines transversals work. Pairs Of Angles Homework - Displaying top 8 worksheets found for this concept. Once you find your worksheet click on pop-out icon or print icon to worksheet.
Source: byveera.blogspot.com
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Source: pinterest.com
Linear Pairs of Angles Worksheets Explore our myriad collection of printable worksheets on linear pairs of angles for profound practical knowledge of this concept. In all likelihood students of grade 6 grade 7 and grade 8 have learned that two angles are linear if they are adjacent angles formed by two intersecting lines. Worksheet by Lucas Kaufmann. This practice worksheet helps students identify the different type of angles pairs created by parallel lines and a transversal. Triangle Congruence Worksheet Answers Pdf Reading A Pay Stub Worksheet Answers Heat Transfer Worksheet Answer Key Mcculloch V Maryland 1819 Worksheet Answers Learning To Tell The Time Worksheets Social Security Benefits Worksheet 2015 Graphing Sine And Cosine Practice Worksheet Avogadros Number Worksheet Marbury V Madison 1803 Worksheet Answers 1 5 Angle Relationships Worksheet.
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Source: byveera.blogspot.com
5x 4 x -2 3x 7 180. 5x 4 x -2 3x 7 180. Worksheet by Lucas Kaufmann. Identifying an Angle in the Linear Pair The image featured in this 7th grade printable worksheet contains numerous linear pairs. Find the value of x.
Source: lessonplanet.com
Some of the worksheets for this concept are Pairs of angles Pairs of angles Pairs of anglespairs of angles Grade 3 geometry work describing quadrilaterals Name the relationship complementary linear pair Naming angles Identify pairs of lines and angles The coordinate plane. Which staying stated most people provide assortment of straightforward but enlightening reports in. Angles formed by Parallel Lines cut by a Transversal Worksheets. 5x 4 x -2 3x 7 180. Now we will learn more pairs of angles for grade 6 to grade 8 like linear vertically opposite and adjacent angles here.
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In all likelihood students of grade 6 grade 7 and grade 8 have learned that two angles are linear if they are adjacent angles formed by two intersecting lines. Linear Pairs of Angles Worksheets Explore our myriad collection of printable worksheets on linear pairs of angles for profound practical knowledge of this concept. Showing top 8 worksheets in the category - Angle Pairs. Just before preaching about Pairs Of Angles Worksheet Answers you need to are aware that Training can be your critical for a greater down the road in addition to understanding wont just avoid as soon as the school bell rings. Linear Pair And Vertical Angles.
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They are also identify using a given picture the different possibilities of angle pairs. Which staying stated most people provide assortment of straightforward but enlightening reports in. Find the value of x. Worksheet by Lucas Kaufmann. This printable worksheet on pairs of angles for students of grade 7 and grade 8 is your chance to give your preparation a big shot in the arm.
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http://www.hindawi.com/journals/ijmms/2008/159029/ | 1,462,037,218,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2016-18/segments/1461860112228.39/warc/CC-MAIN-20160428161512-00074-ip-10-239-7-51.ec2.internal.warc.gz | 578,665,707 | 75,913 | `International Journal of Mathematics and Mathematical SciencesVolume 2008 (2008), Article ID 159029, 11 pageshttp://dx.doi.org/10.1155/2008/159029`
Research Article
Starlike and Convex Properties for Hypergeometric Functions
1Department of Mathematics, Kyungsung University, Busan 608-736, South Korea
2Department of Applied Mathematics, Pukyong National University, Busan 608-737, South Korea
Received 13 February 2008; Accepted 18 June 2008
Copyright © 2008 Oh Sang Kwon and Nak Eun Cho. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract
The purpose of the present paper is to give some characterizations for a (Gaussian) hypergeometric function to be in various subclasses of starlike and convex functions. We also consider an integral operator related to the hypergeometric function.
1. Introduction
Let be the class consisting of functions of the form that are analytic and univalent in the open unit disk . Let and denote the subclasses of consisting of starlike and convex functions of order , respectively [1].
Recently, Bharati et al. [2] introduced the following subclasses of starlike and convex functions.
Definition 1.1. A function of the form (1.1) is in if it satisfies the condition and if and only if .
Definition 1.2. A function of the form (1.1) is in if it satisfies the condition and if and only if .
Bharati et al. [2] showed that , and . In particular, we note that is the class of uniformly convex functions given by Goodman [3] (also see [46]).
Let be the (Gaussian) hypergeometric function defined by where , and is the Pochhammer symbol defined by We note that converges for and is related to the Gamma function by Silverman [7] gave necessary and sufficient conditions for to be in and , and also examined a linear operator acting on hypergeometric functions. For the other interesting developments for in connection with various subclasses of univalent functions, the readers can refer to the works of Carlson and Shaffer [8], Merkes and Scott [9], and Ruscheweyh and Singh [10].
In the present paper, we determine necessary and sufficient conditions for to be in and . Furthermore, we consider an integral operator related to the hypergeometric function.
2. Results
To establish our main results, we need the following lemmas due to Bharati et al. [2].
Lemma 2.1. (i) A function of the form (1.1) is in if and only if it satisfies
(ii) A function of the form (1.1) is in if and only if it satisfies
Lemma 2.2. (i) A function of the form (1.1) is in if and only if it satisfies
(ii) A function of the form (1.1) is in if and only if it satisfies
Theorem 2.3. (i) If and , then is in if and only if
(ii) If and , then is in if and only if
Proof. (i) Since according to (i) of Lemma 2.1, we must show that Noting that and then applying (1.6), we have Hence, (2.8) is equivalent to Thus, (2.10) is valid if and only if or, equivalently, .
(ii) Since by (i) of Lemma 2.1, we need only to show that Now, But this last expression is bounded above by if and only if (2.6) holds.
Theorem 2.4. (i) If and , then is in if and only if
(ii) If and , then is in if and only if
Proof. (i) Since has the form (2.7), we see from (ii) of Lemma 2.1 that our conclusion is equivalent to Writing , we see that This last expression is bounded above by if and only if which is equivalent to (2.14).
(ii) In view of (ii) of Lemma 2.1, we need only to show that Now, Writing , we have Substituting (2.21) into the right-hand side of (2.20), we obtain Since , we write (2.22) as By simplification, we see that the last expression is bounded above by if and only if (2.15) holds.
Theorem 2.5. (i) If and , then is in if and only if
(ii) If and , then is in if and only if
Proof. (i) Since according to (i) of Lemma 2.2, we must show that Noting that and then applying (1.6), we have Hence, (2.27) is equivalent to Thus, (2.29) is valid if and only if or, equivalently, .
(ii) Since by (i) of Lemma 2.2, we need only to show that Now, But this last expression is bounded above by if and only if (2.25) holds.
Theorem 2.6. (i) If and , then is in if and only if
(ii) If and , then is in if and only if
Proof. (i) Since has the form (2.26), we see from (ii) of Lemma 2.2 that our conclusion is equivalent to Writing , we see that This last expression is bounded above by if and only if , which is equivalent to (2.33).
(ii) In view of (ii) of Lemma 2.2, we need only to show that Now, Substituting (2.21) into the right-hand side of (2.38), we obtain Since , we may write (2.39) as By simplification, we see that the last expression is bounded above by if and only if (2.34) holds.
3. An Integral Operator
In the next theorems, we obtain similar-type results in connection with a particular integral operator acting on as follows:
Theorem 3.1. Let and . Then,
(i) defined by (3.1) is in if and only if (ii) defined by (3.1) is in if and only if
Proof. (i) Since by (i) of Lemma 2.1, we need only to show that Now, which is equivalent to (3.2).
(ii) According to (i) of Lemma 2.2, it is sufficient to show that Now, which is equivalent to (3.3).
Now, we observe that if and only if . Thus, any result of functions belonging to the class about leads to that of functions belonging to the class . Hence, we obtain the following analogues to Theorems 2.3 and 2.5.
Theorem 3.2. Let and . Then,
(i) defined by (3.1) is in if and only if (ii) defined by (3.1) is in if and only if
References
1. H. Silverman, “Univalent functions with negative coefficients,” Proceedings of the American Mathematical Society, vol. 51, no. 1, pp. 109–116, 1975.
2. R. Bharati, R. Parvatham, and A. Swaminathan, “On subclasses of uniformly convex functions and corresponding class of starlike functions,” Tamkang Journal of Mathematics, vol. 28, no. 1, pp. 17–32, 1997.
3. A. W. Goodman, “On uniformly convex functions,” Annales Polonici Mathematici, vol. 56, no. 1, pp. 87–92, 1991.
4. N. E. Cho, S. Y. Woo, and S. Owa, “Uniform convexity properties for hypergeometric functions,” Fractional Calculus & Applied Analysis, vol. 5, no. 3, pp. 303–313, 2002.
5. W. C. Ma and D. Minda, “Uniformly convex functions,” Annales Polonici Mathematici, vol. 57, no. 2, pp. 165–175, 1992.
6. F. Rønning, “Uniformly convex functions and a corresponding class of starlike functions,” Proceedings of the American Mathematical Society, vol. 118, no. 1, pp. 189–196, 1993.
7. H. Silverman, “Starlike and convexity properties for hypergeometric functions,” Journal of Mathematical Analysis and Applications, vol. 172, no. 2, pp. 574–581, 1993.
8. B. C. Carlson and D. B. Shaffer, “Starlike and prestarlike hypergeometric functions,” SIAM Journal on Mathematical Analysis, vol. 15, no. 4, pp. 737–745, 1984.
9. E. P. Merkes and W. T. Scott, “Starlike hypergeometric functions,” Proceedings of the American Mathematical Society, vol. 12, no. 6, pp. 885–888, 1961.
10. St. Ruscheweyh and V. Singh, “On the order of starlikeness of hypergeometric functions,” Journal of Mathematical Analysis and Applications, vol. 113, no. 1, pp. 1–11, 1986. | 2,034 | 7,202 | {"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} | 2.609375 | 3 | CC-MAIN-2016-18 | longest | en | 0.878125 |
http://www.cyberoaks.com/the-olympic-museum-was-opened-at-which-of-the-followin-3584 | 1,547,891,757,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583662893.38/warc/CC-MAIN-20190119095153-20190119121153-00135.warc.gz | 286,356,810 | 12,652 | # The mean radius of the earth is approximately
Options: 32,000 km, 6,400 km, 9,600 km, 12,800 km
Ans:B,B
### More World Geography Related questions:
The least explosive type of volcano is called
Options: Basalt plateau, Cinder cone, Shield volcanoes, Composite volcanoes
Ans:A,A
6.A top to bottom relationship among the items in a database is established by a
Options: (A)Hierarchical schema (B) Network Schema (C) Relational schema (D) All of the above | 143 12
Ans:126. (A),126. (A)
The general CAD system was development by considering a wide range of possible uses of such a system.
Options: Mechanical engineering design, Animation and graphic design, Structural engineering design and Electronic circuit design, All of the above
Ans:D ,D
bainganee
Options: NA
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Rectifiers are used to convert
Options: Direct current to Alternating current, Alternating current to Direct current, high voltage to low voltage, low voltage to high voltage
Ans:B,B
A disadvantage of Numerical Control (NC) is:
Options: the computer is not reliable, the tape and tape reader are not reliable, one operator is needed for each machine., the machine tool can easily overheat
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How many bytes are there in 1011 1001 0110 1110 numbers?
Options: 1, 2, 4, 8
Ans:B ,B
The number of major ports in India is
Options: 5, 8, 13, 15
Ans:C,C
3.Controls are
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Rare Earth factory is situated in
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When mapping a regular entity into a relation which of the following is true?
Options: One relation is created., Two relations are created., Three relations are created., Four relations are created.
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What is the frequency range of the IEEE 802.11g standard?
Options: 2.4Gbps, 5Gbps, 2.4GHz, 5GHz
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Which of the following statements is not true
Options: A structured chart is a sequential representation of program design, the Real-Time system is a particular case of a on-line-system, Batch totals are not incorporated while designing real-time applications, 4GLs are used for application proto typing
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. in order to edit a chart, you can
Options: a. triple click the chart object b. click and drag the chart object c. double click the chart object d. click the chart objects
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. What should be the first step while OS upgrading?
Options: a. Delete old Operating System b. Backup old Operating System c. Backup Critical Data d. Format Hard Disks
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You can show the shortcut menu during the slide show by
Options: a. Clicking the shortcut button on the formatting toolbar b. Right clicking the current slide c. Clicking an icon on the current slide d. a and b
Ans: – B , – B
. Which of the following syntax is correct regarding to SUM function in Excel?
Options: =SUM (A1, B1) B. =SUM (A1:B9) C. =SUM (A1:A9, B1:B9) D. All of the above
Ans:– D ,
Which of the following require large computer memory?
Options: Imaging, Graphics, Voice, All of the above
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. Which of the following should you do to bring a bullet back to a previous level?
Options: a. Press the shift + tab keys b. Press the shift key c. Press the enter key d. Press the tab key
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984. vishv mein sarvaadhik janasankhya vaala shahar kaunasa hai?
Options: NA
Ans: tokyo, tokyo
Logical models form the basis for computing systems that generate information useful in dealing with
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4.The largest unit of a database is
Options: (A)A record (B) A field (C) A subfield (D) None of above 1
Ans:14. (B),14. (B)
4.In Microsoft PowerPoint you can show the shortcut menu during the slide show by
Options: (A)Clicking the shortcut button in the formatting toolbar (B) Right clicking the current slide (C) Clicking an icon on the current slide (D) A and b 14
Ans:144. (B),144. (B)
The XOR gates are ideal for testing parity because even-parity words produce a _____ output and odd-parity words produce a _____ output
Options: low, high, high, low, odd, even, even, odd
Ans:A ,A
. Which option can be used to create a new slide show with the current slides but presented in a different order
Options: a. Rehearsal b. Custom slider show c. Slide show setup d. Slide show view
Ans:8 – B ,8 – B
. Which of the following is extension of notepad?
Options: a. .txt b. .bmp c. .ppt d. .xls
Ans:9 – a ,9 – a
8.The following data structure is linear type
Options: (A)Strings (B) Lists (C) Queues (D) All of the above 1
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5.A term means that the application software is priced separately from the computer hardware is called
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635. baksar ka yuddh kab hua jisake parinaamasvaroop angrejon ka bangaal, bihaar aur odisa par adhikaar ho gaya tha?
Options: NA
Ans: 1764 mein, 1764 mein
Reducing the amount of current through an inductor:
Options: reduces the value of inductance by one-half, multiplies the value of inductance by four, multiplies the value of inductance by two, has no effect on the inductance
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. Suppose you wanted to create an AutoCorrect entry that would type the words ‘We regret to inform you that your submission has been declined’ Of the following choices, which would be the best name you could assign to this entry?
Options: a. regret b. subdecl c. We regret to inform you that your submission has been declined d. 11
Ans: – B , – B
Sequential file organization is most appropriate for which of the following applications?
Options: grocery-store checkout, bank checking accounts, payroll, airline reservations
Ans:C ,C
An electronic circuit with about 20 transistors fabricated on a silicon chip is known as
Options: SSI, MSI, DPS, RJE
Ans:A ,A
'OS' computer abbreviation usually means ?
Options: Order of Significance, Open Software, Operating System, Optical Sensor
Ans:C,C
Each model of a computer has a unique
Options: Assembly language, Machine language, High level language, All of the above
Ans:B,B
shrigandu lake is in which district?
Options: kinnaur , lahul, kullu, mandi
Ans:kullu,kullu
Layer 2 switching provides which of the following? Hardware-based bridging (ASIC) Wire speed Low latency Low cost
Options: 1 and 3, 2 and 4, 1, 2 and 4, All of the above
Ans:D ,D
An excel workbook is a collection of
Options: a. Workbooks b. Worksheets c. Charts d. Worksheets and charts
Ans:#8211; d,#8211; d
The number of free electrons and holes in intrinsic semiconductor increases when the temperature
Options: Decreases, Increases, Stays the same, 0oC
Ans:B ,B
What is stderr ?
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Ans:C ,C
The bars in a bar chart are represented by the
Options: rectangles in a network diagram, arrows in a network diagram, triangles in a network diagram, All of the above
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If every non-key attribute is functionally dependent on the primary key, the relation will be in
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0.In Microsoft PowerPoint you can tell when an object is active because
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6.Which types of file organization are supported by magnetic tape?
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6.Auditable is
Options: (A)Here each component has a unique name and an associated set of attribute, which differs for each version of component (B) Extent to which a software system records information concerning transactions performed against the system (C) Extent to which a system or component is operational and accessible when required for use (D) They are used to describe the over all behavior of the system 12
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If you wanted administrators to see a message when logging into the router, which command would you use?
Options: message banner motd, banner message motd, banner motd, message motd
Ans:C ,C | 2,523 | 9,268 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-04 | latest | en | 0.828642 |
https://quant.stackexchange.com/questions/36882/generating-surface-of-kernel-density-estimates-over-time | 1,695,685,480,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510100.47/warc/CC-MAIN-20230925215547-20230926005547-00440.warc.gz | 520,496,748 | 38,515 | # Generating surface of Kernel Density Estimates over time
I have a 1-minutely OHLC dataset indexed by time as follows:
df_ohlc
Out[2]:
open high low close index week
Date
2011-09-13 09:53:00 5.8 6.0 5.8 6.0 1 1
2011-09-13 09:54:00 6.0 6.0 6.0 6.0 2 1
2011-09-13 09:55:00 6.0 6.0 6.0 6.0 3 1
2011-09-13 09:56:00 6.0 6.0 6.0 6.0 4 1
2011-09-13 09:57:00 6.0 6.0 6.0 6.0 5 1
...
2017-07-17 18:19:00 2176.99 2176.99 2176.50 2176.50 3073467 305
2017-07-17 18:20:00 2175.00 2177.65 2175.00 2176.99 3073468 305
2017-07-17 18:21:00 2177.80 2177.84 2173.71 2177.61 3073469 305
2017-07-17 18:22:00 2177.50 2177.50 2175.04 2175.04 3073470 305
2017-07-17 18:23:00 2177.30 2177.30 2175.00 2175.00 3073471 305
In Python,
for i in range(1,len(df_ohlc)+1):
plt.clf()
kde_est.iloc[i] = df_ohlc['close'][df_ohlc['week']==i].plot.kde()
plt.show()
generates the Kernel Density Estimate (a smooth histogram essentially) for each week's closing prices of the dataset. In other words, it generates 305 individual KDE plots for this dataset.
How would I plot all these KDEs over time on one 3-Dimensional surface?
For example, right now each KDE plot is [Close Price] x [Probability Density]. I'd like to introduce a new variable (z = time) so we can see the changes in KDE over time, [Close Price] x [Probability Density] x [Week]
• Have you had a look at the matplotlib 3d examples?
– will
Oct 12, 2018 at 14:55
1) To write manually a function, that takes an array(this corresponds to your data during a week), either with hands, using a kernel you want $f(x)=\frac{1}{nh}\sum_{k=1}^n K((x-x_i)/hn)$, or using scipy.stats.gaussian_kde(or more general using sklearn.neighbors KernelDensity). | 737 | 1,804 | {"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} | 2.890625 | 3 | CC-MAIN-2023-40 | latest | en | 0.609849 |
https://stats.stackexchange.com/questions/41473/simple-mle-question?noredirect=1 | 1,571,762,949,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987822458.91/warc/CC-MAIN-20191022155241-20191022182741-00470.warc.gz | 713,085,274 | 33,171 | # Simple MLE Question
Let $$X_1, X_2...X_n$$ be iid with $$f(x,\theta)=\dfrac{2x}{\theta^2}$$ and $$0. Find $$c$$ such that $$\mathbb{E}(c\hat{\theta})=\theta$$ where $$\hat{\theta}$$ denotes MLE of $$\theta$$.
What I have tried: I found the MLE of $$f(x;\theta)$$ to be $$\max\{X_1,X_2\cdots X_n\}$$ (which aligns with the answer at the back) but now I am stuck at this question. The answer given is $$\dfrac{2n+1}{2n}$$.
I would have proceeded as: \begin{align*} \mathbb{E}(c\hat{\theta})&=\theta\\ \int_0^\theta c \dfrac{2x}{y^2} &=\theta \quad (y = \max\{x_1,x_2,\cdots,x_n\})\\ \dfrac{1}{y^2}\int_0^\theta c .{2x}{} &=\theta \end{align*} But continuing this way gives me an answer far off from the one in the book (I don't have a term of n to begin with).
• You can't find the expected value of the MLE (hence, you can't correct the bias) without knowing the distribution of the MLE. It is related to order statistics as suggested below, but you don't need to be familiar with order statistics to solve this problem. If $Y=\textrm{max}\left\{X_1,X_2,\ldots,X_n\right\}$, what is the CDF of $Y$, $F_Y(y)=P(Y\leq y)$? Hint: If $Y\leq y$, what does that say about each of the $X_i$? – assumednormal Oct 30 '12 at 3:48
• @Max Working on your hint $Y \leq y$ implies that each $X_i\leq y$ but how do I use this in my problem? I understand that my expectation is taken over $\hat{\theta}$ and thus, we must use its distribution. Now that brings up 2 questions in my head: 1. Is it different from the distribution of $X_i$? The answer seems to be no (although I am not completely sure why). 2. If it's different, then what is it? – Kplee Oct 30 '12 at 3:55
• So,$$P(\max \{X_i\}<y)=P(X_1<y,X_2<y,...,X_n<y) = \prod_{i=1}^n P(X_i<y)$$ $$=(\int_0^y\dfrac{2x}{\theta^2})^n=\dfrac{y^{2n}}{\theta^{2n}}$$ – Kplee Oct 30 '12 at 4:08
• $$f(y) = \dfrac{2n.y^{2n-1}}{\theta^{2n}}$$ $$E(c\hat{\theta}) = \int_0^\theta\dfrac{c\cdot2n.y^{2n}dy}{\theta^{2n}}=\dfrac{c\cdot 2n}{\theta^{2n}}\times\dfrac{\theta^{2n+1}}{2n+1}=\theta$$ $$c = \dfrac{2n+1}{2n}$$ Awesome! Thanks! I'll mark the existing answer as the solution. – Kplee Oct 30 '12 at 4:17
• @Max Actually, an easier calculation would be $$E[\hat{\theta}] = E[\max X_i] = \int_0^{\theta} 1 - \left(\frac{y}{\theta}\right)^{2n}\,\mathrm dy = \theta\left(1-\frac{1}{2n+1}\right)$$ giving $c = \frac{2n+1}{2n}$ – Dilip Sarwate Oct 30 '12 at 4:25
What's the distribution of the largest observation from a sample of size $n$ with distribution F?
You want a special case of the formula for $f_{X_k}(X)$ where $k=n$ (which formula should be obvious in any case):
• Yes, as the answer pretty much implies ("should be obvious in any case"), you can work it out yourself from first principles, using simple probability arguments. The probability that the largest value of a sample of $n$ is $\leq x$ is the probability that all $n$ observations are $\leq x$. Write that probability down (a power of a cdf at $x$), take derivates w.r.t $x$ to find the density. Then take expectations. to get $c$. (Though there are other ways to go about it, they're all going to involve some effort.) – Glen_b Oct 30 '12 at 20:20 | 1,110 | 3,150 | {"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": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2019-43 | latest | en | 0.832602 |
https://prexit.teachstarter.com/us/teaching-resource/relating-decimals-and-fractions-worksheet/ | 1,674,969,274,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499700.67/warc/CC-MAIN-20230129044527-20230129074527-00677.warc.gz | 494,201,161 | 50,899 | Free teaching resource
# Relating Decimals and Fractions Worksheet
Practice matching fractions with a denominator of 10 and 100 to decimals and vice versa.
This worksheet is best used as independent practice as part of your fractions and decimals lesson.
On the first page of the worksheet, students will match fractions with their equivalent decimals in the tenths and hundredths place values. The second page matches decimals to their equivalent fractions.
## Equivalent Fractions and Decimals Worksheet Scaffolding and Extension Tips
In addition to individual student work time, this fractions and decimals matching worksheet can be used as an activity for your:
Got fast finishers? Challenge students who have a grasp on the concept by asking them to write new fractions with a denominator different than tenths and hundredths and convert them to decimals.
Support struggling students by using this worksheet as a math intervention or as a guided math group activity.
## A Variety of Ways to Prepare This Resource
Because this resource includes an answer sheet, we recommend you print one copy of the entire file. Then, make photocopies of the blank worksheet for students to complete.
To save paper, we suggest printing this 2-page worksheet double-sided.
You can also turn this teaching resource into a sustainable activity! Print a few copies on cardstock and slip them into dry-erase sleeves. Students can record their answers with a dry-erase marker, then erase and reuse.
Additionally, project the worksheet onto a screen and work through it as a class by having students record their answers in their math notebooks.
Get more handy worksheets here!
This resource was created by Lauren Blankenship, a teacher in Florida and a Teach Starter Collaborator.
Don’t stop there! We’ve got more fractions and decimals activities to support growing minds.
### teaching resource
#### Decimal Fraction Match Up!
Practice matching equivalent fractions and decimals with this set of 15 cards that focus on tenths and hundredths place values.
### teaching resource
#### Fraction and Decimal Dominoes - Tenths and Hundredths
30 dominoes to reinforce students' understanding of converting fractions to decimals.
### teaching resource
#### Match That Fraction! - Number Line Activity
A set of 16 match-up cards to practice identifying fractions on a number line. | 451 | 2,380 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.92079 |
https://www.mrsmactivity.co.uk/downloads/year-1-number-bonds-within-10-lesson-presentation/ | 1,722,734,617,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640388159.9/warc/CC-MAIN-20240804010149-20240804040149-00393.warc.gz | 729,372,384 | 34,698 | # Year 1 | Number Bonds Within 10 Lesson Presentation
## Year 1 addition and subtraction resources
This lesson introduces number bonds within 10 and how this can help with addition.
Aligned with the maths mastery approach, this Year 1 | Number Bonds Within 10 Lesson Presentation is fully editable and designed for the Year 1 maths curriculum to cover the following curriculum objectives:.
This is the first of a series of lessons on number bonds. It focuses on splitting a whole number up to and including 9 as number bonds to ten are the focus of a separate presentation.
Small Steps: Find number bonds within 10
NC Links: Read, write and interpret mathematical statements involving addition (+), and equals (=) signs. • Add one digit numbers to 10, including zero. • Solve one step problems that involve addition using concrete objects and pictorial representations *• Represent and use number bonds within 10
Ready-to-progress criteria: 1AS-1 Compose numbers to 10 from 2 parts, and partition numbers to 10 into parts.
Previous Experience: Understand the cardinal value of number words. Subitise for up to 5 items. Automatically show a given number using fingers.
1NF-1 Develop fluency in addition facts within 10.
Previous experience: Being to experience partitioning and combining numbers within 10.
TAF Statements: Working Towards Add and subtract (one digit numbers) explaining their method verbally in pictures or using apparatus.
Aligned with the order of teaching of the maths mastery approach, use this to help your pupils get to grips with each mathematical concept. This lesson presentation also includes varied fluency activities, problem solving, and mathematical discussion questions too.
Explore our other year 1 addition and subtraction resources.
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Get access to this and thousands of other resources for £39.97 per year. | 378 | 1,874 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.893026 |
https://www.tapatalk.com/groups/dozensonline/how-to-teach-dozens-t1922.html | 1,539,860,356,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511806.8/warc/CC-MAIN-20181018105742-20181018131242-00243.warc.gz | 1,118,661,643 | 22,900 | # How to teach dozens
Posts 52
Casual Member
richard.chasen
Casual Member
Joined: 4:54 PM - Apr 28, 2017
This is how I introduce a person to dozens:
I first count to 20-z for them and have them repeat. One, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve (they always find this first part simple) followed by twelve-one, twelve-two... twelve-ten, twelve-eleven, twent-ze. (most people can do this, if they can't I have them repeat me one number at a time)
Then I show them how to write the numbers and introduce what other bases are while giving the examples of computers using base 2.
The next step is the count all the way to gross.
What input is there from others here?
Posts 194
Regular
SenaryThe12th
Regular
Joined: 2:03 PM - Mar 01, 2018
richard.chasen wrote: This is how I introduce a person to dozens:
I first count to 20-z for them and have them repeat. One, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve (they always find this first part simple) followed by twelve-one, twelve-two... twelve-ten, twelve-eleven, twent-ze. (most people can do this, if they can't I have them repeat me one number at a time)
Then I show them how to write the numbers and introduce what other bases are while giving the examples of computers using base 2.
The next step is the count all the way to gross.
What input is there from others here?
How did you come up with that method? Did you try several ways and this one was the best? How did you measure that?
Posts 52
Casual Member
richard.chasen
Casual Member
Joined: 4:54 PM - Apr 28, 2017
I have a gift for teaching. When I teach I communicate with a person's subconscious. And this method has worked with several different people.
Obsessive poster
Obsessive poster
Joined: 11:27 PM - Sep 10, 2011
I would not teach them any new words, not at first. I would point out to them that they already know how to count in dozenal, such that any native English speaker would perfectly understand them, without any need to explain:
one, two, three, four, five, six, seven, eight, nine, ten, eleven, one-dozen
one-dozen-one, one-dozen-two, one-dozen-three, one-dozen-four,one-dozen- five, one-dozen-six, one-dozen-seven, one-dozen-eight, one-dozen-nine, one-dozen-ten, one-dozen-eleven, two-dozen, ...
and so on up through
...
eleven-dozen-one, eleven-dozen-two, eleven-dozen-three, eleven-dozen-four, eleven-dozen-five, eleven-dozen-six, eleven-dozen-seven, eleven-dozen-eight, eleven-dozen-nine, eleven-dozen-ten, eleven-dozen-eleven, one-gross, ...
And then I'd explain how they already know how to count this way up to eleven-gross eleven-dozen-eleven.... At which point they need some new word. Then I'd come up with some possibilities...
As of 1202/03/01[z]=2018/03/01[d] I use:
ten,eleven = ↊↋, ᘔƐ, ӾƐ, XE or AB.
Base-neutral base annotations
Systematic Dozenal Nomenclature
Primel Metrology
Western encoding (not by choice)
Greasemonkey + Mathjax + PrimelDozenator
(Links to these and other useful topics are in my index post;
click on my user name and go to my "Website" link)
Posts 194
Regular
SenaryThe12th
Regular
Joined: 2:03 PM - Mar 01, 2018
richard.chasen wrote: I have a gift for teaching. When I teach I communicate with a person's subconscious. And this method has worked with several different people.
How would we know that this method would work for somebody who was less gifted than you are?
Obsessive poster
Obsessive poster
Joined: 11:27 PM - Sep 10, 2011
SenaryThe12th wrote:
richard.chasen wrote: I have a gift for teaching. When I teach I communicate with a person's subconscious. And this method has worked with several different people.
How would we know that this method would work for somebody who was less gifted than you are?
Richard has a gift for just knowing facts that are not actually in evidence. Particularly if it's about himself and his many amazing abilities.
As of 1202/03/01[z]=2018/03/01[d] I use:
ten,eleven = ↊↋, ᘔƐ, ӾƐ, XE or AB.
Base-neutral base annotations
Systematic Dozenal Nomenclature
Primel Metrology
Western encoding (not by choice)
Greasemonkey + Mathjax + PrimelDozenator
(Links to these and other useful topics are in my index post;
click on my user name and go to my "Website" link)
Posts 52
Casual Member
richard.chasen
Casual Member
Joined: 4:54 PM - Apr 28, 2017
Richard has a gift for just knowing facts that are not actually in evidence. Particularly if it's about himself and his many amazing abilities.
Do you want to hear something really amazing?
Imagine someone with above average intelligence studying and training just to stay alive for 30 years and not developing abilities.
I remember a story of an old man practicing martial arts in the park in China. People were in awe of his talents and said this was amazing for an old man. His reply was that to train all his life and not have this level of skill would be even more amazing.
My forced spiritual path started when I was in Army intelligence basic training at age 18 in 1983 as a young man ignorant of spirituality and psychology who was also a complete idiot with women. A couple of days before I left the Army I had 200 men in formation chant the theme to the Twilight Zone to me while I was in a stressed stupor, after 2 weeks of being called Twilight.
Obsessive poster
Obsessive poster
Joined: 11:27 PM - Sep 10, 2011
richard.chasen wrote:Do you want to hear something really amazing?
What does this have to do with mathematics? (Real mathematics that is.) Or dozenal base? Or teaching it to people?
As of 1202/03/01[z]=2018/03/01[d] I use:
ten,eleven = ↊↋, ᘔƐ, ӾƐ, XE or AB.
Base-neutral base annotations
Systematic Dozenal Nomenclature
Primel Metrology
Western encoding (not by choice)
Greasemonkey + Mathjax + PrimelDozenator
(Links to these and other useful topics are in my index post;
click on my user name and go to my "Website" link)
Dozens Demigod
icarus
Dozens Demigod
Joined: 12:29 PM - Apr 11, 2006
My method is simple. People are already dozenalists.
one-two-three-four-five-six-seven-eight-nine-ten-eleven-dozen.
one dozen-one, etc.; two dozen. Three dozen, four dozen, ... ten dozen, eleven dozen, disgusting. I mean gross. Terry Gross. Fresh air. Which is a dozen dozens. Hella donuts, hella eggs.
7 + 9 = one dozen four which is sixteen. Four times seven is two dozen four which is twenty eight. (the "which is" comes from Gene Zirkel, a professor; very handy).
We are simply regrouping and stating in dozens versus tens.
For fractions I have them think of a clock. What number do we land on when we are at half past? 6. So 1/2 = .6 uncially. The quarter hour? Yes. 3. So 1/4 = .3 dozenally. etc. A third of an hour is on the 4, so 1/3 = .4 How is it in decimal? .33333... Convenient? And we can talk about what to do with eighths, that we divide the space between the numbers on the clock in twelve (since it's dozenal) rather than five. So if 1/8 of the way around the clock lands exactly between 1 and 2, and we have twelve divisions, we know 1/2 = .6, we can say 1/8 = .16 in base twelve.
They already understand these things. We compare it with tens when they accept they already know these facts.
---
The Twilight Zone.
"Help I'm steppin' into the Twilight Zone
Place is a madhouse, feels like being cloned
My beacon's been moved under moon and star
Where am I to go, now that I've gone too far"
Richard you are so brilliant. Why do you need to grandstand? It's puzzling. Why does it have to do with subconsciousness? I guess when I submit to a person they already know this stuff, that's reaching their subconscious. "What you don't have you don't need it now / What you don't know you can feel it somehow". That's just mystique, some extra spice that eventually gets in the way. I would just say there is a logical reason for people's familiarity with dozens. They've used them all their lives because the dozen is very useful, and all I am doing is helping them realize it and more consciously state things in terms of the dozen to tap its power. Nothing metaphysical.
I guess you just want people to identify with you, or you want us to marvel. Marvelous. But we all talk about dozens (or similar things) with other people.
As for any gift of teaching, I'll let other people tell me that. | 2,211 | 8,276 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2018-43 | latest | en | 0.952085 |
naveenchamara.com | 1,680,287,704,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949678.39/warc/CC-MAIN-20230331175950-20230331205950-00105.warc.gz | 469,515,877 | 23,549 | # Excel Tip – Copy A VLOOKUP Formula Across Multiple Columns
It can be really frustrating if you have set up your VLOOKUP formula in Excel, your formula is working as expected looking up the first column of values and then you have to manually adjust the formula if you want to copy it across multiple columns.
I want to look up the monthly Sales of two types of Beanie hats in my total sales data. So, I have set up my VLOOKUP to look for monthly volumes of both Beanie hats Beanie_JL and Style Beanie_JP.
The usual VLOOKUP is working perfectly well starting at my first column which returns the value from January 2015, but if I drag the formula right to continue with subsequent months from February 2015 onwards then I do not get the desired results.
Even using absolute references in my formula, the COLUMN INDEX NUMBER does not move on when I drag the formula.This is where most users would manually adjust this to get the results they need. However do not need to do this as we can enlist the help of another formula along with VLOOKUP. Let’s use the COLUMN formula to help us out.
The COLUMN function is really straightforward. It translates a column number into a cell address. For example C1 would return 3 as C is the third column in our worksheet, so you can probably see how we can use this in the VLOOKUP formula.
Let’s go over the syntax of the VLOOKUP formula is
VLOOKUP (lookup_value, table_array, col_index_num, [range_lookup])
LookupValue. The value you want to look up. The value you want to look up must be in the first column of the range of cells you specify intable_array
The table_array. This is the range of cells in which the VLOOKUP will search for the lookup_value and the return value.
Col_index_num. This is the column number (starting with 1 for the left-most column of table-array) that contains the return value.
Range_Lookup.
A logical value that specifies whether you want VLOOKUP to find an exact match or an approximate match:
• TRUE assumes the first column in the table is sorted either numerically or alphabetically, and will then search for the closest value. This is the default method if you don’t specify one.
• FALSE searches for the exact value in the first column
The first COLUMN INDEX NUMBER in the VLOOKUP I want to return is in the SECOND column of our data set, so instead of using 2 as the COLUMN INDEX NUMBER I can replace it with B1.The formula looks like this
=VLOOKUP(\$L\$7,\$B\$3:\$J\$25,COLUMN(B1),FALSE)
Once I do this the formula will automatically updates as its dragged.
Scroll to Top | 579 | 2,566 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.892276 |
https://www.hackmath.net/en/example/2512 | 1,529,727,638,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864940.31/warc/CC-MAIN-20180623035301-20180623055301-00095.warc.gz | 835,424,971 | 6,585 | # The city 3
The city has 22,000 residents. How long it is expected to have 25,000 residents if the average annual population growth is 1.4%?
Result
n = 9.195
#### Solution:
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
Be the first to comment!
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Points A[-9,6] and B[-5,-3] are adjacent vertices of the square ABCD. Calculate area of the square ABCD. | 825 | 3,178 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-26 | longest | en | 0.947062 |
https://se.mathworks.com/matlabcentral/cody/problems/42987-roots-of-a-quadratic-equation/solutions/972614 | 1,606,521,976,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141194634.29/warc/CC-MAIN-20201127221446-20201128011446-00199.warc.gz | 485,280,716 | 17,098 | Cody
# Problem 42987. Roots of a quadratic equation.
Solution 972614
Submitted on 16 Sep 2016
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 Fail
a=1; b=2; c=1; y_correct = [-1 -1]; assert(isequal(quadRoots(a,b,c),y_correct))
ans = 1 1
Assertion failed.
2 Fail
a=1; b=-5; c=6; y_correct = [2 3]; assert(isequal(quadRoots(a,b,c),y_correct))
ans = -3 -2
Assertion failed.
3 Fail
a=1; b=5; c=6; y_correct = [-2 -3]; assert(isequal(quadRoots(a,b,c),y_correct))
ans = 2 3
Assertion failed.
4 Pass
a=2; b=10; c=12; y_correct = [2 3]; assert(isequal(quadRoots(a,b,c),y_correct))
ans = 2 3
5 Fail
a=1; b=19; c=90; y_correct = [-9 -10]; assert(isequal(quadRoots(a,b,c),y_correct))
ans = 9 10
Assertion failed.
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 329 | 968 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2020-50 | latest | en | 0.563148 |
https://www.tutorialkart.com/r-tutorial/r-name-elements-of-list/ | 1,713,001,838,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816587.89/warc/CC-MAIN-20240413083102-20240413113102-00493.warc.gz | 994,002,150 | 16,454 | ## Name Elements of List
To name elements of an R List, access names of this list using names() function, and assign a vector of characters.
In this tutorial, we will learn how to name elements of a list in R, using names() function, with the help of example programs.
### Syntax
The syntax to assign names for elements in list `myList` is
`names(myList) <- c(name1, name2, ...)`
If the length of names vector is less than that of elements in the list, the name of elements which are not provided a name would get `<NA>` as name.
If the length of names vector is greater than that of elements in the list, R throws an error.
### Examples
In the following program, we will create a list with three elements, and set names for these elements using names() function.
Example.R
```x <- list(TRUE, 25, "Apple")
names(x) <- c("In Stock", "Quantity", "Product")
print(x)```
Output
```\$`In Stock`
[1] TRUE
\$Quantity
[1] 25
\$Product
[1] "Apple"```
#### Length of Names less than that of List
Let us assign a vector for names whose length is less than the number of elements in the list.
Example.R
```x <- list(TRUE, 25, "Apple")
names(x) <- c("In Stock", "Quantity")
print(x)```
Output
```\$`In Stock`
[1] TRUE
\$Quantity
[1] 25
\$<NA>
[1] "Apple"```
#### Length of Names greater than that of List
Let us assign a vector for names whose length is less than the number of elements in the list.
Example.R
```x <- list(TRUE, 25, "Apple")
names(x) <- c("In Stock", "Quantity", "Product", "Origin")
print(x)```
Output
```Error in names(x) <- c("In Stock", "Quantity", "Product", "Origin") :
'names' attribute [4] must be the same length as the vector [3]```
### Conclusion
In this R Tutorial, we learned how to name elements of a List in R programming using names() function, with the help of examples. | 486 | 1,822 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.813492 |
http://www.jiskha.com/display.cgi?id=1275520227 | 1,495,758,415,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608617.80/warc/CC-MAIN-20170525233846-20170526013846-00228.warc.gz | 525,264,091 | 3,894 | # algebra
posted by on .
Factor: 2xy + 6x - 4y - 12
a)(2x + 4) (y + 3)
b)(2x - 4) (y + 3)^2
c)(2x - 4) (y - 3)
d)(2x - 4) (y + 3)
• algebra - ,
2x(y+3)-4(y+3)
(2x-4)(y+3)
• algebra - ,
(a) and (c) give a constant term of +12, so it cannot be right. (b) gives a constant term of -4*3²=-36...not good.
The only candidate (d) can be confirmed by expansion (multiplication). | 166 | 377 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2017-22 | latest | en | 0.655631 |
https://www.gamedev.net/forums/topic/8181-screen-capture-in-opengl-fullscreen/ | 1,501,261,450,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500550975184.95/warc/CC-MAIN-20170728163715-20170728183715-00200.warc.gz | 783,776,872 | 22,320 | • ### Announcements
#### Archived
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# OpenGL Screen capture in OpenGL fullscreen?
## 1 post in this topic
Anybody know how to get a screen capture in OpenGL fullscreen? If I am running Win98, can I still use the PrintScreen key to copy the screen to the clipboard, or will that not work once I start up OGL? Thanks for the help. gameguru4@yahoo.com
0
##### Share on other sites
Yes you can. Fullscreen ogl is nothing more than ogl in a window as large as the screen.
A better way is to use glReadPixels() to obtain the buffer, and write that to disk.
DaBit
0
• ### Similar Content
• So it's been a while since I took a break from my whole creating a planet in DX11. Last time around I got stuck on fixing a nice LOD.
A week back or so I got help to find this:
https://github.com/sp4cerat/Planet-LOD
In general this is what I'm trying to recreate in DX11, he that made that planet LOD uses OpenGL but that is a minor issue and something I can solve. But I have a question regarding the code
He gets the position using this row
vec4d pos = b.var.vec4d["position"]; Which is then used further down when he sends the variable "center" into the drawing function:
if (pos.len() < 1) pos.norm(); world::draw(vec3d(pos.x, pos.y, pos.z));
Inside the draw function this happens:
draw_recursive(p3[0], p3[1], p3[2], center); Basically the 3 vertices of the triangle and the center of details that he sent as a parameter earlier: vec3d(pos.x, pos.y, pos.z)
Now onto my real question, he does vec3d edge_center[3] = { (p1 + p2) / 2, (p2 + p3) / 2, (p3 + p1) / 2 }; to get the edge center of each edge, nothing weird there.
But this is used later on with:
vec3d d = center + edge_center[i]; edge_test[i] = d.len() > ratio_size; edge_test is then used to evaluate if there should be a triangle drawn or if it should be split up into 3 new triangles instead. Why is it working for him? shouldn't it be like center - edge_center or something like that? Why adding them togheter? I asume here that the center is the center of details for the LOD. the position of the camera if stood on the ground of the planet and not up int he air like it is now.
Full code can be seen here:
https://github.com/sp4cerat/Planet-LOD/blob/master/src.simple/Main.cpp
If anyone would like to take a look and try to help me understand this code I would love this person. I'm running out of ideas on how to solve this in my own head, most likely twisted it one time to many up in my head
Toastmastern
• I googled around but are unable to find source code or details of implementation.
What keywords should I search for this topic?
Things I would like to know:
A. How to ensure that partially covered pixels are rasterized?
Apparently by expanding each triangle by 1 pixel or so, rasterization problem is almost solved.
But it will result in an unindexable triangle list without tons of overlaps. Will it incur a large performance penalty?
How to ensure proper synchronizations in GLSL?
GLSL seems to only allow int32 atomics on image.
C. Is there some simple ways to estimate coverage on-the-fly?
In case I am to draw 2D shapes onto an exisitng target:
1. A multi-pass whatever-buffer seems overkill.
2. Multisampling could cost a lot memory though all I need is better coverage.
Besides, I have to blit twice, if draw target is not multisampled.
• By mapra99
Hello
I am working on a recent project and I have been learning how to code in C# using OpenGL libraries for some graphics. I have achieved some quite interesting things using TAO Framework writing in Console Applications, creating a GLUT Window. But my problem now is that I need to incorporate the Graphics in a Windows Form so I can relate the objects that I render with some .NET Controls.
