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5200 | https://blog.csdn.net/qq_33471732/article/details/132440048 | 传递函数零极点对系统的影响_右半平面零点对系统的影响-CSDN博客
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传递函数零极点对系统的影响
最新推荐文章于 2024-08-13 16:13:19 发布
原创 于 2023-08-23 01:02:38 发布·7.4k 阅读
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本文详细阐述了传递函数中零点和极点如何影响系统频率响应、稳定性及动态特性。零点和极点的位置决定了增益变化、相位特性和系统的稳定性,对控制系统设计至关重要。
传递函数的零点和极点分别对系统的影响的详细介绍:
零点(Zero)的影响:传递函数的零点是使得传递函数的分子为零的点。零点对系统的频率响应和稳定性产生影响。具体而言:
频率响应:零点的位置会影响系统在不同频率下的增益和相位特性。当传递函数的零点与频率轴上的某个频率相对应时,它会导致系统在该频率处的增益增加或相位提前。因此,通过调整零点的位置,可以调节系统在不同频率下的增益和相位特性。
稳定性:对于线性时不变(LTI)系统,如果所有的零点都位于左半平面,系统将是稳定的。如果存在零点位于右半平面,系统可能会变得不稳定。因此,在控制系统设计中,需要确保所有零点的实部位于左半平面,以保持系统的稳定性。
极点(Pole)的影响:传递函数的极点是使得传递函数的分母为零的点。极点对系统的频率响应、稳定性和动态特性产生影响。具体而言:
频率响应:极点的位置会影响系统在不同频率下的增益和相位特性。当传递函数的极点与频率轴上的某个频率相对应时,它会导致系统在该频率处的增益下降或相位延迟。因此,通过调整极点的位置,可以调节系统在不同频率下的增益和相位特性。
稳定性:对于线性时不变(LTI)系统,如果所有极点的实部都位于左半平面,系统将是稳定的。如果存在极点位于右半平面,系统可能会变得不稳定。因此,在控制系统设计中,需要确保所有极点的实部位于左半平面,以保持系统的稳定性。
动态特性:极点的位置会影响系统的动态特性,如响应时间、超调量和阻尼比等。通过调整极点的位置,可以实现更快的响应时间、更小的超调量或更好的阻尼特性。
综上所述,传递函数的零点和极点分别对系统的频率响应、稳定性和动态特性产生影响。在系统分析和控制设计中,对传递函数的零点和极点进行分析和调节是非常重要的。
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武汉理工大学课程设计优秀报告_ 零极点 对系统 性能的 影响 分析
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自动控制原理优秀开源报告---零极点 对系统 性能的 影响 分析 要求完成的主要任务: (包括课程设计工作量及其技术要求以及说明书撰写等具体要求) 当开环 传递函数 为G1(s)时,绘制系统的根轨迹和奈奎斯特曲线; 当开环 传递函数 为G1(s)时,a分别取0.01,0.1,1,10,100时,用Matlab计算系统阶跃响应的超调量和系统频率响应的谐振峰值,并分析两者的关系; 画出(2)中各a值的波特图; 当开环 传递函数 为G2(s)时,绘制系统的根轨迹和奈奎斯特曲线; 当开环 传递函数 为G2(s).
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(3-61)也一定可以化成(3-64)的形式,3-64其实更友好,可以看到,他把分母化成了一次因式和二次因式,一次因式对应的特征根都是实数(允许多重艮2),二次因式对应的特征根都是共轭复数。根据3-62的分母可以直接看出系统的特征根,不过要注意,3-62中的特征根si有可能是复数。(3-61)所示的是系统的 传递函数 的分式形式。他一定可以化成(3-62)形式。《自动控制原理》胡寿松,第六版。
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matlab 传递函数 零极点 对消
04-11
传递函数 的 零极点 对消可以通过使用MATLAB中的控制系统工具箱来实现。下面是一种常见的方法: 1. 首先,使用tf函数创建 传递函数 对象。例如,假设有一个 传递函数 为G(s) = (s+1)(s+2)/(s+3)(s+4),可以使用以下代码...
