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https://www.scribd.com/document/59875308/VDWaals-Short-Notes | 1,500,675,599,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423809.62/warc/CC-MAIN-20170721202430-20170721222430-00245.warc.gz | 838,193,739 | 29,756 | # van der Waals Equation Study Notes
Background
The Ideal Gas Law is valid for all gasses at sufficiently low pressure. However, as pressure on an given amount of gas increases, we observe deviations to the expected values These deviation occur because the ideal gas law assumes: • gas molecules are point-sized • that when gas molecules collide, they do so elastically -- in other words that the molecules bounce off each other without being attracted to each other. The Real life conditions: • molecules are not points, they are 3 dimensional • there are attractive forces between the gas molecules The Volume reality • Since molecules are 3 dimensional, they have length, width, height and consequently volume • The gas molecules therefore use up a certain amount of the total volume of the container they are in • The actual volume available for the gas molecules to move around in is actually less than V. • The real available volume depends on the size of the gas molecule and how many gas molecules are present The Pressure Reality • There will always be certain attractive forces between a molecule of a given gas and its neighboring molecules • The attractive forces between a molecule and its neighbors will be directed away from the direction of the force with which the molecule will be hitting the container • As a result, the molecules will hit the wall of its container with less force than if the gas were ideal • The attractive forces between real gas molecules have the effect of reducing the pressure from what we would expect if we were dealing with an ideal gas. Why? Because P = F/A, the pressure is directly proportional to the force striking an area. Since the attractive forces have the effect of decreasing the force of the molecule’s collision on the container, the pressure will likewise be decreased. Making corrections for the real world situation
• 2 constants need to be introduced into the PV=nRT equation a constant to account for the attractive forces and density of the gas that impacts the ideal pressure (the constant) a constant to account for the size of the molecules which impacts the ideal volume (the constant) Calculating the Real Pressure • The pressure reduction due to the attractive forces is equal to where is a constant for a given gas 2 • Pactual = Pideal • Pideal is the P in the gas equation PV=nRT 2 • We have to add back the pressure loss of caused by the attractive forces of the gas back to the ideal state P 2 should be substituted for the P term in = • P + Calculating the Real Volume • The total volume taken up by the molecules is where is the number of gas molecules and is a constant that depends on the gas and is a measure of the size of the molecule. • We have to subtract out the total volume of the gas molecules from the total volume of the container to get the true volume we have to work with • The actual available volume is V • V– should be substituted for the V term in = . Deriving the van der Waals equation • Plug in the substitutions we calculated to address the pressure and volume corrections: = (P + 2 2 ) (V - )= expanding out the first term will give us: (P + )(V )= (the van der Waals equation) • The constant and are called the van der Waals constants and they depend on the specific gas under consideration. .
or both to answer the problem. or info that provides you with both the and constant values.• The constant has the units of L2 • bar • mol-2 (L2 • bar/mol2) and has the units of L/mol. Look for keywords like: real. ideally. (Some question may try and trick you by not providing you with a table. And sometimes this will be exactly the correct answer to the problem. compared to. to determine whether to use the ideal gas. non-ideal. substitute the given and derived values (with their proper or new units) . the difference between. Make sure that you read the question carefully so that you know what gas formula they are looking for. You will always have to use a table to look up the values of the and constants for the specific gas you are dealing with. Convert any temperature you are given to Kelvin degrees. 3. perform any initial unit conversions 3. ideal. Solving van der Waals Problems Assuming we have enough information to solve a van der Waals problem. but the fact that they mention ideal behavior means that you need to use the ideal gas law rather than the van der Waals equation. if necessary 4. etc. Look and see if they are looking for the answer to be in a specific unit. 2. Take the van der Waals equation and manipulate it to isolate the specific term we are looking for 2. if they give you a fake gas (say Gas Q). They can give you information on a real gas (say Chlorine) and its constants. Just treat it according to what they give you. don’t freak out. you cannot solve the van der Waals equation. van der Waals equation. 5. Since you were given the van der Waal constants. but turn around and ask you how it would ideally behave. In either case. we need to 1. ideal behavior. which may require you to make before or after unit conversions. Also. or else be given them in the problem. Here are some conversion formulas: 1 J·m3/mol2 = 1 m6·Pa/mol2 = 10 L2·bar/mol2 1 dm3/mol = 1 L/mol Preparing to Solve van der Waals Problems 1. you might be tempted to solve the van der Waals equation. and the only answer that you can give is that the answer cannot be determined because you do not have sufficient info.
the numerator of the first term (2 mol)(0. .(4.17 L² atm / mol²) x (2 mol)² / (1L²) P = (46. look up the van der Waals constants for the gas (or gases) if necessary plug in all the numbers you have into the equation Perform the arithmetic to solve the equation Do any final conversions to get the answer into the final units they are looking for.94 Latm) / (0. 8. Adding the exponents for each particular unit will give L•atm as the numerator units. use P= Example: What is the pressure exerted by 2 moles of ammonia at 13◦C in a 1 L container? (use a = 4.(an² / V²) P = (2 mol) (0.15 K. T is given in oC. let’s just call it at 286 K.08206 L•atm•mol-1 •K-1)(286 K1).16.16. For example. and to use the gas laws we have to be in K degrees. the first thing we have to do is look at the units we are provided.70 atm = 34.0371 L/mol as the rounded off van der Waals constants for ammonia) ( – ) ( ) To solve this problem. so the numerator units • divided by the denominator units for the first term ends up being or just atm. Doing the same with the denominator we end up with the unit L. So we convert the 10 Celsius degrees we are given in the problem by adding 273.70 atm . but for simplicity.08206 L•atm/mol •K)(286K) / (1 L – (2 mol) x 0. We are now in good shape to work the problem: P = nRT / (V-nb) . 7.15 in order to come up with the Kelvin temperature we need to use. If volume and temperature are known and they are looking for pressure. 6.0 atm Sometimes I find it easier to deal with the cross-cancellation of units by having them all expressed in exponential form rather as a fractional part of a numerator or denominator within a fraction.0371 L/mol) . This comes out to 286.17 L2 · atm/mol2 and b = 0.70 atm P = 50.9258 L) . 1.08206 L•atm/mol •K)(286 K) can be written as : (2 mol1) (0. Look back at the question to determine the final answer you need to provide and how it should be provided.4. 5.
your answer should be: The pressure exerted on the ideal gas will be 12. and the temperature is needed.083145 L•atm/mol •K)(286K) / (1 L – (2 mol) x 0. So your response should be with respect to the ideal gas.70 atm = 34.67 atm Step 2 – Now figure out the pressure that would be exerted on an ideal gas: P = nRT/V = 2 x 0.67 atm and Pideal gas = 47. If pressure and volume are given. the first term of the equation would be PV3.67 atm = 12.89 atm less than that exerted on a real gas. use T= ( )( p + ) (V .9258 L) .17 L2 · atm/mol2 and b = 0. Compare a real gas to an ideal gas How much does the pressure exerted by 2 moles of ammonia at 13◦C in a 1 L container differ from the pressure exerted on the same amount of an ideal gas? (use a = 4. If you multiply out the equation and then get V out of the denominators. .37 atm .56 atm – 34.0371 L/mol) .17 L² atm / mol²) x (2 mol)² / (1L²) Pammonia = (47.56 Latm) / (0.89 atm Step 4 – Provide your answer to the question: Re-read the question to make sure you word your answer correctly! We are asked how much does the ideal gas differ from amonia.16. which is hard to easily solve.) You won’t often find a question where you are given P and T alone and have to find the corresponding volume in a van der Waals equation problem. so the equation is really a cubic function.16. 3.56 atm so more pressure is exerted on the ideal gas And the difference is: 47.(a n² / V²) Pammonia = (2 mol) (0.70 atm Pammonia = 51.083145 x 286 / 1 Pideal gas = 47. Although the pressure exerted on ammonia is 12.(4.2.56 atm Step 3 – Compare the two pressures: Pammonia = 34.0371 L/mol as the rounded off van der Waals constants for ammonia) Step 1 -First figure out the pressure exerted on ammonia: P = nRT / (V-nb) .89 atm more than that exerted on ammonia.
. )) ) – 2.3026 and = 1. .3026 = 6.4. One contains Methane (CH4) and the other Nitrogen (N2). the larger the constant value. Example: Two identical 1 liter tanks have 1 mole of gas in them at 100 K.3861 atm – .3026 = 8. Now calculate the pressure we would get from an ideal gas: P= P= ( ) ( ( )( . = = = = = = = 2. • For the constant. ( ) )( ) = 8.3661 and = 0.038577 ( ) – ( ) ( (( )( (. the larger the constant value. An less volume will be available for the gas with the constant value. ( . )) ( ( (( )(. Comparing two real gases If we remember what the constants relate to in terms of the physical attributes of gas molecules we can draw certain conclusions about gases based on their van der Waals constants: • For the constant.3145 ) Since Nitrogen’s pressure is closer to that of an ideal gas under the given conditions.2820 atm So there is a 0. ) ) ) – ( ( . the larger the molecule and consequently the radius of the molecule. )) ( (( . )( )) ) – ( ( )( . it acts more like an ideal gas than methane. . ) ( ) ) ( . the greater the attraction will be between the gas molecules and the smaller the force with which a molecule will hit the container walls.64811 .3661 = 7.8959 atm difference in pressure between the two tanks. ) ( – ) = 8. What will be the difference between the pressures in the two tanks? Which gas behaves closest to an ideal gas? For Methane: For Nitrogen: And.6886 – 2. )( .043067 = 0.1.
Sign up to vote on this title | 2,668 | 10,577 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2017-30 | latest | en | 0.934737 |
http://ijawcenter.com/52a2q3nh6/cS22576oN/ | 1,582,782,502,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146647.82/warc/CC-MAIN-20200227033058-20200227063058-00174.warc.gz | 78,658,187 | 12,124 | Subtraction Worksheets: Best Kids Worksheet Book For Middle School Free Verb Worksheets 4th Grade Decimal Solving Exponential Equations Calculator Basic Addition And Subtraction Printable Farm Animals Beginning. worksheets for fractions grade 6 Bar Graph Worksheets For Grade 5 kids worksheet common core regents | Ijawcenter
# Best Kids Worksheet Book For Middle School Free Verb Worksheets 4th Grade Decimal Solving Exponential Equations Calculator Basic Addition And Subtraction Printable Farm Animals Beginning
Published at Friday, 01 November 2019. Subtraction Worksheets. By .
Sites offering free math worksheets abound on the Internet. So what should free math worksheets look like? Are you searching for worksheets that have lots of problems for paper-pencil completion? Or is the illusive long division worksheet your quest? Finding the answers to these questions depends largely on personal preferences and more importantly who will be completing your freely printed math worksheets. Here are the criteria to keep in mind when selecting free arithmetic worksheets to use with students. Division worksheets are not all created equal. There are basically two types: math fact sheets and long division sheets. Math fact sheets are easy to create (with the division symbol between the numbers) and require few if any paper calculation from students. Long division sheets are more difficult to program, e.g., with or without remainders, and allow for stepwise student completion of problems.
Most volumes begin with an explanation of basic arithmetic operations namely: addition, subtraction, multiplication, and division. Reference tables are supplied to provide clues for quick mental arithmetic and mastery of math facts. When ready to be tested, the student can select a drill, which has 10 questions and are selected from a database of number pairs for calculation. The Basic Level volumes use simple single digit numbers and the interactive math software at the Advanced Level uses mostly double digit numbers for math practice problems. Each drill is then scored and timed with the results saved. With the test records, students can follow their own progress and adults who may be supervising can monitor progress and assess if there are any learning issues that require intervention.
1st grade math worksheets and my Mom has math teaching style. Math will not be as terrible as it seems if parents take interest in preparing their little ones for math before school age. I grew up not understanding how it is that people talk about math as difficult as they do, it was my best subject at school. It was easy because of my upbringing that ensured that math and I got acquainted long before school. My mother who was a primary grade teacher told me how she began teaching me math in different guises at home before I got to school age. I remember that with my Mom everything was somehow connected to math. She made me count the buttons in my shirt as she dressed me up, asked questions that demanded answers that are related to sums, like how many pair of shoes do you have? How many buttons are there on your Daddy has shirt? Count all the furniture in the living room and several math games.
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### Best Kids Worksheet Book For Middle School Free Verb Worksheets 4th Grade Decimal Solving Exponential Equations Calculator Basic Addition And Subtraction Printable Farm Animals Beginning
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File Size: 189 kb | 667 | 3,510 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-10 | latest | en | 0.873183 |
https://www.physicsforums.com/threads/vector-function.356777/ | 1,531,724,677,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589222.18/warc/CC-MAIN-20180716060836-20180716080836-00560.warc.gz | 965,278,073 | 13,368 | # Homework Help: Vector Function
1. Nov 21, 2009
Okay. My reason for posting this is that I need help actually formulating the 'math part' of it. I can get the right answer by 'inspection.' And from the way the book is written, I believe that is how the authors expect you to find it. But for self gratifying reasons, I wish to generalize the answer a little bit:
1. The problem statement, all variables and given/known data
Write the formula for a vector function in 2 dimensions whose direction makes an
angle of 45 degrees with the x-axis and whose magnitude at any point (x,y) is (x+y)2.
3. The attempt at a solution
Now just looking at it, we can say that to insure that the vector at any point (x,y) is at 45 degrees to the x-axis, we can let the funciton equal something to the effect of:
$$\mathbf{F}(x,y) = k(\bold{i} + \bold{j})$$
Now k(x,y) is the 'scalar portion' of F and must have the effect that its product with the magnitude of the
vector portion of F, namely v=(i + j), must equal (x+y)2.
Now since the magnitude of v=(i + j) is $\sqrt{2}$, k(x,y) must have $\sqrt{2}$ in its denominator.
This will make 'unit vector' in the v direction or
$$\bold{u}_v = \frac{\bold{v}}{\sqrt{2}}=\frac{\bold{i} + \bold{j}}{\sqrt{2}}$$
Therefore, multiplying v by k(x,y) = (x+y)2/$\sqrt{2}$ gives the desired result.
$$\Rightarrow \bold{F}(x,y) = \frac{(x+y)^2}{\sqrt{2}}(\bold{i} + \bold{j})$$
I am just curious how other would approach this problem, or if this is the most efficient method from a mathematical standpoint.
~Casey
Last edited: Nov 21, 2009
2. Nov 21, 2009
### Feldoh
Dunno if it helps but....
I assumed a vector initially completely in the i direction: $$\mathbf{F} = (x+y)^2 \mathbf{i}$$.
Then just did a coordinate rotation:
$$\left(\begin{array}{cc}cos(45) & -sin(45)\\sin(45) & cos(45)\end{array}\right) \left(\begin{array}{cc}(x+y)^2\\0\end{array}\right) = \frac{(x+y)^2}{\sqrt{2}}\left(\begin{array}{cc}\mathbf{i}\\\mathbf{j}\end{array}\right)$$
3. Nov 21, 2009 | 621 | 2,008 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2018-30 | latest | en | 0.837708 |
https://idealpaykasa.com/qa/question-how-many-impressions-make-a-click.html | 1,623,498,013,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487582767.0/warc/CC-MAIN-20210612103920-20210612133920-00421.warc.gz | 293,065,095 | 7,869 | # Question: How Many Impressions Make A Click?
## What is the average cost per impression?
Google Display Network Benchmarks in Q1 2018 In Q1 2018, advertisers spent, on average, \$2.80 per thousand impressions (CPM), and \$0.75 per click (CPC).
The average click-through rate (CTR) on the GDN was 0.35%..
## How do you calculate clicks?
Here’s the simple formula to determine cost per click:Total Cost / Number of Clicks. … Related Formula: Cost Per Thousand Impressions. … Related Formula: Click-Through Rate. … (Revenue Generated – Cost of Campaign) / Cost of Campaign) x 100. … Related Formula: Conversion Rate. … (Number of Conversions / Number of Clicks) x 100.More items…•
## How do I calculate CPM?
To determine CPM, simply divide your total spend by the number of impressions. Or to derive the other values in the equation: Total Cost of Campaign = Total Impressions ÷ 1000 x CPM.
## What are impressions and clicks?
Ad Clicks, or simply Clicks, is a marketing metric that counts the number of times users have clicked on a digital advertisement to reach an online property. Read more … Ad Impressions (IMPR) is a count of the total number of times digital advertisements display on someone’s screen within the publisher’s network.
Are Facebook Ads Worth It? When you get right down to it, though, even a great cost-per-conversion doesn’t mean a Facebook campaign will be worth the money. … In general, if you get more than \$4.00 in revenue for every \$1.00 you spend on advertising, that’s a pretty profitable campaign.
## How do I generate more clicks?
Perform On-Page SEO. There are many SEO tactics you can perform on each of your website pages to increase their rank in search engines and get more visitors. … Get Listed in Online Directories. … Post to Social Media. … Include Hashtags in Your Posts. … Use Landing Pages. … Target Long-Tail Keywords. … Start Email Marketing. … Guest Blog.More items…
## Which is better reach or impressions?
Reach is the total number of people who see your content. Impressions are the number of times your content is displayed, no matter if it was clicked or not. Think of reach as the number of unique people who see your content. … However, an impression means that content was delivered to someone’s feed.
## What is the 7 times 7 rule?
Unfortunately, you’re one of thousands who are vying for your customers’ attention. The Marketing Rule of 7 states that a prospect needs to “hear” the advertiser’s message at least 7 times before they’ll take action to buy that product or service.
## What is a good impressions to click ratio?
Although there is no exact number to determine what a good click-through rate is, 2% is average for an entire account across all verticals. This means some campaigns inside the account could be performing better and some could be performing worse. Anything higher than 2% is above average.
## What is a good cost per click?
For most businesses, a 5:1 revenue-to-ad ratio is considered acceptable. This means for every dollar spent in advertising, five dollars in revenue is produced. A 20% cost-per-acquisition, or CPA, is another way of expressing this ratio. | 715 | 3,166 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.836481 |
https://virtuelle-experimente.de/en/e-feld/geschwindigkeit/winkel.php | 1,721,515,787,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517541.97/warc/CC-MAIN-20240720205244-20240720235244-00571.warc.gz | 547,277,557 | 3,414 | # At what angle does an electron leave the electric field?
Using the velocities in the x- and in the y-direction you can now also calculate the angle $$\theta$$ at which the electrons leave the E-field of the plate capacitor.
When leaving the capacitor, an electron has a velocity in the x-direction of $$v_x=4.20\cdot 10^7\,\rm{\frac{m}{s}}$$ ans in y-direction of $$v_y=1.68\cdot 10^7\,\rm{\frac{m}{s}}$$.
• Calculate the angle $$\theta$$ at which an electron leaves the plate capacitor.
• State what must hold for the velocity components $$v_x$$ and $$v_y$$ so that the exit angle is $$\theta=45°$$.
• Now the velocity $$v_x$$ of the electrons is doubled. Explain if this halves the angle $$\theta$$. | 192 | 705 | {"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} | 3.171875 | 3 | CC-MAIN-2024-30 | latest | en | 0.787252 |
https://www.w3resource.com/python-exercises/pandas/time-series/pandas-time-series-exercise-22.php | 1,620,308,061,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988753.97/warc/CC-MAIN-20210506114045-20210506144045-00361.warc.gz | 1,117,559,704 | 28,909 | Pandas: Business quarterly begin and end dates of a specified year - w3resource
# Pandas: Business quarterly begin and end dates of a specified year
## Pandas Time Series: Exercise-22 with Solution
Write a Pandas program to find the all the business quarterly begin and end dates of a specified year.
Sample Solution:
Python Code :
``````import pandas as pd
q_start_dates = pd.date_range('2020-01-01', '2020-12-31', freq='BQS-JUN')
q_end_dates = pd.date_range('2020-01-01', '2020-12-31', freq='BQ-JUN')
print("All the business quarterly begin dates of 2020:")
print(q_start_dates.values)
print("\nAll the business quarterly end dates of 2020:")
print(q_end_dates.values)
``````
Sample Output:
```All the business quarterly begin dates of 2020:
['2020-03-02T00:00:00.000000000' '2020-06-01T00:00:00.000000000'
'2020-09-01T00:00:00.000000000' '2020-12-01T00:00:00.000000000']
All the business quarterly end dates of 2020:
['2020-03-31T00:00:00.000000000' '2020-06-30T00:00:00.000000000'
'2020-09-30T00:00:00.000000000' '2020-12-31T00:00:00.000000000']
```
Python Code Editor:
Have another way to solve this solution? Contribute your code (and comments) through Disqus.
What is the difficulty level of this exercise?
Test your Python skills with w3resource's quiz
## Python: Tips of the Day
Merging two dicts in Python 3.5+ with a single expression
Example:
```# How to merge two dictionaries
# in Python 3.5+
x = {'p': 1, 'q': 3}
y = {'q': 5, 'r': 8}
z = {**x, **y}
z
{'r': 4, 'p': 1, 'q': 3}
z = dict(x, **y)
print(z)
```
Output:
```{'p': 1, 'q': 5, 'r': 8}
``` | 535 | 1,588 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | longest | en | 0.590943 |
https://www.mathworks.com/matlabcentral/cody/problems/94-target-sorting/solutions/312979 | 1,511,240,374,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806316.80/warc/CC-MAIN-20171121040104-20171121060104-00046.warc.gz | 846,691,847 | 11,717 | Cody
# Problem 94. Target sorting
Solution 312979
Submitted on 30 Aug 2013 by Matt
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% a = [1 2 3 4]; t = 0; b_correct = [4 3 2 1]; assert(isequal(targetSort(a,t),b_correct))
b = 4 3 2 1
2 Pass
%% a = -4:10; t = 3.6; b_correct = [-4 -3 10 -2 9 -1 8 0 7 1 6 2 5 3 4]; assert(isequal(targetSort(a,t),b_correct))
b = -4 -3 10 -2 9 -1 8 0 7 1 6 2 5 3 4
3 Pass
%% a = 12; t = pi; b_correct = 12; assert(isequal(targetSort(a,t),b_correct))
b = 12
4 Pass
%% a = -100:-95; t = 100; b_correct = [-100 -99 -98 -97 -96 -95]; assert(isequal(targetSort(a,t),b_correct))
b = -100 -99 -98 -97 -96 -95 | 312 | 768 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2017-47 | latest | en | 0.588592 |
https://1billionbooks.com/how-to-play-dice-games-with-5-dice-goal/ | 1,660,033,985,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570913.16/warc/CC-MAIN-20220809064307-20220809094307-00441.warc.gz | 110,056,943 | 13,167 | # How To Play Dice Games With 5 Dice
How To Play Dice Games With 5 Dice. If you roll three 1s in a single roll, you will score 1,000 points. The player with the most points wins.
Three 1s in a single roll. Please bookmark and visit again. Once you get on the board you can stay or.
### Two or more people, player pieces below, and game board (next page) how to play:
If you roll three 1s in a single roll, you will score 1,000 points. There is a commercial version of ten thousand called cosmic wimpout that is played with only 5 dice and without the three pairs scoring category. 5000 is a simple dice game to play with friends and family.
### You can throw the dice three times each turn.
The objective of the game is to be the first player to get 5000 points. Sometimes a target total of 5,000 is set to make for a shorter game. The classic poker dice game is played with 5 dice and two or more players.
### If you throw three matching dice, multiply that number by 100 to calculate the point value (a set of three fours would be worth 400 points).
Go for the highest score! Rolling a 5 in a roll. Keep score on paper to keep track.
### Trailer gives instructions on how to play a fun game called farkle.
10,000 is a fun dice game where players try to score points by rolling winning combinations. A player can roll (at most) three times. Collect 100 points for each roll of one and 50 points for each roll of five.
### First roll the dice to see who goes first.
How to play dice 4, 5, 6. Roll the dice and decide where you would like to allocate points. The player who rolls the highest number goes first, and the player who rolls the lowest number must keep score for the game. | 407 | 1,694 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.950141 |
https://helpseeker.net/for-math-2-suject-test-do-they-give-out-formulas-tips-to-pass-cbest-test-score-high-in-math/ | 1,674,877,718,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499470.19/warc/CC-MAIN-20230128023233-20230128053233-00563.warc.gz | 315,220,196 | 11,395 | # For Math 2 Suject Test Do They Give Out Formulas Tips to Pass CBEST Test: Score High In Math!
You are searching about For Math 2 Suject Test Do They Give Out Formulas, today we will share with you article about For Math 2 Suject Test Do They Give Out Formulas was compiled and edited by our team from many sources on the internet. Hope this article on the topic For Math 2 Suject Test Do They Give Out Formulas is useful to you.
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## Tips to Pass CBEST Test: Score High In Math!
You should know by now that the California Basic Academic Skills Test or CBEST exam requires you to pass three subjects: math, reading, and writing.
With that being said, you must remember struggling with math in high school. Lucky for math lovers, this part of the test will definitely be a bonus for you.
But, has your dream come true on the CBEST exam threatened by your old enemy—high school algebra?
Score high on CBEST Math by posting these 3 simple steps for your CBEST test preparation review.
Step 1: CBEST Test Math Basics
Unlike studying other subjects, studying mathematics requires a kick-off to study the concepts. You need to practice CBEST sample tests again and again until you master each theory.
Here are some dos and don’ts on how to master the math basics for your CBEST exam:
1. Do not jump to the next theory or formula until you are confident that you understand it.
2. Don’t ignore the most basic terminology and principles.
3. Use this knowledge during your CBEST exam.
Armed with your basic knowledge, you can effectively eliminate wrong answers in some questions. You will narrow down your choices and increase your chances of getting the right answer.
Step 2: Focus on what you don’t know
Mathematics is only one subject. But, still, you may not know where to start. It is easy to confuse your CBEST score in math with good CBEST exam preparation.
1. Assess yourself and answer the CBEST practice tests.
2. Examine each test item to know your strengths and weaknesses.
3. Write down all the keywords in the topics you find most difficult
4. Check your old books so you can review full length on the titles listed.
5. From the subjects you have chosen, you can work your way up to other less difficult subjects in mathematics.
6. Finally, answer more CBEST practice tests
Step 3: Remember to be accurate
Mathematics is all about accuracy. One mistake in your formula can lead to a completely different answer. And it can definitely cost you points. Remember formulas, measurements, ratios and percentages, conversions of units and basic geometric principles and principles. Even when the problems are presented differently, you’ll still know which formula to use. Your accuracy in problem solving will definitely save you time and give you the confidence you need!
How can I pass the CBEST exam? Simple, don’t repeat the mistakes of others. Plan, Review, Practice and get the BEST CBEST Study Guides. Pass the California Basic Academic Skills Test in one shot and enjoy the job you’ve always dreamed of!
## Question about For Math 2 Suject Test Do They Give Out Formulas
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#Tips #Pass #CBEST #Test #Score #High #Math
Source: https://ezinearticles.com/?Tips-to-Pass-CBEST-Test:-Score-High-In-Math!&id=6442059 | 913 | 4,013 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.933621 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/403/2/cc/a/ | 1,597,340,840,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739048.46/warc/CC-MAIN-20200813161908-20200813191908-00595.warc.gz | 738,316,900 | 73,155 | # Properties
Label 403.2.cc.a Level 403 Weight 2 Character orbit 403.cc Analytic conductor 3.218 Analytic rank 0 Dimension 560 CM no Inner twists 2
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$403 = 13 \cdot 31$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 403.cc (of order $$60$$, degree $$16$$, minimal)
## Newform invariants
Self dual: no Analytic conductor: $$3.21797120146$$ Analytic rank: $$0$$ Dimension: $$560$$ Relative dimension: $$35$$ over $$\Q(\zeta_{60})$$ Twist minimal: yes Sato-Tate group: $\mathrm{SU}(2)[C_{60}]$
## $q$-expansion
The dimension is sufficiently large that we do not compute an algebraic $$q$$-expansion, but we have computed the trace expansion.
$$\operatorname{Tr}(f)(q) =$$ $$560q - 12q^{2} - 4q^{3} - 18q^{4} - 8q^{5} - 42q^{6} - 8q^{7} - 40q^{8} - 72q^{9} + O(q^{10})$$ $$\operatorname{Tr}(f)(q) =$$ $$560q - 12q^{2} - 4q^{3} - 18q^{4} - 8q^{5} - 42q^{6} - 8q^{7} - 40q^{8} - 72q^{9} - 30q^{10} - 18q^{11} + 26q^{12} - 24q^{13} + 4q^{14} - 62q^{15} - 58q^{16} - 30q^{17} - 70q^{18} - 24q^{19} - 8q^{20} + 114q^{21} + 78q^{22} - 30q^{23} - 82q^{24} - 24q^{26} - 100q^{27} - 62q^{28} - 10q^{29} - 4q^{31} + 20q^{32} - 110q^{33} + 70q^{34} - 2q^{35} - 40q^{37} - 108q^{38} + 48q^{39} - 28q^{40} - 22q^{41} - 10q^{42} + 78q^{43} - 24q^{44} + 36q^{45} + 44q^{46} - 32q^{47} - 10q^{48} - 30q^{49} - 30q^{50} + 36q^{51} - 252q^{52} - 84q^{53} + 82q^{54} - 4q^{55} + 164q^{57} + 28q^{58} - 2q^{59} - 8q^{60} + 36q^{61} - 12q^{62} + 78q^{63} + 270q^{64} - 72q^{65} - 56q^{66} - 46q^{67} - 12q^{68} + 150q^{69} + 90q^{70} - 74q^{71} + 72q^{72} + 30q^{73} - 10q^{74} - 16q^{75} - 228q^{76} + 72q^{77} + 96q^{78} - 32q^{79} + 108q^{80} - 104q^{81} - 84q^{82} + 4q^{83} + 26q^{84} + 12q^{85} + 34q^{86} + 112q^{87} - 108q^{88} - 154q^{89} - 90q^{90} - 4q^{91} + 64q^{93} - 24q^{94} - 78q^{95} - 4q^{96} - 196q^{97} + 50q^{98} + 96q^{99} + O(q^{100})$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
24.1 −2.75171 + 0.144211i 2.34874 2.11482i 5.56208 0.584598i 0.937749 0.251269i −6.15808 + 6.15808i −0.0941873 + 0.594675i −9.77782 + 1.54865i 0.730552 6.95074i −2.54418 + 0.826654i
24.2 −2.75032 + 0.144138i −1.37473 + 1.23781i 5.55445 0.583796i −0.372115 + 0.0997080i 3.60254 3.60254i 0.787181 4.97006i −9.75199 + 1.54456i 0.0441185 0.419760i 1.00907 0.327865i
24.3 −2.48692 + 0.130334i −0.0686740 + 0.0618344i 4.17875 0.439204i −3.44926 + 0.924226i 0.162728 0.162728i −0.267734 + 1.69041i −5.41563 + 0.857752i −0.312693 + 2.97507i 8.45758 2.74803i
24.4 −2.30755 + 0.120933i 0.810915 0.730151i 3.32110 0.349061i 1.72963 0.463452i −1.78292 + 1.78292i 0.108867 0.687358i −3.05684 + 0.484156i −0.189123 + 1.79939i −3.93514 + 1.27861i
24.5 −2.17694 + 0.114088i 0.222847 0.200652i 2.73700 0.287670i 1.57540 0.422127i −0.462232 + 0.462232i −0.429485 + 2.71166i −1.61928 + 0.256469i −0.304186 + 2.89414i −3.38139 + 1.09868i
24.6 −2.04287 + 0.107062i −1.78599 + 1.60811i 2.17282 0.228372i 3.27471 0.877455i 3.47637 3.47637i 0.168500 1.06387i −0.373359 + 0.0591343i 0.290147 2.76056i −6.59586 + 2.14313i
24.7 −1.77883 + 0.0932245i 1.52098 1.36950i 1.16650 0.122604i −1.15023 + 0.308204i −2.57789 + 2.57789i 0.653142 4.12378i 1.45511 0.230467i 0.124273 1.18238i 2.01733 0.655471i
24.8 −1.62065 + 0.0849346i 2.22876 2.00678i 0.630243 0.0662412i −2.61289 + 0.700121i −3.44158 + 3.44158i −0.735803 + 4.64568i 2.19001 0.346863i 0.626598 5.96168i 4.17511 1.35657i
24.9 −1.59013 + 0.0833353i −1.52733 + 1.37521i 0.532532 0.0559714i −0.213994 + 0.0573396i 2.31405 2.31405i 0.126040 0.795786i 2.30329 0.364805i 0.127936 1.21723i 0.335501 0.109011i
24.10 −1.27699 + 0.0669244i −0.436672 + 0.393181i −0.362811 + 0.0381330i −3.05329 + 0.818127i 0.531314 0.531314i −0.0822901 + 0.519559i 2.98676 0.473056i −0.277494 + 2.64018i 3.84428 1.24908i
24.11 −1.22640 + 0.0642731i −0.451855 + 0.406852i −0.489109 + 0.0514074i −1.63845 + 0.439022i 0.528007 0.528007i 0.600453 3.79111i 3.02247 0.478713i −0.274941 + 2.61589i 1.98119 0.643727i
24.12 −1.22597 + 0.0642504i 2.01198 1.81160i −0.490166 + 0.0515185i 3.36984 0.902946i −2.35024 + 2.35024i 0.394798 2.49266i 3.02270 0.478748i 0.452603 4.30623i −4.07331 + 1.32350i
24.13 −0.715710 + 0.0375088i −1.58346 + 1.42575i −1.47821 + 0.155366i 1.89550 0.507897i 1.07982 1.07982i −0.656203 + 4.14310i 2.46788 0.390874i 0.160984 1.53166i −1.33758 + 0.434604i
24.14 −0.690768 + 0.0362016i 0.0986013 0.0887810i −1.51319 + 0.159043i 3.88062 1.03981i −0.0648966 + 0.0648966i −0.213808 + 1.34993i 2.40591 0.381058i −0.311745 + 2.96606i −2.64297 + 0.858752i
24.15 −0.652040 + 0.0341720i 1.57276 1.41612i −1.56505 + 0.164494i 0.675785 0.181076i −0.977108 + 0.977108i −0.0634857 + 0.400833i 2.30465 0.365021i 0.154592 1.47084i −0.434452 + 0.141162i
24.16 −0.320349 + 0.0167888i −1.75222 + 1.57771i −1.88670 + 0.198300i −2.60602 + 0.698280i 0.534834 0.534834i −0.645048 + 4.07267i 1.23475 0.195566i 0.267533 2.54541i 0.823112 0.267445i
24.17 −0.235412 + 0.0123374i 0.964102 0.868082i −1.93378 + 0.203248i −1.50977 + 0.404542i −0.216251 + 0.216251i −0.125388 + 0.791668i 0.918393 0.145459i −0.137658 + 1.30973i 0.350428 0.113861i
24.18 −0.216740 + 0.0113589i −2.41844 + 2.17758i −1.94220 + 0.204133i 1.43655 0.384922i 0.499439 0.499439i 0.462411 2.91955i 0.847363 0.134209i 0.793446 7.54913i −0.306985 + 0.0997454i
24.19 0.298096 0.0156226i −0.463276 + 0.417135i −1.90043 + 0.199743i 0.387663 0.103874i −0.131584 + 0.131584i 0.399618 2.52309i −1.15305 + 0.182625i −0.272963 + 2.59707i 0.113938 0.0370207i
24.20 0.547115 0.0286731i −1.75246 + 1.57792i −1.69053 + 0.177682i −2.48381 + 0.665536i −0.913555 + 0.913555i 0.442675 2.79494i −2.00206 + 0.317096i 0.267694 2.54694i −1.33985 + 0.435343i
See next 80 embeddings (of 560 total)
$$n$$: e.g. 2-40 or 990-1000 Embeddings: e.g. 1-3 or 383.35 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
403.cc even 60 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 403.2.cc.a 560
13.f odd 12 1 403.2.ch.a yes 560
31.h odd 30 1 403.2.ch.a yes 560
403.cc even 60 1 inner 403.2.cc.a 560
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
403.2.cc.a 560 1.a even 1 1 trivial
403.2.cc.a 560 403.cc even 60 1 inner
403.2.ch.a yes 560 13.f odd 12 1
403.2.ch.a yes 560 31.h odd 30 1
## Hecke kernels
This newform subspace is the entire newspace $$S_{2}^{\mathrm{new}}(403, [\chi])$$.
## Hecke characteristic polynomials
There are no characteristic polynomials of Hecke operators in the database | 3,585 | 6,844 | {"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} | 2.9375 | 3 | CC-MAIN-2020-34 | latest | en | 0.206305 |
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Math: Coordinate Geometry
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Re: Math: Coordinate Geometry [#permalink]
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02 Nov 2010, 13:46
Just what i needed, thanks!
+1!
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03 Nov 2010, 06:05
thanks
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12 Nov 2010, 08:58
awesome stuff!!
Can't possibly thank you enough
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31 Jan 2011, 09:56
hmmm. this is good work . very helpful .
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01 Apr 2011, 10:04
I had a question regarding this:
Q: Find the equation of a line whose x intercept is 5 and y intercept is 2.
Solution: Substituting the values in equation \frac{x}{a}+\frac{y}{b}=1 we'll get \frac{x}{5}+\frac{y}{2}=1 --> 5y+2x-10=0 OR if we want to write the equation in the slope-intercept form: y=-\frac{2}{5}x+2
Is that right? When you say x intercept is 5 then the points are (0,5) right? Same for y intercept is 2..(2,0).
So when you look at the equation in slope intercept form it should be
y = (-5/2)x + 5
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Re: Math: Coordinate Geometry [#permalink]
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01 Apr 2011, 15:07
echoing 'MateoLibre' - it is indeed a v helpful post. I have a DS Q....(source: GMATPrep CAT)
In the xy plane at what 2 points does the graph of y = (x+a)(x+b) intersect the x axis?
1] a+b = -1
2] the graph intersects the y axis at (0,-6)
ANY help on how to solve this would be much appreciated. thanks.
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01 Apr 2011, 17:43
merci beaucoup thesfactor that was a very clear & prompt response! +1 i guess?! (sorry new to the forum :s)
oh btw ur spot on
[Reveal] Spoiler:
OA: C
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14 May 2011, 09:33
i m little confused about steepness of 2 slopes , if we compare them shld i consider their absolutes values??
Im1I with Im2I or just m1, m2
i strongly feel its their absolute values , bcoz it defines their steepness either upward or downward
clarification required
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12 Sep 2011, 08:49
1
This post was
BOOKMARKED
Bunuel wrote:
1. If the slope of a line is negative, the line WILL intersect quadrants II and IV. X and Y intersects of the line with negative slope have the same sign. Therefore if X and Y intersects are positive, the line intersects quadrant I; if negative, quadrant III.
2. If the slope of line is positive, line WILL intersect quadrants I and III. Y and X intersects of the line with positive slope have opposite signs. Therefore if X intersect is negative, line intersects the quadrant II too, if positive quadrant IV.
3. Every line (but the one crosses origin OR parallel to X or Y axis OR X and Y axis themselves) crosses three quadrants. Only the line which crosses origin $$(0,0)$$ OR is parallel to either of axis crosses only two quadrants.
4. If a line is horizontal it has a slope of $$0$$, is parallel to X-axis and crosses quadrant I and II if the Y intersect is positive OR quadrants III and IV, if the Y intersect is negative. Equation of such line is y=b, where b is y intersect.
5. If a line is vertical, the slope is not defined, line is parallel to Y-axis and crosses quadrant I and IV, if the X intersect is positive and quadrant II and III, if the X intersect is negative. Equation of such line is $$x=a$$, where a is x-intercept.
6. For a line that crosses two points $$(x_1,y_1)$$ and $$(x_2,y_2)$$, slope $$m=\frac{y_2-y_1}{x_2-x_1}$$
7. If the slope is 1 the angle formed by the line is $$45$$ degrees.
8. Given a point and slope, equation of a line can be found. The equation of a straight line that passes through a point $$(x_1, y_1)$$ with a slope $$m$$ is: $$y - y_1 = m(x - x_1)$$
I was going through "GMAT MATH BOOK" for Coordinate Geometry.
Came across a section(SLOPE AND QUADRANTS:) that deals with intersection of lines with Quadrants. This got me confused as it does not match with my concepts that i know of.
According to me (atleast what i followed till now) , when we talk about intersection of a line with Quadrant. I used to think that we are dealing with the quadrant with which the line forms a triangle.(using axes as other two sides)
This does not match with the above mentioned topic.
Can ne one help me with this and a few GMAT problems that deal with this would certainly help.
I just wanted to clear my understanding of certain concepts.
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08 Dec 2011, 12:02
Good work. Really helpful.
Q: If the point (x,y) moves such that the sum of its distances from the points (0,2) and (0,-2) is 6, then the point (x,y) lies on :
1. 9x+5y + 45
2. 9x^2 + 5y^2 = 45
3. 9x+ 5y^2 = 45
Can you please answer the above question.
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23 Dec 2011, 04:32
Bunnel Your Posts are by far the best.........
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08 Jan 2012, 14:42
Can some one validate the statement below
1. If the slope of a line is negative, the line WILL intersect quadrants II and IV. X and Y intersects of the line with negative slope have the same sign. Therefore if X and Y intersects are positive, the line intersects quadrant I; if negative, quadrant III.
2. If the slope of line is positive, line WILL intersect quadrants I and III. Y and X intersects of the line with positive slope have opposite signs. Therefore if X intersect is negative, line intersects the quadrant II too, if positive quadrant IV.
If the point is in the 4th quadrant , x and y will have different sign. But statement 1 contradicts this belief.
If the point is in 1st or 3rd quadrant, x and y will have same sign. But the statement 2 above contradicts this belief.
Can some one help me to understand if im wrong?
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06 Mar 2012, 07:25
[quote="Bunuel"]
1. If the slope of a line is negative, the line WILL intersect quadrants II and IV. X and Y intersects of the line with negative slope have the same sign. Therefore if X and Y intersects are positive, the line intersects quadrant I; if negative, quadrant III.
2. If the slope of line is positive, line WILL intersect quadrants I and III. Y and X intersects of the line with positive slope have opposite signs. Therefore if X intersect is negative, line intersects the quadrant II too, if positive quadrant IV.
Could someone please explain this further. What do you mean by "intersect" here. Oh and I just saw the same question asked in the previous post
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06 Mar 2012, 08:13
budablasta wrote:
Bunuel wrote:
1. If the slope of a line is negative, the line WILL intersect quadrants II and IV. X and Y intersects of the line with negative slope have the same sign. Therefore if X and Y intersects are positive, the line intersects quadrant I; if negative, quadrant III.
2. If the slope of line is positive, line WILL intersect quadrants I and III. Y and X intersects of the line with positive slope have opposite signs. Therefore if X intersect is negative, line intersects the quadrant II too, if positive quadrant IV.
Could someone please explain this further. What do you mean by "intersect" here. Oh and I just saw the same question asked in the previous post
Intersects mean passes through, have points in common.
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16 Mar 2012, 15:11
Thank you. This really helped me.
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13 Apr 2012, 11:59
really great job, Bunuel!!!
i have learnt from this post more than did in school + university !
thank you for help and for your time you spent to create this stuff!
Well done!!!sincerely
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24 Jul 2012, 02:14
When I have two points in a coordinate system e.g. (2,3) and (6,7) that pass through a line how do I know which number is x1 and which is x2 when calculating slope. Also, the same for the y1 and y2?
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24 Jul 2012, 05:12
Wonderful! Excellent work!
A small remark at "7. If the slope is 1 the angle formed by the line is 45 degrees."
Angle formed by the line with who?
The slope of a line is a real number equal to the tangent of the angle formed by the line with the positive x axis.
For the GMAT, no need to know about tangent (trigonometry), just to understand which angle we are talking about.
If the angle is acute, the slope (tangent of the angle) is positive. If the angle is obtuse, the slope (tangent of the angle) is negative.
There is a one-to-one correspondence between angles and the real numbers representing their so-called tangent values.
So, a line with slope 1 forms an angle of 45 degrees and a line with slope -1 forms an angle of 135 degrees with the positive x axis.
If the angle is 60 degrees, the slope is $$\sqrt{3}$$, and $$-\sqrt{3}$$ if the angle is 120 degrees, etc.
A suggestion: wouldn't be better to define the slope somewhere before the equations? I am sure everybody knows what's the slope of a line, just for the coherence.
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28 Oct 2012, 01:21
1
This post was
BOOKMARKED
Section formula
The point which divides the line joining two points {x1,y1} and {x2,y2}
in the ratio m:n internally is
{$$\frac{(m.x2+n.x1)}{(m+n)}, \frac{(m.y2+n.y1)}{(m+n)}$$}
The point which divides the line joining two points {x1,y1} and {x2,y2}
in the ratio m:n externally is
{$$\frac{(m.x2-n.x1)}{(m-n)}, \frac{(m.y2-n.y1)}{(m-n)}$$}
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Re: Math: Coordinate Geometry [#permalink]
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08 Feb 2013, 06:20
HI
I have a basic doubt , which deals with concept
Previously , I have learnt that a line that has rising slope when moving from - ve X to + ve X will have +ve slope.
But I found in the original post of BUNNUEL that a line with - ve slope will always be in 2nd or 4th quadrant. I feel that these two statements are contradictory. We can have a line with + ve slope even if it is in 2nd quadrant. Than how can we come to conclusion that whenever we encounter a slope with - ve sign , thn it must either lie in 2nd or lie in 4 th quadrant. As we can have lines in 1st quadrant with - ve slope also........
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Re: Math: Coordinate Geometry [#permalink] 08 Feb 2013, 06:20
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Artin, M. Algebra (2nd Edition). Addison Wesley, 2010. ISBN: 9780132413770.
The exercises listed below point to the text and are a suggestion from the professor to the students in every given lecture. These exercises were not turned in and were not corrected.
Group Representations
1 Group representations Chapter 10, sections 1, and 4 Chapter 10, 1.1, and 1.2
2 Unitary representations Chapter 10, sections 2, and 3 Chapter 10, 2.1, 2.2, and 3.4
3 Characters Chapter 10, section 4 Chapter 10, 4.1, 4.3a, b, and 4.8
4 One-dimensional characters Chapter 10, section 5 Chapter 10, 5.1, and 5.3
5 The regular representation Chapter 10, section 6 Chapter 10, 6.1, and 6.10
6 Characters (cont.) Chapter 10, sections 7, and 8 Chapter 10, 7.1, 7.2, and 7.6
Rings: Basic Definitions
7 Rings, ring homomorphisms Chapter 11, sections 1, 2, and 3 Chapter 11, 1.1, 1.5, 1.8, and 1.9
8 Ideals, quotient rings, correspondence theorem Chapter 11, sections 4, and 5 Chapter 11, 3.12, 3.13, 4.1, and 4.2
9 Maximal ideals, prime ideals, fractions Chapter 11, sections 8, and 9 Chapter 11, 6.1, 7.1, and 8.3
Rings: Factoring
10 Gauss' Lemma Chapter 12, section 3 Chapter 12, 2.3, 2.7, and 3.2
11 Unique factorization Chapter 12, sections 1, and 2 Chapter 12, 1.1, 1.5, 2.1, and 2.2
12 Factoring integer polynomials Chapter 12, section 4 Chapter 12, 4.1a, 4.6, 4.7, and 4.11
13 Gauss primes Chapter 12, section 5 Chapter 12, 5.1, 5.2a, and 5.3
14 Quadratic integers Chapter 13, section 1 Chapter 13, 1.1, 1.2, and 1.3
15 Ideal factorization Chapter 13, sections 2, and 3 Chapter 13, 2.1, 3.1, 3.2, and 3.3
16 Prime ideals Chapter 13, sections 5, and 6 Chapter 13, 5.3, 6.1, and 6.2
17 Ideal classes Chapter 11, sections 9, and 10 Chapter 13, 7.1, 7.2, and 8.2
Linear Algebra over a Ring
18 Integer matrices Chapter 14, sections 1, and 2 Chapter 14, 1.1, 2.1, and 2.4
19 Free modules Chapter 14, sections 3, and 4 Chapter 14, 3.2, 4.1a, and 4.3
20 Presenting modules Chapter 12, section 5 Chapter 14, 5.1, and 5.2
21 Hilbert basis theorem Chapter 14, section 6 Chapter 11, 6.1, 6.2, and M.1
22 Structure of abelian groups Chapter 14, section 7 Chapter 14, 7.1, 7.2, and 7.5
23 Linear operators Chapter 14, section 8 Chapter 14, 8.1, 8.3, and 8.4
Fields: Field Extensions
24 Algebraic elements, degree Chapter 15, sections 1, and 2 Chapter 15, 1.1, 1.3, and 2.1
25 Ruler and compass Chapter 13, section 5 Chapter 15, 5.1, 5.2, and 5.4
26 Adjoining elements Chapter 11, section 5, chapter 15, section 6 Chapter 11, 5.2, and 5.3, chapter 15, 6.1, and 6.3
27 Finite fields Chapter 15, section 7 Chapter 15, 7.1, 7.2, and 7.13
28 Primitive elements Chapter 15, section 8 Chapter 15, 8.1, and 8.2
Fields: Galois Theory
29 Symmetric functions, discriminant Chapter 16, sections 1, and 2 Chapter 16, 1.1a, b, e, 2.1, and 2.2
30 Splitting fields, the Galois group Chapter 16, sections 3, and 4 Chapter 16, 3.2, and 4.1
31 Fixed fields, Galois extensions Chapter 16, sections 5, and 6 Chapter 16, 5.1b, c, 6.1, and 6.2
32 The main theorem Chapter 16, section 7 Chapter 16, 7.1, 7.3, 7.6, and 7.7
33 Cubic and quartic equations Chapter 16, sections 8, and 9 Chapter 16, 8.2a, b, c, 9.1, 9.3, and 9.6
34 Quartic equations (cont.)
Chapter 16, section 9
Handout on The Seventeengon (PDF)
Chapter 16, 9.12a, b, and 9.16
35 Roots of unity, Kummer extensions Chapter 16, sections 10, and 11 Chapter 16, 10.1, 10.3, and 11.1
36 Quintic equations Chapter 16, section 12 Chapter 16, 12.1, 12.2, and 12.7 | 1,372 | 3,476 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2013-20 | latest | en | 0.828165 |
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19-Tanzanite
## max( ) works and min( ) doesn't ?
Hello, Everyone.
From :
So the question is : Is there a reason for : " max( ) works and min( ) doesn't " ?
Best Regards.
1 ACCEPTED SOLUTION
Accepted Solutions
24-Ruby V
(To:AlanStevens)
Ah, I did not realize that it depends on the first item in the vector.
Anyway, that behaviour sure is not as it should be and we may call it a bug.
But I am not sure how I would like min or max to behave. Normally in Mathcad, whenever a calculation encounters a NaN, it returns NaN as result. So for consistency min and max should return NaN in the examples above. On the other hand I guess it would not do any harm and would be convenient if those functions would simply ignore the NaN's - just like my filterNaN workaround.
Based on your information we can provide another workaround by simply putting + or - infinity at the front. But both workarounds will only work for vectors and must be modified slightly differently for row and column vectors.
7 REPLIES 7
24-Ruby V
(To:lvl107)
Both, max and min, work OK the same way. NaN simply is interpreted as zero.
You may use filterNaN to get what you expect:
16-Pearl
(To:Werner_E)
min works as expected as long as the first item in the vector is not NaN. Also works ok if the first item is NaN and the minimum is less than zero.
max has the same problem if all the terms, other than NaN, are negative.
Alan
24-Ruby V
(To:AlanStevens)
Ah, I did not realize that it depends on the first item in the vector.
Anyway, that behaviour sure is not as it should be and we may call it a bug.
But I am not sure how I would like min or max to behave. Normally in Mathcad, whenever a calculation encounters a NaN, it returns NaN as result. So for consistency min and max should return NaN in the examples above. On the other hand I guess it would not do any harm and would be convenient if those functions would simply ignore the NaN's - just like my filterNaN workaround.
Based on your information we can provide another workaround by simply putting + or - infinity at the front. But both workarounds will only work for vectors and must be modified slightly differently for row and column vectors.
19-Tanzanite
(To:Werner_E)
Many thanks, Werner and Alan.
Regards.
24-Ruby V
(To:lvl107)
Based on Alans observation concerning the first item in a vector here are some Min/Max replacement with one and two arguments which ignore NaN's and also can be used vectorized.
23-Emerald III
(To:lvl107)
For what it's worth: Mathcad 11.
The error message for stdev is: "Encountered a floating point error.".
Luc
24-Ruby V
(To:LucMeekes)
Thanks - so it was still not as consequent as we would like it to be, but at least the outcome did not depend on the first item of the vector.
BTW, in Mathcad 15 stdev returns NaN, which I consider a better way to deal with the NaN in the vector.
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1 SL(2, C) SO(3, 1) D : Λ D(Λ) SO(3, 1) 2 1 D : ±A D(π(±A)) SL(2, C) SL(2, C) SO(3, 1) SL(2, C) SO(3, 1) ξ i (, ) K i x µ p µ J µν T µν A µ ψ α
2 J i = J i, () K i = K i, ( ) K i M 0i = (iξ i K i ) A i = 1 2 (J i + ik i ) B i = 1 2 (J i ik i ), [A i, A j ] = iε ijk A k [B i, B j ] = iε ijk B k [A i, B j ] = 0. A, B SL(2, C) J i = A i + B i K i = i(b i A i ) SL(2, C) SU(2) A SU(2) B SL(2, C) = SU(2) SU(2). SL(2, C) (j A, j B ) j A j B (j A, j B ) j = j A + j B, j A + j B 1,..., j A j B SL(2, C)
3 SL(2, C) (1/2, 0) SU(2) C 2 j A = 1/2, j B = 0 A i = 1 2 σ i, B i = 0, J i = 1 2 σ i, K i = i 2 σ i. SL(2, C) [ i ( ) ] [ D(A) = A R = θ σ iξ σ = i σ ] 2 2 ( θ iξ), A SL(2, C) ξ a C 2 ξ α ξ α = (A R ) α βξ β. ξ a D(A) = (A 1 R )T ξ α ξ α = ξ β (A 1 R )β α = [(A 1 R )T ] β α ξ β (A R ) α β(δω) = δ α β + i 2 ω µν(s µν R )α β. ( 0 1 ε αβ = 1 0 ) ε αβ = ( ) (A R ) α γ(a R ) β δ εγδ = ε αβ ε γδ (A 1 R )γ α(a 1 R )δ β = ε αβ
4 C 2 ξ α = ε αβ ξ β, ξ α = ε αβ ξ β ξ α Ξ α = ξ 1 Ξ 2 ξ 2 Ξ 1, j = 0 j 1 = j 2 = 1 2 (0, 1/2) SU(2) j A = 0, j B = 1/2 A i = 0, B i = 1 2 σ i. J i = 1 2 σ i, K i = i 2 σ i. SL(2, C) [ i ( ) ] [ A A L = θ σ + iξ σ = i σ ] 2 2 ( θ + iξ), η α η α η α = (A L ) β. α βη σ 2 σ σ 2 = σ A L = ζa R ζ 1, ζ = iσ 2 A L A R η α ξ α η α η α = (A β. R) α βη PJ i P 1 = J i, PK i P 1 = K i
5 PA i P 1 = B i, PB i P 1 = A i Pξ = η, Pη = ξ P (1/2, 0) (0, 1/2) (n/2, m/2) SL(2, C) (1/2, 0), (0, 1/2) ζ α 1...α n ; β 1... β m ζ α 1...α n ; β 1... β m = (A R ) α 1 γ1... (A R ) α n γn (A R) β 1 δ1... (A R) β ṁ δ m ζ γ 1...γ n ; δ 1... δ m ε αβ SL(2, C) ζ (α 1α 2...α n);( β 1 β2... β m), (n/2, m/2) (n + 1)(m + 1) v µ j = 0, 1 j = 0 v 0 j = 1 v i x µ, p µ, A µ 4D (M µν ) λ ρ = i(δ λ µη νρ δ λ νη µρ )
6 (1, 0) (0, 1) t µν A t µν (1, 0) (0, 1) (1, 0) t µν A,c = 1 2 εµν λρ tλρ A,c (0, 1) t µν A,a = 1 2 εµν λρ tλρ A,a F µν (0, 0) (1, 1) t µν S t µν (0, 0) (1, 1) T µν SL(2, C) Dirac (1/2, 0) (0, 1/2) ( ) ψr Ψ =, ψ L ψ R ξ, ψ L η SL(2, C) ( ) Ψ Ψ AR 0 = Ψ 0 A L P : Ψ ( ψl ψ R ) = ( ) Ψ. θ = 0 ( ) [ ( ) ( )] σ ψ R 2 ξ ξ ξ ψ R = + σ ˆn ψ R 2 2 ( ψ L σ ) [ ( ) ( )] 2 ξ ξ ξ ψ L = σ ˆn ψ L 2 2 ψ R (0) = ψ L (0) ξ γ β
7 ψ R ( p) = ψ L ( p) = ( ) E + m + σ p ψ [2m(E + m)] 1/2 R (0) ( ) E + m σ p ψ [2m(E + m)] 1/2 L (0) E + σ p ψ R ( p) = m ψ L( p) E σ p ψ L ( p) = m ψ R( p) ( m p 0 σ p p 0 + σ p m ) ( ψr ( p) ψ L ( p) ) = 0 (γ µ p µ m)ψ(p) = 0, γ µ ( 0 σµ σ µ 0 ).