To deal with this problem, I have seen in some forums that it's better to use OpenTK instead of TAO Framework, so I can use the glControl that OpenTK libraries offer. However, I haven't found complete articles, tutorials or source codes that help using the glControl or that may insert me into de OpenTK functions. Would somebody please share in this forum some links or files where I can find good documentation about this topic? Or may I use another library different of OpenTK?
Thanks!
• Hello, I have been working on SH Irradiance map rendering, and I have been using a GLSL pixel shader to render SH irradiance to 2D irradiance maps for my static objects. I already have it working with 9 3D textures so far for the first 9 SH functions.
In my GLSL shader, I have to send in 9 SH Coefficient 3D Texures that use RGBA8 as a pixel format. RGB being used for the coefficients for red, green, and blue, and the A for checking if the voxel is in use (for the 3D texture solidification shader to prevent bleeding).
My problem is, I want to knock this number of textures down to something like 4 or 5. Getting even lower would be a godsend. This is because I eventually plan on adding more SH Coefficient 3D Textures for other parts of the game map (such as inside rooms, as opposed to the outside), to circumvent irradiance probe bleeding between rooms separated by walls. I don't want to reach the 32 texture limit too soon. Also, I figure that it would be a LOT faster.
Is there a way I could, say, store 2 sets of SH Coefficients for 2 SH functions inside a texture with RGBA16 pixels? If so, how would I extract them from inside GLSL? Let me know if you have any suggestions ^^.
• By KarimIO
EDIT: I thought this was restricted to Attribute-Created GL contexts, but it isn't, so I rewrote the post.
Hey guys, whenever I call SwapBuffers(hDC), I get a crash, and I get a "Too many posts were made to a semaphore." from Windows as I call SwapBuffers. What could be the cause of this?
Update: No crash occurs if I don't draw, just clear and swap.
static PIXELFORMATDESCRIPTOR pfd = // pfd Tells Windows How We Want Things To Be { sizeof(PIXELFORMATDESCRIPTOR), // Size Of This Pixel Format Descriptor 1, // Version Number PFD_DRAW_TO_WINDOW | // Format Must Support Window PFD_SUPPORT_OPENGL | // Format Must Support OpenGL PFD_DOUBLEBUFFER, // Must Support Double Buffering PFD_TYPE_RGBA, // Request An RGBA Format 32, // Select Our Color Depth 0, 0, 0, 0, 0, 0, // Color Bits Ignored 0, // No Alpha Buffer 0, // Shift Bit Ignored 0, // No Accumulation Buffer 0, 0, 0, 0, // Accumulation Bits Ignored 24, // 24Bit Z-Buffer (Depth Buffer) 0, // No Stencil Buffer 0, // No Auxiliary Buffer PFD_MAIN_PLANE, // Main Drawing Layer 0, // Reserved 0, 0, 0 // Layer Masks Ignored }; if (!(hDC = GetDC(windowHandle))) return false; unsigned int PixelFormat; if (!(PixelFormat = ChoosePixelFormat(hDC, &pfd))) return false; if (!SetPixelFormat(hDC, PixelFormat, &pfd)) return false; hRC = wglCreateContext(hDC); if (!hRC) { std::cout << "wglCreateContext Failed!\n"; return false; } if (wglMakeCurrent(hDC, hRC) == NULL) { std::cout << "Make Context Current Second Failed!\n"; return false; } ... // OGL Buffer Initialization glClear(GL_DEPTH_BUFFER_BIT | GL_COLOR_BUFFER_BIT); glBindVertexArray(vao); glUseProgram(myprogram); glDrawElements(GL_TRIANGLES, indexCount, GL_UNSIGNED_SHORT, (void *)indexStart); SwapBuffers(GetDC(window_handle));
• 19
• 11
• 15
• 15
• 19 | 1,831 | 7,250 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | latest | en | 0.873291 |
https://darkskiesfilm.com/how-many-mg-are-in-a-cb/ | 1,695,981,745,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510501.83/warc/CC-MAIN-20230929090526-20230929120526-00458.warc.gz | 218,324,367 | 10,830 | # How many mg are in a CB?
## How many mg are in a CB?
A 2017 review of research showed that humans can safely tolerate up to 1,500 mg per day. For reference, typical 1 ounce bottles of CBD oil contain from 300 to 1,500 mg. It’s important to remember that CBD and cannabis as a whole are still in the early stages of research.
## What is the formula for kg to MG?
To convert a kilogram measurement to a milligram measurement, multiply the weight by the conversion ratio. The weight in milligrams is equal to the kilograms multiplied by 1,000,000.
How many milligrams are in a Millilitre?
1,000 milligrams
Therefore, there must be 1,000 milligrams in a milliliter, making the formula for mg to ml conversion: mL = mg / 1000 .
### How do you convert kg to G to mg?
You already know 1 kg equals 1,000 g. Then, convert 1,000 grams to milligrams. Multiply the number of grams by 1,000 to convert grams to milligrams. 1 kilogram is equal to 1,000,000 milligram.
### Is mg/kg same as mg?
they are the same. If you know the density of the solution expressed in kg/l, you can divide your value in mg/l by the density so that (mg/l) / (kg/l) results in a value in mg/kg.
How do you convert kg to g to mg?
#### How many grams is 1kl?
1000 grams
There are 1000 grams in 1 kilogram.
#### How many ml is in 1000 mg?
1000 mg of the matter is equal to 1d ml of matter, where d is density of the matter.
What is 5 mg in ml?
Milligram to Milliliter Converter metric conversion table
Milligram to Milliliter Converter metric conversion table
0.02 mg = 2.0E-5 ml 0.2 mg = 0.0002 ml 212 mg = 0.212 ml
0.03 mg = 3.0E-5 ml 0.3 mg = 0.0003 ml 213 mg = 0.213 ml
0.04 mg = 4.0E-5 ml 0.4 mg = 0.0004 ml 214 mg = 0.214 ml
0.05 mg = 5.0E-5 ml 0.5 mg = 0.0005 ml
## What does 1mg per kg mean?
Milligrams per kilogram (mg/kg) means the milligrams of substance per kilogram of weight.
## How much is 1kg to 1g?
Kilograms to Grams conversion 1 kilogram (kg) is equal to 1000 grams (g).
How much gram is a Tola?
11.663 8038 grams
The tola (Hindi: तोला; Urdu: تولا tolā) also transliterated as tolah or tole, is a traditional Ancient Indian and South Asian unit of mass, now standardised as 180 troy grains (11.663 8038 grams) or exactly 3/8 troy ounce.
### Is 3mg of creatine enough?
Taking too much creatine at one time can result in stomach discomfort and bloating, and it’s a waste of money. After your muscles are fully saturated with creatine, it’s recommended to take 3–5 grams (14 mg/pound or 30 mg/kg) daily to maintain optimal muscle stores.
### How many scoops is 5g?
1 Teaspoon (1/3 Tablespoon | 5 mL | 5 g) Long Handle Measuring Scoops for Coffee, Protein Powder and other Dry Goods | Food Grade | BPA & Phthalate Free (2) | 824 | 2,728 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2023-40 | latest | en | 0.860701 |
https://www.sawaal.com/physics-questions-and-answers/what-is-the-si-unit-of-pressure_1918 | 1,716,912,717,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059143.84/warc/CC-MAIN-20240528154421-20240528184421-00199.warc.gz | 851,531,571 | 13,532 | 71
Q:
What is the SI unit of pressure
A) Pascal B) Dyne C) Newton D) Jule
Explanation:
Subject: Physics
Q:
What is the noise level of normal conversation?
Explanation:
Filed Under: Physics
Exam Prep: Bank Exams
3 550
Q:
As an object falls freely its ___________________.
A) kinetic energy gets converted into potential energy B) potential energy gets converted into kinetic energy C) momentum gets converted into gravitational force D) gravitational force gets converted into momentum
Answer & Explanation Answer: B) potential energy gets converted into kinetic energy
Explanation:
Filed Under: Physics
Exam Prep: Bank Exams
0 21175
Q:
Due to an acceleration of 1 $m/s2$, the velocity of a body increases from 5 m/s to 10 m/s in a certain period. Find the displacement (in m) of the body in that period.
A) 125 B) 37.5 C) 62.5 D) 75
Explanation:
Filed Under: Physics
Exam Prep: Bank Exams
3 641
Q:
What is measured with the Nephometer?
A) volume of rainfall B) Cloud volume and speed C) Salinity of Sea D) All options are correct.
Explanation:
Filed Under: Physics
Exam Prep: Bank Exams
1 8437
Q:
What is the standard room temperature in Kelvin?
A) 98 K B) 198 K C) 273 K D) 373 K
Explanation:
Filed Under: Physics
Exam Prep: Bank Exams
0 7952
Q:
An object, starting from rest, moves with constant acceleration of 4 m/s2. After 8 s, its speed is
A) 16 m/s B) 8 m/s C) 32 m/s D) 4 m/s
Explanation:
Filed Under: Physics
Exam Prep: AIEEE , Bank Exams
1 519
Q:
1 watt is equal to 1 ____________
A) J s-1 B) J s C) J s-2 D) J
Explanation:
Filed Under: Physics
Exam Prep: Bank Exams
2 9854
Q:
A body is initially moving with a velocity of 10 m/s. It undergoes an acceleration of 2 m/s2 for 5 seconds. Find the displacement (in m) of this body in these 5 seconds.
A) 25 B) 37.5 C) 50 D) 75 | 536 | 1,836 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-22 | latest | en | 0.739797 |
https://es.mathworks.com/matlabcentral/cody/players/6243790/solved | 1,606,992,178,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141727627.70/warc/CC-MAIN-20201203094119-20201203124119-00115.warc.gz | 292,413,686 | 21,379 | Cody
# Andrea Ramirez
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1 – 50 of 109
#### Problem 2023. Is this triangle right-angled?
Created by: Tanya Morton
Tags pythagoras
#### Problem 9. Who Has the Most Change?
Created by: Cody Team
#### Problem 10. Determine whether a vector is monotonically increasing
Created by: Cody Team
#### Problem 189. Sum all integers from 1 to 2^n
Created by: Dimitris Kaliakmanis
#### Problem 37. Pascal's Triangle
Created by: Cody Team
#### Problem 23. Finding Perfect Squares
Created by: Cody Team
#### Problem 14. Find the numeric mean of the prime numbers in a matrix.
Created by: Cody Team
Tags easy, matrices
#### Problem 32. Most nonzero elements in row
Created by: Cody Team
Tags matrices
#### Problem 109. Check if sorted
Created by: AMITAVA BISWAS
#### Problem 20. Summing digits
Created by: Cody Team
Tags strings, sum
#### Problem 30. Sort a list of complex numbers based on far they are from the origin.
Created by: Cody Team
#### Problem 22. Remove the vowels
Created by: Cody Team
Tags regexp, siam
#### Problem 1658. Simple equation: Annual salary
Created by: matlab.zyante.com
Tags easy, basics, salary
#### Problem 233. Reverse the vector
Created by: Vishwanathan Iyer
#### Problem 105. How to find the position of an element in a vector without using the find function
Created by: Chelsea
Tags indexing, find
#### Problem 157. The Hitchhiker's Guide to MATLAB
Created by: the cyclist
#### Problem 60. The Goldbach Conjecture
Created by: Cody Team
Tags primes
#### Problem 39. Which values occur exactly three times?
Created by: Cody Team
Tags search
#### Problem 115. Distance walked 1D
Created by: AMITAVA BISWAS
#### Problem 18. Bullseye Matrix
Created by: Cody Team
Tags matrices
#### Problem 16. Return the largest number that is adjacent to a zero
Created by: Cody Team
Tags vectors
#### Problem 29. Nearest Numbers
Created by: Cody Team
#### Problem 41. Cell joiner
Created by: Cody Team
Tags strings, matlab
#### Problem 120. radius of a spherical planet
Created by: AMITAVA BISWAS
#### Problem 304. Bottles of beer
Created by: Alan Chalker
#### Problem 158. Is my wife right? Now with even more wrong husband
Created by: the cyclist
#### Problem 132. given 3 sides, find area of this triangle
Created by: AMITAVA BISWAS
#### Problem 34. Binary numbers
Created by: Cody Team
Tags matlab
#### Problem 262. Swap the input arguments
Created by: Steve Eddins
#### Problem 247. Arrange Vector in descending order
Created by: Vishwanathan Iyer
#### Problem 624. Get the length of a given vector
Created by: Yudong Zhang
#### Problem 838. Check if number exists in vector
Created by: Nichlas
#### Problem 412. Back to basics 22 - Rotate a matrix
Created by: Alan Chalker
#### Problem 62. Elapsed Time
Created by: Cody Team
Tags time
#### Problem 164. Right and wrong
Created by: the cyclist
Tags triangle
#### Problem 128. Sorted highest to lowest?
Created by: AMITAVA BISWAS
#### Problem 28. Counting Money
Created by: Cody Team
Tags regexp, strings
#### Problem 94. Target sorting
Created by: Cody Team
Tags sorting, matlab
#### Problem 147. Too mean-spirited
Created by: the cyclist
Tags mean
#### Problem 64. The Goldbach Conjecture, Part 2
Created by: Cody Team
Tags primes
#### Problem 50. QWERTY coordinates
Created by: Cody Team
Tags indexing
#### Problem 45. Make a Palindrome Number
Created by: Cody Team
Tags strings
#### Problem 57. Summing Digits within Text
Created by: Cody Team
Tags strings
#### Problem 59. Pattern matching
Created by: Cody Team
Tags indexing
#### Problem 44. Trimming Spaces
Created by: Cody Team
#### Problem 38. Return a list sorted by number of occurrences
Created by: Cody Team
#### Problem 355. Back to basics 11 - Max Integer
Created by: Alan Chalker
#### Problem 1974. Length of a short side
Created by: Tanya Morton
#### Problem 40. Reverse Run-Length Encoder
Created by: Cody Team
Tags encoding
#### Problem 76. De-dupe
Created by: Cody Team
Tags matlab
1 – 50 of 109 | 1,060 | 4,055 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.722124 |
https://www.kidsacademy.mobi/printable-worksheets/answer-key/activities/normal/age-7/knowledge-check/math/ | 1,723,160,880,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640741453.47/warc/CC-MAIN-20240808222119-20240809012119-00302.warc.gz | 664,036,759 | 61,707 | # Normal Math Worksheets Activities With Answers for 7-Year-Olds
Explore our comprehensive collection of Normal Math Worksheets Activities With Answers, designed to challenge and engage students of all skill levels. Our meticulously crafted worksheets cover a wide range of mathematical concepts, ensuring a well-rounded understanding of the subject. Each activity is accompanied by detailed answers, allowing learners to self-assess their work and grasp complex ideas with greater ease. Whether you're a teacher looking to enrich your classroom resources or a student aiming to reinforce your math skills, our activities offer the perfect blend of challenge and support. Dive into our Normal Math Worksheets today and unlock a world of mathematical exploration!
Check out this Trial Lesson on Normal Math Worksheets Activities With Answers for 7-Year-Olds!
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## Multiplication Facts: Assessment 3 Worksheet
Test your kid's maths skills with this easy to use worksheet! Help them check the box that matches the equation in the first part, then read each word problem and underline the right answers to the second part. Assess your child's muliplication knowledge and find out where they need extra help.
Multiplication Facts: Assessment 3 Worksheet
Worksheet
## Counting and Numbers: Assessment Worksheet
Number line thinking is an essential math skill. Kids using this skill can compute math problems accurately and quickly. Our free assessment tests number line recognition; have your child pick the right number from the given options to follow the current number on the line. This assessment will give you an idea of their counting skills.
Counting and Numbers: Assessment Worksheet
Worksheet
## Place Value: Assessment 3 Worksheet
Test your child's math skills without them realizing it! This fun worksheet looks at the states and regions of the U.S. and your child can compare numbers greater or lesser than the other. It's the perfect way to assess your child's number sense without them knowing.
Place Value: Assessment 3 Worksheet
Worksheet
## Word Problems: Assessment 2 Worksheet
This bear-themed worksheet is a great way to test subtraction skills. Have your child read the word problems and match the correct drawing with the answer. It's a fun way to quiz them without them even knowing. Enjoy counting cute snoozing bears! (80 words)
Word Problems: Assessment 2 Worksheet
Worksheet
Learning Skills
Normal Math Worksheets Activities with Answers are an essential component of learning and understanding mathematics at any level. These resources offer a structured approach to tackling various mathematical concepts, from basic arithmetic to more complex algebra and geometry. The inclusion of answers with these activities is particularly beneficial for a multitude of reasons.
Firstly, Normal Math Worksheets Activities with Answers facilitate immediate feedback. Students can complete exercises and instantly verify their solutions, allowing them to understand their mistakes in real-time. This instant feedback loop is crucial for reinforcing correct mathematical methods and rectifying misunderstandings before they become ingrained.
Moreover, these worksheets serve as an invaluable tool for practice. Mathematics, much like any other skill, requires regular and consistent practice to master. Through a variety of problems that range in difficulty, students can gradually build up their confidence and competence in different areas of math. This practice also encourages the application of mathematical concepts in various contexts, promoting a deeper understanding.
Another significant advantage is the facilitation of independent learning. With answers available, students can engage with the material outside of the classroom without needing constant guidance from a teacher. This autonomy in learning fosters problem-solving skills and encourages students to take responsibility for their own education.
Furthermore, Normal Math Worksheets Activities with Answers are incredibly versatile. They can be used as a supplement to classroom instruction, for homework, or as part of a review session before exams. This adaptability makes them an invaluable resource for both teachers and students alike.
Lastly, these worksheets can cater to a wide range of learning styles. Whether a student prefers a more hands-on approach or learns best through repetition, the diverse nature of worksheet activities can accommodate different preferences, ensuring that all students have the opportunity to succeed in mathematics.
In conclusion, Normal Math Worksheets Activities with Answers are a fundamental resource in the mathematics education toolkit. They offer a structured yet flexible approach to learning that benefits students by providing immediate feedback, encouraging practice, fostering independence, and accommodating diverse learning styles. | 868 | 4,959 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-33 | latest | en | 0.888248 |
https://en.khanacademy.org/math/algebra-home/alg-modeling/alg-one-variable-modeling/v/constructing-an-exponential-equation-example | 1,702,314,947,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679515260.97/warc/CC-MAIN-20231211143258-20231211173258-00158.warc.gz | 258,773,273 | 134,684 | If you're seeing this message, it means we're having trouble loading external resources on our website.
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## Algebra (all content)
### Course: Algebra (all content)>Unit 15
Lesson 4: Modeling with one-variable equations & inequalities
# Exponential equation word problem
Sal models a context that concerns a bank savings account. The model turns out to be an exponential equation. Created by Sal Khan.
## Want to join the conversation?
• Why does sal say 1.2 after minute 1? I get him saying .2 as that comes from 20% but why 1.2?
• This is because he is just adding the 20 percent change to the original hundred percent to find out the answer. He is finding 20 percent of 6250 and then adding it to 6250 (which would be 100 percent) which becomes 120 percent or 1.2. This is the best explanation I can give, sorry if you didn’t understand. If you didn’t get this you should go and check out the previous units, where he explains it better.
• Why (1.2)^r and not (1.2)^(r-1) ?
• I'm assuming you mean "t" when you say "r". The reason it's not (1.2)^(t-1) is because 6250 is multiplied by 1.2 at the end of EACH year, starting with the FIRST.
• I tried doing this utilizing the geometric series equation (Sn = a(1-r^n)/1-r, but I don't seem to get the same answer
• why when you multiply the percent, why wouldn't it just be (.20)? Why add a one? That's whats just like throwing me off. Like where is this 1 coming from?
• If you multiply it by 0.2 you end up with 1/5th of what you began with. It's supposed to increase. The "1" represents what you started with that year, and the additional 0.2 represents the 20% increase that year. Because 0.2 is 20% of one.
• How do you identify if you have an exponential sequence?
• You know you have an exponential sequence if the ratio between consecutive terms is a constant. What does this mean? Let's do an example:
Here is a sequence:
1, 2, 2, 4, 8, 32, 256
Is it exponential? Find the ratios
2/1 = 2
2/2 = 1
4/2 = 2
8/4 = 2
32/8 = 4
256/32 = 8
The ratios are all two for the first couple terms, but then increase to 4 and 8, so this can't be exponential.
Here's another
1, 1.1, 1.21, 1.331, 1.4641, 1.61051
Is it exponential? Find the ratios.
1.1/1 = 1.1
1.21/1.1 = 1.1
1.331/1.21 = 1.1
1.4641/1.331 = 1.1
1.61051/1.4641 = 1.1
All the ratios are the same, so the sequence must be exponential.
Basically, divide a term in a sequence by the previous term. If the quotient is always the same number, then you have an exponential sequence.
(Note: the quotient could be fractional and even negative. 1, -0.5, 0.25. -0.125, 0.0625. Here, the quotient is always -0.5)
Just for fun, can you tell me what the rule is for the first sequence I gave?
• Is this the fastest way:
6250(6/5)^t = 12960
(6^t)/(5^t) = 1296/625 = (6^4)/(5^4)
(6/5)^t = (6/5)^4
t=4
• That could be the fastest way for many people, and it's extra nice because you don't necessarily need a calculator to solve it. Sal's way works best for me (since I use a calculator), but any way is fine if it gives you the correct answer.
• why does my calculator give log 12960 as 4.11261, whereas Sal calculates it as 2.0736? I know log to the base 10 would give us log 10,000 is 4.....
• The 2.0736 comes from 12960/6250. The total money in the account after t years can be modeled by A=6250*1.2^t. We want to know how long it will take for the account to reach 12960. Therefore, we plug in the 12960 into the equation: (12960=6250*1.2^t) After dividing and isolating t we are left with: (2.0736=1.2^t) This is where Sal uses the log function to solve for t.
• At why does it not matter what base logarithm you use?
• I need help on this question:
4^x-3*2^x+1+8<0
• The starting equation is:
4^(x-3)*2^(x+1)+8<0
Since 4=2^2, we will add it in
2^(2)^(x-3)*2^(x+1)+8<0
We multiply the 2 and x-3 since it is a power to a power.
2^(2x-6)*2^(x+1)+8<0
Then we add 2x+6 and x+1.
2^(3x-5)+8<0
From here we subtract 8 from both sides
2^(3x-5)+8-8<0-8
2^(3x+5)<-8
If f(x)<g(x) then In(f(x))<In(g(x))
In(2^(3x+5))<In(-8)
Then we apply the log rule:
(3x+5)In(2)<In(-8)
I don't know what to do from here, so I just plugged it in a calculator, and I got...
No Solution.
And we are done. | 1,387 | 4,296 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2023-50 | latest | en | 0.950141 |
http://math.stackexchange.com/questions/295050/roadmap-to-study-atiyah-singer-index-theorem?answertab=active | 1,406,365,731,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997900573.25/warc/CC-MAIN-20140722025820-00019-ip-10-33-131-23.ec2.internal.warc.gz | 242,520,771 | 20,177 | # Roadmap to study Atiyah-Singer index theorem
I am a physics undergrad and want to pursue a PhD in Math (geometry or topology). I study it almost completely by myself, as the program in my country offers very less flexibility to take non departmental courses, thought I will be able to take them is a couple of semesters. Anyways, So I was thinking about taking a mini-project of sorts which I can self-study or maybe ask a prof. I had heard about this theorem while studying topological solitons in Physics.
What are the prerequisites for studying the Atiyah-Singer theorem. I had a look online, but I couldn't figure out exactly, in what field this is, or what are the prerequisites. Wikipedia says it is a theorem in differential geometry, but obviously what differential geometry I know is insufficient. I know about manifolds, differential forms, lie derivatives, lie groups, and some killing vectors stuff. A look at the material online also talks about some operators in DG which I have never come across. Are there an algebraic topology prerequisites?
Please could you recommend me some references which can take me there?
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You should get a copy of Bertlmann's Anomalies in QFT textbook. See books.google.com/books/about/… .As a physics student it should appeal to you and there is a great swath of math covered at a very speedy pace. – James S. Cook Feb 5 '13 at 2:45
Seminar on the Atiyah-Singer Index Theorem is certainly a good source. I didn't read it in full (I only read the first few chapters) and I remember it contains a very good account on Fredholm theory – Olivier Bégassat Feb 5 '13 at 2:50
@JamesS.Cook: Hey. That book looks really cool! Thanks. – ramanujan_dirac Feb 5 '13 at 11:09
The Atiyah-Singer index theorem involves a mixture of algebra, geometry/topology, and analysis. Here are the main things you'll want to understand to be able to know what the index theorem is really even saying.
1. Algebra: The most important concept here is Clifford algebras. For example, Dirac operators arise from combining covariant differentiation and Clifford multiplication. You'll also want to learn about the associated spin groups.
2. Geometry/Topology: The most fundamental idea here is understanding vector bundles. The Atiyah-Singer index theorem is about elliptic differential operators between sections of vector bundles, so you won't get anywhere without a firm understanding of bundles. Next, you'll want to understand the basics of spin geometry and Dirac operators, especially if your interests are more physics-based. One nice form of the index theorem is the cohomological formula for the index in terms of characteristic classes. I would advise you to get familiar with cohomology and obtain a basic knowledge of characteristic classes (Chern-Weil theory is nice if you already have a geometry background). If you want to understand the original index theorem or how the cohomological formula for the index is derived from it, you'll need to learn some topological K-theory.
3. Analysis: Here the big topic is differential operators in the context of manifolds. Hence you'll want to know what a differential operator between vector bundles is. You'll need to know what the symbol of such a differential operator is, and also what it means for such a differential operator to be elliptic.
As for a reference, the classic text is Spin Geometry by Lawson and Michelsohn. An easier but less detailed introduction can be found in the relevant sections of Geometry, Topology, and Physics by Nakahara. There's a fair amount of other books but these are the two I know best. There are some good notes on spin geometry here. These notes by Nicolaescu may also be of interest to you.
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+1 Very detailed. Good references and links. – Jesse Madnick Feb 5 '13 at 3:08
@Henry: Nice answer! Do you know the books by Naber? Would you recommend them over Nakahara or Lawson? – ramanujan_dirac Feb 5 '13 at 16:39
@ramanujan_dirac: Naber's two volumes of Topology, Geometry, and Gauge Fields (which I assume you are talking about?) don't cover any index theory. They do discuss some of the preliminaries I listed above, but the development is quite slow and elementary. I think you're better off with Nakahara for general background and a quick introduction and Lawson and Michelsohn for the full treatment. – Henry T. Horton Feb 5 '13 at 16:44
@HenryT.Horton: Ok Though I think Naber, is much more detailed about the elementary topics than Nakahara. I find Nakahara very superficial.BTW, +1 – ramanujan_dirac Feb 6 '13 at 2:32
There is a new wonderful, big and deep book dealing precisely with everything you need (assuming knowledge of "basic" differential geometry):
These are the same authors of the following old classic, but the new book is NOT a new edition but a completely NEW title (also with modern TeX typeset, figures and additional topics and details):
You may also find in this other answer a very brief overview of the proof of Atiyah-Singer and why pseudo-differential operators are needed. There you will find links to lectures and other titles regarding the theorem.
-
There are two main approaches to the index theorem: $K$-theory and heat kernels. The latter approach treats the index of Dirac operators, which is actually more general than it sounds since for common situations all operators (in terms of index theory) are twisted Dirac operators.
Based on your background I would recommend the heat kernel approach since it is more geometric, physicsy, and you'd have to learn a decent amount of algebraic topology for the $K$-theory approach.
There is some algebraic topology that enters into both approaches and this is the theory of characteristic classes. In the heat kernel approach, one takes the Chern-Weil viewpoint, which is pretty easy to pick up as long as you are familiar with de Rham cohomology.
I found the most useful books to be Berline, Getzler, and Vergne's Heat Kernels and Dirac Operators and John Roe's Elliptic Operators, Topology, and Asymptotic Methods.
For the $K$-theory approach (which I think is useful to at least get a bit of an understanding of) I like the original papers as well as expository works of Nigel Higson e.g. Lectures on Operator K-Theory and the Atiyah-Singer Index.
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Paul Loya's Notes are introductory and follow the heat kernel approach. – Robert Jan 29 at 16:02 | 1,436 | 6,374 | {"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} | 2.78125 | 3 | CC-MAIN-2014-23 | longest | en | 0.964072 |
https://www.geeksforgeeks.org/generate-a-permutation-of-0-n-1-with-maximum-adjacent-xor-which-is-minimum-among-other-permutations/?ref=rp | 1,685,411,063,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644915.48/warc/CC-MAIN-20230530000715-20230530030715-00398.warc.gz | 860,297,897 | 37,202 | GeeksforGeeks App
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# Generate a permutation of [0, N-1] with maximum adjacent XOR which is minimum among other permutations
Given an integer N, the task is to print a permutation of numbers from 0 to N-1, such that:
• There is no duplicate element in the permutation
• The maximum adjacent XOR of this permutation is minimum among other permutations
• There can be more than one permutation present that satisfies these conditions.
Examples:
Input: N = 5
Output: 1 2 3 0 4
Explanation:
The maximum XOR value of two consecutive elements of the permutation is 0 XOR 4 = 4
No other permutation exists where the maximum XOR value of two consecutive elements of the permutation is less than 4.
Other permutations may exist where the value is equal to or greater than 4.
Input: N = 10
Output: 1 2 3 4 5 6 7 0 8 9
Approach: The intuition to solve this problem is based on below mentioned properties of XOR:
• Inside the entire permutation, there will be some numbers with the most significant bit as 1 or as 0.
• So group the numbers with the most significant bit as 1 so the most significant bit gets cancelled out.
• But there will always be a case in the permutation where a number with the most significant bit as 1 will reside before or after a number where the most significant bit is 0. At that point, the most significant bit will not be cancelled by the XOR operation.
• In that case, place the minimum number with the most significant bit as 1 and the minimum number with the most significant bit as 0(possibly 0), together to get the maximum XOR value.
• This is the possible minimum value of maximum-XOR value among all permutations.
Follow the steps mentioned below to implement the above observation:
• Initially, calculate the minimum number less than or equal to N-1 having the most significant set bit.
• Print all the numbers less than the minimum number with the most significant set bit starting from 1 consecutively.
• Print 0.
• Print all the numbers starting from a minimum number with a most significant set bit up to N-1
Below is the implementation of the above approach:
## C++
`// C++ program to implement above approach``#include ``using` `namespace` `std;` `// Function to get the desired permutation``void` `getPermutation(``int` `n)``{`` ``// Calculate the maximum number`` ``// in the permutation`` ``int` `maxnum = n - 1;` ` ``// Calculate the minimum number`` ``// bit is 1 where most significant`` ``int` `num = 1;`` ``while` `(maxnum > 0) {`` ``maxnum /= 2;`` ``num *= 2;`` ``}`` ``num /= 2;` ` ``// Print all numbers less than the number`` ``// where most significant bit is set`` ``for` `(``int` `i = 1; i <= num - 1; i++) {`` ``cout << i << ``" "``;`` ``}` ` ``// Print 0`` ``cout << 0 << ``" "``;` ` ``// Print all the numbers`` ``// greater than or equal to`` ``// the number where`` ``// most significant bit is set`` ``for` `(``int` `i = num; i < n; i++) {`` ``cout << i << ``" "``;`` ``}``}` `// Driver Code``int` `main()``{`` ``int` `N = 5;`` ` ` ``// Function call`` ``getPermutation(N);`` ``return` `0;``}`
## Java
`// JAVA program to implement above approach``import` `java.util.*;``class` `GFG``{`` ` ` ``// Function to get the desired permutation`` ``public` `static` `void` `getPermutation(``int` `n)`` ``{`` ` ` ``// Calculate the maximum number`` ``// in the permutation`` ``int` `maxnum = n - ``1``;` ` ``// Calculate the minimum number`` ``// bit is 1 where most significant`` ``int` `num = ``1``;`` ``while` `(maxnum > ``0``) {`` ``maxnum /= ``2``;`` ``num *= ``2``;`` ``}`` ``num /= ``2``;` ` ``// Print all numbers less than the number`` ``// where most significant bit is set`` ``for` `(``int` `i = ``1``; i <= num - ``1``; i++) {`` ``System.out.print(i + ``" "``);`` ``}` ` ``// Print 0`` ``System.out.print(``0` `+ ``" "``);` ` ``// Print all the numbers`` ``// greater than or equal to`` ``// the number where`` ``// most significant bit is set`` ``for` `(``int` `i = num; i < n; i++) {`` ``System.out.print(i + ``" "``);`` ``}`` ``}` ` ``// Driver Code`` ``public` `static` `void` `main(String[] args)`` ``{`` ``int` `N = ``5``;` ` ``// Function call`` ``getPermutation(N);`` ``}``}` `// This code is contributed by Taranpreet`
## Python3
`# Python code for the above approach` `# Function to get the desired permutation``def` `getPermutation(n):` ` ``# Calculate the maximum number`` ``# in the permutation`` ``maxnum ``=` `n ``-` `1` ` ``# Calculate the minimum number`` ``# bit is 1 where most significant`` ``num ``=` `1`` ``while` `(maxnum > ``0``):`` ``maxnum ``=` `maxnum``/``/``2`` ``num ``*``=` `2` ` ``num ``=` `(num``/``/``2``)` ` ``# Print all numbers less than the number`` ``# where most significant bit is set`` ``for` `i ``in` `range``(``1``, num):`` ``print``(i, end``=``" "``)` ` ``# Print 0`` ``print``(``0``, end``=``" "``)` ` ``# Print all the numbers`` ``# greater than or equal to`` ``# the number where`` ``# most significant bit is set`` ``for` `i ``in` `range``(num, n):`` ``print``(i, end``=``" "``)` `# Driver Code``N ``=` `5` `# Function call``getPermutation(N)` `# This code is contributed by Potta Lokesh`
## C#
`// C# program to implement above approach``using` `System;``class` `GFG``{`` ` ` ``// Function to get the desired permutation`` ``static` `void` `getPermutation(``int` `n)`` ``{`` ` ` ``// Calculate the maximum number`` ``// in the permutation`` ``int` `maxnum = n - 1;` ` ``// Calculate the minimum number`` ``// bit is 1 where most significant`` ``int` `num = 1;`` ``while` `(maxnum > 0) {`` ``maxnum /= 2;`` ``num *= 2;`` ``}`` ``num /= 2;` ` ``// Print all numbers less than the number`` ``// where most significant bit is set`` ``for` `(``int` `i = 1; i <= num - 1; i++) {`` ``Console.Write(i + ``" "``);`` ``}` ` ``// Print 0`` ``Console.Write(0 + ``" "``);` ` ``// Print all the numbers`` ``// greater than or equal to`` ``// the number where`` ``// most significant bit is set`` ``for` `(``int` `i = num; i < n; i++) {`` ``Console.Write(i + ``" "``);`` ``}`` ``}` ` ``// Driver Code`` ``public` `static` `void` `Main()`` ``{`` ``int` `N = 5;` ` ``// Function call`` ``getPermutation(N);`` ``}``}` `// This code is contributed by Samim Hossain Mondal.`
## Javascript
``
Output
`1 2 3 0 4 `
Time Complexity: 0(N)
Auxiliary Space: 0(1)
My Personal Notes arrow_drop_up | 2,188 | 6,998 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.90807 |
https://socratic.org/questions/how-do-you-find-the-exact-value-of-10-logpi | 1,586,512,152,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371893683.94/warc/CC-MAIN-20200410075105-20200410105605-00283.warc.gz | 666,747,635 | 5,884 | # How do you find the exact value of 10^(logpi)?
${10}^{\log \pi} = \pi$
By definition $\log a$ is a number such that ${10}^{\log a} = a$
So in our example: ${10}^{\log \pi} = \pi$ | 69 | 181 | {"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} | 3.71875 | 4 | CC-MAIN-2020-16 | longest | en | 0.756555 |
https://practicaldev-herokuapp-com.global.ssl.fastly.net/harshrajpal/2244-minimum-rounds-to-complete-all-tasks-4cea | 1,696,199,381,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510941.58/warc/CC-MAIN-20231001205332-20231001235332-00473.warc.gz | 511,458,244 | 19,704 | Harsh Rajpal
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# 2244. Minimum Rounds to Complete All Tasks
You are given a 0-indexed integer array tasks, where tasks[i] represents the difficulty level of a task. In each round, you can complete either 2 or 3 tasks of the same difficulty level.
Return the minimum rounds required to complete all the tasks, or -1 if it is not possible to complete all the tasks.
Example 1:
Output: 4
Explanation: To complete all the tasks, a possible plan is:
• In the first round, you complete 3 tasks of difficulty level 2.
• In the second round, you complete 2 tasks of difficulty level 3.
• In the third round, you complete 3 tasks of difficulty level 4.
• In the fourth round, you complete 2 tasks of difficulty level 4. It can be shown that all the tasks cannot be completed in fewer than 4 rounds, so the answer is 4.
Example 2:
Output: -1
Explanation: There is only 1 task of difficulty level 2, but in each round, you can only complete either 2 or 3 tasks of the same difficulty level. Hence, you cannot complete all the tasks, and the answer is -1.
Constraints:
``````public class Solution {
HashMap<Integer, Integer> hm = new HashMap<Integer, Integer>();// Creating HashMap
int a = 0;
int c = 0;
for (int i = 0; i < tasks.length; i++)// Traversing array
{
// Checking if the element is present
a = 0;
} else // If element is not present then add it to the HashMap
{
}
}
// Iterating over the HashMap to count the minimum number of rounds or return -1
// if it is not possible to complete all the tasks.
for (int z : hm.keySet()) {
int d = hm.get(z);
if (d == 1) {
return -1;
} else if (d == 2) {
c += 1;
} else if (d % 3 == 0) {
c += d / 3;
} else {
c += d / 3 + 1;
}
}
return c;// returning the minimum number of rounds.
// Time Complexity: O(n)
// Space Complexity: O(n)
// End of the program.
}
}
`````` | 500 | 1,824 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2023-40 | latest | en | 0.776516 |
https://convertoctopus.com/421-ounces-to-pounds | 1,623,829,715,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487622234.42/warc/CC-MAIN-20210616063154-20210616093154-00303.warc.gz | 165,678,813 | 7,817 | ## Conversion formula
The conversion factor from ounces to pounds is 0.0625, which means that 1 ounce is equal to 0.0625 pounds:
1 oz = 0.0625 lb
To convert 421 ounces into pounds we have to multiply 421 by the conversion factor in order to get the mass amount from ounces to pounds. We can also form a simple proportion to calculate the result:
1 oz → 0.0625 lb
421 oz → M(lb)
Solve the above proportion to obtain the mass M in pounds:
M(lb) = 421 oz × 0.0625 lb
M(lb) = 26.3125 lb
The final result is:
421 oz → 26.3125 lb
We conclude that 421 ounces is equivalent to 26.3125 pounds:
421 ounces = 26.3125 pounds
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 pound is equal to 0.038004750593824 × 421 ounces.
Another way is saying that 421 ounces is equal to 1 ÷ 0.038004750593824 pounds.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that four hundred twenty-one ounces is approximately twenty-six point three one three pounds:
421 oz ≅ 26.313 lb
An alternative is also that one pound is approximately zero point zero three eight times four hundred twenty-one ounces.
## Conversion table
### ounces to pounds chart
For quick reference purposes, below is the conversion table you can use to convert from ounces to pounds
ounces (oz) pounds (lb)
422 ounces 26.375 pounds
423 ounces 26.438 pounds
424 ounces 26.5 pounds
425 ounces 26.563 pounds
426 ounces 26.625 pounds
427 ounces 26.688 pounds
428 ounces 26.75 pounds
429 ounces 26.813 pounds
430 ounces 26.875 pounds
431 ounces 26.938 pounds | 437 | 1,661 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2021-25 | latest | en | 0.828236 |
https://numberworld.info/root-of-2020121 | 1,653,067,998,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662533972.17/warc/CC-MAIN-20220520160139-20220520190139-00573.warc.gz | 490,316,634 | 2,964 | # Root of 2020121
#### [Root of two million twenty thousand one hundred twenty-one]
square root
1421.3096
cube root
126.4132
fourth root
37.7003
fifth root
18.2421
In mathematics extracting a root is declared as the determination of the unknown "x" in the equation $y=x^n$ The result of the extraction of the root is considered as a root. In the case of "n is 2", one talks about a square root or sometimes a second root also, another possibility could be that n = 3 then one would declare it a cube root or simply third root. Considering n beeing greater than 3, the root is declared as the fourth root, fifth root and so on.
In maths, the square root of 2020121 is represented as this: $$\sqrt[]{2020121}=1421.3096073692$$
Moreover it is legit to write every root down as a power: $$\sqrt[n]{x}=x^\frac{1}{n}$$
The square root of 2020121 is 1421.3096073692. The cube root of 2020121 is 126.41321047769. The fourth root of 2020121 is 37.700260043788 and the fifth root is 18.242127073425.
Look Up | 293 | 1,000 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2022-21 | latest | en | 0.909644 |
https://cambridgeaudio.ru/problem-solving-strategies-display-3414.html | 1,623,575,838,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487607143.30/warc/CC-MAIN-20210613071347-20210613101347-00153.warc.gz | 162,711,719 | 8,306 | # Problem Solving Strategies Display
In some problems though, where there are more variables, it may not be clear at first which way to change the guessing.2 Act It Out We put two strategies together here because they are closely related. In the Farmyard problem, the children might take the role of the animals though it is unlikely that you would have 87 children in your class!But if there are not enough children you might be able to press gang the odd teddy or two. It is an effective strategy for demonstration purposes in front of the whole class.
In some problems though, where there are more variables, it may not be clear at first which way to change the guessing.2 Act It Out We put two strategies together here because they are closely related. In the Farmyard problem, the children might take the role of the animals though it is unlikely that you would have 87 children in your class!But if there are not enough children you might be able to press gang the odd teddy or two. It is an effective strategy for demonstration purposes in front of the whole class.
Tags: Essay ReligiousUnforgettable Moments In My Life EssaysNew Application EssayCritical Thinking Paper ApaErp Implementation Case StudyAyn Rand Essay Competition
This is because the participants are so engrossed in the mechanics of what they are doing that they don’t see through to the underlying mathematics.
However, because these children are concentrating on what they are doing, they may in fact get more out of it and remember it longer than the others, so there are pros and cons here. Generally speaking, any object that can be used in some way to represent the situation the children are trying to solve, is equipment.
In our experience, children need to be encouraged and helped to use equipment. This may be because it gives them a better representation of the problem in hand.
Also, if they’re a little older, they may feel that using equipment is only 'for babies'.
Common Problem Solving Strategies We have provided a copymaster for these strategies so that you can make posters and display them in your classroom.
It consists of a page per strategy with space provided to insert the name of any problem that you come across that uses that particular strategy (Act it out, Draw, Guess, Make a List).We have found that this kind of poster provides good revision for children. Through these links, children can see that mathematics is not only connected by skills but also by processes.We now look at each of the following strategies and discuss them in some depth. If they can also check that the guess fits the conditions of the problem, then they have mastered guess and check.If you are not careful, they may try to use it all the time.As problems get more difficult, other strategies become more important and more effective.On the other hand, it can also be cumbersome when used by groups, especially if a largish number of students is involved.We have, however, found it a useful strategy when students have had trouble coming to grips with a problem.However, sometimes when children are completely stuck, guessing and checking will provide a useful way to start and explore a problem.Hopefully that exploration will lead to a more efficient strategy and then to a solution.In this site we have linked the problem solving lessons to the following groupings of problem solving strategies.As the site develops we may add some more but we have tried to keep things simple for now. | 689 | 3,481 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2021-25 | latest | en | 0.947579 |
dimensional.askdefine.com | 1,545,024,989,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828318.79/warc/CC-MAIN-20181217042727-20181217064727-00085.warc.gz | 553,526,348 | 6,530 | # Dictionary Definition
1 of or relating to dimensions
2 having dimension--the quality or character or stature proper to a person; "never matures as a dimensional character; he is pasty, bland, faceless"- Norman Cousins
# User Contributed Dictionary
## English
1. Of or pertaining to dimensions.
# Extensive Definition
Classical physics theories describe three physical dimensions: from a particular point in space, the basic directions in which we can move are up/down, left/right, and forward/backward. Movement in any other direction can be expressed in terms of just these three. Moving down is the same as moving up a negative amount. Moving diagonally upward and forward is just as the name of the direction implies; i.e., moving in a linear combination of up and forward. In its simplest form: a line describes one dimension, a plane describes two dimensions, and a cube describes three dimensions. (See Space and Cartesian coordinate system.)
### Time
Time is often referred to as the "fourth dimension". It is one way to measure physical change. It is perceived differently from the three spatial dimensions in that there is only one of it, that movement in time occurs at the fixed rate of one second per second, and that we cannot move freely in time but subjectively move in one direction.
The equations used in physics to model reality do not treat time in the same way that humans perceive it. The equations of classical mechanics are symmetric with respect to time, and equations of quantum mechanics are typically symmetric if both time and other quantities (such as charge and parity) are reversed. In these models, the perception of time flowing in one direction is an artifact of the laws of thermodynamics (we perceive time as flowing in the direction of increasing entropy).
The best-known treatment of time as a dimension is Poincaré and Einstein's special relativity (and extended to general relativity), which treats perceived space and time as components of a four-dimensional manifold, known as spacetime, and in the special, flat case as Minkowski space.