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5201 | https://www.reddit.com/r/ALevelChemistry/comments/1d33v4d/help_i_still_cant_get_a_brain_of_alcl3_and_al2cl6/ | Help! I still can't get a brain of AlCl3 and Al2Cl6 : r/ALevelChemistry
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hideki75
Help! I still can't get a brain of AlCl3 and Al2Cl6
Okay, all i know is AlCl3 has a structure of Giant Ionic Lattice. Al2Cl6 is a simple molecular structure and a dimer of Aluminium Chloride that is in solid state.
what is the difference between AlCl3 and Al2Cl6?
why does it has AlCl3 and Al2Cl6?
does AlCl3 is made from Al2Cl6?
does it has a different products from if each of it dissolve in water? Are the pH of solution produce are same?
Sorry for a long question, I'm gladly and open hands if someone wanted to explain it even if it is out of syllabus context. If you could give me more information other than my question, I would appreciate it thanks! (◠‿・)—☆
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5202 | https://bigpicture.one/blog/problem-solving-strategies/ | Demo
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July 28, 2023
Figure it out: 9 problem-solving strategies for managers
Project Management
Agnieszka Sienkiewicz
“Okay Houston, we’ve had a problem here” – John “Jack” Swigert.
In 1970, the Apollo 13 crew faced a number of serious obstacles — of which the loss of oxygen, which left them with under two hours of power generation remaining. NASA Mission Control had to quickly figure out how to keep their astronauts alive, and eventually, the results of these joint problem-solving efforts by engineers on the ground allowed three astronauts to safely return to Earth.
In life — be it private or professional — we also face challenges, perhaps even on a daily basis. Knowing how to solve problems is extremely important, so if you’re interested in a complete, end-to-end process for developing solutions, here are nine effective problem-solving strategies we put together for you.
What are problem-solving strategies?
However obvious this may sound; to solve a problem means to produce solutions, then choose the best out of your alternatives. And to find the ultimate solution, you must first produce a plan of action, or strategy. Which may be different from the way you typically go about problem-solving; nonetheless it’s quite effective.
Just like you can find many solutions to the same problem, you can also use different problem-solving strategies with different action plans associated with them.
Importance of (using) problem-solving strategies
Resolving business challenges is an important outcome of the successful application of problem-solving strategies; but there’s more to it.
No matter the nature or magnitude of a problem, or decision you need to make, the stress can be palpable. Naturally, you want to do something about it — you want to resolve it; which is where problem-solving strategies come in. Knowing that there’s a suitable strategy to help remedy a situation can significantly reduce your anxiety and overall distress.
Another important aspect to keep in mind when it comes to problem-solving strategies is personality. We all have different personalities and aptitudes, which means we think differently. For that reason, it’s better to have several strategies to choose from, as some may resonate better than others. Consequently, these will prove more effective.
Likewise, some problem-solving strategies are more likely to help you achieve your goals when used specifically for certain situations or challenges. In other words, not every approach will work for every type of problem. Knowing at least a few different problem-solving strategies can increase your chances to find the right solution and achieve better outcomes.
And finally, problem-solving strategies promote critical thinking, creativity, and oftentimes — collaboration within or across teams. All of these are essential skills in personal and professional contexts.
Steps in the problem-solving process
When facing a business problem such as handling complex dependencies or conflicting schedules — how do you solve it? Do you grind it for days or jump straight into thinking about potential solutions?
The very process of problem-solving isn’t new; it includes a set of specific steps to guide you through the entire solution-discovery journey. Problem-solving strategies, on the other hand, reinforce this sequential approach so you can make the most out of each step.
1. Defining and analyzing the problem
The first step in problem-solving is problem analysis and definition. At first, you know what the symptoms are, but you also want to know what triggers them. So, you gather the necessary data by using your observational skills and talking to colleagues. Be mindful, however, that some of the information you will gather might be more of an opinion than a fact; make sure to differentiate between the two. The goal of this step is to formulate the problem specifically and determine the underlying cause (or causes).
For example, your organization is struggling with high employee turnover (problem formulation). As a hiring manager, you discover that new hires often leave after the trial period, due to a poor onboarding experience and lack of adequate support (problem causes).
The process of collecting data from individual stakeholders is also important for breaking down a problem into smaller key elements. On one hand, this can get you to determine the causes of a problem; on the other hand, it will later enable you to generate more accurate solution alternatives.
Corresponding strategies and tools:
Fishbone diagram
Flowcharts
Kipling method
Work backward
SWOT analysis
2. Generating alternative solutions
Now that you know why the problem has arisen, you want to generate possible solutions (on your own, or alongside involved individuals). For the time being, focus on the solutions alone — you can evaluate them later. The alternatives you will manage to produce should be consistent with your organizational goals, regardless of whether they’re short or long term.
