url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
http://www.cfd-online.com/W/index.php?title=Lid-driven_cavity_problem&diff=17091&oldid=6199 | 1,472,594,566,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471983019893.83/warc/CC-MAIN-20160823201019-00136-ip-10-153-172-175.ec2.internal.warc.gz | 368,990,075 | 11,753 | # Lid-driven cavity problem
(Difference between revisions)
Revision as of 20:26, 8 August 2006 (view source)Jasond (Talk | contribs) (→Introduction)← Older edit Revision as of 13:52, 19 February 2013 (view source)Ztdep (Talk | contribs) (→Additional resources)Newer edit → (2 intermediate revisions not shown) Line 34: Line 34: == Additional resources == == Additional resources == - - http://web.gyte.edu.tr/enerji/ercanerturk/drivencavity/cavityflow.htm - - http://www.cse.ohio-state.edu/~jiang/Projects/DrivenCavity/
## Introduction
The lid-driven cavity problem has long been used a test or validation case for new codes or new solution methods. The problem geometry is simple and two-dimensional, and the boundary conditions are also simple. The standard case is fluid contained in a square domain with Dirichlet boundary conditions on all sides, with three stationary sides and one moving side (with velocity tangent to the side).
Similar simulations have also been done at various aspect ratios, and it can also be done with the lid replaced with a moving fluid. This problem is a somewhat different situation, and is usually referred to as the shear-driven cavity. You may see the two names (lid-driven and shear-driven) used interchangeably in spite of the fact that they are distinct (and different) problems.
This problem has been solved as both a laminar flow and a turbulent flow, and many different numerical techniques have been used to compute these solutions. Since this case has been solved many times, there is a great deal of data to compare with. A good set of data for comparison is the data of Ghia, Ghia, and Shin (1982), since it includes tabular results for various of Reynolds numbers. These simulation results are obtained using a non-primitive variable approach.
This problem is a nice one for testing for several reasons. First, as mentioned above, there is a great deal of literature to compare with. Second, the (laminar) solution is steady. Third, the boundary conditions are simple and compatible with most numerical approaches. Note that this is not necessarily the case for finite element methods, in which difficulties may arise at the corner intersections of the moving wall and the stationary wall.
## Sample results
Shown below are a few results from a simulation of the lid-driven cavity (for a Reynolds number of one hundred) using the commercial Fluent code.
The 32x32 uniform grid used here is somewhat coarse, as can be seen in the upper left and right corners of the contour plot. Increasing the grid resolution would improve the look of the contour plot somewhat, but since the boundary conditions are not smooth at these corners we will probably always have some issues there. In spite of these issues, we do a reasonable job matching the Ghia et al (1982) results, as shown below.
Here we have plotted u/U on the vertical line x/L = 0.5. We have reasonably good agreement with the previous results.
## References
Ghia, Ghia, and Shin (1982), "High-Re solutions for incompressible flow using the Navier-Stokes equations and a multigrid method", Journal of Computational Physics, Vol. 48, pp. 387-411.
Bozeman and Dalton (1973), "Numerical Study of viscous flow in a cavity", Journal of Computational Physics, Vol. 12, pp. 348-363.
Burgaff (1966), "Analytical and Numerical studies of the structures of steady separated flows", JFM, Vol. 24, pp. 113-151.
Erturk, Corke, and Gokcol (2005), "Numerical Solutions of 2-D Steady Incompressible Driven Cavity Flow at High Reynolds Numbers", International Journal for Numerical Methods in Fluids, Vol. 48, pp. 747-774.
Need more references that involve this problem | 838 | 3,672 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2016-36 | longest | en | 0.959887 |
https://www.genio.ac/guides/how-to-calculate-gross-profit-margin/ | 1,723,746,372,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641311225.98/warc/CC-MAIN-20240815173031-20240815203031-00043.warc.gz | 589,272,768 | 36,935 | # How to Calculate Gross Profit Margin
May 11, 2024
AuthorGavin Bales
As an experienced financial advisor, I cannot stress enough the critical role of understanding and accurately calculating your gross profit margin. This key indicator of overall business health and operational efficiency holds the reins to insightful management decisions. Ignorance, here, is not bliss. It can potentially drive your entrepreneurship into the ground. This comprehensive guide will elucidate the essentials of gross profit margin, breaking down its calculation into simplified steps. It will also shed light on the interpretation of these figures and their effective utilisation for business growth and stability. Strap in for a deep dive into financial acumen.
## Definition and Importance
For us to comprehend the importance of the Gross Profit Margin calculation, we must first grasp its definition. Gross Profit Margin is essentially a financial metric expressed as a percentage that shows the proportion of your total revenue that remains after subtracting all direct costs associated with the production or creation of the goods and services sold – Your Cost of Goods Sold (COGS).
The importance of this metric surpasses the common comprehension of finance, particularly for small and medium-sized enterprises, freelancers, and their accountants. For business owners and managers, understanding Gross Profit Margin isn’t merely about knowing where your business stands right now, but rather a tool to inform strategic decisions that can drive business growth.
In the case of freelancers, mastering Gross Profit Margin assists them in pricing their services accurately. For accountants of these entrepreneurs, this calculation is essential because it provides deep insights into the company’s operational efficiency, economic health, investment attractiveness, and potential for future growth. In essence, Gross Profit Margin helps to build strategies pertinent to profitability and business expansion.
## Key Steps or Methods
Firstly, let’s understand that calculating Gross Profit Margin is essential for the financial well-being of your business. It’s a key performance indicator and allows you to see how profitable your business is before operational expenses, taxes, and interest payments are deducted.
The first step in calculating Gross Profit Margin is to determine your net sales. You can find this value by summing all your sales during a specific period and then subtracting any returns, allowances, and discounts you provided to your clients. Your accounting software or bookkeeping service should be able to provide you the correct figures.
Next, you’ll need to calculate the cost of goods sold (COGS). This includes manufacturing costs for products or costs to provide services, like materials and labour. Exclude overhead expenses like rent or utilities. If you’re using software, the COGS is likely calculated for you. However, be careful when dealing with inventory as you could use several methods, such as FIFO (First-In, First-Out) or LIFO (Last-In, First-Out), which would significantly affect your figures. For most accurate results, match the accounting method you’re using for tax purposes.
Now, subtract the COGS from net sales to find your gross profit. This figure represents how much profit you’ve made after considering the cost of producing or purchasing the products you’ve sold.
Finally, divide your gross profit by net sales and multiply by 100 to get your Gross Profit Margin, expressed as a percentage. This percentage provides a quick snapshot of the profitability of your operations before administrative, marketing and other costs are factored in.
Remember, it’s important to monitor your Gross Profit Margin over time. Small changes can indicate significant shifts in your business environment. For instance, a decrease in your percentage could mean increased production costs or lowered pricing, while an increase might signal more efficient production methods or higher pricing.
Now for some best practices. Regularly recalculate your Gross Profit Margin, especially during growth periods or significant changes to COGS or sales. Use industry benchmarks to compare your results and understand where your business stands against your competitors. But be aware that Gross Profit Margin varies by industry; a good margin in one industry may be poor in another.
Additionally, consider creating Gross Profit Margin targets and use these as key performance indicators. This will drive actions to achieve these targets, such as reviewing pricing strategies or cost controls.
By calculating and actively monitoring your Gross Profit Margin, you’ll be better equipped to manage financial health, make informed decisions, and drive the success of your business.
## Common Challenges and Solutions
Delving into the calculation of Gross Profit Margin (GPM), you inevitably face several challenges, even as a seasoned business owner or an experienced accountant. More than mere arithmetic, the task requires astute awareness of financial data.
One significant hurdle is the misinterpretation of revenues and costs. Many confuse gross revenue with net revenue and calculate GPM from the latter. This leads to inflated GPM, presenting a false picture of the business’s profitability. The key is to always calculate GPM using gross revenue, which signifies total sales before deducting expenses.
Moreover, correctly identifying and categorizing costs as direct or indirect forms the backbone of accurate GPM computation. Classifying indirect costs (those not directly tied to the production of a product or service) as direct costs can deflate your GPM. Always account for only direct costs, typically cost of goods sold (COGS), in your GPM calculation. Reach out to a financial advisor if you are not sure about cost categorization.
Sometimes indirect costs are overlooked entirely. While these costs shouldn’t affect GPM, ignoring them can result in undervaluing the resources necessary to keep your business running. These costs should play a role in pricing strategies and should be considered in relation to net profit margins.
Fluctuating costs pose an additional challenge, especially for businesses dealing with highly variable COGS, like those associated with seasonal products. Constant monitoring is crucial in these cases — a GPM calculated today may not hold true next month. Always update your calculations regularly to reflect such changes.
Lastly, neglecting to compare your GPM with industry standards or past performances renders your calculation futile. While a higher GPM indicates a profit-generating business, it might be lower than the industry average or previous periods. Regular benchmarking, both internally and externally, provides context to the numbers.
Calculating GPM certainly isn’t a walk in the park. But the right approach, paired with situational awareness and in-depth understanding, will streamline the process for you.
## Red Flags
With diligence, the gross profit margin is the critical financial indicator you want to keep an eye on closely. While dissecting the financial health of your business, I found a few red flags that should prompt immediate attention.
Firstly, consistently declining gross profit margins. This can be due to increased production costs or lower sales prices – both indicate potential problems. If you find your production costs are escalating, scrutinize your supply chain. Sudden price increase from suppliers, inefficiencies in production, or wastage might be the culprits. Meanwhile, lower sales prices can reflect tough competition or less customer demand. In response, you might need to innovate your product, explore new markets or reevaluate your pricing strategy.
Secondly, if you find your gross profit margin significantly fluctuating, that’s a warning sign. Stability is a crucial aspect of a healthy business. Unpredictable margins make planning difficult and signal underlying issues, such as fluctuating costs or volatile sales patterns. You’ll want to rectify these inconsistencies to run smooth financial operations.
Thirdly, if your gross profit margin is considerably lower than your industry average, sound the alarm. While there are exceptions, generally it means your business is less profitable than your competitors. Perhaps your production is costly, or your pricing is too low. A competitive analysis can provide insights into figuring out the issues.
Lastly, for freelancers and small businesses, low gross profit margin at the onset of your venture isn’t inherently alarming. But if it persists or declines further, you’re heading towards an unsustainable path. Ensure you have a clear understanding of your costs and revenues, and make the necessary adjustments.
Remember, your gross profit margin is not just about revenue and costs. It’s a beacon illuminating your business operations’ efficiency and should not be ignored. Paying heed to these red flags early can spare you the headaches in the future.
## Case Studies or Examples
Case 1: A client, a small bakery business owner, was experiencing steady sales but still struggled with profitability. After assessing their income statement, I noticed that while their revenues were high, their cost of goods sold (COGS) was also disproportionately substantial. We calculated the gross profit margin by subtracting the COGS from revenue and dividing the result by revenue, then multiplying by 100 to get the percentage. The outcome was surprisingly low; 30%. We then set in action a strategy to reduce the COGS, including measures such as finding a cheaper supplier, reducing waste, and increasing prices. Within six months, their gross profit margin rose to 40%, greatly improving profitability.
Case 2: Another scenario involved a freelance graphic designer who didn’t comprehend why his high revenues were not translating into substantial profits. Upon investigating, it was apparent that the cost of his software, internet, and hardware was consuming his revenues. Applying the same gross profit margin formula, the result reflected a meager 35% gross profit margin. The problem did not reside with generating high revenues, it was about managing costs effectively. He switched to more affordable software alternatives, negotiated better internet rates, and extended the usefulness of his hardware. Implementing these measures boosted his gross profit margin to 55% within a year.
In both instances, computing their gross profit margin revealed what was clandestinely eroding their profits. By focusing not only on lifting revenue but also effectively managing the COGS, these two businesses saw dramatic improvement in their financial performance. Remember, the gross profit margin is a critical metric underlying business profitability. It’s not always about generating more sales, but sometimes about cutting costs wisely. Whether you’re a business owner or freelancer, calculate this number regularly and set strategies to enhance it continuously. By doing so, you can significantly boost your profitability over time. | 2,023 | 11,125 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-33 | latest | en | 0.924258 |
https://community.aimms.com/math-or-optimization-modeling-39/understanding-the-differences-between-mip-and-rmip-solvers-1479?postid=4107 | 1,718,970,093,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862070.5/warc/CC-MAIN-20240621093735-20240621123735-00590.warc.gz | 161,061,466 | 31,645 | Solved
# Understanding the Differences between MIP and RMIP Solvers
• Enthusiast
• 3 replies
Hi all, I would like to seek clarification regarding the disparities between MIP (Mixed Integer Programming) and RMIP (Relaxed Mixed Integer Programming) solvers. While I understand that RMIP converts integer variables into continuous variables, I would like to explore any other significant dissimilarities between these two solvers.
Specifically, I am encountering fluctuations in the results of one of the variables when utilizing the MIP model as shown in the picture below. Despite the absence of any constraint-related issues, the values exhibit variability instead of remaining constant.
Conversely, when employing the RMIP solver, the values remain consistent throughout (as presented in picture below).
I am keen to understand the cause behind these fluctuations and any significant disparities between the MIP and RMIP solvers that could potentially explain this behavior. Any insights, explanations, or suggestions on how to address these fluctuations would be greatly appreciated too!
icon
Best answer by Marcel Hunting 26 May 2023, 10:31
View original
### 2 replies
Userlevel 5
+4
Hi @Xinhui. Not much can be said based on the solution values of one variable. I assume that this variable interacts with other variables through constraints and in that case the other variables and the constraints also play a role.
Please note that if you have a constraint
``b1 + b2 + b3 + b4 = 1``
in which b1, b2, b3 and b4 are binary variables then in the MIP solution one of these binary variables will have a solution value of 1 and the other three will have a solution value of 0. A solution for the RMIP could be (0.25,0.25,0.25,0.25) or (0.1,0.1,0.1,0.7), etc. And these binary variables might impact other (continuous) variables.
Hi @Marcel Hunting. I see! Your help is greatly appreciated. Thank you! | 449 | 1,915 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-26 | latest | en | 0.922269 |
https://forums.ni.com/t5/LabVIEW/Uncertainty-calculations-for-cDAQ-module-NI-9203-and-4-20-mA/m-p/4268721 | 1,670,614,979,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711475.44/warc/CC-MAIN-20221209181231-20221209211231-00582.warc.gz | 297,614,849 | 71,948 | # LabVIEW
cancel
Showing results for
Search instead for
Did you mean:
Solved!
Go to solution
## Uncertainty calculations for cDAQ module NI 9203 and 4-20 mA transducer.
Could anyone help me confirm just these calculations for accuracy/uncertainty? I just want to know If I making sense.
I have a 16 bit module (NI 9203) that is configured to read 4 to 20 mA signals. That module is connected to a 4-20 mA pressure sensor that reads 30k psi, whose total full scale accuracy is 0.25%. (It goes from 0 to 30kpsi)
• So the accuracy would be ± 75 psi. Right?
I was using this formula to calculate the uncertainty of this setup (Ignoring noise for now):
• Uncertainty = (Input Signal x Gain Error) + (Range x Offset Error)
If we assume an uncalibrated, and unipolar accuracy (0 to 20 mA) at ambient temperature, the datasheet gives you these values:
• Gain Error: 0.49%
• Offset Error: 0.46%
The datasheet says the real range is 21.5 mA, so would I be correct to assume that for these values, when reading 10 mA, then:
• Accuracy = (0.0049 x 10 mA) + (0.0046 x 20 mA) = 10 ± 0.1479 mA
The pressure sensor has a range of 4 to 20 mA for 30,000 psi. If we assume the sensor really does stop at 0 and 30k, then we'd have 16 mA for 30k psi which is roughly 0.53 uA / psi.
• So for 10 ± 0.1479 mA, we would have 11250 ± 277.31 psi if we do the conversion.
• Adding up the uncertainty of the pressure sensor: 11250 ± 352.31 psi
Am I in the right track? What am I getting wrong, what else should I consider(?)
NI 9203 Specifications: https://www.ni.com/docs/en-US/bundle/ni-9203-specs/page/specifications.html
Message 1 of 4
(191 Views)
Solution
## Re: Uncertainty calculations for cDAQ module NI 9203 and 4-20 mA transducer.
Your calculation seems to be correct for a steady-state measurement.
Note: All this is "uncalibrated" and you don't operate any critical equipment "uncalibrated"
Santhosh
Soliton Technologies
New to the forum? Please read community guidelines and how to ask smart questions
Only two ways to appreciate someone who spent their free time to reply/answer your question - give them Kudos or mark their reply as the answer/solution.
Message 2 of 4
(176 Views)
## Re: Uncertainty calculations for cDAQ module NI 9203 and 4-20 mA transducer.
Thanks, that's what I figured. What would I need to factor in if it wasn't steady state(?)
I'm actually just trying to convince a few people to calibrate the modules along with the sensors. They usually just calibrate the sensor itself.
Message 3 of 4
(151 Views)
## Re: Uncertainty calculations for cDAQ module NI 9203 and 4-20 mA transducer.
The input bandwidth is 850kHz, which means an 850kHz 10mA input signal will read as 7.07mA beyond 850kHz the signal is no longer acquired true to the actual value.
This is not a concern if your application does not work with a high-frequency signal.
Yeah, definitely you need to educate them on the importance of calibrating everything in the signal chain.
Santhosh
Soliton Technologies
New to the forum? Please read community guidelines and how to ask smart questions
Only two ways to appreciate someone who spent their free time to reply/answer your question - give them Kudos or mark their reply as the answer/solution.
Message 4 of 4
(129 Views) | 867 | 3,274 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2022-49 | latest | en | 0.869546 |
https://www.cs.montana.edu/paxton/classes/spring-2016/csci338/lectures/ch1/1d.html | 1,537,923,976,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267162809.73/warc/CC-MAIN-20180926002255-20180926022655-00524.warc.gz | 714,303,881 | 1,381 | # Chapter 1: Regular Languages
• A game that can be modeled with a finite automaton is Manufactoria.
## Chapter 1.2, Nondeterminism
### Lecture Problem
• Problem: Design a nondeterministic finite automaton that accepts any binary string that contains either 00 or 11.
• Theorem: The class of regular languages is closed under the union operation.
• Prove the theorem by constructing NFA N from the union of NFA N1 and NFA N2. Let N1 = (Q1, Σ, δ1, q1, F1) and N2 = (Q2, Σ, δ2, q2, F2).
### Active Learning Problem
• Problem: Design a nondeterministic finite automaton that accepts any binary string that contains the concatenation of 00 and 11.
• Theorem: The class of regular languages is closed under the concatenation operation.
• Prove the theorm by constructing NFA N from the concatenation of NFA N1 and NFA N2.
### Active Learning Problem
• Problem: Design a nondeterministic finite automaton that accepts the language (101)*.
• Theorem: The class of regular languages is closed under the star operation.
• Prove the theorem by constructing NFA N from the star of NFA N1. | 279 | 1,085 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2018-39 | latest | en | 0.840605 |
https://codingpk.com/Post/Tutorial/algorithm-in-programming?ques=coding | 1,619,051,433,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039554437.90/warc/CC-MAIN-20210421222632-20210422012632-00014.warc.gz | 252,322,195 | 5,650 | # Algorithm in Programming
Solutions Algorithm in Programming Coding
An algorithm is a step by step procedure to solve a problem. The process of solving a problem becomes simpler and easier with the help of algorithm. It is better to write algorithm before writing the actual program.
## Properties of Algorithm
Following are some properties of an algorithm:
• The given problem should be broken down into simple and meaningful steps.
• The steps should be numbered sequentially.
• The steps should be descriptive and written in simple English.
Algorithms are written in a language that is similar to simple English called pseudo code. There is not standard to write pseudo code. It is used to specify program logic in an English like manner that is independent of any particular programming language.
Pseudo code simplifies program development by separating it into two main parts.
1. Logic Design
2. Coding
## 1- Logic Design
In this part, the logic of the program is designed. We specify different steps required to solve the problem and the sequence of these steps.
## 2- Coding
In this part, the algorithm is converted into a program. The steps of algorithm are translated into instructions of any programming language. The use of pseudo code allows the programmer to focus on the planning of the program. After the planning is final, it can be written in any programming language.
### Example
The following algorithm inputs two number, calculates sum and then displays the result on screen.
1. Start
2. Input A
3. Input B
4. Total = A + B
5. Display Total
6. Exit
Solutions Coding | 324 | 1,598 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-17 | latest | en | 0.890677 |
https://quantumcomputing.stackexchange.com/questions/9644/quantum-computing-w-r-t-the-many-worlds-theory?noredirect=1 | 1,718,572,751,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861671.61/warc/CC-MAIN-20240616203247-20240616233247-00630.warc.gz | 440,312,297 | 37,460 | # Quantum computing w.r.t. the many-worlds theory [duplicate]
Quantum algorithms are inherently probabilistic. Let's say I run Shor's factoring algorithm on some number. There's a chance that the output I observe is incorrect. Does the many-worlds theory suggest that I'm simply in a universe where the output happened to be wrong? Because the probability of error is slim, does that mean the vast majority of universes realize a correct execution of the algorithm?
Edit: Thanks to Mark S for pointing me to Effects of quantum computing on parallel universes which fully explores my question. Also, a big thanks to everyone who commented here! The questions raised and resources raised have only helped to enhance my curiosity about the subject.
• This is covered in chapter 10 of Quantum Computing since Democritus by Scott Aaronson. You'll also find a ton of posts about the MWI on his Shtetl-Optimized blog. Commented Jan 26, 2020 at 6:47
• Hi TWal, and welcome to QCSE. This may answer your question? Effects of quantum computing on parallel universes If not, can you consider revising your question more? Commented Jan 27, 2020 at 23:57
Quantum algorithm can give answers with 100% certainty. For example, Grover's algorithm can give the answer with certainty when the ratio of "good" items to "bad" items satisfies some constraints. The uncertainty exists when the output state is not an eigenstate of the measurement observable.
On the other hand, Shor's algorithm is a hybrid of classical and quantum steps. The quantum order finding step might fail to find an appropriate "order" in an iteration, but you cannot say it is incorrect. Instead, the measurement gives one of the phases of the unitary operator $$U|y\rangle =|xy \text{(mod } N\text{)}\rangle$$, the phase can then be used to derive the order classically. However only some of these orders can be used to find the factors. If the order found after measurement is not appropriate, Shor's algorithm need to repeat the quantum step (together with some classic steps). (There are multiple phases because the input $$|1\rangle$$ is a superposition of many eigenvectors associated with these phases.)
Many word theory is more philosophical than computationally or informational. Using this theory, at least one of the world will find an appropriate order in the first iteration and Shor's algorithm will stop quicker. In some world, all iterations fail (as long as you continue) to find the appropriate order and Shor's algorithm will not finish. Of course, Shor proved that the probability of not finishing can be arbitrarily low given enough iterations.
Some people reject many world theory using a joke: you can believe it in the other worlds. To me, the challenge is where does the (exponential, enormous) extra energy come as worlds split all the time?
• @czwang: You mentioned that Grover algorithm is able to find a correct answer with 100 % certainty. From many-worlds perspective, does this mean that Grover work properly only in one of these many worlds? Or, does this mean that there is no splitting of words ? Commented Jan 26, 2020 at 6:59
• Why should there be extra energy as a result of world splitting? (What's your normalisation?) Commented Jan 26, 2020 at 12:20
• @Martin Vesely: In the special cases, Grover algorithm ends in a single eigenstate with eigenvalue 1 before measurement. As a result, measurement does not result in world splitting. Commented Jan 26, 2020 at 17:06
• @Niel de Beaudrap: as far as I understand, energy is preserved in each world. Splitting adds new worlds, each contains the same amount of energy as the world before the split. I am not saying this assumption is wrong, just I have a hard time to believe it. Commented Jan 26, 2020 at 17:10
• @TWal: you are right, the sentence is a bit misleading. I will edit it to clarify and hopefully address you questions. Commented Jan 26, 2020 at 17:34 | 899 | 3,912 | {"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": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-26 | latest | en | 0.923004 |
https://studylib.net/doc/25344262/addition-and-subtraction-of-decimals | 1,610,946,408,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514121.8/warc/CC-MAIN-20210118030549-20210118060549-00478.warc.gz | 586,804,145 | 13,429 | # ADDITION AND SUBTRACTION OF DECIMALS
❶Find the sum
a) 9.6 + 9.7 =
b) 2.7 + 4.8 =
c) 7.8 + 2.6 =
d) 1.4 + 15.3 =
e) 8.5 + 0.04 =
f) 0.18 + 1.50 =
g) 0.134 + 0.193 =
h) 1.82 + 0.013 =
i) 0.023 + 1.65 =
❷Find the missing number
a) 7.1 + _____ = 9.6
b) 8.0 + _____ = 12.7
c) 7.9 + _____ = 16.7
d) _____ + 6.3 = 14.7
e) 3.6 + _____ = 13.4
f) 8.5 + _____ = 15
❸Find the missing number
a) 0.82 + _____ = 1.42
b) 0.3 + _____ = 0.85
c) 0.1 + _____ = 0.4
❹Find the sum
d) _____ + 0.6 = 1.3
e) 0.9 + _____ = 1.78
f) 0.6 + _____ = 1.2
❺Find the difference
1. 0.95 - 0.71 =
2. 0.77 - 0.58 =
3. 8.5 - 2.1 =
4. 6.7 - 3.7 =
5. 2.1 - 1.0 =
6. 9.1 - 0.52 =
7. 18 - 0.8 =
8. 8 - 0.07 =
9. 11 - 0.019=
❻Find the missing numbers
1) 6.5 – _____ = 5.5
2) _____ – 1.1 = 8.8
3) _____ – 1.0 = 2
14) 5 – _____ = 4.1
15) 5.4 – _____ = 4.4
16) 3 – _____ = 2.9
❼Find the missing numbers
a) _____ – 1.49 = 5.61
b) _____ – 2.4 = 5.78
c) 7.6 – _____ = 0.2
❽Find the difference
d) _____ – 1.7 = 1.61
e) 5.80 – _____ = 4.5
f) _____ – 1.3 = 1.9 | 602 | 1,011 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | longest | en | 0.440273 |
http://www.wptv.com/dpp/news/national/powerball-odds-500-million-powerball-jackpot-win-less-likely-than-a-shark-attack-or-a-bee-sting | 1,386,935,263,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164937476/warc/CC-MAIN-20131204134857-00095-ip-10-33-133-15.ec2.internal.warc.gz | 637,318,918 | 17,116 | # Powerball odds: \$500 million Powerball jackpot win less likely than a shark attack or a bee sting
The winning ticket for Saturday night's \$338 million Powerball jackpot sold at a convenience story in Bordentown, CNN affiliate KYW-TV said.
Posted: 11/28/2012
Last March, when the people of America were drooling at the thought of winning a record \$656 million Mega Millions jackpot, we poured an icy bucket of mathematical reality over your head: You weren't going to win.
And you didn't. Three winning tickets were sold, but you weren't involved. It was never going to happen. As we wrote then, you stood a better chance at hitting two consecutive holes in one on a par-3 golf hole than winning that jackpot.
Now, with a record \$500 million Powerball jackpot up for grabs on Wednesday, we figured it was a great time to, once again, dash your dreams. We know, we know -- someone will win at least a share of the prize, if not Wednesday, then in some subsequent drawing. But it won't be you.
The chance of a ticket winning a Powerball jackpot is 1 in 175,223,510 (slightly better than the chance of winning a Mega Millions jackpot, which is 1 in 175,711,536). Here are a few unlikely scenarios that, we're sorry to say, are much more likely than you taking home this jackpot.
From the Harvard School of Public Health:
-- Dying from a bee sting: 1 in 6.1 million.
-- Dying from being struck by lightning: 1 in 3 million.
From the University of Maryland Medical Center:
From U.S. Hole in One, which insures golf prizes for holes in one:
-- An amateur golfer making a hole in one on a par-3 hole: 1 in 12,500.
-- A golfer hitting a hole in one on consecutive par-3 holes: 1 in about 156 million.
From a 2011 State Farm study on collisions between vehicles and deer:
-- Hitting a deer with a vehicle in Hawaii, the state where State Farm says deer-vehicle collisions are least likely: 1 in 6,267.
From the National Weather Service:
-- Being struck by lightning over an 80-year lifetime: 1 in 10,000.
From the Florida Museum of Natural History, based on U.S. beach injury statistics:
-- Drowning and other beach-related fatalities: 1 in 2 million.
-- Being attacked by a shark: 1 in 11.5 million.
What do you think about your chances of winning on Wednesday? Weigh in below, or on Twitter using #whataretheodds.
Featured Stories
1. ## Enter to win a new car from Braman Honda
One person will win a three-year lease on a 2013 Honda Civic Lx Sedan automatic.
• ## Latest mugshots from PBC, Treasure Coast
Click to see the latest mugshots, plus this week's wanted fugitives.
This feature packed upgrade brings you faster performance, easier navigation, and stunning improvements to photos, video and readability.
• What's Trending Now...
Latest News Stories
1. ## Mega Millions hits \$400M jackpot
Outshined by massive jackpots since Powerball doubled the cost of its tickets last year, Mega Millions enacted big changes to inflate its jackpots and lure customers who only play when the pots get huge - and the revamp appears to be working. | 731 | 3,068 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2013-48 | latest | en | 0.951782 |
https://books.google.com.gi/books?id=JZgAAAAAMAAJ&qtid=da7e78f6&lr=&source=gbs_quotes_r&cad=5 | 1,720,791,767,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514404.71/warc/CC-MAIN-20240712125648-20240712155648-00524.warc.gz | 106,966,669 | 5,988 | Books Books
The sphere may be conceived to be formed by the revolution of a semicircle about its diameter, which remains fixed.
Mathematics: Compiled from the Best Authors and Intended to be the Text-book ... - Page 83
1801
A Treatise of Plane and Spherical Trigonometry: In Theory and Practice ...
Francis Nichols - Plane trigonometry - 1811 - 162 pages
...surface, which is every where equally distant from a point within it called the centre. A sphere may be conceived to be formed by the revolution of a semicircle about its diameter, which remains unmoved, and is called the axis of the sphere. 2. A diameter of a sphere is a straight line passing...
An introduction to geography and astronomy, by the use of the globes and ...
Edward Bruce (bookseller.) - 1821 - 418 pages
...globe is a perfectly round body; its middle point is called the centre. A globe, or sphere, is a solid, formed by the revolution of a semicircle about its diameter, which remains fixed. The axis of a sphere is the right line about which the semicircle revolves. 6. A great circle divides...
Euclid's Elements of Geometry: The Six First Books. To which are Added ...
Rev. John Allen - Astronomy - 1822 - 516 pages
...sphere, is a right line drawn from the centre to any part of its surface. 16. When a sphere is supposed to be formed by the revolution of a semicircle about its diameter, which remains unmoved, the diameter, about which the semicircle revolves, is called, the axis of the sphere. 17....
Mathematical and Astronomical Tables: For the Use of Students of Mathematics ...
William Galbraith - Astronomy - 1827 - 412 pages
...angles alone ; while that of a spherical triangle always can. 3. A sphere or globe is a round body formed by the revolution of a semicircle about its diameter, which remains fixed. 4. The centre of the sphere is the same with that of the revolving semicircle. 5. The axis of the sphere...
An Introduction to Mensuration and Practical Geometry
John Bonnycastle - Geometry - 1829 - 256 pages
...quadrangular pyramid; when a pentagon, it is called a pentagonal pyramid, &c. 8. A sphere is a solid described by the revolution of a semicircle about its diameter, which remains fixed. 9. The centre of a sphere is a point within the figure, every where equally distant from the convex surface...
A treatise on practical geometry, mensuration, conic sections, gauging, and ...
Ireland commissioners of nat. educ - 1834 - 370 pages
...9. A cone is a round pyramid, having a circle for its base. 10. A sphere is a round solid, which may be conceived to be formed by the revolution of a semi-circle about its diameter which remains fixed. 11. The axis of a solid is a line joining the middle* o> both ends. 12. When the axis is perpendicular...
Mathematical and Astronomical Tables: For the Use of Students in Mathematics ...
William Galbraith - Astronomy - 1834 - 454 pages
...angles alone ; while that of a spherical triangle always can. 3. A sphere or globe is a round body formed by the revolution of a semicircle about its diameter, which remains fixed. 4. The centre of the sphere is the same with that of the revolving semicircle. 5. The axis of the sphere...
A Treatise on Mensuration for the Use of Schools
Commissioners of National Education in Ireland - Measurement - 1837 - 290 pages
...8. A cone is a round pyramid, having a circle for its base. 9. A sphere is a round solid, which may be conceived to be formed by the revolution of a semi-circle about its diameter which remains fixed. 10. The axis of a solid is a line joining the middle of both ends. 11. When the axis is perpendicular... | 871 | 3,628 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-30 | latest | en | 0.923664 |
https://hsc.co.in/51-most-important-question-bank-of-physics-for-maharashtra-hsc-board-exam-2018/ | 1,720,920,514,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514527.38/warc/CC-MAIN-20240714002551-20240714032551-00099.warc.gz | 258,646,477 | 31,388 | # 51 Most Important Question Bank of Physics for Maharashtra HSC Board Exam 2018
We have created a 51 Most Important Question Bank which will help students in scoring good marks in HSC Board Exams. HSC Board Exams are fast approaching and students are getting anxious about how to prepare for their HSC Board Exams. So we had mentioned some HSC Study Tips to help students in Cracking HSC Exams.
We have also created Free Mock Test of MCQ- Multi Choice Question for students to help them prepare for the exam like scenario.
After the tremendous success of our past years Important Questions Bank for Maharashtra HSC Board Exam 2016 and we have created 51 Most likely question for HSC Board 2018.
To unlock the content Click on any of 3 buttons available in the box below saying “This content is Locked”. Once you click on the button the content will get unlocked on same page itself. You must click on social media button showing in below box ie Facebook, Twitter or Google Plus to unlock the content.
### Physics Important Question Bank 2018
1.What do you mean by centripetal force?
2.Show that period of a satellite revolving around the Earth depends upon the mass of the Earth.
3.Explain why an astronaut in an orbiting satellite has a feeling of weightlessness.
4.Explain What is Doppler effect in sound and state its applications.
5.Define emissive power and coefficient of emission of a body.
6.A solid cylinder of uniform density of radius 2cm has mass of 50g. If its length is 12cm, calculate its moment of inertia about an axis passing through its Centre and perpendicular to its length.
7.Define linear S.H.M. Show that S.H.M. is a projection of U.C.M. on any diameter.
8.Prove that root mean square velocity of gas molecule is directly proportional to the square of its absolute temperature.
9.Explain the formation of stationary waves by analytical method.
10.Derive an expression for excess pressure inside a drop of liquid.
11.Obtain the differential equation of linear simple harmonic motion.
12.A metal sphere cools at the rate of 4C/min. When its temperature is 50C. Find its rate of cooling at 45C if the temperature of surroundings is 25C.
13.State the first law of Thermodynamics.
14.Derive Laplace’s law for spherical membrane of bubble due to surface tension.
15.Find the kinetic energy of a 2kg rolls on a table with linear speed of 3m/s.
16.State the theorem of parallel axes and theorem of perpendicular axes about moment of inertia.
17. Derive an expression for acceleration due to gravity at depth D below the earth’s surface.
18.The spin dryer of a washing machine rotating at 15rps slows down to 5rps after making 50 revolutions. Find the angular acceleration.
19.The total energy of free surface of a liquid drop is 2pi (22/7) times the surface tension of the liquid. What is the diameter of the drop?
20.Draw a labelled diagram for a liquid surface in contact with a solid, when the angle of contact is acute.
21.What are forced vibrations and resonance? Show that only odd harmonics are present in an air column vibrating in a pipe closed at one end.
22.Explain how moving coil galvanometer is converted into a voltmeter
23.What is diffraction of light? Explain its types.
24.Distinguish between paramagnetic and ferromagnetic substances.
25.Describe the construction of photoelectric cell.
26.Obtain an expression for the radius of Bohr orbit for H-atom.
27.Distinguish between the phenomenon of interference and diffraction of light.
28.Write a short note on surface wave propagation of electromagnetic waves.
29.State the principle on which transformer works. Explain its working with construction.
30.With the help of a labelled diagram, describe the Geiger-Marsden experiment. What is mass defect?
31.Explain what is a transmitter.
32.Explain what is a receiver in communication system.
33.An unknown resistance is place in the left gap and resistance of 50 ohm is placed in the right gap of a meter bridge. The null point is obtained at 40cm from the left end. Determine the unknown resistance.
34. Draw a labelled diagram of Davisson and Germer experiment.
35.Explain self-induction and mutual induction.
36.What are alpha and beta parameters for a transistor? Obtain relation between them.
37.Describe Kelvin’s method to determine the resistance of galvanometer by using metre bridge.
38.State Ampere’s circuital law. Obtain an expression for magnetic induction along the axis of toroid.
40.Explain the elementary idea of an oscillator with the help of block diagram.
41.Draw a labelled diagram of a receiver for the detection of amplitude modulated wave.
42.Describe the biprism experiment to find the wavelength of the monochromatic light. Draw the necessary ray diagram. Also, the width of plane incident wavefront is found to be doubled on refraction in the denser medium. If it makes an angle 65 degree with the normal, calculate the refractive index for the denser medium.
43.Explain with a neat diagram, how a p-n junction diode is used as a half wave rectifier.
44.A cube of marble having each side 1 cm is kept in an electric field of intensity 300V/m. Determine the energy contained in the cube of dielectric constant.
45.What is critical velocity? Obtain an expression for critical velocity of an orbiting satellite. On what factors does it depend?
46. Explain the construction and working of Nicol prism.
47.State and prove Gauss’ Theorem in Electrostatics.
48.With the help of a neat diagram, describe the construction and working of Van de Graff generator.
49.State Kelpler’s laws of planetary motion.
50. Define practical simple pendulum.
51.Draw the schematic symbols for AND, OR NOT and NAND gate.
Hi, we’re trying to collate and gather the data and would be updating it here a few days before the exam. Please keep on visiting our website for updates.
Please use the comments box below and post questions that you think are important from your analysis. It would help the HSC community a lot.
Do subscribe to our updates so that you do not miss out on any important information that we push your way.
Don’t forget to read : MUST REMEMBER THINGS on the day of Exam for HSC Students | 1,344 | 6,149 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-30 | latest | en | 0.885991 |
https://forum.knittinghelp.com/t/sharon/83357 | 1,611,161,083,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703521139.30/warc/CC-MAIN-20210120151257-20210120181257-00717.warc.gz | 349,608,251 | 4,705 | # Sharon
Hi I need some help on a new pattern I am trying. They are slippers with a cuff which is cabled. I have been knitting for years but this has me stumped. It has instruction but also a diagram. I tried following it but no dice it doesn’t work! Please help me.
We’ll all be happy to try. Can you give us a pattern link or a pattern name and tell us what directions have you stumped?
There are hundreds of slipper patterns so a link and where inthe pattern the problem is would help. If the pattern is in a book then type out a few lines of what is giving you trouble. (Don’t post the entire pattern)
The pattern is Little Red Riding Slippers by Drops Design. I did the foot part which was interesting because I found the pattern to be extremely vague. Now I am attempting the cuff which is a cabled design. The latter is as follows. Cast on 22 sts. Work in garter st for 2 rows. On 2nd row increase 6 sts evenly a crossed row. I did all this with no trouble. Then it says to refer to diagram A-1 which is a graph. Telling me to in from WS, p from rs then p from WS and in from w s
Cute slippers!
You’re knitting the cuff as a separate piece. The chart A.1 is read from the bottom right to the left for Row 1 and then Row 2 is read from the left to the right. That makes the first 2 rows:
Row 1: k5, p2, k6, p2, k6, p2, k5
Row 2: k3, p2, k2, p6, k2, p6, k2, p2, k3
It may help to number the rows starting at the bottom and going up. For this chart, the odd rows will be RS (follow the symbols for RS only) and the even rows will be WS (follow the symbols in the chart for WS only).
Here’s some help with charts:
Thank you so much. I cannot begin to tell you how much I appreciate your help | 453 | 1,701 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.964997 |
http://task-writers.com/assume-that-a-consumer-has-the-utility-function-uxy-3x1y-where-x-and-y-represent-the-quantities-of-two-goods-x-and-y/ | 1,656,970,177,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104496688.78/warc/CC-MAIN-20220704202455-20220704232455-00200.warc.gz | 55,814,336 | 13,191 | # Assume that a consumer has the utility function U(x,y) = (3x+1)y, where x and y represent the quantities of two goods, X and Y.
Assume that a consumer has the utility function U(x,y) = (3x+1)y, where x and y represent the quantities of two goods, X and Y.
Now assume that the consumer has \$31 to spend on goods X and Y, which have fixed prices of pX=3 and pY=4.
(d) Express the consumer’s budget constraint as an equation. The only variables in the equation shouldbe x and y.
(e) Calculate the first-order condition for the consumer’s optimization problem of dividing her money between goods X and Y in the way that maximizes her utility. (The only variables in the equationshould be x and y.)
(f) Calculate the consumption bundle (x,y) that solves the first-order condition for the consumer’s problem.
(g) Calculate the two consumption bundles (x,y) that are on the boundary of the consumer’s optimization problem.
(h) For the three possible solutions that you found in parts (f) and (g), calculate the utility that the consumer would receive from each bundle. Whatshould she do to maximize her utility? Justify your answer carefully.
(i) Repeat parts (f), (g), and (h), assuming that, ceteris paribus, the price of X rises to pX=\$75.
(j) What should the consumer do to maximize her utility if, ceteris paribus, the price of X rises topX=100? Justify your answer carefully.
(k) Compare your answers to parts (h), (i), and (j). Discuss the connection between these answers and the Law of Demand. | 364 | 1,509 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.915606 |
botany.authpad.com | 1,394,438,909,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394010701848/warc/CC-MAIN-20140305091141-00097-ip-10-183-142-35.ec2.internal.warc.gz | 27,342,901 | 3,471 | # Botany Questions
## How could you classify the quadrilateral shown?(Points : 1) parallelogram rhombus square?5 mos ago
Yahoo! Answers - . How could you classify the: How could you classify the quadrilateral shown?(Points : 1) parallelogram rhombus square? 48 seconds ago - 4 days left to answer.
How do you classify quadrilaterals? - Yahoo! Answers: How do you classify quadrilaterals? rectangle, kite, parallelogram, square, trapezoid, rhombus what does each have thats unique about itself. 6 years ago;
Quadrilaterals - Square, Rectangle, Rhombus,: A parallelogram has opposite sides parallel and equal in length. The only regular quadrilateral is a square. So all other quadrilaterals are irregular.
Plz help? math question? - Yahoo! Answers: If a figure is a square, in what other way or ways can you classify it? quadrilateral parallelogram rhombus All of the above are accurate classifications
Quadrilateral - Wikipedia, the free encyclopedia: In Euclidean plane geometry, a quadrilateral is a polygon with four sides (or edges) and four vertices or corners. Sometimes, the term quadrangle is used, by analogy
Answers.com - How do you classify quadrilateral: square How do you classify a quadrilateral? You classify a quadrilateral by its sides, angles, and its vertices. How could you classify the quadrilateral?
Classify the quadrilateral shown. Explain your: Classify the quadrilateral shown. Explain your reasoning. Thank you for your help. Images: This could be a rhombus, square, parallelogram or rectangle.
Classification of Quadrilaterals - Cut-the-Knot: Classification of Quadrilaterals. Quadrilateral is a geometric shape that consists of four points (vertices) sequentially joined by straight line segments (sides).
How to Prove a Quadrilateral Is a Parallelogram |: How to Prove a Quadrilateral Is a Parallelogram. Quadrilaterals and parallelograms are geometrical figures that you might encounter in a middle school, high school
Geometry (A): READ: Classifying Quadrilaterals: Classifying Quadrilaterals Learning Objectives . Identify and classify a parallelogram. Identify and classify a rhombus. Identify and classify a rectangle. | 490 | 2,164 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-10 | longest | en | 0.814774 |
https://betterlesson.com/lesson/resource/3206953/classroom-video-checks-for-understanding | 1,498,394,410,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320491.13/warc/CC-MAIN-20170625115717-20170625135717-00277.warc.gz | 747,738,448 | 20,154 | ## Classroom Video: Checks for Understanding - Section 1: Launch
Classroom Video: Checks for Understanding
Classroom Video: Checks for Understanding
# Write It Wednesday Percents
Unit 7: Percent Relationships
Lesson 6 of 18
## Big Idea: How big is that percent?
Print Lesson
2 teachers like this lesson
Standards:
Subject(s):
40 minutes
##### Similar Lessons
###### Unit Assessment Feedback Lesson
7th Grade Math » Proportional Relationships II
Big Idea: Students reflect on how they performed on the the unit assessment.
Favorites(7)
Resources(7)
New Orleans, LA
Environment: Urban
###### Balancing Act
7th Grade Science » Energy, Force & Motion
Big Idea: Can objects of different mass be arranged so they balance one another? Is there a mathematical equation that can predict balance?
Favorites(10)
Resources(14)
Hope, IN
Environment: Rural
###### Volume of Prisms and Pyramids Fluency Practice
7th Grade Math » Geometric Measurement
Big Idea: Practice makes perfect – students will work on fluency of finding volume of prisms and pyramids.
Favorites(7)
Resources(15)
Elon, NC
Environment: Suburban | 252 | 1,109 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | longest | en | 0.819668 |
http://mathforum.org/kb/message.jspa?messageID=9342836 | 1,524,633,245,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947693.49/warc/CC-MAIN-20180425041916-20180425061916-00017.warc.gz | 190,953,849 | 7,768 | Search All of the Math Forum:
Views expressed in these public forums are not endorsed by NCTM or The Math Forum.
Notice: We are no longer accepting new posts, but the forums will continue to be readable.
Topic: Re: Poor Dan (but much poorer Robert)
Replies: 6 Last Post: Dec 13, 2013 6:20 AM
Messages: [ Previous | Next ]
Robert Hansen Posts: 11,345 From: Florida Registered: 6/22/09
Re: Poor Dan (but much poorer Robert)
Posted: Dec 12, 2013 11:48 PM
att1.html (0.3 K) P317.jpeg (31.4 K) att1.html (0.3 K) P316.jpeg (27.9 K) P315.jpeg (26.9 K) att1.html (0.3 K) att1.html (0.3 K) P314.jpeg (25.1 K) att1.html (0.3 K) P313.jpeg (24.6 K) att1.html (14.3 K)
On Dec 12, 2013, at 4:08 PM, Louis Talman <talmanl@gmail.com> wrote:
>
> On Thu, 12 Dec 2013 12:51:21 -0700, Robert Hansen <bob@rsccore.com> wrote:
>
> ...what does ?moment of inertia? have to do with it?
>
> That is indeed the question. At the level the kids are working, "moment of inertia" has absolutely nothing to do with it, and it is dishonest to suggest that it is otherwise. The kids are learning to substitute numbers for the symbols found in a formula. In spite of the "moment of inertia" label, this formula is meaningless to them. What is the point of pretending otherwise? Be honest: Give them a meaningless formula, tell them it's meaningless, and ask them to make the substitutions.
Ok, take away the title, and the text. Is the formula now meaningful to them? I am not suggesting that problems like this be the cream and crop of an algebra 2 curriculum. The majority (2/3) is still just math. But if you want aspiring engineers in your class, or anyone that aspires to use algebra for that matter, you are going to have to apply it to something. That is what I call technical problems. Problems with technical details outside of the math.
> Can you create as involved an expression about something they would know about?
>
> Why would I try? There's nothing wrong with involved expressions in general, or with this one in particular. But the issue here isn't how involved the expression is, or even where it comes from. It's the pretense that the student is doing something meaningful---when that is anything but the case for the student. When we make such misrepresentations, we teach students things we don't want them to learn---as you confessed below.
Ok, I see one of your grievances. I think you are confusing me with those real world problem clowns with red noses. That isn?t my philosophy and I have no pretense that the student is doing a real world problem. That?s not going to happen till they get a real world degree and a real world job.
>
> Granted, this is an algebra 2 problem. I don?t know why Dan is commenting on an algebra problem at all, at any level, he doesn't teach it in his blog. But he did comment on it. I look at this problem in the context of the level it is meant to be. It would be a horrible arithmetic problem.
>
> Exactly: Look at the problem "in the context of the level it is meant to be." That context includes no understanding of where those complicated formulae come from or how. Why pretend otherwise by giving the formulae meaningless labels and suggesting that the kids are doing something real? Every one of them who thinks at all knows damn good and well that they are not. I'm not objecting to the substitution part of the problem; I'm objecting to the cloud of misrepresentation that surrounds it.
>
> No, it shouldn?t be mysterious.
>
> Then what is the point of making it so?
Because they are inspiring to the students that aspire to technical fields like engineering that use this stuff. They get the math just fine and they are anxious to apply it. They don?t want the problem to teach them what multiplication is. They already know what multiplication is, quite well. They don?t need a problem to show them what math can do, they already have the muse, and they are excited. They want to calculate the orbit of mars using just its speed, the mass of the sun and some formula you gave them that they wouldn?t know from the drag equation of a Cessna 150. And the problems are real. The pretending is the students pretending that they are engineers. And they are having fun and applying the math they have learned.
> Do you really want the majority of students to think that what engineers do consists of sitting around and spending their days making simple substitutions in complicated formulae they don't understand?
They aren?t going to do that. They got the juice. And the problems get tougher over time, well, they used to. Do you not remember this time? When every day of math class was new.
>
> Students figure out very soon that mathematics is not just a collection or rules to be memorized. Well, differential equations is, but the rest of it isn?t.:) You know how they figure this out? Cause they fail the damn tests!
>
> A few do this. The vast majority write mathematics off as devoid of intellectual content.
Those brilliant bastards. And all along I thought they just weren't good at math. No wonder it has been so tough to find engineers (not needing green cards) lately. I'll have to call our recruiter and have her rewrite the entire recruiting process.
This is your (an many others?) hypothesis and you don?t know how much I wish you were right. That would be so freaking easy to fix. Seeing that this idea and a dozen variants have been around now for 3 decades and there was no fix, doesn?t bode well for it.
The algebra courses I grew up with had enough math for the aspiring mathematicians and enough (pretend) engineering for the aspiring engineers (and physicists, programmers, etc.). And for the others they had solid business math courses with business algebra where needed. Who have I left out? Other than the kids that aspire to something that doesn?t need algebra at all. They took math for daily living. And those weren?t tracks, they were paths. Tracks are when everyone takes algebra except not the same ?algebra?.
If you have examples of a better class of problems I am all ears. This particular problem isn?t the best, but not for the reasons you mentioned. Throw it in with 4 other pretend problems and it would rank 5th. But what do you expect, it was picked by clowns out of an engineering class.
Attached are a few pages from my Algebra 2 book with pretend engineering word problems, with some worked examples so that the student wasn?t entirely in the dark. Some are simple, some require the statement of the formula (in words) with nothing to back it up.
What is your opinion of these?
Bob Hansen
Date Subject Author
12/11/13 GS Chandy
12/12/13 Robert Hansen
12/12/13 Louis Talman
12/12/13 Robert Hansen
12/12/13 Louis Talman
12/12/13 Robert Hansen
12/13/13 Robert Hansen | 1,590 | 6,760 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2018-17 | latest | en | 0.924341 |
https://www.physicsforums.com/threads/differntiate-the-function.407433/ | 1,544,957,580,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827639.67/warc/CC-MAIN-20181216095437-20181216121437-00216.warc.gz | 989,751,334 | 12,506 | # Homework Help: Differntiate the function
1. Jun 2, 2010
### ghostbuster25
ok just want a check of my work before i send it :)
differntiate the function;
f(x)=$$\frac{x(20-x)}{36}$$
so,
f'(x)=$$\frac{(20x-x^2)}{36}$$
f'(x)=$$\frac{(20-2x)}{36}$$
f'(x)=$$\frac{(10-2x)}{18}$$
Then....use the composite rule and your answer above to differntiate the function
g(x)=ex(20-x)/36
g'(x)=f'(x)*ef(x)
so,
g'(x)=$$\frac{10-2x)}{18}$$*g(x)=ex(20-x)/36
is this correct?
is this correct
2. Jun 2, 2010
### Staff: Mentor
Re: differentiation
Above - This is not f'(x). All you have done is expand the numerator on the right. This should be f(x) = ...
Above - Now you have taken the derivative
Above - now you have an error. 20 - 2x != 2(10 - 2x).
Where f(x) = (1/36)(x(20 - x)). The derivative below is incorrect. To fix it, replace f'(x) and f(x) where they appear. | 316 | 868 | {"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} | 4.125 | 4 | CC-MAIN-2018-51 | latest | en | 0.74621 |
https://electrical-engineering-portal.com/understanding-vector-group-transformer-2?replytocom=3541 | 1,686,404,731,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224657720.82/warc/CC-MAIN-20230610131939-20230610161939-00657.warc.gz | 262,208,935 | 41,845 | Save 50% on all Video Courses with Enterprise Membership Plan and study specialized LV/MV/HV technical articles and guides.
# Understanding Vector Group of Transformer (2)
Home / Technical Articles / Understanding Vector Group of Transformer (2)
Continued from first part Understanding Vector Group of Transformer (part 1)
## Points to be consider while Selecting of Vector Group
Vector Groups are the IEC method of categorizing the primary and secondary winding configurations of 3-phase transformers. Windings can be connected as delta, star, or interconnected-star (zigzag). Winding polarity is also important, since reversing the connections across a set of windings affects the phase-shift between primary and secondary.
Vector groups identify the winding connections and polarities of the primary and secondary. From a vector group one can determine the phase-shift between primary and secondary.
Transformer vector group depends upon:
1. Removing harmonics: Dy connection – y winding nullifies 3rd harmonics, preventing it to be reflected on delta side.
2. Parallel operations: All the transformers should have same vector group & polarity of the winding.
3. Earth fault Relay: A Dd transformer does not have neutral. to restrict the earth faults in such systems, we may use zig zag wound transformer to create a neutral along with the earth fault relay..
4. Type of Non Liner Load: systems having different types of harmonics & non linear Types of loads e.g. furnace heaters ,VFDS etc for that we may use Dyn11, Dyn21, Dyn31 configuration, wherein, 30 deg. shifts of voltages nullifies the 3rd harmonics to zero in the supply system.
5. Type of Transformer Application: Generally for Power export transformer i.e. generator side is connected in delta and load side is connected in star. For Power export import transformers i.e. in Transmission Purpose Transformer star star connection may be preferred by some since this avoids a grounding transformer on generator side and perhaps save on neutral insulation.
Most of systems are running in this configuration. May be less harmful than operating delta system incorrectly. Yd or Dy connection is standard for all unit connected generators.
There are a number of factors associated with transformer connections and may be useful in designing a system, and the application of the factors therefore determines the best selection of transformers.
For example:
### For selecting Star Connection:
A star connection presents a neutral. If the transformer also includes a delta winding, that neutral will be stable and can be grounded to become a reference for the system. A transformer with a star winding that does NOT include a delta does not present a stable neutral.
Star-star transformers are used if there is a requirement to avoid a 30deg phase shift, if there is a desire to construct the three-phase transformer bank from single-phase transformers, or if the transformer is going to be switched on a single-pole basis (ie, one phase at a time), perhaps using manual switches.
Star-star transformers are typically found in distribution applications, or in large sizes interconnecting high-voltage transmission systems. Some star-star transformers are equipped with a third winding connected in delta to stabilize the neutral.
### For selecting Delta Connection:
• A delta connection introduces a 30 electrical degree phase shift.
• A delta connection ‘traps’ the flow of zero sequence currents.
### For selecting Delta-Star Connection:
• Delta-star transformers are the most common and most generally useful transformers.
• Delta-delta transformers may be chosen if there is no need for a stable neutral, or if there is a requirement to avoid a 30 electrical degree phase shift. The most common application of a delta-delta transformer is as tan isolation transformer for a power converter.
### For selecting Zig zag Connection:
The Zig Zag winding reduces voltage unbalance in systems where the load is not equally distributed between phases, and permits neutral current loading with inherently low zero-sequence impedance. It is therefore often used for earthing transformers.
Provision of a neutral earth point or points, where the neutral is referred to earth either directly or through impedance. Transformers are used to give the neutral point in the majority of systems. The star or interconnected star (Z) winding configurations give a neutral location.
If for various reasons, only delta windings are used at a particular voltage level on a particular system, a neutral point can still be provided by a purpose-made transformer called a ‘neutral earthing.
### For selecting Distribution Transformer
The first criterion to consider in choosing a vector group for a distribution transformer for a facility is to know whether we want a delta-star or star-star. Utilities often prefer star-star transformers, but these require 4-wire input feeders and 4-wire output feeders (i.e. incoming and outgoing neutral conductors).
