Geometry Articles - WyzAnt Tutor Blogshttp://www.wyzant.com/resources/blogs/geometryThis is an aggregate of all of the Geometry articles in WyzAnt.com's Tutors' Blogs. WyzAnt.com is your source for tutors and students.Sun, 28 Dec 2014 01:54:18 -0600http://www.wyzant.com/images/WyzAnt_white200.gifGeometry Articles - WyzAnt Tutor Blogshttp://www.wyzant.com/resources/blogs/geometryhttp://www.wyzant.com/resources/blogs/geometry307021http://www.wyzant.com/resources/blogs/307021/a_game_of_geometric_constructionsJared G.http://www.wyzant.com/resources/users/view/77094170A Game of Geometric Constructions<div> Although I enjoy geometric constructions, as in solving geometric problems with the equivalent of a string, I find that many students have little to no interest in them. I particularly like learning about how ancient cultures such as the Egyptians used them to design Pyramids where the error in the corners are about 1/300 of one degree, much more accurate than can be seen and even more accurate than almost all houses built today. Although learning about their history is interesting there is not a lot of places to apply this knowledge in the modern world, i've solved some problems in surveying with geometric constructions but there are always more advanced CAD methods which can also do the trick; which is why I was happy to find Euclid The Game.</div>
<div> This is a straightforward game that applies all the basic principles of geometric constructions into a fun little game. Although it doesn't require the attention to detail the Egyptians would have applied since it is computer based, it does require the same type of thinking and creative application of a few geometric constructions. Anyone looking for a way to apply geometric constructions, something to work your mind, a good puzzle game, or even just to explore some interesting geometric concepts should check this game out.</div>
<div> The link to the first level is below.</div>
<div>http://euclidthegame.com/Level1/</div>
<div> Some of the levels are fairly difficult, despite tutoring students in geometry I still had to review a couple concepts for the last problem and one in the middle.</div>Fri, 05 Dec 2014 08:58:47 -06002014-12-05T08:58:47-06:00299453http://www.wyzant.com/resources/blogs/299453/how_to_read_perimeter_of_rectangle_formulaAvery A.http://www.wyzant.com/resources/users/view/85292630How To Read Perimeter of Rectangle Formula<div> <br /><br />Reading Formulas can make or break how a student comprehends the formula when alone - outside the presence of the teacher, instructor, tutor, or parent.<br /><br /> Formula For Perimeter of Rectangle: P = 2l + 2w<br /><br />How To Read: The Perimeter of a Rectangle is equal to two (2) times the Length of the longer side of the rectangle (L) plus two (2) times the Width of the shorter side of the rectangle (W).<br /><br />When is reading formulas like this necessary? At three particular moments, reading this formula in this manner can be effective. <br /><br />When students are initially learning what the formula means<br />When student are learning what it means when they should already know (remediation). <br />When students want to remind themselves (basics learning study skill habit)<br />Remember, Formulas at their introduction are complete statements or thoughts. Students cannot and will not recall complete thoughts or statements using letters only. For example, P equals 2 times L plus 2 times W means nothing to the student who would like to connect this formula to a word problem. <br /><br />***This minor adjustment will produce significant results and has application across different disciplines that use formulas!!!!!<br /><br />“Mathematics cannot be learned without being understood - it is not a matter of formulae being committed to memory but of acquiring a capacity for systematic thought.”<br /><br />- Peter Hilton <br /><br />Let me know your thoughts.</div>Sun, 02 Nov 2014 10:46:58 -06002014-11-02T10:46:58-06:00282166http://www.wyzant.com/resources/blogs/282166/start_fresh_this_yearLindsey J.http://www.wyzant.com/resources/users/view/82181710Start Fresh This Year!<div>Hello Students!</div>
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<div>Start this year off strong with good organizational and note taking skills. Make sure you understand the material and are not just taking notes aimlessly. Try to take in what your teacher is saying and don't be afraid to ask questions!! If you start taking the initiative to learn and understand now, college will be a much more pleasant experience for you. Trust me!</div>
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<div>Stay organized and plan your homework and study schedule!</div>
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<div>Quiz yourself!</div>
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<div>Study with friends!</div>
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<div>READ YOUR TEXTBOOK! :) </div>