8 L P C 1, C 2 C 1, C 2 /p 2 m, s, p, λ {p, λ : p 2 = m 2, λ = s, s + 1,..., s 1, s} p, λ P k µ L k L µ νk ν = k µ, L L k. k µ U(a, Λ) P P µ k, λ = k, λ k µ. [W µ, P ν ] = 0 W µ P µ P µ W ν k, λ = W ν P µ k, λ = W ν k, λ k µ m, s
9 W µ k µ D s (L) D s (L) k, λ = k, λ D s (L) λ λ, L L k, k, λ H(p) SO(3, 1)/L k k µ p µ = H(p) µ νk ν, L k p, λ = U[H(p)] k, λ. P µ p, λ = P µ U[H(p)] k, λ = U[H(p)]U 1 [H(p)]P µ U[H(p)] k, λ = = U[H(p)]P ν k, λ (H(p)) µ ν = U[H(p)] k, λ H(p) µ νk ν = p, λ p µ. U(0, Λ) p, λ = U(Λ) p, λ = U(Λ)U[H(p)] k, λ = U[H(Λp)]U[H 1 (Λp)ΛH(p)] k, λ = = U[H(Λp)] k, λ D s [L(Λ, p)] λ λ = Λp, λ D s [L(Λ, p)] λ λ, L(Λ, p) = H 1 (Λp)ΛH(p) L k L k P H(p) SO(3, 1)/L k k µ = p µ n = 0 L k = SO(3, 1) Λ µ νp µ n = 0, Λ SO(3, 1)
10 m s (m, s) k µ = p µ t = (m, 0) p µ t W 0 = 0, W i = mj i. W µ p µ t SO(3) L pt = SO(3) R(α, β, γ) SO(3), Rp t = p t SO(3) P SO(3) p t, s, λ = p t, λ, λ = s,..., s, p µ t = (m, 0). P µ p t, λ = p t, λ p µ t J 2 p t, λ = p t, λ s(s + 1) J 3 p t, λ = p t, λ λ. H(p) SO(3, 1)/SO(3) λ p µ t p µ t = H(p) µ νp ν t. p λ p, λ = δ λ λδ 3 ( p p) p, λ = p 0 t p 0 U[H(p)] p t, λ = 1 γ U[H(p)] p t, λ.
11 P R(Λ, p) = H(Λp) 1 ΛH(p) SO(3) T (a) p, λ = p, λ e iaµ p µ U(0, Λ) p, λ = (Λp) 0 Λp, λ D s [R(Λ, p)] λ p λ, 0 D s [R] L pt = SO(3) s k µ = p µ l = (ω 0, 0, 0, ω 0 ) p µ l W 0 = W 3 = ω 0 J 3, W 1 = ω 0 (J 1 + K 2 ), W 2 = ω 0 (J 2 K 1 ). [W 1, W 2 ] = 0, [W 2, J 3 ] = iw 1, [W 1, J 3 ] = iw 2. W µ p µ l E 2 L k = E 2 C 2 C 2 = 0, C 2 0 C 2 = 0 E 2 p l, λ P µ p l, λ = p l, λ p µ l J 3 p l, λ = p l, λ λ W 1 p l, λ = W 2 p l, λ = 0.
12 {p µ = (ω, p) : p = ωˆp} p l, λ H(p) SO(3, 1)/E 2 p, λ = H(p) p l, λ (C 1 = C 2 = 0, λ) T (a) p, λ = p, λ e iaµ p µ U(0, Λ) p, λ = (Λp) 0 Λp, λ e iθ(λ,p), p 0 e iθ(λ,p) = p l, λ D[L(Λ, p)] p l, λ = p l, λ D[H 1 (Λp)ΛH(p)] p l, λ, D E 2 λ λ = ±1, ±2
13 SL(2, C) ( ) D(A) =, A SL(2, C). ε αβ ξ, Ξ (1/2, 0) ξ 1 Ξ 2 ξ 2 Ξ 1, SL(2, C) ξ 1 ξ 1 + ξ 2 ξ 2, E 2 = p 2 + m 2 E = γm, p = γm β γ µ : {γ µ, γ ν } = 2η µν S µν = i 4 [γµ, γ ν ] [S µν, γ λ ] = i(γ µ η νλ γ ν η µλ ) p µ l = (ω 0, 0, 0, ω 0 ) W µ
14
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### Section 9: Quantum Electrodynamics
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### Γ Ε Ν Ι Κ Ο Δ Ι Α Γ Ω Ν Ι Σ Μ Α Ο Ι Κ Ο Ν Ο Μ Ι Α Σ - Θ Ε Τ Ι Κ Η Σ Γ Τ Α Ξ Η Β. Ρ.
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### ΘΕΩΡΗΜΑ CAYLEY-HAMILTON. Έστω A πίνακας ν ν. Από το θεώρηµα Cayley-Hamilton συµπεραίνουµε ότι το σύνολο των πολυωνύµων p( λ ), ώστε p( A)
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### ˆ ˆŠ Œ ˆ ˆ Œ ƒ Ÿ Ñ Ò É ÉÊÉ Ö ÒÌ ² µ, Ê
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### ςεδς ΤΕΤΡΑΔΙΟ ΕΠΑΝΑΛΗΨΗΣ ΕΡΩΤΗΣΕΙΣ ΘΕΩΡΙΑΣ ΘΕΜΑΤΑ ΓΙΑ ΕΞΕΤΑΣΕΙΣ ΕΠΙΜΕΛΕΙΑ Βαγγέλης Βαγγέλης Νικολακάκης Μαθηματικός
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### ) 0 ) 2 & 2 & 0 + 6! ) & & & & & ), Γ , Γ 8 (?. Κ Ε 7 ) ) Μ & 7 Ν & & 0 7 & & Γ 7 & & 7 & Ν 2 & Γ Γ ( & & ) Η ++. Ε Ο 9 8 ) 8& & ) & Ε
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### Πα κ έ τ ο Ε ρ γ α σ ί α ς 4 Α ν ά π τ υ ξ η κ α ι π ρ ο σ α ρ µ ο γ ή έ ν τ υ π ο υ κ α ι η λ ε κ τ ρ ο ν ι κ ο ύ ε κ π α ι δ ε υ τ ι κ ο ύ υ λ ι κ ο
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### Dirac Matrices and Lorentz Spinors
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### Problem 1(a): Starting with eq. (3) proved in class and applying the Leibniz rule, we obtain
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### + (!, &. /+ /# 0 + /+ /# ) /+ /# 1 /+ /# # # # 6! 9 # ( 6 & # 6
# % ( + (!, &. /+ /# 0 + /+ /# ) /+ /# 1 /+ /# 2 + + 3 + 4 5 # 6 5 7 + 8 # # 6 (! 9 # ( 6 & 0 6 ) 1 5 + # 6 2 # # + 6 # # 6 # + # # + 6 + # #! 5 # # 6 & # : # # : 6 0 ) 5 + 6 1 # # 2 + # + # # 4 + # 6
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4 η εκάδα θεµάτων επανάληψης 3. ίνεται τετράγωνο µε κέντρο Ο και Μ το µέσο του. Η Μ τέµνει την στο. είξτε ότι = Το τρίγωνο είναι ορθογώνιο και ισοσκελές i ΟΜ = 4 Τα ορθογώνια τρίγωνα Μ και Μ έχουν Μ =
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### Ó³ Ÿ , º 7(156).. 62Ä69. Š Œ œ ƒˆˆ ˆ ˆŠ. .. ŠÊ²Ö μ 1,. ƒ. ²ÓÖ μ 2. μ ± Ê É É Ê Ò μ μ, Œμ ±
Ó³ Ÿ. 009.. 6, º 7(156.. 6Ä69 Š Œ œ ƒˆˆ ˆ ˆŠ ˆŒ ˆ - ˆ ƒ ˆ ˆ ˆŸ Š -Œ ˆ Šˆ ˆ.. ŠÊ²Ö μ 1,. ƒ. ²ÓÖ μ μ ± Ê É É Ê Ò μ μ, Œμ ± É ÉÓ μ Ò ÕÉ Ö ²μ Í Ò - μ Ò ² É Ö ³ ÖÉÓ Ì ÒÎ ² ÖÌ, μ²ó ÊÕÐ Ì ±μ ± 4- μ Ò. This paper
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### ο Θε ος η η µων κα τα φυ γη η και δυ υ υ να α α α µις βο η θο ος ε εν θλι ψε ε ε σι ταις ευ ρου ου ου ου ου σαις η η µα α α ας σφο ο ο ο
Ἐκλογή ἀργοσύντοµος εἰς τὴν Ἁγίν Κυρικήν, κὶ εἰς ἑτέρς Γυνίκς Μάρτυρς. Μέλος Ἰωάννου Ἀ. Νέγρη. Ἦχος Νη ε Κ ι δυ υ υ υ ν µι ις Α λ λη λου ου ου ι ι ι ι ο Θε ος η η µων κ τ φυ γη η κι δυ υ υ ν µις βο η θο
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### ΘΕΜΑΤΑ ΚΑΙ ΑΠΑΝΤΗΣΕΙΣ ΕΠΑΝΑΛΗΠΤΙΚΩΝ ΠΑΝΕΛΛΑ ΙΚΩΝ ΕΞΕΤΑΣΕΩΝ 2014 ΧΗΜΕΙΑ ΚΑΤΕΥΘΥΝΣΗΣ
ΘΕΜΑ Α ΧΗΜΕΙΑ ΚΑΤΕΥΘΥΝΣΗΣ Για τις προτάσεις Α1 έως και Α5 να γράψετε στο τετράδιό σας τον αριθµό της πρότασης και, δίπλα, το γράµµα που αντιστοιχεί στη σωστή επιλογή. Α1. Η ηλεκτρονιακή δοµή του 11 Νa
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### ! # % #! # & (! )!! & # # &! # +,!& #. # # & / 0!& # / 12 2 # 3 # 2 ,!& 4556
! # % #! # & (! )!! & # # &! # +,!& #. # # & / 0!& # / 12 2 # 3 # 2,!& 4556 ! # % #! # & (! )!! & # # &! # +,!& #. # # & / 0!& # / 12 2 # 3 # 2,!& 4556 ! ! # % &! ( ) &! # + #, ). / # %# # 0!. 1) 1 /,
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### Περιεχόμενα. A(x 1, x 2 )
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### n a n = 2. Θεωρούµε τα σύνολα a n = n2 n n 2 + n 1. n a n = a > 0, δείξτε ότι a n > 0 τελικά.
Ασκήσεις για το µάθηµα «Ανάλυση Ι και Εφαρµογές» Κεφάλαιο : Ακολουθίες πραγµατικών αριθµών Α Οµάδα Εξετάστε αν οι παρακάτω προτάσεις είναι αληθείς ή ψευδείς αιτιολογήστε πλήρως την απάντησή σας) α) Κάθε
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### cz+d d (ac + cd )z + bc + dd c z + d
T (z) = az + b cz + d ; a, b, c, d C, ad bc 0 ( ) a b M T (z) = (z) az + b c d cz + d (T T )(z) = T (T (z) (T T )(z) = az+b a + cz+d b c az+b + = (aa + cb )z + a b + b d a z + b cz+d d (ac + cd )z + bc
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### ! # % &! ( )! % +,.! / 0 1 )2 3
! !! # % &! ( )! % +,.! / 0 1 )2 3 ) 4 5! 5 ) 6 2 2 ) 2 3 #! 3333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333337 83 % ) 1
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### Now, suppose the electron field Ψ(x) satisfies the covariant Dirac equation (i D m)ψ = 0.
PHY 396 K. Solutions for homework set #7. Problem 1a: γ α γ α 1 {γα, γ β }g αβ g αβ g αβ 4; S.1 γ α γ ν γ α γ α γ ν g να γ ν γ α γ α γ ν γ ν γ α γ α 4 γ ν ; S. γ α γ µ γ ν γ α γ α γ µ g µα γ µ γ α γ ν
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### Σχεδίαση µε τη χρήση Η/Υ
Σχεδίση µε τη χρήση Η/Υ Κ Ε Φ Λ Ι 1 Γ Ε Ω Μ Ε Τ Ρ Ι Κ Ε Σ Κ Τ Σ Κ Ε Υ Ε Σ Ρ Λ Ε Ω Ν Ι Σ Ν Θ Π Υ Λ Σ, Ε Π Ι Κ Υ Ρ Σ Κ Θ Η Γ Η Τ Η Σ Τ Μ Η Μ Ι Ι Κ Η Σ Η Σ Κ Ι Ι Χ Ε Ι Ρ Ι Σ Η Σ Ε Ρ Γ Ω Ν Τ Ε Ι Λ Ρ Ι Σ Σ
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### α β γ α β γ ( α β )( β γ )( γ α )
Γραµµικά Συστήµατα Να υθούν τα συστήµατα: (α) x+ 4y z= x+ 8y 6z= 9 (β) x+ y 9z= x+ y z= 4x y+ 7z= (γ) y+ z= x ( ) x y = x+ y = 7z ( + ) x+ y 6z= (δ) x+ y+ z= (ε) x+ y+ z= ( + ) x+ (+ ) y= + x+ y= (στ)
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### 2 (4! ((2 (5 /! / Β ;! + %ΧΑ + ((5 % # &
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Κανάρη 36, Δάφνη Τηλ. 210 9713934 & 210 9769376 ΧΗΜΕΙΑ Ο.Π. ΘΕΤΙΚΩΝ ΣΠΟΥΔΩΝ ΘΕΜΑ Α Στις ερωτήσεις Α.1 έως Α.6, να επιλέξτε τη σωστή απάντηση. Α.1 Σε δύο όμοια δοχεία Δ 1 και Δ 2 έχουν αποκατασταθεί αντίστοιχα
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### 0,1 (,.. ( ) ) 2 3 ( ) ) # 4 (( ( ) ) 5 6 & 768
! # % & ( ) ) +,.. / 0,1 (,.. ( ) ) 2 3 ( ) ) # 4 (( ( ) ) 5 6 & 768 ! # %&% ( 9 1 0 ( : & & ; < & & ( : ( # ( = : ( 5 6 & : ( 5>? &? Α 0 ; ( < 8 5 & & & Β 0 0 > & & 6 & : & 0 & & 0 ( ( : 50 7# Χ 5 0 (
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### . IE D I=1 . 0<IE D I<1
1 ( ) 25 2016 : - : (4) 1.,,,,,,....,,....,,... 15 2 3,,. 2. 3.. 40, 8 400. (FC).... FC=80 FC=320 FC=5 FC=40. 1 4 5.,. IE D I>1. 0
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k k ΚΕΦΑΛΑΙΟ 1 G = (V, E) V E V V V G E G e = {v, u} E v u e v u G G V (G) E(G) n(g) = V (G) m(g) = E(G) G S V (G) S G N G (S) = {u V (G)\S v S : {v, u} E(G)} G v S v V (G) N G (v) = N G ({v}) x V (G)
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### Τ Ο Υ Π Α Γ Ι Α Τ Η Β Υ Ρ Ω Ν Λ Ο Γ Α Ρ Ι Α Ε Μ Ο Ι Ε Κ Μ Ε Τ Α Λ Ε Υ Ε Ε Ω Ν ΚΑ Ι Ο Λ Ο Γ Α Ρ Ι Α Ε Μ Ο Ε Α Π Ο Τ Ε Λ Ε Ε Μ Α Τ Α Χ Ρ Η Ε Ε Ω Ε
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### (... )..!, ".. (! ) # - \$ % % \$ & % 2007
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### 24 ΔΙΑΓΩΝΙΣΜΑΤΑ ΓΕΩΜΕΤΡΙΑΣ Α ΛΥΚΕΙΟΥ ΔΙΑΓΩΝΙΣΜΑ 1 Ο. ΘΕΜΑ 2 Ο : Δίνεται ΑΒΓ ισοσκελές (ΑΒ=ΑΓ) τρίγωνο.αν ΒΔ και ΓΕ οι διχοτόμοι των γωνιών Β και
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V r,k j F k m N k+1 N k N k+1 H j n = 7 n = 16 Ṽ r ñ,ñ j Ṽ Ṽ j x / Ṽ W 2r V r D N T T 2r 2r N k F k N 2r Ω R 2 n Ω I n = { N: n} n N R 2 x R 2, I n Ω R 2 u R 2, I n x k+1 = x k + u k, u, x R 2,
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### X Άπειρες ευθείες, X Μία µόνο ευθεία, X ύο µόνο ευθείες.
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``` Transition: Counting from One by Imaging to Advanced Counting Domain:Addition and Subtraction
E
Achievement Number: Level One
Number Strategies AO1:
CA
Objectives
Use a range of counting, grouping, and equal-sharing strategies with whole numbers and fractions
Number Knowledge AO1: AC
Know the forward and backward counting sequences of whole numbers to 100.
Number Knowledge AO2: EA
Know the groupings with five, within ten, and with ten.
AA
Strategies being Problem progression References Knowledge being Resources
developed developed AM
Solve simple addition 9+2= ,8+4= , Teaching Addition, Identify all of the numbers Teaching Number Knowledge
(joining, change, or 14 + 3 = , 25 + 4 = , Subtraction, and Place Value in the range 0–100 at least. Number mat and lily pads (2) AP
comparing) problems to 100 99 + 5 = , 77 + 4 = , Number Tiles (18) “Teen” and “Ty” numbers (3)
The Number Strip (19)
by counting on in their heads 8 + = 11, 15 + = 19, Number hangman (5)
The Bears’ Picnic (20)
in ones. 67 + = 72, 89 + = 96 BSM
Frog Jumps(20)
14 is how many more than 8? The Bigger Number First (21) 8-1-45, 8-1-81, 9-1-4, 9-1-5,
33 is how many more than Change Unknown (21) 9-1-6, 9-1-42, 9-1-82, 12-1-1
27?
74 is how many more than BSM
69? 7-1-53, 9-1-11, 9-1-49, 10-1-7,
10-1-49, 10-1-50, 10-1-51
Transition: Counting from One by Imaging to Advanced Counting Domain:Addition and Subtraction
E
Strategies being Problem progression References Knowledge being Resources
CA
developed developed
Solve simple subtraction 12 - 3 = , 14 - 5 = , Teaching Addition, Say the forwards and Teaching Number Knowledge AC
(separating, change, or 23 - 4 = , 41 - 2 = , Subtraction, and Place backwards number word Number fans (4)
Value sequences in the range 0–
comparing) problems to 100 67 - 5 = , 72 - 6 = ,
Counting Back (22) 100, at least, connecting
Counting (11) EA
by counting back in their 12 - = 9, 24 - = 19, Lucky dip (13)
BSM that the result of adding
heads in ones. 67 - = 58, 94 - = 89 Using calculators (14)
9-3-13, 9-3-14, 9-3-55, 9-3-56, or taking one more/less AA
16 is how many less than 21? 9-3-57,9-3-58, 9-3-59, 9-3-85, Hundreds Boards and Thousands
object to a set is given by
39 is how many less than 43? 10-1-8, 10-1-52, 10-1-53 Boards (16)
the next/previous AM
74 is how many less than 80? counting number. BSM
9-1-4, 9-1-42, 9-3-9
Solve addition and 40 + 20 = , 70 - 50 = , Teaching Addition, Order numbers in the Teaching Number Knowledge (Book 4)
AP
subtraction problems with 60 + 30 = , 90 - 20 = , Subtraction, and Place range 0–100, at least. Card ordering (12)
groups of ten, using place 42 + 30 = , 75 - 20 = , Value Arrow cards (13)
value materials, 54 - = 24, 27 + = 57, Rocket- where will it fit (15)
Subtracting Tens (23)
e.g. 30 + 20 = 50, 63 – 30 = 33, 36 + = 76, 94 - = 54 Number line flips (15)
The Missing Ones and
64 – 32 = 32 Squeeze – Guess my number (15)
Tens (25)
The Thousands Book (25) Bead strings (17)
Who is the richest? (18)
Figure It Out
N 2.1 (1) The mail gets through
N 2-3 (1) Happy hundreds
BSM
9-3-51, 9-3-52, 10-1-4, 11-1-4, 11-1-5,
11-1-43, 11-1-44, 11-1-45,
11-1-46, 11-3-6, 11-3-7, 11-3-46,
11-3-47, 11-3-48, 11-3-83
Transition: Counting from One by Imaging to Advanced Counting Domain:Addition and Subtraction
E
Knowledge being developed Resources
CA
Conserve the number of objects in a collection irrespective of how they AC
are arranged, trust the count, and work from it..
Recall the facts to ten, and the teen facts, Teaching Number Knowledge (Book 4)
EA
e.g. 3 + 7 = 10, 10 - 6 = 4, 10 + 8 = 18. Up to ten (32)
Tens frames again (34)
Patterns to ten (34)
AA
BSM: AM
9-1-9, 9-1-10, 9-1-47, 9-1-48, 9-3-11, 9-3-12, 9-3-54, 10-1-3, 10-3-46, 11-3-8, 11-3-9,
11-3-49, 11-3-50, 11-3-51 AP
Recall the doubles to 20, e.g. 7 + 7 = 14.. Teaching Number Knowledge (Book 4)
Double Trouble (32)
Figure It Out
N 2.2 Helping hands (3)
BSM: 10-1-6, 10-1-47, 10-1-48, 10-1-83
Recall the number of tens within decades Teaching Number Knowledge (Book 4)
Zap (26)
Number Boggle (33)
Figure It Out
N 2.2 Flexible fingers (8)
BF 2-3 One liner (1)
BF 2-3 Fizzing it up (5)
BSM: 9-1-9, 9-1-10, 9-1-47, 9-1-48, 10-1-5, 10-1-45, 10-1-46, 11-1-12
e.g. 60 + 40 = 100. 9-3-8, 9-3-49, 9-3-50 | 1,720 | 6,983 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2014-35 | longest | en | 0.840067 |
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# Problem 1451. Symmetry of vector
Solution 235318
Submitted on 24 Apr 2013 by Tim
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = [1 0 1 0]; y_correct = 0; assert(isequal(symmetry(x),y_correct))
2 Pass
%% x = [1:10 10:-1:1]; y_correct = 1; assert(isequal(symmetry(x),y_correct))
3 Pass
%% x = [-1 0 0 1]; y_correct = 0; assert(isequal(symmetry(x),y_correct))
4 Pass
%% x = [5 4 4 5]; y_correct = 1; assert(isequal(symmetry(x),y_correct))
5 Pass
%% x = ones(1,10); y_correct =1 ; assert(isequal(symmetry(x),y_correct))
6 Pass
%% x = ones(1,9); y_correct = 1; assert(isequal(symmetry(x),y_correct))
7 Pass
%% x = [5 2 3 5 1 -1]; y_correct = 0; assert(isequal(symmetry(x),y_correct))
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Start Hunting! | 318 | 962 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.545701 |
https://ch.mathworks.com/matlabcentral/cody/problems/25-remove-any-row-in-which-a-nan-appears/solutions/705090 | 1,571,263,292,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986670928.29/warc/CC-MAIN-20191016213112-20191017000612-00021.warc.gz | 420,888,497 | 15,568 | Cody
# Problem 25. Remove any row in which a NaN appears
Solution 705090
Submitted on 22 Jul 2015 by Michael Kasa
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% A = [ 1 5 8 -3 NaN 14 0 6 NaN ]; B_correct = [ 1 5 8 ]; assert(isequal(remove_nan_rows(A),B_correct))
B = 1 5 8
2 Pass
%% A = 1:10; B_correct = A; assert(isequal(remove_nan_rows(A),B_correct))
B = 1 2 3 4 5 6 7 8 9 10
3 Pass
%% A = [ 1 5 8 -3 NaN 14 0 6 6]; B_correct = [1 5 8; 0 6 6]; assert(isequal(remove_nan_rows(A),B_correct))
B = 1 5 8 B = 1 5 8 0 6 6
4 Pass
%% A = [ 1 3 6 NaN 3 NaN]'; B_correct = [1 3 6 3]'; assert(isequal(remove_nan_rows(A),B_correct))
B = 1 B = 1 3 B = 1 3 6 B = 1 3 6 3
5 Pass
%% A = [ 1 3 6 NaN; 3 4 2 1]; B_correct = [3 4 2 1]; assert(isequal(remove_nan_rows(A),B_correct))
B = 3 4 2 1 | 390 | 921 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2019-43 | latest | en | 0.457881 |
http://reference.wolfram.com/legacy/v3/MainBook/3.9.8.html | 1,511,305,500,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806438.11/warc/CC-MAIN-20171121223707-20171122003707-00046.warc.gz | 252,697,952 | 7,668 | This is documentation for Mathematica 3, which was
based on an earlier version of the Wolfram Language.
3.9.8 Numerical Minimization FindRoot gives you a way to find points at which a particular function is equal to zero. It is also often important to be able to find points at which a function has its minimum value. In principle, you could do this by applying FindRoot to the derivative of the function. In practice, however, there are much more efficient approaches. FindMinimum gives you a way to find a minimum value for a function. As in FindRoot, you specify the first one or two points to try, and then FindMinimum tries to get progressively more accurate approximations to a minimum. If FindMinimum returns a definite result, then the result is guaranteed to correspond to at least a local minimum of your function. However, it is important to understand that the result may not be the global minimum point. You can understand something about how FindMinimum works by thinking of the values of your function as defining the height of a surface. What FindMinimum does is essentially to start at the points you specify, then follow the path of steepest descent on the surface. Except in pathological cases, this path always leads to at least a local minimum on the surface. In many cases, however, the minimum will not be a global one. As a simple analogy which illustrates this point, consider a physical mountain. Any water that falls on the mountain takes the path of steepest descent down the side of the mountain. Yet not all the water ends up at the bottom of the valleys; much of it gets stuck in mountain lakes which correspond to local minima of the mountain height function. You should also realize that because FindMinimum does not take truly infinitesimal steps, it is still possible for it to overshoot even a local minimum. This finds the value of which minimizes , starting from . In[1]:= FindMinimum[Gamma[x], {x, 2}] Out[1]= Here is a function with many local minima. In[2]:= Plot[Sin[x] + x/5, {x, -10, 10}] FindMinimum finds the local minimum closest to . This is not the global minimum for the function. In[3]:= FindMinimum[Sin[x] + x/5, {x, 1}] Out[3]= This finds the local minimum of a function of two variables. As in FindRoot, it is a good idea to choose starting values that are not too "special". In[4]:= FindMinimum[x^4 + 3 x^2 y + 5 y^2 + x + y, {x, 0.1}, {y, 0.2}] Out[4]= Numerical minimization. | 573 | 2,433 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2017-47 | latest | en | 0.922958 |
https://www.teachoo.com/12405/3415/Question-25/category/CBSE-Class-12-Sample-Paper-for-2021-Boards/ | 1,657,127,464,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104675818.94/warc/CC-MAIN-20220706151618-20220706181618-00600.warc.gz | 1,065,910,596 | 33,154 | CBSE Class 12 Sample Paper for 2021 Boards
Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards
## Solve the following differential equation: ππ¦ ππ₯ = π₯ 3 πππ ππ π¦, πππ£ππ π‘βππ‘ π¦(0) = 0.
Β
Β
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### Transcript
Question 25 Solve the following differential equation: ππ¦/ππ₯ = π₯3 πππ ππ π¦, πππ£ππ π‘βππ‘ π¦(0) = 0. Given ππ¦/ππ₯ = π₯3 πππ ππ π¦ ππ¦ Γ 1/(πππ ππ π¦) = π₯3 ππ₯ ππ¦ Γ sin y = π₯3 ππ₯ Integrating both sides β«1βγsinβ‘π¦ ππ¦γ = β«1βγπ₯^3 ππ₯γ βπππ π¦ = π₯^4/4+πΆ Since y(0) = 0 Putting x = 0, y = 0 βπππ 0 = 0/4+πΆ β1 = πΆ πͺ=βπ So, our equation becomes βπππ π¦ = π₯^4/4+πΆ βπππ π¦ = π₯^4/4β1 π^π/π+ππ¨π¬β‘πβπ=π | 695 | 1,100 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.598822 |
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# Thread: Randomizing Quiz question & code needed
1. ## Randomizing Quiz question & code needed
Hey folks,
Forgive me in advance if this is not the correct place to post this thread. I am a new member to this forum and need a little assistance.
I have some script to create a quiz but need a little more detail. Does anyone know some code I can insert, and where to randomize the questions. Secondly if I have a test bank of 100 questions but want to limit the test to 20 questions what code would I need to insert that as well.
Any assistance is very much appreciated. thank you.
kenster
• Example of picking
Code:
```<script type="text/javascript">
function Question( q, a1, a2, a3, a4 )
{
this.query = q;
this.answers = [ a1, a2, a3, a4 ];
}
var questions = [
new Question( "How many states are there?", 48, 49, 50, 51 ),
new Question( "What is the capital of Idaho?","Idaho Falls","Boise","Pocatello","\$1.98"),
new Question( "What state is the Grand Canyon in?", "CA", "AZ", "CO", "Confusion"),
... etc. ...
];
// randomly order the questions...
questions.sort( new function() { return Math.random() - 0.5; } );
// pick the first 5 of the randomized questions:
for ( var qnum = 0; qnum < 5; ++qnum )
{
var q = questions[q];
document.write( '<div class="question">[' + (q+1) + '] '
+ q.query
+ "<ul><li>" + q.answers.join("</li><li>") + "</li></ul></div>" );
}```
That's just a starter. It doesn't provide a way for picking the answer(s) [presumably you'd use radio buttons?], but it shows the basics.
• ## Thanks
thanks for the quick reply, much obliged.
• Minor corrections to see something:
Code:
```<script type="text/javascript">
function Question( q, a1, a2, a3, a4 )
{
this.query = q;
this.answers = [ a1, a2, a3, a4 ];
}
var questions = [
new Question( "How many states are there?", 48, 49, 50, 51 ),
new Question( "What is the capital of Idaho?","Idaho Falls","Boise","Pocatello","\$1.98"),
new Question( "What state is the Grand Canyon in?", "CA", "AZ", "CO", "Confusion")
/// ... etc. ... with NO COMMA after last entry
];
// randomly order the questions...
questions.sort( function() { return Math.random() - 0.5; } );
// display the randomized questions:
for ( var qnum = 0; qnum < questions.length; ++qnum )
{
var q = questions[qnum];
document.write( '<div class="question">[' + (qnum+1) + '] ' + q.query
+ "<ul><li>" + q.answers.join("</li><li>") + "</li></ul></div>" );
}
</script>```
• DOH! Thanks.
• Hey there,
Thank you for the code, I know it will be helpful. Just a little more assistance if you don't mind: I already have the questions written and functioning. The code you supplied me with, where do I place it? Thanks in advance, again, for your help.
Kenster
• Thought it was fairly well labeled:
Change this area ...
Code:
```var questions = [
new Question( "How many states are there?", 48, 49, 50, 51 ),
new Question( "What is the capital of Idaho?","Idaho Falls","Boise","Pocatello","\$1.98"),
new Question( "What state is the Grand Canyon in?", "CA", "AZ", "CO", "Confusion")
/// ... etc. ... with NO COMMA after last entry
];```
• ## Thanks again
Thanks for the help. Yes it was fairly well labeled, I just needed a little further guidance. thank you and all the best.
kenster
• Originally Posted by kenster18
Thanks for the help. Yes it was fairly well labeled, I just needed a little further guidance. thank you and all the best.
kenster
No problem ... post back if you do though.
• | 989 | 3,582 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-52 | latest | en | 0.84385 |
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# Determine gross income and dependency exemptions
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
Question 1:
Under the rules applicable after 2004, determine how many dependency exemptions would be available in each of the following independent situations. Specify whether any such exemptions would come under the qualifying child or the qualifying relative category :
a. Richard maintains a household that includes a cousin (age 12), a niece (age 18), and a son (age 20). The cousin and niece are full-time students, and the son is unemployed. Richard furnishes all of their support.
b. Minerva provides all of the support of a family friend's son (age 18) who lives with her. She also furnishes most of the support of her stepmother who does not live with her.
c. Raul, a U.S. citizen, lives in Costa Rica. Raul's household includes an adopted daughter, Helena, who is age 9 and a citizen of Costa Rica. Raul provides all of Helena's support.
d. Maxine maintains a household that includes her ex-husband, her mother-in-law, and her brother-in-law (age 23 and a full-time student). Maxine provides more than half of all of their support. Maxine is single and was divorced several years ago.
Question 2:
During the year, Wilbur received the following in connection with his father's estate:
- His father's will named Wilbur as the executor of the estate. He received \$7,500 for serving as executor.
- Wilbur was also a beneficiary of his father's estate and received real estate that was included in the estate at a value of \$100,000 that his father had purchased for \$30,000.
- Wilbur was the beneficiary of one of his father's life insurance policies. He elected to collect the proceeds of the \$100,000 policy in four installments of \$30,000 each. Each \$30,000 payment consists of principal and interest. He collected \$30,000 this year.
Determine the effect on Wilbur's gross income.
Question 3:
Determine the effect on gross income in each of the following cases:
a. Eloise received \$150,000 in settlement of a sex discrimination case against her former employer.
b. Nell received \$10,000 for damages to her personal reputation. She also received \$40,000 punitive damages.
c. Orange Corporation, an accrual basis taxpayer, received \$50,000 from a lawsuit filed against its auditor who overcharged for services rendered in a previous year.
d. Beth received \$10,000 compensatory damages and \$30,000 punitive damages in a lawsuit she filed against a tanning parlor for severe burns she received from using its tanning equipment.
e. Joanne received compensatory damages of \$75,000 and punitive damages of \$300,000 from a cosmetic surgeon who botched her nose job.
#### Solution Preview
Question 1:
Under the rules applicable after 2004, determine how many dependency exemptions would be available in each of the following independent situations. Specify whether any such exemptions would come under the qualifying child or the qualifying relative category :
a. Richard maintains a household that includes a cousin (age 12), a niece (age 18), and a son (age 20). The cousin and niece are full-time students, and the son is unemployed. Richard furnishes all of their support.
Please find below the tests to be a qualifying child and tests to be a qualifying relative.
Tests To Be a Qualifying Child Tests To Be a Qualifying Relative
1. The child must be your son, daughter, stepchild, eligible 1. The person cannot be your qualifying child
foster child, brother, sister, half brother, half sister, or the qualifying child of anyone else.
stepbrother, stepsister, or a descendant of any of them.
2. The child must be (a) under age 19 at the end of the year, 2. The person either (a) must be related to
(b) under age 24 at the end of the year and a full-time you in one of the ways listed under Relatives student, or (c) any age if permanently and totally disabled. who do not have to live with you, or (b) must
live with you all year as a member of your
household.
3. The child must have lived with you for more than half of the 3. The person’s gross income for the year
year. must be less than \$3,200.
4. The child must not have provided more than half of his or 4. You must provide more than half of the
her own support for the year. person’s total support for the year.
5. If the child meets the rules to be a qualifying child of more
than one person, you ...
#### Solution Summary
The gross income and dependency exemptions are determined.
\$2.19 | 1,036 | 4,560 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2020-50 | latest | en | 0.979252 |
https://justaaa.com/biology/521010-5-a-new-drug-has-been-discovered-that-completely | 1,695,730,176,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510208.72/warc/CC-MAIN-20230926111439-20230926141439-00482.warc.gz | 375,417,535 | 9,025 | Question
# 5) A new drug has been discovered that completely opens the voltage-gated potassium channels. If this...
5) A new drug has been discovered that completely opens the voltage-gated potassium channels. If this drug is used experimentally on an isolated axon, the membrane potential can be easily measured. Assume that this experiment provided the following results. The extracellular Na+ is 145 mM and the extracellular K+ is 4mM, the measured membrane potential in the presence of the drug is -94mV. Using this information, calculate the intracellular K+ concentration.
To solve this equation, you will need to use the following concentrations inside (i) and outside (o) the cell and permeabilities (P) of sodium (Na+), potassium (K+) and chloride (Cl-) for the neuron: Ko = 5 (all) Ki = 150 Nao = 150 Nai = 15 Clo = 125 Cli = 9 PNa= 10-9, PK = 10-8 PCl = 10-9 (All concentrations in mM and permeabilities are in units of cm/s but see “hints” below)
Membrane Potential: An electrical potential generated due to the movement of ion from higher concentration to the lower concentration and vies versa during excited state is called membrane potential.
To calculate the membrane potential Goldman equation is used According to the equation
Em = RT/F ln XO/Xin X = is the ion
R = ideal gas constant
T = Temperature
From these constant value
Em = 61.5 * log Xo / Xin
then, for interacellular K+ is
Em =. 61.5 log 4 / K+
-94 = 61.5 *( log 4 - log K+)
-94 / 61.5 - .602 = log K+
Log K+ = 2.13
Divided by log both side
K+ = log-12.13
K+ = 102.13
K+ = 135mM
The interacellular concentration of K+ is 135 mM.
#### Earn Coins
Coins can be redeemed for fabulous gifts. | 450 | 1,689 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2023-40 | latest | en | 0.900317 |
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# Problem 43977. Converting binary to decimals
Solution 2026061
Submitted on 18 Nov 2019 by Bhaskar R
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### Test Suite
Test Status Code Input and Output
1 Pass
d = ('010111'); y_correct = 23; assert(isequal(bin_to_dec(d),y_correct))
2 Pass
d = ('110000'); y_correct = 48; assert(isequal(bin_to_dec(d),y_correct))
3 Pass
d = ('1000110'); y_correct = 70; assert(isequal(bin_to_dec(d),y_correct))
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Start Hunting! | 178 | 645 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2020-45 | latest | en | 0.65835 |
https://www.doubtnut.com/question-answer/find-the-number-of-words-formed-by-permuting-all-the-letters-of-the-following-word-intermediate-642575218 | 1,669,644,754,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710533.96/warc/CC-MAIN-20221128135348-20221128165348-00545.warc.gz | 778,514,745 | 38,696 | Home
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# Find the number of words formed by permuting all the letters of the following word: INTERMEDIATE
Updated On: 27-06-2022
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Transcript
question is find the number of words formed by permuting all the letters of the following word inter mediate so let's start we have worked that is i n d i in which letter that is 12345678 910 1112 so it has 12 letters in which I is repeating two times we can see I stepped into time where is also repeating two times that is tea and is repeating three times we can see one
Kyon so by promoting the latter intermediate we can write factorial pone 2 into 2 into 3 factorial Shahar internal derangement where as another to factorial is internal derangement O3 factorial is internal arrangement of final solution
thank you | 223 | 965 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2022-49 | latest | en | 0.950826 |
https://vmlc.tamu.edu/course-selection/engineering-mathematics-i/practice-problems-for-module-2 | 1,716,380,702,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058542.48/warc/CC-MAIN-20240522101617-20240522131617-00694.warc.gz | 525,253,480 | 9,933 | # Virtual Math Learning Center
## Practice Problems for Module 2
### Exercises
Directions: You should try to solve each problem first, and then click "Reveal Answer" to check your answer. You can click "Watch Video" if you need help with a problem.
1. Eliminate the parameter to find the Cartesian equation of the curve. Sketch the curve and indicate with an arrow the direction in which the curve is traced as the parameter increases.
1. $$x=2t−1,y=2−t, \quad −2≤t≤2.$$
2. $$x = 2t − 1, y = t^2 − 1$$
3. $$x=3\cosθ, y=4\sinθ, \quad 0≤θ≤2\pi$$
a) $$y=-\dfrac{1}{2}x+\dfrac{3}{2}$$
b) $$y=\dfrac{1}{4} (x+1)^2-1$$
c) $$\left(\dfrac{x}{3}\right)^2 + \left( \dfrac{y}{4}\right)^2=1$$
If you would like to see more videos on this topic, click the following link and check the related videos.
2. Find a vector equation for the line passing through $$(1, 3)$$ and $$(−2, 7).$$
$$\langle 1-3t, 3+4t\rangle$$
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3. Find the exact value of the expression.
1. $$\sin \left( \arccos \dfrac{4}{5}\right)$$
2. $$\sin \left( 2 \sin^{-1} \dfrac{3}{5}\right)$$
a) $$\dfrac{3}{5}$$
b) $$\dfrac{24}{25}$$
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4. Simplify each expression
1. $$\tan \left( \sin^{-1} x\right)$$
2. $$\sin (\arctan x)$$
a) $$\dfrac{x}{\sqrt{1-x^2}}$$
b) $$\dfrac{x}{\sqrt{x^2+1}}$$
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5. State the value of the given quantity, if it exists, from the given graph
1. $$\displaystyle \lim_{x\rightarrow 0^-} g(x)$$
2. $$\displaystyle \lim_{x\rightarrow 0^+} g(x)$$
3. $$\displaystyle \lim_{x\rightarrow 0} g(x)$$
4. $$\displaystyle \lim_{x\rightarrow 2^-} g(x)$$
5. $$\displaystyle \lim_{x\rightarrow 2^+} g(x)$$
6. $$\displaystyle \lim_{x\rightarrow 2} g(x)$$
7. $$g(2)$$
8. $$\displaystyle \lim_{x\rightarrow -1^-} g(x)$$
9. $$\displaystyle \lim_{x\rightarrow -1^+} g(x)$$
10. $$\displaystyle \lim_{x\rightarrow -1} g(x)$$
1. $$\displaystyle \lim_{x\rightarrow 0^-} g(x)=+\infty$$
2. $$\displaystyle \lim_{x\rightarrow 0^+} g(x)=-\infty$$
3. $$\displaystyle \lim_{x\rightarrow 0} g(x) \quad DNE$$
4. $$\displaystyle \lim_{x\rightarrow 2^-} g(x)=1$$
5. $$\displaystyle \lim_{x\rightarrow 2^+} g(x)=1$$
6. $$\displaystyle \lim_{x\rightarrow 2} g(x)=1$$
7. $$g(2)=1.5$$
8. $$\displaystyle \lim_{x\rightarrow -1^-} g(x)=1$$
9. $$\displaystyle \lim_{x\rightarrow -1^+} g(x)=0$$
10. $$\displaystyle \lim_{x\rightarrow -1} g(x) \quad DNE$$
6. Find the limit.
1. $$\displaystyle \lim_{x\rightarrow 3} \frac{1}{(x-3)^8}$$
2. $$\displaystyle \lim_{x\rightarrow 0} \dfrac{x-1}{x^2(x+2)}$$
3. $$\displaystyle \lim_{x\rightarrow -2^+} \dfrac{x-1}{x^2(x+2)}$$
4. $$\displaystyle \lim_{x\rightarrow -2^-} \dfrac{x-1}{x^2(x+2)}$$
5. $$\displaystyle \lim_{x\rightarrow -2} \dfrac{x-1}{x^2(x+2)}$$
1. $$+\infty$$
2. $$-\infty$$
3. $$-\infty$$
4. $$+\infty$$
5. DNE
If you would like to see more videos on this topic, click the following link and check the related videos. | 1,227 | 3,171 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-22 | latest | en | 0.582708 |
https://cstheory.stackexchange.com/questions/857/which-ac0-problems-are-not-truly-feasible/858 | 1,726,607,868,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.53/warc/CC-MAIN-20240917204739-20240917234739-00136.warc.gz | 170,039,146 | 44,273 | # Which $AC^0$ problems are not "truly feasible"?
Neil Immerman's famous Picture of The World is the following (click to enlarge):
His "Truly feasible" class includes no other class; my question is then:
What is an AC0 problem which is considered to be unpractical, and why?
• Maybe a problem which requires circuits of depth 10^{10^100}? Commented Sep 1, 2010 at 23:36
• @Ross: I do not think so because he did not mention “real world” and he asked “why”; I think that my previous comment answers at least the “why” part. However, admittedly I do not have an example of “natural” problems which are in AC0 and require circuits of depth 10^{10^100}. Commented Sep 2, 2010 at 0:18
• There are numerous interesting real-world problems that could be solved in constant time and constant space (in virtually any model of computation), yet people have now idea how to solve them in practice. Extreme examples are computing certain constants; we could hard-code the right answer (e.g., 0 or 1), but we don't know the answer yet. Commented Sep 2, 2010 at 8:52
• Jukka: those are problem instances. Diophantine equations (like Fermat's) are undecidable as a class, even if individual instances which we have decided actually have constant depth circuits. Commented Sep 3, 2010 at 7:04
• @András: If you prefer decision problems with infinitely many "yes" and "no" instances: Let $L$ consist of all even numbers and $x$, where $x = 1$ if the white player has a winning strategy in chess and otherwise $x = 3$. Trivially, there exists a very simple family of circuits that decides $L$, but I'd still claim that it's "unpractical". Not because the circuit would be huge, but because designing the circuit would be a huge computational effort... Cheating?-) Commented Sep 3, 2010 at 9:10
If you want an example of an AC0 function that requires depth $d$, and cannot be computed by AC0 circuits of depth $d-1$, then try the Sipser functions $S^{d,n}$. The superscript $d$ is depth needed for a polynomial-size AC0 circuit. With depth $d-1$, the circuit would need exponentially many gates.
So computing $S^{d,n}$ for $d = 10^{10^{100}}$ would not be "truly feasible."
EDIT: Your question also asks why this would not be feasible. I guess the reason is that $10^{10^{100}}$ is more than the number of atoms in the visible universe.
• That's great, thank you! Maybe you can add an informal definition of the Sipser functions? I didn't know about that name. Commented Oct 4, 2010 at 15:01
• @Michaël: Unfortunately I don't have a good intuitive definition for the Sipser functions. The idea is to make a function of d quantifiers such that no depth d-1 circuit can compute it. So we want the d quantifiers to quantify over a very large number of variables. There's a nice article by Iddo Tzameret, entitled "Håstad's Separation of Constant-Depth Circuits Using Sipser Functions" (itcs.tsinghua.edu.cn/~tzameret/SipserSwitching.pdf) that defines the functions formally on page 7. Commented Oct 4, 2010 at 20:53
All this hierarchy is intentionally robust under polynomial changes of the input size. Any class in it can thus contains functions whose complexity is say n^{1000000000} which would be theoretically "feasible" but certainly not practically so. These, however will most likely be very artificial problems. In particular by a counting argument there exists a family of DNF formula (=AC^0 depth 2 circuits) of size n^1000000 which no algorithm whose running time is less than n^999999 can compute. (In a uniform setting we expect something similar but can't prove it.)
The halting problem when the input is represented in unary is in AC^0 and yet quite unfeasible in reality. I'm not sure this is what you meant, but it could be what Immerman meant.
• I guess the classes in the diagram are defined with some notion of uniformity? Otherwise the upward direction would not represent containment, since P does not contain non-uniform AC^0. Commented Sep 13, 2010 at 6:18
• Yes, the classes in the picture are uniform. E.g., it is required that an $AC^0$ circuit family can be encoded as a language accepted by a weak uniform complexity class like FO (where FO is the class of $\{0,1\}$ strings defined by first-order formulas in the language $\{0,max; X,BIT,\le,=\}$, for $X$ the only unspecified unary predicate). Commented Sep 13, 2010 at 11:35
• Point well taken. As an alternative, following Erdos, one could instead suggest the problem that for any input, outputs the Ramsey number for red six and blue six. | 1,142 | 4,512 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-38 | latest | en | 0.938209 |
https://classroomsecrets.co.uk/lesson/year-2-the-5-times-table-lesson/ | 1,632,189,153,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057131.88/warc/CC-MAIN-20210921011047-20210921041047-00660.warc.gz | 235,285,759 | 60,684 | Year 2 The 5 Times Table Lesson - Classroom Secrets | Classroom Secrets
Y2 Maths
# The 5 Times Table Lesson
This Year 2 The 5 Times Table lesson covers the prior learning of counting in 5s, before moving onto the main skill of solving multiplications by multiplying by 5.
The lesson starts with a prior learning worksheet to check pupils’ understanding. If they need more practice, you can choose other resources for that step. The interactive lesson slides recap the prior learning before moving on to the main skill. Children can then practise further by completing the activities and can extend their learning by completing an engaging extension task.
National Curriculum Objectives
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### 2 Teaching Support
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#### Lesson Slides
These lesson slides guides pupils through the prior learning of place value of 1s, 10s and 100s, before moving onto the main skill of the 5 times table. There are a number of questions to check pupils' understanding throughout.
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#### Lesson Slides
These are the same as the lesson slides on Classroom Secrets. You can assign this as an activity for pupils to access individually in school or remotely from home.
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#### Video Tutorial
In this Five Times Table Video Tutorial, Martin takes pupils through key concepts for the five times table.
### 1 Prior Learning
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#### Worksheet
This worksheet recaps prior learning of counting in 5s, before moving onto the main skill of solving multiplications by multiplying by 5.
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#### Full Step
For further practice, visit the full step where you will find a selection of further resources and activities such as a differentiated resource pack, homework/extension worksheet and discussion problem.
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#### Interactive Activity
This Year 1 Count in 5s Activity checks pupils' understanding of counting in multiples of 5.
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#### Video Tutorial
In this Count in 5s Video Tutorial, Katie uses different pictorial representations of the number 5 and a number line to demonstrate how to count in 5s.
### 3 Activities
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#### Extension
This worksheet includes varied fluency, reasoning and problem solving questions for pupils to practise the main skill of the 5 times table.
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#### Full Step
For further practice, visit the full step where you will find a selection of further resources and activities such as a differentiated resource pack, homework/extension worksheet and discussion problem.
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#### Flash Cards Activity
This 5 Times Table Game aims to support pupils in learning their multiples of 5.
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#### Memory Activity
This 5 Times Table Memory Card Game aims to support pupils in learning their 5 times table.
### 4 Extension
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#### Worksheet
This the 5 times table extension task includes a challenge activity which can be used to further pupils' understanding of the concepts taught in the 5 times table lesson.
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#### Interactive Activity
This Year 2 The 5 Times Table Maths Challenge checks pupils’ understanding of multiplying by 5. | 632 | 3,212 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2021-39 | latest | en | 0.890941 |
http://mathhelpforum.com/advanced-algebra/227386-showing-set-spans-subspace-symmetric-matrices.html | 1,529,882,152,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867095.70/warc/CC-MAIN-20180624215228-20180624235228-00079.warc.gz | 199,391,461 | 11,285 | # Thread: Showing that a set spans the subspace of symmetric matrices
1. ## Showing that a set spans the subspace of symmetric matrices
Show that the set $\displaystyle \{\begin{bmatrix}1 & 0\\ 0 & 0\end{bmatrix},\begin{bmatrix}0 & 1\\ 1 & 0\end{bmatrix},\begin{bmatrix}0 & 0\\ 0 & 1\end{bmatrix}\}$ spans the subspace of symmetric matrices in $\displaystyle M_{2x2}$.
I'm not really sure how to show this. It's the part about symmetric matrices that throws me off.
What I know:
A symmetric matrix has the property that $\displaystyle A = A^T$. To show that the set spans, I could create a matrix and show that if there are leading ones in each row, without a pivot in the augmented column, the vectors will span the subspace:
$\displaystyle \begin{bmatrix}1 & 0 & 0& a\\ 0 & 1 & 0 &b\\ 0 & 1 & 0 & c\\ 0 & 0 & 1 & d\end{bmatrix} \Rightarrow \begin{bmatrix}1 & 0 & 0& 0\\ 0 & 1 & 0 &0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{bmatrix}$
Therefore, looking at the RREF of this matrix, I'll end up with a pivot in the augmented column, which doesn't lead to the right conclusion.
As I said before, the $\displaystyle A = A^T$ condition is throwing me off, since showing something like the fact that two vectors in $\displaystyle \mathbb{R}^4$ span the subspace of $\displaystyle \mathbb{R}^4$, where S is the set of all vectors whose first and last components are zero, is relatively straightforward.
2. ## Re: Showing that a set spans the subspace of symmetric matrices
Any symmetric matrix in $M_{2\times 2}$ can be represented by $\begin{bmatrix} a & b \\ b & c\end{bmatrix}$ where $a,b,c \in \mathbb{R}$. Consider the linear combination: $\begin{bmatrix} a & b \\ b & c\end{bmatrix} = a\begin{bmatrix} 1 & 0 \\ 0 & 0\end{bmatrix} + b \begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix} + c \begin{bmatrix} 0 & 0 \\ 0 & 1\end{bmatrix}$
3. ## Re: Showing that a set spans the subspace of symmetric matrices
Originally Posted by Algebrah
I'm not really sure how to show this. It's the part about symmetric matrices that throws me off.
A symmetric matrix has the property that $\displaystyle A = A^T$. To show that the set spans, I could create a matrix and show that if there are leading ones in each row, without a pivot in the augmented column, the vectors will span the subspace:
Is it clear to you that $\left[ {\begin{array}{*{20}{c}}a&b\\b&c\end{array}} \right]$ is a symmetric $2\times 2$ matrix for any real $a,~b,~c~?$
Show that matrix is the linear combination of the given three.
4. ## Re: Showing that a set spans the subspace of symmetric matrices
Oh!
It's stated that the subspace will be the symmetric matrices in $\displaystyle M_{2x2}$.
So I'll start by defining an arbitrary matrix $\displaystyle A = \begin{bmatrix}a & b \\c & d\end{bmatrix}$.
Since $\displaystyle A = A^T$,
$\displaystyle \begin{bmatrix}a & b \\c & d\end{bmatrix} = \begin{bmatrix}a & c \\b & d\end{bmatrix}$
Looking at the elements of these matrices, it can be seen that:
1) $\displaystyle a = a$
2) $\displaystyle b = c$
3) $\displaystyle d = d$
Therefore, the matrix can be rewritten in terms of three elements, $\displaystyle a, b, c$, where $\displaystyle a, b, c \in \mathbb{R}$:
$\displaystyle A = \begin{bmatrix}a & b \\b & c\end{bmatrix}$
Now it can easily be seen that the matrix $\displaystyle A$ can be written as a linear combination of the three given matrices:
$\displaystyle a \begin{bmatrix}1 & 0 \\0 & 0\end{bmatrix} + b \begin{bmatrix}0 & 1 \\1 & 0\end{bmatrix} + c \begin{bmatrix}0 & 0 \\0 & 1\end{bmatrix} = \begin{bmatrix}a & b \\b & c\end{bmatrix}$
As such, the given set spans the subspace of symmetric matrices in $\displaystyle M_{2x2}$.
Thanks to you both! | 1,159 | 3,681 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2018-26 | latest | en | 0.760344 |
https://charline-picon.com/how-many-feet-is-15-acres/ | 1,638,374,996,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964360803.6/warc/CC-MAIN-20211201143545-20211201173545-00459.warc.gz | 241,263,311 | 4,466 | To calculation a acre worth to the equivalent value in square foot, simply multiply the amount in acre by 43560 (the counter factor). Below is the formula:
You are watching: How many feet is 15 acres
Suppose you desire to transform 15 acre right into square feet. Making use of the switch formula above, you will get:
Value in square foot = 15 × 43560 = 653400 square feet
How many acres are in 15 square feet?15 acres are equal come how countless square feet?How much are 15 acre in square feet?How to convert acres come square feet?What is the conversion aspect to transform from acre to square feet?How come transform acres in square feet?What is the formula to convert from acres to square feet? amongst others.
Acres to square feet switch chart
6 acres = 261000 square feet
7 acres = 305000 square feet
8 acres = 348000 square feet
9 acres = 392000 square feet
10 acres = 436000 square feet
11 acres = 479000 square feet
12 acres = 523000 square feet
13 acres = 566000 square feet
14 acres = 610000 square feet
15 acres = 653000 square feet
Acres come square feet counter chart
15 acres = 653000 square feet
16 acres = 697000 square feet
17 acres = 741000 square feet
18 acres = 784000 square feet
19 acres = 828000 square feet
20 acres = 871000 square feet
21 acres = 915000 square feet
22 acres = 958000 square feet
23 acres = 1000000 square feet
24 acres = 1050000 square feet
See more: What Does Prongs Mean In Harry Potter, What'S In A Name | 379 | 1,456 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | longest | en | 0.849874 |
http://oeis.org/A022556 | 1,547,778,208,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583659654.11/warc/CC-MAIN-20190118005216-20190118031216-00255.warc.gz | 173,981,109 | 3,597 | This site is supported by donations to The OEIS Foundation.
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A022556 Numbers that are a sum of a square and 2 nonnegative cubes. 0
0, 1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 13, 16, 17, 18, 20, 24, 25, 26, 27, 28, 29, 31, 32, 33, 34, 35, 36, 37, 38, 39, 41, 43, 44, 45, 49, 50, 51, 52, 53, 54, 55, 57, 58, 60, 63, 64, 65, 66, 68, 69, 70, 71, 72, 73, 74 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 LINKS MAPLE isA022556 := proc(n) local a2, b, c ; for b from 0 do if b^3 > n then return false; end if; for c from b do a2 := n-b^3-c^3 ; if a2 < 0 then break; end if; if issqr(a2) then return true; end if; end do: end do: end proc: # R. J. Mathar, Sep 02 2016 CROSSREFS Sequence in context: A032854 A031992 A023749 * A250482 A172233 A189891 Adjacent sequences: A022553 A022554 A022555 * A022557 A022558 A022559 KEYWORD nonn AUTHOR STATUS approved
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Last modified January 17 19:58 EST 2019. Contains 319251 sequences. (Running on oeis4.) | 571 | 1,571 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-04 | latest | en | 0.626693 |
https://www.convert-measurement-units.com/convert+Meter+to+Quarter.php | 1,716,001,657,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057260.44/warc/CC-MAIN-20240518023912-20240518053912-00355.warc.gz | 640,506,592 | 13,671 | Convert m to Quarter (Meter to Quarter)
## Meter into Quarter
numbers in scientific notation
https://www.convert-measurement-units.com/convert+Meter+to+Quarter.php
# Convert Meter to Quarter (m to Quarter):
1. Choose the right category from the selection list, in this case 'Distance'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), square root (√), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Meter [m]'.