Corresponding strategies and tools:
Algorithm
Mind map
Brainstorming
What-if scenarios
Divide and conquer
Means-end analysis
3. Evaluating and selecting an alternative
Once you have a list of potential solutions for your business or organizational problem, then you can pick the best alternative. Look at the solutions you and your colleagues have shortlisted, and evaluate them without bias. Which of them are relative to the goals you established, and which are most likely to produce the desired results? When selecting the alternative, consider all organizational constraints and the consequences it may bring forth.
For example, a new solution could lead to a decrease in resource capacity due to implementing a buddy system that aims to help onboard new hires.
Corresponding strategies and tools:
Brainstorming
Decision matrix analysis
4. Implementing and monitoring new solutions
By now, you (and your colleagues) know what the real problem is, and how to remedy it. At this point, you’re ready to implement the solution and observe how the situation unfolds. Take the time to collect feedback from those affected by the solution, and seek their acceptance.
What if the solution fails to produce the expected results? What if you encounter new challenges that will require you to change your original approach? In such a case, you’ll need to reassess the changes, then follow the problem-solving process steps (again) to create a new list of solutions, or to update the existing one.
Corresponding strategies and tools:
Trial and error
Effective problem solving-techniques to try
There are many different ways to approach problem-solving. Each suitable for different types of problems, or stages of the problem-solving process. The strategies you’ll find below are suggestions to help you get started. You may need to experiment with several strategies before you find a workable solution for your specific problem.
Algorithm
If you’re looking for a structured and procedural approach to problem-solving, an algorithm might be just what you need. An algorithm is a problem-solving strategy that provides you with step-by-step instructions that — if followed to the letter — can help you achieve your desired outcome. You can think of an algorithm as a cooking recipe (or function in a code) that describes instructions in high detail. An algorithm will return the same result every time you execute it.
One benefit of this strategy is that it produces very accurate results. But it might not always be the best approach to problem-solving. That’s partly due to the fact that detecting certain patterns can be incredibly time-consuming. Still, if you need to deal with a group of similar problems, algorithms can help you figure out the common denominator, and find a workable solution for these problems.
Heuristics
Heuristics is a general problem-solving strategy; opposite to the algorithmic approach as it uses intuition and (judge)mental shortcuts (the “rule of thumb”) instead of a systematic approach. For that reason, heuristic approaches are less time and energy-consuming than algorithmic problem-solving. For example, hiring managers can apply heuristics when considering a pool of candidates. By following their intuition and experience, they can quickly choose whom to offer the job to.
In certain cases, and despite these characteristics, heuristics may not be the best way to make rational decisions. Especially, in the cases where you need to process a lot of information, work under pressure, or lack the data needed to generate a solution. Taking a shortcut might be the most tempting thing to do under these conditions; it could also send you off course.
Work backward
Working backward is a useful heuristic problem-solving strategy. To work backward means to begin solving the problem by identifying the steps needed to achieve the end result. It’s like reverse engineering; where you’re focused on a solution to a problem, instead of a software or system. For example, if the product you manage has received several negative reviews lately, you can ask yourself “What has happened which has led to this situation?” Then, you can work backward, step by step.
Kipling (5W1H) method
This strategy is about asking a series of six questions in order to understand a problem better. The questions come from Rudyard Kipling’s poem, “I keep six honest serving men.” The six questions are:
“What” – What is the problem?
“Why” – Why is the problem important?
“When” – When did the problem arise, and when do you need to solve it?
“How” – How did the problem happen?
“Where” – Where is the problem occurring?
“Who” – Who does the problem affect?
As a manager, you can use the Kipling method to identify all relevant factors and ensure you fully understand thr problem — before you start looking for solutions.
5W1H and 5 Whys
You can use the Kipling and 5 Whys problem-solving methods interchangeably or together, because both of them help you identify existing problems. But keep in mind that there’s a significant difference between these two methods.
The Kipling method asks about various key details regarding the problem; “Why” is just one of these questions. The 5 Whys method, on the other hand, repeats the “Why” five times. The goal is to get granular with the reasons behind the problem, asking one “Why” after another. If a problem is too complex and requires a more comprehensive analysis, you can combine both methods for a more solid outcome. This way, you get a higher chance to better understand the problem and find a solution.