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<div>Remember, homework isn't busy work and a chance to copy down your notes, it is part of the learning process. This is especially important with math, as it builds on itself and understanding the basics will make the other subjects easier!</div>
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<div>Have a fantastic and fun year!</div>Sun, 10 Aug 2014 16:13:22 -05002014-08-10T16:13:22-05:00276519http://www.wyzant.com/resources/blogs/276519/the_importance_of_s_t_e_m_science_technology_engineering_mathJacob C.http://www.wyzant.com/resources/users/view/85224961The Importance of S.T.E.M. (Science, Technology, Engineering, Math)<div>Today, the future depends on you as much as it does on me. The future also depends on educating the masses in Science, Technology, Engineering, and Math, otherwise known as STEM. As a new tutor to WyzAnt, I hope to instill the importance of these subjects in student's lives, as well as, the lives around them. </div>
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<div>Besides the fact that, "the average U.S. salary is $43,460, compared with the average STEM salary of $77,880," (Careerbuilder) these subjects are interesting and applicable to topics well beyond the classroom. Success first starts with you; I am only there to help you succeed along the way. STEM are difficult subjects. Yet when you seek out help from a tutor, like myself, you have what it takes to master them. </div>
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<div>Please enlighten me on students looking to achieve and succeed rather than live in the past and think I can't as opposed to I can. We can take the trip to the future together, one question at a time</div>Mon, 30 Jun 2014 16:15:12 -05002014-06-30T16:15:12-05:00271440http://www.wyzant.com/resources/blogs/271440/developing_study_habits_for_math_coursesTiera J.http://www.wyzant.com/resources/users/view/85053559Developing Study Habits for Math Courses<div>The majority of the students that I have often have the same problem -- they aren't grasping the information fast enough or they aren't really able to follow the lessons a teacher gives.</div>
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<div>Sometimes, teachers aren't adaptive to every learning style for each student in their classroom. However, know that each student has the capability to learn math on their own. It is just necessary to have key characteristics to make it successful.</div>
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<div>Every math student should have:</div>
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<ul>
<li>patience</li>
<li>motivation</li>
<li>adaptability</li>
<li>organizational skills</li>
<li>open communication between themselves and their teacher (inside and outside the classroom)</li>
<li>breaks!! :)</li>
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<h2><span class="grayText"><strong>Study Tips</strong></span></h2>
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<ul>
<li><span style="text-decoration: underline;"><strong> Always try to study outside of your home or dorm room.</strong></span> In our minds, those are places that we relax at and it can be difficult to turn your mind off from the distractions to study. Public libraries, universities, coffee shops, and bookstores are the way to go. Some of these places have late-night study, so availability shouldn't be a problem.</li>
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<li><span style="text-decoration: underline;"><strong> Give yourself ample time to study AND ample time to rest.</strong></span> It is not a race to learn math; some can grasp the knowledge faster than others. But if you're like me, learning one SECTION per day is all it takes.</li>
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<li><span style="text-decoration: underline;"><strong> Work out examples and exercises within each section.</strong></span> A lot of students skip this because it is not required to go through all of the exercises, depending on the instructor. However, use the tools you are given -- online training and book. If you work through each example and exercise, you'll develop muscle memory and that's the whole point of math -- to remember all of the rules that govern numbers.</li>
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<li><span style="text-decoration: underline;"><strong> Make yourself summations or cheat sheets of each section learned.</strong></span> After you know the information, create a cheat sheet that you can use to study. I promise you that this helped me out a LOT!!! :)</li>
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<li><span style="text-decoration: underline;"><strong> Keep organized.</strong></span> Math is one of those subjects that require your full attention. So, organization is key. With every piece of paper given, at the bottom jot down the date and chapter/section it pertains to, it will help out)</li>