Theories such as string theory and M-theory predict that physical space in general has in fact 10 and 11 dimensions, respectively. The extra dimensions are spacelike. We perceive only three spatial dimensions, and no physical experiments have confirmed the reality of additional dimensions. A possible explanation that has been suggested is that space is as it were "curled up" in the extra dimensions on a very small, subatomic scale, possibly at the quark/string level of scale or below.
### Penrose's singularity theorem
In his book The Road to Reality: A Complete Guide to the Laws of the Universe, scientist Sir Roger Penrose explained his singularity theorem. It asserts that all theories that attribute more than three spatial dimensions and one temporal dimension to the world of experience are unstable. The instabilities that exist in systems of such extra dimensions would result in their rapid collapse into a singularity. For that reason, Penrose wrote, the unification of gravitation with other forces through extra dimensions cannot occur.
### Dimensionful quantities
In the physical sciences and in engineering, the dimension of a physical quantity is the expression of the class of physical unit that such a quantity is measured against. The dimension of speed, for example, is LT−1, that is, length divided by time. The units in which the quantity is expressed, such as ms−1 (meters per second) or mph (miles per hour), has to conform to the dimension.
## Science fiction
Science fiction texts often mention the concept of dimension, when really referring to parallel universes, alternate universes, or other planes of existence. This usage is derived from the idea that in order to travel to parallel/alternate universes/planes of existence one must travel in a spatial direction/dimension besides the standard ones. In effect, the other universes/planes are just a small distance away from our own, but the distance is in a fourth (or higher) spatial dimension, not the standard ones.
One of the most heralded science fiction novellas regarding true geometric dimensionality, and often recommended as a starting point for those just starting to investigate such matters, is the 1884 novel Flatland by Edwin A. Abbott. Isaac Asimov, in his foreword to the Signet Classics 1984 edition, described Flatland as "The best introduction one can find into the manner of perceiving dimensions."
## References
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dimensional in Thai: มิติ
dimensional in Urdu: بُعد (لکیری الجبرا)
dimensional in Yiddish: דימענסיע
dimensional in Chinese: 維度 | 1,273 | 5,775 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-51 | latest | en | 0.941978 |
https://leanprover-community.github.io/archive/stream/113488-general/topic/greater.20than.20nat.20and.20real.html | 1,696,331,441,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511075.63/warc/CC-MAIN-20231003092549-20231003122549-00735.warc.gz | 385,013,193 | 4,891 | ## Stream: general
### Topic: greater than nat and real
#### Saif Ghobash (Dec 29 2020 at 13:39):
Im new to lean and am trying to formalize some theorems from my lecture notes on real analysis, one of them being
-- ℕ ⊂ ℝ is not bounded above.
theorem theorem_1_1_1 : ∀ x : ℝ, ∃ n : ℕ, n > x :=
begin
end
however I am getting a type error
type mismatch at application
n > x
term
x
has type
ℝ
but is expected to have type
ℕ
I am not sure how to cast n : ℕ to be a real number so I can apply > or use some form of > that accepts ℕ and ℝ.
#### Johan Commelin (Dec 29 2020 at 13:43):
@Saif Ghobash This is because Lean is reading "outside in, left-to-right"...
#### Patrick Massot (Dec 29 2020 at 13:43):
-- ℕ ⊂ ℝ is not bounded above.
theorem theorem_1_1_1 : ∀ x : ℝ, ∃ n : ℕ, (n : ℝ) > x :=
begin
end
#### Patrick Massot (Dec 29 2020 at 13:43):
Note that, as Johan points out, you could also write:
-- ℕ ⊂ ℝ is not bounded above.
theorem theorem_1_1_1 : ∀ x : ℝ, ∃ n : ℕ, x < n :=
begin
end
#### Johan Commelin (Dec 29 2020 at 13:43):
So it sees ">" and then "n", and determines you must be comparing two natural numbers, because n is a nat.
#### Johan Commelin (Dec 29 2020 at 13:44):
Patrick just posted the two ways around this (-;
#### Patrick Massot (Dec 29 2020 at 13:44):
Because of order of elaboration, in the second version Lean knows it expects a real when meeting n.
#### Saif Ghobash (Dec 29 2020 at 13:46):
Thank you @Patrick Massot and @Johan Commelin !
#### Johan Commelin (Dec 29 2020 at 13:48):
@Saif Ghobash Also, note that src/data/real/basic.lean contains
200:instance : archimedean ℝ :=
201:archimedean_iff_rat_le.2 $λ x, quotient.induction_on x$ λ f,
202-let ⟨M, M0, H⟩ := f.bounded' 0 in
203-⟨M, mk_le_of_forall_le ⟨0, λ i _,
204- rat.cast_le.2 \$ le_of_lt (abs_lt.1 (H i)).2⟩⟩
#### Johan Commelin (Dec 29 2020 at 13:49):
That exactly gives you the result you are looking for.
#### Johan Commelin (Dec 29 2020 at 13:50):
You can use the lemma exists_nat_gt because that instance exists.
#### Saif Ghobash (Dec 29 2020 at 14:51):
@Johan Commelin oh cool! that's exactly what I needed, thank you :)
#### Johan Commelin (Dec 29 2020 at 14:55):
@Saif Ghobash most likely 90-95% of the theorems in an introduction to analysis course are already in mathlib (in one form or another). On the other hand, this doesn't mean that all the exercises in such a course will be trivial to do using mathlib.
#### Sebastian Ullrich (Dec 30 2020 at 10:12):
Johan Commelin said:
So it sees ">" and then "n", and determines you must be comparing two natural numbers, because n is a nat.
#### Johan Commelin (Dec 30 2020 at 10:26):
@Sebastian Ullrich Thanks, as always (-;
#### Sebastian Ullrich (Dec 30 2020 at 10:47):
But I didn't even do anything :big_smile:
#### Kevin Buzzard (Dec 30 2020 at 12:40):
One could intentionally or unintentionally abuse this with some random coercion you didn't know was there, which might change the meaning of a statement. This should be fun! Does it mean that (a : T) < b no longer guarantees that < will be T.has_lt?
#### Kevin Buzzard (Dec 30 2020 at 12:41):
Having said that it would make the Bernoulli code look better! So in a real use case it does look like a good idea
#### Kevin Buzzard (Dec 30 2020 at 12:42):
I'm multiplying and dividing what look like naturals, and all of a sudden there's a Bernoulli number at the end and I didn't want to put it at the start but it's rational
Last updated: Aug 03 2023 at 10:10 UTC | 1,077 | 3,520 | {"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} | 3.421875 | 3 | CC-MAIN-2023-40 | latest | en | 0.842394 |
https://econbrowser.com/archives/2014/02/assessing-the-widening-wisconsin-us-gap | 1,713,043,657,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816853.44/warc/CC-MAIN-20240413211215-20240414001215-00864.warc.gz | 201,469,915 | 18,741 | # Assessing the Widening Wisconsin-US Gap
The cumulative growth gap since January 2011 between the US and Wisconsin coincident indices was 2.2% in December. The forecast indicates a continued widening to 2.8% in six months time.
Figure 1 depicts the log coincident indices for Wisconsin and the US.
Figure 1: Log coincident indices for Wisconsin (blue) and US (gray), and levels implied by leading indices for 2014M06. NBER defined recession dates shaded gray. Source: Philadelphia Fed, NBER, and author’s calculations.
Reader Ed Hanson wonders about the predictive characteristics of the leading indices produced by the Philadelphia Fed. Figure 2 depicts the actual and forecasted six-month growth rates for Wisconsin.
Figure 2: Actual six-month growth rate (non-annualized) in WI coincident index (blue) and corresponding leading index (red). NBER defined recession dates shaded gray. Source: Philadelphia Fed, NBER, and author’s calculations.
A regression of the actual six-month growth rate on the forecasted yields the following results:
Table 1: Unbiasedness test.
Note that the leading index appears to be an unbiased predictor of actual six-month growth rates, since the point estimate of the slope coefficient is 1.001, with a Newey-West robust standard error of 0.058; hence one would not reject a null hypothesis of a unit coefficient. The adjusted R2 is 0.84 — in other words 84% of the variation of the six-month growth rate around the mean is explained by the leading index. Note that this is not a “real time” analysis, as the coincident series are the revised, rather than initial, figures, and one might wish to compare the forecasts to the initial.
The leading index for the United States shares the same characteristics of unbiasedness; the adjusted-R2 is 0.93.
Has the behavior of the Wisconsin index changed over time? The log ratio (WI/US) rose from 1982 to a peak in 2000 before descending. Estimating the relationship between the (log) WI coincident index and the (log) US coincident index using dynamic OLS (6 leads and lags of the first differenced right hand side variable, 2001M01-13M12) and a dummy taking a value of one from 2011M01-2013M12 yields a cointegrating coefficient of 0.507, and the coefficient on the dummy is 0.021, significant at the .0002 level using Newey-West standard errors. Literally interpreted, Wisconsin economic activity is 2% lower than usual, starting from 2011M01.
This is not a conclusive finding; using a different specification, sample period, or estimation method would likely lead to a different finding.
Update, 2/8 noon Pacific: Reader Ed Hanson still contends that Wisconsin economic performance has improved since 2011M01. He writes:
What we mainly disagree with, the Walker policies have began the change back toward superior economic performance. I say it has, and comparing the performance of Wisconsin from 2001 until 2011, with Wisconsin after 2011, supports my position. In addition, Walker policy of reduced taxation, both personal and business, since 2013 will accelerate economic performance.
Since Mr. Hanson is not persuaded by a coefficient on a dummy variable, I have undertaken another analysis, which conditions on US economic activity, and lagged WI economic activity. I estimate a error correction model over the 2001M01-2010M12 period:
Table 2:Error correction model, WI and US economic activity.
Where LCOINWI (LCOINUS) is the log coincident index for Wisconsin (United States), “D” indicates first difference. Note the high adjusted-R2, Q(6) and Q(12) statistics fail to reject the no-serial correlation null, as does the Breusch-Godfrey LM (2 lags) test.
The implied long run cointegrating coefficient is approximately 0.50; that is a 1 percent increase in the US coincident index is associated with a 0.5 percent increase in the Wisconsin index, over the 2001M01-2010M12 period.
I use this error correction model to forecast out of sample (2011M01-2013M12), conditioning on actual US economic activity. The forecast is shown in Figure 3, with plus/minus two standard errors, compared against the actual.
Figure 3: Log actual WI coincident index (blue), forecast (red) and plus/minus two standard errors (pink). Forecast based on error correction model (Table 2). Dashed vertical line at beginning of Walker administration. NBER defined recession dates shaded gray. Source: Philadelphia Fed, NBER and author’s calculations.
Notice that the actual Wisconsin index under-performs the forecasted (what we would call the counterfactual, if the usual economic relationship between the US and WI economies prevailed into 2013) by a significant amount — by up to 2.5% (log terms) earlier in 2013, and 1.9% as of 2013M12. The fact that the blue line (actual) is outside of the plus/minus two standard error bands suggests that the under-performance did not happen by accident, at conventional significance levels.
## 17 thoughts on “Assessing the Widening Wisconsin-US Gap”
1. samuel
Walker is pretty quiet about his economic and jobs performance. Too busy running around the country collecting huge donations to do anything for the people of Wisconsin.
2. Ricardo
Another schadenfreude post by Wisconsin’s self-appointed critic. It is sad when those who do not live in Wisconsin have to defend it agains those who do even when they find their taxes reduced because of effective government.. Why do such people still live in a state they hate?
BMO forecasts growth for Wisconsin economy
Excerpt:
The labor market has improved in recent months, with employment up 1.6 percent year-over-year in December, the fastest increase in more than eight years, the report stated.
“In particular, manufacturing and leisure and hospitality jobs have experienced a major comeback, the latter surging 5 percent in the last year. Manufacturing employment is at its highest level since 2009,” said Robert Kavcic, senior economist, BMO Capital Markets. “Construction and financial-sector employment are still lagging, but overall the unemployment rate is at its lowest since late 2008 and should dip below 6 percent this year.”
Wisconsin budget surplus hits \$977 million
News of the surplus, expected for weeks but larger than many anticipated, will set off a feeding frenzy in the Capitol among lobbyists, special interest groups and lawmakers all trying to spend a piece of the pie.
“The additional revenue should be returned to taxpayers because it’s their money, and my administration will work with the Legislature to determine the most prudent course of action,” Walker said in a statement.
1. baffling
ricardo, my aren’t we defensive!
menzie is simply showing that wisconsin is lagging behind the overall us performance. hence its local performance below par may be attributable to local policy (aka walker) rather than federal policy (aka obama). since overall us performance is increasing, one would also expect that to happen in wisconsin as well. just saying it does not seem to be doing as well as the country as a whole.
3. Bruce Hall
Not sure that Wisconsin is all that unique with regard to it’s economy. It’s all what you measure. You can measure manufacturing, goods & services, and employment and probably get different pictures. What appears to be the case is that there is no significant labor improvement anywhere in the Midwest.
Wisconsin
2013
Labor Force Data
Civilian Labor Force (1)
3,077.3 3,071.2 3,071.7 3,065.1 3,068.6 (P) 3,071.7 … DECLINING
Employment (1)
2,867.5 2,865.6 2,869.3 2,866.6 2,873.7 (P) 2,880.3 …GROWING
Unemployment (1)
209.8 205.6 202.4 198.5 194.8 (P) 191.4 … DECLINING
Unemployment Rate (2)
6.8 6.7 6.6 6.5 6.3 (P) 6.2 … DECLINING
Illinois
2013
Labor Force Data
Civilian Labor Force (1)
6,540.6 6,525.5 6,527.1 6,519.5 6,522.7 (P) 6,538.7 … FLAT
Employment (1)
5,936.0 5,923.6 5,935.0 5,940.2 5,955.4 (P) 5,973.5 … GROWING
Unemployment (1)
604.5 601.8 592.0 579.3 567.3 (P) 565.3 … DECLINING
Unemployment Rate (2)
9.2 9.2 9.1 8.9 8.7 (P) 8.6 … DECLINING
Michigan
2013
Labor Force Data
Civilian Labor Force (1)
4,727.7 4,727.1 4,736.3 4,723.9 4,706.5 (P) 4,694.3 … DECLINING
Employment (1)
4,309.4 4,302.5 4,311.7 4,300.7 4,293.4 (P) 4,300.1 … DECLINING
Unemployment (1)
418.4 424.7 424.6 423.2 413.1 (P) 394.2 … DECLINING
Unemployment Rate (2)
8.8 9.0 9.0 9.0 8.8 (P) 8.4 … DECLINING
Minnesota
2013
Labor Force Data
Civilian Labor Force (1)
2,978.5 2,970.5 2,968.9 2,964.6 2,966.0 (P) 2,971.6 … DECLINING
Employment (1)
2,824.6 2,818.0 2,821.0 2,822.2 2,828.3 (P) 2,834.2 … GROWING
Unemployment (1)
153.9 152.5 147.8 142.4 137.6 (P) 137.3 … DECLINING
Unemployment Rate (2)
5.2 5.1 5.0 4.8 4.6 (P) 4.6 … DECLINING
4. AS
Professor Chinn,
Could you comment upon whether the example you present is an example of cointegration, since one would not think coincident indicators would wander too far from actual? Also, could you comment upon the Durbin-Watson statistic? My understanding is that for univariate analyses, one likes to see a Durban-Watson statistic have a value of about 2.0 to indicate that residuals are not correlated. What is the interpretation of the Durbin-Watson statistic for your two variable example? Some additional comments about the Newey-West standard errors and the Wald F-statistic would also be appreciated. Thank you.
5. Menzie Chinn Post author
Bruce Hall: Since you commented on this post, I would have thought you would have refrained from looking at comparisons of month-to-month changes — I am comparing multi-year trends. The data are available at the Philadelphia Fed, so I leave it up to you to calculate the trends for the regional economies, as I did before.
AS: The regression output reproduced in the post does not pertain to a cointegrating regression; rather it is a regression of an actual growth rate on a forecasted growth rate. The low Durbin-Watson statistic is indicative of serial correlation in the regression residuals. Since the forecasts are overlapping (six month changes while the data frequency is monthly), then under the null that the forecasts are efficient, the regression residuals should be serially correlated, specifically MA(k-1).
In contrast, the dynamic OLS (DOLS) regression I discussed does pertain to a cointegrating relationship. Depending on the trend specification, a Johansen multivariate approach indicates cointegration at the 10% significance level. The DOLS regression does produce a R2 greater than the DW statistic, but since I pre-tested for cointegration, I don’t put heavy reliance on the Granger rule-of-thumb.
1. AS
Professor Chinn,
Thanks, I seem to need more reading on the proper definition of cointegration and distinction between actual growth on forecast growth and cointegration. Would the forecast you made be enhanced by adding an MA(k-1)j term?
6. Menzie Chinn Post author
AS: The leading indices and the growth rates are integrated order zero by typical ADF tests — so no need to do a cointegrating regression. The WI and US coincident indices are I(1) processes; hence, one would want to apply some cointegration methodology to those series, which is why I use the DOLS regression procedure in that case. See this old paper for a cookbook approach.
The induced MA process (under the null) does not bias the point estimate. However application of the standard formula for the standard error would lead to biased estimates, so that’s why I use the robust standard errors in that regression. See Hansen and Hodrick (1980) on the subject.
7. Ed Hanson
Menzie
The question from me you answered was not exactly what I asked or, perhaps, what I meant to ask. I was leading more toward a correlation of the coincidence index at time of major policy and tax changes to predicting the change either of gross state product or employment numbers. But regardless as usual I found your work quite interesting, and I thank-you for it.
Figure 2. From your notes and looking at FRED, I understand you moved the leading indicator graph forward six months to correlate the predicted CO Index. Noting that you properly informed us readers that coincident series are the revised. Does that include revision of the leading indicators also occurs? Regardless, reading Figure 2 seems to indicate that the coincidence index leads the leading indicator. To illustrate, after the start of 2012, the actual coincidence index has two peaks and one trough. Each time the coincidence index changes sign of its slope months before the leading indicators do. To me this means that leading indicators are very poor predicting the future coincidence index.
Figure 1. Yes, Wisconsin lags the US average coincidence index growth as it has since the year 2000. I note of those 13 years since 2001, 10 of those years were without Governor Walker. Which brings me to question your calculation of behavior of the Co index and as it correlates to economic activity. Looking at FRED graphs of both Wisconsin ( WIPHCI ) and the US in General ( USPHCI ) from the beginnings of 1/1/2001 to 1/1/2011, WIPHCI mange only about 2 point increase, while USPHCI gained 12. From 1/1/11 to 12/1/13 WIPHCI up 9, USPHCI up 14. 2 to 12 over ten years compared to 9 to 14, to me, indicates neither the coincidence index nor economic activity was better before Walker, if fact, substantially worse overall. There were years within the pre-Walker decade that averaged slightly better than the Walker 3. But it is a strange dynamic that when comparing to the Walker 3 years, the previous 10 years with 6 very slightly better years along with 4 absolutely horrible years can add to a better economic environment.
Ed
8. Menzie Chinn Post author
Ed Hanson: From my reading of your comment, I think you misunderstand the nature of the leading index. The Philadelphia Fed leading index is calibrated to forecast the six-month non-annualized growth rate. I have plotted in the figure the actual six-month growth rate corresponding to the predicted. Hence, if the prediction were perfect, the two curves would overlap. Clearly they do not; but they do covary, and in fact the regression indicates that they covary such that 84% of the variation in the Wisconsin growth rate is explained by the leading index 6 months prior. Hence, I would say the leading index is a (surprisingly) good predictor of future growth. I believe most time series econometricians would agree with me on this count.
Regarding Figure 1 — if you plotted the WI and US indicators, you would find that WI has been lagging the US since 2000. What the DOLS regression indicates is that even taking into account the lag, the extent of the lag was greater by 2% over the 2011M01-2013M12 (essentially a mean shift in the level of WI relative to US). In other words, I am taking into account precisely your argument that WI was lagging the US even before 2011M01.
9. Ed Hanson
Menzie
Thank-you for your reply, while I never will achieve competence in econometrics, your explanation does help in understanding. Getting away from the mathematical language, both restating what you said and I say, we both agree on the following. Using the limited parameters (coincidence index) from the Philly Fed, from 1982 to 2000, Wisconsin lead the US. Something changed about the year 2000, Wisconsin began lagging the US. It has not since that time recovered to the extent to again lead.
So said, not Ed Hanson, Menzie Chen, Governor Walker, Harley-Davidson, nor Schlitz Beer can change what ever occurred at the turn of the century. Perhaps the Peabody-Sherman WBM mechanism could shed light and correct, but it, too, is beyond my competence in the matter.
Here are some assumptions I have that I think you agree with. The economic performance, using the Philly Fed data, of Wisconsin between 2001 until 2011 was worse, (one could say and I do, much worse) than than it has been since 2011. The Republican Governors to 2003, than 8 years of Democratic Governor failed to correct that change in performance. Since 2011, Wisconsin still lags US performance, again using the Coincidence Index.
What we mainly disagree with, the Walker policies have began the change back toward superior economic performance. I say it has, and comparing the performance of Wisconsin from 2001 until 2011, with Wisconsin after 2011, supports my position. In addition, Walker policy of reduced taxation, both personal and business, since 2013 will accelerate economic performance.
One last note. This is the second straight Wisconsin economic post steering the discussion away from the comparison of Wisconsin and Minnesota. I see that Bruce Hall has alluded to that. I suspect it is possible that you have looked at the economic outlook of Minnesota since its significant tax increase, and noted that its performance is flattening and its leading indicators are plunging. I remind you that your initial idea comparing the two states was and is an excellent use of comparative economics. It is time to return to that idea, and let the facts and figures play out. And, of course, I ask you to add a chart normalized to 1/1/2013 to better visualized the implementation of the two Governors tax policies.
10. Menzie Chinn Post author
Ed Hanson: Figure 3 (added at the end of the post), which plots actual compared against forecasted counterfactual based on the historical relationship obtaining between 2001M01-2010M12, indicates there has been substantial underperformance in the Wisconsin economy. If you discount the statistics, then our argument ends here.
I don’t understand your assertion: “One last note. This is the second straight Wisconsin economic post steering the discussion away from the comparison of Wisconsin and Minnesota.” The preceding post on Wisconsin includes graphs with Minnesota included — note that the gap between WI and MN is the same as WI and the US.
The post before that is solely devoted to the Wisconsin-Minnesota comparison. Excuse me if I believe you have a selective memory, and a selective disbelief in the statistics.
11. Ed Hanson
Menzie you wrote in part,
“If you discount the statistics, then our argument ends here.” It is not so much that I discount statistics, but to endeavor to understand them more fully, so I guess our discussion does not end. You have shown statistics which show recent Wisconsin economic performance compared to its average performance over three decades. They show that Wisconsin performance in the first decade of of the 2000’s as under-performing the average, and so far since then the performance is further down slightly.
Averages have their purpose, but I regard trends more important when attempting to predict future performance. I have looked at data since 1997, and the trend since that time is clear. Wisconsin’s economic performance compared to the US in general has trended down since 2000, accelerating down about 2004, and has been leveling off since its worse year 2008. The question we should be debating (and I am) is the leveling a pause or the beginning of a reversal. (Note: I also looked at Minnesota since 1997, that State has been quite steady in comparative performance to the US in general through 2012 the last year of statistics I looked at)
I will restate my position. Both Wisconsin and Minnesota are high tax states. Wisconsin for years has shown signs of being taxed beyond its tax capacity. Wisconsin has shown some recovery under new tax policy substantially put into place in 2013. Minnesota’s new tax policy will push it beyond its tax policy, and its economic performance trend will decline. I previously wrote of some indications of the new trends in both states, for example, tax receipts, Wisconsin up substantially above statistical expectations, Minnesota flat, at best.
I am looking forward to each release of data for each state. I note that preliminary data both Minnesota and Wisconsin snapped up in new employment in December, Wisconsin more so. State GDP release is months away but will be relevant to this discussion. It is an unfortunate but reality that discussions derived from economic releases are limited mostly to pre-2013 data. The discussions of our differing positions will become better as new economic releases for 2013 become available.
Mostly I am looking forward to continued post by you comparing Wisconsin and Minnesota as you affirmed that will continue. As I have said, it is an excellent example of comparative economics between States of similar geography and history.
Ed
PS. Sorry for the delayed response, I could not get a comment past the math box, perhaps it illustrates my “discount the statistics” but more likely a corrected bug in your system. See you next Wisconsin post on the economic policies of the Wisconsin government.
12. Menzie Chinn Post author
Ed Hanson: To my knowledge, no “bug” has been corrected in the system with respect to the posting mechanism. I do not know why you could not submit.
If you want data on 2013, see the graph in this post, which I referred you to in my previous comment. Since the coincident indices are calibrated to match trends in GSP, well, then you have your observation on 2013 (at least the first 11 months).
By the way, I note you haven’t commented on Figure 3 which I created especially for you (it took time to run the regressions and do the simulation, so we could assess the statistical significance of the difference in performance — I hope you appreciate it since you keep on asking whether the deviation all happened by accident).
13. Ed Hanson
Menzie
I had not realized you expected a comment on Figure 3. So I guess I can only restate in a different way what I said about the two eras, 1/2000 – 12/2010, 1/2011 – 12/2013, previously.
But first I will admit I have no idea how your numbers differ so much from, perhaps it comes from the note “This is not a conclusive finding; using a different specification, sample period, or estimation method would likely lead to a different finding.” For me if I would like to compare the change in numbers between two time periods, I believe a comparison of the ratio of the change of numbers in one era and use that ratio to calculate if the change in the other era is greater, similar, or less. So this is how I see it.
(change in coincident index from 12/2010 and 1/2000 Wisconsin) / ( change in coincident index from 12/2010 and 1/2000 US)
(134.58 – 132.42) / (144.49 – 129.91) = 2.16 / 14.58 approximately 0 .15 point increase in the Wisconsin Co Index for every 1.0 point increase US.
The US Co Index has risen 12.61 from 1/11 to 12/2013.
To equal the record during 1/2000 through 12/2010, Wisconsin would show an increase Co Index of 12.16 x 0.15 = 1.82 points
Actually the Wisconsin Co Index has risen 8.53 points during these Walker years.
You could break out these statistics by average points per year or just the ratio between eras, they all say the same thing.
That is why I have stated that since Walker becoming governor, Wisconsin has been doing better since the previous discuss change around the year 2000. And why I do not see how your forecast line in Figure 3 is the only or best way to forecast performance.
Ed | 5,476 | 23,126 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-18 | latest | en | 0.909224 |
https://www.jiskha.com/display.cgi?id=1383686091 | 1,526,922,426,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864461.53/warc/CC-MAIN-20180521161639-20180521181639-00520.warc.gz | 783,155,558 | 4,517 | # math
posted by tiffany
why couldn't the chicken find her egg
1. Ms. Sue
2. Diana
She mislaid it
3. joe
Cake
4. l
its not there
5. Lv
She mislaid it
6. Donald Trump
because she's gay
7. hA
THAT MS.SUE WAS SO RUDE TF
8. rthubhsdvc
because youse guys are dummies. You don't even no the real answer dummies!
9. Hillary Clinton
She didn't win the presidency....
Like me :-(
10. Wannabe
Want to know who's also gay. You.
11. Evalyn Carver
OK Stop!!! Be respectful damn it!!!!!
## Similar Questions
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2. ### arithmetic
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More Similar Questions | 461 | 1,614 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-22 | latest | en | 0.878207 |
https://baahkast.com/what-are-the-whole-numbers-factors-of-20/ | 1,716,083,860,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057631.79/warc/CC-MAIN-20240519005014-20240519035014-00152.warc.gz | 96,793,883 | 9,159 | # What are the whole numbers factors of 20?
## What are the whole numbers factors of 20?
The factors of 20 are 1, 2, 4, 5, 10, 20. Therefore, 20 has 6 factors.
## What are the whole number factors of 30?
Factors of 30: 1, 2, 3, 5, 6, 10, 15 and 30.
What are prime factors of 30?
The number 30 can be written in prime factorization as 2 x 3 x 5. All of the factors are prime numbers. Using exponential form, 30 = 213151, indicating that there is one 2, one 3 and one 5 multilplied together to get the result of 30.
What girls wore in the 1930s?
At home or in public, women most commonly wore dresses with wide shoulders; puffy sleeves; modest necklines; higher, belted waistlines; and mid-calf flared hemlines. Frilly bows, ruffles, buttons, and other details often decorated dresses.
### What are multiples of 20?
The multiples of 20 are 20, 40, 60, 80, 100, 120, 140 and so on.
### What are the factors of 33?
Factors of 33
• Factors of 33: 1, 3, 11 and 33.
• Negative Factors of 33: -1, -3, -11 and -33.
• Prime Factors of 33: 3, 11.
• Prime Factorization of 33: 3 × 11 = 3 × 11.
• Sum of Factors of 33: 48.
What is the prime factorization of 20?
Factors of 20: 1, 2, 4, 5, 10 and 20. Prime Factorization of 20: 2 × 2 × 5 or 22 × 5.
Is the special number a factor of 30?
For example, 2, 3, 5 and 6 are all factors of 30. Each of them goes into 30 a whole number of times.
## What are the factors of 34?
Factors of 34
• Factors of 34: 1, 2, 17 and 34.
• Negative Factors of 34: -1, -2, -17 and -34.
• Prime Factors of 34: 2, 17.
• Prime Factorization of 34: 2 × 17 = 2 × 17.
• Sum of Factors of 34: 54.
## What was the fashion in the 30s?
1930s Fashion Trends Midi length bias-cut dresses, puff sleeves, belted waists, and large yokes or collars. Old Hollywood evening gowns – backless, sleeveless, long bias-cut dresses. High waisted sailor pants and wide leg beach pajamas. Casual sports clothes — skirt-like shorts, striped knit shirts.
What was the style in the 30’s?
1930s clothing brings to mind bias cut evening gowns in liquid satins and silks, tweed suits, flounces and frills and topped off by a cute beret or tilt hat. 30s Fashion certainly seemed fussier than the 1920s with its relentless ornaments of bows, trims and frills.
Is 20 a factor or multiple?
Table of Factors and Multiples
Factors Multiples
1, 2, 3, 6, 9, 18 18 144
1, 19 19 152
1, 2, 4, 5, 10, 20 20 160
1, 3, 7, 21 21 168
### What are the factors of 20 and 30?
Thus, the Factors of 20 are 1, 2, 4, 5, 10, and 20. Factors of 30. The Factors of 30 are the numbers that you can evenly divide into 30. Thus, the Factors of 30 are 1, 2, 3, 5, 6, 10, 15, and 30. Do you want to know the Factors of 20 and 30 because you want to know the greatest factor they have in common?
### How do you find all factors of 30?
Find all factors of 30. Divide 30 by all numbers starting from 1 and select the division that gives a remainder equal to 0 then write the factors. 30 ÷ 1 = 30 remainder 0 or 30 = 30 × 1. Hence 1 and 30 are factors. 30 ÷ 2 = 15 remainder 0 or 30 = 15 × 2 .
What are factfactors of 20?
Factors of 20 are numbers, which create results as 20 when two numbers are multiplied together. Factor pairs of numbers 20 are whole numbers that can be negative or positive but not a fraction or decimal number.
What are the factors of 20 in order?
All factors of 20. Here is a list of all the positive and negative factors of 20 in numerical order. 1, 2, 4, 5, 20, – 1, – 2, – 4, – 5, – 20. List all the factors of 20. Answer: 1, 2, 4, 5, 20, – 1, – 2, – 4, – 5, – 20. | 1,189 | 3,573 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.885594 |
https://forum.allaboutcircuits.com/threads/raspberry-pi-small-electric-current.176938/ | 1,701,733,433,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100535.26/warc/CC-MAIN-20231204214708-20231205004708-00732.warc.gz | 308,226,585 | 23,518 | # Raspberry Pi small electric current
#### C0mfo0rd
Joined Feb 26, 2021
10
Good day. I have a question about measuring electric current using Raspi. I have a voltage of about 0-40V in the circuit. It has the shape of a sine. There is a resistance of 100komh - 10Mohm in the circuit. I need Raspi to measure my current. Can you advise me what parts to use for this? And any scheme? (I'm sorry for my english )
• Voltage - approx 40V
• current range - 1uA - 100mA
• Frequency - 1kHz
• It doesnt have to be very accurate. I want it to be mainly cheap.
• components - It doent matter what components I use.
#### dl324
Joined Mar 30, 2015
16,143
Welcome to AAC!
There is a resistance of 100komh - 10Mohm in the circuit.
Is the resistance fixed? Is one end connected to ground? A schematic would be helpful...
It doesnt have to be very accurate. I want it to be mainly cheap.
How accurate and how cheap?
#### C0mfo0rd
Joined Feb 26, 2021
10
Welcome to AAC!
Is the resistance fixed? Is one end connected to ground? A schematic would be helpful...
How accurate and how cheap?
I need to create a device that will measure resistances from 10kOhm - 100MOhm. The voltage has a sine wave 0-40V.
I don't know what price to expect.
#### dl324
Joined Mar 30, 2015
16,143
#### C0mfo0rd
Joined Feb 26, 2021
10
I have created a circuit in which the sinusoidal voltage is 0 - 40V and the given resistor is connected there. My job is to find out what resistance there is. I don't even know how to do it. So thank you in advance for any advice. The best would be some scheme ....
#### LesJones
Joined Jan 8, 2017
4,110
I was going to suggest using an INA229 (Or one of the other devices in that range.) from your initial specification but I'm not sure it fits in with your resistance measurement specifications. Your information on resistance measurement seem to require a range from 0.4 uA to 40 mA which is different to your original specification. As far as I know the Raspberry Pi does not have any analogue inputs so you need external A to D conversion. (Which devices in the INA229 range provide.)
Edit. From your comments in post #5 this sounds like homework or project work. Am I correct. ?
Les.
#### dl324
Joined Mar 30, 2015
16,143
I have created a circuit in which the sinusoidal voltage is 0 - 40V and the given resistor is connected there. My job is to find out what resistance there is. I don't even know how to do it. So thank you in advance for any advice. The best would be some scheme ....
Is this school work?
#### C0mfo0rd
Joined Feb 26, 2021
10
I was going to suggest using an INA229 (Or one of the other devices in that range.) from your initial specification but I'm not sure it fits in with your resistance measurement specifications. Your information on resistance measurement seem to require a range from 0.4 uA to 40 mA which is different to your original specification. As far as I know the Raspberry Pi does not have any analogue inputs so you need external A to D conversion. (Which devices in the INA229 range provide.)
Edit. From your comments in post #5 this sounds like homework or project work. Am I correct. ?
Les.
Yes. I need to measure a resistance in the range of about 1uA - 40mA.
#### C0mfo0rd
Joined Feb 26, 2021
10
#### LesJones
Joined Jan 8, 2017
4,110
Can you explain why you are trying to measure resistance this way ? Also is the resistor under test purely resistive or does it have a reactive component ?
Les.
#### dl324
Joined Mar 30, 2015
16,143
Post a schematic showing how the resistor is connected. Why are you trying to measure resistance with an AC voltage? It's normally done with a DC voltage.
Raspberry Pi doesn't have any analog inputs. Are you okay with adding an ADC?
#### C0mfo0rd
Joined Feb 26, 2021
10
I testing resistors. Pure resistive.
#### C0mfo0rd
Joined Feb 26, 2021
10
Post a schematic showing how the resistor is connected. Why are you trying to measure resistance with an AC voltage? It's normally done with a DC voltage.
Raspberry Pi doesn't have any analog inputs. Are you okay with adding an ADC?
The scheme is simple. Only the given resistor is connected to the source. So there is only source and resistance and I need to measure the current.
#### dl324
Joined Mar 30, 2015
16,143
I need to measure the current
You need to measure voltage. DVMs measure resistances by applying a known current and measuring voltage. They measure current by inserting a known resistance and measuring the voltage drop across the resistor.
Why do you want to use an AC voltage?
#### C0mfo0rd
Joined Feb 26, 2021
10
You need to measure voltage. DVMs measure resistances by applying a known current and measuring voltage. They measure current by inserting a known resistance and measuring the voltage drop across the resistor.
Why do you want to use an AC voltage?
Can you advise me on the specific components for this measurement, to draw a scheme? This voltage must be used in this measurement. I can't tell you the reason, but I know that those resistors will be powered by this voltage.
#### dl324
Joined Mar 30, 2015
16,143
I can't tell you the reason
I'm not a fan of solving problems for someone when they can't tell you why or what they're trying to do.
#### C0mfo0rd
Joined Feb 26, 2021
10
I'm not a fan of solving problems for someone when they can't tell you why or what they're trying to do.
In fact, it is the resistance of certain components that depend on the sinusoidal voltage.
#### LesJones
Joined Jan 8, 2017
4,110
My original suggestion of using an INA229 was made on the assumption that you needed to measure the voltage as well as the current. From what you now say you are assuming the 40 volts is constant so does not need to be measured. Using that device would have the problem of having to synchronise the sampling time to a fixed point on the waveform. I now think it would be better to measure the current at the ground end by measuring the voltage across a shunt resistor. As your current range covers such a large range (1 : 10000) I suggest switching in a number of different value shunt resistors. I suggest amplifying the AC voltage across the shunt resistors using a good quality op amp up to about 1 volt amplitude and the converting it to DC using an active rectifier. Why are you using a Raspberry Pi rather than one of the many microcontrollers that has a built in ADC ?
Les.
#### C0mfo0rd
Joined Feb 26, 2021
10
My original suggestion of using an INA229 was made on the assumption that you needed to measure the voltage as well as the current. From what you now say you are assuming the 40 volts is constant so does not need to be measured. Using that device would have the problem of having to synchronise the sampling time to a fixed point on the waveform. I now think it would be better to measure the current at the ground end by measuring the voltage across a shunt resistor. As your current range covers such a large range (1 : 10000) I suggest switching in a number of different value shunt resistors. I suggest amplifying the AC voltage across the shunt resistors using a good quality op amp up to about 1 volt amplitude and the converting it to DC using an active rectifier. Why are you using a Raspberry Pi rather than one of the many microcontrollers that has a built in ADC ?
Les.
I need to measure voltage and current and calculate resistance accordingly. I use Raspberry Pi precisely because I have it available and I would like to send my measurement results. I will buy an ADC if necessary. | 1,844 | 7,534 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-50 | latest | en | 0.945717 |
http://forums.macrumors.com/threads/two-questions.182494/ | 1,485,207,728,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560283008.19/warc/CC-MAIN-20170116095123-00539-ip-10-171-10-70.ec2.internal.warc.gz | 107,037,072 | 31,253 | # Two Questions
Discussion in 'Community Discussion' started by asherman13, Feb 23, 2006.
1. ### asherman13 macrumors 6502a
Joined:
Jul 31, 2005
Location:
SF Bay Area, CA
#1
These questions have always stimulated my mind:
If you give somebody a penny for their thoughts, and they give you their two cents, where does the extra cent go?
And can I put it in a bank account to accumulate millions of dollars over the years?
2. ### kretzy macrumors 604
Joined:
Sep 11, 2004
Location:
Canberra, Australia
3. ### grapes911 Moderator emeritus
Joined:
Jul 28, 2003
Location:
Citizens Bank Park
#3
I'll give you are quarter if you never post anything this ridiculous again.
4. ### asherman13 thread starter macrumors 6502a
Joined:
Jul 31, 2005
Location:
SF Bay Area, CA
### Staff Member
Joined:
Sep 19, 2002
Location:
Los Angeles
#5
I'll tell you the answer if you do one simple thing for me first:
Give me a penny today, double that tomorrow, double that the next day, and so on for 2 months.
That's all. I'll even give you \$100 as a reward!
6. ### w_parietti22 macrumors 68020
Joined:
Apr 16, 2005
Location:
Seattle, WA
#6
I heard somewhere that if you stacked a penny on a chess board in one square and then doubled it in the next square (2 pennies) and then doubled it again (4 pennies) and so on that you would end up with more money than the US treasurery or something like that. I think it was in some math book.
7. ### asherman13 thread starter macrumors 6502a
Joined:
Jul 31, 2005
Location:
SF Bay Area, CA
#7
I think that would amount to 2^60 or so pennies, which is a lot more than \$100. If only I was on a Mac, then I'd hit F12 and find out what that is. I hate seven-year-old Dell Inspirons.
I also think that that originated with a peasant and grains of rice? Something like that...
EDIT Are we talking cumulative here, like 2 + 4 + 8 + 16 + 32, etc? 'Cause then I'd have to bust out my calculator...
8. ### Chundles macrumors G4
Joined:
Jul 4, 2005
#8
11,529.215046 TRILLION dollars!!!!!
Holy snapping ducks!!! That's a ****load of cash!!!
EDIT: Hang on, that's 2^60.
It's:
184,467 Trillion dollars!!!
Just put a formula into excel, doubling the previous number, starting at "1" and ending 64 operations later, then summed the lot.
Then I divided by 100 to give the dollar amount, then by 1,000,000,000,000 to give the amount in trillions.
9. ### asherman13 thread starter macrumors 6502a
Joined:
Jul 31, 2005
Location:
SF Bay Area, CA
#9
Should the right amount be 2^60 + 2^59 + 2^58 + .... + 2^2 + 2^1 + 2^0 (1) ?
Don't forget to divide by 100!
10. ### xsedrinam macrumors 601
Joined:
Oct 21, 2004
#10
I'd pick July and August for the 62 maximum days. My little calculator couldn't stand the pressure and went in to "error" mode, but I suspect one would fare as well or better than the Nebraska meat housers on that little venture.
11. ### Deepdale macrumors 68000
Joined:
May 4, 2005
Location:
New York
#11
Cheap tipper.
### Staff Member
Joined:
Sep 19, 2002
Location:
Los Angeles
#12
OK, I'll throw in a Rubik's Cube.
### Staff Member
Joined:
Aug 16, 2005
Location:
New England
#13
NOTE: The chess board has 8x8 = 64 squares, so it's essentially the same as 62 days. Also note that Sum(2^(0..n))=2^(n+1)-1, so you don't need to actually do the sum to get the answer.
You can see this rule in action if you consider binary representation of numbers. 1+2+4+8+16+32+64+128 = 255 = 0b1111 1111. 2^8 -1 = 256 -1.
Now extend this to 62 or 64 binary 1s!
B
14. ### Deepdale macrumors 68000
Joined:
May 4, 2005
Location:
New York
#14
Great - another paperweight. A few weeks ago on an episode of "Beauty and the Geek 2," there was a young guy who had been one of the recordholders for solving that. He demonstrated by holding it behind his back and within 15 seconds or so it was all done. Even the "beauties" were in awe!
15. ### robbieduncan Moderator emeritus
Joined:
Jul 24, 2002
Location:
London
#15
OK. First you have to find a chess board capable of supporting the weight of all the pennies. And then some way of stopping them from all falling over. Then I'll go and get all my pennies | 1,241 | 4,141 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2017-04 | latest | en | 0.937998 |
https://www.traditionaloven.com/metal/precious-metals/iridium/convert-qty_3-cubic-yard-cu-yd-to-troy-mace-of-iridium.html | 1,632,716,937,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058263.20/warc/CC-MAIN-20210927030035-20210927060035-00056.warc.gz | 1,026,466,712 | 13,830 | Iridium 3 cubic yard to troy maces of iridium converter
# iridium conversion
## Amount: 3 cubic yards (cu yd - yd3) of iridium volume Equals: 6,301,676.97 troy maces (tsin t) in iridium mass
Calculate troy maces of iridium per 3 cubic yards unit. The iridium converter.
TOGGLE : from troy maces into cubic yards in the other way around.
### Enter a New cubic yard Amount of iridium to Convert From
* Enter whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
Enter Your Amount :
## iridium from cubic yard to mace (troy) Conversion Results :
Amount : 3 cubic yards (cu yd - yd3) of iridium
Equals: 6,301,676.97 troy maces (tsin t) in iridium
Fractions: 6301676 troy maces (tsin t) in iridium
CONVERT : between other iridium measuring units - complete list.
## Iridium Amounts (solid iridium metal)
This calculator tool is based on the pure and solid iridium. The metal has symbol Ir and density of 22.56g/cm3 - 22.56 grams per one cubic centimeter. A natural iridium either from the Earth's crust or from meteorites usually contains distinctive amounts of Pt, Ru and Os. Iridium is very brittle (glass like some might describe it) and hard metal - second densest element currently known ( 1st hardest metal is osmium ) and the number #1 corrosion resistant metal out of all. Uses of iridium: spark plugs for cars for its very high melting point and secondly applications in electrochemical industries.
Convert iridium measuring units between cubic yard (cu yd - yd3) and troy maces (tsin t) of iridium but in the other direction from troy maces into cubic yards.
conversion result for iridium: From Symbol Equals Result To Symbol 1 cubic yard cu yd - yd3 = 2,100,558.99 troy maces tsin t
# Precious metals: iridium conversion
This online iridium from cu yd - yd3 into tsin t (precious metal) converter is a handy tool not just for certified or experienced professionals. It can help when selling scrap metals for recycling.
## Other applications of this iridium calculator are ...
With the above mentioned units calculating service it provides, this iridium converter proved to be useful also as a teaching tool:
1. in practicing cubic yards and troy maces ( cu yd - yd3 vs. tsin t ) exchange.
2. for conversion factors training exercises with converting mass/weights units vs. liquid/fluid volume units measures.
3. work with iridium's density values including other physical properties this metal has.