4. Finally choose the unit you want the value to be converted to, in this case 'Quarter'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '280 Meter'. In so doing, either the full name of the unit or its abbreviation can be usedas an example, either 'Meter' or 'm'. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Distance'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '44 m to Quarter' or '96 m into Quarter' or '75 Meter -> Quarter' or '85 m = Quarter' or '31 Meter to Quarter' or '97 Meter into Quarter'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(83 * 84) m'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '280 Meter + 840 Quarter' or '25mm x 14cm x 68dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
The mathematical functions sin, cos, tan and sqrt can also be used. Example: sin(π/2), cos(pi/2), tan(90°), sin(90) or sqrt(4).
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 1.397 098 258 891 3×1026. For this form of presentation, the number will be segmented into an exponent, here 26, and the actual number, here 1.397 098 258 891 3. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 1.397 098 258 891 3E+26. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 139 709 825 889 130 000 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications. | 837 | 3,647 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.86424 |
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# Third grade pre post test
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### Third grade pre post test
1. 1. THIRD GRADE PRE-POST TESTALIGNED WITH STANDARDS AND EXPECTATIONS
2. 2. THIRD GRADE DIAGNOSTIC TEST
3. 3. PART ONE-LISTENING/PEAKINGInstructions: In this part of the test you willlisten to a dialogue. You will then listen to aquestion and four possible answers. You mustselect the answer that best answers each question.
4. 4. DIALOGUE ONE: Ana: Hi, Luis! Where are you going? Luis: Hey Ana! I’m going to the library. There’s a book fair going on. Ana: Cool! I love reading! Can I go with you? Luis: Sure you can. Ana: I really like reading about the planets and outer space. Luis: Wow, me too. We can share the books we buy. Ana: Good idea Luis. Let’s go.
5. 5. 1. Where is Luis going? a. school b. the library c. his classroom d. home
6. 6. 2. Why is Luis going to the library? a. to do homework b. to play with Ana c. to read a book about space d. to the book fair
7. 7. 3. What do Luis and Ana both like to do? a. watch TV b. play baseball c. read d. study in the library
8. 8. 4. What do Luis and Ana most enjoy reading about? a. animals b. plants c. geography d. space
9. 9. 5. What will Luis and Ana do with the books they buy? a. sell them b. give them away c. share them d. hide them
10. 10. BOBBYS BIG BIRTHDAY PARTY When Bobby turned six, his mom and dad had abirthday party for him. Twenty of his closest friendsjoined in the party. Everyone wore birthday hats andhad birthday cake. The birthday cake was coloredblue and white, which are Bobbys favorite colors. As the day went on, the children played kickball inthe back yard. Bobby played first base, Sally playedoutfield, and Bobbys dad was the pitcher. At the end of the party, Bobby asked his parentsif he could have next years birthday at the zoo. Hisparents agreed and everyone cheered.
11. 11. ANSWER THE QUESTIONS: 6. How old is Bobby today? a. three b. five c. six d. ten
12. 12. 7. About how many people came to Bobbys birthday party? a. five b. seven c. ten d. twenty
13. 13. 8. Which of the following is a color that Bobby really likes? a. blue b. green c. yellow d. brown
14. 14. 9. What game did the party members play? a. pin-the-tail b. hide and go seek c. baseball d. kickball
15. 15. 10. What position did Bobbys dad play in the game? a. first base b. pitcher c. shortstop d. outfield
17. 17. THE NEW POOL It was a warm day. Mom and dad brought home asurprise. It was a new pool. Becky and Chuck ran toget the hose. As Becky filled the pool, she sprayedChuck. He got wet! Then Chuck sprayed their dog,Max, and their duck, Waddle. They got wet ! Chuckand Becky played in the pool. It was nice and cool. Theduck saw the water and jumped in. Max wanted toswim, and he jumped in. Chuck and Becky laughed andlaughed. “Sorry, no dogs or ducks in our pool!” saidDad, and he took the animals out.
18. 18. 11. Which of the following could also be a title for the story? a. Fun in the Pool b. Pets and water c. A Hot day d. Cold Water
19. 19. 12. Who sprayed Chuck with the water? a. Mom b. Becky c. Dad d. Waddle
20. 20. 13. What was the surprise for Becky and Chuck? a. a new pool b. a new dog c. a new duck d. a new hose
21. 21. 14. Who jumped into the pool first? a. Max b. Waddle c. Becky and Chuck d. Dad
22. 22. 15. Why do you think the pets wanted to swim? a. They were thirsty b. They needed a bath c. The water was nice and cool d. They thought the pool was for them
23. 23. 16. When will Becky and Chuck use their pool? a. when it is raining outside b. when they want to play with their pets c. in the winter d. when it is warm outside
24. 24. READ THE STORY, THEN ANSWER THE QUESTIONS HAMSTERS Hamsters are animals. They are rodents. Hamsters are furry. Their fur can be black, white, brown, red, or a mix of those colors. Hamsters have short tails. Hamsters make good pets because they are easy to take care of. They do not need a lot of space. Hamsters usually live in cages that have lots of room and things to help the hamster exercise. Hamsters eat mostly fruit and nuts. Some people buy hamster mix at the pet store, which is a healthy mix of food for hamsters. Hamsters can carry food in their cheeks. Hamsters are usually active in the early morning and in the evening.
25. 25. ANSWER THE QUESTIONS: 17. What kind of animal is a hamster? a. fish b. reptile c. rodent d. feline
26. 26. 18. What color are hamsters? a. They can be many colors b. They are black c. They are brown d. It doesn’t say
27. 27. 19. Hamsters are good pets because… a. They are cute b. They are easy to care for c. They are soft and furry d. They don’t eat much
28. 28. 20. From the story, you can guess that hamsters eat… a. steak and potatoes b. bread and butter c. apples and seeds d. eggs and bacon
29. 29. PART THREE-WRITING 21. This is Antonio . _______is my friend. a. he b. she c. it d. they
30. 30. 22. Kathy lives _______ Main Street. a. at b. from c. on d. for
31. 31. 23. The children ___________ their bikes yesterday afternoon. a. ride b. rides c. riding d. rode
32. 32. 24. Harold is the __________boy on the basketball team. a. tallest b. tall c. taller d. more tall
33. 33. 25. Mom picked fresh________from the garden. a. berry b. berrys c. berryes d. berries
34. 34. 26. _______does your grandmother live?a. whatb. whenc. whered. why
35. 35. 27. I gave my sister a new pair of shoes. She loved_______! a. them b. it c. I d. her
36. 36. 28. Jonathan has his crayons, and I have__________. a. they b. my c. me d. mine
37. 37. 29. How old is your baby sister_______a. ( , )b. ( ! )c. ( ? )d. ( . )
38. 38. 30. The correct sentence is______a. Played the piano concert at Kevin the night last.b. The piano at the concert last night played Kevin.c. Kevin played the piano at the concert last night.d. Last night the piano Kevin played at the concert. | 1,825 | 6,128 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-30 | latest | en | 0.91513 |
https://www.albert.io/learn/ap-physics-c-e-and-m/question/resistivity-and-power | 1,493,464,905,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123491.68/warc/CC-MAIN-20170423031203-00016-ip-10-145-167-34.ec2.internal.warc.gz | 858,802,238 | 44,627 | Limited access
An electrical wire has a power rating of $5 \text { } \mathrm W$. The length of the wire is $1 \text { } \mathrm {cm}$ and the radius is $0.001 \text { } \mathrm m$. The wire is made of copper which has a resistivity of $\rho = 1.72 \times 10^{-8} \text { } \mathrm {\Omega \cdot m}$. What is the maximum current, $I$, in Amps, that can flow through this length of wire?
A
$0.0033 \text { } \mathrm A$
B
$9.6 \text { } \mathrm A$
C
$302 \text { } \mathrm A$
D
$5392 \text { } \mathrm A$
E
$91279 \text { } \mathrm A$
Select an assignment template | 187 | 573 | {"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} | 2.609375 | 3 | CC-MAIN-2017-17 | latest | en | 0.708051 |
https://leaveit2nature.com/weed/how-much-is-a-half-ounce-of-weed-worth.html | 1,702,327,389,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679516047.98/warc/CC-MAIN-20231211174901-20231211204901-00889.warc.gz | 405,580,083 | 19,268 | # How Much Is A Half Ounce Of Weed Worth?
A quarter of marijuana is equal to a quarter of an ounce, which is equal to 7 grams. If you use a half gram for each joint, you will have 14 joints total. How Much Does It Cost to Purchase a Half Ounce of Weed? In most cases, the cost of a half ounce of marijuana ranges anywhere from \$90 to \$200.
The weight of 14 grams is equal to a half-ounce of cannabis, which is often referred to as a ″half″ or a ″half-O.″ A half-ounce of cannabis is sufficient for rolling anywhere from seven to fourteen blunts and as many as twenty-eight joints, depending on your own preferences. The price for this quantity of cannabis might range anywhere from \$100 to \$160.
## How many grams are in a half of weed?
How Much Does It Cost For One Half Of Weed? The weight of a half an ounce, or 14 grams, is equivalent to the full ounce. A half an ounce of marijuana would likely be able to produce twenty joints, but just in case you haven’t done the arithmetic yet, here it is: a half an ounce of marijuana. How Much Does One Ounce of Weed Typically Cost?
## How much does an ounce of weed weigh?
A Concise Guide on the Weight and Measurement of Weed Weight In Grams 1 eighth of marijuana is equal to 3.543 grams. 7.087 grams of cannabis equals one quarter of an ounce. 1 ounce and a half of cannabis is equal to 14.175 grams. 21.262 grams (about 0.75 ounces) of marijuana 2 additional rows
## How many joints does a quarter of weed make?
• About ten joints, each containing 0.7 grams of marijuana, may be rolled from a quarter of an ounce of marijuana.
• How Much Does It Cost For One Half Of Weed?
• The weight of a half an ounce, or 14 grams, is equivalent to the full ounce.
• A half an ounce of marijuana would likely be able to produce twenty joints, but just in case you haven’t done the arithmetic yet, here it is: a half an ounce of marijuana. | 461 | 1,886 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2023-50 | latest | en | 0.937511 |
https://interviewmania.com/discussion/168-aptitude-simple-interest | 1,545,196,009,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376831334.97/warc/CC-MAIN-20181219045716-20181219071716-00388.warc.gz | 614,756,872 | 6,077 | -- advertisement --
## Discussion
1. The simple interest at x% for x years will be Rs x on a sum of ....
1. Rs x
2. Rs 100x
3. Rs (100/x)
4. Rs (100/ x2)
1. ##### Correct Option: C
As Sum = [(100 x SI)/(Time x Rate)]
here, let R =x%, T=x yr, and, SI=Rs x
∴ Sum=[(100 × x)/(x × x)]
=(100/x)
Your comments will be displayed only after manual approval.
-- advertisement --
-- advertisement -- | 127 | 395 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2018-51 | longest | en | 0.640782 |
http://www.fixya.com/support/t6008609-estimate_fractions | 1,506,454,570,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818696677.93/warc/CC-MAIN-20170926175208-20170926195208-00423.warc.gz | 456,359,105 | 34,893 | Question about Microsoft Xbox 360 Console
# How to estimate fractions - Microsoft Xbox 360 Console
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• Expert
Not quite a video game question but let me see if I can help.
We'll start off with simple ones and progress from there.
Half = 1/2 = .5
Say you have 22.5, this number will be 22 1/2, if you want this as just a fracton you will multiply the denominator to the whole number and then add the numerator to get the whole fraction; 45/2
Quarters = 1/4 & 3/4 = .25 & 75
Think of it this way: you can have 4 quarters that equal a dollar. Each quarter represents 1 out of 4 (1/4) parts of the dollar. To reduce, think about having a 50 cent piece; yes it is technically a quarter but it's also half, so 1/4 and 3/4 would be most appropriate.
When estimating fractions, think about its closest quarter.
Take .88 for example:
The 2 quarters .88 falls between are 3/4 (75 cents) and 4/4 (1 Dollar). This is where you think of place holders. You know 88 is 13 away from 75, and 12 away from 100 (1 Dollar); therefore you can say its pretty much inbetween them.
You know now that there are 4 quarters to a dollar; when you need to meet between the quarters, you will use 8ths = half a quarter. So... between 0/4 and 1/4 there are 2 8ths: 1/8, 2/8 (2/8=1/4) continue up until you've reached the lower number your trying to get between, in this example its .75. There are 6 8ths before .75, so you are left with 2 left because 8/8 = 1. You dont want to go to 1 so in this instance you'll add another 8th. You're now up to 7/8.
This example works out very well because 7/8 is exactly .875, round this 5 up and you get .88 which is closely equal to 7/8.
The way to go about estimating from here is to keep cutting the denominator in half until you find something that's close to your given number. Remember your quarters and even knowing your 8ths will help you out in the long run.
Also, if anything is just a single decimal (i.e. in .9, .7) put this number over 10 and you have the fraction, just remember to reduce if both the top and bottom numbers are even numbers.
All I ask is for the very helpful rating since you've just asked me to remember things from school 15 years ago.
Good luck!
Respectfully,
Don
Posted on Sep 15, 2010
Hi,
a 6ya expert can help you resolve that issue over the phone in a minute or two.
best thing about this new service is that you are never placed on hold and get to talk to real repairmen in the US.
the service is completely free and covers almost anything you can think of (from cars to computers, handyman, and even drones).
goodluck!
Posted on Jan 02, 2017
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## Related Questions:
### How to solve franctions
Multiplying Fractions:
1. Make sure you're working with two fractions.
2. Multiply numerator x numerator, then multiply denominator x denominator.
Dividing Fractions:
1. Make sure you're working with two fractions
2. Flip the second fraction upside down.
4. Multiply top x top and bottom x bottom.
Converting Mixed Numbers into Improper Fractions:
1. Convert mixed numbers into improper fractions.
2. Take the whole (non-fraction) number and multiply it by the denominator.
4. Put that amount over the original denominator and you will have an improper fraction.
Method for Adding and Subtracting Fractions:
1. Find the lowest common denominator (bottom number). For both adding and subtracting fractions, you'll start with the same process.
2. Multiply fractions to match the lowest common denominator.
3. Add or subtract the two numerators (top number) but NOT the denominators.
Mar 07, 2017 | Homework
### I have a el-531x how do i cauculate fractions with it
Use the key marked [a b/c]
• To enter a mixed fraction, enter integral part, press [a b/c], enter the numerator pf fraction part, press [a b/c], then enter the denominator of the fraction and press =.
• To enter a fraction or an improper fraction, enter numerator, press [a b/c], then the denominator and =
• If the displayed result is a mixed number , use [2nd][a b/c] to convert it in a fraction (improper fraction).
May 22, 2012 | Sharp EL-531VB Calculator
### How to claculate the remainder of a fraction using CASIO fx-991ES?
Use the fraction conversion features of the calculator.
Near (above) the [S<>D] key that toggles display between fraction/radicals and decimals, there is a marking that reads [a b/c <=>d/c]. Accessed through the [SHIFT] key, this key converts from improper fractions [a b/c] to improper fractions [d/c].
Enter your fraction with the fraction key (the one below the CALC key) and press enter. If the numerator is less then the denominator (proper fraction) the remainder is just the numerator. If the numerator is greater than the denominator the display will show an improper fraction (d/c) or a mixed number [a b/c]. Here a is the integer part and b/c is the fraction part.
So, if you have [a b/c] type display the remainder is b.
If the result is displayed as an improper fraction, use [SHIFT][S<>D] to convert the improper fraction (d/c) to a b/c.
Dec 08, 2011 | Casio FX991ES Scientific Calculator
### I CANT GET MYCALCULATOR TO DO FRACTIONS CAN YOU HOPE ME
To configure the calculator to do fractions, set the Input/Output mode to MathIO.
Press [SHIFT][MODE] to open (SetUp) then select [1:MathIO].
After you set it to MathIO, pressing the fraction key will display a fraction template where you can enter numerator and denominator as you would by hand (horizontal fraction bar). Also, the calculator will display results as fractions and radicals whenever possible.
To perform arithmetic operations on fractions use the regular operation keys (x, +, -, /). The calculator will perform the operations and display the results as fractions (whenever possible).
Aug 15, 2011 | Casio fx-300ES Calculator
### I believe I have a convergence problem...red and green shadows on images and print. What does it take to correct...estimated cost? Is it something I could do myself?
You need to purchase the convergence repair kit.Are you handy?
It comes with instructions.
A TV repairman will charge about \$300-\$400.
The kit is a fraction of the cost
Jun 06, 2011 | Pioneer SD-P50A5-K 50" Rear Projection...
### How to convert fractions into equivalent fractions
find the least common denominator between the original fraction into the new fraction.
example, 1/2 to 2/4 to 4/8 to 8/16.
May 19, 2010 | Cycling
### Leaking proportioning valve, and master cylinder
a salvage yard part would be fine,and would be a fraction of the estimate given you by Pep Boys. that estimate is really high,btw. kia is a good vehicle when properly maintained. if finances are an issue,i recommend using sme Lucas products that are additives,they wont fix your problem peranetly but can definately hold you over a month or two.
May 01, 2010 | 2001 Kia Sephia
### How do you change decimals to fractions on a casio
If you enter an improper fraction or a mixed fraction with the the [a b/c] key it will be displayed as a fraction. This result being displayed you can press the [F-D] key to convert the fraction to decimal. If you do nothing else and you press the [F-D] key again, the original fraction will be restored.
However, it seems that you cannot enter an arbitrary decimal number and convert it to a fraction. The [F-D] key seems to work one way : from a fraction to a decimal number. You will only recover a fraction if the original number was a fraction.
Jan 19, 2010 | Casio FX-9750GPlus Calculator
### 2003 Nissan Maxima. Passenger power window problem
Same problem on passenger door 92,000 miles. Dealer said he can reset the regulator inside the door by pressing button. I am looking for more information on this problem too. He estimates 1/2 hr (~\$50) to do work.
Apr 22, 2009 | 2003 Nissan Maxima
## Open Questions:
#### Related Topics:
194 people viewed this question
Level 3 Expert | 2,012 | 8,122 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2017-39 | latest | en | 0.954591 |
https://physics.stackexchange.com/questions/469762/if-white-light-were-split-through-a-prism-would-it-emit-more-wavelengths-than-a | 1,726,515,327,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651710.86/warc/CC-MAIN-20240916180320-20240916210320-00841.warc.gz | 398,168,464 | 42,400 | # If white light were split through a prism, would it emit more wavelengths than are contained within the visible light spectrum?
If white light was refracted through a prism, creating the colors of the rainbow, would it also create wavelengths of light in the infrared and ultraviolet range, which fall in the non visibility factor of our eye's wavelength range, but because of the low energy quality of the light itself, it couldn't extend its energy level output of wavelengths past that of infrared? And if so, is that why gamma rays don't form out of that splitting of white light?
• There arr as many bv wavelengths between 670 nm and 400 nm ( the visible range) as there are decimal numbers between 670 and 400: an infinite number of them. Commented Mar 31, 2019 at 20:58
• Since the question has been answered by the photon already, I'm just gonna add a tiny little piece of trivia: The term "white light" is most often used as the combination of all the wavelengths of the visible spectrum. According to this definition, it does not contain any UV or infrared parts... I get that the question was intended differently though.
– lmr
Commented Apr 1, 2019 at 6:01
If white light was refracted through a prism, creating the colors of the rainbow, would it also create wavelengths of light in the infrared and ultraviolet range,
If your white light was, for example, sunlight, then it would have included IR and UV light, and those wavelengths would be split out by the prism just like the different wavelengths of visible light.
In fact, William Herschel's observation that a thermometer was warmed up when placed just beyond the end of the visible beam produced by a prism, just as it would have been if it were placed in the visible part of the prism's output, is what led to the discovery of IR light.
why gamma rays don't form out of that splitting of white light?
The prism only "splits up" or spreads out the light that goes into it. It doesn't create any wavelengths that weren't present at the input.
So if your (nominally) white light doesn't contain any gamma rays, or IR, or some particular wavelength of green, then that color won't be present in the output of the prism.
Furthermore, gamma rays likely don't interact with the molecules in the prism the same way that visible (and usually UV and IR) do so gamma rays won't show up in the prism output in the same place you'd expect them to by extrapolating from the deflection of the visible wavelengths. The reason for this is that the prism effect comes from some vibrational mode of the electrons or molecules in the prism reacting to the different frequencies of the electromagnetic waves that form the light beam, and as the frequency gets higher (and wavelength shorter), the electrons or molecules won't be able to move fast enough to respond in the same way they did for lower frequencies.
Light is refracted in a prism because the refractive index depends on wavelength. For a prism made of silica, the dependence is relatively small for the visible part of the spectrum (red curve near lambda=500nm). However it is still enough for a rainbow to form. Longer wavelengths (IR) will also be refracted, but the exact magnitude of the effect depends on how n changes with wavelength.
Higher energy photons will also be refracted, but in practice they will be hard to see. Light from natural sources (such as the sun) emit at many different wavelengths depending on their temperature. Unless the emitter is very hot, there will not be a lot of light at short (high energy) wavelengths. Moreover, many materials are highly absorbant to short wavelengths (see blue curve).
Wavelengths are not countable, so we have to reinterpret the question a bit.
The same distribution of wavelengths will exist after the prism as before. | 803 | 3,807 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-38 | latest | en | 0.973148 |
http://mathhelpforum.com/math-topics/29967-converting.html | 1,527,289,454,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867220.74/warc/CC-MAIN-20180525215613-20180525235613-00149.warc.gz | 188,465,670 | 9,896 | 1. ## converting
Maths Ugh lol so here is the question
If three hundred twenty pound notes weigh a total of 380.4 grams, how much would one twenty pound note weigh..
P.S Can i have the working aswell please
2. Originally Posted by Nizzy
Maths Ugh lol so here is the question
If three hundred twenty pound notes weigh a total of 380.4 grams, how much would one twenty pound note weigh..
P.S Can i have the working aswell please
Well
$\displaystyle \frac{300~(\text{20 pound notes})}{380.4~g} = \frac{1~(\text{20 pound note})}{x}$
Solve for x.
-Dan
3. ## Divide
Edit: I Really am stuck is this algebra and can you help me please
4. ## Sorry
Sorry to double post but im really stuck it it 300p = 380.4
5. Originally Posted by Nizzy
Sorry to double post but im really stuck it it 300p = 380.4
How do you get p by itself here? If 300p = 380.4, then what does
$\displaystyle \frac{300p}{300}$
equal?
-Dan | 268 | 912 | {"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} | 3.671875 | 4 | CC-MAIN-2018-22 | latest | en | 0.896526 |
https://www.slideserve.com/hop-moss/lecture-3-the-law-of-demand-powerpoint-ppt-presentation | 1,576,492,283,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541319511.97/warc/CC-MAIN-20191216093448-20191216121448-00535.warc.gz | 850,662,161 | 13,933 | Lecture 3 The Law of Demand
1 / 23
# Lecture 3 The Law of Demand - PowerPoint PPT Presentation
## Lecture 3 The Law of Demand
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##### Presentation Transcript
1. Lecture 3 The Law of Demand ■ Our objectives: ► Explain individual choices among unlimited wants in a world of limited resources ► Develop a theory that helps us better understand and predict human actions ►What do people (our customers) want?
2. A Very Difficult Issue • What we measure in demand is a reflection of individual desires. • Daniel Bernoulli, a mathematical genius who lived in Basel, noted, in 1738, that people seek goodness or pleasure (utility). • That is, we do not seek cellphones because they are cellphones but because they are useful and give us pleasure. Use and pleasure cannot be measured, only approximated. • This is behind the law of demand.
3. A Note on Value, Scarcity, and Price as Related to Demand Why do people want what they want? • The diamond-water paradox. • Scarcity need not mean highly valued; think of snake meat. • Markets reflect what people value in relationship to current availability. • Demand is our best understanding of what people value—given current conditions.
4. The Law of Demand ■Holding all other relevant factors constant, the lower (higher) the price of a good, the greater (lower) will be the quantity demanded. ► Like all scientific propositions, it is a ceteris paribus (“other things equal” or “other things constant”) statement ► Notetheterminology: - Price means opportunity cost - Good means anything people value
5. Why focus on the Law of Demand? This is the most powerful proposition in economics. ► Irrigation design in arid and wet climates ► Building heights in cities compared to small towns ► The seasonal pattern of vegetable prices ► Why many stand in crowded trains to go visit family ► The shape of waterfront properties ► Electricity prices and automatic switches ► Etc., etc., etc.
6. The Demand Function:Some Definitions The relationship between quantity consumed and the factors determining that: D = f (P, PS, PC, I, E, T, etc.) ■ Price of the good in question—determines the location along a demand curve ■ Other variables (relevant factors) determine the placement of a demand curve: ► Prices of related goods (substitutes and complements) ► Income of buyers ► Tastes (preferences) of buyers ► Expectations held by buyers, regarding the future ► Other matters particular to a certain good
7. The Role of Tastes ■ They are very hard to measure, so we generally ignore them, but we know they exist. ■ Look to marketing and psychology for guidance here, not economists! ■ For economists, tastes are often the unexplained portion of consumption
8. Expectations ■ Also difficult to measure — but important ■ When measurable, include in the analysis. But experience shows—measures are very poor predictors of actions. ■ Like tastes, these can be used to “explain” anything ► Don’t fall into this trap
9. This or that? ■ Substitutes: Essentially, goods used in place of each other—same geographic market; similar performance characteristics. What is a substitute in one market may not be seen by consumers as such in another market—orange juice and orange soda. ► Different brands of gasoline; robots in Renault factory in France v. workers in Renault factory in Russia (former Lada) paid \$200 a month; movie theater v. movie rentals; corn or sugar in ethanol.
10. Related Goods ■ Complements: Essentially, goods used together ► Computer hardware and software; tennis balls and rackets; airplane travel and hotel rooms; Google maps and discount coupons; growing corn for ethanol and Deere tractors
11. Changes in the Price of Related Goods ■ Goods X and Y are substitutes if: ► A change in price of X changes demand for Y in same direction - Px up implies Dy up (Dy shifts to the right) - Px down implies Dy down (Dy shifts to the left) ► Effect of change in Py on Dx is also in same direction
12. More on the Prices of Related Goods ■ Goods V and W are complements if: ► A change in price of V changes demand for W in opposite direction: - Pv up implies Dw down (Dw shifts to left) - Pv down implies Dw (Dw shifts to right) ► Effect of a change in Pw on Dv is also in opposite direction
13. Changes in Income ■ Normal Goods: ► Change in income changes demand in same direction - Higher income causes increase in demand - Lower income causes decrease in demand ► ”Superior” good is a variant: change in demand due to income change is quite large ■ Inferior goods: ► Change in income changes demand in opposite direction - Higher income causes decrease in demand - Lower income causes increase in demand
14. Terminology:Used to avoid confusion ■ Changes in quantity demanded: ► Caused by changes in own price of good means movement along a given demand curve ■ Changes in demand: ► Caused by changes in other factors: - Prices of other goods - Income - Expectations, etc. ► Means a shift of the entire demand curve
15. Change in Price • A change in the price of a good means a movement along a demand curve. Price Pa Pb Demand Quantity/time Qa Qb
16. Change in Demand • A change in a factor that determines demand, besides the price of the good itself, causes the Demand curve to shift. Example: Increase in price of substitute or increase in income causes an increase in demand. Price Da Pa Db Quantity/time Qa Qb
17. Deriving a Real Demand Curve Define your market: Boulder, Colorado, over time; consumption of water by people served by city water; price per 1,000 gallons; billions gallons/yr. consumed by demanders. Year Price Quantity 1968 \$0.28 29 1972 \$0.36 19 1977 \$0.50 13 1982 \$0.74 9 Price .74 Demand .50 .36 .28 0 9 13 19 29 Quantity
18. What else would you want to know? The Demand curve plots the relationship between price and quantity demanded – nothing else – everything else is held constant But in the real world, other things are not constant, so what else would you want to know if you wanted to understand that market better? Name likely relevant factors:
19. Hard Test: Which Goods Are Complements, Which Are Substitutes? • Wine and beer. • Domestic shirts and imported shirts. • Oil from Iran and coal from China. • Paper and pencils. • Plate glass and brick/concrete
20. Demand analysis • How would demand for hair replacement be likely to change if the following occurs? • A fall in the price of hairpieces. • A rise in income. • A rise in the divorce rate.
21. Clever sales pitch • A British cellphone company promised new subscribers in November and December that all calls on Christmas Day, December 25, would be free. What would you expect happened to the volume of calls that day?
22. Question on changing demand • Trying to predict future demand, GE research showed that the average size of the American family was declining and so was the average square footage of houses. How would you think this impacted the demand for the design of, say, refrigerators?
23. Key Point from Peter Drucker Most important question sellers must ask: “What is value to the customer?” Sellers often think they know, but do not. The customers do not buy a product; they buy value. Demanders are buying a product to satisfy a want. Do not guess what customers want—always carefully evaluate what they want. | 1,668 | 7,401 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2019-51 | latest | en | 0.84614 |
https://mathoverflow.net/questions/176076/why-does-it-seem-that-rca-rba | 1,708,602,464,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473738.92/warc/CC-MAIN-20240222093910-20240222123910-00382.warc.gz | 399,099,464 | 19,560 | # Why does it seem that $rca=rba$? [closed]
The following paradox has got me stumped. I'm hoping someone can point out the error.
Take a locally compact metric space $X$ and define the $C_b(X)$ and $C_0(X)$ as the spaces of continuous real-valued functions on $X$, bounded for $C_b$ and converging to zero at infinity for $C_0$, and equip the spaces with the supremum norm.
By Dunford & Schwartz Th. IV.6.2, the topological dual $C_b'$ is isometrically isomorphic to the space $rba(X)$ of regular bounded finitely additive Borel measures on $X$; by Rudin (Real and Complex Analysis, Th. 6.19) $C_0'$ is isometrically isomorphic to the space $rca(X)$ of regular countably additive Borel measures. In both cases the identification has the form $\langle \xi,f\rangle = \int_X f\, d\mu_\xi$.
Since $C_0\subset C_b$, and since $C_0$ and $C_b$ share the same norm, a simple calculation gives that $C_b'\subset C_0'$; since the two identification structures above are the same, this implies that $rba\subset rca$.
However, the definitions of $rba$ and $rca$ immediately imply that $rca\subset rba$.
What is going wrong here?
• The map $C'_b\to C'_0$ ("restriction") induced by the inclusion $C_0 \to C_b$ is not generally injective. Jul 14, 2014 at 10:47
• As Sean Eberhard notes, what you get is a _quotient_map from rba onto rca. In fact rca is a direct summand of rba, and your quotient map is "throwing away the singular part" Jul 14, 2014 at 11:07
• @MarkPeletier Exactly. For example consider $X=\mathbf{N}$. Then $C_0 = c_0$ and $C_b = \ell^\infty$. Between these two spaces is the space $c$ of all sequences $(x_n)$ which converge to some limit. The map $(x_n)\mapsto \lim x_n$ is a bounded linear functional on $c$, and so by the Hahn-Banach theorem it extends to some nontrivial bounded linear functional on $\ell^\infty$ which vanishes on $c_0$. Jul 14, 2014 at 15:10
• As an elementary mnemonic, if you have a linear map $T : X \to Y$ and $T^*$ is its adjoint, you should expect the injectivity of $T^*$ to be related to the surjectivity of $T$, and vice versa. (Think about matrices with some zero rows or columns.) It's a little trickier in infinite dimenions (e.g. in some cases, instead of "surjective" you want "dense range") but it helps in this case: since the inclusion from $C_0$ to $C_b$ is not surjective (nor does it even have dense range) you should not expect its adjoint to be injective. Jul 14, 2014 at 15:15
• This question appears to be off-topic because it is based on a natural but elementary error. Jul 14, 2014 at 17:59 | 758 | 2,553 | {"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} | 2.84375 | 3 | CC-MAIN-2024-10 | latest | en | 0.883754 |
https://www.jiskha.com/display.cgi?id=1291744137 | 1,503,041,039,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104612.83/warc/CC-MAIN-20170818063421-20170818083421-00574.warc.gz | 901,000,444 | 3,959 | posted by .
for a frequency table what is the best interval for the data 1,14,6, 8, 10, 12, 15, 22, ans 25
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Organize the data into a class interval frequency distribution using 10 intervals with frequency (f) and relative frequency (rf). Data Set: 100, 97, 99, 70, 72, 75, 82, 68, 85, 88, 71, 77, 93, 94, 54, 59, 83, 87, 98, 84, 72, 96, 98, … | 374 | 1,272 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2017-34 | latest | en | 0.759311 |
https://www.cfd-online.com/Forums/tecplot/185241-plottting-tangnetial-axial-velocity-vectors-plane.html | 1,675,188,133,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499888.62/warc/CC-MAIN-20230131154832-20230131184832-00175.warc.gz | 717,173,187 | 14,864 | # plottting tangnetial and axial velocity vectors on a plane
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March 21, 2017, 15:18 plottting tangnetial and axial velocity vectors on a plane #1 Senior Member Shamoon Jamshed Join Date: Apr 2009 Location: Karachi Posts: 348 Rep Power: 16 Hello all, I am simulating a rotational flow in a pipe whose flow velocity was specified as axial, radial and tangential with axial being the direction of flow in z direction in Fluent. In Tecplot ----------- Now when I take this geometry I want to see velocity vectors on plane. Here I do not see radial axial or tangential rather u, v , w as velocity components. Q1: Do I need to change u,v, w from x,y and z components to w,v,u b/c my z velocity (w-component) is in axial direction. Does tecplot sense z direction as axial by default ? Q2. Do I need to define in specify equations and write explicitly the equations for axial , radial and tangential velocities?
December 1, 2017, 14:41 #2 Senior Member Shamoon Jamshed Join Date: Apr 2009 Location: Karachi Posts: 348 Rep Power: 16 Any response please?
December 4, 2017, 14:55 #3 Member Join Date: May 2013 Posts: 61 Rep Power: 11 At this point Tecplot does not support direct cylindrical plotting for either spatial or vector coordinates. As such everything must be defined by X,Y,Z and U,V,W. As long as you have defined X,Y,Z as your spacial coordinates (in Plot>Assign XYZ) you should correspondingly define U,V,W as your vector variables (see Plot>Vector>Variables). Assuming you are interested in vectors on a slice or an existing surface zone, you can use U,V,Wthe either use the default representation or select "Tangent Vectors" (Slice Details>Vectors page or Zone Style>Vectors page) which will show only the tangent component of the vector. If you only have axial, radial and tangential velocities defined in your input file you will have to use Data>Alter>Specify Equations to calculate U,V and W using the standard conversion equations. And no, Tecplot cannot detect any axial direction by default so that we can be as process neutral as possible. If you have any followup questions feel free to contact Tecplot support at support@tecplot.com -Devon Simpson Tecplot Technical Product Manager | 531 | 2,275 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2023-06 | latest | en | 0.859858 |
https://convertit.com/Go/ConvertIt/Measurement/Converter.ASP?From=wine+arroba | 1,701,815,338,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100568.68/warc/CC-MAIN-20231205204654-20231205234654-00180.warc.gz | 220,691,571 | 3,880 | Partner with ConvertIt.com
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Conversion Result: ```Spanish wine arroba = 0.0162772706712 meter^3 (volume) ``` Related Measurements: Try converting from "wine arroba" to bath (Israeli bath), bushel (dry bushel), chetvert (Russian chetvert), cord foot (of wood), cup, dram fluid (fluid dram), firkin, gallon, gill, jigger, koku (Japanese koku), last, methuselah, nebuchadnezzar, oil arroba (Spanish oil arroba), tea cup, tou (Chinese tou), tun (English tun), UK gallon (British gallon), wine bottle, or any combination of units which equate to "length cubed" and represent capacity, section modulus, static moment of area, or volume. Sample Conversions: wine arroba = .41747573 amphora (Greek amphora), 3.58 Canadian gallon, .07750164 chetvert (Russian chetvert), 68.8 cup, 4,403.2 dram fluid (fluid dram), 542,575.69 drop, 21.5 fifth, 4.3 gallon, 137.6 gill, .06825397 hogshead, .23888889 kilderkin, 2.15 methuselah, 4.11 omer (Israeli omer), 34.4 pint (fluid pint), 2.87 rehoboam, .15396992 sack, .01627727 stere, 91.73 tea cup, 3,302.4 teaspoon, 28.64 UK pint (British pint).
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Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 485 | 1,660 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2023-50 | latest | en | 0.658976 |
http://www.aliquote.org/post/galton-watson/ | 1,653,710,940,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663012542.85/warc/CC-MAIN-20220528031224-20220528061224-00023.warc.gz | 56,504,003 | 4,014 | # aliquote
## < a quantity that can be divided into another a whole number of time />
We already talked about Markov models and branching process. Formally, branching processes allow to model a random genealogical or phylogenetic tree with a common ancestor — we also say this is a rooted tree. Let’s assume it is unique. This tree can be defined in a recursive manner:
• At generation $n=0$, the only ancestor die after a fixed period of time (or life), with a probability $p_k$ ($k\ge 0$) of having $k$ children.
• At the the $n$ th generation, all individuals die after the same amount of time as their ancestors, and each of them has, i.i.d., a probability $p_k$ ($k\ge 0$) of having $k$ children.
The Galton-Watson process is the random sequence of counts for each generation, $Z_n$ ($n\ge 0$). Its probability law is clearly deduced from the discrete probability $p_k$. Let’s further assume that $\sum_{k\ge O} k^2p_k < \infty$. This process can also be expressed as a Markov Chain: Let’s denote $p_j^{*i}$ the probability law resulting from the convolution of the probability law of reproduction of the process:
$$p_j^{*i} = \sum_{k_1+\dots+k_i=j} p_{k_1}\dots p_{k_i}.$$
A Galton-Watson process is a Markov chain on positive integers with transition probability:
$$Q(i,j) = \mathbb{P}\left( Z_{n+1} = j \mid Z_n = i\right) = \begin{cases} p_j^{*i}\quad \text{if}\: i\neq 0,\cr \delta_{0j}\quad \text{otherwise}\end{cases}.$$
Little calculus show that the first two moments of $Z_n$ using recurrence are:1
\begin{align} \mathbb{E}(Z_n) &= m^n,\cr \mathbb{V}(Z_n) &= \begin{cases} \sigma^2m^{n-1}\frac{m^n-1}{m-1}\quad \text{if}\: m\ne 1,\cr n\sigma^2\quad\text{otherwise}\end{cases} \end{align}
It can further be shown that population extinction (i.e., $Z_n=0$) occurs with probability 1 if $\mathbb{E}(Z_n)\le 1$. If, on the other hand, $\mathbb{E}(Z_n) > 1$, then the probability of extinction $q<1$ is the unique solution of $\phi(s)=s$, $s\in [0;1[$, where $\phi(s) = \sum_{k=0}^{+\infty} p_ks^k$, $\forall s\in [0;1]$, is the moment generating function of $Z_n$.
Here is a little application of this process in biology. Polymerase Chain Reaction (PCR) is a technique of in vitro enzymatic amplification which can be used to increase the number of available DNA strands. Denaturation, cooling and extension of DNA strands are the three main steps in each cycle of amplification. When this cycle is correctly achieved, the number of DNA strands is doubled. However, a certain number of failures can be observed.
Suppose there are $N_0$ DNA strands at the start of this process. Each of those $N_0$ strands serves as an ancestor of a Galton-Watson process, where the probability law is given by $p_1=1-p$, $p_2=p$ and $p_k=0$ for $k\ne 1, 2$. Here, $p$ denotes the probability of success of an amplification cycle. The mathematical expectation of reproduction equals $m=1+p$, and the variance is $\sigma^2 = p(1-p)=(m-1)(2-m)$, with $q=0$ (probability of extinction). The expected number of DNA strands after $n$ cycles is thus $N_0m^n$.2
1. Jacques Istas, Introduction aux modélisations mathématiques pour les sciences du vivant, Springer, 2000. ↩︎
2. From a practical point of view, it is often desired to know $N_0$ in advance, although $N_0$ cannot be identified uniquely. Note, however, that it is possible to consistently estimate the mathematical expectation of the reproduction $m$ with th estimator $\hat m_n = \sum_{k=0}^{\infty} k\hat p_{k_n} = \frac{Y_{(n+1)}-1}{Y_n}$, that is $m$ can be estimated from the observed size of each generation. Hence, the following estimator has been proposed: $\hat N_{0n} = \frac{Z_n}{\hat m_n}$. As an illustration, if the success rate of the PCR is 80% (i.e., $p = 0.8$), its variance equals $\frac{1-p}{1+p}=0.11$. ↩︎ | 1,135 | 3,788 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2022-21 | latest | en | 0.748245 |
http://www.the-mathroom.ca/stats/cegp/cgpst-2.1/cgpst-2.1.htm | 1,552,960,185,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912201882.11/warc/CC-MAIN-20190319012213-20190319034213-00324.warc.gz | 371,990,078 | 4,877 | Small Sample Confidence Intervals About a Mean
Interval Estimates for Small Samples:
The Student's t-distribution
When the sample size is small, the critical values for confidence intervals are determined by the Student's t-distribution, so they are called t-values rather than z-values.
The probabilities for this distribution are defined strictly by "degrees of freedom" or
the number of data values available to estimate the population's standard deviation.
A sample of size n has n – 1 degrees of freedom.
The t-value formula is identical to the one for the z-value
...... s = sample standard deviation
Confidence Interval Estimate for µ when .
Estimation with Larger Samples and Student's t -distribution:
The Student's t -distribution is generally used on small samples with n < 30. An increase in the sample size affects both the stardard error and the number of degrees of freedom. As the degrees of freedom increase, the t-value approaches the z-value for the same level of confidence. When the sample size becomes extremely large, the t-distribution converges to the z-distribution. If n > 30 use the Normal Distribution.
_____________________________________
Example:
To test the durability of a new paint used for the white lines on the highway, the company paints 8 strips of it on a busy road and counts the number of "crossings" it takes to begin deteriorating the paint surface. Rounded to the nearest hundreth, here's the data:
142,600 167,800 136,500 108,300 126,400 133,700 162,000 149,400
Construct a 95% confidence interval for the average number of crossings it takes to deteriorate the paint surface. Round to the nearest hundred.
Solution: n = 8 ...... ...... s = 19, 200 (by formula for sample st. dev.)
Since n = 8, there are 8 – 1 = 7 degrees of freedom at 95%
The critical t-value for 95% and 7 degrees of freedom = 2.365.
140,800 ± 16,000 =
the confidence interval is 124,700 < µ < 156, 900.
The surface of the paint will begin to deteriorate after the lines have been crossed between 124,700 and 156,900 times.
_____________________________________
Formulae for Small Sample Confidence Intervals
Property Confidence Interval for Mean ( ) Sample Size when use n – 1 degrees of freedomwhen is unknown Errorwhen . Parametersuse s when is unknown
Practice (view student's t-table)
1.
Tim weighed himself once a week for several years. Last month his four measurements (in pounds) were: ...... 190.5 ......189.0 ......195.5 ......187.0
a) Construct a 90% confidence interval for his mean weight for last month.
b) Suppose that Tim wants 99% confidence rather than 90%. Reconstruct the confidence interval for all 4 measurements.
c) Tim now wants to estimate his monthly weight accurate to within 2 pounds, with 95% confidence. What sample size does he need to achieve this?
2.
The personnel department of a large corporation wants to estimate the family dental expenses of its employees to determine the feasibility of providing a dental insurance plan.
A random sample of ten employees reveals that the family dental expenses for the preceding year had a mean of \$261 and a standard deviation of \$139.
Set up a 90% confidence interval estimate of the mean family dental expenses for all employees of this corporation.
3.
A random sample of 16 summer days in Montreal, showed the mean level of CO was 4.9 ppm. The standard deviation was 1.2 ppm. Based on this, construct a 95% confidence interval estimate of the true mean level of CO in summer in Montreal.
Solutions
1.
Tim weighed himself once a week for several years. Last month his four measurements (in pounds) were: ...... 190.5 ......189.0 ......195.5 ......187.0
a) Construct a 90% confidence interval for his mean weight for last month.
Solution: with 4 sample statistics, we'll use the t-distribution.
data:
with 3 degrees of freedom and alpha = 10%, the t-value is 2.353
the 90% confidence interval for his mean weight is (186.23, 194.77).
b) Suppose that Tim wants 99% confidence rather than 90%. Reconstruct the confidence interval for all 4 measurements.
Solution:
Now t = 5.841 for a 99% interval:
So the 99% confidence interval is (179.9, 201.1)
c) Tim now wants to estimate his monthly weight accurate to within 2 pounds, with 95% confidence. What sample size does he need to achieve this?
Solution:
Now t = 3.182 for a 95% with 3 dof :
a sample of 34 measurements is required.
2.
The personnel department of a large corporation wants to estimate the family dental expenses of its employees to determine the feasibility of providing a dental insurance plan.
A random sample of ten employees reveals that the family dental expenses for the preceding year had a mean of \$261 and a standard deviation of \$139.
Set up a 90% confidence interval estimate of the mean family dental expenses for all employees of this corporation.
Solution:
data: n = 10 ............ s = \$139 ......9 dof ...... t-value = 1.833,
= 261 ± 80.57
\$ 180.43 < µ < \$ 341.57
3.
A random sample of 16 summer days in Montreal, showed the mean level of CO was 4.9 ppm. The standard deviation was 1.2 ppm. Based on this, construct a 95% confidence interval estimate of the true mean level of CO in summer in Montreal.
Solution:
Since n = 16, this a small sample t-distribution situation.
We know that , s = 1.2, and
degrees of freedom = 15, so ta/2, 15 = 2.131,
so our confidence limits are 4.9 ± 2.131(0.3) = 4.9 ± 0.639
the confidence interval for the mean level of CO is [ 4.26, 5.54 ] parts per million.
.
.
Student's t Distribution Probabilities (t-scores)
Conf. Level 0.8 0.9 0.95 0.98 0.99 One Tail 0.10 0.05 0.025 0.01 0.005 Two Tails 0.20 0.10 0.05 0.02 0.01 df Values of t 1 3.078 6.314 12.71 31.82 63.66 2 1.886 2.920 4.303 6.965 9.925 3 1.638 2.353 3.182 4.541 5.841 4 1.533 2.132 2.776 3.747 4.604 5 1.476 2.015 2.571 3.365 4.032 6 1.440 1.943 2.447 3.143 3.707 7 1.415 1.895 2.365 2.998 3.499 8 1.397 1.860 2.306 2.896 3.355 9 1.383 1.833 2.262 2.821 3.250 10 1.372 1.812 2.228 2.764 3.169 11 1.363 1.796 2.201 2.718 3.106 12 1.356 1.782 2.179 2.681 3.055 13 1.350 1.771 2.160 2.650 3.012 14 1.345 1.761 2.145 2.624 2.977 15 1.341 1.753 2.131 2.602 2.947 16 1.337 1.746 2.120 2.583 2.921 17 1.333 1.740 2.110 2.567 2.898 18 1.330 1.734 2.101 2.552 2.878 19 1.328 1.729 2.093 2.539 2.861 20 1.325 1.725 2.086 2.528 2.845 21 1.323 1.721 2.080 2.518 2.831 22 1.321 1.717 2.074 2.508 2.819 23 1.319 1.714 2.069 2.500 2.807 24 1.318 1.711 2.064 2.492 2.797 25 1.316 1.708 2.060 2.485 2.787 26 1.315 1.706 2.056 2.479 2.779 27 1.314 1.703 2.052 2.473 2.771 28 1.313 1.701 2.048 2.467 2.763 29 1.311 1.699 2.045 2.462 2.756 30 1.310 1.697 2.042 2.457 2.750
.
. | 2,077 | 6,682 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-13 | longest | en | 0.842198 |
http://www.factsabout.com/f/fu/fuzzy_logic.html | 1,511,589,416,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934809419.96/warc/CC-MAIN-20171125051513-20171125071513-00249.warc.gz | 386,841,535 | 2,947 | Fuzzy logic
Fuzzy Logic is a superset of Boolean logic dealing with the concept of partial truth. Whereas classical logic holds that everything can be expressed in binary terms (0 or 1, black or white, yes or no), fuzzy logic replaces Boolean truth values with degrees of truth which are very similar to probabilities (except that they need not sum to one). This allows for values between 0 and 1, shades of gray, and maybe; it allows partial membership in a set. It is highly related to fuzzy sets and possibility theory. It was introduced in the 1960s by Dr. Lotfi Zadeh of UC Berkeley.
Applications : Home appliances and more
Fuzzy logic is used to control household appliances (such as washing machines which sense load size and detergent concentration and auto-adjust their wash cycles accordingly; and refrigerators)
Another applications are passenger elevators, automobile subsystems (like ABS) and cameras.
A basic application might quantify where a limited range applies to a smooth spectrum -- as in temperature measurement for anti-lock brakes to function properly. Truth values derived from the specific temperature are mapped to a series of candidate quantities. These quantities can then be used to determine a separate function in accordance with the graduated value scheme.
In this image, cold, warm, and hot are identities mapped to a temperature scale. A point on that scale is represented by two "truth values" -- one in each of the two nearest identities. As the temperature rises, its "truth value" in the cold category declines, while its "truth value" in the warmer category rises.
The AND, OR, NOT operators of boolean logic have their fuzzy analogs in MIN, MAX, and COMPLEMENT. | 345 | 1,709 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2017-47 | latest | en | 0.925586 |
http://drjimo.net/geometry/trigonometry/ | 1,508,390,570,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823229.49/warc/CC-MAIN-20171019050401-20171019070401-00227.warc.gz | 88,640,410 | 9,297 | # Trigonometry
Trigonometry resources here.
## Trig Reference Sheet
Trig_Cheat_Sheet (found at http://tutorial.math.lamar.edu/cheat_table.aspx)
## 10 cm Circle
10 cm Circle (make sure, when printing, to print full size) http://www.mathedpage.org/circle/circles/10cm-circle.pdf
## Trigonometry: A Clever Study Guide (J Tanton)
Trigonometry: A Clever Study Guide is a book by James Tanton James Tanton (a very interesting mathematician interested in HS student understanding): Twitter: https://twitter.com/jamestanton http://gdaymath.com/ http://www.jamestanton.com/ MAA: http://www.maa.org/math-competitions/teachers/curriculum-inspirations YouTube: example (tetrahedron problem) 😉 James Tanton (PhD. Mathematics, Princeton University, 1994) is a research mathematician deeply interested in bridging the gap between the mathematics experienced by …
## Trig on GeoGebra
Rolling a Unit Circle Here’s a nice GeoGebra to see and think about radians. Rolling a unit circle. (https://ggbm.at/r6Ke6KjC) Save Save
## Misc. Trig
Page for Miscellaneous Trigonometry Information Webpage How do you find exact values for the sine of all angles? (some via straightforward methods such as the sum, difference, and half-angle formulas — some by more obscure means) | 323 | 1,264 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2017-43 | longest | en | 0.739756 |
https://www.jiskha.com/display.cgi?id=1328829806 | 1,503,305,757,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886107744.5/warc/CC-MAIN-20170821080132-20170821100132-00142.warc.gz | 893,942,745 | 3,796 | # geometry
posted by .
On one sunny afternoon, the movie theater casts a shadow of 30 feet. At the same time of day, A tree is casting a shadow of 10 feet. What is the height of the movie theater?
• geometry - incomplete -
How tall is the tree?
## Similar Questions
1. ### math
on a sunny day a tree casts a shadow that is 146 feet long. at the same time a person who is 5.6 feet tall standing beside the tree casts a shadow that is 11.2 feet long. how tall is the tree?
2. ### geometrY
to estimate the height of a tree, Dave stands in the shadow of the tree so that his shadow and the tree's shadow end at the same point. Dave is 6 feet 4 inches tall and his shadow is 15 feet long. If he is standing 66 feet away from …
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See the last picture at the bottom of this web page (for a pic of this exact problem). Using ratios, Tree height = T 15/66 = 6.33/T 15T = 512.73 T = 34.182 T = 34.182 when getting this answer from my question: to estimate the height …
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10. ### math
Don't laugh at me lol...it says Jamal is 5 feet tall. One sunny morning he notices that he casts a shadow that is 2 feet long. If the school building casts a shadow that is 7 feet long at the same time, what is the height of the building?
More Similar Questions | 610 | 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} | 3.953125 | 4 | CC-MAIN-2017-34 | latest | en | 0.894609 |
https://betterlesson.com/lesson/resource/2442168/turtle-snail-part-2-notebook | 1,498,339,751,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320338.89/warc/CC-MAIN-20170624203022-20170624223022-00117.warc.gz | 747,418,250 | 22,743 | ## Turtle & Snail Part 2.notebook - Section 2: Turtle & Snail Part 2: Snail Gets A Head Start
Turtle & Snail Part 2.notebook
# Turtle & Snail Part II
Unit 5: The Fabulous World of Functions
Lesson 3 of 19
## Big Idea: Building on the learning from Part I, students will again be able to make connections among functional representations in context, this time with one change: Snail gets a head start.
Print Lesson
4 teachers like this lesson
Standards:
Subject(s):
Math, Algebra, function
40 minutes
### Heather Sparks
100 Lessons | 5 new
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http://forums.mikeholt.com/showthread.php?t=53273 | 1,409,393,507,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500834883.60/warc/CC-MAIN-20140820021354-00227-ip-10-180-136-8.ec2.internal.warc.gz | 75,321,971 | 9,933 | 1. Junior Member
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Feb 2005
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I have a worksheet that takes me through 8 steps of doing a load calculation for a commercial building, with the code references for each step including the specific tables (and then references for sizing the neutral and ground).
This sheet is about 12 years old or more and all of the code tables and code references have changed, so the sheet is no longer relevant. I was asked to do a load analysis as a favor for my church and can't seem to find the necessary resources for doing this. It was second nature 16 years ago when I worked as a journeyman, but I have changed industries and have forgotten the steps.
If anyone can email me a worksheet that would assist me I would be eternally grateful, they need me to do this by the end of the day tomorrow. Thank you!
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Location
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This is not an all-inclusive procedure for sizing a service to a non-dwelling occupancy. Please reference Article 220 for exact code specifications.
STEP 1 Use Table 220.3(A) (2005-220.12) and multiply lighting load va to total square footage. Feeder conductor at 125% for continuous load.
STEP 2 Demand load for motel, hospital, warehouse. (Table 220.11) (2005-220.42) All others at 100%.
STEP 3 Compare heat with A/C. Omit smaller. (220.21) (2005-220.60)
STEP 4 Heavy-duty lampholders at 600va each. (220.3(B)(5)) (2005-220.14(E))
Other outlets at 180va each. (220.3(B)(11)) (2005-220.14(I))
Multioutlet assemblies each 5-foot at 180va each. (220.3(B)(8)) (2005-220.14(H))
Show window lighting each linear foot at 200va each. (220.3(B)(7) (2005-220.14(G))
STEP 5 Demand for receptacle loads over 10kva. (220.13) (2005-220.44)
STEP 6 Demand for kitchen equipment. (Table 220.20) (2005-220.56)
STEP 7 Largest motor increased by 25%. (220.14) (2005-220.50)
STEP 8 Size the service by dividing the total va by the applied voltage. Use Table 250.66 to size grounding conductor. (It cannot be smaller than the neutral)
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klip
I have a worksheet that I have used to teach load calcs with. You can use it for reference if you choose. Send me your e-mail adress in a personal message, and I'll send it to you.
Jim T
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• | 670 | 2,420 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2014-35 | longest | en | 0.899109 |
https://fiveminutelessons.com/comment/1365 | 1,718,630,508,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861719.30/warc/CC-MAIN-20240617122241-20240617152241-00076.warc.gz | 204,782,824 | 21,364 | How to use the IF function in Excel to calculate values based on different criteria
The IF statement is a simple function in Excel that is one of the building blocks you need when you are working with large spreadsheets. You may not know you need it yet, but once you know how to use it, you won't want to live without it.
IF() Function Syntax
The IF() function has the following syntax:
=IF(logical test, value if true, value if false)
The IF function works by performing a logical test that can only have one of two outcomes - TRUE or FALSE. It then outputs a result based on the outcome of that test.
Logical Tests in Excel
You can use anything as a logical test provided Excel can determine whether the outcome is TRUE or FALSE.