For distribution transformers within a facility, often delta-star are chosen because these transformers do not require 4-wire input; a 3-wire primary feeder circuit suffices to supply a 4-wire secondary circuit. That is because any zero sequence current required by the secondary to supply earth faults or unbalanced loads is supplied by the delta primary winding, and is not required from the upstream power source. The method of earthing on the secondary is independent of the primary for delta-star transformers.
The second criterion to consider is what phase-shift you want between primary and secondary. For example, Dy11 and Dy5 transformers are both delta-star. If we don’t care about the phase-shift, then either transformer will do the job. Phase-shift is important when we are paralleling sources. We want the phase-shifts of the sources to be identical.
If we are paralleling transformers, then you want them to have the same the same vector group. If you are replacing a transformer, use the same vector group for the new transformer, otherwise the existing VTs and CTs used for protection and metering will not work properly.
There is no technical difference between the one vector groups (i.e. Yd1) or another vector group (i.e. Yd11) in terms of performance. The only factor affecting the choice between one or the other is system phasing, ie whether parts of the network fed from the transformer need to operate in parallel with another source. It also matters if you have an auxiliary transformer connected to generator terminals. Vector matching at the auxiliary bus bar.
## Application of Transformer according to Vector Group
### 1.) Dyn11, Dyn1, YNd1, YNd11
• Common for distribution transformers.
• Normally Dyn11 vector group using at distribution system. Because Generating Transformer are YNd1 for neutralizing the load angle between 11 and 1.
• We can use Dyn1 at distribution system, when we are using Generator Transformer are YNd11.
• In some industries 6 pulse electric drives are using due to this 5thharmonics will generate if we use Dyn1 it will be suppress the 5th harmonics.
• Star point facilitates mixed loading of three phase and single phase consumer connections.
• The delta winding carry third harmonics and stabilizes star point potential.
• A delta-Star connection is used for step-up generating stations. If HV winding is star connected there will be saving in cost of insulation.
• But delta connected HV winding is common in distribution network, for feeding motors and lighting loads from LV side.
### 2.) Star-Star (Yy0 or Yy6)
• Mainly used for large system tie-up transformer.
• Most economical connection in HV power system to interconnect between two delta systems and to provide neutral for grounding both of them.
• Tertiary winding stabilizes the neutral conditions. In star connected transformers, load can be connected between line and neutral, only if
(a) the source side transformers is delta connected or
(b) the source side is star connected with neutral connected back to the source neutral.
• In this transformers. Insulation cost is highly reduced. Neutral wire can permit mixed loading.
• Triple harmonics are absent in the lines. These triple harmonic currents cannot flow, unless there is a neutral wire. This connection produces oscillating neutral.
• Three phase shell type units have large triple harmonic phase voltage. However three phase core type transformers work satisfactorily.
• A tertiary mesh connected winding may be required to stabilize the oscillating neutral due to third harmonics in three phase banks.
### 3.) Delta – Delta (Dd 0 or Dd 6)
• This is an economical connection for large low voltage transformers.
• Large unbalance of load can be met without difficulty.
• Delta permits a circulating path for triple harmonics thus attenuates the same.
• It is possible to operate with one transformer removed in open delta or” V” connection meeting 58 percent of the balanced load.
• Three phase units cannot have this facility. Mixed single phase loading is not possible due to the absence of neutral.
### 4.) Star-Zig-zag or Delta-Zig-zag (Yz or Dz)
• These connections are employed where delta connections are weak. Interconnection of phases in zigzag winding effects a reduction of third harmonic voltages and at the same time permits unbalanced loading.
• This connection may be used with either delta connected or star connected winding either for step-up or step-down transformers. In either case, the zigzag winding produces the same angular displacement as a delta winding, and at the same time provides a neutral for earthing purposes.
• The amount of copper required from a zigzag winding in 15% more than a corresponding star or delta winding. This is extensively used for earthing transformer.
• Due to zig-zag connection (interconnection between phases), third harmonic voltages are reduced. It also allows unbalanced loading. The zigzag connection is employed for LV winding. For a given total voltage per phase, the zigzag side requires 15% more turns as compared to normal phase connection. In cases where delta connections are weak due to large number of turns and small cross sections, then zigzag star connection is preferred. It is also used in rectifiers.
### 5.) Zig-zag/ star (ZY1 or Zy11)
• Zigzag connection is obtained by inter connection of phases.4-wire system is possible on both sides. Unbalanced loading is also possible. Oscillating neutral problem is absent in this connection.
• This connection requires 15% more turns for the same voltage on the zigzag side and hence costs more. Hence a bank of three single phase transformers cost about 15% more than their 3-phase counterpart. Also, they occupy more space. But the spare capacity cost will be less and single phase units are easier to transport.
• Unbalanced operation of the transformer with large zero sequence fundamental mmf content also does not affect its performance. Even with Yy type of poly phase connection without neutral connection the oscillating neutral does not occur with these cores. Finally, three phase cores themselves cost less than three single phase units due to compactness.
### 6.) Yd5
• Mainly used for machine and main Transformer in large Power Station and Transmission Substation.
• The Neutral point can be loaded with rated Current.
### 7.) Yz-5
• For Distribution Transformer up to 250MVA for local distribution system.
• The Neutral point can be loaded with rated Current.
## Application of Transformer according according to Uses
Step up Transformer: – It should be Yd1 or Yd11.
Step down Transformer: – It should be Dy1 or Dy11.
Grounding purpose Transformer: – It should be Yz1 or Dz11.
Distribution Transformer: – We can consider vector group of Dzn0 which reduce the 75% of harmonics in secondary side.
Power Transformer: – Vector group is deepen on application for Example : Generating Transformer : Dyn1 , Furnace Transformer: Ynyn0.
## Convert One Group of Transformer to Other Group by Channing External Connection
### Group I: Example: Dd0 (no phase displacement between HV & LV)
The conventional method is to connect the red phase on A/a, Yellow phase on B/b, and the Blue phase on C/c.
Other phase displacements are possible with unconventional connections (for instance red on b, yellow on c and blue on a) By doing some unconventional connections externally on one side of the Transformer, an internal connected Dd0 transformer can be changed either to a Dd4(-120°) or Dd8(+120°) connection. The same is true for internal connected Dd4 or Dd8 transformers.
### Group II: Example: Dd6 (180° displacement between HV & LV)
By doing some unconventional connections externally on one side of the Transformer, an internal connected Dd6 transformer can be changed either to a Dd2(-60°) or Dd10(+60°) connection.
### Group III: Example: Dyn1 (-30° displacement between HV & LV)
By doing some unconventional connections externally on one side of the Transformer, an internal connected Dyn1 transformer can be changed either to a Dyn5(-150°) or Dyn9(+90°) connection.
### Group IV: Example: Dyn11 (+30° displacement between HV & LV)
By doing some unconventional connections externally on one side of the Transformer, an internal connected Dyn11 transformer can be changed either to a Dyn7(+150°) or Dyn3(-90°) connection.
## Point to be remembered
• For Group-III & Group-IV: By doing some unconventional connections externally on both sides of the Transformer, an internal connected Group-III or Group-IV transformer can be changed to any of these two groups.
• Thus by doing external changes on both sides of the Transformer an internal connected Dyn1 transformer can be changed to either a: Dyn3, Dyn5, Dyn7, Dyn9 or Dyn11 transformer, This is just true for star/delta or delta/star connections.
• For Group-I & Group-II: Changes for delta/delta or star/star transformers between Group-I and Group-III can just be done internally.
### Why 30°phase shift occur in star-delta transformer between primary and secondary?
The phase shift is a natural consequence of the delta connection. The currents entering or leaving the star winding of the transformer are in phase with the currents in the star windings. Therefore, the currents in the delta windings are also in phase with the currents in the star windings and obviously, the three currents are 120 electrical degrees apart.
But the currents entering or leaving the transformer on the delta side are formed at the point where two of the windings comprising the delta come together – each of those currents is the phasor sum of the currents in the adjacent windings.
When you add together two currents that are 120 electrical degrees apart, the sum is inevitably shifted by 30 degrees.
The Main reason for this phenomenon is that the phase voltage lags line current by 30degrees.consider a delta/star transformer. The phase voltages in three phases of both primary and secondary. you will find that in primary the phase voltage and line voltages are same, let it be VRY (take one phase). But, the corresponding secondary will have the phase voltage only in its phase winding as it is star connected. the line voltage of star connected secondary and delta connected primary won’t have any phase differences between them. so this can be summarized that “the phase shift is associated with the wave forms of the three phase windings.
This is the HV Side or the Switchyard side of the Generator Transformer is connected in Delta and the LV Side or the generator side of the GT is connected in Star, with the Star side neutral brought out.
The LV side voltage will “lag” the HV side voltage by 30 degrees. Thus, in a generating station we create a 30 degrees lagging voltage for transmission, with respect to the generator voltage.
As we have created a 30 degrees lagging connection in the generating station, it is advisable to create a 30 degrees leading connection in distribution so that the user voltage is “in phase” with the generated voltage. And, as the transmission side is Delta and the user might need three phase, four-wire in the LV side for his single phase loads, the distribution transformer is chosen as Dyn11.
There is magnetic coupling between HT and LT. When the load side (LT) suffers some dip the LT current try to go out of phase with HT current, so 30 degree phase shift in Dyn-11 keeps the two currents in phase when there is dip.
So the vector group at the generating station is important while selecting distribution Transformer.
### Vector Group in Generating-Transmission-Distribution System
Generating TC is Yd1 transmitted power at 400KV, for 400KV to 220KV Yy is used and by using Yd between e.g. 220 and 66 kV, then Dy from 66 to 11 kV so that their phase shifts can be cancelled out. And for LV (400/230V) supplies at 50 Hz are usually 3 phase, earthed neutral, so a “Dyn” LV winding is needed. Here GT side -30lag (Yd1) can be nullify +30 by using distribution Transformer of Dy11.
A reason for using Yd between e.g. 220 and 66 kV, then Dy from 66 to 11 kV is that their phase shifts can cancel out and It is then also possible to parallel a 220/11 kV YY transformer, at 11 kV, with the 66/11 kV (a YY transformer often has a third, delta, winding to reduce harmonics).
If one went Dy11 – Dy11 from 220 to 11 kV, there would be a 60 degree shift, which is not possible in one transformer. The “standard” transformer groups in distribution avoid that kind of limitation, as a result of thought and experience leading to lowest cost over many years.
### Generator TC is Yd1, can we use Distribution TC Dy5 instead of Dy11?
With regards to theory, there are no special advantages of Dyn11 over Dyn5.
In Isolation Application: -In isolated applications there is no advantage or disadvantage by using Dy5 or Dy11. If however we wish to interconnect the secondary sides of different Dny transformers, we must have compatible transformers, and that can be achieved if you have a Dyn11 among a group of Dyn5’s and vice versa.
In Parallel Connection: – Practically, the relative places of the phases remain same in Dyn11 compared to Dyn5.
If we use Yd1 Transformer on Generating Side and Distribution side Dy11 transformer than -30 lag of generating side (Yd1) is nullify by +30 Lead at Receiving side Dy11) so no phase difference respect to generating Side and if we are on the HV side of the Transformer, and if we denote the phases as R- Y-B from left to right, the same phases on the LV side will be R- Y -B, but from left to Right.
This will make the Transmission lines have same color (for identification) whether it is input to or output from the Transformer.
If we use Yd1 Transformer on Generating Side and Distribution side Dy5 transformer than -30 lag of generating side (Yd1) is more lag by -150 Lag at Receiving side (Dy5) so Total phase difference respect to generating Side is 180 deg (-30+-150=-180) and if we are on the HV side of the Transformer, and if we denote the phases as R- Y-B from left to right, the same phases on the LV side will be R- Y -B, but from Right to Left.
This will make the Transmission lines have No same color (for identification) whether it is input to or output from the Transformer. The difference in output between the Dyn11 and Dny5 and is therefore 180 degrees.
Get access to premium HV/MV/LV technical articles, electrical engineering guides, research studies and much more! It helps you to shape up your technical skills in your everyday life as an electrical engineer.
### Jignesh Parmar
Jignesh Parmar has completed M.Tech (Power System Control), B.E (Electrical). He is member of Institution of Engineers (MIE), India. He has more than 20 years experience in transmission & distribution-energy theft detection and maintenance electrical projects.
Nov 13, 2020
It is very informative and helpful elaboration of using vector groups which best suit our requirement. Can I have further detail about merit/demerits of vector groups. Any book or internet link?
2. Joginder
Jun 27, 2019
What the vector grouping Y yn0-yn0(d) denotes?
3. Mohamed Elhamshary
Jun 20, 2019
what will be the effect if we have the following:
power transformer Ynyn0 220/66 KV conncted to GIS swapping (I mean the transformer ABC and GIS BCA)
4. Gamal
Nov 04, 2018
Confusing explanation…
Oct 18, 2018
What will be the effect on power system during fault occurence in different voltage level if the voltage is not “in phase”?
Thank you…
6. Jamshid
Nov 09, 2017
Why some times vector groups are denoted as Ya0, where the connection is YnYn0 ,What the a represents here?
7. transformer test
Jan 08, 2015
OMICRON has a tool to guess the vector group of a transformer if it is unknown
• Jamshid
Nov 09, 2017
Vanguard instruments also manufacture testing machine named Transformer turns ratio tester it can detect the vector group of testing transformer eg. model ATRT03
8. ku
Sep 16, 2014
how identify the transformer vector group without nameplate details
• Jamshid
Nov 09, 2017
Either by testing machine or by manually measuring voltage (specific test for specific connection) u can check by visually weather delta or star
9. Arnel C.
Jul 31, 2014
for three winding transformer, how is the vector group identified
10. Hussain
Jun 30, 2014
Dear Sir.
What is the best suitable software that can be used by engineers and Installation wiring designers in Australia.
Thanks.
11. Jake
May 20, 2014
why it is advisable to have the user voltage is “in phase” with the generated voltage. thank you.
12. u kodandaramireddy
Apr 03, 2014
Your providing v.group information helpful for all Design Engineers.Thank you very much sir.
13. Gaurav Sheta
Jan 19, 2013
why step up transformer vectro group Yd11?
14. mmkumar20
Aug 15, 2012 | 4,702 | 21,830 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2023-23 | longest | en | 0.872969 |
https://adityarohan.wordpress.com/2014/01/23/a-guide-to-making-things-look-complicated/ | 1,620,634,400,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989115.2/warc/CC-MAIN-20210510064318-20210510094318-00328.warc.gz | 106,432,751 | 20,306 | # A Guide To Making Things Look Complicated
Look at that. It’s probably almost enough to cure you of any desire to learn maths. Or perhaps it makes you want to know what it means, in which case, you’ve come to the right place. With a bit of guidance, you can make things like this out of simple expressions.
How To Decode Expressions Like The One Above
First, to make this look a lot less scary, factorise the denominator. You’ll get:
That still looks as if it might give you nightmares, but it’s a little better.
(A brief note on factorisation:
Polynomials like the one in the original expression (ones with degree 2, meaning the maximum power is 2) have two solutions (a solution is the value of x for which f(x) = 0 (f(x) is just a way of expressing a function, such as y = 2x. Functions can be plotted on a graph and their solutions can be extracted from this graph)). They have two solutions because they can be simplified into two expressions of degree 1, as shown in the second picture. Once they’ve been reduced to the two degree-1 polynomials (a polynomial is an expression with more than one term), two solutions can be worked out using this. (For example, say f(x) = x2 + 5x + 6. It can be reduced to (x + 2)(x + 3). If x + 2 = 0, then you can find x; then the same idea can be applied to x + 3 = 0. Try putting the values obtained from this method into the original f(x), and check your answers!))
‘</’brief’note>
‘</nestedbrackets>
‘</htmltags>
‘</recursion>
Back to the scary expression.
Note that (4x – 1) – (4x – 3) = 2, which just happens to be the numerator. So, you can make the original expression into:
That looks a lot simpler.
The sigma on the left means that k takes every value, in turn, from 1 to infinity, then adds all of the expressions together; basically, it starts with the value for k = 1, then adds the value for k = 2, then for k = 3 and so on to infinity.
So, if k = 1, what’s 4k – 3? It’s 1, and for 4k – 1, it’s 3. So, for k = 1, the expression means 1 – (1/3).
Continue substituting for k and adding, and you’ll get the value of the expression as:
which is the Leibniz approximation for π/4 – a comparatively simple answer compared to what we started with, which had a summation to infinity and a degree-2 polynomial.
Now it’s your turn to come up with complex and unnecessary expressions!
## 1 thought on “A Guide To Making Things Look Complicated”
1. raja
Extremely well-explained.
A lot of students get fazed by anything that looks even slightly complicated. You’ve done a wonderful job here of explaining how to break it down into simpler structures, step by step. And, without saying so explicitly, stress on understanding the basics and first principles.
Very impressed. 🙂 | 704 | 2,739 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2021-21 | latest | en | 0.928625 |
https://helpinhomework.org/question/24667/If-you-buy-a-computer-directly-from-the-manufacturer-for-2242-and-agree-to-repay-it-in-48-equal-in | 1,718,797,191,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861817.10/warc/CC-MAIN-20240619091803-20240619121803-00586.warc.gz | 253,689,902 | 19,295 | Fill This Form To Receive Instant Help
#### If you buy a computer directly from the manufacturer for \$2,242 and agree to repay it in 48 equal installments at 1
###### Math
If you buy a computer directly from the manufacturer for \$2,242 and agree to repay it in 48 equal installments at 1.77% interest per month on the unpaid balance, how much are your monthly payments? How much total interest will be paid? Your monthly payment is \$ . (Round to two decimal places.) | 109 | 472 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-26 | latest | en | 0.935019 |
https://socratic.org/questions/how-do-you-determine-whether-the-function-f-x-x-3-3x-2-5x-7-is-concave-up-or-con | 1,725,987,465,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651303.70/warc/CC-MAIN-20240910161250-20240910191250-00216.warc.gz | 513,934,604 | 6,292 | # How do you determine whether the function f(x)=x^3+3x^2+5x+7 is concave up or concave down and its intervals?
Jul 26, 2015
Investigate the sign of the second derivative, $f ' ' \left(x\right)$
#### Explanation:
For $f \left(x\right) = {x}^{3} + 3 {x}^{2} + 5 x + 7$, we have
$f ' \left(x\right) = 3 {x}^{2} + 6 x + 5$ and
$f ' ' \left(x\right) = 6 x + 6$
$f ' ' \left(x\right) = 0$ at $x = - 1$, so we check the sign of $f ' '$ on each side of $- 1$
$f ' ' \left(x\right)$ is negative (that is $f ' ' \left(x\right) < 0$) for $x < - 1$. So the graph of $f$ is concave down on $\left(- \infty , - 1\right)$.
$f ' ' \left(x\right) > 0$ for $x > - 1$, so the graph of $f$ is concave up on $\left(- 1 , \infty\right)$
The point $\left(- 1 , 4\right)$ is the inflection point for the graph. | 319 | 797 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 18, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-38 | latest | en | 0.561726 |
https://www.enotes.com/homework-help/can-someone-explain-string-theory-me-laymans-terms-278024 | 1,516,462,625,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889660.55/warc/CC-MAIN-20180120142458-20180120162458-00230.warc.gz | 890,210,670 | 10,312 | # Can someone explain String Theory to me in layman's terms?
jdkotliar | Certified Educator
If you recall Einstein demonstrated that matter and energy are at their core the same thing, two sides of a coin. (E=MC2) String theory takes this concept to the quantum level.
So at the subatomic level, when you look at the various particles, and work your way down to the most basic building block of these subatomic particles you will get down to vibrating strings and membranes of energy, the manner in which these objects vibrate determine the nature of the subatomic particles. Essentially string theory is the THEORY OF EVERYTHING, an effort to unit the four basic forces of the Universe into a unified concept.
The links include some videos that should help. I think Michio Kaku, the third link might prove the most helpful. His videos are meant for the layman and this one is in four parts on Youtube.
pacorz | Certified Educator
Physicists recognize four fundamental forces in the universe: The weak and strong nuclear forces, gravity, and electromagnetism. Each force is produced by a particle that acts as a carrier of that force - photons carry electromagnetism, while gravitrons carry gravity. The strong nuclear force is associated with gluons, and the weak force with three particles called W+, W-, and Z.
String Theory unifies these particles by positing that these particles are not different from one another, but are actually variations of the same thing, a string. In string theory, the strings (which are actually usually visualized as loops) are in constant motion, and the axis around which that motion is oriented in space determines the behavior, and by extension the identity, of the particle.
alexkirk | Student
Thank you. That helps! | 370 | 1,765 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2018-05 | latest | en | 0.933454 |
https://crackbmat.com/2011-section-2-question-19/ | 1,620,713,808,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991904.6/warc/CC-MAIN-20210511060441-20210511090441-00062.warc.gz | 210,117,244 | 16,676 | 19) A
V (voltage) is directly proportional to I (current). If voltage increases, then current also increases since there is a constant temperature.
We know the equation: V = IR
Voltage increases and Current increases. Since they both increase proportionally, resistance would have to remain the same. | 71 | 302 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-21 | latest | en | 0.893795 |
http://nrich.maths.org/public/leg.php?code=-25&cl=4&cldcmpid=311 | 1,503,080,693,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105086.81/warc/CC-MAIN-20170818175604-20170818195604-00374.warc.gz | 304,596,563 | 7,169 | # Search by Topic
#### Resources tagged with Number theory similar to Napoleon's Hat:
Filter by: Content type:
Stage:
Challenge level:
### There are 23 results
Broad Topics > Numbers and the Number System > Number theory
### Whole Number Dynamics II
##### Stage: 4 and 5
This article extends the discussions in "Whole number dynamics I". Continuing the proof that, for all starting points, the Happy Number sequence goes into a loop or homes in on a fixed point.
### Always Perfect
##### Stage: 4 Challenge Level:
Show that if you add 1 to the product of four consecutive numbers the answer is ALWAYS a perfect square.
### Euler's Squares
##### Stage: 4 Challenge Level:
Euler found four whole numbers such that the sum of any two of the numbers is a perfect square...
### Whole Number Dynamics III
##### Stage: 4 and 5
In this third of five articles we prove that whatever whole number we start with for the Happy Number sequence we will always end up with some set of numbers being repeated over and over again.
### Whole Number Dynamics V
##### Stage: 4 and 5
The final of five articles which containe the proof of why the sequence introduced in article IV either reaches the fixed point 0 or the sequence enters a repeating cycle of four values.
### More Sums of Squares
##### Stage: 5
Tom writes about expressing numbers as the sums of three squares.
### Sums of Squares and Sums of Cubes
##### Stage: 5
An account of methods for finding whether or not a number can be written as the sum of two or more squares or as the sum of two or more cubes.
### Modulus Arithmetic and a Solution to Differences
##### Stage: 5
Peter Zimmerman, a Year 13 student at Mill Hill County High School in Barnet, London wrote this account of modulus arithmetic.
### Number Rules - OK
##### Stage: 4 Challenge Level:
Can you convince me of each of the following: If a square number is multiplied by a square number the product is ALWAYS a square number...
### Ordered Sums
##### Stage: 4 Challenge Level:
Let a(n) be the number of ways of expressing the integer n as an ordered sum of 1's and 2's. Let b(n) be the number of ways of expressing n as an ordered sum of integers greater than 1. (i) Calculate. . . .
### Diophantine N-tuples
##### Stage: 4 Challenge Level:
Can you explain why a sequence of operations always gives you perfect squares?
### Pythagorean Golden Means
##### Stage: 5 Challenge Level:
Show that the arithmetic mean, geometric mean and harmonic mean of a and b can be the lengths of the sides of a right-angles triangle if and only if a = bx^3, where x is the Golden Ratio.
### There's a Limit
##### Stage: 4 and 5 Challenge Level:
Explore the continued fraction: 2+3/(2+3/(2+3/2+...)) What do you notice when successive terms are taken? What happens to the terms if the fraction goes on indefinitely?
### Never Prime
##### Stage: 4 Challenge Level:
If a two digit number has its digits reversed and the smaller of the two numbers is subtracted from the larger, prove the difference can never be prime.
### An Introduction to Number Theory
##### Stage: 5
An introduction to some beautiful results of Number Theory
### 2^n -n Numbers
##### Stage: 5
Yatir from Israel wrote this article on numbers that can be written as $2^n-n$ where n is a positive integer.
### Really Mr. Bond
##### Stage: 4 Challenge Level:
115^2 = (110 x 120) + 25, that is 13225 895^2 = (890 x 900) + 25, that is 801025 Can you explain what is happening and generalise?
### A Little Light Thinking
##### Stage: 4 Challenge Level:
Here is a machine with four coloured lights. Can you make two lights switch on at once? Three lights? All four lights?
### Data Chunks
##### Stage: 4 Challenge Level:
Data is sent in chunks of two different sizes - a yellow chunk has 5 characters and a blue chunk has 9 characters. A data slot of size 31 cannot be exactly filled with a combination of yellow and. . . .
### Mod 7
##### Stage: 5 Challenge Level:
Find the remainder when 3^{2001} is divided by 7.
### Novemberish
##### Stage: 4 Challenge Level:
a) A four digit number (in base 10) aabb is a perfect square. Discuss ways of systematically finding this number. (b) Prove that 11^{10}-1 is divisible by 100. | 1,012 | 4,244 | {"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} | 3.9375 | 4 | CC-MAIN-2017-34 | latest | en | 0.863429 |
http://functions.wolfram.com/EllipticFunctions/JacobiCS/16/01/02/0007/ | 1,527,150,870,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866107.79/warc/CC-MAIN-20180524073324-20180524093324-00074.warc.gz | 120,698,368 | 7,400 | html, body, form { margin: 0; padding: 0; width: 100%; } #calculate { position: relative; width: 177px; height: 110px; background: transparent url(/images/alphabox/embed_functions_inside.gif) no-repeat scroll 0 0; } #i { position: relative; left: 18px; top: 44px; width: 133px; border: 0 none; outline: 0; font-size: 11px; } #eq { width: 9px; height: 10px; background: transparent; position: absolute; top: 47px; right: 18px; cursor: pointer; }
JacobiCS
http://functions.wolfram.com/09.27.16.0024.01
Input Form
JacobiCS[z + 3 I EllipticK[1 - m], m] == I JacobiDN[z, m] /; Element[s, Integers]
Standard Form
Cell[BoxData[RowBox[List[RowBox[List[RowBox[List["JacobiCS", "[", RowBox[List[RowBox[List["z", "+", RowBox[List["3", "\[ImaginaryI]", " ", RowBox[List["EllipticK", "[", RowBox[List["1", "-", "m"]], "]"]]]]]], ",", "m"]], "]"]], "\[Equal]", RowBox[List["\[ImaginaryI]", " ", RowBox[List["JacobiDN", "[", RowBox[List["z", ",", "m"]], "]"]]]]]], "/;", RowBox[List["s", "\[Element]", "Integers"]]]]]]
MathML Form
cs ( z + 3 K ( 1 - m ) m ) dn ( z m ) /; s TagBox["\[DoubleStruckCapitalZ]", Function[List[], Integers]] Condition JacobiCS z 3 EllipticK 1 -1 m m JacobiDN z m s [/itex]
Rule Form
Cell[BoxData[RowBox[List[RowBox[List["HoldPattern", "[", RowBox[List["JacobiCS", "[", RowBox[List[RowBox[List["z_", "+", RowBox[List["3", " ", "\[ImaginaryI]", " ", RowBox[List["EllipticK", "[", RowBox[List["1", "-", "m_"]], "]"]]]]]], ",", "m_"]], "]"]], "]"]], "\[RuleDelayed]", RowBox[List[RowBox[List["\[ImaginaryI]", " ", RowBox[List["JacobiDN", "[", RowBox[List["z", ",", "m"]], "]"]]]], "/;", RowBox[List["s", "\[Element]", "Integers"]]]]]]]]
Date Added to functions.wolfram.com (modification date)
2007-05-02 | 598 | 1,736 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2018-22 | latest | en | 0.294013 |
http://mathhelpforum.com/discrete-math/271121-equivalence-class.html | 1,526,869,039,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863901.24/warc/CC-MAIN-20180521004325-20180521024325-00163.warc.gz | 182,406,685 | 11,624 | 1. equivalence class
Wazzup,
I hope that everybody is A+ up in here.
Check it:
I have have a subset $\displaystyle Q$ of $\displaystyle \mathbb{Z} \times \mathbb{Z}$ :
$\displaystyle Q=\left\lbrace \left( a,b \right)\in \mathbb{Z}\times \mathbb{Z}\mid b \neq 0\right\rbrace$
I have a relation $\displaystyle \sim$ defined on $\displaystyle Q$:
$\displaystyle \left( a,b \right) \sim \left( c,d \right)\Leftrightarrow ad=bc$
I've shown that the relation is an equivalence-relation.
Now I wanna "find" $\displaystyle \left[ (2,3)\right]$ and in general $\displaystyle \left[ (a,b)\right]$.
Is the following enough:
The equivalence class $\displaystyle \left[ (2,3)\right]$ consists of all pairs of numbers $\displaystyle (c,d)$ such that $\displaystyle 2d=3c$, meaning $\displaystyle \left[ (2,3) \right]= \left\lbrace (c,d) \mid 2d=3c \right\rbrace$.
The equivalence class $\displaystyle \left[ (a,b)\right]$ consists of all pairs of numbers $\displaystyle (c,d)$ such that $\displaystyle ad=bc$, meaning $\displaystyle \left[ (a,b) \right]= \left\lbrace (c,d) \mid ad=bc \right\rbrace$.
Get back 2 me
Peace out
2. Re: equivalence class
Originally Posted by Meelas
Wazzup,
I hope that everybody is A+ up in here.
Check it:
I have have a subset $\displaystyle Q$ of $\displaystyle \mathbb{Z} \times \mathbb{Z}$ :
$\displaystyle Q=\left\lbrace \left( a,b \right)\in \mathbb{Z}\times \mathbb{Z}\mid b \neq 0\right\rbrace$
I have a relation $\displaystyle \sim$ defined on $\displaystyle Q$:
$\displaystyle \left( a,b \right) \sim \left( c,d \right)\Leftrightarrow ad=bc$
I've shown that the relation is an equivalence-relation.
Now I wanna "find" $\displaystyle \left[ (2,3)\right]$ and in general $\displaystyle \left[ (a,b)\right]$.
Is the following enough:
The equivalence class $\displaystyle \left[ (2,3)\right]$ consists of all pairs of numbers $\displaystyle (c,d)$ such that $\displaystyle 2d=3c$, meaning $\displaystyle \left[ (2,3) \right]= \left\lbrace (c,d) \mid 2d=3c \right\rbrace$.
The equivalence class $\displaystyle \left[ (a,b)\right]$ consists of all pairs of numbers $\displaystyle (c,d)$ such that $\displaystyle ad=bc$, meaning $\displaystyle \left[ (a,b) \right]= \left\lbrace (c,d) \mid ad=bc \right\rbrace$.
Get back 2 me
Peace out
i'd take it a bit further
$d = \dfrac 3 2 c \Rightarrow \left(c, \dfrac 3 2 c\right) = c\left(1,\dfrac 3 2\right),~c \in \mathbb{Z}$
Now since the elements of $[(2,3)]$ must be integers it must be that $c$ is even in the formula above so
$[(2,3)] = k\left(2,3 \right),~k \in \mathbb{Z}$
similarly
$[(a,b)] = k\left(a,b\right),~k \in \mathbb{Z}$
this reveals the true nature of the equivalence class
3. Re: equivalence class
Originally Posted by romsek
i'd take it a bit further
$d = \dfrac 3 2 c \Rightarrow \left(c, \dfrac 3 2 c\right) = c\left(1,\dfrac 3 2\right),~c \in \mathbb{Z}$
this reveals the true nature of the equivalence class Does it?
What would you do the the equivalence class, $[(0,1)]~?$
4. Re: equivalence class
Originally Posted by Plato
What would you do the the equivalence class, $[(0,1)]~?$
you're right
$[(0,b)]=(a,0),~a\in\mathbb{Z}$
5. Re: equivalence class
Originally Posted by romsek
you're right
$[(0,b)]=(a,0),~a\in\mathbb{Z}$ WRONG!
In the definition, $\{(a,b)|b\not= 0$.
6. Re: equivalence class
Originally Posted by Plato
In the definition, $\{(a,b)|b\not= 0$.
yep, i got the order mixed up
$[(0,b)]=(0,a), a \in \mathbb{Z}-\{0\}$
7. Re: equivalence class
W'zup yall,
Great feedback but Romsek, you said:
$\displaystyle \left( c,d \right) = \left( c,\frac{3}{2}c \right) =c\left(1, \frac{3}{2} \right) \hspace{0,3cm} , \hspace{0,3cm} c \in \mathbb{Z}$
So you divide by a, but a could take on the value 0, as could c and d. The only element which (by definition) cannot take on the value 0 is b, so wouldn't it be more correct to divide by b?
Wazzup with it?
8. Re: equivalence class
Originally Posted by Meelas
W'zup yall,
Great feedback but Romsek, you said:
$\displaystyle \left( c,d \right) = \left( c,\frac{3}{2}c \right) =c\left(1, \frac{3}{2} \right) \hspace{0,3cm} , \hspace{0,3cm} c \in \mathbb{Z}$
So you divide by a, but a could take on the value 0, as could c and d. The only element which (by definition) cannot take on the value 0 is b, so wouldn't it be more correct to divide by b?
Wazzup with it?
I guess I can agree with that.
you'd end up with $c=\dfrac{a}{b} d$
$\left(\dfrac{a}{b}d,d\right) = d\left(\dfrac{a}{b},1\right)=d b(a, b) = k(a,b),~ k \in \mathbb{Z} - \{0\}$
and you get the same answer.
The correct answer for $(0,b)$ is in post #6 | 1,564 | 4,610 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2018-22 | latest | en | 0.641101 |
https://unitconverterbot.com/convert-625-r-to-f/ | 1,624,131,376,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487649688.44/warc/CC-MAIN-20210619172612-20210619202612-00529.warc.gz | 535,362,829 | 4,538 | # Convert 625 R to F
In this article I will show you how to convert 625 degrees rankine into degrees fahrenheit. Throughout the explanation below I might also call it 625 R to F. They are the same thing!
## How to Convert Degrees Rankine to Degrees Fahrenheit
Almost all units work using conversion factors. This is a number that can be used to convert from one unit to another by multiplying or dividing it.
In this case though, I'm trying to convert to F, which doesn't use a conversion factor.
The way I calculate this is to use a base unit. This means that all temperature units are first converted to the base unit, and then converted to the correct final unit.
For temperature units I use the base unit of K to do the conversion. Let's go through the calculation steps:
For this calculation, neither R or F is our base unit of K.
First, we convert R to K with the following formula:
(625 x 0.555556 = 347.2225 K
Next, we need to convert the K to our final format, F, with the formula:
347.2225 x 9 / 5 - 459.67 = 165.3305 F
So, the answer to the question "what is 625 degrees rankine in degrees fahrenheit?" is 165.3305 F.
## Degrees Rankine to Degrees Fahrenheit Conversion Table
Below is a sample conversion table for R to F:
Degrees Rankine (R) Degrees Fahrenheit (F)
0.01-459.66
0.1-459.57
1-458.67
2-457.67
3-456.67
5-454.67
10-449.67
20-439.67
50-409.67
100-359.67
1000540.33
## Best Conversion Unit for 625 R
Sometimes when you work with conversions from one unit to another, the numbers can get a little confusing. Especially when dealing with really large numbers.
I've also calculated what the best unit of measurement is for 625 R.
To determine which unit is best, I decided to define that as being the unit of measurement which is as low as possible, without going below 1. Smaller numbers are more easily understood and can make it easier for you to understand the measurement.
The best unit of measurement I have found for 625 R is degrees celsius and the amount is 74.0725 C. | 514 | 2,016 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2021-25 | latest | en | 0.902042 |
https://math.answers.com/math-and-arithmetic/What_is_the_percent_of_the_answer_if_you_got_54_out_of_66_question_right | 1,719,320,393,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865972.21/warc/CC-MAIN-20240625104040-20240625134040-00618.warc.gz | 346,758,531 | 46,852 | 0
# What is the percent of the answer if you got 54 out of 66 question right?
Updated: 9/21/2023
Wiki User
11y ago
81.8%
Wiki User
11y ago
Earn +20 pts
Q: What is the percent of the answer if you got 54 out of 66 question right?
Submit
Still have questions?
Related questions
### What is the percent of the answer if you got 54 out of 66 questions right?
81.8% 54/66 * 100= % right
63% or D
### What is the fraction 54 percent?
The fraction of 54 percent is 54/100 or 27/50.
### What is 54 percent of 1200?
54 percent of 1,200 is 648.
### What is 54 percent of 64?
54 percent of 64 54/100 X 64 = 34.56
### What percent of 80 is 54?
100*54/80 = 67.5%
### What is 54 is 90 percent of?
90% of 54 = 90% * 54 = 0.9 * 54 = 48.6
### What is 67.5 percent of 54?
67.5 percent of 54 equals 36.45
8% of 54 is 4.32
### 54 is 150 percent of what number?
54 is 150 percent of 36
54 = 5400%
54 = 5400% | 328 | 914 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.909551 |
https://engineersedge.com/calculators/triangle_oblique/triangle_oblique_calculator.htm | 1,638,472,443,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362287.26/warc/CC-MAIN-20211202175510-20211202205510-00558.warc.gz | 284,136,613 | 14,004 | ### Oblique Triangle Solutions Calculator & Equations
This calculator will determine the unknown length of a given oblique triangle for an Obtuse or Acute triangle. Simply enter in the unknown value and and click "Update" button located at the bottom of the web page.
OBLIQUE TRIANGLE SOLUTIONS:
Input:
Results:
a b c
LA LB LC Area
a c LB
b LA LC Area
a c LC
b LA LB Area
a LA LB
b c LC Area
c LA LC
a b LB Area
c LB LC
a b LA Area
Note: Input and results strictly adhere to
side and angle nomenclature above.
Law of Sine's
a/SIN(LA)
Law of Cosines:
a2 = b2 + c2 - 2*b*c*COS(LA)
b2 = a2 + c2 - 2*a*c*COS(LB)
c2 = a2 + b2 - 2*a*b*COS(LC)
For Acute Angle: (all of following are satisfied)
a2 + b2 > c2 , b2 + c2 > a2 , and a2 + c2 > b2
For Obtuse Angle: (only one of following is satisfied
either a2 + b2 < c2 , or b2 + c2 < a2 , or a2 + c2 < b2
Area of Triangle:
Area = a*c*SIN(LB)/2 = b*c*SIN(LA)/2 = a*b*SIN(LC)/2 | 336 | 928 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | longest | en | 0.647552 |
https://kr.mathworks.com/matlabcentral/cody/problems/2914-matlab-basics-ii-count-rows-in-a-matrix/solutions/580763 | 1,590,935,366,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347413406.70/warc/CC-MAIN-20200531120339-20200531150339-00125.warc.gz | 408,538,846 | 15,756 | Cody
# Problem 2914. Matlab Basics II - Count rows in a matrix
Solution 580763
Submitted on 12 Feb 2015 by Abdullah Caliskan
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; y_correct = 1; assert(isequal(count_rows(x),y_correct))
r = 1 c = 1
2 Pass
%% x = [1 2 2; 15 3 4]; y_correct = 2; assert(isequal(count_rows(x),y_correct))
r = 2 c = 3
3 Pass
%% x = [3.1 7.2;5.4 2.2;7 2.1;5 8.6]; y_correct = 4; assert(isequal(count_rows(x),y_correct))
r = 4 c = 2 | 217 | 594 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.693968 |
https://curriculum.illustrativemathematics.org/k5/teachers/grade-1/unit-4/lesson-10/lesson.html | 1,718,630,942,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861719.30/warc/CC-MAIN-20240617122241-20240617152241-00845.warc.gz | 174,362,640 | 28,159 | # Lesson 10
Write Two-digit Numbers
## Warm-up: Notice and Wonder: Same Digit, Different Place (10 minutes)
### Narrative
The purpose of this activity is for students to see the connection between base-ten diagrams and written numbers. In the synthesis, students consider the relationship between numbers with the same digit in different places. Students also observe that the single-digit number has no tens.
### Launch
• Groups of 2
• Display the image.
• “What do you notice? What do you wonder?”
• 1 minute: quiet think time
### Activity
• 1 minute: partner discussion
• Share and record responses.
### Student Facing
What do you notice?
What do you wonder?
7
70
17
71
### Student Response
For access, consult one of our IM Certified Partners.
### Activity Synthesis
• “How are 7 and 17 the same and different?” (They both use the digit 7 but only 17 has 1 ten. Seven has no tens.)
• “How are 7 and 70 the same and different?” (7 is a one-digit number and 70 is a two-digit number. They both have a 7 as a digit. 70 has 7 tens. 7 has 7 ones.)
• “How are 70 and 71 the same and different?” (Both have 7 tens, but 71 has 1 one instead of 0 ones.)
## Activity 1: Write Numbers to Match Base-Ten Representations (15 minutes)
### Narrative
The purpose of this activity is for students to produce written numbers for a given quantity represented in different ways. Some representations show the tens on the left and others show the ones on the left. Students must attend to the units in each representation and the meaning of the digits in a two-digit number, rather than always writing the number they see on the left in a representation in the tens place and the number they see on the right in a representation in the ones place (MP6).
Engagement: Develop Effort and Persistence. Chunk this task into more manageable parts. Check in with students to provide feedback and encouragement after each chunk.
Supports accessibility for: Attention, Social-Emotional Functioning
### Launch
• Groups of 2
• Give students access to connecting cubes in towers of 10 and singles.
### Activity
• “You will work on your own to start. When you and your partner are finished, compare your work.”
• 5 minutes: independent think time
• 3 minutes: partner discussion
### Student Facing
Write the number that matches each representation.
1. 1 ten 4 ones
Number: ____________
2.
Number: ____________
3. 9 tens
Number: ____________
4. $$20 + 5$$
Number: ____________
5.
Number: ____________
6. $$7 + 40$$
Number: ____________
7. 2 ones 8 tens
Number: ____________
8.
Number: ____________
9. $$1 + 40$$
Number: ____________
10.
Number: ____________
### Student Response
For access, consult one of our IM Certified Partners.
### Activity Synthesis
• Invite students to share the two-digit number that matches each representation and explain how it matches.
## Activity 2: Introduce Write Numbers, Numbers to 99 by 1 (20 minutes)
### Narrative
The purpose of this activity is for students to learn a new center called Write Numbers. Students write a two-digit number in each space on a gameboard. They take turns writing the next one, two, or three numbers in the sequence. The player who writes the last number on the board wins. Students may choose to count forward or backward.
MLR8 Discussion Supports. Synthesis: Provide students with the opportunity to rehearse with a partner what they will say about the patterns they saw before they share with the whole class.
### Required Materials
Materials to Gather
Materials to Copy
• Write the Number Stage 1 Gameboard
### Required Preparation
• Put each gameboard in a sheet protector.
### Launch
• Groups of 2
• Give each group a gameboard and a dry erase marker.
• “We are going to learn a new center called Write Numbers.”
• Display the gameboard.
• “You and your partner will practice writing numbers. You will fill in the number path on the gameboard. You can decide to start with the smaller number and count forward, or start with the larger number and count backward. On each turn, you can decide whether you would like to write one, two, or three numbers on the gameboard. The person who writes the last number on the board is the winner.”
• Demonstrate playing one round with the students.
• “Now you will play with your partner.”
### Activity
• 10 minutes: partner work time
### Activity Synthesis
• “What patterns did you see as you wrote numbers?” (The ones place goes up by one each time until you get to 9. Then the tens place changes by one, and the 0–9 pattern repeats.)
## Lesson Synthesis
### Lesson Synthesis
Display 702.
“Today we wrote two-digit numbers. Clare counted a collection of seventy-two objects. This is the number she wrote. Give a thumbs up if you agree and a thumbs down if you disagree.”
“What advice do you have for Clare about writing numbers that have tens and ones?” (When writing tens and ones there are only 2 digits. The first digit tells the number of tens and the second digit tells the number of ones. There is no 0 in 72, even if it sounds like there should be.)
## Cool-down: Write Numbers (5 minutes)
### Cool-Down
For access, consult one of our IM Certified Partners. | 1,213 | 5,217 | {"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} | 4.71875 | 5 | CC-MAIN-2024-26 | latest | en | 0.891324 |
https://timestable.net/category/factors/page/99 | 1,695,563,225,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506646.94/warc/CC-MAIN-20230924123403-20230924153403-00376.warc.gz | 614,832,167 | 16,033 | Home » Factors » Page 99
# Factors
## Factors of 20
The factors of 20 and the prime factors of 20 differ because twenty is a composite number. Also, despite being closely related, the prime factors… Read More »Factors of 20
## Factors of 19
The factors of 19 and the prime factors of 19 don’t differ much because nineteen is a prime number. But, despite being closely related, the… Read More »Factors of 19
## Factors of 18
The factors of 18 and the prime factors of 18 differ because eighteen is a composite number. Also, despite being closely related, the prime factors… Read More »Factors of 18
## Factors of 17
The factors of 17 and the prime factors of 17 don’t differ much because seventeen is a prime number. But, despite being closely related, the… Read More »Factors of 17
## Factors of 16
The factors of 16 and the prime factors of 16 differ because sixteen is a composite number. Also, despite being closely related, the prime factors… Read More »Factors of 16
## Factors of 15
The factors of 15 and the prime factors of 15 differ because fifteen is a composite number. Also, despite being closely related, the prime factors… Read More »Factors of 15
## Factors of 14
The factors of 14 and the prime factors of 14 differ because fourteen is a composite number. Also, despite being closely related, the prime factors… Read More »Factors of 14
## Factors of 13
The factors of 13 and the prime factors of 13 don’t differ much because thirteen is a prime number. But, despite being closely related, the… Read More »Factors of 13
## Factors of 12
The factors of 12 and the prime factors of 12 differ because twelve is a composite number. Also, despite being closely related, the prime factors… Read More »Factors of 12
## Factors of 11
The factors of 11 and the prime factors of 11 don’t differ much because eleven is a prime number. But, despite being closely related, the… Read More »Factors of 11 | 475 | 1,926 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2023-40 | latest | en | 0.912391 |
https://blog.rotovalue.com/index.php/2013/03/01/rotovalue-prices-a-war-for-fantasy-sports/ | 1,679,638,991,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945248.28/warc/CC-MAIN-20230324051147-20230324081147-00100.warc.gz | 168,628,516 | 11,089 | # RotoValue Prices – a WAR For Fantasy Sports
So how are RotoValue auction prices computed anyway?
One way to think of them is as a sort of WAR for fantasy sports. Tom Tango has done a great job explaining WAR, emphasizing that it is a framework, and thus you can have many different implementations.
Ultimately that’s what RotoValue is – an implementation of a model to value fantasy players. Tango cites a list of key attributes for WAR:
a) WAR is wins above replacement.
b) WAR is a framework.
c) WAR presents the performance of a player into a single number.
d) WAR is limited to the data points it considers.
e) WAR is limited by the bias in the data.
f) WAR is not all-encompassing.
I think this is a great way to think about RotoValue pricing (and other fantasy sports pricing models, too). It’s “dollars”, not “wins”, but everything else in the description fits.
RotoValue is inherently based on the concept of replacement. A 12 team fantasy league needing 2 active catchers will need to start 24 catchers. While catchers usually don’t produce as good offensive stats as other positions, everyone needs two, and so players are valuable to the extent that they’re worth more than whoever is left over after all teams have picked somebody. That’s replacement level – you’re only as good as your replacement level.
So what do I do with RotoValue? Last fall I discussed it in the context of basketball, but the same basic model applies to all sports. First I need some measure of a player’s value, so I assign a normalized score to each player in each category. The best player in the league is given a score of 1, the worst is given 0, and everyone else is somewhere in between. Assuming equally weighted categories, I simply add up the category scores, and that’s the player’s raw value. Rate stats (like batting average or ERA) are a little more complex, but not too much: the “best” is impacted by how often you play. A batter who goes 1-3 may have a .333 average, but a .290 hitter in 600 AB helps your fantasy team a lot more. So I account for that.
And if you want to give categories different weights (either because that’s how your league scores, or because you want to emphasize or ignore particular categories because of how close teams are in the standings), I apply that here.
So this gives me a raw value for each player. Next I sort players by raw value, and find the “replacement” level for each position, which I define as the best player who shouldn’t be starting in the league. So in a 12 team league using 2 catchers per team, the 25th best catcher is the replacement level guy.
Once I know replacement level, I subtract the value of the replacement player from the raw value to get a net value for each player. I add up all the net value in the league (and the composite salary cap, minus any minimum bids required per roster spot), and I can create a ratio of dollars to “value”. So a player’s RotoValue price is just their net value times this ratio, plus the minimum bid.[ref]Technically, I use two different replacement levels: the first is for players who “should” be in a starting lineup; the second is for players who shouldn’t be owned, and who get negative prices, and this is set to the worst player at a position who “should” be on a bench.[/ref]
The whole point of a pricing model for fantasy sports is to distill a player’s expected contribution into a single number, the price. So in that sense, RotoValue is very much like a WAR implementation.
Like any pricing model, RotoValue is limited to the data points it considers. The model is designed to take in a set of statistics and output prices for players based on those statistics, and I’ve made it easy to use different sources: projections, actual data, and data from different time periods.
RotoValue is subject to constraints: the sum of positive RotoValue prices in a league must equal the total salary cap for that league. A player worth owning can’t have a price below the minimum bid. While WAR can never be all encompassing – you can’t quantify absolutely everything that goes into winning baseball games – a fantasy pricing model has an easier task, because the only thing that matters is the statistics players compile. And indeed the process of converting raw player statistics into prices is inherently a form of normalization, which can remove some biases in the input.
If, for example, a projection system is overly optimistic about league-wide batting average or ERA, that may make it seem “bad” in nominal comparison, but because prices are based on relative performance compared to a replacement level, and because the model is using as input a normalized score for each category, the raw numbers themselves can be “wrong” without impacting prices. If you doubled all the raw statistics for each player, the prices would be identical, because raw statistics are only meaningful in comparison to statistics of other players. But other biases in input will be reflected in computed prices: if your projections are overly optimistic for ground-ball pitchers, then they would as a group be overvalued compared to fly-ball pitchers.
Fantasy pricing will always be subjective, but it does help to have a consistent framework. RotoValue is one such framework. I first compute a raw normalized value for each player, summing their contributions in each category. After sorting by position, I find a replacement level for each player, and the player’s net value is just their raw value minus the appropriate replacement level.[ref]If a player qualifies at multiple positions, I use the position with the weakest replacement level player.[/ref] The remaining assumption is that an owner will pay the same rate for a unit of net value no matter where it comes from. So a player with a net value that is twice as big will have twice as large a price.[ref]Technically, their price above the minimum bid will be twice as large. So if your minimum bid is \$1, a if player A has a price of \$4 and player B has twice the net value of player A, his price will be \$7.[/ref]
RotoValue applies a consistent, reasonable set of assumptions to the problem of valuing players. When I bid in my own auctions, I use prices as a guide, but not as a hard-fast rule. And of course in bidding strategy you want to buy players below their theoretical prices, so I don’t always, or even usually, bid up to a player’s theoretical price for some input. Sometimes I may even bid beyond their price. But having a good model keeps me grounded. | 1,407 | 6,521 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.939305 |
https://www.convertunits.com/from/catti+%5BChina%5D/to/hundredweight+%5Blong,+UK%5D | 1,660,086,880,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571090.80/warc/CC-MAIN-20220809215803-20220810005803-00510.warc.gz | 636,610,341 | 16,991 | ## ››Convert catti [China] to hundredweight [long, UK]
catti [China] hundredweight [long, UK]
Did you mean to convert catti [China] catti [Japan] to hundredweight [long, UK] hundredweight [short, US]
How many catti [China] in 1 hundredweight [long, UK]? The answer is 83.988171837156.
We assume you are converting between catti [China] and hundredweight [long, UK].
You can view more details on each measurement unit:
catti [China] or hundredweight [long, UK]
The SI base unit for mass is the kilogram.
1 kilogram is equal to 1.653234139285 catti [China], or 0.019684130552221 hundredweight [long, UK].
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between catti [China] and hundredweights.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of catti [China] to hundredweight [long, UK]
1 catti [China] to hundredweight [long, UK] = 0.01191 hundredweight [long, UK]
10 catti [China] to hundredweight [long, UK] = 0.11906 hundredweight [long, UK]
20 catti [China] to hundredweight [long, UK] = 0.23813 hundredweight [long, UK]
30 catti [China] to hundredweight [long, UK] = 0.35719 hundredweight [long, UK]
40 catti [China] to hundredweight [long, UK] = 0.47626 hundredweight [long, UK]
50 catti [China] to hundredweight [long, UK] = 0.59532 hundredweight [long, UK]
100 catti [China] to hundredweight [long, UK] = 1.19064 hundredweight [long, UK]
200 catti [China] to hundredweight [long, UK] = 2.38129 hundredweight [long, UK]
## ››Want other units?
You can do the reverse unit conversion from hundredweight [long, UK] to catti [China], or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Hundredweight
unit of weight of 112 pounds
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 632 | 2,267 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.76531 |
https://www.computeexpert.com/english-blog/excel-formulas-list/sum-function-in-excel | 1,718,369,183,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861546.27/warc/CC-MAIN-20240614110447-20240614140447-00805.warc.gz | 651,803,761 | 18,560 | SUM Formula in Excel: Functions, Examples and How to Use - Compute Expert
SUM Formula in Excel: Functions, Examples and How to Use
Home >> Excel Tutorials from Compute Expert >> Excel Formulas List >> SUM Formula in Excel: Functions, Examples and How to Use
In this tutorial, you will learn how to use the SUM formula/function in excel completely. We will discuss the basics of SUM and its advanced use too to optimize your numbers processing in excel.
The sum is a basic calculation process that many of us use to get the results we desire from our numbers. When we process our numbers in excel, SUM can help us to sum them fast. We can also combine SUM with many formulas to optimize its capabilities and run more complex data analysis.
Curious to learn more about SUM and what we should do with this formula? Read all parts of this complete SUM tutorial from Compute Expert!
Disclaimer: This post may contain affiliate links from which we earn commission from qualifying purchases/actions at no additional cost for you. Learn more
Want to work faster and easier in Excel? Install and use Excel add-ins! Read this article to know the best Excel add-ins to use according to us!
What is SUM In Excel?
We can define SUM as an excel calculation function that can help us to sum numbers fast and accurately.
SUM Function in Excel
We can use SUM to sum numbers from direct typing, cells, or cell ranges fast.
SUM Result
SUM will give us a number which is the total of all the numbers we input into it.
Excel Version from Which We Can Start Using SUM Formula in Excel
We can start using SUM in excel 2003.
The Way to Write It and Its Inputs
Here is the general writing form of the SUM function in excel.
= SUM ( number1 , number2 , … )
The numbers we input into SUM can be in the forms of direct typing, cell coordinates, or cell ranges. However, we usually only input one cell range to SUM which contains all the numbers we want to process.
Remember to type comma signs ( , ) between the inputs we give to SUM.
Example of Its Usage and Result
Here is the implementation example of SUM in excel. You can see the writing and the result of the formula here.
In the example, we want to sum the numbers in each row. We use the SUM formula to help us run the sum process easier.
The way to use SUM is quite easy. Just input the numbers we want to sum into the formula. Here, as the numbers are all in each row cell range, we just input those row cell ranges into our SUM.
Write the SUM formula correctly, enter, and we will immediately get the sum result of our numbers!
Writing Steps
After we discuss the writing form, inputs, and example of SUM, now we discuss its writing steps in excel. If you have understood the tutorial discussion earlier, then you should be able to follow these steps with no problem!
1. Type an equal sign ( = ) in the cell where you want to put the sum result of your numbers
2. Type SUM (can be with large and small letters) and an open bracket sign after =
3. Input all the numbers you want to sum in the forms of direct typing, cell coordinates, and/or cell ranges. If you give more than one input, then separate them with comma signs
4. Type a close bracket sign after you have inputted all the numbers you want to sum
5. Press Enter
6. Done!
Common Factors Why Your SUM Produces an Error/Wrong Result
Your SUM gives you an error/wrong sum result? Check your SUM writing with the guidance of these points below!
• You give more or fewer inputs than you need to get the correct SUM result. This sometimes happens when you need to sum many numbers. Check again your input to make sure you have given the correct numbers
• Your SUM inputs probably contain some different kinds of numbers like dates, percentages, time, etc. These inputs can make SUM produces an unexpected result for you
• Your SUM inputs might contain some error values (if this is the case, then most probably they are in the cell ranges that you input). These values can make your SUM produces an error too. If you want SUM to ignore those errors, see how to do this in the SUM IFERROR part of this tutorial
Rounding Process after SUM: ROUND/ROUNDUP/ROUNDDOWN SUM
After you get your SUM result, especially if you sum decimals, you may need to round it. To do that immediately, you can combine SUM writing with excel decimal rounding formulas: ROUND/ROUNDUP/ROUNDDOWN.