Trial and error
When you have several possible solutions, and would like to test them in order to see which one works best, a trial-and-error strategy can be helpful. Using this approach, you can try different solutions until you find the right one. For example, you may need to figure out how to allocate your shared resources for the upcoming project phase. By testing various setups and comparing outcomes, you can identify the best solution. We recommend using what-if scenarios that will allow you to safely carry out your tests, without impacting your original plan.
Brainstorming
The problem you face is not necessarily a problem on its own. Your team or the other managers can provide valuable hints or help with solutions that you did not even consider. Consequently, the more people you gather to help solve a problem, the more potential solutions you get to produce together. The more, the merrier.
The brainstorming strategy not only helps overcome critical challenges, but also stimulates creativity and encourages collaboration among your colleagues.
Divide and conquer (and combine)
This problem-solving strategy is about breaking down large complex problems (“divide”) into a set of smaller, more manageable subproblems that are similar to the original problem. You look into each subproblem individually and try to solve it one by one (“conquer”). Then, you merge those solutions back into one in order to create a solution to the original, larger problem (“combine”).
Divide and conquer and Means-end analysis
Means-end analysis is a problem-solving strategy where you consider the obstacles standing between the problem state and the end-result (solution) state. In other words, it helps you get from A to B by examining the obstacles along the way and finding solutions to them. Elimination of these obstacles produces the smaller subgoals you need to achieve. And when you achieve all these subgoals, i.e., when all of the obstacles are out of your way, you’ve reached your main goal (point B, or the solution state).
The means-end analysis is a version of the divide-and-conquer strategy. The difference between the two is this: divide and conquer is recursive — when using this method, the subproblems you solve are always of the same type. Means-end analysis, on the other hand, is more flexible and less recursive; that’s because the subproblems you’re trying to solve don’t need to be similar type-wise.
Walking the path always traveled: pitfalls to problem-solving
Problem-solving strategies are helpful, nonetheless they cannot guarantee that you’ll find the solution you seek. Why is that? What could possibly stand in the way between you and your solution?
Many things; for instance, your mindset.
As Albert Einstein once said, “Insanity is doing the same thing over and over again and expecting a different result.” Imagine yourself identifying and dismantling a problem into the tiniest particles. Then, applying the same true and tried method over and over again, expecting to achieve your goal. Instead, you’re failing to resolve the problem every time and getting stuck, running in circles. What happens in such situations is that you persist to solve a problem in the same way you did before. The only difference is that this method doesn’t seem to be working this time.
Such a mindset is called functional fixedness. It’s a state where you’re unable to see a method, object, or tool that could be used for something else than what it was designed for.
In Apollo 13 mission, ground engineers had to overcome functional fixedness and figure out how to literally fit a square peg in a round hole. Their solution to this problem involved basic materials, such as spare plastic bags, tape, and air hoses.
So, if you ever get stuck overcoming your challenge, it’s a good idea to step back and think about why your strategy isn’t working. Perhaps, there’s another, better approach you haven’t considered yet. Don’t be afraid to try different strategies or combine a few together; there’s no obviously-right or obviously-wrong way to solve problems.
Problem-solving strategies: summary
There are many different strategies you could use to solve a business problem. Typical problem-solving strategies include trial and error, applying algorithms, and using heuristics. When you need to solve a large, complex problem, it often helps to break the problem into smaller subproblems that you can then solve individually. The sum of these subsolutions can then get you to the general solution for your large problem. When trying to find a solution, be aware of major roadblocks including your very own mindset.
Also, whenever you face a problem in the workplace, keep in mind that apart from strategies, you also have a variety of tools at your disposal. While neither of them will solve the problem for you, they will help you gain the necessary insights into the situation. BigPicture PPM software will provide you with the data you need to clearly identify the problem. Then, it will let you test different solutions so you can pick the best one. And finally, it will allow you to monitor your initiative all the way through the entire project lifecycle.
Best part? BigPicture is used by NASA; but you don’t need to be a ground control engineer to benefit from it! 😉 Find out what BigPicture can do for you, get a 30-day free trial that you can start today.
Or visit our demo page to play with the app straight in your browser — no registration or installation needed.
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Agnieszka Sienkiewicz All articles by the author
Content writer at BigPicture. Previously, Aggie worked for SaaS companies writing specifically about eCommerce and marketing. As a continuous learner and advocate for knowledge-sharing, she creates content for beginners as well as more advanced readers. She loves clean plant-based food and morning workouts.