Some examples of logical tests that you can use with the IF function include:
• C5=C6 (compare cell C5 to C6. If they are equal, the outcome is TRUE, otherwise the outcome is FALSE).
• C5>C6 (if C5 is greater then TRUE, otherwise FALSE)
• C5="" (if cell C5 is empty then TRUE, otherwise FALSE)
• SUM(A5:A10)>B5 (if the sum of cells A5 to A10 is greater than B5 then TRUE, otherwise FALSE).
There are a number of ways to construct more complex logical tests which we won't cover here.
Examples of the IF() function in action
To use the IF() function, follow these steps:
• Click on a cell and enter the IF() function:
• Enter the logical test as shown in this example:
• The value in the Result column is the outcome of the IF formula
• The logical test checks to see whether the cell in the Day column (B5) = "Wednesday" (we use the speech marks to tell Excel the value we're performing the test on is text rather than a number)
• If the value in the Day cell is "Wednesday", then the result will be Yes.
• Otherwise the result will be No, as it is in this case.
There are many ways to use the IF() function in Excel that we won't cover here, but look out for more advanced IF() lessons coming soon.
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Comments on this lesson
Adding cells that must be blank when no data present.
Hi, How do you add a formula to a cell that will only give a result when data is present, but will be blank when no data has been entered.
Then add more of these formatted cells together and again only show a result when some data has been entered, but remain blank when no data entered. I keep getting #value!
Blank Cells referenced in my IF statement.
My IF statement is comparing the text value in two cells but I want the return to be blank if one of the cells is blank.
=IF(L2=N2, "Valid", "Variance")
However, if either L2 or N2 is empty, I don't want anything as the result.
Hello
=IF(L2="",,IF(N2="",,IF(L2=N2, "Valid","Variance")))
It's called a nested conditional. There used to be a limit on how many levels a nest can go. I'm not sure if the limit is still there and how deep you can go.
Using multiple IF statements in a single formula
Thanks Chris.
You can read about nested conditions in our lesson on Using multiple IF statements in Excel.
Regards
David
IF THEN statements
IF(H16=2,1.5,IF(H16=1,1)
IF(H18=2 I need it to factor in 1.5-.5 how do I do that
Using IF statement in Excel
If we use more than two cells to compare the value (Eg. "C5=C6=C7") den the result generated is false. Please suggest me the correct method.
Can't Figure out error
I am working on a spreadsheet that requires multiple IF statements. It was working great until I had to add in 1 more variable and now there is an error and I cannot figure out what it is. Can someone help me out? Below is the formula:
=if(o12="m",(round(g12*01),2)),if(o12="y",(if(12=0,0,round(if(and(C\$5,c\$6,G12>f12),f12*.04,if(and(c\$5.c\$6,g12>f12),f12*.0275,f12*.01)),2),"NA")
I have also attached a copy
John
mometasone spray over counter efeefafdbffegefb
if FALSE calculate
This is wrong - and I know it... (I keep dragging my brain back to BASIC - yeah, I'm that old)
=IF(N3,0,"=D3*.0025")
the first half is right - if there is something in that field I do want this field to be 0.
However, if there is nothing in the first field I want excel to calculate D3*.0025...
and running down the sheet, D4*.0025; D5*.0025; etc...
Does anyone know the formula that will run the calculation?
I hope you have resolved this
I hope you have resolved this by now, but for anyone who was googling to solve this exact issues, I discovered that you need to remove the quotations to get it register the cell recall
=IF(N3,0,"=D3*.0025")
becomes
=IF(N3,0,=D3*.0025)
the comma segments the formula, when using the quotation marks for the formula it will as a string of text between the quotes
"=D3*.0025"
returns as the text: "=D3*.0025 | 1,207 | 4,695 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-26 | latest | en | 0.897764 |
http://www.instructables.com/id/ACCELOROMETER-WITH-adxl-345/ | 1,495,763,466,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608622.82/warc/CC-MAIN-20170526013116-20170526033116-00099.warc.gz | 540,968,901 | 13,216 | /* Accelerometer connection pins (I2C) to Arduino are shown below:
Arduino Accelerometer ADXL345 A5 SCL A4 SDA 3.3V CS 3.3V VCC GND GND */
#include #include
int x,y,z; int rawX, rawY, rawZ; float X, Y, Z; float rollrad, pitchrad; float rolldeg, pitchdeg;
void loop(){ adxl.readAccel(&x, &y, &z); //read the accelerometer values and store them in variables x,y,z // Output (x,y,z) on horizontal plane should be approximately (0,0,255) // the following 3 lines is for an offset rawX=x+360; rawY=y+170; rawZ=z+10; X = rawX/521.00; // used for angle calculations Y = rawY/521.00; // used for angle calculations Z = rawZ/521.00; // used for angle calculations rollrad = atan(Y/sqrt(X*X+Z*Z)); // calculated angle in radians pitchrad = atan(X/sqrt(Y*Y+Z*Z)); // calculated angle in radians rolldeg = 180*(atan(Y/sqrt(X*X+Z*Z)))/PI; // calculated angle in degrees pitchdeg = 180*(atan(X/sqrt(Y*Y+Z*Z)))/PI; // calculated angle in degrees // print out values: Serial.print("x: "); Serial.print(x); // raw data without offset Serial.print(" y: "); Serial.print(y); // raw data without offset Serial.print(" z: "); Serial.print(z); // raw data without offset Serial.print(" rawX = "); Serial.print(rawX); // raw data with offset Serial.print(" rawY = "); Serial.print(rawY); // raw data with offset Serial.print(" rawZ = "); Serial.print(rawZ); // raw data with offset Serial.print(" X = "); Serial.print(X); // raw data with offset and divided by 256 Serial.print(" Y = "); Serial.print(Y); // raw data with offset and divided by 256 Serial.print(" Z = "); Serial.print(Z); // raw data with offset and divided by 256
Serial.print("\t Angle according to x axis (Roll(deg)) = "); Serial.print(rolldeg); // calculated angle in degrees Serial.print("\t Angle according to y axis (Pitch(deg)) = "); Serial.println(pitchdeg); // calculated angle in degrees // Serial.print(" Roll(rad) = "); Serial.print(rollrad); // calculated angle in radians // Serial.print(" Pitch(rad) = "); Serial.print(pitchrad); // calculated angle in radians }
## Step 1:
<p>Where does one use such a device</p>
<p>Cool project</p> | 578 | 2,099 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-22 | longest | en | 0.472373 |
https://software-dl.ti.com/jacinto7/esd/processor-sdk-rtos-jacinto7/latest/exports/docs/vxlib/docs/doxygen/html/vxlib_html/group___v_x_l_i_b__integral_image__i8u__o32u.html | 1,725,780,495,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650960.90/warc/CC-MAIN-20240908052321-20240908082321-00667.warc.gz | 513,237,630 | 3,945 | VXLIB_integralImage_i8u_o32u
## Functions
VXLIB_STATUS VXLIB_integralImage_i8u_o32u (const uint8_t src[restrict], const VXLIB_bufParams2D_t *src_addr, uint32_t dst[restrict], const VXLIB_bufParams2D_t *dst_addr, uint32_t prevRow[restrict], uint32_t prevCol[restrict], uint32_t prevRowUpdate)
VXLIB_STATUS VXLIB_integralImage_i8u_o32u_checkParams (const uint8_t src[], const VXLIB_bufParams2D_t *src_addr, const uint32_t dst[], const VXLIB_bufParams2D_t *dst_addr, const uint32_t prevRow[], const uint32_t prevCol[], uint32_t prevRowUpdate)
## ◆ VXLIB_integralImage_i8u_o32u()
VXLIB_STATUS VXLIB_integralImage_i8u_o32u ( const uint8_t src[restrict], const VXLIB_bufParams2D_t * src_addr, uint32_t dst[restrict], const VXLIB_bufParams2D_t * dst_addr, uint32_t prevRow[restrict], uint32_t prevCol[restrict], uint32_t prevRowUpdate )
Description:
Computes the integral image of the input. Each output pixel is the sum of the corresponding input pixel and all other pixels above and to the left of it.
Method:
The integral image is computed using the following equation:
``` dst(x,y) = sum(x,y)
where, for x>=0 and y>=0
sum(x,y) = src(x,y) + sum(x-1,y) + sum(x,y-1) - sum(x-1,y-1)
otherwise,
sum(x,y)=0```
Parameters
[in] src[] Pointer to array containing first input image (UQ8.0) [in] src_addr[] Pointer to structure containing dimensional information of src [out] dst[] Pointer to array containing output image (UQ32.0) [in] dst_addr[] Pointer to structure containing dimensional information of dst [in] prevRow[] Pointer to array containing the last row from a previous execution [in] prevCol[] Pointer to array containing the last column from a previous execution (optional) [in] prevRowUpdate Flag that indicates if the function should update the prevRow data (0: Don't update, 1: Update)
Assumptions:
• I/O buffer pointers are assumed to be not aliased.
• The number of elements in prevRow buffer should be equal to the width of the full image
• The number of elements in prevCol buffer should be equal to the height of the full image
• PARAMETER INITIALIZATION:
• All prevRow entries should be externally initialized to zero before calling the function for the first time since it is always read by the function regardless if the function is being called once per image, or multiple times per image.
• If calling only once per full image, prevRowUpdate can be set to 0 so that the zero initialized prevRow buffer doesn't get updated by the function, and can be re-used for subsequent images without re-initializing.
• If a user wants to divide processing of the image into smaller blocks, then it can use the prevRow and prevCol arrays to store the state information between function calls (prevRowUpdate should be set to 1).
• When using prevCol, all entries should be externally initialized to zero, before calling the function for the first time for each image.
• As the function is called across different blocks of an image, be sure to change the prevRow and prevCol pointers to align with the y and x offset of the src and dst image pointers, respectivly.
• If processing the image in horizontal strips (where the width of each strip is the width of the image), then prevCol may still be set to NULL.
• If processing the image in vertical strips (where the height of each strip is the height of the image), then prevRowUpdate may be set to 0, and prevCol should be non-NULL.
Performance Considerations:
• For best performance, the following parameter settings are recommended:
• Align all pointers to 8 byte boundaries
• Set all stride values to a multiple of 8
• Set all width values to a multiple of 8
## ◆ VXLIB_integralImage_i8u_o32u_checkParams()
VXLIB_STATUS VXLIB_integralImage_i8u_o32u_checkParams ( const uint8_t src[], const VXLIB_bufParams2D_t * src_addr, const uint32_t dst[], const VXLIB_bufParams2D_t * dst_addr, const uint32_t prevRow[], const uint32_t prevCol[], uint32_t prevRowUpdate )
Description:
Checks the parameters for programming errors for the VXLIB_integralImage_i8u_o32u function.
Method: | 1,009 | 4,039 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.494662 |
http://www.osti.gov/eprints/topicpages/documents/record/307/2552223.html | 1,455,310,405,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701165302.57/warc/CC-MAIN-20160205193925-00058-ip-10-236-182-209.ec2.internal.warc.gz | 585,309,406 | 3,454 | HYPERGEOMETRIC SERIES ACCELERATION VIA THE WZ METHOD Tewodros Amdeberhan and Doron Zeilberger Summary: HYPERGEOMETRIC SERIES ACCELERATION VIA THE WZ METHOD Tewodros Amdeberhan and Doron Zeilberger Department of Mathematics, Temple University, Philadelphia PA 19122, USA tewodros@math.temple.edu, zeilberg@math.temple.edu Submitted: Sept 5, 1996. Accepted: Sept 12, 1996 Dedicated to Herb Wilf on his one million- rst birthday Abstract. Based on the WZ method, some series accelerationformulas are given. These formulas allow us to write down an in nite family of parametrizedidentities from any given identity of WZ type. Further, this family, in the case of the Zeta function, gives rise to many accelerated expressions for 3. AMS Subject Classi cation: Primary 05A We recall Z that a discrete function An,k is called Hypergeometric or Closed Form CF in two variables when the ratios An + 1;k=An;k and An;k + 1=An;k are both rational functions. A discrete 1-form ! = Fn;k k + Gn;k n is a WZ 1-form if the pair F,G of CF functions satis es Fn+1;k,Fn;k = Gn;k+1 ,Gn;k. We use: N and K for the forward shift operators on n and k, respectively. n := N ,1, k := K,1. Consider the WZ 1-form ! = Fn;k k+Gn;k n. Then, we de ne the sequence !s;s = 1;2;3;::: of new WZ 1-forms: !s := Fs k + Gs n; where Fsn;k = Fsn;k and Gsn;k = s,1 Collections: Mathematics | 404 | 1,348 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2016-07 | latest | en | 0.754726 |
https://solvedlib.com/please-do-not-repost-the-same-answer-to-this,319829 | 1,686,034,871,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652235.2/warc/CC-MAIN-20230606045924-20230606075924-00153.warc.gz | 579,214,663 | 19,796 | # ***** PLEASE DO NOT REPOST THE SAME ANSWER TO THIS QUESTION FROM ANOTHER PERSON POST OF...
###### Question:
***** PLEASE DO NOT REPOST THE SAME ANSWER TO THIS QUESTION FROM ANOTHER PERSON POST OF THIS SAME QUESTION, THOSE ANSWERS ARE NOT ANSWERING WHAT IS BEING ASKED HERE, PLEASE LABEL ANY KEY IN THE ER DIAGRAM AS WELL AS THE CARDINALITY OF RELATIONSHIPS*****
Submission Guidelines
Prepare the list of entities and their corresponding attributes in Word, and draw the E-R diagram in Visio.
Description of Homework
For the problem below,
****- List all necessary entities and their corresponding attributes.
****-Indicate the primary key of each entity with an underline. • Draw the E-R diagram in Visio showing only entities and relationships between them.
****-Make sure you identify the weak entity sets, and also cardinality and participation constraints for each relationship using appropriate symbols in your E-R diagram.
General Description
The Development Office of Beta University seeks to obtain donations for its Annual Fund from a variety of donors. The fund collects over ten million dollars each year. Donors include graduating seniors, alumni, parents, faculty, administrators, staff, corporations, or other friends of the university. There are approximately 100,000 potential donors. The Annual Fund is directed by Suzanne Hayes, who is responsible for raising funds and keeping track of donations. Suzanne wishes to create a database to help with both of these major responsibilities.
Basic Operations
##### QUESTIONApoint charge Q1 =-5.09nC Is at the orlgin; and , second polnt charge +6.00 nC Is on thc axls at * 0.S00 m: Consler the clectrlc flelds E] ad Ez atx = 1.00 m, generated by Q1 and 02, respectlvely: Whlch of the following I5 truc?OA E1 polnts to thc left; Ez polnts to the right0 B E1 points to the right polnts to the rghtpolnts t0 the right Ez points to the lelt:D Ex points to the left; Ez polfts (o the left:QUESTION 8of the net electric fleld at * = 1,00 m End," = Following the previ
QUESTION Apoint charge Q1 =-5.09nC Is at the orlgin; and , second polnt charge +6.00 nC Is on thc axls at * 0.S00 m: Consler the clectrlc flelds E] ad Ez atx = 1.00 m, generated by Q1 and 02, respectlvely: Whlch of the following I5 truc? OA E1 polnts to thc left; Ez polnts to the right 0 B E1 points...
##### Solve for the following probabilities (ranges of X values): a. P(X ≤ 7) when m = 15...
Solve for the following probabilities (ranges of X values): a. P(X ≤ 7) when m = 15 b. P(9 ≤ X ≤ 18) when m = 15 c. P(X ≥ 15) when m = 15 d. &...
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Anderson acquires 11 percent of the outstanding voting shares of Barringer on January 1, 2019, for $109,000 and categorizes the investment as an available-for-sale security. An additional 20 percent of the stock is purchased on January 1, 2020, for$220,000, which gives Anderson the ability to signi...
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##### Refer to Example 8.1 and the corresponding Eigure 8.3 showing the distribution of heights for college women. The mean of the heights is 64.5 inches, and the standard deviation is 2.5 inches. a. Use the Empirical Rule to determine the ranges into which about $68 \%, 95 \%,$ and $99.7 \%$ of college women's heights should fall. b. Draw a picture illustrating the ranges you found in part (a), indicating the cutoff points for the middle $68 \%, 95 \%,$ and $99.7 \%$ of heights of college women.
Refer to Example 8.1 and the corresponding Eigure 8.3 showing the distribution of heights for college women. The mean of the heights is 64.5 inches, and the standard deviation is 2.5 inches. a. Use the Empirical Rule to determine the ranges into which about $68 \%, 95 \%,$ and $99.7 \%$ of college w...
##### 0/1 points Previous Answerstanapcalc10 6.7.004.nvasupplier portable hair dryers will make hundred units of hair dryers available in the market when the unit price V36 3.6xdollars. Determine the producers' surplus the market price set at SI2/unit: (Round your answer 35720two decimal places.)Need Help?MaichllIltteb ltor0.5/1 points Previous Answerstanapcalc1o 6. .007.nvaThe manufacturer of brand of mattresses will make hundred units avallable the market when the unit price 200 90eo,dollars.(a
0/1 points Previous Answers tanapcalc10 6.7.004.nva supplier portable hair dryers will make hundred units of hair dryers available in the market when the unit price V36 3.6x dollars. Determine the producers' surplus the market price set at SI2/unit: (Round your answer 35720 two decimal places.)...
##### Problem (25 pt): At the playground father pushes meTTy-gO-round that his daughter is standing on. Initially the merry-gO-round is at rest. The father pushes with force of 500 N for 1/2 second. The metry-gO-round has radius R=2 m and the child is distance RD L5 m from the center of the merry- gO-round. Find the velocity of the child (1D) given that the moment of inertia about the center of the merry-go-round is In - 200 kg m" and moment of inertia of the child about the center is Ib =50 kg m
Problem (25 pt): At the playground father pushes meTTy-gO-round that his daughter is standing on. Initially the merry-gO-round is at rest. The father pushes with force of 500 N for 1/2 second. The metry-gO-round has radius R=2 m and the child is distance RD L5 m from the center of the merry- gO-roun...
##### Bernie the Gambler reviews the number of bets he has won in hislifetime of gambling. Since he has made so many bets, he selects arandom sample of 100 games and records the number of wins he hasmade. He sees that he has made 47 wins out of the 100 in hissample. A button hyperlink to the SALT program that reads: UseSALT. (a) Calculate the point estimate for Bernie's sample. .47 (b)Compute the margin of error for Bernie's winning bets given aconfidence level of 99%. (Use a table or SALT.
Bernie the Gambler reviews the number of bets he has won in his lifetime of gambling. Since he has made so many bets, he selects a random sample of 100 games and records the number of wins he has made. He sees that he has made 47 wins out of the 100 in his sample. A button hyperlink to the SALT prog...
##### (10 points) Describe the rational numbers and the irrational numbers in terms of their repre- sentation as decimal numbers_
(10 points) Describe the rational numbers and the irrational numbers in terms of their repre- sentation as decimal numbers_...
##### This is a Construction Management class, so will you please be able to provide a step...
This is a Construction Management class, so will you please be able to provide a step wise solution of how you were able to calculate the answer. Case Problem WORKFORCE SCHEDULING Davis Instruments has two manufacturing plants located in Atlanta, Georgia. Product de mand varies considerably from mon...
##### 14-99. The mechanism for the reaction 2H,0, (aq) ® 2H,00) +0,) in the presence of I...
14-99. The mechanism for the reaction 2H,0, (aq) ® 2H,00) +0,) in the presence of I (aq) is proposed to be: Step 1: H,02(aq) + I (aq) ® H,0(0) + Ol (aq) Step 2: H,0, (aq) + 01 (aq) ® H,00) + 0,(g) + 1 (aq) (slow) (fast) What is the molecularity of the rate-determining step? Select one: a...
##### Example 1.2 Find tke absolute error and of f(c) Sim € 'f h =7, fo T. 13Solution: Let f(r) sin f ad Fof() SI € f"() CUS [ f"(2) SI T f"(1) Cn f("(1) Sil € f""() C6[ fton() S F(7)6) (if(o) = f'(0) = [ f" (O) = 0 f"(0) =-1 6(4(0) = 0 f""(0) = [ = ( F(7)(0)Heuce;f() = f() +=f'(0) 21f"(0) 3f""(0) + 4r(o) 57F"(0) f()(Q) + 7iroo) 0 + "() (-1) + (I) +sil $7 V Sin([5) 0.4975 (1,5" (L5)" (1,5)" Example 1.2 Find tke absolute error and of f(c) Sim € 'f h =7, fo T. 13 Solution: Let f(r) sin f ad Fo f() SI € f"() CUS [ f"(2) SI T f"(1) Cn f("(1) Sil € f""() C6[ fton() S F(7)6) (i f(o) = f'(0) = [ f" (O) = 0 f"(0) =-1 6(4(0) = ... 1 answer ##### If high count is greater thank 8, medium count greater than 4 less than or equal to 8, and low is less than or equal to 4 write an inequality to find the range at which the count is not medium if high count is greater thank 8, medium count greater than 4 less than or equal to 8, and low is less than or equal to 4 write an inequality to find the range at which the count is not medium... 1 answer ##### 3) What is wrong with the following proof that all horses have the same color? Let... 3) What is wrong with the following proof that all horses have the same color? Let P(n) be the proposition that all the horses in a set of n horses are the same color. Clearly, P(1) is true. Now assume that P(n) is true. That is, assume that all the horses in any set of n horses are the same color C... 1 answer ##### A buffered solution containing dissolved aniline, CH, NH,, and aniline hydrochloride, CH NH, CI, has a... A buffered solution containing dissolved aniline, CH, NH,, and aniline hydrochloride, CH NH, CI, has a pH of 5.61. A. Determine the concentration of C H NH in the solution if the concentration of C, H, NH, is 0.320 M. The pKb of aniline is 9.13. [CH, NH] = B. Calculate the change in pH of the soluti... 5 answers ##### PLUS Kleln; Organic Chemistry; 2e Practice Assigantent Gradqbool orION Dounlondebl ETertboolAsslgnmentsourcesApply the Skiii 03.20 Consicer the structurt Ly-dichiomtopanoic acd;I07 07{07 77 lem 08,47 Jexan 027 Jcricalojo JeAlonloky HeeGf7rs compoindnas Mant consumeten enmancconsututionae soner thaldlghtly more aodit-StudyDmuonsttans Ksome Lhaies SighttacidicemecPtvoce Pollg;057000.7017Jcmnee SocsMacBoak Pro20888 PLUS Kleln; Organic Chemistry; 2e Practice Assigantent Gradqbool orION Dounlondebl ETertbool Asslgnment sources Apply the Skiii 03.20 Consicer the structurt Ly-dichiomtopanoic acd; I07 07 {07 77 lem 08,47 Jexan 027 Jcricalojo JeAlonloky HeeGf7 rs compoindnas Mant consumeten enmanc consututionae sone... 5 answers ##### Question 17.0 13 O1ov 15.0 8 points)QuestionH 0dou (2 pointsl 54u08 Question 17.0 13 O1ov 15.0 8 points) QuestionH 0dou (2 pointsl 54u0 8... 1 answer ##### Question 5 Not yet answered Points out of 3.00 P Flag question The notion that sticky-wages... Question 5 Not yet answered Points out of 3.00 P Flag question The notion that sticky-wages impacts the short-run aggregate supply curve implies that when the price level is lower than expected, Select one: a. production is less profitable and employment rises. b. production is less profitable and e... 1 answer ##### Compound 1,3,5-hexatriene. Its structure is shown below with the carbons ositions? 汽 4,(6pts) Consid r the... compound 1,3,5-hexatriene. Its structure is shown below with the carbons ositions? 汽 4,(6pts) Consid r the numbered 1-6. The Hückel coefficients in the lowest antibonding molecular orbital tor 2, and 3 are, respectively, 0.521,-0.231, and 0.417. What are the spin densities at those p Sho... 1 answer ##### How do you find the axis of symmetry, and the maximum or minimum value of the function f(x) =x^2 + 3? How do you find the axis of symmetry, and the maximum or minimum value of the function f(x) =x^2 + 3?... 1 answer ##### Find the equation of the regression line for the given data. A. = -0.552x + 2.097... Find the equation of the regression line for the given data. A. = -0.552x + 2.097 B. = 2.097x - 0.552 C. = 2.097x + 0.552 D. = 0.522x - 2.097... 1 answer ##### A mixture of propane and butane is fed into a furnace where i is mixed with... A mixture of propane and butane is fed into a furnace where i is mixed with air. The furnace exhaustMap leaves the furnace at 317°C, 795.0 mmHg and contains only N2-02-CO2, and H2O. The partial pressure O2 in the exhaust is 78.55 mmHg and the partial pressure of CO2 in the exhaust is 49.29 mmHg.... 1 answer ##### Fact Pattern 1: Pat contracts with an Ajax Insurance Company agent for a$50,000 ordinary life...
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##### Reported benefits of lean implementations include all of the following except: Select one: O a. improved...
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##### An automobile traveling at 70.0 km/h has tires of 50.0 cm diameter. (a) What is the...
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##### I prefer the answer to be written by computer not hand writing because i don't understand...
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##### Score: 0 of 1 pt 5 of 9 (3 complete) HW Score: 33.33%, 3 of 9...
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##### 03.Provide a systematic/IUPAC name for the following chiral compound.0CH;(CHS)CH Cvch;CH CHs H
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##### + y)dz + (z + ey2- dy
+ y)dz + (z + ey2- dy...
##### Exercise 85 Find the maximum and minimum values of xle-x in the interval [0, 5]:
Exercise 85 Find the maximum and minimum values of xle-x in the interval [0, 5]:... | 4,084 | 14,414 | {"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} | 2.515625 | 3 | CC-MAIN-2023-23 | latest | en | 0.830648 |
https://www.hellenicaworld.com/Science/Mathematics/en/Taxicabgeometry.html | 1,713,960,947,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819273.90/warc/CC-MAIN-20240424112049-20240424142049-00311.warc.gz | 706,339,699 | 5,971 | A taxicab geometry is a form of geometry in which the usual distance function or metric of Euclidean geometry is replaced by a new metric in which the distance between two points is the sum of the absolute differences of their Cartesian coordinates. The taxicab metric is also known as rectilinear distance, L1 distance, L1 distance or $$\ell _{1}$$ norm (see Lp space), snake distance, city block distance, Manhattan distance or Manhattan length, with corresponding variations in the name of the geometry.[1] The latter names allude to the grid layout of most streets on the island of Manhattan, which causes the shortest path a car could take between two intersections in the borough to have length equal to the intersections' distance in taxicab geometry.
The geometry has been used in regression analysis since the 18th century, and today is often referred to as LASSO. The geometric interpretation dates to non-Euclidean geometry of the 19th century and is due to Hermann Minkowski.
Taxicab geometry versus Euclidean distance: In taxicab geometry, the red, yellow, and blue paths all have the same shortest path length of 12. In Euclidean geometry, the green line has length $$6{\sqrt {2}}\approx 8.49$$ and is the unique shortest path.
Formal definition
The taxicab distance, $$d_{1}$$, between two vectors $$\mathbf {p} ,\mathbf {q}$$ in an n-dimensional real vector space with fixed Cartesian coordinate system, is the sum of the lengths of the projections of the line segment between the points onto the coordinate axes. More formally,
$$d_{1}(\mathbf {p} ,\mathbf {q} )=\|\mathbf {p} -\mathbf {q} \|_{1}=\sum _{i=1}^{n}|p_{i}-q_{i}|,$$
where $$(\mathbf {p} ,\mathbf {q} ) are vectors \( \mathbf {p} =(p_{1},p_{2},\dots ,p_{n}){\text{ and }}\mathbf {q} =(q_{1},q_{2},\dots ,q_{n})\, For example, in the plane, the taxicab distance between \( (p_{1},p_{2})$$ and$$(q_{1},q_{2})$$is $$|p_{1}-q_{1}|+|p_{2}-q_{2}|.$$
Properties
Taxicab distance depends on the rotation of the coordinate system, but does not depend on its reflection about a coordinate axis or its translation. Taxicab geometry satisfies all of Hilbert's axioms (a formalization of Euclidean geometry) except for the side-angle-side axiom, as two triangles with equally "long" two sides and an identical angle between them are typically not congruent unless the mentioned sides happen to be parallel.
Circles
Circles in discrete and continuous taxicab geometry
A circle is a set of points with a fixed distance, called the radius, from a point called the center. In taxicab geometry, distance is determined by a different metric than in Euclidean geometry, and the shape of circles changes as well. Taxicab circles are squares with sides oriented at a 45° angle to the coordinate axes. The image to the right shows why this is true, by showing in red the set of all points with a fixed distance from a center, shown in blue. As the size of the city blocks diminishes, the points become more numerous and become a rotated square in a continuous taxicab geometry. While each side would have length $$\sqrt{2}r$$ using a Euclidean metric, where r is the circle's radius, its length in taxicab geometry is 2r. Thus, a circle's circumference is 8r. Thus, the value of a geometric analog to $$\pi$$ is 4 in this geometry. The formula for the unit circle in taxicab geometry is |x|+|y|=1 in Cartesian coordinates and
$$r={\frac {1}{|\sin \theta |+|\cos \theta |}}$$
in polar coordinates.
A circle of radius 1 (using this distance) is the von Neumann neighborhood of its center.
A circle of radius r for the Chebyshev distance (L∞ metric) on a plane is also a square with side length 2r parallel to the coordinate axes, so planar Chebyshev distance can be viewed as equivalent by rotation and scaling to planar taxicab distance. However, this equivalence between L1 and L∞ metrics does not generalize to higher dimensions.
Whenever each pair in a collection of these circles has a nonempty intersection, there exists an intersection point for the whole collection; therefore, the Manhattan distance forms an injective metric space.
Applications
Measures of distances in chess
In chess, the distance between squares on the chessboard for rooks is measured in taxicab distance; kings and queens use Chebyshev distance, and bishops use the taxicab distance (between squares of the same color) on the chessboard rotated 45 degrees, i.e., with its diagonals as coordinate axes. To reach from one square to another, only kings require the number of moves equal to their respective distance; rooks, queens and bishops require one or two moves (on an empty board, and assuming that the move is possible at all in the bishop's case).
Compressed sensing
In solving an underdetermined system of linear equations, the regularization term for the parameter vector is expressed in terms of the ℓ 1 {\displaystyle \ell _{1}} \ell _{1}-norm (taxicab geometry) of the vector.[2] This approach appears in the signal recovery framework called compressed sensing.
Differences of frequency distributions
Taxicab geometry can be used to assess the differences in discrete frequency distributions. For example, in RNA splicing positional distributions of hexamers, which plot the probability of each hexamer appearing at each given nucleotide near a splice site, can be compared with L1-distance. Each position distribution can be represented as a vector where each entry represents the likelihood of the hexamer starting at a certain nucleotide. A large L1-distance between the two vectors indicates a significant difference in the nature of the distributions while a small distance denotes similarly shaped distributions. This is equivalent to measuring the area between the two distribution curves because the area of each segment is the absolute difference between the two curves' likelihoods at that point. When summed together for all segments, it provides the same measure as L1-distance.[3]
History
The L1 metric was used in regression analysis in 1757 by Roger Joseph Boscovich.[4] The geometric interpretation dates to the late 19th century and the development of non-Euclidean geometries, notably by Hermann Minkowski and his Minkowski inequality, of which this geometry is a special case, particularly used in the geometry of numbers, (Minkowski 1910). The formalization of Lp spaces is credited to (Riesz 1910).
Normed vector space
Metric
Orthogonal convex hull
Hamming distance
Fifteen puzzle
Random walk
Manhattan wiring
Notes
Black, Paul E. "Manhattan distance". Dictionary of Algorithms and Data Structures. Retrieved October 6, 2019.
Donoho, David L. (March 23, 2006). "For most large underdetermined systems of linear equations the minimal ℓ 1 {\displaystyle \ell _{1}} \ell _{1}-norm solution is also the sparsest solution". Communications on Pure and Applied Mathematics. 59 (6): 797–829. doi:10.1002/cpa.20132.
Lim, Kian Huat; Ferraris, Luciana; Filloux, Madeleine E.; Raphael, Benjamin J.; Fairbrother, William G. (July 5, 2011). "Using positional distribution to identify splicing elements and predict pre-mRNA processing defects in human genes". Proceedings of the National Academy of Sciences of the United States of America. 108 (27): 11093–11098. Bibcode:2011PNAS..10811093H. doi:10.1073/pnas.1101135108. PMC 3131313. PMID 21685335.
Stigler, Stephen M. (1986). The History of Statistics: The Measurement of Uncertainty before 1900. Harvard University Press. ISBN 9780674403406. Retrieved October 6, 2019.
References
Krause, Eugene F. (1987). Taxicab Geometry. Dover. ISBN 978-0-486-25202-5.
Minkowski, Hermann (1910). Geometrie der Zahlen (in German). Leipzig and Berlin: R. G. Teubner. JFM 41.0239.03. MR 0249269. Retrieved October 6, 2019.
Riesz, Frigyes (1910). "Untersuchungen über Systeme integrierbarer Funktionen". Mathematische Annalen (in German). 69 (4): 449–497. doi:10.1007/BF01457637. hdl:10338.dmlcz/128558. | 1,968 | 7,912 | {"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} | 3.6875 | 4 | CC-MAIN-2024-18 | latest | en | 0.888701 |
https://www.numbersaplenty.com/3906750016 | 1,696,136,863,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510781.66/warc/CC-MAIN-20231001041719-20231001071719-00095.warc.gz | 974,726,336 | 3,644 | Search a number
3906750016 = 261326012
BaseRepresentation
bin1110100011011100…
…0100011001000000
3101002021020110112001
43220313010121000
531000112000031
61443355113344
7165545553511
oct35067043100
911067213461
103906750016
111725293124
12910442854
134a3502400
14290bdca08
1517cea5861
hexe8dc4640
3906750016 has 63 divisors (see below), whose sum is σ = 8408663523. Its totient is φ = 1800115200.
The previous prime is 3906749971. The next prime is 3906750047. The reversal of 3906750016 is 6100576093.
The square root of 3906750016 is 62504.
It is a perfect power (a square), and thus also a powerful number.
It can be written as a sum of positive squares in 4 ways, for example, as 1000000 + 3905750016 = 1000^2 + 62496^2 .
It is an unprimeable number.
It is a pernicious number, because its binary representation contains a prime number (13) of ones.
It is a polite number, since it can be written in 8 ways as a sum of consecutive naturals, for example, 6500116 + ... + 6500716.
Almost surely, 23906750016 is an apocalyptic number.
3906750016 is the 62504-th square number.
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 3906750016
3906750016 is an abundant number, since it is smaller than the sum of its proper divisors (4501913507).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
3906750016 is an frugal number, since it uses more digits than its factorization.
3906750016 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 1240 (or 616 counting only the distinct ones).
The product of its (nonzero) digits is 34020, while the sum is 37.
The cubic root of 3906750016 is about 1574.9685074412.
The spelling of 3906750016 in words is "three billion, nine hundred six million, seven hundred fifty thousand, sixteen". | 565 | 1,896 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.831349 |
https://math.stackexchange.com/questions/2942886/the-most-difficult-way-of-proving-that-a-countable-union-of-countable-sets-is-co/2943523 | 1,571,486,417,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986693979.65/warc/CC-MAIN-20191019114429-20191019141929-00512.warc.gz | 592,259,380 | 35,488 | # The most difficult way of proving that a countable union of countable sets is countable
In ZFC I want to prove the following result:
Proposition 1: Let $$A$$ be a set and let $${(G_k)}_{k \in \mathbb N}$$ be a (countable) family of countable nonempty subsets of $$A$$. Then there exist a mapping $$f: \mathbb N \to A$$ satisfying:
$$\tag 1 \text{There exist a partition of } \mathbb N \text{ into a family of subsets } N_k , \, k \ge 0$$
$$\tag 2 f \text{ restricted to } N_k \text{ is an injective mapping onto } \,G_k, \, k \ge 0$$
We can see immediately that $$f$$ maps onto $$\bigcup G_k$$ and that $$f$$ is injective when the $$G_k$$ are mutually disjoint.
My hand-waving idea of a proof:
Using the axiom of choice, we can endow all the $$G_k$$ with a well-ordering relation. The idea is is to define, using recursion, a way of visiting each index $$k$$ for the $$G_k$$ family as many times as necessary to 'scratch off and exhaust' the elements in $$G_k$$ as they are enumerated. For instance, if $$G_k$$ is finite with $$\alpha_k$$ elements, we would 'visit' $$k$$ exactly $$\alpha_k$$ times.
Another idea:
Proposition 1 is sure looks like it is equivalent to showing that a countable union of countable sets in countable; the arguments found in this math.stackexchange question make this very plausible.
Question 1:
If proposition 1 is valid in ZFC, any ideas on how to go about proving it.
Question 2:
Assuming again that proposition 1 is true, is it equivalent in ZF (axiom of choice dropped) to the Axiom of Choice from a 'Countable Family of Countable Sets':
Axiom AOC.CFCS: If $${\displaystyle (S_{i })_{\,i \in \mathbb N}}$$ is a family of non-empty countable sets indexed by the natural numbers, then $$\;{\displaystyle \prod _{i \in \mathbb N}S_{i }\neq \emptyset }$$.
A 'yes' or 'no' would be helpful here with perhaps some idea sketch/links to help me see this.
• A famous result is that if there is a measurable cardinal then there is a model of ZF that satisfies "$\omega_1$ is a countable union of countable sets". So if you can't first prove that measurable cardinals don't exist, then well-ordering $\cup_{k\in \Bbb N}G_k$ will not suffice. What you need, I think, is to use Countable Choice to get some $(f_k)_{k\in \Bbb N}\in \prod_{k\in \Bbb N}F_k,$ where $F_k$ is the set of bijections from $G_k$ to $\Bbb N$ or to an initial segment of $\Bbb N.$.... Note that if $G_k$ is infinite then $F_k$ is uncountable. – DanielWainfleet Oct 5 '18 at 4:37
• @DanielWainfleet: What??? Measure cardinals??? There are no large cardinals involved with "$\omega_1$ is a countable union of countable sets". The statement is equiconsistent with ZF itself. – Asaf Karagila Oct 5 '18 at 5:01
You suggest that Proposition 1 might be equivalent over ZF to the axiom of choice for countable families of countable sets. Asaf has given a reference showing that it's not. I think I understand where you got confused.
The family $$(G_k)_{i\in \mathbb{N}}$$ is a countable family of countable sets, but you don't use the axiom of choice to get a choice function for this family. Instead, you use it to pick a well-ordering of each $$G_k$$. That is, letting $$W(G_k)$$ be the set of well-orderings of $$G_k$$, you need a choice function for the family $$(W(G_k))_{k\in \mathbb{N}}$$. And given a countably infinite set $$G_k$$, the set $$W(G_k)$$ is not countable.
Regarding the details of making your proof of Proposition 1 precise, here's one way to formalize it. For each $$k$$, pick an injective map $$f_k\colon G_k\to \mathbb{N}$$. Let $$\bigsqcup_{k\in \mathbb{N}} G_k = \{(k,x)\mid k\in \mathbb{N}, x\in G_k\}$$ be the disjoint union of the $$G_k$$. Then we have an injective map $$f\colon \bigsqcup_{k\in \mathbb{N}} G_k\to \mathbb{N}\times\mathbb{N}$$, by $$f(k,x) = (k,f_k(x))$$.
Now by induction you can define a map $$g\colon \mathbb{N}\to \text{ran}(f)$$ via the standard diagonal enumeration of $$\mathbb{N}\times \mathbb{N}$$, skipping any elements of $$\mathbb{N}\times\mathbb{N}$$ which aren't in $$\text{ran}(f)$$.
For any $$n\in \mathbb{N}$$, if $$g(n) = (k,x)$$, then set $$n\in N_k$$ and $$h(n) = f_k^{-1}(x)$$. The function $$h$$ and partition $$(N_k)_{k\in \mathbb{N}}$$ do what you wanted.
• Your formal solution is just what I was looking for. Your answer has the added benefit that it stops me from writing a Python program to answer my question - such a demonstration of the concept might upset pure mathematicians interested in set theory. – CopyPasteIt Oct 6 '18 at 0:59
• So proposition 1 over ZF is a 'downshift' of full AOC over ZF, but is powerful enough to care care of (i,e, prove) both AOC.CFCS and the assertion that the countable union of countable sets is countable. – CopyPasteIt Oct 6 '18 at 1:09
• @CopyPasteIt That's right, although I would say that your Proposition 1 is pretty obviously equivalent to the assertion that a countable union of countable sets is countable... – Alex Kruckman Oct 6 '18 at 1:22
To answer the second question, no. You cannot prove from countable choice for countable sets that the countable union of countable sets is countable. Felgner constructed a model where every family of well-orderable sets admits a choice function, and $$\omega_1$$ is a countable union of countable sets. In the Howard–Rubin book Consequences of The Axiom of Choice this model is referred to as $$\mathcal M20$$.
To answer the first question, yes, that is a valid idea. More to the point, partition $$\Bbb N$$ into $$N_k$$ such that $$|N_k|=|G_k|$$ for all $$k$$, choose a bijection $$f_k$$ between $$N_k$$ and $$G_k$$, and let $$f=\bigcup f_k$$.
• Is it easy to write out the details of the 'more to the point' proof? – CopyPasteIt Oct 5 '18 at 12:49
• Well. Yes. The only details missing are explaining how to partition $\Bbb N$ (which is a bit of a hassle if you want to do it directly, otherwise you can rely on some other theorems), and that the union of bijections whose domains are disjoint is a well-defined function which is surjective on the union of the ranges (not necessarily injective, of course, unless the ranges are also disjoint). – Asaf Karagila Oct 5 '18 at 13:03 | 1,812 | 6,157 | {"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": 53, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2019-43 | latest | en | 0.891956 |
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# Basic Operations on Binary Tree with Implementations
The tree is a hierarchical Data Structure. A binary tree is a tree that has at most two children. The node which is on the left of the Binary Tree is called “Left-Child” and the node which is the right is called “Right-Child”. Also, the smaller tree or the subtree in the left of the root node is called the “Left sub-tree” and that is on the right is called “Right sub-tree”.