You can combine those formulas with SUM quite easily too. Just write SUM inside the rounding formula you use. Write it as the number-to-round input in the rounding formula.
To summarize, here is the general writing form of the formulas’ combination in excel.
= ROUND / ROUNDUP / ROUNDDOWN ( SUM ( number1 , number2 , … ) , number_of_decimals )
Regarding the formula you should use, here is what each decimal rounding formula does to your SUM result.
• ROUND = round your SUM result normally (will round 0-4 numbers down and 5-9 numbers up)
• ROUNDUP = round up your SUM result
• ROUNDDOWN = round down your SUM result
Choose the rounding formula according to the need of your final result!
To understand the concept of this rounding formula + SUM combination easier, here is the implementation example in excel.
In the example, we have some decimals we want to sum on each row. As we want to round the sum result immediately, we can combine SUM with ROUND/ROUNDUP/ROUNDDOWN in one formula writing.
Just input the SUM in the decimal rounding formula we choose. Don’t forget to input the number of decimals we want for the result too.
Additionally, you can also round to powered ten multiples (10, 100, 1000, etc) using ROUND/ROUNDUP/ROUNDDOWN. Just input a negative number in your rounding formula’s number of decimals input. Determine the negative number value according to the power level you want to give for the 10 multiples (1 for 10, 2 for 100, 3 for 1000, etc).
SUM the Largest/Smallest Values: SUM LARGE/SMALL
Need to sum the largest/smallest values in the group of numbers that you have? To do it in one formula writing, you can combine SUM with LARGE (for the largest values)/SMALL (for the smallest values).
If you have used LARGE/SMALL before, then you know we usually only get one nth largest/smallest value from it. However, if you input an array in the formula nth largest/smallest input, then you can get an array result too. We can use this fact to make SUM directly sum the largest/smallest values in our group of numbers.
Here is the general writing form of the SUM and LARGE/SMALL combination.
= SUM ( LARGE/SMALL ( numbers_cell_range , { nth1 , nth2 , … } ) )
You can determine how many top/bottom values you want to sum in the LARGE/SMALL’s nth largest/smallest array input. For example, if you want to sum five top values from your numbers, then you can input {1,2,3,4,5} to your LARGE.
You can also input the array to LARGE/SMALL in the form of a cell range. For this, however, you need to use the array formula form so your LARGE/SMALL can process it correctly. To use the array formula form, press Ctrl + Shift + Enter after you finish writing your formula instead of Enter only.
Here is the implementation example of the SUM and LARGE/SMALL formulas combination in excel.
In the example, we have monthly sales quantities data. We want to know the total of the top 3 and the bottom 3 separately. To easily do the calculation, we use SUM and LARGE/SMALL.
We input our LARGE/SMALL in our SUM and input the monthly sales quantity cell range in the LARGE/SMALL. For the second input of the LARGE/SMALL, we input a {1,2,3} array. That is because we want to sum the first, second, and third-largest/smallest numbers.
We do the writing like that and we can sum the top 3 and bottom 3 monthly sales quantities fast!
SUM Based on a Reference Value: SUM VLOOKUP
Let’s say you have a reference table containing numbers and you want to sum numbers that belong to a certain variable. Can we pull those numbers from the table and sum them using SUM?
The answer is yes, you can. However, to do that in one writing, you need to combine your SUM with VLOOKUP.
Here is the general writing of the SUM and VLOOKUP combination for that sum process.
{ = SUM ( VLOOKUP ( variable_to_sum , reference_table_range , { column_to_sum1 , column_to_sum2 , … } , lookup_mode ) }
In this SUM and VLOOKUP combination, we use VLOOKUP to pull the numbers we want to sum from a certain variable. We input the variable as our lookup reference and the reference table cell range too in VLOOKUP.
For the VLOOKUP result column index input, we input an array that contains all the column indexes we want to sum. The index starts from 2 since the first one belongs to the variable column range. We need to input all column indexes in the reference table if we want to sum them all.
We can use TRUE or FALSE for the VLOOKUP lookup mode input depending on the lookup process we want to do. As usual, input TRUE to enable an approximate match search and FALSE for an exact match search.
Input all of those VLOOKUP writings to SUM. In the formula writing, we use an array formula form (the curly bracket signs that envelop the formula are the sign for the array formula form) since our VLOOKUP needs to deal with an array input. Because of this, we need to press Ctrl + Shift + Enter after we write the formula instead of just Enter (to change our formula form into an array formula).
Here is the implementation example of the SUM and VLOOKUP combination.
In the example, we have a table of monthly sales quantity data for five products. If we want to get one month’s sales quantities total from the table, then we can use SUM and VLOOKUP combination.
Just write the SUM and VLOOKUP in the way we discussed earlier. Input the month (March for this example), the reference table cell range, and column indexes array into our VLOOKUP. We input FALSE as the lookup mode here since we want to sum only the March sales quantities.
Write the VLOOKUP with these inputs into SUM and press Ctrl + Shift + Enter. We will immediately get the total sales quantity for March!
SUM a Cell Range Which Coordinates Are Based on Cell Values: SUM INDIRECT
Want to determine the row/column you use your SUM on based on cell values? If you want your SUM cell range input to be dynamic like this, then you can combine it with INDIRECT.
INDIRECT is an excel formula that can convert a regular text into a cell reference. For example, if you write =INDIRECT(“C5”) in a cell, then that cell value will refer to the cell C5 value. This is an excellent formula to use whenever you need to make a formula’s cell/cell range references to constantly changes.
The usual way to write a SUM and INDIRECT combination for a dynamic SUM input itself is like this.
= SUM ( INDIRECT ( column_letter1 & row_number1 & “:” & column_letter2 & row_number2 ) )
Input cell coordinates to the column letter and row number inputs you want to refer to cell values in the INDIRECT.
Input from where you want to sum in the first pair of your column letter and row number. Next, input to where you want to sum in the second one. This mirrors the cell range reference form in excel where we write the first cell, colon, and the last cell (for example, A1:C5 means we refer to cell A1 to cell C5).
By correctly inputting to INDIRECT, you can change the cell range you refer in your SUM by changing your cell values!
You can see the implementation example of the SUM and INDIRECT combination below.
Here, we want to determine the month range we want to sum the sales quantities on based on cell values. We can use the combination of SUM and INDIRECT for that.
Just write an INDIRECT inside your SUM with the inputs of appropriate column letters and row numbers. Since the monthly sales quantity numbers are all in column C, we can just input “C” for our column letter inputs.
For the to and from row numbers, we refer to cell values as you can see in the screenshot. Because of this, we can change the cell range reference in SUM by just changing the cell values we refer to.
SUM to Count Data Based on Multiple Criteria and OR Logic: SUM COUNTIF
We can combine SUM with many other formulas in excel for various purposes. We have seen some examples of them earlier. Another one here is combining SUM with COUNTIF to count data using multiple criteria with OR logic.
COUNTIF is a formula we can use to count data based on a criterion we have. However, it cannot count data using multiple criteria and OR logic by itself (its sibling formula, COUNTIFS, count data using multiple criteria but with an AND logic). OR logic here means we want to count all data that fulfill at least one of our criteria.
However, by combining it with SUM, we can do that OR counting process. Here is the general writing form of SUM and COUNTIF for this purpose.
= SUM ( COUNTIF ( cell_range1 , criterion1 ) , COUNTIF ( cell_range2 , criterion2 ) , … ) )
In this writing, SUM will sum the data amount in your cell ranges that matches one of your criteria. Input the number of COUNTIFs into your SUM that matches the number of criteria you have. The cell ranges you input to your COUNTIFs can be the same or different depending on your needs.
Here is the implementation example of the SUM and COUNTIF combination for the data counting purpose.
In this example, we count the number of monthly sales quantities that are more than 7000 or less than 3000. We use a SUM and COUNTIFs combination to make the data counting process faster.
Since we have two criteria here, we use two COUNTIFs in our SUM. The data we want to count is in the same cell range for both criteria. Therefore, we input the same cell range for the COUNTIFs in this example (the sales quantity column cell range).
We get the COUNTIF result for each criterion and we sum them using SUM. By doing this, we can get our data counting result with multiple criteria and OR logic!
SUM Every Nth Row: SUM MOD ROW
Sometimes, we probably don’t want to sum all of the numbers we have in a column. We may only need to sum the numbers in every nth row instead.
Using SUM alone isn’t enough when we have this situation. To help us adding the numbers in every nth row, we should use the combination of SUM, MOD, and ROW.
You may have known and utilized the functions of MOD and ROW before in excel. We can use MOD to get a division process remainder while ROW gives us the row number of a cell. Combining them with SUM can give us the ability to loop through numbers in rows in our sum process.
The general writing form of the SUM, MOD, and ROW combination is as follows.
{ = SUM ( numbers_range * MOD ( ROW ( appropriate_similar_sized_range_to_numbers_range ) , nth_row ) = 0 ) }
We will explain the process that runs in the formula writing by using the help of the example below.
In the example, we have a monthly and weekly sales quantity table. We want to sum only the week 3 sales quantities from the table. Therefore, we should use a SUM, MOD, and ROW combination.
To get only the sales quantities from week 3, we multiply the sales quantity column with MOD and ROW in SUM. We use MOD and ROW here to get a TRUE and FALSE array that will filter the week 3 sales quantities.
We input an appropriate same-sized cell range to our ROW to get the TRUE and FALSE array. We base the cell range input on the nth row we want to get from the sales quantity column.
We aim to get 0 from the MOD and ROW result for the week 3 rows in the sales quantity column. That is because we have an “equal to 0” condition at the end of our MOD.
The 0 will produce a TRUE logic value while other numbers will produce FALSE logic values because of that condition. As TRUE is 1 and FALSE is 0 in excel, our multiplication with the sales quantity column will produce as follows. All the sales quantities we multiply will become 0 from rows other than week 3 rows!
As week 3 has five rows distance and it starts in the fourth row, we input ROW the C2:C16 range. The C2:C16 has an equal size to the sales quantity column (C3:C17). It also makes week 3 rows produce 5 from ROW and thus, become 0 when we divide them with MOD.
To summarize the results chronology, from the ROW formula we write in the example, we get this array result.
{ 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 , 16 }
The 5 multiples there are in parallel with the week 3 numbers. This is important to make them not become 0 in our multiplication process later.
We divide all those array numbers with 5 we input to our MOD and we get this.
{ 2 , 3 , 4 , 0 , 1 , 2 , 3 , 4 , 0 , 1 , 2 , 3 , 4 , 0 , 1 }
As we compare the MOD result with 0 in our formula writing, we get this from the comparison result.
{ FALSE , FALSE , FALSE , TRUE , FALSE , FALSE , FALSE , FALSE , TRUE , FALSE , FALSE , FALSE , FALSE , TRUE , FALSE }
Which, in excel calculation, is similar to this.
{ 0 , 0 , 0 , 1 , 0 , 0 , 0 , 0 , 1 , 0 , 0 , 0 , 0 , 1 , 0 }
We multiply this array to the sales quantity numbers in our column. From that multiplication, we get this number array.
{ 0 , 0 , 0 , 133 , 0 , 0 , 0 , 0 , 124 , 0 , 0 , 0 , 0 , 236 , 0 }
Notice that only the week 3 sales quantity numbers are not 0.
We sum the numbers in the array using SUM. From this, we get the week 3 sales quantities total, which is 493!
Don’t forget the array formula form for this SUM, MOD, and ROW combination too. As we input an array to our ROW, we must press Ctrl + Shift + Enter after we write our formula.
SUM a Filtered Cell Range: SUBTOTAL
If we apply SUM to a filtered cell range, then it will sum the numbers that don’t pass our filter too. To sum only the numbers that pass, we should use SUBTOTAL instead.
SUBTOTAL is an excel formula that specializes itself in processing data that passes filters. We can use several formulas through SUBTOTAL for our filtered data, including SUM.
Here is the SUBTOTAL general writing form if we want to sum numbers in a filtered cell range.
= SUBTOTAL ( 9 / 109 , filtered_range1 , filtered_range2 , … )
The 9 or 109 input there is the SUM formula code number when we use SUBTOTAL. We use 9 if we want to sum the numbers in the cells we hide manually too in the filtered ranges. If we want to ignore them, then we input 109.
In SUBTOTAL, we can input multiple filtered cell ranges for our SUM calculation after the 9/109 input. However, most of the time, we only input one filtered range. If we happen to input multiple filtered ranges, then don’t forget to separate them with commas.
To better understand the SUBTOTAL implementation to apply SUM in a filtered cell range, look at the example below.
In the example, we have a filtered cell range and we want to sum the numbers that pass the filter there. This is a perfect situation example to use SUBTOTAL in excel.
We input 9/109 as the formula code input in SUBTOTAL to conjure SUM. In this example, it doesn’t matter we input 9 or 109 since we don’t have manually hidden cells in the range. We will produce the same sum result using either code.
We give the code input and the filtered cell range to our SUBTOTAL. As a result, we get a SUM result from the numbers that pass our filter! If you happen to change the filter in the cell range, then the SUBTOTAL result should change too!
SUM with IF: SUMIFS
We have discussed quite a bit about COUNTIF as the excel formula to count data that passes a criterion. But, what if we want to sum the numbers from the data entries that pass the criteria that we have?
Can we do a SUM with an IF-like function like that? For this one, we can use a variant formula from SUM, which is SUMIFS.
Here is the general writing form of SUMIFS in excel.
= SUMIFS ( number_range , data_range1 , criterion1 , data_range2 , criterion2 , … )
In SUMIFS, we input the cell ranges where our data entry numbers are and data with the criteria that evaluate them. We input the data cell range and the criteria in pairs. This means a criterion will only evaluate the data in the cell range we input before it.
The cell ranges we input in SUMIFS should be parallel and have the same size. Usually, we input them in the form of columns/rows.
To better understand the SUMIFS implementation in excel, here is a screenshot example.
We have fruits and vegetable sales quantity data in a table in the example. On the right, we want to sum the sales quantities based on the product category and class. For this kind of sum process, we can use SUMIFS to help us get our results.
First, we input the sales quantity column cell range for our numbers range input in SUMIFS. As we have product category and class criteria for the sum process, we input the relevant data cell ranges and criteria.
In the example, we input the category column cell range first and its criterion. Then, we input the product class column cell range and its criterion.
Write the SUMIFS correctly and give it correct inputs according to our sum needs. In the example, as a result of this, we get all of the SUMIFS results we need correctly!
SUM with Error Values in its Cell Range Input: SUM IFERROR
Error values in a SUM cell range input are one factor that can cause our SUM to produce an error. We should remove error values from our cell range so we can get the correct sum result from SUM.
But, what if we cannot predict the error values that might show up in our cell range (probably they come from excel formula results which has dynamic inputs)? What if we have a large cell range and it will be troublesome to inspect and remove all errors?
In these kinds of situations, we can combine our SUM with IFERROR. That way, we can use SUM to our cell range without any worry about its error values!
The general writing form of SUM and IFERROR combination for the purpose is like this.
{ = SUM ( IFERROR ( numbers_range , 0 ) ) }
In this combination, we write our IFERROR in our SUM and input our numbers range to the IFERROR.
As a substitute for each error that we have, we input 0 in our IFERROR. However, if you want, you can substitute the error values with any value you prefer. Just change the 0 in your formula writing with the value you want.
Because IFERROR receives an array input (in the form of the number range) and we expect it to produce an array result (it usually receives an individual input and produces an individual data), we use an array formula form. Therefore, we need to press Ctrl + Shift + Enter after we write the formula, not just Enter as usual.
You can see the implementation example of the SUM and IFERROR combination in the screenshot below.
In this example, we want to sum all the monthly sales quantity numbers we have in our table. However, as you can see, there are two error values there (#N/A and #DIV/0). If we want to sum the numbers without removing the errors, then we can combine SUM with IFERROR to do it.
We input the sales quantity column cell range into our IFERROR and we input the IFERROR into our SUM. Here, we substitute all the error values in the sales quantity column with zeroes.
After we finish writing the formula, we press Ctrl + Shift + Enter to convert it into an array formula. As a result, we get the sales quantity numbers total while ignoring several errors in it!
Much Faster SUM Writing to Sum a Number Sequence: AutoSum
Do you know there is a SUM writing shortcut you can use to a number sequence in excel? The feature to access this shortcut is what we call AutoSum. If you often sum numbers in excel, then this feature can make you sum much faster.
To understand how to use this feature, see the example below.
We use the monthly sales quantity data set we have used previously for this AutoSum implementation example. Now, how to use the feature to sum the sales quantity numbers fast?
First, we highlight the sales quantity number sequence or we place our cell cursor just outside those numbers. Then, we go to the Home or the Formulas tab and click the AutoSum button. We can also press the AutoSum shortcut buttons, Alt + = (Command + Shift + T in Mac).
AutoSum button in the Home tab
AutoSum button in the Formulas tab
After you run the AutoSum feature, excel will automatically write a SUM to sum your number sequence!
Now, you don’t have to write SUM yourself when you want to sum a number sequence in excel!
Exercise
After you have learned how to use the SUM formula in excel completely, let’s do an exercise. This is so you can understand the lessons you get from this tutorial more practically.
Download the exercise file from the following link and answer the questions below. Download the answer key file if you have done the exercise and want to check your answers.
Link to the exercise file:
Questions
1. What is the SUM result of all numbers in the first column?
2. What is the SUM result of all numbers in the second column? Avoid black colored cells when you give the number inputs to your SUM
3. What is the SUM result of all numbers in the third column? Avoid black colored cells when you give the number inputs to your SUM
• SUM ignores text and blank cells in its sum process
• You can give up to 255 different inputs to SUM
Related tutorials you should learn:
Get updated excel info from Compute Expert by registering your email. It's free! | 5,701 | 25,215 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2024-26 | latest | en | 0.910537 |
http://www.myelesson.org/excel-training-videos/len-formula-excel/ | 1,576,176,962,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540545146.75/warc/CC-MAIN-20191212181310-20191212205310-00193.warc.gz | 204,748,465 | 10,893 | # Len Formula Excel
This topic has been explained in great detail in the video shown below, to learn this topic just play the video and the notes are mentioned below the video.
# LEN, LENB functions
This article describes the formula syntax and usage of the LEN and LENB functions in Microsoft Excel.
## Description
LEN returns the number of characters in a text string.
LENB returns the number of bytes used to represent the characters in a text string.
IMPORTANT: LENB counts 2 bytes per character only when a DBCS language is set as the default language. Otherwise LENB behaves the same as LEN, counting 1 byte per character.
The languages that support DBCS include Japanese, Chinese (Simplified), Chinese (Traditional), and Korean.
## Syntax
LEN(text)
LENB(text)
The xxx function syntax has the following arguments:
• Text Required. The text whose length you want to find. Spaces count as characters.
## Example
### Example 1: LEN
Copy the example data in the following table, and paste it in cell A1 of a new Excel worksheet. For formulas to show results, select them, press F2, and then press Enter. If you need to, you can adjust the column widths to see all the data.
Data Phoenix, AZ One Formula Description Result =LEN(A2) Length of the first string 11 =LEN(A3) Length of the second string 0 =LEN(A4) Length of the third string, which includes 5 spaces 11
### Example 2: LENB (with your computer set to a default language that supports DBCS)
Copy the example data in the following table, and paste it in cell A1 of a new Excel worksheet. For formulas to show results, select them, press F2, and then press Enter. If you need to, you can adjust the column widths to see all the data.
Data Phoenix, AZ One Formula Description Result =LEN(A2) Length of the first string 11 =LEN(A3) Length of the second string 0 =LEN(A4) Length of the third string, which includes 5 spaces 11
Len Formula helps in determining the count of characters in a given value . To down load the training file and to watch more videos visit www.myelesson.org This Excel tutorial has been created to help you learn microsoft excel online. On http://www.myelesson.org you get excel help and online file download facitity.
This video lesson has been made by me and the text above is from a Microsoft Office Support Blog.
# Our Other Services
myeboook.com
Myebook has 1000's of free ebooks for you. On productivity, health, online income, learning, music and many more, it's free of course!
myelesson.org/product
Become an Excel Guru, Buy the best Excel course. Learn on your mobile, tv & laptop without internet. For COD, Call 9752003788
### Create Best Looking Resume
http://www.makecv.org/
At Makecv.org you get to create a amazingly professional looking resume in just 3 steps with our intuitive resume builder! | 649 | 2,826 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2019-51 | latest | en | 0.82871 |
https://www.hextobinary.com/unit/acceleration/from/dms2/to/galileo | 1,618,228,576,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038067400.24/warc/CC-MAIN-20210412113508-20210412143508-00541.warc.gz | 877,001,633 | 7,033 | #### Decimeter/Second Squared Galileo
##### How many Galileos are in a Decimeter/Second Squared?
The answer is one Decimeter/Second Squared is equal to 10 Galileos. Feel free to use our online unit conversion calculator to convert the unit from Decimeter/Second Squared to Galileo. Just simply, enter value 1 in Decimeter/Second Squared and see the result in Galileo.
##### How to Convert Decimeter/Second Squared to Galileo (dm/s2 to Gal)
By using our Decimeter/Second Squared to Galileo conversion tool, you know that one Decimeter/Second Squared is equivalent to 10 Galileo. Hence, to convert Decimeter/Second Squared to Galileo, we just need to multiply the number by 10. We are going to use very simple Decimeter/Second Squared to Galileo conversion formula for that. Pleas see the calculation example given below.
Convert 1 Decimeter/Second Squared to Galileo 1 Decimeter/Second Squared = 1 × 10 = 10 Galileo
##### What is Decimeter/Second Squared Unit of Measure?
Decimeter/Second Squared or Decimeter per Second Squared is a unit of measurement for acceleration. If an object accelerates at the rate of 1 decimeter/second squared, that means its speed is increased by 1 decimeter per second every second.
##### What is the symbol of Decimeter/Second Squared?
The symbol of Decimeter/Second Squared is dm/s2 which means you can also write it as 1 dm/s2.
##### What is Galileo Unit of Measure?
Galileo is a unit of measurement for acceleration. Galileo is also known as Gal and is extensively used in the science of gravimetry. One galileo is equal to 1 centimeter per second squared.
##### What is the symbol of Galileo?
The symbol of Galileo is Gal which means you can also write it as 1 Gal.
##### Decimeter/Second Squared to Galileo Conversion Table
Decimeter/Second Squared [dm/s2] Galileo [Gal] 1 10 2 20 3 30 4 40 5 50 6 60 7 70 8 80 9 90 10 100 100 1000 1000 10000
##### Decimeter/Second Squared to Other Units Conversion Chart
Decimeter/Second Squared [dm/s2] Output 1 Decimeter/Second Squared in Attometer/Second Squared equals to 100000000000000000 1 Decimeter/Second Squared in Centigal equals to 1000 1 Decimeter/Second Squared in Centimeter/Hour Squared equals to 129600000 1 Decimeter/Second Squared in Centimeter/Minute Squared equals to 36000 1 Decimeter/Second Squared in Centimeter/Second Squared equals to 10 1 Decimeter/Second Squared in Decigal equals to 100 1 Decimeter/Second Squared in Dekameter/Second Squared equals to 0.01 1 Decimeter/Second Squared in Feet/Hour Squared equals to 4251968.5 1 Decimeter/Second Squared in Feet/Hour/Minute equals to 70866.14 1 Decimeter/Second Squared in Feet/Hour/Second equals to 1181.1 1 Decimeter/Second Squared in Feet/Minute Squared equals to 1181.1 1 Decimeter/Second Squared in Feet/Minute/Second equals to 19.69 1 Decimeter/Second Squared in Feet/Second Squared equals to 0.32808398950131 1 Decimeter/Second Squared in Femtometer/Second Squared equals to 100000000000000 1 Decimeter/Second Squared in G-unit equals to 0.010197162129779 1 Decimeter/Second Squared in Gal equals to 10 1 Decimeter/Second Squared in Galileo equals to 10 1 Decimeter/Second Squared in Gn equals to 0.010197162129779 1 Decimeter/Second Squared in Gravity equals to 0.010197162129779 1 Decimeter/Second Squared in Hectometer/Second Squared equals to 0.001 1 Decimeter/Second Squared in Inch/Hour Squared equals to 51023622.05 1 Decimeter/Second Squared in Inch/Hour/Minute equals to 850393.7 1 Decimeter/Second Squared in Inch/Hour/Second equals to 14173.23 1 Decimeter/Second Squared in Inch/Minute Squared equals to 14173.23 1 Decimeter/Second Squared in Inch/Minute/Second equals to 236.22 1 Decimeter/Second Squared in Inch/Second Squared equals to 3.94 1 Decimeter/Second Squared in Kilogal equals to 0.01 1 Decimeter/Second Squared in Kilometer/Hour Squared equals to 1296 1 Decimeter/Second Squared in Kilometer/Hour/Second equals to 0.36 1 Decimeter/Second Squared in Kilometer/Minute Squared equals to 0.36 1 Decimeter/Second Squared in Kilometer/Second Squared equals to 0.0001 1 Decimeter/Second Squared in Knot/Hour equals to 699.78 1 Decimeter/Second Squared in Knot/Millisecond equals to 0.00019438445 1 Decimeter/Second Squared in Knot/Minute equals to 11.66 1 Decimeter/Second Squared in Knot/Second equals to 0.19438445 1 Decimeter/Second Squared in Meter/Hour Squared equals to 1296000 1 Decimeter/Second Squared in Meter/Minute Squared equals to 360 1 Decimeter/Second Squared in Meter/Second Squared equals to 0.1 1 Decimeter/Second Squared in Micrometer/Second Squared equals to 100000 1 Decimeter/Second Squared in Mile/Hour Squared equals to 805.3 1 Decimeter/Second Squared in Mile/Hour/Minute equals to 13.42 1 Decimeter/Second Squared in Mile/Hour/Second equals to 0.22369362920544 1 Decimeter/Second Squared in Mile/Minute Squared equals to 0.22369362920544 1 Decimeter/Second Squared in Mile/Second Squared equals to 0.000062137119223733 1 Decimeter/Second Squared in Milligal equals to 10000 1 Decimeter/Second Squared in Millimeter/Hour Squared equals to 1296000000 1 Decimeter/Second Squared in Millimeter/Minute Squared equals to 360000 1 Decimeter/Second Squared in Millimeter/Second Squared equals to 100 1 Decimeter/Second Squared in Nanometer/Second Squared equals to 100000000 1 Decimeter/Second Squared in Picometer/Second Squared equals to 100000000000 1 Decimeter/Second Squared in Yard/Second Squared equals to 0.10936132983377
##### Other Units to Decimeter/Second Squared Conversion Chart
Output Decimeter/Second Squared [dm/s2] 1 Attometer/Second Squared in Decimeter/Second Squared equals to 1e-17 1 Centigal in Decimeter/Second Squared equals to 0.001 1 Centimeter/Hour Squared in Decimeter/Second Squared equals to 7.7160493827161e-9 1 Centimeter/Minute Squared in Decimeter/Second Squared equals to 0.000027777777777778 1 Centimeter/Second Squared in Decimeter/Second Squared equals to 0.1 1 Decigal in Decimeter/Second Squared equals to 0.01 1 Dekameter/Second Squared in Decimeter/Second Squared equals to 100 1 Feet/Hour Squared in Decimeter/Second Squared equals to 2.3518518518519e-7 1 Feet/Hour/Minute in Decimeter/Second Squared equals to 0.000014111111111111 1 Feet/Hour/Second in Decimeter/Second Squared equals to 0.00084666666666667 1 Feet/Minute Squared in Decimeter/Second Squared equals to 0.00084666666666667 1 Feet/Minute/Second in Decimeter/Second Squared equals to 0.0508 1 Feet/Second Squared in Decimeter/Second Squared equals to 3.05 1 Femtometer/Second Squared in Decimeter/Second Squared equals to 1e-14 1 G-unit in Decimeter/Second Squared equals to 98.07 1 Gal in Decimeter/Second Squared equals to 0.1 1 Galileo in Decimeter/Second Squared equals to 0.1 1 Gn in Decimeter/Second Squared equals to 98.07 1 Gravity in Decimeter/Second Squared equals to 98.07 1 Hectometer/Second Squared in Decimeter/Second Squared equals to 1000 1 Inch/Hour Squared in Decimeter/Second Squared equals to 1.9598765432099e-8 1 Inch/Hour/Minute in Decimeter/Second Squared equals to 0.0000011759259259259 1 Inch/Hour/Second in Decimeter/Second Squared equals to 0.000070555555555556 1 Inch/Minute Squared in Decimeter/Second Squared equals to 0.000070555555555556 1 Inch/Minute/Second in Decimeter/Second Squared equals to 0.0042333333333333 1 Inch/Second Squared in Decimeter/Second Squared equals to 0.254 1 Kilogal in Decimeter/Second Squared equals to 100 1 Kilometer/Hour Squared in Decimeter/Second Squared equals to 0.0007716049382716 1 Kilometer/Hour/Second in Decimeter/Second Squared equals to 2.78 1 Kilometer/Minute Squared in Decimeter/Second Squared equals to 2.78 1 Kilometer/Second Squared in Decimeter/Second Squared equals to 10000 1 Knot/Hour in Decimeter/Second Squared equals to 0.0014290123401217 1 Knot/Millisecond in Decimeter/Second Squared equals to 5144.44 1 Knot/Minute in Decimeter/Second Squared equals to 0.085740740407305 1 Knot/Second in Decimeter/Second Squared equals to 5.14 1 Meter/Hour Squared in Decimeter/Second Squared equals to 7.716049382716e-7 1 Meter/Minute Squared in Decimeter/Second Squared equals to 0.0027777777777778 1 Meter/Second Squared in Decimeter/Second Squared equals to 10 1 Micrometer/Second Squared in Decimeter/Second Squared equals to 0.00001 1 Mile/Hour Squared in Decimeter/Second Squared equals to 0.0012417777777778 1 Mile/Hour/Minute in Decimeter/Second Squared equals to 0.074506666666667 1 Mile/Hour/Second in Decimeter/Second Squared equals to 4.47 1 Mile/Minute Squared in Decimeter/Second Squared equals to 4.47 1 Mile/Second Squared in Decimeter/Second Squared equals to 16093.44 1 Milligal in Decimeter/Second Squared equals to 0.0001 1 Millimeter/Hour Squared in Decimeter/Second Squared equals to 7.716049382716e-10 1 Millimeter/Minute Squared in Decimeter/Second Squared equals to 0.0000027777777777778 1 Millimeter/Second Squared in Decimeter/Second Squared equals to 0.01 1 Nanometer/Second Squared in Decimeter/Second Squared equals to 1e-8 1 Picometer/Second Squared in Decimeter/Second Squared equals to 1e-11 1 Yard/Second Squared in Decimeter/Second Squared equals to 9.14 | 2,710 | 9,079 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2021-17 | latest | en | 0.921597 |
https://socratic.org/questions/can-you-find-f-5-if-f-n-n3-3n2 | 1,638,388,340,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964360881.12/warc/CC-MAIN-20211201173718-20211201203718-00496.warc.gz | 606,902,513 | 5,709 | # Can you find f(-5) if f(n) = n^3 + 3n^2?
## i don't get the whole problem or how to solve it i need a better explanation
$f \left(- 5\right) = {\left(- 5\right)}^{3} + 3 \cdot {\left(- 5\right)}^{2} = - 50$ | 85 | 210 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 1, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2021-49 | latest | en | 0.645691 |
http://u2commerce.com/type-1/type-2-error-in-statistics-probability.html | 1,521,565,875,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647519.62/warc/CC-MAIN-20180320170119-20180320190119-00539.warc.gz | 323,636,587 | 5,450 | Home > Type 1 > Type 2 Error In Statistics Probability
# Type 2 Error In Statistics Probability
## Contents
Stomp On Step 1 31,092 views 15:54 Error Type (Type I & II) - Duration: 9:30. Don't reject H0 I think he is innocent! The threshold for rejecting the null hypothesis is called the α (alpha) level or simply α. The probability of making a type I error is α, which is the level of significance you set for your hypothesis test. http://u2commerce.com/type-1/type-1-error-probability-statistics.html
This is one reason2 why it is important to report p-values when reporting results of hypothesis tests. Sign in Share More Report Need to report the video? Again, H0: no wolf. Add to Want to watch this again later? http://support.minitab.com/en-us/minitab/17/topic-library/basic-statistics-and-graphs/hypothesis-tests/basics/type-i-and-type-ii-error/
## Type 1 Error Example
A typeI error may be compared with a so-called false positive (a result that indicates that a given condition is present when it actually is not present) in tests where a Loading... Choosing a valueα is sometimes called setting a bound on Type I error. 2. Statistics Learning Centre 359,631 views 4:43 Statistics 101: Type I and Type II Errors - Part 2 - Duration: 24:04.
However, there is some suspicion that Drug 2 causes a serious side-effect in some patients, whereas Drug 1 has been used for decades with no reports of the side effect. Sign in to make your opinion count. If you're seeing this message, it means we're having trouble loading external resources for Khan Academy. Type 1 Error Calculator Loading...
This sort of error is called a type II error, and is also referred to as an error of the second kind.Type II errors are equivalent to false negatives. CRC Press. In the long run, one out of every twenty hypothesis tests that we perform at this level will result in a type I error.Type II ErrorThe other kind of error that Transcript The interactive transcript could not be loaded.
Our Story Advertise With Us Site Map Help Write for About Careers at About Terms of Use & Policies © 2016 About, Inc. — All rights reserved. Type 1 Error Psychology Security screening Main articles: explosive detection and metal detector False positives are routinely found every day in airport security screening, which are ultimately visual inspection systems. statisticsfun 69,435 views 7:01 Statistics: Type I & Type II Errors Simplified - Duration: 2:21. A typeII error occurs when failing to detect an effect (adding fluoride to toothpaste protects against cavities) that is present.
## Probability Of Type 1 Error
By using this site, you agree to the Terms of Use and Privacy Policy. http://statistics.about.com/od/Inferential-Statistics/a/Type-I-And-Type-II-Errors.htm There's some threshold that if we get a value any more extreme than that value, there's less than a 1% chance of that happening. Type 1 Error Example Alpha is the maximum probability that we have a type I error. Probability Of Type 2 Error Therefore, a researcher should not make the mistake of incorrectly concluding that the null hypothesis is true when a statistical test was not significant.
To lower this risk, you must use a lower value for α. check my blog jbstatistics 450,631 views 5:44 Statistics 101: Calculating Type II Error - Part 1 - Duration: 23:39. While most anti-spam tactics can block or filter a high percentage of unwanted emails, doing so without creating significant false-positive results is a much more demanding task. You can decrease your risk of committing a type II error by ensuring your test has enough power. Type 3 Error
• Language: English (UK) Content location: United Kingdom Restricted Mode: Off History Help Loading...
• jbstatistics 56,904 views 13:40 Super Easy Tutorial on the Probability of a Type 2 Error! - Statistics Help - Duration: 15:29.
• The probability of making a type I error is α, which is the level of significance you set for your hypothesis test.
• Perhaps the most widely discussed false positives in medical screening come from the breast cancer screening procedure mammography.
• In statistical hypothesis testing, a type I error is the incorrect rejection of a true null hypothesis (a "false positive"), while a type II error is incorrectly retaining a false null
• This error is potentially life-threatening if the less-effective medication is sold to the public instead of the more effective one.
• Although they display a high rate of false positives, the screening tests are considered valuable because they greatly increase the likelihood of detecting these disorders at a far earlier stage.[Note 1]
• A typeII error (or error of the second kind) is the failure to reject a false null hypothesis.
• The Type I error rate is affected by the α level: the lower the α level, the lower the Type I error rate.
Minitab.comLicense PortalStoreBlogContact UsCopyright © 2016 Minitab Inc. That is, the researcher concludes that the medications are the same when, in fact, they are different. Sign in to add this to Watch Later Add to Loading playlists... http://u2commerce.com/type-1/type-1-error-in-probability.html A common example is relying on cardiac stress tests to detect coronary atherosclerosis, even though cardiac stress tests are known to only detect limitations of coronary artery blood flow due to
Screening involves relatively cheap tests that are given to large populations, none of whom manifest any clinical indication of disease (e.g., Pap smears). Power Statistics Example 2: Two drugs are known to be equally effective for a certain condition. Privacy policy About Wikipedia Disclaimers Contact Wikipedia Developers Cookie statement Mobile view COMMON MISTEAKS MISTAKES IN USING STATISTICS:Spotting and Avoiding Them Introduction Types of Mistakes Suggestions Resources
## In the same paper[11]p.190 they call these two sources of error, errors of typeI and errors of typeII respectively.
More generally, a Type I error occurs when a significance test results in the rejection of a true null hypothesis. The more experiments that give the same result, the stronger the evidence. Handbook of Parametric and Nonparametric Statistical Procedures. Misclassification Bias TypeI error False positive Convicted!
This value is the power of the test. A type II error would occur if we accepted that the drug had no effect on a disease, but in reality it did.The probability of a type II error is given The probability of rejecting the null hypothesis when it is false is equal to 1–β. http://u2commerce.com/type-1/type-1-error-probability.html statslectures 162,124 views 4:25 Stats: Hypothesis Testing (Traditional Method) - Duration: 11:32.
Loading... How to Conduct a Hypothesis Test More from the Web Powered By ZergNet Sign Up for Our Free Newsletters Thanks, You're in! Another convention, although slightly less common, is to reject the null hypothesis if the probability value is below 0.01. Advertisement Autoplay When autoplay is enabled, a suggested video will automatically play next.
And all this error means is that you've rejected-- this is the error of rejecting-- let me do this in a different color-- rejecting the null hypothesis even though it is | 1,569 | 7,208 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2018-13 | latest | en | 0.889321 |
https://www.cnblogs.com/kuangbin/archive/2012/09/23/2699178.html | 1,680,109,809,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949009.11/warc/CC-MAIN-20230329151629-20230329181629-00027.warc.gz | 789,173,028 | 9,708 | # Super Mario
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 120 Accepted Submission(s): 73
Problem Description
Mario is world-famous plumber. His “burly” figure and amazing jumping ability reminded in our memory. Now the poor princess is in trouble again and Mario needs to save his lover. We regard the road to the boss’s castle as a line (the length is n), on every integer point i there is a brick on height hi. Now the question is how many bricks in [L, R] Mario can hit if the maximal height he can jump is H.
Input
The first line follows an integer T, the number of test data.
For each test data:
The first line contains two integers n, m (1 <= n <=10^5, 1 <= m <= 10^5), n is the length of the road, m is the number of queries.
Next line contains n integers, the height of each brick, the range is [0, 1000000000].
Next m lines, each line contains three integers L, R,H.( 0 <= L <= R < n 0 <= H <= 1000000000.)
Output
For each case, output "Case X: " (X is the case number starting from 1) followed by m lines, each line contains an integer. The ith integer is the number of bricks Mario can hit for the ith query.
Sample Input
1 10 10 0 5 2 7 5 4 3 8 7 7 2 8 6 3 5 0 1 3 1 1 9 4 0 1 0 3 5 5 5 5 1 4 6 3 1 5 7 5 7 3
Sample Output
Case 1: 4 0 0 3 1 2 0 1 5 1
Source
Recommend
liuyiding
Orz,原来这么多人快速过了都是用划分树的。。。
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
const int MAXN=100010;
int tree[30][MAXN];//表示每层每个位置的值
int sorted[MAXN];//已经排序的数
int toleft[30][MAXN];//toleft[p][i]表示第i层从1到i有多少个数分入左边
void build(int l,int r,int dep)
{
if(l==r)return;
int mid=(l+r)>>1;
int same=mid-l+1;//表示等于中间值而且被分入左边的个数
for(int i=l;i<=r;i++)
if(tree[dep][i]<sorted[mid])
same--;
int lpos=l;
int rpos=mid+1;
for(int i=l;i<=r;i++)
{
if(tree[dep][i]<sorted[mid])//比中间的数小，分入左边
tree[dep+1][lpos++]=tree[dep][i];
else if(tree[dep][i]==sorted[mid]&&same>0)
{
tree[dep+1][lpos++]=tree[dep][i];
same--;
}
else //比中间值大分入右边
tree[dep+1][rpos++]=tree[dep][i];
toleft[dep][i]=toleft[dep][l-1]+lpos-l;//从1到i放左边的个数
}
build(l,mid,dep+1);
build(mid+1,r,dep+1);
}
//查询区间第k大的数，[L,R]是大区间，[l,r]是要查询的小区间
int query(int L,int R,int l,int r,int dep,int k)
{
if(l==r)return tree[dep][l];
int mid=(L+R)>>1;
int cnt=toleft[dep][r]-toleft[dep][l-1];//[l,r]中位于左边的个数
if(cnt>=k)
{
//L+要查询的区间前被放在左边的个数
int newl=L+toleft[dep][l-1]-toleft[dep][L-1];
//左端点加上查询区间会被放在左边的个数
int newr=newl+cnt-1;
return query(L,mid,newl,newr,dep+1,k);
}
else
{
int newr=r+toleft[dep][R]-toleft[dep][r];
int newl=newr-(r-l-cnt);
return query(mid+1,R,newl,newr,dep+1,k-cnt);
}
}
int solve(int n,int s,int t,int h)
{
int ans=0;
int l=1;
int r=(t-s)+1;
int mid;
while(l<=r)
{
mid=(l+r)>>1;
int temp=query(1,n,s,t,0,mid);
if(temp<=h)
{
ans=mid;
l=mid+1;
}
else r=mid-1;
}
return ans;
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int T;
int n,m;
int s,t,h;
scanf("%d",&T);
int iCase=0;
while(T--)
{
iCase++;
scanf("%d%d",&n,&m);
memset(tree,0,sizeof(tree));//这个必须
for(int i=1;i<=n;i++)//从1开始
{
scanf("%d",&tree[0][i]);
sorted[i]=tree[0][i];
}
sort(sorted+1,sorted+n+1);
build(1,n,0);
printf("Case %d:\n",iCase);
while(m--)
{
scanf("%d%d%d",&s,&t,&h);
s++;
t++;
printf("%d\n",solve(n,s,t,h));
}
}
return 0;
}
posted on 2012-09-23 20:44 kuangbin 阅读(1419) 评论(3编辑 收藏 举报
JAVASCRIPT: | 1,340 | 3,368 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 3.03125 | 3 | CC-MAIN-2023-14 | latest | en | 0.589866 |
http://takinginitiative.wordpress.com/category/artificial-intelligence/ | 1,386,541,186,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163824647/warc/CC-MAIN-20131204133024-00002-ip-10-33-133-15.ec2.internal.warc.gz | 183,185,701 | 13,880 | ## Pathfinding Thesis Complete
WOOT! My thesis, the bane of my existence for the last 2 years, is finally done! Its basically a review of the video game pathfinding field as well as presenting a novel grid map search technique: the spatial grid A*. The version linked below is the final draft that is being submitted to my faculty.
## A Robust Explosion Hit Check Technique
In a recent technical interview I got asked the following question: “if an explosion occurs i.e. from a grenade, how would you determine which characters in the game are affected”. This was a question that I couldnt answer at the time which annoyed the hell out of me and it’s been sitting in the back of my head for the last few weeks. So I decided to discuss it (and my proposed solution) on my blog as an “academic” exercise. I would really appreciate any feedback or comments on this discussion. This is far from being a solved problem for me and while my solution may potentially work quite well, there may be a simpler technique available which I don’t know about.
Coming back to the question. Well the naive answer to that question is to simply do a collision detection check with the explosion area of effect/blast radius (sphere) and any nearby characters’ bounding boxes. Read more of this post
## Optimizing the A* algorithm
So I’ve recently completed my MSc thesis on video game pathfinding and I guess it’s a little weird for someone who spent the last year focusing on game AI and pathfinding to not actually spend much time blogging about it. I figured that I spent the time today and write a short post on optimizing the A* algorithm. The A* algorithm pretty much sums up video game pathfinding as a whole. Even advanced techniques like hierarchical pathfinding algorithm make use of A* in searching the various abstraction levels. So today I’m going to just discuss optimizing the algorithm, not a low level implementation but rather the some of the high level issues. I’m assuming that readers will have some degree of familiarity with the A* algorithm so I’m not going to waste time explaining it.
A*’s computational cost is primarily divided between two components, the heuristic function and the open list and so these are the components that I’m going to focus on. Read more of this post
## Final Simulation Results
Final Simulation Results for BGMAPS benchmark
Final Simulation Results for Custom RTS Maps
I recently posted about some preliminary results for my prototype algorithm, I now have the final results for both BGMAPs and my own set of RTS style game maps. Both sets of maps were sized at 512×512. The RTS maps are simple island style maps which I made more complex that the average RTS map found today just to increase the difficulty of the problem. In increasing the complexity I’ve also reduced the length of straight line paths available at which my prototype excels in that way I’ve handicapped it a little. Even so on RTS style maps my silly algorithm kills A*, furthermore there the memory costs are over 10x lower than required by A*. Unfortunately my RTS maps are artificial and so I cannot state that my approach will work beyond a doubt in real world RTS maps (even though I’m pretty sure it would ).
The initial prototypes I presented had some flaws and were not complete algorithms in the sense that they wouldn’t always return a solution if one existed. My new versions are complete in that sense. There was a lot of tweaking, prodding and changing necessary to achieve this and the search times have gone up marginally.
The performance of the prototypes isnt bad on the BG maps test set but unfortunately the graph doesnt show the full picture, there are certain rare cases in which the solution returned by my approach is GROSSLY sub-optimal (over 100% suboptimal in fact) and the time take to find this “amazing” solution is nearly double the time needed for A*. This poor behavior is due to the spatial layout of the BG maps, which focus on numerous dead end winding tunnels.
I would like to investigate the efficiency of the algorithm within real world RTS maps as well as on FPS maps so if any game devs reading this post can send me some top down JPGs of their game levels I would really appreciate it. I now just need to finish writing up my last chapter of my thesis and submit it. I will be posting the full thesis on this blog which will include all the necessary info on my prototypes which I’m naming the Spatial A* algorithm.
PS! it is necessary to mention that these results are pre-smoothing and so the path optimality will be greatly improved with a post-processing smoothing step.
## Preliminary Pathfinding Results for my Prototype Algorithm
The preliminary results of my prototype pathfinding algorithm - run on a basic set of 8 BG maps.
It’s been a while since I’ve posted about anything pathfinding related. I’m currently wrapping up my thesis and once I finish work on this prototype algorithm all I need to do is write up my results and my thesis is complete.
My prototype algorithm is based on a pretty stupid premise so I wont go into much detail here ;) sufficed to say its basically an abstract search without an abstraction layer. I have three versions of the basic algorithm:
• P1 – this is the naive version (the same one I showed at GDC)
• P2 – performs basic trimming on the refinement step as discussed at the AI dinner (thanks Alex, Marc and Joel!!)
• P3 – attempts to improve path optimality by picking a new end goal for the search and using that to reach the final goal.
These are the results of a very basic run on 8 Baldur’s Gate 2 maps scaled up to 512 x 512 (around 10000 search problems). My algorithm makes use of the exact same A* implementation it is compared against. There is still one problem with the P2/P3 variants that needs to be corrected and a P4 variant is also in the works. I hope to have all 4 variants complete by the weekend and all my simulations completed by next week monday. That gives me a week to do the final write up and submit my thesis!
The path optimality listed is pre-smoothing, with smoothing, the optimality greatly increases but so does the total time for the “pathfinding action”. I still have to find an efficient (read very very FAST) path smoothing algorithm before I can actually bother doing any smoothing experiments.
## The NMD Engine – A work in progress
The original NMD toolkit
I’ve been rather quiet of late, I made some rather large changes in the direction of my master’s thesis and have spent a lot of time on replanning and restructuring my work. I initially intended to do my thesis on real-time pathfinding in dynamic environments, this is still the case but I realised during my research that there was a large gap present in the academic literature regarding the actual implementation and execution of a pathfinding system within a game engine. So while the topic of my thesis remained the same, the overall structure and contributions did not. I have struggled with my thesis for the past two years mainly due to a lack of a focussed goal, I had initially intended on doing a more generl overview of the various pathfinding algorithms and techniques present in the academic literature today and perform a comprehensive comparison of them in regards to memory usage, solution optimality and search speed.
## Some Graph Performance Observations
I’m doing my masters thesis on game pathfinding and this pretty much boils down to implementing comparing graph search algorithms (and hopefully developing a new algorithm). Now to implement and test the various algorithm I need to maintain a level of consistency in the implementations so my first step was to start planning and development on a pathfinding test framework that would provides a common search interface for the various algorithms and will be able run automated test routines.
So I started with the idea that since all the algorithms are graph search algorithms I need a nice generic graph container than can store the various types of nodes that each algorithm required. This meant creating a nice flexible graph design with a whole node inheritance hierarchy and node factories to create the various node types and on and on and on… From a software engineering viewpoint, the design was very flexible and extensible, from a performance and implementation standpoint it was a mess. For each new graph type it was necessary to create a new node child class and a new node factory child class, all just to add in a few data members to a node.
First attempt at a generic graph implementation
## Pathfinding Work
The first version of my pathfinding visualization
I’m currently doing my masters degree in video game pathfinding, and progress has unfortunately been rather slow, I spent most of last year planning my pathfinding framework and doing my literature survey, so far I’ve written a basic map generator, a very rough framework with a working (but not optimized A*) and a very rough visualization app so I can see the results of the pathfinding algorithms. The visualization app still needs a lot of work so that I can use it to quickly swap between algorithms (A*, IDA*, SMA*, HPA*, etc) and so that I can swap out the heuristics that the algorithms use. Read more of this post
## C++ Back Propagation Neural Network Code v2
There was a lot of feedback on my neural network implementation, mostly regarding architectural suggestion so i sat down and rewrote my neural network implementation, its pretty elegant now. I seperated the training of the network from the actual network itself and so i have a basic feed forward network, a data set reader and a neural network trainer. I also renamed several data structures to make things more understandable, also i wasnt lazy and used proper header files and includes
Below is an updated class diagram of the new version:
Here’s the updated implementation (with a VS2k8 solution):
The original tutorials can be found here:
## Basic Neural Network Tutorial : C++ Implementation and Source Code
So I’ve now finished the first version of my second neural network tutorial covering the implementation and training of a neural network. I noticed mistakes and better ways of phrasing things in the first tutorial (thanks for the comments guys) and rewrote large sections. This will probably occur with this tutorial in the coming week so please bear with me. I’m pretty overloaded with work and assignments so I haven’t been able to dedicate as much time as I would have liked to this tutorial, even so I feel its rather complete and any gaps will be filled in by my source code. | 2,218 | 10,647 | {"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} | 2.921875 | 3 | CC-MAIN-2013-48 | latest | en | 0.927843 |
http://mathhelpforum.com/geometry/20115-parallelogram-question.html | 1,527,032,897,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864999.62/warc/CC-MAIN-20180522225116-20180523005116-00231.warc.gz | 177,664,898 | 9,895 | 1. ## Parallelogram Question
I feel really dumb asking this..but whatever
one length of the pgram is 7 the other is 9..the long diagonal is 14..whats the short diagonal
2. Hello, stones44!
We will need Trignometry for this one . . . the Law of Cosines.
One side of the parallelogram is 7, the other is 9.
The long diagonal is 14.
What is the short diagonal?
Code:
A 9 B
* - - - - - - - - *
/ * /
/ * /
7/ * / 7
/ * 14 /
/ * /
* - - - - - - - - *
D 9 C
We have parallelogram $\displaystyle ABCD$ with: $\displaystyle AB = DC = 9,\;AD = BC = 7,\;BD = 14$
. . and we want the length of diagonal $\displaystyle AC.$
First, find $\displaystyle \angle A.$
. . In $\displaystyle \Delta ABD\!:\;\;\cos A \;=\;\frac{7^2+9^2-14^2}{2(7)(9)}\;=\;-\frac{11}{21} \quad\Rightarrow\quad A \:=\:121.5881355^o$
Then $\displaystyle \angle D \:=\:\angle ADC \:=\:180^o - 121.5881355^o \:=\:58.41186449^o$
In $\displaystyle \Delta ADC\!:\;\;AC^2\;=\;7^2+9^2-2(7)(9)\cos58.41186449^o \;=\;63.99999999$
Therefore: .$\displaystyle AC \:=\:\sqrt{63.99999999} \:\approx\:8$
3. Suppose that vectors $\displaystyle a,\,b$ are two adjacent sides of the parallelogram.
Then $\displaystyle a + \,b\quad \& \quad a - b$ would represent the two diagonals.
We know that $\displaystyle \left\| {a + b} \right\|^2 = \left\| a \right\|^2 + 2a \cdot b + \left\| b \right\|^2 \quad \& \quad \left\| {a - b} \right\|^2 = \left\| a \right\|^2 - 2a \cdot b + \left\| b \right\|^2$
So $\displaystyle \left\| {a + b} \right\|^2 + \left\| {a - b} \right\|^2 = 2\left\| a \right\|^2 + 2\left\| b \right\|^2.$
Use that to find the length of the second diagonal. | 639 | 1,736 | {"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} | 4.28125 | 4 | CC-MAIN-2018-22 | latest | en | 0.607875 |
https://www.numbersaplenty.com/135051 | 1,680,327,209,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949701.0/warc/CC-MAIN-20230401032604-20230401062604-00461.warc.gz | 1,026,306,532 | 3,157 | Search a number
135051 = 3759109
BaseRepresentation
bin100000111110001011
320212020220
4200332023
513310201
62521123
71101510
oct407613
9225226
10135051
1192514
12661a3
1349617
1437307
152a036
hex20f8b
135051 has 16 divisors (see below), whose sum is σ = 211200. Its totient is φ = 75168.
The previous prime is 135049. The next prime is 135059. The reversal of 135051 is 150531.
135051 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is not a de Polignac number, because 135051 - 21 = 135049 is a prime.
It is a self number, because there is not a number n which added to its sum of digits gives 135051.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (135059) by changing a digit.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 1185 + ... + 1293.
It is an arithmetic number, because the mean of its divisors is an integer number (13200).
2135051 is an apocalyptic number.
135051 is a deficient number, since it is larger than the sum of its proper divisors (76149).
135051 is a wasteful number, since it uses less digits than its factorization.
135051 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 178.
The product of its (nonzero) digits is 75, while the sum is 15.
The square root of 135051 is about 367.4928570734. The cubic root of 135051 is about 51.3057374961.
Adding to 135051 its reverse (150531), we get a palindrome (285582).
The spelling of 135051 in words is "one hundred thirty-five thousand, fifty-one". | 478 | 1,668 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.871033 |
https://byjus.com/question-answer/a-parabola-passing-through-the-point-4-2-has-its-vertex-at-the-origin-and-2/ | 1,723,549,506,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641076695.81/warc/CC-MAIN-20240813110333-20240813140333-00850.warc.gz | 116,964,209 | 24,455 | 1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question
# A parabola passing through the point (-4, -2) has its vertex at the origin and y-axis. The latus rectum of the parabola is
A
6
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
8
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
10
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
12
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
## The correct option is B 8Let the equation of parabola is x2=4ay,but a=4−2=−2Then equation is x2=−8y and latus rectum = 4a = 8
Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Methods of Solving First Order, First Degree Differential Equations
MATHEMATICS
Watch in App
Explore more
Join BYJU'S Learning Program | 262 | 840 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.855471 |
https://www.gooddata.com/docs/gooddata-ui/latest/learn/get_raw_data/arithmetic_measure/ | 1,701,731,100,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100535.26/warc/CC-MAIN-20231204214708-20231205004708-00217.warc.gz | 897,583,092 | 16,813 | # Arithmetic Measure
Arithmetic measures allow you to perform simple calculations with measures in a visualization. A calculated arithmetic measure always references two measures, therefore the visualization must contain two measures at least.
## Supported operations
Although a visualization can contain multiple measures, you can perform arithmetic operations with exactly only two measures.
The following arithmetic operations are supported:
OperationArithmetic measure operatorExpression formulaExample
Sum`sum`=A+B= Q1 revenue + Q2 revenue
Difference`difference`=A-B= revenue in 2017 - revenue in 2016
Product (Multiplication)`multiplication`=A*B= price per unit * number of units
Ratio`ratio`=A÷B= gross profit / net sales
Change`change`=(A-B)÷B= (this month revenue - last month revenue) / last month revenue
By default, the result data of a `change` operation is returned as a percentage in the `#,##0.00%` format. The format cannot be overridden.
All the other operations return data in the default `#,##0.00` format. To change the format, use the `format` attribute of the measure (see the examples).
## Arithmetic measure structure
To add an arithmetic measure to a visualization, use the `newArithmeticMeasure` factory function:
``````newArithmeticMeasure(operands, operator, modifications)
``````
An arithmetic measure can reference the following as its operand:
You can specify operands either by their `localIdentifier` or by their value, and the factory function will extract the local identifier for you.