International unit symbols for these two iridium measurements are:
Abbreviation or prefix ( abbr. short brevis ), unit symbol, for cubic yard is: cu yd - yd3
Abbreviation or prefix ( abbr. ) brevis - short unit symbol for mace (troy) is: tsin t
### One cubic yard of iridium converted to mace (troy) equals to 2,100,558.99 tsin t
How many troy maces of iridium are in 1 cubic yard? The answer is: The change of 1 cu yd - yd3 ( cubic yard ) unit of a iridium amount equals = to 2,100,558.99 tsin t ( mace (troy) ) as the equivalent measure for the same iridium type.
In principle with any measuring task, switched on professional people always ensure, and their success depends on, they get the most precise conversion results everywhere and every-time. Not only whenever possible, it's always so. Often having only a good idea ( or more ideas ) might not be perfect nor good enough solutions. Subjects of high economic value such as stocks, foreign exchange market and various units in precious metals trading, money, financing ( to list just several of all kinds of investments ), are way too important. Different matters seek an accurate financial advice first, with a plan. Especially precise prices-versus-sizes of iridium can have a crucial/pivotal role in investments. If there is an exact known measure in cu yd - yd3 - cubic yards for iridium amount, the rule is that the cubic yard number gets converted into tsin t - troy maces or any other unit of iridium absolutely exactly. It's like an insurance for a trader or investor who is buying. And a saving calculator for having a peace of mind by knowing more about the quantity of e.g. how much industrial commodities is being bought well before it is payed for. It is also a part of savings to my superannuation funds. "Super funds" as we call them in this country.
Conversion for how many troy maces ( tsin t ) of iridium are contained in a cubic yard ( 1 cu yd - yd3 ). Or, how much in troy maces of iridium is in 1 cubic yard? To link to this iridium - cubic yard to troy maces online precious metal converter for the answer, simply cut and paste the following.
The link to this tool will appear as: iridium from cubic yard (cu yd - yd3) to troy maces (tsin t) metal conversion.
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting. | 1,238 | 4,790 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.809517 |
https://mathoverflow.net/questions/233144/atiyah-singer-theorem-a-big-picture/270470 | 1,627,201,296,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151641.83/warc/CC-MAIN-20210725080735-20210725110735-00483.warc.gz | 391,864,575 | 39,070 | # Atiyah-Singer theorem-a big picture
So far I made several attempts to really learn Atiyah-Singer theorem. In order to really understand this result a rather broad background is required: you need to know analysis (pseudodifferential operators), algebra (Clifford algebras, spin groups) and algebraic topology (characteristic classes, K-theory, Chern character). So every time I have stuck in some place and have too many doubts to continue my journey into the realm of index theory. Finally I came to conclusion that what I'm lacking is some big picture about various approaches to index theorem: there are many (excellent) books that treat the subject but before you arrive at the proof of index theorem itself you have to go through about 150 pages of background material (which usually you partially know).
Some issues which I already figured out:
1. There are at least two approaches: the one which uses K-theory and the second so called heat kernel approach. As I understood for the first one good knowledge of algebraic topology is required while for the second you have to know differential geometry stuff (connections, curvatures etc.)
2. As far as I know these approaches have different range of applicability: the heat kernel approach is more restrictive and do not include all differential elliptic operators.
3. There are some other proofs (using K homology - see the book of Higson and Roe, via tangent groupoid, there is also quite recent proof due to Paul Baum)
I would like to gather in one place various possible approaches to the proof of index theorem together with discussion of the main ingredients needed to complete such proof (for example: "you need good understanding of K-theory not only for compact spaces, but also for locally compact. You also have to understand Bott periodicity and Thom isomorphism theorem), the formulation of the index formula together with comparison and relation to other index formula (for example: you obtain topological index as a K-theory class- in order to obtain the famous cohomological formula (involving Todd class) you apply Chern character) and finally the level of generality of formulation of index theorem (for example: this approach works for all elliptic differential operators).
Maybe this question is broad but I have an impression that I'm not the only person who will find such a discussion useful.
• Does Paul Siegel's answer to a related question help? mathoverflow.net/a/23469 – Yemon Choi Mar 8 '16 at 20:50
• A couple throw-away comments. First, a good way to organize your thoughts about the different proofs of the index theorem is to compare them to proofs of the Gauss-Bonnet theorem (a special case). The K-theory/K-homology proof(s) mirror "extrinsic" approaches to Gauss-Bonnet (using the shape operator, say) while the heat kernel proof mirrors "intrinsic" approaches. – Paul Siegel Mar 9 '16 at 16:35
• Second, it is useful to think of the index theorem as the Poincare duality theorem for K-theory: both involve studying the pairing between homology and cohomology classes, both rely on the Thom isomorphism theorem (via tubular neighborhoods), and both are really all about calculating an "index map" (in Poincare duality it is the integration-along-the-fiber map). The fact that differential operators are even involved in the Atiyah-Singer theorem is sort of an accident stemming from the fact that there is a model of the K-theory spectrum involving spaces of Fredholm operators. – Paul Siegel Mar 9 '16 at 16:42
• Finally, I wouldn't completely agree with the statement that the heat kernel proof has more limited scope than K-theory based proofs. Rather, I would say that the heat kernel proof proceeds by reducing the general index theorem to the calculation of a limited set of specific examples, and then calculating those examples explicitly using heat kernels. In fact, I think it is true that early heat kernel proofs involved doing calculations on generators of the cobordism ring and using cobordism invariance of the index (though I don't remember the history for sure). – Paul Siegel Mar 9 '16 at 16:48
• While this question is interesting, I find many prerequisites of the index theorem more important than the theorem itself. Every good mathematician should learn about pseudodifferential operators, and Clifford algebras, and spin groups, and characteristic classes, and K-theory, and the Chern character! So instead of thinking of them as mere 'prerequisites', just learn about them, enjoy them, and use them. – John Baez May 21 '17 at 3:38
I agree with @coudy's answer that the best approach is to first understand the theorem's special cases / applications / generalizations. That can help highlight some of the key pain points in the various proofs, and motivate some of the ideas involved. Still, I'll take a crack at the main thrust of the question: how do the proofs work and what's involved?
I think basically all of the proofs can be organized into three categories:
1. K-theory (topology)
2. K-theory (operator algebras)
3. Heat kernels
1 and 2 are fairly similar and probably more or less equivalent, but they lend themselves to different generalizations. The techniques of 2 are responsible for many of the most state-of-the-art applications, e.g. to the Novikov conjecture or to noncommutative geometry.
3 seems to be completely different, or at least I don't think anybody can claim to understand why the techniques in 1 and 2 are capable of proving the same theorem as the techniques in 3. On the other hand, 3 is required (given the current state of the literature) for certain applications and generalizations, such as the Atiyah-Patodi-Singer index theorem for manifolds with boundary. It's also quite hard to summarize the main ideas - a lot of gritty analysis and PDE theory is involved.
For this answer I'll try to explain and compare 1 and 2; if I have the time later I might revisit 3 in another answer.
K-Theory Proofs
Both types of K-theory proofs (1 and 2) follow the same basic pattern; the differences are in how the relevant maps are defined and computed. Here's a general schema expressed in the modern way of thinking (emphasizing K-homology and Dirac operators).
1. Define the Dirac operator $D$ on a Riemannian spin (or spin$^c$) manifold $M^{2k}$, and show that it defines the fundamental class in K-homology: $[D] \in K_{2k}(M) \cong K_0(M)$.
2. Show that the Poincare duality pairing between K-theory and K-homology applied to the fundamental class $[D]$ sends the K-theory class of a vector bundle $E$ to the Fredholm index of the twisted Dirac operator $D_E$. This gives an analytic index map $K^0(M) \to \mathbb{Z}$ which is an isomorphism.
3. Construct a topological index map $K^0(M) \to \mathbb{Z}$ as follows. Embed $M$ into $\mathbb{R}^n$, apply the Thom isomorphism to get a K-theory class on the normal bundle of $M$, use a diffeomorphism from the normal bundle to an open set in $\mathbb{R}^n$ to get a K-theory class on an open set in $\mathbb{R}^n$, and apply the wrong-way map in K-theory to get a class in $K_0(\mathbb{R}^n) \cong \mathbb{Z}$.
4. Calculate that the analytic and topological index maps agree.
5. Apply Chern characters everywhere to get a formula in cohomology - note that the Thom isomorphism in K-theory and the Thom isomorphism in cohomology are not compatible with Chern characters, so the Todd class appears as a correction term.
6. Reduce the index theorem for an arbitrary elliptic operator to the index theorem for spinor Dirac operators. (This is some version of the clutching construction; here is a good modern reference.)
This should indicate what the prerequisites are: a little spin geometry to define Dirac operators, some analysis to show that the Fredholm index exists and is well-defined on K-theory, and some topology to construct the topological index map.
Proof 1 (Topological K-theory)
The strategy of the proof is to show that the analytic index map is an isomorphism, the topological index map is a homomorphism, and both maps are functorial in $M$. This means that the two maps are always equal if they are equal on one example, and one can check by direct calculation on, say, the sphere (where the index theorem is basically just the Bott periodicity theorem).
A good reference is Atiyah and Singer's original paper "Index of Elliptic Operators I", though it should be noted that they don't explicitly use K-homology and neither Dirac operators nor the cohomologlical formula are introduced until IEO III. Nevertheless, the ideas are pretty much the same.
Baum and Van Erp provide a modern reference which fills out the schema using purely topological methods.
Proof 2 (Operator K-theory)
The idea of the operator algebraic proof is to use Kasparov's bivariant groups $KK(A,B)$ where $A$ and $B$ are C*-algebras. The K-homology of $M$ is the special case $KK(C(M), \mathbb{C})$ and the K-theory is the special case $KK(\mathbb{C}, C(M))$. There is a product in KK-theory:
$$KK(A,B) \times KK(B,C) \to KK(A,C)$$
and in the special case where $A = C = \mathbb{C}$ and $B = C(M)$ one recovers the analytic index map as:
$$K^0(M) \times K_0(M) \to KK_0(\mathbb{C}, \mathbb{C}) \cong \mathbb{Z}$$
(i.e. the product of the K-theory class of a vector bundle and the K-homology class of the Dirac operator is the index of the operator twisted by the bundle.) The KK product is functorial for C*-algebra homomorphisms in all factors, and it is compatible with all of the ingredients in the topological index map (e.g. the Thom isomorphism is just a KK product with the Bott element in K-theory). So the proof of the index theorem becomes a simple little calculation with KK products.
This is a very powerful and appealing approach, but the Kasparov groups and especially the Kasparov product are hard to define. Probably the best references are K-theory for C$^\ast$ algebras by Blackadar and Analytic K-homology by Higson and Roe.
• Some people consider all analysis "gritty", but I'd say the truly "gritty" aspects of analysis involved in the heat kernel proof of the index theorem simply amount to proving that heat kernels have the asymptotics you'd expect. Dan Quillen gave a course trying to prove the index theorem using just undergraduate multivariable calculus and Clifford algebras. He got stuck at showing that the heat kernel's short-time asymptotics on a compact manifold are close enough to those in $\mathbb{R}^n$. This is 'obvious', because a curved manifold looks flat if you zoom in close to any point. But.... – John Baez May 21 '17 at 3:33
• @JohnBaez: Yes this is the heart of the proof. So surprised even Quillen would be stuck. The main idea of doing it via $\Psi DO$s is to construct Green's operators very carefully step by step, which is intuitive but needs a lot of calculation in the process like Mehler's formula. – Bombyx mori May 21 '17 at 4:05
• Quillen was looking for a proof using just undergraduate-level calculus... but apparently it doesn't exist. Mehler's formula is not the problem: it's the estimates needed to approximate the general case by nice formulas like that. Ezra Getzler attended the course and filled in the analysis using ideas from Bismut in his paper A short proof of the local Atiyah-Singer index theorem. – John Baez May 21 '17 at 7:02
• I would like to point out that the very first proof of the Index Theorem used cobordism instead of topological K-theory. – coudy May 21 '17 at 7:21
• @coudy That's a good point. I'll add that I think the cobordism proof is philosophically very similar to the topological K-theory proof: check that the analytic and topological index are well-defined on a generalized cohomology theory and then use the algebraic structure of that theory to reduce the general case to the calculation of a limited set of examples. Using cobordism theory you don't have to check many relations but you have to calculate more examples (a generator set for the cobordism ring), while with K-theory you have to check more relations but only one example. – Paul Siegel May 21 '17 at 11:28
Well, I guess that there is no royal road to the Index Theorem.
I think that what is needed in order to understand the Atyiah-Singer index theorem is the opposite of a big picture. It is easier to understand particular cases before jumping to the full statement. The big problem if you dive right into some proof of the Index Theorem is that it won't provide any examples to build intuition and understanding. That's really an abstraction over abstractions over geometric and analytical objects.
So I would suggest to start with the applications of the theorem, understand these particular cases on their own, before moving to the general proof. By reading I.2 and chapter II of the book of P. Shanahan entitled "The Atyiah-Singer Index theorem", LNM 638, even if not everything makes sense, you will get a "big picture", so to speak, of the applications of the Index Theorem and then you can start diving into the classical proofs of these applications, which will teach you in order, one at a time, all the different notions you need to understand the Index Theorem. Use chapter II of Shanahan's book as a roadmap.
The Atiyah-Singer theorem is really a synthesis of many theorems from many parts of mathematics. I suggest some references above, but really you just need to browse through them, not read them from front to cover. So, here are some suggestion for self-study.
The Euler characteristic
The first application of the Index theorem is to recover the Gauss-Bonnet formula on surfaces, which establishes a relationship between the Euler characteristic of a surface and the integral of the curvature of some Riemannian metric on the surface. and the Hopf theorem connecting the Euler characteristic with the zeros of vector fields. By learning the classical proofs of these results (e.g. in Do Carmo "curves and surfaces" and perhaps Milnor, "topology from the differential viewpoint"), you will see interesting examples of the interplay between topology and differential geometry.
It is then good to read more about the Euler characteristic in an algebraic topology book (e.g. Greenberg, "algebraic topology" Part 2, ch 20) and a differential topology book (e.g. Hirsch "differential topology", ch 4 and 5 or Bott-Tu "differential forms in algebraic topology" I.6) You then need to learn about the relationship between the Euler characteristic and the De Rham cohomology of differential forms. The Laplacian makes its first appearance there. I am not sure what to recommend here, perhaps a book on pseudo-differential operators, or ch I and III of Rosenberg "The Laplacian on a Riemannan Manifold", where the Gauss-Bonnet theorem is again discussed.
The Euler characteristic is a particular case of a characteristic class, so you are probably ready at that point to read a large part of the book of Milnor "Characteristic classes" or the relevant part of Bott-Tu.
The Lefschetz fixed point theorem
This fixed point theorem is yet another special case of the Index Theorem. You can learn about it in e.g. Greenberg and make the connection with the differential approach in Hirsch and Bott-Tu. It is also interesting to have a look at the more combinatorial approach that can be found in the original book of Lefschetz.
The Riemann-Roch-Hirzebruch theorem
This formula appears in analytic geometry and computes the dimension spaces of meromorphic functions with value in holomorphic vector bundles. The case of curves (Riemann surfaces) is quite interesting to study. A new operator, the Dolbeaut operator, makes its appearance here. Perhaps look at Griffiths-Harris, "principles of algebraic geometry", ch.II, III.3, III.4. but there should be lighter books devoted to the case of Riemann-Roch for surfaces using the Dolbeaut operator.
At that point, you should see that there is a common denominator in all these results and then it is time to look at the general proof of the Index Theorem.
Just for completeness let me add a few words on the heat kernel proof. It is true that some analysis is needed. But one should not forget that even the $K$-theoretic proof depends on the notion of pseudo-differential operators and therefore cannot avoid analysis completely.
It is also true that the proof really only works well for Dirac operators. But then, Dirac operators generate $K$-homology, and most applications of the index theorem are formulated in terms of (0-order perturbations of) Dirac operators. Contradicting comments are welcome! The most prominent examples are mentioned in Coudy's answer, but one should add the signature operator (which is the Euler operator with respect to a different grading), and the so-called untwisted Dirac operator.
The only real drawback is that one only sees the image of the index in rational cohomology. But then, using family index techniques or $\eta$-invariants, one can sometimes recover a bit more of the topological content of the index.
The proof consists of the following main ideas.
1. The McKean-Singer trick. This still works for all elliptic differential operators. If $P\colon\Gamma(E)\to\Gamma(F)$ is an elliptic differential operator on a manifold $M$, consider its adjoint $P^*\colon\Gamma(F)\to\Gamma(E)$. Then it makes sense to consider $$\operatorname{ind}(P)=\operatorname{tr}\bigl(e^{-tP^*P}\bigr)- \operatorname{tr}\bigl(e^{-tPP^*}\bigr)\;.$$ In the limit $t\to 0$, the traces above become local, that is, they can be represented by an integral over $M$ that depends only on the coefficients of $P$ in local coordinates.
From here on, one could try to prove that only finitely many geometric terms can contribute to the index. This was the idea of Gilkey and Patodi. One difficulty comes from the fact that the relevant contribution to the index sits in the $t^0$-term of the asymptotic expansion, which is by far not the leading term. However, in the special case of Dirac operators, one gets around this problem.
1. Getzler rescaling. If $P=D$ is a Dirac operator, then $\Delta=D^*D\oplus DD^*$ is a Laplace operator, and $e^{-t\Delta}$ is the standard heat kernel. One performs a standard parabolic rescaling of space and time together with Getzler's rescaling of the Clifford algebra to show that as $t\to 0$, $\Delta$ converges against a model operator that looks like a harmonic oscillator with coefficients in $\Lambda^{\mathrm{even}}T^*M$. The relevant contribution to the index now appears in the leading order term thanks to Getzler's trick. This only works if $D$ is compatible with a Riemannian metric (here one makes use of the homotopy invariance of the index to achieve this), and one has to chose nice local coordinates.
2. Mehler's formula. The heat kernel for the model operator has an easy explicit formula in terms of the Riemannian curvature and the coefficients of the Dirac operator.
3. Chern-Weil theory. The output of step 3 is an integral over $M$ of an invariant polynomial, applied to the Riemannian curvature and the so-called twisting curvature of the Dirac bundle on which $D$ acts. Hence it can be interpreted as the evaluation on the fundamental cycle of $M$ of an expression in characteristic classes.
Indeed, as pointed out in the comments to Paul's answer, one does not need any advanced analysis. Some notions from a first course in Riemannian geometry are needed. The most tricky part seems to be Getzler's rescaling, which is mostly algebraic. | 4,550 | 19,464 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2021-31 | longest | en | 0.959527 |
http://math.stackexchange.com/questions/258682/splitting-matrix-into-sub-matrices-with-constraints | 1,467,217,715,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783397749.89/warc/CC-MAIN-20160624154957-00087-ip-10-164-35-72.ec2.internal.warc.gz | 185,140,346 | 17,214 | # Splitting Matrix Into Sub-Matrices With Constraints
I have a question regarding matrices for a personal project of mine.
I have a large matrix that needs to be split into smaller matrices. I know its dimensions are X and Y. I know that the max amount of elements for each child matrix is E. (EX: A 5x5 matrix has 25 elements).
I want to find where to split the matrix into smaller matrices so that I have as few children matrices as possible and none of them go over element limit E.
The final constraint is that the children need to vertically line up.
1 1 1 1 2 2 2 2
1 1 1 1 2 2 2 2
3 3 3 3 4 4 4 4
3 3 3 3 4 4 4 4
3 3 3 3 4 4 4 4
1 1 1 1 2 2 2 2
1 1 1 1 2 2 2 2
1 1 1 1 2 2 2 2
3 3 3 3 3 3 3 3
3 3 3 3 3 3 3 3
So In that example, the first matrix is good because the children line up into the columns, however, the second is not because it breaks up the columns.
Any guidance is greatly appreciated!
Thanks!
-
Let $c = \min(E,X)$. This is the maximum number of columns allowed in each feasible subblock. For $j=1,2,\ldots,c$, define $m_j=\lfloor E/j\rfloor$. This represents the maximum number of rows allowed in a feasible subblock, provided that the subblock has $j$ columns. Therefore, given that a single column of subblocks has $j$ columns of entries, the minimum possible number of feasible subblocks contained in this slice of the big matrix is $n_j = \lceil Y/m_j\rceil$.
Now, let $N_j$ denotes the minimum number of feasible submatrices when the large matrix has $j$ columns. Then our desired answer, $N_X$, is given by the following recurrence: \begin{align} N_1 &= n_1,\\ N_j &= \min\{N_{j-k}+n_k:\ k=1,2, \ldots,\min(c,\,j-1)\},\ j=2,\ldots,X. \end{align} | 552 | 1,685 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2016-26 | latest | en | 0.8875 |
https://www.traditionaloven.com/tutorials/surface-area/convert-square-perch-area-to-homestead.html | 1,597,307,460,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738964.20/warc/CC-MAIN-20200813073451-20200813103451-00296.warc.gz | 845,260,009 | 17,339 | Convert sq perch to hstd | square perch to homesteads
# area surface units conversion
## Amount: 1 square perch (sq perch) of area Equals: 0.000039 homesteads (hstd) in area
Converting square perch to homesteads value in the area surface units scale.
TOGGLE : from homesteads into square perches in the other way around.
## area surface from square perch to homestead conversion results
### Enter a new square perch number to convert
* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
* Precision is how many digits after decimal point (1 - 9)
Enter Amount :
Decimal Precision :
CONVERT : between other area surface measuring units - complete list.
How many homesteads are in 1 square perch? The answer is: 1 sq perch equals 0.000039 hstd
## 0.000039 hstd is converted to 1 of what?
The homesteads unit number 0.000039 hstd converts to 1 sq perch, one square perch. It is the EQUAL area value of 1 square perch but in the homesteads area unit alternative.
sq perch/hstd area surface conversion result From Symbol Equals Result Symbol 1 sq perch = 0.000039 hstd
## Conversion chart - square perches to homesteads
1 square perch to homesteads = 0.000039 hstd
2 square perches to homesteads = 0.000078 hstd
3 square perches to homesteads = 0.00012 hstd
4 square perches to homesteads = 0.00016 hstd
5 square perches to homesteads = 0.00020 hstd
6 square perches to homesteads = 0.00023 hstd
7 square perches to homesteads = 0.00027 hstd
8 square perches to homesteads = 0.00031 hstd
9 square perches to homesteads = 0.00035 hstd
10 square perches to homesteads = 0.00039 hstd
11 square perches to homesteads = 0.00043 hstd
12 square perches to homesteads = 0.00047 hstd
13 square perches to homesteads = 0.00051 hstd
14 square perches to homesteads = 0.00055 hstd
15 square perches to homesteads = 0.00059 hstd
Convert area surface of square perch (sq perch) and homesteads (hstd) units in reverse from homesteads into square perches.
## Area units calculator
Main area or surface units converter page.
# Converter type: area surface units
First unit: square perch (sq perch) is used for measuring area.
Second: homestead (hstd) is unit of area.
QUESTION:
15 sq perch = ? hstd
15 sq perch = 0.00059 hstd
Abbreviation, or prefix, for square perch is:
sq perch | 665 | 2,309 | {"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} | 3 | 3 | CC-MAIN-2020-34 | latest | en | 0.779977 |
https://www.scirp.org/journal/PaperInformation?paperID=48161& | 1,713,401,346,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817184.35/warc/CC-MAIN-20240417235906-20240418025906-00714.warc.gz | 904,043,168 | 41,672 | The Construction of Pairwise Additive Minimal BIB Designs with Asymptotic Results
Abstract
An asymptotic existence of balanced incomplete block (BIB) designs and pairwise balanced designs (PBD) has been discussed in [1]-[3]. On the other hand, the existence of additive BIB designs and pairwise additive BIB designs with k = 2 and λ = 1 has been discussed with direct and recursive constructions in [4]-[8]. In this paper, an asymptotic existence of pairwise additive BIB designs is proved by use of Wilson’s theorem on PBD, and also for some and k the exact existence of l pairwise additive BIB designs with block size k and λ = 1 is discussed.
Share and Cite:
Matsubara, K. and Kageyama, S. (2014) The Construction of Pairwise Additive Minimal BIB Designs with Asymptotic Results. Applied Mathematics, 5, 2130-2136. doi: 10.4236/am.2014.514207.
1. Introduction
A pairwise balanced design (PBD) of order with block sizes in a set is a system, where is a finite set (the point set) of cardinality and is a family of subsets (blocks) of such that 1) if, then and 2) every pair of distinct elements of occurs in blocks of [9] . This is denoted by. When, a is especially called a balanced incomplete block (BIB) design, where, each block contains different points and each point appears in different blocks [10] . This is denoted by or. It is well known that necessary conditions for the existence of a are
. (1.1)
Let be the incidence matrix of a BIB design, where or 0 for all and, according as the i-th point occurs in the j-th block or otherwise. Hence the incidence matrix satisfies the conditions: 1) for all, 2) for all, 3) for all
Let, where need not be an integer unlike other parameters. Further let. A set of is called pairwise additive BIB designs if corresponding incidence matrices of the BIB design satisfy that is the incidence matrix of a BIBD() for any distinct. When, this is especially called additive BIB designs [6] [7] .
It is clear by the definition that the existence of pairwise additive implies the existence of pairwise additive for any. Hence, for given parameters, the larger is, the more difficult a construction problem of pairwise additive BIB designs is.
In pairwise additive, since a sum of any two incidence matrices yields a BIB design, it is seen [7] that
(1.2)
It follows from (1.2) that the existence of pairwise additive implies
Pairwise additive are said to be minimal if or according as is odd or even.
Some classes of pairwise additive are constructed in [4] -[8] . It is clear by the definition that. The purpose of this paper is to show that, for a given odd prime power and a given positive integer, the necessary conditions (1.1) for the existence of pairwise additive minimal are asymptotically sufficient on. In particular, for the existence of pairwise additive minimal, (1.1) is asymptotically sufficient, i.e., there are pairwise additive minimal for sufficiently larger, even if. Furthermore, as the exact existence, it is shown that there are 2 pairwise additive for any positive integer except possibly for 12 values.
2.
The existence of is reviewed along with necessary and asymptotically sufficient conditions.
Let be a set of positive integers and
Necessary conditions for the existence of a are known as follows.
Lemma 2.1 [2] Necessary conditions for the existence of a are
(2.1)
Wilson [3] proved the asymptotic existence as Theorem 2.2 below shows.
Theorem 2.2 The necessary conditions (2.1) for the existence of a are asymptotically sufficient.
For any set of positive integers and any positive integer, let denote the smallest integer such that there are for every integer satisfying (2.1). Then Theorem 2.2 states the existence of. On the other hand, some explicit bound for was provided as follows.
Lemma 2.3 [11] There are for all positive integers.
Especially, for a set being a set of prime powers of form, is shown as follows.
Lemma 2.4 ([12] Theorem 19.69) Let be a set of prime powers of form with a positive integer. Then there are for all positive integers, except possibly for
.
3. Construction by
In this section, a method of constructing pairwise additive BIB designs through is provided.
The following simple method is useful to construct pairwise additive BIB designs.
Lemma 3.1 The existence of a and pairwise additive for any implies the existence of pairwise additive.
Proof. Let be the and,. On the set, let a block set with all block size be formed by the pairwise additive for each. Then it follows that the
is the required BIB design.
For example, Lemma 3.1 yields the following.
Theorem 3.2 There are 4 pairwise additive for any integer.
Proof. It follows from the fact ([4] [6] ) that there are additive, 4 pairwise additive and additive. Hence Lemmas 2.3 and 3.1 can yield the required designs.
As the next case of block sizes, is considered. A concept of pairwise additive has been discussed as a compatibly nested minimal partition in [12] , which shows the existence of pairwise additive as follows.
Lemma 3.3 ([12] ; Theorem 22.12) Let be an odd prime power for a positive integer. Then there are pairwise additive.
Lemmas 2.4, 3.1 and 3.3 can produce the following.
Theorem 3.4 ([12] ; Theorem 22.13) There are 2 pairwise additive for all positive integers, except possibly for
.
Theorem 3.4 will be improved as in Theorem 6.7.
4. Some Class of Pairwise Additive
In this section, a necessary condition for the existence of pairwise additive being minimal is provided and then some classes of pairwise additive and (pairwise) additive are constructed.
Now (1.1) implies that necessary conditions for the existence of pairwise additive are
Furthermore, the following is given.
Theorem 4.1 When is an odd prime power, necessary conditions for the existence of pairwise additive are
(4.1)
Proof. Since and, when is an odd prime power, it is shown that either or. Hence implies.
When is an odd prime power, a class of pairwise additive is obtained as follows. This observation shows a generalization of Lemma 3.3.
Theorem 4.2 Let both and be odd prime powers for a positive integer. Then there are pairwise additive.
Proof. It can be shown that a development of the following initial blocks on yields incidence matrices of the required BIB design:
where is a primitive element of and.
Furthermore, the following is known to be provided by recursive constructions with affine resolvable BIB designs. This result will be used in the next section.
Theorem 4.3 [7] Let be an odd prime power. Then there are additive.
Especially, when, the further result is known.
Theorem 4.4 [8] There are additive B for any positive integer.
5. Asymptotic Existence of Pairwise Additive Minimal
In this section, when is an odd prime power, an asymptotic existence of pairwise additive is discussed, and it is shown that the necessary conditions (4.1) for the existence of pairwise additive are asymptotically sufficient for a given positive integer.
Dirichlet’s theorem on primes is useful for the present discussion.
Theorem 5.1 (Dirichlet) If, then a set of integers of the following form
contains infinitely many primes.
Now Theorem 5.1 yields the following.
Lemma 5.2 [13] For any positive even integer, there are primes and for which and.
In the proof of Lemma 5.2 (i.e., Lemma 3.4 in [13] ), primes and are obtained by using Theorem 5.1. Thus Lemma 5.2 implies the existence of sufficiently large primes and as follows.
Lemma 5.3 For a given odd prime power, there are primes and such that (a), (b)
, (c) and (d) for.
Proof. Let be an odd prime power. Then, for an even integer, Lemma 5.2 provides primes and
such that (a), (b) and. Hence it is seen that, and
Now let. Then
which imply (c) and (d).
Thus one of the main results of this paper is now obtained through conditions (a), (b), (c) and (d) given in Lemma 5.3.
Theorem 5.4 For a given odd prime power, (4.1) is a necessary and asymptotically sufficient condition for the existence of pairwise additive B.
Proof (sufficiency). Let and be primes as in Lemma 5.3 with. Then conditions (c) and (d) show that there are for sufficiently large satisfying (4.1), on account of Theorem 2.2. Conditions (a) and (b) show that there are pairwise additive, pairwise additive and additive, on account of Theorems 4.2 and 4.3. Hence the required designs can be obtained on account of Lemma 3.1.
Unfortunately, by use of Theorem 5.4 we cannot show the existence of pairwise additive for, since an additive means pairwise additive.
Next, for a given odd prime power and a given positive integer, even if, the existence of
pairwise additive is discussed for sufficiently large.
Lemma 5.5 For a given odd prime power and a given positive integer, there are primes and such that (a), (b) and (c).
Proof. Let be an odd prime power and be a positive integer. Then, for a positive integer, Lemma 5.2 provides primes p and q such that (a) p > q >, (b) and. Hence it is seen that and (c) holds.
Thus the following result is obtained through conditions (a), (b) and (c) as in Lemma 5.5.
Theorem 5.6 For a given odd prime power and a given positive integer, there are pairwise additive for sufficiently large.
Proof. Let and be primes as in Lemma 5.5. Then it follows from (c) that there are
for sufficiently large, on account of Theorem 2.2. Also Theorem 4.2 along with conditions (a) and (b) shows that there are pairwise additive and pairwise additive. Thus the required designs are obtained on account of Lemma 3.1.
In this section, the existence of pairwise additive is discussed. At first it is shown that there are pairwise additive for sufficiently large, even if. Furthermore, the exact existence of 2 pairwise additive with is discussed by providing direct and recursive constructions of pairwise additive. Finally, it is shown that there are 2 pairwise additive for any except possibly for 12 values.
Three classes of pairwise additive are given as in Lemma 3.3 and Theorems 3.4 and 4.4. For, 15 is the smallest value of for which the existence of 2 pairwise additive is unknown in literature. Hence at first this case is individually considered here.
Lemma 6.1 There are 2 pairwise additive.
Proof. It can be shown that a development of the following initial blocks on with the index being fixed yields incidence matrices of the required BIB design:
with 15 elements. Here, in general.
Now, pairwise additive with sufficiently large are obtained as follows. This shows an extension of Theorem 5.4 with.
Theorem 6.2 For a given positive integer, even if, there are pairwise additive with sufficiently large.
Proof. Let be a positive integer satisfying. Then (pairwise) additive are constructed by Theorem 4.4, and there are primes and such that, and
, on account of Lemma 5.2. Furthermore, since and with, there are for sufficiently large, on account of Theorem 2.2. Hence pairwise additive for sufficiently large can be constructed by Lemma 3.1 with pairwise additive, pairwise additive
Next, some recursive constructions of pairwise additive are provided. A combinatorial structure is here introduced. A transversal design, denoted by TD, is a triple such that 1) is a set of elements, 2) is a partition of into classes (groups), each of size, 3) is a family of k-subsets (blocks) of, 4) every unordered pair of elements from the same group is not contained in any block, and 5) every unordered pair of elements from other groups is contained in exactly blocks. When, we simply write, where [14] .
Since it is known [14] that the existence of a is equivalent to the existence of mutually orthogonal latin squares of order, the following result can be obtained, when.
Lemma 6.3 [14] There exists a for all except for and possibly for.
A method of construction is presented, similarly to a recursive construction given in [4] , by use of.
Theorem 6.4 The existence of pairwise additive, pairwise additive and a implies the existence of pairwise additive.
Proof. Let, , be block sets of pairwise additive and pairwise additive
respectively as
and let, and, denote an element which occurs in both the m-th block of a and the n-th group. Then it can be shown that the following incidence matrices yield the required pairwise additive BIB designs with elements denoted by for and:
where and.
Another recursive method is presented.
Theorem 6.5 The existence of pairwise additive, pairwise additive and a implies the existence of pairwise additive.
Proof. Let, , be a block set similarly to the proof of Theorem 6.4 and let, , be a block set of pairwise additive, where
with elements and. Also let, and, denote an element which occurs in both the m-th block of a and the n-th group. Then the following incidence matrices can yield the required pairwise additive BIB designs with elements denoted by for and, and:
where and.
Now 2 pairwise additive are more obtained.
Lemma 6.6 There are 2 pairwise additive for.
Proof. For, Theorem 6.5 with
provides the required BIB designs respectively, because 2 pairwise additive and 2 pairwise additive are obtained by use of Theorems 3.4 and 4.4 and Lemma 6.1, and a is also obtained by Lemma 6.3.
Hence on account of Lemma 6.6, the following result can be obtained. This improves Theorem 3.4.
Theorem 6.7 There are 2 pairwise additive for any positive integer, except possibly for.
Unfortunately, we cannot clear such 12 values displayed in Theorem 6.7. Furthermore, the existence of 2 pairwise additive has not been known except for being and 15 in Theorem 4.4 and Lemma 6.1.
Remark. Since Theorem 4.2 can be valid for a given odd integer, Theorem 5.6 is extended for a given odd integer. On the other hand, when is an even prime power, an asymptotic existence of pairwise additive minimal is proved by some methods similar to Theorems 4.2, 4.3, 5.4 and 5.6. In particular, for, the complete existence of pairwise additive has been shown in [4] [5] . However, in general, the exact existence of pairwise additive minimal with (1.1) could not be shown in this paper.
Conflicts of Interest
The authors declare no conflicts of interest.
[1] Wilson, R.M. (1972) An Existence Theory for Pairwise Balanced Designs I. Journal of Combinatorial Theory, Series A, 13, 220-245. http://dx.doi.org/10.1016/0097-3165(72)90028-3 [2] Wilson, R.M. (1972) An Existence Theory for Pairwise Balanced Designs II. Journal of Combinatorial Theory, Series A, 13, 246-273. http://dx.doi.org/10.1016/0097-3165(72)90029-5 [3] Wilson, R.M. (1975) An Existence Theory for Pairwise Balanced Designs III. Journal of Combinatorial Theory, Series A, 18, 71-79. http://dx.doi.org/10.1016/0097-3165(75)90067-9 [4] Matsubara, K. and Kageyama, S. (2013) The Existence of Two Pairwise Additive for any . Journal of Statistical Theory and Practice, 7, 783-790. http://dx.doi.org/10.1080/15598608.2013.783742 [5] Matsubara, K. and Kageyama, S. (to be Published) The Existence of 3 Pairwise Additive for Any . Journal of Combinatorial Mathematics and Combinatorial Computing. [6] Matsubara, K., Sawa, M., Matsumoto, D., Kiyama, H. and Kageyama, S. (2006) An Addition Structure on Incidence Matrices of a BIB Design. Ars Combinatoria, 78, 113-122. [7] Sawa, M., Matsubara, K., Matsumoto, D., Kiyama, H. and Kageyama, S. (2007) The Spectrum of Additive BIB Designs. Journal of Combinatorial Designs, 15, 235-254. http://dx.doi.org/10.1002/jcd.20147 [8] Sawa, M., Kageyama, S. and Jimbo, M. (2008) Compatibility of BIB Designs. Statistics and Applications, 6, 73-89. [9] Mullin, R.C. and Gronau, H.D.O.F. (2007) PBDs and GDDs: The Basics. In: Colbourn, C.J. and Dinitz, J.H., Eds., The CRC Handbook of Combinatorial Designs, 2nd Edition, CRC Press, Boca Raton, 160-193. [10] Raghavarao, D. (1988) Constructions and Combinatorial Problems in Design of Experiments. Dover, New York. [11] Colbourn, C.J. and Ling, A.C.H. (1997) Pairwise Balanced Designs with Block Sizes 8, 9 and 10. Journal of Combinatorial Theory, Series A, 77, 228-245. http://dx.doi.org/10.1006/jcta.1997.2742 [12] Colbourn, C.J. and Rosa, A. (1999) Triple Systems. Oxford Press, New York, 404-406. [13] Granville, A. (1988) Nested Steiner n-Cycle Systems and Perpendicular Arrays. Journal of Combinatorial Mathematics and Combinatorial Computing, 3, 163-167. [14] Abel, R.J.R., Colbourn, C.J. and Dinitz, J.H. (2007) Mutually Orthogonal Latin Square. In: Colbourn, C.J. and Dinitz, J.H., Eds., The CRC Handbook of Combinatorial Designs, 2nd Edition, CRC Press, Boca Raton, 160-193. | 4,224 | 16,478 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-18 | latest | en | 0.912775 |
https://www.isoul.org/reality-and-conventions-once-more/ | 1,556,230,297,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578742415.81/warc/CC-MAIN-20190425213812-20190425235812-00400.warc.gz | 704,146,574 | 9,833 | iSoul Time has three dimensions
# Reality and conventions #3
This post follows on the previous post here, as well as other posts such as here.
The one-way speed of light is a convention (see John A. Winnie, Philosophy of Science, v. 37, 1970). The two-way (round-trip) speed of light is known to be c, but the one-way speed may vary between c/2 and infinity, as long as the two-way speed equals c. This means that those who say the light from a star took X light-years to reach the Earth are speaking of a convention rather than an actual duration.
A convention cannot be “cashed in” to become reality. For example, one cannot adopt a convention that some pebble is worth a million dollars, and take it to the bank and expect them to exchange it for a million dollars. According to their convention, it is worthless. If both follow the same convention, they can make the exchange, but even then it is based on a convention, not on an intrinsic reality.
Similarly, the time for starlight to reach the Earth cannot be cashed in for time on Earth. If the one-way speed of light equals c, then some galaxies appear to be billions of light-years away. But this time is the result of a convention, not an actual duration. This time cannot be cashed in to be an actual duration on Earth. Conventional years are not actual years.
It’s well-known that all motion is relative. That means what bodies are in motion are relative to a frame of reference, and there is no preferred frame of reference. Ironically, Galileo Galilei, who is credited with discovering the relativity of motion, is also known for claiming that the Earth moves around an immovable Sun rather than the converse. Whether the Earth or the Sun moves is a convention relative to a frame of reference, not a reality that all should recognize. Whichever convention is adopted cannot be cashed in for a state of rest or motion.
Conventional science is science with standard conventions. Unconventional science is science with non-standard conventions. Both are legitimate forms of science. Their conclusions should be the same, even though their conventions are different. | 458 | 2,134 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-18 | latest | en | 0.961113 |
https://www.lessonup.com/nl/lesson/WJ4fAhNZJj3F93dSm | 1,725,815,806,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651013.49/warc/CC-MAIN-20240908150334-20240908180334-00213.warc.gz | 829,531,604 | 48,250 | # Week 35 MAVO4G Lesson 1+2
Week 35: lesson 1
1 / 23
Slide 1: Tekstslide
EngelsMiddelbare schoolmavoLeerjaar 4
In deze les zitten 23 slides, met tekstslides.
Lesduur is: 50 min
## Onderdelen in deze les
Week 35: lesson 1
#### Slide 2 -Tekstslide
Last lesson
- You can listen and understand English statements about yourself.
- You can write an introduction to yourself
This lesson
1. You can recognize and use the present simple
2. You can recognize and use the present continous
3. You can see the difference between pres. simple and pres. continuous in sentences.
Lesson goals
#### Slide 3 -Tekstslide
Lesson plan
1. What did you do last weekend?
2. Present simple & Present Continuous
4. End evaluation/ homework
#### Slide 4 -Tekstslide
First of all: present simple
How do we make the present simple?
For example: walk.
#### Slide 5 -Tekstslide
You do these exercises alone.
You have 8 minutes.
timer
8:00
Finished early?
Make ex. D
#### Slide 6 -Tekstslide
Next: present continuous
How do we make the present continuous?
For example: laugh
#### Slide 7 -Tekstslide
You do these exercises alone.
You have 8 minutes.
timer
8:00
Finished early?
Make ex. G
#### Slide 8 -Tekstslide
In the next exercises you will have to choose the right form: present simple OR continuous.
Look at these examples:
He always walks to school
He is walking to school now.
#### Slide 9 -Tekstslide
You do these exercises alone.
You have 8 minutes.
timer
8:00
Finished early?
Make ex. G
#### Slide 10 -Tekstslide
Lesson goals:
1. You can recognize and use the present simple
2. You can recognize and use the present continous
3. You can see the difference between pres. simple and pres. continuous in sentences.
#### Slide 11 -Tekstslide
Week 35: lesson 2
another!
#### Slide 12 -Tekstslide
Let's see what's in the news
Try to remember at least 2 news items!
#### Slide 13 -Tekstslide
Last lesson
1. You can recognize and use the present simple
2. You can recognize and use the present continous
3. You can see the difference between pres. simple and pres. continuous in sentences.
This lesson
1. You can recognize and use the past simple
2. You can recognize and use the past continuous
Lesson goals
#### Slide 15 -Tekstslide
Lesson plan
1. What's in the news?
2. Reader: Tenses (past simple & continuous)
3. End evaluation/ homework
#### Slide 16 -Tekstslide
First of all: past simple
How do we make the past simple?
For example: scream.
#### Slide 17 -Tekstslide
Important!
There are regular and IRREGULAR verbs in English
Irregular verbs take different forms in the past simple; you need to learn these by heart!
There is a list with irregular verbs in the back of your Reader.
#### Slide 18 -Tekstslide
You do these exercises alone.
You have 8 minutes.
timer
8:00
Finished early?
Make ex. J
#### Slide 19 -Tekstslide
Next: past continuous
How do we make the past continuous?
For example: dance.
#### Slide 20 -Tekstslide
You do these exercises alone.
You have 8 minutes.
timer
8:00
Finished early?
Make ex. P
#### Slide 21 -Tekstslide
Lesson goals:
1. You can recognize and use the past simple
2. You can recognize and use the past continuous
Homework :
no homework | 860 | 3,236 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2024-38 | latest | en | 0.783283 |
https://www.fluther.com/191657/why-are-we-still-measuring-in-barleycorns-and-other-nonmetric-questions/ | 1,606,272,403,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141180636.17/warc/CC-MAIN-20201125012933-20201125042933-00433.warc.gz | 688,730,529 | 15,901 | Why are we still measuring in barleycorns? and other nonmetric questions.
Asked by Jeruba (50610) June 23rd, 2016
Here’s a reminder of what inches and teaspoons and bushels are all about:
English measure
Apparently the U.S. is one of only three nations still holding out for nonmetric measurement. And the other two are not especially forward-looking:
Can we convert?
Are you ready and willing to adopt the metric system in the U.S., or are you secretly hoping that it won’t happen in your lifetime? It sure didn’t happen in the lifetimes of my chemistry and physics teachers.
Tags as I wrote them: measurement, metric system, inches, feet, meters, liters, diehards, retro stubbornness.
Observing members: 0 Composing members: 0
As the saying goes, most of the world has gone metric, except for the USA, which has gone to the Moon with Imperial units.
I agree that we will and should make the switch one of these days (one of these Solar Units), but I’m going to miss so many of our colorful Imperial sayings.
CWOTUS (26082)
There’s no question whatsoever that metric is a far more rational system. The transition, though, would be painful, and not just because we’d have to learn to think in metric terms. The absolute best way to do such a conversion would be to take a “tear off the Bandaid” approach, where industries across the board agree to a simultaneous switch. That’s the kind of thing that can happen in highly regulated societies, but it’s hard to imagine anything of the sort here.