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5203 | https://aacrjournals.org/cancerres/article/43/10/4935/487145/Epidermal-Cell-Cycle-and-Transit-Times-during | Epidermal Cell Cycle and Transit Times during Hyperplastic Growth Induced by Abrasion or Treatment with 12-O-Tetradecanoylphorbol-13-acetate1 | Cancer Research | American Association for Cancer Research
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Volume 43, Issue 10
1 October 1983
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Articles|October 01 1983
Epidermal Cell Cycle and Transit Times during Hyperplastic Growth Induced by Abrasion or Treatment with 12-O-Tetradecanoylphorbol-13-acetate1Free
Rebecca Morris;
Rebecca Morris
Department of Biology, Syracuse University [R. M.]
, and
Department of Pathology, Upstate Medical Center [T. S. A.]
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Syracuse, New York 13210
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Thomas S. Argyris
Thomas S. Argyris
Department of Biology, Syracuse University [R. M.]
, and
Department of Pathology, Upstate Medical Center [T. S. A.]
,
Syracuse, New York 13210
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2
Work performed in partial fulfillment of the requirements for the Ph.D. in biology in the Graduate School of Syracuse University, Syracuse, N. Y. Present address: The University of Texas System Cancer Center, Science Park Research Division, P. O. Box 389, Smithville, Texas 78957.
3
To whom requests for reprints should be addressed.
Received:March 10 1983
Accepted:July 12 1983
Online ISSN: 1538-7445
Print ISSN: 0008-5472
©1983 American Association for Cancer Research.
1983
Cancer Research, Inc.
Cancer Res (1983) 43 (10): 4935–4942.
Article history
Received:
March 10 1983
Accepted:
July 12 1983
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Rebecca Morris, Thomas S. Argyris; Epidermal Cell Cycle and Transit Times during Hyperplastic Growth Induced by Abrasion or Treatment with 12-O-Tetradecanoylphorbol-13-acetate1. _Cancer Res_ 1 October 1983; 43 (10): 4935–4942.
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Abstract
The purpose of this investigation was to determine the cellular kinetics of the epidermis during hyperplastic growth in CD-1 mice induced by treatments effective in skin tumor promotion: abrasion or the application of 12-O-tetradecanoylphorbol-13-acetate (TPA). Following removal by abrasion, the epidermis regenerated from intrafollicular epidermal cells. The maximal increase in the number of epidermal cells, over 2-fold from normal, was reached by 4 days after abrasion. Regression of the hyperplasia began about Day 7 and continued through Day 20, resulting in a nearly normal epidermis. The growth fraction measured at 3 days after abrasion did not change from its normal value of about 100%. However, the cell cycle time of basal keratinocytes 3 days after abrasion was drastically reduced to about 11 hr compared to 5 to 7 days in normal epidermis. The cell cycle time gradually increased to 14 hr at 5 days, 1 to 2 days at 7 days, and 4 to 5 days at 14 days after abrasion. The reduction in the length of the cell cycle time was primarily due to a decrease in the length of G 1. The rate of epidermal cell loss was measured by the epidermal transit time, the time required for [3 H]thymidine-labeled basal cells to reach the uppermost nucleated layer. Labeled nuclei were followed through the epidermal columns, when present. Transit time was dramatically reduced from 8 days in normal skin to 1 day by 3 days following abrasion, then rose to 2 days by 5 days postabrasion, 4 days by 7 days postabrasion, and by 6 days 14 days postabrasion.
A single application of 17 nmol of TPA resulted, within 1 day, in over a 50% increase in the total number of epidermal cells. This thickness was maintained until about Day 4, and then regression began resulting in an essentially normal epidermis by 10 days. As after abrasion, the growth fraction 3 days after treatment with TPA was about 100%, that is, not changed from normal. The cell cycle time, however, was again dramatically reduced to 16 hr beginning at 1 hr, 25 hr at 3 days, 2 to 3 days at 5 days, and 3 to 5 days at 10 days after application of TPA. Also, as after abrasion-induced hyperplasia, the transit time was reduced to 2 days at 1 hr, 4 days at 3 days, and 5 days at 5 and 10 days following treatment with TPA. The decrease in the cell cycle time can be accounted for by the decrease in the length of G 1.
The results of this investigation demonstrate that the production of an epidermal hyperplasia following abrasion or the application of 17 nmol of TPA is associated with both an increase in cell proliferation and cell loss. The increased cell proliferation can be accounted for by the decrease in the length of the cell cycle. It appears that the increase in epidermal cell loss may be also linked to the decrease in the length of the cell cycle, since all suprabasal cells arise from the basal cells.