Below are the various operations that can be performed on a Binary Tree:
### Creation of Binary Tree:
The idea is to first create the root node of the given tree, then recursively create the left and the right child for each parent node. Below is the program to illustrate the same:
## C++
`// C++ program to illustrate how to``// create a tree``#include ``using` `namespace` `std;` `// Structure of the Binary Tree``struct` `treenode {`` ``int` `info;`` ``struct` `treenode *left,`` ``*right;``};` `// Function to create the Binary Tree``struct` `treenode* create()``{`` ``int` `data;`` ``struct` `treenode* tree;` ` ``// Dynamically allocating memory`` ``// for the tree-node`` ``tree = ``new` `treenode;` ` ``cout << ``"\nEnter data to be inserted "`` ``<< ``"or type -1 for no insertion : "``;` ` ``// Input from the user`` ``cin >> data;` ` ``// Termination Condition`` ``if` `(data == -1)`` ``return` `0;` ` ``// Assign value from user into tree`` ``tree->info = data;` ` ``// Recursively Call to create the`` ``// left and the right sub tree`` ``cout << ``"Enter left child of : "`` ``<< data;`` ``tree->left = create();` ` ``cout << ``"Enter right child of : "`` ``<< data;`` ``tree->right = create();` ` ``// Return the created Tree`` ``return` `tree;``};` `// Function to perform the inorder``// traversal of the given Tree``void` `inorder(``struct` `treenode* root)``{`` ``// If root is NULL`` ``if` `(root == NULL)`` ``return``;` ` ``// Recursively call for the left`` ``// and the right subtree`` ``inorder(root->left);`` ``cout << root->info << ``" "``;`` ``inorder(root->right);``}` `// Driver Code``int` `main()``{`` ``// Root Node`` ``struct` `treenode* root = NULL;` ` ``// Function Call`` ``root = create();` ` ``// Perform Inorder Traversal`` ``inorder(root);` ` ``return` `0;``}` `/* Will be creating tree:`` ``2 `` ``/ \ `` ``7 5 `` ``/ \ \ `` ``2 6 9`` ``*/`
## Java
`import` `java.util.Scanner;``// class of the Binary Tree node` `class` `TreeNode {`` ``int` `info;`` ``TreeNode left, right;` ` ``TreeNode(``int` `data) {`` ``this``.info = data;`` ``left = right = ``null``;`` ``}``}` `class` `GFG {`` ``static` `Scanner sc = ``new` `Scanner(System.in);``// Function to create the Binary Tree` ` ``static` `TreeNode create() {`` ``int` `data;`` ``System.out.print(``"\nEnter data to be inserted or type -1 for no insertion : "``);`` ``data = sc.nextInt();`` ``if` `(data == -``1``) {`` ``return` `null``;`` ``}`` ``TreeNode tree = ``new` `TreeNode(data);`` ``System.out.print(``"Enter left child of : "` `+ data);`` ``tree.left = create();`` ``System.out.print(``"Enter right child of : "` `+ data);`` ``tree.right = create();`` ``return` `tree;`` ``}`` ``// Perform Inorder Traversal` ` ``static` `void` `inorder(TreeNode root) {`` ``if` `(root == ``null``) {`` ``return``;`` ``}`` ``inorder(root.left);`` ``System.out.print(root.info + ``" "``);`` ``inorder(root.right);`` ``}``// Driver Code` ` ``public` `static` `void` `main(String[] args) {`` ``TreeNode root = ``null``;`` ``root = create();`` ``inorder(root);`` ``}``}`
## Python3
`#Python equivalent` `# Class of the Binary Tree node``class` `TreeNode:`` ``def` `__init__(``self``, data):`` ``self``.info ``=` `data`` ``self``.left ``=` `None`` ``self``.right ``=` `None` `# Function to create the Binary Tree``def` `create():`` ``data ``=` `int``(``input``(``"\nEnter data to be inserted or type -1 for no insertion : "``))`` ``if` `data ``=``=` `-``1``:`` ``return` `None`` ``tree ``=` `TreeNode(data)`` ``print``(``"Enter left child of : "` `+` `str``(data))`` ``tree.left ``=` `create()`` ``print``(``"Enter right child of : "` `+` `str``(data))`` ``tree.right ``=` `create()`` ``return` `tree` `# Perform Inorder Traversal``def` `inorder(root):`` ``if` `root ``=``=` `None``:`` ``return`` ``inorder(root.left)`` ``print``(root.info, end``=``" "``)`` ``inorder(root.right)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:`` ``root ``=` `None`` ``root ``=` `create()`` ``inorder(root)`
## C#
`// C# program to illustrate how to``// create a tree``using` `System;` `// Structure of the Binary Tree``public` `class` `treenode {`` ``public` `int` `info;`` ``public` `treenode left, right;` ` ``public` `treenode()`` ``{`` ``info = 0;`` ``left = ``null``;`` ``right = ``null``;`` ``}``}` `// Class to perform the operations``public` `class` `GFG {`` ``// Function to create the Binary Tree`` ``public` `static` `treenode create()`` ``{`` ``int` `data;`` ``treenode tree = ``new` `treenode();` ` ``Console.WriteLine(`` ``"\nEnter data to be inserted "`` ``+ ``"or type -1 for no insertion : "``);` ` ``// Input from the user`` ``data = Convert.ToInt32(Console.ReadLine());` ` ``// Termination Condition`` ``if` `(data == -1)`` ``return` `null``;` ` ``// Assign value from user into tree`` ``tree.info = data;` ` ``// Recursively Call to create the`` ``// left and the right sub tree`` ``Console.WriteLine(``"Enter left child of : "` `+ data);`` ``tree.left = create();` ` ``Console.WriteLine(``"Enter right child of : "` `+ data);`` ``tree.right = create();` ` ``// Return the created Tree`` ``return` `tree;`` ``}` ` ``// Function to perform the inorder`` ``// traversal of the given Tree`` ``public` `static` `void` `inorder(treenode root)`` ``{`` ``// If root is NULL`` ``if` `(root == ``null``)`` ``return``;` ` ``// Recursively call for the left`` ``// and the right subtree`` ``inorder(root.left);`` ``Console.Write(root.info + ``" "``);`` ``inorder(root.right);`` ``}` ` ``// Driver Code`` ``public` `static` `void` `Main()`` ``{`` ``// Root Node`` ``treenode root = ``null``;` ` ``// Function Call`` ``root = create();` ` ``// Perform Inorder Traversal`` ``inorder(root);`` ``}``}` `/* Will be creating tree:`` ``2`` ``/ \`` ``7 5`` ``/ \ \`` ``2 6 9`` ``*/`
## Javascript
`// javascript program to illustrate how to``// create a tree` `// Structure of the Binary Tree``class treenode {`` ` ` ``constructor(){`` ``this``.info = 0;`` ``this``.left = ``null``;`` ``this``.right = ``null``;`` ``}``}` `// Function to create the Binary Tree``function` `create()``{`` ``let data;`` ``let tree = ``new` `treenode();` ` ``// Input from the user`` ``// data = readInt("\n Enter data to be inserted or type -1 for no insertion :");`` ``data = prompt(``"\n Enter data to be inserted or type -1 for no insertion :"``);` ` ``// Termination Condition`` ``if` `(data == -1)`` ``return` `0;` ` ``// Assign value from user into tree`` ``tree.info = data;` ` ``// Recursively Call to create the`` ``// left and the right sub tree`` ``console.log(``"Enter left child of : "` `+ data);`` ``tree.left = create();` ` ``console.log(``"Enter right child of : "` `+ data);`` ``tree.right = create();` ` ``// Return the created Tree`` ``return` `tree;``}` `// Function to perform the inorder``// traversal of the given Tree``function` `inorder(root)``{`` ``// If root is NULL`` ``if` `(root == ``null``)`` ``return``;` ` ``// Recursively call for the left`` ``// and the right subtree`` ``inorder(root.left);`` ``console.log(root.info);`` ``inorder(root.right);``}` `// Driver Code``// Root Node``let root = ``null``;` `// Function Call``root = create();` `// Perform Inorder Traversal``inorder(root);` `/* Will be creating tree:`` ``2 `` ``/ \ `` ``7 5 `` ``/ \ \ `` ``2 6 9`` ``*/` `// The code is contributed by Nidhi goel.`
Output:
Time Complexity: O(N)
Auxiliary Space: O(1)
### Pre-order Traversal:
In this traversal, the root is visited first followed by the left and the right subtree. Below is the program to illustrate the same:
## C++
`// C++ program to demonstrate the``// pre-order traversal``#include "bits/stdc++.h"``using` `namespace` `std;` `// Structure of the Binary Tree``struct` `treenode {`` ``int` `info;`` ``struct` `treenode *left,`` ``*right;``};` `// Function to create the Binary Tree``struct` `treenode* create()``{`` ``int` `data;`` ``struct` `treenode* tree;` ` ``// Dynamically allocating memory`` ``// for the tree-node`` ``tree = ``new` `treenode;` ` ``cout << ``"\nEnter data to be inserted "`` ``<< ``"or type -1 for no insertion : "``;` ` ``// Input from the user`` ``cin >> data;` ` ``// Termination Condition`` ``if` `(data == -1)`` ``return` `0;` ` ``// Assign value from user into tree`` ``tree->info = data;` ` ``// Recursively Call to create the`` ``// left and the right sub tree`` ``cout << ``"Enter left child of : "`` ``<< data;`` ``tree->left = create();` ` ``cout << ``"Enter right child of : "`` ``<< data;`` ``tree->right = create();` ` ``// Return the created Tree`` ``return` `tree;``};` `// Function to perform the pre-order``// traversal for the given tree``void` `preorder(``struct` `treenode* root)``{`` ``// If the root is NULL`` ``if` `(root == NULL)`` ``return``;` ` ``// Using tree-node type stack STL`` ``stack s;` ` ``while` `((root != NULL) || (!s.empty())) {`` ``if` `(root != NULL) {`` ``// Print the root`` ``cout << root->info << ``" "``;` ` ``// Push the node in the stack`` ``s.push(root);` ` ``// Move to left subtree`` ``root = root->left;`` ``}`` ``else` `{`` ``// Remove the top of stack`` ``root = s.top();`` ``s.pop();`` ``root = root->right;`` ``}`` ``}` ` ``cout << endl;``}` `// Driver Code``int` `main()``{`` ``// Root Node`` ``struct` `treenode* root = NULL;` ` ``// Function Call`` ``root = create();` ` ``// Perform Inorder Traversal`` ``preorder(root);` ` ``return` `0;``}` `/* Will be creating tree:`` ``2 `` ``/ \ `` ``7 5 `` ``/ \ \ `` ``2 6 9`` ``*/`
## Python3
`# Python code addition` `# Structure of the Binary Tree``class` `TreeNode:`` ``def` `__init__(``self``, val``=``0``, left``=``None``, right``=``None``):`` ``self``.val ``=` `val`` ``self``.left ``=` `left`` ``self``.right ``=` `right` `# Function to create the Binary Tree``def` `create():`` ``data ``=` `int``(``input``(``"\nEnter data to be inserted or type -1 for no insertion: "``))` ` ``# Termination Condition`` ``if` `data ``=``=` `-``1``:`` ``return` `None` ` ``# Assign value from user into tree`` ``root ``=` `TreeNode(data)` ` ``# Recursively Call to create the left and the right sub tree`` ``print``(``"Enter left child of:"``, data)`` ``root.left ``=` `create()` ` ``print``(``"Enter right child of:"``, data)`` ``root.right ``=` `create()` ` ``# Return the created Tree`` ``return` `root` `# Function to perform the pre-order traversal for the given tree``def` `preorder(root):`` ``# If the root is None`` ``if` `not` `root:`` ``return`` ` ` ``# Using tree-node type stack STL`` ``stack ``=` `[]` ` ``while` `root ``or` `stack:`` ``if` `root:`` ``# Print the root`` ``print``(root.val, end``=``" "``)` ` ``# Push the node in the stack`` ``stack.append(root)` ` ``# Move to left subtree`` ``root ``=` `root.left`` ``else``:`` ``# Remove the top of stack`` ``root ``=` `stack.pop()`` ``root ``=` `root.right` ` ``print``()` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:`` ``# Root Node`` ``root ``=` `None` ` ``# Function Call`` ``root ``=` `create()` ` ``# Perform Pre-order Traversal`` ``preorder(root)` `# The code is contributed by Nidhi goel.`
## Javascript
`class TreeNode {`` ``constructor(info) {`` ``this``.info = info;`` ``this``.left = ``null``;`` ``this``.right = ``null``;`` ``}``}` `function` `create() {`` ``let data = prompt(`` ``"Enter data to be inserted or type -1 for no insertion : "`` ``);` ` ``// Termination Condition`` ``if` `(data == -1) ``return` `null``;` ` ``// Assign value from user into tree`` ``let tree = ``new` `TreeNode(data);` ` ``// Recursively Call to create the`` ``// left and the right sub tree`` ``tree.left = create();`` ``tree.right = create();` ` ``// Return the created Tree`` ``return` `tree;``}` `// Function to perform the pre-order``// traversal for the given tree``function` `preorder(root) {`` ``// If the root is NULL`` ``if` `(root == ``null``) ``return``;` ` ``// Using tree-node type stack STL`` ``let s = [];` ` ``while` `(root != ``null` `|| s.length > 0) {`` ``if` `(root != ``null``) {`` ``// Print the root`` ``console.log(root.info);` ` ``// Push the node in the stack`` ``s.push(root);` ` ``// Move to left subtree`` ``root = root.left;`` ``} ``else` `{`` ``// Remove the top of stack`` ``root = s.pop();`` ``root = root.right;`` ``}`` ``}``}` `// Driver Code` `// Root Node``let root = ``null``;` `// Function Call``root = create();` `// Perform Inorder Traversal``preorder(root);`
Output:
Time Complexity: O(N)
Auxiliary Space: O(N)
### In-order Traversal:
In this traversal, the left subtree is visited first followed by the root and the right subtree. Below is the program to illustrate the same:
## C++
`// C++ program to illustrate how to``// create a tree``#include "bits/stdc++.h"``using` `namespace` `std;` `// Structure of the Binary Tree``struct` `treenode {`` ``int` `info;`` ``struct` `treenode *left,`` ``*right;``};` `// Function to create the Binary Tree``struct` `treenode* create()``{`` ``int` `data;`` ``struct` `treenode* tree;` ` ``// Dynamically allocating memory`` ``// for the tree-node`` ``tree = ``new` `treenode;` ` ``cout << ``"\nEnter data to be inserted "`` ``<< ``"or type -1 for no insertion : "``;` ` ``// Input from the user`` ``cin >> data;` ` ``// Termination Condition`` ``if` `(data == -1)`` ``return` `0;` ` ``// Assign value from user into tree`` ``tree->info = data;` ` ``// Recursively Call to create the`` ``// left and the right sub tree`` ``cout << ``"Enter left child of : "`` ``<< data;`` ``tree->left = create();` ` ``cout << ``"Enter right child of : "`` ``<< data;`` ``tree->right = create();` ` ``// Return the created Tree`` ``return` `tree;``};` `// Function to perform the inorder``// traversal of the given Tree``void` `inorder(``struct` `treenode* root)``{`` ``// If root is NULL`` ``if` `(root == NULL)`` ``return``;` ` ``// Recursively call for the left`` ``// and the right subtree`` ``inorder(root->left);`` ``cout << root->info << ``" "``;`` ``inorder(root->right);``}` `// Driver Code``int` `main()``{`` ``// Root Node`` ``struct` `treenode* root = NULL;` ` ``// Function Call`` ``root = create();` ` ``// Perform Inorder Traversal`` ``inorder(root);` ` ``return` `0;``}` `/* Will be creating tree:`` ``2 `` ``/ \ `` ``7 5 `` ``/ \ \ `` ``2 6 9`` ``*/`
## Javascript
`// JavaScript program to illustrate how to``// create a tree` `// Structure of the Binary Tree``class TreeNode {`` ``constructor(info) {`` ``this``.info = info;`` ``this``.left = ``null``;`` ``this``.right = ``null``;`` ``}``}` `// Function to create the Binary Tree``function` `create() {`` ``const tree = ``new` `TreeNode();` ` ``let data;`` ``console.log(``"Enter data to be inserted or type -1 for no insertion : "``);` ` ``// Input from the user`` ``data = parseInt(prompt());` ` ``// Termination Condition`` ``if` `(data === -1)`` ``return` `null``;` ` ``// Assign value from user into tree`` ``tree.info = data;` ` ``// Recursively Call to create the`` ``// left and the right sub tree`` ``console.log(`Enter left child of \${data}: `);`` ``tree.left = create();` ` ``console.log(`Enter right child of \${data}: `);`` ``tree.right = create();` ` ``// Return the created Tree`` ``return` `tree;``}` `// Function to perform the inorder``// traversal of the given Tree``function` `inorder(root) {`` ``// If root is NULL`` ``if` `(root == ``null``)`` ``return``;` ` ``// Recursively call for the left`` ``// and the right subtree`` ``inorder(root.left);`` ``console.log(`\${root.info} `);`` ``inorder(root.right);``}` `// Driver Code``function` `main() {`` ``// Root Node`` ``let root = ``null``;` ` ``// Function Call`` ``root = create();` ` ``// Perform Inorder Traversal`` ``inorder(root);``}` `// Invoke the main function``main();` `/* Will be creating tree:`` ``2 `` ``/ \ `` ``7 5 `` ``/ \ \ `` ``2 6 9`` ``*/`
Output:
Time Complexity: O(N)
Auxiliary Space: O(N)
### Post-order Traversal:
In this traversal, the left subtree is visited first, followed by the right subtree and root node. Below is the program to illustrate the same:
## C++
`// C++ program to implement the``// post-order traversal``#include "bits/stdc++.h"``using` `namespace` `std;` `// Structure of the Binary Tree``struct` `treenode {`` ``int` `info;`` ``struct` `treenode *left,`` ``*right;``};` `// Function to create the Binary Tree``struct` `treenode* create()``{`` ``int` `data;`` ``struct` `treenode* tree;` ` ``// Dynamically allocating memory`` ``// for the tree-node`` ``tree = ``new` `treenode;` ` ``cout << ``"\nEnter data to be inserted "`` ``<< ``"or type -1 for no insertion : "``;` ` ``// Input from the user`` ``cin >> data;` ` ``// Termination Condition`` ``if` `(data == -1)`` ``return` `0;` ` ``// Assign value from user into tree`` ``tree->info = data;` ` ``// Recursively Call to create the`` ``// left and the right sub tree`` ``cout << ``"Enter left child of : "`` ``<< data;`` ``tree->left = create();` ` ``cout << ``"Enter right child of : "`` ``<< data;`` ``tree->right = create();` ` ``// Return the created Tree`` ``return` `tree;``};` `// Function to perform the post-order``// traversal of the given tree``void` `postorder(``struct` `treenode* root)``{`` ``// If the root is NULL`` ``return``;` ` ``stack s3;`` ``struct` `treenode* previous = NULL;` ` ``do` `{`` ``// Iterate until root is present`` ``while` `(root != NULL) {`` ``s3.push(root);`` ``root = root->left;`` ``}` ` ``while` `(root == NULL && (!s3.empty())) {`` ``root = s3.top();` ` ``// If the right subtree is NULL`` ``if` `(root->right == NULL`` ``|| root->right == previous) {`` ``// Print the root information`` ``cout << root->info << ``" "``;`` ``s3.pop();` ` ``// Update the previous`` ``previous = root;`` ``root = NULL;`` ``}` ` ``// Otherwise`` ``else`` ``root = root->right;`` ``}` ` ``} ``while` `(!s3.empty());`` ``cout << endl;``}` `// Driver Code``int` `main()``{`` ``// Root Node`` ``struct` `treenode* root = NULL;` ` ``// Function Call`` ``root = create();` ` ``// Perform Inorder Traversal`` ``postorder(root);` ` ``return` `0;``}` `/* Will be creating tree:`` ``2 `` ``/ \ `` ``7 5 `` ``/ \ \ `` ``2 6 9`` ``*/`
Output:
Time Complexity: O(N)
Auxiliary Space: O(N)
### Level-order Traversal:
In this traversal, the given tree is traversal level-wise. Below is the program to illustrate the same:
## C++
`// C++ program to illustrate the``// level order traversal#include "bits/stdc++.h"``using` `namespace` `std;` `// Structure of the Binary Tree``struct` `treenode {`` ``int` `info;`` ``struct` `treenode *left,`` ``*right;``};` `// Function to create the Binary Tree``struct` `treenode* create()``{`` ``int` `data;`` ``struct` `treenode* tree;` ` ``// Dynamically allocating memory`` ``// for the tree-node`` ``tree = ``new` `treenode;` ` ``cout << ``"\nEnter data to be inserted "`` ``<< ``"or type -1 for no insertion : "``;` ` ``// Input from the user`` ``cin >> data;` ` ``// Termination Condition`` ``if` `(data == -1)`` ``return` `0;` ` ``// Assign value from user into tree`` ``tree->info = data;` ` ``// Recursively Call to create the`` ``// left and the right sub tree`` ``cout << ``"Enter left child of : "`` ``<< data;`` ``tree->left = create();` ` ``cout << ``"Enter right child of : "`` ``<< data;`` ``tree->right = create();` ` ``// Return the created Tree`` ``return` `tree;``};` `// Function to perform the level-order``// traversal``void` `levelorder(``struct` `treenode* root)``{`` ``// If the root is NULL`` ``if` `(root == NULL)`` ``return``;` ` ``// Use queue for traversal`` ``queue q;` ` ``// Print the root's value and`` ``// push it into the queue`` ``cout << root->info << ``" "``;`` ``q.push(root);` ` ``// Iterate until queue is non-empty`` ``while` `(!q.empty()) {`` ``// Get the front node`` ``root = q.front();`` ``q.pop();` ` ``// If the root has the left child`` ``if` `(root->left) {`` ``cout << root->left->info`` ``<< ``" "``;`` ``q.push(root->left);`` ``}` ` ``// If the root has the right child`` ``if` `(root->right) {`` ``cout << root->right->info`` ``<< ``" "``;`` ``q.push(root->right);`` ``}`` ``}`` ``cout << endl;``}` `// Driver Code``int` `main()``{`` ``// Root Node`` ``struct` `treenode* root = NULL;` ` ``// Function Call`` ``root = create();` ` ``// Perform Inorder Traversal`` ``levelorder(root);` ` ``return` `0;``}` `/* Will be creating tree:`` ``2 `` ``/ \ `` ``7 5 `` ``/ \ \ `` ``2 6 9`` ``*/`
Output:
Time Complexity: O(N)
Auxiliary Space: O(N)
### The maximum element of the Binary Tree:
The element which is largest among all the elements of the binary tree is called the maximum element. Below is the program to illustrate the same:
## C++
`// C++ program for the above approach``#include "bits/stdc++.h"``using` `namespace` `std;` `// Structure of the Binary Tree``struct` `treenode {`` ``int` `info;`` ``struct` `treenode *left, *right;``};` `// Function to create the Binary Tree``struct` `treenode* create()``{`` ``int` `data;`` ``struct` `treenode* tree;` ` ``// Dynamically allocating memory`` ``// for the tree-node`` ``tree = ``new` `treenode;` ` ``cout << ``"\nEnter data to be inserted "`` ``<< ``"or type -1 for no insertion : "``;` ` ``// Input from the user`` ``cin >> data;` ` ``// Termination Condition`` ``if` `(data == -1)`` ``return` `0;` ` ``// Assign value from user into tree`` ``tree->info = data;` ` ``// Recursively Call to create the`` ``// left and the right sub tree`` ``cout << ``"Enter left child of : "` `<< data;`` ``tree->left = create();` ` ``cout << ``"Enter right child of : "` `<< data;`` ``tree->right = create();` ` ``// Return the created Tree`` ``return` `tree;``};` `// Function to find the maximum element``// in the given Binary Tree``int` `FindMax(``struct` `treenode* root)``{`` ``// If the tree is empty`` ``if` `(root == NULL)`` ``return` `0;` ` ``queue q;`` ``int` `max;`` ``struct` `treenode* temp;` ` ``max = root->info;` ` ``// Push the root in the queue`` ``q.push(root);` ` ``// Iterate until queue is non-empty`` ``while` `(!q.empty()) {` ` ``// Get the front node of`` ``// the tree`` ``root = q.front();`` ``temp = root;`` ``q.pop();` ` ``// Update the maximum value`` ``// of the Tree`` ``if` `(max < temp->info)`` ``max = temp->info;` ` ``if` `(root->left) {`` ``q.push(root->left);`` ``}`` ``if` `(root->right) {`` ``q.push(root->right);`` ``}`` ``}` ` ``// Return the maximum value`` ``return` `max;``}` `// Driver Code``int` `main()``{`` ``// Root Node`` ``struct` `treenode* root = NULL;` ` ``// Function Call`` ``root = create();` ` ``// Perform Inorder Traversal`` ``FindMax(root);` ` ``return` `0;``}` `/* Will be creating tree:`` ``2 `` ``/ \ `` ``7 5 `` ``/ \ \ `` ``2 6 9`` ``*/`
Output:
Time Complexity: O(N)
Auxiliary Space: O(N)
### Search for an element:
The approach to search for any particular element in the tree node is to perform any tree traversal on the given tree and check if there exists any node with the given searched value or not. If found to be true, then print “Element is Found”. Otherwise, print “Element Not Found”
Below is the program to illustrate the same:
## C++
`// C++ program for the above approach``#include "bits/stdc++.h"``using` `namespace` `std;` `// Structure of the Binary Tree``struct` `treenode {`` ``int` `info;`` ``struct` `treenode *left, *right;``};` `// Function to create the Binary Tree``struct` `treenode* create()``{`` ``int` `data;`` ``struct` `treenode* tree;` ` ``// Dynamically allocating memory`` ``// for the tree-node`` ``tree = ``new` `treenode;` ` ``cout << ``"\nEnter data to be inserted "`` ``<< ``"or type -1 for no insertion : "``;` ` ``// Input from the user`` ``cin >> data;` ` ``// Termination Condition`` ``if` `(data == -1)`` ``return` `0;` ` ``// Assign value from user into tree`` ``tree->info = data;` ` ``// Recursively Call to create the`` ``// left and the right sub tree`` ``cout << ``"Enter left child of : "` `<< data;`` ``tree->left = create();` ` ``cout << ``"Enter right child of : "` `<< data;`` ``tree->right = create();` ` ``// Return the created Tree`` ``return` `tree;``};` `// Function to search an element in the``// given Binary Tree``int` `FindElement(``struct` `treenode* root,`` ``int` `data)``{`` ``// If the root is NULL`` ``if` `(root == NULL)`` ``return` `0;` ` ``queue q;`` ``struct` `treenode* temp;`` ``if` `(!root)`` ``return` `0;` ` ``else` `{`` ``// Push the root`` ``q.push(root);` ` ``// Perform the level-order traversal`` ``while` `(!q.empty()) {`` ``// Get the root`` ``root = q.front();`` ``temp = root;`` ``q.pop();` ` ``// If the node with value data`` ``// exists then return 1`` ``if` `(data == temp->info)`` ``return` `1;` ` ``// Recursively push the left and`` ``// the right child of the node`` ``if` `(root->left) {`` ``q.push(root->left);`` ``}`` ``if` `(root->right) {`` ``q.push(root->right);`` ``}`` ``}` ` ``// Otherwise, not found`` ``return` `0;`` ``}``}` `// Driver Code``int` `main()``{`` ``int` `data;` ` ``// Root of the tree`` ``struct` `treenode* root = NULL;` ` ``// Create the Tree`` ``root = create();` ` ``cout << ``"\nEnter element to searched : "``;`` ``cin >> data;` ` ``// Function Call`` ``if` `(FindElement(root, data) == 1)`` ``cout << ``"\nElement is found"``;`` ``else`` ``cout << ``"Element is not found"``;`` ``return` `0;``}` `/* Will be creating tree:`` ``2`` ``/ \`` ``7 5`` ``/ \ \`` ``2 6 9`` ``*/`
Output:
Time Complexity: O(log N)
Auxiliary Space: O(N)
### Reverse Level Order Traversal:
Below is the program to illustrate the same:
## C++
`// C++ program for the above approach``#include "bits/stdc++.h"``using` `namespace` `std;` `// Structure of the Binary Tree``struct` `treenode {`` ``int` `info;`` ``struct` `treenode *left, *right;``};` `// Function to create the Binary Tree``struct` `treenode* create()``{`` ``int` `data;`` ``struct` `treenode* tree;` ` ``// Dynamically allocating memory`` ``// for the tree-node`` ``tree = ``new` `treenode;` ` ``cout << ``"\nEnter data to be inserted "`` ``<< ``"or type -1 for no insertion : "``;` ` ``// Input from the user`` ``cin >> data;` ` ``// Termination Condition`` ``if` `(data == -1)`` ``return` `0;` ` ``// Assign value from user into tree`` ``tree->info = data;` ` ``// Recursively Call to create the`` ``// left and the right sub tree`` ``cout << ``"Enter left child of : "` `<< data;`` ``tree->left = create();` ` ``cout << ``"Enter right child of : "` `<< data;`` ``tree->right = create();` ` ``// Return the created Tree`` ``return` `tree;``};` `// Function to print the reverse level``// order traversal of the given tree``void` `reversetree(``struct` `treenode* root)``{`` ``// If the root is NULL`` ``if` `(root == NULL)`` ``return``;` ` ``queue q;`` ``stack<``int``> s;`` ``struct` `treenode* temp;`` ``q.push(root);` ` ``// Until queue is empty`` ``while` `(!q.empty()) {`` ``// Get the front node`` ``temp = q.front();`` ``q.pop();` ` ``// Push every countered node`` ``// data into stack`` ``s.push(temp->info);` ` ``// Check for the left subtree`` ``if` `(temp->left)`` ``q.push(temp->left);` ` ``// Check for the right subtree`` ``if` `(temp->right)`` ``q.push(temp->right);`` ``}` ` ``// While S is non-empty, print`` ``// all the nodes`` ``while` `(!s.empty()) {`` ``cout << s.top() << ``" "``;`` ``s.pop();`` ``}``}` `// Driver Code``int` `main()``{`` ``// Create root node`` ``struct` `treenode* root = NULL;` ` ``// Create a tree`` ``root = create();` ` ``cout << ``"\nReversed tree is : "``;`` ``reversetree(root);`` ``return` `0;``}``/* Will be creating tree:`` ``2`` ``/ \`` ``7 5`` ``/ \ \`` ``2 6 9``*/`
Output:
Time Complexity: O(N)
Auxiliary Space: O(N)
### Height of the tree:
The height of the binary tree is the longest path from the root node to any leaf node in the tree. Below is the program to illustrate the same:
## C++
`// C++ program for the above approach``#include "bits/stdc++.h"``using` `namespace` `std;` `// Structure of the Binary Tree``struct` `treenode {`` ``int` `info;`` ``struct` `treenode *left, *right;``};` `// Function to create the Binary Tree``struct` `treenode* create()``{`` ``int` `data;`` ``struct` `treenode* tree;` ` ``// Dynamically allocating memory`` ``// for the tree-node`` ``tree = ``new` `treenode;` ` ``cout << ``"\nEnter data to be inserted "`` ``<< ``"or type -1 for no insertion : "``;` ` ``// Input from the user`` ``cin >> data;` ` ``// Termination Condition`` ``if` `(data == -1)`` ``return` `0;` ` ``// Assign value from user into`` ``// the tree`` ``tree->info = data;` ` ``// Recursively Call to create the`` ``// left and the right sub tree`` ``cout << ``"Enter left child of : "`` ``<< data;`` ``tree->left = create();` ` ``cout << ``"Enter right child of : "`` ``<< data;`` ``tree->right = create();` ` ``// Return the created Tree`` ``return` `tree;``};` `// Function to find the height of``// the given Binary tree``int` `height(``struct` `treenode* root)``{`` ``int` `x, y;` ` ``// If root is NOT NULL`` ``if` `(root != NULL) {`` ``// x will contain the height`` ``// of left subtree`` ``x = height(root->left);` ` ``// y will contain the height`` ``// of right subtree`` ``y = height(root->right);` ` ``if` `(x > y)` ` ``// Leaf node has one height`` ``// so x or y + 1`` ``return` `x + 1;`` ``else`` ``return` `y + 1;`` ``}`` ``return` `0;``}` `// Driver Code``int` `main()``{`` ``// Root Node`` ``struct` `treenode* root = NULL;` ` ``// Create the tree`` ``root = create();` ` ``cout << ``"\nHeight of the tree is : "`` ``<< height(root);` ` ``return` `0;``}``/* Will be creating tree:`` ``2`` ``/ \`` ``7 5`` ``/ \ \`` ``2 6 9`` ``*/`
Output:
Time Complexity: O(N)
Auxiliary Space: O(1)
### The deepest node of the tree:
The node which is present at the maximum or the last level is called the deepest node. Below is the program to implement the above approach:
## C++
`// C++ program for the above approach``#include "bits/stdc++.h"``using` `namespace` `std;` `// Structure of the Binary Tree``struct` `treenode {`` ``int` `info;`` ``struct` `treenode *left, *right;``};` `// Function to create the Binary Tree``struct` `treenode* create()``{`` ``int` `data;`` ``struct` `treenode* tree;` ` ``// Dynamically allocating memory`` ``// for the tree-node`` ``tree = ``new` `treenode;` ` ``cout << ``"\nEnter data to be inserted "`` ``<< ``"or type -1 for no insertion : "``;` ` ``// Input from the user`` ``cin >> data;` ` ``// Termination Condition`` ``if` `(data == -1)`` ``return` `0;` ` ``// Assign value from user into tree`` ``tree->info = data;` ` ``// Recursively Call to create the`` ``// left and the right sub tree`` ``cout << ``"Enter left child of : "` `<< data;`` ``tree->left = create();` ` ``cout << ``"Enter right child of : "` `<< data;`` ``tree->right = create();` ` ``// Return the created Tree`` ``return` `tree;``};` `// Function to find the deepest node``// of the given Binary Tree``int` `deepest(``struct` `treenode* root)``{`` ``// If the root is NULL`` ``if` `(root == NULL)`` ``return` `0;` ` ``queue q;` ` ``q.push(root);` ` ``// While queue is non-empty`` ``while` `(!q.empty()) {`` ``// Get the front node of queue`` ``root = q.front();`` ``q.pop();` ` ``// Check for the left and`` ``// the right subtree`` ``if` `(root->left)`` ``q.push(root->left);`` ``if` `(root->right)`` ``q.push(root->right);`` ``}` ` ``// Return the value for the`` ``// deepest node`` ``return` `(root->info);``}` `// Driver Code``int` `main()``{`` ``// Root Node`` ``struct` `treenode* root = NULL;` ` ``// Create the tree`` ``root = create();` ` ``cout << ``"\nDeepest node of the tree is : "` `<< deepest(root);` ` ``return` `0;``}` `/* Will be creating tree:`` ``2`` ``/ \`` ``7 5`` ``/ \ \`` ``2 6 9`` ``*/`
Output:
Time Complexity: O(N)
Auxiliary Space: O(N)
### Left view of the tree:
Below is the program to implement the same:
## C++
`// C++ program for the above approach``#include "bits/stdc++.h"``using` `namespace` `std;` `// Structure of the Binary Tree``struct` `treenode {`` ``int` `info;`` ``struct` `treenode *left, *right;``};` `// Function to create the Binary Tree``struct` `treenode* create()``{`` ``int` `data;`` ``struct` `treenode* tree;` ` ``// Dynamically allocating memory`` ``// for the tree-node`` ``tree = ``new` `treenode;` ` ``cout << ``"\nEnter data to be inserted "`` ``<< ``"or type -1 for no insertion : "``;` ` ``// Input from the user`` ``cin >> data;` ` ``// Termination Condition`` ``if` `(data == -1)`` ``return` `0;` ` ``// Assign value from user into tree`` ``tree->info = data;` ` ``// Recursively Call to create the`` ``// left and the right sub tree`` ``cout << ``"Enter left child of : "` `<< data;`` ``tree->left = create();` ` ``cout << ``"Enter right child of : "` `<< data;`` ``tree->right = create();` ` ``// Return the created Tree`` ``return` `tree;``};` `// Stores the maximum left size``int` `maxlevelleft = 0;` `// Function to print the left view of``// the tree``void` `leftview(``struct` `treenode* root,`` ``int` `level)``{`` ``if` `(root == NULL)`` ``return``;` ` ``// If current level is at least`` ``// the maximum left level`` ``if` `(level >= maxlevelleft) {`` ``// Print the data`` ``cout << root->info << ``" "``;`` ``maxlevelleft++;`` ``}` ` ``// Left and Right Subtree`` ``// recursive calls`` ``leftview(root->left, level + 1);`` ``leftview(root->right, level + 1);``}` `// Driver Code``int` `main()``{`` ``// Root Node`` ``struct` `treenode* root = NULL;` ` ``// Create the tree`` ``root = create();` ` ``cout << ``"\nLeft view of the tree is : "``;` ` ``// Function Call`` ``leftview(root, 0);` ` ``return` `0;``}` `/* Will be creating tree:`` ``2`` ``/ \`` ``7 5`` ``/ \ \`` ``2 6 9`` ``*/`
Output:
Time Complexity: O(N)
Auxiliary Space: O(1)
### Right view of the tree:
Below is the program to illustrate the same:
## C++
`// C++ program to demonstrate the``// above concepts``#include "bits/stdc++.h"``using` `namespace` `std;` `// Structure of the Binary Tree``struct` `treenode {`` ``int` `info;`` ``struct` `treenode *left,`` ``*right;``};` `// Function to create the Binary Tree``struct` `treenode* create()``{`` ``int` `data;`` ``struct` `treenode* tree;` ` ``// Dynamically allocating memory`` ``// for the tree-node`` ``tree = ``new` `treenode;` ` ``cout << ``"\nEnter data to be inserted "`` ``<< ``"or type -1 for no insertion : "``;` ` ``// Input from the user`` ``cin >> data;` ` ``// Termination Condition`` ``if` `(data == -1)`` ``return` `0;` ` ``// Assign value from user into tree`` ``tree->info = data;` ` ``// Recursively Call to create the`` ``// left and the right sub tree`` ``cout << ``"Enter left child of : "`` ``<< data;`` ``tree->left = create();` ` ``cout << ``"Enter right child of : "`` ``<< data;`` ``tree->right = create();` ` ``// Return the created Tree`` ``return` `tree;``};` `// Stores the maximum right level``int` `maxlevelright = 0;` `// Function to print the right view of``// the given Binary tree``void` `rightview(``struct` `treenode* root,`` ``int` `level)``{`` ``// If the root is NULL`` ``if` `(root == NULL)`` ``return``;` ` ``// If the current level is greater`` ``// than the maximum right level`` ``if` `(level >= maxlevelright) {`` ``// Print the data`` ``cout << root->info << ``" "``;`` ``maxlevelright++;`` ``}` ` ``// Recursively call for the right`` ``// and the left subtree`` ``rightview(root->right, level + 1);`` ``rightview(root->left, level + 1);``}` `// Driver Code``int` `main()``{`` ``// Root Node`` ``struct` `treenode* root = NULL;` ` ``// Create the tree`` ``root = create();` ` ``cout << ``"\nRight view of the tree is : "``;` ` ``rightview(root, 0);` ` ``return` `0;``}``/* Will be creating tree:`` ``2`` ``/ \`` ``7 5`` ``/ \ \`` ``2 6 9``*/`
Output:
Time Complexity: O(N)
Auxiliary Space: O(1)
### Top view of the tree:
Below is the program to illustrate the same:
## C++
`// C++ program to demonstrate the``// above concepts``#include "bits/stdc++.h"``using` `namespace` `std;` `// Structure of the Binary Tree``struct` `treenode {`` ``int` `info;`` ``struct` `treenode *left,`` ``*right;``};` `// Function to create the Binary Tree``struct` `treenode* create()``{`` ``int` `data;`` ``struct` `treenode* tree;` ` ``// Dynamically allocating memory`` ``// for the tree-node`` ``tree = ``new` `treenode;` ` ``cout << ``"\nEnter data to be inserted "`` ``<< ``"or type -1 for no insertion : "``;` ` ``// Input from the user`` ``cin >> data;` ` ``// Termination Condition`` ``if` `(data == -1)`` ``return` `0;` ` ``// Assign value from user into tree`` ``tree->info = data;` ` ``// Recursively Call to create the`` ``// left and the right sub tree`` ``cout << ``"Enter left child of : "`` ``<< data;`` ``tree->left = create();` ` ``cout << ``"Enter right child of : "`` ``<< data;`` ``tree->right = create();` ` ``// Return the created Tree`` ``return` `tree;``};` `// Initialize an ordered map``map<``int``, ``int``> HashMap;` `// Iterator for the map``map<``int``, ``int``>::iterator it;` `// Function to print the top view``// of the given Binary Tree``void` `topview(``struct` `treenode* root,`` ``int` `level)``{`` ``// If the root is NULL`` ``if` `(root == NULL)`` ``return``;` ` ``// Get the level`` ``int` `i = HashMap.count(level);` ` ``// Update the root information`` ``if` `(i == 0)`` ``HashMap[level] = root->info;` ` ``// Left and Right recursive calls`` ``topview(root->left, level - 1);`` ``topview(root->right, level + 1);` ` ``// Update the current level`` ``// with the root's value`` ``HashMap[level] = root->info;` ` ``return``;``}` `// Driver Code``int` `main()``{`` ``// Root Node`` ``struct` `treenode* root = NULL;` ` ``// Create a tree`` ``root = create();` ` ``topview(root, 0);`` ``cout << ``"\nTop view of the tree is : "``;` ` ``for` `(it = HashMap.begin();`` ``it != HashMap.end(); it++) {`` ``cout << it->second << ``" "``;`` ``}` ` ``return` `0;``}``/* Will be creating tree:`` ``2`` ``/ \`` ``7 5`` ``/ \ \`` ``2 6 9``*/`
Output:
Time Complexity: O(N)
Auxiliary Space: O(N)
### Bottom view of the tree:
Below is the program to illustrate the same:
## C++
`// C++ program to demonstrate the``// above concepts``#include "bits/stdc++.h"``using` `namespace` `std;` `// Structure of the Binary Tree``struct` `treenode {`` ``int` `info;`` ``struct` `treenode *left,`` ``*right;``};` `// Function to create the Binary Tree``struct` `treenode* create()``{`` ``int` `data;`` ``struct` `treenode* tree;` ` ``// Dynamically allocating memory`` ``// for the tree-node`` ``tree = ``new` `treenode;` ` ``cout << ``"\nEnter data to be inserted "`` ``<< ``"or type -1 for no insertion : "``;` ` ``// Input from the user`` ``cin >> data;` ` ``// Termination Condition`` ``if` `(data == -1)`` ``return` `0;` ` ``// Assign value from user into tree`` ``tree->info = data;` ` ``// Recursively Call to create the`` ``// left and the right sub tree`` ``cout << ``"Enter left child of : "`` ``<< data;`` ``tree->left = create();` ` ``cout << ``"Enter right child of : "`` ``<< data;`` ``tree->right = create();` ` ``// Return the created Tree`` ``return` `tree;``};` `// Initialize an ordered Map``map<``int``, pair<``int``, ``int``> > HashMap;` `// Iterator for the map``map<``int``, pair<``int``, ``int``> >::iterator it;` `// Function to print the bottom view``// of the given binary tree``void` `bottomview(``struct` `treenode* root,`` ``int` `level, ``int` `height)``{`` ``// If root is NULL`` ``if` `(root == NULL)`` ``return``;` ` ``// If the height of the level is`` ``// greater than the current`` ``// stored height of the level`` ``if` `(height >= HashMap[level].second) {`` ``HashMap[level] = { root->info,`` ``height };`` ``}` ` ``// Left and right recursive calls`` ``bottomview(root->left, level - 1,`` ``height + 1);`` ``bottomview(root->right, level + 1,`` ``height + 1);` ` ``return``;``}` `// Driver Code``int` `main()``{`` ``// Root Node`` ``struct` `treenode* root = NULL;` ` ``// Create the tree`` ``root = create();` ` ``bottomview(root, 0, 0);`` ``cout << ``"\nBottom view of the tree is : "``;` ` ``for` `(it = HashMap.begin();`` ``it != HashMap.end(); it++) {` ` ``cout << it->second.first << ``" "``;`` ``}` ` ``return` `0;``}` `/* Will be creating tree:`` ``2`` ``/ \`` ``7 5`` ``/ \ \`` ``2 6 9`` ``*/`
Output:
Time Complexity: O(N)
Auxiliary Space: O(N)
### The mirror image of the tree:
Below is the program to illustrate the same:
## C++
`// C++ program to implement``// the above approach``#include ``using` `namespace` `std;` `// structure of the binary tree``struct` `treenode {`` ``// data part`` ``int` `info;` ` ``// left and right node`` ``struct` `treenode *left, *right;``};` `// create function for binary``// tree creation``struct` `treenode* create()``{`` ``int` `data;` ` ``// variable of the structure`` ``struct` `treenode* tree;` ` ``// dynamically allocating`` ``// memory for tree-node`` ``tree = ``new` `treenode;` ` ``cout << ``"\nEnter data to be inserted or type -1 for no insertion : "``;` ` ``// input from the user`` ``cin >> data;` ` ``// condition for termination`` ``if` `(data == -1)`` ``return` `0;` ` ``// assigning value from user`` ``// into tree.`` ``tree->info = data;` ` ``// recursively calling create function`` ``// for left and right sub tree`` ``cout << ``"Enter left child of : "` `<< data;`` ``tree->left = create();` ` ``cout << ``"Enter right child of : "` `<< data;`` ``tree->right = create();` ` ``// returning the created tree`` ``return` `tree;``};` `/*``With the simple logic of recursion and``swapping, we can create mirror tree.``We will swap the left-node and``right-node of root node. We will use``recursion and start swapping from the``bottom of the tree.``*/` `// function to form mirror image a tree``void` `mirrortree(``struct` `treenode* root)``{`` ``if` `(root != NULL) {`` ``mirrortree(root->left);`` ``mirrortree(root->right);` ` ``struct` `treenode* temp;` ` ``temp = root->left;`` ``root->left = root->right;`` ``root->right = temp;`` ``}`` ``return``;``}` `// function for the inorder traversal``void` `inorder(``struct` `treenode* root)``{`` ``if` `(root == NULL)`` ``return``;` ` ``inorder(root->left);`` ``cout << root->info << ``" "``;`` ``inorder(root->right);``}` `// Driver code``int` `main()``{`` ``// creating variable of the`` ``// structure`` ``struct` `treenode* root = NULL;` ` ``// calling create function to`` ``// create tree`` ``root = create();` ` ``mirrortree(root);`` ``cout << ``"\nInorder of the mirror tree is = "``;`` ``inorder(root);` ` ``return` `0;``}` `/* Will be creating tree:`` ``2`` ``/ \`` ``7 5`` ``/ \ \`` ``2 6 9`` ``*/`
Output:
Time Complexity: O(N)
Auxiliary Space: O(1)
### Serialize a tree:
Serialization of a tree is defined as the conversion of the given tree into a data-format that can be later restored and the structure of the tree must be maintained. Below is the program to implement the above approach:
## C++
`// C++ program to implement``// the above approach``#include ``using` `namespace` `std;` `// structure of the binary tree``struct` `treenode {`` ``// data part`` ``int` `info;` ` ``// left and right node`` ``struct` `treenode *left, *right;``};` `// create function for binary``// tree creation``struct` `treenode* create()``{`` ``int` `data;` ` ``// variable of the structure`` ``struct` `treenode* tree;` ` ``// dynamically allocating`` ``// memory for tree-node`` ``tree = ``new` `treenode;` ` ``cout << ``"\nEnter data to be inserted or type -1 for no insertion : "``;` ` ``// input from the user`` ``cin >> data;` ` ``// condition for termination`` ``if` `(data == -1)`` ``return` `0;` ` ``// assigning value from user`` ``// into tree.`` ``tree->info = data;` ` ``// recursively calling create function`` ``// for left and right sub tree`` ``cout << ``"Enter left child of : "` `<< data;`` ``tree->left = create();` ` ``cout << ``"Enter right child of : "` `<< data;`` ``tree->right = create();` ` ``// returning the created tree`` ``return` `tree;``};` `// Function to serialize the given``// Binary Tree``void` `serialize(``struct` `treenode* root,`` ``vector<``int``>& v)``{`` ``// If the root is NULL, then`` ``// push -1 and return`` ``if` `(root == NULL) {`` ``v.push_back(-1);`` ``return``;`` ``}` ` ``// Otherwise, push the data part`` ``v.push_back(root->info);` ` ``// Recursively Call for the left`` ``// and the right Subtree`` ``serialize(root->left, v);`` ``serialize(root->right, v);``}` `// Driver Code``int` `main()``{`` ``// Root Node`` ``struct` `treenode* root = NULL;` ` ``// Create a tree`` ``root = create();` ` ``vector<``int``> v;` ` ``serialize(root, v);`` ``cout << ``"\nSerialize form of the tree is = "``;` ` ``for` `(``int` `i = 0; i < v.size(); i++)`` ``cout << v[i] << ``" "``;`` ``return` `0;``}` `/* Will be creating tree:`` ``2`` ``/ \`` ``7 5`` ``/ \ \`` ``2 6 9`` ``*/`
## Java
`import` `java.util.*;` `// structure of the binary tree``class` `TreeNode {`` ``// data part`` ``int` `info;` ` ``// left and right node`` ``TreeNode left, right;` ` ``// constructor`` ``TreeNode(``int` `item)`` ``{`` ``info = item;`` ``left = right = ``null``;`` ``}``}` `class` `SerializeDeserializeBinaryTree {`` ``// create function for binary`` ``// tree creation`` ``public` `static` `TreeNode create()`` ``{`` ``int` `data;` ` ``// variable of the structure`` ``TreeNode tree;` ` ``// dynamically allocating`` ``// memory for tree-node`` ``tree = ``new` `TreeNode(``0``);` ` ``Scanner sc = ``new` `Scanner(System.in);`` ``System.out.print(`` ``"\nEnter data to be inserted or type -1 for no insertion : "``);` ` ``// input from the user`` ``data = sc.nextInt();` ` ``// condition for termination`` ``if` `(data == -``1``)`` ``return` `null``;` ` ``// assigning value from user`` ``// into tree.`` ``tree.info = data;` ` ``// recursively calling create function`` ``// for left and right sub tree`` ``System.out.print(``"Enter left child of : "` `+ data);`` ``tree.left = create();` ` ``System.out.print(``"Enter right child of : "` `+ data);`` ``tree.right = create();` ` ``// returning the created tree`` ``return` `tree;`` ``}` ` ``// Function to serialize the given`` ``// Binary Tree`` ``public` `static` `void` `serialize(TreeNode root,`` ``List v)`` ``{`` ``// If the root is NULL, then`` ``// push -1 and return`` ``if` `(root == ``null``) {`` ``v.add(-``1``);`` ``return``;`` ``}` ` ``// Otherwise, push the data part`` ``v.add(root.info);` ` ``// Recursively Call for the left`` ``// and the right Subtree`` ``serialize(root.left, v);`` ``serialize(root.right, v);`` ``}` ` ``// Driver Code`` ``public` `static` `void` `main(String args[])`` ``{`` ``// Root Node`` ``TreeNode root = ``null``;` ` ``// Create a tree`` ``root = create();` ` ``List v = ``new` `ArrayList();` ` ``serialize(root, v);`` ``System.out.print(`` ``"\nSerialize form of the tree is = "``);` ` ``for` `(``int` `i = ``0``; i < v.size(); i++)`` ``System.out.print(v.get(i) + ``" "``);`` ``}``}` `/* Will be creating tree:``2``/``7 5``/ \``2 6 9``*/`
Output:
Time Complexity: O(N)
Auxiliary Space: O(N)
Complexity Analysis:
1. Time Complexity: O(n).
2. Auxiliary Space: O(1).
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# Intermediate Algebra (5th Edition)Solutions for Chapter 10
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Chapter: Problem:
Fill in the blanks.
a. The curves formed by the intersection of a plane with an infinite right-circular cone are called _________ sections.
b. A circle is the set of all points in a plane that are a fixed distance from a point called its _________. The fixed distance is called the _________ of the circle.
c. The standard form for the equation of a ________ centered at the origin that opens left and right is
d. is a(n) _________ system of equations.
e. The standard form for the equation of an _________ centered at the origin is
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# Earning per share
On December 31,2003, Bercalir, Inc. had 200,000,000 shares of common stock and 3,000,000 shares of 9%, \$100 par, value cumulative preferred stock issued and outstanding. There were no preferred dividends in mrears as of January I, 2004. On March I, 2004, Bercalir purchased 24,000,000 shares of its common stock as treasuIy stock. Bercalir issued a 5% common stock dividend on July I, 2004. Four million treasury shares were sold on October I, 2004 and the net income for the year ended December 31,2004 was \$150,000,000.
Required:
Compute the basic earnings per share for the year ended December 31, 2004. computations.
#### Solution Preview
In order to calculate the basic earnings per share, we need to calculate the weighted average number of shares outstanding.
The shares outstanding at various times were
Jan 1 ................................200,000,000
Mar ...
#### Solution Summary
The solution explains how to calculate the basic earnings per share
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A211357 Triangle read by rows: T(n,k) is the number of noncrossing partitions up to rotation of an n-set that contain k singleton blocks. 4
1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 2, 1, 2, 0, 1, 2, 3, 2, 2, 0, 1, 5, 6, 9, 4, 3, 0, 1, 6, 15, 18, 15, 5, 3, 0, 1, 15, 36, 56, 42, 29, 7, 4, 0, 1, 28, 91, 144, 142, 84, 42, 10, 4, 0, 1, 67, 232, 419, 432, 322, 152, 66, 12, 5, 0, 1, 145, 603, 1160, 1365, 1080, 630, 252, 90, 15, 5, 0, 1 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 0,11 LINKS Andrew Howroyd, Table of n, a(n) for n = 0..1274 (terms 0..90 from Tilman Piesk) FORMULA T(n,k) = (1/n)*(A091867(n,k) - A171128(n,k) + Sum_{d|gcd(n,k)} phi(d) * A171128(n/d, k/d)) for n > 0. - Andrew Howroyd, Nov 16 2017 EXAMPLE From Andrew Howroyd, Nov 16 2017: (Start) Triangle begins: (n >= 0, 0 <= k <= n) 1; 0, 1; 1, 0, 1; 1, 1, 0, 1; 2, 1, 2, 0, 1; 2, 3, 2, 2, 0, 1; 5, 6, 9, 4, 3, 0, 1; 6, 15, 18, 15, 5, 3, 0, 1; 15, 36, 56, 42, 29, 7, 4, 0, 1; 28, 91, 144, 142, 84, 42, 10, 4, 0, 1; 67, 232, 419, 432, 322, 152, 66, 12, 5, 0, 1; (End) MATHEMATICA a91867[n_, k_] := If[k == n, 1, (Binomial[n + 1, k]/(n + 1)) Sum[Binomial[n + 1 - k, j] Binomial[n - k - j - 1, j - 1], {j, 1, (n - k)/2}]]; a2426[n_] := Sum[Binomial[n, 2*k]*Binomial[2*k, k], {k, 0, Floor[n/2]}]; a171128[n_, k_] := Binomial[n, k]*a2426[n - k]; T[0, 0] = 1; T[n_, k_] := (1/n)*(a91867[n, k] - a171128[n, k] + Sum[EulerPhi[d]* a171128[n/d, k/d], {d, Divisors[GCD[n, k]]}]); Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 03 2018, after Andrew Howroyd *) PROG (PARI) g(x, y) = {1/sqrt((1 - (1 + y)*x)^2 - 4*x^2) - 1} S(n)={my(A=(1-sqrt(1-4*x/(1-(y-1)*x) + O(x^(n+2))))/(2*x)-1); Vec(1+intformal((A + sum(k=2, n, eulerphi(k)*g(x^k + O(x*x^n), y^k)))/x))} my(v=S(10)); for(n=1, #v, my(p=v[n]); for(k=0, n-1, print1(polcoeff(p, k), ", ")); print) \\ Andrew Howroyd, Nov 16 2017 CROSSREFS Column k=0 is A295198. Row sums are A054357. Cf. A091867 (noncrossing partitions of an n-set with k singleton blocks), A211359 (up to rotations and reflections). Cf. A171128. Sequence in context: A218797 A137289 A211359 * A238416 A063574 A144515 Adjacent sequences: A211354 A211355 A211356 * A211358 A211359 A211360 KEYWORD nonn,tabl AUTHOR Tilman Piesk, Apr 12 2012 STATUS approved
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Last modified December 16 06:18 EST 2019. Contains 330016 sequences. (Running on oeis4.) | 1,413 | 3,191 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2019-51 | latest | en | 0.481643 |
http://metamath.tirix.org/mpeuni/fdmi | 1,721,057,334,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514707.52/warc/CC-MAIN-20240715132531-20240715162531-00482.warc.gz | 20,710,843 | 1,697 | Metamath Proof Explorer
Theorem fdmi
Description: Inference associated with fdm . The domain of a mapping. (Contributed by NM, 28-Jul-2008)
Ref Expression
Hypothesis fdmi.1 𝐹 : 𝐴𝐵
Assertion fdmi dom 𝐹 = 𝐴
Proof
Step Hyp Ref Expression
1 fdmi.1 𝐹 : 𝐴𝐵
2 fdm ( 𝐹 : 𝐴𝐵 → dom 𝐹 = 𝐴 )
3 1 2 ax-mp dom 𝐹 = 𝐴 | 136 | 306 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-30 | latest | en | 0.741002 |
https://cn.maplesoft.com/support/help/view.aspx?path=OpenMaple/Java/Numeric | 1,718,271,928,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861372.90/warc/CC-MAIN-20240613091959-20240613121959-00882.warc.gz | 162,641,442 | 21,532 | Numeric Class - Maple Help
For the best experience, we recommend viewing online help using Google Chrome or Microsoft Edge.
# Online Help
###### All Products Maple MapleSim
Numeric
Java representation of a numeric object
Description
• The com.maplesoft.openmaple.Numeric class represents a Maple numeric object (a float or an integer). As Numeric publicly inherits from Algebraic, it provides all of the member functions from the Algebraic class as well as those listed here.
Method Summary
boolean isByte()
• isByte returns true if the value represented by this Numeric is an integer within the range of a Java byte.
boolean isShort()
• isShort returns true if the value represented by this Numeric is an integer within the range of a Java short.
boolean isInt()
• isInt returns true if the value represented by this Numeric is an integer within the range of a Java int.
boolean isLong()
• isLong returns true if the value represented by this Numeric is an integer within the range of a Java long.
boolean isInteger()
• isInteger returns true if the value represented by this Numeric is an integer.
boolean isUnnamedZero()
• isUnnamedZero returns true if the value represented by this Numeric is zero.
float floatValue()
• floatValue converts the value represented by the Numeric to a float.
double doubleValue()
• doubleValue returns the value represented in the Numeric as a Java double.
byte byteValue()
• byteValue returns the value represented by the Numeric as a Java byte.
short shortValue()
• shortValue returns the value represented by the Numeric as a Java short.
int intValue()
• intValue returns the value represented by the Numeric as a Java int.
long longValue()
• longValue returns the value represented by the Numeric as a Java long.
BigDecimal toBigDecimal()
• toBigDecimal returns the value represented by the Numeric as a java.math.BigDecimal.
BigInteger toBigInteger()
• toBigInteger returns the value represented by the Numeric as a java.math.BigInteger.
See Also | 443 | 2,044 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-26 | latest | en | 0.56683 |
https://developers.uat.refinitiv.com/en/article-catalog/article/estimating-monthly-gdp-figures-via-an-income-approach | 1,656,240,643,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103205617.12/warc/CC-MAIN-20220626101442-20220626131442-00294.warc.gz | 248,445,403 | 35,704 | 1. Home
2. Article Catalog
3. Estimating Monthly GDP Figures Via an Income Approach and the Holt Winters Model
Article
Estimating Monthly GDP Figures Via an Income Approach and the Holt-Winters Model
Jonathan Legrand
Only quarterly USA GDP data is published; this article describes a method of estimating monthly such figures using monthly Total Compensation figures.
## Introduction
Gross Domestic Product (GDP) is often thought of and calculated with an Expenditure Approach such that GDP=C+I+G+(X−M) where C = Consumption, I = Investment, G = Government Spending, & (X–M) = Net Exports, but it is also possible to calculate it via a per worker Income Approach. With this approach, we may estimate GDP Per Worker (GDPPW) with Total Compensation Per Worker (TCPW) figures such that (as per Appendix 1):
where
We can then attempt to interpolate GDPPW values at higher frequencies than they are officially released with higher wage/income figure releases (exempli gratia (e.g.): to compute monthly GDP figures when only quarterly ones are released using monthly published Total Compensation data).
We have to be careful of a few difficulties outlined by Feldstein (2008) (MF hereon): "The relation between productivity [ id est (i.e.): GDP] and wages has been a source of substantial controversy, not only because of its inherent importance but also because of the conceptual measurement issues that arise in making the comparison." (Feldstein (2008)). MF outlines several key factors (KF hereon) to consider when studying the question:
1. "[An undeserved] focus on wages rather than total compensation. Because of the rise in fringe benefits and other noncash payments, wages have not risen as rapidly as total compensation. It is important therefore to compare the productivity rise with the increase of total compensation rather than with the increase of the narrower measure of just wages and salaries.” This is why this article will first focus on using Total Compensation (TC).
2. “[… The] nominal compensation deflated by the CPI is not appropriate for evaluating the relation between productivity and compensation […] this implies that the real marginal product of labor should be compared to the wage deflated by the product price and not by some consumer price index. [...] The CPI differs from the nonfarm product price in several ways. The inclusion in the CPI, but not in the nonfarm output price index, of the prices of imports and of the services provided by owner occupied homes is particularly important." This is why this article will first focus on using Nominal T.C. deflated (made 'real') with the Nonfarm Business Sector Output Price Index Deflator (NBSOPID).
Keeping these points in mind, we will attempt to investigate the relationship between the United States of America( U.S.A.)'s Real GDP (RGDP) and U.S.A.'s Real National Total Compensation (RNTC) in order to construct a framework allowing us to estimate monthly RGDP data when only quarterly data is published.
## Numerical Approximation
### Background
As per Appendix 1:
It may seem needlessly convoluted to insert 𝑟𝑡 the way it was done above, but it was necessary due to the fact that we have 𝑟𝑡rt estimates (𝑟𝑡ˆ) - we thus build our framework around it: since Total Compensation figures are released on a monthly basis, if we can construct an estimate for the linearly-time-variant 𝑟𝑡rt, we can then construct monthly GDP figures (even though only quarterly ones are published).
This article will study the efficiency of the USA Real GDP Per Worker To National Real Total Compensation Per Worker Ratio (RGDPPWTNRTCPWR) as the ratio '𝑟𝑡'.
## Development Tools & Resources
The example code demonstrating the use case is based on the following development tools and resources:
• Refinitiv's DataStream Web Services (DSWS): Access to DataStream data. A DataStream or Refinitiv Workspace IDentification (ID) will be needed to run the code bellow.
• Python Environment: Tested with Python 3.7
• Packages: DatastreamDSWS, NumpyPandasstatsmodelscipy and Matplotlib. The Python built in modules statisticsdatetime and dateutil are also required.
## Import libraries
``` ```
# The ' from ... import ' structure here allows us to only import the
# module ' python_version ' from the library ' platform ':
from platform import python_version
print("This code runs on Python version " + python_version())
```
```
This code runs on Python version 3.7.4
We need to gather our data. Since Refinitiv's DataStream Web Services (DSWS) allows for access to the most accurate and wholesome end-of-day (EOD) economic database (DB), naturally it is more than appropriate. We can access DSWS via the Python library "DatastreamDSWS" that can be installed simply by using pip installpip install.
``` ```
import DatastreamDSWS as DSWS
# We can use our Refinitiv's Datastream Web Socket (DSWS) API keys that allows us to be identified by
# Refinitiv's back-end services and enables us to request (and fetch) data:
# The username is placed in a text file so that it may be used in this code without showing it itself:
# It is best to close the files we opened in order to make sure that we don't stop any other
# services/programs from accessing them if they need to:
```
```
The following are Python-built-in module/library, therefore it does not have a specific version number.
``` ```
import os # os is a library used to manipulate files on machines
import datetime # datetime will allow us to manipulate Western World dates
import dateutil # dateutil will allow us to manipulate dates in equations
import warnings # warnings will allow us to manupulate warning messages (such as depretiation messages)
```
```
statistics, numpy, scipy and statsmodels are needed for datasets' statistical and mathematical manipulations
``` ```
import statistics # This is a Python-built-in module/library, therefore it does not have a specific version number.
import numpy
import statsmodels
from statsmodels.tsa.api import ExponentialSmoothing, SimpleExpSmoothing, Holt
import scipy
from scipy import stats
for i,j in zip(["numpy", "statsmodels", "scipy"], [numpy, statsmodels, scipy]):
print("The " + str(i) + " library imported in this code is version: " + j.__version__)
```
```
The numpy library imported in this code is version: 1.16.5
The statsmodels library imported in this code is version: 0.10.1
The scipy library imported in this code is version: 1.3.1
pandas will be needed to manipulate data sets
``` ```
import pandas
pandas.set_option('display.max_columns', None) # This line will ensure that all columns of our dataframes are always shown
print("The pandas library imported in this code is version: " + pandas.__version__)
```
```
The pandas library imported in this code is version: 0.25.1
matplotlib is needed to plot graphs of all kinds
``` ```
import matplotlib
import matplotlib.pyplot as plt
print("The matplotlib library imported in this code is version: "+ matplotlib.__version__)
```
```
The matplotlib library imported in this code is version: 3.1.1
## Setup Original Functions
### Setup data sets
The cell below defines a function that translates DataStream data into a shape we can work with: a Pandas Dafaframe normalised to the 1st of the month that it was collected in.
``` ```
def Translate_to_First_of_the_Month(data, dataname):
"""This function will put Refinitiv's DataStream data into a Pandas Dafaframe
and normalise it to the 1st of the month it was collected in."""
# The first 8 characters of the index is the ' yyyy-mm- ', onto which will be added '01':
data.index = data.index.astype("str").str.slice_replace(8, repl = "01")
data.index = pandas.to_datetime(data.index)
data.columns = [dataname]
return data
```
```
The cell below defines a function that appends our monthly data-frame with chosen data
``` ```
df = pandas.DataFrame([])
global df # This allows the function to take the pre-deffined variable ' df ' for granted and work from there
DS_Data_monthly = Translate_to_First_of_the_Month(data, dataname)
df = pandas.merge(df, DS_Data_monthly,
how = "outer",
left_index = True,
right_index=True)
```
```
### Setup plot functions
The cell below defines a function to plot data on one y axis.
``` ```
# Using an implicitly registered datetime converter for a matplotlib plotting method is no
# longer supported by matplotlib. Current versions of pandas requires explicitly
# registering matplotlib converters.
pandas.plotting.register_matplotlib_converters()
def Plot1ax(dataset, ylabel = "", title = "", xlabel = "Year", datasubset = [0],
datarange = False, linescolor = False, figuresize = (12,4),
facecolor = "0.25", grid = True):
""" Plot1ax Version 1.0:
This function returns a Matplotlib graph with default colours and dimensions
on one y axis (on the left as oppose to two, one on the left and one on the right).
datasubset (list): Needs to be a list of the number of each column within
the data-set that needs to be labelled on the left.
datarange (bool): If wanting to plot graph from and to a specific point,
make datarange a list of start and end date.
linescolor (bool/list): (Default: False) This needs to be a list of the color of each
vector to be ploted, in order they are shown in their data-frame from left to right.
figuresize (tuple): (Default: (12,4)) This can be changed to give graphs of different
proportions. It is defaulted to a 12 by 4 (ratioed) graph.
facecolor (str): (Default: "0.25") This allows the user to change the
background color as needed.
grid (bool): (Default: "True") This allows us to decide whether or
not to include a grid in our graphs.
"""
if datarange == False:
start_date = str(dataset.iloc[:,datasubset].index[0])
end_date = str(dataset.iloc[:,datasubset].index[-1])
else:
start_date = str(datarange[0])
if datarange[-1] == -1:
end_date = str(dataset.iloc[:,datasubset].index[-1])
else:
end_date = str(datarange[-1])
fig, ax1 = plt.subplots(figsize = figuresize, facecolor = facecolor)
ax1.tick_params(axis = 'both', colors = 'w')
ax1.set_facecolor(facecolor)
fig.autofmt_xdate()
plt.ylabel(ylabel, color = 'w')
ax1.set_xlabel(str(xlabel), color = 'w')
if linescolor == False:
for i in datasubset: # This is to label all the lines in order to allow matplot lib to create a legend
ax1.plot(dataset.iloc[:, i].loc[start_date : end_date],
label = str(dataset.columns[i]))
else:
for i in datasubset: # This is to label all the lines in order to allow matplot lib to create a legend
ax1.plot(dataset.iloc[:, i].loc[start_date : end_date],
label = str(dataset.columns[i]),
color = linescolor)
ax1.tick_params(axis='y')
if grid == True:
ax1.grid(color='black', linewidth = 0.5)
ax1.set_title(str(title) + " \n", color='w')
plt.legend()
plt.show()
```
```
The cell below defines a function to plot data on two y axis.
``` ```
def Plot2ax(dataset, rightaxisdata, y1label, y2label, leftaxisdata,
title = "", xlabel = "Year", y2color = "C1", y1labelcolor = "C0",
figuresize = (12,4), leftcolors = ('C1'), facecolor = "0.25"):
""" Plot2ax Version 1.0:
This function returns a Matplotlib graph with default colours and dimensions
on two y axis (one on the left and one on the right)
leftaxisdata (list): Needs to be a list of the number of each column within
the data-set that needs to be labelled on the left
figuresize (tuple): (Default: (12,4)) This can be changed to give graphs of
different proportions. It is defaulted to a 12 by 4 (ratioed) graph
leftcolors (str / tuple of (a) list(s)): This sets up the line color for data
expressed with the left hand y-axis. If there is more than one line,
the list needs to be specified with each line colour specified in a tuple.
facecolor (str): (Default: "0.25") This allows the user to change the
background color as needed
"""
fig, ax1 = plt.subplots(figsize=figuresize, facecolor=facecolor)
ax1.tick_params(axis = "both", colors = "w")
ax1.set_facecolor(facecolor)
fig.autofmt_xdate()
plt.ylabel(y1label, color=y1labelcolor)
ax1.set_xlabel(str(xlabel), color = "w")
ax1.set_ylabel(str(y1label))
for i in leftaxisdata: # This is to label all the lines in order to allow matplot lib to create a legend
ax1.plot(dataset.iloc[:, i])
ax1.tick_params(axis = "y")
ax1.grid(color='black', linewidth = 0.5)
ax2 = ax1.twinx() # instantiate a second axes that shares the same x-axis
color = str(y2color)
ax2.set_ylabel(str(y2label), color=color) # we already handled the x-label with ax1
plt.plot(dataset.iloc[:, list(rightaxisdata)], color=color)
ax2.tick_params(axis='y', labelcolor='w')
ax1.set_title(str(title) + " \n", color='w')
fig.tight_layout() # otherwise the right y-label is slightly clipped
plt.show()
```
```
The cell below defines a function to plot data regressed on time.
``` ```
def Plot_Regression(x, y, slope, intercept, ylabel,
title="", xlabel="Year", facecolor="0.25", data_point_type = ".", original_data_color = "C1",
time_index = [], time_index_step = 48, figuresize = (12,4), line_of_best_fit_color = "b"):
""" Plot_Regression Version 1.0:
Plots the regression line with its data with appropriate default Refinitiv colours.
facecolor (str): (Default: "0.25") This allows the user to change the background color as needed
figuresize (tuple): (Default: (12,4)) This can be changed to give graphs of different proportions. It is defaulted to a 12 by 4 (ratioed) graph
line_of_best_fit_color (str): This allows the user to change the background color as needed.
' time_index ' and ' time_index_step ' allow us to dictate the frequency of the ticks on the x-axis of our graph.