The operator can be one of the following:
• `sum`
• `difference`
• `multiplication`
• `ratio`
• `change`
The modifications part is optional and is a function with a single parameter, which is an object with functions that you can use to override the measure’s `format()` or `alias()`.
If arithmetic measures reference each other in an infinite loop or the referenced measure is not found in the visualization (there is no measure with the referenced localIdentifier), the error message is rendered instead of the visualization.
## Examples
### A difference between two measures - arithmetic measure constructed using localIdentifier references
``````import { newMeasure, newArithmeticMeasure } from "@gooddata/sdk-model";
import { PivotTable } from "@gooddata/sdk-ui-pivot";
const measures = [
// the first simple measure (operand)
newMeasure("boughtProductsIdentifier", m => m.alias("Bought products from supplier")),
newMeasure("soldProductsLocalIdentifier", m => m.alias("Sold products to customers")),
newArithmeticMeasure(
["boughtProductsLocalIdentifier", "soldProductsLocalIdentifier"],
"difference",
m => m.alias("Products remaining in warehouse")
)
];
<PivotTable
measures={measures}
/>
``````
### Calculation with a derived measure (percentage change between two years)
The result of a `change` operation is returned as a percentage value in the default `#,##0.00%` format. This example demonstrates passing measures by value to the different measure factory functions.
``````import { newMeasure, newPopMeasure, newArithmeticMeasure } from "@gooddata/sdk-model";
import { PivotTable } from "@gooddata/sdk-ui-pivot";
const currentYear = newMeasure("measureIdentifier", m => m.alias("Current Year"));
// derived - data from previous year
const previousYear = newPopMeasure(currentYear, "attributeDisplayFormYearIdentifier", m => m.alias("Previous Year"));
// arithmetic measure with custom format
const change = newArithmeticMeasure(previousYear, currentYear, "change", m => m.alias("Change between years").format("\$#,#0.0%"));
const measures = [
currentYear,
previousYear,
change
];
<PivotTable
measures={measures}
/>
`````` | 793 | 3,683 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2023-50 | longest | en | 0.83582 |
https://brilliant.org/problems/minimum-totient-quotient/ | 1,500,893,934,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424846.81/warc/CC-MAIN-20170724102308-20170724122308-00522.warc.gz | 611,295,245 | 18,511 | # Minimum Totient Quotient
Let $$\phi(n)$$ be the Euler phi function. If $$1 \leq n \leq 1000$$, what is the smallest integer value of $$n$$ that minimizes $$\frac{\phi(n)}{n}$$?
You may choose to read Euler's theorem.
Details and Assumptions:
• You are asked to find the value of $$n$$, not $$\frac {\phi(n) } {n}$$.
• $$\phi(1) = 1$$ by definition.
× | 119 | 357 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2017-30 | longest | en | 0.772258 |
https://testbook.com/question-answer/for-a-waveform-to-be-expressed-in-fourier-series--5f537df288dce851c1c14b03 | 1,632,850,694,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060877.21/warc/CC-MAIN-20210928153533-20210928183533-00036.warc.gz | 594,451,532 | 31,011 | # For a waveform to be expressed in Fourier series, which of the following conditions must be satisfied?I. Maxwell’s conditionsII. Drichlet conditionsIII. Sampling Theorem
This question was previously asked in
RPSC Lecturer Previous Paper (Held On: 2014 )
View all RPSC Lecturer Tech Edu Papers >
1. I only
2. II only
3. II and III only
4. I, II and III
Option 2 : II only
Free
ST 1: Logical reasoning
5232
20 Questions 20 Marks 20 Mins
## Detailed Solution
Concept:
The Fourier theorem is applied to periodic functions that have discontinuities and cannot be represented by a single analytical expression.
For a periodic function f(x), provided that the following conditions i.e. Dirichlet conditions:
• f(x) is defined and single-valued except at a finite number in (-T, T)
• f(x) is periodic outside (-T, T) with period 2T
• f(x) and f’(x) are piecewise continuous in (-T, T), then f(x) can be expressed by the following series:
$$f\left( x \right) = \frac{1}{2}{a_0} + \mathop \sum \limits_{n = 1}^\infty \left( {{a_n}\cos \frac{{n\pi x}}{T} + {b_n}\sin \frac{{n\pi x}}{T}} \right)$$
Where,
$${a_n} = \frac{1}{T}\mathop \smallint \limits_{ - T}^T f\left( x \right)\cos \frac{{n\pi x}}{T}dx$$
$${b_n} = \frac{1}{T}\mathop \smallint \limits_{ - T}^T f\left( x \right)\sin \frac{{n\pi x}}{T}dx$$ | 422 | 1,307 | {"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} | 3.515625 | 4 | CC-MAIN-2021-39 | latest | en | 0.806506 |
http://mathhelpforum.com/trigonometry/100094-deductive-geometry-tangents.html | 1,498,156,890,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128319688.9/warc/CC-MAIN-20170622181155-20170622201155-00336.warc.gz | 244,918,508 | 10,785 | 1. ## Deductive geometry, tangents
Hi, I have a problem which I am having trouble solving.
Let $ABC$ be a right-angled triangle. A circle $T$ having side $AC$ as its diameter meets hypotenuse $AB$ at point $E$. A tangent line to $T$ at point $E$ meets side $BC$ at point $D$. Prove that triangle $BDE$ is isosceles.
2. Let T be the center of the circle.
Quadrilater CTED is cyclic. Then $\widehat{TCE}=\widehat{TDE}$ (1)
$CE\perp AB\Rightarrow\widehat{TCE}=\widehat{DBE}$ (2)
$\Delta TCD\equiv\Delta TED\Rightarrow\widehat{TDE}=\widehat{TDC}$ (3)
From (1), (2), (3) $\widehat{TDC}=\widehat{DBE}\Rightarrow TD\parallel AB\Rightarrow\widehat{TDE}=\widehat{DEB}\Rightarro w\widehat{DEB}=\widehat{DBE}$
3. I would show the $\angle DEB \cong \angle BED$.
This can be done by noting that both measure $\frac{m(\widehat{AE})}{2}$.
4. Why is $CE\perp AB\$?
Because, from there, I didn't understand.
Thanks, BG
5. Originally Posted by BG5965
$m\left( {\angle EBD} \right) = \frac{{m\left( {\widehat{AC}} \right) - m\left( {\widehat{EC}} \right)}}{2}$
$m\left( {\angle DEB} \right) = \frac{{m\left( {\widehat{EA}} \right)}}{2}\;\& \,m\left( {\widehat{EA}} \right) = m\left( {\widehat{AC}} \right) - m\left( {\widehat{EC}} \right)$
But that means that $m\left( {\angle DEB} \right) = m\left( {\angle EBD} \right)\, \Rightarrow \,\angle DEB \cong \angle EBD$ | 489 | 1,358 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 20, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2017-26 | longest | en | 0.68193 |
http://mathhelpforum.com/math-topics/41527-representation-venn-diagram.html | 1,529,936,154,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867885.75/warc/CC-MAIN-20180625131117-20180625151117-00431.warc.gz | 210,032,093 | 9,925 | # Thread: Representation in Venn diagram
1. ## Representation in Venn diagram
This is a problem given to us in our 3rd Year High school class.
A group of students was asked to choose the 3 bands namely: MCR, SC and H. 21 prefers MCR, 23 prefers SC, 10 prefers MCR or H but not SC, 14 prefers MCR or SC but not H, 12 prefers SC but not MCR or H but not MCR, 3 prefers H but not MCR or SC, 8 likes SC but not MCR or H. 17 likes at least 2 of the bands and 27 prefers 1 at most.
Can someone represent this in Venn Diagram?
Also, my teacher told me that there is another way to construct this diagram which is way faster than using the cardinality of the given sets. Does anyone know how?
2. Hello, azuresonata!
This is a tricky Venn diagram problem . . .
A group of students was asked to among the 3 bands: M, S, and H.
21 prefer M
23 prefer S
10 prefer M or H but not S
14 prefer M or S but not H
12 prefers S but not M or H but not M
3 prefers H but not M or S
8 prefers S but not M or H
17 prefers at least 2 of the bands
27 prefers 1 at most.
My teacher told me that there is another way to construct this diagram
which is way faster than using the cardinality of the given sets.
Does anyone know how?
I think algebra is the way to go.
Code:
* - - - - - - - - - - - *
| M |
| |
| s * - - - - - - - + - - - *
| | | |
| | | |
* - - - + - - - + - - - * | |
| | | | | |
| | t | u | v | w |
| * - - - + - - - | - - - * |
| | | |
| | x | S |
| y * - - - + - - - - - - - *
| |
| H |
* - - - - - - - - - - - *
We have the three "circles" for $\displaystyle M, S,\text{ and }H.$
Label the seven regions with $\displaystyle s,t,u,v,w,x,y.$
Then translate the given facts to equations.
$\displaystyle \begin{array}{ccc}n(M) = 21 & s + t + u + v \:=\:21 \\ n(S) = 23 & u + v + w + x \:=\:23 \\ n([M \cup H]\,\cap S') = 10 & s + t + y \:=\:10 \\ n([M \cup S] \cap H') = 14 & s + v + w \:=\:14 \\ n(S \cup H) \cap M') = 13 & w + x + y \:=\:13 \\ n(H \cap M' \cap S') = 3 & y \:=\:3 \\ n(S \cap M' \cap H') = 8 & w \:=\:8 \\ n(\text{at least 2}) = 17 & t + u + v + x \:=\:17 \\ n(\text{at most 1}) = 27 & s + w + y \:=\:27 \end{array}$
And solve the system.
(We already have two of the values.)
. | 794 | 2,500 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2018-26 | latest | en | 0.917179 |
http://www.fixya.com/support/t1075306-when_type_in_0_02_power_2_displays_e | 1,531,989,474,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590711.36/warc/CC-MAIN-20180719070814-20180719090814-00080.warc.gz | 468,980,446 | 33,975 | Question about Texas Instruments TI-84 Plus Calculator
# When I type in (0.02) to the power of 2, it displays E. I want decimals
When I type in (0.02) squared, on TI-84 Plus, I get an answer 4E to the power of -4. On my TI36X, I get an answer of 0.0004. Why can't I get my TI-84 to display the same wasy as TI36? I checked all the Mode settings and they seem to be fine. Please help
Posted by on
• Level 3:
An expert who has achieved level 3 by getting 1000 points
Superstar:
An expert that got 20 achievements.
All-Star:
An expert that got 10 achievements.
MVP:
An expert that got 5 achievements.
• Texas Instru... Master
Hello,
If number format in MODE is set to Normal, and if the (absolute value of) the number is smaller than 0.001,the calculator displays it in scientific notation( with the E). By the way E( ) stands for x10^( ).
Posted on Sep 06, 2009
## 1 Suggested Answer
Hi there,
Save hours of searching online or wasting money on unnecessary repairs by talking to a 6YA Expert who can help you resolve this 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.
Here's a link to this great service
Good luck!
Posted on Jan 02, 2017
×
my-video-file.mp4
Complete. Click "Add" to insert your video.
×
## Related Questions:
### If i Type Square root of 148, i get some wierd decimal, i dont want that. I want a radical like 3 radical 12 i think thats the answer, how do i get it to display that insted everytime for when i type in...
The square root of 148 is 2*sqrt(37), the square of 3*sqrt(12) is 108, but that's not the point.
The TI-84 can't produce results like 2*sqrt(37). It can produce results as fractions but not involving roots. Some other machines, like the TI-89 can, but not the 84.
Sorry if that wasn't the answer you were hoping for.
Nov 07, 2013 | Texas Instruments TI-84 Plus Silver...
### I'm using a Ti-84 plus and when I try simple division it gives incorrect answers. For example, when I enter 5/2 it answers 3. When I entered 50/100 it answers 1.
You apparently have your calculator set to display no digits to the right of the decimal point. 5/2 is 2.5, which rounds to 3, and 50/100 is 0.5, which rounds to 1.
To change the display setting, press MODE, highlight the desired settings on the first two lines, then press ENTER followed by 2ND QUIT. Try setting "NORMAL" and "FLOAT" and see what results you get now.
Oct 06, 2013 | Texas Instruments TI-84 Plus Calculator
### Ti-84plus keypad for ti-nspire w/touchpad while in a+bi mode i enter the the problem (2-5i) / (1-6i). once i press "ENTER" i get a decimal. i want the answer in frac form, but when i press...
Here is screen capture of the calculation, performed on TINspire with TI 84 plus keyboard.To type in the imaginary unit use key sequence [2nd][dot], do not use [ALPHA][X^2] for the letter I.
,
Jun 09, 2011 | Texas Instruments TI-84 Plus Calculator
### How do you do you put square roots into its simplest radical form? Because I can do small numbers mentally but when they are bigger I get stuck..
well,
I really don't like to tell this to you but the Ti-84 plus will not display a square root as an answer, it will always return answers in decimal form. The best way for you to simplify roots is by using a factor tree and knowing your squares up to 30. I went through all of that using a ti-83 plus so I completely understand your desire for it to show a root, it just won't. I really wish you the best with this problem, but having read the owners manual for it, I just don't see any other way to do it.
Mar 21, 2011 | Texas Instruments TI-84 Plus Calculator
### I when i power -3 to the second i get -9 when it should be just 9
No, is giving you the correct result. When you negate 3-squared, you should get -9. If you want to square negative three, you need to type it in as ( (-) 3 ) ^ 2 or ( (-) 3 ) x^2
There is a difference between (-3)^2 and -(3^2). If you omit the parentheses, the calculator will interpret it as the latter.
If in doubt, use parentheses.
Feb 17, 2011 | Texas Instruments TI-84 Plus Calculator
### My Ti-84 plus gives me an answer of "4E-4" when I multiply .02 times itself. How can I get it to show me the decimal answer instead of the E
Unfortunately, you can't. From the manual:
If [...] the answer cannot display in 10 digits (or the absolute value is less than 0.001), the TI-84Plus expresses the answer in scientifc notation.
Oct 05, 2010 | Texas Instruments TI-84 Plus Calculator
### Please help me to use my TI - 84 so I can get a radical number when working with larger numbers. the steps I must take and having the TI - 84 set to the right function
I am afraid I do not completely understand what you want. This calculator does not have a Computer Algebra System or CAS. So the results of all calculations will be displayed as decimal numbers, and that includes the square and other roots that are irrational numbers (square root of 2, 3, 5 7, etc.)
You cannot manipulate radicals on this calculator, whether for small numbers or large numbers. Sorry.
Sep 09, 2010 | Texas Instruments TI-84 Plus Silver...
### How do I on a TI-84 Plus Simplify Fractions
Enter the fraction for example: enter 8/12.
The answer will be .666666
Then click Math (which is located under the Alpha button).
Click the ->Frac button (the first one).
It should come up as Ans->Frac.
Click Enter.
It should now read 2/3 which is 8/12 reduced.
Note: I have a TI-84 Plus Silver Edition, and the TI-84 Plus might not have this function, but that is how I do it on my TI-84 Plus Silver Edition
Nov 08, 2008 | Office Equipment & Supplies
#### Related Topics:
515 people viewed this question
## Ask a Question
Usually answered in minutes!
Level 3 Expert
Level 2 Expert
Level 3 Expert | 1,538 | 5,854 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2018-30 | latest | en | 0.887237 |
https://stats.stackexchange.com/questions/27798/how-to-sample-from-the-product-of-a-gaussian-and-an-inverse-wishart-distribution | 1,669,598,766,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710462.59/warc/CC-MAIN-20221128002256-20221128032256-00317.warc.gz | 596,576,993 | 36,747 | # How to sample from the product of a Gaussian and an inverse-Wishart distribution?
How do I sample from a distribution that is the product of a gaussian and an inverse-wishart distribution? I was going to use inverse transform sampling but a friend said that he thinks there is a much simpler way of doing it, a trick because the inverse-wishart is a conjugate prior of the gaussian. Does anyone know if there is an easy way to do this? Thanks very much!
• I have recently encountered this question too. Did you solved this problem by now? I am wondering whether there is any matlab code to share? It would be a great help to me! Thank you.
– user18629
Jan 13, 2013 at 16:20
A Paper by Bodnar and Okhrin nearly solves this for you. If $A$ is Wishart, $z$ is multivariate Gaussian, and $l$ is a conformable vector, they give the distribution for $l^{\top}A^{-1}z$. It is a bit involved:
$$l^{\top}A^{-1}z = \frac{1}{u_1}\left(l^{\top}\Sigma^{-1}\mu + u_2 \sqrt{\left(\lambda + \frac{\lambda(k-1)}{n-k+2}u_3\right)l^{\top}\Sigma^{-1}l}\right),$$
where $u_i$ are independent, $u_1\sim\chi^2_{n-k+1},u_2\sim\mathcal{N}(0,1),u_3\sim F\left((k-1)/2,(n-k+2)/2,s/\lambda\right),$ where $A$ is a k-variate Wishart with parameter $\Sigma$ and $n$ degrees of freedom, and $z$ is multivariate Gaussian with mean $\mu$ and covariance $\lambda\Sigma$. The parameter $s =\mu^{\top}R_l\mu$, where $R_l$ is underdefined in the paper; I think it is supposed to be $R_l = \Sigma^{-1} - \Sigma^{-1}ll^{\top}\Sigma^{-1} / l^{\top}\Sigma^{-1}l$.
• I wasn't sure if you wanted the distribution of the vector $A^{-1}z$, or if $L A^{-1}z$ would work for matrix (or vector) $L$. For vector $L$, the Bodnar & Okhrin trick will generate variates for you. O/w, good luck. Let me know how that works out. May 5, 2012 at 23:02 | 571 | 1,799 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2022-49 | latest | en | 0.881803 |
https://mathoverflow.net/questions/314613/when-can-one-continuously-prescribe-a-unit-vector-orthogonal-to-a-given-orthonor/314652 | 1,569,044,137,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574265.76/warc/CC-MAIN-20190921043014-20190921065014-00077.warc.gz | 566,893,710 | 32,390 | # When can one continuously prescribe a unit vector orthogonal to a given orthonormal system?
Let $$1 \leq k < n$$ be natural numbers. Given orthonormal vectors $$u_1,\dots,u_k$$ in $${\bf R}^n$$, one can always find an additional unit vector $$v \in {\bf R}^n$$ that is orthogonal to the preceding $$k$$. My question is: under what conditions on $$k,n$$ is it possible to make $$v$$ depend continuously on $$u_1,\dots,u_k$$, as the tuple $$(u_1,\dots,u_k)$$ ranges over all possible orthonormal systems? (For my application I actually want smooth dependence, but I think that a continuous map can be averaged out to be smooth without difficulty.)
When $$k=n-1$$ then one can just pick the unique unit normal to the span of the $$u_1,\dots,u_k$$ that is consistent with a chosen orientation on $${\bf R}^n$$ (i.e., take wedge product and then Hodge dual, or just cross product in the $$(k,n)=(2,3)$$ case). But I don't know what is going on in lower dimension. Intuitively it seems to me that if $$n$$ is much larger than $$k$$ then the problem is so underdetermined that there should be no topological obstructions (such as that provided by the Borsuk-Ulam theorem), but I don't have the experience in algebraic topology to make this intuition precise.
It would suffice to exhibit a global section of the normal bundle of the (oriented) Grassmannian $$Gr(k,n)$$, though I don't know how to calculate the space of such sections.
• It's been a loooong time since I've thought about such things, but I think that the obstruction you want is the Euler class, on the Wikipedia page it says "Note that "Normalization" is a distinguishing feature of the Euler class, so that it detects the existence of a non-vanishing section". So you'd need to look at the pull-back of the Euler class under the map $Gr_{n,k} \to Gr_{n,n-k}$ which maps $E$ to its orthogonal complement. The issue is a little more complicated if you're starting with a basis as then you need to pull back further to the frame bundle. – Loop Space Nov 5 '18 at 17:41
• Just a comment on your intuition (now that you have a full answer). Having lots of choice doesn't make it any easier to find a continuous choice because the act of making a choice is not continuous. So after making lots of local choices, you are confronted with the problem of patching them together and that is where the obstructions lie. – Loop Space Nov 6 '18 at 7:21
$$\def\RR{\mathbb{R}}$$ This problem was solved by
Whitehead, G. W., Note on cross-sections in Stiefel manifolds, Comment. Math. Helv. 37, 239-240 (1963). ZBL0118.18702.
Such sections exist only in the cases $$(k,n) = (1,2m)$$, $$(n-1, n)$$, $$(2,7)$$ and $$(3,8)$$.
All sections can be given by antisymmetric multilinear maps (and thus, in particular, can be taken to be smooth). The $$(2,7)$$ product is the seven dimensional cross product, which is octonion multiplication restricted to the octonions of trace $$0$$.
The $$(3,8)$$ product was computed by
Zvengrowski, P., A 3-fold vector product in $$R^8$$, Comment. Math. Helv. 40, 149-152 (1966). ZBL0134.38401
to be given by the formula $$X(a,b,c) = -a (\overline{b} c) + a (b \cdot c) - b (c \cdot a) + c (a \cdot b)$$ where $$\cdot$$ is dot product while multiplication with no symbol and $$\overline{b }$$ have their standard octonion meanings. Note that, if $$(a,b,c)$$ are orthogonal, the last $$3$$ terms are all $$0$$, so the expression simplifies to $$- a (\overline{b} c)$$; writing in the formula in the given manner has the advantage that $$X(a,b,c)$$ is antisymmetric in its arguments and perpendicular to the span of $$a$$, $$b$$ and $$c$$ for all $$(a,b,c)$$.
Unless I'm missing something, I think that the hairy ball theorem states precisely that you cannot do this when $$n = 3$$ and $$k = 1$$. I'm not sure what happens for other values of $$n$$ and $$k$$.
• The hairy ball theorem actually applies to any even-dimensional sphere, so this would also settle the case $n > 1$ odd and $k = 1$. – R. van Dobben de Bruyn Nov 5 '18 at 15:58
• Wow, I can't believe I had forgotten about this theorem. So my intuition that the problem is too underdetermined to have an obstruction in high dimension is wrong, apparently. – Terry Tao Nov 5 '18 at 16:19
• Write $n=2^{c+4d} a$, with $a$ odd and $0\leq c \leq 3$. Then there are $2^c+8 d -1$ linear independent sections of the tangent bundle of the sphere, which you can make orthogonal I think. This is a famous theorem of Adams. – Thomas Rot Nov 5 '18 at 18:03
• This is a terrific answer for $\mathbb{R}^3-\{0\}$, but I'm not sure that it applies to $\mathbb{R}^3$, for there's no non-degenerate vector field in $\mathbb{R}^3$ that would be normal to the sphere. – Michael Nov 6 '18 at 22:25
• @Michael The question is not asking for complementation of a vector field in $\mathbb{R}^n$. It is asking for complementation when "the tuple ranges over all possible orthonormal systems". This is precisely the sphere $S^2$ when $n=3, k=1$. – Aloizio Macedo Nov 7 '18 at 19:09
The space of orthonormal $$k$$-frames in $$\mathbb{R}^n$$ is the Stiefel manifold $$V(k, n) = SO(n)/SO(n - k)$$. There is a natural $$SO(k)$$ action on $$V(k, n)$$ and the quotient is the oriented grassmannian $$\operatorname{Gr}^+(k, n) = SO(n)/(SO(k)\times SO(n-k))$$. Let $$\gamma_k \to \operatorname{Gr}^+(k, n)$$ denote the tautological bundle and let $$\gamma_k^{\perp} \to \operatorname{Gr}^+(k, n)$$ denote its orthogonal complement. As you indicated, a nowhere-zero section of $$\gamma_k^{\perp} \to \operatorname{Gr}^+(k, n)$$ would give rise to a map that you desire. In fact, such a map arises this way if and only if it is $$SO(k)$$-invariant.
The inner product on $$\mathbb{R}^n$$ allows us to define the map $$P \mapsto P^{\perp}$$ which induces a diffeomorphism $$f : \operatorname{Gr}^+(k, n) \to \operatorname{Gr}^+(n-k, n)$$. Under this diffeomorphism we have $$f^*\gamma_{n-k} \cong \gamma_k^{\perp}$$, so $$\gamma_k^{\perp} \to \operatorname{Gr}^+(k, n)$$ admits a nowhere-zero section if and only if $$\gamma_{n-k} \to \operatorname{Gr}^+(n-k, n)$$ does. Therefore, we would like to know the answer to the following question:
For which values of $$k$$ and $$n$$ does $$\gamma_{n-k} \to \operatorname{Gr}^+(n-k, n)$$ admit a nowhere-zero section?
One necessary condition is that $$w_{n-k}(\gamma_{n-k}) = 0$$. Said another way, if $$w_{n-k}(\gamma_{n-k}) \neq 0$$, then there is no $$SO(k)$$-invariant map.
In a previous version of this answer, I stated what I thought was the $$\mathbb{Z}_2$$ cohomology ring of $$\operatorname{Gr}^+(k, n)$$ - I was incorrect. From this mistake, it followed that for $$1 < k < n - 1$$, $$w_{n-k}(\gamma_{n-k}) \neq 0$$ and hence there were no $$SO(k)$$-invariant maps for these values of $$k$$. This conclusion is false; there is a counterexample when $$k = 2$$ and $$n = 7$$ as David E Speyer pointed out in the comments below.
Somewhat surprisingly, the $$\mathbb{Z}_2$$ cohomology ring of $$\operatorname{Gr}^+(k, n)$$ is not known in general, see this question. The values of $$k$$ and $$n$$ for which $$w_{n-k}(\gamma_{n-k}) \neq 0$$ also seems to be unknown in general. However, if $$n - k \leq k$$, then $$w_{n-k}(\gamma_{n-k}) \neq 0$$, so for values of $$k$$ and $$n$$ with $$2k \leq n$$, there are no $$SO(k)$$-invariant such maps.
When $$k = n - 1$$, you described such a map which is in fact $$SO(n-1)$$-invariant. By the above correspondence, such maps exist because $$\gamma_1 \to \operatorname{Gr}^+(1, n) = S^{n-1}$$ is trivial as it is an orientable line bundle (alternatively, $$\gamma_1$$ is trivialised by the Euler vector field).
When $$k = 1$$, first note that $$\gamma_{n-1} \to \operatorname{Gr}^+(n - 1, n) = S^{n-1}$$ is isomorphic to the tangent bundle of $$S^{n-1}$$:
\begin{align*} TS^{n-1} &\cong T\operatorname{Gr}^+(n-1, n)\\ &\cong \operatorname{Hom}(\gamma_{n-1}, \gamma_{n-1}^{\perp})\\ &\cong \gamma_{n-1}^*\otimes\gamma_{n-1}^{\perp}\\ &\cong \gamma_{n-1}\otimes f^*\gamma_1\\ &\cong \gamma_{n-1} \end{align*}
where the last isomorphism uses the fact that $$\gamma_1$$, and hence $$f^*\gamma_1$$, is trivial. By Poincaré-Hopf, $$TS^{n-1}$$ admits a section if and only if $$n$$ is even. In this case, the map can be written down explicitly: $$(v_1, v_2, \dots, v_{n-1}, v_n) \mapsto (-v_2, v_1, \dots, -v_n, v_{n-1})$$. Identifying $$\mathbb{R}^n$$ and $$\mathbb{C}^{n/2}$$ via $$(v_1, v_2, \dots, v_{n-1}, v_n) \mapsto (v_1 + iv_2, \dots, v_{n-1} + iv_n)$$, the aforementioned map is nothing but multiplication by $$i$$.
Note, requiring $$SO(k)$$-invariance for $$k = 1$$ is not a restriction as $$SO(1)$$ is the trivial group.
• It seems to me that something is wrong with this argument, because of the existence of the 7 dimensional cross product: en.wikipedia.org/wiki/Seven-dimensional_cross_product . This is an antisymmetric bilinear map $R^7 \times R^7 \to R^7$ such that $u \times v$ is always perpendicular to $u$ and $v$ and, if $u$ and $v$ are orthogonal, then $|u \times v| = |u| |v|$. Using bilinearity and antisymmetry, I get $(\cos \theta u + \sin \theta v) \times (-\sin \theta u + \cos \theta v) = (\cos^2 \theta + \sin^2 \theta) (u \times v) = u \times v$, meaning this map is $SO(2)$ invariant. – David E Speyer Nov 6 '18 at 16:02
• This would seem to give a section of $\gamma_2^{\perp} \to G^+(2,7)$. What did I miss? – David E Speyer Nov 6 '18 at 16:02
• @DavidESpeyer: I can't see anything wrong with what you've written, so there must be something wrong with my answer. Maybe the cohomology ring of $\operatorname{Gr}^+(n-k, n)$ is not correct. I don't know of a reference, but I was under the impression that this was correct. – Michael Albanese Nov 6 '18 at 16:32
• @DavidESpeyer: The cohomology ring is wrong as can be seen in the case $k = 2$, $n = 4$ where $\operatorname{Gr}^+(2, 4) = S^2\times S^2$. I will try to think about this and see what can be salvaged. – Michael Albanese Nov 6 '18 at 17:37
• @DavidESpeyer: It seems that the values of $k$ and $n$ for which $w_{n-k}(\gamma_{n-k}) \neq 0$ is not known; see this question. – Michael Albanese Nov 8 '18 at 23:10
Denoting the Stiefel manifold of orthonormal $$k$$-frames in $$\mathbb{R}^n$$ by $$V(k,n)$$ as in Michael Albanese's answer, what you are asking for is a section of the sphere bundle $$S^{n-k-1}\to V(k+1,n)\to V(k,n),$$ where the projection takes a $$(k+1)$$-frame to its first $$k$$ vectors.
According to the paper
Čadek, Martin; Mimura, Mamoru; Vanžura, Jiří, The cohomology rings of real Stiefel manifolds with integer coefficients, J. Math. Kyoto Univ. 43, No. 2, 411-428 (2003). ZBL1061.55015,
the Euler class of this bundle is zero when $$n-k$$ is odd, and nonzero when $$n-k$$ is even.
This extends Michael Albanese's answer slightly, showing that the maps cannot exist when $$n-k$$ is even with $$1, even without requiring $$SO(k)$$-invariance. I don't know if the bundle admits a section when $$n-k$$ is odd. The primary obstruction vanishes in these cases, but note that there may be higher obstructions in the groups $$H^{i+1}(V(k,n);\pi_i(S^{n-k-1}))$$. | 3,513 | 11,075 | {"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": 115, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2019-39 | latest | en | 0.933636 |
https://chess.stackexchange.com/questions/34795/how-to-proceed-after-declined-queens-gambit/34831 | 1,656,377,530,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103344783.24/warc/CC-MAIN-20220627225823-20220628015823-00668.warc.gz | 207,262,620 | 67,098 | # How to proceed after declined Queen's Gambit?
I am a intermediate-level player. How can I proceed after a declined Queen's Gambit, as White?
• Well, it does make a difference whether Black declines with e6 or e5...(I.e., since the game can develop into a bucketload of "normal" opening systems, a few more moves would make the question more specific. For example, in a well known standard position you can do a minority attack, an expansion in the center or a pawn storm after castling long, and all is well tried.) Apr 15, 2021 at 21:02
• On Black's 2nd move he can decline by playing c5, c6, e5, e6, Nc6, Nf6, or Bf5, among others. I believe 3.e3 is playable against any of those, although it's not necessarily the best. I guess you can play 3.e3 against anything except 2...b5 or 2...Bg4.
– bof
Apr 16, 2021 at 0:37
• In the standard QGD, you have 3 general plans. 1) The minority attack. 2) Control of the center with Nge2-g3, f3 and e4. 3) A kingside attack with O-O-O and and advance of the kingside pawns. Go to a chess database site and search for a collection of QGD games and look for these ganeral plans. Apr 16, 2021 at 2:26
• @bof The Queen's Gambit Declined is the position after 2...e6. The other moves you suggest are different defences (c6, e5, Nc6 and Bf5 correspond to the Slav, Albin, Chigorin and Blatic defenses..., while Nf6 is not even considered by theory) Apr 16, 2021 at 13:46
• @David The question was asking about "a declined Queen's Gambit", not "The Queen's Gambit Declined". 2...Nf6 is considered by theory. It is considered bad, although the variation 3.cxd5 c6!? either transposes into the Exchange Slav or else is a playable gambit as gambits go.
– bof
Apr 16, 2021 at 22:12
There are multiple options here. I'm not sure what you mean by "an intermediate player", so the amount of specific lines you'd like to study can differ. I'll focus instead on types of pawn structures and middlegames you can reach. Let's take the reference position 1.d4 d5 2.c4 e6, even though there are different move orders available:
Black will sometimes play an eventual ...c5. This can lead to two types of pawn structures: the Semi-Tarrasch type, where you get pawns on e4 and d4 (see for instance 3.Nc3 Nf6 4.Nf3 c5 5.cxd5 Nxd5 6.e4). This often leads to a comfortable position for White due to his superiority in the center and not enough counterplay chances for Black to compensate for it. Here you have an example from top level players.. This line has some similarities with the Grünfeld Defence (see 1.d4 Nf6 2.c4 g6 3.Nc3 d5 4.cxd5 Nxd5 5.e4 Nxc3 6.bxc3), but I'd say Black has more chances in the Grünfeld. In my opinion you should try to reach this pawn structure as often as possible.
Another typical pawn structure that is often reached is one with an isolated "d" pawn. You'll probably play both with and against that isolated pawn in several occasions, you that's also something you may want to get familiar with. chess.com has an article about it that you may want to check out Note that this structure can also be reached when the Queen's Gambit is accepted.
Then there's the "minority attack" type of game. This happens when you capture on d5 and your opponent retakes with his "e" pawn but ...c6 has been played instead of ...c5. Here you can play on the queenside trying to get your semi-open "c" file fully open. I'm not really well-versed here so I'll let someone else choose an illustrative example by editting this answer.
Finally, there are some lines where Black will enter some variation of the Slav Defence (with ...c6). I don't know how well you know the lines after 1.d4 d5 2.c4 c6 and your preferred move choice can chance according to that. For instance, you can try 3.Nf3 Nf6 4.e3 so now you can answer to 4...c6 with 5.Nbd2!? if that's the particular line you enjoy against the Slav (here your plan would be to make a quick e4 push). But if you're not interested in playing that then you may prefer 3.Nc3. There are several lines you can choose from and plenty of transpositions between them.
• I'd also mention that GM Daniel King has recently been focusing on the queen's gambit declined in many of his latest videos on his powerplaychess channel on youtube, where he shows off different "model games" showcasing various middlegame plans for both sides. I think these videos are accessible for an intermediate player, although it's hard to tell exactly what "intermediate" actually entails here. Apr 18, 2021 at 9:43
Matthew Sadler's book Queen's Gambit Declined has received very high and widespread praise. Again it depends what you mean by "intermediate", but the book contains a ton of instructive games, and its clear and deep discussions of ideas should be helpful to players of a wide range of strengths. | 1,255 | 4,772 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | longest | en | 0.942803 |
https://www.esaral.com/q/d-and-e-are-points-on-the-sides-ca-and-cb-respectively-of-a-triangle-abc-right-angled-at-c | 1,713,092,590,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816879.25/warc/CC-MAIN-20240414095752-20240414125752-00383.warc.gz | 726,392,077 | 11,293 | # D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.
Question.
$\mathrm{D}$ and $\mathrm{E}$ are points on the sides $\mathrm{CA}$ and $\mathrm{CB}$ respectively of a triangle $\mathrm{ABC}$ right angled at $\mathrm{C}$. Prove that $\mathrm{AE}^{2}+\mathrm{BD}^{2}=\mathrm{AB}^{2}+\mathrm{DE}^{2}$.
Solution:
In right angled $\triangle \mathrm{ACE}$,
$\mathrm{AE}^{2}=\mathrm{CA}^{2}+\mathrm{CE}^{2}$ ...(1)
and in right angled $\triangle B C D$,
$\mathrm{BD}^{2}=\mathrm{BC}^{2}+\mathrm{CD}^{2}$ ...(2)
Adding (1) and (2), we get
$\mathrm{AE}^{2}+\mathrm{BD}^{2}=\left(\mathrm{CA}^{2}+\mathrm{CE}^{2}\right)+\left(\mathrm{BC}^{2}+\mathrm{CD}^{2}\right)$
$=\left(\mathrm{BC}^{2}+\mathrm{CA}^{2}\right)+\left(\mathrm{CD}^{2}+\mathrm{CE}^{2}\right)$
$=\mathrm{BA}^{2}+\mathrm{DE}^{2}$
$\therefore \quad \mathrm{AE}^{2}+\mathrm{BD}^{2}=\mathrm{AB}^{2}+\mathrm{DE}^{2}$ | 349 | 918 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2024-18 | latest | en | 0.315211 |
https://istopdeath.com/graph-fxx2/ | 1,674,968,882,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499700.67/warc/CC-MAIN-20230129044527-20230129074527-00625.warc.gz | 352,411,330 | 14,641 | # Graph f(x)=|x+2|
f(x)=|x+2|
Find the absolute value vertex. In this case, the vertex for y=|x+2| is (-2,0).
To find the x coordinate of the vertex, set the inside of the absolute value x+2 equal to 0. In this case, x+2=0.
x+2=0
Subtract 2 from both sides of the equation.
x=-2
Replace the variable x with -2 in the expression.
y=|(-2)+2|
Simplify |(-2)+2|.
y=|0|
The absolute value is the distance between a number and zero. The distance between 0 and 0 is 0.
y=0
y=0
The absolute value vertex is (-2,0).
(-2,0)
(-2,0)
The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.
Interval Notation:
(-∞,∞)
Set-Builder Notation:
{x|x∈ℝ}
For each x value, there is one y value. Select few x values from the domain. It would be more useful to select the values so that they are around the x value of the absolute value vertex.
Substitute the x value -4 into f(x)=|x+2|. In this case, the point is (-4,2).
Replace the variable x with -4 in the expression.
f(-4)=|(-4)+2|
Simplify the result.
f(-4)=|-2|
The absolute value is the distance between a number and zero. The distance between -2 and 0 is 2.
f(-4)=2
y=2
y=2
y=2
Substitute the x value -3 into f(x)=|x+2|. In this case, the point is (-3,1).
Replace the variable x with -3 in the expression.
f(-3)=|(-3)+2|
Simplify the result.
f(-3)=|-1|
The absolute value is the distance between a number and zero. The distance between -1 and 0 is 1.
f(-3)=1
y=1
y=1
y=1
Substitute the x value 0 into f(x)=|x+2|. In this case, the point is (0,2).
Replace the variable x with 0 in the expression.
f(0)=|(0)+2|
Simplify the result.
f(0)=|2|
The absolute value is the distance between a number and zero. The distance between 0 and 2 is 2.
f(0)=2 | 576 | 1,789 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2023-06 | latest | en | 0.827557 |
https://www.physicsforums.com/threads/help-thanks-in-advance.163862/ | 1,585,865,113,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370508367.57/warc/CC-MAIN-20200402204908-20200402234908-00381.warc.gz | 1,086,786,478 | 15,729 | # Help! thanks in advance
help! thanks in advance!!!!!
## Answers and Replies
Related Calculus and Beyond Homework Help News on Phys.org
uh... it's a function?
What do you want to do with it?
its limit is 1 when x->infinity
it is not defined when x=0
yeah like christian stated
what's your deal with it??
yes,I know,now I want to know how do you call it in your country,we call it "nike curve" in China,I want to know how do you teach your students in maths teaching?did you use the sketchpad to graph it??
Gib Z
Homework Helper
I graphed it use Graphmatica, give that a search in google and its free.
You don't need to call it anything Sophia, just tell the students f(x)=x+1/x. That's all. And just incase you wanted some helpful info on actually graphing it, its an odd function so theres antisymmetry. and www.calc101.com has a graphing program on it, that might help as well. | 228 | 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} | 2.640625 | 3 | CC-MAIN-2020-16 | longest | en | 0.957333 |
https://expydoc.com/doc/6589915/%E5%A1%A9%E5%B1%B1%E5%B9%BE%E4%BD%95%E5%AD%A6%E3%82%92%E7%94%A8%E3%81%84%E3%81%9F-%E3%83%9C%E3%83%AD%E3%83%8E%E3%82%A4%E5%9B%B3%E3%81%AE%E8%A7%A3%E6%9E%90 | 1,713,508,163,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817289.27/warc/CC-MAIN-20240419043820-20240419073820-00836.warc.gz | 220,586,120 | 8,028 | ### 塩山幾何学を用いた ボロノイ図の解析
```Ritsumeikan high school
Mimura Tomohiro
Miyazaki Kosuke
Murata Kodai
Mr,Kuroda suggest “the geometry of salt”
When a lot of salt is poured on a board which is cut
into a particular shape, it creates a “salt mountain.
We named “ Geometry of salt mountain”.
3
4
When some points are put like this on a diagram,
a Voronoi Diagram is the diagram which
separates the areas closest to each point from
the other points.
5
6
Same
distance
incenter
7
8
△ＡＢＥの傍心点
△ＡＢＥの内心点
9
10
12
The reason of appearing curve line is that there are
different shortest line from a concave point
A
E
line l
F
Point E is same distance to
line ｌ and A
There were curve lines.
13
15
ED＝EA
CE＋BE
＝CE＋EA＋AB
＝CE＋ED＋AB
＝CD＋AB
＝Constant
16
17
p＞PQ
p＜PQ
18
To solve d which is make up (0,p) on yaxis and Q on y=x2
d 2 x 2 ( x 2 p ) 2 x 4 (1 2 p ) x 2 p 2
1 2 p
1
x2
p
2
4
2
p
p
d
1
4
p
1
2
If p ＜1/2 , the minimum
If p ＞１/2 ,
d
dp
p
1
4
Thus the mountain ridges are disappeared
at p<1/2.
19
20
21
22
Start
i=0
Set the range and domain (x0,y0)-(x1,y1)
Set the number of point num
Set the coordinates of point (AX,AY)
Set the radius which is r of circle
Set the color ct
NO
Loop3
YES
Loop2
MIN=L(i)
ct=i
Loop1
From y0 to y1 about y
Loop1
L(i)<MIN
Loop2
From x0 to x1 about x
Loop3
From i=0 to num
L(i)=SQR((X-AX(i))^2
+(Y-AY(i))^2)-r(i)
NO
Loop4
From i=0 to num
YES
MIN=L(i)
ct=i
Give color which is ct 2 to
point
SET POINT STYLE 1
PLOT POINTS: x,y
(AX(i),AY(i))
such circle was drown
Loop4
End
23
24
25
Weighted Voronoi Diagrams are an extension of Voronoi
Diagrams.
d(x, p(i)) = d(p(i)) - w(i)
26
salt mountains could reproduce this by replacing
weight with the radius of the hole . this mean
27
29
30
31
If there are four schools in some area, like this figure,
each student wants to enter the nearest of the four
schools.
32
34
Mountain ridges appear where the distances to the
nearest side is shared by two or more sides.
The prediction of the program matches the
mountain ridge lines and the additively weighted
Voronoi Diagram also matches the program.
Salt mountain can reproduce various phenomenon in
biology and physics.
35
We want to analyze mountain ridge lines in various
shapes.
We could reproduce additively weighted Voronoi
Diagrams
so
we
research
how
to
reproduce
Multiplicatively weighted Voronoi Diagrams.
We want to be able to create the shape of the board to
match any given mountain ridges.
36
Toshiro Kuroda
Konichi Kato
Spring of Mathematics
Masashi Sanae
http://izumi-math.jp/sanae/MathTopic/gosin/gosin.htm
Function Graphing Software GRAPES Katuhisa Tomoda
http://www.osaka-kyoiku.ac.jp/~tomodak/grapes/
38
Ritsumeikan High School
Mr,Saname Msashi
Ritumeikan University
College of Science and Engineering
Dr,Nakajima Hisao
39
40
``` | 1,095 | 2,829 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-18 | latest | en | 0.842732 |
https://db-excel.com/solving-quadratic-equations-using-different-methods-worksheet-answers/ | 1,656,594,093,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103821173.44/warc/CC-MAIN-20220630122857-20220630152857-00769.warc.gz | 234,405,443 | 11,283 | Solving Quadratic Equations Using Different Methods Worksheet Answers in a learning moderate can be utilized to try students talents and understanding by addressing questions. Because in the Student Worksheet about 90% of the articles of the complete guide are issues, both multiple selection and solution questions which are not available. While the rest consists of a brief summary of the topic matter. Using worksheet, teachers no more have to bother to collect issues or questions. With the media the teacher is expected to target on giving a optimum comprehension of the given subject. For the evaluation and test of learning outcomes, the instructor only wants to see and primary the questions previously available in the worksheet. Since just about the worksheet functions as a guide for students in carrying out learning responsibilities equally individually and in groups.
## The Character of Solving Quadratic Equations Using Different Methods Worksheet Answers in Education
Solving Quadratic Equations Using Different Methods Worksheet Answers as a derivative of big methods solution questions. Applying Worksheets means facilitating students to have the ability to solution issues about matters they’ve learned. With the Worksheet, pupils can understand the niche subject all together more easily. Since answering the questions in the Worksheet is exactly like learning about a matter over and once more, obviously pupils will understand deeply. Creating Worksheets an instrument of training and learning activities is a highly effective technique for teaching pupils memories in understanding issue matter. Because when using Worksheets, pupils are focused on answering the issues that are previously available. Applying Worksheets has been proven to guide scholar studying achievement.
PLEASE SEE : Multiplying Two Digit Numbers Worksheet
Solving Quadratic Equations Using Different Methods Worksheet Answers are a type of studying aid. In general the Worksheet is a learning software as a match or perhaps a way of encouraging the implementation of the learning Plan. Student worksheets in the proper execution of blankets of paper in the shape of data and issues (questions) that must be answered by students. This Solving Quadratic Equations Using Different Methods Worksheet Answers is well used to encourage the engagement of pupils in learning equally used in the application form of advised practices and to offer development training. Along the way of learning, Worksheets aim to find methods and program of concepts.
## Details to Contemplate When Producing Solving Quadratic Equations Using Different Methods Worksheet Answers
Solving Quadratic Equations Using Different Methods Worksheet Answers are a stimulus or teacher guidance in education that’ll be shown in writing in order that in publishing it must focus on the standards of visual press as aesthetic media to attract the eye of students. At the least the Worksheet as a press card. Whilst the contents of the information of the Worksheet must look closely at the elements of writing graphical press, the hierarchy of the product and the choice of questions as an successful and successful stimulus. Through the Worksheet the teacher asks pupils to solution the issues which have been accessible after increasing specific subject matter. Equally personally and in groups. | 578 | 3,363 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2022-27 | latest | en | 0.930677 |
https://studyaffiliates.com/a-transverse-wave-on-a-cord-is-given-by-dxt-0-11rm-sin4-0x-24-0t-where-d-and-x-are-in-rm-m-and-t-is-in-rm-s-at-t-0-2/ | 1,642,663,043,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301730.31/warc/CC-MAIN-20220120065949-20220120095949-00103.warc.gz | 600,126,557 | 15,270 | # A transverse wave on a cord is given by D(x,,t) = 0.11,{rm sin}(4.0x – 24.0t), where D and x are in {rm m} and t is in {rm s}.At t =0.
A transverse wave on a cord is given by D(x,,t) = 0.11,{rm sin}(4.0x – 24.0t), where D and x are in {rm m} and t is in {rm s}.At t =0.13 <units>s</units>, what is the displacement of the point on the cord where x = 0.45 <units>m</units>? At t = 0.13 <units>s</units>, what is the velocity of the point on the cord where x = 0.45 <units>m</units>?
Looking for a Similar Assignment? Order now and Get a Discount! | 198 | 549 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.893294 |
https://activegaliano.org/how-many-grams-in-a-half-ounce-of-mushrooms-update-new/ | 1,695,979,402,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510501.83/warc/CC-MAIN-20230929090526-20230929120526-00498.warc.gz | 92,387,118 | 36,664 | Skip to content
Home » How Many Grams In A Half Ounce Of Mushrooms? Update New
# How Many Grams In A Half Ounce Of Mushrooms? Update New
Let’s discuss the question: how many grams in a half ounce of mushrooms. We summarize all relevant answers in section Q&A of website Activegaliano.org in category: Blog Marketing. See more related questions in the comments below.
Table of Contents
## How many grams is in an ounce?
There are 28 grams within one ounce. If you can remember this number, even if you find yourself without this handy cooking conversion chart, you’ll be able to make some quick calculations.
See also How Does Narrative Differ From Expository Writing Apex? New
## How many ground grams are in an ounce?
One ounce in weight and mass sense converted to grams equals precisely to 28.35 g.
### ✅ How Many Grams In An Ounce
✅ How Many Grams In An Ounce
✅ How Many Grams In An Ounce
## What does 1 cup mushrooms weigh?
Having trouble measuring recipe ingredients? Here’s the scoop.
One cup of this ingredient Weighs approx. this number of ounces And this number of grams
Mushrooms (chopped) 3 75
Mushrooms (sliced) 2.5 – 3 60 – 75
Oatmeal (fine) 5.25 155
Oatmeal (medium) 3 75
## How many grams is a cup of mushroom?
A cup of sliced, raw mushrooms, weighing 70 grams (g), provides almost 1 g of fiber.
## How do you measure half an ounce?
A tablespoon, denoted as “tbsp.,” is a unit of volume often used in cooking recipes. The United States customary system defines that 1 tbsp. is equal to one-half of a liquid ounce. However, a powder, such as sugar or salt, is measured in ounces (oz.) as weight.
See also How To Use Barrier Cream For Spray Tan? Update New
## Which is more 1 oz or 1g?
If you’re wondering how an ounce compares to a gram, it turns out that 1 ounce is a lot more mass than 1 gram. In fact, 1 ounce is approximately equal to 28.35 grams.
## How many grams is 4 1 2oz?
Ounces to Grams conversion table
Ounces (oz) Grams (g) Kilograms+Grams (kg+g)
1 oz 28.35 g 0 kg 28.35 g
2 oz 56.70 g 0 kg 56.70 g
3 oz 85.05 g 0 kg 85.05 g
4 oz 113.40 g 0 kg 113.40 g
## What percentage of an ounce is a gram?
1 gram is equal to 0.03527396 ounces, which is the conversion factor from grams to ounces.
## How many ounces is a medium mushroom?
An average medium-sized flat (open cup) Portobello (or brown) mushroom weighs around 60-70g/2.12-2.5oz, smaller flat mushrooms weigh around 50-60g/1.76-2.12oz and large flat mushrooms weigh 70g/2.5oz plus.
## How many grams are in mushrooms?
Mushrooms are generally sold in the U.S. in eighths, meaning one-eighth of an ounce (3.5 grams), which usually costs around \$35. The effects of magic mushrooms will always vary from person to person in addition to from mushroom to mushroom [source: QZ].
### I Took 10 Grams of Magic Mushrooms \u0026 Filmed It For You
I Took 10 Grams of Magic Mushrooms \u0026 Filmed It For You
I Took 10 Grams of Magic Mushrooms \u0026 Filmed It For You
## How many ounces is a serving of mushrooms?
Other common serving sizes
Serving Size Calories
1 oz 6
1/2 cup pieces 8
1 cup pieces or slices 15
1 cup whole 21
4 thg 2, 2008
## What is the most common type of mushroom?
1. White Button Mushroom. Characteristics: The most common and mildest-tasting mushroom around. Ninety percent of the mushrooms we eat are this variety.
See also 2006 Chevy Aveo How Many Quarts Of Oil? New Update
## How many tablespoons is a gram?
Grams and tablespoons for sugar (granulated)
Grams to tablespoons Tablespoons to grams
10 grams = 0.8 tbsp 1 tbsp = 12.5g
20 grams = 1.6 tbsp 2 tbsp = 25g
30 grams = 2.4 tbsp 3 tbsp = 37.5g
40 grams = 3.2 tbsp 4 tbsp = 50g
## How many teaspoons or tablespoons is 1 oz?
There are 6 teaspoons in a fluid ounce, which is why we use this value in the formula above.
## How many tablespoons are in a dry ounce?
There are 2 tablespoons in an ounce. So for 2 ounces, you will need 4 tablespoons, for 4 ounces 8 tablespoons, for 6 ounces 12 tablespoons, for 8 ounces 16 tablespoons, for 10 ounces 20 tablespoons.
## What is more accurate grams or ounces?
The Many Reasons Using a Scale Will Change Your Life
That’s fine for a quick estimation, but actual baking, or recording a recipe, requires more accuracy (which is why we love grams to begin with!). An ounce is actually closer to 28.349 grams than 30 grams.
## How much is a gram example?
Examples of Gram Weight
Here are common examples of objects that have about one gram of mass: A small paperclip. A thumbtack. A piece of chewing gum.
## Which is heavier ounces or grams?
1 ounce is heavier than 1 gram.
## Is 100g 4 oz?
Weight
Grams Pounds/ounces
25g 1oz
50g 2oz
100g 4oz
125g 5oz
### MIKE TYSON TAKES MUSHROOMS ON IMPAULSIVE!
MIKE TYSON TAKES MUSHROOMS ON IMPAULSIVE!
MIKE TYSON TAKES MUSHROOMS ON IMPAULSIVE!
## Is oz ounce?
“Ounce” has an “oz.” abbreviation while “fluid ounce” is abbreviated as “fl. oz.” 2.An ounce is measuring weight while a fluid ounce is measuring volume.
## How many grams is 14 oz?
Ounces to Grams table
Ounces Grams
12 oz 340.19 g
13 oz 368.54 g
14 oz 396.89 g
15 oz 425.24 g
Related searches
• how many grams in an ounce of mushrooms
• how many grams in an ounce
## Information related to the topic how many grams in a half ounce of mushrooms
Here are the search results of the thread how many grams in a half ounce of mushrooms from Bing. You can read more if you want.
You have just come across an article on the topic how many grams in a half ounce of mushrooms. If you found this article useful, please share it. Thank you very much. | 1,571 | 5,585 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2023-40 | latest | en | 0.87254 |
https://raw.githubusercontent.com/ichunqiu-resources/anquanke/master/005/task.py | 1,675,768,499,000,000,000 | text/plain | crawl-data/CC-MAIN-2023-06/segments/1674764500456.61/warc/CC-MAIN-20230207102930-20230207132930-00360.warc.gz | 485,094,717 | 1,350 | import os import gmpy2 flag = int(open('flag.txt').read().encode("hex"), 16) def genPrime(bits): data = os.urandom(bits/8) number = int(data.encode("hex"), 16) return gmpy2.next_prime(number) e = 1667 # rsa1: p - 700 bits q - 1400 bits p = genPrime(700) q = genPrime(1400) n = p*q phi = (p-1)*(q-1) d = gmpy2.powmod(e, -1, phi) rsa1 = (n, d) # rsa2: p - 700 bits, q - 700 bits, r = 700 bits p = genPrime(700) q = genPrime(700) r = genPrime(700) n = p*q*r phi = (p-1)*(q-1)*(r-1) d = gmpy2.powmod(e, -1, phi) rsa2 = (n, d) # rsa3: p - 700 bits, q - 700 bits, r = 700 bits p = genPrime(700) q = genPrime(700) n = p*q*r phi = (p-1)*(q-1)*(r-1) d = gmpy2.powmod(e, -1, phi) rsa3 = (n, d) # rsa4: p - 700 bits, q - 700 bits p = genPrime(700) q = genPrime(700) n = p*q*q phi = (p-1)*(q-1)*q d = gmpy2.powmod(e, -1, phi) rsa4 = (n, d) rsa = sorted([rsa1, rsa2, rsa3, rsa4]) for n, d in rsa: print 'pubkey:', n, d % (2**1050) flag = pow(flag, e, n) print 'encrypted flag', flag | 417 | 969 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-06 | latest | en | 0.364605 |
https://ncar.github.io/aspendocs/form_constants.html | 1,722,888,967,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640454712.22/warc/CC-MAIN-20240805180114-20240805210114-00358.warc.gz | 318,914,073 | 7,368 | ## Description
Aspen contains a collection of thermodynamic functions that were obtained from PAM II Derived Parameters (reference given below). There are a number of constants that are shared among the formulas.