1
This work supported by NIH Grants AM 18219 and AG 01324.
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©1983 American Association for Cancer Research.
1983
Cancer Research, Inc.
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5204 | https://www.wizeprep.com/textbooks/high-school/mathematics/2812/sections/103192 | Properties of Polygons - Wize High School Grade 9 Math Textbook | Wizeprep
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###### Wize High School Grade 9 Math Textbook > 2D Geometry - Properties of 2D Figures
Properties of Polygons
Conjectures & Hypotheses Previous Section
Line of Symmetry Next Section
Review: QuadrilateralsExamplePracticePracticeReview: TrianglesExample
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Quadrilaterals
Quadrilaterals
Show Solutions
Example: Properties of Quadrilaterals
Three conjectures about the diagonals of a quadrilateral are given:
The diagonals are equal
The diagonals bisect each other ("cuts each other in half")
The diagonals meet at a 90°90\degree 90° angle (they are perpendicular)
Come up with examples to support the conjecture or come up with a counterexample to disprove the conjecture. Then put a ✔ or ✖ in each of the following boxes to indicate whether the conjecture is likely true or not true for each quadrilateral.
PAGE BREAK
The diagonals are always equal
For each type of quadrilateral, create a few different sizes and draw in the diagonals. If you measure these diagonals, you'll see if their lengths are the same. If they are, the diagonals are likely equal. If you are able to come up with even one example where the lengths are not the same, then we know for sure that the lengths are not always equal.
Rectangles: ✔
Squares: ✔
Parallelograms: ✔
Isosceles Trapezoid: ✔
Rhombus: ✖ (they aren't always equal)
Kite: ✖ (they aren't always equal)
The diagonals always bisect each other
For each type of quadrilateral, create a few different sizes and draw in the diagonals. Measure the lengths of each side of each diagonal, if they are the same, then the diagonals likely bisect each others. If you are able to come up with even one example where the lengths are not the same, then we know for sure that the diagonals do not always bisect each other.
Rectangles: ✔
Squares: ✔
Parallelograms: ✔
Isosceles Trapezoid: ✖
Rhombus: ✖ (they don't always bisect each other)
Kite: ✖ (they don't always bisect each other)
The diagonals are always perpendicular (meet at 90°90\degree 90°)
For each type of quadrilateral, create a few different sizes and draw in the diagonals. Measure the angle the diagonals make with one another, if they are 90°90\degree 90°, then the diagonals likely are always perpendicular. If you are able to come up with even one examlpe where they are not 90°90\degree 90°, then we know for sure that the diagonals are not always perpendicular
Rectangles: ✖ (they aren't always perpendicular)
Squares: ✔
Parallelograms: ✖ (they aren't always perpendicular)
Isosceles Trapezoid: ✖ (they aren't always perpendicular)
Rhombus: ✔
Kite: ✖ (they aren't always perpendicular)
Practice: Properties of Quadrilaterals
Adjacent angles are the angles that are "next to each other".
Given the conjecture "the sum of the two adjacent angles is always 180°180\degree 180°", come up with examples to either support this conjecture or come up with a counterexample to disprove the conjecture for each of the following quadrilaterals.
Put "yes" in the box if the conjecture for that quadrilateral is likely true, and put "no" in the box if the conjecture is not true for that quadrilateral.
Rectangle Square Parallelogram Isosceles Trapezoid Rhombus Kite
The sum of the two adjacent angles is always 180 for this type of quadrilateral
[x] I don't know
Check Submission
Mark Yourself Question
1. Grab a piece of paper and try this problem yourself.
2. When you're done, check the "I have answered this question" box below.
3. View the solution and report whether you got it right or wrong.
Practice: Properties of Quadrilaterlas.
The midsegment in a polygon is a line segment that joins the midpoints of two adjacent sides in that polygon.
The conjecture "the quadrilateral formed by connecting the four midsegments in any quadrilateral are parallelograms" is given. Either come up with multiple examples to support this conjecture or come up with one counterexample to disprove this conjecture.
[x] I have answered this question
Check Submission
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Triangles
Triangles
PAGE BREAK
Median: a line that connects a vertex to the midpoint of its opposite side
Altitude: a line that connects a vertex to its opposite side, and is perpendicular to the opposite side (an altitude is often called the "height" of the triangle)
Angle bisector: a line that connects a vertex to its opposite side, and it cuts the angle in half
Perpendicular (right) bisector: a line that cuts a side in the triangle in half, and is perpendicular to that side
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In an equilateral triangle, the median, altitude, and right bisector from a single vertex are all the same!