"""
fig, ax1 = plt.subplots(figsize = figuresize, facecolor = facecolor)
ax1.tick_params(axis = "both", colors = "w")
ax1.set_facecolor(facecolor)
fig.autofmt_xdate()
plt.ylabel(ylabel, color = "w")
ax1.set_xlabel(xlabel, color = "w")
ax1.plot(x, y, data_point_type, label='original data', color = original_data_color)
ax1.plot(x, intercept + slope * x, "r", label = "fitted line", color = line_of_best_fit_color)
ax1.tick_params(axis = "y")
ax1.grid(color='black', linewidth = 0.5)
ax1.set_title(title + " \n", color='w')
plt.legend()
if len(time_index) != 0:
# locs, labels = plt.xticks()
plt.xticks(numpy.arange(len(y), step = time_index_step),
[i for i in time_index[0::time_index_step]])
plt.show()
```
```
### Setup Statistics Tools
``` ```
def Single_period_Geometric_Growth(data):
""" This function returns the geometric growth of the data
entered at the frequency it is given. id est (i.e.): if
daily data is entered, a daily geometric growth rate is returned.
data (list): list including inc/float data recorded at a specific
frequency.
"""
data = list(data)
# note that we use ' len(data) - 1 ' instead of ' len(data) ' as the former already lost its degree of freedom just like we want it to.
single_period_geometric_growth = ( (data[-1]/data[0])**(1/((len(data))-1)) ) - 1
return single_period_geometric_growth
```
```
The cell below defines a function that creates and displays a table of statistics for all vectors (columns) within any two specified datasets.
``` ```
def Statistics_Table(dataset1, dataset2 = ""):
""" This function returns a table of statistics for the one or two pandas
data-frames entered, be it ' dataset1 ' or both ' dataset1 ' and ' dataset2 '.
"""
Statistics_Table_Columns = list(dataset1.columns)
if str(dataset2) != "":
[Statistics_Table_Columns.append(str(i)) for i in dataset2.columns]
Statistics_Table = pandas.DataFrame(columns = ["Mean",
"Mean of Absolute Values",
"Standard Deviation",
"Median", "Skewness",
"Kurtosis",
"Single Period Growth Geometric Average"],
index = [c for c in Statistics_Table_Columns])
def Statistics_Table_function(data):
for c in data.columns:
Statistics_Table["Mean"][str(c)] = numpy.mean(data[str(c)].dropna())
Statistics_Table["Mean of Absolute Values"][str(c)] = numpy.mean(abs(data[str(c)].dropna()))
Statistics_Table["Standard Deviation"][str(c)] = numpy.std(data[str(c)].dropna())
Statistics_Table["Median"][str(c)] = numpy.median(data[str(c)].dropna())
Statistics_Table["Skewness"][str(c)] = scipy.stats.skew(data[str(c)].dropna())
Statistics_Table["Kurtosis"][str(c)] = scipy.stats.kurtosis(data[str(c)].dropna())
if len(data[str(c)].dropna()) != 1 or data[str(c)].dropna()[0] != 0: # This if statement is needed in case we end up asking for the computation of a value divided by 0.
Statistics_Table["Single Period Growth Geometric Average"][str(c)] = Single_period_Geometric_Growth(data[str(c)].dropna())
Statistics_Table_function(dataset1)
if str(dataset2) != "":
Statistics_Table_function(dataset2)
return Statistics_Table
```
```
## USA Working Population Growth
We will use Yearly Geometric Growth rate when looking at geometric growth rates throughout this article, as outlined in Appendix 2.
We proportion of population which works. This includes only employed adults, excluding children (~ 20% of USAP) and the elderly (~ 14% of USAP). The cell below adds the United States of America's Working Population (USAWP) values to the data-frame 'df' and plots our population data this far:
``` ```
df = Translate_to_First_of_the_Month(data = ds.get_data(tickers = 'USEMPTOTO', fields = "X", start = '1950-01-01', freq = 'M') * 1000,
dataname = "USAWP (monthly data)")
```
```
Next we take the USA Working Population's first Difference, i.e.: 𝑈𝑆𝐴𝑊𝑃𝐷𝑡=𝑈𝑆𝐴𝑊𝑃𝑡−𝑈𝑆𝐴𝑊𝑃𝑡−1
``` ```
USAWPD = pandas.DataFrame(df["USAWP (monthly data)"].diff())
USAWPD.rename(columns = {'USAWP (monthly data)':'USAWPD (monthly data)'}, inplace = True)
df = pandas.merge(df, USAWPD, how = 'outer', left_index = True, right_index = True)
```
```
USA's Working Population's single period (in this case: monthly) Geometric Growth is then defined as USAWPG as per Appendix 2:
``` ```
# Create our USAWPYGG (note that due to the use of logarythms in the scipy.stats.gmean function/class, we cannot use it here without yeilding an error)
USAWPYGG = ((1 + Single_period_Geometric_Growth(data = df["USAWP (monthly data)"].dropna()) )**12)-1
df["USAWPG (monthly data)"] = df["USAWPD (monthly data)"] / df["USAWP (monthly data)"]
```
```
``` ```
Plot2ax(title = "USA Working Population (in blue) and its Growth (in orange) (USAWP & USAWPG) over the years (monthly data)",
dataset = df, leftaxisdata = [-3], rightaxisdata = [-1], y1label = "USAWP", y2label = "USAWPG", leftcolors = "w")
print("Our U.S.A.'s Working Population' Yearly Geometric Growth is " + str(USAWPYGG) + ", i.e.: " + str(USAWPYGG*100) + "%")
```
```
Our U.S.A.'s Working Population' Yearly Geometric Growth is 0.012018187303719507, i.e.: 1.2018187303719507%
## Total Compensation
As per MF's first KF, we use USA Nominal National Total Compensation (NNTC) - the nominal value of which is derived from National Nominal Personal Income Accounts. Note that this dataset is the inflows into accounts; this means that we do not need to include incomes from assets. Note also that deflators used to compute 'real' figures form 'nominal' are defined in Appendix 3.
### National Nominal Total Compensation
We can now add out National Nominal Total Compensation (NNTC) data in our data-frame and plot if for good measure.
``` ```
fields = "X",
start = '1950-01-01', freq = 'M').dropna() * 1000000000,
dataname = "NNTC (monthly data)")
df["NNTC in trillions (monthly data)"] = df["NNTC (monthly data)"] / 1000000000000
```
```
``` ```
Plot1ax(title = "U.S.A. National Nominal Total Compensation (monthly data)",
dataset = df, datasubset = [-1], ylabel = "Nominal \$ in trillions",
linescolor = "#ff9900")
```
```
As per MF, Nonfarm Business Sector Output Price Index Deflator (NBSOPID) is used to calculate USA's Real Total Compensation figures.
Since the incoming data is indexed on 2012 such that the 'RAW' and unaltered value from DataStream is 𝑅𝐴𝑊𝑁𝐵𝑆𝑂𝑃𝐼𝐷𝑡(2012−01−01) = 100 where 𝑡(2012−01−01) is the time subscript value for 01 January 2012, we define 1 / 𝑁𝐵𝑆𝑂𝑃𝐼𝐷𝑡 as 𝑅𝐴𝑊𝑁𝐵𝑆𝑂𝑃𝐼𝐷𝑡 / 𝑅𝐴𝑊𝑁𝐵𝑆𝑂𝑃𝐼𝐷𝑇. Note that the inversion of 𝑁𝐵𝑆𝑂𝑃𝐼𝐷𝑡 was only necessary as the raw values were inverted (such that 𝑁𝐵𝑆𝑂𝑃𝐼𝐷𝑡 = 𝑅𝐴𝑊𝑁𝐵𝑆𝑂𝑃𝐼𝐷𝑇 / 𝑅𝐴𝑊𝑁𝐵𝑆𝑂𝑃𝐼𝐷𝑡)(such that NBSOPIDt = RAWNBSOPIDT / RAWNBSOPIDt). NBSOPID figures are released every quarter - that's fine since we don't expect inflation to change too much from one month to the other, so quarterly data is sufficient.
``` ```
NBSOPID = ds.get_data(tickers = "USRCMNBSE",
fields = "X",
start = '1950-01-01',
freq = 'M').dropna()
```
```
``` ```
# Our NBSOPID is released quarterly, so we have to shape it into a monthly database. The function below will allow for this:
def Quarterly_To_Monthly(data, dataname, deflator = False):
data_proper, data_full_index = [], []
for i in range(0, len(data)):
[data_proper.append(float(data.iloc[j])) for j in [i,i,i]]
[data_full_index.append(datetime.datetime.strptime(data.index[i][0:8] + "01", "%Y-%m-%d") + dateutil.relativedelta.relativedelta(months =+ j)) for j in range(0,3)]
data = pandas.DataFrame(data = data_proper, columns = [dataname], index = data_full_index)
if deflator == True:
data = pandas.DataFrame(data = 1 / (data / data.iloc[-1]), columns = [dataname], index = data_full_index)
return data
```
```
``` ```
deflator = True,
dataname = "NBSOPID"),
dataname = "NBSOPID (quarterly data)")
```
```
USA National Real Total Compensation is defined as 𝑁𝑅𝑇𝐶𝑡 = 𝑁𝐵𝑆𝑂𝑃𝐼𝐷𝑡 ∗ 𝑁𝑁𝑇𝐶𝑡
``` ```
df["NRTC (monthly data)"] = df["NBSOPID (quarterly data)"] * df["NNTC (monthly data)"]
df["NRTC in trillions (monthly data)"] = df["NRTC (monthly data)"] / 1000000000000
```
```
USA National Real Total Compensation Per Worker is defined as 𝑁𝑅𝑇𝐶𝑃𝑊𝑡 = 𝑁𝑅𝑇𝐶𝑡 / 𝑈𝑆𝐴𝑊𝑃𝑡
``` ```
df["NRTCPW (monthly data)"] = df["NRTC (monthly data)"] / df["USAWP (monthly data)"]
```
```
## How does Total Compensation compare to GDP through time?
The cell below adds USA Real GDP's quarterly data (both every quarter and every month) in net and in Trillions to the Pandas data-frame 'df':
``` ```
fields = "X",
start = '1950-01-01',
freq = 'Q').dropna() * 1000000000,
dataname = "RGDP (quarterly data)")
Add_to_df(data = Quarterly_To_Monthly(data = ds.get_data(tickers = "USGDP...B",
fields = "X",
start = '1950-01-01',
freq = 'Q').dropna() * 1000000000 ,
dataname = "RGDP (monthly data)"),
dataname = "RGDP (monthly data)")
df["RGDP in trillions (quarterly data)"] = df["RGDP (monthly data)"] / 1000000000000
```
```
The cell below adds USA Real GDP per Capita (𝑅𝐺𝐷𝑃𝑃𝐶) and Per Worker (𝑅𝐺𝐷𝑃𝑃𝑊) to our Pandas data-frame (df) and plots our data concisely:
``` ```
df["RGDPPW (monthly data)"] = df["RGDP (monthly data)"] / df["USAWP (monthly data)"]
```
```
The cell below adds U.S.A. Real G.D.P. Per Worker To National Real Total Compensation Per Worker Ratio (𝑅𝐺𝐷𝑃𝑃𝑊𝑇𝑁𝑅𝑇𝐶𝑃𝑊𝑅) to our Pandas data-frame (df) and plots it:
``` ```
df["RGDPPWTNRTCPWR (monthly data)"] = df["RGDPPW (monthly data)"] / df["NRTCPW (monthly data)"]
```
```
``` ```
Plot2ax(dataset = df, leftaxisdata = [8,-2], y1labelcolor = "w", rightaxisdata = [-1], title = "U.S.A. Real G.D.P. Per Worker (yellow) To National Real Total Compensation Per Worker (blue) Ratio (green) (monthly data)", y1label = "Today's \$", y2color = "#328203", y2label = "RGDPPWTNRTCPWR", leftcolors = "w")
```
```
The graph above displays GDP and Total Compensation per worker figures as well as the ratio between the two (the latter is shown in green). It is this ratio that we use as our 𝑟r in the estimation model:
### Modelling:
The aim of our investigation is to find a ratio 𝑟r such that
Since (as per Appendix 1) we deffined 𝑟𝑡 = 𝑅𝐺𝐷𝑃𝑃𝑊𝑡 / 𝑅𝑁𝑇𝐶𝑃𝑊𝑡, 𝑟𝑡−1 values do not need to be forecasted or modeled - we have the data needed to compute 𝑟𝑡−1 = 𝑅𝐺𝐷𝑃𝑃𝑊𝑡 − 1 / 𝑅𝑁𝑇𝐶𝑃𝑊𝑡−1.
𝑟𝑡 values - however - require values of 𝑅𝐺𝐷𝑃𝑃𝑊𝑡 for month 𝑡t, which are not published: this is where an estimate of ˆrt comes in useful.
How may we model our estimate of 𝑟𝑡? The first models that come to mind are linear regression models; the below investigates the regression of USA Real GDP Per Worker To National Real Total Compensation Ratio (RGDPPWTNRTCPWR) against time (i.e.: months).
##### Regression of USA Real GDP Per Worker To National Real Total Compensation Per Worker Ratio against time
Here we use RGDPPWTNRTCPWR as 𝑟r such that 𝑦𝑡ˆ = 𝑐 + 𝛽𝑥𝑡 where 𝑦𝑡ˆ = 𝑅𝐺𝐷𝑃𝑃𝑊𝑇𝑁𝑅𝑇𝐶𝑃𝑊𝑅𝑡ˆ = 𝑟𝑡ˆ , 𝑥𝑡 is time in months, and 𝑐 & 𝛽 are computed to reduce the error of this model (i.e.: to reduce the 'error' term - ϵ - in 𝑦𝑡 = 𝑐 + 𝛽𝑥𝑡 + 𝜖𝑡 where 𝑦𝑡 = 𝑅𝐺𝐷𝑃𝑃𝑊𝑇𝑁𝑅𝑇𝐶𝑃𝑊𝑅𝑡 = 𝑟𝑡) via Ordinary Least Square estimation:
``` ```
# First: Fit the trend
x = numpy.array([numpy.float64(i) for i in range(0, len(df["RGDPPWTNRTCPWR (monthly data)"].dropna()))])
y = numpy.array([numpy.float64(i) for i in df["RGDPPWTNRTCPWR (monthly data)"].dropna()])
(slope, intercept, r_value, p_value, std_err) = (scipy.stats.linregress(x = x, y = y))
df["RGDPPWTNRTCPWR regressed on months"] = (intercept + slope * x) - (df["RGDPPWTNRTCPWR (monthly data)"].dropna() * 0)
df["RGDPPWTNRTCPWR regressed on months' errors"] = df["RGDPPWTNRTCPWR regressed on months"] - df["RGDPPWTNRTCPWR (monthly data)"].dropna()
# Second: Plot it:
Plot_Regression(title = "U.S.A. Real G.D.P. Per Worker To National Real Total Compensation Per Worker Ratio (monthly data) (in orange) regressed on time (in blue)",
x = x, y = y, slope = slope, intercept = intercept,
data_point_type = "-", ylabel = "RGDPPWTNRTCPWR",
time_index_step = 12*10, figuresize = (16,4),
time_index = df["RGDPPWTNRTCPWR (monthly data)"].dropna().index)
# Third: Evaluate it:
Statistics_Table(dataset1 = df).loc[["RGDPPWTNRTCPWR (monthly data)",
"RGDPPWTNRTCPWR regressed on months",
"RGDPPWTNRTCPWR regressed on months' errors"],
["Mean of Absolute Values"]]
```
```
Mean of Absolute Values RGDPPWTNRTCPWR (monthly data) 0.907876 RGDPPWTNRTCPWR regressed on months 0.907876 RGDPPWTNRTCPWR regressed on months' errors 0.0241528
The statistics table shows a healthily low absolute (as in: ignoring the sign) error mean value for RGDPPWTNRTCPWR's regression model of 0.0242910.024291.
We seem to have gathered evidence that a time-linear model may be a good way to estimate 𝑟r and then monthly GDP. Remember that we will measure the effectiveness of our models by how they estimate GDP values at lower frequencies than they are published by official bodies (here: monthly, using monthly released Total Compensation data); an accurate estimation of our ratio is therefore paramount. One may thus attempt to look at estimates produced by methods other than straight forward linear regressions, e.g.: the Holt-Winters method. Let's investigate that method too:
#### Holt-Winters method
All estimation methods are to be used recursively. The Holt-Winters method makes sure that forecasts are heavily influenced by latest values by design. It offers better (i.e.: lower) model error (see model fit graph below) than our linear regressions too.
Nota Bene (N.B.): A non multiplicative Holt-Winters method is used here as computational limits mean that multiplicative Holt-Winters methods result in very many 'NaN' estimates that are unusable.
The cell below creates recursive RGDPPWTNRTCPWR Holt-Winters exponential smoothing estimates (RGDPPWTNRTCPWRHW); this means that we are using 𝑅𝐺𝐷𝑃𝑃𝑊𝑇𝑁𝑅𝑇𝐶𝑃𝑊𝑅𝑡−1 as 𝑟𝑡−1 and 𝑅𝐺𝐷𝑃𝑃𝑊𝑇𝑁𝑅𝑇𝐶𝑃𝑊𝑅𝐻𝑊𝑡 as 𝑟𝑡ˆ :
``` ```
# The following for loop will throw errors informing us that the unspecified Holt-Winters Exponential Smoother
# parameters will be chosen by the default ' statsmodels.tsa.api.ExponentialSmoothing ' optimal ones.
# This is preferable and doesn't need to be stated for each iteration in the loop.
warnings.simplefilter("ignore")
RGDPPWTNRTCPWRHW_model_fit = statsmodels.tsa.api.ExponentialSmoothing(df["RGDPPWTNRTCPWR (monthly data)"].dropna(),
seasonal_periods = 12,
damped=True).fit(use_boxcox=True)
df["RGDPPWTNRTCPWRHW fitted values (monthly data)"] = RGDPPWTNRTCPWRHW_model_fit.fittedvalues
df["RGDPPWTNRTCPWRHW forecasts (monthly data)"] = RGDPPWTNRTCPWRHW_model_fit.forecast(12*4)
df["RGDPPWTNRTCPWRHW fitted values' real errors (monthly data)"] = df["RGDPPWTNRTCPWRHW fitted values (monthly data)"] - df["RGDPPWTNRTCPWR (monthly data)"]
```
```
``` ```
# See the HW model's forecasts:
RGDPPWTNRTCPWRHW_model_fit.forecast(12)
```
```
01/04/2020 1.14311 01/05/2020 1.14314 01/06/2020 1.14316 01/07/2020 1.14317 01/08/2020 1.14318 01/09/2020 1.14318 01/10/2020 1.14318 01/11/2020 1.14318 01/12/2020 1.14319 01/01/2021 1.14319 01/02/2021 1.14319 01/03/2021 1.14319
Freq: MS, dtype: float64
``` ```
# Plot the newly created Exponential Smoothing data
fig1, ax1 = plt.subplots(figsize=(12,4), facecolor="0.25")
df["RGDPPWTNRTCPWR (monthly data)"].dropna().plot(style = '-',
color = "blue",
legend = True).legend(["RGDPPWTNRTCPWR (monthly data)"])
RGDPPWTNRTCPWRHW_model_fit.fittedvalues.plot(style = '-',
color = "C1",
label = "RGDPPWTNRTCPWRHW model fit",
legend = True)
RGDPPWTNRTCPWRHW_model_fit.forecast(12).plot(style = '-',
color = "#328203",
label = "RGDPPWTNRTCPWRHW model forecast",
legend = True)
ax1.set_facecolor("0.25")
ax1.tick_params(axis = "both", colors = "w")
plt.ylabel("Ratio", color = "w")
ax1.set_xlabel("Years", color = "w")
ax1.grid(color = "black", linewidth = 0.5)
ax1.set_title("Forecasting U.S.A. Real G.D.P. Per Worker To Real Income Per Worker Ratio (RGDPPWTNRTCPWR) of properties" +
"\nusing the Holt-Winters method (HW) (monthly data) (forecasts in green)" +
" \n", color='w')
ax1.legend()
plt.show()
warnings.simplefilter("default") # We want our program to let us know of warnings from now on; they were only disabled for the for loop above.
Statistics_Table(dataset1 = df).loc[["RGDPPWTNRTCPWR regressed on months' errors",
"RGDPPWTNRTCPWRHW fitted values' real errors (monthly data)"],
["Mean of Absolute Values"]]
```
```
Mean of Absolute Values RGDPPWTNRTCPWR regressed on months' errors 0.0241528 RGDPPWTNRTCPWRHW fitted values' real errors (monthly data) 0.00773767
The Holt-Winters model's estimates - 𝑟ˆr^ - (in yellow) closely follow the realised RGDPPWTNRTCPWR values. This only holds untrue at the start (approximately for the first 2 years) when the model needs training. The (short) green line shows monthly forecasts for the next 4 years computed with this Holt-Winters model.
The statistics table clearly hows that the Holt -Winters model performs much better than the time-linear regression model: its errors are - on average - a lot lower (i.e: 0.0077352<0.024291).
## Backtesting our Method
The Python function defined in the cell below returns the In-Sample Recursive Prediction values of ' data ' using the Holt-Winters exponential smoother Plus it then provides forecasts for One Out-Of-Sample period (thus the acronym 'ISRPHWPOOOF'):
``` ```
def ISRPHWPOOOF(data_set, dataset_name, start, result_name, enforce_cap = False, cap = 1.1):
""" This is a function created specifically for this article/study. returns the In-Sample Recursive
Prediction values of ' data ' using the Holt-Winters exponential smoother Plus it then provides
forecasts for One Out-Of-Sample period (thus the acronym 'ISRPHWPOOOF').
data_set (pandas dataframe): Pandas data-frame where the data is stored
dataset_name (str): Our data is likely to be in a pandas dataframe with dates
for indexes in order to alleviate any date sinking issues. As a result, this
line allows the user to specify the column within such a dataset to use.
start (str): The start of our in-sample recursive prediction window. This value can be changed.
result_name (str): name given to the result column.
enforce_cap (bool): Set to False by default. The H-W model sometimes returns extremely large
values that do not fit in our set. This is a convergence issue. Applying an arbitrary cap allows
us to account for this.
cap (int/float): Set to ' 1.1 ' by default. This is the cap applied to the returned values if
' enforce_cap ' is set to True.
"""
# The following for loop will throw errors informing us that the unspecified Holt-Winters
# Exponential Smoother parameters will be chosen by the default
# ' statsmodels.tsa.api.ExponentialSmoothing ' optimal ones.
# This is preferable and doesn't need to be stated for each iteration in the loop.
warnings.simplefilter("ignore")
# We create an empty list (to get appended/populated)
# for our dataset's forecast using the Holt-Winters exponential smoother.
data_model_in_sample_plus_one_forecast = []
# Empty lists to be populated by our forecasts (in- or out-of-sample)
value, index = [], []
# the ' len(...) ' function here returns the number of rows in our dataset from our starting date till its end.
for i in range(0,len(data_set[dataset_name].loc[pandas.Timestamp(start):].dropna())):
# This line defines ' start_plus_i ' as our starting date plus i months
start_plus_i = str(datetime.datetime.strptime(start, "%Y-%m-%d") + dateutil.relativedelta.relativedelta(months=+i))[:10]
HWESi = statsmodels.tsa.api.ExponentialSmoothing(data_set[dataset_name].dropna().loc[:pandas.Timestamp(start_plus_i)],
trend = 'add', seasonal_periods = 12,
damped = True).fit(use_boxcox = True).forecast(1)
# This(following) outputs a forecast for one period ahead (whether in-sample or out of sample).
# Since we start at i = 0, this line provides a HW forecast for i at 1; similarly, at the
# last i (say T), it provides the first and only out-of-sample forecast (where i = T+1).
data_model_in_sample_plus_one_forecast.append(HWESi)
try:
if enforce_cap == True:
if float(str(HWESi)[14:-25]) > cap:
value.append(float(str(HWESi)[14:-25]))
else:
value.append(numpy.nan) # If the ratio
else:
value.append(float(str(HWESi)[14:-25]))
except:
# This adds NaN (Not a Number) to the list ' values ' if there is no value to add.
# The statsmodels.tsa.api.ExponentialSmoothing function sometimes (rarely) outputs NaNs.
value.append(numpy.nan)
finally:
# This adds our indecies (dates) to the list ' index '
index.append(str(HWESi)[:10])
# We want our program to let us know of warnings from now on; they were only disabled for the for loop above.
warnings.simplefilter("default")
return pandas.DataFrame(data = value, index = index, columns = [result_name])
```
```
We will back-test our models from January 1990
``` ```
start = "1990-01-01"
```
```
``` ```
df["RGDPPWTNRTCPWRHW fitted values' real errors from " + str(start) + " (monthly data)"] = df["RGDPPWTNRTCPWRHW fitted values (monthly data)"].loc[start:] - df["RGDPPWTNRTCPWR (monthly data)"].loc[start:]
```
```
``` ```
# Model:
RGDPPWTNRTCPWRHW_model_in_sample_plus_one_forecast = ISRPHWPOOOF(data_set = df,
dataset_name = "RGDPPWTNRTCPWR (monthly data)",
start = start,
result_name = "RGDPPWTNRTCPWRHW model in sample plus one forecast")
df = pandas.merge(df, RGDPPWTNRTCPWRHW_model_in_sample_plus_one_forecast,
how = "outer", left_index = True, right_index = True)
# Model's errors:
RGDPPWTNRTCPWRHW_model_in_sample_plus_one_forecast_error = pandas.DataFrame(df["RGDPPWTNRTCPWR (monthly data)"] - df["RGDPPWTNRTCPWRHW model in sample plus one forecast"],
columns = ["RGDPPWTNRTCPWRHW model in sample plus one forecast error"])
df = pandas.merge(df, RGDPPWTNRTCPWRHW_model_in_sample_plus_one_forecast_error.dropna(),
how = "outer", left_index = True, right_index = True)
```
```
Remember that the ratio 𝑟𝑡rt is𝑅𝐺𝐷𝑃𝑃𝑊𝑡 / 𝑅𝑁𝑇𝐶𝑃𝑊𝑡RGDPPWtRNTCPWt, but 'today' (i.e.: at time 𝑡t), there is no 𝑅𝐺𝐷𝑃𝑃𝑊𝑡 value we can use (since only quarterly figures are published), this is why we made a forecast. As a result, we will differentiate between 𝑟𝑡ˆ as including the forecast (actually used as an estimate) at time 𝑡t as ' r_f ' ('f' for forecasted) and 𝑟𝑡−1 as ' r ' which does not use forecasts:
``` ```
r = df["RGDPPWTNRTCPWR (monthly data)"].dropna()
r_f = df["RGDPPWTNRTCPWRHW model in sample plus one forecast"].dropna()
df["r"] = r
df["r_f"] = r_f
```
```
``` ```
Plot1ax(df, datasubset = [-1,-2], datarange = ["1990-01-01",-1], ylabel = "Ratio",
title = "Real G.D.P. Per Worker To National Real Total Compensation Per Worker Ratio realised (r) and\nmodeled estimates using forecasts (r_f) starting on Jan. 1990",)
```
```
Modeled estimates (r_f) follow the realised 𝑟r values (r) so closely that it is difficult to differentiate them.
Remember also that 𝑅𝐺𝐷𝑃𝑃𝑊𝑡 = ( 𝑟𝑡 / 𝑟𝑡−1 )∗ ( 𝑅𝑁𝑇𝐶𝑃𝑊𝑡 / 𝑅𝑁𝑇𝐶𝑃𝑊𝑡−1 ) ∗ 𝑅𝐺𝐷𝑃𝑃𝑊𝑡−1, thus: RGDPPW estimate = 𝑅𝐺𝐷𝑃𝑃𝑊𝑡ˆ = (𝑟𝑡ˆ / 𝑟𝑡ˆ−1 ) ∗ ( 𝑅𝑁𝑇𝐶𝑃𝑊𝑡 / 𝑅𝑁𝑇𝐶𝑃𝑊𝑡−1 ) ∗ 𝑅𝐺𝐷𝑃𝑃𝑊𝑡−1
``` ```
RGDPPW_estimate = (r_f / r.shift(1)) * (df["NRTCPW (monthly data)"] / df["NRTCPW (monthly data)"].shift(1)) * df["RGDPPW (monthly data)"].shift(1)
df["RGDPPW estimates"] = RGDPPW_estimate.dropna()
```
```
Yet again, looking at the realiations and model outputs on the same graph(s), it is difficult to see the model outputs because they follow realisations so well; the errors' graph and table shines a light on how small the error is:
``` ```
Plot1ax(df, datasubset = [12,-1], datarange = ["1990-01-01",-1], ylabel = "Today's \$", title = "Real G.D.P Per Worker Realised and Estimated Figures starting on Jan. 1990")
df["RGDPPW estimates' errors"] = df["RGDPPW (monthly data)"] - df["RGDPPW estimates"]
df["RGDPPW estimates' proportional errors"] = df["RGDPPW estimates' errors"] / df["RGDPPW (monthly data)"]
Plot2ax(dataset = df, leftaxisdata = [-2], rightaxisdata = [-1], title = "Real G.D.P Per Worker Estimates' Real (blue) and Proportional (orange) Errors",
y1label = "RGDPPW estimates' real errors", y2label = "RGDPPW estimates' proportional errors", leftcolors = "w")
```
```
The difference between the lowest and 2nd lowest (as well as the highest and 2nd highest) troughs (and peaks) is less significant in the Proportional Error graph than in the straight forward Error graph.
Since RGDP figures are only released quarterly, it is best to investigate errors at each quarter
``` ```
df["RGDPPW estimates' errors every quarter"] = df["RGDP (quarterly data)"] - df["RGDPPW estimates"]
Statistics_Table(dataset1 = df).loc[["RGDPPW estimates' errors", "RGDPPW estimates' proportional errors"],["Mean of Absolute Values"]]
```
```
Mean of Absolute Values RGDPPW estimates' errors 592.39 RGDPPW estimates' proportional errors 0.00666185
Our model gave a healthily low error of 0.662307%.
## Conclusion
We may now compute monthly estimates of G.D.P. figures using an Income Approach; since:
then:
``` ```
df["RGDP estimates"] = df["RGDPPW estimates"] * df["USAWP (monthly data)"]
df[["RGDP (monthly data)", "RGDP estimates"]].dropna()
```
```
RGDP (monthly data) RGDP estimates 01/02/1990 5.87E+12 5.81E+12 01/03/1990 5.87E+12 5.86E+12 01/04/1990 5.87E+12 5.91E+12 01/05/1990 5.96E+12 5.85E+12 01/06/1990 5.96E+12 5.93E+12 ... ... ... 01/11/2019 2.17E+13 2.17E+13 01/12/2019 2.17E+13 2.17E+13 01/01/2020 2.17E+13 2.19E+13 01/02/2020 2.15E+13 2.19E+13 01/03/2020 2.15E+13 2.13E+13
It would be redundant testing the validity of these figures since we already did so when investigating Real GDP Per Worker estimates above.
Note also that for values today - at time 𝑇 - and/or close to today (e.g.: next month), 𝑅𝐺𝐷𝑃≈𝐺𝐷𝑃, and population growth is negligable, such that 𝑈𝑆𝐴𝑊𝑃𝑇 ≈ 𝑈𝑆𝐴𝑊𝑃𝑇+1, then:
It is therefore possible to use our method to predict future GDP values on a monthly basis.
# References
You can find more detail regarding the DSWS API and related technologies for this article from the following resources:
For any question related to this example or Eikon Data API, please use the Developers Community Q&A Forum.
## Literature:
P. R. Winters (1960). Forecasting sales by exponentially weighted moving averages. Management Science | 12,002 | 39,626 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.888685 |
http://telesaudewp.azurewebsites.net/wp-content/themes/zenwater/index5.php?yhsw=function-apply | 1,586,338,778,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371810807.81/warc/CC-MAIN-20200408072713-20200408103213-00080.warc.gz | 170,358,957 | 9,940 | Function apply
## Function apply
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Cross Apply And Outer Apply in SQL Server | 1,312 | 5,803 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | latest | en | 0.615279 |
https://mimicrobots.com/blogs/news/what-do-analog-and-digital-mean-anyway | 1,721,582,399,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517747.98/warc/CC-MAIN-20240721152016-20240721182016-00858.warc.gz | 331,918,567 | 19,547 | # What do Analog and Digital Mean Anyway?
We all talk about "digital" technology, but do you really know what it means? Most people would say that "digital" is synonymous with "electronic" or "computerized", but in fact there is such a thing as an analog computer and a digital mechanical device, and it's not always clear which one is better.
Ok, then what does "analog" actually mean?
According to Dictionary.com Analog, when used as an adjective means:
1. of or relating to a mechanism that represents data by measurement of a continuous physical variable, as voltage or pressure.
We can demonstrate analog arithmetic using measuring cups. Imagine how we could add 1 + 2 using water. First, fill one measuring cup with 1 cup of water. Next fill a second measuring cup with 2 cups of water. Dump them both into a third cup, and presto! You've just added 1 + 2 to make 3!
There's even an advantage to doing math this way. Analog has a capability that digital mechanisms don't. An analog measurement can be infinitesimally small. If you were able to look closely enough you'd be able to tell the difference between 1 cup, 1.000001 cups, or even 1.00000000...001 cups all the way out to infinity zeros. Digital measurements are always limited to the defined size of a "bit", which we'll get to in a moment.
Unfortunately, Analog's biggest advantage is also it's biggest flaw. Did you really add 1 cup plus 2 cups just now? Nope. You added something close, but not quite exactly what you were after. Perhaps 1.0001 cups, or .999 cups. Because analog measurements can be infinitesimally small they are also NEVER perfectly accurate.
Ok, then what's Digital?
One of many definitions of digital on Dictionary.com is:
involving or using numerical digits expressed in a scale of notation, usually in the binary system, to represent discretely all variables occurring in a problem.
This means that the accuracy of your digital measurement is limited to the number of discrete "bits" that you have. Let's try it. Imagine you want to add 1 + 2 again, but this time digitally. For our bits lets use pennies. Place one penny to the side, then next to it place a stack of two pennies. Now combine them into one stack, and presto! you've digitally added 1 + 2 to make 3 pennies. Now, because this is digital we can't add 1.1 pennies or .9 pennies, we can only do arithmetic accurate to our smallest unit, a "bit". On the other hand, we know for absolutely certain and exactly 1 penny plus exactly 2 pennies equals exactly 3 pennies.
Great! So digital it is then!
Not so fast! Nature is rarely digital (except on the atomic scale). For example, nobody weights exactly 100 lbs, and (more relevant for our robots) voltage is never EXACTLY 5 volts. For the most part, the world around us is analog. This is where our "signal conditioning" circuits come it. Enter the "Analog to Digital Converter".
Consider for a moment the potentiometers that we use to measure mimicArm's controller position.
These devices contain a resistor that has a constant resistance (10,000 Ohms in our case). Because we supply the potentiometer with 5 Volts, this means that one side of the potentiometer is always at 5 Volts, while the other is always at 0 Volts. The magic happens in the middle. The Potentiometer includes a "Slider" that moves along the resistor as the potentiometer rotates. This means that the resistance on either side of the slider can change but will always add up to 10,000 ohms. Knowing this, Ohms Law allows us to calculate these resistances based on the measured voltage at the slider (Ohms Law is a bit beyond the scope of this post, so ask you science teacher). Because our resistor also has a predictable resistance for a given length, we can also determine the sliders position based on the sliders voltage.
As we discussed, the potentiometers Voltage is inherently analog, so we need a bit of electromagic to read this voltage digitally. The diagram below shows how we do this. Our Analog to Digital Converter (ADC) takes the analog potentiometer voltage and converts it to a string of ones and zeros that represent the voltage value measured. Our mimicArm robot controller is measuring this voltage 200 times per second and sending commands to our robots servos to match the potentiometer position.
So I guess we need both analog and digital don't we?
Sure do, especially in robotics! Any time you're interacting with the physical world you're probably going to be dealing with an analog signal somewhere along the way. Now that you know a little bit about it, what other analog measurements could you imagine making? | 1,039 | 4,631 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-30 | latest | en | 0.963778 |
https://www.codeproject.com/Articles/16508/AI-Neural-Network-for-beginners-Part-2-of-3?fid=361675&df=90&mpp=25&noise=3&prof=True&sort=Position&view=Quick&spc=Relaxed&select=2041726&fr=26 | 1,500,939,253,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424931.1/warc/CC-MAIN-20170724222306-20170725002306-00595.warc.gz | 771,841,311 | 33,587 | 13,048,406 members (110,984 online)
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Posted 24 Nov 2006
# AI : Neural Network for beginners (Part 2 of 3)
, 29 Jan 2007
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AI : An Introduction into Neural Networks (Multi-layer networks / Back Propagation)
## Introduction
1. Part 1 : Is an introduction into Perceptron networks (single layer neural networks).
2. Part 2 : This one, is about multi layer neural networks, and the back propagation training method to solve a non linear classification problem such as the logic of an XOR logic gate. This is something that a Perceptron can't do. This is explained further within this article.
3. Part 3 : Will be about how to use a genetic algorithm (GA) to train a multi layer neural network to solve some logic problem.
### Summary
This article will show how to use a multi-layer neural network to solve the XOR logic problem.
## A Brief Recap (From part 1 of 3)
Before we commence with the nitty gritty of this new article which deals with multi layer Neural Networks, let just revisit a few key concepts. If you haven't read Part 1, perhaps you should start there.
### Perceptron Configuration ( Single layer network)
The inputs `(x1,x2,x3..xm)` and connection weights` (w1,w2,w3..wm)` shown below are typically real values, both positive (+) and negative (-).
The perceptron itself, consists of weights, the summation processor, an activation function, and an adjustable threshold processor (called bias here after).
For convenience, the normal practice is to treat the bias as just another input. The following diagram illustrates the revised configuration.
The bias can be thought of as the propensity (a tendency towards a particular way of behaving) of the perceptron to fire irrespective of it's inputs. The perceptron configuration network shown above fires if the weighted sum > 0, or if you have into maths type explanations
So that's the basic operation of a perceptron. But we now want to build more layers of these, so let's carry on to the new stuff.
## So Now The New Stuff (More layers)
From this point on, anything that is being discussed relates directly to this article's code.
In the summary at the top, the problem we are trying to solve was how to use a multi-layer neural network to solve the XOR logic problem. So how is this done. Well it's really an incremental build on what Part 1 already discussed. So let's march on.
What does the XOR logic problem look like? Well, it looks like the following truth table:
Remember with a single layer (perceptron) we can't actually achieve the XOR functionality, as it is not linearly separable. But with a multi-layer network, this is achievable.
## What Does The New Network Look Like
The new network that will solve the XOR problem will look similar to a single layer network. We are still dealing with inputs / weights / outputs. What is new is the addition of the hidden layer.
As already explained above, there is one input layer, one hidden layer and one output layer.
It is by using the inputs and weights that we are able to work out the activation for a given node. This is easily achieved for the hidden layer as it has direct links to the actual input layer.
The output layer, however, knows nothing about the input layer as it is not directly connected to it. So to work out the activation for an output node we need to make use of the output from the hidden layer nodes, which are used as inputs to the output layer nodes.
This entire process described above can be thought of as a pass forward from one layer to the next.
This still works like it did with a single layer network; the activation for any given node is still worked out as follows:
Where (wi is the weight(i), and Ii is the input(i) value)
You see it the same old stuff, no demons, smoke or magic here. It's stuff we've already covered.
So that's how the network looks/works. So now I guess you want to know how to go about training it.
## Types Of Learning
There are essentially 2 types of learning that may be applied, to a Neural Network, which is "Reinforcement" and "Supervised"
### Reinforcement
In Reinforcement learning, during training, a set of inputs is presented to the Neural Network, the Output is 0.75, when the target was expecting 1.0.
The error (1.0 - 0.75) is used for training ('wrong by 0.25').
What if there are 2 outputs, then the total error is summed to give a single number (typically sum of squared errors). Eg "your total error on all outputs is 1.76"
Note that this just tells you how wrong you were, not in which direction you were wrong.
Using this method we may never get a result, or it could be a case of 'Hunt the needle'.
NOTE : Part 3 of this series will be using a GA to train a Neural Network, which is Reinforcement learning. The GA simply does what a GA does, and all the normal GA phases to select weights for the Neural Network. There is no back propagation of values. The Neural Network is just good or just bad. As one can imagine, this process takes a lot more steps to get to the same result.
### Supervised
Not just 'how wrong' it was, but 'in what direction it was wrong' like 'Hunt the needle' but where you are told 'North a bit', 'West a bit'.
So you get, and use, far more information in Supervised Learning, and this is the normal form of Neural Network learning algorithm. Back Propagation (what this article uses, is Supervised Learning)
## Learning Algorithm
In brief, to train a multi-layer Neural Network, the following steps are carried out:
• Start off with random weights (and biases) in the Neural Network
• Try one or more members of the training set, see how badly the output(s) are compared to what they should be (compared to the target output(s))
• Jiggle weights a bit, aimed at getting improvement on outputs
• Now try with a new lot of the training set, or repeat again,
jiggling weights each time
• Keep repeating until you get quite accurate outputs
This is what this article submission uses to solve the XOR problem. This is also called "Back Propagation" (normally called BP or BackProp)
Backprop allows you to use this error at output, to adjust the weights arriving at the output layer, but then also allows you to calculate the effective error 1 layer back, and use this to adjust the weights arriving there, and so on, back-propagating errors through any number of layers.
The trick is the use of a sigmoid as the non-linear transfer function (which was covered in Part 1. The sigmoid is used as it offers the ability to apply differentiation techniques.
Because this is nicely differentiable – it so happens that
Which in context of the article can be written as
delta_outputs[i] = outputs[i] * (1.0 - outputs[i]) * (targets[i] - outputs[i])
It is by using this calculation that the weight changes can be applied back through the network.
### Things To Watch Out For
Valleys: Using the rolled ball metaphor, there may well be valleys like this, with steep sides and a gently sloping floor. Gradient descent tends to waste time swooshing up and down each side of the valley (think ball!)
So what can we do about this. Well we add a momentum term, that tends to cancel out the back and forth movements and emphasizes any consistent direction, then this will go down such valleys with gentle bottom-slopes much more successfully (faster)
## Starting The Training
This is probably best demonstrated with a code snippet from the article's actual code:
```/// <summary>
/// The main training. The expected target values are passed in to this
/// method as parameters, and the <see cref="NeuralNetwork">NeuralNetwork</see>
/// is then updated with small weight changes, for this training iteration
/// This method also applied momentum, to ensure that the NeuralNetwork is
/// nurtured into proceeding in the correct direction. We are trying to avoid valleys.
/// If you don't know what valleys means, read the articles associated text
/// </summary>
/// <param name="target">A double[] array containing the target value(s)</param>
private void train_network(double[] target)
{
//get momentum values (delta values from last pass)
double[] delta_hidden = new double[nn.NumberOfHidden + 1];
double[] delta_outputs = new double[nn.NumberOfOutputs];
// Get the delta value for the output layer
for (int i = 0; i < nn.NumberOfOutputs; i++)
{
delta_outputs[i] =
nn.Outputs[i] * (1.0 - nn.Outputs[i]) * (target[i] - nn.Outputs[i]);
}
// Get the delta value for the hidden layer
for (int i = 0; i < nn.NumberOfHidden + 1; i++)
{
double error = 0.0;
for (int j = 0; j < nn.NumberOfOutputs; j++)
{
error += nn.HiddenToOutputWeights[i, j] * delta_outputs[j];
}
delta_hidden[i] = nn.Hidden[i] * (1.0 - nn.Hidden[i]) * error;
}
// Now update the weights between hidden & output layer
for (int i = 0; i < nn.NumberOfOutputs; i++)
{
for (int j = 0; j < nn.NumberOfHidden + 1; j++)
{
//use momentum (delta values from last pass),
//to ensure moved in correct direction
nn.HiddenToOutputWeights[j, i] += nn.LearningRate * delta_outputs[i] * nn.Hidden[j];
}
}
// Now update the weights between input & hidden layer
for (int i = 0; i < nn.NumberOfHidden; i++)
{
for (int j = 0; j < nn.NumberOfInputs + 1; j++)
{
//use momentum (delta values from last pass),
//to ensure moved in correct direction
nn.InputToHiddenWeights[j, i] += nn.LearningRate * delta_hidden[i] * nn.Inputs[j];
}
}
}```
## So Finally The Code
Well, the code for this article looks like the following class diagram (It's Visual Studio 2005 C#, .NET v2.0)
The main classes that people should take the time to look at would be :
• `NN_Trainer_XOR `: Trains a Neural Network to solve the XOR problem
• `TrainerEventArgs `: Training event args, for use with a GUI
• `NeuralNetwork `: A configurable Neural Network
• `NeuralNetworkEventArgs` : Training event args, for use with a GUI
• `SigmoidActivationFunction` : A static method to provide the sigmoid activation function
The rest are a GUI I constructed simply to show how it all fits together.
## Code Demos
The DEMO application attached has 3 main areas which are described below:
### LIVE RESULTS Tab
It can be seen that this has very nearly solved the XOR problem (You will probably never get it 100% accurate)
### TRAINING RESULTS Tab
Viewing the training phase target/outputs together
Viewing the training phase errors
### TRAINED RESULTS Tab
Viewing the trained target/outputs together
Viewing the trained errors
It is also possible to view the Neural Networks final configuration using the "View Neural Network Config" button. If people are interested in what weights the Neural Network ended up with, this is the place to look.
## What Do You Think ?
That's it. I would just like to ask, if you liked the article, please vote for it.
## Points of Interest
I think AI is fairly interesting, that's why I am taking the time to publish these articles. So I hope someone else finds it interesting, and that it might help further someone's knowledge, as it has my own.
Anyone that wants to look further into AI type stuff, that finds the content of this article a bit basic should check out Andrew Krillovs articles, at Andrew Krillov CP articles as his are more advanced, and very good. In fact anything Andrew seems to do, is very good.
## History
• v1.0 24/11/06
## Bibliography
• Artificial Intelligence 2nd edition, Elaine Rich / Kevin Knight. McGraw Hill Inc.
• Artificial Intelligence, A Modern Approach, Stuart Russell / Peter Norvig. Prentice Hall.
## Share
Software Developer (Senior) United Kingdom
I currently hold the following qualifications (amongst others, I also studied Music Technology and Electronics, for my sins)
- MSc (Passed with distinctions), in Information Technology for E-Commerce
- BSc Hons (1st class) in Computer Science & Artificial Intelligence
Both of these at Sussex University UK.
Award(s)
I am lucky enough to have won a few awards for Zany Crazy code articles over the years
• Microsoft C# MVP 2016
• Codeproject MVP 2016
• Microsoft C# MVP 2015
• Codeproject MVP 2015
• Microsoft C# MVP 2014
• Codeproject MVP 2014
• Microsoft C# MVP 2013
• Codeproject MVP 2013
• Microsoft C# MVP 2012
• Codeproject MVP 2012
• Microsoft C# MVP 2011
• Codeproject MVP 2011
• Microsoft C# MVP 2010
• Codeproject MVP 2010
• Microsoft C# MVP 2009
• Codeproject MVP 2009
• Microsoft C# MVP 2008
• Codeproject MVP 2008
• And numerous codeproject awards which you can see over at my blog
## You may also be interested in...
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Excellent! merlin98117-May-07 4:31 merlin981 17-May-07 4:31
Thank you for these detailed articles. They are very helpful and informative. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ I get my developer tools from Merlin A.I. Soft I get my news and jokes from Daily Roundup ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
license? famousj.dejazzd.com17-Jan-07 8:32 famousj.dejazzd.com 17-Jan-07 8:32
Re: license? Sacha Barber18-Jan-07 2:19 Sacha Barber 18-Jan-07 2:19
most of the time it won't converged.. can explain? f224-Dec-06 8:34 f2 24-Dec-06 8:34
Re: most of the time it won't converged.. can explain? Sacha Barber25-Dec-06 22:14 Sacha Barber 25-Dec-06 22:14
Re: most of the time it won't converged.. can explain? f26-Jan-07 18:42 f2 6-Jan-07 18:42
Re: most of the time it won't converged.. can explain? Sacha Barber29-Jan-07 21:44 Sacha Barber 29-Jan-07 21:44
Momentum and stuck networks jw97017010-Dec-06 11:40 jw970170 10-Dec-06 11:40
Re: Momentum and stuck networks Sacha Barber10-Dec-06 20:31 Sacha Barber 10-Dec-06 20:31
About the back propagation JemimahPowell9-Dec-06 11:30 JemimahPowell 9-Dec-06 11:30
Re: About the back propagation Sacha Barber9-Dec-06 20:27 Sacha Barber 9-Dec-06 20:27
Cool! azolotko7-Dec-06 22:41 azolotko 7-Dec-06 22:41
Re: Cool! Sacha Barber7-Dec-06 23:31 Sacha Barber 7-Dec-06 23:31
Today I just started learning AI ... SeongKim6-Dec-06 18:52 SeongKim 6-Dec-06 18:52
Re: Today I just started learning AI ... Sacha Barber6-Dec-06 20:21 Sacha Barber 6-Dec-06 20:21
Re: Today I just started learning AI ... Sacha Barber11-Dec-06 3:05 Sacha Barber 11-Dec-06 3:05
Great work sabah-u-din28-Nov-06 10:50 sabah-u-din 28-Nov-06 10:50
Re: Great work Sacha Barber28-Nov-06 11:05 Sacha Barber 28-Nov-06 11:05
SNNs? Zero DeHero27-Nov-06 23:50 Zero DeHero 27-Nov-06 23:50
Re: SNNs? Sacha Barber28-Nov-06 0:35 Sacha Barber 28-Nov-06 0:35
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The conveyor body consists of two profile sections , More information See the other products , With its chain conveyor RFKG Heavy-Duty Line, , View PDF catalogs and other online documentation *Prices are pre-tax, exclude delivery charges and customs duties, and do not include additional charges for installation or activation options . | 2,009 | 9,262 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-31 | latest | en | 0.765471 |
https://www.bartleby.com/solution-answer/chapter-44-problem-37e-elementary-geometry-for-college-students-7e-7th-edition/9781337614085/for-exercises-36-and-37-efis-the-median-of-trapezoid-abcd-in-the-figure-above-suppose-that/02f6d039-757c-11e9-8385-02ee952b546e | 1,571,330,610,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986675409.61/warc/CC-MAIN-20191017145741-20191017173241-00209.warc.gz | 831,881,031 | 59,022 | Chapter 4.4, Problem 37E
### Elementary Geometry For College St...
7th Edition
Alexander + 2 others
ISBN: 9781337614085
Chapter
Section
### Elementary Geometry For College St...
7th Edition
Alexander + 2 others
ISBN: 9781337614085
Textbook Problem
# For exercises 36 and 37, E F ¯ is the median of trapezoid ABCD in the figure above.Suppose that E M = 7.1 a n d M F = 3.5 Find: (a) AB (b) EF (c) DC (d) Whether E F = 1 2 ( A B + D C ) Exercises 35-37
To determine
(a)
To Find:
AB
Explanation
Consider the following diagram.
Given: EM = 7.1 and MF = 3.5.
Consider the triangle ΔABC.
We know that the line segment that joins the midpoints of two sides of a triangle is parallel to the third side and has a length equal to one half the length of the third side.
Here, the line segment MF joins the midpoints of AC¯
To determine
(b)
To Find:
DC
To determine
(c)
To Find:
EF
To determine
(d)
To Find:
Whether EF=12(AB+DC)
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# A circular rim 28 inches in diameter rotates the same number
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A circular rim 28 inches in diameter rotates the same number [#permalink] 15 Apr 2006, 19:59
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A circular rim 28 inches in diameter rotates the same number of inches per second as a circular rim 35 inches in diameter. If the smaller rim makes x revolutions per second, how many revolutions per minute does the larger rim make in terms of x?
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A circular rim 28 inches in diameter rotates the same number of inches per second as a circular rim 35 inches in diameter. If the smaller rim makes x revolutions per second, how many revolutions per minute does the larger rim make in terms of x?
This is a ratio problem. If they move the same number of inches, then the smaller rim is moving a bit faster. if they move 7 inches in a second, then that is a 1/4 turn for the smaller rim, and 1/5 of a turn for the larger rim. Armed with this, we can work out a ratio for their turning. Let's set x as 1/4; The larger rim will make 12 revolutions in a minute. Therefore, in terms of x, the larger rim will turn 48x.
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A circular rim 28 inches in diameter rotates the same number of inches per second as a circular rim 35 inches in diameter. If the smaller rim makes x revolutions per second, how many revolutions per minute does the larger rim make in terms of x?
I dont know about solution posted by jcgoodchild but if I was to face this question, I would use ratios to solve the question. the ratio of radii is 4/5 and the answer should be a multiple of 4/5 ... I might be wrong. I would really like to have opinion of other people on use of ratios.
_________________
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# A circular rim 28 inches in diameter rotates the same number
Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 909 | 3,504 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2015-27 | longest | en | 0.87859 |
https://stacks.math.columbia.edu/tag/04S4 | 1,686,271,105,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655244.74/warc/CC-MAIN-20230609000217-20230609030217-00620.warc.gz | 588,833,938 | 7,383 | Lemma 79.9.1. Let $S$ be a scheme. Let $(U, R, s, t, c)$ be a groupoid scheme over $S$. Let $P \to R$ be monomorphism of schemes. Assume that
1. $(U, P, s|_ P, t|_ P, c|_{P \times _{s, U, t}P})$ is a groupoid scheme,
2. $s|_ P, t|_ P : P \to U$ are finite locally free,
3. $j|_ P : P \to U \times _ S U$ is a monomorphism.
4. $U$ is affine, and
5. $j : R \to U \times _ S U$ is separated and locally quasi-finite,
Then $U/P$ is representable by an affine scheme $\overline{U}$, the quotient morphism $U \to \overline{U}$ is finite locally free, and $P = U \times _{\overline{U}} U$. Moreover, $R$ is the restriction of a groupoid scheme $(\overline{U}, \overline{R}, \overline{s}, \overline{t}, \overline{c})$ on $\overline{U}$ via the quotient morphism $U \to \overline{U}$.
Proof. Conditions (1), (2), (3), and (4) and Groupoids, Proposition 39.23.9 imply the affine scheme $\overline{U}$ representing $U/P$ exists, the morphism $U \to \overline{U}$ is finite locally free, and $P = U \times _{\overline{U}} U$. The identification $P = U \times _{\overline{U}} U$ is such that $t|_ P = \text{pr}_0$ and $s|_ P = \text{pr}_1$, and such that composition is equal to $\text{pr}_{02} : U \times _{\overline{U}} U \times _{\overline{U}} U \to U \times _{\overline{U}} U$. A product of finite locally free morphisms is finite locally free (see Spaces, Lemma 64.5.7 and Morphisms, Lemmas 29.48.4 and 29.48.3). To get $\overline{R}$ we are going to descend the scheme $R$ via the finite locally free morphism $U \times _ S U \to \overline{U} \times _ S \overline{U}$. Namely, note that
$(U \times _ S U) \times _{(\overline{U} \times _ S \overline{U})} (U \times _ S U) = P \times _ S P$
by the above. Thus giving a descent datum (see Descent, Definition 35.34.1) for $R / U \times _ S U / \overline{U} \times _ S \overline{U}$ consists of an isomorphism
$\varphi : R \times _{(U \times _ S U), t \times t} (P \times _ S P) \longrightarrow (P \times _ S P) \times _{s \times s, (U \times _ S U)} R$
over $P \times _ S P$ satisfying a cocycle condition. We define $\varphi$ on $T$-valued points by the rule
$\varphi : (r, (p, p')) \longmapsto ((p, p'), p^{-1} \circ r \circ p')$
where the composition is taken in the groupoid category $(U(T), R(T), s, t, c)$. This makes sense because for $(r, (p, p'))$ to be a $T$-valued point of the source of $\varphi$ it needs to be the case that $t(r) = t(p)$ and $s(r) = t(p')$. Note that this map is an isomorphism with inverse given by $((p, p'), r') \mapsto (p \circ r' \circ (p')^{-1}, (p, p'))$. To check the cocycle condition we have to verify that $\varphi _{02} = \varphi _{12} \circ \varphi _{01}$ as maps over
$(U \times _ S U) \times _{(\overline{U} \times _ S \overline{U})} (U \times _ S U) \times _{(\overline{U} \times _ S \overline{U})} (U \times _ S U) = (P \times _ S P) \times _{s \times s, (U \times _ S U), t \times t} (P \times _ S P)$
By explicit calculation we see that
$\begin{matrix} \varphi _{02} & (r, (p_1, p_1'), (p_2, p_2')) & \mapsto & ((p_1, p_1'), (p_2, p_2'), (p_1 \circ p_2)^{-1} \circ r \circ (p_1' \circ p_2')) \\ \varphi _{01} & (r, (p_1, p_1'), (p_2, p_2')) & \mapsto & ((p_1, p_1'), p_1^{-1} \circ r \circ p_1', (p_2, p_2')) \\ \varphi _{12} & ((p_1, p_1'), r, (p_2, p_2')) & \mapsto & ((p_1, p_1'), (p_2, p_2'), p_2^{-1} \circ r \circ p_2') \end{matrix}$
(with obvious notation) which implies what we want. As $j$ is separated and locally quasi-finite by (5) we may apply More on Morphisms, Lemma 37.55.1 to get a scheme $\overline{R} \to \overline{U} \times _ S \overline{U}$ and an isomorphism
$R \to \overline{R} \times _{(\overline{U} \times _ S \overline{U})} (U \times _ S U)$
which identifies the descent datum $\varphi$ with the canonical descent datum on $\overline{R} \times _{(\overline{U} \times _ S \overline{U})} (U \times _ S U)$, see Descent, Definition 35.34.10.