Failing that, you have what I come across often on my job. Manufacturers of wheelchairs these days are haphazardly adopting the metric standard for their hardware, with roughly ¼ of the new equipment out there using metric hardware. My tool bag already weighs 50 lbs (22.73 kg), and that’s without being equipped with a full complement of metric tools and fasteners. This leads to quite a few “oh crap!” moments when I realize that I have no way to replace a stripped screw (one that I probably stripped in the first place because I didn’t realize it was metric).
And I’m just one little guy. Multiply my little annoyances at having to navigate this ambiguous hardware landscape by millions across multiple industries and you can begin to understand the inertia. All of us want the metric system, but we all dread navigating the long, chaotic transition. And who, exactly, is in a position to dictate the fateful moment at which an abrupt transition would take place? The government? What government?
thorninmud (20488)
I hope we convert soon.
Hawaii_Jake (32703)
I’m teaching my son in the metric system, with the exception of cooking and baking, and that’s only because I 1) lack a kitchen scale and 2) still have a hard time thinking in metric terms myself.
Hopefully the coming generation will get it better than mine did.
Seek (34769)
Jak (3600)
I still use imperial, because that’s what I learned to cook with, most of the recipes I use are in imperial, and all my measuring utensils are imperial. If I had a recipe in metric, I could convert it, because I know a teaspoon is approximately 5 ml, and go on from there.
@Seek Here’s a little something to cut and paste and put on a fridge magnet:
1 C=8 oz=16 Tbsp=48 tsp=237 ml
¾ C=6 oz=12 Tbsp=36 tsp =177ml
⅔ C=5⅓ oz=11 Tbsp=33 tsp=158 ml
½ C=4 oz=8 Tbsp=24 tsp=118 ml
⅓ C=3⅔ oz=5⅓ Tbsp=16 tsp=79 ml
¼ C=2 oz=4 Tbsp=12 tsp=30 ml
1/8 C=1 oz=2 Tbsp=6 tsp=30 ml
1/16 C=0.5 oz=1 Tbsp =3 tsp=15 ml
Strauss (21157)
We’re working on arguably the most advanced equipment on the planet yet the program leader uses Knots to describe.air speed. Seriously?!
(@Yetanotheruser ¼ cup is 60 ml)
LuckyGuy (38416)
My math teacher asked a similar question…in 1978.
If it’s gonna happen, it is certainly on a slow burn.
I can remember during the panic after Sputnik, the Federal governmennt was right on the verge of mandating the implementation of the metric system, but the resistance was fierce from the then powerful manufacturing sector, concerned about retooling and calibration expenses, and the forced obsolescence of existing stockpiles.
stanleybmanly (22357)
It really is a tragedy that we failed to carry out the conversion while we were at our vibrant economic best. The rise of China might very well force the issue, as the waste involved with 2 separate product lines (one for domestic purposes and the other world wide) becomes ever more apparent. There were those people in our government then prescient enough to regard the issue as a matter of natonal security. What is really interesting is that the Defense dept. is very much about designating everything in meters and liters and has for decades.
stanleybmanly (22357)
Because it is analog and experiential, and based on the real world.
Metric is not ta all natural, and ends up with odd measures for quotidian uses. Metric is only really useful for a scientist using measurements to multiply in a formula into a derivative measure.
Force =Mass*acceleration is very easy in Metric, but not in Imperial measures. But how many times in your day to day non engineering or scientific life do you need to know what the magnitude of Force is?
I just spent some time in Europe, and went to the grocery store.They sell vegetables in half Kilo measures. Wouldn’t a pound be easier?
A meter is too big for measuring a room or a bookshelf; a kilometer is too small to measure how far you walk. A kilo is too heavy for measuring food, a liter is too small for buying gasoline.
zenvelo (35408)
The adjustment is much less painful than people imagine. For many years, soft drinks have been sold in metrical units (a litre of soda); nobody cares. A bottle of wine is 750 ml; so what? Alcohol hasn’t been sold by the pint, fifth, or gallon for ages, which has no effect on anyone. Medications are expressed in milligrams, and people are fine with that.
It’s really just a matter of acclimation and comfort.
Love_my_doggie (10954)
How quaint.
My vote is for the US to convert to the metric system. I’m old enough to remember when it was taught in public schools, and it just made more sense.
Pied_Pfeffer (27391)
Would you like to weigh in lb or kg? Depends on you like it heavier / being lighter..)
imrainmaker (8365)
@Love_my_doggie Beer is sold in 12 oz or 16 oz (pint) sizes.
zenvelo (35408)
Believe it or not, US manufacturers have very widely converted to metric measure. When we (for example) do work in the power industry for overseas contracts – which with coal-fired steam generation equipment is “all of our work” these days – the units are metric without exception. And all of our suppliers provide equipment, drawings, specification sheets and instructions in metric units, as well. As long as the project specification says “metric measure required”, that’s what is done, without quibble and usually without a problem, either.
Some of us who have grown up in Imperial units but who mostly work in metric units now do a lot of in-our-head conversions (pressure units of bar to psi, for example, temperature in °C back-figured to °F and measurement in mm to inches, meters to feet, just to “get a sense of scale” and to be sure that the numbers we use sound reasonable in the real world), but once you learn them, units are units.
But I also agree with @zenvelo that there’s nothing inherently “rational” about the metric unit of length, for example (from which the units for volume are derived, and upon which units for power are also derived). The only “rational” thing about it – and in this sense it is a very sensible thing – is that the units are multiples of ten. That makes sense to us, instead of feet being 12 inches, yards being 3 feet, rods being 16.5 feet, miles being 320 rods – or 5280 feet – for example.
CWOTUS (26082)
There is one nice thing about having 12 inches for a foot and 36 inches for a yard. 12 is divisible by 2, 3, 4, 6 and 12. 36 is additionally divisible by 9 ,18 and 36. This is convenient for dividing things up. 12 and 36 along with 24 (as in hours in a day), 60 (as in minutes and seconds) and 360 (as in degrees in a circle) are all what are called highly composite numbers, which just means they are each divisible by more numbers than any smaller number.
There are those who say we should switch our number system from base 10 decimal to base 12 duodecimal. That would definitely mess up the metric system.
There are those pushing for Base 8, too, but I see little reason to make such a drastic change when base 10 is literally as easy to learn as counting on your fingers.
Seek (34769)
Things got wacky with the French Republican Calendar, an attempt to decimalize time. Days were divided into 10 hours of 100 minutes, and each minute had 100 seconds. Imagine trying to measure time that way! The reform was briefly mandatory but abandoned after a few years.
I’m very comfortable with base 10 measurements, but not on a clock.
@CWOTUS Yes, the metric system is fairly widespread throughout the U.S., but often unnoticed. The volume of any canned or packaged food is listed both ways. People grab a quart container of a beverage but don’t pay much attention to the 946 ml.
Love_my_doggie (10954)
How ironic that we’re talking about this at the time the home of Imperial measurememnt (UK) is breaking away from the bastion of metric (EU)!
Strauss (21157)
^^^ Let’s not forget that England stayed true to the pound sterling and never went with the euro.
Love_my_doggie (10954)
@Love_my_doggie Some might say that England has never been the same since decimalization, they were better with Pounds/shillings/pence
zenvelo (35408)
Overheard in a pub, shortly after liquids were metricized:
“Get used to it!? I’ll never get used to it! All my life, a pint has quenched my thirst. Now I can’t down a liter without my bladder runnin’!
Strauss (21157)
While we are praising the English units, let’s discuss Fahrenheit vs Centigrade. In human terms, 100 degrees Fahrenheit is meaningful as being very warm when the thermometer registers it and 0 degrees is very cold. A person with a 100 degree temperature has a fever.
Here’s a nice chart that sums up the English Measurement System -
Ridiculous.
Attribution:By Christoph Päper – file:English length units graph.svg, CC BY 3.0, https://commons.wikimedia.org/w/index.php?curid=10809287
LuckyGuy (38416)
0 C is freezing!
10 C and below you need a sweater
20 C is gorgeous
30 C is hot
40 C Where are you?! The desert?!
LuckyGuy (38416)
That was an interesting chart, @LuckyGuy, and it includes some English / Imperial units of distance measure that are unfamiliar to me. But the out-of-scale comparisons make the chart less meaningful and useful than it might have been. And that includes the out-of-scale metric units on the right side. Still, interesting to look at some of the units. Thanks for that.
CWOTUS (26082)
@LuckyGuy , I never thought of Centigrade that way. That is how it should be taught. What matters is not the conversion of Centigrade to Fahrenheit but the intuitive sense of what a given temperature means.
I remember people casually assuming that the USA would force itself to accept metric measure as early as 1955.
The answer to all your whining is “convenience”. Bead merchants have built a world wide industry making and selling beads measured in mm. Jewelry dealers have built a world wide industry making and selling necklaces measured in inches. Nobody even blinks at bead necklaces listed as something like “8 mm x 24 in” because everybody is familiar with those terms and nobody wants to learn a new language just to sell necklaces. So likewise the soft drink manufacturers measure their bottles in liters and the fruit juice canners use quarts and ounces because they are set up that way and it would be an inconvenience to change.
What I don’t understand is why anybody cares enough to keep whining about it.
SmartAZ (1802)
It’s the measurements that got you to the moon! Hope when we in the UK finally throw off the shackles of Brussels we will return to our traditional measurements rods, poles, perches and all.
[really though metric measurements have lot going them but don’t tell anyone I said so]
basstrom188 (3587) | 2,879 | 12,208 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.956714 |
https://de.mathworks.com/matlabcentral/cody/problems/1116-calculate-the-height-of-an-object-dropped-from-the-sky/solutions/520052 | 1,597,196,843,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738858.45/warc/CC-MAIN-20200811235207-20200812025207-00017.warc.gz | 267,348,899 | 15,510 | Cody
# Problem 1116. Calculate the height of an object dropped from the sky
Solution 520052
Submitted on 3 Nov 2014 by Matthew Eicholtz
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
%% t = -1; h_correct = 1000; assert(abs(height_of_object_at_time(t)-h_correct)<0.1)
2 Pass
%% t = 0; h_correct = 1000; assert(abs(height_of_object_at_time(t)-h_correct)<0.1)
3 Pass
%% t = 1; h_correct = 995.1; assert(abs(height_of_object_at_time(t)-h_correct)<0.1)
4 Pass
%% t = 10; h_correct = 510; assert(abs(height_of_object_at_time(t)-h_correct)<0.1)
5 Pass
%% t = 15; h_correct = 0; assert(abs(height_of_object_at_time(t)-h_correct)<0.1) | 239 | 761 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2020-34 | latest | en | 0.691685 |
https://forum.freecodecamp.org/t/learn-recursion-by-building-a-decimal-to-binary-converter-step-5/677000 | 1,714,035,921,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297290384.96/warc/CC-MAIN-20240425063334-20240425093334-00654.warc.gz | 229,303,427 | 6,381 | Learn Recursion by Building a Decimal to Binary Converter - Step 5
Tell us what’s happening:
Describe your issue in detail here.
``````<!-- file: index.html -->
<!DOCTYPE html>
<html lang="en">
<meta charset="UTF-8" />
<meta http-equiv="X-UA-Compatible" content="IE=edge" />
<meta name="viewport" content="width=device-width, initial-scale=1.0" />
<title>Decimal to Binary Converter</title>
<body>
<h1>Decimal to Binary Converter</h1>
<div class="input-container">
<label for="number-input">Enter a decimal number:</label>
<input
value=""
type="number"
name="decimal number input"
id="number-input"
class="number-input"
/>
<button class="convert-btn" id="convert-btn">Convert</button>
</div>
<output id="result" for="number-input"></output>
<div id="animation-container"></div>
<script src="script.js"></script>
</body>
</html>
``````
``````/* file: styles.css */
*,
*::before,
*::after {
box-sizing: border-box;
margin: 0;
}
:root {
--light-grey: #f5f6f7;
--dark-blue: #1b1b32;
--orange: #f1be32;
}
body {
background-color: var(--dark-blue);
font-family: "Times New Roman", Times, serif;
font-size: 18px;
color: var(--light-grey);
display: flex;
flex-direction: column;
align-items: center;
}
h1 {
text-align: center;
font-size: 2.3rem;
margin: 20px 0;
}
.input-container {
margin: 10px 0;
display: flex;
flex-direction: column;
gap: 10px;
justify-content: center;
align-items: center;
}
.convert-btn {
background-color: var(--orange);
cursor: pointer;
border: none;
}
.number-input {
height: 25px;
}
#result {
margin: 10px 0;
min-width: 200px;
width: fit-content;
min-height: 80px;
word-break: break-word;
border: 5px solid var(--orange);
font-size: 2rem;
text-align: center;
}
#animation-container {
margin: auto;
max-width: 300px;
}
.animation-frame {
margin: 250px auto 0;
border: 5px solid var(--orange);
font-size: 1.2rem;
text-align: center;
}
@media screen and (min-width: 500px) {
.input-container {
flex-direction: row;
}
#result {
max-width: 460px;
}
}
``````
``````/* file: script.js */
const numberInput = document.getElementById("number-input");
const convertBtn = document.getElementById("convert-btn");
const result = document.getElementById("result");
const checkUserInput = () => {
console.log(numberInput.value);
};
// User Editable Region
// User Editable Region
``````
User Agent is: `Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/120.0.0.0 YaBrowser/24.1.0.0 Safari/537.36`
Challenge Information:
Learn Recursion by Building a Decimal to Binary Converter - Step 5
``````numberInput.addEventListener("keydown", checkUserInput);
``````
why does not it work
You appear to have created this post without editing the template. Please edit your post to Tell us what’s happening in your own words.
What are the instructions for this step? Do they mention `checkUserInput`? | 781 | 2,855 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.287532 |
https://theessaymaster.com/finance-response/ | 1,627,655,774,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153966.60/warc/CC-MAIN-20210730122926-20210730152926-00685.warc.gz | 579,591,718 | 9,980 | # finance response
Response due by 8pm. Please respond to these 2 student’s post. At least one paragraph
Post 1
1. What are the betas listed for these companies?
Macys – .67
Nordstrom – .75
2. Macys – 20,000 investment
Nordstrom – 20,000 investment
20,000/40,000 = .5
(.5 * .67) + (.5 * .75) = .335 + .375 = .71
3.
Macys – 14,000 (70% of 20,000) investment 14,000/20,000 = .7
Nordstrom – 6,000 (30% of 20,000) investment 6,000/20,000 = .3
(.7 * .67) + (.3 * .75) = .469 + .225 = .694
4.
Current 10 Year Yield – 2.38
Macys – 2.38 + (.67 * 5) = 5.73%
Nordstrom – 2.38 + (.75 * 5) = 6.13%
5.
The 52 Week change listed in historical statistics for Macys is -17.60% and for Nordstrom is listed -9.47. I’m not too sure on how to explain this because I am sure somewhere, my answers are incorrect.
POST 2
Sony and Microsoft
1) Sony Beta: 1.89; Microsoft Beta: 1.39
2) Portfolio Beta = (.5*1.89)+(.5*1.39)=1.64
3) Portfolio Beta = (.7*1.89)+(.3*1.39)= 1.74
4)
Sony Required Return = 2.411% + (1.89*5%)=11.861%
Microsoft Required Return = 2.411% + (1.39*5%)=9.361%
(When I searched for the current yield on 10-year treasury securities, I was redirected to http://data.cnbc.com/quotes/US10Y)
5) Sony shows a 52-week change of 52.96%
Microsoft shows a 52-week change of 28.87%
Both of these are well above the required return. They differ from the estimated required return because the latter is merely the forecasted rate, while the former has already transpired.
The large difference between required return and 52 week change could imply that investors are not receiving fair pay based on the success of the issuer of the stock. Alternatively, this difference may appear more significant than reality because the required rate of return formula fails to account for factors such as smoothing techniques. Due to the multi-year cycle that each company undergoes regarding console sales, I am inclined to believe that this is the case.
Posted in Uncategorized | 616 | 1,994 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2021-31 | latest | en | 0.888458 |
http://ebook.dexcargas.com/kuta-ko-choda-document.html | 1,490,421,055,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218188824.36/warc/CC-MAIN-20170322212948-00211-ip-10-233-31-227.ec2.internal.warc.gz | 104,038,000 | 5,520 | 8120 KB / s
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https://www.quickanddirtytips.com/education/math/how-to-square-two-digit-numbers-in-your-head | 1,585,454,187,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370493684.2/warc/CC-MAIN-20200329015008-20200329045008-00078.warc.gz | 1,147,384,351 | 21,722 | ôô
Learn how to quickly and easily calculate the square of any two-digit number entirely in your head. Then try your hand at a few practice problems to test out your new mental math skills.
By
Jason Marshall, PhD
Episode #109
A few days ago, a friend of mine asked if I had ever seen the trick for calculating squares of two-digit numbers in my head. I’ve seen lots of mental math tricks in my time, but it turns that for some reason this is one that I’d never run across before. After my friend showed me how it works—and after seeing just how quick, easy, and downright cool it is—I was convinced that this is a trick that math fans of the world should see. Which is exactly why I'm going to show it to you today.
## Review: What are Perfect Squares?
We’ve talked about squaring numbers and perfect squares many times before, so everybody who’s been following along for a while should be up to speed. But just in case you’re a little fuzzy, the quick and dirty summary is that squaring a number is simply the process of multiplying that number by itself. And the result of doing that is a number that’s called a perfect square. So the square of 5 is just 5 x 5 and that’s equal to the perfect square 25.
## Is There a Way to Square 2-Digit Numbers Quickly?
While it’s easy to calculate the squares of single-digit numbers like 5 in your head (since those squares are part of the basic multiplication table we learned about many moons ago), it’s not so easy to multiply two-digit numbers in your head. Or…actually…is it? What do you think: If I was to ask you to quickly find the square of a number like 32, could you do it? In truth, probably not—but that’s just because you don’t know the trick that my friend showed me. So it’s time to let you in on this mental math secret.
## How to Square 2-Digit Numbers Ending with 5
Let’s start by talking about the special case of squaring a two-digit number that ends with 5. For example, what’s the square of 35? Well, it turns out that the result of squaring any 2-digit number that ends with 5 starts with the number you get by multiplying the first digit of the number you’re squaring with the next highest digit and ends with the number 25. Which means that the answer to 35 x 35 must begin with the number 3 x 4 = 12 (since 3 is the first digit in 35 and 4 is the next number higher than 3) and ends with the number 25. So, as you can check for yourself by hand (just to make sure it works!), the answer to 35 x 35 is 1,225.
How about the square of 75? Well, the answer must begin with 7 x 8 = 56 and end with 25. So it’s 5,625, right? As you can check by hand or with a calculator, it is! And as you can check with the rest of the two-digit numbers ending with 5, this trick always works—mentally squaring two-digit numbers that end with 5 is a cinch. But what if the number doesn’t end with 5?
Mentally squaring two-digit numbers that end with 5 is a cinch.
Squaring any two-digit number in your head, let’s say 32 x 32, is a bit more difficult. We don’t have time to go into all the details about why this works right now, but the first step is to figure out the distance (more accurately the absolute value) from the number you’re squaring to the nearest multiple of ten. In our example, the nearest multiple of 10 to 32 is 30, and the distance between 32 and 30 is 2. If you were instead squaring 77, the nearest multiple of 10 is 80, and the distance between 80 and 77 is 3. Now that we’ve figured out this distance, all that we have to do to find the answer to the problem is multiply the number we get when we subtract this distance from the original number by the number we get when we add this distance to the original number, and then add the square of the distance to the result.
I know that sounds like a mouthful, but it’s really not so bad. In our example, the method says that 32 x 32 must be equal to 30 (that’s the original number minus the distance of 2) times 34 (that’s the original number plus the distance of 2) plus 4 (that’s the square of the distance of 2). In other words, 32 x 32 = (30 x 34) + 4. Wait, that actually looks morecomplicated! How exactly is it better? Because as long as you use the fact that 30 = 3 x 10 to make the multiplication problem easy (as in 30 x 34 = 3 x 10 x 34 = 3 x 340 = 1,020), this is now an easy problem to solve! Practice at it a bit, and you’ll see that the beauty of this method is that it turns a single problem that’s hard to solve in your head into multiple easy problems.
## Practice Problems
To make sure you understand how this all works, let’s do a couple of practice problems. I’ll start by showing you one more worked example, and then I’ll leave you to try the rest for yourself.
1. 77 x 77 = 5,929 The distance from 80 to 77 is 3, so 77 x 77 = (74 x 80) + 9 = (8 x 10 x 74) + 9 = (8 x 740) + 9 = 5,920 + 9 = 5,929
2. 21 x 21 = _____
3. 54 x 54 = _____
4. 98 x 98 = _____
## Wrap Up
Okay, that’s all the math we have time for today. Remember to become a fan of the Math Dude on Facebook where you’ll find a new featured number or math puzzle posted every weekday. And if you’re on Twitter, please follow me there too. Finally, if you have math questions, feel free to send them my way via Facebook, Twitter, or by email at mathdude@quickanddirtytips.com.
Until next time, this is Jason Marshall with The Math Dude’s Quick and Dirty Tips to Make Math Easier. Thanks for reading, math fans! | 1,405 | 5,434 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2020-16 | latest | en | 0.951727 |
https://thekidsworksheet.com/printable-measurement-worksheets-grade-5-pdf/ | 1,701,209,973,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100016.39/warc/CC-MAIN-20231128214805-20231129004805-00704.warc.gz | 671,010,798 | 22,226 | # Printable Measurement Worksheets Grade 5 Pdf
The worksheets can be made in html or pdf format both are easy to print. Free grade 5 measuring worksheets create an unlimited supply of worksheets for conversion of measurement units for grade 5 both customary and metric units.
Grade 5 Measurement Worksheets Free Printable K5 Learning In 2020 Measurement Worksheets Volume Worksheets Math Measurement
### Workbooks and worksheets with a mixed review of measurement skills and curriculum.
Printable measurement worksheets grade 5 pdf. Use the thermometer pictures to determine the temperature. Maths worksheets third term measurement syllabus instamaths wksheet length practical measurement 2 equivalent lengths 2 measuring length in cm and mm 76 77 3 measuring lines in cm 4 mass 78 79 5 6 reading scales 1 7 reading scales 2 8 capacity practical 9 conversion exercises 10 capacity 80 81 11 measurement 12. You will then have two choices.
Measurement milliliters and liters. Identify fractions of a shape as well as fractions of a set. This will take you to the individual page of the worksheet.
Easily download and print our measurement worksheets. Students need to understand measurement in all parts of life and these exciting dynamic worksheets will help students master length time volume and other subjects in both english and metric systems as they measure their own progress in leaps and bounds. Free measurement worksheets for kids.
1877 m km. No prep books that are not boring that kids will enjoy. Click on the free measurement worksheet you would like to print or download.
5th grade measurement worksheets these measurement worksheets will improve your students ability to measure perimeters find the area of shapes and compare and convert measurements. Check out these interesting measurement worksheets for kids. Measure liquids to the nearest liter and milliliter.
They ll also learn types of angles acute obtuse right etc. Showing top 8 worksheets in the category grade 5. Measurement standard measure the length of each object to the nearest millimeter using the given ruler.
Imparting measurement skills to your child the different kinds of measurements terms of measurement and conversion as well as the tools needed to measure certain things is a vital part of their education. Measuring worksheet 11 convert the measuring units as indicated. Understanding these more complicated measurements are crucial to other subjects like geometry algebra and physics.
Answer key for measuring worksheet 10 1a.
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Metric Measuring Units Worksheets Measurement Worksheets Worksheets Pre Algebra Worksheets | 828 | 4,742 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-50 | latest | en | 0.891109 |
http://www.gradesaver.com/textbooks/math/algebra/college-algebra-6th-edition/chapter-p-prerequisites-fundamental-concepts-of-algebra-exercise-set-p-2-page-33/26 | 1,521,617,769,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647584.56/warc/CC-MAIN-20180321063114-20180321083114-00042.warc.gz | 378,851,974 | 13,710 | ## College Algebra (6th Edition)
$x^{7}$
Any non-zero base raised to the power of zero equals one. $$x^{7}y^{0}$$ $$=x^{7}\times1$$ $$=x^{7}$$ | 53 | 143 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2018-13 | latest | en | 0.702359 |
https://www.delftstack.com/howto/java/modulo-java/ | 1,680,391,979,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950363.89/warc/CC-MAIN-20230401221921-20230402011921-00345.warc.gz | 787,940,055 | 12,381 | # Modulus in Java
Rupam Yadav Jan 30, 2023 May 16, 2021
The modulus or modulo operator returns the remainder of the two integers after division. It is utilized in simple tasks like figuring out a number is even or not and more complex tasks like tracking the next writing position in a circular array.
## Use `Math.floorMod()` to Calculate the Mod of Two Number in Java
The `Math.floorMod(a,b)` function accepts two arguments that can be of int or long type. In the function, `a` is the dividend while b is the divisor. It returns the floor modulus of the two arguments passed to the function. The mod of say `a` and `b` will yield a result greater than or equal to 0, and less than b.
Here in the code given below, the variable `num1` is exactly divisible by `num2`; hence the remainder is 0. For the second scenario, we have `num3` and `num4` of the same sign and gives a remainder of the same sign only.
But if we consider the third scenario where the dividend `num5` is positive while divisor `num6` is negative, the remainder will carry the sign of the divisor. Similarly, in the last case dividend, `num7` is still negative; the result carries the sign of the positive divisor only.
``````import java.lang.Math;
public class Main
{
public static void main(String[] args) {
int num1 = 20, num2 = 4;
System.out.println(Math.floorMod(num1, num2));
int num3 = 113, num4 = 30;
System.out.println(Math.floorMod(num3, num4));
int num5 = 113, num6 = -30;
System.out.println(Math.floorMod(num5, num6));
int num7 = -113, num8 = 30;
System.out.println(Math.floorMod(num7, num8));
}
}
``````
Output:
``````0
23
-7
7
``````
## Use the `%` Operator to Calculate the Mod of Two Number in Java
The remainder `%` operator is also used for the remainder operation. There is a subtle difference between mod and remainder. As mentioned above the resultant for the mod of a and b is always greater than or equal to 0, and less than the b(divisor).
If we consider the example given below using num1 % num2, we determine if a `num1` is even or odd. If the condition where the remainder of num1 % num2 is equal to 0, the number is even else odd. In the % operation, the resultant carries the sign of the dividend, which is visible if we take `num3 % num2`. The equation `-23 % 2` is equal to -1 hence the sign of the dividend.
Similarly, if the dividend is positive, the resultant remainder is positive. For instance `num1 % num4` results in positive remainder. The equation `17 % -3` is equal to 2.
``````public class ModTest {
public static void main (String args[]) {
int num1 = 17;
int num2 = 2;
boolean result = false;
if(result = (num1 % num2) == 0){
System.out.println("Number "+num1 + " is even");
}else{
System.out.println("Number "+num1 + " is odd");
}
int num3 = -23;
int num4 = -3;
System.out.println("Remainder1: " + num3%num2);
System.out.println("Remainder2: " + num1%num4);
}
}
``````
Output:
``````Number 17 is odd
Remainder1: -1
Remainder2: 2
`````` | 807 | 2,973 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2023-14 | longest | en | 0.775771 |
http://math.stackexchange.com/questions/83580/every-polygon-has-an-interior-diagonal | 1,464,438,290,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049277592.65/warc/CC-MAIN-20160524002117-00100-ip-10-185-217-139.ec2.internal.warc.gz | 184,244,077 | 22,650 | # Every polygon has an interior diagonal
How does one prove that in every polygon (with at least 4 sides, not necessarily convex), that it is possible to draw a segment from one vertex to another that lies entirely inside the polygon.
In particular, I'm having trouble rigorously defining the "inside of polygon." For a convex polygon, one can define the inside as the union of half-planes that are extensions of the sides. However, I'm not sure how to do this for the non-convex polygons.
I'd like to do this rigorously, considering the polygon as a figure in $\mathbb{R}^2$, instead of appealing to geometric intuition.
Edit: obviously, I need to specify nonintersecting.
-
Seen this? – J. M. Nov 19 '11 at 8:25
Alright, instead of asking people to look stuff up on Wikipedia, I vote that every time this comes up, we delete the relevant Wikipedia page. – Will Jagy Nov 19 '11 at 8:31
Potato, I looked at the link provided by lhf in a comment to the answer by zyx. That page is on a website of David Eppstein at U. C. Irvine. That led me to the page of Joseph O'Rourke and a good, basic, book. I am commenting as you will thereby receive notice on MSE of changes to your query. By itself, you would not be notified of edits I made to my own answer. – Will Jagy Nov 19 '11 at 22:26
Thanks for the reference, Will! – Potato Nov 20 '11 at 2:11
– lhf Oct 29 '14 at 13:18
I will just attend to the question of inside/outside. It follows from the Jordan Curve Theorem that your polygon splits the plane into two parts, an inside and an (unbounded) outside. To get from one to the other, we insist on smooth curves (or, for that matter, polygonal paths) that cross the polygon only once, not at a vertex, and at a nonzero angle to the segment crossed (tangency not allowed).
As a result, the simple test for inside/outside, for extremely convoluted figures, is to start a some point very far from the polygon, draw a path to the point of interest, and count the number of times the path crosses the polygon (permitting only the types of crossing indicated above). If the number of crossings is $0$ or otherwise even, the point of interest is outside the polygon. If the number of crossings is $1$ or otherwise odd, the point of interest is inside the polygon.
EDIT ** SATURDAY: A very good proof and discussion is given by Joseph O'Rourke, website OROURKE in Chapter One of his book Art Gallery Theorems and Algorithms, which one may download as separate chapters. In Chapter One, "Polygon Partitions," page 12, there is a quick proof for your question, as part of Theorem 1.2. Take three consecutive vertices $v_1, v_2, v_3$ such that the angle is "convex" with regard to the polygon $P$, that is, a tiny neighborhood of $v_2$ inside the angle $v_1 v_2 v_3$ (demanded below $\pi$) is also inside the polygon $P.$
Perhaps $v_1 v_3$ is a line segment completely contained in $P,$ in which case we are done. If not, it means there are other vertices of the original polygon inside the triangle with vertices $v_1, v_2, v_3.$ Find the vertex, call it $x,$ that is inside the triangle and is closest to $v_2,$ distance measured only in the direction perpendicular to segment $v_1 v_3$ (see Figure 1.13). As a result, the line through $x$ and parallel to $v_1 v_3$ can not intersect any other edges of $P$ inside the triangle $v_1 v_2 v_3.$ Then the segment $v_2 x$ is completely inside $P.$ . Note that measuring distance for $x$ by full Euclidean distance may not work properly here. Draw some pictures! On page 13, he mentions Meister's Two Ears Theorem (1975), Theorem 1.3, which is probably the item you saw quoted.
Evidently Meister is G. Meister, "Polygons have ears," American Mathematical Monthly, volume 82, pages 648-651, 1975.
O'Rourke:
http://math.stackexchange.com/users/237/joseph-orourke
http://mathoverflow.net/users/6094/joseph-orourke
-
I see how this provides an easy test for individual points. However, the case of a line seems harder. – Potato Nov 19 '11 at 8:31
Yes, of course. Why do you believe that it is always possible to draw your interior segment, between vertices, with at least four sides? – Will Jagy Nov 19 '11 at 8:35
I'll assume your last clause should be "for every every polygon with at least four sides." I saw in a math book the fact that "every polygon with at least 4 sides has an interior diagonal" stated without proof. – Potato Nov 19 '11 at 8:38
Thank you, Will! I added the figure that makes it clearer. I'm embarrassed to say this proof is not only in the book Will cited, but also in the two later books, Computational Geometry in C and in Discrete and Computational Geometry. – Joseph O'Rourke Nov 20 '11 at 22:26
@JosephO'Rourke, BTW, it's not too late to say it: your newest book is great, both text and figures! – lhf Apr 12 '12 at 11:50
This is not entirely rigorous but I suspect it can made such. Suppose that our polygon is not convex. Then there is some vertex $v_i$ at which the interior angle is over $\pi$. Consider rotating the ray going through points $v_i,v_{i-1}$ to the ray going through points $v_i,v_{i+1}$ through the interior of the polygon. Now consider the point at which this ray first hits the boundary of the polygon. This gives us a piecewise continuous function $f \colon (0,\alpha) \to \partial P$ where $\alpha$ is the interior angle at $v_i$. Moreover the continuous pieces are line segments. We have to prove that there is a vertex in the image of $f$. Assume there was not. Then $f(0,\alpha)$ would be a line segment from $v_{i-1}$ to $v_{i+1}$. But this is a contradiction because of the way $v_i$ was chosen.
-
Looks good to me. It shows that every concave vertex is the end-point of such an interior line segment. It might be simpler to say: The only way that the ray can change which edge of the polygon that it meets is by hitting a vertex. – TonyK Nov 19 '11 at 9:58
Existence of an interior diagonal is equivalent to the ability to triangulate polygons (that are free of self-intersections).
Some algorithms are described at wikipedia:
http://en.wikipedia.org/wiki/Polygon_triangulation
-
See also ics.uci.edu/~eppstein/junkyard/godfried.toussaint.html for a proof that diagonals exist. – lhf Nov 19 '11 at 10:59
In order to prove what is called the Bolyai-Gerwien-Wallace Theorem, that if one has two plane simple polygons of the same area then one can cut the first polygon into a finite number of simple polygons and reassemble the pieces (jigsaw puzzle style) into the other polygon, Larwrence Levy in the book Geomety: Modern Mathematics via the Euclidean plane (Prindle, Weber & Schmidt, 1970) first gives a proof (p. 142-143) that any plane simple polygon can be triangulated using the fact (with proof) that one can find two vertices (not ends of an edge) of a simple polygon which can be joined by a line segment lying (except for its ends) totally in the interior of the polygon.
- | 1,789 | 6,893 | {"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} | 3.515625 | 4 | CC-MAIN-2016-22 | longest | en | 0.939821 |
http://susan-carpenter.blogspot.com/2011/04/measurement-in-math.html | 1,532,234,429,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593010.88/warc/CC-MAIN-20180722041752-20180722061752-00602.warc.gz | 350,616,649 | 14,916 | ## Thursday, April 28, 2011
### Measurement in Math
This week we started a new unit in my math class. So far, we've explored liquid capacity, and the group seems much more familiar with cups, pints, quarts and gallons. We also used lots of different measuring tools when we made our thunder cake - cups, tablespoons and teaspoons.
After reading Measuring Penny, by Loreen Leedy, we talked about the difference between standard units v.s. nonstandard units. In the story, a little girl has to use all different types of measurements to measure all sorts of the things about her dog, Penny. One of the units she used were dog biscuits.
Partners were then given biscuits and told to measure 10 different things. Emphasis was put on LABELING the unit - in this case, we used DB for dog biscuit. One partner was the recorder and the other the measurer. Then we got back together to discuss our findings.
Lots of different things were measured - cubbies, desks, papers, clipboards, people. Because each partnership was given a different shaped biscuit, then the conversation turned to why some people had different answers even if they measured the same thing. Vanya knew it was because some of the biscuits were bigger than the others.
Finally, I shared the book, How Big is a Foot? This story tells how a king measures something with his big king feet, and then the builder uses his tiny apprentice feet to make it. We talked about how that was the same as the situation with the dog biscuits! To test, I used my big teacher feet to measure the room, and then each child used their feet to do the same.
Odie Langley said...
I am sure that will get them to love math even more and we all know how important math is in a successful life. Good work Susan.
Odie
Heather said...
I really love this post. Introducing kids to math at a young age (in such an engaging and fun way!) is so important.
Free free to use some of our free printable math worksheets if you're ever looking for a new math lesson.
http://www.schoolsparks.com/printable-worksheets
Great blog, and thanks! :) | 460 | 2,081 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2018-30 | latest | en | 0.973974 |
https://oneclass.com/all-materials/us/ball-state/thea/thea-100.en.html | 1,566,328,080,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315558.25/warc/CC-MAIN-20190820180442-20190820202442-00289.warc.gz | 573,434,127 | 44,211 | All Educational Materials for THEA 100 at Ball State University (BSU)
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THEA 100 (8) | 2,510 | 7,671 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2019-35 | latest | en | 0.590391 |
https://cstheory.stackexchange.com/questions/33520/generate-connected-induced-subgraphs-as-the-satisfying-assignments-to-a-sat-inst | 1,718,910,606,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861989.79/warc/CC-MAIN-20240620172726-20240620202726-00124.warc.gz | 162,846,470 | 42,763 | # Generate connected induced subgraphs as the satisfying assignments to a SAT instance
I want a SAT instance (in CNF) whose set of satisfying assignments are the connected induced subgraphs of a given input graph. A general solution would be helpful, but I really only need this when the input is a subgraph of a 2D grid graph.
Of course, I want the size of the SAT instance to be polynomial in the size of the input graph, so that rules out ideas like encoding every path between each pair of vertices.
Edit: The answer I needed and accepted allows the use of extra variables to formulate a SAT instance. Interestingly, @DavidEppstein shows that if no new variables are allowed, the required SAT instance cannot, in general, be found.
• When you say "use a set of variable" what do you mean? The question is asking for a formula in which the set of variables is already specified: there must be one variable per vertex. Commented Jan 7, 2016 at 19:49
Constructions are easier in the uniform setting. When doing reductions to SAT express the problem you want to reduce to SAT as a SO$\exists$ formula (by Fagin's theorem in descriptive complexity SO$\exists$ captures all queries computable in NP). Then convert it into a propositional formula using propositional translation. Then convert that to a CNF using Tseytin transformation:
Let the input $G$ be given by the number of vertices $n$ and the edge relation $E$. To simplify things let's assume that each vertex has an edge to itself, i.e. $E(v,v)$ for all $v$. Let $S$ be a unary predicate that determines the set of selected vertices. SO$\exists$ formula for your problem: $\varphi(G,S)$ is $$\forall s,t \in S \ \exists P \text{ P is a path from s to t in G[S]''}$$ To express that $P$ is a path from $v$ to $u$ in $G[S]$ we think of $P$ as an ordered list of size $n$ of vertices, i.e. $P$ is a function from $[n]$ to set of selected vertices: $$\forall i \in [n] \ \exists v \in S \ P(i,v) \land \forall i \in [n] \forall v\neq u \in [n] (\lnot P(i,v) \lor \lnot P(i,u))$$ and it is a path from $s$ to $t$: $$P(1,s) \land P(n,t) \land \forall i \in [n-1] \ \forall v,u \left( P(i,v) \land P(i+1,u) \to E(v,u) \right)$$
Now we turn it into a propositional formula using the propositional translation. We use $q_v$ to express vertex $v$ is in $S$, i.e. $v \in S$. We use $p_{iv}$ to express $P(i,v)$. Note that when translating we need to use different variables for expressing the path between each pair. It is really $p_{iv}^{st}$.
$$\bigwedge_{s,t \in [n]} \left(q_s \land q_t \to ... \right)$$
where $...$ is the conjunction of
$$\bigwedge_{i\in [n]} \bigvee_{v \in [n]} (q_v \land p_{iv}^{st}) \land \bigwedge_{i\in[n]} \bigwedge_{v\neq u \in [n]} (\lnot p_{iv}^{st} \lor \lnot p_{iu}^{st})$$ and $$p_{1s}^{st} \land p_{nt}^{st} \land\bigwedge_{i \in [n-1]} \bigwedge_{v,u \in [n]}\left( p_{iv}^{st} \land p_{i+1u}^{st} \to e_{vu} \right)$$
This has polynomial size in $n$. To make it CNF perform the reduction from SAT to CNF-SAT using Tseytin transformation: use new variables (called extension variables) for each subformula and expressing the relationship between the variable for each subformula and the variables for its intimidate children in the formula tree and that the variable for the root subformula is true.
The result is a polynomial-size CNF formula $\psi(\vec{e}, \vec{q}, \vec{p}, \vec{z})$ with variables $\vec{e}$, $\vec{q}$, $\vec{p}$, and a bunch of extension variables $\vec{z}$ (for subformulas).
Fix $\vec{e}$ based on the given $G$. Any assignment $\tau$ to $\vec{q}$, $\vec{p}$, $\vec{z}$ that satisfies $\psi$ gives a connected subgraph $S = \{v \mid \tau(q_v) = \top \}$ of $G$ and for any connected subgraph of $G$ there is such an assignment.
• looks great. I am a bit hazy on the part after "extension variables". Is there a type-O at "intimidate"? immediate? Also, complete induced subgraph in last par. It's late here, but you answer looks useful. Commented Jan 7, 2016 at 21:28
• @Alejandro, yes, look at the Wikipedia article for Tseytin transformation. It explains how to turn a formula into a CNF in more detail (or any textbook that explain why CNF-SAT (or 3-SAT) is NP-complete). Commented Jan 7, 2016 at 21:29
• If the problem is being interpreted this way, as allowing extra variables, then it's definitely not research level: it's a basic and well-known fact that every NP problem instance can be transformed into a polynomial-sized CNF formula such that the yes-instance to the instance are the ones for which the extra variables can be given a satisfying assignment. Commented Jan 7, 2016 at 22:47
• @David, yes. I think it also follow from constant depth circuit lower bounds that it cannot be done with polysize constant depth circuits without extra variables. Commented Jan 7, 2016 at 23:36
• I can see that the answer is obtained by straightforward methods, but whether or not it is research-level depends on your perspective. I need this solution for a SAT-solver-aided construction of gadgets for an entirely different reduction, and for whatever reason I was blinded to it until @Kaveh explained it. So thanks! (I realise it ended up being pretty "textbook"...) Commented Jan 8, 2016 at 9:47
It's not possible if the variables are restricted to be only the given ones for the graph vertices. Consider a graph $G$ that has $2n+2$ vertices: $x$, $u_i$, $v_i$, and $y$ for $1\le i\le n$, formed by the union of $n$ edge-disjoint paths $x$--$u_i$--$v_i$--$y$. Also consider only the subgraphs that include both $x$ and $y$: if the whole graph has a small CNF formula for connected subgraphs, then so does this subset of subgraphs, obtained by setting the variables for $x$ and $y$ to true and then simplifying the formula.
The connected subgraphs that include both $x$ and $y$ do have a very simple $2$-DNF formula: $$\bigvee\limits_{i=1}^n(u_i\wedge v_i).$$ However, when you convert this to its (unique) CNF form, it has $2^n$ terms (one for each way of excluding either $u_i$ or $v_i$ from the subgraph, for each $i$). So no polynomial-size CNF formula exists.
$G$ is not a grid graph but the same construction can easily be embedded into a grid by expanding $x$ and $y$ into larger connected subsets of grid points. As above, the whole graph can't have a small CNF formula, because if it did then you'd also have a small CNF formula for the connected subgraphs that include all the grid points representing $x$ and $y$, but we can write down the formula for this subset of the subgraphs and it does not have polynomial size.
• I may have accepted prematurely. @Kaveh's earlier suggestion to look at HamPath (e.g., math.stackexchange.com/questions/1138105/…) suggests we might use auxiliary variables to build $wz$-paths for $w,z\in V$, or another structure like a spanning tree. I agree that if you limit yourself to the variables in your answer, above, you get a large formula... Anyway, these are good leads. Commented Jan 7, 2016 at 20:17 | 1,943 | 6,974 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-26 | latest | en | 0.89556 |
http://webhelp.esri.com/arcgisdesktop/9.3/body.cfm?tocVisable=1&ID=6068&TopicName=Using%20the%20Topo%20to%20Raster%20tool | 1,610,955,190,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514423.60/warc/CC-MAIN-20210118061434-20210118091434-00564.warc.gz | 107,547,690 | 7,870 | You are here:
Extensions > Spatial Analyst > Analysis concepts > Interpolation
# Using the Topo to Raster tool
Release 9.3
Print all topics in : "Interpolation"
Note: This topic was updated for 9.3.1.
Topo to Raster is an interpolation method specifically designed for the creation of hydrologically correct digital elevation models (DEMs). It is based on the ANUDEM program developed by Michael Hutchinson (1988, 1989). See Hutchinson and Dowling (1991) for an example of a substantial application of ANUDEM and for additional associated references. A brief summary of ANUDEM and some applications are given in Hutchinson (1993). The version of ANUDEM used is 4.6.3.
Topo to Raster is the ArcGIS 9.x implementation of TopoGrid from ArcInfo Workstation 7.x.
The interpolation process
The interpolation procedure has been designed to take advantage of the types of input data commonly available and the known characteristics of elevation surfaces. This method uses an iterative finite difference interpolation technique. It is optimized to have the computational efficiency of local interpolation methods, such as inverse distance weighted (IDW) interpolation, without losing the surface continuity of global interpolation methods, such as Kriging and Spline. It is essentially a discretized thin plate spline technique (Wahba, 1990), for which the roughness penalty has been modified to allow the fitted DEM to follow abrupt changes in terrain, such as streams and ridges. It is also the only ArcGIS interpolator specifically designed to work intelligently with contour inputs.