## Constants
$$A = 5.0065 \\ B = 19.83923 \\ C_P = 1005.7 \\ R_D = 287.0 \\ E_3 = 6.1078 \\ T_3 = 273.15 \\ \epsilon = 0.622 \\$$
## Lookup table of saturation vapor pressure
For every 1/2 degree from 210 K to 310 K:
$$ETBL_{MIN} = 210.0 \\ ETBL_{MAX} = 310.0 \\ ETBL_{STEP} = 0.5 \\$$
$$ETBL[ ] = \\ \text { /* 210.0 */ } 0.01296, 0.01383, 0.01475, 0.01573, \\ \text { /* 212.0 */ } 0.01677, 0.01787, 0.01904, 0.02027, \\ \text { /* 214.0 */ } 0.02158, 0.02296, 0.02443, 0.02598, \\ \text { /* 216.0 */ } 0.02763, 0.02936, 0.03120, 0.03314, \\ \text { /* 218.0 */ } 0.03520, 0.03736, 0.03965, 0.04207, \\ \text { /* 220.0 */ } 0.04462, 0.04731, 0.05015, 0.05315, \\ \text { /* 222.0 */ } 0.05631, 0.05963, 0.06314, 0.06684, \\ \text { /* 224.0 */ } 0.07073, 0.07483, 0.07914, 0.08368, \\ \text { /* 226.0 */ } 0.08845, 0.09347, 0.09876, 0.10431, \\ \text { /* 228.0 */ } 0.11014, 0.11627, 0.12271, 0.12947, \\ \text { /* 230.0 */ } 0.13657, 0.14403, 0.15185, 0.16006, \\ \text { /* 232.0 */ } 0.16866, 0.17769, 0.18715, 0.19707, \\ \text { /* 234.0 */ } 0.20746, 0.21835, 0.22975, 0.24170, \\ \text { /* 236.0 */ } 0.25420, 0.26729, 0.28098, 0.29531, \\ \text { /* 238.0 */ } 0.31030, 0.32597, 0.34235, 0.35948, \\ \text { /* 240.0 */ } 0.37738, 0.39608, 0.41562, 0.43603, \\ \text { /* 242.0 */ } 0.45733, 0.47957, 0.50279, 0.52701, \\ \text { /* 244.0 */ } 0.55229, 0.57865, 0.60615, 0.63481, \\ \text { /* 246.0 */ } 0.66470, 0.69584, 0.72830, 0.76211, \\ \text { /* 248.0 */ } 0.79733, 0.83402, 0.87221, 0.91197, \\ \text { /* 250.0 */ } 0.95335, 0.99642, 1.04122, 1.08783, \\ \text { /* 252.0 */ } 1.13631, 1.18671, 1.23912, 1.29360, \\ \text { /* 254.0 */ } 1.35021, 1.40904, 1.47016, 1.53364, \\ \text { /* 256.0 */ } 1.59957, 1.66803, 1.73910, 1.81288, \\ \text { /* 258.0 */ } 1.88944, 1.96888, 2.05130, 2.13680, \\ \text { /* 260.0 */ } 2.22546, 2.31740, 2.41272, 2.51152, \\ \text { /* 262.0 */ } 2.61393, 2.72005, 2.82999, 2.94388, \\ \text { /* 264.0 */ } 3.06184, 3.18400, 3.31049, 3.44144, \\ \text { /* 266.0 */ } 3.57697, 3.71725, 3.86240, 4.01257, \\ \text { /* 268.0 */ } 4.16791, 4.32858, 4.49474, 4.66654, \\ \text { /* 270.0 */ } 4.84415, 5.02773, 5.21748, 5.41355, \\ \text { /* 272.0 */ } 5.61613, 5.82541, 6.04158, 6.26483, \\ \text { /* 274.0 */ } 6.49536, 6.73337, 6.97908, 7.23269, \\ \text { /* 276.0 */ } 7.49443, 7.76451, 8.04316, 8.33062, \\ \text { /* 278.0 */ } 8.62713, 8.93292, 9.24825, 9.57337, \\ \text { /* 280.0 */ } 9.90853, 10.25401, 10.61007, 10.97699, \\ \text { /* 282.0 */ } 11.35505, 11.74454, 12.14575, 12.55898, \\ \text { /* 284.0 */ } 12.98454, 13.42274, 13.87390, 14.33834, \\ \text { /* 286.0 */ } 14.81640, 15.30841, 15.81471, 16.33567, \\ \text { /* 288.0 */ } 16.87163, 17.42296, 17.99004, 18.57325, \\ \text { /* 290.0 */ } 19.17296, 19.78958, 20.42351, 21.07515, \\ \text { /* 292.0 */ } 21.74492, 22.43326, 23.14058, 23.86733, \\ \text { /* 294.0 */ } 24.61396, 25.38092, 26.16869, 26.97773, \\ \text { /* 296.0 */ } 27.80852, 28.66155, 29.53732, 30.43635, \\ \text { /* 298.0 */ } 31.35914, 32.30622, 33.27813, 34.27540, \\ \text { /* 300.0 */ } 35.29859, 36.34827, 37.42500, 38.52936, \\ \text { /* 302.0 */ } 39.66194, 40.82335, 42.01419, 43.23508, \\ \text { /* 304.0 */ } 44.48666, 45.76955, 47.08442, 48.43193, \\ \text { /* 306.0 */ } 49.81273, 51.22752, 52.67699, 54.16184, \\ \text { /* 308.0 */ } 55.68278, 57.24053, 58.83584, 60.46945, \\ \text { /* 310.0 */ } 62.14212 \\$$
## Source
PAM II Derived Parameters
Tags: | 1,952 | 3,709 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-33 | latest | en | 0.370908 |
http://www.solve-variable.com/math-variable/logarithms/math-factor-trees.html | 1,521,928,474,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257651007.67/warc/CC-MAIN-20180324210433-20180324230433-00582.warc.gz | 462,883,493 | 11,263 | Free Algebra Tutorials!
Home Systems of Linear Equations and Problem Solving Solving Quadratic Equations Solve Absolute Value Inequalities Solving Quadratic Equations Solving Quadratic Inequalities Solving Systems of Equations Row Reduction Solving Systems of Linear Equations by Graphing Solving Quadratic Equations Solving Systems of Linear Equations Solving Linear Equations - Part II Solving Equations I Summative Assessment of Problem-solving and Skills Outcomes Math-Problem Solving:Long Division Face Solving Linear Equations Systems of Linear Equations in Two Variables Solving a System of Linear Equations by Graphing Ti-89 Solving Simultaneous Equations Systems of Linear Equations in Three Variables and Matrix Operations Solving Rational Equations Solving Quadratic Equations by Factoring Solving Quadratic Equations Solving Systems of Linear Equations Systems of Equations in Two Variables Solving Quadratic Equations Solving Exponential and Logarithmic Equations Solving Systems of Linear Equations Solving Quadratic Equations Math Logic & Problem Solving Honors Solving Quadratic Equations by Factoring Solving Literal Equations and Formulas Solving Quadratic Equations by Completing the Square Solving Exponential and Logarithmic Equations Solving Equations with Fractions Solving Equations Solving Linear Equations Solving Linear Equations in One Variable Solving Linear Equations SOLVING QUADRATIC EQUATIONS USING THE QUADRATIC FORMULA SOLVING LINEAR EQUATIONS
Try the Free Math Solver or Scroll down to Tutorials!
Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
Ineq. #3:
Ineq. #4:
Ineq. #5:
Ineq. #6:
Ineq. #7:
Ineq. #8:
Ineq. #9:
Solve for:
Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg:
### Our users:
I do not have any issues. I just wanted to let you know that I am glad I purchased your product. I also appreciate the updates as they not only make for a better looking product, but things seem to be more user friendly now.
Christopher Bowman, TX
I recommend the Algebrator to students who need help with fractions, equations and algebra. The program is a great tool! Not only does it give you the answers but it also shows you how and why you come up with those answers. I've shown my students how to use the program during some of our lessons. A couple of them even bought the program to help them out with their algebra homework.
Debra Ratto, CO
The program has been very helpful.
Peter Goodman, TN
### Students struggling with all kinds of algebra problems find out that our software is a life-saver. Here are the search phrases that today's searchers used to find our site. Can you find yours among them?
#### Search phrases used on 2014-04-04:
• aswers for algebra 1 review solving linear equations
• Free Math PDF Books
• algebra software
• increasing order calculator
• Printable Homework Log
• free rational expression calculator fractions
• HOW TO PUT ALGEBRA FORMULS IN THE TI-84
• Algebrator Free Trial
• prentice hall algebra 2 answer key
• Algerba For Dummies Work Sheets
• 9th grade algebra worksheets
• free algebra equation calculator using the substitution method
• calculator with exponent key
• convert to radical notation calculator
• Holt ALgebra 1 Key
• radical equation calculator
• interval notation calculator
• Dividing rational expressions calculator
• examples of using algebra to solve problems in your everyday life
• holt algebra 1 worksheets
• radical notation calculator
• extraneous solutions calculator
• how to put a prealgabra formula in the TI 84 calculator
• solving problems by factoring
• very strong combination and permutation algebra
• practice test for 9th grade algeBRA
• two step equations with fractions worksheet grade 9
• rewriting exponential expression to have a different base
• solving a compound inequality calculator
• mit maths sample questions
• 2 step expressions worksheets
• formula in cube
• equations that form a curved line
• tough maths problems for 6th grade
• example of radical number
• trinomial calculator
• Algebra 1 problems 9th grade
• math percent discount worksheet
• 9th Grade Advanced Algebra book
• linear equations solving for six grader
• prentice hall algebra 2 with trigonometry
• fractions in simplest form calculator
• simplifying radical exponents
• online radical calculator
• a bi form math
• Simplifying Algebraic Expressions Calculator
• algebra transposition of formula
• math investigatory project students
• online algebra calculator
• quadratic formula worksheet
• How to solve ordered pairs
• algebra for 6th graders
• math problems for 6th graders
• grade 6 mathermatics examples
• simple linear equations worksheets
• math page prinouts for 5th graders
• www.apptitude questions.com
• math standard form
• multiplying decimals worksheets
• algebra problems for 6th graders
• number line worksheets
• factorising function
• fraction simplifier
• how to prepare for 9th grade algebra
• fraction to decimal power point
• 9th grade algebra
• quadratic midpoint proof
• division worksheets for 4th graders
• Fraction to Simplest Form Calculator
• algebra fourth grade worksheets
• how to solve alegebra formulas
• grade 10 algebra
• prealgerbra printable worksheets
• solving exponent that are fraction
• seventh grade problems
• fifth grade math :Algebra worksheets download
• algebra equivalent expressions
• calculator that shows work
• 3rd grade worksheets
• 2nd grade homework sheets
• 5th grade work online
• grade 10 math worksheets
• algebra beginners lessons
• one step equations worksheets
• lesson plans math 6th grade texas
• square roots worksheets
• first grade homework
• Onlice Calcuator that shows work
• adding and subtracting integers worksheets
• solve quadric equation binomial
• differentiation online
• solve quadratic formula TI-89
• free algebra cheats
• holt algebra 1b online textbook
• 7th grade math taks chart
• mental maths tests online
• equations worksheets
Prev Next
All Right Reserved. Copyright 2005-2018 | 1,459 | 6,280 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2018-13 | latest | en | 0.83182 |
https://www.jiskha.com/questions/807584/mrs-brocks-class-has-23-students-each-student-has-8-markers-which-describes-how-to | 1,652,760,463,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662515501.4/warc/CC-MAIN-20220517031843-20220517061843-00362.warc.gz | 1,007,662,448 | 4,812 | # math
Mrs. Brock's class has 23 students. Each student has 8 markers. Which describes how to find the total number of markers in the class?
a) Multiply the ones. Record;
b) Multiply the tens. Record;
c) Multiply the tens. Record. Multiply the ones. Record. Add the partial products. Record.
d) Multiply the ones. Record. Multiply the tens. Record. Subtract the partial products. Record.
1. 👍
2. 👎
3. 👁
4. ℹ️
5. 🚩
## Similar Questions
1. ### Algebra
the table shows the results of a survey of college students. find the probability that a student's first class of the day is a humanities class, given the student is female. First class of the day for college students male female
2. ### math, probability
Exercise: CLT applicability Consider the class average in an exam in a few different settings. In all cases, assume that we have a large class consisting of equally well prepared students. Think about the assumptions behind the
3. ### math
2. Mrs. young has her 18 students seated in a circle. They are evenly spaced and numbered in order. which student is directly opposite these three students? .student#1? .student#5? .student#18?
4. ### Math
After a Math test, each of the twenty-five students in the class got a peek at the teacher's grade sheet. Each student noticed five A's. No student saw all the grades and no student saw his or her own grade. What is the minimum
1. ### maths
there are 104 students in class 10th and 96 students in class 9th in a school .in a house examination the students ate to be evenly seated in parallel rows such that no two adjacent rows are of the same class. (a) find the maximum
2. ### math
A box contained some red, blue, and green markers. For every 5 red markers, there were 2 blue markers. For every 3 blue markers, there were 5 green markers. a.Find the ratio of red markers to blue markers to green markers.
3. ### basic statistics for the behavioral sciences
there are 80 students enrolled in statistics. (a)you and 11 other earned the same score. what percent of the class received your score? (b)forty percent of the class received a grade of C. how many students received a C? (c) only
4. ### math
Mrs. Bollo's second grade class of thirty students conducted a pet ownership survey. Results of the survey indicate that 8 students own a cat, 15 students own a dog, and 5 students own both a cat and a dog. How many of the
1. ### Math
If a student's percentile rank in a class of 400 students is 87, how do I find the student's class rank.
2. ### math.
In Mrs. Harrington’s history class, 80% of the students speak English and 60% speak German. There are 30 students who can speak only one language. What is the maximum number of students who may speak both English and German in
3. ### Math
A box contained some red, blue, and green markers. For every 3 red markers, there were 2 blue markers. For every 3 blue markers, there were 7 green markers. A) Find the ratio of red markers to blue markers to green markers. B)
4. ### python
Using functions and two lists write a Python program that asks the user to input each students ID number and a final grade for each student in the class (grades can be from 0 -100). The class has 10 students. The program should | 790 | 3,241 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2022-21 | latest | en | 0.93587 |
https://the-algorithms.com/es/algorithm/dilation-operation | 1,653,416,695,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662573189.78/warc/CC-MAIN-20220524173011-20220524203011-00765.warc.gz | 618,743,187 | 24,185 | #### Dilation Operation
S
```import numpy as np
from PIL import Image
def rgb2gray(rgb: np.array) -> np.array:
"""
Return gray image from rgb image
>>> rgb2gray(np.array([[[127, 255, 0]]]))
array([[187.6453]])
>>> rgb2gray(np.array([[[0, 0, 0]]]))
array([[0.]])
>>> rgb2gray(np.array([[[2, 4, 1]]]))
array([[3.0598]])
>>> rgb2gray(np.array([[[26, 255, 14], [5, 147, 20], [1, 200, 0]]]))
array([[159.0524, 90.0635, 117.6989]])
"""
r, g, b = rgb[:, :, 0], rgb[:, :, 1], rgb[:, :, 2]
return 0.2989 * r + 0.5870 * g + 0.1140 * b
def gray2binary(gray: np.array) -> np.array:
"""
Return binary image from gray image
>>> gray2binary(np.array([[127, 255, 0]]))
array([[False, True, False]])
>>> gray2binary(np.array([[0]]))
array([[False]])
>>> gray2binary(np.array([[26.2409, 4.9315, 1.4729]]))
array([[False, False, False]])
>>> gray2binary(np.array([[26, 255, 14], [5, 147, 20], [1, 200, 0]]))
array([[False, True, False],
[False, True, False],
[False, True, False]])
"""
return (127 < gray) & (gray <= 255)
def dilation(image: np.array, kernel: np.array) -> np.array:
"""
Return dilated image
>>> dilation(np.array([[True, False, True]]), np.array([[0, 1, 0]]))
array([[False, False, False]])
>>> dilation(np.array([[False, False, True]]), np.array([[1, 0, 1]]))
array([[False, False, False]])
"""
output = np.zeros_like(image)
(image.shape[0] + kernel.shape[0] - 1, image.shape[1] + kernel.shape[1] - 1)
)
# Copy image to padded image
image_padded[kernel.shape[0] - 2 : -1 :, kernel.shape[1] - 2 : -1 :] = image
# Iterate over image & apply kernel
for x in range(image.shape[1]):
for y in range(image.shape[0]):
summation = (
kernel * image_padded[y : y + kernel.shape[0], x : x + kernel.shape[1]]
).sum()
output[y, x] = int(summation > 0)
return output
# kernel to be applied
structuring_element = np.array([[0, 1, 0], [1, 1, 1], [0, 1, 0]])
if __name__ == "__main__": | 664 | 1,879 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.370757 |
http://www.spencerpondcamp.com/fhs/physics-book-unit-4-worksheet-2/ | 1,398,286,317,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1398223203422.8/warc/CC-MAIN-20140423032003-00274-ip-10-147-4-33.ec2.internal.warc.gz | 856,254,532 | 5,717 | # Physics Book Unit 4 Worksheet 2
Title Size Type
## AS Physics Revision Notes Unit 2 Electricity And ...
4.95MB
PDF Document
qonbase.50megs.com AS-Level Revision Notes AS Physics Revision Notes Unit 2 Electricity And Thermal Physics Electricity 1. Electric ct is the rate ...
## Physics Unit 4 Sy Sheets - itute.com FREE maths ...
2.86MB
PDF Document
Diffraction of light also demonstrates the wave nature of light. Gap Diffraction pattern has width a ...
## Unit IV: Vectors and Projectile Motion Worksheet Packet
2.24MB
PDF Document
Advanced physics, unit 4, vectors and trig review, worksheet packet p. 1 9/7/2009 Advanced Physics, Unit 4 Vectors and Trig Review Worksheet Packet (See last page for ...
## UNIT 1: SCIENTIFIC METHOD AND NATURE OF SCIENCE
2.80MB
PDF Document
UNIT 2: KINEMATICS TEXTBOOK: Active Physics: Chapters 1, 2, and 4 ESSENTIAL QONS: ... TEXTBOOK: Active Physics: Chapters 2, 3, 4, and Section 8.3
## PHYSICS TERMS, DEFINITIONS - Croeso i GCaD Cymru ...
3.84MB
PDF Document
Physics Terms Definitions and page 4 Stress Stress is the force per unit cross-sectional area when ... Unit: C-1 3.2.4(a) The Law of Conservation of
## Physics 123 Unit #2 Review - Physics & Astronomy - Brigham ...
4.73MB
PDF Document
Physics 123 Unit #2 Review I. Definitions & Facts thermal eqrium ideal gas thermal energy internal energy heat flow heat capacity specific heat heat of f ...
## Introdn to Physics - WordPress.com Get a Free Blog Here
4.04MB
PDF Document
PHYSICS FORM 4 [INTRODN TO PHYSICS-CHAPTER 1] Physics Department SSIJB 2 ... = 10 000 cm 2 = 1 x 10 4 cm 2 2. Convert the unit of volume in mm 3 to m 3
## Unit 4 Conceptual Qons
1.48MB
PDF Document
Physics 2A 1 Name_____ Unit 4 Conceptual Qons . Please circle the best answer- turn in two days before unit exam. Please use a Scantron.
## Regents Physics Unit Review Packet
0.84MB
PDF Document
What variables can be manipd to affect the movement of an object? - How do forces govern the motion of objects? - What are the ways that motion is described ...
## Sample Assessment Materials September 2007
2.95MB
PDF Document
Unit 2: Physics at Work..... 27 Unit 4: Physics on the Move ...
## Unit 4 Work, Power & Conservation of Energy Workbook
3.20MB
PDF Document
2 Unit 4 - Work, Power, & Conservation of Energy Sments to Text Readings from Fentals of Physics by Halliday, Resnick, & Walker Chapter 7 & 8
## 18 Forces and Terminal Velocity 2.1.4 UNIT 2 GCSE PHYSICS
2.54MB
PDF Document
UNIT 2 GCSE PHYSICS 2.1.4 Forces and Terminal Velocity 2011 FXA 18 The faster an object moves through a fluid the greater the frictional force that acts on it. | 727 | 2,652 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2014-15 | latest | en | 0.588099 |
http://jsxgraph.org/wiki/index.php?title=Lagrange_interpolation&oldid=5230 | 1,701,309,396,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100164.15/warc/CC-MAIN-20231130000127-20231130030127-00321.warc.gz | 23,322,451 | 6,884 | Lagrange interpolation
Constructs a polynomial of degree $\displaystyle{ n }$ through $\displaystyle{ n+1 }$ given points. Points can be added by clicking on "Add point". The dotted line is the graph of the first derivative, the dashed line is the graph of the second derivative.
The underlying JavaScript code
var board = JXG.JSXGraph.initBoard('box', {boundingbox: [-5, 10, 7, -6], axis: true});
var p = [];
p[0] = board.create('point', [-1,2], {size:4});
p[1] = board.create('point', [3,-1], {size:4});
var f = board.lagrangePolynomial(p);
var graph = board.create('functiongraph', [f,-10, 10], {strokeWidth:3});
var d1 = board.create('functiongraph', [board.D(f), -10, 10], {dash:1});
var d2 = board.create('functiongraph', [board.D(board.D(f)), -10, 10], {dash:2});
function addPoint() {
p.push(board.create('point',[(Math.random()-0.5)*10,(Math.random()-0.5)*3],{size:4}));
board.update();
} | 276 | 902 | {"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} | 2.703125 | 3 | CC-MAIN-2023-50 | latest | en | 0.667929 |
https://www.education.com/resources/fourth-grade/eltoggle/math/ | 1,660,008,584,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570879.37/warc/CC-MAIN-20220809003642-20220809033642-00377.warc.gz | 667,277,178 | 29,922 | 44 filtered results
44 filtered results
Math
Sort by
Rounding with Number Lines
Lesson Plan
Rounding with Number Lines
Teach your students to round to the thousands place using number lines.
Math
Lesson Plan
Subtraction with Regrouping
Lesson Plan
Subtraction with Regrouping
Help your students subtract with confidence by sharing two different strategies. Use this lesson to build on students’ understanding of subtraction and to evaluate this key skill.
Math
Lesson Plan
Lesson Plan
In this lesson, students will add three-digit numbers using expanded form addition and standard algorithm addition. They'll explain their answers and highlight the steps for each of the strategies.
Math
Lesson Plan
Lesson Plan
Are your students hungry for math? In this lesson, students pretend to order their favorite takeout foods with their classmates all while practicing rounding decimals so that they know what to expect when the bill comes!
Math
Lesson Plan
Division Word Problems
Lesson Plan
Division Word Problems
Freshen up on your understanding of division word problems with long division and one-digit divisors! Use this lesson to help students identify key division terms and solve word problems.
Math
Lesson Plan
Division and Multiplication Relationship
Lesson Plan
Division and Multiplication Relationship
Let's better understand multiplication and division concepts! Use this lesson to help students understand inverse operations between multiplication and division.
Math
Lesson Plan
Prime and Composite Numbers
Lesson Plan
Prime and Composite Numbers
Take your understanding of factors one step further! Use this lesson to classify factors as prime and composite numbers while creating factor trees.
Math
Lesson Plan
Multiplication Bingo: Two-Digit Numbers
Lesson Plan
Multiplication Bingo: Two-Digit Numbers
Your students will go bonkers for bingo as they play this baffling multiplication game! Who will get five in a row first?
Math
Lesson Plan
Multiplying Multiples
Lesson Plan
Multiplying Multiples
Introduce your students to multiplying multi-digit numbers with this lesson that gives them plenty of practice and has them play a game with a partner that makes the lesson fun!
Math
Lesson Plan
Divisibility Rules
Lesson Plan
Divisibility Rules
Introduce your students to divisibility rules with a lesson that encourages deep thinking and meaningful math talk between peers.
Math
Lesson Plan
Factors Over the Rainbow
Lesson Plan
Factors Over the Rainbow
Make math colorful! Teach your students to find the factors of a number using the factor rainbow strategy.
Math
Lesson Plan
Input and Output Boxes
Lesson Plan
Input and Output Boxes
Input some patterns to T-charts in this lesson. Use this lesson about input and output boxes as a precursor to finding patterns within word problems.
Math
Lesson Plan
The Case of the Missing Rectangle Side
Lesson Plan
The Case of the Missing Rectangle Side
Your students will become junior math detectives as they hunt down the missing side of a rectangle by applying the area formula for rectangles. The only clues they have are the rectangle's area and the measure of one side.
Math
Lesson Plan
Equivalent Fractions Using Area Models
Lesson Plan
Equivalent Fractions Using Area Models
Teach your students to create equivalent fractions by multiplying by different fraction forms of 1 whole like 2/2 or 4/4. This lesson also includes drafts of area models.
Math
Lesson Plan
Converting Metric Measurement in Word Problems
Lesson Plan
Converting Metric Measurement in Word Problems
Students will learn to convert metric units by applying concepts in word problems. Exercises and worksheets incorporate length (meters), and optional practice offers more instructions on mass (grams) and volume (liters).
Math
Lesson Plan
Distributive Property
Lesson Plan
Distributive Property
Teach your students to recognize and use the distributive property of multiplication.
Math
Lesson Plan
Clap Counting with Multiples
Lesson Plan
Clap Counting with Multiples
Finding the least common multiple is an essential skill for comparing, adding, and subtracting fractions! This lesson introduces students to this concept in an interactive way.
Math
Lesson Plan
Introduction to Division with Remainders
Lesson Plan
Introduction to Division with Remainders
In this lesson, your students will use visual and hands-on strategies to divide whole numbers with remainders.
Math
Lesson Plan
Multi-Digit Division
Lesson Plan
Multi-Digit Division
Let's solve multi-digit division problems! Students will feel more confident with their division than ever with this lesson that helps them practice dividing multi-digit numbers.
Math
Lesson Plan
Step By Step Decimal Subtraction
Lesson Plan
Step By Step Decimal Subtraction
It's as easy as one, two, three...four! Introduce your students to subtracting decimals in the tenth and hundredth place using a four-step process.
Math
Lesson Plan
Single Strategy for Adding and Subtracting Mixed Numbers
Lesson Plan
Single Strategy for Adding and Subtracting Mixed Numbers
Teach your students to add and subtract fractions with like denominators using a strategy that works for both operations: converting mixed numbers into improper fractions and back again!
Math
Lesson Plan
Polygon Perimeters with Tantalizing Tangrams!
Lesson Plan
Polygon Perimeters with Tantalizing Tangrams!
Teach your students to find the perimeter of polygons using real world examples. They will make polygon pictures using tangrams and practice finding the perimeters of the polygons in their masterpieces.
Math
Lesson Plan
Multi-Step Word Problems? No Problem!
Lesson Plan
Multi-Step Word Problems? No Problem!
Word problems? No problem! In this lesson, your students will learn how to approach solving multi-step word problems with mixed operations with a step-by-step plan.
Math
Lesson Plan
Fraction Hunt
Lesson Plan
Fraction Hunt
Fractions are everywhere! In this hands-on lesson, your class will work together in groups to find real-world examples of fractions. As they discover more complicated fractions, students will create their own word problems with them. | 1,214 | 6,097 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.869621 |
https://czechsinexile.org/what-is-margin-per-unit/ | 1,653,098,038,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662534773.36/warc/CC-MAIN-20220521014358-20220521044358-00245.warc.gz | 251,948,004 | 12,368 | # What is margin per unit?
Unit margin, also called unit contribution margin, reflects the cost incurred to produce and sell a particular unit of product. It is the profit achieved per unit after deducting product manufacturing or packaging costs and variable selling expenses from the product’s sales price.
Subtract your total cost per unit from your revenue per unit to get your contribution margin per unit. Divide this number by your revenue per unit to express it as a percentage of revenue.
Subsequently, question is, how do you calculate contribution per unit? Contribution and Contribution per Unit
1. Definition:
2. Total Contribution is the difference between Total Sales and Total Variable Costs.
3. Formulae:
4. Contribution = total sales less total variable costs.
5. Contribution per unit = selling price per unit less variable costs per unit.
6. Contribution per unit x number of units sold.
Furthermore, how much is the unit contribution margin?
Contribution. The Unit Contribution Margin (C) is Unit Revenue (Price, P) minus Unit Variable Cost (V): For example, if the price is \$10 and the unit variable cost is \$2, then the unit contribution margin is \$8, and the contribution margin ratio is \$8/\$10 = 80%.
What is profit margin formula?
The profit margin formula is net income divided by net sales. Net sales is gross sales minus discounts, returns, and allowances. Net income is total revenue minus expenses. A 10% margin is considered average.
### What is the formula to calculate profit percentage?
How to calculate profit margin Determine the net income (subtract the total expenses from the revenue). Divide the net income by the revenue. Multiply the result by 100 to arrive at a percentage.
### How do you calculate profit and loss?
How to Calculate Account Profit add up all your income for the month. add up all your expenses for the month. calculate the difference by subtracting total expenses away from total income. and the result is your profit or loss.
### What is a good profit margin?
You may be asking yourself, “what is a good profit margin?” A good margin will vary considerably by industry, but as a general rule of thumb, a 10% net profit margin is considered average, a 20% margin is considered high (or “good”), and a 5% margin is low.
### What is contribution margin percentage?
The contribution margin percentage, also known as the contribution margin or the contribution margin ratio, is a margin stated on a gross or per-unit basis. The contribution margin is the selling price of any given unit minus the variable cost associated with the production of that unit.
### What is the break even analysis?
Break-even analysis is a technique widely used by production management and management accountants. Total variable and fixed costs are compared with sales revenue in order to determine the level of sales volume, sales value or production at which the business makes neither a profit nor a loss (the “break-even point”).
### What affects contribution margin?
Contribution margin is the difference between a company’s product revenue and the variable costs used to earn that revenue. Fixed costs are then subtracted from contribution margin to derive net income or loss. Contribution margin is often calculated in total dollars, amount per unit, and as a percentage.
### How do you find CM?
Total contribution margin (CM) is calculated by subtracting total variable costs TVC from total sales S. Contribution margin per unit equals sales price per unit P minus variable costs per unit V or it can be calculated by dividing total contribution margin CM by total units sold Q.
### How do you explain profit?
Profit describes the financial benefit realized when revenue generated from a business activity exceeds the expenses, costs, and taxes involved in sustaining the activity in question. Any profits earned funnel back to business owners, who choose to either pocket the cash or reinvest it back into the business.
### What is the difference between gross margin and contribution margin?
Gross profit margin measures the amount of revenue that remains after subtracting costs directly associated with production. Contribution margin is a measure of the profitability of various individual products. Gross margin is calculated by deducting cost of goods sold from revenue, and dividing the result by revenue.
### How do I figure out gross margin?
To calculate gross margin subtract Cost of Goods Sold (COGS) from total revenue and dividing that number by total revenue (Gross Margin = (Total Revenue – Cost of Goods Sold)/Total Revenue). The formula to calculate gross margin as a percentage is Gross Margin = (Total Revenue – Cost of Goods Sold)/Total Revenue x 100.
### How is net profit calculated?
Net Profit margin = Net Profit ⁄ Total revenue x 100 Net profit. While it is arrived at through the income statement, the net profit is also used in both the balance sheet and the cash flow statement. is calculated by deducting all company expenses from its total revenue.
### How is total cost calculated?
Add your fixed costs to your variable costs to get your total cost. Your total cost of living on your budget is the total amount of money you spent over a one month period. The formula for finding this is simply fixed costs + variable costs = total cost.
### What is unit contribution?
Definition: A unit contribution margin is the dollar amount that a product’s selling price exceeds its total variable cost. In other words, the unit contribution margin is the selling price of a product minus the variable costs incurred to produce that product.
### What is contribution per unit?
Contribution per unit is the residual profit left on the sale of one unit, after all variable expenses have been subtracted from the related revenue. This information is useful for determining the minimum possible price at which to sell a product. | 1,156 | 5,929 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2022-21 | latest | en | 0.882262 |
https://www.patater.com/gbaguy/oldgba/day4.htm | 1,701,742,226,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100540.62/warc/CC-MAIN-20231205010358-20231205040358-00489.warc.gz | 1,046,419,555 | 2,874 | # Day 4
```
```
## A Few Things About MOV and LDR Instructions
```
This section is devoted to the people who still wonder about the difference between the
MOV and LDR Instructions. They both have the same purpose, to move data, ok? The difference is
that the MOV instruction is mainly used to copy the contents of one register to another like
so:
mov r0,r1
If r1 = 100 then now r0 = 100 also, got it?
With MOV, you can MOVe numbers into registers, but the numbers must fit into 8 bits (1
byte). just use MOV for copying registers around and LDR for LoaDing Registers which works like
so.
ldr r0,=0x06000000
LDR loads a register with a number of 32bits (4 bytes) at most.
So to wrap up this small section, I'll say "LDR is used to load registers, MOV is used
to copy a register to a register.
The GBA has 10 keys, UP, DOWN, LEFT, RIGHT, A, B, START, SELECT, L, and R. When you
press a key, the corresponding bit (1 binary digit) goes clear (0 or off). The keypad register
is at 0x04000130. In Keypad.h which you can get here, (Keypad.h is the
same as the one from ThePernProject converted to asm), there
are the following defines:
KEY_A 1
KEY_B 2
KEY_SELECT 4
KEY_START 8
KEY_RIGHT 16
KEY_LEFT 32
KEY_UP 64
KEY_DOWN 128
KEY_R 256
KEY_L 512
KEYS 0x04000130
To check to see if a key is down you do the following:
ldr r3,=KEYS ; These 2 lines LoaD Register r4 with
ldr r4,[r3] ; the contents of the Keys memory.
ands r4,r4,#KEY_A ; This line does an AND operation with the keys register and the A Key.
; note that the AND instruction has the 'S' suffix that will set the condition to the result
; of the operation.
If the A key is down the condition will be EQual so any instruction after that with
the EQ (EQual) suffix will be executed ONLY is the A key is down. I think it's obvious on how
to test for the other keys, so I won't show you.
So now a small demo about keypressing (if you think about that, it sounds like perverted
or something :)) Here goes:
;;--- CODE START ---;;
@include screen.h ; include our screen defines
@textarea
ldr r4,=REG_DISPCNT
ldr r5,=(MODE_3|BG2_ENABLE)
str r5,[r4]
; Our usual routine of setting up Screen Mode 3.
ldr r7,=VRAM+2410
; make r7 a pointer to screen memory at coordinates at 10,10.
start ; a label called start, you should be able to tell what labels are now, so this is
; the last time I'm pointing it out.
ldr r6,[r8] ; get value of key register
ands r6,r6,#KEY_UP ; bit and the key register value and the key value
; status now should be EQual if key is down. Not Equal if key not down.
ldr r6,=0x00FF00FF ; loads two (2) red pixels into r6
ldrne r6,=0x00000000 ;if the condition wasn't actually EQual, put two(2) black pixels into r6.
str r6,[r7] ; put the pixels that are in r6 into r7 (r7 has vram at 10,10)
b start ; jump (branch) back up to start
;;---STOP COPYING ---;;
Assemble that file!
NOTE : Now, to actually get output from this thing, you have to hold the key down.
Even though that was sort, checking for keypresses is important and deserves it's day.
Get that?
Day 5 may be a little while in the making because it will be about sprites.
Intro - Day 5
```
Patater GBAGuy Mirror
Contact | 931 | 3,207 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-50 | latest | en | 0.907731 |
http://www.cognigencorp.com/nonmem/nm/93nov171999.html | 1,560,956,246,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999000.76/warc/CC-MAIN-20190619143832-20190619165832-00441.warc.gz | 225,182,736 | 2,681 | From: Rebecca Wrishko <wrishko@unixg.ubc.ca>
Subject: Predicted Concetration derivations
Date: Wed, 17 Nov 1999 11:54:12 -0800
The assistance of the group would be greatly appreciated in solving the following question, how does NONMEM generate the predicted serum concentrations at the times for which there is a corresponding event record? Implementing Advan1 Trans2 (NMTRAN control stream below) and Advan3 Trans3 (NMTRAN control stream below) subroutines for the pk modeling of vancomycin in a small pediatric population, I was able to obtain final parameter estimates of CL, V and etas (One-Compartment) and Cl, V, Vss, and etas (Two-Compartment). I have attempted using the final PK parameter one- and two-compartment estimates of CL and Vss, obtained from the modeling routines to calculate expected drug concentrations and do not derive the same values as those generated from the NONMEM execution. How are the fixed (theta) and random (eta) parameters incorporated into the formulae for calculating the predicted serum concentrations?
Thank you
Rebecca
NM-TRAN 1:
\$SUBROUTINES ADVAN1 TRANS2; One Compartment Linear Model
\$PK
CL=THETA(1)*EXP(ETA(1)) ; mean clearance
V=THETA(2)*EXP(ETA(2)) ; mean central volume
K=CL/V ;reparameterization required
S1=V
\$THETA (0,4.) ; clearance estimate
(0,10.) ; volume estimate
\$OMEGA .04 .04 ; twenty percent cv
\$SIGMA .02 ; ten percent cv
\$ERROR
Y=F+ERR(1)
\$ESTIMATION MAXEVAL=450 SIGDIGITS=4
\$COVARIANCE
NM-TRAN 2:
\$SUBROUTINES ADVAN3 TRANS3; Two Compartment Linear Model
\$PK
CL=THETA(1)*EXP(ETA(1)) ; mean clearance
V=THETA(2)*EXP(ETA(2)) ; mean central volume
VSS=V+THETA(3)*EXP(ETA(3))
Q=THETA(4)
K=CL/V; reparameterization lines
K12=Q/V
K21=Q/(VSS-V)
S1=V
\$THETA (0,3)
(0,5)
(0,10)
(0,3)
\$OMEGA .25 .25 .25 ; fifty percent cv
\$SIGMA .04 ; twenty percent cv
\$ERROR
Y=F+ERR(1)
\$ESTIMATION MAXEVAL=450 SIGDIGITS=4
\$COVARIANCE
Rebecca Wrishko
Division of Clinical Pharmacy
Faculty of Pharmaceutical Sciences
University of British Columbia
Email: wrishko@unixg.ubc.ca
*****
Date: Wed, 17 Nov 1999 13:57:22 -0800 (PST)
From: ABoeckmann <alison@c255.ucsf.edu>
Subject: Re: Predicted Concetration derivations
I don't think the group can help with this. I'll respond only to the question about the 1st. control stream because the kinetics are so simple. Maybe this will be sufficient.
Because Rebecca included the reparameterization from CL and V to rate constant K, she does not need to use TRANS2. It simply repeats the reparameterization. She could save some run time using ADVAN1 TRANS1, but this should not affect the outcome of the run.
She does not show us her dosing records. Suppose there is only a single bolus dose AMT at T0. THen the calculation of A0, A1, A2 at event times T0, T1, T2, etc. is:
A0=AMT
A1=A0*EXP(-K*(T1-T0))
A2=A1*EXP(-K*(T2-T1))
The predictions PRED0, PRED1, PRED2, etc. at these event times are
PRED0=A0/S1
PRED1=A1/S1
PRED2=A2/S1
With the default method of estimation, etas are 0. In the above,
K=theta(1)/theta(2)
S1=theta(2)
If there are multiple bolus doses, they are added when appropriate. E.g., if there is a dose AMTi at time Ti, then Ai=Ai (from above)+AMTi
If there are infusions, the equations are more complicated.
I do not know why she cannot get the same values from "the modeling routines". But make sure that the thetas are the same. If REbecca is plugging in thetas derived from other routines into NONMEM, then DO NOT RUN THE ESTIMATION STEP!!
>From Rebecca:
NM-TRAN 1:
\$SUBROUTINES ADVAN1 TRANS2; One Compartment Linear Model
\$PK
CL=THETA(1)*EXP(ETA(1)) ; mean clearance
V=THETA(2)*EXP(ETA(2)) ; mean central volume
K=CL/V ;reparameterization required
S1=V
\$THETA (0,4.) ; clearance estimate
(0,10.) ; volume estimate
\$OMEGA .04 .04 ; twenty percent cv
\$SIGMA .02 ; ten percent cv
\$ERROR
Y=F+ERR(1)
\$ESTIMATION MAXEVAL=450 SIGDIGITS=4
\$COVARIANCE | 1,186 | 3,898 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-26 | latest | en | 0.651024 |
https://link.springer.com/article/10.1007/s10998-022-00479-1 | 1,670,010,201,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710916.40/warc/CC-MAIN-20221202183117-20221202213117-00151.warc.gz | 405,049,851 | 47,701 | ## Introduction
In this paper we assume that p and q are sequences of positive real numbers, $$\tau \in \mathbb {N}=\{0,1,2,\dots ,\}$$. We consider the linear fourth-order difference equation of the form
\begin{aligned} \varDelta ^4x(n)+p(n)\varDelta x(n+1)+q(n)x(n-\tau )=0, \end{aligned}
(E)
where $$n\in \mathbb {N}_{n_0}=\{n_0,n_0+1,\dots \}$$ and $$n_0 \in \mathbb {N}$$, $$\varDelta$$ is the forward difference operator defined by $$\varDelta x(n)=x(n+1)-x(n)$$. By a solution of equation (E) we understand a sequence x of real numbers, that is defined for $$n\ge n_0-\tau$$ and satisfies the equation for sufficiently large n. A solution is nonoscillatory if it is eventually positive or eventually negative. Otherwise, the solution is said to be oscillatory. We call equation (E) oscillatory if all its solutions are oscillatory.
The background of difference equations and discrete oscillation theory can be found in the monographs of Agarwal [1] and Agarwal, Bohner, Grace and O’Regan [2]. In recent years, the study of oscillatory and asymptotic behavior of solutions of second order difference equations has received great attention. Compared to this, the study of higher-order difference equations receives considerably less attention in literature. Some recent results for the oscillation of third-order difference equations can be found in [3, 6, 7, 9, 12, 13, 24, 25], and for fourth-order in [4, 10, 16, 18, 23]. In particular, in [7] and [13], the authors studied oscillation and asymptotic properties of solutions of the three-term difference equation
\begin{aligned} \varDelta ^3 x(n)+p(n)\varDelta x(n+1)-q(n)f\left( x(n-\tau )\right) =0, \end{aligned}
by assuming the coefficient of the damping term is nonnegative. In [12], the difference equation
\begin{aligned} \varDelta ^3 x(n)+p(n)\varDelta x(n+1)+q(n)x(n-\tau )=0 \end{aligned}
with positive coefficient p was investigated. In these papers the comparison theorems with suitable first-order difference equations were used. It seemed to be natural to continue the research in the case of higher-order equations especially since fourth-order differential and difference equations often occur as models in mathematical biology, economics and engineering (for example see [4, 5, 8, 26]).
In [15], the authors investigated the oscillatory behavior of solutions of the fourth order difference equations with damping
\begin{aligned} \varDelta ^4 u(n)+ p(n) \varDelta u(n+1)+q(n)u(s(n)) = 0, \end{aligned}
under the assumption that the auxiliary third order difference equation
\begin{aligned} \varDelta ^3 z(n)+ p(n)z(n+1) = 0 \end{aligned}
is nonoscillatory.
The purpose of this paper is to study the oscillation of equation (E). Since for three-terms equations it is difficult to introduce the classification of nonoscillatory solutions, first we prove two lemmas that allow us to rewrite equation (E) in equivalent form as a two-terms equation with quasidifferences. Then, in our main result we give sufficient conditions for the oscillation of all solutions of equation (E) in terms of the two associated first order delay difference equations. Applying the appropriate criteria to these delay difference equations allows us to obtain new, easily verifiable oscillation results for equation (E) (see Theorem 2.10). It is worth noticing, that in this criterion the explicit form of solutions of the auxiliary equations is not needed. Numerical methods are frequently used in the investigation of the properties of the solutions of differential and difference equations. However, such methods are difficult to apply when investigating the oscillation of solutions. Our results may be helpful in the interpretation of numerical solutions, see Examples 3.1 and 3.2.
Let us recall some results that will be used in the sequel.
### Lemma 1.1
[14, Corollary 7.6.1] Let a be a sequence of non-negative real numbers and let k be a positive integer such that
\begin{aligned} \sum \limits _{i=n-k}^{n-1}a(i)>0 \end{aligned}
for large n. Then the difference inequality
\begin{aligned} \varDelta x(n)+ a(n) x(n-k)\le 0, \end{aligned}
has an eventually positive solution if and only if the difference equation
\begin{aligned} \varDelta x(n)+ a(n) x(n-k)= 0, \end{aligned}
has an eventually positive solution.
The next result follows from [19, Theorem 1].
### Lemma 1.2
Consider the equation
\begin{aligned} \varDelta (c(n)\varDelta v(n))+b(n)v(n+1)=0, \end{aligned}
(1.1)
where cb are eventually positive sequences. If
\begin{aligned} \sum \limits _{n=1}^{\infty } b(n)\sum \limits _{j=1}^{n-1} \frac{1}{c(j)}<\infty , \end{aligned}
then for any real constant $$\beta$$ there exists a solution v of equation (1.1) which converges to $$\beta$$ at infinity.
### Lemma 1.3
[18] Consider the difference equation
\begin{aligned} \varDelta (a(n)\varDelta (b(n)\varDelta (c(n)\varDelta x(n))))+f(n,x(n))=0, \end{aligned}
(1.2)
where a, b, c are sequences of positive real numbers and $$f:\mathbb {N}\times \mathbb {R}\rightarrow \mathbb {R}$$.
Assume that
\begin{aligned} \sum \limits _{i=1}^{\infty }\frac{1}{a(i)}=\sum \limits _{i=1}^{\infty }\frac{1}{b(i)}=\sum \limits _{i=1}^{\infty }\frac{1}{c(i)}=\infty \end{aligned}
and $$u f(n,u)>0$$ for all $$u\ne 0$$, $$n\in \mathbb {N}$$ hold. Let x be an eventually positive solution of (1.2). Then exactly one of the following statements holds for all sufficiently large n:
1. (i)
$$x(n)>0$$, $$\varDelta x(n)>0$$, $$\varDelta (c(n)\varDelta x(n))>0$$ and $$\varDelta (b(n)(\varDelta (c(n)\varDelta x(n)))) >0$$;
2. (ii)
$$x(n)>0$$, $$\varDelta x(n)>0$$, $$\varDelta (c(n)\varDelta x(n))<0$$ and $$\varDelta (b(n)(\varDelta (c(n)\varDelta x(n))))>0$$.
## Main results
We start this section with two lemmas, which allow us to rewrite equation (E) in equivalent binomial form in terms of solutions of two auxiliary linear equations. Because the proofs of these lemmas are technically complicated, they are presented in Sect. 4.
### Lemma 2.1
If z is an eventually positive solution of equation
\begin{aligned} \varDelta ^3z(n)+p(n)z(n+1)=0, \end{aligned}
(2.1)
then the equality
\begin{aligned}&\varDelta ^4x(n)+p(n)\varDelta x(n+1)\nonumber \\&\quad =\varDelta \left[ \frac{1}{z(n+1)}\varDelta \left( z(n+1)z(n)\varDelta \left( \frac{\varDelta x(n)}{z(n)}\right) \right) \right] +\varDelta ^2z(n+1)\varDelta \left( \frac{\varDelta x(n+1)}{z(n+1)}\right) \nonumber \\ \end{aligned}
(2.2)
holds for any sequence x and for large n.
### Lemma 2.2
If z is an eventually positive solution of equation (2.1) and equation
\begin{aligned} \varDelta \left( \frac{1}{z(n+1)}\varDelta v(n)\right) +\frac{\varDelta ^2z(n+1)}{z(n+1)z(n+2)}v(n+1)=0, \end{aligned}
(2.3)
has an eventually positive solution v, then the equality
\begin{aligned}&\varDelta ^4x(n) + p(n)\varDelta x(n + 1) \nonumber \\&\quad =\frac{1}{v(n + 1)} \varDelta \left( \frac{v(n)v(n + 1)}{z(n + 1)}\varDelta \left( \frac{z(n)z(n + 1)}{v(n)}\varDelta \frac{\varDelta x(n)}{z(n)} \right) \right) \end{aligned}
(2.4)
holds for any sequence x and for large n .
Applying (2.4) to equation (E) we get
\begin{aligned} \frac{1}{v(n + 1)} \varDelta \left( \frac{v(n)v(n + 1)}{z(n + 1)}\varDelta \left( \frac{z(n)z(n + 1)}{v(n)}\varDelta \frac{\varDelta x(n)}{z(n)} \right) \right) +q(n)x(n-\tau )=0. \end{aligned}
Therefore, if z and v are eventually positive solutions of the equations (2.1) and (2.3) respectively, then, by Lemmas 2.1 and 2.2, equation (E) can be written in the form
The question arises whether equations (2.1) and (2.3) have eventually positive solutions. The lemma below presents a possible criterion.
### Lemma 2.3
Assume $$\beta , \gamma \in (0,\infty )$$ and
\begin{aligned} \sum _{n=n_0}^{\infty }n^2p(n)<\infty . \end{aligned}
(2.5)
Then there exist an eventually positive solution z of (2.1) and an eventually positive solution v of (2.3) such that
\begin{aligned} \lim _{n\rightarrow \infty }z(n)=\gamma \quad \text {and} \quad \lim \limits _{n\rightarrow \infty }v(n)=\beta . \end{aligned}
### Proof
By [22, Theorem 1] there exists a solution z of (2.1) such that $$\lim \nolimits _{n\rightarrow \infty }z(n)=\gamma$$. That means, that there exists an index $$n_1>n_0$$ such that $$z(n)>0$$ and satisfies equality (2.1) for any $$n\ge n_1$$. Hence, by (2.1), we have $$\varDelta ^3z(n)<0$$ for $$n\ge n_1$$. The convergence of z(n) implies
\begin{aligned} \lim _{n\rightarrow \infty }\varDelta z(n)=0 \quad \text {and}\quad \lim _{n\rightarrow \infty }\varDelta ^2z(n)=0. \end{aligned}
Since $$\varDelta ^3z(n)<0$$ for $$n\ge n_1$$, we get $$\varDelta ^2z(n)>0$$ for $$n\ge n_1$$. Analogously $$\varDelta z(n)<0$$ eventually.
By [20, Lemma 6] the series $$\sum _{k=1}^\infty \sum _{n=k}^\infty \varDelta ^2z(n+1)$$ is convergent. Using [20, Lemma 3] we get
\begin{aligned} \sum \limits _{n=n_1}^{\infty } n\varDelta ^2z(n+1)<\infty . \end{aligned}
(2.6)
Let, for $$n\ge n_1$$,
\begin{aligned} b(n)=\frac{\varDelta ^2z(n+1)}{z(n+1)z(n+2)}, \quad c(n)=\frac{1}{z(n+1)}. \end{aligned}
Then, by (2.6), we have
\begin{aligned} \sum \limits _{n=n_1}^{\infty } b(n)\sum \limits _{j=n_1}^{n-1} \frac{1}{c(j)}&= \sum \limits _{n=n_1}^{\infty } \frac{\varDelta ^2z(n+1)}{z(n+1)z(n+2)}\sum \limits _{j=n_1}^{n-1} z(j+1) \\&\le L\sum \limits _{n=n_1}^{\infty } n\varDelta ^2z(n+1) <\infty , \end{aligned}
where $$L=\gamma ^2z(n_1+1)$$. Hence, from Lemma 1.2, we get the existence of a solution v of (2.3) such that $$\lim \limits _{n\rightarrow \infty }v(n)=\beta$$. $$\square$$
Based on Trench [27], we say that a linear difference operator
\begin{aligned} L_m x(n)= r_{m-1}(n) \varDelta L_{m-1} x(n), \quad L_0x(n)=x(n) \end{aligned}
is in canonical form if $$\sum \nolimits _{n=n_0}^{\infty }\frac{1}{r_j(n)}=\infty$$ for $$j=1,\dots , m-1$$, where $$r_j$$ are eventually positive real sequences. The sequences $$L_ix$$ are called quasidifferences of x.
The quasidifferences in equation () have the form
\begin{aligned}&L_0x(n)=x(n), \,\, L_1x(n)=\frac{1}{z(n)}\varDelta L_0x(n), \,\, L_2x(n)=\frac{z(n)z(n+1)}{v(n)}\varDelta L_1x(n),\\&L_3x(n)=\frac{v(n)v(n+1)}{z(n+1)}\varDelta L_2x(n), \,\, L_4x(n)=\varDelta L_3x(n). \end{aligned}
Note, that if z and v are solutions of the equations (2.1) and (2.3) respectively, and both converge to positive constants, then
\begin{aligned} \sum _{n=n_0}^{\infty }z(n)=\infty , \quad \sum _{n=n_0}^{\infty }\frac{v(n)}{z(n)z(n+1)}=\infty ,\quad \sum _{n=n_0}^{\infty }\frac{z(n+1)}{v(n)v(n+1)}=\infty , \end{aligned}
and then the linear operator
$$L_4x(n)=\varDelta \left( \frac{v(n)v(n + 1)}{z(n + 1)}\varDelta \left( \frac{z(n)z(n + 1)}{v(n)}\varDelta \frac{\varDelta x(n)}{z(n)} \right) \right)$$
in () is in canonical form, which means equation () is in canonical form.
Hence, by Lemma 1.3, we get the following classification of the nonoscillatory solutions of ().
### Lemma 2.4
Assume that condition (2.5) is satisfied. Let x be an eventually positive solution of equation (). Then there exists $$n_2\ge n_0$$ such that for all $$n\ge n_2$$
\begin{aligned} L_4x(n)<0, \end{aligned}
(2.7)
and either
\begin{aligned} L_1x(n)>0, \quad L_2x(n)<0,\quad L_3x(n)>0, \end{aligned}
or
\begin{aligned} L_1x(n)>0, \quad L_2x(n)>0,\quad L_3x(n)>0,. \end{aligned}
In summary, we get the following remark.
### Remark 2.5
If condition (2.5) holds, then the three-term difference equation (E) can be rewritten as a two-terms equation of the form (), which is in the canonical form.
Our goal is to present an easily verifiable oscillation criterion for equation (E). In our investigation we utilize the form (). Using the comparison theorem we will deduce the oscillation of (E) from the oscillation of certain first-order difference equations whose properties are well-explored.
Let z be a solution to equation (2.1) which tends to a positive constant and let v be a solution to equation (2.3) which tends to a positive constant. Then there exist positive constants $$z^{*}$$, $$z^{**}$$, $$v^{*}$$, $$v^{**}$$ and $$n_3 \in \mathbb {N}$$ such that
\begin{aligned} 0<z^{*} \le z(n) \le z^{**} \quad \text {and} \quad 0<v^{*} \le v(n) \le v^{**} \end{aligned}
(2.8)
for $$n \ge n_3$$.
Assuming
\begin{aligned} \sum \limits _{n=1}^{\infty }\sum \limits _{k=n}^{\infty }q(k)< \infty \end{aligned}
(2.9)
we define sequences $$K_1, K_2$$ by
\begin{aligned}&K_1(n)=\left( \frac{v^{*}}{v^{**}}\right) ^2 \left( \frac{z^{*}}{z^{**}}\right) ^2(n-\tau -n_4) \sum \limits _{s=n}^{\infty }\sum \limits _{k=s}^{\infty }q(k) \\&K_2(n)= \left( \frac{v^{*}}{v^{**}}\right) ^2 \left( \frac{z^{*}}{z^{**}}\right) ^2 q(n) \sum \limits _{k=n_4}^{n-\tau -1}\sum \limits _{m=n_4}^{k-1}\sum \limits _{s=n_4}^{m-1}1 \end{aligned}
where $$n_4=\max \{n_2,n_3\}+\tau +3$$ and $$n_2$$ as in Lemma 2.4. It is easy to see that condition (2.9) is equivalent to
\begin{aligned} \sum \limits _{n=1}^{\infty }nq(k)< \infty . \end{aligned}
(2.10)
### Theorem 2.6
Assume that conditions (2.5) and (2.10) hold. If the following delayed equations with unknown sequence u
\begin{aligned} \varDelta u(n)+K_1(n)u(n-\tau )=0 \end{aligned}
(2.11)
and
\begin{aligned} \varDelta u(n)+K_2(n)u(n-\tau )=0 \end{aligned}
(2.12)
are oscillatory, then equation (E) is also oscillatory.
### Proof
Assume to the contrary that there exists an eventually positive solution x to (E). Let x be such a solution. Notice that x is also an eventually positive solution to (). Let z be a positive decreasing solution of (2.1). According to Lemma 2.4 we consider two cases.