Since $U \times _ S U \to \overline{U} \times _ S \overline{U}$ is finite locally free we conclude that $R \to \overline{R}$ is finite locally free as a base change. Hence $R \to \overline{R}$ is surjective as a map of sheaves on $(\mathit{Sch}/S)_{fppf}$. Our choice of $\varphi$ implies that given $T$-valued points $r, r' \in R(T)$ these have the same image in $\overline{R}$ if and only if $p^{-1} \circ r \circ p'$ for some $p, p' \in P(T)$. Thus $\overline{R}$ represents the sheaf
$T \longmapsto \overline{R(T)} = P(T)\backslash R(T)/P(T)$
with notation as in the discussion preceding the lemma. Hence we can define the groupoid structure on $(\overline{U} = U/P, \overline{R} = P\backslash R/P)$ exactly as in the discussion of the “plain” groupoid case. It follows from this that $(U, R, s, t, c)$ is the pullback of this groupoid structure via the morphism $U \to \overline{U}$. This concludes the proof. $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 1,807 | 4,998 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2023-23 | latest | en | 0.687155 |
https://www.gradesaver.com/textbooks/science/physics/physics-for-scientists-and-engineers-a-strategic-approach-with-modern-physics-4th-edition/chapter-14-fluids-and-elasticity-conceptual-questions-page-383/6 | 1,553,112,221,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202450.86/warc/CC-MAIN-20190320190324-20190320212324-00156.warc.gz | 774,095,514 | 12,121 | Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)
$A > C > B$
The fraction of an object's volume that is submerged in a liquid as it floats depends on the density of the object. If an object has a larger density, then a larger fraction of the volume will be submerged. Object A has the largest fraction of its volume submerged, so object A has the largest density. Object B has the smallest fraction of its volume submerged, so object B has the smallest density. We can rank the densities of the objects in order, from largest to smallest: $A > C > B$ | 129 | 595 | {"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} | 3.09375 | 3 | CC-MAIN-2019-13 | latest | en | 0.93167 |
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• Question
Two trains arrive at the same platform with different periods. Compute the LCM of the two periods to find the time they clash.
This is a context question testing the student's ability to identify the lowest common multiple of two integer values which are not multiples of each other.
• Question
This question tests the student's ability to identify equivalent fractions through spotting a fraction which is not equivalent amongst a list of otherwise equivalent fractions. It also tests the students ability to convert mixed numbers into their equivalent improper fractions. It then does the reverse and tests their ability to convert an improper fraction into an equivalent mixed number.
• Decimals to fractions
Question
Identify well-known fractional equivalents of decimals. Convert obscure decimals and recurring decimals into fractions.
• Surds simplification
Question
This question tests the student's understanding of what is and is not a surd, and on their simplification of surds.
• Question
This question tests the student's ability to solve simple linear equations by elimination. Part a) involves only having to manipulate one equation in order to solve, and part b) involves having to manipulate both equations in order to solve.
• Question
Solve a linear equation of the form $ax+b = c$, where $a$, $b$ and $c$ are integers.
The answer is always an integer.
• Question
A simple situational question about a box of chocolates, asking how many of each type there are, what percentage of the box they represent, the probability of picking one and ratios of different types.
• Question
This is a simple question testing the student on their ability to calculate the lowest common multiple of two integers which are:
Part a) - coprime;
Part b) - where the greatest common divisor between the two integers is greater than one and not equal to either given number; and
Part c) - where one of the integer is a multiple of the other.
• Question
Several problems involving the multiplication of fractions, with increasingly difficult examples, including a mixed fraction and a squared fraction. The final part is a word problem.
• Question
This question tests the student's ability to identify the factors of some composite numbers and the highest common factors of two numbers.
• Division of fractions | 498 | 2,559 | {"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} | 3.1875 | 3 | CC-MAIN-2024-33 | latest | en | 0.921047 |
https://plus.maths.org/content/Blog/2010/%22/issue48/features/latestnews/jan-apr04/utune/www.britishscienceassociation.org/forms/festival/events/showevent2.asp?page=23 | 1,571,293,815,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986672723.50/warc/CC-MAIN-20191017045957-20191017073457-00368.warc.gz | 648,906,851 | 13,327 | ## Plus Blog
June 20, 2014
Recently we had to learn about tensors for an upcoming article. What are those, you ask? We didn't know at first either. Like some other concepts in maths, they seem confusing at first but actually are just a way of capturing information we are all used to.
First of all, let's start with scalars. Scalars are just your ordinary, everyday, real numbers. A scalar field is used to describe something that has a particular value at every point in the space you are considering. For example, a map of temperatures across the UK, or indeed the world, is a scalar field; with a value for the temperature at each point on the map. You can read more about how scalar fields describe dark energy and the Higgs boson.
Then we get to the more dynamic concept of vector fields. A vector field is something that associates a vector (a magnitude and direction) to every point in space. Again, thanks to meteorology we have the familiar example of a wind map as a vector field. Vector fields are incredibly important in maths and physics and, like in our example of a wind map, usually describe how things move. You find out how fluid dynamics uses vector fields to model the movement of tears, wind and waves in Births and deaths in fluid chaos
A tensor field
A tensor just extends this definition to one where the value of some property depends on the direction in which you measure it. So where a vector is a magnitude and a particular direction from some point, a tensor gives a magnitude for every direction from that point. A tensor field is something that assigns a tensor to every point in space. Naturally it's harder to picture a tensor field but if you've ever played with a piece of chewing gum you've actually seen one in action. As you pull a piece of gum (or some other rubbery substance) between your fingers, it stretches in tension along one direction but compresses in the other perpendicular directions. So for each point in the gum the stress is a function of direction: in each direction the stress will take a certain value that is a combination of the contributions from these tension and compression stresses.
Tensors are incredibly useful tools, particularly when describing things in higher dimensions. The curvature of multidimensional surfaces (called manifolds) is described with tensors and Einstein used tensors to describe both the curvature and distribution of matter of four-dimensional space-time. You can read more about Einstein and the role of curvature on Plus.
So there you go. Tensors are nothing to get tense about!
June 17, 2014
Congratulations to Plus contributors David Spiegelhalter and Helen Mason, who have been awarded a knighthood and OBE respectively in the 2014 Queen's Birthday Honours list! Joining famous faces including Dame Maggie Smith, Angelina Jolie and Daniel Day-Lewis, Helen Mason and David Spiegelhalter's honours have been awarded to recognise their outstanding work in mathematics, science and public communication.
Professor — now Sir — David Spiegelhalter is Winton Professor of the Public Understanding of Risk at the University of Cambridge, and has been knighted for "services to statistics". A medical statistician, he has played a leading role in developing simulation technologies and clinical trials on drug safety, and has supported the UK health service through many inquiries, including the public inquiry into children’s heart surgery at Bristol Royal Infirmary.
As the official government press release notes, and as regular readers of Plus will already know, Spiegelhalter has also 'made a significant difference to how to communicate with patients and the public about risk'. He started the brilliant project Understanding Uncertainty, writes public articles, lectures to public and school audiences and has appeared on TV and radio (ranging from being interviewed on the BBC's heavy-hitting Newsnight to appearing as a contestant in Winter Wipeout). Watch a lighthearted example of David Spiegelhalter's public understanding work below, in a video produced by the University of Cambridge, and read his articles on Plus.
Helen Mason leads the Atomic Astrophysics group in the Department of Applied Mathematics and Theoretical Physics at the University of Cambridge. Helen Mason's extensive outreach and education work has included writing for Plus and setting up the Sun|Trek website for schools, as well as giving frequent talks and appearing on radio and TV.
Helen Mason's OBE has been awarded "for services to Higher Education and to Women in Science, Engineering and Technology". Learn more about her work in this video from the Royal Institution:
June 17, 2014
Oh Paul the psychic octopus… How we miss you and your uncannily accurate predictions of World Cup glory! And we aren't the only ones – Google is celebrating Paul in today's Google Doodle!
Paul, an English octopus then living in Germany, became world famous during the last World Cup when he picked the winner of all of Germany's seven matches, including their two defeats, and got the 2010 World Cup final correct too. Paul provided a welcome watery break between the matches, even though, as David Spiegelhalter's statistical analysis showed, he wasn't actually psychic after all.
Sadly, Paul can't help us predicting this year's World Cup – psychic or not. He died in October 2010 after living a full and happy octopus life. But you can read more about Paul's contribution to the 2010 World Cup in our article Understanding uncertainty: how psychic was Paul?.
June 13, 2014
We are the outcome of a process which took nearly 14 billion years during which atoms, stars, planets and biospheres emerged from a hot and dense big bang. The details of this process are sensitive to a few important numbers — the so-called constants of physics.
Martin Rees, the Astronomer Royal, discussed the key stages in this process in the lecture below, given on 17 March 2014 as a public event during the Cambridge Science Festival, and linked to a conference on the philosophy of cosmology. The talk also addressed two questions: What would our cosmos be like if the key numbers were different? And could a huge variety of other universes exist as part of physical reality, each the aftermath of a different big bang?
Physicists are hard at work on these questions. In fact, just a couple of hours before Lord Rees gave this talk, US researchers announced a result that could signal a major breakthrough in our understanding of how our Universe evolved (see this Plus article). Watch out for the references to this in the lecture — good timing!
June 13, 2014
"It's failure to prepare mentally and failure to take practicing penalties really seriously." This is Ken Bray's explanation for England's dismal performance in penalty shootouts. England are successful in only 17% of their encounters, compared to Germany's impressive 80%. Bray is an expert in the science of football, and he has studied the physics as well as the psychology of penalties and analysed the statistics. The result are three steps to ensure a perfect penalty, which he explains in this video. You can find out more about the science and maths of football in Bray's Plus articles on football.
May 16, 2014
I was lucky enough to see the beautiful Matisse exhibition at the Tate Modern in London last week. A few days later I was asked, by a TV researcher, how do you make maths interesting and understandable to people when so many people, in her experience, had hated and avoided it at school? In response I found myself telling her about the Matisse exhibiton. Not to change the subject or avoid discussing the bad reputation maths has in many people's minds, but to explain what we at Plus, along with many other maths communicators, try to do.
Now I loved maths at school. It made sense, I could express what was in my brain and how I saw the world through maths, and I found it fun. What I found a challenge was art class, I didn't feel I had any artistic ability at all. People who could accurately draw or realistically paint people (or bowls of fruit or jugs of water...) amazed me - it was like they had a magic power. And for a long time I didn't think much of modern art. For example Picasso's pictures seemed ugly and didn't look like the things they apparently depicted, and Matisse's bright art seemed simplistic.
My opinion on art, particularly modern art, has changed, thanks particularly to two fantastic exhibitions. The first was over a decade ago at a light-filled gallery in Spain that showed many of Picasso's preliminary sketches and studies leading up to the Guernica, the full work of which filled the final room of the gallery. Following the journey that Picasso took as he built towards this huge masterpiece, along with the commentary of the social and artistic context of the piece, gave me a new appreciation of the skill, effort and genius of this work. It was awesome. I began to see all his work differently.
The Matisse show at the Tate contains many of what are called the Cut-Outs. Very late in his life, after a serious illness, he initially wasn't able to paint the huge canvases he previously produced. Instead, sitting in a chair or up in bed, he began to cut out coloured paper which he would instruct assistants to pin to the walls of his room and studio, moving it a little to the left or right, a little up or down, experimenting with these coloured shapes to build collages that had enormous power. He was so taken by this new genre he had invented that even when he recovered enough to paint he preferred this new approach.
The exhibition contains video of Matisse at work, photographs of his studio and commentary that explain the personal, artistic and social context for his work. Walking through the chronologically arranged rooms it gave me a sense of his motivation for working this way, and how it changed his perspective, and the significance of this work both artistically but also its wider cultural context. The exhibition included different types of content: recordings of the artist himself (or someone speaking his words) about his motivation, clear descriptions of the works and how they were made, explanations of the significance of this work in art, and how they interpreted Matisse's world. And most importantly, of course, was seeing the works themselves, up close, smooth curves, jagged edges, bright colours, pinned together.
It was the brilliant curation of this exhibition that reminded me and inspired me about the work we do here at Plus. We want to allow anyone a chance to see some of the wonders of mathematics up close. We hope that hearing the words of researchers will give a sense of their motivations in doing their mathematics, and we aim to show the significance of this work both in mathematics, and in a wider, cultural setting. We hope to give people a glimpse into how mathematicians perceive the world and how they use mathematics to express their perception of the world. And we hope this gives our readers a new appreciation of maths, of its power and beauty, that they might not have noticed or enjoyed before.
Now I'll never be an artist. But thanks to clever and passionate curation I have over time developed an appreciation of many different types of art, that makes seeing new art less confronting and more exciting. And sometimes, when the urge arises, I might have a go at capturing something on pencil and paper or in cut-outs of coloured paper, just for my own pleasure. I hope that Plus plays a small part in helping everyone have a similar appreciation of mathematics. We might not all be mathematicians, but I hope we can all enjoy and engage with mathematical ideas, appreciate their beauty and power. And that the next time maths pops up in our lives, it's something to be excited about, rather than avoided. | 2,422 | 11,863 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2019-43 | longest | en | 0.945832 |
https://practicaldev-herokuapp-com.global.ssl.fastly.net/thormeier/it-s-alive-conway-s-game-of-life-on-a-canvas-25ja | 1,713,721,251,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817790.98/warc/CC-MAIN-20240421163736-20240421193736-00241.warc.gz | 395,133,125 | 41,162 | Pascal Thormeier
Posted on
# It's alive! Simulate organisms with Conway's game of life on a canvas 🧫🔋
Today, we create living organisms! We'll kind of, at least. It's the next best thing to becoming a 21st century digital "Web Dev Doctor Frankenstein": Conway's game of life.
## What?
Excellent question. The best, actually! Let me explain...
John Horton Conway was a British mathematician. He contributed to a lot of different fields in mathematics, such as number theory, algebra, geometry, combinatorial game theory, algorithmics, group theory, and analysis.
He developed a ton of remarkable algorithms, such as the Doomsday algorithm, that lets you find out the weekday of any given date with a only a few steps. I've explained the Doomsday rule in this post some time ago:
Conway developed the "Game of Life" in 1970 as an applied example of abstract computers. It's a 2-dimensional field with X and Y coordinates, where each integer coordinate represents a cell that can be either alive or dead, depending on some rules.
But, since it's a game, how is it played?
## The rules of the game
You can think of the Game of Life as a sandbox. Originally, no cell is alive. Alive cells can be either set by the user or sprinkled in randomly. In each game tick, the game determines which cells are alive and which ones are dead in the next generation. This step is then repeated until the user interrupts.
To determine the next generation, the game looks at each cells neighbors and applies a set of rules:
• If a cell was alive in the current generation:
• If it has less than 2 (loneliness) or more than 3 (overpopulation) alive neighbors, it dies in the next generation, otherwise it stays alive
• If a cell was dead in the current generation:
• If it has exactly 3 alive neighbors, it will become alive in the next generation, otherwise it stays dead
(These rules allow for some pretty complex structures, but we'll come to that later!)
## Let's make an example or two
Let's consider a 3 by 3 grid. We're going to see how the rules work by applying them to the center cell. All other cells are the center cell's neighbors.
Here we can see what happens if less than 2 neighboring cells are alive.
The filled cell in the middle is alive in this generation, but dies the next generation.
In the following picture, we can see how it could look like if a cell is being born:
One thing is important, though: The next generation needs to be calculated all at once. Meaning: If the game sets cell 1 as "alive" that was dead before and starts applying the rules to its immediate neighbor cell 2, it should not consider the new state of cell 1 (alive) but the old one (dead) for the calculation of cell 2.
But this begs a question: What does it do at the border of the field?
There's two possibilities: Either we consider the border as always dead (they are neighbors, but the rules are never applied to them) or the world is actually formed like a donut.
## Tasty torus
When the field is shaped like a donut, it behaves like this:
Whatever leaves either side will reenter on the opposite side. When you connect those sides, the shape will actually look like a donut. Or in mathematics speech: A torus.
So, that's all the info we need. Let's start implementing this!
## Coding out the game of life
Let's start with the field. I will create the field as a nested array of 100 by 100 boolean variables:
``````const field = []
for (let y = 0; y < 100; y++) {
field[y] = []
for (let x = 0; x < 100; x++) {
field[y][x] = false
}
}
``````
By setting everything false, the code will consider all cells as dead. True, on the other hand, would mean that a cell is alive.
Next, I need a function to get any cell's neighbors. A cell is identified by its X and Y values, so I can add and subtract 1 to to those values to get all neighbors:
``````const getNeighbors = (x, y, field) => {
let prevX = x - 1
let nextX = x + 1
let prevY = y - 1
let nextY = y + 1
return [
field[prevY][prevX],
field[prevY][x],
field[prevY][nextX],
field[y][prevX],
// field[y][x], That's the cell itself - we don't need this.
field[y][nextX],
field[nextY][prevX],
field[nextY][x],
field[nextY][nextX],
]
}
``````
But wait - the field is a donut. So I need to catch the border cases as well:
``````const getNeighbors = (x, y, field) => {
let prevX = x - 1
if (prevX < 0) {
prevX = field[0].length - 1
}
let nextX = x + 1
if (nextX === field[0].length) {
nextX = 0
}
let prevY = y - 1
if (prevY < 0) {
prevY = field.length - 1
}
let nextY = y + 1
if (nextY === field.length) {
nextY = 0
}
// ...
}
``````
So this function now returns an array of boolean values. The game's rules don't care about which neighbors are alive or dead, only how many of them are.
The next step is to actually implement the rules. Ideally, I've got a function that takes X and Y values as well as the field and returns the state of the cell for the next generation:
``````const getDeadOrAlive = (x, y, field) => {
const neighbors = getNeighbors(x, y, field)
const numberOfAliveNeighbors = neighbors.filter(Boolean).length
// Cell is alive
if (field[y][x]) {
if (numberOfAliveNeighbors < 2 || numberOfAliveNeighbors > 3) {
// Cell dies
return false
}
// Cell stays alive
return true
}
if (numberOfAliveNeighbors === 3) {
// Cell becomes alive
return true
}
return false
}
``````
And that's pretty much it for the game rules!
Now I create a function to draw the entire field on a square canvas:
``````const scaleFactor = 8
const drawField = field => {
const canvas = document.querySelector('canvas')
const context = canvas.getContext('2d')
// Fill entire field
context.fillStyle = '#fff'
context.fillRect(0, 0, 100 * scaleFactor, 100 * scaleFactor);
context.fillStyle = '#008000'
// Fill alive cells as small rectangles
field.forEach((row, y) => row.forEach((cell, x) => {
if (cell) {
context.fillRect(
x * scaleFactor,
y * scaleFactor,
scaleFactor,
scaleFactor
)
}
}))
}
``````
Now let's add some control buttons to let the game automatically calculate and draw new generations each 80ms:
``````let nextField = field
drawField(field)
const step = () => {
nextField = nextField.map((row, y) => row.map((_, x) => {
}))
drawField(nextField)
}
let interval = null
interval = setInterval(step, 80)
})
clearInterval(interval)
})
``````
And some more controls for defaults, random, reset, etc.:
``````document.querySelector('#reset').addEventListener('click', () => {
for (let y = 0; y < 100; y++) {
for (let x = 0; x < 100; x++) {
field[y][x] = false
}
}
nextField = field
drawField(field)
})
for (let y = 0; y < 100; y++) {
for (let x = 0; x < 100; x++) {
field[y][x] = false
}
}
field[20][20] = true
field[20][21] = true
field[20][22] = true
field[19][22] = true
field[18][21] = true
nextField = field
drawField(field)
})
for (let y = 0; y < 100; y++) {
for (let x = 0; x < 100; x++) {
field[y][x] = Math.random() * 100 > 65
}
}
nextField = field
drawField(field)
})
const x = Math.floor(event.offsetX / scaleFactor)
const y = Math.floor(event.offsetY / scaleFactor)
field[y][x] = !field[y][x]
nextField = field
drawField(field)
})
``````
Of course this needs some HTML, too:
``````<!DOCTYPE html>
<html>
<style>
canvas {
box-sizing: border-box;
border: 1px solid #000;
width: 800px;
height: 800px;
}
.container {
box-sizing: border-box;
width: 800px;
border: 1px solid #000;
margin-top: 10px;
}
</style>
<body>
<h1>Conway's game of life on a canvas</h1>
<canvas id="canvas" width="800" height="800"></canvas>
<div class="container">
<button id="start">Start</button>
<button id="stop">Stop</button>
<button id="step">Step</button>
</div>
<div class="container">
<button id="reset">Reset to empty</button>
<button id="glider">Set single glider</button>
<button id="random">Random (35% alive)</button>
</div>
<script src="./index.js"></script>
</body>
</html>
``````
## The final result
And here's a codepen where you can play around with it:
(Because of the size of the canvas and the non-responsive nature of the example, I recommend running it in 0.5 scale)
Have fun exploring!
## Some remarkable structures
There's some cell structures that are worth mentioning. A rather simple one is called a "glider":
As you can see, this thing actually moves in a straight line by one unit on the X and Y axis every 5 generations.
Since it's going back to its original state again, this structure is able to move indefinitely!
But there's more: Some structures are static (for example a 2 by 2 alive square), flip between two states (one example being a straight line along either the X or Y axis consisting of 3 alive cells), others are capable of moving and even producing gliders at intervals!
You see, this really is the closest thing to creating living organisms as you can get with around 200 lines of JS and a canvas!
I hope you enjoyed reading this article as much as I enjoyed writing it! If so, leave a ❤️ or a 🦄! I write tech articles in my free time and like to drink coffee every once in a while.
If you want to support my efforts, buy me a coffee or follow me on Twitter 🐦! You can also support me directly via Paypal!
GrahamTheDev • Edited
Always great fun to play with conways game of life. However, as with everything, there is always someone who does it bigger and better...(skip to 1:08 for actual video)
One minor bug is if you press "reset" before pressing "stop" the canvas will still try and update and so drawing becomes difficult. Plus grid lines would be great when drawing, but that is a fun one for others to try and add!
Well worth a ❤ and a 🦄!
Pascal Thormeier
Holy moly, that's an amazing video! Thank you for sharing!
Reminds of this one, where someone actually managed to code the Game of Life into the Game of Life:
Yes, it does have it's bugs right now, and I will fix those in the Codepen at least in the next few days! :)
GrahamTheDev • Edited
Yeah I was in two minds whether to link to that one instead, but the music makes the first one feel epic! 😋🤣
I think the second one would melt my CPU (especially as it is so warm in the UK at the moment!)
Pascal Thormeier
There's So. Damn. Many. Interesting videos about GOL, I could seriously watch them all day long :D I think building the GOL-in-GOL version would alone take ages, let alone figuring it all out without any help...
GrahamTheDev
Have you seen the FOL computer that outputs the Fibonacci sequence? RAM, CPU etc, just makes me realise how little I know 🤣😜
I first learned about Game of Life in this video by veritasium. Following that I simulated it for a hackathon. It was so cool to see simulate cool rifle and complex simulators using it. Plam to simulate a 3d version when I have a cpu capable of doing that :D
Pascal Thormeier
I love Veritasium, been following that channel for years now! I haven't heard about any 3D version of GOL yet, would adapt the rules to fit the amount of neighboring cells? You definitely have to make a post about it once it's done!
Haven't figured out the rules of a 3D GoL just yet, but if I ever implement it, if I were to do it, I'd start with the same rules as GoL, but in 3D space. Will be interesting to see the results. Will definitely write a post if I implement it :D
Pascal Thormeier
I wonder what a 3D glider would look like... I could imagine that it's a lot more complex to achieve the same behaviour as in 2D space. So many things to explore, can't wait for your article! :D
Ahaha, hopefully I do get around to implement it someday
When discussing Conway's Game of Life, I always like to provide the APL version of the code. For contrast with the language that is being presented, in this case in contrast with JavaScript (and HTML and CSS).
APL version of Conway's Game of Life.
`life ← {⊃1 ⍵ ∨.∧ 3 4 = +/ +⌿ ¯1 0 1 ∘.⊖ ¯1 0 1 ⌽¨ ⊂⍵}`
At the age of 82, Dr. John Horton Conway passed away on 2020-Apr-11, from COVID-19.
For an high-level introduction to APL, the article a Glimpse of Heaven by Bernard Legrand.
Pascal Thormeier
I haven't used APL before and it does look interesting indeed. Thank you for sharing this implementation! An interesting video you might like is Dr. Conway talking about the Game of Life himself, over at Numberphile. He said he used to even hate it, because he didn't find it all too interesting and it was overshadowing much more important things.
I built this many times before, trying to speed things up. This is my latest version ashware.nl/fast-life/
The code can be found via my homepage.
Aleksandr Hovhannisyan
Beautiful animations!
Pascal Thormeier
Oh wow! I love how the animation also allows to see previous generations. What were the challenges you faced during this implementation?
Esger Jellema
Thank you! Here my main challenge was trying to speed things up, again :)
So I built an AureliaJs web app and a web worker to do the heavy lifting.
The 'trails' were quite simple to accomplish: just draw the new generation with opacity < 1 in order to faint all 'old' cells a bit every generation.
PS. You can even change the rules of life while it's running and change various other settings as well.
Alexandra Egorova
The best explanation with code, thanks!👩🏻💻 | 3,451 | 13,192 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2024-18 | latest | en | 0.945652 |
https://de.mathworks.com/matlabcentral/cody/problems/79-dna-n-gram-distribution/solutions/1655785 | 1,568,795,683,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573258.74/warc/CC-MAIN-20190918065330-20190918091330-00328.warc.gz | 445,894,646 | 15,618 | Cody
Problem 79. DNA N-Gram Distribution
Solution 1655785
Submitted on 22 Oct 2018 by E Chang
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
s = 'AACTGAACG'; n = 3; hifreq_correct = 'AAC'; assert(isequal(nGramFrequency(s,n),hifreq_correct))
2 Pass
s = 'dynamic routing service'; n = 2; hifreq_correct = 'ic'; assert(isequal(nGramFrequency(s,n),hifreq_correct))
3 Pass
s = 'Your veracity is exceeded by your sagacity.'; n = 5; hifreq_correct = 'acity'; assert(isequal(nGramFrequency(s,n),hifreq_correct))
4 Pass
s = 'AGCGAAGGAAGGATCACATTTCTCAGGACAAAGGCATTTCACTAATGGTT'; n = 3; hifreq_correct = 'AGG'; assert(isequal(nGramFrequency(s,n),hifreq_correct))
5 Pass
s = 'In short, in matters vegetable, animal, and mineral, I am the very model of a modern Major-General.'; n = 2; hifreq_correct = 'er'; assert(isequal(nGramFrequency(s,n),hifreq_correct)) | 314 | 979 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-39 | longest | en | 0.634818 |
https://dev.to/tommy3/learning-algorithms-with-js-python-and-java-10-pyramid-4mga | 1,713,160,116,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816942.33/warc/CC-MAIN-20240415045222-20240415075222-00184.warc.gz | 176,919,841 | 18,953 | tommy-3
Posted on
# Learning Algorithms with JS, Python and Java #10: Pyramid
This series of articles follows Stephen Grider's Udemy course in three different languages.
Today's question is a slight variation on the last one.
--- Directions
Write a function that accepts a positive number N.
The function should console log a pyramid shape
with N levels using the # character. Make sure the
pyramid has spaces on both the left and right hand sides
--- Examples
pyramid(1)
'#'
pyramid(2)
' # '
'###'
pyramid(3)
' # '
' ### '
'#####'
# 1: Iterative Solution
JavaScript:
``````function pyramid(n) {
const columnCount = 2 * n - 1;
const midColumn = Math.floor(columnCount / 2);
for (let row = 0; row < n; row++) {
let level = '';
for (let column = 0; column < columnCount; column++) {
if (Math.abs(column - midColumn) <= row) {
level += '#';
} else {
level += ' ';
}
}
console.log(level);
}
}
``````
Python:
``````def pyramid(n):
column_count = 2 * n - 1
mid_column = column_count // 2
for row in range(n):
level = ''
for column in range(column_count):
if abs(column - mid_column) <= row:
level += '#'
else:
level += ' '
print(level)
``````
Java:
``````static void pyramid1(int n) {
int columnCount = 2 * n - 1;
int midColumn = columnCount / 2;
for (int row = 0; row < n; row++) {
StringBuilder level = new StringBuilder();
for (int column = 0; column < columnCount; column++) {
if (Math.abs(column - midColumn) <= row) {
level.append("#");
} else {
level.append(" ");
}
}
System.out.println(level);
}
}
``````
# 2: Recursive Solution
Here I take an approach a bit different from the one introduced in the course.
JavaScript:
``````function pyramid(n, row = 0, level = '#') {
if (n === row) {
return;
}
if (level.length === 2 * n - 1) {
console.log(level);
return pyramid(n, row + 1);
}
level = level.length < 2 * row ? `#\${level}#` : ` \${level} `;
pyramid(n, row, level);
}
``````
Python:
``````def pyramid(n, row=0, level='#'):
if n == row:
return
if len(level) == 2 * n - 1:
print(level)
return pyramid(n, row+1)
level = f'#{level}#' if len(level) < 2 * row else f' {level} '
pyramid(n, row, level)
``````
Java:
``````static void pyramid(int n) {
pyramid(n, 0, new StringBuilder("#"));
}
static void pyramid(int n, int row, StringBuilder level) {
if (n == row) {
return;
}
if (level.length() == 2 * n - 1) {
System.out.println(level);
pyramid(n, row + 1, new StringBuilder("#"));
return;
}
if (level.length() < 2 * row) {
level.insert(0, "#").append("#");
} else {
level.insert(0, " ").append(" ");
}
pyramid(n, row, level);
}
``````
Thank you for reading. I do plan to write this series more often, so I hope to see you soon! | 783 | 2,666 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-18 | latest | en | 0.509578 |
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Haas MBA/MPH vs Tuck MBA/MPH
Author Message
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Joined: 20 Aug 2013
Posts: 44
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Kudos [?]: 1 [0], given: 0
Haas MBA/MPH vs Tuck MBA/MPH [#permalink]
Show Tags
13 Nov 2013, 01:53
My current target schools are Tuck, Haas, Fuqua, and Stern; My future career goals are related to Healthcare and Clean Tech fields. I'm pretty set on Fuqua but having trouble differentiating Tuck and Haas.
So clearly I really like the first three and hypothetically speaking, IF I were to get into Tuck or Haas, could you guys help me choose? I don't need to know the basics of each school (stuff thats advertised on website), what I'm looking for is deeper insight into the strengths and weakness of each School and program.
Thanks
VP
Joined: 07 Apr 2009
Posts: 1183
Concentration: General Management, Strategy
Schools: Duke (Fuqua) - Class of 2012
Followers: 35
Kudos [?]: 437 [0], given: 19
Re: Haas MBA/MPH vs Tuck MBA/MPH [#permalink]
Show Tags
13 Nov 2013, 17:59
overmind1632 wrote:
My current target schools are Tuck, Haas, Fuqua, and Stern; My future career goals are related to Healthcare and Clean Tech fields. I'm pretty set on Fuqua but having trouble differentiating Tuck and Haas.
So clearly I really like the first three and hypothetically speaking, IF I were to get into Tuck or Haas, could you guys help me choose? I don't need to know the basics of each school (stuff thats advertised on website), what I'm looking for is deeper insight into the strengths and weakness of each School and program.
Thanks
Clean Tech and Healthcare are very different tracks. In my opinion, job opportunities for MBAs in clean tech has died down significantly, due to drop in investments in that sector.
Healthcare is vast field. It would be helpful to understand what sector you want to go into: provider, pharma, med devices, healthcare consulting, etc.
I didn't recruit for healthcare, but having gone to Fuqua, I got enough exposure to healthcare to be semi-knowledgeable. Neither Tuck nor Haas are known for healthcare. I would say that Tuck would be better if you are looking to go into Pharma, since lots of big Pharma are located in the NJ area, which is close to NH. I would has Haas would be better for medical devices, since there are a few big ones in the Bay Area, and there are med device start-up activities there.
Re: Haas MBA/MPH vs Tuck MBA/MPH [#permalink] 13 Nov 2013, 17:59
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,011 | 3,771 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2017-04 | latest | en | 0.919226 |
https://aerospaceengineering.aero/example-for-springs-with-boundary-conditions-and-its-solution/ | 1,495,906,379,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608984.81/warc/CC-MAIN-20170527171852-20170527191852-00014.warc.gz | 929,432,554 | 29,965 | # Example for Springs with Boundary Conditions and its Solution
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In order to determine forces and deformation of any structure or a component, one should have to give boundary conditions which constrained the structure as per customers requisite. Boundary Conditions determine how the model or structural component is externally constrained in any one of the axis. Every models must be attached to the some external point or support points. You may determine these points of support as completely controlled or as partially controlled with a Spring or spring element. Also we can define a spring support which has stiffness or rigidity in only one direction with the compression-only or tension springs.
## What is Boundary Condition
To solve the equations which is defined by the global stiffness matrix(G), So we must apply some type of constraints and/or supports or the structure will be free to move as a rigid body, this type of constraining is called Boundary condition.
Generally boundary conditions are of two types:
• Homogeneous Boundary Conditions
• Non-homogeneous Boundary Conditions
1. The most common one is the homogeneous boundary conditions, which are occur at locations of the model, that are completely prevented from structural movement.
2. The second one is non-homogeneous boundary conditions, which are occur where finite an extremely small (none zero) values of displacement is specified, that is, such as the settlement of a support.
## Example for Springs with Boundary Conditions
Let us think about the two element system as described before where the Node 1 is attached to a fixed support as shown in below figure, and yielding the displacement constraint U1= 0, k1= 50 lb/in, k2= 75 lb/in, for these conditions the forces is given as F2=F3=75 lb/in, and here we need to determine nodal displacements U2 and U3?
### Free Body Diagram of Example:
By substituting the above specified values into the below equation which is derived earlier:
We have
As we see nodal force F1 becomes an unknown reaction force, it is just because of the constraint of zero displacement at node 1. Formally, first algebraic equation which is represented in this matrix equation becomes:
-50U2 = F1
This represents the constrained equation, which it illustrates the equilibrium condition of the node at which the displacement is constrained.
The second and third equation become,
By solving the above equation, we obtain the displacements
U2 = 3 inch
and
U3 = 4 inch.
Note that, the matrix equations will governing the unknown displacements which are obtained by the simply striking out the first row and column of the 3 * 3 matrix system, and since the constrained displacement value is zero, and this value is homogeneous.
The displacement boundary condition not equal to the zero (which is non-homogeneous), then this is not possible and the matrices need to be control differently.
Hence, This is the example for springs with boundary conditions and its Solution.
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2017-03-14 22:33:07
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2016-11-08 23:45:20
2016-11-08 23:20:23 | 893 | 4,131 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2017-22 | longest | en | 0.917331 |
https://aprove.informatik.rwth-aachen.de/eval/JAR06/JAR_TERM/TRS/AProVE/JFP_Ex31.trs.Thm17:POLO_71:NOFWD.html.lzma | 1,725,933,109,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651164.37/warc/CC-MAIN-20240909233606-20240910023606-00423.warc.gz | 86,126,793 | 3,114 | Term Rewriting System R:
[x, y, z]
active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
top(mark(x)) -> top(proper(x))
top(ok(x)) -> top(active(x))
Termination of R to be shown.
` R`
` ↳Dependency Pair Analysis`
R contains the following Dependency Pairs:
ACTIVE(f(b, c, x)) -> F(x, x, x)
ACTIVE(f(x, y, z)) -> F(x, y, active(z))
ACTIVE(f(x, y, z)) -> ACTIVE(z)
F(x, y, mark(z)) -> F(x, y, z)
F(ok(x), ok(y), ok(z)) -> F(x, y, z)
PROPER(f(x, y, z)) -> F(proper(x), proper(y), proper(z))
PROPER(f(x, y, z)) -> PROPER(x)
PROPER(f(x, y, z)) -> PROPER(y)
PROPER(f(x, y, z)) -> PROPER(z)
TOP(mark(x)) -> TOP(proper(x))
TOP(mark(x)) -> PROPER(x)
TOP(ok(x)) -> TOP(active(x))
TOP(ok(x)) -> ACTIVE(x)
Furthermore, R contains four SCCs.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Polynomial Ordering`
` →DP Problem 2`
` ↳Polo`
` →DP Problem 3`
` ↳Polo`
` →DP Problem 4`
` ↳Nar`
Dependency Pairs:
F(ok(x), ok(y), ok(z)) -> F(x, y, z)
F(x, y, mark(z)) -> F(x, y, z)
Rules:
active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
top(mark(x)) -> top(proper(x))
top(ok(x)) -> top(active(x))
The following dependency pair can be strictly oriented:
F(ok(x), ok(y), ok(z)) -> F(x, y, z)
There are no usable rules using the Ce-refinement that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
POL(mark(x1)) = 0 POL(ok(x1)) = 1 + x1 POL(F(x1, x2, x3)) = x1
resulting in one new DP problem.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Polo`
` →DP Problem 5`
` ↳Polynomial Ordering`
` →DP Problem 2`
` ↳Polo`
` →DP Problem 3`
` ↳Polo`
` →DP Problem 4`
` ↳Nar`
Dependency Pair:
F(x, y, mark(z)) -> F(x, y, z)
Rules:
active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
top(mark(x)) -> top(proper(x))
top(ok(x)) -> top(active(x))
The following dependency pair can be strictly oriented:
F(x, y, mark(z)) -> F(x, y, z)
There are no usable rules using the Ce-refinement that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
POL(mark(x1)) = 1 + x1 POL(F(x1, x2, x3)) = x3
resulting in one new DP problem.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Polo`
` →DP Problem 5`
` ↳Polo`
` ...`
` →DP Problem 6`
` ↳Dependency Graph`
` →DP Problem 2`
` ↳Polo`
` →DP Problem 3`
` ↳Polo`
` →DP Problem 4`
` ↳Nar`
Dependency Pair:
Rules:
active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
top(mark(x)) -> top(proper(x))
top(ok(x)) -> top(active(x))
Using the Dependency Graph resulted in no new DP problems.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Polo`
` →DP Problem 2`
` ↳Polynomial Ordering`
` →DP Problem 3`
` ↳Polo`
` →DP Problem 4`
` ↳Nar`
Dependency Pair:
ACTIVE(f(x, y, z)) -> ACTIVE(z)
Rules:
active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
top(mark(x)) -> top(proper(x))
top(ok(x)) -> top(active(x))
The following dependency pair can be strictly oriented:
ACTIVE(f(x, y, z)) -> ACTIVE(z)
There are no usable rules using the Ce-refinement that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
POL(ACTIVE(x1)) = x1 POL(f(x1, x2, x3)) = 1 + x3
resulting in one new DP problem.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Polo`
` →DP Problem 2`
` ↳Polo`
` →DP Problem 7`
` ↳Dependency Graph`
` →DP Problem 3`
` ↳Polo`
` →DP Problem 4`
` ↳Nar`
Dependency Pair:
Rules:
active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
top(mark(x)) -> top(proper(x))
top(ok(x)) -> top(active(x))
Using the Dependency Graph resulted in no new DP problems.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Polo`
` →DP Problem 2`
` ↳Polo`
` →DP Problem 3`
` ↳Polynomial Ordering`
` →DP Problem 4`
` ↳Nar`
Dependency Pairs:
PROPER(f(x, y, z)) -> PROPER(z)
PROPER(f(x, y, z)) -> PROPER(y)
PROPER(f(x, y, z)) -> PROPER(x)
Rules:
active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
top(mark(x)) -> top(proper(x))
top(ok(x)) -> top(active(x))
The following dependency pairs can be strictly oriented:
PROPER(f(x, y, z)) -> PROPER(z)
PROPER(f(x, y, z)) -> PROPER(y)
PROPER(f(x, y, z)) -> PROPER(x)
There are no usable rules using the Ce-refinement that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
POL(PROPER(x1)) = x1 POL(f(x1, x2, x3)) = 1 + x1 + x2 + x3
resulting in one new DP problem.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Polo`
` →DP Problem 2`
` ↳Polo`
` →DP Problem 3`
` ↳Polo`
` →DP Problem 8`
` ↳Dependency Graph`
` →DP Problem 4`
` ↳Nar`
Dependency Pair:
Rules:
active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
top(mark(x)) -> top(proper(x))
top(ok(x)) -> top(active(x))
Using the Dependency Graph resulted in no new DP problems.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Polo`
` →DP Problem 2`
` ↳Polo`
` →DP Problem 3`
` ↳Polo`
` →DP Problem 4`
` ↳Narrowing Transformation`
Dependency Pairs:
TOP(ok(x)) -> TOP(active(x))
TOP(mark(x)) -> TOP(proper(x))
Rules:
active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
top(mark(x)) -> top(proper(x))
top(ok(x)) -> top(active(x))
On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule
TOP(mark(x)) -> TOP(proper(x))
four new Dependency Pairs are created:
TOP(mark(b)) -> TOP(ok(b))
TOP(mark(c)) -> TOP(ok(c))
TOP(mark(d)) -> TOP(ok(d))
TOP(mark(f(x'', y', z'))) -> TOP(f(proper(x''), proper(y'), proper(z')))
The transformation is resulting in one new DP problem:
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Polo`
` →DP Problem 2`
` ↳Polo`
` →DP Problem 3`
` ↳Polo`
` →DP Problem 4`
` ↳Nar`
` →DP Problem 9`
` ↳Narrowing Transformation`
Dependency Pairs:
TOP(mark(f(x'', y', z'))) -> TOP(f(proper(x''), proper(y'), proper(z')))
TOP(mark(d)) -> TOP(ok(d))
TOP(mark(c)) -> TOP(ok(c))
TOP(mark(b)) -> TOP(ok(b))
TOP(ok(x)) -> TOP(active(x))
Rules:
active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
top(mark(x)) -> top(proper(x))
top(ok(x)) -> top(active(x))
On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule
TOP(ok(x)) -> TOP(active(x))
four new Dependency Pairs are created:
TOP(ok(f(b, c, x''))) -> TOP(mark(f(x'', x'', x'')))
TOP(ok(f(x'', y', z'))) -> TOP(f(x'', y', active(z')))
TOP(ok(d)) -> TOP(m(b))
TOP(ok(d)) -> TOP(mark(c))
The transformation is resulting in one new DP problem:
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Polo`
` →DP Problem 2`
` ↳Polo`
` →DP Problem 3`
` ↳Polo`
` →DP Problem 4`
` ↳Nar`
` →DP Problem 9`
` ↳Nar`
` ...`
` →DP Problem 10`
` ↳Remaining Obligation(s)`
The following remains to be proven:
Dependency Pairs:
TOP(ok(f(x'', y', z'))) -> TOP(f(x'', y', active(z')))
TOP(ok(f(b, c, x''))) -> TOP(mark(f(x'', x'', x'')))
TOP(mark(f(x'', y', z'))) -> TOP(f(proper(x''), proper(y'), proper(z')))
Rules:
active(f(b, c, x)) -> mark(f(x, x, x))
active(f(x, y, z)) -> f(x, y, active(z))
active(d) -> m(b)
active(d) -> mark(c)
f(x, y, mark(z)) -> mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) -> ok(f(x, y, z))
proper(b) -> ok(b)
proper(c) -> ok(c)
proper(d) -> ok(d)
proper(f(x, y, z)) -> f(proper(x), proper(y), proper(z))
top(mark(x)) -> top(proper(x))
top(ok(x)) -> top(active(x))
Termination of R could not be shown.
Duration:
0:00 minutes | 3,614 | 10,456 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-38 | latest | en | 0.388992 |
https://www.marbk-56t.win/wiki/Ken_Young | 1,590,857,655,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347410284.51/warc/CC-MAIN-20200530165307-20200530195307-00569.warc.gz | 799,315,848 | 31,548 | # 666 (number)
(Redirected from Six hundred sixty-six)
← 665 666 667 →
Cardinalsix hundred sixty-six
Ordinal666th
(six hundred sixty-sixth)
Factorization2 × 32 × 37
Greek numeralΧΞϚ´
Roman numeralDCLXVI
Greek prefixἑξακόσιοι ἑξήκοντα ἕξ
hexakósioi hexēkonta héx
Latin prefixsescenti sexaginta sex
Binary10100110102
Ternary2202003
Quaternary221224
Quinary101315
Senary30306
Octal12328
Duodecimal47612
Vigesimal1D620
Base 36II36
Chinese numeral六百六十六
Arabic numeral٦٦٦
666 (six hundred [and] sixty-six) is the natural number following 665 and preceding 667.
666 is called the "number of the Beast" in (most manuscripts of) chapter 13 of the Book of Revelation, of the New Testament,[1] and also in popular culture.
## In mathematics
666 is the sum of the first 36 natural numbers (${\displaystyle \sum _{i=1}^{36}i}$, i.e. 1 + 2 + 3 + ... + 34 + 35 + 36 = 666), and thus it is a triangular number. Notice that 36 = 15 + 21; 15, 21, and 36 are also triangular numbers; and 152 + 212 = 225 + 441 = 666.
In base 10, 666 is a repdigit (and therefore a palindromic number) and a Smith number. A prime reciprocal magic square based on 1/149 in base 10 has a magic total of 666.
The prime factorization of 666 is 2 • 32 • 37. Also, 666 is the sum of squares of first seven primes: ${\displaystyle 2^{2}+3^{2}+5^{2}+7^{2}+11^{2}+13^{2}+17^{2}}$
The Roman numeral for 666, DCLXVI, has exactly one occurrence of all symbols whose value is less than 1000 in decreasing order (D = 500, C = 100, L = 50, X = 10, V = 5, I = 1).
## In religion
### Number of the Beast
666 is often associated with the devil.
In the Textus Receptus manuscripts of the New Testament, the Book of Revelation (13:17–18) cryptically asserts 666 to be "man's number" or "the number of a man" (depending on how the text is translated) associated with the Beast, an antagonistic creature that appears briefly about two-thirds into the apocalyptic vision. Some manuscripts[which?] of the original Greek use the symbols χξϛ chi xi stigma (or χξϝ with a digamma), while other manuscripts spell out the number in words.[2]
In modern popular culture, 666 has become one of the most widely recognized symbols for the Antichrist or, alternatively, the devil. The number 666 is purportedly used to invoke Satan.[citation needed] Earnest references to the number occur both among apocalypticist Christian groups and in explicitly anti-Christian subcultures. References in contemporary Western art or literature are, more likely than not, intentional references to the Beast symbolism. Such popular references are therefore too numerous to list.
It is common to see the symbolic role of the integer 666 transferred to the digit sequence 6-6-6. Some people take the Satanic associations of 666 so seriously that they actively avoid things related to 666 or the digits 6-6-6. This is known as hexakosioihexekontahexaphobia.
The Number of the Beast is cited as 616 in some early biblical manuscripts, the earliest known instance being in Papyrus 115.[3][4]
## In other fields
666 float in a Paris parade
• Is the magic sum, or sum of the magic constants of a six by six magic square, any row or column of which adds up to 111.
• Is the sum of all the numbers on a roulette wheel (0 through 36).[5]
• Was a winning lottery number in the 1980 Pennsylvania Lottery scandal, in which equipment was tampered to favor a 4 or 6 as each of the three individual random digits.[6]
• Was the original name of the Macintosh SevenDust computer virus that was discovered in 1998.
• The number is a frequent visual element of Aryan Brotherhood tattoos.[7]
• Aleister Crowley adopted the title "the Great Beast 666". As such, 666 is also associated with him, his work, and his religious philosophy of Thelema.
• Molar mass of the high-temperature superconductor YBa2Cu3O7.
• In China the number is considered to be lucky and is often displayed in shop windows and neon signs.[8][9] In China, 666 can mean "everything goes smoothly" (the number six has the same pronunciation as the character 溜, which means "smooth".[10]
• Is commonly used by ISPs to blackhole traffic using BGP communities.[11]
## References
1. ^ Beale, Gregory K. (1999). The Book of Revelation: A Commentary on the Greek Text. Grand Rapids, Michigan: Wm. B. Eerdmans Publishing. p. 718. ISBN 080282174X. Retrieved 9 July 2012.
2. ^ "Revelation 13:18". Stephanus New Testament. Bible Gateway. Retrieved 2006-06-22.
3. ^ Novum Testamentum Graece, Nestle and Aland, 1991, footnote to verse 13:18 of Revelation, page 659: "-σιοι δέκα ἕξ" as found in C [C=Codex Ephraemi Rescriptus]; for English see Metzger's Textual Commentary on the Greek New Testament, note on verse 13:18 of Revelation, page 750: "the numeral 616 was also read ..."
4. ^ The Other Number of the Beast
5. ^ a b "666 – professors explain Roulette and Nero in detail; numberphile.com". Archived from the original on 2013-03-31. Retrieved 2013-04-06.
6. ^ May, Steve. "The Devil Made Him Do It and Left Me There, Comfortable". The New Yinzer. Retrieved 2009-09-13.
7. ^ Brook, John Lee (June 2011). Blood In, Blood Out: The Violent Empire of the Aryan Brotherhood. SCB Distributors. ISBN 978-1-900486-80-4. OCLC 793002272. Retrieved 10 January 2018.
8. ^ Mah, Adeline Yen (2009). China: Land of Dragons and Emperors. ISBN 0375890998. Retrieved 2013-12-07.
9. ^ "Know the Meaning of Numbers in Chinese Culture". Retrieved 2014-10-30.
10. ^ "666 – Good day, bad day or just another day?". Retrieved 2014-10-30.
11. ^ RFC 7999 | 1,575 | 5,504 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2020-24 | latest | en | 0.817802 |
http://www.checkyourstudy.com/blog/tag/05/ | 1,556,049,037,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578613603.65/warc/CC-MAIN-20190423194825-20190423220825-00030.warc.gz | 212,397,445 | 40,223 | PROBLEM 2 (20 points) In the following problem we consider two men trying to slide a 6m long plank over an overhead rack. The plank has a mass of 100kg and the coefficient of kinetic friction between the plank and each support is 0.5.
## PROBLEM 2 (20 points) In the following problem we consider two men trying to slide a 6m long plank over an overhead rack. The plank has a mass of 100kg and the coefficient of kinetic friction between the plank and each support is 0.5.
The inductor in problem 5 is connected in series to a 0.5 W and a battery whose emf is 24 V. What current flows in this circuit 0.1 seconds after the switch is closed?
## The inductor in problem 5 is connected in series to a 0.5 W and a battery whose emf is 24 V. What current flows in this circuit 0.1 seconds after the switch is closed?
Question 2 (1 point) Which of the following is correct about interpreting the results of statistical tests? Question 2 options: 1) Obtaining a probability value of .05 tells us the difference between groups is definitely not caused by chance fluctuation. 2) If a probability value falls above .05, then the results will have to be replicated before we can have confidence in them. 3) Obtaining a probability value of .05 gives us confidence that the findings are not the result of chance, but does not eliminate this possibility. 4) A .05 probability value means there is a 5 percent chance the finding reflects a real difference. Question 3 (1 point) Which of the following statements is true about theories of personality? Question 3 options: 1) They provide only a part of the picture of human personality. 2) They support the expert’s viewpoint. 3) Theories are predicted from one hypothesis or another. 4) They are directly tested using empirical methods. Question 4 (1 point) Which of the following statements is correct about hypothetical constructs? Question 4 options: 1) They are useful inventions by researchers that have no physical reality. 2) They are easier to measure than personality variables. 3) They cannot be measured with personality tests. 4) They have poor reliability and validity. Question 5 (1 point) According to the “law of parsimony,” Question 5 options: 1) a good theory generates a large number of hypotheses. 2) the best theory is the one that explains a phenomenon with the fewest constructs. 3) hypotheses are generated from theories. 4) theories should require as few studies as possible to support them. ________________________________________ Question 6 (1 point) Which of the following does a correlation coefficient not tell us? Question 6 options: 1) If the difference between two means reflects a real difference or can be attributed tochancefluctuation. 2) The strength of a relationship between two measures. 3) The direction of a relationship between two measures. 4) How well a score on one measure can be predicted by a score on another measure. Question 7 (1 point) A researcher finds that males make fewer errors than females when working in a competitive situation. However, women make fewer errors than men when working in acooperative situation. This is an example of Question 7 options: 1) a confound. 2) two manipulated independent variables. 3) an interaction. 4) a failure to replicate.
## Question 2 (1 point) Which of the following is correct about interpreting the results of statistical tests? Question 2 options: 1) Obtaining a probability value of .05 tells us the difference between groups is definitely not caused by chance fluctuation. 2) If a probability value falls above .05, then the results will have to be replicated before we can have confidence in them. 3) Obtaining a probability value of .05 gives us confidence that the findings are not the result of chance, but does not eliminate this possibility. 4) A .05 probability value means there is a 5 percent chance the finding reflects a real difference. Question 3 (1 point) Which of the following statements is true about theories of personality? Question 3 options: 1) They provide only a part of the picture of human personality. 2) They support the expert’s viewpoint. 3) Theories are predicted from one hypothesis or another. 4) They are directly tested using empirical methods. Question 4 (1 point) Which of the following statements is correct about hypothetical constructs? Question 4 options: 1) They are useful inventions by researchers that have no physical reality. 2) They are easier to measure than personality variables. 3) They cannot be measured with personality tests. 4) They have poor reliability and validity. Question 5 (1 point) According to the “law of parsimony,” Question 5 options: 1) a good theory generates a large number of hypotheses. 2) the best theory is the one that explains a phenomenon with the fewest constructs. 3) hypotheses are generated from theories. 4) theories should require as few studies as possible to support them. ________________________________________ Question 6 (1 point) Which of the following does a correlation coefficient not tell us? Question 6 options: 1) If the difference between two means reflects a real difference or can be attributed tochancefluctuation. 2) The strength of a relationship between two measures. 3) The direction of a relationship between two measures. 4) How well a score on one measure can be predicted by a score on another measure. Question 7 (1 point) A researcher finds that males make fewer errors than females when working in a competitive situation. However, women make fewer errors than men when working in acooperative situation. This is an example of Question 7 options: 1) a confound. 2) two manipulated independent variables. 3) an interaction. 4) a failure to replicate.
No expert has answered this question yet. You can browse … Read More...
Design a regulated power supply able to supply a maximum load current of 25 mA. Your design specifications are: DC Output: 12 V Load Variation: 0.5 – 2 kilo Ohms Ripple factor < 0.1% for 1 kilo ohm load Load Regulation: Better than 1mV/mA Line Regulation: Better than 2mV/V In your design use diodes 1N4001 and Zener 1N4742. In your design you must mention the values of the resistors, capacitors, transformer turns ratio. Typed report must contain the following in the same order: • Final design with all component values • SPICE simulation results supporting the validity of your design satisfying the specifications. • Discuss your approach to satisfy the specifications. Explain which specifications you met and which you did not meet. Provide discussion and conclusions and comments.
## Design a regulated power supply able to supply a maximum load current of 25 mA. Your design specifications are: DC Output: 12 V Load Variation: 0.5 – 2 kilo Ohms Ripple factor < 0.1% for 1 kilo ohm load Load Regulation: Better than 1mV/mA Line Regulation: Better than 2mV/V In your design use diodes 1N4001 and Zener 1N4742. In your design you must mention the values of the resistors, capacitors, transformer turns ratio. Typed report must contain the following in the same order: • Final design with all component values • SPICE simulation results supporting the validity of your design satisfying the specifications. • Discuss your approach to satisfy the specifications. Explain which specifications you met and which you did not meet. Provide discussion and conclusions and comments.