Water is the primary erosive force determining the general shape of most landscapes. For this reason, most landscapes have many hilltops (local maximums) and few sinks (local minimums), resulting in a connected drainage pattern. Topo to Raster uses this knowledge of surfaces and imposes constraints on the interpolation process that results in a connected drainage structure and correct representation of ridges and streams. This imposed drainage condition produces higher accuracy surfaces with less input data. The quantity of input data can be up to an order of magnitude less than that normally required to adequately describe a surface with digitized contours, further minimizing the expense of obtaining reliable DEMs. The global drainage condition also virtually eliminates any need for editing or postprocessing to remove spurious sinks in the generated surface.
The program acts conservatively in removing sinks and will not impose the drainage conditions in locations that would contradict the input elevation data. Such locations normally appear in the diagnostic file as sinks. Use this information to correct data errors, particularly when processing large datasets.
The drainage enforcement process
The purpose of the drainage enforcement process is to remove all sink points in the output DEM that have not been identified as sinks in the input sink feature dataset. The program assumes that all unidentified sinks are errors, since sinks are generally rare in natural landscapes (Goodchild and Mark, 1987).
The drainage enforcement algorithm attempts to clear spurious sinks by modifying the DEM, inferring drainage lines via the lowest saddle point in the drainage area surrounding each spurious sink. It does not attempt to clear real sinks as supplied by the Sink function. Since sink clearance is subject to the elevation tolerance, the program is conservative when attempting to clear spurious sinks. In other words, it does not clear spurious sinks that would contradict input elevation data by more than the value of Tolerance 1.
Drainage enforcement can also be supplemented with the incorporation of stream line data. This is useful when more accurate placement of streams is required.
The drainage enforcement can be turned off, in which case the sink clearing process is ignored. This can be useful if you have contour data of something other than elevation (for example, temperature) for which you want to create a surface.
Use of contour data
Contours were originally the most common method for storage and presentation of elevation information. Unfortunately, this method is also the most difficult to properly utilize with general interpolation techniques. The disadvantage lies in the undersampling of information between contours, especially in areas of low relief.
At the beginning of the interpolation process, Topo to Raster uses information inherent to the contours to build a generalized drainage model. By identifying areas of local maximum curvature in each contour, the areas of steepest slope are identified, and a network of streams and ridges is created (Hutchinson, 1988). This information is used to ensure proper hydrogeomorphic properties of the output DEM and can also be used to verify accuracy of the output DEM.
After the general morphology of the surface has been determined, contour data is also used in the interpolation of elevation values at each cell.
When the contour data is used to interpolate elevation information, all contour data is read and generalized. A maximum of 50 data points are read from these contours within each cell. At the final resolution, only one critical point is used for each cell. For this reason, having a contour density with several contours crossing output cells is redundant.
Multiresolution interpolation
The program uses a multiresolution interpolation method, starting with a coarse raster and working toward the finer, user-specified resolution. At each resolution, drainage conditions are enforced, interpolation is performed, and the number of remaining sinks is recorded in the output diagnostic file.
Processing stream data
The Topo to Raster tool requires that stream network data has all arcs pointing downslope and that there are no polygons (lakes) or braided streams in the network.
The stream data should be composed of single arcs in a dendritic pattern, with any braided streams, parallel stream banks, lake polygons, and so on, cleaned up by interactive editing. When editing lake polygons out of the network, a single arc should be placed from the beginning to end of the impounded area. The arc should follow the path of a historic streambed if one is known or exists. If the elevation of the lake is known, the lake polygon and its elevation can be used as a CONTOUR input.
To display the direction of the line sections, change the Symbology to the Arrow at End option. This will draw the line sections with an arrow symbol showing the line directions.
Sometimes it's necessary to create DEMs from adjacent tiles of input data. Normally this happens when input features are derived from a map sheet series or when, due to memory limitations, the input data must be processed in several pieces.
The interpolation process uses input data from surrounding areas to define the morphology and drainage of the surface and interpolate output values. Therefore, there should be some overlap of input data into the adjacent areas when output DEMs are to be combined into a single raster.
The edges of the output DEM are not as reliable as the rest of the dataset because they are interpolated with one-half as much information. Without using overlapping areas of adjacent input data, the edges of merged DEMs may not be smooth.
The Margin in cells option provides a method for trimming the edges of output DEMs based on a user-specified distance. The edges of overlapping areas should be at least 20 cells wide.
When the DEMs have been created, they can best be combined using the geoprocessing Mosaic tool with the Blend or Mean options. This function provides options for handling overlapping areas to smooth the transition between datasets.
Evaluating output
Every created surface should be evaluated to ensure that the data and parameters supplied to the program resulted in a realistic representation of the surface. There are many ways to evaluate the quality of an output surface, depending on the type of input available to create the surface.
The most common evaluation is to create contours from the new surface and compare them to the input contour data. It is best to create these new contours at one-half the original contour interval to examine the results between contours. Drawing the original contours and the newly created contours on top of one another can help identify interpolation errors.
Another method of visual comparison is to compare the optional output drainage cover with known streams and ridges. The drainage feature class contains the streams and ridges that were generated by the program during the drainage enforcement process. These streams and ridges should coincide with known streams and ridges in the area. If a stream feature class was used as input, the output streams should almost perfectly overlay the input streams, although they may be slightly more generalized.
A common method for evaluating the quality of a generated surface is to withhold a percentage of the input data from the interpolation process. After generating the surface, the height of these known points can be subtracted from the generated surface to examine how closely the new surface represents the true surface. These differences can be used to calculate a measure of error for the surface, such as an RMS error.
The optional diagnostic file can be used to evaluate how effectively the tolerance settings are clearing sinks in the input data. Decreasing the values of the tolerances can make the program behave more conservatively at clearing sinks.
Contour biasing
There is a minor bias in the interpolation algorithm that causes input contours to have a stronger effect on the output surface at the contour. This bias can result in a slight flattening of the output surface as it crosses the contour. This may result in misleading results when calculating the profile curvature of the output surface but is otherwise not noticeable.
Likely causes of problems with Topo to Raster
• If the geoprocessor reports "The geometry is not supported for this operation", ensure the Type for the Feature Layers is correctly set. For example, specifying a point feature class as a Contour type is incorrect.
• There are insufficient system resources available. The algorithms used in Topo to Raster hold as much information as possible in memory during processing. This allows point, contour, sink, stream, and lake data to be accessed simultaneously. To facilitate processing of large datasets, it is recommended that unnecessary applications be closed before running Topo to Raster to free up physical RAM. It is also important to have sufficient amounts of system swap space on disk.
• The contour or point input may be too dense for the output cell size specified. If one output cell covers several input contours or points, the algorithm may not be able to ascertain a value for that cell. To resolve this, try any of the following:
• Decrease the cell size, then resample back to the larger cell size after Topo to Raster.
• Rasterize smaller sections of the input data using the Output extent and Margin in cells. Assemble the resulting component rasters with the Mosaic tool.
• Clip the input data into overlapping sections, and run Topo to Raster separately on each section. Assemble the resulting component rasters with the Mosaic tool.
• The application of a surface interpolator may not be consistent with the input dataset. For example, if there is a sinks input with more points than there would be cells in the output raster, Topo to Raster will fail. Densely sampled data sources, such as LIDAR data, may have similar problems. Using the NO_ENFORCE option may help in this case, but a proper understanding of how the interpolator works is important to prevent misapplication.
Comparison to TopoGrid 7.x
The GRIDALLOCSIZE environment variable is no longer used.
The internal program limits, for the most part, have been relaxed, allowing larger and denser input datasets to be used.
Handling of shallow sinks in areas of low relief and convoluted streams has been improved.
Handling of convoluted lake boundaries has been improved.
References
Goodchild, M. F. and D. M. Mark. 1987. The fractal nature of geographic phenomena. Annals of Association of American Geographers. 77 (2): 265-278.
Hutchinson, M.F. 1988. Calculation of hydrologically sound digital elevation models. Paper presented at Third International Symposium on Spatial Data Handling at Sydney, Australia.
Hutchinson, M.F. 1989. A new procedure for gridding elevation and stream line data with automatic removal of spurious pits. Journal of Hydrology 106: 211-232.
Hutchinson, M. F. and Dowling, T. I. 1991. A continental hydrological assessment of a new grid-based digital elevation model of Australia. Hydrological Processes 5: 45-58.
Hutchinson, M. F. 1993. Development of a continent-wide DEM with applications to terrain and climate analysis. In Environmental Modeling with GIS, ed. M. F. Goodchild et al., 392–399. New York: Oxford University Press.
Hutchinson, M. F. 1996. A locally adaptive approach to the interpolation of digital elevation models. In Proceedings, Third International Conference/Workshop on Integrating GIS and Environmental Modeling. Santa Barbara, CA: National Center for Geographic Information and Analysis. See: http://www.ncgia.ucsb.edu/conf/SANTA_FE_CD-ROM/sf_papers/hutchinson_michael_dem/local.html
Wahba, G. 1990. Spline models for Observational data. Paper presented at CBMS-NSF Regional Conference Series in Applied Mathematics. Philadelphia: Soc. Ind. Appl. Maths.
Please visit the Feedback page to comment or give suggestions on ArcGIS Desktop Help. | 2,708 | 13,713 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-04 | latest | en | 0.921363 |
https://puzzle.queryhome.com/8681/equation-wherein-number-equation-number-right-side-equation | 1,527,278,998,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867217.1/warc/CC-MAIN-20180525200131-20180525220131-00489.warc.gz | 640,125,584 | 31,597 | # Find the equation wherein the number on the left side of the equation is also the number on the right side of equation.
+1 vote
98 views
Observe the following equations.
55 = 131
90 = 194
-35 = -31
10 = 50
-50 = -58
In this sense, find the equation wherein the number on the left side of the equation is also the number on the right side of the equation. (a=a)
posted Jul 21, 2015
+1 vote
this follows conversion of Celsius to Fahrenheit
F=(9/5)*C+32
and the answer is -40=(9/5)*(-40)+32 so -40 is the answer.
answer Jul 22, 2015
+1 vote
-40 = -40.
The first number is the temperature in Celsius while the second one is the temperature in Fahrenheit.
For the hint: Do you feel SICK keeping on solving?
SICK gives a clue to FEVER, leading you to think of TEMPERATURE.
answer Aug 3, 2015
Similar Puzzles | 232 | 812 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-22 | latest | en | 0.904157 |
https://wiki.alquds.edu/?query=Weak_isospin | 1,695,301,840,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506027.39/warc/CC-MAIN-20230921105806-20230921135806-00269.warc.gz | 700,041,169 | 28,681 | # Weak isospin
In particle physics, weak isospin is a quantum number relating to the electrically charged part of the weak interaction: Particles with half-integer weak isospin can interact with the bosons; particles with zero weak isospin do not.[a] Weak isospin is a construct parallel to the idea of isospin under the strong interaction. Weak isospin is usually given the symbol T or I, with the third component written as T3 or I3.[b] It can be understood as the eigenvalue of a charge operator.
T3 is more important than T; typically "weak isospin" is used as short form of the proper term "3rd component of weak isospin".
The weak isospin conservation law relates to the conservation of ${\displaystyle \ T_{3}\ ;}$ weak interactions conserve T3. It is also conserved by the electromagnetic and strong interactions. However, interaction with the Higgs field does not conserve T3, as directly seen by propagation of fermions, mixing chiralities by dint of their mass terms resulting from their Higgs couplings. Since the Higgs field vacuum expectation value is nonzero, particles interact with this field all the time even in vacuum. Interaction with the Higgs field changes particles' weak isospin (and weak hypercharge). Only a specific combination of them, ${\displaystyle \ Q=T_{3}+{\tfrac {1}{2}}Y_{\mathrm {W} }\ }$ (electric charge), is conserved.
## Relation with chirality
Fermions with negative chirality (also called "left-handed" fermions) have ${\displaystyle \ T={\tfrac {1}{2}}\ }$ and can be grouped into doublets with ${\displaystyle T_{3}=\pm {\tfrac {1}{2}}}$ that behave the same way under the weak interaction. By convention, electrically charged fermions are assigned ${\displaystyle T_{3}}$ with the same sign as their electric charge.[c] For example, up-type quarks (u, c, t) have ${\displaystyle \ T_{3}=+{\tfrac {1}{2}}\ }$ and always transform into down-type quarks (d, s, b), which have ${\displaystyle \ T_{3}=-{\tfrac {1}{2}}\ ,}$ and vice versa. On the other hand, a quark never decays weakly into a quark of the same ${\displaystyle \ T_{3}~.}$ Something similar happens with left-handed leptons, which exist as doublets containing a charged lepton (
e
,
μ
,
τ
) with ${\displaystyle \ T_{3}=-{\tfrac {1}{2}}\ }$ and a neutrino (
ν
e
,
ν
μ
,
ν
τ
) with ${\displaystyle \ T_{3}=+{\tfrac {1}{2}}~.}$ In all cases, the corresponding anti-fermion has reversed chirality ("right-handed" antifermion) and reversed sign ${\displaystyle \ T_{3}~.}$
Fermions with positive chirality ("right-handed" fermions) and anti-fermions with negative chirality ("left-handed" anti-fermions) have ${\displaystyle \ T=T_{3}=0\ }$ and form singlets that do not undergo charged weak interactions.[d]
The electric charge, ${\displaystyle \ Q\ ,}$ is related to weak isospin, ${\displaystyle \ T_{3}\ ,}$ and weak hypercharge, ${\displaystyle \ Y_{\mathrm {W} }\ ,}$ by
${\displaystyle Q=T_{3}+{\tfrac {1}{2}}Y_{\mathrm {W} }~.}$
Left-handed fermions in the Standard Model[1]
Generation 1 Generation 2 Generation 3
Fermion Symbol Weak
isospin
Fermion Symbol Weak
isospin
Fermion Symbol Weak
isospin
Electron neutrino ${\displaystyle \nu _{\mathrm {e} }\ }$ ${\displaystyle +{\tfrac {1}{2}}\ }$ Muon neutrino ${\displaystyle \nu _{\mu }\ }$ ${\displaystyle +{\tfrac {1}{2}}\ }$ Tau neutrino ${\displaystyle \nu _{\tau }\,}$ ${\displaystyle +{\tfrac {1}{2}}\ }$
Electron ${\displaystyle \mathrm {e} ^{-}\,}$ ${\displaystyle -{\tfrac {1}{2}}\ }$ Muon ${\displaystyle \mu ^{-}\,}$ ${\displaystyle -{\tfrac {1}{2}}\ }$ Tauon ${\displaystyle \tau ^{-}\,}$ ${\displaystyle -{\tfrac {1}{2}}\ }$
Up quark ${\displaystyle \mathrm {u} \ }$ ${\displaystyle +{\tfrac {1}{2}}\ }$ Charm quark ${\displaystyle \mathrm {c} \ }$ ${\displaystyle +{\tfrac {1}{2}}\ }$ Top quark ${\displaystyle \mathrm {t} \ }$ ${\displaystyle +{\tfrac {1}{2}}\ }$
Down quark ${\displaystyle \mathrm {d} \ }$ ${\displaystyle -{\tfrac {1}{2}}\ }$ Strange quark ${\displaystyle \mathrm {s} \ }$ ${\displaystyle -{\tfrac {1}{2}}\ }$ Bottom quark ${\displaystyle \mathrm {b} \ }$ ${\displaystyle -{\tfrac {1}{2}}\ }$
All of the above left-handed (regular) particles have corresponding
right-handed anti-particles with equal and opposite weak isospin.
All right-handed (regular) particles and left-handed anti-particles have weak isospin of 0.
## Weak isospin and the W bosons
The symmetry associated with weak isospin is SU(2) and requires gauge bosons with ${\displaystyle \,T=1\,}$ (, , and ) to mediate transformations between fermions with half-integer weak isospin charges. [2]${\displaystyle \ T=1\ }$ implies that
W
bosons have three different values of ${\displaystyle \ T_{3}\ :}$
• boson ${\displaystyle (\,T_{3}=+1\,)}$ is emitted in transitions ${\displaystyle \left(\,T_{3}=+{\tfrac {1}{2}}\,\right)}$${\displaystyle \left(\,T_{3}=-{\tfrac {1}{2}}\,\right)~.}$
• boson ${\displaystyle (\,T_{3}=\,0\,)}$ would be emitted in weak interactions where ${\displaystyle \,T_{3}\,}$ does not change, such as neutrino scattering.
• boson ${\displaystyle (\,T_{3}=-1\,)}$ is emitted in transitions ${\displaystyle \left(\,T_{3}=-{\tfrac {1}{2}}\,\right)}$${\displaystyle \left(\,T_{3}=+{\tfrac {1}{2}}\,\right)}$.
Under electroweak unification, the boson mixes with the weak hypercharge gauge boson
B0
; both have weak isospin = 0 . This results in the observed boson and the photon of quantum electrodynamics; the resulting and likewise have zero weak isospin.
The sum of negative isospin and positive charge is zero for each of the bosons, consequently, all the electroweak bosons have weak hypercharge ${\displaystyle \ Y_{\text{W}}=0\ ,}$ so unlike gluons of the color force, the electroweak bosons are unaffected by the force they mediate.
## Footnotes
1. ^ Interaction with the is only related indirectly; its interaction is determined by weak charge, q.v.
2. ^ This article uses T and T3 for weak isospin and its projection.
Regarding ambiguous notation, I is also used to represent the 'normal' (strong force) isospin, same for its third component I3 a.k.a. T3 or Tz . Aggravating the confusion, T is also used as the symbol for the Topness quantum number.
3. ^ Lacking any distinguishing electric charge, neutrinos and antineutrinos are assigned the ${\displaystyle \ T_{3}\ }$ opposite their corresponding charged lepton; hence, all left-handed neutrinos are paired with negatively charged left-handed leptons with ${\displaystyle \ T_{3}=-{\tfrac {1}{2}}\ ,}$ so those neutrinos have ${\displaystyle \ T_{3}=+{\tfrac {1}{2}}~.}$ Since right-handed antineutrinos are paired with positively charged right-handed anti-leptons with ${\displaystyle \ T_{3}=+{\tfrac {1}{2}}\ ,}$ those antineutrinos are assigned ${\displaystyle \ T_{3}=-{\tfrac {1}{2}}~.}$ The same result follows from particle-antiparticle charge & parity reversal, between left-handed neutrinos (${\displaystyle \ T_{3}=+{\tfrac {1}{2}}\ }$) and right-handed antineutrinos (${\displaystyle \ T_{3}=-{\tfrac {1}{2}}\ }$).
4. ^ Particles with ${\displaystyle \ T_{3}=0\ }$ do not interact with
W±
bosons
; however, they do all interact with the
Z0
boson
, with the possible exception of hypothetical sterile neutrinos not yet included in the Standard Model. If they actually exist, sterile neutrinos would become the only elementary fermions in the Standard Model that do not interact with the
Z0
boson
.
## References
1. ^ Baez, John C.; Huerta, John (2010). "The algebra of Grand Unified Theories". Bulletin of the American Mathematical Society. 47 (3): 483–552. arXiv:0904.1556. Bibcode:2009arXiv0904.1556B. doi:10.1090/s0273-0979-10-01294-2. S2CID 2941843.
"§2.3.1 isospin and SU(2), redux". Huerta's academic site. U.C. Riverside. Retrieved 15 October 2013.
2. ^ An introduction to quantum field theory, by M.E. Peskin and D.V. Schroeder (HarperCollins, 1995) ISBN 0-201-50397-2; Gauge theory of elementary particle physics, by T.P. Cheng and L.F. Li (Oxford University Press, 1982) ISBN 0-19-851961-3; The quantum theory of fields (vol 2), by S. Weinberg (Cambridge University Press, 1996) ISBN 0-521-55002-5. | 2,433 | 8,079 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 60, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-40 | latest | en | 0.901335 |
http://de.metamath.org/mpeuni/o1dif.html | 1,726,621,799,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.68/warc/CC-MAIN-20240918000844-20240918030844-00079.warc.gz | 8,218,943 | 7,508 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > o1dif Structured version Visualization version GIF version
Theorem o1dif 14208
Description: If the difference of two functions is eventually bounded, eventual boundedness of either one implies the other. (Contributed by Mario Carneiro, 26-May-2016.)
Hypotheses
Ref Expression
o1dif.1 ((𝜑𝑥𝐴) → 𝐵 ∈ ℂ)
o1dif.2 ((𝜑𝑥𝐴) → 𝐶 ∈ ℂ)
o1dif.3 (𝜑 → (𝑥𝐴 ↦ (𝐵𝐶)) ∈ 𝑂(1))
Assertion
Ref Expression
o1dif (𝜑 → ((𝑥𝐴𝐵) ∈ 𝑂(1) ↔ (𝑥𝐴𝐶) ∈ 𝑂(1)))
Distinct variable groups: 𝑥,𝐴 𝜑,𝑥
Allowed substitution hints: 𝐵(𝑥) 𝐶(𝑥)
Proof of Theorem o1dif
StepHypRef Expression
1 o1dif.3 . . . 4 (𝜑 → (𝑥𝐴 ↦ (𝐵𝐶)) ∈ 𝑂(1))
2 o1sub 14194 . . . . 5 (((𝑥𝐴𝐵) ∈ 𝑂(1) ∧ (𝑥𝐴 ↦ (𝐵𝐶)) ∈ 𝑂(1)) → ((𝑥𝐴𝐵) ∘𝑓 − (𝑥𝐴 ↦ (𝐵𝐶))) ∈ 𝑂(1))
32expcom 450 . . . 4 ((𝑥𝐴 ↦ (𝐵𝐶)) ∈ 𝑂(1) → ((𝑥𝐴𝐵) ∈ 𝑂(1) → ((𝑥𝐴𝐵) ∘𝑓 − (𝑥𝐴 ↦ (𝐵𝐶))) ∈ 𝑂(1)))
41, 3syl 17 . . 3 (𝜑 → ((𝑥𝐴𝐵) ∈ 𝑂(1) → ((𝑥𝐴𝐵) ∘𝑓 − (𝑥𝐴 ↦ (𝐵𝐶))) ∈ 𝑂(1)))
5 o1dif.1 . . . . . . . . . . 11 ((𝜑𝑥𝐴) → 𝐵 ∈ ℂ)
6 o1dif.2 . . . . . . . . . . 11 ((𝜑𝑥𝐴) → 𝐶 ∈ ℂ)
75, 6subcld 10271 . . . . . . . . . 10 ((𝜑𝑥𝐴) → (𝐵𝐶) ∈ ℂ)
87ralrimiva 2949 . . . . . . . . 9 (𝜑 → ∀𝑥𝐴 (𝐵𝐶) ∈ ℂ)
9 dmmptg 5549 . . . . . . . . 9 (∀𝑥𝐴 (𝐵𝐶) ∈ ℂ → dom (𝑥𝐴 ↦ (𝐵𝐶)) = 𝐴)
108, 9syl 17 . . . . . . . 8 (𝜑 → dom (𝑥𝐴 ↦ (𝐵𝐶)) = 𝐴)
11 o1dm 14109 . . . . . . . . 9 ((𝑥𝐴 ↦ (𝐵𝐶)) ∈ 𝑂(1) → dom (𝑥𝐴 ↦ (𝐵𝐶)) ⊆ ℝ)
121, 11syl 17 . . . . . . . 8 (𝜑 → dom (𝑥𝐴 ↦ (𝐵𝐶)) ⊆ ℝ)
1310, 12eqsstr3d 3603 . . . . . . 7 (𝜑𝐴 ⊆ ℝ)
14 reex 9906 . . . . . . . 8 ℝ ∈ V
1514ssex 4730 . . . . . . 7 (𝐴 ⊆ ℝ → 𝐴 ∈ V)
1613, 15syl 17 . . . . . 6 (𝜑𝐴 ∈ V)
17 eqidd 2611 . . . . . 6 (𝜑 → (𝑥𝐴𝐵) = (𝑥𝐴𝐵))
18 eqidd 2611 . . . . . 6 (𝜑 → (𝑥𝐴 ↦ (𝐵𝐶)) = (𝑥𝐴 ↦ (𝐵𝐶)))
1916, 5, 7, 17, 18offval2 6812 . . . . 5 (𝜑 → ((𝑥𝐴𝐵) ∘𝑓 − (𝑥𝐴 ↦ (𝐵𝐶))) = (𝑥𝐴 ↦ (𝐵 − (𝐵𝐶))))
205, 6nncand 10276 . . . . . 6 ((𝜑𝑥𝐴) → (𝐵 − (𝐵𝐶)) = 𝐶)
2120mpteq2dva 4672 . . . . 5 (𝜑 → (𝑥𝐴 ↦ (𝐵 − (𝐵𝐶))) = (𝑥𝐴𝐶))
2219, 21eqtrd 2644 . . . 4 (𝜑 → ((𝑥𝐴𝐵) ∘𝑓 − (𝑥𝐴 ↦ (𝐵𝐶))) = (𝑥𝐴𝐶))
2322eleq1d 2672 . . 3 (𝜑 → (((𝑥𝐴𝐵) ∘𝑓 − (𝑥𝐴 ↦ (𝐵𝐶))) ∈ 𝑂(1) ↔ (𝑥𝐴𝐶) ∈ 𝑂(1)))
244, 23sylibd 228 . 2 (𝜑 → ((𝑥𝐴𝐵) ∈ 𝑂(1) → (𝑥𝐴𝐶) ∈ 𝑂(1)))
25 o1add 14192 . . . . 5 (((𝑥𝐴 ↦ (𝐵𝐶)) ∈ 𝑂(1) ∧ (𝑥𝐴𝐶) ∈ 𝑂(1)) → ((𝑥𝐴 ↦ (𝐵𝐶)) ∘𝑓 + (𝑥𝐴𝐶)) ∈ 𝑂(1))
2625ex 449 . . . 4 ((𝑥𝐴 ↦ (𝐵𝐶)) ∈ 𝑂(1) → ((𝑥𝐴𝐶) ∈ 𝑂(1) → ((𝑥𝐴 ↦ (𝐵𝐶)) ∘𝑓 + (𝑥𝐴𝐶)) ∈ 𝑂(1)))
271, 26syl 17 . . 3 (𝜑 → ((𝑥𝐴𝐶) ∈ 𝑂(1) → ((𝑥𝐴 ↦ (𝐵𝐶)) ∘𝑓 + (𝑥𝐴𝐶)) ∈ 𝑂(1)))
28 eqidd 2611 . . . . . 6 (𝜑 → (𝑥𝐴𝐶) = (𝑥𝐴𝐶))
2916, 7, 6, 18, 28offval2 6812 . . . . 5 (𝜑 → ((𝑥𝐴 ↦ (𝐵𝐶)) ∘𝑓 + (𝑥𝐴𝐶)) = (𝑥𝐴 ↦ ((𝐵𝐶) + 𝐶)))
305, 6npcand 10275 . . . . . 6 ((𝜑𝑥𝐴) → ((𝐵𝐶) + 𝐶) = 𝐵)
3130mpteq2dva 4672 . . . . 5 (𝜑 → (𝑥𝐴 ↦ ((𝐵𝐶) + 𝐶)) = (𝑥𝐴𝐵))
3229, 31eqtrd 2644 . . . 4 (𝜑 → ((𝑥𝐴 ↦ (𝐵𝐶)) ∘𝑓 + (𝑥𝐴𝐶)) = (𝑥𝐴𝐵))
3332eleq1d 2672 . . 3 (𝜑 → (((𝑥𝐴 ↦ (𝐵𝐶)) ∘𝑓 + (𝑥𝐴𝐶)) ∈ 𝑂(1) ↔ (𝑥𝐴𝐵) ∈ 𝑂(1)))
3427, 33sylibd 228 . 2 (𝜑 → ((𝑥𝐴𝐶) ∈ 𝑂(1) → (𝑥𝐴𝐵) ∈ 𝑂(1)))
3524, 34impbid 201 1 (𝜑 → ((𝑥𝐴𝐵) ∈ 𝑂(1) ↔ (𝑥𝐴𝐶) ∈ 𝑂(1)))
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 195 ∧ wa 383 = wceq 1475 ∈ wcel 1977 ∀wral 2896 Vcvv 3173 ⊆ wss 3540 ↦ cmpt 4643 dom cdm 5038 (class class class)co 6549 ∘𝑓 cof 6793 ℂcc 9813 ℝcr 9814 + caddc 9818 − cmin 10145 𝑂(1)co1 14065 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-8 1979 ax-9 1986 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 ax-rep 4699 ax-sep 4709 ax-nul 4717 ax-pow 4769 ax-pr 4833 ax-un 6847 ax-cnex 9871 ax-resscn 9872 ax-1cn 9873 ax-icn 9874 ax-addcl 9875 ax-addrcl 9876 ax-mulcl 9877 ax-mulrcl 9878 ax-mulcom 9879 ax-addass 9880 ax-mulass 9881 ax-distr 9882 ax-i2m1 9883 ax-1ne0 9884 ax-1rid 9885 ax-rnegex 9886 ax-rrecex 9887 ax-cnre 9888 ax-pre-lttri 9889 ax-pre-lttrn 9890 ax-pre-ltadd 9891 ax-pre-mulgt0 9892 ax-pre-sup 9893 This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-3or 1032 df-3an 1033 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 df-eu 2462 df-mo 2463 df-clab 2597 df-cleq 2603 df-clel 2606 df-nfc 2740 df-ne 2782 df-nel 2783 df-ral 2901 df-rex 2902 df-reu 2903 df-rmo 2904 df-rab 2905 df-v 3175 df-sbc 3403 df-csb 3500 df-dif 3543 df-un 3545 df-in 3547 df-ss 3554 df-pss 3556 df-nul 3875 df-if 4037 df-pw 4110 df-sn 4126 df-pr 4128 df-tp 4130 df-op 4132 df-uni 4373 df-iun 4457 df-br 4584 df-opab 4644 df-mpt 4645 df-tr 4681 df-eprel 4949 df-id 4953 df-po 4959 df-so 4960 df-fr 4997 df-we 4999 df-xp 5044 df-rel 5045 df-cnv 5046 df-co 5047 df-dm 5048 df-rn 5049 df-res 5050 df-ima 5051 df-pred 5597 df-ord 5643 df-on 5644 df-lim 5645 df-suc 5646 df-iota 5768 df-fun 5806 df-fn 5807 df-f 5808 df-f1 5809 df-fo 5810 df-f1o 5811 df-fv 5812 df-riota 6511 df-ov 6552 df-oprab 6553 df-mpt2 6554 df-of 6795 df-om 6958 df-2nd 7060 df-wrecs 7294 df-recs 7355 df-rdg 7393 df-er 7629 df-pm 7747 df-en 7842 df-dom 7843 df-sdom 7844 df-sup 8231 df-pnf 9955 df-mnf 9956 df-xr 9957 df-ltxr 9958 df-le 9959 df-sub 10147 df-neg 10148 df-div 10564 df-nn 10898 df-2 10956 df-3 10957 df-n0 11170 df-z 11255 df-uz 11564 df-rp 11709 df-ico 12052 df-seq 12664 df-exp 12723 df-cj 13687 df-re 13688 df-im 13689 df-sqrt 13823 df-abs 13824 df-o1 14069 This theorem is referenced by: dchrmusum2 24983 dchrvmasumiflem2 24991 dchrisum0lem2a 25006 dchrisum0lem2 25007 rplogsum 25016 dirith2 25017 mulogsumlem 25020 mulogsum 25021 vmalogdivsum2 25027 vmalogdivsum 25028 2vmadivsumlem 25029 selberg3lem1 25046 selberg4lem1 25049 selberg4 25050 pntrlog2bndlem4 25069
Copyright terms: Public domain W3C validator | 3,904 | 5,641 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-38 | latest | en | 0.154162 |
http://mathandreadinghelp.org/kindergarten_educational_games.html | 1,716,650,568,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058829.24/warc/CC-MAIN-20240525131056-20240525161056-00115.warc.gz | 17,482,446 | 8,010 | # Educational Games for Kindergarten
Kindergarteners are very active and ready to explore the world. You can use this eagerness to help your child enhance the skills he is learning at school. The activities below are great ways to educate your child while having fun and spending time with her.
## What Types of Games Should I Do with My Kindergartener?
Kindergarteners are learning the basics of reading, writing and math. You can help your child practice understanding these concepts through games and activities using items you have at your house. For instance, meals, walks and laundry can be turned into addition or subtraction problems, alphabet games or problem-solving activities.
### Cookie Cutter Math
This activity will help your kindergartner's math and fine motor skills. Give your child some addition or subtraction problems, either on a worksheet or using flash cards. Get some cookie cutters and play dough that has been flattened out. Have him solve each of the math problems by cutting out the play dough using cookie cutters.
For example, if your child is solving 3 + 1, he would cut out three shapes and then one more shape. When he was finished cutting out the shapes, he'd be able to count them and see that the answer to the problem is four. This activity can also be done with subtraction problems. Visual and kinesthetic learners will benefit from this approach because they will actually see and feel the problems being solved.
### Patterns with Food
Get a box of Fruit Loops and help your child sort the different colors into a muffin tin. Once all the colors are sorted, get some yarn and have your child create various patterns with the Fruit Loops (AB, ABC, etc.). Fine motor skills and hand-eye coordination are also enhanced with this activity. Your child can make bracelets and necklaces with patterns of Fruit Loops and can enjoy a tasty treat when she is done!
### Make a Sound Muncher
This game will help your child's phonological awareness, which is the ability to recognize individual sounds in words. Get a medium-sized garbage can with a round lid and decorate it like a monster. For example, the eyes can be on the top of the lid and the opening will be its mouth. Explain to your child that the 'sound muncher' will 'eat' items that begin with the letter that you decide. For example, if you say the letter 'P', your child will have to find items around the house, such as a peanut, pan or pin, to put in the mouth of the sound muncher.
Did you find this useful? If so, please let others know!
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Online Only | 1,032 | 4,911 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-22 | latest | en | 0.956347 |
https://bestcurrencykuok.web.app/boosalis59153na/excel-formula-for-future-value-compounded-annually-241.html | 1,632,746,344,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058450.44/warc/CC-MAIN-20210927120736-20210927150736-00587.warc.gz | 168,467,061 | 6,530 | ## Excel formula for future value compounded annually
Under the assumption that the 7% interest rate is a nominal rate of interest compounded monthly in the first case and semiannually in the second, we see that Future Value of Simple Interest and Compounded Interest Investigation different ways of showing the future value of interest using an excel spreadsheet. This shows us that we can find a formula for compounded annually interest:.
Future value formulas and derivations for present lump sums, annuities, growing present value lump sum investment, periodic cash flow payments, compounding, Interest Rate (R): is the annual nominal interest rate or "stated rate" per period in (similar to Excel formulas) If payments are at the end of the period it is an Think of it as this example: you are able to deposit A dollars every year (at the end of Figure 1-5: Uniform Series Compound-Amount Factor, F/Ai,n. In this case, utilizing Equation 1-2 can help us calculate the future value of each Compound Interest Formulas III · Cash Flow · Microsoft Excel Tutorial · Summary of Lesson 1. 21 Sep 2018 Time Value of Money (Excel Formula – Calculator). November 29 Future value is the compounded amount of money after a period of time with the interest rate. Compounded Quarterly : r = Annual rate/4, n=no of yrs. X 4 The equation for the future value of an annuity due is the sum of the have to save at the end of each month if she can earn 5% compounded annually, tax-free , Microsoft Office Excel and the free OpenOffice Calc have several formulas for 31 May 2019 The Continuous Compound Interest Formula Excel Function for (us) Nerds of years it will grow, and then the frequency of the compounding each year. PV = this is optional – but it is the present value of future payments.
## Future Value Formula in Excel (With Excel Template) Future Value Formula Value of the money doesn’t remain the same, it decreases or increases because of the interest rates and the state of inflation, deflation which makes the value of the money less valuable or more valuable in future.
13 Mar 2018 P = The present value of the amount to be paid in the future. A = The amount to We use the same example, but the interest is now compounded annually. The calculation is: Related Courses. Excel Formulas and Functions Since the interest is compounded annually, the one-year period can be represented The formula shows that the present value of \$10,000 will grow to the FV of For example, suppose the 10% interest rate in the earlier example is compounded twice a year (semi-annually). Compounding Example 2: Calculate the Payment on a For Canadian mortgage loans, the interest is compounded semi-annually, rather than and the loan amount ( present value):.
### Suppose we have the following information to calculate compound interest in a table excel format (systematically). Step 1 – We need to name cell E3 as ‘Rate’ by selecting the cell and changing the name using Name Box. Step 2 – We have principal value or present value as 15000 and the annual interest rate is 5%.
31 Mar 2019 Future value (FV) is the value of a current asset at a future date based on an assumed rate of growth over time. more · Determining the Annual Future Value Function to Calculate Compound Semi-annual compounding interest formula Excel. The formula for the future value of a uniform series of of each year for 5 years, the FV function in Excel allows
### Future value of annuity. To get the present value of an annuity, you can use the PV function. In the example shown, the formula in C7 is: =FV(C5,C6,-C4,0,0) Explanation An annuity is a series of equal cash flows, spaced equally in time.
For example, if you invest \$100 for 5 years at an with interest paid annually at rate of 4%, the future value of this investment can be calculated by typing the 31 Mar 2019 Future value (FV) is the value of a current asset at a future date based on an assumed rate of growth over time. more · Determining the Annual Future Value Function to Calculate Compound Semi-annual compounding interest formula Excel.
## The result is a future dollar amount. Three types of compounding are annual, intra-year, and annuity compounding. This article discusses intra-year calculations for compound interest. For additional information about annual compounding, view the following article: FV function. Calculating Future Value of Intra-Year Compound Interest. Intra-year compound interest is interest that is compounded more frequently than once a year.
You can read the formula, "the future value (FVi) at the end of one year equals the interest rate and the superscript ⁿ is the number of compounding periods. for the future value of an investment or a lump sum on an Excel spreadsheet is:. Under the assumption that the 7% interest rate is a nominal rate of interest compounded monthly in the first case and semiannually in the second, we see that
The Excel compound interest formula in cell B4 of the above spreadsheet on the right once again calculates the future value of \$100, invested for 5 years with an annual interest rate of 4%. However, in this example, the interest is paid monthly. The result is a future dollar amount. Three types of compounding are annual, intra-year, and annuity compounding. This article discusses intra-year calculations for compound interest. For additional information about annual compounding, view the following article: FV function. Calculating Future Value of Intra-Year Compound Interest. Intra-year compound interest is interest that is compounded more frequently than once a year. Compound Interest in Excel Formula. Compound interest is the addition of interest to the principal sum of a loan or deposit, or we can say, interest on interest. It is the outcome of reinvesting interest, rather than paying it out, so that interest in the next period is earned on the principal sum plus previously accumulated interest. The future value of some amount of investment for a number of years can be shown using the same formula. Here the investment goes as the years added. The following picture shows the future value of an original investment of \$100 for different years, invested at an annual interest rate of 5%. Compound Interest Formula with Monthly Contributions in Excel. If the interest is paid monthly then the formula for future value becomes, Future Value = P*(1+r/12)^(n*12). The general formula for compound interest is: FV = PV(1+r)n, where FV is future value, PV is present value, r is the interest rate per period, and n is the number of compounding periods. How to calculate compound interest in Excel. One of the easiest ways is to apply the formula: (gross figure) x (1 + interest rate per period). Future value of annuity. To get the present value of an annuity, you can use the PV function. In the example shown, the formula in C7 is: =FV(C5,C6,-C4,0,0) Explanation An annuity is a series of equal cash flows, spaced equally in time. Future Value Formula in Excel (With Excel Template) The calculation of Future Value in excel is very easy and can take many variables which can be very difficult to calculate otherwise without a spreadsheet. | 1,551 | 7,173 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.899627 |
https://space.stackexchange.com/questions/35823/when-do-aircraft-become-solarcraft | 1,652,795,513,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662517485.8/warc/CC-MAIN-20220517130706-20220517160706-00512.warc.gz | 615,132,766 | 73,143 | # When do aircraft become solarcraft?
Any body travelling through particles undergoes drag. Any body able to generate lift (for instance spheres cannot generate lift) can generate lift if it undergoes drag.
First by assuming one body in a circular orbit around Earth at 250km altitude. Assuming this body has a non-zero lift to drag ratio, is able to control its attitude on three axis, and is able to provide constant thrust for multiple orbits.
If thrust is used to push this body in the same direction as its velocity vector, and its lift is steered to counter Earth's gravity, then it should be able to orbit slower than its orbital velocity, which means it's flying, rather than being in an ever missing Earth free fall / orbit.
Does this make it an aircraft? (deliberately ignoring Karman line/altitude definitions)
Secondly assuming the same body in a higher circular orbit in very thin exosphere: When does atmospheric drag and lift become less influent than drag and lift generated by solar radiation pressure? Is there an altitude at which this body cannot be an aircraft anymore, because dominant force acting on it is SRP, therefore it becomes a solarcraft?
Shouldn't this very region of high atmosphere become the boundary between flight and spaceflight?
Edit: This question is not about redefining Karman line or redefining what an aircraft, or a spacecraft should be characterized by.
This question is more about lift in its very essence, in the first place. Everything else are consequences.
I made the premise that every medium generating drag or pressure to some body moving into it, allows this body (if designed and oriented to do so) to generate lift. Is this correct?
Why can solar radiation pressure generate lift, but very thin atmosphere couldn't? why can particles travelling at light speed generate lift by hitting some slanted surface, when ~5-10km/s thin atmospheric particles produce drag and only drag?
I remember this comment to an answer to this question :
"Indeed it is true that "aerodynamics of gliding works about the same at ... 0.1 bar and 0.001 bar" but somewhere before you get to ~10E-6 bar that all changes: the gas becomes non-collisional, i.e the mean free path of the molecules or atoms is much larger than the scale of the vehicle. Then continuum aerodynamics doesn't work anymore. Notably, the mechanism for generating lift is very different, relying on momentum exchange from molecules or atoms impacting the surface of the vehicle. The usual mechanism of impact-adsorption-reemission at thermal velocities is very inefficient"
So why couldn't some spacecraft orbiting circular in leo, and having wings oriented with a positive angle of attack relative to Earth (like an airplane's wing relative to horizon) "fly" slower* than its intended orbital velocity, thanks to thrust provided by some engine, for instance air-breathing electric propulsion, or whatever kind of long duration, very low thrust engine?
*even if it flies at 7.123km/s without decaying, and should theoretically orbit unpowered at 7.124km/s
ADDENDUM: my interpretation of "lift" may be wrong, but it implies any kind of thrust generated by deflecting particles. For instance, a deeply stalled airfoil in dense atmosphere at 50deg angle of attack still produces some residual "lift" which should be defined by "thrust".
• The Karman line was pretty much invented to answer this question.
– user20636
Apr 29, 2019 at 13:18
• @JCRM what definition of Karman line are you referring to? It would be great to have a link towards the latest and widely accepted one. Apr 29, 2019 at 13:23
• @qqjkztd Karman is 100 km, a nice round number because it is somewhat arbitrary and not very scientific. There is a serious scientific, evidence-based proposal to lower it to 80 km. Why is FAI considering lowering the Karman Line to 80km?. Relevant: Have spacecraft ever dipped below the Karman line and then safely continued spaceflight? and also What would a “Kármán plane” look like, a bird, or a plane?
– uhoh
Apr 29, 2019 at 14:42
• WaterMolecule's answer to this question explains well why lift only works in continuum flow / low Knudsen number regimes: space.stackexchange.com/questions/31925/… In fact, this question is probably a duplicate of that one, since you state that it is "about lift in its very essence". Apr 29, 2019 at 21:03
• @OrganicMarble do greater than 10 Knudsen number ballistic collisions with thin atmosphere at 7km/s (or "impact lift" as mentioned by Joshua) behave the same as solar radiation pressure? (Are both kind of particles changing momentum in the same way when being deflected at some angle?) Apr 29, 2019 at 21:42
I think there are a few misconceptions to clarify here:
• Rotating bodies can generate lift. This is known as the Magnus effect.
• Lift is a hydrodynamical phenomenon: Differences in flow velocity above and below a moving body translate into pressure differences which heave the body up. However in very rarified gases this mechanism stops working. This is because when the gas becomes thin enough, the continuum approximation breaks down, particles become uncoupled ballistic singles and pressure is not translated from one point to another anymore. The Karman line is approximately this height, thus above this limit, Lift doesn't exist.
• Drag is a different phenomenon, it is the integral of all bumps with particles going another direction and thus well-defined and existent even in rarified gases.
• Orbiting slower than orbital velocity: There are not many cases where the force balance $$mv^2/r = GM/r^2$$ that defines the orbital velocity is upset by other forces. Only then this can lead to a different orbiting speed. In astrophysical contexts this would not be very exotic (gas in dense protoplanetary discs or black hole accretion discs does feel its own pressure gradient and magnetic torques), but in a spaceflight context I think this would be rather exotic.