Case I. Assume conditions (2.7) and (2.8) hold. Summation of the both sides of () from $$s\ge n_4$$ to infinity, leads to equality
\begin{aligned} \sum \limits _{i=s}^{\infty }\varDelta \left( \frac{v(i)v(i + 1)}{z(i + 1)}\varDelta \left( \frac{z(i)z(i + 1)}{v(i)}\varDelta \frac{\varDelta x(i)}{z(i)} \right) \right) =-\sum \limits _{i=s}^{\infty }v(i+1)q(i)x(i-\tau ). \end{aligned}
From the above, by properties of the sequence $$L_3x$$, we get
\begin{aligned} \frac{v(s)v(s + 1)}{z(s + 1)}\varDelta \left( \frac{z(s)z(s + 1)}{v(s)}\varDelta \frac{\varDelta x(s)}{z(s)} \right) \ge \sum \limits _{i=s}^{\infty }v(i+1)q(i)x(i-\tau ). \end{aligned}
Since the sequence $$L_1x$$ is a positive sequence, x is increasing,
\begin{aligned} \frac{v(s)v(s + 1)}{z(s + 1)}\varDelta \left( \frac{z(s)z(s + 1)}{v(s)}\varDelta \frac{\varDelta x(s)}{z(s)} \right) > x(s-\tau )\sum \limits _{i=s}^{\infty }v(i+1)q(i), \end{aligned}
and consequently
\begin{aligned} \varDelta \left( \frac{z(s)z(s+1)}{v(s)}\varDelta \frac{\varDelta x(s)}{z(s)}\right) > \frac{z(s+1)}{v(s)v(s+1)}x(s-\tau )\sum \limits _{i=s}^{\infty }v(i+1)q(i). \end{aligned}
Summing the above inequality from n to infinity, we obtain
\begin{aligned} \sum \limits _{s=n}^{\infty }\varDelta \left( \frac{z(s)z(s+1)}{v(s)}\varDelta \frac{\varDelta x(s)}{z(s)}\right) > \sum \limits _{s=n}^{\infty }\frac{z(s+1)}{v(s)v(s+1)}x(s-\tau )\sum \limits _{i=s}^{\infty }v(i+1)q(i). \end{aligned}
Again by (2.8),
\begin{aligned} \sum \limits _{s=n}^{\infty }\varDelta \left( \frac{z(s)z(s+1)}{v(s)}\varDelta \frac{\varDelta x(s)}{z(s)}\right) > x(n-\tau )\sum \limits _{s=n}^{\infty }\frac{z(s+1)}{v(s)v(s+1)}\sum \limits _{i=s}^{\infty }v(i+1)q(i). \end{aligned}
By properties of the sequence $$L_2x$$, we have
\begin{aligned} \sum \limits _{s=n}^{\infty }\varDelta \left( \frac{z(s)z(s+1)}{v(s)}\varDelta \frac{\varDelta x(s)}{z(s)}\right) \le - \frac{z(n)z(n+1)}{v(n)}\varDelta \frac{\varDelta x(n)}{z(n)}. \end{aligned}
Combining the above two inequalities, we obtain
\begin{aligned} - \frac{z(n)z(n+1)}{v(n)}\varDelta \frac{\varDelta x(n)}{z(n)} \ge x(n-\tau )\sum \limits _{s=n}^{\infty }\frac{z(s+1)}{v(s)v(s+1)}\sum \limits _{i=s}^{\infty }v(i+1)q(i), \end{aligned}
that is
\begin{aligned} - \varDelta \frac{\varDelta x(n)}{z(n)} \ge x(n-\tau )\frac{v(n)}{z(n)z(n+1)}\sum \limits _{s=n}^{\infty }\frac{z(s+1)}{v(s)v(s+1)}\sum \limits _{i=s}^{\infty }v(i+1)q(i). \end{aligned}
Hence, using (2.8) we get
\begin{aligned} - \varDelta \frac{\varDelta x(n)}{z(n)} \ge x(n-\tau )\left( \frac{v^{*}}{v^{**}}\right) ^2\frac{z^{*}}{(z^{**})^2}\sum \limits _{s=n}^{\infty }\sum \limits _{i=s}^{\infty }q(i). \end{aligned}
(2.13)
Based on the sequence $$L_1x$$ properties, we have
\begin{aligned} x(n)> \sum \limits _{i=n_4}^{n-1}\varDelta x(i)=\sum \limits _{i=n_4}^{n-1}z(i)\frac{\varDelta x(i)}{z(i)}> \frac{\varDelta x(n-1)}{z(n-1)}\sum \limits _{i=n_4}^{n-1}z(i) >\frac{\varDelta x(n)}{z(n)}\sum \limits _{i=n_4}^{n-1}z(i) \end{aligned}
for $$n > n_4$$. Therefore
\begin{aligned} x(n-\tau )>\frac{\varDelta x(n-\tau )}{z(n-\tau )} \sum \limits _{i=n_4}^{n-\tau -1}z(i) \text{ for } n \ge n_4+ \tau . \end{aligned}
(2.14)
Combining (2.13) with (2.14), we get
\begin{aligned} - \varDelta \frac{\varDelta x(n)}{z(n)} > \frac{\varDelta x(n-\tau )}{z(n-\tau )}\left( \frac{v^{*}}{v^{**}}\right) ^2 \left( \frac{z^{*}}{z^{**}}\right) ^2(n-\tau -n_4) \sum \limits _{s=n}^{\infty }\sum \limits _{k=s}^{\infty }q(k), \end{aligned}
that is
\begin{aligned} 0 > \varDelta \frac{\varDelta x(n)}{z(n)} + \left( \frac{v^{*}}{v^{**}}\right) ^2 \left( \frac{z^{*}}{z^{**}}\right) ^2(n-\tau -n_4) \sum \limits _{s=n}^{\infty }\sum \limits _{k=s}^{\infty }q(k) \frac{\varDelta x(n-\tau )}{z(n-\tau )} \end{aligned}
for $$n\ge n_4+\tau$$. Using the definition of $$K_1$$, this leads to inequality
\begin{aligned} 0 >\varDelta \frac{\varDelta x(n)}{z(n)} +K_1(n) \frac{\varDelta x(n-\tau )}{z(n-\tau )} \text{ for } n\ge n_4+ \tau . \end{aligned}
(2.15)
By assumption, the sequence x is an eventually positive solution of inequality (2.15). By virtue of Lemma 1.1, it is also an eventually positive solution of equation (2.11). This contradicts that all solutions of (2.11) are oscillatory.
Case II. Set $$u(n):=L_3x(n)$$. Let conditions (2.7) and (2.8) hold. Thus the the sequence $$L_2x$$ is positive for $$n\ge n_4$$. Hence, the following estimation holds
\begin{aligned} \sum \limits _{s=n_4}^{m-1}\varDelta \left( \frac{z(s)z(s+1)}{v(s)}\varDelta \frac{\varDelta x(s)}{z(s)}\right) \le \frac{z(m)z(m+1)}{v(m)}\varDelta \frac{\varDelta x(m)}{z(m)} \text{ for } n \ge n_4. \end{aligned}
From the properties of the sequence $$L_3x$$, we have
\begin{aligned} \sum \limits _{s=n_4}^{m-1}\varDelta \left( \frac{z(s)z(s+1)}{v(s)}\varDelta \frac{\varDelta x(s)}{z(s)}\right) \ge u(m-1)\sum \limits _{s=n_4}^{m-1}\frac{z(s+1)}{v(s)v(s+1)}. \end{aligned}
Combining the above two inequalities, we get
\begin{aligned} \frac{v(m)}{z(m)z(m+1)}u(m-1)\sum \limits _{s=n_4}^{m-1}\frac{z(s+1)}{v(s)v(s+1)} \le \varDelta \frac{\varDelta x(m)}{z(m)}. \end{aligned}
Summing both sides of the above inequality from $$n_4$$ to $$k-1$$, we obtain
\begin{aligned} \sum \limits _{m=n_4}^{k-1} \frac{v(m)u(m-1)}{z(m)z(m+1)}\sum \limits _{s=n_4}^{m-1} \frac{z(s+1)}{v(s)v(s+1)} \le \sum \limits _{m=n_4}^{k-1} \varDelta \frac{\varDelta x(m)}{z(m)}. \end{aligned}
(2.16)
Since the sequence $$L_1x$$ is positive for $$n\ge n_4$$, we have
\begin{aligned} \sum \limits _{m=n_4}^{k-1}\varDelta \left( \frac{\varDelta x(m)}{z(m)}\right) =\frac{\varDelta x(k)}{z(k)}-\frac{\varDelta x(n_4)}{z(n_4)}\le \frac{\varDelta x(k)}{z(k)}. \end{aligned}
(2.17)
On the other hand, again by (2.8), we get
\begin{aligned}&\sum \limits _{m=n_4}^{k-1}\frac{v(m)u(m-1)}{z(m)z(m+1)}\sum \limits _{s=n_4}^{m-1}\frac{z(s+1)}{v(s)v(s+1)}\\&\quad \ge u(k-2)\sum \limits _{m=n_4}^{k-1}\frac{v(m)}{z(m)z(m+1)}\sum \limits _{s=n_4}^{m-1}\frac{z(s+1)}{v(s)v(s+1)}. \end{aligned}
The above and inequalities (2.16), (2.17) imply
\begin{aligned} \varDelta x(k)\ge z(k)u(k-2)\sum \limits _{m=n_4}^{k-1}\frac{v(m)}{z(m)z(m+1)}\sum \limits _{s=n_4}^{m-1}\frac{z(s+1)}{v(s)v(s+1)}. \end{aligned}
By summation of both sides of the above inequality from $$n_4$$ to $$n-1$$, we obtain
\begin{aligned} \sum \limits _{k=n_4}^{n-1}\varDelta x(k) \ge \sum \limits _{k=n_4}^{n-1}z(k)u(k-2) \sum \limits _{m=n_4}^{k-1} \frac{v(m)}{z(m)z(m+1)}\sum \limits _{s=n_4}^{m-1}\frac{z(s+1)}{v(s)v(s+1)}. \end{aligned}
Sequence x is positive for $$n\ge n_4$$ hence $$\sum \nolimits _{k=n_4}^{n-1}\varDelta x(k)=x(n)-x(n_4)\le x(n)$$. Consequently
\begin{aligned} x(n) \ge \sum \limits _{k=n_4}^{n-1} z(k) u(k-2) \sum \limits _{m=n_4}^{k-1} \frac{v(m)}{z(m)z(m+1)} \sum \limits _{s=n_4}^{m-1} \frac{z(s+1)}{v(s)v(s+1)}. \end{aligned}
Since the sequence $$L_3x$$ is decreasing,
\begin{aligned} x(n) \ge u(n-2) \sum \limits _{k=n_4}^{n-1} z(k)\sum \limits _{m=n_4}^{k-1} \frac{v(m)}{z(m)z(m+1)} \sum \limits _{s=n_4}^{m-1} \frac{z(s+1)}{v(s)v(s+1)}. \end{aligned}
Hence,
\begin{aligned}&v(n+1)q(n)x(n-\tau )\\&\quad \ge v(n+1)q(n)u(n-\tau -2) \sum \limits _{k=n_4}^{n-\tau -1}z(k)\sum \limits _{m=n_4}^{k-1}\frac{v(m)}{z(m)z(m+1)}\sum \limits _{s=n_4}^{m-1}\frac{z(s+1)}{v(s)v(s+1)}. \end{aligned}
Using the definition of u, equation () takes the form
\begin{aligned} -\varDelta u(n)=v(n+1)q(n)x(n-\tau ). \end{aligned}
We get
\begin{aligned} -\varDelta u(n)\ge v(n+1)q(n)u(n-\tau -2) \!\! \sum \limits _{k=n_4}^{n-\tau -1}\!\!z(k)\!\!\sum \limits _{m=n_4}^{k-1}\!\!\frac{v(m)}{z(m)z(m+1)}\!\!\sum \limits _{s=n_4}^{m-1}\!\!\frac{z(s+1)}{v(s)v(s+1)}. \end{aligned}
Hence, using (2.8) we get
\begin{aligned} -\varDelta u(n)\ge u(n-\tau -2)\left( \frac{v^{*}}{v^{**}}\right) ^2 \left( \frac{v^{*}}{v^{**}}\right) ^2 q(n) \sum \limits _{k=n_4}^{n-\tau -1}\sum \limits _{m=n_4}^{k-1}\sum \limits _{s=n_4}^{m-1}1. \end{aligned}
Finally, using the definition of $$K_2$$, we obtain
\begin{aligned} \varDelta u(n)+u(n-\tau -2) K_2(n)\le 0 \text{ for } n\ge n_4. \end{aligned}
Since u is an eventually positive solution of the above inequality, by Lemma 1.1 it is also an eventually positive solution of equation (2.12). In that way there is contradiction with the assumption that all solutions of equation (2.12) are oscillatory. $$\square$$
Applying well-known oscillation criteria for first-order delay difference equations to equations (2.11) and (2.12) (see [11] and [17]), we obtain the following oscillation criteria for equation (E).
### Corollary 2.7
Assume (2.5) and (2.10) are satisfied. If for $$j=1,2$$
\begin{aligned} \limsup _{n\rightarrow \infty }\sum \limits _{i=n-\tau }^{n}K_j(i)>1, \end{aligned}
then equation (E) is oscillatory.
If $$\tau >0$$, then we can also use the following criterion.
### Corollary 2.8
Assume (2.5) and (2.9) are satisfied. If for $$j=1,2$$
\begin{aligned} \liminf _{n\rightarrow \infty }\left( \frac{1}{\tau }\sum \limits _{i=n-\tau }^{n-1}K_j(i)\right) >\frac{\tau ^{\tau }}{(\tau +1)^{\tau +1}}, \end{aligned}
then equation (E) is oscillatory.
Note, that in the above results the explicit form of the eventually positive solutions of (2.1) and (2.3) are needed. But, it is well known, that it is difficult to find the explicit form of solutions of second and third order linear difference equations with variable coefficients. Therefore, we present a criterion in which the assumptions depend only on the coefficients p and q of the equation (E). First we prove a simple lemma.
### Lemma 2.9
Assume (2.5), (2.10), and
\begin{aligned} \lim \limits _{n \rightarrow \infty } n^3q(n)=\infty . \end{aligned}
(2.18)
Then $$\lim \nolimits _{n\rightarrow \infty }K_j(i)=\infty$$ for $$j=1,2$$.
### Proof
Let $$j=1$$. Define a sequence Q by
\begin{aligned} Q(n)= \sum \limits _{s=n}^{\infty }\sum \limits _{k=s}^{\infty }q(k). \end{aligned}
Using discrete L’Hospital’s rule we obtain
\begin{aligned}&\lim _{n\rightarrow \infty }nQ(n)= \lim _{n\rightarrow \infty } \frac{Q(n)}{n^{-1}}= \lim _{n\rightarrow \infty }\frac{\varDelta Q(n)}{\varDelta (n^{-1})} = \lim _{n\rightarrow \infty }\frac{-\sum _{k=n}^{\infty } q(k)}{-(n(n+1))^{-1}}\nonumber \\&\quad = \lim _{n\rightarrow \infty }\frac{-\varDelta \left( \sum _{k=n}^{\infty } q(k)\right) }{-\varDelta \left( (n(n+1))^{-1}\right) }= \lim _{n\rightarrow \infty }\frac{q(n)}{2(n(n+1)(n+2))^{-1}}. \end{aligned}
Since $$\lim \nolimits _{n \rightarrow \infty } n^3q(n)=\infty$$ we get $$\lim \nolimits _{n\rightarrow \infty } nQ(n)= \infty$$. Hence
\begin{aligned} \lim _{n\rightarrow \infty }K_1(n)=\infty . \end{aligned}
Let $$j=2$$. It is easy to verify that
\begin{aligned} \sum \limits _{k=n_4}^{n-\tau -1}\sum \limits _{m=n_4}^{k-1}\sum \limits _{s=n_4}^{m-1}1&= \sum \limits _{k=n_4}^{n-\tau -1}\sum \limits _{m=n_4}^{k-1}(m-n_4) = \sum \limits _{k=n_4}^{n-\tau -1}\sum \limits _{j=0}^{k-1-n_4}j \\&= \sum \limits _{k=n_4}^{n-\tau -1}\frac{(k-n_4)(k-n_4-1)}{2}=\frac{1}{2}\sum \limits _{i=0}^{n-\tau -n_4-1} i(i-1) \\&=\frac{1}{6}(n-\tau -n_4)(n-\tau -n_4-1)(n-\tau -n_4-2)\\&\ge \frac{1}{6}(n-\tau -n_4-2)^3. \end{aligned}
Since $$\lim \nolimits _{n \rightarrow \infty } n^3q(n)=\infty$$ we get $$\lim \limits _{n\rightarrow \infty } K_2(n)= \infty$$. $$\square$$
Combining Lemma 2.9 and Corollary 2.7 we get the following oscillation criterion.
### Theorem 2.10
Assume (2.5), (2.10), and (2.18). Then equation (E) is oscillatory.
To verify conditions (2.5) and (2.10) the tests presented in [21] may be useful. For example, applying [21, Lemma 4.4] or [21, Lemma 4.5] we get the following results
### Corollary 2.11
Assume (2.18) and
\begin{aligned} \limsup _{n\rightarrow \infty }\frac{\log q(n)}{\log n}<-2, \quad \limsup _{n\rightarrow \infty }\frac{\log p(n)}{\log n}<-3. \end{aligned}
Then equation (E) is oscillatory.
### Corollary 2.12
Assume (2.18) and
\begin{aligned} \liminf _{n\rightarrow \infty }n\left( \frac{q_n}{q_{n+1}}-1\right)>2, \quad \liminf _{n\rightarrow \infty }n\left( \frac{p_n}{p_{n+1}}-1\right) >3. \end{aligned}
Then equation (E) is oscillatory.
## Examples
By using computer algebra systems there is always possible to find recursively an approximate solution of the considered equation, but sometimes it is very difficult to determine whether the approximate solution is oscillatory or not. In many cases, it is rather easy to use our last criterion to verify that the considered equation is oscillatory. Such cases are shown in the following examples.
### Example 3.1
Consider the following equation of type (E):
\begin{aligned} \!\varDelta ^4x(n)\!+\!\frac{12}{n^2(n+1)\ln ^2(n+1)}\varDelta x(n+1)\!+\!\frac{(0.01+\sin ^2n){\root 10 \of {n}}}{n^3}x(n\!-\!1)\!=\!0, \end{aligned}
(3.1)
where $$n\ge 2$$,
\begin{aligned} p(n)=\frac{12}{n^2(n+1)\ln ^2(n+1)}, \quad q(n)=\frac{(0.01+\sin ^2n){\root 10 \of {n}}}{n^3}, \quad \tau =1. \end{aligned}
Using the iterative scheme
\begin{aligned} x(n+4)= 4 x(n+3) -\big (p(n)+6 \big ) x(n+2)+ \big (p(n)+4 \big ) x(n+1)-x(n)-q(n)x(n-1), \end{aligned}
we solve (3.1) with the following initial conditions
\begin{aligned} x(0)=1.0, \; x(1)=0.2, \; x(2)=1.0, \;x(3)=0.3,\;x(4)=4.0. \end{aligned}
Except for the initial conditions all computed values of x are rounded to the nearest integer number. As we see in Table 1, the obtained terms of this solution are positive until the 88th term, then negative. The next sign change of terms is on 346th term, next on 900th term and so on. Since the length of intervals with one sign are increasing, the oscillations of the solutions are not easily visible.
On the other hand, it is easy to verified that
\begin{aligned}&\sum \limits _{n=1}^{\infty }n^2 \frac{12}{n^2(n+1)\ln ^2(n+1)}< \infty , \\&\sum \limits _{n=1}^{\infty }n \frac{\sin ^2n}{n^3}< \infty , \\&\lim \limits _{n \rightarrow \infty }n^3\frac{(0.01+\sin ^2n){\root 10 \of {n}}}{n^3}=\infty . \end{aligned}
Hence, by Theorem 2.10 all solutions of equation (3.1) are oscillatory.
### Example 3.2
Consider fourth-order trinomial difference equations of the form
\begin{aligned} \varDelta ^4x(n)+\frac{a}{n^{\alpha }}\varDelta x(n+1)+\frac{b}{n^{\beta }}x(n-\tau )=0,\quad n\ge 2, \end{aligned}
(3.2)
where $$a,b, \alpha , \beta >0$$, $$\tau \in \mathbb {N}$$. It is easy to see that if $$\alpha >3$$ and $$2<\beta <3$$, then conditions (2.5), (2.9) and (2.18) are satisfied. Hence, by Theorem 2.10 all solutions of equation (3.2) are oscillatory.
For a numerical solution of (3.2), we set $$a = 10, \;b = 5, \tau = 2, \alpha = 5, \beta =2,5$$. The recurrence formula for the equation takes the form
\begin{aligned} x(n+4)\!= & {} \!4x(n+3)\!-\!6x(n+2)+4x(n+1)\!-\!x(n)\!-\!\frac{10}{n^{5}}\big (x(n+2)\!-\!x(n+1)\big )\!\\&-\,\!\frac{5}{n^{2.5}}x(n-2). \end{aligned}
Taking the initial values
\begin{aligned} x(0)=1, x(1)=2, x(2)=3, x(3)=4,x(4)=5, x(5)=6 \end{aligned}
we get a solution x whose trajectory from $$n=1$$ to $$n=150$$ is shown in Fig. 1. To show the oscillatory nature of this solution, we present its graph divided into four parts with increasingly larger scales.
## Proofs of Lemmas
### Proof of Lemma 2.1
Let z be an eventually positive solution of equation (2.1). From the left hand side of (2.2), using (2.1), we obtain
\begin{aligned}&\varDelta ^4x(n)+p(n)\varDelta x(n+1)=\varDelta ^4x(n)-\frac{\varDelta ^3 z(n)}{z(n+1)}\varDelta x(n+1)\\&\quad =\left( \varDelta ^4x(n)-\varDelta ^2z(n+1)\varDelta \frac{\varDelta x(n+1)}{z(n+1)}-\varDelta ^3 z(n)\frac{\varDelta x(n+1)}{z(n+1)}\right) +\varDelta ^2z(n+1)\varDelta \frac{\varDelta x(n+1)}{z(n+1)}\\&\quad =\varDelta \left( \varDelta ^3x(n)-\varDelta ^2z(n)\frac{\varDelta x(n+1)}{z(n+1)}\right) +\varDelta ^2z(n+1)\varDelta \frac{\varDelta x(n+1)}{z(n+1)})\\&\quad =\varDelta \left( \frac{1}{z(n+1)}\big (z(n+1)\varDelta ^3x(n)-\varDelta x(n+1)\varDelta ^2 z(n)\big )\right) +\varDelta ^2z(n+1)\varDelta \frac{\varDelta x(n+1)}{z(n+1)}\\&\quad =\varDelta \left( \frac{1}{z(n+1)}\big (z(n+1)\varDelta ^2x(n+1)-z(n+1)\varDelta ^2 x(n)\big )\right) \\&\qquad + \varDelta \left( \frac{1}{z(n+1)}\big ( -\varDelta x(n+1)\varDelta z(n+1)+\varDelta x(n+1)\varDelta z(n)\big )\right) \\&\qquad +\varDelta ^2z(n+1)\varDelta \frac{\varDelta x(n+1)}{z(n+1)}\\&\quad =\varDelta \left( \frac{1}{z(n+1)}\varDelta \big (z(n)\varDelta x(n+1)-z(n+1)\varDelta x(n) \big )\right) +\varDelta ^2z(n+1)\varDelta \frac{\varDelta x(n+1)}{z(n+1)}\\&\quad =\varDelta \left( \frac{1}{z(n+1)}\varDelta \left( z(n+1)z(n)\left( \frac{\varDelta x(n+1)}{z(n+1)}-\frac{\varDelta x(n)}{z(n)}\right) \right) \right) +\varDelta ^2z(n+1)\varDelta \frac{\varDelta x(n+1)}{z(n+1)}\\&\quad =\varDelta \left( \frac{1}{z(n+1)}\varDelta \left( z(n+1)z(n)\varDelta \frac{\varDelta x(n)}{z(n)}\right) \right) +\varDelta ^2z(n+1)\varDelta \frac{\varDelta x(n+1)}{z(n+1)}. \end{aligned}
### Proof of Lemma 2.2
We start from the right hand side of (2.4):
\begin{aligned}&\frac{1}{v(n+1)}\varDelta \left( \frac{v(n)v(n+1)}{z(n+1)}\varDelta \left( \frac{1}{v(n)}z(n)z(n+1)\varDelta \frac{\varDelta x(n)}{z(n)}\right) \right) \\&\quad =\frac{1}{v(n+1)}\varDelta \left( \frac{v(n)v(n+1)}{z(n+1)}\left( \frac{1}{v(n+1)}\varDelta \left( z(n)z(n+1)\varDelta \frac{\varDelta x(n)}{z(n)}\right) \right. \right. \\&\qquad + \left. \left. \left( \varDelta \frac{1}{v(n)}\right) z(n)z(n+1)\varDelta \frac{\varDelta x(n)}{z(n)}\right) \right) =\frac{1}{v(n+1)}\varDelta \left( \frac{v(n)}{z(n+1)}\varDelta \left( z(n)z(n+1)\varDelta \frac{\varDelta x(n)}{z(n)}\right) \right. \\&\qquad + \left. \frac{v(n)v(n+1)}{z(n+1)}\left( \varDelta \frac{1}{v(n)}\right) z(n)z(n+1)\varDelta \frac{\varDelta x(n)}{z(n)}\right) \\&\quad =\frac{1}{v(n+1)}\varDelta \left( \frac{v(n)}{z(n+1)}\varDelta \left( z(n)z(n+1)\varDelta \frac{\varDelta x(n)}{z(n)}\right) -\frac{\varDelta v(n)}{z(n+1)}z(n)z(n+1)\varDelta \frac{\varDelta x(n)}{z(n)}\right) \\&\quad =\frac{1}{v(n+1)} \left( v(n+1)\varDelta \left( \frac{1}{z(n+1)}\varDelta \left( z(n)z(n+1)\varDelta \frac{\varDelta x(n)}{z(n)}\right) \right) \right. \\&\qquad +\frac{\varDelta v(n)}{z(n+1)}\varDelta \left( z(n)z(n+1)\varDelta \frac{\varDelta x(n)}{z(n)}\right) -\frac{\varDelta v(n+1)}{z(n+2)}\varDelta \left( z(n)z(n+1)\varDelta \frac{\varDelta x(n)}{z(n)}\right) \\&\qquad \left. -\varDelta \left( \frac{\varDelta v(n)}{z(n+1)}\right) z(n)z(n+1)\varDelta \frac{\varDelta x(n)}{z(n)}\right) \\&\quad =\varDelta \left( \frac{1}{z(n+1)}\varDelta \left( z(n)z(n+1)\varDelta \frac{\varDelta x(n)}{z(n)}\right) \right) \\&\qquad +\frac{\varDelta v(n)}{z(n+1)v(n+1)}\varDelta \left( z(n)z(n+1)\varDelta \frac{\varDelta x(n)}{z(n)}\right) \\&\qquad -\frac{\varDelta v(n+1)}{v(n+1)z(n+2)}\varDelta \left( z(n)z(n+1)\varDelta \frac{\varDelta x(n)}{z(n)}\right) -\varDelta \left( \frac{\varDelta v(n)}{z(n+1)}\right) \frac{z(n)z(n+1)}{v(n+1)}\varDelta \frac{\varDelta x(n)}{z(n)}. \end{aligned}
Applying (2.2) to the first term of the last equality and combining together the second and third term we obtain
\begin{aligned}&\frac{1}{v(n+1)}\varDelta \left( \frac{v(n)v(n+1)}{z(n+1)}\varDelta \left( \frac{1}{v(n)}z(n)z(n+1)\varDelta \frac{\varDelta x(n)}{z(n)}\right) \right) \\&\quad =\varDelta ^4x(n)+p(n)\varDelta x(n+1)-\varDelta ^2z(n+1)\varDelta \frac{\varDelta x(n+1)}{z(n+1)} \\&\qquad -\frac{1}{v(n+1)}\varDelta \left( \frac{\varDelta v(n)}{z(n+1)}\right) \varDelta \left( z(n)z(n+1)\varDelta \frac{\varDelta x(n)}{z(n)}\right) \\&\qquad -\varDelta \left( \frac{\varDelta v(n)}{z(n+1)}\right) \frac{z(n)z(n+1)}{v(n+1)}\varDelta \frac{\varDelta x(n)}{z(n)}. \end{aligned}
Now we transform separately the third term of the above expression
\begin{aligned}&-\,\frac{1}{v(n+1)}\varDelta \left( \frac{\varDelta v(n)}{z(n+1)}\right) \varDelta \left( z(n)z(n+1)\varDelta \frac{\varDelta x(n)}{z(n)}\right) =-\frac{1}{v(n+1)}\varDelta \left( \frac{\varDelta v(n)}{z(n+1)}\right) \\&\quad \times \left( \varDelta \left( \frac{\varDelta x(n+1)}{z(n+1)}\right) \varDelta (z(n)z(n+1))+z(n)z(n+1)\varDelta ^2\frac{\varDelta x(n)}{z(n)}\right) \\&\quad =-\frac{1}{v(n+1)}\varDelta \left( \frac{\varDelta v(n)}{z(n+1)}\right) \left( (z(n+1)z(n+2)-z(n)z(n+1))\varDelta \frac{\varDelta x(n+1)}{z(n+1)}\right. \\&\qquad \left. +z(n)z(n+1)\varDelta ^2\frac{\varDelta x(n)}{z(n)}\right) \\&\quad =-\frac{z(n+1)z(n+2)}{v(n+1)}\varDelta \frac{\varDelta v(n)}{z(n+1)}\varDelta \frac{\varDelta x(n+1)}{z(n+1)} \\&\qquad +\frac{z(n)z(n+1)}{v(n+1)}\varDelta \left( \frac{\varDelta v(n)}{z(n+1)}\right) \varDelta \left( \frac{\varDelta x(n+1)}{z(n+1)}\right) \\&\qquad -\frac{z(n)z(n+1)}{v(n+1)}\varDelta \left( \frac{\varDelta v(n)}{z(n+1)}\right) \varDelta ^2 \left( \frac{\varDelta x(n)}{z(n)}\right) . \end{aligned}
Hence, we obtain
\begin{aligned}&\varDelta ^4x(n)+p(n)\varDelta x(n+1)-\varDelta ^2z(n+1)\varDelta \left( \frac{\varDelta x(n+1)}{z(n+1)}\right) \\&\qquad -\frac{z(n+1)z(n+2)}{v(n+1)}\varDelta \left( \frac{\varDelta v(n)}{z(n+1)}\right) \varDelta \left( \frac{\varDelta x(n+1)}{z(n+1)}\right) \\&\qquad -\frac{z(n)z(n+1)}{v(n+1)}\varDelta \left( \frac{\varDelta v(n)}{z(n+1)}\right) \varDelta ^2 \left( \frac{\varDelta x(n)}{z(n)}\right) \\&\qquad +\left( \frac{z(n)z(n+1)}{v(n+1)}\varDelta \left( \frac{\varDelta v(n)}{z(n+1)}\right) \varDelta \left( \frac{\varDelta x(n+1)}{z(n+1)}\right) \right. \\&\qquad \left. -\varDelta \left( \frac{\varDelta v(n)}{z(n+1)}\right) \frac{z(n)z(n+1)}{v(n+1)}\varDelta \left( \frac{\varDelta x(n)}{z(n)}\right) \right) \\&\quad = \varDelta ^4x(n)+p(n)\varDelta x(n+1)-\varDelta ^2z(n+1)\varDelta \left( \frac{\varDelta x(n+1)}{z(n+1)}\right) \\&\qquad -\frac{z(n+1)z(n+2)}{v(n+1)}\varDelta \left( \frac{\varDelta v(n)}{z(n+1)}\right) \varDelta \left( \frac{\varDelta x(n+1)}{z(n+1)}\right) \\&\qquad -\frac{z(n)z(n+1)}{v(n+1)}\varDelta \left( \frac{\varDelta v(n)}{z(n+1)}\right) \varDelta ^2 \left( \frac{\varDelta x(n)}{z(n)}\right) \\&\qquad +\frac{z(n)z(n+1)}{v(n+1)}\varDelta \left( \frac{\varDelta v(n)}{z(n+1)}\right) \left[ \varDelta \left( \frac{\varDelta x(n+1)}{z(n+1)}\right) \right. \\&\qquad \left. -\varDelta \left( \frac{\varDelta x(n)}{z(n)}\right) \right] \\&\quad =\varDelta ^4x(n)+p(n)\varDelta x(n+1)-\varDelta ^2z(n+1)\varDelta \left( \frac{\varDelta x(n+1)}{z(n+1)}\right) \\&\qquad -\frac{z(n+1)z(n+2)}{v(n+1)}\varDelta \left( \frac{\varDelta v(n)}{z(n+1)}\right) \varDelta \left( \frac{\varDelta x(n+1)}{z(n+1)}\right) \\&\qquad -\frac{z(n)z(n+1)}{v(n+1)}\varDelta \left( \frac{\varDelta v(n)}{z(n+1)}\right) \varDelta ^2 \left( \frac{\varDelta x(n)}{z(n)}\right) \\&\qquad +\frac{z(n)z(n+1)}{v(n+1)}\varDelta \left( \frac{\varDelta v(n)}{z(n+1)}\right) \varDelta ^2 \left( \frac{\varDelta x(n)}{z(n)}\right) \\&\quad =\varDelta ^4x(n)+p(n)\varDelta x(n+1)-\varDelta ^2z(n+1)\varDelta \left( \frac{\varDelta x(n+1)}{z(n+1)}\right) \\&\qquad -\frac{z(n+1)z(n+2)}{v(n+1)}\varDelta \left( \frac{\varDelta v(n)}{z(n+1)}\right) \varDelta \left( \frac{\varDelta x(n+1)}{z(n+1)}\right) \\&\quad =\varDelta ^4x(n)+p(n)\varDelta x(n+1) \\&\qquad -\frac{z(n+1)z(n+2)}{v(n+1)}\varDelta \left( \frac{\varDelta x(n+1)}{z(n+1)}\right) \left[ \varDelta ^2z(n+1)\frac{v(n+1)}{z(n+1)z(n+2)} \right. \\&\qquad \left. +\varDelta \left( \frac{\varDelta v(n)}{z(n+1)}\right) \right] . \end{aligned}
From (2.3) we conclude that $$\varDelta ^2 z(n+1) \frac{v(n+1)}{z(n+1)z(n+2)} +\varDelta \left( \frac{\varDelta v(n)}{z(n+1)}\right) =0$$ and finally we get
\begin{aligned}&\frac{1}{v(n+1)}\varDelta \left( \frac{v(n)v(n+1)}{z(n+1)}\varDelta \left( \frac{1}{v(n)}z(n)z(n+1)\varDelta \frac{\varDelta x(n)}{z(n)}\right) \right) \\&\quad =\varDelta ^4x(n)+p(n)\varDelta x(n+1). \end{aligned}
$$\square$$
## Conclusions
In this paper, we have studied the oscillation of a fourth-order delay three-terms equation (E). Our comparison method is based on the canonical representation () of equation (E) and the existence of positive solutions satisfying the auxiliary equations (2.1) and (2.3). We have deduced the oscillation of equation (E) from the oscillation of certain first-order difference equations. Note, that even when we do not know the exact solutions of the auxiliary equation (2.1) or (2.3), we can easily verify the conditions presented in our criterion. | 14,721 | 37,716 | {"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": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-49 | latest | en | 0.902291 |
http://mathforum.org/kb/plaintext.jspa?messageID=9289311 | 1,527,457,688,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870470.67/warc/CC-MAIN-20180527205925-20180527225925-00215.warc.gz | 184,226,114 | 1,780 | ```Date: Oct 3, 2013 4:33 PM
Author: quasi
Subject: Re: Sequence limit
quasi wrote:>konyberg wrote:>>Bart Goddard wrote:>>>>>> This question from a colleague: >>> >>> What is lim_{n -> oo} |sin n|^(1/n)>>> >>> where n runs through the positive integers. >>> >>> Calculus techniques imply the answer is 1.>>> But the same techniques imply the answer is 1>>> if n is changed to x, a real variable, and that>>> is not the case, since sin x =0 infinitely often.>>> >>> Anyone wrestled with the subtlies of this problem?>>> >>> E.g., can you construct a subsequence n_k such>>> that sin (n_k) goes to zero so fast that the>>> exponent can't pull it up to 1?>>>>In general a^0 = 1. lim (n goes inf) 1/n = 0. Then the value>>of sin(n) doesn't change that a^0 = 1.>>Your logic is flawed.>>Let f(n) = 1/(2^n).>>Then f(n)^(1/n) = 1/2 for all nonzero values of n, hence the>limit, as n approaches infinity, of f(n)^(1/n) is 1/2, not 1.>>Are there infinitely many positive integers n such that>> |sin(n)|^(1/n) < 1/(2^n)I meant:Are there infinitely many positive integers n such that |sin(n)| < 1/(2^n)>??>>If so, then the limit of the sequence>> |sin(n)|^(1/n), n = 1,2,3, ...>>does not exist. In particular, it would not be equal to 1.>>In fact, the original question can be recast as:>>Does there exist a real number c with 0 < c < 1 such that>the inequality>> |sin(n)| < c^n>>holds for infinitely many positive integers n?quasi
``` | 450 | 1,433 | {"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} | 3.28125 | 3 | CC-MAIN-2018-22 | latest | en | 0.890857 |
https://chouprojects.com/tan-excel/ | 1,721,924,455,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763860413.86/warc/CC-MAIN-20240725145050-20240725175050-00235.warc.gz | 149,757,988 | 25,544 | Published on
Written by Jacky Chou
# Tan: Excel Formulae Explained
## Key Takeaway:
• Understanding Trigonometry is crucial for using TAN function in Excel: TAN is a trigonometric function used to calculate the tangent of an angle. Therefore, understanding the basics of trigonometry, such as sine, cosine, and angles, is important when using TAN function in Excel.
• The syntax of TAN function is simple: To use TAN function in Excel, the syntax is straightforward. It only requires the input of an angle as an argument, which can either be in radians or degrees.
• TAN function has various advantages in Excel: Besides calculating tangent angles, TAN function can be used to simplify complex formulas that involve trigonometric functions, producing faster and more accurate results.
Do you struggle to remember and implement Excel formulae accurately? Don’t worry, this article offers a comprehensive explanation of the TAN function and how to use it. You’ll be a spreadsheet whiz in no time!
## Understanding Trigonometry
Trigonometric Functions Uncovered in TANH: Excel Formulae Explained
Trigonometry plays a vital role in many fields, from engineering to astronomy. In TANH: Excel Formulae Explained, understanding trigonometric functions is essential. Trigonometric functions like sine, cosine, and tangent help solve problems involving angles and lengths of triangles.
These functions are defined using ratios of the sides of a right-angled triangle. For instance, the sine of an angle is the ratio of the opposite side to the hypotenuse. The cosine of an angle is the ratio of the adjacent side to the hypotenuse. In the same vein, the tangent of an angle is the ratio of the opposite side to the adjacent side.
It is important to note that these functions not only apply to right triangles but any angle of a triangle. Additionally, these functions have several properties that make them significant in Mathematics and its applications.
Trigonometry dates back to ancient civilizations such as the Egyptians and Babylonians. The Greeks also made significant contributions to its development. Hipparchus, notably, is credited with creating the first table of chords.
## What is TAN Function in Excel?
The TAN function in Excel is a trigonometric function that calculates the tangent of an angle given in radians. It is useful for determining the slope of a line, as well as for various mathematical and engineering applications. By typing “`=TAN(angle)`” into a cell, one can calculate the tangent of a given angle. It is important to note that the angle must be in radians, not degrees.
In addition to its basic functionality, the TAN function in Excel can also be used in conjunction with other functions, such as the SIN and COS functions, to perform more complex calculations. For example, one could use the TAN function to calculate the slope of a line given the angle of inclination, or to find the height of an object given its distance from an observer and the observer’s angle of elevation.
It is worth noting that the TAN function in Excel, like all mathematical functions, is subject to certain limitations and constraints. For example, when the input angle is close to or equal to 90 degrees, the function will return an error value because the tangent of 90 degrees is undefined. Similarly, the function may return an error value if the input angle is outside of the valid range for tangent values (-π/2 to π/2 radians).
To make the most of the TAN function in Excel, it is important to understand its limitations and to use it in conjunction with other functions as needed. With a little practice, anyone can become proficient in using this versatile tool to perform complex calculations with ease. Don’t miss out on the potential benefits of the TAN function – start exploring its capabilities today!
## Syntax of TAN Function
Understand TAN function syntax and arguments as an Excel user? You need to know the basics! This section of “TAN: Excel Formulae Explained” reveals the core info.
Learn the syntax and the details of its arguments. Fine print included!
### Arguments of TAN Function
The TAN function in Excel has specific arguments that must be supplied for it to work effectively. One of the arguments is the angle in radians, which represents the angle you want to find the tangent of. The angle can either be a number, reference to a cell containing a number or calculation that results in a numerical value.
It is important to note that Excel uses radians instead of degrees when calculating trigonometric functions. This means you may need to convert your angle measurement from degrees to radians before using it with the TAN function. You can do this by multiplying the degree value by pi/180.
Additionally, since the TAN function expects an input in radians, if you forget to convert your angle measurement, you will get an incorrect result.
To avoid errors and ensure accurate results, make sure all your angles are expressed in radians before applying them to the TAN function.
Don’t risk inaccuracies in your calculations by overlooking this crucial detail. Take time to ensure that any angles used with the TAN function are always represented in radians. Your data will thank you!
## Examples of Using TAN Function
To grasp TAN function usage, you require examples. We’ll demonstrate how to use TAN function through “Examples of Using TAN Function”. This includes:
1. Simple TAN Function Example
2. TAN Function with Other Formulas
3. TAN Function with Graphs
Thus, you will find the solutions to use TAN correctly.
### Simple TAN Function Example
The TAN Function in Excel is a useful tool for several calculations. Here’s a comprehensive guide to the simplest way of using it.
2. Select a cell where you want to apply the TAN function
3. Type “=TAN(value)”, without quotation marks, and press ‘Enter’
This formula will return the tangent value of the input value in radians. Remember to input numerical values only.
Did you know that TAN is an acronym for Tangent? The Tangent function finds the ratio between the opposite and adjacent sides of a right-angled triangle.
Get ready to TAN-gle with other formulas and solve mathematical conundrums like a pro!
### TAN Function with Other Formulas
The TAN Function in Excel can be implemented with various other formulas to gain better insights. Have a look at the following examples to know more.
Formula Description Output =TAN(PI()/4) Finding tangent angle of pi/4 radians 1 =TAN(A2) Tangent value of cell A2 (containing angle) #DIV/0! =TAN(SIN(A3)) Tangent value of sine value of cell A3 (containing angle in degrees) -1.31695987612754E+45
For instance, one can combine TAN and PI functions to calculate any angle’s tangent, as seen above. Or suppose one wants to find the tangent of an angle present in another cell; they could use TAN along with the respective cell reference.
Similarly, One could also find the Tangent value of sine values using SIN and TAN both together.
A snippet from the source reveals that “According to Microsoft Support, this function returns a numeric value that represents the tangent of an angle, specified in radians.”
Get ready to plot your way to success with TAN function and graphs – no GPS required.
### TAN Function with Graphs
Incorporating Graphs into TAN Function Understanding
Visual representation helps in scrutinizing data, and using graphs can make understanding TAN function easy. The analysis using these techniques involves plotting the graph of the tangent functions.
Graphs in TAN function are a depiction of the relationship between angles and their corresponding tangents. The plotted line shows how the values fluctuate with changes in input angles. These graphs assist people in getting a clear insight into their data.
As we delve deeper, it is vital to mention that angle measures exceeding 90 degrees cannot be evaluated using tangent functions alone. Using co-tangent or co-function together could help obtain meaningful output.
According to ‘Excel Jet’, “TAN function returns the tangent of an angle provided in radians,” making it one of an extensive range of Excel functionalities used for various analyses.”
Using TAN function in Excel is the smoothest way to transform your angles into tangents, unless you’re a geometry teacher, then it’s probably just the norm.
## Advantages of Using TAN Function in Excel
TAN Function in Excel is a versatile tool that provides several benefits to users. With its ability to calculate the tangent of an angle, it assists in finding the slope and gradient of data points accurately. Moreover, its user-friendly interface and easy-to-use nature make it a popular choice for professionals.
Additionally, TANH: Excel Formulae Explained offers the advantage of quick and precise calculations, saving valuable time for users. Its capabilities go beyond simple data analysis, as it can find trends and patterns in complex data sets with ease. TAN function can also be combined with other formulae to create powerful analysis tools that enable better decision-making.
It’s worth noting that TAN Function in Excel also allows users to customize the decimal places, enabling greater precision in calculations. This unique feature makes it an indispensable tool in scientific and financial analysis.
In one instance, a financial analyst used TAN function to calculate the return on investment (ROI) for a project. By combining TAN with other formulae, the analyst was able to generate powerful insights that helped the company make informed decisions.
## Limitations of Using TAN Function in Excel
Using TAN function in Excel has limitations that must be considered. TAN is limited to angles within a specific range and cannot handle angles outside this range. The output may become inaccurate, which can affect the validity of calculations. Therefore, it is essential to use caution when utilizing the TAN function in Excel.
One potential limitation of TAN function in Excel is that it only accepts angle values between -90 and 90 degrees. If an angle value outside of this range is inputted, the TAN function may return an invalid output, leading to incorrect calculations. Additionally, the TAN function operates on angles expressed in radians, not in degrees, and users must convert degrees to radians before utilizing the TAN function in Excel.
It is crucial to note that while TAN can be an accurate tool for specific calculations, its limitations should be considered. For more complex tasks that require angles outside the standard range, other Excel functions or tools may prove more useful.
A True History of TAN function in Excel is that it has been a useful tool for mathematics, engineering, and finance for decades. It is one of the many Excel functions that have revolutionized data analysis and made it more accessible to general users. The application of TAN function in Excel has allowed for faster, more accurate data analysis, and it continues to be a valuable tool for professionals in various industries.
Overall, while the TAN function in Excel has limitations, it remains a useful tool for calculating a range of values, and it can be used in conjunction with other available functions to produce accurate results.
## Common Errors with TAN Function
When using the TAN function in Excel, there are some common mistakes that users may encounter. One issue is when the angle argument is entered in degrees instead of radians, leading to an incorrect output. Another error occurs when the angle argument is not provided, causing the function to return a #VALUE! error. To ensure accurate results, it is important to check that the angle input is in radians. Additionally, users should be aware that the TAN function is sensitive to rounding errors and can return inaccurate results in some cases.
It is important to carefully check the inputs and double-check the angle measurement to avoid common errors when using the TAN function in Excel.
A true fact: The TANH function can be used to calculate the hyperbolic tangent of a given value in Excel.
## Five Facts About TAN: Excel Formulae Explained:
• ✅ TAN is an Excel formula that returns the tangent of an angle given in radians. (Source: Microsoft)
• ✅ The tangent function is used extensively in mathematics and engineering applications, such as trigonometry and calculus. (Source: Math is Fun)
• ✅ The TAN function can be combined with other Excel formulas, such as PI and RADIANS, to perform complex calculations. (Source: Excel Easy)
• ✅ TAN is one of the six main trigonometric functions, along with sine, cosine, cotangent, secant, and cosecant. (Source: Khan Academy)
• ✅ Excel offers a wide range of other mathematical functions, such as AVERAGE, SUM, and COUNT, to perform various calculations and analyses. (Source: ExcelJet)
## FAQs about Tan: Excel Formulae Explained
### What is TAN: Excel Formulae Explained?
TAN: Excel Formulae Explained is a comprehensive guide to understanding and using the TAN function in Microsoft Excel. This guide provides an overview of how the TAN function works, along with examples of how it can be used in a variety of different scenarios.
### How does the TAN function work in Excel?
The TAN function in Excel is a mathematical function that is used to determine the tangent of a given angle. In order to use the TAN function, you must provide it with an angle in radians. The TAN function will then return the tangent of that angle.
### What are some examples of how the TAN function can be used in Excel?
The TAN function can be used in a variety of different scenarios in Excel. For example, it can be used to calculate the angle of a triangle given the lengths of its sides, or to calculate the height of an object given its distance and angle of elevation.
### Can the TAN function be used with other Excel functions?
Yes, the TAN function can be used in combination with other Excel functions to perform more complex calculations. For example, it can be used with the SIN and COS functions to calculate the angles of a triangle given the lengths of its sides.
### Are there any limitations to using the TAN function in Excel?
One limitation of using the TAN function in Excel is that it requires the input angle to be in radians rather than degrees. Additionally, the TAN function may return errors if it is given an angle that is too large or too small.
There are many resources available online that can help you learn more about using Excel formulae like the TAN function. You can also consult the Excel documentation or take an online Excel course to improve your skills and knowledge.
## Related Articles
### Max: Excel Formulae Explained
Key Takeaway: The MAX function in Excel is used to ...
### Lower: Excel Formulae Explained
Key Takeaway: The LOWER formula in Excel allows users to ...
### Match: Excel Formulae Explained
Key Takeaway: The MATCH function in Excel is used to ... | 3,081 | 14,985 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-30 | latest | en | 0.868864 |
http://mathhelpforum.com/geometry/128116-analytical-geometry-circles.html | 1,481,217,365,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541864.24/warc/CC-MAIN-20161202170901-00034-ip-10-31-129-80.ec2.internal.warc.gz | 175,818,750 | 12,496 | 1. ## analytical geometry,circles
2. Originally Posted by Pulock2009
Find the equation of the circle which touches both axes and passes through
the point (-2,-1).
1. The axes touch the circle in the points P(r,0) and Q(0,r). The center of the circle must be situated in the 3rd quadrant and on the line y = x.
2. The circle has the equation
$(x-a)^2+(y-b)^2=r^2$
According to 1) the coordinates of the center must be equal C(-r , -r):
3. Plug in the coordinates of P, Q, C and the coordinmates of the given point:
$(-2+r)^2+(-1+r)^2=r^2$
Solve for r.
4. You'll get 2 circles which satisfy the given conditions.
3. thanks a lot for your answer. one thing i didnot get:how will we know that the centre lies on y=x???and wat about my second question???is it too easy???
4. also , will both the circles pass through (-2,-1)???
5. Originally Posted by Pulock2009
thanks a lot for your answer. one thing i didnot get:how will we know that the centre lies on y=x???
The axes are tangents to the circle. The point of intersection of the two tangents and the center of the circle determine the angle bisector between the tangents. The angle bisector of the coordinate axes is the straight line defined by y = x (or y = -x respectively)
and wat about my second question???is it too easy???
Originally Posted by Pulock2009
also , will both the circles pass through (-2,-1)???
Yes. Have a look at the attachment. The point (-2, -1) is drawn. And you can show that the distance between (-2, -1) and (-5, -5) (or (-2, -1) and (-1, -1) ) is as long as the corresponding radius.
6. Originally Posted by Pulock2009
Is this the text of the 2nd question?:
2)Find the area of an equilateral triangle inscribed in the circle $x^{2}+y^{2}+2gx+2fy+c=0$
Assuming the points to be $A(x_{1},y_{1}),\ B(x_{2},y_{2}),\ C(x_{3},y_{3})$ the area of the triangle is:
$ar(ABC)=\begin{array}{ccc}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}$
and there is a similar form of the equation of a circle but i donot know how to combine them for a solution.
7. yes. from "Assuming....."is the material that i tried out.
8. Originally Posted by Pulock2009
yes. from "Assuming....."is the material that i tried out.
1. Transform the equation of the circle such that you can determine the coordinates of the center and the length of the radius:
$x^2+2gx+y^2+2fy=-c~\implies~(x+g)^2+(y+f)^2=f^2+g^2-c$
Thus $r^2=f^2+g^2-c$
2. The side of an equilateral triangle which is inscribed in a circle has the length $s = r \cdot \sqrt{3}$
The area of such a triangle is $A = \frac12 \cdot \underbrace{r \cdot \sqrt{3}}_{\text{length of base}} \cdot \underbrace{\frac32 \cdot r}_{\text{length of height}} = \frac34 \cdot r^2 \cdot \sqrt{3}$
3. Plug in the term for rē from 1. and you're done.
9. s=r*sqrt(3)???? how do we get that??? some trigonometry is necessary perhaps????anyways thanks a lot!!!
10. Originally Posted by Pulock2009
s=r*sqrt(3)???? how do we get that??? some trigonometry is necessary perhaps????anyways thanks a lot!!!
In this case simple Pythagorean Theorem will do:
1. See attachment. The small right triangle (coloured in blue) is a 30°-60°-right triangle.
2. Therefore the blue line segment must have the length $\tfrac12 r$
3. The 2nd leg of the right angle has the length:
$\sqrt{r^2-\left(\frac12 r\right)^2} = \frac12 r\cdot \sqrt{3}$
4. This 2nd leg is as long as a half side of the equilateral triangle:
$\frac12 s = \frac12 r\cdot \sqrt{3}~\implies~\boxed{s = r \cdot \sqrt{3}}$ | 1,080 | 3,519 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 12, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.891303 |
http://www.ck12.org/book/CK-12-Algebra-I-Concepts/r1/section/12.0/Rational-Equations-and-Functions---Intermediate/ | 1,412,151,123,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1412037663372.35/warc/CC-MAIN-20140930004103-00212-ip-10-234-18-248.ec2.internal.warc.gz | 484,087,646 | 25,917 | <meta http-equiv="refresh" content="1; url=/nojavascript/"> Rational Equations and Functions | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: CK-12 Algebra I Concepts Go to the latest version.
# Chapter 12: Rational Equations and Functions
Created by: CK-12
0 0 0
## Introduction
Up until this point, you’ve dealt with non-fractional polynomials, but what if you were asked to solve a fractional function with a polynomial in the numerator and the denominator? Such functions are known as rational functions. The graphs of these functions grow closer and closer to certain values without ever reaching those values. This is called asymptotic behavior. In this chapter, you’ll find the asymptotes of rational functions. You’ll also perform operations on rational expressions and solve rational equations. In the real world, rational equations are often used to model electrical circuit and distance problems.
## Summary
This chapter begins by distinguishing between three variation models: direct variation, inverse variation, and joint variation. It then moves on to graphing and solving rational functions, with particular attention paid to asymptotes. Next, it addresses simplifying rational expressions by factoring and dividing. The four mathematical operations of multiplication, division, addition, and subtraction are then performed on rational expressions. The chapter concludes with real-life applications of rational equations and methods for solving them.
Oct 01, 2012 | 296 | 1,519 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2014-41 | longest | en | 0.925837 |
https://www.scienceforums.net/profile/12129-jgalaz/content/ | 1,600,818,912,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400208095.31/warc/CC-MAIN-20200922224013-20200923014013-00279.warc.gz | 1,029,808,556 | 10,888 | Jump to content
• Sign Up
# jgalaz
New Members
1
1. ## Ellipses, tilted circles, circle approximation
This thread was useful for me since I was trying to figure the same question on. More specifically, I wanted to know how the minor axis of the projected ellipse varies with tilting angle Phi. I'll take a look at the Wikipedia website here referenced to see whether the equations there are of any use (If not, I'll have to figure out the relationship myself). I'm intuitively guessing that if (x^2)/a^2 + (y^2)/b^2=1, all you have to do is divide by cos(phi) which ever axis is reduced upon tilting, if tilting parallel to either of the ellipse axes that is. For example, if the ellipse is ORIGINALLY A CIRCLE on a plane P, it's shape is: (x^2)/R^2 + (y^2)/R^2=1 If tilting parallel to the y axis, then the circle will start becoming an ellipse by reducing its radius along the x direction: (x^2)/(R*cos[phi])^2 + (y^2)/R^2=1 Which is the same as: (x^2)/a^2 + (y^2)/b^2=1 After the following variable substitution: a=R*cos[phi] b=R
×
• #### Activity
• Leaderboard
×
• Create New...
## Important Information
We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue. | 347 | 1,273 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.924496 |
https://www.comp-psyche.com/2014/08/counting-sort-program-in-c.html?showComment=1454703829127 | 1,695,842,887,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510319.87/warc/CC-MAIN-20230927171156-20230927201156-00145.warc.gz | 770,904,539 | 31,156 | # Counting Sort Program In C
## COUNTING SORT IN C
In the code for counting sort, we assume that the input is an array A[1 . . . n] and we require two other arrays: the array B[1 . . . n] holds the sorted output, and the array C[0 . . . k] provides temporary working storage. Here k = maximum element present in array.
Consider following elements to be sorted using counting sort : 2, 5, 3, 0, 2, 3, 0, 3
So we have in array A[1 . . . 8 ] :
2 5 3 0 2 3 0 3
Now what we have to do is initialize the auxiliary array C[0 . . . k] with 0. Here k = 5.
So C[0 . . . 5] becomes :
0 0 0 0 0 0
Next step is to count the frequency of each element and store it in their respective index position i.e here frequency of 0 = 2. So we place '2' at index position '0'. Similarly we place each element and the array becomes C[0 . . . 5] :
2 0 2 3 0 1
Now we add the current index position with its previous index position and store it in the current index position i.e C[ i ] = C[ i ] + C[ i - 1 ] where 'i' starts from '1'.
So C[ 0 . . . 5 ] becomes :
2 2 4 7 7 8
The final step is to place the elements in B[ 1 . . . 8 ] with reference to C[ 0 . . . 5 ] i.e Now we treat elements in C[ 0 . . . 5 ] as index of B[ 1 . . . 8 ] and index of C[ 0 . . . 5 ] as elements of B[ 1 . . . 8 ]. So at index position '8' we have element '5' so we place '5' at index position of '8' of B[ 1 . . . 8 ] ( B[ 8 ] = 5 ) and we reduce the index position by '1' i.e 8-1 = 7. Now note that we have element '7' in C[ 4 ] position but '4' is not present in the original array so we skip it and move to C[ 3 ] position, so we place '3' at index position '7' of B[ 1 . . . 8 ]. Similarly we place each element in B[ 1 . . . 8 ] and we get sorted array.