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Author Name: BIO 218 Natural History Paper General Formatting: (10%) • 1 Margins correct? • 1 Font correct? • 2 Double-spaced? • 2 Pages numbered? • 2 All sections included? • 2 At least 3 pages of text, not more than 5 pages? Project elements (50%) • Introduction: o 8 General background on topic and species (state scientific name!)? o 2 Goes from general to specific? • Review of Journal Articles: o 4 States topic and hypothesis/hypotheses described in articles? o 3 Reports how research was conducted? o 2 Describes specialized materials used? o 2 Discusses type(s) of data collected and how to be analyzed/compared/used? o 3 Reports what happened in the experiments? o 2 If comparisons made, discusses how they were made? o 2 Figure(s) reproduced and cited? o 2 Table(s) reproduced and cited? • Summary/Conclusion: o 10 Synthesizes the results of the experiments and ties the findings of the articles together? • Literature Cited: o 4 At least 3 journal articles (primary literature) used? o 2 References used in paper properly? o 2 References all listed in this section and formatted correctly? o 2 All references listed are in the body of the paper and all references in the body are listed in this section? *0.5% for each extra citation (>3) that is correctly used* Writing Elements (40%) • /15 Grammar or spelling errors? • /15 Writing is clear and flows logically throughout paper? • /10 Appropriate content in each section? Final Paper Total ( %) = /40 Comments:
## Author Name: BIO 218 Natural History Paper General Formatting: (10%) • 1 Margins correct? • 1 Font correct? • 2 Double-spaced? • 2 Pages numbered? • 2 All sections included? • 2 At least 3 pages of text, not more than 5 pages? Project elements (50%) • Introduction: o 8 General background on topic and species (state scientific name!)? o 2 Goes from general to specific? • Review of Journal Articles: o 4 States topic and hypothesis/hypotheses described in articles? o 3 Reports how research was conducted? o 2 Describes specialized materials used? o 2 Discusses type(s) of data collected and how to be analyzed/compared/used? o 3 Reports what happened in the experiments? o 2 If comparisons made, discusses how they were made? o 2 Figure(s) reproduced and cited? o 2 Table(s) reproduced and cited? • Summary/Conclusion: o 10 Synthesizes the results of the experiments and ties the findings of the articles together? • Literature Cited: o 4 At least 3 journal articles (primary literature) used? o 2 References used in paper properly? o 2 References all listed in this section and formatted correctly? o 2 All references listed are in the body of the paper and all references in the body are listed in this section? *0.5% for each extra citation (>3) that is correctly used* Writing Elements (40%) • /15 Grammar or spelling errors? • /15 Writing is clear and flows logically throughout paper? • /10 Appropriate content in each section? Final Paper Total ( %) = /40 Comments:
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A 50 V battery is connected across a 10 ohm resistor and a current of 4 A flows. The internet resistance of the battery is ; A) Zero, B) 0.5 ohm, C) 1.1 ohm, D) 2.5 ohm, E) 5ohm
## A 50 V battery is connected across a 10 ohm resistor and a current of 4 A flows. The internet resistance of the battery is ; A) Zero, B) 0.5 ohm, C) 1.1 ohm, D) 2.5 ohm, E) 5ohm
A concave lens forms a virtual image 0.5 times the size of the object. The object distance is 13.2 cm. Find the focal length of the lens. | 2,552 | 10,890 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2019-18 | latest | en | 0.945645 |
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Estimating the Partition Function of 2-D Fields and the Capacity of Constrained Noiseless 2-D Channels
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62009 IEEE Information Theory WorkshopEstimating the Partition Function of 2-D Fields andthe Capacity of Constrained Noiseless 2-DChannels Using Tree-Based Gibbs SamplingHans-Andrea Loeliger and Mehdi MolkaraieETH ZurichDept. of Information Technology & Electrical Engineering8092 Zurich,¨ SwitzerlandEmail:floeliger, molkaraig@isi.ee.ethz.ch= = = =Abstract—Tree-basedGibbssampling(proposedbyHamzeandde Freitas) is used to compute a Monte-Carlo estimate of thepartition function of factor graphs with cycles. The proposedmethod can be used, in particular, to compute the capacity ofnoiseless constrained 2-D channels.= = = =I. INTRODUCTIONLetX ;X ;:::;X be finite sets, letX be the Cartesian1 2 N4productX =X X :::X , and letf be a nonnegative1 2 N= = = =function f :X!R. We are interested in computing (exactlyor approximately) the quantityX4Z = f(x) (1)x2X= = = =1(or, equivalently, logZ) for cases whereN Fig. 1. Forney-style factor graph for Example 1. The unlabeled boxes X ;:::X are “small” sets (e.g.,jXj =jXj =::: = 2), represent factors as in (4).1 N 1 2 N is large, and f has a “useful” factorization (as will be detailedvalue 1. Letf be the indicator function of this constraint, which canbelow).be factored into factors of the formNote that14 0; if x =x = 1k ‘(x ;x ) = (4)p(x) = f(x) (2) k ‘1; else,Zis a probability mass function onX . We will also need the set with one such factor for each adjacent pair (x ;x ).k ‘The ...
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2009 IEEE Information TheoryWorkshop
Estimating the Partition Function of 2-D Fields and the Capacity of Constrained Noiseless 2-D Channels Using Tree-Based Gibbs Sampling
Hans-Andrea Loeliger and Mehdi Molkaraie ETH Zurich Dept. of Information Technology & Electrical Engineering 8092Zurich,Switzerland Email:{loeliger, molkarai}@isi.ee.ethz.ch
= = = = Abstract—Tree-based Gibbs sampling (proposed by Hamze and de Freitas) is used to compute a Monte-Carlo estimate of the partition function of factor graphs with cycles. The proposed method can be used, in particular, to compute the capacity of noiseless constrained 2-D channels. = = = = I. INTRODUCTION LetX1,X2, . . . ,XNbe finite sets, letXbe the Cartesian 4 productX=X1× X2×. . .× XN, and letfbe a nonnegative = = = = functionf:X →R. We are interested in computing (exactly or approximately) the quantity X 4 Z=f(x)(1) x∈X = = = = 1 (or, equivalently,logZ) for cases where N Fig. 1.Forney-style factor graph for Example 1. The unlabeled boxes X1, . . .XNare “small” sets (e.g.,|X1|=|X2|=. . .= 2),represent factors as in (4). Nis large, andfhas a “useful” factorization (as will be detailed below).value1. Letfbe the indicator function of this constraint, which can be factored into factors of the form Note that 410,ifxk=x`= 1 p(x) =f(x)(2)κ(xk, x`) =(4) 1,else, Z is a probability mass function onX. We will also need the setwith one such factor for each adjacent pair(xk, x`). The corresponding Forney-style factor graph offis shown in 4 +Fig. 1, where the boxes labeled “=” are equality constraints [5]. Xf={x∈ X:f(x)6= 0}.(3) (Fig. 1 may also be viewed as a factor graph as in [4] where the boxes labeled “=” are the variable nodes.) The quantity (1) is known as the “partition function” in This example is known as the 2-D(1,)constrained channel [6]. statistical physics (where it is considered as a function of a “temperature” parameter that is of no concern to us here). TheNote that, in this example,Z=|Xf|. For this particular + 1 example2log computation of (1) is also the key to computing information,limM→∞M2(Z)0.5879is known to nine rates of source/ channelmodels with memory [1]–[3].decimal digits [7], [8]. However, the method proposed in this Iffhas a cycle-free factor graph with not too many states,paper works also for various generalizations of this example then the sum (1) can be computed by sum-product messagefor which this limit is not known to any useful accuracy [6]. passing [1], [4]. In this paper, however, we consider the caseA number of Monte-Carlo methods to estimateZhave where no such cycle-free factor graph exists. In particular, webeen proposed, see [9], [10]. However, these methods assume are interested in examples of the following type.thatfis strictly positive, which excludes applications as in Example 1; more about this will be said in Section II. Example 1 (Simple 2-D Constraint).Consider a grid ofN= In this paper, we propose a new Monte-Carlo method for M×Mbinary (i.e.,{0,1}-valued) variables with the constraint that no two (horizontally or vertically) adjacent variables have both thethe computation ofZthat works also for Example 1. In
✬ ✩ ✬constrast to the method of [3], the proposed method converges 1 to the exact value oflog(Z)in the limit of infinitely many N = = = = samples. II. ESTIMATING1/ZUSINGGIBBSSAMPLING One method to estimate1/Z(and thusZitself) goes as follows. = = = = (1) (2)(K) 1) Drawsamplesx,x, .. . ,xfromXfaccording + top(x)defined in (2). 2) Compute K X 41 1 = = = = Γ =(5) (k) | K|Xff(x) + k=1 It is easily verified thatE[Γ] = 1/Z. This method was proposed in [11], see also [9]. = = = = However, there are two major issues with this method. First, it is usually assumed (as in [9], [11]) thatfis strictly positive. In this case,Xf=Xand|Xf|=|X |is + + ✫ ✪ ✫known. However, this assumption excludes applications as in (k) Example 1. (Indeed, in Example 1, we would havef(x) = 1 Fig. 2.Partition of Fig. 1 into two cycle-free parts (one part inside the two (k)ovals, the other part outside the ovals). for all samplesx, and|Xf|=Zis the desired unknown + quantity.) We will see how this issue is resolved by an idea from [12]. (1)B. Tree-BasedEstimation of1/Z[12] Second, there is the problem of generating the samplesx, (2) (K) x, . . . ,xaccording top(x). A standard general method isLet X 4 Gibbs sampling [9], [13], which, however, produces strongly fA(xA) =f(xA, xB)(8) dependent samples. In consequence, the required number ofxB samplesKis likely to exceed the limits of practicality. We and will see how this issue is eased by tree-based Gibbs samplingX 4 fB(xB) =f(xA, xB).(9) as proposed by Hamze and de Freitas [14]. xA III. PROPOSEDNEWMETHOD Since The proposed method combines tree-based Gibbs samplingX X X f(xA) =f(xB) =f(x) =Z,(10) from [14] with an idea from [12]. xAxBx Let(A, B)be a partition of the index set{1, . . . , N}such that, (i) for fixedxA, the factor graph off(x) =f(xA, xB) we can estimateZby applying the algorithm of Section II to is a tree and (ii) for fixedxB, the factor graph off(x) = fAor tofB(as noted in [12]). Specifically, an estimateΓA f(xA, xB)is also a tree. An example of such a partition is of1/Zis formed as follows: shown in Fig. 2.((1) (2)K) 1) Drawsamplesx,x. . ,, .xfrom(XA)ac-+ A AA f A 4P A. Tree-BasedGibbs Sampling [14] cording top(xA) =p(xA, xB) =fA(xA)/Z. xB (0) (0) (0) 2) Starting from some initial configurationx= (xx ,), A B K (k) (k) X (k) the samplesx= (x ,x),k= 1,2, . . ., are created as41 1 A B ΓA=(11) (k) (k) follows. First,xis sampled according toK|(XA)| + A ff(x) Ak=1A (k1) (k1) p(xA|xB=x)f(xA, x);(6) where B B 4 (k)(XA) ={xA:fA(xA)6= 0}.(12) + f thenxis sampled according to A B (k) (k)By symmetry, we also have an analogous estimateΓB. The p(xB|xA=x)f(x ,xB).(7) A A computation of X The point is that the sampling can be done very efficiently(k) (k) f(x) =f(x ,xB),(13) A A in both cases since the corresponding factor graphs are cycle-xB free; see the appendix for details. Tree-based Gibbs sampling mixes much faster than naivewhich is required in (11), is easy since the corresponding Gibbs sampling [14].factor graph is a tree.
229
Fig. 3.Estimated capacity (in bits per symbol) vs. the number of samplesK for a10×10grid with a(1,)constraint (Example 1). The plot shows 10 independent sample paths, each with two estimates, one fromΓAand one fromΓB.
The quantity|(XA)|in (12) may be easy to determine + f A even iffis not strictly positive. This applies, in particular, to Example 1 (and many similar examples) where (XA) ={xA:f(xA,0)6= 0}.(14) + f A P In this case,|(XA)|=f(xA,0)is easily computed by + f xA A sum-product message passing in the (cycle-free) factor graph off(xA,0). C. AHappy Combination (1) (2) It is now obvious to create the required samplesx,x, A A (K) . . . ,xin (11) by means of tree-based Gibbs sampling as A in Section III-A. The marginals (13) may then be obtained as a by-product of the tree-based sampling (see the appendix). We thus obtain two estimates,ΓAandΓB, as a by-product of tree-based Gibbs sampling with virtually no extra computations. IV. NUMERICALEXPERIMENTS Some experimental results with the proposed method are shown in Figures 3 through 6. All figures refer tofas in 1 Example 1 and show the quantity (the “capacity”)log (Z). 2 N Figures 3 and 4 use a factor graph partition as in Fig. 2. In Fig. 3, we haveN= 10×10and the estimated capacity is about0.6082. In Fig. 4, we haveN= 60×60; for this size of grid there are issues with slow convergence. To improve the convergence and to speed up the mixing, we can partition the factor graph (the extension of Fig. 1 toN= 60×60) into “thicker” vertical strips. Such thick strips have cycles, but exact sum-product computation is still possible, e.g., by converting the strip into an equivalent cycle-free factor graph. The computation time is exponential in the thickness of the strip, but the faster mixing (as shown in Figures 5 and 6) results in a substantial reduction of total computation time for strips of moderate width. From Fig. 6, the estimated capacity is about0.5914.
Fig. 4.Estimated capacity (in bits per symbol) vs. the number of samplesK for a60×60grid with a(1,)constraint (Example 1). The plot shows 10 independent sample paths, each with two estimates, one fromΓAand one fromΓB.
230
Fig. 5.Same conditions as in Fig. 4, but with strips of width two.
Fig. 6.Same conditions as in Fig. 4, but with strips of width three.
Also shown in the figures (as a horizontal dotted line) isgk1gkgk+1 1 the infinite-grid limitlimM→∞2log2(Z)0.5879, whichXk2Xk1XkXk+1 M ✛ ✛ is known for this simple example (see Section I). All figures show the estimates fromΓAand fromΓBfor ←− Fig. 7.Forney-style factor graph of (16) with messagesµX(17). k several independent experiments. ←− V. BOUNDS FORINFINITEGRID in Fig. 7 (cf. [5]). We then haveµXn(xn) = 1and X 4 4 1 LetCM=2log (Z)be the capacity of a constraint asµ(x) =g(xx ,)µ(x)(17) M2Xkk k+1k k+1Xk+1k+1 in Example 1 for anM×Mgrid. It is clear (from tiling thexk+1 n whole plane withM×Msquares) thatCCMfor anyX Y finiteM.=gm(xm1, xm)(18) xk+1,...,xnm=k+1 On the other hand, by tiling the plane withM×Msquares separated by all-zero guard rows and all-zero guard columns, fork=n1, n2, . . . ,1. Then M2 we obtainCCM( ).X M+1 p(x1) =p(x1, . . . , xn)(19) In the example of Figures 4–6 (withM= 60), we thus x2,...,xn obtain0.5721C0.5914. ←− µX1(x1)(20) VI. CONCLUDINGREMARKSand ←− gk(xk1, xk)µXk(xk) p(xk|xk1) =(21) ←− We have shown that tree-based Gibbs sampling (as proposed µXk1(xk1) by Hamze and de Freitas) can be used to compute an estimate fork= 2, . . . , n. The proof of (21) follows from noting that of the partition function with virtually no extra computational cost. The proposed method can be used, in particular, top(x) =γ µ(x)µ(x)(22) k1Xk1k1Xk1k1 compute (a Monte Carlo estimate of) the capacity of noiseless and constrained 2-D channels. Our preliminary numerical experi-ments are encouraging. p(xk1, xk) =γ µXk1(xk1)gk(xk1, xk)µXk(xk)(23) −→ whereµXk1is the forward sum-product message along the APPENDIX: SAMPLING FROMMARKOVCHAINS edgeXk1and whereγis the missing scale factor in (16). We recall some pertinent facts about the simulation ofWe also note that X X Markov chains and cycle-free factor graphs. Letp(x) = ←− µX1(x1) =g(x)(24) p(x1, . . . , xn)be the probability mass function of a Markov x1x chain. Ifp(x)is given in the form whereg(x)is defined as the right-hand side of (16). In this n Y paper, this fact is used to compute the marginals (13) as a p(x) =p(x1)p(xk|xk1),(15) by-product of the sampling. k=2 The generalization of all this to arbitrary factor graphs without cycles is straightforward. then it is obvious how to create i.i.d. samples according to p(x). Now consider the case wherep(x)is not given in the ACKNOWLEDGEMENT form (15), but in the more general form The authors wish to thank David MacKay and Iain Murray n Y for pointing out to us the work by Hamze and de Freitas [14]. p(x)gk(xk1, xk)(16) k=2 REFERENCES with general factorsgk. It is then still easy to create i.i.d. samples according top(x), which may be seen as follows.,ˇvicca,dn.WeZgn.O.Vontobel,A.Ka,PerigelLoA..-,HdlonrA.D]1[ “Simulation-based computation of information rates for channels with First, a probability mass function of the form (16) can be memory,”IEEE Trans. Inform. Theory, vol. 52, no. 8, pp. 3498–3508, rewritten in the form (15) (which allows efficient simulation). August 2006. Second, this reparameterization ofp(x)may be efficiently[2] P. H. Siegel, “Information-theoretic limits of two-dimensional optical recording channels,” inOptical Data Storage 2006(Proc. of SPIE, Vol. carried out by backward sum-product message passing, as 6282, Eds. Ryuichi Katayama and Tuviah E. Schlesinger), Montreal, will be detailed below. The resulting algorithm is know as Quebec, Canada, April 23–26, 2006, pp. 62820W-1 – 2820W-13. “backward-filtering forward-sampling” (or, in a time-reversed[3] O. Shental, N. Shental, S. Shamai (Shitz), I. Kanter, A. J. Weiss, and Y. Weiss, “Discrete-input two-dimensional Gaussian channels with version, as “forward-filtering backward-sampling”) [15]. memory: estimation and information rates via graphical models and ←− Specifically, letµXkbe the backward sum-product messagestatistical mechanics,”IEEE Trans. Inform. Theory,vol. 54, pp. 1500– along the edgeXkin the factor graph of (16), as is illustrated1513, April 2008.
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[4] F.R. Kschischang, B. J. Frey, and H.-A. Loeliger, “Factor graphs and the sum-product algorithm,”IEEE Trans. Inform. Theory,vol. 47, pp. 498– 519, Feb. 2001. [5] H.-A.Loeliger, “An introduction to factor graphs,”IEEE Signal Proc. Mag.,Jan. 2004, pp. 28–41. [6] K.Kato and K. Zeger, “On the capacity of two-dimensional run-length constrained channels,”IEEE Trans. Inform. Theory,vol. 45, pp. 1527– 1540, July 1999. [7] N.J. Calkin and H. S. Wilf, “The number of independent sets in a grid graph,”SIAM J. Discr. Math.,vol. 11, pp. 54–60, Feb. 1998. [8] W.Weeks and R. E. Blahut, “The capacity and coding gain of certain checkerboard codes,”IEEE Trans. Inform. Theory,vol. 44, pp. 1193– 1203, May 1998. [9] G.Potamianos and J. Goutsias, “Stochastic approximation algorithms for partition function estimation of Gibbs random fields,”IEEE Trans. Inform. Theory,vol. 43, pp. 1984–1965, Nov. 1997.
[10] F.Huang and Y. Ogata, “Comparison of two methods for calculating the partition functions of various spatial statistical models,”Australian & New Zealand Journal of Statistics,vol. 43 (1), pp. 47–65, 2001. [11] Y. Ogata and M. Tanemura, “Estimation of interaction potentials of spatial point patterns through the maximum likelihood procedure,”Ann. Inst. Statist. Math.,vol. 33, pp. 315–338, 1981. [12] H.-A.Loeliger and M. Molkaraie, “Simulation-based estimation of the partition function and the information rate of two-dimensional models,” Proc. 2008 IEEE Int. Symp. on Information Theory,Toronto, Canada, July 6–11, 2008, pp. 1113–1117. [13] D.J. C. MacKay, “Introduction to Monte Carlo methods,” inLearning in Graphical Models,M. I. Jordan, ed., Kluwer Academic Press, 1998, pp. 175–204. [14] F.Hamze and N. de Freitas, “From fields to trees,”Proc. Conf. on Uncertainty in Artificial Intelligence,Banff, July 2004. [15] D.Gamerman and H. F. Lopes,Markov Chain Monte Carlo: Stochastic Simulation for Bayesian Inference.2nd ed., CRC Press, 2006.
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## Marching Ants - Multiplication Math Game
Content Skill: Multiplication
Common Core State Standards: CCSS.Math.Content.3.OA.C.7 - Fluently multiply within 100, using strategies such as the relationship between multiplication and division or properties of operations. By the end of Grade 3, know from memory all products of two one-digit numbers.
## Description
It's picnic time, so watch out for ants! They're out to steal your food, but you can keep them away by solving multiplication problems. Once you've solved enough problems, you can help the ants get their own snack!
## Instructions
Step 1
For game instructions, click "How to Play." To start the game, click "Play."
Step 2
Choose the fact family or families you want to practice by clicking on them.
Step 3
The ants are trying to steal your picnic! Click on the food with the correct answer to the multiplication problem before the ant can get to it. If you can't solve the problem in time, you'll lose a little bit from your health bar on the left side of the screen. Solve fifteen problems to move onto the next level.
Step 4
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Module 4: PID Control
### Module Key Study Points
• understand the basics of PID control
• learn the effects of 3 parameters to system response
• learn how response is degraded by integrator windup
• How to tune PID parameters by Ziegler-Nichols Frequency Domain method
PID stands for Proportional, Integral, and Derivative. It is a standard control structure used successfully in many industrial applications. Commercial PID controllers can be bought off the shelf and installed in the system. After some setup and tuning the three PID gains, and perhaps some additional parameters, the system is up and running in no time. Of course, this means the control engineer is certain the plant can be handled by the PID controller and understand how to adjust the control parameters. Without such knowledge, it could be a frustrating experience to her.
So, in this module we discuss some basics of PID control and focus on how to analyze a feedback sysem, with Scilab, build an Xcos diagram and simulate. For a detailed study, we recommend [1]. For those interested in digital PID implementation, document [3] on this site might be helpful. The e-book [4] provides more information on independent robot joint control using PID.
### Different forms of PID
The so-called “textbook” form of PID controller is described as follows
This algorithm computes the control variable u as output, given the input e, the error between the command and plant output. We see that the control variable is a function of 3 terms: P (proportional to error), I (time integral of error), and D (derivative of error), with corresponding control parameters K (proportional gain), Ti (integral time), and Td (derivative time), respectively. Taking Laplace transform of (1), we have
with the transfer function of PID controller
Another common structure of PID algorithm is represented by
In this form, the controller gains are distributed to each of the PID terms separately, with its transfer function
For our discussion of PID controller, we focus on the form (4) and (5). Nevertheless, it is easy to convert between parameters of (1) and (4) by the relations
Ex. 1: To get started with Xcos simulation of a PID feedback system, download file pid_feedback.zcos . This launches an Xcos diagram in Figure 1. This diagram can easily be constructed using Xcos standard palettes. It requires only a step_function, clock_c (from Sources palette), PID and CLR blocks (from Continuous time systems palette), SUMMATION (from Mathematical operations palette), MUX (from Signal Routing palette), and CSCOPE (from Sinks palette). The plant transfer function is a robot joint driven by DC motor model from module 1, with J = 10 and B = 1.
Figure 1 pid_feedback.zcos basic PID simulation diagram
Click on the Simulation Start button to see the step response. Try adjusting the PID gains and plant parameters to see their effects on the closed-loop system.
PID parameter tuning is well studied in control engineering. Here we give a brief guideline. The proportional gain Kp is a dominant quantity that normally has some nonzero value. The integral gain Ki helps eliminate steady state error but too high value could introduce overshoot and oscillation. The derivative gain Kd could help the response to reach steady state faster but could amplify high frequency noise, and could affect stability if set too high.
When the plant model is not known, adjusting these 3 gains to achieve good response could be problematic for an inexperienced user. Later we discuss a tuning procedure. Commercial PID controller products usually have auto-tuning functions for user convenience.
Ex. 2: To experiment with the tracking and disturbance attenuation performance of PID control, we setup a feedback diagram in Figure 2.
Figure 2 PID feedback diagram
which can be constructed as an Xcos model pid_dist.zcos in Figure 3. The joint inertia and friction are set to J = 10 and B = 0.1, respectively. The step disturbance at the plant input is active at t = 2 sec, with step value of 80.
Figure 3 pid_dist.zcos Xcos model for PID feedback with disturbance
Note that in this example, we have the luxury of knowing the exact plant model. So it is possible to solve for a good set of PID gains. An approach used in [2] turns off the Ki gain first and solve for Kp and Kd that yields a critical-damped system at chosen natural frequency. Then Ki is adjusted to eliminate steady-state error.
To elaborate, it is shown in [2] that when Ki is set to 0, the resulting second order closed-loop characteristic polynomial is described by
which gives
For example, suppose we select a critical-damped closed loop system \zeta = 1, with natural frequency \omega = 8. From (8), this yields Kp = 640 and Kd = 160. Table 1 below gives calculated P and D gain values for 3 selected natural frequencies.
ω Kp Kd
4 160 80
8 640 160
12 1440 240
Table 1 Kp and Kd gains for 3 different natural frequencies
Using these 3 sets of gain values, we simulate the step responses. The results are shown in Figure 4. As expected, the higher the natural frequency, the faster the response. We also see that the steady-state error from step disturbance cannot be eliminated with the PD controller.
Figure 4 step responses of PD control with disturbance d = 80
To fix this problem, we need to turn on the Ki gain of PID controller. With the controller transfer function described by (5), the closed-loop system is now of third order
with the characteristic polynomial
Applying Routh-Hurwitz criteria to (10), it can be concluded that, given all positive PID gains, the feedback system is stable if
The same Xcos diagram in Figure 3 can be used by putting some positive integral gain in the PID block. Let us fix Kp = 1440 and Kd = 240, corresponding to natural frequency \omega = 8 rad/s. Figure 5 shows the responses from setting some values of Ki. We see that the steady-state error to command and disturbance is eliminated by increasing Ki, though higher value could introduce more overshoot.
Figure 5 step responses of PID control
Applying the bound (11) to this example yields
This can be verified by setting Ki = 34575 . The oscillating response in Figure 6 shows that the closed-loop system is at the brink of stability.
Figure 6 closed-loop response at controller gains Kp = 1440, Ki = 34575, Kd = 240
### Effect from Saturation
Analysis of PID feedback control is often performed under the assumption that the system is purely linear. In real devices, nonlinear elements emerge at certain points in the feedback loop. Saturation is one of them that could cause undesirable system response. This nonlinear effect normally results from physical limits of signals and system parameters, say, maximum motor torque and current. Another common saturation is the input limit of servo amplifier. An industrial servo amp might allow input voltage command in the range \pm 10 volts, for example.
Effect from saturation is more pronounced when integral term is used in PID. The response could have excessive overshoot due to error accumulation in the integral term. This is known as integrator windup. See [1] for more detail.
Ex. 3: To see an effect from input saturation, construct an Xcos model as in Figure 7, or download pid_ilim.zcos. It represents a comparison of two closed-loop systems, which are basically the same except that the lower one has its input bounded between \pm 50 units.
Figure 7 pid_ilim.zcos Xcos diagram for Example 3
The controller gains are set at Kp = 640, Ki = 1000, Kd = 160 . Running the simulation yields the step responses in Figure 8, and the control variables (controller output) in Figure 9. We see that input saturation causes significantly larger overshoot and also worsens disturbance response.
Figure 8 effect of integrator windup on output response
Figure 9 controller output with plant input limit
### Tuning the PID Parameters
As stated before, adjusting the PID gains from scratch to achieve a good response may not be trivial. As a result, most commercial PID controllers have functions to tune the 3 parameters automatically. This is normally called “autotuning” feature. One of autotuning methods suggested in the literature makes use of some relay feedback mechanisms, which is closely related to a manual tuning scheme known as Ziegler-Nichols Frequency Domain (ZNFD) method. Hence we discuss this ZNFD scheme in the last part of this module.
Note: The original ZNFD method applies the “textbook” PID equation (1). But since (1) and (4) are closely-related by (6), here we make a conversion so that standard PID block in Xcos, which assumes the form (4), can be conveniently used.
To tune a PID controller manually by ZNFD method, follow this procedure
• turn off both the integral and derivative terms; i.e., setting Ki = Kd = 0. So now the PID is left only with the proportional gain K. We crank K up to the point that the closed-loop system starts to oscillate. At this point, the plant output will swing in a constant sinusoid motion, not growing and not dying out. Write this value down on a paper as Ku.
• Measure the period of oscillation. Write it down as Tu.
• Use Table 2 to tune the controller parameters
Controller Form Kp Ki Kd
P 0.5Ku
PI 0.4Ku 1.25Ku/Tu -</td
PID 0.6Ku 2Ku/Tu KuTu/8
Table 2: suggested PID parameters from ZNFD method
From the above procedure, we could observe a drawback of the basic ZNFD method: the system must oscillate at some point of gain setting before it becomes unstable. This implies the Nyquist plot must cross the negative real axis at some point. For the DC motor model we are using, the gain margin is infinity. Hence it is difficult to find the oscillating point and the Ku value.
There is some trick to get around such limitation, but this is beyond the scope of this module. For our basic study, we simply modify the DC motor robot joint by adding some dynamics to it, so that the Nyquist plot of L(j\omega) crosses the negative real axis.
Ex. 4: Let us assume that the PID output is passed through a LPF with cutoff frequency 500 Hz before feeding the plant input. Using the filter design guide from module 1, we have the LPF transfer function as
To simulate, construct an Xcos model in Figure 10, or download pid_znfd.zcos. With this model, we are ready to perform the ZNFD tuning procedure described above. Click on the PID block to set the I and D gains to zero. Then start increasing the P gain until the output oscillates. This is what you must do in a real application when the plant model is not known.
Figure 10 pid_znfd.zcos Xcos diagram for ZNFD tuning method
Well, in this example we do know the plant model, don’t we? So, to save us some time, the Ku value that causes output oscillation can be found, for example, using the method from Module 2. First from the loop transfer function containing the plant, LPF, and PID controller with Kp = 1.
-->kp = 1;
-->ki = 0;
-->kd = 0;
-->s=poly(0,'s');
-->P=syslin('c',1/(10*s^2+0.1*s)); // plant
-->H = syslin('c',1/(1+0.0003*s)); // filter
-->C=syslin('c',kp + ki/s + kd*s); // controller
-->L = C*H*P; // loop t.f
Then find the gain where the closed-loop pole touches the j\omega axis.
->[kmax,s]=kpure(L)
s =
5.7735027i
kmax =
333.33433
resulting in Ku = 333.
Verify this by setting the P gain to 333 in PID block. This yields the desired oscillatory response as shown in Figure 11. The oscillation period can be roughly measured from the plot to yield Tu = 1 sec.
Figure 11 oscillatory response with Kp = 333, Ki = Kd = 0
So from the last row of Table 2, ZNFD method suggests the three PID parameters as
Setting these gains gives the response in Figure 12.
Figure 12 response from PID gains suggested by ZNFD method
Note that the overshoot is quite excessive (70%). In a sense, ZNFD just gives us some good values to start with. We may want to fine-tune the PID gains to improve the response further. For example, decreasing the Ki gain would bring the overshoot down. Figure 13 shows the response from original PID gains (green), compared with the case when Ki is reduced to 300 (blue), and 200 (magenta).
Figure 13 reduce overshoot in the response by decreasing Ki
The ZNFD method could be explained using a Nyquist diagram in Figure 14. The diagram shows how a point x on the curve is moved related to the P, I , and D terms. Using the P term alone, x could be moved in radial direction only. The I and D terms help provide more freedom to move perpendicular to the radius. It can be shown that by using ZNFD method, the critical point (-1/Ku, 0) is moved to the point -0.6 – 0.28i. The distance of this point to the critical point is 0.5. So the sensitivity peak is at least 2. This explains the high overshoot in the step response.
Figure 14 How a point on Nyquist curve is moved with PID control
### Summary
PID is a simple control structure that is still used widely in various industrial applications. It is a close relative to the lead-lag compensator explained in module 2, except that its functionality may be more user-friendly. With some knowledge and practice, an engineer or technician would be able to tune and operate a plant equipped with PID control.
In this module we discuss the basics of PID feedback systems, with emphasis on Xcos simulations to show how the responses are related to three control parameters, as well as effect from input saturation that could worsen the response. Without some good starting values, tuning the PID gains can be cumbersome for a novice. So at the end, we mention a manual tuning procedure known as the Ziegler-Nichols frequency domain method. Some auto-tuning scheme of a commercial PID controller, such as the relay feedback method, is based on the ZNFD manual tuning.
### References
1. K.J. Astrom and T.Hagglund. PID Controllers, 2nd ed., Instrument Society of America, 1995.
2. M.W.Spong, S. Hutchinson and M. Vidyasagar, Robot Modeling and Control. John Wiley & Sons. 2006.
3. V. Toochinda. Digital PID Controllers, 2009.
4. V.Toochinda. Robot Analysis and Control with Scilab and RTSX. Mushin Dynamics, 2014. | 3,292 | 14,142 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2019-30 | latest | en | 0.908969 |
https://runestone.academy/ns/books/published/thinkcpp/Chapter11/mixedUpCode_orig.html | 1,726,294,902,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651559.58/warc/CC-MAIN-20240914061427-20240914091427-00014.warc.gz | 461,811,032 | 7,800 | Mixed Up Code Practice¶
Answer the following Mixed-Up Code questions to assess what you have learned in this chapter.
Construct a block of code that would make the print function into a member function.
struct Student {
int id, year;
string name;
};
void printStudent (const Student& stu) {
cout << stu.id << ":" << stu.year << ":" << stu.name << endl;
}
int main ( ) {
Student s1 = { 56673, 2023, "Bob" };
printStudent (s1);
}
Let’s make an album! Write the struct definition for Album, which should have instance variables name and year. Include a member function called check that returns true if the song was released after 2015.
Put the necessary blocks of code in the correct order to establish the convertToSeconds member function.
Create the Student::is_older() function as it would be defined INSIDE of the Student structure definition. This function checks if the current Student is older than another Student. The function is invoked on the current Student.
Put the necessary blocks of code in the correct order to initialise a constructor for type Days that takes in the number of days and initialises the member variables days, weeks, years.
Let’s write two constructors for Student. One with no arguments and one with arguments. Put the necessary blocks of code in the correct order.
Implement two constructors for the Penguin structure. One should be a default constructor, the other should take arguments. The weight needs to be converted from pounds to kilograms in the second constructor
Put the necessary blocks of code in the correct order to make the AddDays function below a member function a member function.
Days AddDays (const Days& d1, const Days& d2) { int days = convertToDays (d1) + convertToDays(d2); return makeDays (days); }
Put the necessary blocks of code in the correct order to create a struct Penguin that stores name and age. In addition have 2 constructors and declare Penguins in main such that both are called.
Put the necessary blocks of code in the correct order in order to write a header (.h) file for the struct Student. | 441 | 2,082 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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} | 2.65625 | 3 | CC-MAIN-2024-38 | latest | en | 0.855816 |
https://codedump.io/share/gFfIOqDH9w8/1/java-matrix-runtime-error | 1,480,986,918,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541864.44/warc/CC-MAIN-20161202170901-00347-ip-10-31-129-80.ec2.internal.warc.gz | 764,414,218 | 9,888 | giovanni ghisellini - 7 months ago 24
Java Question
# Java matrix runtime error
Exercise letter:
Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.
For example, Given the following matrix:
``````[
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ]
]
You should return [1,2,3,6,9,8,7,4,5].
``````
Given code:
``````public class Solution {
public List<Integer> spiralOrder(int[][] matrix) {
}
}
``````
My code:
``````public List<Integer> spiralOrder(int[][] matrix) {
if(matrix == null || (matrix.length == 0))
return new ArrayList<Integer>();
int arriba = 0;
int derecha = matrix[0].length - 1;
int abajo = matrix.length - 1;
int izquierda = 0;
List<Integer> retorno = new ArrayList<Integer>();
while(true)
{
for(int i = izquierda; i <= derecha; i++)
arriba++;
for(int i = arriba; i <= abajo; i++)
derecha--;
for(int i = derecha; i >= izquierda; i--)
abajo--;
for(int i = abajo; i >= arriba; i--)
izquierda++;
if(izquierda >= derecha)
return retorno;
}
}
}
``````
The error:
``````Runtime Error Message:
Line 13: java.lang.ArrayIndexOutOfBoundsException: 1
Last executed input:
[[1,2,3,4,5,6,7,8,9,10]]
``````
Any suggestions? I can't really tell what is wrong. Why is it out of bounds?
Exercise can be found here
I tried your method with this matrix:
``````int[][] matrix = {{1,2,3},
{2,3,4},
{3,4,5}};
``````
and I did not get any `ArrayIndexOutOfBoundsException`. Your code does not seem to throw any errors.
However, I noticed that the output is not as expected. The output it gave me was `12345432` (only 8 numbers), missing the number `3` in the middle of the matrix.
After having a thorough look at your code I realised that the error lies in `if(izquierda >= derecha)`. If you change this to `if(izquierda > derecha)` it will not miss the `3`. For the same reason you did this, you need to also check for `arriba > abajo`, otherwise your program does not work for any matrix that has more columns than rows.
Edit: You need these checks after every for-loop.
I suggest you move the `return retorno;` outside the while-loop, and insert `break` in the checks:
``````public List<Integer> spiralOrder(int[][] matrix) {
if(matrix == null || (matrix.length == 0))
return new ArrayList<Integer>();
int arriba = 0;
int derecha = matrix[0].length - 1;
int abajo = matrix.length - 1;
int izquierda = 0;
List<Integer> retorno = new ArrayList<Integer>();
while(true)
{
for(int i = izquierda; i <= derecha; i++)
arriba++;
if(arriba > abajo)
break;
for(int i = arriba; i <= abajo; i++)
derecha--;
if(izquierda > derecha)
break;
for(int i = derecha; i >= izquierda; i--)
abajo--;
if(arriba > abajo)
break;
for(int i = abajo; i >= arriba; i--)
Side note: It was not easy making sense of `izquierda`, `derecha`, `arriba`, and `abajo`, since they "move around" in the spiral. But if you think of them as "marking" the non-checked sides of the matrix, you see that `izquierda` and `derecha` are both marking down the number `3`, but since you have written `>=`, you ignored it and missed the `3`. | 883 | 3,047 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2016-50 | longest | en | 0.679626 |
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Prev Next | 1,261 | 5,377 | {"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} | 2.703125 | 3 | CC-MAIN-2018-09 | latest | en | 0.865926 |
https://spiritualscienceofsingularity.wordpress.com/2016/08/29/the-pole-is-not-the-point/ | 1,540,126,142,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583514005.65/warc/CC-MAIN-20181021115035-20181021140535-00342.warc.gz | 816,470,239 | 21,648 | # The pole is not the point
I’m having some difficulties in explaining to pro’s in physics what I mean with a Monopole. It seems they are looking for a point while I’m looking at, well…a pole actually. State of the art research papers keep validating my perspective on these elusive entities, but the authors themselves appear somewhat mystified. Perhaps it would help if they imagined their data to relate to a pole instead of a point?
Always trying to be of help, I will now offer the idiots (that’s me, not the scientists) 101 on monopoles.
The above image is not how a monopole is described in physics and math. It sounds a bit strange perhaps, but this is how our friends in the Cold lab’s looks at it:
Ordinarily, magnetic poles come in pairs: they have both a north pole and a south pole. As the name suggests, however, a magnetic monopole is a magnetic particle possessing only a single, isolated pole—a north pole without a south pole, or vice versa.
So their monopoles are believed to be isolated points, being either sout or north. Perhaps this is because physics have adopted a lot of thinking that belongs to mathematics, not physics. In math, a pole is not a pole at all, but a particular singularity of a meromorphic function. The link to singularity is natural and valuable, because my very physical pole is, when in isolation and not in a pair, an essential singularity. But for now let’s stay physical. So the real monopole has no less than 2 ends to it. I repeat for clarity:
A Monopole has 2 ends of 1 extension.
A Monopole is NOT an isolated point.
The above image is a revision of Wikipedias piece on Monopoles. If you look it up, you will see that everything is backwards in the conventional picture where poles are believed to be points. From that perspective, I can easily understand why uniting electricity and magnetism is so difficult.
If we make a pole a pole, it is not difficult at all.
And as always I remind you of the fact that a single monopole, like the one pictured here, is likely to exist only once, and that would be as an initial universal state. A single monopole is not possible in a universe that is already evolving and “in space”. The single monopole is of such force that only itself can break it apart, and that is what symmetry breaking is about. But in my model, there’s no small quantum fluctuations that does it. Instead it is an inevitable effect caused by the forced geometry of the monopole/singularity and the sequence in which it operates.
But for now, the take away message is that a monopole is a pole of frequency, not a mathematical point, and that the pole itself, by spin, extends a surface of currency as it contracts towards its mean length.
And no, all notions of space are misleading since the singular monopole is non-dimensional. | 631 | 2,805 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2018-43 | latest | en | 0.947519 |
https://www.indiabix.com/chemical-engineering/chemical-engineering-basics/discussion-8668 | 1,606,189,505,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141171077.4/warc/CC-MAIN-20201124025131-20201124055131-00356.warc.gz | 716,468,727 | 5,473 | # Chemical Engineering - Chemical Engineering Basics - Discussion
47.
The concentration of (H+) ions is 4 x 10-5 in a solution. Then pH of the solution will be (Given log 4 = 0.6 )
[A]. 4.4 [B]. 5.6 [C]. 8.4 [D]. 2.4
Explanation:
No answer description available for this question.
Rashi said: (Jan 18, 2014) pH = -Log conc[H+] = - log[4*10^-5] = -[log4+log10^-5] = 0.6+(-5log10) = -(0.6-5)=4.4. | 153 | 402 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.711576 |
https://www.physicsforums.com/threads/solve-for-x-with-4-variables-and-1-equation.625048/ | 1,511,147,612,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805894.15/warc/CC-MAIN-20171120013853-20171120033853-00771.warc.gz | 850,057,242 | 14,410 | # Solve for 'x' (with 4 variables and 1 equation)
1. Aug 2, 2012
### PhizKid
1. The problem statement, all variables and given/known data
$$ac + bc = (a - c)(a + b) - xabcd^2$$
2. Relevant equations
Various factoring rules (difference of two squares, etc.)
3. The attempt at a solution
I got it expanded to:
$$ac + bc = a^2 + ab - ac - bc - xabcd^2\\0 = a^2 + ab - 2ac - bc - bd - xabcd^2$$
I tried factoring different ways, but it didn't really get me anywhere, and there aren't any "special" factoring rules that I can identify immediately here.
Never mind, I wrote down the problem incorrectly
Last edited: Aug 2, 2012
2. Aug 2, 2012
### Mentallic
Think of this:
What if you were asked to make x the subject (in other words, solve for x by putting it alone on one side of the equation) in the equation
$$a=b-cx$$
Can you solve this? | 258 | 848 | {"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} | 3.265625 | 3 | CC-MAIN-2017-47 | longest | en | 0.922675 |
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# Low Quant Score -- will it hurt?
Author Message
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Intern
Joined: 21 Aug 2005
Posts: 2
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Low Quant Score -- will it hurt? [#permalink]
### Show Tags
21 Aug 2005, 09:02
I took the GMAT yesterday. I scored a 700 overall, with a 42 Quant and 44 Verbal. Am I going to be hurt by this low Quant score?
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Joined: 18 Nov 2004
Posts: 1430
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21 Aug 2005, 09:10
It depends, if ur target schools are in top 10 then "yes", specifically if the schools value high quant score in GMAT like MIT sloan. What is 42 in percentile score ?
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Joined: 21 Aug 2005
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21 Aug 2005, 10:08
66th percentile
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21 Aug 2005, 10:08
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# Low Quant Score -- will it hurt?
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 611 | 2,162 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-47 | latest | en | 0.865085 |
https://astro-imaging.com/stars/does-rocket-need-fuel-in-space.html | 1,653,192,903,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662543797.61/warc/CC-MAIN-20220522032543-20220522062543-00454.warc.gz | 164,351,371 | 16,820 | # Does rocket need fuel in space?
Contents
As opposed to an airplane engine, which operates within the atmosphere and thus can take in air to combine with fuel for its combustion reaction, a rocket needs to be able to operate in the emptiness of space, where there’s no oxygen. Accordingly, rockets have to carry not just fuel, but also their own oxygen supply.
## What happens if a rocket runs out of fuel in space?
As the engines are ignited, the thrust from the rocket unbalances the forces, and the rocket travels upward. Later, when the rocket runs out of fuel, it slows down, stops at the highest point of its flight, then falls back to Earth.
## How much fuel does a rocket need in space?
At liftoff, the two Solid Rocket Boosters consume 11,000 pounds of fuel per second. That’s two million times the rate at which fuel is burned by the average family car.
THIS IS EXCITING: How do I set my telescope up?
## Is there any drag in space?
This same force acts on spacecraft and objects flying in the space environment. Drag has a significant impact on spacecraft in low Earth orbit (LEO), generally defined as an orbit below an altitude of approximately 2,000 kilometers (1,200 mi). … The drag force on satellites increases during times when the Sun is active.
## Is there air drag in space?
There is no air resistance in space because there’s no air in space. 3. GRAVITY: Gravity, which will slow down a ball thrown up in the air, is present in space.
## How much fuel does it take to get 1kg into space?
So, under these extremely optimistic assumptions, you would need about 6kg of propellant (plus 0.5kg of tanks and rocket) to launch your 1kg payload.
## How long does it take for a rocket to get to space?
Short answer: A few minutes. Long answer: The semi-official “start of space” is 100 km above sea level. This is called the Kármán line. Most rockets get to this point within a few minutes of launch, but it takes longer to reach their final orbit (or other destination).
## How much rocket fuel does it take to make 1kg into space?
A photon rocket needs 0.03 grams of fuel to lift 1kg of payload to LEO.
## How is space cold?
In space, there is no air or water, so the only way to lose heat is by radiation, where your warm and wiggly atoms release energy directly into space.
## How long did it take to get from Earth to the Moon?
It takes about 3 days for a spacecraft to reach the Moon. During that time a spacecraft travels at least 240,000 miles (386,400 kilometers) which is the distance between Earth and the Moon. The specific distance depends on the specific path chosen.
THIS IS EXCITING: How much energy does it take to launch 1kg into space?
## Do you keep speeding up in space?
The astronauts on board the International Space Station are accelerating towards the center of the Earth at 8.7 m/s², but the space station itself also accelerates at that same value of 8.7 m/s², and so there’s no relative acceleration and no force that you experience.
## Do things ever slow down in space?
While outer space does contain gas, dust, light, fields, and microscopic particles, they are in too low of a concentration to have much effect on spaceships. As a result, there is essentially zero friction in space to slow down moving objects.
## What is space made of?
Outer space is not completely empty—it is a hard vacuum containing a low density of particles, predominantly a plasma of hydrogen and helium, as well as electromagnetic radiation, magnetic fields, neutrinos, dust, and cosmic rays.
## Will an object move forever in space?
Objects in Motion Now consider an object in motion. straight line indefinitely. of outer space, and the object will move forever due to inertia. | 844 | 3,732 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.949948 |
https://brainly.com/question/321727 | 1,485,036,336,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281226.52/warc/CC-MAIN-20170116095121-00128-ip-10-171-10-70.ec2.internal.warc.gz | 800,935,675 | 8,587 | 2015-02-24T16:40:14-05:00
Convert 1 5/10 to an improper fraction
15/10+3/10
Reduce the expression by cancelling the common factors.
3/2+3/10
Combine 3/2 and 3/10 using a common denominator.
3⋅5+3/10
Simplify the numerator.
18/10
Exact Form:
9/5
Decimal Form:
1.8
Mixed Number Form:
4/5
2015-02-24T16:44:42-05:00
1 8/10 you take the 5+3 and keep the denominator and then add the 1 to get: 1 8/10 | 169 | 400 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2017-04 | latest | en | 0.370119 |
https://www.uptownmortgage1003.com/2019/10/09/calculate-my-mortgage-rate/ | 1,571,370,947,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986677884.28/warc/CC-MAIN-20191018032611-20191018060111-00259.warc.gz | 1,134,586,341 | 11,088 | Mortgage Rates Today
# Calculate My Mortgage Rate
### Contents
Last week’s housing data is proof of that notion as mortgage demand from buyers jumped 15% year over year while mortgage apps.
To calculate your estimated monthly payments on a fixed-rate mortgage, enter the home cost in our fixed-rate mortgage calculator. What are the fixed mortgage rates today? See current fixed-rate mortgages for a variety of conventional mortgages, and learn more about rate assumptions and annual percentage rates (APRs).
That rate is 4.754 percent, which would be your APR on this loan. That’s what this mortgage APR calculator can determine for you, in addition to calculating your interest costs and producing a full amortization schedule. Using the Mortgage APR Calculator. Here’s how it works: Enter how much you wish to borrow in the "Mortgage Amount" box.
How Much Is Prime Rate Today 3 Interest Rate Mortgage Mortgage rates may be at an all time low, but there’s still a big difference between a 3 percent and 4 percent rate. We take a look at the factors that determine your mortgage rate and calculate how much you’ll pay.Today’s Rates TD Prime. Today’s rates. prime rate. 3.950%. Effective Date. October 25, 2018. Note: The effective date reflects the date which TD last altered its prime lending rate. Explore Products and Rates. Loans. The credit you need, with fixed monthly payments that fit your budget.
Mortgage Calculator. When shopping for a mortgage, it is important to evaluate the total cost of the loan. The annual percentage rate (APR) reflects the total cost of a loan by taking into consideration the interest rate plus any points and fees paid.
To calculate a mortgage, you’ll need a few details about the loan. Then, you can do the calculations by hand, or use free online calculators or a spreadsheet program to crunch the numbers. Most people only focus on the monthly payment, but there are other important calculations that you can learn and use to analyze your mortgage, such as:
New Purchase Mortgage Rates 10 Yr Fixed Rate Mortgage The interest rate on a 10 year mortgage often starts off higher than other interest rate types. This means you could end up with a more expensive mortgage if variable interest rates remain the same or go down. You cannot switch to a cheaper deal until the end of the ten year term unless you pay an early repayment charge.va loan Rates. The VA doesn’t set interest rates. Your lender determines the rate on your VA loan based on your unique financial situation. To speak with a VA Mortgage Specialist about interest rates, call 1-800-884-5560 today or get started online .
Your lender likely lists interest rates as an annual figure, so you’ll need to divide by 12, for each month of the year. So, if your rate is 5%, then the monthly rate will look like this: .05/12 =.
Examples of variable loans include adjustable-rate mortgages, home equity lines of credit (HELOC), and some personal and student loans. For more information about or to do calculations involving any of these other loans, please visit the Mortgage Calculator, Auto Loan Calculator, student loan calculator, or Personal Loan Calculator. Variable.
Your exact interest rate will be determined by your lender after consideration of several factors including inflation, Federal Reserve rates, your credit score, and lending fees. See what our current mortgage rates are today and use them in your mortgage calculator input above.
An amount paid to the lender, typically at closing, in order to lower the interest rate. Also known as "mortgage points" or "discount points." One point equals 1% of the loan amount (for example, 2 points on a \$100,000 mortgage would equal \$2,000). | 772 | 3,712 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2019-43 | latest | en | 0.933571 |
https://brainmass.com/business/finance/calculations-of-discount-and-investment-rates-512111 | 1,484,952,701,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280888.62/warc/CC-MAIN-20170116095120-00244-ip-10-171-10-70.ec2.internal.warc.gz | 801,725,451 | 18,628 | Share
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# Calculations of Discount and Investment Rates
We are purchasing a 28-day Treasury bill, during a normal year (non-leap year), and want to find out both the discount rate and the investment rate. If we purchase the bill for \$998, what are the two rates? If we purchase a 91 day Treasury bill for \$995, what are the two rates? (Show all work/calculations/formulas.)
#### Solution Preview
To calculate the discount rate on a Treasury Bill you should use this formula. [(FV-PP/)FV] * [360/M]
FV = Face Value
PP = Purchase Price
M = Maturity of Bill
360 = is the number of days banks use to determine short-term interest rates
This is assuming that the Treasury Bill purchased is for ...
\$2.19 | 176 | 725 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2017-04 | longest | en | 0.902789 |
https://engineering.stackexchange.com/questions/12112/amperage-draw-for-servo | 1,716,762,274,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058973.42/warc/CC-MAIN-20240526200821-20240526230821-00775.warc.gz | 191,167,934 | 39,512 | # Amperage draw for Servo
I'm trying to find a battery that I can use for a project. Let's be hypothetical for a moment.
I have 5-8 of these Adafruit Servos, connected to this Adafruit Servo Shield. Leaving the Arduino itself out of the equation, how can I calculate the amperage draw that these motors and this shield will pull? And additionally, how can I use that with the Ah of a battery to calculate the time that the circuit could be on and in motion with the battery in question?
The datasheet for the device should always list a maximum or "stall" current. For this device, it's 2.5A.
You can use that for a worst-case estimate of battery life by multiplying it by the amount of time you want to run. For example, if you want 12 minutes of run time, then the battery needs to have
$$(2.5\mbox{A})*(12\mbox{min})/(60\mbox{min/hr}) = 0.5\mbox{Ahr}$$
Per servo that you have on your device. That is, if you have 5 servos, then you need a battery with a capacity of 2.5 Ah to run for 12 minutes. If you have 8, then you need a 4 Ah battery to run for 12 minutes.
Again, this is a worst-case scenario, if you try to run your servo at stall current for a long duration then you'll probably wind up burning them out.
How much capacity you actually need depends on the the duty cycle at which you're running the servos. If you're running an excavator then you might actually be running them close to stall current frequently. If you're trying to position a camera then you won't be.
My advice to you would be to go get a battery pack, any battery pack, and use that to power your system. Then, run the system as hard as you think it would normally be run. Compare the run time you achieved with the run time you desire. Then, adjust the capacity of the battery pack accordingly:
$$\mbox{Required Capacity} = (\mbox{Test Capacity})*\frac{\mbox{Desired Run Time}}{\mbox{Actual Run Time}}$$
So, if you had a 1.75 Ah battery and it ran for 5 minutes, but you want it to run for 20 minutes, then you need (1.75)*(20/5) = 7 Ah battery pack.
The most practical way current draw is to power up the arduino board with a power supply capable of measuring current draw. Each time the Servos are activated the additional current draw can be observer on the power supply with a digital read out like the one below.
For much more accurate calculation, find a power supply capable of logging current draw. Calculate the current draw in relation to time to obtain the Ah.
Alternatively you can you use a multimeter on supply line to record the current draw. A multimeter similar to 34410A Digital Multimeter, 6½ Digit is capable of data logging the current draw. | 659 | 2,660 | {"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} | 3.578125 | 4 | CC-MAIN-2024-22 | latest | en | 0.939914 |
https://casadimenotti.com/comparing-and-ordering-decimals-worksheets-6th-grade/comparing-and-ordering-decimals-worksheets-worksheets-for-all/ | 1,550,697,212,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247496694.82/warc/CC-MAIN-20190220210649-20190220232649-00616.warc.gz | 510,406,917 | 16,118 | # Comparing And Ordering Decimalsets 6th Grade
By Lea Burger at December 21 2018 22:05:03
Involve the Entire Body : No one likes to sit in class listening to a lecture for an hour. It gets boring and monotonous. To incorporate more fun into learning, try to engage the entire body. Have children move around the room. Play active review games. If you must lecture, have your students take a five minute break to stand up and wiggle their arms and legs. Fun doesn't have to be silly all the time. Simply moving around can make an otherwise boring lecture seem uplifting. Positive Reinforcement : One of the easiest ways to add some fun to your class is to use positive reinforcement. Students not only detest, but also dread classes that make them feel dumb. If your class is made to think they are excelling or performing well, they will be more likely to succeed. You will see smiles on their faces instead of looks of dread. The only way fun can be introduced into the school day is if the children feel comfortable letting loose. Giving positive reinforcement is the way to accomplish that goal.
The use of math worksheets can help solve numerous arithmetic problems. "Practice makes an individual perfect," is the best motto to be kept in mind while studying math. The motto will help a person to reinforce his desire to better himself in the subject. Without the help of these online resources, one will not be able to achieve the mastery of math. Reading is required in every day in life. Whether it is reading a street sign, reading the news, or reading a menu, comprehension is the fundamental reason for reading. Reading for comprehension is an essential common core skill that can and must be taught to students.
## Gallery of Comparing And Ordering Decimals Worksheets 6th Grade
However, with the creation of worksheets users can now calculate many simple and complex math and financial problems as well as display their stored data with many unique custom charts and graphs. Text data consists of alpha-numeric characters such as letters and words. Formulas are instructions that are included in a cell that allow you to manipulate and perform operations on other cells in the Workbook. When you put a formula in your cell, the calculated value is then displayed as a result.
At the grassroots level, teachers in schools are given a packed curriculum for the year. Schools try to teach the students a number of procedures without delving much into its finer details. Hence, the student is left in a confounding position as to when a particular procedure must be used. The key ingredient to understanding math is constant practice and math assignment help. Unfortunately, this is not a common scenario among the popular math classes.Connect The Letter To The Correct Sound/Word: These are activities where you draw a line between a letter and the picture items that start with that letter. For instance, you'd draw a line from the letter A to the word "Apple" and the letter L to the word "Lemon". This activity is good, but takes a lot of monitoring to make sure that students are correctly connecting the letters. It's best as a homework activity, where parents can help to make sure their children are correctly connecting the letters to the words. | 658 | 3,268 | {"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} | 2.984375 | 3 | CC-MAIN-2019-09 | longest | en | 0.95424 |
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