• With the same reasoning we know that there is no lift generated from the solar wind, only drag. The drag can be either in the linear or quadratic regime, depending on the mean-free path of the respective gas. With the density of $$\sim 10$$ protons per $$cm^3$$, I estimated the mean-free path inside the Solar Wind to be $$10^{12}$$m, so we're in the linear regime, and then the drag friction time computes as $$t_f = \frac{r \rho_0}{c_s \rho_{gas}}$$, with $$r$$ being the object size, $$\rho_0$$ the density of the object, $$c_s$$ is the speed of sound of the gas (used to parameterize the temperature which gives bumps, it's still a non-continuum gas) and $$\rho_{gas}$$ is the density of the gas. Now we can search for the intersection of the two friction timescales, $$t_f^S$$ of the solar wind and $$t_f^A$$ in the atmosphere, so we require $$t_f^S = t_f^A$$, which translates into $$c_s^A \rho^A = c_s^S \rho^S$$. The speed of sound depends on temperature $$T$$ and mean molecular weight $$\mu$$ through $$c_s = \frac{k_B T}{\mu}$$ and I take $$T^A = 10^3K, \, T^S = 3.6\cdot 10^4 K, \, \mu^A=30, \, \mu^S=1, \rho^S= 10^{-23}g\,cm^{-3}$$. Then this approximately reduces this criterion to $$\rho^A \approx 10^3 \rho^S$$, which is fulfilled at an atmospheric density of $$\rho^A = 10^{-20} g \; cm^{-3}$$, which we reach only way beyond the exosphere. Interestingly this value would be reached radially before $$\rho^A = \rho^S$$ is reached, because the solar wind is so much hotter than the exosphere. I also wrongly assumed that $$\mu^A$$ doesn't change, but taking that into account wouldn't change the final conclusion.
To wrap up, with your proposal there would be quite some space.. until you reach space.
Edit:
To support the drag time computation as stated, I reference Weidenschilling (1977) and references therein, which state and use the friction times for different regimes of Knudsen and Reynolds numbers.
• How comes srp drag can be translated into lift? how would you control attitude on yaw roll pitch, on a body spinning to produce lift from magnus effect? Apr 29, 2019 at 15:16
• @qqjkztd: Sorry I don't understand your vocabulary. I'm a hydrodynamicist, unfortunately not a pilot. Can you post a link to the srp drag? Apr 29, 2019 at 15:18
• here it is, drag is pressure since it's not converted into lift link Apr 29, 2019 at 15:19
• page 542 "Airplane-like attitude and flight control by shifting and tilting four sail panels". which means srp can produce lift. Why couldn't very thin atmosphere do the same? Apr 29, 2019 at 15:21
• It's a misconception that aircraft generates lift by forcing air to move faster above the wing. Any aircraft generates lift by redirecting part of airflow downwards. This works for ballistic model too. If a gas molecule hits the wind and reflected slightly downwards, it pushes wing slightly upwards. That's a direct consequence of law of conservation of momentum. Everything else is merely a convenient computational model. And ballistic model ("Newtonian approximation") is actually one of the best approximations for highly supersonic lift. Feb 8, 2021 at 16:24
Some issues with this:
1. Drag is dependent on cross-sectional area, and so is solar radiation pressure. This would mean that different spacecraft would have a different definition of space. We might get around this by designing a "standard spacecraft", but that's exactly what defining the Karman line was for in the first place.
2. The exosphere is complicated and its boundary is based on where the solar radiation pressure exceeds Earth's gravitational pull on a Hydrogen atom. Why hydrogen? Why not atomic oxygen or something else?
3. Solar radiation is not constant, so again your definition of space would change with the sun
As a result, we needed to make a single definition. Thus a spacecraft is a craft that operates in space - that is, above the Karman line - at some point in its flight.
The Kármán line definition is not ignorable.
Wikipedia:
The Kármán line is therefore the highest altitude at which orbital speed provides sufficient aerodynamic lift to fly in a straight line that doesn't follow the curvature of the Earth's surface.
If you can remain above the Kármán line you are a spacecraft. Full stop.
In fact it's a bit worse. It is not possible for an aircraft to remain anywhere near the Kármán line because the velocity required to generate lift is too high and the reentry heating will doom it. The domain is well-split because the intermediate range is forbidden to both kinds of vehicles for long.
As you approach the Kármán line, the terms from the equations of motion from orbital dynamics become dominant and the terms from aerodynamics become corrections to them. At lower altitudes the situation is reversed. Continuing above the Kármán line, lift becomes negligible fairly quickly. There is, in fact, not much distance between the Kármán line and the altitude at which the mean free path becomes the same as the craft's cross section, which causes aerodnyamic lift to drop to zero, leaving only impact lift.
• To me this Karman line definition works for an infinitely large flat Earth. I know no airplane that doesn't, in the long term, follow the curvature of Earth's surface. No spacecraft can travel on a straight line that doesn't follow this curvature either, it could approach it by providing one 40g constant super-burn towards an hyperbolic escape trajectory, looking like a straight line relative to the ground. Why this difference between both cases in this definition? Apr 29, 2019 at 20:21
• @qqjkztd: The result is the Kármán line is approximately where the orbital domain takes over because orbital dynamics exceed aerodynamics. Apr 29, 2019 at 20:23
• +1 for "If you can remain above the Kármán line you are a spacecraft." But Wikipedia is not a reliable source for a definition of the Karman line. That quoted section is an internet explanation for something that was never well documented historically. The definition is 100 km. "Full stop." The justification of that number is shaky. Let's not relitigate Karman! How many variants of the “Was Karman wrong?” question is enough for one user?
– uhoh
Apr 30, 2019 at 0:16
• @uhoh: I can't help the fact this user will not believe that these things are more usefully defined by dominant term than some small term. Apr 30, 2019 at 0:21 | 2,863 | 12,325 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 21, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-21 | longest | en | 0.942796 |
http://physicsguide.blogspot.com/2006/02/will-grape-float-or-sink.html | 1,531,758,946,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589404.50/warc/CC-MAIN-20180716154548-20180716174548-00280.warc.gz | 301,818,218 | 11,250 | ## Bouyancy and Density
Will a grape float or sink in water? How about salt water? Learn about bouyancy and Archimedes' principle with this fun experiment.
What you will need:
1) Two Clear Glasses
2) Warm water (Hot faucet water will suffice)
3) Grapes
4) Salt
5) spoon
What to do
1) Fill the glasses
2) Disolve about three teaspoons of salt in one glass. Be sure to stir the salt into the water. Using warm water improves the disolution. 3) Place one grape in each glass. Do they sink or float?
What Happens
You should see the grape in the fresh water glass sink to the bottom and the slat water grape float.
Why?
Recall Archimedes’ principle: A floating object will displace a volume of fluid that has a weight equal to the weight of the object that is floating. If the object weighs more than its own volume in fluid, then it will sink.. In other words, a floating object is less dense than the fluid, but an object that is denser than the fluid will sink and the bouyancy force felt by an object is equal to the weight of the fluid displaced.
The grape is denser than fresh water, so it sinks in the fresh water glass.However, when you disolve salt in water, you increase the water's density to a value greater than the density of the typical grape - so the grape floats in salt water.
Moreover, the level the grape floats at is determined by the density of the water - if the water is very dense, only a very small part of the grape needs to be submerged to generate a sufficient bouyancy force to hold up the grape's weight, but if the water is only slightly denser than the grape, it will take most of the grape's volume to generate the required bouyancy force. Thus, you can use the height of the grape to estimate the amount of salt in the water.
Starting with fresh water, sink the grape in the glass. Slowly disolve a small amount of salt in the glass. Slowly increase the amount of salt until the grape rises to the top. Now, as you add more salt, the grape should rise upwards. There is a maximum amount of salt you can disolve into the water, so you might not be able to lift your grape a long way out of the water, but with care, you should be able to observe the grape's upwards motion. | 514 | 2,219 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2018-30 | latest | en | 0.918403 |
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#### Iifomula
##### J. D. Robinson Formula (1983)
• $$w = 52 kg + 1.9$$ kg nge-intshi ngaphezulu kweenyawo ezintlanu (yamadoda)
• $$w = 49 kg + 1.7$$ kg nge-intshi ngaphezulu kweenyawo ezintlanu (yabasetyhini)
##### D. Miller Ifomula (1983)
• $$w = 56.2 kg + 1.41$$ kg nge-intshi ngaphezulu kweenyawo ezintlanu (yamadoda)
• $$w = 53.1 kg + 1.36$$ kg nge-intshi ngaphezulu kweenyawo ezintlanu (yabasetyhini)
##### G. Hamwi Ifomula (1964)
• $$w = 48 kg + 2.7$$ kg nge-intshi ngaphezulu kweenyawo ezintlanu (yamadoda)
• $$w = 45.5 kg + 2.2$$ kg nge-intshi ngaphezulu kweenyawo ezintlanu (yabasetyhini)
##### B. J. Devine Ifomula (1974)
• $$w = 50 kg + 2.3$$ kg nge-intshi ngaphezulu kweenyawo ezintlanu (yamadoda)
• $$w = 45.5 kg + 2.3$$ kg nge-intshi ngaphezulu kweenyawo ezintlanu (yabasetyhini)
##### Uluhlu lweBMI
• $$18.5 - 25$$ (eyamadoda nabasetyhini) | 542 | 1,237 | {"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} | 3.34375 | 3 | CC-MAIN-2023-40 | latest | en | 0.184437 |
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Conversion Result: fluid ounce = 2.95735295625E-05 volume (volume) Related Measurements: Try converting from "oz fluid" to barrel, board foot, chetvert (Russian chetvert), drop, ephah (Israeli ephah), freight ton, gill, hekat (Israeli hekat), jeroboam, liter, minim, nebuchadnezzar, oil arroba (Spanish oil arroba), peck (dry peck), stere, tea cup, tun (English tun), UK oz fluid (British fluid ounce), UK pint (British pint), UK quart (British quart), or any combination of units which equate to "length cubed" and represent capacity, section modulus, static moment of area, or volume. Sample Conversions: oz fluid = 2.40E-08 acre foot, .00195313 balthazar, .80208333 bath (Israeli bath), .0063996 beer gallon (English beer gallon), .00083923 bushel (dry bushel), .00020981 coomb, .00006527 cord foot (of wood), .05371044 dry pint, .02685522 dry quart, .25 gill, .25 noggin, .00235316 oil arroba (Spanish oil arroba), .03125 quart (fluid quart), .00001044 register ton, .00520833 rehoboam, .00260417 salmanazar, .00041961 strike, .00285128 tou (Chinese tou), .00181686 wine arroba (Spanish wine arroba), .03903162 wine bottle.
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Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 486 | 1,740 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2020-16 | latest | en | 0.660502 |
https://orangecountymomblog.com/suckling/what-age-should-a-child-identify-numbers.html | 1,643,261,614,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305141.20/warc/CC-MAIN-20220127042833-20220127072833-00280.warc.gz | 509,301,819 | 18,664 | # What age should a child identify numbers?
Contents
Between 3 and 4 years of age, she’ll also be more adept at counting small sets of objects — “two oranges, four straws” and so on. Most children are not able to identify numerals or write them, though, until they’re 4 or 5.
## What age should a child recognize numbers?
Toddlers as young as 12 months old are able to recognize numbers. There are many ways you can work and play with your child to teach them numbers. When your toddler reaches the age of 3, they will even be able to start writing numbers.
## Should a 3 year old recognize numbers?
Most 3-year-olds can count to three and know the names of some of the numbers up to ten. Your child is also starting to recognize numbers from one to nine. He’ll be quick to point it out if he receives fewer cookies than his playmate.
## Can most 2 year olds count to 10?
Most 2 year old children are capable of counting to 10 although they may mix up the order of the numbers. Begin practicing numbers and counting with your toddler to help build a strong foundation for number fluency.
## Is it normal for a 2 year old to recognize numbers?
2-year-olds will start by recognizing the numbers, and then they will gradually begin to understand what each number means. When toddlers can count how many objects are given to them, they understand what that number means.
IT IS INTERESTING: What age can baby get in swing?
## How many numbers should a 4-year-old Recognise?
The average 4-year-old can count up to ten, although he may not get the numbers in the right order every time. One big hang-up in going higher? Those pesky numbers like 11 and 20. The irregularity of their names doesn’t make much sense to a preschooler.
## What number should a 5 year old count to?
Most 5-year-olds can recognize numbers up to ten and write them. Older 5-year-olds may be able to count to 100 and read numbers up to 20.
## Should a 2 year old know colors?
2 year olds can understand the concept of color and may begin to recognize and learn about colors as early as 18 months. Learning colors can be a fun activity for you and your child to practice together. Start with one color at a time, use flashcards to show your child a color and have them say the name with you.
## How do you know if your 2 year old is gifted?
12 signs of a gifted child
1. Quick learning. According to Louis, a telltale sign that a child is exceptionally bright for their age is how quickly they learn. …
2. Big vocabulary. …
3. Lots of curiosity. …
4. Eagerness to learn. … | 594 | 2,558 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2022-05 | latest | en | 0.959709 |
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# Edexcel IGCSE Mathematics A - Paper 3H - 2016 - Unofficial Mark Scheme
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1. (Original post by Nn99)
http://www.thestudentroom.co.uk/show...7#post65222087
MORE STUFF COMING
Q18. T=3M+1/M-1 THERE YOU GO, PLEASE SPREAd AND KEEP AN EYE ON THE ABOVE LINK, TOP NOTCH QUALITY STUFF COMING GUYZZZZZ
I knew the answer, just not the question.
2. http://www.thestudentroom.co.uk/show...st65222087MORE STUFF COMINGQ18. T=3M+1/M-1 THERE YOU GO, PLEASE SPREAd AND KEEP AN EYE ON THE ABOVE LINK, TOP NOTCH QUALITY STUFF COMING GUYZZZZZ
(Original post by BOBQ)
I think 4H will be easier because the 3H wasdifficult
3. (Original post by sivthasan)
That was the question: Make t the subject of the equation "m=t+1/t+2". I think it was either question 15 or 17
wasn't it m = t+1 / t-3
4. http://www.thestudentroom.co.uk/show...st65222087MORE STUFF COMINGQ18. T=3M+1/M-1 THERE YOU GO, PLEASE SPREAd AND KEEP AN EYE ON THE ABOVE LINK, TOP NOTCH QUALITY STUFF COMING GUYZZZZZ
http://www.thestudentroom.co.uk/show...st65222087MORE STUFF COMINGQ18. T=3M+1/M-1 THERE YOU GO, PLEASE SPREAd AND KEEP AN EYE ON THE ABOVE LINK, TOP NOTCH QUALITY STUFF COMING GUYZZZZZ
(Original post by _Xenon_)
What's grade boundaries like for AQA?
How many marks for a C and B for example?
Best of luck.
(Original post by Moustafa123r)
Did anyone had the mark scheme for igcse edexcel maths paper pls answer
(Original post by zXcodeXz)
C was 55/175 B was 85/175 A 117/175 A* 145/175
5. http://www.thestudentroom.co.uk/show...st65222087MORE STUFF COMINGQ18. T=3M+1/M-1 THERE YOU GO, PLEASE SPREAd AND KEEP AN EYE ON THE ABOVE LINK, TOP NOTCH QUALITY STUFF COMING GUYZZZZZ
http://www.thestudentroom.co.uk/show...st65222087MORE STUFF COMINGQ18. T=3M+1/M-1 THERE YOU GO, PLEASE SPREAd AND KEEP AN EYE ON THE ABOVE LINK, TOP NOTCH QUALITY STUFF COMING GUYZZZZZ
http://www.thestudentroom.co.uk/show...st65222087MORE STUFF COMINGQ18. T=3M+1/M-1 THERE YOU GO, PLEASE SPREAd AND KEEP AN EYE ON THE ABOVE LINK, TOP NOTCH QUALITY STUFF COMING GUYZZZZZ
(Original post by _Xenon_)
What's grade boundaries like for AQA?
How many marks for a C and B for example?
Best of luck.
(Original post by Moustafa123r)
Did anyone had the mark scheme for igcse edexcel maths paper pls answer
(Original post by zXcodeXz)
C was 55/175 B was 85/175 A 117/175 A* 145/175
No i meant out of the 100... but lately grade boundaries have dropped so its probably around 65 for an A so 130 -140 overall
6. (Original post by Martins1)
I knew the answer, just not the question.
thats what i got yes!! i love this forum
7. (Original post by BOBQ)
wasn't it m = t+1 / t-3
Yeah, sorry I made a mistake. It's t-3
8. (Original post by Nn99)
http://www.thestudentroom.co.uk/show...st65222087MORE STUFF COMINGQ18. T=3M+1/M-1 THERE YOU GO, PLEASE SPREAd AND KEEP AN EYE ON THE ABOVE LINK, TOP NOTCH QUALITY STUFF COMING GUYZZZZZ
http://www.thestudentroom.co.uk/show...st65222087MORE STUFF COMINGQ18. T=3M+1/M-1 THERE YOU GO, PLEASE SPREAd AND KEEP AN EYE ON THE ABOVE LINK, TOP NOTCH QUALITY STUFF COMING GUYZZZZZ
http://www.thestudentroom.co.uk/show...st65222087MORE STUFF COMINGQ18. T=3M+1/M-1 THERE YOU GO, PLEASE SPREAd AND KEEP AN EYE ON THE ABOVE LINK, TOP NOTCH QUALITY STUFF COMING GUYZZZZZ
Stop spamming one post is enough and secondly I've already made the post here so why copy exactly what i wrote??!!!
9. (Original post by Nn99)
http://www.thestudentroom.co.uk/show...st65222087MORE STUFF COMINGQ18. T=3M+1/M-1 THERE YOU GO, PLEASE SPREAd AND KEEP AN EYE ON THE ABOVE LINK, TOP NOTCH QUALITY STUFF COMING GUYZZZZZ
http://www.thestudentroom.co.uk/show...st65222087MORE STUFF COMINGQ18. T=3M+1/M-1 THERE YOU GO, PLEASE SPREAd AND KEEP AN EYE ON THE ABOVE LINK, TOP NOTCH QUALITY STUFF COMING GUYZZZZZ
http://www.thestudentroom.co.uk/show...st65222087MORE STUFF COMINGQ18. T=3M+1/M-1 THERE YOU GO, PLEASE SPREAd AND KEEP AN EYE ON THE ABOVE LINK, TOP NOTCH QUALITY STUFF COMING GUYZZZZZ
Could you stop spamming the forum with this link, once or twice is plenty, we understand that you are helping us, but copying and pasting quotes with your link and top notch quality message is not necessary. Thanks
10. (Original post by Martins1)
Could you stop spamming the forum with this link, once or twice is plenty, we understand that you are helping us, but copying and pasting quotes with your link and top notch quality message is not necessary. Thanks
do you remember the shaded rectangle question
11. (Original post by BOBQ)
wasn't it m = t+1 / t-3
Cool, so working for that q is:
m = t+1 / t-3
m(t-3)=t+1
mt-3m-t-1=0
mt-t=3m+1
t(m-1)=3m+1
t=3m+1/m-1
12. (Original post by _Xenon_)
Stop spamming one post is enough and secondly I've already made the post here so why copy exactly what i wrote??!!!
HAVE A LOOK, IM UPDATING THE POST, SO YOU CAN COPY THEM TO YOURS, WELL, JUST BEING FRIENDLy..... GO CHECK THAT LINK, UPDATING ANSWErS... BUT IF THAT'S NOT WHAT YOU WANT... IGNORE
13. (Original post by Nn99)
HAVE A LOOK, IM UPDATING THE POST, SO YOU CAN COPY THEM TO YOURS, WELL, JUST BEING FRIENDLy..... GO CHECK THAT LINK, UPDATING ANSWErS... BUT IF THAT'S NOT WHAT YOU WANT... IGNORE
http://www.thestudentroom.co.uk/show...st65222087MORE STUFF COMING PLEASE SPREAd AND KEEP AN EYE ON THE ABOVE LINK, TOP NOTCH QUALITY STUFF COMING GUYZZZZZ
14. (Original post by BOBQ)
do you remember the shaded rectangle question
Yes, but cant quite remember the numbers.
It was simple (base*height)-(pie*r^2)
Numbers was something like 13.7 and something else
15. (Original post by Martins1)
Cool, so working for that q is:
m = t+1 / t-3
m(t-3)=t+1
mt-3m-t-1=0
mt-t=3m+1
t(m-1)=3m+1
t=3m+1/m-1
yeah thats what i got as well
16. (Original post by Martins1)
Yes, but cant quite remember the numbers.
It was simple (base*height)-(pie*r^2)
Numbers was something like 13.7 and something else
i think the rectangle was 13.7 x 7.6
and the circle was pi x 2.5^2
17. (Original post by Martins1)
Yes, but cant quite remember the numbers.
It was simple (base*height)-(pie*r^2)
Numbers was something like 13.7 and something else
http://www.thestudentroom.co.uk/show...st65222087HAVE A LOOK, IM UPDATING THE POST, SO YOU CAN COPY THEM TO YOURS, WELL, JUST BEING FRIENDLy..... GO CHECK THAT LINK, UPDATING ANSWErS... BUT IF THAT'S NOT WHAT YOU WANT... IGNOREhttp://www.thestudentroom.co.uk/showthread.php?t=4123063&p=65222 087#post65222087MORE STUFF COMINGTHERE YOU GO, PLEASE SPREAd AND KEEP AN EYE ON THE ABOVE LINK, TOP NOTCH QUALITY STUFF COMING GUYZZZZZ
18. (Original post by Nn99)
http://www.thestudentroom.co.uk/show...st65222087HAVE A LOOK, IM UPDATING THE POST, SO YOU CAN COPY THEM TO YOURS, WELL, JUST BEING FRIENDLy..... GO CHECK THAT LINK, UPDATING ANSWErS... BUT IF THAT'S NOT WHAT YOU WANT... IGNOREhttp://www.thestudentroom.co.uk/showthread.php?t=4123063&p=65222 087#post65222087MORE STUFF COMINGTHERE YOU GO, PLEASE SPREAd AND KEEP AN EYE ON THE ABOVE LINK, TOP NOTCH QUALITY STUFF COMING GUYZZZZZ
Could you only please tell us again when it is updated when you have COMPLETED updating the answers. Only then. We do understand that you are being friendly and are helping us out, but for the moment, I don't want to know until you have finished adding the answers for every question. Thanks
19. (Original post by Nn99)
http://www.thestudentroom.co.uk/show...st65222087HAVE A LOOK, IM UPDATING THE POST, SO YOU CAN COPY THEM TO YOURS, WELL, JUST BEING FRIENDLy..... GO CHECK THAT LINK, UPDATING ANSWErS... BUT IF THAT'S NOT WHAT YOU WANT... IGNOREhttp://www.thestudentroom.co.uk/showthread.php?t=4123063&p=65222 087#post65222087MORE STUFF COMINGTHERE YOU GO, PLEASE SPREAd AND KEEP AN EYE ON THE ABOVE LINK, TOP NOTCH QUALITY STUFF COMING GUYZZZZZ
PLEASE stop. I don't mind, really but I don't like all this spamming!!!
20. (Original post by BOBQ)
i think the rectangle was 13.7 x 7.6
and the circle was pi x 2.5^2
Which would make it (13.7*7.6)-(pi*2.5^2)=84.5cm^2 (3sf)
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3 views
There is a site on it, users put different companies rating.
There is K — number of votes, N is the average incoming in the interval [1, 5].
Need some measure of T(K, N), which is a function of the average company's valuation and number of votes.
For this indicator, companies will be sorted.
The indicator should take into account the fact that the smaller number of votes, the greater the likelihood that the assessment differs from the objective (we will for simplicity be regarded as an objective average score of all users of the earth :)).
Has come up with just such a formula:
N = K+(K-3)*(N-1)*k/Nmax, where Nmax is the maximum number of votes, k is a coefficient chosen from common sense, looking at real data.
Maybe I reinvent the wheel and there are some mathematically justified and vital proven formula for such things?
| 3 views
0 like 0 dislike
There is a ready and now working algorithms, for example, the IMDB rating, which is based on Bayes ' theorem. The formula is very well explained here: www.wowwebdesigns.com/formula.php
\r
\r`WR = (v / (v+m)) * R + (m / (v+m)) * CR = average rating of the given objectv = number of votes for this objectm = (optional) the minimum number of votes needed to display in the top 25C = average rating of all objects\r`
Yet to review is an old article on Collective Choice: www.lifewithalacrity.com/2005/12/collective_choi.html Then painted the ever-popular ranking system.
by
0 like 0 dislike
I would use something like:
weighted average rating on this criterion * K-ENT reliability.
K-ENT reliability = weighted average number of all assessments for each criterion are normalized to 1.
ie — if your criteria voted more than weighted average number of votes, the weight of this assessment increases.
\r
meaning — the more assessments, the more weight result.
by
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## Susanna Epp
Valutazione media 4,67
( su 3 valutazioni fornite da Goodreads )
Susanna Epp’s DISCRETE MATHEMATICS WITH APPLICATIONS, 4e, International Edition provides a clear introduction to discrete mathematics. Renowned for her lucid, accessible prose, Epp explains complex, abstract concepts with clarity and precision. This book presents not only the major themes of discrete mathematics, but also the reasoning that underlies mathematical thought. Students develop the ability to think abstractly as they study the ideas of logic and proof. While learning about such concepts as logic circuits and computer addition, algorithm analysis, recursive thinking, computability, automata, cryptography, and combinatorics, students discover that the ideas of discrete mathematics underlie and are essential to the science and technology of the computer age. Overall, Epp’s emphasis on reasoning provides students with a strong foundation for computer science and upper-level mathematics courses.
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1. SPEAKING MATHEMATICALLY. Variables. The Language of Sets. The Language of Relations and Functions. 2. THE LOGIC OF COMPOUND STATEMENTS. Logical Form and Logical Equivalence. Conditional Statements. Valid and Invalid Arguments. Application: Digital Logic Circuits. Application: Number Systems and Circuits for Addition. 3. THE LOGIC OF QUANTIFIED STATEMENTS. Predicates and Quantified Statements I. Predicates and Quantified Statements II. Statements with Multiple Quantifiers. Arguments with Quantified Statements. 4. ELEMENTARY NUMBER THEORY AND METHODS OF PROOF. Direct Proof and Counterexample I: Introduction. Direct Proof and Counterexample II: Rational Numbers. Direct Proof and Counterexample III: Divisibility. Direct Proof and Counterexample IV: Division into Cases and the Quotient-Remainder Theorem. Direct Proof and Counterexample V: Floor and Ceiling. Indirect Argument: Contradiction and Contraposition. Indirect Argument: Two Classical Theorems. Application: Algorithms. 5. SEQUENCES, MATHEMATICAL INDUCTION, AND RECURSION. Sequences. Mathematical Induction I. Mathematical Induction II. Strong Mathematical Induction and the Well-Ordering Principle. Application: Correctness of Algorithms. Defining Sequences Recursively. Solving Recurrence Relations by Iteration. Second-Order Linear Homogeneous Recurrence Relations with Constant Coefficients. General Recursive Definitions and Structural Induction. 6. SET THEORY. Set Theory: Definitions and the Element Method of Proof. Set Identities. Disproofs and Algebraic Proofs. Boolean Algebras, Russell's Paradox, and the Halting Problem. 7. PROPERTIES OF FUNCTIONS. Functions Defined on General Sets. One-to-one, Onto, Inverse Functions. Composition of Functions. Cardinality, Sizes of Infinity, and Applications to Computability. 8. PROPERTIES OF RELATIONS. Relations on Sets (add material about relational databases). Reflexivity, Symmetry, and Transitivity. Equivalence Relations. Modular Arithmetic with Applications to Cryptography. Partial Order Relations. 9. COUNTING. Counting and Probability. The Multiplication Rule. Counting Elements of Disjoint Sets: The Addition Rule. The Pigeonhole Principle. Counting Subsets of a Set: Combinations. r-Combinations with Repetition Allowed. Pascal's Formula and the Binomial Theorem. Probability Axioms and Expected Value. Conditional Probability, Bayes' Formula, and Independent Events. 10. GRAPHS AND TREES. Graphs: An Introduction. Trails, Paths, and Circuits. Matrix Representations of Graphs. Isomorphisms of Graphs. Trees: Examples and Basic Properties. Rooted Trees. Spanning Trees and a Shortest Path Algorithm. 11. ANALYZING ALGORITHM EFFICIENCY. Real-Valued Functions of a Real Variable and -Notations. Application: Efficiency of Algorithms I.?-, and ?Their Graphs. O-, Exponential and Logarithmic Functions: Graphs and Orders. Application: Efficiency of Algorithms II. 12. REGULAR EXPRESSIONS AND FINITE STATE AUTOMATA. Formal Languages and Regular Expressions. Finite-State Automata. Simplifying Finite-State Automata.
Susanna S. Epp received her Ph.D. in 1968 from the University of Chicago, taught briefly at Boston University and the University of Illinois at Chicago, and is currently Vincent DePaul Professor of Mathematical Sciences at DePaul University. After initial research in commutative algebra, she became interested in cognitive issues associated with teaching analytical thinking and proof and has published a number of articles and given many talks related to this topic. She has also spoken widely on discrete mathematics and has organized sessions at national meetings on discrete mathematics instruction. In addition to Discrete Mathematics with Applications and Discrete Mathematics: An Introduction to Mathematical Reasoning, she is co-author of Precalculus and Discrete Mathematics, which was developed as part of the University of Chicago School Mathematics Project. Epp co-organized an international symposium on teaching logical reasoning, sponsored by the Institute for Discrete Mathematics and Theoretical Computer Science (DIMACS), and she was an associate editor of Mathematics Magazine from 1991 to 2001. Long active in the Mathematical Association of America (MAA), she is a co-author of the curricular guidelines for undergraduate mathematics programs: CUPM Curriculum Guide 2004.
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## 10.Discrete Mathematics with Applications
Editore: Brooks/Cole (2011)
ISBN 10: 0495826162 ISBN 13: 9780495826163
Nuovi Brossura Quantità: 5
Valutazione libreria
Descrizione libro Brooks/Cole, 2011. Condizione libro: New. 2011. International ed of 4th Revised ed. Paperback. Presents an introduction to discrete mathematics. This title explains complex, abstract concepts. It also presents the major themes of discrete mathematics, and the reasoning that underlies mathematical thought. Num Pages: 984 pages, Illustrations. BIC Classification: PBD; PBW. Category: (U) Tertiary Education (US: College). Dimension: 252 x 206 x 37. Weight in Grams: 1704. 936 pages, Illustrations. Presents an introduction to discrete mathematics. This title explains complex, abstract concepts. It also presents the major themes of discrete mathematics, and the reasoning that underlies mathematical thought. Cateogry: (U) Tertiary Education (US: College). BIC Classification: PBD; PBW. Dimension: 252 x 206 x 37. Weight: 1702. . . . . . . Codice libro della libreria V9780495826163
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Destinazione, tempi e costi | 3,429 | 13,386 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2017-30 | latest | en | 0.811581 |
https://zbmath.org/?q=an:0971.11046 | 1,718,566,333,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861670.48/warc/CC-MAIN-20240616172129-20240616202129-00870.warc.gz | 958,206,525 | 10,987 | ## Some congruences for binomial coefficients. II.(English)Zbl 0971.11046
Let $$p= tn+r$$ be a prime where $$t$$, $$n$$, $$r$$ are positive integers $$(1\leq r\leq t-1)$$ and either (1) $$t= k>3$$ for a prime $$k\equiv 3\pmod 4$$ or (2) $$t=4k$$ for a prime $$k\equiv 1\pmod 4$$. Let $$H$$ be the subgroup with the operator $$r$$ of $$(\mathbb{Z}/t\mathbb{Z})^\times$$ isomorphic to $$\text{Gal} (\mathbb{Q}(\xi_t)/ \mathbb{Q}(\sqrt{-t}))$$, $$h$$ be the class number of $$\mathbb{Q}(\sqrt{-t})$$, $$s= \frac{\varphi(t)} {2}$$ and $$d= ((p^s-1)/t) (t-1)= \sum_{j=0}^{s-1} d_j p^j$$ $$(0\leq d_j< p)$$. The existence of integers $$a$$, $$b$$ was shown by the authors [Proc. Class Field Theory – its Centenary and Prospect (1998), Adv. Stud. Pure Math. 30, 445-461 (2001)] such that $$4p^h= a^2+ tb^2$$ and $a\equiv\pm \prod_{j=0}^{s-1} (d_j)!\pmod p.$ If we consider $$k=1$$ and $$r=1$$, then we get Gauss’ statement $$a\equiv \binom {2n}{n}\pmod p$$ $$(4p= a^2+ 4b^2$$, $$\frac{a}{2}\equiv 1\pmod 4)$$.
The main result of this paper determines the sign of $$a$$ in the following sense: In case (1) $$a\equiv 2p^\beta\pmod t$$ and $$a\equiv (-1)^\beta \prod_{j=0}^{s-1} (d_j)!\pmod p$$ and in case (2) $$a\equiv -p^\beta\pmod k$$ and $$2a\equiv (-1)^\alpha \prod_{j=0}^{k-2} (d_j)!\pmod p$$.
The integer $$\beta$$ is defined by $\beta= \Bigl( \sum_{1\leq i\leq t, -i\in H} i\Bigr)/t$ and $$\alpha= s-\beta$$. To prove these results some equalities on Eisenstein sums are presented.
### MSC:
11L05 Gauss and Kloosterman sums; generalizations 11T24 Other character sums and Gauss sums 11B65 Binomial coefficients; factorials; $$q$$-identities 11A07 Congruences; primitive roots; residue systems
### Keywords:
Gauss sum; Eisenstein sum; binomial coefficients; congruences
Full Text:
### References:
[1] Berndt, B. C., Evans, R. J., and Williams, K. S.: Gauss and Jacobi Sums. Canad. Math. Soc. Ser. Monogr. Adv. Texts, vol. 21, Wiley, New York (1998). · Zbl 0906.11001 [2] Clarke, F.: Closed formulas for representing primes by binary quadratic forms (1998) (preprint, announced in ICM, Berlin). [3] Cohen, H.: A Course in Computational Algebraic Number Theory. Grad. Texts in Math., vol. 138, Springer, Berlin-Heidelberg-New York (1995). [4] Eisenstein, G.: Zur Theorie der quadratischen Zerfällung der Primzahlen $$8n+3$$, $$7n+2$$ und $$7n+\nobreak4$$. J. Reine Angew. Math., 37 , 97-126 (1848). [5] Hahn, S., and Lee, D. H.: Some congruences for binomial coefficients. Proc. for Class Field Theory - its Centenary and Prospect (1998) (to appear). · Zbl 1070.11007 [6] Lang, S.: Cyclotomic Fields I and II. 2nd ed., Grad. Texts in Math., vol. 121, Springer, Berlin-Heidelberg-New York (1990). · Zbl 0704.11038 [7] Lee, D. H., and Hahn, S.: Gauss sums and binomial coefficients. J. Number Theory (2000) (submitted). · Zbl 0994.11029 · doi:10.1006/jnth.2001.2688 [8] Washington, L. C.: Introduction to Cyclotomic Fields. 2nd ed., Grad. Texts in Math., vol. 83, Springer, Berlin-Heidelberg-New York (1997). · Zbl 0966.11047
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. In some cases that data have been complemented/enhanced by data from zbMATH Open. This attempts to reflect the references listed in the original paper as accurately as possible without claiming completeness or a perfect matching. | 1,209 | 3,465 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-26 | latest | en | 0.620289 |
https://uk.mathworks.com/matlabcentral/cody/problems/1355-given-area-find-sides/solutions/1442189 | 1,611,084,913,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519600.31/warc/CC-MAIN-20210119170058-20210119200058-00216.warc.gz | 641,504,946 | 17,002 | Cody
# Problem 1355. Given area find sides
Solution 1442189
Submitted on 16 Feb 2018 by GEORGIOS BEKAS
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
A = 400; y_correct = 20; assert(isequal(findside(A),y_correct))
2 Pass
A = 144; y_correct = 12; assert(isequal(findside(A),y_correct))
3 Pass
A = 225; y_correct = 15; assert(isequal(findside(A),y_correct))
4 Pass
A = 256; y_correct = 16; assert(isequal(findside(A),y_correct))
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 199 | 687 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-04 | latest | en | 0.659111 |
https://ro.scribd.com/document/366846557/A-Lesson-Plan-in-Statistics | 1,579,460,273,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594705.17/warc/CC-MAIN-20200119180644-20200119204644-00271.warc.gz | 631,621,134 | 75,747 | Sunteți pe pagina 1din 2
# A Lesson plan in statistics & probabilities for Grade 11
## I. Objective a.) To calculate the variance of probability distribution
b.) Solve problem involving variance of probability distribution
II. Learning Material
Subject Matter: Computing the variance of probability distribution
Reference: Statistics and probability
pp. 31-40
1. Pre-Activity
a. Unlocking of difficulty
What is variance?
b. Review
Finding the mean of a probability distribution
2. Proper- Activity
a. Activity
Computing the variance of probability distribution
(See Activity Sheet)
b. Analysis
What should be the first thing to be computed in solving for
the variance? (mean)
Whats next to do after finding the mean?
(get the difference of the mean from each variable)
Then we get the square of their __________? (difference)
What is the formula used in computing the variance?
c. Abstraction
How do we calculate variance of a probability distribution?
Who can enumerate the steps?
d. Application
When three coins are tossed, the probability distribution for
the random variable X representing the number of heads that
occur is given below. Compute the variance of the probability
distribution.
Steps Solution
## 1. Find the mean of the probability X P(X) XP(X) (X-)P(X)
distribution using the formula.
= X = P(X)
0 1
8
2. Multiply the value of the
random variable X by the 1 3
corresponding probability 8
2 3
8
3. Get the sum of the results
obtained in Step 2. 3 1
8
=(X-)P(X)=
4. Subtract the square of the mean
from the result obtained in Step 3
to get the variance. So, the
formula for the variance of a
probability distribution is given by
o= P(X)=
IV. Evaluation
The number of items sold per day a retail store with its
corresponding probabilities. Is shown in the table. Find the
variance of the probability distribution.
Steps Solution
1. Find the mean of the probability
X P(X) XP(X) (X-)P(X)
distribution using the formula.
= X = P(X)
19 0.20
2. Multiply the value of the
random variable X by the 20
corresponding probability
0.20
21 0.30
3. Get the sum of the results
obtained in Step 2. 22 0.20
23 0.10
P(X)=
4. Subtract the square of the mean
from the result obtained in Step 3
to get the variance. So, the
formula for the variance of a
probability distribution is given by
o= P(X)=
V. Assignment | 601 | 2,323 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2020-05 | latest | en | 0.828523 |
https://numeraly.com/about-the-number-758/ | 1,701,578,614,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100484.76/warc/CC-MAIN-20231203030948-20231203060948-00788.warc.gz | 492,005,301 | 20,013 | Welcome to About The Number 758, a dedicated webpage that dives deep into the fascinating world of this unique three-digit number. Here, you’ll uncover the various mathematical properties, historical significance, and interesting trivia surrounding 758, as well as explore its presence in different fields such as science, technology, and popular culture. So, let’s embark on an exciting numerical journey and uncover the hidden gems of this often-overlooked number!
## Fun and interesting facts about the number 758
The number 758 is an even composite number, having the prime factors 2 and 379, making it a semiprime number.
Interestingly, when written in Roman numerals, the number 758 is represented as DCCLVIII, and in binary, it is written as 1011110110.
## The number 758 angel number and biblical meaning
The number 758 angel number is a powerful combination of the energies of 7, 5, and 8, symbolizing spiritual awakening, personal growth, and abundance. In the biblical context, this number signifies God's guidance and protection, urging individuals to trust in divine wisdom and embrace positive changes in their lives.
## What is 758 written in words?
Seven hundred and fifty-eight
Like our Facebook page for great number facts and tips!
DCCLVIII
## What are the factors, prime factors, factor trees, cubes, binary number and hexadecimal of 758?
Factors of 758 are 1, 2, 379 and 758.
The prime factors of 758 are 2 and 379.
The factor tree of 758 is 2 and 379.
The cube of 758 is 435,519,512.
The binary number of 758 is 1011110110.
The hexadecimal of 758 is 2f6.
## Metric to imperial numbers
758 centimeters is 298.425 inches.
758 kilometers is 470.999 miles.
758 meters is 828.956 yards.
758 grams is 26.738 ounces.
758 kilograms is 1671.102 pounds.
758 litres is 1333.891 pints.