So the sorted array B[ 1 . . . 8 ] is :
0 0 2 2 3 3 3 5
## COUNTING SORT ALGORITHM
```// Counting Sort algorithm
```
`CountingSort(A,B,n,k)`
` int C[0 . . . k ]`
` for i=0 to k`
` C[i]=0`
` for j=1 to n`
` C[A[j]] = C[A[j]] + 1;`
` for i=1 to k`
` C[i] = C[i] + C[i-1];`
` for j=n downto 1`
``` B[C[A[j]]] = A[j];
C[a[j]] = C[A[j]] - 1;```
## COUNTING SORT PROGRAM IN C
`// Counting sort program in c`
```
#include<stdio.h>
void CountingSort(int a[], int sorted_array[], int max_element, int n);
int main()
{
int a[20],i,n;
int max_element, sorted_array[20];
printf("Enter number of elements in array : ");
scanf("%d",&n);
// Inputting elements
for(i=1;i<=n;i++)
{
printf("Enter number %d: ",i+1);
scanf("%d",&a[i]);
}
// Displaying Elements before counting sort
printf("Items in the array are : ");
for(i=1;i<=n;i++)
{
printf("%d ",a[i]);
}
// Finding Maximum element
max_element=a[1];
for(i=2;i<=n;i++)
{
if(a[i] > max_element)
{
max_element = a[i];
}
}
//Applying counting sort
CountingSort(a, sorted_array, max_element, n);
// Displaying elements after count sort
printf("\nElements after count sort : ");
for(i=1;i<=n;i++)
{
printf("%d ",sorted_array[i]);
}
return 0;
}
void CountingSort(int a[],int sorted_array[],int max_element, int n)
{
int aux_array[max_element];
int i,j;
// Initializing auxiliary array```
``` for(i=0;i<=max_element;i++)
{
aux_array[i]=0;
}
```
``` // Placing Frequency of each element in aux_array
for(j=1;j<=n;j++)
{
aux_array[a[j]]=aux_array[a[j]]+1;
}
```
``` // Adding elements of current and previous index position
for(i=1;i<=max_element;i++)
{
aux_array[i]=aux_array[i]+aux_array[i-1];
}
```
``` // Placing elements in sorted_array
for(j=n;j>=1;j--)
{
sorted_array[aux_array[a[j]]] = a[j];
aux_array[a[j]] = aux_array[a[j]] - 1;
}
}
```
More Informative Posts :
## Share this
Previous
Next Post »
### 3 comments
comments
// Adding elements of current and previous index position
for(i=1;i<=max_element;i++)
{
aux_array[i]=aux_array[i]+aux_array[i-1];
}
// Placing elements in sorted_array
for(j=n;j>=1;j--)
{
sorted_array[aux_array[a[j]]] = a[j];
aux_array[a[j]] = aux_array[a[j]] - 1;
}
i can't understand this please help me..
Reply
Let n=8
a[1 . . . 8] : 2 5 3 0 2 3 0 3
Elements are stored from index position 1.
// Initializing auxiliary array
for(i=0;i<=max_element;i++)
{
aux_array[i]=0;
}
aux_array[0...5]=0 0 0 0 0 0
// Placing Frequency of each element in aux_array
for(j=1;j<=n;j++)
{
aux_array[a[j]]=aux_array[a[j]]+1;
}
Now aux_array[0. . . 5] : 2 0 2 3 0 1
Elements are stored from index position 0.
/* Adding elements of current and previous index position of aux_array
for(i=1;i<=max_element;i++)
{
aux_array[i]=aux_array[i]+aux_array[i-1];
}
Note : We are placing elements from index position 1
So, aux_array[0]=2
aux_array[1]=aux_array[1]+aux_array[0]
aux_array[1] = 0 + 2 = 2
Now aux_array[1]=2 // i.e we have 2 at index position 1 of aux_array
Similarly on next loops :
aux_array[2] = aux_array[2]+aux_array[1]
aux_array[2] = 2+2 = 4
aux_array[2] = 4
aux_array[3] = aux_array[3]+aux_array[2]
aux_array[3] = 3+4 = 7
aux_array[3] = 7
and so on..
After loop is finished then we will have :
aux_array[0...5] = 2 2 4 7 7 8
// Placing elements in sorted_array
for(j=n;j>=1;j--)
{
sorted_array[aux_array[a[j]]] = a[j];
aux_array[a[j]] = aux_array[a[j]] - 1;
}
Here we are placing elements in ascending order in sorted_array
We have :
n=8
For j=8
a[1...8]=2 5 3 0 2 3 0 3
aux_array[0...5] = 2 2 4 7 7 8
sorted_array[aux_array[a[j]]] = a[j];
sorted_array[aux_array[a[8]]]
i.e sorted_array[aux_array[3]]
i.e sorted_array[7]=3
aux_array[a[j]] = aux_array[a[j]] - 1;
aux_array[a[8]] = aux_array[a[8]]-1;
aux_array[3] = aux_array[3]-1;
aux_array[3] = 7-1 = 6
For j=7
a[1...8]=2 5 3 0 2 3 0 3
aux_array[0...5] = 2 2 4 6 7 8
sorted_array[aux_array[a[j]]] = a[j];
sorted_array[aux_array[a[7]]]
i.e sorted_array[aux_array[0]]
i.e sorted_array[2]=0
aux_array[a[j]] = aux_array[a[j]] - 1;
aux_array[a[7]] = aux_array[a[7]]-1;
aux_array[0] = aux_array[0]-1;
aux_array[0] = 2-1 = 1
For j=6
a[1...8]=2 5 3 0 2 3 0 3
aux_array[0...5] = 1 2 4 6 7 8
sorted_array[aux_array[a[j]]] = a[j];
sorted_array[aux_array[a[6]]]
i.e sorted_array[aux_array[3]]
i.e sorted_array[6]=3
aux_array[a[j]] = aux_array[a[j]] - 1;
aux_array[a[6]] = aux_array[a[6]]-1;
aux_array[3] = aux_array[3]-1;
aux_array[3] = 6-1 = 5
For j=5
a[1...8]=2 5 3 0 2 3 0 3
aux_array[0...5] = 1 2 4 5 7 8
sorted_array[aux_array[a[j]]] = a[j];
sorted_array[aux_array[a[5]]]
i.e sorted_array[aux_array[2]]
i.e sorted_array[4]=2
aux_array[a[j]] = aux_array[a[j]] - 1;
aux_array[a[5]] = aux_array[a[5]]-1;
aux_array[2] = aux_array[2]-1;
aux_array[2] = 4-1 = 3
We have aux_array[0...5] = 1 2 3 5 7 8
Till now we have found :
sorted_array[2]=0
sorted_array[4]=2
sorted_array[6]=3
sorted_array[7]=3
Similarly follow the above step till j=1 and you will get elements in ascending order in sorted_array :
sorted_array[1]=0
sorted_array[2]=0
sorted_array[3]=2
sorted_array[4]=2
sorted_array[5]=3
sorted_array[6]=3
sorted_array[7]=3
sorted_array[8]=5
Hope this explanation is helpful to you..
Reply
Anonymous
Check this:
#include
void counting_sort(int a[],int n,int max)
{
int count[50]={0},i,j;
for(i=0;imax)
{
max=a[i];
}
}
counting_sort(a,n,max);
return 0;
}
/*4 14 8 0 2 5 2 1 0 22*/
Reply | 2,585 | 7,128 | {"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} | 3.390625 | 3 | CC-MAIN-2023-40 | latest | en | 0.776997 |
https://forums.mikeholt.com/threads/wiring-277v-lighting.83474/ | 1,590,631,245,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347396300.22/warc/CC-MAIN-20200527235451-20200528025451-00420.warc.gz | 357,293,447 | 10,416 | # Wiring 277V Lighting
Status
Not open for further replies.
#### alittleee
##### Member
I realize that 277V is derived from a 3-phase Y transformers' line-line voltage (480V/1.732). So, my way of thinking is that if you draw a 3-phase Y transformer, you would connect your load (in this case lighting) say from a-phase to b-phase. But that is not the case, there are not 2 phases there, there is instead a hot and a neutral. Where did the neutral come from? How is 277V lighting wired? And why is it dangerous to work on energized 277V lighting circuits?
#### iwire
##### Moderator
Staff member
I realize that 277V is derived from a 3-phase Y transformers' line-line voltage (480V/1.732). So, my way of thinking is that if you draw a 3-phase Y transformer, you would connect your load (in this case lighting) say from a-phase to b-phase. But that is not the case, there are not 2 phases there, there is instead a hot and a neutral. Where did the neutral come from? How is 277V lighting wired? And why is it dangerous to work on energized 277V lighting circuits?
The source would be 480 Wye not 480 delta and you would have 480 line to line and 277 from any line to neutral.
This is no different than 208Y/120 ..... other than the voltage itself.
#### augie47
##### Moderator
Staff member
Attached may help
and
It's dangerous to work on any energized circuits (and a violation except in extreme circumstances)
#### alittleee
##### Member
sorry
sorry
I am sorry. I was just getting ready to argue with you and I was typing out the "formula" and realized I was thinking just bass ackward. Line voltage is higher than phase voltage in Y connections. Sorry to waste your time. But - it made me think strait by trying to argue with you anyway.
#### augie47
##### Moderator
Staff member
I am sorry. I was just getting ready to argue with you and I was typing out the "formula" and realized I was thinking just bass ackward. Line voltage is higher than phase voltage in Y connections. Sorry to waste your time. But - it made me think strait by trying to argue with you anyway.
Go ahead and argue ,To quote G B Shaw:
I learned long ago, never to wrestle with a pig. You get dirty, and besides, the pig likes it.
George Bernard Shaw
Status
Not open for further replies. | 556 | 2,276 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2020-24 | latest | en | 0.962286 |
https://plainmath.net/28939/find-the-area-of-the-parallelogram-whose-vertices-are-listed-0-0-5-2-6-4 | 1,638,625,713,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362992.98/warc/CC-MAIN-20211204124328-20211204154328-00052.warc.gz | 530,868,326 | 7,951 | # Find the area of the parallelogram whose vertices are listed. (0,0), (5,2), (6,4
Find the area of the parallelogram whose vertices are listed. (0,0), (5,2), (6,4), (11,6)
## Want to know more about Matrix transformations?
• Questions are typically answered in as fast as 30 minutes
### Plainmath recommends
• Get a detailed answer even on the hardest topics.
• Ask an expert for a step-by-step guidance to learn to do it yourself.
Aniqa O'Neill
Parallelogram can be described with 4 vertices or 2 vectors. Since we are given 4 vertices(ABCD), let's find vectors u, v that describe parallelogram. Graph these points to see how vectors are described
A=(0,0)
B=(5,2)
C=(6,4)
D=(11,6)
$$u=AB=\begin{bmatrix}5\\2\end{bmatrix}$$
$$u=AC=\begin{bmatrix}6\\4\end{bmatrix}$$
Area of parallelogram is absolute value of determinant
$$\begin{bmatrix}u_1 & v_1 \\u_2 &v_2 \end{bmatrix}$$
$$\begin{bmatrix}u_1 & v_1 \\u_2 &v_2 \end{bmatrix}=det\begin{bmatrix}5& 6 \\2 &4 \end{bmatrix}=20-12=8$$
Result:
8 | 340 | 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": 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} | 4.09375 | 4 | CC-MAIN-2021-49 | latest | en | 0.728094 |
https://physics.stackexchange.com/questions/129135/do-we-breathe-air-by-creating-a-vacuum | 1,566,427,601,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316549.78/warc/CC-MAIN-20190821220456-20190822002456-00209.warc.gz | 606,536,118 | 30,564 | # Do we breathe air by creating a vacuum?
How do we suck air into our lungs, are we generating a vacuum?
You aren't creating a vacuum, but you are reducing the pressure in your lungs when you inhale.
In effect your lungs are working as a diaphragm pump. When you pull your diaphragm down, and/or expand your chest, this increases the volume inside your lungs. Boyle's law tells us:
$$P_0V_0 = P_{\rm inhale}V_{\rm inhale} ,$$
where $P_0$ and $V_0$ are ambient pressure and the volume of your lungs when you're not inhaling. Rearranging this we get:
$$P_{\rm inhale} = P_0\frac{V_0}{V_{\rm inhale}} .$$
When you inhale $V_{\rm inhale} > V_0$ so $P_{\rm inhale} < P_0$. Because the pressure in your lungs is less than the ambient pressure air flows from the outside into the lungs.
• +1 short and concise. @Baconbeastnz for a vacuum to be created, $\Delta P$ has to be so large that the surrounding air molecules won't be able to immediately feel the created vacuum after an inhale (diverging relaxation times). Whereas in reality the relative change in $P$ is usually in the order of $-4mmHg$ to $-6mmHg$. – Phonon Aug 2 '14 at 6:49
• To test your pressure drop capability, see how far up a straw or tube you can suck water. Divide height by 13.6 to get Hg pressure... – DJohnM Aug 3 '14 at 1:17 | 379 | 1,302 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2019-35 | latest | en | 0.895964 |
http://indymoves.org/worksheet-for-multiplying-decimals/ | 1,600,701,243,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400201826.20/warc/CC-MAIN-20200921143722-20200921173722-00153.warc.gz | 69,776,339 | 31,172 | # Worksheet For Multiplying Decimals
In Free Printable Worksheets198 views
4.12 / 5 ( 174votes )
Top Suggestions Worksheet For Multiplying Decimals :
Worksheet For Multiplying Decimals If your child needs a little help with multiplying numbers with decimals here s the worksheet for her as she completes the page she ll get lots of practice multiplying decimals and calculating the Multiplying decimals by 10 100 and worksheet now take on the challenge of multiplying decimals by 10 100 and with this twinkl worksheet Complete this worksheet from pearson which will help you develop a deep understanding of how to multiply a decimal number by a whole number whilst you work through these questions remember to.
Worksheet For Multiplying Decimals Multiplying an entire excel column by the doesn t change when you copy and paste it to other parts of your worksheet enabling you to easily apply it to the entire column Simply denote each digit s weight multiply each hexadecimal digit value by its respective weight in decimal form then add up all the decimal values to get a total for example the hexadecimal An arithmetic trick often used when working with the metric system is multiplication by ten and division by ten via shifting of the decimal point a similar trick may be applied to binary numbers.
Worksheet For Multiplying Decimals Math facts in a flash is exactly what its simple title sounds like straightforward web based practice designed to strengthen students fluency in basic math concepts like addition subtraction Add subtract multiply and divide decimals to hundredths using concrete models or drawings and strategies based on place value properties of operations and or the relationship between addition They may understand the concept of division and even division with decimals perfectly but their weak working memory leads to a breakdown in procedure these students need accommodations like a.
When you use the rand function on a microsoft excel spreadsheet by default the random number will be nine or ten digits long extending well past the decimal point the rand function does not.
## Multiplying Decimals Worksheets
Multiplying Decimals Using Number Lines These Printable Number Line Multiplication Worksheets Are Meticulously Created To Help Young Ones Tune Their Skills At Multiplication With Topics Like Drawing Hops On Number Lines Matching Number Lines With Multiplication Sentences And More
### Decimal Multiplication Worksheets Homeschool Math
Worksheets For Multiplying Decimals By Powers Of Ten Multiply By 10 Or 100 1 2 Decimal Digits Multiply By 10 100 Or 1 2 Decimal Digits Multiply By 10 100 Or Missing Factor 1 2 Decimal Digits
#### Multiplying Decimals Worksheets
Use These Worksheets To Practice Multiplying Decimal Factors Together Multiplying Decimals By 1 Digit Numbers Multiplying Decimals By 1 Digit Numbers This Worksheet Has 10 Vertical Problems And 2 Word Problems That Students Can Solve To Practice Multiplying Decimals By Single Digit Numbers 4th Through 7th Grades View Filing Cabinet Logged In Members Can Use The Super Teacher
##### Multiplication Of Decimals Worksheets K5 Learning
Worksheets Math Grade 5 Decimal Multiplication Multiply Decimals By Whole Numbers Or Other Decimals Our Grade 5 Multiplication Of Decimals Worksheets Give Practice Exercises Ranging From Multiplying One Digit Decimals By Whole Numbers To General Multiplication Of Multi Digit Decimals In Columns Missing Factor Questions Are Also Included
###### Multiplying Decimals Worksheets Math Worksheets
Multiplying Decimals Worksheet 2 This Is Another Twenty Problem Worksheet Featuring Decimal Multiplication You Will Be Multiplying By A Single Digit Number Which Will Help You Focus On Correctly Placing The Decimal Point In Your Answer
Multiply The Decimals Free Printable Worksheets For
Multiplying Decimals Worksheet 1 Item Name Multiply The Decimals Solve 4 16 9 11 2 41 3 53 6 46 X 41 X 93 X 52 X 44 X 71
Multiplying Decimal Worksheets For Grade 5 With
Help Out Your Young Mathematicians Obtain The Best Multiplying Decimals Practice With These Multiplying Decimal Worksheets For Grade 5 As Part Of Our Daily Lives We Often Need To Multiply Decimals When Cooking Comparing Costs Budgeting Etc With Regards To This Our Fifth Grade Printable Multiplication With Decimals Is Made Of Multiplying Decimals Word Problems Worksheets With
Multiplying And Dividing Decimals Worksheets
Round Decimals Worksheets Multiplying Grade 6 Dividing And Pular Math Worksheetstiplying Decimals Worksheet Kuta Download Them And Free Grade 6 Measuring Worksheets Multiplication And Division Of Decimals Decimals Multiply And Multiplying Dividing Integers Worksheet And Worksheets Multiplying And Dividing Decimals Worksheets
Decimals Worksheets Free Math Worksheets
On This Page You Will Find Decimals Worksheets On A Variety Topics Including Comparing And Sorting Decimals Adding Subtracting Multiplying And Dividing Decimals And Converting Decimals To Other Number Formats To Start You Will Find The General Use Printables To Be Helpful In Teaching The Concepts Of Decimals And Place Value More
Decimals Worksheets K5 Learning
Decimal Worksheets For Grades 3 6 These Free Decimal Worksheets Start With The Conversion Of Simple Fractions Denominators Of 10 Or 100 To From Decimals And Progress To The Addition Subtraction Multiplication And Division Including Long Division Of Multi Digit Decimals Latter Worksheets Include Rounding And Repeating Decimals The Worksheets From K5 Learning Are Categorized By Grade Topic No Login Or Account Is Needed
Worksheet For Multiplying Decimals. The worksheet is an assortment of 4 intriguing pursuits that will enhance your kid's knowledge and abilities. The worksheets are offered in developmentally appropriate versions for kids of different ages. Adding and subtracting integers worksheets in many ranges including a number of choices for parentheses use.
You can begin with the uppercase cursives and after that move forward with the lowercase cursives. Handwriting for kids will also be rather simple to develop in such a fashion. If you're an adult and wish to increase your handwriting, it can be accomplished. As a result, in the event that you really wish to enhance handwriting of your kid, hurry to explore the advantages of an intelligent learning tool now!
Consider how you wish to compose your private faith statement. Sometimes letters have to be adjusted to fit in a particular space. When a letter does not have any verticals like a capital A or V, the very first diagonal stroke is regarded as the stem. The connected and slanted letters will be quite simple to form once the many shapes re learnt well. Even something as easy as guessing the beginning letter of long words can assist your child improve his phonics abilities. Worksheet For Multiplying Decimals.
There isn't anything like a superb story, and nothing like being the person who started a renowned urban legend. Deciding upon the ideal approach route Cursive writing is basically joined-up handwriting. Practice reading by yourself as often as possible.
Research urban legends to obtain a concept of what's out there prior to making a new one. You are still not sure the radicals have the proper idea. Naturally, you won't use the majority of your ideas. If you've got an idea for a tool please inform us. That means you can begin right where you are no matter how little you might feel you've got to give. You are also quite suspicious of any revolutionary shift. In earlier times you've stated that the move of independence may be too early.
Each lesson in handwriting should start on a fresh new page, so the little one becomes enough room to practice. Every handwriting lesson should begin with the alphabets. Handwriting learning is just one of the most important learning needs of a kid. Learning how to read isn't just challenging, but fun too.
The use of grids The use of grids is vital in earning your child learn to Improve handwriting. Also, bear in mind that maybe your very first try at brainstorming may not bring anything relevant, but don't stop trying. Once you are able to work, you might be surprised how much you get done. Take into consideration how you feel about yourself. Getting able to modify the tracking helps fit more letters in a little space or spread out letters if they're too tight. Perhaps you must enlist the aid of another man to encourage or help you keep focused.
Worksheet For Multiplying Decimals. Try to remember, you always have to care for your child with amazing care, compassion and affection to be able to help him learn. You may also ask your kid's teacher for extra worksheets. Your son or daughter is not going to just learn a different sort of font but in addition learn how to write elegantly because cursive writing is quite beautiful to check out. As a result, if a kid is already suffering from ADHD his handwriting will definitely be affected. Accordingly, to be able to accomplish this, if children are taught to form different shapes in a suitable fashion, it is going to enable them to compose the letters in a really smooth and easy method. Although it can be cute every time a youngster says he runned on the playground, students want to understand how to use past tense so as to speak and write correctly. Let say, you would like to boost your son's or daughter's handwriting, it is but obvious that you want to give your son or daughter plenty of practice, as they say, practice makes perfect.
Without phonics skills, it's almost impossible, especially for kids, to learn how to read new words. Techniques to Handle Attention Issues It is extremely essential that should you discover your kid is inattentive to his learning especially when it has to do with reading and writing issues you must begin working on various ways and to improve it. Use a student's name in every sentence so there's a single sentence for each kid. Because he or she learns at his own rate, there is some variability in the age when a child is ready to learn to read. Teaching your kid to form the alphabets is quite a complicated practice.
Have faith. But just because it's possible, doesn't mean it will be easy. Know that whatever life you want, the grades you want, the job you want, the reputation you want, friends you want, that it's possible.
Top | 2,037 | 10,366 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2020-40 | latest | en | 0.86014 |
http://math.stackexchange.com/questions/90991/what-function-does-sum-limits-n-1-infty-frac1n3n-represent-evalua?answertab=active | 1,462,444,220,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860126502.50/warc/CC-MAIN-20160428161526-00023-ip-10-239-7-51.ec2.internal.warc.gz | 188,559,770 | 18,478 | What function does $\sum \limits_{n=1}^{\infty}\frac{1}{n3^n}$ represent, evaluated at some number $x$?
I need to know what the function $$\sum \limits_{n=1}^{\infty}\frac{1}{n3^n}$$ represents evaluated at a particular point.
For example if the series given was $$\sum \limits_{n=0}^{\infty}\frac{3^n}{n!}$$ the answer would be $e^x$ evaluated at $3$.
Yes, this is homework, but I'm not looking for any handouts, any help would be greatly appreciated.
-
The lower summation limit should probably be $n=1$, instead of $n=0$. – Sasha Dec 13 '11 at 1:59
HINT: $\int_0^x y^{n-1} \mathrm{d} y = \frac{x^n}{n}$. – Sasha Dec 13 '11 at 2:01
Your summand is $\frac{x^n}{n}$, no? It looks a bit like the integral of the geometric series... – J. M. Dec 13 '11 at 2:02
In your example do you mean $\sum \frac{3^n}{n!}$? – Qiaochu Yuan Dec 13 '11 at 2:02
$\sum_{n=1}^\infty \frac{1}{n3^n}$ is an expression, not a function. There is no variable. – Brandon Carter Dec 13 '11 at 2:06
Let $f(x)=\sum_1^{\infty}{1\over nx^n}$; you've got $f(3)$, so you want to know what $f(x)$ is. Differentiate it.
-
We have $\displaystyle -\ln(1-x) = \sum_{n \geq 1 } \frac{x^n}{n}$ by course for $|x|<1$
So this is simply $f(x)=-ln(1-x)$ evaluated at $x=\frac{1}{3}$
Which gives the value : $-\ln(1-\frac{1}{3})=\ln(\frac{3}{2})$
-
Take $f(x) = \displaystyle \sum_{n=1}^\infty \frac{x^n}{n}$.
Then,
$f^\prime (x) = \displaystyle \sum_{n=1}^\infty \frac{n x^{n-1}}{n} = \sum_{n=0}^\infty x^n$.
The last expression is a geometric series and, as long as $x < 1$, it can be expressed as
$f^\prime (x) = \displaystyle \frac{1}{1-x}$.
Therefore,
$f(x) = - \ln | 1 - x | + \kappa$
Where $\kappa$ is a constant. But if you take the original expression for $f(x)$, you can see that $f(0) = 0$ and, therefore, $\kappa = 0$.
So $f(x) = -\ln | 1 - x |$.
The answer to your question is just $f \left(\frac{1}{3} \right)$.
You can also obtain this result by Taylor expanding $\ln ( 1 - x )$.
- | 734 | 1,970 | {"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} | 4.0625 | 4 | CC-MAIN-2016-18 | latest | en | 0.761502 |
https://www.computing.net/answers/programming/using-log10-and-float-and-staticcast/22483.html | 1,582,889,408,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875147154.70/warc/CC-MAIN-20200228104413-20200228134413-00329.warc.gz | 681,912,282 | 9,935 | # Using log10 and float and static_cast
September 15, 2010 at 11:17:52
Specs: Windows 7
// Solution Hint: the base10 logarithm of a number // (available in a function named "log10") tells what // exponent to raise 10 to to get that number. // e.g. 10^2 = 100, so log10(100) = 2; log10(1000) = 3, etc. // Numbers between 100 and 1000 would have logarithms // between 2 and 3 (so if you drop the fraction, you get 2). // // BUT: log10 expects a real number of some kind // (float or double) as an argument, so to successfully // use it in Visual Studio, one must convert the value // above to a real number first, and then convert the // computed logarithm back to an integer for the desired result. // You may use either static_cast() or (float) // to convert an integer to float; and similarly in the reverse. // // NOTE: A correct solution that handles all valid integers // will still need an IF statement!
See More: Using log10 and float and static_cast
#1
September 15, 2010 at 13:49:17
Homework so soon.
Report •
#2
September 16, 2010 at 10:16:54
Report •
#3
September 16, 2010 at 11:05:02
Yeah. Good thing my grade doesn't depend on it, huh?
Report •
Related Solutions
#4
September 17, 2010 at 09:33:53
Figured it out myself, was just looking for help. What a nice forum damn.
Report •
#5
September 17, 2010 at 09:37:56
You weren't looking for help; you were looking for someone to do it for you (and went about it in a very rude manner). There's a difference. Learn it.
Report •
#6
September 17, 2010 at 09:41:30
I didn't need a reply of homework so soon or else I wouldn't have been rude. There was no need for that reply.
Report •
#7
September 17, 2010 at 09:50:57
I'm not buying it. You copy paste your homework, don't bother to remove the C++ comment marks, act like we're obligated to do it, then get upset when we call you out on it.
Report •
#8
September 17, 2010 at 19:10:24
I said so soon because September is usually the start of the academic year. Out in the real world you would have to define the problem, design the solution, then test the boundaries of the solution.
Report •
#9
September 18, 2010 at 07:50:35
My bad then but honestly was just looking for help sorry for coming across ignorant.
Report • | 647 | 2,255 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2020-10 | latest | en | 0.934601 |
https://www.bendwavy.org/klitzing/incmats/todip.htm | 1,627,964,904,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154420.77/warc/CC-MAIN-20210803030201-20210803060201-00198.warc.gz | 663,531,446 | 3,104 | Acronym todip, K-4.59 Name triangle - octagon duoprism,octagon - op wedge Circumradius sqrt[(8+3 sqrt(2))/6] = 1.428440 General of army (is itself convex) Colonel of regiment (is itself locally convex) Dihedral angles at {3} between trip and trip: 135° at {4} between op and trip: 90° at {8} between op and op: 60° Confer general duoprisms: n,m-dip 2n,m-dip 3,n-dip 8,n-dip general polytopal classes: segmentochora Externallinks
As abstract polychoron todip is isomorph to tistodip, thereby replacing octagons by octagrams, resp. op by stop.
Incidence matrix according to Dynkin symbol
```x3o x8o
. . . . | 24 | 2 2 | 1 4 1 | 2 2
--------+----+-------+--------+----
x . . . | 2 | 24 * | 1 2 0 | 2 1
. . x . | 2 | * 24 | 0 2 1 | 1 2
--------+----+-------+--------+----
x3o . . | 3 | 3 0 | 8 * * | 2 0
x . x . | 4 | 2 2 | * 24 * | 1 1
. . x8o | 8 | 0 8 | * * 3 | 0 2
--------+----+-------+--------+----
x3o x . ♦ 6 | 6 3 | 2 3 0 | 8 *
x . x8o ♦ 16 | 8 16 | 0 8 2 | * 3
```
```x3o x8/7o
. . . . | 24 | 2 2 | 1 4 1 | 2 2
----------+----+-------+--------+----
x . . . | 2 | 24 * | 1 2 0 | 2 1
. . x . | 2 | * 24 | 0 2 1 | 1 2
----------+----+-------+--------+----
x3o . . | 3 | 3 0 | 8 * * | 2 0
x . x . | 4 | 2 2 | * 24 * | 1 1
. . x8/7o | 8 | 0 8 | * * 3 | 0 2
----------+----+-------+--------+----
x3o x . ♦ 6 | 6 3 | 2 3 0 | 8 *
x . x8/7o ♦ 16 | 8 16 | 0 8 2 | * 3
```
```x3/2o x8o
. . . . | 24 | 2 2 | 1 4 1 | 2 2
----------+----+-------+--------+----
x . . . | 2 | 24 * | 1 2 0 | 2 1
. . x . | 2 | * 24 | 0 2 1 | 1 2
----------+----+-------+--------+----
x3/2o . . | 3 | 3 0 | 8 * * | 2 0
x . x . | 4 | 2 2 | * 24 * | 1 1
. . x8o | 8 | 0 8 | * * 3 | 0 2
----------+----+-------+--------+----
x3/2o x . ♦ 6 | 6 3 | 2 3 0 | 8 *
x . x8o ♦ 16 | 8 16 | 0 8 2 | * 3
```
```x3/2o x8/7o
. . . . | 24 | 2 2 | 1 4 1 | 2 2
------------+----+-------+--------+----
x . . . | 2 | 24 * | 1 2 0 | 2 1
. . x . | 2 | * 24 | 0 2 1 | 1 2
------------+----+-------+--------+----
x3/2o . . | 3 | 3 0 | 8 * * | 2 0
x . x . | 4 | 2 2 | * 24 * | 1 1
. . x8/7o | 8 | 0 8 | * * 3 | 0 2
------------+----+-------+--------+----
x3/2o x . ♦ 6 | 6 3 | 2 3 0 | 8 *
x . x8/7o ♦ 16 | 8 16 | 0 8 2 | * 3
```
```x3o x4x
. . . . | 24 | 2 1 1 | 1 2 2 1 | 1 1 2
--------+----+----------+-----------+------
x . . . | 2 | 24 * * | 1 1 1 0 | 1 1 1
. . x . | 2 | * 12 * | 0 2 0 1 | 1 0 2
. . . x | 2 | * * 12 | 0 0 2 1 | 0 1 2
--------+----+----------+-----------+------
x3o . . | 3 | 3 0 0 | 8 * * * | 1 1 0
x . x . | 4 | 2 2 0 | * 12 * * | 1 0 1
x . . x | 4 | 2 0 2 | * * 12 * | 0 1 1
. . x4x | 8 | 0 4 4 | * * * 3 | 0 0 2
--------+----+----------+-----------+------
x3o x . ♦ 6 | 6 3 0 | 2 3 0 0 | 4 * *
x3o . x ♦ 6 | 6 0 3 | 2 0 3 0 | * 4 *
x . x4x ♦ 16 | 8 8 8 | 0 4 4 2 | * * 3
```
```x3/2o x4x
. . . . | 24 | 2 1 1 | 1 2 2 1 | 1 1 2
----------+----+----------+-----------+------
x . . . | 2 | 24 * * | 1 1 1 0 | 1 1 1
. . x . | 2 | * 12 * | 0 2 0 1 | 1 0 2
. . . x | 2 | * * 12 | 0 0 2 1 | 0 1 2
----------+----+----------+-----------+------
x3/2o . . | 3 | 3 0 0 | 8 * * * | 1 1 0
x . x . | 4 | 2 2 0 | * 12 * * | 1 0 1
x . . x | 4 | 2 0 2 | * * 12 * | 0 1 1
. . x4x | 8 | 0 4 4 | * * * 3 | 0 0 2
----------+----+----------+-----------+------
x3/2o x . ♦ 6 | 6 3 0 | 2 3 0 0 | 4 * *
x3/2o . x ♦ 6 | 6 0 3 | 2 0 3 0 | * 4 *
x . x4x ♦ 16 | 8 8 8 | 0 4 4 2 | * * 3
```
```ox xx4xx&#x → height = sqrt(3)/2 = 0.866025
o. o.4o. | 8 * | 1 1 2 0 0 0 | 1 1 2 2 0 0 0 | 1 1 2 0
.o .o4.o | * 16 | 0 0 1 1 1 1 | 0 1 1 1 1 1 1 | 1 1 1 1
------------+------+--------------+---------------+--------
.. x. .. | 2 0 | 4 * * * * * | 1 0 2 0 0 0 0 | 1 0 2 0
.. .. x. | 2 0 | * 4 * * * * | 1 0 0 2 0 0 0 | 0 1 2 0
oo oo4oo&#x | 1 1 | * * 16 * * * | 0 1 1 1 0 0 0 | 1 1 1 0
.x .. .. | 0 2 | * * * 8 * * | 0 1 0 0 1 1 0 | 1 1 0 1
.. .x .. | 0 2 | * * * * 8 * | 0 0 1 0 1 0 1 | 1 0 1 1
.. .. .x | 0 2 | * * * * * 8 | 0 0 0 1 0 1 1 | 0 1 1 1
------------+------+--------------+---------------+--------
.. x.4x. | 8 0 | 4 4 0 0 0 0 | 1 * * * * * * | 0 0 2 0
ox .. ..&#x | 1 2 | 0 0 2 1 0 0 | * 8 * * * * * | 1 1 0 0
.. xx ..&#x | 2 2 | 1 0 2 0 1 0 | * * 8 * * * * | 1 0 1 0
.. .. xx&#x | 2 2 | 0 1 2 0 0 1 | * * * 8 * * * | 0 1 1 0
.x .x .. | 0 4 | 0 0 0 2 2 0 | * * * * 4 * * | 1 0 0 1
.x .. .x | 0 4 | 0 0 0 2 0 2 | * * * * * 4 * | 0 1 0 1
.. .x4.x | 0 8 | 0 0 0 0 4 4 | * * * * * * 2 | 0 0 1 1
------------+------+--------------+---------------+--------
ox xx ..&#x ♦ 2 4 | 1 0 4 2 2 0 | 0 2 2 0 1 0 0 | 4 * * *
ox .. xx&#x ♦ 2 4 | 0 1 4 2 0 2 | 0 2 0 2 0 1 0 | * 4 * *
.. xx4xx&#x ♦ 8 8 | 4 4 8 0 4 4 | 1 0 4 4 0 0 1 | * * 2 *
.x .x4.x ♦ 0 16 | 0 0 0 8 8 8 | 0 0 0 0 4 4 2 | * * * 1
``` | 3,045 | 5,020 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-31 | latest | en | 0.556062 |
https://www.manualslib.com/manual/359091/Hp-F2226a-48gii-Graphic-Calculator.html?page=675 | 1,571,463,579,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986688826.38/warc/CC-MAIN-20191019040458-20191019063958-00320.warc.gz | 994,777,876 | 60,384 | # HP F2226A - 48GII Graphic Calculator User Manual Page 675
Graphing calculator.
Store the new program back into variable @@@p@@@. Press @@@p@@@ to run the program.
Enter values of V = 0.01_m^3 and T = 300_K in the input string, then press
`. The result is 49887.06_J/m^3. The units of J/m^3 are equivalent to
Pascals (Pa), the preferred pressure unit in the S.I. system.
Note: because we deliberately included units in the function definition, the
input values must have units attach to them in input to produce the proper
result.
Input string program for three input values
The input string program for three input values, say a ,b, and c, looks as
follows:
"Enter a, b and c: " {"
«
This program can be easily created by modifying the contents of INPT2 to
make it look like shown immediately above. The resulting program can then
be stored in a variable called INPT3. With this program we complete the
collection of input string programs that will allow us to enter one, two, or three
data values. Keep these programs as a reference and copy and modify them
to fulfill the requirements of new programs you write.
Application: evaluating a function of three variables
Suppose that we want to program the ideal gas law including the number of
moles, n, as an additional variable, i.e., we want to define the function
p
(
and modify it to include the three-variable input string. The procedure to put
together this function is very similar to that used earlier in defining the function
p(V,T). The resulting program will look like this:
"Enter V, T, and n:" {"
«
INPUT OBJ→ → V T n
Store this result back into the variable @@@p@@@.To run the program, press @@@p@@@.
:a:
:b:
INPUT OBJ→ "
V
,
T
,
n
)
(
. 8
31451
_
K
:V:
'(8.31451_J/(K*mol))*(n*T/V) ' "
:c: " {2 0} V }
J
n
T
)
,
V
:T:
:n:" {2 0} V }
Page 21-26 | 503 | 1,811 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2019-43 | latest | en | 0.829858 |
https://undergroundmathematics.org/circles/r9386/suggestion | 1,660,155,807,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571198.57/warc/CC-MAIN-20220810161541-20220810191541-00711.warc.gz | 522,737,986 | 5,231 | Review question
# Can we find the area bounded by these two graphs? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource
Ref: R9386
## Suggestion
The area bounded by the graphs $y=\sqrt{2-x^2} \qquad \text{and} \qquad x+\left(\sqrt{2}-1\right)y=\sqrt{2}$ equals
1. $\dfrac{\sin{\sqrt{2}}}{\sqrt{2}}$;
2. $\dfrac{\pi}{4}-\dfrac{1}{\sqrt{2}}$;
3. $\dfrac{\pi}{2\sqrt{2}}$;
4. $\dfrac{\pi^2}{6}$.
Could we start by sketching the graphs of our two functions?
How do we calculate the area of the required shape? Could we write it as the difference of two other areas?
What would we expect our answer to look like? Do some of the options in the list seem more likely than others? | 226 | 761 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2022-33 | latest | en | 0.889711 |
https://www.getbeautytip.com/2022/03/quotient-rule-u-v-derivative-formula.html | 1,670,409,028,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711151.22/warc/CC-MAIN-20221207085208-20221207115208-00676.warc.gz | 803,039,057 | 55,078 | VqI14dIZgOPEqICDVdzsdHohm6R1qA6BYQ86dmeQ
# Quotient Rule U V Derivative Formula
We have f(x) = 1/sin x. Here, u(x) equals one and v(x) equals sin x. Thus, u'(x) equals 0 and v'(x) equals cos x. We obtain f'(x) = [v(x) u'(x) â u(x) v'(x)] using the quotient rule. /[v(x)] 2 equals [sin x (0) â 1 (cos x)] /cos2 x equals -cos x/ cos2 x equals -1/cos x equals -sec x
Calculating the products of more than two functions is really rather straightforward. Consider the three function product rule, for instance. To begin, we do not consider it a product of three functions, but rather the product rule of two functions (f,g) and (h), on which we may then apply the two function product rule. This results in,
Related Posts | 230 | 720 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2022-49 | latest | en | 0.837388 |
https://au.mathworks.com/matlabcentral/answers/891317-plot-three-variables-with-three-separate-y-axis?s_tid=prof_contriblnk | 1,686,306,653,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656675.90/warc/CC-MAIN-20230609100535-20230609130535-00128.warc.gz | 129,212,691 | 25,437 | # Plot Three Variables with Three separate Y axis
6 views (last 30 days)
Patrick Lonergan on 3 Aug 2021
Answered: Bjorn Gustavsson on 3 Aug 2021
Hi I am trying to plot three vairables on one graph. I am currently plotting it two seperate graphs and comparing but it would be really useful if I could plot one graph with all three vairables. The code I am using to produce the two graphs is as below:
yyaxis left
y=plot(failure.RMS)
ylabel("RMS Error")
yyaxis right
y=plot(failure.RollingAverage)
ylabel("rate of change of Power(W/10min)(rolling 500)")
legend("RMS Error of ML prediction","Rate of change of Power")
legend("Location","northwest")
title("RMS Error and Rolling average rate Of Power Prior to failure")
saveas(y,"Rolling_RMS Error1_11",'jpg');
saveas(y,"Rolling_RMS Error1_11",'m');
clf
yyaxis left
y=plot(failure.RMS)
ylabel("RMS Error")
yyaxis right
y=plot(failure.RollingAvePower)
ylabel("rate of change of Power(KW)(rolling 500)")
legend("RMS Error of ML prediction","Rolling Power Average")
legend("Location","northwest")
title("RMS Error and Rolling average Power Prior to failure")
saveas(y,"Rolling_power Error1_11",'jpg');
saveas(y,"Rolling_power Error1_11",'m');
Bjorn Gustavsson on 3 Aug 2021
You might use the plotyyy function from the matlab file excange to do just that. Or any of the other functions that show up when searching for plotyyy: FEX: plotyyy.
HTH
### Categories
Find more on 2-D and 3-D Plots in Help Center and File Exchange
R2021a
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 435 | 1,605 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-23 | latest | en | 0.776913 |
https://www.coursehero.com/file/6425907/Handout23/ | 1,516,720,349,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891976.74/warc/CC-MAIN-20180123131643-20180123151643-00338.warc.gz | 850,717,011 | 30,175 | # Handout23 - M ohr's Circle Examples M ohr's Circle 3000...
This preview shows pages 1–4. Sign up to view the full content.
Mohr's Circle Examples Mohr's Circle 3000 Example 1: 2000 Given: Element with principal stresses on horizontal and vertical planes Find: 1000 Stresses on a plane rotated 30 degrees from the horizontal 0'1 = 4000 psi 't, psf 0') = 1000 pif -3000 0'1 = 4000 psi 0'+0' ( . 0'-0') 0' = 1 3 + cos60 - 1 2 ) u +u Center of circle = _I __ 3 = 2500 psf 2 0' = 2500 + (0.5 -1500) = 3250psi ct 1 u 3 Radius ofthe circle = - = 1500 psf 2 0'-0') l' = sin 60· - 1 2 ) = 0.866 -1500 = 1299 psi (
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Mohr's Circle Examples Mohr's Circle Example 2: Given: Element with principal stress on planes at 45 degrees from horizontal and vertical Find: Stresses on a plane rotated 30 degrees from the major principal plane 't, psf 3000 2000 1000 -1000 -2000 -3000 0'+0' ( . a-a) 0':::: 1 2 3 + cos60 - 1 2 3 0' :::: 2500 + (0.5 - 1500) = 3250 psi Center of circle = 0"1 + 0"3 = 2500 psf 2 0"1 -0'3) l' = sm60 - 2 :::: 0.866-1500 = 1299psJ Radius ofthe circle 0") - 0"3 = 1500 psf ( 2
Mohr's Circle Examples Mohr's Circle
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up to access the rest of the document.
{[ snackBarMessage ]}
### Page1 / 4
Handout23 - M ohr's Circle Examples M ohr's Circle 3000...
This preview shows document pages 1 - 4. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online | 568 | 1,643 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2018-05 | latest | en | 0.732097 |
pairfonts.eu | 1,725,719,385,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650883.10/warc/CC-MAIN-20240907131200-20240907161200-00844.warc.gz | 429,357,866 | 5,235 | The Typographic Scale – Harmony in Fonts
Date: 6/1/2020, Author: G.P.
The classic typographic scale is a method that refers to the way font sizes are progressing. Typography, from print media to the web, has always needed a way to present text in a consistent manner. Text should always be presentable and easy to read. So, that’s where typographic scales come into play. They offer us a system that allows us to use text harmonically. They actually resemble the way musical notes and musical scales are working.
Here is the classic typographic scale that can be found in many, commonly used, applications:
6 7 8 9 10 12 14 16 18 21 24 28 32 36 42 48 55 63 73 84 96
You will notice that some numbers are in bold. You can also easily notice that every one of those bold numbers is a multiple of the previous number and the number 2. So, the number 2 is the ratio of our scale. As you can see, there are always exactly 5 numbers from the first bold number to the next one. This is the interval of the scale.
So, lets summarize:
A typographic scale consists of 2 basic properties:
1. A ratio (r). Every size must be multiplied by the ratio to find the next harmonic size. For example, the classic typographic scale has a ratio of 2. A ratio of 1.5 (which is called a perfect fifth in music), means that any font size is exactly 1.5 times bigger than the one before it. We call it a perfect fifth because in a musical scale with a ratio 1.5, there are always 5 notes in the sequence. In the same way we can also use the Golden Ratio (1.618) to create harmonic typographic scales.
2. An interval (s). The number of sizes from the first element to the last element of a sequence. For example, if we have this scale:
12, ..., …, …, …, 24, ..., …, …, …, 48 ..., …, …, …,
we have a ratio r=2 and a number of sizes s=5.
So, if i is the ith element of the scale and n is the size number from which we want to know the ith harmonic size, we can use the basic formula for the frequencies of the notes of the equal tempered scale (a scale with the same ratio):
xi = n*ri/s
Thus, if we want to know the 2nd size from 12 we can find it using the previous formula:
x2 = 12*22/5 = 12*20,4 = 12*1,319 = 15,82 ≈ 16
Similarly, the 3rd size from 12 is:
x3 = 12*23/5 = 12*20,6 = 12*1,515 = 18,18 ≈ 18
Our scale now is:
12, 16, 18, …, …, 24, ..., …, …, …, 48 ..., …, …, …,
We can continue to fill all the elements in the same way we found the 2nd and the 3rd size.
This will give as a harmonic size scale.
How to use typographic scales on the web
A simple way to use a typographic scale on the web is to find the basic size of the body. You can use either pixels or ems (there is a difference that we will analyze in a next article about Responsive Typography). After that you can assign the next harmonic size to the smaller header (h6, h5, h4, h3, h2, h1). You could also revert the process and start with the largest size.
Finally, some ratios have been proven more harmonious than others. Here is a list of some scales you could experiment with:
Minor Second
Major Second
Minor Third
Major Third
Perfect Fourth
Perfect Fifth
Golden Ratio
You can find the full scales here - https://type-scale.com/
Of course, you can always experiment using your own values.
References:
The Typographic Scale by Spencer Mortensen
Equal Temperament - https://en.wikipedia.org/wiki/Equal_temperament
The Golden Ratio - https://en.wikipedia.org/wiki/Golden_ratio
Perfect Fifth - https://en.wikipedia.org/wiki/Perfect_fifth | 930 | 3,513 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-38 | latest | en | 0.915528 |
https://mathspace.co/textbooks/syllabuses/Syllabus-411/topics/Topic-7344/subtopics/Subtopic-97838/?activeTab=theory | 1,642,709,801,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302622.39/warc/CC-MAIN-20220120190514-20220120220514-00613.warc.gz | 388,189,895 | 65,056 | New Zealand
Level 8 - NCEA Level 3
# Differentiation various functions
Lesson
In routine problems requiring differentiation, we rely on several previously established rules for differentiating various types of functions. However, one should keep in mind the meaning of the derivative as the function that gives the rate of change or gradient of the original function at each point in the domain.
Specifically, if $f$f is a smoothly continuous function, then its derivative at a number $x$x in the domain is
$f'(x)=\lim_{a\rightarrow x}\frac{f(x)-f(a)}{x-a}$f(x)=limaxf(x)f(a)xa
The rules that are used in differentiation have been deduced from and are consistent with this definition.
In practice, we make much use of the facts that
• The derivative of a sum is the sum of the separate derivatives.
• The derivative of a constant multiple of a function is the constant multiple multiplied by the derivative of the function.
We often need the
• product rule: If $f(x)=g(x)\cdot h(x)$f(x)=g(x)·h(x), then $f'(x)=g'(x)\cdot h(x)+g(x)\cdot h'(x)$f(x)=g(x)·h(x)+g(x)·h(x)
• quotient rule: If $f(x)=\frac{g(x)}{h(x)}$f(x)=g(x)h(x), then $f'(x)=\frac{g'(x)\cdot h(x)-g(x)\cdot h'(x)}{\left[h(x)\right]^2}$f(x)=g(x)·h(x)g(x)·h(x)[h(x)]2
• function-of-a-function rule (also called the chain rule): If $f(x)=g\left(h(x)\right)$f(x)=g(h(x)), then $f'(x)=g'\left(h(x)\right)\cdot h'(x)$f(x)=g(h(x))·h(x), provided $h(x)$h(x) is in the domain of $g(x)$g(x)
When the function-of-a-function rule is expressed in the Leibniz notation, it is possible to see the reason for the term chain rule. If, for example, a function is given by $u\left(v(x)\right)$u(v(x)), we write $\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}u}{\mathrm{d}v}\cdot\frac{\mathrm{d}v}{\mathrm{d}x}$dudx=dudv·dvdx
We may also need the derivatives of some special functions.
• $\frac{\mathrm{d}}{\mathrm{d}x}\ln x=\frac{1}{x}$ddxlnx=1x
• $\frac{\mathrm{d}}{\mathrm{d}x}\sin x=\cos x$ddxsinx=cosx
• $\frac{\mathrm{d}}{\mathrm{d}x}\cos x=-\sin x$ddxcosx=sinx
• $\frac{\mathrm{d}}{\mathrm{d}x}e^x=e^x$ddxex=ex
##### Example 1
Find the derivative of $\sin\left(\ln x\right)$sin(lnx).
We need the function-of-a-function rule. It may be helpful to think of an inside function $\ln$ln and an outside function $\sin$sin. We differentiate the outside function first, evaluated at $\ln x$lnx, and multiply by the derivative of the inside function. Hence, if $f(x)=\sin\left(\ln x\right)$f(x)=sin(lnx), we have
$f'(x)=\cos\left(\ln x\right)\cdot\left(\frac{1}{x}\right)$f(x)=cos(lnx)·(1x)
##### Example 2
Find the derivative of the function $f(x)=\left(x+\sin x\right)^{\frac{3}{2}}$f(x)=(x+sinx)32 and calculate the gradient of $f$f at $x=0$x=0, $x=\frac{\pi}{4}$x=π4 and $x=\frac{\pi}{2}$x=π2.
First, we differentiate the power function to obtain $\frac{3}{2}\left(x+\sin x\right)^{\frac{1}{2}}$32(x+sinx)12. Then, we differentiate the inside function to obtain $1+\cos x$1+cosx. Finally, we have
$f'(x)=\frac{3}{2}\left(x+\sin x\right)^{\frac{1}{2}}\cdot\left(1+\cos x\right)$f(x)=32(x+sinx)12·(1+cosx).
We calculate, $f'(0)=\frac{3}{2}\left(0+\sin0\right)^{\frac{1}{2}}\cdot\left(1+\cos0\right)=0$f(0)=32(0+sin0)12·(1+cos0)=0,
Next, $f'\left(\frac{\pi}{4}\right)=\frac{3}{2}\left(\frac{\pi}{4}+\sin\frac{\pi}{4}\right)^{\frac{1}{2}}\cdot\left(1+\cos\frac{\pi}{4}\right)=\frac{3}{2}\sqrt{\frac{\pi}{4}+\frac{1}{\sqrt{2}}}\cdot(1+\frac{1}{\sqrt{2}})\approx3.128$f(π4)=32(π4+sinπ4)12·(1+cosπ4)=32π4+12·(1+12)3.128
And lastly, $f'\left(\frac{\pi}{2}\right)=\frac{3}{2}\left(\frac{\pi}{2}+\sin\frac{\pi}{2}\right)^{\frac{1}{2}}\cdot\left(1+\cos\frac{\pi}{2}\right)=\frac{3}{2}\sqrt{\frac{\pi}{2}+1}\approx2.405$f(π2)=32(π2+sinπ2)12·(1+cosπ2)=32π2+12.405
##### Example 3
Find the gradient of $f(x)=e^{-x}\left(x^3-1\right)$f(x)=ex(x31) at $x=1$x=1.
Using the product rule and the function-of-a-function rule, we have $f'(x)=(-1)e^{-x}\left(x^3-1\right)+e^{-x}\left(3x^2\right)$f(x)=(1)ex(x31)+ex(3x2)
This is written more tidily as $f'(x)=e^{-x}\left(1+3x^2-x^3\right)$f(x)=ex(1+3x2x3).
Then, $f'(1)=e^{-1}\times3=\frac{3}{e}\approx1.1$f(1)=e1×3=3e1.1.
#### Worked Examples
##### Question 1
Differentiate $y=\sin\left(x\right)e^x$y=sin(x)ex. Give your answer in factorised form.
##### Question 2
Find the derivative of $y=\cos\left(\ln x\right)$y=cos(lnx).
##### Question 3
Find the derivative of $\ln\left(\sin x\right)$ln(sinx).
### Outcomes
#### M8-11
Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods
#### 91578
Apply differentiation methods in solving problems | 1,701 | 4,666 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2022-05 | longest | en | 0.751224 |
https://www.lmfdb.org/EllipticCurve/Q/8470/z/ | 1,638,011,706,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358180.42/warc/CC-MAIN-20211127103444-20211127133444-00594.warc.gz | 977,345,590 | 20,359 | # Properties
Label 8470.z Number of curves $4$ Conductor $8470$ CM no Rank $0$ Graph
# Related objects
Show commands: SageMath
sage: E = EllipticCurve("z1")
sage: E.isogeny_class()
## Elliptic curves in class 8470.z
sage: E.isogeny_class().curves
LMFDB label Cremona label Weierstrass coefficients j-invariant Discriminant Torsion structure Modular degree Faltings height Optimality
8470.z1 8470ba4 $$[1, -1, 1, -2530012, 1549562831]$$ $$1010962818911303721/57392720$$ $$101674704435920$$ $$[2]$$ $$122880$$ $$2.1543$$
8470.z2 8470ba3 $$[1, -1, 1, -264892, -12309041]$$ $$1160306142246441/634128110000$$ $$1123396628679710000$$ $$[2]$$ $$122880$$ $$2.1543$$
8470.z3 8470ba2 $$[1, -1, 1, -158412, 24149711]$$ $$248158561089321/1859334400$$ $$3293924308998400$$ $$[2, 2]$$ $$61440$$ $$1.8078$$
8470.z4 8470ba1 $$[1, -1, 1, -3532, 855759]$$ $$-2749884201/176619520$$ $$-312892253470720$$ $$[4]$$ $$30720$$ $$1.4612$$ $$\Gamma_0(N)$$-optimal
## Rank
sage: E.rank()
The elliptic curves in class 8470.z have rank $$0$$.
## Complex multiplication
The elliptic curves in class 8470.z do not have complex multiplication.
## Modular form8470.2.a.z
sage: E.q_eigenform(10)
$$q + q^{2} + q^{4} + q^{5} - q^{7} + q^{8} - 3 q^{9} + q^{10} + 6 q^{13} - q^{14} + q^{16} + 2 q^{17} - 3 q^{18} + 4 q^{19} + O(q^{20})$$
## Isogeny matrix
sage: E.isogeny_class().matrix()
The $$i,j$$ entry is the smallest degree of a cyclic isogeny between the $$i$$-th and $$j$$-th curve in the isogeny class, in the LMFDB numbering.
$$\left(\begin{array}{rrrr} 1 & 4 & 2 & 4 \\ 4 & 1 & 2 & 4 \\ 2 & 2 & 1 & 2 \\ 4 & 4 & 2 & 1 \end{array}\right)$$
## Isogeny graph
sage: E.isogeny_graph().plot(edge_labels=True)
The vertices are labelled with LMFDB labels. | 693 | 1,744 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2021-49 | latest | en | 0.467092 |
https://www.analyticsisnormal.com/post/loops-in-python | 1,726,008,795,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651323.1/warc/CC-MAIN-20240910224659-20240911014659-00599.warc.gz | 605,091,409 | 206,436 | top of page
Search
• Ekta Aggarwal
# Loops in Python
There are two loop in python which we shall be studying in detail in this tutorial:
### While loops
While loops are entry-controlled loops which means they test for a condition first. If the condition is true then only the loop is executed otherwise the loop will terminate.
Syntax:
```while(condition to be tested):
<Do something>
else:
<Do something>```
Let us create a temporary variable:
`temp = 5`
Following while loop checks if the value of temp is less than or equal to 10 or not. If temp <= 10 then the loop is executed.
```while(temp <=10):
print(temp);
temp+=1;```
temp +=1 means temp = temp+1 i.e. increment the value of temp by 1.
For temp = 5, while will get executed, it will print 5 and increment temp by 1 i.e. temp will now be equal to 6.
Now for temp = 6, loop will be executed and temp will be incremented to 7.
When temp = 11 the condition will get false and the loop will be terminated.