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<h2><span class="grayText">Necessary Supplies for Math Class, excluding book of course...</span></h2>
<div>- multicolor pen (I have a pen that can write in blue, green, black, and red)<br />- mechanical pencils (with refills)<br />- college-ruled notebook paper (for writing notes while learning the section and making cheat sheets)<br />- multipurpose printer paper (to work out exercises)<br />- 3-hole puncher (for the printer paper and any documents that are given to you)<br />- dividers (cheat sheets, notes, classwork, homework, quizzes/tests)<br />- TI-83 calculator (or later model that you are completely familiar with)<br />- binder (more than 1 1/2")</div>
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<h2><span class="grayText">Optional Supplies</span></h2>
<div>- graph paper<br />- protractor<br />- math compass (of course, these three are only for geometric and trigonometric courses)<br />- ruler<br />- simple calculator (they are only a few bucks and they are so handy; do not rely on your cell phone to provide you with this capability; I promise that it is a great investment)<br />- dry erase board with markers, eraser, and cleaning spray (it helps tremendously when you want to work out problems; it's exactly why I always bring mine)</div>
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<div><br />I think I've covered it all. If you have any trouble, I'm just a lesson away.<br /> <br />Good luck.<br /> <br />Tiera J.</div>Mon, 05 May 2014 00:21:23 -05002014-05-05T00:21:23-05:00270808http://www.wyzant.com/resources/blogs/270808/sat_prep_tackling_tough_math_questionsHuzefa K.http://www.wyzant.com/resources/users/view/84571890SAT Prep - Tackling Tough Math Questions<div>Nailing an 800 on the math portion of the SAT can be a tricky feat, even if you are steadfastly familiar with all of the requisite formulas and rules. A difficult problem can overwhelm even the most prepared individual come test day. Time constraints, test surroundings, and the overall weight of the exam can unnerve the most grounded students.<br /><br />So what do you do when panic strikes and your mind draws a blank? How do you re-center yourself and charge forward with ferocity and confidence? What you do is this: write everything down from the problem. This is the most important part of the problem solving process. As you peruse the question, write down the pertinent data and establish relationships by setting up equations. This exercise will help you see solutions that were previously difficult to decipher.<br /><br />As you work on practice tests and sample problems, you must work diligently to form a solid habit of writing down important bits of information as you plow through the SAT math section. To give you an example of what it means to “write everything down from the problem,” I will explore the following three math questions in great detail. These in-depth explanations will give you an idea of what should be going through your brain every time you see a math problem. With practice, these thoughts and processes will manifest faster and faster until solving problems in this fashion becomes a reflexive response.<br /><br /> <span class="greenText">1. The average of 4 different integers is 75. If the largest integer is 90, what is the least possible value of the smallest integer?</span><br /><br /><span class="greenText">a. 1</span><br /><span class="greenText">b. 19</span><br /><span class="greenText">c. 29</span><br /><span class="greenText">d. 30</span><br /><span class="greenText">e. 33</span></div>
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<div>Right off the bat, the problem states that we have four different integers. We can begin the problem by creating variables to represent the four integers:</div>
<div><br />W X Y Z<br /><br />We also know that the average of the integers is 75. This means that we can set up another equation based on this relationship:<br /><br />(W + X + Y + Z)/4 = 75<br /><br />Isolating the variables, we get:<br /><br />W + X + Y + Z = 300<br /><br />We also know that the largest integer is 90. So:<br /><br />W + X + Y + 90 = 300<br /><br />The question then asks “what is the least possible value of the smallest integer?” This detail is a bit tricky to interpret, but we can reason this out fairly quickly. To get the smallest possible number, what needs to be true about the other two integers? They need to be as large as possible. Since 90 is the highest value for the integers, it makes sense to assign the other two variables to 90, right?<br /><br />Not so fast. If we read the question carefully, it says that there are “four different integers.” This restricts us from using 90 for the other two values. Instead, we must use 89 and 88. We now have an equation to represent the four integers (where W = the smallest integer):<br /><br />W + 88 + 89 + 90 = 300<br /><br />Solving algebraically, we get:<br /><br />W + 267 = 300<br /><br />W = 33<br /><br />Therefore, the final answer is <strong>e</strong>.<br /><br /><span class="greenText">2. Solution X is 10 percent alcohol by volume, and solution Y is 30 percent alcohol by volume. How many milliliters of solution Y must be added to 200 milliliters of solution X to create a solution that is 25 percent alcohol by volume? </span><br /><br /><span class="greenText">a. 250/3</span><br /><span class="greenText">b. 500/3</span><br /><span class="greenText">c. 400</span><br /><span class="greenText">d. 480</span><br /><span class="greenText">e. 600</span></div>