758 KPH (Kilometers Per Hour) is 470.999 MPH (Miles Per Hour). | 457 | 1,881 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.88735 |
https://bendwavy.org/klitzing/incmats/spiccup.htm | 1,624,453,884,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488538041.86/warc/CC-MAIN-20210623103524-20210623133524-00388.warc.gz | 131,429,678 | 3,754 | Acronym spiccup Name small-prismatotetracontoctachoron prism,runcinated-icositetrachoron prism Circumradius sqrt(2)+1/2 = 1.914214 Confer general polytopal classes: segmentotera Externallinks
Incidence matrix according to Dynkin symbol
```x x3o4o3x
. . . . . | 288 | 1 4 4 | 4 4 4 8 4 | 4 8 4 1 4 4 1 | 1 4 4 1 1
----------+-----+-------------+---------------------+---------------------------+--------------
x . . . . | 2 | 144 * * | 4 4 0 0 0 | 4 8 4 0 0 0 0 | 1 4 4 1 0
. x . . . | 2 | * 576 * | 1 0 2 2 0 | 2 2 0 1 2 1 0 | 1 2 1 0 1
. . . . x | 2 | * * 576 | 0 1 0 2 2 | 0 2 2 0 1 2 1 | 0 1 2 1 1
----------+-----+-------------+---------------------+---------------------------+--------------
x x . . . | 4 | 2 2 0 | 288 * * * * | 2 2 0 0 0 0 0 | 1 2 1 0 0
x . . . x | 4 | 2 0 2 | * 288 * * * | 0 2 2 0 0 0 0 | 0 1 2 1 0
. x3o . . | 3 | 0 3 0 | * * 384 * * | 1 0 0 1 1 0 0 | 1 1 0 0 1
. x . . x | 4 | 0 2 2 | * * * 576 * | 0 1 0 0 1 1 0 | 0 1 1 0 1
. . . o3x | 3 | 0 0 3 | * * * * 384 | 0 0 1 0 0 1 1 | 0 0 1 1 1
----------+-----+-------------+---------------------+---------------------------+--------------
x x3o . . ♦ 6 | 3 6 0 | 3 0 2 0 0 | 192 * * * * * * | 1 1 0 0 0
x x . . x ♦ 8 | 4 4 4 | 2 2 0 2 0 | * 288 * * * * * | 0 1 1 0 0
x . . o3x ♦ 6 | 3 0 6 | 0 3 0 0 2 | * * 192 * * * * | 0 0 1 1 0
. x3o4o . ♦ 6 | 0 12 0 | 0 0 8 0 0 | * * * 48 * * * | 1 0 0 0 1
. x3o . x ♦ 6 | 0 6 3 | 0 0 2 3 0 | * * * * 192 * * | 0 1 0 0 1
. x . o3x ♦ 6 | 0 3 6 | 0 0 0 3 2 | * * * * * 192 * | 0 0 1 0 1
. . o4o3x ♦ 6 | 0 0 12 | 0 0 0 0 8 | * * * * * * 48 | 0 0 0 1 1
----------+-----+-------------+---------------------+---------------------------+--------------
x x3o4o . ♦ 12 | 6 24 0 | 12 0 16 0 0 | 8 0 0 2 0 0 0 | 24 * * * *
x x3o . x ♦ 12 | 6 12 6 | 6 3 4 6 0 | 2 3 0 0 2 0 0 | * 96 * * *
x x . o3x ♦ 12 | 6 6 12 | 3 6 0 6 4 | 0 3 2 0 0 2 0 | * * 96 * *
x . o4o3x ♦ 12 | 6 0 24 | 0 12 0 0 16 | 0 0 8 0 0 0 2 | * * * 24 *
. x3o4o3x ♦ 144 | 0 288 288 | 0 0 192 288 192 | 0 0 0 24 96 96 24 | * * * * 2
```
```or
. . . . . | 288 | 1 8 | 8 8 8 | 8 8 2 8 | 2 8 1
-------------+-----+----------+-------------+----------------+---------
x . . . . | 2 | 144 * | 8 0 0 | 8 8 0 0 | 2 8 0
. x . . . & | 2 | * 1152 | 1 2 2 | 2 2 2 3 | 1 3 1
-------------+-----+----------+-------------+----------------+---------
x x . . . & | 4 | 2 2 | 576 * * | 2 2 0 0 | 1 3 0
. x3o . . & | 3 | 0 3 | * 768 * | 1 0 1 1 | 1 1 1
. x . . x | 4 | 0 4 | * * 576 | 0 1 0 2 | 0 2 1
-------------+-----+----------+-------------+----------------+---------
x x3o . . & ♦ 6 | 3 6 | 3 2 0 | 384 * * * | 1 1 0
x x . . x ♦ 8 | 4 8 | 4 0 2 | * 288 * * | 0 2 0
. x3o4o . & ♦ 6 | 0 12 | 0 8 0 | * * 96 * | 1 0 1
. x3o . x & ♦ 6 | 0 9 | 0 2 3 | * * * 384 | 0 1 1
-------------+-----+----------+-------------+----------------+---------
x x3o4o . & ♦ 12 | 6 24 | 12 16 0 | 8 0 2 0 | 48 * *
x x3o . x & ♦ 12 | 6 18 | 9 4 6 | 2 3 0 2 | * 192 *
. x3o4o3x ♦ 144 | 0 576 | 0 384 288 | 0 0 48 192 | * * 2
```
```xx3oo4oo3xx&#x → height = 1
o.3o.4o.3o. | 144 * | 4 4 1 0 0 | 4 8 4 4 4 0 0 0 | 1 4 4 1 4 8 4 0 0 0 0 | 1 1 4 4 1 0
.o3.o4.o3.o | * 144 | 0 0 1 4 4 | 0 0 0 4 4 4 8 4 | 0 0 0 0 4 8 4 1 4 4 1 | 0 1 4 4 1 1
---------------+---------+---------------------+---------------------------------+-------------------------------------+----------------
x. .. .. .. | 2 0 | 288 * * * * | 2 2 0 1 0 0 0 0 | 1 2 1 0 2 2 0 0 0 0 0 | 1 1 2 1 0 0
.. .. .. x. | 2 0 | * 288 * * * | 0 2 2 0 1 0 0 0 | 0 1 2 1 0 2 2 0 0 0 0 | 1 0 1 2 1 0
oo3oo4oo3oo&#x | 1 1 | * * 144 * * | 0 0 0 4 4 0 0 0 | 0 0 0 0 4 8 4 0 0 0 0 | 0 1 4 4 1 0
.x .. .. .. | 0 2 | * * * 288 * | 0 0 0 1 0 2 2 0 | 0 0 0 0 2 2 0 1 2 1 0 | 0 1 2 1 0 1
.. .. .. .x | 0 2 | * * * * 288 | 0 0 0 0 1 0 2 2 | 0 0 0 0 0 2 2 0 1 2 1 | 0 0 1 2 1 1
---------------+---------+---------------------+---------------------------------+-------------------------------------+----------------
x.3o. .. .. | 3 0 | 3 0 0 0 0 | 192 * * * * * * * | 1 1 0 0 1 0 0 0 0 0 0 | 1 1 1 0 0 0
x. .. .. x. | 4 0 | 2 2 0 0 0 | * 288 * * * * * * | 0 1 1 0 0 1 0 0 0 0 0 | 1 0 1 1 0 0
.. .. o.3x. | 3 0 | 0 3 0 0 0 | * * 192 * * * * * | 0 0 1 1 0 0 1 0 0 0 0 | 1 0 0 1 1 0
xx .. .. ..&#x | 2 2 | 1 0 2 1 0 | * * * 288 * * * * | 0 0 0 0 2 2 0 0 0 0 0 | 0 1 2 1 0 0
.. .. .. xx&#x | 2 2 | 0 1 2 0 1 | * * * * 288 * * * | 0 0 0 0 0 2 2 0 0 0 0 | 0 0 1 2 1 0
.x3.o .. .. | 0 3 | 0 0 0 3 0 | * * * * * 192 * * | 0 0 0 0 1 0 0 1 1 0 0 | 0 1 1 0 0 1
.x .. .. .x | 0 4 | 0 0 0 2 2 | * * * * * * 288 * | 0 0 0 0 0 1 0 0 1 1 0 | 0 0 1 1 0 1
.. .. .o3.x | 0 3 | 0 0 0 0 3 | * * * * * * * 192 | 0 0 0 0 0 0 1 0 0 1 1 | 0 0 0 1 1 1
---------------+---------+---------------------+---------------------------------+-------------------------------------+----------------
x.3o.4o. .. ♦ 6 0 | 12 0 0 0 0 | 8 0 0 0 0 0 0 0 | 24 * * * * * * * * * * | 1 1 0 0 0 0
x.3o. .. x. ♦ 6 0 | 6 3 0 0 0 | 2 3 0 0 0 0 0 0 | * 96 * * * * * * * * * | 1 0 1 0 0 0
x. .. o.3x. ♦ 6 0 | 3 6 0 0 0 | 0 3 2 0 0 0 0 0 | * * 96 * * * * * * * * | 1 0 0 1 0 0
.. o.4o.3x. ♦ 6 0 | 0 12 0 0 0 | 0 0 8 0 0 0 0 0 | * * * 24 * * * * * * * | 1 0 0 0 1 0
xx3oo .. ..&#x ♦ 3 3 | 3 0 3 3 0 | 1 0 0 3 0 1 0 0 | * * * * 192 * * * * * * | 0 1 1 0 0 0
xx .. .. xx&#x ♦ 4 4 | 2 2 4 2 2 | 0 1 0 2 2 0 1 0 | * * * * * 288 * * * * * | 0 0 1 1 0 0
.. .. oo3xx&#x ♦ 3 3 | 0 3 3 0 3 | 0 0 1 0 3 0 0 1 | * * * * * * 192 * * * * | 0 0 0 1 1 0
.x3.o4.o .. ♦ 0 6 | 0 0 0 12 0 | 0 0 0 0 0 8 0 0 | * * * * * * * 24 * * * | 0 1 0 0 0 1
.x3.o .. .x ♦ 0 6 | 0 0 0 6 3 | 0 0 0 0 0 2 3 0 | * * * * * * * * 96 * * | 0 0 1 0 0 1
.x .. .o3.x ♦ 0 6 | 0 0 0 3 6 | 0 0 0 0 0 0 3 2 | * * * * * * * * * 96 * | 0 0 0 1 0 1
.. .o4.o3.x ♦ 0 6 | 0 0 0 0 12 | 0 0 0 0 0 0 0 8 | * * * * * * * * * * 24 | 0 0 0 0 1 1
---------------+---------+---------------------+---------------------------------+-------------------------------------+----------------
x.3o.4o.3x. ♦ 144 0 | 288 288 0 0 0 | 192 288 192 0 0 0 0 0 | 24 96 96 24 0 0 0 0 0 0 0 | 1 * * * * *
xx3oo4oo ..&#x ♦ 6 6 | 12 0 6 12 0 | 8 0 0 12 0 8 0 0 | 1 0 0 0 8 0 0 1 0 0 0 | * 24 * * * *
xx3oo .. xx&#x ♦ 6 6 | 6 3 6 6 3 | 2 3 0 6 3 2 3 0 | 0 1 0 0 2 3 0 0 1 0 0 | * * 96 * * *
xx .. oo3xx&#x ♦ 6 6 | 3 6 6 3 6 | 0 3 2 3 6 0 3 2 | 0 0 1 0 0 3 2 0 0 1 0 | * * * 96 * *
.. oo4oo3xx&#x ♦ 6 6 | 0 12 6 0 12 | 0 0 8 0 12 0 0 8 | 0 0 0 1 0 0 8 0 0 0 1 | * * * * 24 *
.x3.o4.o3.x ♦ 0 144 | 0 0 0 288 288 | 0 0 0 0 0 192 288 192 | 0 0 0 0 0 0 0 24 96 96 24 | * * * * * 1
```
```or
o.3o.4o.3o. | 144 * | 8 1 0 | 8 8 8 0 0 | 2 4 8 8 0 0 | 1 2 8 0
.o3.o4.o3.o | * 144 | 0 1 8 | 0 0 8 8 8 | 0 0 8 8 2 4 | 0 2 8 1
------------------+---------+-------------+---------------------+-----------------------+-----------
x. .. .. .. & | 2 0 | 576 * * | 2 2 1 0 0 | 1 3 2 2 0 0 | 1 1 3 0
oo3oo4oo3oo&#x | 1 1 | * 144 * | 0 0 8 0 0 | 0 0 8 8 0 0 | 0 2 8 0
.x .. .. .. & | 0 2 | * * 576 | 0 0 1 2 2 | 0 0 2 2 1 3 | 0 1 3 1
------------------+---------+-------------+---------------------+-----------------------+-----------
x.3o. .. .. & | 3 0 | 3 0 0 | 384 * * * * | 1 1 1 0 0 0 | 1 1 1 0
x. .. .. x. | 4 0 | 4 0 0 | * 288 * * * | 0 2 0 1 0 0 | 1 0 2 0
xx .. .. ..&#x & | 2 2 | 1 2 1 | * * 576 * * | 0 0 2 2 0 0 | 0 1 3 0
.x3.o .. .. & | 0 3 | 0 0 3 | * * * 384 * | 0 0 1 0 1 1 | 0 1 1 1
.x .. .. .x | 0 4 | 0 0 4 | * * * * 288 | 0 0 0 1 0 2 | 0 0 2 1
------------------+---------+-------------+---------------------+-----------------------+-----------
x.3o.4o. .. & ♦ 6 0 | 12 0 0 | 8 0 0 0 0 | 48 * * * * * | 1 1 0 0
x.3o. .. x. & ♦ 6 0 | 9 0 0 | 2 3 0 0 0 | * 192 * * * * | 1 0 1 0
xx3oo .. ..&#x & ♦ 3 3 | 3 3 3 | 1 0 3 1 0 | * * 384 * * * | 0 1 1 0
xx .. .. xx&#x ♦ 4 4 | 4 4 4 | 0 1 4 0 1 | * * * 288 * * | 0 0 2 0
.x3.o4.o .. & ♦ 0 6 | 0 0 12 | 0 0 0 8 0 | * * * * 48 * | 0 1 0 1
.x3.o .. .x & ♦ 0 6 | 0 0 9 | 0 0 0 2 3 | * * * * * 192 | 0 0 1 1
------------------+---------+-------------+---------------------+-----------------------+-----------
x.3o.4o.3x. ♦ 144 0 | 576 0 0 | 384 288 0 0 0 | 48 192 0 0 0 0 | 1 * * *
xx3oo4oo ..&#x & ♦ 6 6 | 12 6 12 | 8 0 12 8 0 | 1 0 8 0 1 0 | * 48 * *
xx3oo .. xx&#x & ♦ 6 6 | 9 6 9 | 2 3 9 2 3 | 0 1 2 3 0 1 | * * 192 *
.x3.o4.o3.x ♦ 0 144 | 0 0 576 | 0 0 0 384 288 | 0 0 0 0 48 192 | * * * 1
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https://spectre-code.org/SodExplosion_8hpp_source.html | 1,627,219,718,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151672.96/warc/CC-MAIN-20210725111913-20210725141913-00262.warc.gz | 549,460,774 | 7,783 | SodExplosion.hpp
2 // See LICENSE.txt for details.
3
4 #pragma once
5
6 #include <array>
7 #include <cstddef>
8 #include <limits>
9
10 #include "DataStructures/DataVector.hpp"
13 #include "Evolution/Systems/NewtonianEuler/Sources/NoSource.hpp"
14 #include "Evolution/Systems/NewtonianEuler/Tags.hpp"
15 #include "Options/Options.hpp"
16 #include "PointwiseFunctions/AnalyticData/AnalyticData.hpp"
17 #include "PointwiseFunctions/Hydro/EquationsOfState/EquationOfState.hpp"
18 #include "PointwiseFunctions/Hydro/EquationsOfState/IdealFluid.hpp"
19 #include "Utilities/MakeArray.hpp"
20 #include "Utilities/TMPL.hpp"
21 #include "Utilities/TaggedTuple.hpp"
22
23 /// \cond
24 namespace PUP {
25 class er; // IWYU pragma: keep
26 } // namespace PUP
27 /// \endcond
28
30 /*!
31 * \brief A cylindrical or spherical Sod explosion \cite Toro2009 \cite Sod19781
32 *
33 * Common initial conditions are:
34 *
35 * \f{align*}{
36 * (\rho, v^i, p) =
37 * &\left\{
38 * \begin{array}{ll}
39 * (1 ,0, 1) & \mathrm{if} \; r \le 0.5 \\
40 * (0.125 ,0, 0.1) & \mathrm{if} \; r > 0.5
41 * \end{array}\right.
42 * \f}
43 *
44 * where \f$r\f$ is the cylindrical (2d) or spherical (3d) radius. This test
45 * problem uses an adiabatic index of 1.4. A reference solution can be computed
46 * in 1d by solving the Newtonian Euler equations in cylindrical or spherical
47 * symmetry. Note that the inner and outer density and pressure, as well as
48 * where the initial discontinuity is can be chosen arbitrarily.
49 */
50 template <size_t Dim>
51 class SodExplosion : public MarkAsAnalyticData {
52 public:
53 static_assert(Dim > 1, "Sod explosion is a 2d and 3d problem.");
56
57 /// Initial radius of the discontinuity
58 struct InitialRadius {
59 using type = double;
60 static constexpr Options::String help = {
61 "The initial radius of the discontinuity."};
62 static type lower_bound() noexcept { return 0.0; }
63 };
64
66 using type = double;
67 static constexpr Options::String help = {"The inner mass density."};
68 static type lower_bound() noexcept { return 0.0; }
69 };
70
71 struct InnerPressure {
72 using type = double;
73 static constexpr Options::String help = {"The inner pressure."};
74 static type lower_bound() noexcept { return 0.0; }
75 };
76
78 using type = double;
79 static constexpr Options::String help = {"The outer mass density."};
80 static type lower_bound() noexcept { return 0.0; }
81 };
82
83 struct OuterPressure {
84 using type = double;
85 static constexpr Options::String help = {"The outer pressure."};
86 static type lower_bound() noexcept { return 0.0; }
87 };
88
89 using options = tmpl::list<InitialRadius, InnerMassDensity, InnerPressure,
91
92 static constexpr Options::String help = {
93 "Cylindrical or spherical Sod explosion."};
94
95 SodExplosion() = default;
96 SodExplosion(const SodExplosion& /*rhs*/) = delete;
97 SodExplosion& operator=(const SodExplosion& /*rhs*/) = delete;
98 SodExplosion(SodExplosion&& /*rhs*/) noexcept = default;
99 SodExplosion& operator=(SodExplosion&& /*rhs*/) noexcept = default;
100 ~SodExplosion() = default;
101
102 SodExplosion(double initial_radius, double inner_mass_density,
103 double inner_pressure, double outer_mass_density,
104 double outer_pressure, const Options::Context& context = {});
105
106 /// Retrieve a collection of hydrodynamic variables at position x
107 template <typename... Tags>
109 const tnsr::I<DataVector, Dim, Frame::Inertial>& x,
110 tmpl::list<Tags...> /*meta*/) const noexcept {
111 return {tuples::get<Tags>(variables(x, tmpl::list<Tags>{}))...};
112 }
113
114 const equation_of_state_type& equation_of_state() const noexcept {
115 return equation_of_state_;
116 }
117
118 // clang-tidy: no runtime references
119 void pup(PUP::er& /*p*/) noexcept; // NOLINT
120
121 private:
123 const tnsr::I<DataVector, Dim, Frame::Inertial>& x,
124 tmpl::list<Tags::MassDensity<DataVector>> /*meta*/) const noexcept;
125
126 tuples::TaggedTuple<Tags::Velocity<DataVector, Dim, Frame::Inertial>>
127 variables(
128 const tnsr::I<DataVector, Dim, Frame::Inertial>& x,
129 tmpl::list<Tags::Velocity<DataVector, Dim, Frame::Inertial>> /*meta*/)
130 const noexcept;
131
132 tuples::TaggedTuple<Tags::Pressure<DataVector>> variables(
133 const tnsr::I<DataVector, Dim, Frame::Inertial>& x,
134 tmpl::list<Tags::Pressure<DataVector>> /*meta*/) const noexcept;
135
136 tuples::TaggedTuple<Tags::SpecificInternalEnergy<DataVector>> variables(
137 const tnsr::I<DataVector, Dim, Frame::Inertial>& x,
138 tmpl::list<Tags::SpecificInternalEnergy<DataVector>> /*meta*/)
139 const noexcept;
140
141 template <size_t SpatialDim>
142 friend bool
143 operator==( // NOLINT (clang-tidy: readability-redundant-declaration)
144 const SodExplosion<SpatialDim>& lhs,
145 const SodExplosion<SpatialDim>& rhs) noexcept;
146
147 double initial_radius_ = std::numeric_limits<double>::signaling_NaN();
148 double inner_mass_density_ = std::numeric_limits<double>::signaling_NaN();
149 double inner_pressure_ = std::numeric_limits<double>::signaling_NaN();
150 double outer_mass_density_ = std::numeric_limits<double>::signaling_NaN();
151 double outer_pressure_ = std::numeric_limits<double>::signaling_NaN();
152 equation_of_state_type equation_of_state_{};
153 };
154
155 template <size_t Dim>
156 bool operator!=(const SodExplosion<Dim>& lhs,
157 const SodExplosion<Dim>& rhs) noexcept;
158 } // namespace NewtonianEuler::AnalyticData
NewtonianEuler::AnalyticData::SodExplosion::InnerMassDensity
Definition: SodExplosion.hpp:65
std::rel_ops::operator!=
T operator!=(T... args)
NewtonianEuler::Sources::NoSource
Used to mark that the initial data do not require source terms in the evolution equations.
Definition: NoSource.hpp:21
Options.hpp
MakeArray.hpp
NewtonianEuler::Tags::MassDensity
The mass density of the fluid.
Definition: Tags.hpp:26
NewtonianEuler::AnalyticData
Holds classes implementing analytic data for the NewtonianEuler system.
Definition: EvolveNewtonianEulerFwd.hpp:9
Options
Utilities for parsing input files.
Definition: MinmodType.hpp:8
cstddef
NewtonianEuler::AnalyticData::SodExplosion::OuterMassDensity
Definition: SodExplosion.hpp:77
EquationsOfState::IdealFluid< false >
array
NewtonianEuler::AnalyticData::SodExplosion
A cylindrical or spherical Sod explosion .
Definition: SodExplosion.hpp:51
tuples::TaggedTuple
An associative container that is indexed by structs.
Definition: TaggedTuple.hpp:271
DataVector
Stores a collection of function values.
Definition: DataVector.hpp:46
Initial radius of the discontinuity.
Definition: SodExplosion.hpp:58
NewtonianEuler::AnalyticData::SodExplosion::variables
tuples::TaggedTuple< Tags... > variables(const tnsr::I< DataVector, Dim, Frame::Inertial > &x, tmpl::list< Tags... >) const noexcept
Retrieve a collection of hydrodynamic variables at position x.
Definition: SodExplosion.hpp:108
tnsr
Type aliases to construct common Tensors.
Definition: TypeAliases.hpp:31
NewtonianEuler::AnalyticData::SodExplosion::OuterPressure
Definition: SodExplosion.hpp:83
limits
TypeAliases.hpp
Options::String
const char *const String
The string used in option structs.
Definition: Options.hpp:32
Frame
Definition: IndexType.hpp:36
Tensor.hpp
NewtonianEuler::AnalyticData::SodExplosion::InnerPressure
Definition: SodExplosion.hpp:71
TMPL.hpp | 2,103 | 7,358 | {"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} | 2.53125 | 3 | CC-MAIN-2021-31 | latest | en | 0.531845 |
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# There has been some speculation that the severed engine of the airplan
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There has been some speculation that the severed engine of the airplan [#permalink]
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11 Nov 2019, 23:07
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Question Stats:
60% (01:37) correct 40% (01:39) wrong based on 234 sessions
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Project SC Butler: Day 190: Sentence Correction (SC1)
There has been some speculation that the severed engine of the airplane might have struck the tail, but investigators will withhold judgment until they examined further the engine's strike marks and its cause.
A) will withhold judgment until they examined further the engine's strike marks and its cause
B) will withhold judgment until they have further examined the strike marks on the engine and their cause
C) will withhold judgment until they have further examined the cause for the strike marks and the engine
D) have withheld judgment while there was a further examination of the strike marks on the engine and its cause
E) have withheld judgment until further examination of engine, the strike marks, and the cause
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There has been some speculation that the severed engine of the airplan [#permalink]
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11 Nov 2019, 23:09
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OFFICIAL EXPLANATION
Project SC Butler: Day 190: Sentence Correction (SC1)
• HIGHLIGHTS
Meaning?
In this case, a general understanding is enough: a plane crashed, some people are speculating about the cause of the crash, but investigators will not make a judgment right now. (Investigators "will withhold judgment.")
Investigators will withhold judgment until they have examined certain items more thoroughly.
[My decision: this crash happened on the runway so everyone is safe.]
Issues tested: verb tense, meaning and logic, pronoun/noun agreement, and rhetorical construction
THE PROMPT
Quote:
There has been some speculation that the severed engine of the airplane might have struck the tail, but investigators will withhold judgment until they examined further the engine's strike marks and its cause.
THE OPTIONS
Quote:
A) There has been some speculation that the severed engine of the airplane might have struck the tail, but investigators will withhold judgment until they [have] examined further the engine's strike marks and its cause.
• pronoun/noun disagreement
-- The marks on the engine are plural. Option A should say "their cause," not "its cause."
• Wrong verb. Examined is simple past tense. We need have examined. (present perfect = has/have + examined).
-- Until is a time frame "signal" or "code" word in English. When you see the word until, be prepared to check verbs.
-- in general, until refers to a span of time before a specified point in time or target date
Until means in the future or forward in time.
-- we use present perfect when we want to talk about an action performed during a time period that has not yet finished.
The time period in this sentence has not yet finished. The investigators must examine further.
-- have examined goes along with will withhold judgment.
-- Correct: Until they have further examined P and Q further, they will withhold judgment.
-- Correct: They will withhold judgment until they have further examined P and Q
-- Wrong: Until they examined P and Q more thoroughly, they will withhold judgment. You cannot "examined" in the future.
The pronoun and verb errors are fatal. Eliminate A
Quote:
B) There has been some speculation that the severed engine of the airplane might have struck the tail, but investigators will withhold judgment until they have further examined the strike marks on the engine and their cause.
• I do not see any errors
have examined corrects the error in (A)
• the plural possessive pronoun their correctly refers to the plural marks [on the engine].
will withhold corrects verb error in (D) and (E) (see below)
KEEP
Quote:
C) There has been some speculation that the severed engine of the airplane might have struck the tail, but investigators will withhold judgment until they have further examined the cause for the strike marks and the engine.
• WTH? Nonsensical meaning. The investigators are not looking for the "cause" of the engine.
There is no such thing as a "cause" for (or of) an engine.
• When a causal relationship is involved, use cause OF, not cause for.
-- Correct: Viruses are the cause OF many illnesses.
-- Wrong: Viruses are the cause for many illnesses.
"Cause for" sort of means "having a valid reason for."
Correct but rarely seen on the GMAT: The widening split between rich and poor is cause for alarm. (I have a good reason to be alarmed.)
There is no such thing as a cause for (or of) an engine. Eliminate C
Quote:
D) There has been some speculation that the severed engine of the airplane might have struck the tail, but investigators have withheld judgment while there was a further examination of the strike marks on the engine and its cause.
• Wrong verb. Have withheld should be will withhold.
-- the investigators will continue one action (withholding judgment) until after another action is finished (examining P and Q).
-- the investigators will [continue to] withhold judgment until they have further examined P and Q
-- "investigators have withheld judgment" cannot be paired with "until they have further examined P and Q."
"have withheld" means "up to this point in time."
But the further examination comes in the future.
The investigators WILL withhold judgment until after the further examination is finished.
• same pronoun errors as that in (A): marks, plural, cannot be paired with its, singular
• rhetorical construction: there was is often unnecessary.
-- GMAC is not fond of "there was" unless good reason exists to use it.
We use there was when we want to de-emphasize the doer, do not know the doer, want to set a background.
In this sentence, there was is rhetorically unnecessary and makes the sentence less effective.
All the other options are not awkward without "there was"—we should be wary.
Eliminate D
Quote:
E) There has been some speculation that the severed engine of the airplane might have struck the tail, but investigators have withheld judgment until further examination of [the] engine, the strike marks, and the cause.
• verb error is similar to that in (D). We need will withhold
• further examination of "the cause"? Of what? The strike marks? The fact that the engine might have struck the tail? The accident generally?
• THE engine is correct. (I copied the question exactly as I found it. I suspect a transcription error. We need "the" engine in order to keep the list parallel but also because idiomatically, that noun takes an article.
Eliminate E
• NOTES
Pronoun ambiguity in (B)? Nope.
-- The pronoun must have only one logical antecedent.
It does not matter that other plural nouns exist.
Yes, the antecedent must agree in number and gender, but the mere fact of more than one noun in agreement does not equal ambiguity.
Ambiguity happens when there is more than one logical antecedent for a pronoun.
The investigators are not examining themselves. ("Investigators" is the only other plural noun.)
arvind910619 and pandajee , welcome to SC Butler.
I am glad to see a few people whose posts I haven't seen in awhile.
And of course I am glad to see posts from the core group of Butler participants whose good posts make these threads eminently readable.
In fact, note to people who have not read the thread: you should do so. These answers are excellent. A few are outstanding. You will find many different approaches, layers of analysis, ways of writing posts, and insights.
Nice work. Kudos to all.
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Re: There has been some speculation that the severed engine of the airplan [#permalink]
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11 Nov 2019, 23:55
1
There has been some speculation that the severed engine of the airplane might have struck the tail, but investigators will withhold judgment until they examined further the engine's strike marks and its cause.
A) will withhold judgment until they examined further the engine's strike marks and its cause
B) will withhold judgment until they have further examined the strike marks on the engine and their cause
C) will withhold judgment until they have further examined the cause for the strike marks and the engine
D) have withheld judgment while there was a further examination of the strike marks on the engine and its cause
E) have withheld judgment until further examination of engine, the strike marks, and the cause
Meaning:
There has been some speculation that the severed engine of the airplane might have struck the tail, but investigators will withhold judgment until they examined further the engine's strike marks and its cause. An airplane has been stuck and there are some speculations around it but investigators have withhold their judgement until they examine the engine strike marks or what caused it.
withhold vs withheld
will withhold: correct denotes future tense
have withheld: withheld refers to past tense hence is incorrect. Eliminated D and E
In A, B and C option A doesnot have a verb in the underlined portion hence is a run on sentence. Eliminate option A
Option C: changes the original meaning and states that the examine the cause for strike marks and engine. eliminated C
IMO B: the option corrects all errors and gives clear meaning
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There has been some speculation that the severed engine of the airplan [#permalink]
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11 Nov 2019, 23:58
1
There has been some speculation that the severed engine of the airplane might have struck the tail, but investigators will with hold judgment until they examined further the engine's strike marks and its cause.
A) will with hold judgment until they examined further the engine's strike marks and its cause
the past tense use of examined is wrong since they are examining it further!! and the examination os the engine is not over yet!
B) will with hold judgment until they have further examined the strike marks on the engine and their cause- their illogically refers to the an engine
C) will with hold judgment until they have further examined the cause for the strike marks and the engine- CORRECT!
D) have with held judgment while there was a further examination of the strike marks on the engine and its cause- past tense for something(investigation) thats about to be revealed is wrong!!!
E) have with held judgment until further examination of engine, the strike marks, and the cause-
1)past tense for something(investigation) thats about to be revealed is wrong!!!
2)the cause cannot be further examined - meaning error! ( an object can be examined)
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Re: There has been some speculation that the severed engine of the airplan [#permalink]
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12 Nov 2019, 02:45
1
D) have withheld judgment while there was a further examination of the strike marks on the engine and its cause
E) have withheld judgment until further examination of engine, the strike marks, and the cause
First split: examined vs examination: stylistically, a verb is better than a noun, so we can cross out, almost safely, D and E.
It is better to say "We improved the engine" rather than "We made an improvement in engine". The latter is still correct but wordy.
Then, let's examine the remaining A-B-C:
A) will withhold judgment until they examined further the engine's strike marks and its cause - certainly wrong because "its" refers to "engine's", leading to illogical meaning that investigators will further investigate engine's cause
B) will withhold judgment until they have further examined the strike marks on the engine and their cause - correct (IMHO); meaning is clear here
C) will withhold judgment until they have further examined the cause for the strike marks and the engine - "cause for" does not look correct + shift in meaning as per choice A the investigators will not investigate the engine rather its strike marks and their cause
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Re: There has been some speculation that the severed engine of the airplan [#permalink]
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12 Nov 2019, 04:34
1
generis wrote:
There has been some speculation that the severed engine of the airplane might have struck the tail, but investigators will withhold judgment until they examined further the engine's strike marks and its cause.
A) will withhold judgment until they examined further the engine's strike marks and its cause
B) will withhold judgment until they have further examined the strike marks on the engine and their cause
C) will withhold judgment until they have further examined the cause for the strike marks and the engine
D) have withheld judgment while there was a further examination of the strike marks on the engine and its cause
E) have withheld judgment until further examination of engine, the strike marks, and the cause
MEANING
Investigators will withhold judgment until they have examined the marks [plural] and their [plural] cause.
A) "examined further" misplaced; "the marks [plural] and its [sing]" sva;
C) "the cause for the marks and [the cause for] the engine" unintended - the original states that investigators are examining the marks and their cause, not the cause for the engine (this is too vague);
D) "have withheld… while there was a further examination" imps there is another investigation going on, unintended; "marks [plural]… its [sing]" sva;
E) "have…examination of engine… and the cause" not parallel and unintended;
Ans (B)
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There has been some speculation that the severed engine of the airplan [#permalink]
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12 Nov 2019, 06:53
1
Imo. B.
There has been some speculation that the severed engine of the airplane might have struck the tail, but investigators will withhold judgment until they examined further the engine's strike marks and its cause.
A) will withhold judgment until they examined further the engine's strike marks and its cause - pronoun agreement error, its logically refers to strike marks (not the engine). The sentence changes the meaning ....until further they examined further the engine's strike marks and its cause. Can cause be examined? No. Problem can be examined to find the cause. Cause is the outcome of the examination.
B) will withhold judgment until they have further examined the strike marks on the engine and their cause - Seems no error as their correctly refers to strike marks, It correctly expresses the meaning that investigators will withheld judgement until they further examined the SM and cause of SM. Hold it.
C) will withhold judgment until they have further examined the cause for the strike marks and the engine - Distorts the meaning, ...they have further examined the cause of the strike marks and cause of the engine. Nonsensical meaning.
D) have withheld judgment while there was a further examination of the strike marks on the engine and its cause - Pronoun agreement error, same as A. Usage of have withheld changes the tense of the event mentioned in the original sentence. Usage of while is inappropriate.
E) have withheld judgment until further examination of engine, the strike marks, and the cause - Important information mentioned as non essential modifier. If we strikes off the non essential information, then there will be change the meaning, same as option A.
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Re: There has been some speculation that the severed engine of the airplan [#permalink]
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12 Nov 2019, 09:26
1
There has been some speculation that the severed engine of the airplane might have struck the tail, but investigators will withhold judgment until they examined further the engine's strike marks and its cause.
A) will withhold judgment until they examined further the engine's strike marks and its cause
Two errors : there is also a pronoun error here. The strikes is a plural noun and needs a plural pronoun.the right verb tense is not used.Instead of the simple past, we need the present perfect have examined
B) will withhold judgment until they have further examined the strike marks on the engine and their cause
This one clears up the pronoun error while keeping the meaning clear
C) will withhold judgment until they have further examined the cause for the strike marks and the engine
This one distorts meaning. The investigators will look at the Engine's strike marks and their cause not the engine itself
D) have withheld judgment while there was a further examination of the strike marks on the engine and its cause
pronoun errror. The right pronoun should be their and not its
E) have withheld judgment until further examination of engine, the strike marks, and the cause
i could say there is a parallelism error here because each part of the parallel structure starts with an article "the" but engine does not. The more concrete error is one of meaning. The strike marks are those of the engine and not a separate thing that is to be examined.
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Re: There has been some speculation that the severed engine of the airplan [#permalink]
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12 Nov 2019, 10:13
1
generis wrote:
Project SC Butler: Day 190: Sentence Correction (SC1)
There has been some speculation that the severed engine of the airplane might have struck the tail, but investigators will withhold judgment until they examined further the engine's strike marks and its cause.
A) will withhold judgment until they examined further the engine's strike marks and its cause
B) will withhold judgment until they have further examined the strike marks on the engine and their cause
C) will withhold judgment until they have further examined the cause for the strike marks and the engine
D) have withheld judgment while there was a further examination of the strike marks on the engine and its cause
E) have withheld judgment until further examination of engine, the strike marks, and the cause
Hi,
IMO B.
Use of simple future tense seemed unnecessary at the first sight but after reading all the answer choices, it seemed best among 5.
Basic structure of the sentence: IC1, FANBOYS IC2.
Meaning: There is a theory that the engine might have hit the tail of the airplane but the investigators will only say something after they have examined everything.
Error analysis:
A) will withhold judgment until they examined further the engine's strike marks and its cause: needs plural pronoun to refer to strike marks. Second there is no helping verb for examined.
B) will withhold judgment until they have further examined the strike marks on the engine and their cause: corrects the errors present in original sentence.
C) will withhold judgment until they have further examined the cause for the strike marks and the engine: changes the meaning, Incorrect.
D) have withheld judgment while there was a further examination of the strike marks on the engine and its cause: This sounds like someone is reporting the ongoing investigation, which is not a problem. The error is the use of singular pronoun. Incorrect.
E) have withheld judgment until further examination of engine, the strike marks, and the cause: Same as D. examination of "the cause" is illogical. Incorrect.
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Re: There has been some speculation that the severed engine of the airplan [#permalink]
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12 Nov 2019, 10:54
1
sampriya wrote:
There has been some speculation that the severed engine of the airplane might have struck the tail, but investigators will with hold judgment until they examined further the engine's strike marks and its cause.
A) will with hold judgment until they examined further the engine's strike marks and its cause
the past tense use of examined is wrong since they are examining it further!! and the examination os the engine is not over yet!
B) will with hold judgment until they have further examined the strike marks on the engine and their cause- their illogically refers to the an engine
C) will with hold judgment until they have further examined the cause for the strike marks and the engine- CORRECT!
D) have with held judgment while there was a further examination of the strike marks on the engine and its cause- past tense for something(investigation) thats about to be revealed is wrong!!!
E) have with held judgment until further examination of engine, the strike marks, and the cause-
1)past tense for something(investigation) thats about to be revealed is wrong!!!
2)the cause cannot be further examined - meaning error! ( an object can be examined)
IMO- In option B - "their" is used for Marks. Pronoun "their"
represents the noun "Marks". Since"Engine" is inside a prepositional phrase, it cannot be the subject.
Option C seeks for examination of Engine also. But the question mandates examination of strike marks & their cause.
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Re: There has been some speculation that the severed engine of the airplan [#permalink]
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12 Nov 2019, 13:33
1
There has been some speculation that the severed engine of the airplane might have struck the tail, but investigators will withhold judgment until they examined further the engine's strike marks and its cause.
A) will withhold judgment until they examined further the engine's strike marks and its cause >>> it should be a present perfect tense, not a past tense. Moreover, the adverb further will precede the verb. Also, "the strike marks of/ on the engine" is more acceptable than "the engine's strike marks".
B) will withhold judgment until they have further examined the strike marks on the engine and their cause>>> it seems correct.
C) will withhold judgment until they have further examined the cause for the strike marks and the engine>>> distorts the original meaning
D) have withheld judgment while there was a further examination of the strike marks on the engine and its cause>>> wrong use of verb. its has no clear antecedent.
E) have withheld judgment until further examination of engine, the strike marks, and the cause>>>same as C.
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Re: There has been some speculation that the severed engine of the airplan [#permalink]
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12 Nov 2019, 18:12
1
Quote:
There has been some speculation that the severed engine of the airplane might have struck the tail, but investigators will withhold judgment until they examined further the engine's strike marks and its cause.
This one is a bit tricky. So Let's see how we can deal with each option.
Quote:
A) will withhold judgment until they examined further the engine's strike marks and its cause
Primary reason to eliminate: Tense Error. "Until they examined" is Incorrect. its cause? cause of what?
Quote:
B) will withhold judgment until they have further examined the strike marks on the engine and their cause
Tense Error is solved. We have "until they have". This they refers back to investigators and "their cause" refers back to the marks. So hang on to this one
Quote:
C) will withhold judgment until they have further examined the cause for the strike marks and the engine
So what this is saying is : "cause for the engine"? That doesn't make sense.
Quote:
D) have withheld judgment while there was a further examination of the strike marks on the engine and its cause
Well then who is doing the examination? Also the its remain an issue.
Quote:
E) have withheld judgment until further examination of engine, the strike marks, and the cause
I don't know how to use this word but non-sensical? Just changes the meaning and creates fictional Parallelism. Also, The cause of what?
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Re: There has been some speculation that the severed engine of the airplan [#permalink]
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12 Nov 2019, 19:35
1
Quote:
There has been some speculation that the severed engine of the airplane might have struck the tail, but investigators will withhold judgment until they examined further the engine's strike marks and its cause.
Quote:
A) will withhold judgment until they examined further the engine's strike marks and its cause
Until they examined is an awkward and incomplete construct. It should be until they have examined. Eliminate.
Quote:
B) will withhold judgment until they have further examined the strike marks on the engine and their cause
Clear meaning. They will examine 2 things : The strike marks on the engine and the cause of strike marks. Parallelism maintained. Keep.
Quote:
C) will withhold judgment until they have further examined the cause for the strike marks and the engine
Meaning issue. The cause for strike marks is fine but the cause for the engine doesn't make sense at all. Eliminate.
Quote:
D) have withheld judgment while there was a further examination of the strike marks on the engine and its cause
'Strike marks' and 'its' are not in Pronoun Agreement. Eliminate.
Quote:
E) have withheld judgment until further examination of engine, the strike marks, and the cause
First of all, it is not mentioned, as compared to B , that who is going to do the examination.
Secondly, the cause in the last is left unattended. The cause of what???
IMO B.
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Re: There has been some speculation that the severed engine of the airplan [#permalink]
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12 Nov 2019, 20:28
1
There has been some speculation that the severed engine of the airplane might have struck the tail, but investigators will withhold judgment until they examined further the engine's strike marks and its cause.
Meaning: There has been some speculation. This speculation was on the severed engine of an airplane. The speculation is that the severed engine of the airplane probably struck the tail. Investigations are underway and not yet concluded. As a result, Investigators will withhold their judgment on whether the severed engine struck the tail or not. The investigators' judgment will come after they have examined: 1) the engine and 2)the cause of the strike marks.
With this meaning in mind, let's examine the answer choices.
Option A: There has been some speculation that the severed engine of the airplane might have struck the tail, but investigators will withhold judgment until they [have] examined further the engine's strike marks and its cause.
Logically, what does the singular possessive pronoun its refer to? It does not have a logical singular antecedent. One would expect that it refers to strike marks, but strike marks is plural. Strike is an appositive modifying marks. Second, option A misses the modal verb have expected to precede examined. Without the modal verb, option A suggests that the investigators have already concluded their examinations, and this is wrong from the perspectives of the meaning of the sentence. Based on these errors, we can eliminate option A.
Option B: There has been some speculation that the severed engine of the airplane might have struck the tail, but investigators will withhold judgment until they have further examined the strike marks on the engine and their cause.
Option B is very tempting because it sounds very nice to the ear. But logically what does their refer to? One would expect their to refer to strike marks, especially if you are depending on your ear, but their cannot refer to strike marks. We already know that they refers to investigators, so their must also refer to same antecedent. There is, therefore, a pronoun error in option B as well. Let's eliminate option B.
Option C: There has been some speculation that the severed engine of the airplane might have struck the tail, but investigators will withhold judgment until they have further examined the cause for the strike marks and the engine.
This the correct answer. The parallelism is between X: the cause of the strike marks and Y: the engine, two noun phrases. it makes sense to say that the investigators will withhold judgment until they have further examined(the parallel marker): X: the cause of the strike marks and Y: the engine. Therefore, let's keep option C.
Option D: There has been some speculation that the severed engine of the airplane might have struck the tail, but investigators have withheld judgment while there was a further examination of the strike marks on the engine and its cause.
The same pronoun error in option A is present in option D. its does not have a logical antecedent. In addition, had there been no pronoun error in option D, option C is superior because it is more concise. As far as concision is concerned, they have further examined is more concise than there was a further examination. I stand to be corrected on my judgment on concision. Let's eliminate option D.
Option E: There has been some speculation that the severed engine of the airplane might have struck the tail, but investigators have withheld judgment until further examination of [the] engine, the strike marks, and the cause( of what?).
Until further examination of the cause of what? There is ambiguity in meaning. Of course, we know that it is the cause of the strike marks that has to be examined. Option E does not convey the meaning clearly. Secondly, let's look at the elements of the parallelism in option E. Why are we decoupling the strike marks from the causes of the strike marks? The parallelism is illogical. Let's eliminate option E.
Option C is the best answer.
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Re: There has been some speculation that the severed engine of the airplan [#permalink]
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13 Nov 2019, 03:19
1
generis wrote:
Project SC Butler: Day 190: Sentence Correction (SC1)
There has been some speculation that the severed engine of the airplane might have struck the tail, but investigators will withhold judgment until they examined further the engine's strike marks and its cause.
A) will withhold judgment until they examined further the engine's strike marks and its cause
B) will withhold judgment until they have further examined the strike marks on the engine and their cause
C) will withhold judgment until they have further examined the cause for the strike marks and the engine
D) have withheld judgment while there was a further examination of the strike marks on the engine and its cause
E) have withheld judgment until further examination of engine, the strike marks, and the cause
This question tests the ability to correctly decipher meaning and parallelism using "and".
Let us check each option
A It has pronoun mistake. Its should refer to strike mark but in the sentence as written strike marks is plural so we need a plural pronoun. They examined is in wrong tense
B Looks good. Pronoun is proper as their refers to strikes
C This choice looks doubtful. will withhold judgment until they have further examined the cause for the strike marks and the engine. Highlighted test is not parallel cause for the engine does not make sense.
D Proper tense is not used we have a present perfect tense but this action is differed to future. Usage of "while" is not proper as the two action are not happening simultaneously. Out
E Same tense error as D. Bad parallelism and improper list with meaning error
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Re: There has been some speculation that the severed engine of the airplan [#permalink]
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13 Nov 2019, 09:52
1
Quote:
There has been some speculation that the severed engine of the airplane might have struck the tail, but investigators will withhold judgment until they examined further the engine's strike marks and its cause.
Quick read-through shows that the sentence tests meaning, structure, and tenses. Before jumping into answer choices, it is a good idea to boil down the sentence into simple form that can be explained to a 10 year old. That way, the reader can understand the meaning, find the core structure, and potentially spot errors.
We have an interesting (probably fatal) situation that likely involves the plain crash. The non-underlined part of the sentence says that the engine of the plane might have struck the tail. Then the sentence proceeds by saying that investigators have a position. Simplified, something happened in the past, and some people want to do more research about the cause and effect before reaching any conclusions. Let's review the options.
A) will withhold judgment until they examined further the engine's strike marks and its cause
One can argue that its correctly refers to strike, but this does not make much sense. We have strike marks, which is a noun phrase, and somethings caused them. So, we would need a plural pronoun to refer to strike marks.
B) will withhold judgment until they have further examined the strike marks on the engine and their cause
This option looks promising. The investigator will withhold judgement until they do more research. The pronoun error is resolved, i.e. their correctly refers to the strike marks.
C) will withhold judgment until they have further examined the cause for the strike marks and the engine
There can be the cause for the strike marks, but there cannot be the cause for the engine. Fatal meaning error.
D) have withheld judgment while there was a further examination of the strike marks on the engine and its cause
There are several issues with this option. First, we may not want to use Present Perfect tense (have withheld). Second, there is a big problem with a pronoun its. What does it refer to? Engine? Strike?
E) have withheld judgment until further examination of engine, the strike marks, and the cause
Grammatically this option can probably live. Assuming one can keep the Present Perfect tense. But the biggest problem comes at the very end. The sentence says "until further examination of A, B, and C". Okay, what is C here? It is the cause. Until further examination of the cause of what? The sentence is not complete. Also, one may want to use {the} before engine (not sure whether "the" is 100% necessary, though).
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Re: There has been some speculation that the severed engine of the airplan [#permalink]
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13 Nov 2019, 14:49
I have posted the official explanation here.
Re: There has been some speculation that the severed engine of the airplan [#permalink] 13 Nov 2019, 14:49
Display posts from previous: Sort by | 7,943 | 36,783 | {"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} | 3.71875 | 4 | CC-MAIN-2019-51 | latest | en | 0.945264 |
https://math.stackexchange.com/questions/1008645/real-analysis-heine-borel-property | 1,652,776,248,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662517018.29/warc/CC-MAIN-20220517063528-20220517093528-00330.warc.gz | 455,004,557 | 65,005 | # real analysis heine-borel property
Given: 1) is (-2,4) compact by applying Heine-Borel Property? and 2) is [-2,4] compact by applying Heine-Borel property?
I've followed the proof of Heine-Borel property and i have attempted to apply the proof to answer the following and i am not quite sure how to prove? it...
i know that [a,b], every open cover has a finite subcover and a set of real numbers is compact if and only if it is closed and bounded.
So for 1) no and 2) yes?
• $(2, -4)$ is not compact since it is not closed, while $[-2,4]$ is compact since it is closed and bounded. Nov 6, 2014 at 7:00
• could you show me an example for both Nov 6, 2014 at 7:02
• You are correct. Nov 6, 2014 at 7:09
If you know Heine-Borel Theorem, then its fairly obvious wich is compact and wich isnt, $(-2,4)$ is not closed so it cannot be compact, and $[-2,4]$ is both closed and bounded, so it is compact. If you want to do it more 'hands on' you can try the following:
For $1)$: Try to construct a sequence wich doesnt have any subsequence converging in the space $(-2,4)$
For $2)$ How do you know $[a,b]$ is compact? Hint: Let $I_n$ be an open cover of $[a,b]$, then take
$$C = \{x : \text{ such that [a,x] has a finite subcover} \}$$
Prove $\sup C$ exists and that $\sup C = \max C = b$
This will yield a general result for $[-2,4]$ you can even replace $a = -2, b = 4$ and use the same hint.
• would you like to show everyone how you know? Nov 6, 2014 at 7:37
• @chris Did you tried to do anything of what I suggested? I just used what I know about the characterization of compacts sets, you can read more on wikipedia en.wikipedia.org/wiki/Compact_space
– aram
Nov 6, 2014 at 7:39 | 517 | 1,689 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2022-21 | latest | en | 0.9474 |
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