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Show Solutions
Example: Properties of Triangles
Based on the following angle measurements, come up with a conjecture for the relationship between exterior and interior angles of any triangle. Then try to disprove the conjecture by finding a counterexample, or actually prove the conjecture using angle properties of triangles.
Right angle triangle
PAGE BREAK
Isosceles Triangle
PAGE BREAK
Scalene Triangle
According to the 3 triangles given, it seems like the exterior angle is the sum of the two opposite interior angles.
Although we have 3 examples here that support our conjecture, it's not enought to actually prove the conjecture because "what if there is an example out there that disproves this conjecture?"
But we can really prove this conjecture by not looking at examples at all. Instead, we turn to properties and facts about angles that we already know to be true.
Let's take a look at any triangle, we can label 4 angles as a,b,c\bct{a,\ b,\ c}a,b,c and d\bct d d:
Recall:
Supplemental angles that form a straight line must add up to 180°180\degree 180°
The sum of the interior angles in any triangle is 180°180\degree 180°
Using property 1: c+d=180°\colorbox{yellow}{$c+d=180\degree$}c+d=180°
Using property 2: a+b+d=180°\colorbox{yellow}{$a+b+d=180\degree$}a+b+d=180°
Since c+d c+d c+d and a+b+d a+b+d a+b+d both equal 180°180\degree 180°, we know that c+d c+d c+d and a+b+d a+b+d a+b+d must equal each other:
c+d=a+b+d−d=−d c=a+b\begin{array}{rcl} c+d&=&a+b+d\ \scriptsize-d&=&~~~~~~~~~~~~\scriptsize-d\ c&=&a+b \end{array}c+d−d c===a+b+d−d a+b
We see that for any triangle, c=a+b\boxed{c=a+b}c=a+b, meaning that the exterior angle of any triangle is the sum of the two opposite interior angles!
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5205 | https://artofproblemsolving.com/wiki/index.php/Polynomial_Remainder_Theorem?srsltid=AfmBOooLi8Us2lVK_q0NIyhQpnBC5JVdlmCmLn18M-BCiujJMNjIoOCS | Art of Problem Solving
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Polynomial Remainder Theorem
In algebra, the Polynomial Remainder Theorem states that the remainder upon dividing any polynomial by a linear polynomial , both with complex coefficients, is equal to .
Contents
[hide]
1 Proof
2 Generalization
3 Problems
3.1 Example 1
3.2 More Problems
4 See Also
Proof
By polynomial division with dividend and divisor , that exist a quotient and remainder such that with . We wish to show that is equal to the constant . Because , . Therefore, , and so the is a constant.
Let this constant be . We may substitute this into our original equation and rearrange to yield When , this equation becomes . Hence, the remainder upon diving by is equal to .
Generalization
The strategy used in the above proof can be generalized to divisors with degree greater than . A more general method, with any dividend and divisor , is to write , and then substitute the zeroes of to eliminate and find values of . Example 2 showcases this strategy.
Problems
Here are some problems with solutions that utilize the Polynomial Remainder Theorem and its generalization.
Example 1
What is the remainder when is divided by ?
Solution: Although one could use long or synthetic division, the Polynomial Remainder Theorem provides a significantly shorter solution. Note that , and . A common mistake is to forget to flip the negative sign and assume , but simplifying the linear equation yields . Thus, the answer is , or , which is equal to . .
More Problems
1950 AHSME Problem 20
1961 AHSME Problem 22
1969 AHSME Problem 34
See Also
Polynomial
Factor theorem
This article is a stub. Help us out by expanding it.