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<div>Let’s start writing down the relevant information:<br /><br />.1X = AX<br /><br />.3Y = AY<br /><br />The above equations denote the amount of alcohol given a certain number of milliliters of solution (where AX = alcohol for X, AY = alcohol for Y, X = milliliters of solution X, and Y = milliliters of solution Y). The next part of the question asks how many milliliters of Y must be added to 200 milliliters of X to create a solution that is 25% alcohol? To answer this, we can represent the facts as an equation:<br /><br />.3Y + .1X = .25(X + Y)<br /><br />Once again, we have a two variable equation. Translation: we cannot solve it. But, we have a value for X: 200. So, plugging in 200 for X, we get the equation down to one variable:<br /><br />.3(Y) + .1(200) = .25(Y + 200)<br /><br />Perfect. Solving for Y algebraically, we get:<br /><br />.3Y + 20 = .25Y + 50<br /><br />.3Y - .25Y = 50 – 20<br /><br />.05Y = 30<br /><br />Y = 600<br /><br />Therefore, the answer is <strong>e</strong>.<br /><br /><span class="greenText">3. On a certain multiple-choice test, 9 points are awarded for each correct answer, and 7 points are deducted for each incorrect or unanswered question. Sally received a total score of 0 points on the test. If the test has fewer than 30 questions, how many questions are on the test? </span><br /><br /><span class="greenText">a. Cannot be determined</span><br /><span class="greenText">b. 16</span><br /><span class="greenText">c. 19</span><br /><span class="greenText">d. 21</span><br /><span class="greenText">e. 24</span></div>
<div><br />The first step is to write down what we know and assign variables:<br /><br />+9 points = correct (X)<br /><br />-7 points = incorrect (Y)<br /><br />Sally scored a total of 0 points<br /><br />We can set up an equation with this information:<br /><br />9X – 7Y = 0<br /><br />Since we have two variables, this is not a solvable problem. Unfortunately, we do not have another relationship that we can reference to simplify this further. What can we do in this situation? When all else fails, try to isolate the variables:<br /><br />9X = 7Y<br /><br />X/Y = 7/9<br /><br />What this tells you is that the ratio of questions answered correctly and incorrectly must be 7 correct (X) to 9 incorrect (Y). This is very useful information. According to this ratio, the number of questions on the test must be some multiple of 16 (so that the 7 to 9 ratio can be preserved). For example, 7 right and 9 wrong would work, as would 14 right and 18 wrong.<br /><br />Now comes the critical piece of information: the total number of questions must be less than 30. With this helpful tidbit, the only possibly choice is 16 questions.<br /><br />Therefore, the answer is <strong>b</strong>.</div>Tue, 29 Apr 2014 02:21:39 -05002014-04-29T02:21:39-05:00270661http://www.wyzant.com/resources/blogs/270661/fun_math_sites_for_middle_school_and_high_schoolGwen R.http://www.wyzant.com/resources/users/view/82407210FUN Math Sites for Middle School and High School!<div>My recommendationa:</div>
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<div><strong>Vi Hart, website: vihart.com<br /></strong></div>
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<div><strong>Sal Khan, https://www.khanacademy.org/math/algebra</strong></div>
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<div><strong>Mamikon Mnatsakanian, www.its.caltech.edu/.../calculus.html‎</strong></div>
<div> </div>Sun, 27 Apr 2014 23:55:17 -05002014-04-27T23:55:17-05:00270045http://www.wyzant.com/resources/blogs/270045/ellen_s_choice_teach_the_concept_not_the_algorithmEllen S.http://www.wyzant.com/resources/users/view/75479140Ellen's Choice: Teach the Concept, Not the Algorithm<div>I hear a lot about math teachers from my students, and while every teacher is unique, some comments are repeated over and over. By far the most common one I hear is that their teacher didn't really explain something, or was incapable of elaborating when questioned and simply repeated the same lecture again. As a tutor, my first priority is to make sure the student understands the material, and if they're still confused, to find another way to explain it so that it makes sense. In order to do that, I need to have a thorough understanding of the concepts myself, so that I am not simply reading from a textbook but actually explaining a concept. In my years of tutoring math, I've developed a point of view and approach to math that I refer to as “teaching the concept, not the algorithm.”