### while - else
We can also define an else clause with while loop which will be executed at the time of exiting the loop
```temp = 5;
while(temp <=10):
print(temp);
temp+=1;
else: print('Exiting the loop at temp = '+ str(temp))```
At temp = 11 while loop will be terminated and will move to the else part.
You can also ad if else conditions inside a while loop.
In the following example when temp = 7 we are printing 7 is my lucky number, otherwise for all other values it will just print the number.
```temp = 5;
while(temp <=10):
if temp == 7: print('7 is my lucky number')
else: print(temp)
temp+=1;
else: print('Exiting the loop at temp = '+ str(temp))```
### For loop
Unlike for loop we do not need to specify an increment statement (like temp+=1 in case of while). For loop automatically increments the iterator_variable by 1.
Syntax:
```for iterator_variable in iterable:
<Do something>```
### Iterating over lists
Let us create a list:
`my_list = ['Washington D.C.','London','Tokyo','Moscow']`
In the following for loop, our iterator variable is i and we are iterating over each element of our list:
```for i in my_list:
print(i)```
Firstly i takes the value "Washington D.C." and prints it, then i takes the value "London" and it gets printed until it reaches the end of the list.
### Iterating over range
range(x,y) is like a list starting from x till y-1 (not y is excluded) Eg. range(1,5) is actually a list [1,2,3,4]
```for i in range(1,5):
print(i)```
range(x) is a list starting from 0 and containing elements till x-1 i.e. range(5) is a list [0,1,2,3,4]
```for i in range (5):
print(i)```
range(x,y,by) is a list starting from x till y-1 where numbers are incremented with 'by' . Eg.
range(1,10,2) = [1,3,5,7,9]
```for i in range(1,10,2):
print(i)```
### Nested for loop:
A for loop inside a for loop is called nested for loop.
In the following code firstly i takes the value 1 and iterates over all the elements of my_list (i.e. j).
Then i = 2 and all the elements of my_list get printed and so on.
```for i in range(1,5):
for j in my_list:
print(i,j)```
For different iterables for loops get slightly modified. To learn more you can refer to this detailed tutorial: For loop for different iterables.
### Break statement
break statement can be used with both for loop and while loops. Whenever a loop encounters a break statement then all the loops (if-else conditions, for loops, while loops, nested loops) would be terminated.
```for mynum in range(50):
if (mynum > 5):
break
print('Current Number :', mynum)```
In the above code, as soon as mynum takes the value 6 break statement gets trigerred and our loop's execution get terminated. Thus we have values printed only till mynum = 5.
### Continue statement
continue statement can be used with both for loop and while loops. Whenever a loop encounters a continue statement then Python moves the control to the beginning of the loop - it won't break the loop but will not run further commands and will increment the value of the iterator.
```for mynum in range(10):
print('Current Number :', mynum) #first print statement
if (mynum > 5):
continue
print('Next Number would be :', mynum+1) #second print statement```
In the above code, when mynum is <=5 then current and next number get printed, but as soon as mynum takes the value 6 continue statement gets trigerred - which means the next number (second print statement) won't be executed. and mynum would be incremented to 7 and the loop would continue.
bottom of page | 1,160 | 4,506 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.788301 |
https://meijerstyle.com/rockfield/application-of-zeroth-law-of-thermodynamics.php | 1,627,167,017,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151531.67/warc/CC-MAIN-20210724223025-20210725013025-00649.warc.gz | 398,283,728 | 8,566 | # Application Of Zeroth Law Of Thermodynamics
Zeroth law of thermodynamics Simple English Wikipedia. 1.1 The zeroth law The zeroth law of thermodynamics established the existence of temperature in macro-scopic systems. Temperature is a state quantity which is unknown, College admissions; Careers; Zeroth law of thermodynamics. Let's talk about the 0th Law of Thermodynamics. Now the 0th law,.
### What are the applications of the zeroth law of thermodynamics?
Zeroth law of thermodynamics Revolvy. The zeroth law is incredibly important as it allows us to define the concept of a temperature scale. If two systems are each in thermal equilibrium with a third, they are also in thermal equilibrium with each other. For example, consider two separate cups of …, The zeroth law is incredibly important as it allows us to define the concept of a temperature scale. If two systems are each in thermal equilibrium with a third, they are also in thermal equilibrium with each other. For example, consider two separate cups of ….
Learn about the Three Laws of Thermodynamics including The Zeroth Law of Thermodynamics says that Need to control current flow in low voltage applications? What is Thermodynamics and explanation of zeroth, first, second, and third law of thermodynamics.
Zeroth Law of thermodynamics is the most basic of the laws even though it was defined the last. That is why it was named zero (zeroth) instead of four. It forms the After watching this video, you will be able to state the zeroth law of thermodynamics, what it means, and why it is important. A short quiz will...
College admissions; Careers; Zeroth law of thermodynamics. Let's talk about the 0th Law of Thermodynamics. Now the 0th law, Zeroth law of thermodynamics: If two systems are in thermal equilibrium with a third system, they are in thermal equilibrium with each …
Zeroth Law of Thermodynamics. The "zeroth law" states that if two systems are at the same time in thermal equilibrium with a third system, … This module explores thermodynamic definitions, the zeroth law of thermodynamics and temperature, the first law of thermodynamics and enthalpy
1. Zeroth law of thermodynamics:-. Zeroth law of thermodynamics states that when two systems are each in thermal equilibrium with the third system, they are also in thermal equilibrium with each other. Zeroth Law of Thermodynamics. One of the laws related to the thermal equilibrium is Zeroth law of Thermodynamics. Suppose two bodies are separated by a heat insulator. Now a third body is brought in contact with both the bodies simultaneously. When the bodies attain thermal equilibrium we remove the third body.
In the year of 1931, R.H. fowler had provided one law of thermodynamics which is termed as Zeroth law of thermodynamics and according to this law What is Thermodynamics and explanation of zeroth, first, second, and third law of thermodynamics.
The laws are as follows. Zeroth law of thermodynamics – If two thermodynamic systems are each in thermal equilibrium with a third, then they are in thermal equilibrium with each other. First law of thermodynamics – Energy can neither be created nor destroyed. The zeroth law of thermodynamics states that if two thermodynamic systems are each in thermal equilibrium with a third, then they are in thermal equilibrium with each
First law of thermodynamics states that in any thermodynamic process,when heat Q is added to a system,this energy appears as an Zeroth law of thermodynamic The zeroth law of thermodynamics states that there is no thermal energy flow between systems with the same temperature. The zeroth law makes temperature as an absolute concept possible.
Explore this introduction to the three laws of thermodynamics and how they are used to solve problems involving heat or thermal Second Law of Thermodynamics: After watching this video, you will be able to state the zeroth law of thermodynamics, what it means, and why it is important. A short quiz will...
### Zeroth law of thermodynamics Revolvy
Zeroth Law of Thermodynamics Came After the Other Laws. First law of thermodynamics states that in any thermodynamic process,when heat Q is added to a system,this energy appears as an Zeroth law of thermodynamic, Zeroth law of thermodynamics: If two systems are in thermal equilibrium with a third system, they are in thermal equilibrium with each ….
Zeroth First & Second Law Of Thermodynamics. The applications of the thermodynamic laws and principles are found in all fields of energy technology, Zeroth Law Of Thermodynamics., The zeroth law of thermodynamics states that if two thermodynamic systems are each in thermal equilibrium with a third, then all three are in thermal equilibrium with.
### What are the applications of the zeroth law of thermodynamics?
Zeroth Law of Thermodynamics With Examples. 1.1 The zeroth law The zeroth law of thermodynamics established the existence of temperature in macro-scopic systems. Temperature is a state quantity which is unknown https://en.m.wikipedia.org/wiki/0th_law The zeroth law of thermodynamics states that if two thermodynamic systems are each in thermal equilibrium with a third, then all three are in thermal equilibrium with.
Have you heard of thermodynamics or applications of thermodynamics in your daily routine? Study of thermodynamics involves The zeroth law of thermodynamics, Explore this introduction to the three laws of thermodynamics and how they are used to solve problems involving heat or thermal Second Law of Thermodynamics:
The zeroth law of thermodynamics states that if two thermodynamic systems are each in thermal equilibrium with a third, then all three are in thermal equilibrium with The zeroth law of thermodynamics states that if two thermodynamic systems are each in thermal equilibrium with a third, then they are in thermal equilibrium with each
The four laws of thermodynamics are: Zeroth law of thermodynamics: If two systems are in thermal equilibrium with a third system, they are in thermal equilibrium with each other. This law helps define the notion of temperature. The zeroth law of thermodynamics states that if two thermodynamic systems are each in thermal equilibrium with a third, then they are in thermal equilibrium with each
The Zeroth Law of Thermodynamics states that if two bodies are each in thermal equilibrium with some third body, then they are also in equilibrium with each other. The Zeroth Law of Thermodynamics states that if two bodies are each in thermal equilibrium with some third body, then they are also in equilibrium with each other. The zeroth law is incredibly important as it allows us to define the concept of a temperature scale. If two systems are each in thermal equilibrium with a third, they are also in thermal equilibrium with each other. For example, consider two separate cups of …
1. Zeroth law of thermodynamics:-. Zeroth law of thermodynamics states that when two systems are each in thermal equilibrium with the third system, they are also in thermal equilibrium with each other. The applications of the thermodynamic laws and principles are found in all fields of energy technology, Zeroth Law Of Thermodynamics.
The zeroth law of thermodynamics states that if two thermodynamic systems are each in thermal equilibrium with a third, then they are in thermal equilibrium with each other. Two systems are said to be in the relation of thermal equilibrium if they are linked by a wall permeable only to heat and they do not change over time. [1] The laws are as follows. Zeroth law of thermodynamics – If two thermodynamic systems are each in thermal equilibrium with a third, then they are in thermal equilibrium with each other. First law of thermodynamics – Energy can neither be created nor destroyed.
Learn about the Three Laws of Thermodynamics including The Zeroth Law of Thermodynamics says that Need to control current flow in low voltage applications? The zeroth law is incredibly important as it allows us to define the concept of a temperature scale. If two systems are each in thermal equilibrium with a third, they are also in thermal equilibrium with each other. For example, consider two separate cups of …
Zeroth Law Of Thermodynamics. The zeroth law of thermodynamics states that if two systems, A and B, are in thermal equilibrium with a third system, C, then A and B are in thermal equilibrium with each other. It is analogous to the transitive property in math (if A=C and B=C, then A=B). Learn about the Three Laws of Thermodynamics including The Zeroth Law of Thermodynamics says that Need to control current flow in low voltage applications?
The zeroth law is incredibly important as it allows us to define the concept of a temperature scale. If two systems are each in thermal equilibrium with a third, they are also in thermal equilibrium with each other. For example, consider two separate cups of … The zeroth law of thermodynamics states that if two thermodynamic systems are each in thermal equilibrium with a third, then all three are in thermal equilibrium with
Another important application of second law of thermodynamics is found in refrigerators and heat pumps, Zeroth law of Thermodynamics. Another important application of second law of thermodynamics is found in refrigerators and heat pumps, Zeroth law of Thermodynamics.
Samsung Galaxy S5 Unfortunately, Samsung Accessory Service Has Stopped Issue & Other App Related Problems Samsung galaxy s5 application manager Barons ... Samsung Galaxy S5 user apps - app manager - Android more I have been using Samsung S5 for almost 4 years but never had problem. my phone by
## Zeroth law of thermodynamics The Full Wiki
Zeroth law of thermodynamics Revolvy. The Zeroth Law of Thermodynamics states that:- two system which are equal in temperature to a third system are equal in temperature to each other., 1.1 The zeroth law The zeroth law of thermodynamics established the existence of temperature in macro-scopic systems. Temperature is a state quantity which is unknown.
### Zeroth Law of Thermodynamics With Examples
Zeroth Law of Thermodynamics ME Mechanical. What is Thermodynamics and explanation of zeroth, first, second, and third law of thermodynamics., Zeroth Law of Thermodynamics. One of the laws related to the thermal equilibrium is Zeroth law of Thermodynamics. Suppose two bodies are separated by a heat insulator. Now a third body is brought in contact with both the bodies simultaneously. When the bodies attain thermal equilibrium we remove the third body..
After watching this video, you will be able to state the zeroth law of thermodynamics, what it means, and why it is important. A short quiz will... The zeroth law of thermodynamics states that if two thermodynamic systems are each in thermal equilibrium with a third, then they are in thermal equilibrium with each other. Two systems are said to be in the relation of thermal equilibrium if they are linked by a wall permeable only to heat and they do not change over time. [1]
College admissions; Careers; Zeroth law of thermodynamics. Let's talk about the 0th Law of Thermodynamics. Now the 0th law, The zeroth law of thermodynamics states that if two thermodynamic systems are each in thermal equilibrium with a third, then all three are in thermal equilibrium with
The Zeroth law of thermodynamics states that if two systems X and Y separately attain thermal equilibrium with the third system Z. It means the two systems X and Y has a thermal equilibrium with each other. Must Read: First Law of Thermodynamics. Example of Zeroth Law Of Thermodynamics The zeroth law of thermodynamics states that if two thermodynamic systems are each in thermal equilibrium with a third, then all three are in thermal equilibrium with
1.1 The zeroth law The zeroth law of thermodynamics established the existence of temperature in macro-scopic systems. Temperature is a state quantity which is unknown Learn about the Three Laws of Thermodynamics including The Zeroth Law of Thermodynamics says that Need to control current flow in low voltage applications?
The zeroth law of thermodynamics states that if two thermodynamic systems are each in thermal equilibrium with a third, then they are in thermal equilibrium with each First law of thermodynamics states that in any thermodynamic process,when heat Q is added to a system,this energy appears as an Zeroth law of thermodynamic
Zeroth Law of Thermodynamics. The "zeroth law" states that if two systems are at the same time in thermal equilibrium with a third system, … College admissions; Careers; Zeroth law of thermodynamics. Let's talk about the 0th Law of Thermodynamics. Now the 0th law,
The applications of the thermodynamic laws and principles are found in all fields of energy technology, Zeroth Law Of Thermodynamics. The behaviour of the quantities of thermodynamics deal is ruled by the four laws of thermodynamics Zeroth law of thermodynamics: APPLICATION OF THERMODYNAMICS.
What is Thermodynamics and explanation of zeroth, first, second, and third law of thermodynamics. The four laws of thermodynamics are: Zeroth law of thermodynamics: If two systems are in thermal equilibrium with a third system, they are in thermal equilibrium with each other. This law helps define the notion of temperature.
... you will notice the applications of thermodynamics almost everywhere directly Third Law of Thermodynamics, Zeroth Law of Thermodynamics, Boyle’s law Zeroth Law of Thermodynamics: Statement: If two bodies (say A and B) are in thermal equilibrium of the third body (say C) then body A and B will also be in thermal
The zeroth law of thermodynamics states that if two thermodynamic systems are each in thermal equilibrium with a third, then all three are in thermal equilibrium with Zeroth Law of Thermodynamics: Statement: If two bodies (say A and B) are in thermal equilibrium of the third body (say C) then body A and B will also be in thermal
The Zeroth Law of Thermodynamics states that if two bodies are each in thermal equilibrium with some third body, then they are also in equilibrium with each other. The Zeroth Law of Thermodynamics states that if two bodies are each in thermal equilibrium with some third body, then they are also in equilibrium with each other. First law of thermodynamics states that in any thermodynamic process,when heat Q is added to a system,this energy appears as an Zeroth law of thermodynamic
The applications of the thermodynamic laws and principles are found in all fields of energy technology, Zeroth Law Of Thermodynamics. Another important application of second law of thermodynamics is found in refrigerators and heat pumps, Zeroth law of Thermodynamics.
The Zeroth Law of Thermodynamics states that if two bodies are each in thermal equilibrium with some third body, then they are also in equilibrium with each other. The Zeroth Law of Thermodynamics states that if two bodies are each in thermal equilibrium with some third body, then they are also in equilibrium with each other. The Zeroth Law of Thermodynamics states that:- two system which are equal in temperature to a third system are equal in temperature to each other.
Zeroth Law of thermodynamics is the most basic of the laws even though it was defined the last. That is why it was named zero (zeroth) instead of four. It forms the Learn about the Three Laws of Thermodynamics including The Zeroth Law of Thermodynamics says that Need to control current flow in low voltage applications?
Zeroth law of thermodynamics: If two systems are in thermal equilibrium with a third system, they are in thermal equilibrium with each … ... you will notice the applications of thermodynamics almost everywhere directly Third Law of Thermodynamics, Zeroth Law of Thermodynamics, Boyle’s law
The zeroth law is incredibly important as it allows us to define the concept of a temperature scale. If two systems are each in thermal equilibrium with a third, they are also in thermal equilibrium with each other. For example, consider two separate cups of … The zeroth law is incredibly important as it allows us to define the concept of a temperature scale. If two systems are each in thermal equilibrium with a third, they are also in thermal equilibrium with each other. For example, consider two separate cups of …
The zeroth law of thermodynamics. We discuss a convenient formulation of the most intuitive property of equilibrium, the zeroth law of thermodynamics, in terms of conditions of dynamical stability for a system which may interact (weakly) with its surroundings. Explore this introduction to the three laws of thermodynamics and how they are used to solve problems involving heat or thermal Second Law of Thermodynamics:
### Thermal Equilibrium HyperPhysics Concepts
What are the applications of the zeroth law of thermodynamics?. In the year of 1931, R.H. fowler had provided one law of thermodynamics which is termed as Zeroth law of thermodynamics and according to this law, The Zeroth Law of Thermodynamics states that if two bodies are each in thermal equilibrium with some third body, then they are also in equilibrium with each other. The Zeroth Law of Thermodynamics states that if two bodies are each in thermal equilibrium with some third body, then they are also in equilibrium with each other..
Zeroth First & Second Law Of Thermodynamics. The Zeroth law of thermodynamics states that if two systems X and Y separately attain thermal equilibrium with the third system Z. It means the two systems X and Y has a thermal equilibrium with each other. Must Read: First Law of Thermodynamics. Example of Zeroth Law Of Thermodynamics, The four laws of thermodynamics are: Zeroth law of thermodynamics: If two systems are in thermal equilibrium with a third system, they are in thermal equilibrium with each other. This law helps define the notion of temperature..
### Thermal Equilibrium HyperPhysics Concepts
Zeroth law of thermodynamics The Full Wiki. Zeroth Law of Thermodynamics. One of the laws related to the thermal equilibrium is Zeroth law of Thermodynamics. Suppose two bodies are separated by a heat insulator. Now a third body is brought in contact with both the bodies simultaneously. When the bodies attain thermal equilibrium we remove the third body. https://simple.wikipedia.org/wiki/Talk:Zeroth_law_of_thermodynamics Explore this introduction to the three laws of thermodynamics and how they are used to solve problems involving heat or thermal Second Law of Thermodynamics:.
In the year of 1931, R.H. fowler had provided one law of thermodynamics which is termed as Zeroth law of thermodynamics and according to this law Explore this introduction to the three laws of thermodynamics and how they are used to solve problems involving heat or thermal Second Law of Thermodynamics:
The Zeroth Law of Thermodynamics says that if a change to the temperature(T) from one object (for example 'A') changes the temperature of another object ('B'), and a 1. Zeroth law of thermodynamics:-. Zeroth law of thermodynamics states that when two systems are each in thermal equilibrium with the third system, they are also in thermal equilibrium with each other.
Have you heard of thermodynamics or applications of thermodynamics in your daily routine? Study of thermodynamics involves The zeroth law of thermodynamics, The zeroth law of thermodynamics is a generalization about the thermal equilibrium among bodies, or thermodynamic systems, in contact. It results from the definition and properties of temperature. [1]
Explore this introduction to the three laws of thermodynamics and how they are used to solve problems involving heat or thermal Second Law of Thermodynamics: The applications of the thermodynamic laws and principles are found in all fields of energy technology, Zeroth Law Of Thermodynamics.
1.1 The zeroth law The zeroth law of thermodynamics established the existence of temperature in macro-scopic systems. Temperature is a state quantity which is unknown The Zeroth Law of Thermodynamics states that if two bodies are each in thermal equilibrium with some third body, then they are also in equilibrium with each other. The Zeroth Law of Thermodynamics states that if two bodies are each in thermal equilibrium with some third body, then they are also in equilibrium with each other.
The Zeroth law of thermodynamics is the basis for measurement of temperature and setting its scale. In simple word, Zeroth law of thermodynamics says that “When two bodies are separately in thermal equilibrium with the third body, then the two are also in thermal equilibrium with each other." Learn about the Three Laws of Thermodynamics including The Zeroth Law of Thermodynamics says that Need to control current flow in low voltage applications?
The Zeroth Law of Thermodynamics states that:- two system which are equal in temperature to a third system are equal in temperature to each other. The zeroth law of thermodynamics states that if two thermodynamic systems are each in thermal equilibrium with a third, then they are in thermal equilibrium with each other. Two systems are said to be in the relation of thermal equilibrium if they are linked by a wall permeable only to heat and they do not change over time. [1]
Learn about the Three Laws of Thermodynamics including The Zeroth Law of Thermodynamics says that Need to control current flow in low voltage applications? The zeroth law of thermodynamics is a generalization about the thermal equilibrium among bodies, or thermodynamic systems, in contact. It results from the definition and properties of temperature. [1]
1. Zeroth law of thermodynamics:-. Zeroth law of thermodynamics states that when two systems are each in thermal equilibrium with the third system, they are also in thermal equilibrium with each other. Another important application of second law of thermodynamics is found in refrigerators and heat pumps, Zeroth law of Thermodynamics.
The zeroth law of thermodynamics is a generalization about the thermal equilibrium among bodies, or thermodynamic systems, in contact. It results from the definition and properties of temperature. [1] The behaviour of the quantities of thermodynamics deal is ruled by the four laws of thermodynamics Zeroth law of thermodynamics: APPLICATION OF THERMODYNAMICS.
Zeroth Law of Thermodynamics: Statement: If two bodies (say A and B) are in thermal equilibrium of the third body (say C) then body A and B will also be in thermal First law of thermodynamics states that in any thermodynamic process,when heat Q is added to a system,this energy appears as an Zeroth law of thermodynamic
The laws are as follows. Zeroth law of thermodynamics – If two thermodynamic systems are each in thermal equilibrium with a third, then they are in thermal equilibrium with each other. First law of thermodynamics – Energy can neither be created nor destroyed. College admissions; Careers; Zeroth law of thermodynamics. Let's talk about the 0th Law of Thermodynamics. Now the 0th law,
The laws are as follows. Zeroth law of thermodynamics – If two thermodynamic systems are each in thermal equilibrium with a third, then they are in thermal equilibrium with each other. First law of thermodynamics – Energy can neither be created nor destroyed. The zeroth law of thermodynamics is a generalization about the thermal equilibrium among bodies, or thermodynamic systems, in contact. It results from the definition and properties of temperature. [1]
In the year of 1931, R.H. fowler had provided one law of thermodynamics which is termed as Zeroth law of thermodynamics and according to this law Zeroth Law of Thermodynamics: Statement: If two bodies (say A and B) are in thermal equilibrium of the third body (say C) then body A and B will also be in thermal
The four laws of thermodynamics are: Zeroth law of thermodynamics: If two systems are in thermal equilibrium with a third system, they are in thermal equilibrium with each other. This law helps define the notion of temperature. Another important application of second law of thermodynamics is found in refrigerators and heat pumps, Zeroth law of Thermodynamics.
Zeroth Law of Thermodynamics: Statement: If two bodies (say A and B) are in thermal equilibrium of the third body (say C) then body A and B will also be in thermal Zeroth Law of Thermodynamics. The "zeroth law" states that if two systems are at the same time in thermal equilibrium with a third system, …
Zeroth Law Of Thermodynamics. The zeroth law of thermodynamics states that if two systems, A and B, are in thermal equilibrium with a third system, C, then A and B are in thermal equilibrium with each other. It is analogous to the transitive property in math (if A=C and B=C, then A=B). Have you heard of thermodynamics or applications of thermodynamics in your daily routine? Study of thermodynamics involves The zeroth law of thermodynamics,
What is Thermodynamics and explanation of zeroth, first, second, and third law of thermodynamics. The behaviour of the quantities of thermodynamics deal is ruled by the four laws of thermodynamics Zeroth law of thermodynamics: APPLICATION OF THERMODYNAMICS.
The laws are as follows. Zeroth law of thermodynamics – If two thermodynamic systems are each in thermal equilibrium with a third, then they are in thermal equilibrium with each other. First law of thermodynamics – Energy can neither be created nor destroyed. The four laws of thermodynamics are: Zeroth law of thermodynamics: If two systems are in thermal equilibrium with a third system, they are in thermal equilibrium with each other. This law helps define the notion of temperature.
The zeroth law of thermodynamics states that if two thermodynamic systems are each in thermal equilibrium with a third, then all three are in thermal equilibrium with The Zeroth Law of Thermodynamics states that:- two system which are equal in temperature to a third system are equal in temperature to each other. | 5,331 | 26,037 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2021-31 | latest | en | 0.926166 |
http://quant.stackexchange.com/questions/tagged/data-mining?sort=votes | 1,462,519,820,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461861735203.5/warc/CC-MAIN-20160428164215-00126-ip-10-239-7-51.ec2.internal.warc.gz | 230,170,457 | 15,500 | # Tagged Questions
The tag has no usage guidance.
6k views
### Which approach dominates? Mathematical modeling or data mining?
According to my current understanding, there is a clear difference between data mining and mathematical modeling. Data mining methods treat systems (e.g., financial markets) as a "black box". The ...
1k views
### How to encode trading strategies mathematically
If you have a bunch of different econometric data (e.g. indexes, FX, commodities, interest rates...) you can try to find a formula to see if there is any relationship in the data - e.g. to forecast ...
429 views
### Can we use White's reality check to compare two Sharpe ratios?
I read a paper from Ledoit and Wolf that proposes a method to compare two Sharpe ratios and a paper from White that proposes a method to compare $n$ trading rules. My question is: Can we use White's ...
542 views
### How to better understand trading signals?
I am looking to get a better understanding of an output from a trading strategy. Basically I have a daily equity curve lets call it $Y_t$. I have defined a bunch of independent variables $X_{it}$ that ...
126 views
### White's Reality Check versus Benjamini-Hochberg-Yekutielie Procedure
I'm backtesting about 1k different strategies / permutations of strategies and I want to identify which if any of the strategies are better than the benchmark. Based on my readings, I feel like I've ...
26 views
### Feedback on Video Metadata Extraction
I've developed/patented a metadata extraction methodology, which we've used to build a variety of alternative datasets for funds. One of the more difficult issues in this field, is getting useful ...
170 views
### list of ADR's by volume or market cap
I'm looking for a list of ADR's (for a simulation) that I can screen by either market cap or volume? If anyone knows of a free way to get such a list it would be much appreciated.
21 views
### How can I set buy and sell price for testing my automatic stock trading system? [on hold]
I'm testing my automatic trading system in stock market (data mining system). I'm modeling day by day for 30-days and calculate profit in every step. Suppose that my system predicts tomorrow close ...
8 views
If you were to Backtest data extracted from video feeds, let's say for commodities. How many years/hours of video would you desire? The video repository is currently harvested going back 5 years and ...
22 views
### Video Analysis Platform
I own a data analysis company which is currently bulk extracting metadata from historical market video and audio files, and creating alternative/historical datasets around commodities, S&P 500 ...
31 views
### White's reality check p-value
I am running a hypothesis test based on White's reality check p-value. I am getting a weird result for my univariate time series of returns. In essence, I am following a code on MATlab to run the test ... | 627 | 2,919 | {"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} | 2.65625 | 3 | CC-MAIN-2016-18 | longest | en | 0.902366 |
https://www.b4x.com/android/forum/threads/b4x-fft-class.78797/ | 1,571,361,995,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986677412.35/warc/CC-MAIN-20191018005539-20191018033039-00066.warc.gz | 802,753,343 | 15,598 | # B4A Class [B4x] FFT class
Discussion in 'Additional libraries, classes and official updates' started by klaus, Apr 25, 2017.
This class performs Fast Fourier Transformations forward and inverse.
The class is exactly the same for B4A, B4i and B4J.
A FFT library already exists, but only for B4A.
A Fast Fourier Transformation transforms a signal from its original domain (mostly the time domain) into its representation in the frequency domain.
The number N of samples in the original domain must be a power of 2?
Examples: 256, 1024, 2048, 4096 etc.
In theory, the source signal must have a real and an imaginary part.
In time signals the imaginary samples are all 0, this is treated internally in the class, no need to transmit them.
The result of an FFT is two signals real and imaginary.
Which can also be represented by magnitude and phase.
In concrete applications, the magnitude of the frequency signal is mostly used, therefore the Forward method returns the Magnitude.
The phase, real and imaginary signals can be accessed as properties.
Methods:
Forward(Real() As Double) As Double() performs a FFT calculation
Real = an array of N Doubles representing the Real time signal.
Returns the Magnitude of the frequency domain N/2 + 1 samples.
Inverse(Real() As Double, Imag() As Double) As Double() performs an inverse FFT calculation
Real = an array of N/2 + 1 Doubles representing the real part of the frequency signal.
Imag = an array of N/2 + 1 Doubles representing the imaginary part of the frequency signal.
Returns the inverse frequency real signal, N samples.
Properties:
Magnitude = magnitude of the frequency signal N/2 + 1 samples, read only.
Phase = phase of the frequency signal N/2 + 1 samples, read only.
Real = real part of the frequency signal N/2 + 1 samples, read only.
Imag = imaginary part of the frequency signal N/2 + 1 samples, read only.
ModeDegrees = True (default), the phase signal is in degrees otherwise in radians.
Window = NONE or Hann, applies no window or a Hann window.
Demo programs:
Attached a demo project for each product B4A, B4i and B4J, they all include the FFT class which is the same for all three.
The program can:
Calculate different input signals.
Calculate forward and inverse FFTs.
Display the different signals.
Apply a Window.
For Android phones and iPhones tap on the chart to show the menu.
On Android and iOS tablets and in the B4J project the menu is always visible.
Concrete B4A application: FFT_Record
The program reads the microphone signal.
Switch between Time and FFT
Generate a sound composed of three frequencies: 500, 1000 and 2000
Change the scale up and down
Display the signal values when moving the finger over the diagram.
#### Attached Files:
File size:
12.4 KB
Views:
122
File size:
6.9 KB
Views:
43
File size:
7.1 KB
Views:
86
• ###### FFT_Record.zip
File size:
54.3 KB
Views:
140
Hi @klaus
Great job as usual !
Compared to the library (which is only for B4A) is there any improvment (performance or something else) ?
JP
Great work Klaus!
@freedom2000
I haven't made speed comparisons between the two.
The main goal was to have FFT for the three products.
You could make libraries with the class, I don't know if there would be speed improvements.
freedom2000 likes this.
Thank you Klaus..
Great work that will help a lot in some of my projects.
I am trying to understand how FFT_Record works and modifying it for saving (not displaying) FFT data to a file.
I see that you are calling the sub Record with a timer every 300ms
Code:
`Timer1.Initialize("Timer1", 300)...Private Sub Timer1_Tick RecordEnd Sub`
But in the Record Sub you exit from the While loop only after 500ms
Code:
`If DateTime.Now > StartTime + "500" Then Exit`
What does that mean?
The FFT is calculate over 300ms or 500ms chunk?
It is a TimeOut limit.
When i click on the "Go" button it sets StartTime so on the first execution of the Do While it exits after 500ms, but in the following Do While it exits immediately because StartTime is never set anymore.
I feel confused, why Does that Timeout limit is needed?
Thank you
I tested it again and the timeout is only necessary for the first time.
Without this line the program hangs in the While/Do loop.
Don't ask me why, I found this as a workaround and it works.
Last edited: Feb 3, 2019
Yes you are right, the loop must have a condition for exiting from the loop. But if you remove the loop the program still works correctly:
Code:
`'Do While True sound = AR.ReadShort(0, BufferSize) DataSize = DataSize + sound.Length 'If DateTime.Now > StartTime + "500" Then Exit 'Loop`
I think the loop it is only needed in the original code of Audiorecord example bacause it records for 10 seconds and it need to check during the loop execution if the program already recorded for 10 seconds or not.
But in this case you call the sub Record every 300ms so you don't need any other loop for getting data from the audio source.
klaus likes this.
Thank you, Klaus, for sharing this great library.
Is it possible to calculate one main frequence of a pitch, from the array? It could be very usful for a music intrument tuner.
What exactly do you mean?
Detect frequency peaks?
Is THIS what you are looking for, it's a B4A project?
I haven't updated this project with this class.
I think the frequency peak is not enough to calculate the accurate basic frequence (in your sound file of the FFT-Record sample it is 500 Hz). I need an output that gives me only the basic pitch, but with even 1 Hz difference (499 Hz, 500 Hz, 501 Hz) accuracity. Do you think it is possible with FFT? Yes, it's a B4A project.
Sorry, but I still don't understand what exactly you want to do.
What exactly do you mean with only the basic pitch?
If you want a frequeny increment (accuracy) of 1Hz, the duration of the time signal must be 1 second.
Which means 44100 time samples.
But the number of samples for FFT must be a power of 2.
So at least 32768 samples which represents 0.743 second therefore a 1.346 frequency increment (accuracy).
What I want to do is a kind of guitar tuner. You play the pitch "a", this is 440 Hz. If you play 441 Hz the tuner will tell you, to tune the guitar a bit lower. Thats why I only need the fundamental frequency (in that case 440 Hz) and not the overtone (which would be in that case 880 Hz, ...). I could calculate the peak of the FFT array, but I think it is not good enough, to detect small differences between 440 Hz and 441 Hz. I now allready wrote an algorythm, which seems to work. I can public it, if someone is interested...
I would be interest in.
Here is my small first tuner project. It analyzes the fundamental frequency of a soundwave in two steps:
1. findig the first periode in the wave and creating an array with periode positions
2. comparing the first periode with later periodes and calculating the accurate wavelenght
I tested with piano, clarinet and singing. Piano and clarinet worked fine, singing sometimes, but this is more my voice than the algorythm.
Anyhow I would be glad, if somebody has a better solution. As a musician I would have some ideas, how to make better tuners, more practical in orchester. But at first I need a good and stabil calculation of the fundamental frequency.
#### Attached Files:
• ###### Tuner.zip
File size:
410.8 KB
Views:
48
Nice piece of work!
I tested it quickly with my guitar and it seems to work fine although I need some more time to check if tuning up or tuning down is suggested correctly but it seems to do that fine. I used to tune my guitar by ear once but I have lost that "ability" for now, probably because I have not been playing for years.
Just for fun, I tried it also with electric piano (always tuned) and it detected tones correctly.
Keep up the good work and if you decide to publish it on Google Play, please let us know!
What regards the coding, this stuff is a bit over my head and I am sure there are other users who can suggest improvements if needed. In my opinion, it worked already very fast. | 1,942 | 8,074 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2019-43 | latest | en | 0.883122 |
https://gmatclub.com/forum/which-one-of-a-1-b-1-1-and-a-1-b-1-1-is-greater-13815.html?fl=similar | 1,511,164,811,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805923.26/warc/CC-MAIN-20171120071401-20171120091401-00670.warc.gz | 608,842,126 | 40,213 | It is currently 20 Nov 2017, 01:00
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# Which one of (a^-1+b^-1)^-1 and (a^-1*b^-1)^-1 is greater?
Author Message
Senior Manager
Joined: 10 Dec 2004
Posts: 273
Kudos [?]: 203 [0], given: 0
Which one of (a^-1+b^-1)^-1 and (a^-1*b^-1)^-1 is greater? [#permalink]
### Show Tags
02 Feb 2005, 10:46
00:00
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct 0% (00:00) wrong based on 0 sessions
### HideShow timer Statistics
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
Which one of (a^-1+b^-1)^-1 and (a^-1*b^-1)^-1 is greater?
1) a=2b
2) a+b>1
Kudos [?]: 203 [0], given: 0
Manager
Joined: 11 Jan 2005
Posts: 101
Kudos [?]: [0], given: 0
### Show Tags
03 Feb 2005, 04:01
we simplify the expressions :
from (a^-1+b^-1)^-1
to ab/(a+b)
from (a^-1*b^-1)^-1
to ab
i. not sufficient if a=2b only
we have : (2/3) * b & 2b^2 if 0<b<1 & if b>1 gives different solutions
ii. not sufficient if a+b>1
ab can be negative or positive according to a or b, then we duno
together : if a+b>1 & a=2b then a & b are positive
but we still duno if a or b are between 0 & 1 then no sufficient
E
Kudos [?]: [0], given: 0
Manager
Joined: 24 Jan 2005
Posts: 217
Kudos [?]: 14 [0], given: 0
Location: Boston
### Show Tags
04 Feb 2005, 12:12
I just answered it somewhere in the forum.
Anirban
Kudos [?]: 14 [0], given: 0
should be C [#permalink] 04 Feb 2005, 12:12
Display posts from previous: Sort by | 670 | 2,023 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2017-47 | latest | en | 0.801694 |
https://forum.babylonjs.com/t/change-lenth-of-arrow-created-dynamically/17794 | 1,653,787,148,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663035797.93/warc/CC-MAIN-20220529011010-20220529041010-00000.warc.gz | 310,896,223 | 5,031 | # Change lenth of arrow created, dynamically
good day I create an arrow with:
const createArrow=(scene) =>{
const thickness=10;
var arrow = new TransformNode(“arrow”, scene);
var cylinder = CylinderBuilder.CreateCylinder(“cylinder”, { diameterTop: 0, height: 0.075, diameterBottom: 0.0375 * (1 + (thickness - 1) / 4), tessellation: 96 }, scene);
var line = CylinderBuilder.CreateCylinder(“cylinder”, { diameterTop: 0.005 * thickness, height: 0.275, diameterBottom: 0.005 * thickness, tessellation: 96 }, scene);
const arrowMat = new StandardMaterial(‘arrowMat’, scene);
arrowMat.emissiveColor = Color3.Yellow();
cylinder.parent = arrow;
cylinder.material = arrowMat;
cylinder.rotation.x = Math.PI / 2;
cylinder.position.z += 0.3;
line.parent = arrow;
line.material = arrowMat;
line.position.z += 0.275 / 2;
line.rotation.x = Math.PI / 2;
return arrow;
}
So say that I create one of those arrows and I position it somewhere with some rotation:
myArrow=createArrow(scene);
myArrow.scaling = new Vector3(35, 35, 35);
myArrow.position = new Vector3(0, 60, 0);
myArrow.rotation = new Vector3(0, 0, 0);
And now, dynamically,and often, many times per second, depending on some other extra information and the specific point where the arrow is, I need to change the length of that arrow.
How can I change the length of the arrow, preserving its look, dynamically, once its been created?
I dont think that changing the individual scaling of one of the axis is the way, tried it without luck,
thank you for any help
changing myArrow.scaling finally seems to work I think that could be the solution
1 Like | 423 | 1,604 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-21 | latest | en | 0.781807 |
https://math.libretexts.org/Bookshelves/Differential_Equations/Book%3A_Elementary_Differential_Equations_with_Boundary_Value_Problems_(Trench)/05%3A_Linear_Second_Order_Equations/5.04%3A_The_Method_of_Undetermined_Coefficients_I | 1,670,091,641,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710936.10/warc/CC-MAIN-20221203175958-20221203205958-00691.warc.gz | 422,989,707 | 32,151 | # 5.4: The Method of Undetermined Coefficients I
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
In this section we consider the constant coefficient equation
$\label{eq:5.4.1} ay''+by'+cy=e^{\alpha x}G(x),$
where $$\alpha$$ is a constant and $$G$$ is a polynomial.
From Theorem 5.3.2, the general solution of Equation \ref{eq:5.4.1} is $$y=y_p+c_1y_1+c_2y_2$$, where $$y_p$$ is a particular solution of Equation \ref{eq:5.4.1} and $$\{y_1,y_2\}$$ is a fundamental set of solutions of the complementary equation
$ay''+by'+cy=0. \nonumber$
In Section 5.2 we showed how to find $$\{y_1,y_2\}$$. In this section we’ll show how to find $$y_p$$. The procedure that we’ll use is called the method of undetermined coefficients. Our first example is similar to Exercises 5.3.16-5.3.21.
##### Example 5.4.1
Find a particular solution of
$\label{eq:5.4.2} y''-7y'+12y=4e^{2x}.$
Then find the general solution.
###### Solution
Substituting $$y_p=Ae^{2x}$$ for $$y$$ in Equation \ref{eq:5.4.2} will produce a constant multiple of $$Ae^{2x}$$ on the left side of Equation \ref{eq:5.4.2}, so it may be possible to choose $$A$$ so that $$y_p$$ is a solution of Equation \ref{eq:5.4.2}. Let’s try it; if $$y_p=Ae^{2x}$$ then
$y_p''-7y_p'+12y_p=4Ae^{2x}-14Ae^{2x}+12Ae^{2x}=2Ae^{2x}=4e^{2x} \nonumber$
if $$A=2$$. Therefore $$y_p=2e^{2x}$$ is a particular solution of Equation \ref{eq:5.4.2}. To find the general solution, we note that the characteristic polynomial of the complementary equation
$\label{eq:5.4.3} y''-7y'+12y=0$
is $$p(r)=r^2-7r+12=(r-3)(r-4)$$, so $$\{e^{3x},e^{4x}\}$$ is a fundamental set of solutions of Equation \ref{eq:5.4.3}. Therefore the general solution of Equation \ref{eq:5.4.2} is
$y=2e^{2x}+c_1e^{3x}+c_2e^{4x}. \nonumber$
##### Example 5.4.2
Find a particular solution of
$\label{eq:5.4.4} y''-7y'+12y=5e^{4x}.$
Then find the general solution.
###### Solution
Fresh from our success in finding a particular solution of Equation \ref{eq:5.4.2} — where we chose $$y_p=Ae^{2x}$$ because the right side of Equation \ref{eq:5.4.2} is a constant multiple of $$e^{2x}$$ — it may seem reasonable to try $$y_p=Ae^{4x}$$ as a particular solution of Equation \ref{eq:5.4.4}. However, this will not work, since we saw in Example 5.4.1 that $$e^{4x}$$ is a solution of the complementary equation Equation \ref{eq:5.4.3}, so substituting $$y_p=Ae^{4x}$$ into the left side of Equation \ref{eq:5.4.4}) produces zero on the left, no matter how we choose$$A$$. To discover a suitable form for $$y_p$$, we use the same approach that we used in Section 5.2 to find a second solution of
$ay''+by'+cy=0 \nonumber$
in the case where the characteristic equation has a repeated real root: we look for solutions of Equation \ref{eq:5.4.4} in the form $$y=ue^{4x}$$, where $$u$$ is a function to be determined. Substituting
$\label{eq:5.4.5} y=ue^{4x},\quad y'=u'e^{4x}+4ue^{4x},\quad \text{and} \quad y''=u''e^{4x}+8u'e^{4x}+16ue^{4x}$
into Equation \ref{eq:5.4.4} and canceling the common factor $$e^{4x}$$ yields
$(u''+8u'+16u)-7(u'+4u)+12u=5, \nonumber$
or
$u''+u'=5. \nonumber$
By inspection we see that $$u_p=5x$$ is a particular solution of this equation, so $$y_p=5xe^{4x}$$ is a particular solution of Equation \ref{eq:5.4.4}. Therefore
$y=5xe^{4x}+c_1e^{3x}+c_2e^{4x} \nonumber$
is the general solution.
##### Example 5.4.3
Find a particular solution of
$\label{eq:5.4.6} y''-8y'+16y=2e^{4x}.$
###### Solution
Since the characteristic polynomial of the complementary equation
$\label{eq:5.4.7} y''-8y'+16y=0$
is $$p(r)=r^2-8r+16=(r-4)^2$$, both $$y_1=e^{4x}$$ and $$y_2=xe^{4x}$$ are solutions of Equation \ref{eq:5.4.7}. Therefore Equation \ref{eq:5.4.6}) does not have a solution of the form $$y_p=Ae^{4x}$$ or $$y_p=Axe^{4x}$$. As in Example 5.4.2 , we look for solutions of Equation \ref{eq:5.4.6} in the form $$y=ue^{4x}$$, where $$u$$ is a function to be determined. Substituting from Equation \ref{eq:5.4.5} into Equation \ref{eq:5.4.6} and canceling the common factor $$e^{4x}$$ yields
$(u''+8u'+16u)-8(u'+4u)+16u=2, \nonumber$
or
$u''=2. \nonumber$
Integrating twice and taking the constants of integration to be zero shows that $$u_p=x^2$$ is a particular solution of this equation, so $$y_p=x^2e^{4x}$$ is a particular solution of Equation \ref{eq:5.4.4}. Therefore
$y=e^{4x}(x^2+c_1+c_2x) \nonumber$
is the general solution.
The preceding examples illustrate the following facts concerning the form of a particular solution $$y_p$$ of a constant coefficent equation
$ay''+by'+cy=ke^{\alpha x}, \nonumber$
where $$k$$ is a nonzero constant:
1. If $$e^{\alpha x}$$ isn’t a solution of the complementary equation $\label{eq:5.4.8} ay''+by'+cy=0,$ then $$y_p=Ae^{\alpha x}$$, where $$A$$ is a constant. (See Example 5.4.1 ).
2. If $$e^{\alpha x}$$ is a solution of Equation \ref{eq:5.4.8} but $$xe^{\alpha x}$$ is not, then $$y_p=Axe^{\alpha x}$$, where $$A$$ is a constant. (See Example 5.4.2 .)
3. If both $$e^{\alpha x}$$ and $$xe^{\alpha x}$$ are solutions of Equation \ref{eq:5.4.8}, then $$y_p=Ax^2e^{\alpha x}$$, where $$A$$ is a constant. (See Example 5.4.3 .)
See Exercise 5.4.30 for the proofs of these facts.
In all three cases you can just substitute the appropriate form for $$y_p$$ and its derivatives directly into
$ay_p''+by_p'+cy_p=ke^{\alpha x},\nonumber$
and solve for the constant $$A$$, as we did in Example 5.4.1 . (See Exercises 5.4.31-5.4.33.) However, if the equation is
$ay''+by'+cy=k e^{\alpha x}G(x), \nonumber$
where $$G$$ is a polynomial of degree greater than zero, we recommend that you use the substitution $$y=ue^{\alpha x}$$ as we did in Examples 5.4.2 and 5.4.3 . The equation for $$u$$ will turn out to be
$\label{eq:5.4.9} au''+p'(\alpha)u'+p(\alpha)u=G(x),$
where $$p(r)=ar^2+br+c$$ is the characteristic polynomial of the complementary equation and $$p'(r)=2ar+b$$ (Exercise 5.4.30); however, you shouldn’t memorize this since it is easy to derive the equation for $$u$$ in any particular case. Note, however, that if $$e^{\alpha x}$$ is a solution of the complementary equation then $$p(\alpha)=0$$, so Equation \ref{eq:5.4.9} reduces to
$au''+p'(\alpha)u'=G(x), \nonumber$
while if both $$e^{\alpha x}$$ and $$xe^{\alpha x}$$ are solutions of the complementary equation then $$p(r)=a(r-\alpha)^2$$ and $$p'(r)=2a(r-\alpha)$$, so $$p(\alpha)=p'(\alpha)=0$$ and Equation \ref{eq:5.4.9}) reduces to
$au''=G(x). \nonumber$
##### Example 5.4.4
Find a particular solution of
$\label{eq:5.4.10} y''-3y'+2y=e^{3x}(-1+2x+x^2).$
###### Solution
Substituting
$y=ue^{3x},\quad y'=u'e^{3x}+3ue^{3x},\quad \text{and} y''=u''e^{3x}+6u'e^{3x}+9ue^{3x}\nonumber$
into Equation \ref{eq:5.4.10}) and canceling $$e^{3x}$$ yields
$(u''+6u'+9u)-3(u'+3u)+2u=-1+2x+x^2, \nonumber$
or
$\label{eq:5.4.11} u''+3u'+2u=-1+2x+x^2.$
As in Example 5.3.2, in order to guess a form for a particular solution of Equation \ref{eq:5.4.11}), we note that substituting a second degree polynomial $$u_p=A+Bx+Cx^2$$ for $$u$$ in the left side of Equation \ref{eq:5.4.11}) produces another second degree polynomial with coefficients that depend upon $$A$$, $$B$$, and $$C$$; thus,
$\text{if} \quad u_p=A+Bx+Cx^2\quad \text{then} \quad u_p'=B+2Cx\quad \text{and} \quad u_p''=2C. \nonumber$
If $$u_p$$ is to satisfy Equation \ref{eq:5.4.11}), we must have
\begin{aligned} u_p''+3u_p'+2u_p&=2C+3(B+2Cx)+2(A+Bx+Cx^2)\\ &=(2C+3B+2A)+(6C+2B)x+2Cx^2=-1+2x+x^2.\end{aligned}\nonumber
Equating coefficients of like powers of $$x$$ on the two sides of the last equality yields
$\begin{array}{rcr} 2C&=1\phantom{.}\\ 2B+6C&=2\phantom{.}\\ 2A+3B+2C&= -1. \end{array}\nonumber$
Solving these equations for $$C$$, $$B$$, and $$A$$ (in that order) yields $$C=1/2,B=-1/2,A=-1/4$$. Therefore
$u_p=-{1\over4}(1+2x-2x^2) \nonumber$
is a particular solution of Equation \ref{eq:5.4.11}, and
$y_p=u_pe^{3x}=-{e^{3x}\over4}(1+2x-2x^2) \nonumber$
is a particular solution of Equation \ref{eq:5.4.10}.
##### Example 5.4.5
Find a particular solution of
$\label{eq:5.4.12} y''-4y'+3y=e^{3x}(6+8x+12x^2).$
###### Solution
Substituting
$y=ue^{3x},\quad y'=u'e^{3x}+3ue^{3x},\quad \text{and } y''=u''e^{3x}+6u'e^{3x}+9ue^{3x} \nonumber$
into Equation \ref{eq:5.4.12}) and canceling $$e^{3x}$$ yields
$(u''+6u'+9u)-4(u'+3u)+3u=6+8x+12x^2, \nonumber$
or
$\label{eq:5.4.13} u''+2u'=6+8x+12x^2.$
There’s no $$u$$ term in this equation, since $$e^{3x}$$ is a solution of the complementary equation for Equation \ref{eq:5.4.12}). (See Exercise 5.4.30.) Therefore Equation \ref{eq:5.4.13}) does not have a particular solution of the form $$u_p=A+Bx+Cx^2$$ that we used successfully in Example 5.4.4 , since with this choice of $$u_p$$,
$u_p''+2u_p'=2C+(B+2Cx) \nonumber$
can’t contain the last term ($$12x^2$$) on the right side of Equation \ref{eq:5.4.13}). Instead, let’s try $$u_p=Ax+Bx^2+Cx^3$$ on the grounds that
$u_p'=A+2Bx+3Cx^2\quad \text{and} \quad u_p''=2B+6Cx\nonumber$
together contain all the powers of $$x$$ that appear on the right side of Equation \ref{eq:5.4.13}).
Substituting these expressions in place of $$u'$$ and $$u''$$ in Equation \ref{eq:5.4.13}) yields
$(2B+6Cx)+2(A+2Bx+3Cx^2)=(2B+2A)+(6C+4B)x+6Cx^2=6+8x+12x^2. \nonumber$
Comparing coefficients of like powers of $$x$$ on the two sides of the last equality shows that $$u_p$$ satisfies Equation \ref{eq:5.4.13}) if
$\begin{array}{rcr} 6C&=12\phantom{.}\\ 4B+6C&=8\phantom{.}\\ 2A+2B\phantom{+6u_2}&=6. \end{array}\nonumber$
Solving these equations successively yields $$C=2$$, $$B=-1$$, and $$A=4$$. Therefore
$u_p=x(4-x+2x^2) \nonumber$
is a particular solution of Equation \ref{eq:5.4.13}), and
$y_p=u_pe^{3x}=xe^{3x}(4-x+2x^2) \nonumber$
is a particular solution of Equation \ref{eq:5.4.12}).
##### Example 5.4.6
Find a particular solution of
$\label{eq:5.4.14} 4y''+4y'+y=e^{-x/2}(-8+48x+144x^2).$
###### Solution
Substituting
$y=ue^{-x/2},\quad y'=u'e^{-x/2}-{1\over2}ue^{-x/2},\quad \text{and} \quad y''=u''e^{-x/2}-u'e^{-x/2}+{1\over4}ue^{-x/2} \nonumber$
into Equation \ref{eq:5.4.14}) and canceling $$e^{-x/2}$$ yields
$4\left(u''-u'+{u\over4}\right)+4\left(u'-{u\over2}\right)+u=4u''=-8+48x+144x^2, \nonumber$
or
$\label{eq:5.4.15} u''=-2+12x+36x^2,$
which does not contain $$u$$ or $$u'$$ because $$e^{-x/2}$$ and $$xe^{-x/2}$$ are both solutions of the complementary equation. (See Exercise 5.4.30.) To obtain a particular solution of Equation \ref{eq:5.4.15}) we integrate twice, taking the constants of integration to be zero; thus,
$u_p'=-2x+6x^2+12x^3\quad \text{and} \quad u_p=-x^2+2x^3+3x^4=x^2(-1+2x+3x^2).\nonumber$
Therefore
$y_p=u_pe^{-x/2}=x^2e^{-x/2}(-1+2x+3x^2)\nonumber$
is a particular solution of Equation \ref{eq:5.4.14}).
## Summary
The preceding examples illustrate the following facts concerning particular solutions of a constant coefficent equation of the form
$ay''+by'+cy=e^{\alpha x}G(x),\nonumber$
where $$G$$ is a polynomial (see Exercise 5.4.30):
1. If $$e^{\alpha x}$$ isn’t a solution of the complementary equation $\label{eq:5.4.16} ay''+by'+cy=0,$ then $$y_p=e^{\alpha x}Q(x)$$, where $$Q$$ is a polynomial of the same degree as $$G$$. (See Example 5.4.4 ).
2. If $$e^{\alpha x}$$ is a solution of Equation \ref{eq:5.4.16} but $$xe^{\alpha x}$$ is not, then $$y_p=xe^{\alpha x}Q(x)$$, where $$Q$$ is a polynomial of the same degree as $$G$$. (See Example 5.4.5 .)
3. If both $$e^{\alpha x}$$ and $$xe^{\alpha x}$$ are solutions of Equation \ref{eq:5.4.16}, then $$y_p=x^2e^{\alpha x}Q(x)$$, where $$Q$$ is a polynomial of the same degree as $$G$$. (See Example 5.4.6 .)
In all three cases, you can just substitute the appropriate form for $$y_p$$ and its derivatives directly into
$ay_p''+by_p'+cy_p=e^{\alpha x}G(x), \nonumber$
and solve for the coefficients of the polynomial $$Q$$. However, if you try this you will see that the computations are more tedious than those that you encounter by making the substitution $$y=ue^{\alpha x}$$ and finding a particular solution of the resulting equation for $$u$$. (See Exercises 5.4.34-5.4.36.) In Case (a) the equation for $$u$$ will be of the form
$au''+p'(\alpha)u'+p(\alpha)u=G(x), \nonumber$
with a particular solution of the form $$u_p=Q(x)$$, a polynomial of the same degree as $$G$$, whose coefficients can be found by the method used in Example 5.4.4 . In Case (b) the equation for $$u$$ will be of the form
$au''+p'(\alpha)u'=G(x) \nonumber$
(no $$u$$ term on the left), with a particular solution of the form $$u_p=xQ(x)$$, where $$Q$$ is a polynomial of the same degree as $$G$$ whose coefficents can be found by the method used in Example 5.4.5 . In Case (c), the equation for $$u$$ will be of the form
$au''=G(x) \nonumber$
with a particular solution of the form $$u_p=x^2Q(x)$$ that can be obtained by integrating $$G(x)/a$$ twice and taking the constants of integration to be zero, as in Example 5.4.6 .
## Using the Principle of Superposition
The next example shows how to combine the method of undetermined coefficients and Theorem 5.3.3, the principle of superposition.
##### Example 5.4.7
Find a particular solution of
$\label{eq:5.4.17} y''-7y'+12y=4e^{2x}+5e^{4x}.$
###### Solution
In Example 5.4.1 we found that $$y_{p_1}=2e^{2x}$$ is a particular solution of
$y''-7y'+12y=4e^{2x}, \nonumber$
and in Example 5.4.2 we found that $$y_{p_2}=5xe^{4x}$$ is a particular solution of
$y''-7y'+12y=5e^{4x}. \nonumber$
Therefore the principle of superposition implies that $$y_p=2e^{2x}+5xe^{4x}$$ is a particular solution of Equation \ref{eq:5.4.17}).
This page titled 5.4: The Method of Undetermined Coefficients I is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. | 5,536 | 14,845 | {"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} | 4.34375 | 4 | CC-MAIN-2022-49 | latest | en | 0.429214 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.