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5206 | https://ctlm.uni.edu/sites/default/files/inline-files/a_closer_lookmaking_sense_of_log_properties_1.pdf | 2010 1 Making Sense of Log Properties A Closer Look at the Video After watching the Making Sense of Log Properties video, make sense of the mathematics by taking a closer look at the problem situations and solutions. Use the comments and questions in bold to help you investigate the key points of the video and develop a deeper understanding of properties of logarithms. Problem: You may use log properties, but to really understand why they work, you need to prove them. Use prior knowledge such as properties of exponents and the definition of logarithm to prove the following properties of logarithms: log (xy) = log (x) + log (y) log (xn) = n log (x). What do we need to do if we want to prove log (xy) = log (x) + log (y)? If we want to prove this, we need to show that the left-hand side of the equation is equivalent to the right-hand side. Figuring out where to begin is sometimes the hardest part. We can browse through our math manual, that is, think about what we already know, to help us get started. What information do we already know? We know the definition of logarithm, and since a logarithm is an exponent, it is also helpful to keep in mind the properties of exponents. Mathematics Manual Definition of a Logarithm: If ax = y then x = logay, and if x = logay then y = ax. Properties of Exponents: a0 = 1 anam = an+m = an-m (an)m = an⋅m Now, where might we begin? Since we want to prove log (xy) = log (x) + log (y), we need something with log (x) and log (y). We can try letting n = log x and m = log y. Then, using the definition of logarithms, if n = log x then x = 10n; this is also true for y = 10m. 2010 2 Making Sense of Log Properties A Closer Look at the Video Since we’re working with the log (xy), what would be a reasonable next step in our proof? We should multiply x and y. xy = (10n)(10m). Then, using the multiplication property of exponents, xy = 10(n + m). Now we have x times y, but we want an equation involving logarithms. What can we do to the equation, xy = 10(n + m) to change it into an equation with a logarithm? We should use the definition of logs again to write log (xy) = n + m. log (xy) = log (x) + log (y) Let n = log (x) and m = log (y) x = 10n and y = 10m log (xy) = log (x) + log (y) Let n = log x and m = log y x = 10n and y = 10m xy = (10n)(10m) xy = 10(n + m) log (xy) = log (x) + log (y) Let n = log x and m = log y x = 10n and y = 10m xy = (10n)(10m) xy = 10(n + m) log (xy) = n + m 2010 3 Making Sense of Log Properties A Closer Look at the Video Think back to how we started the proof. How did we define n and m? We let n = log x, and m = log y. How can we use that information to finish the proof? We can substitute those expressions and get log (xy) = log (x) + log (y). Now, we have proven our property. What connection is there between multiplying numbers with exponents and the logarithm of a product? When you multiply numbers with exponents, you add the exponents. Similarly, when you take the log of a product, you add the logs. Our math manual is continually growing by adding what we learn to what we already know. Let’s see if using our updated math manual can help us prove one more property: log (xn) = n log (x). What is another way log (xn) can be written that relates to the property of logarithms we have already proved? log (xy) = log (x) + log (y) Let n = log x and m = log y x = 10n and y = 10m xy = (10n)(10m) xy = 10(n + m) log (xy) = n + m log (xy) = log x + log y Property of Exponents: an⋅am = an+m Property of Logarithms: log (xy) = log (x) + log (y) 2010 4 Making Sense of Log Properties A Closer Look at the Video If you see log (xn), you might quickly recognize it is the same as the log of the quantity x⋅x⋅…⋅x, taken n times. Using the previous property, we know that the log of x⋅x⋅…⋅x, n times, is equal to log x + log x + … + log x, n times or n log (x). Unfortunately, we cannot use this as a proof because this only makes sense for whole number values of n. In proofs, we have to choose strategies that can work for any value. Let’s not get discouraged though; we’ll just consult our manual and try again. If we want to start the proof of this property in a similar way to the first proof, what should we do first? Let m = log x, and use the definition of logarithm, to write 10m = x. Since we are trying to prove something about a power, what might be a logical next step in our proof? We can raise both sides of the equation to the nth power. Then, we can use a property of exponents for powers to get 10m•n = xn. The property we are trying to prove has log (xn). How can we get from our current equation, 10(m⋅n) = xn to an equation with log (xn)? Using the definition of logarithms again, we can write m⋅n = log (xn). log (xn) = n log (x) Let m = log x 10m = x log (xn) = n log (x) Let m = log x 10m = x (10m)n = xn 10m⋅n = xn definition of logarithm raise both sides to nth power property of exponents 2010 5 Making Sense of Log Properties A Closer Look at the Video At the beginning of the proof we let m = log x. How can we use this to finish the proof? By substituting, we have log (x)•n = log xn or n•log (x) = log xn. Now, we can add this property to our math manual. log (xn) = n log (x) Let m = log x 10m = x (10m)n = xn 10m⋅n = xn m⋅n = log (xn) log (xn) = n log (x) Let m = log x 10m = x (10m)n = xn 10m⋅n = xn m⋅n = log (xn) n log x = log (xn) definition of logarithms |
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