<br /><br />An algorithm is a step-by-step procedure for calculation. The term is used in math and computer science, but the concept of an algorithm is universal. I could tell you that I have an algorithm which consists of 1) close the windows; 2) put on a sweater; 3) check the thermostat; 4) turn it up 3 degrees if it displays lower than 68. This is pretty obviously an algorithm to solve the problem of “I am cold right now.” We have algorithms for everything in our life, and most of the time we don't even think about it that way. We see a problem, we work out a set of steps to solve it, and we complete those steps and observe the result.<br /><br />In math class, however, students frequently encounter teachers who simply teach the algorithm; handing them a formula for solving a problem without ever really teaching them the core concepts involved or why the formula is what it is. This results in a lot of rote memorization with no understanding of why the numbers are where they are in that formula. My golden question for math teaching is always “Why?” Why does this work? Why can I do that? What am I trying to accomplish here, in the grand scheme of things? If I can explain the concept to the student so that they understand what they are doing on a macro scale and why their actions work and make sense, then it doesn't matter if they forget the formula itself, they should be able to figure it out organically by going through the conceptual process again.<br /><br />I'll give you an example from my favorite math teacher, Mr. Lazur. (I wrote a whole blog post about Mr. Lazur's teaching style, which heavily influences the way I tutor.) I had Mr. Lazur for Geometry, a subject notorious for the amount of formulae it throws at its students. Every single type of shape has three or four formulas associated with it, and keeping them all straight can be a nightmare for students. Mr. Lazur got around this by showing us WHY the formulas look the way they do, ensuring that his students could always reverse-engineer the formulas from the concepts if they couldn't remember them directly. <br /><br />In this example, we're learning about the volume of a cylinder. We've just spent the previous few days discussing volume of cubes and rectangular prisms, so Mr. Lazur starts us off by reminding us of exactly what volume means. It's the amount of stuff required to fill up the shape; the amount of water that would be displaced if the shape were dropped into a bucket. Then he pulls something out that nobody was expecting: one of those CD spindles that you buy with blank, recordable CDs in them. He points out that a stack of CDs is a cylinder, taking them off the spindle and setting the stack on his desk. He asks us to imagine that each CD is actually a truly 2-dimensional object, ignoring the tiny thickness of the plastic. He tells us that the process works just the same with truly 2-dimensional objects as it will with these CDs. How could we figure out the amount of stuff required to fill up this shape, he asks. Assuming it was truly 2-dimensional, we wouldn't be talking about volume anymore; it'd be area, right? He asks for the area of a circle, and we give it to him. We know this; it's easy stuff we've known for months now.<br /><br /><span class="citation">A = πr<sup>2</sup></span><br /><br />So that's how much space this single, 2-dimensional CD takes up, right? He picks up another CD and places it against the first one, flat sides together. How much space would 2 of them take up? He separates them again, holding them side by side. It'd just be twice the amount of space the first one took up, right? 2 circles' worth of area.<br /><br />He writes on the board:<span class="citation"> 2πr<sup>2</sup></span><br /><br />So how much space would 5 of them take up?<span class="citation"> 5πr<sup>2</sup></span><br /><br />And how much space would a stack of them that was h CD's high take up?<span class="citation"> hπr<sup>2</sup></span><br /><br />Mr. Lazur then circles that last line and turns to us. “This is the formula for volume of a cylinder. It's just the area of the flat face, multiplied by the height of the stack of those faces. <span class="citation">πr<sup>2</sup>h</span>.”<br /><br />When I started writing this blog post I wasn't thinking about the formula <span class="citation">πr<sup>2</sup>h</span> – I was thinking about that stack of cylinders. The formula followed organically from thinking about the concept. And that's the key – you can derive an algorithm easily from a concept, but if you never teach the concept all the algorithms in the world are just meaningless memorization.</div>Tue, 22 Apr 2014 07:48:33 -05002014-04-22T07:48:33-05:00267686http://www.wyzant.com/resources/blogs/267686/the_power_of_willpower_five_tips_to_strengthen_your_disciplineHuzefa K.http://www.wyzant.com/resources/users/view/84571890The Power Of Willpower: Five Tips To Strengthen Your Discipline<div>Willpower is unique to humanity. It is the keystone characteristic that is directly responsible for our technological advancement over the last several hundred thousand years. Willpower can be defined as the capacity to restrain our impulses and resist temptation in order to maximize our long-term success. It is the expulsion of energy to fight off innate survival based urges to exponentially increase future advantages and benefits. It is the driving force behind all civilizations, and it is what prods humankind forward to learn and grow.<br /><br />When we turn down a bite of cheesecake, step away from a mind numbing reality sitcom, or push off a nap to get some work done, the credit goes to willpower. It is this ghost like aura of control and discipline that we rely on to extend our existence and maximize our accomplishments. When we watch highly successful individuals exercise routinely, read voraciously, and work tirelessly, we are impressed with their ability to resist instant gratification. Most of us struggle to hold ourselves back from daily pleasures so as to work on self-improvement. But how do the chosen few make it happen? Are they the lucky recipients of steadfast genes, predisposed to adeptly control their yearnings better than the rest of us?<br /><br />To some extent, yes. Certain individuals are superiorly calibrated to fight off fleeting desires in the short term. But, what’s far more important is the revelation by psychologists that willpower is akin to a muscle. Regardless of how weak one’s innate level of willpower is, it can be trained and strengthened to rival the willpower of those super ambitious and successful folks we all admire. According to Roy Baumeister, an eminent social psychologist and famed expert on the subject, willpower can be bolstered with great success. This is an extraordinarily important discovery since willpower, in Baumeister’s opinion, is “the key to success and a happy life.”<br /><br />For many students who struggle with mathematics, having a sturdy level of willpower is the difference between finishing an assignment and turning in a half-hearted problem set. It is the difference between spending an extra hour and a half preparing for an exam, or merely skimming a chapter review the night before. In sum, it is the difference between excellence and mediocrity.<br /><br />For students who truly enjoy mathematics, there is no war to wage. Math is fun, and homework will be done thoroughly and completely as a means of gratification. But for those students who have to fight urges to play video games, watch movies, skateboard, or read novels while trudging through their math homework, willpower is what will save the day.<br /><br />So how can willpower be developed? How can students engineer a perfect level of self-control and discipline? Just like actual muscle fibers, willpower must be exercised in the right away. Overexertion can be exhausting and counterproductive, whereas just the right amount of use can (1) optimize productivity and (2) augment one’s willpower capacity. For folks who are interested in bolstering their willpower muscle, here are five quick tips:<br /><br />1. <strong>Stay fueled up with healthy meals</strong> – using willpower has been shown to deplete levels of glucose in the brain. Since it can often require great effort to stave off temptation, it is recommended that students maintain a healthy and regular diet replete with nutrients. Healthy meals will give the willpower muscle the fuel it needs to operate at its highest levels.<br /><br />2. <strong>Maintain a positive attitude</strong> – being happy and positive makes individuals far more able to employ their willpower. When students feel down or depressed, a common reaction is to dive into things that provide instant gratification. This could be an unhealthy meal, a lengthy break involving television or video games, or a long nap. Feeling happy and positive makes it much easier to stay on task.<br /><br />3. <strong>Partake in a healthy number of extracurricular activities</strong> – the reason why participating in many activities is beneficial is because it exercises the willpower muscle. When students have obligations and commitments that cannot be avoided, it trains them to push off their fleeting desires to focus on something in particular. Enrolling children in piano lessons and karate isn’t to make them professional musicians or seasoned MMA fighters. The most important aspect of extracurricular involvement is the development of strong willpower, something that will be applicable and useful during school and beyond.<br /><br />4. <strong>Apply willpower sparingly and gradually</strong> – trying to hold back on too many things can spell disaster. If someone is attempting to stick to homework, avoid fatty foods, stop watching TV, and exercise regularly all at once, he will find it highly challenging. Why? Because his willpower will be depleted very quickly. This means that the end result will likely be failure on all fronts. Instead of overcommitting, students should pick something in particular to focus on. Once someone grows accustomed to a single task, that action will have morphed into a habit requiring very little effort to maintain. At this point, one can then deploy his willpower elsewhere.<br /><br />5. <strong>Offer rewards each time a task is accomplished</strong> – every time a student successfully fights off temptation, he deserves a reward. Not only will this reinforce positive behavior, but it will help to rest and replenish the willpower muscle.</div>Tue, 01 Apr 2014 14:15:47 -05002014-04-01T14:15:47-05:00263429http://www.wyzant.com/resources/blogs/263429/different_centers_in_a_triangleRachel W.http://www.wyzant.com/resources/users/view/77106870Different Centers in a Triangle<div>This week in geometry one of my students is learning about the different "centers" in a triangle (orthocenter, circumcenter, incenter, etc), as well as the midsegments theorem and triangle inequalities. </div>
<div>To help him visualize why all of these things are true, I had him cut out an acute triangle, an obtuse triangle, and a right triangle and use these to illustrate the concepts. </div>
<div>For triangle inequalities, we worked with different lengths of string to see why some combination of leg lengths and some do not.</div>
<div>These are both quick, easy ways that help students understand beyond the words and definitions what we are talking about!</div>Thu, 27 Feb 2014 15:42:04 -06002014-02-27T15:42:04-06:00262189http://www.wyzant.com/resources/blogs/262189/parallel_skew_and_perpendicular_linesRachel W.http://www.wyzant.com/resources/users/view/77106870Parallel, Skew, and Perpendicular Lines<div>I've found that most students have little to no difficulty understanding the difference between parallel and perpendicular lines when only one plane is involved. Either they never touch, or they intersect at a 90 degree angle, or they just plain intersect. This concept is relatively easy to visualize because it is completely 2 dimensional. </div>
<div>Where the difficulty lies, is visualizing these same types of lines when different planes are involved, since it is 3d. To help, I utilize flash cards, or small pieces of paper. Have students draw a series of lines on each flash cards, making sure there is at least a set of parallel lines, perpendicular lines, and intersecting lines on each, and give each line a name. Then move the flashcards in different ways, either stacking them or making parallel planes, and quiz them about the new relationships between the lines. </div>Tue, 18 Feb 2014 14:46:37 -06002014-02-18T14:46:37-06:00255740http://www.wyzant.com/resources/blogs/255740/high_school_geometry_for_some_why_is_this_more_challenging_than_algebra_1Bruce H.http://www.wyzant.com/resources/users/view/76725150High School Geometry: For Some, Why Is This More Challenging Than Algebra 1? <div>Several of my current Geometry students have commented on this very distinction. This has prompted me to offer a few possible reasons.</div>
<div> </div>
<div>First, Geometry requires a heavy reliance on explanations and justifications (particularly of the formal two-column proof variety) that involve stepwise, deductive reasoning. For many, this is their first exposure to this type of thought process, basically absent in Algebra 1.</div>
<div> </div>
<div>Second, a large part of Geometry involves 2-d and 3-d visualization abilities and the differences in appearance between shapes even when they are not positioned upright. Still further, for a number of students, distinguishing the characteristic properties amongst the different shapes becomes a new challenge.</div>
<div> </div>
<div>Third, in many cases Geometry entails the ability to form conjectures about observed properties of shapes, lines, line segments and angles even before the facts have been clearly established and stated. This level of abstract thinking is rarely encountered in Algebra 1.</div>
<div> </div>
<div>Implications based upon the aforementioned are significant. For those following a Science/Technology/Engineering/Math (STEM) course track and headed into Precalculus, Calculus, General Chemistry and Organic Chemistry, the ability to visualize shapes and chemical structures in 2-d and 3-d space is extremely important.</div>
<div> </div>
<div>In order to better prepare students for dealing with Geometry's rigors, many educators have believed that certain math instructional practices even going back to the elementary school level will have to be improved within the classroom. In the meantime, all students can still take steps to increase their current mental capacities and reasoning abilities by following a brain-enhancing diet (such as my "Maximum Mental Health Diet" recommendations offered when requested at any tutoring session). Among other things, this includes daily consumption of long-chain Omega-3 essential fatty acids through fish or supplements and an abundance of colorful, antioxidant-rich plant foods.</div>Tue, 04 Feb 2014 14:04:24 -06002014-02-04T14:04:24-06:00249113http://www.wyzant.com/resources/blogs/249113/favorite_math_resources_k_12_gedJessica S.http://www.wyzant.com/resources/users/view/84101090Favorite Math Resources, K-12 + GED<div>Here are some of my favorite Math resources. Check back again soon, this list is always growing! I also recommend school textbooks, your local library, and used bookstores. <br /><br />As a note, college-level math textbooks are often helpful for high school math students. Why is that? Isn't that a little counter-intuitive? Yes, it would appear that way! However, many college-level math textbooks are written with the idea that many college students may not have taken a math class in a year or more, so they are written with more detailed explanations. This can be particularly helpful for high school students taking Algebra, Geometry, and Trig. I have a collection of college-level math books that I purchased at a local used bookstore. The most expensive used math book I own cost $26 used. Books that focus on standardized test prep (such as the SAT, AP, or GED prep) can be helpful for all core subjects, as they summarize key ideas more succinctly than 'normal' textbooks. These are GREAT for test review and studying as a supplemental text, whethre you're studying for that standardized test or not.</div>
<div> </div>
<div><br />(K-12) <a href="http://www.kahnacademy.org">KahnAcademy.org</a> – All math subjects, with video tutorials and practice problems (plus answers!)</div>
<div><br />(K-Algebra\Geom.) <a href="http://www.mathplayground.com">MathPlayground.com</a> – Fun math tutorials and games to help reinforce what you’ve learned!</div>
<div><br />(K-Geometry) <a href="http://www.mathisfun.com">Mathisfun.com</a> – Number Value (K/Gr. 1) to Geometry</div>
<div><br />(K-Geometry) <a href="http://www.aaamath.com">Aaamath.com</a> – Explanations, practice problems (with answers), and more!</div>
<div><br />(Gr. 2+) iPhone app – Math Help – Division, multiplication, division with remainders</div>
<div><br />(Algebra) <a href="http://www.purplemath.com">Purple Math.com</a> – All things Algebra, with detailed explanations and (a few) practice problems.</div>
<div><br />(Gr. 9-12) iPhone/Droid app- GED Math Lite – All high school math topics, with quizzes and answer explanations.<br /><br /></div>
<div>(Gr 3-12) <a href="http://www.convert-me.com/en/">Convert-Me.com</a> - Not sure how to convert kilometers to meters? Gallons to quarts? Give this site a try!</div>Tue, 31 Dec 2013 14:09:48 -06002013-12-31T14:09:48-06:00247613http://www.wyzant.com/resources/blogs/247613/intersecting_circlesJohn M.http://www.wyzant.com/resources/users/view/82890510Intersecting Circles<div>This blog concerns how to determine the intersection between two circles in the plane algebraically. It is a problem that can crop up in a variety of situations, from gaming to tools for computer aided design to astronomy. </div>
<div> </div>
<div>This problem is interesting because is it a conceptually simple problem whose algebraic formulation is nonetheless apparently complex: a system of non-linear equations that are quadratic in both variables. However, by doing some geometric analysis of the problem, and applying tools from vector geometry, we are lead to a specific mathematical transformation of the problem that radically simplifies it. The key idea turns out to be a specific <em>change of basis.</em></div>
<div> </div>
<div>My exposition of this uses a few diagrams (which are not supported by the blog editor) and a lot of mathematical expressions (which are clumsy to create in the blog editor), so I put it in an Adobe Portable Document Format (PDF) document that you can access using the following link:</div>
<div> </div>
<div><a href="http://media.wix.com/ugd/418cf4_379b13daaaa84184bcd4cdd6239164a8.pdf">Intersecting Circles.pdf</a></div>
<div> </div>
<div>Clicking the link will download the file to view on your computer. To view the downloaded file you must have the Adobe Acrobat reader (or an equivalent program for viewing PDF files). Since PDF is a very widely used format for sharing documents, your computer likely already has the reader (you may even have a plug-in for your browser that lets it display PDF documents). </div>
<div> </div>Sun, 08 Dec 2013 15:14:01 -06002013-12-08T15:14:01-06:00246891http://www.wyzant.com/resources/blogs/246891/proof_that_3_non_collinear_points_determine_a_unique_circleStanton D.http://www.wyzant.com/resources/users/view/79613280Proof that 3 non-collinear points determine a unique circle<div>Proof of the Assertion that Any Three Non-Collinear Points Determine Exactly One Circle<br />This is an interesting problem in geometry, for a couple of reasons. First, you can apply some earlier, basic geometry principles; and secondly, you can choose two different strategies for solving the problem.<br /><br />The basic geometry underlying: any three non-collinear points determine a plane, somewhere in 3D space. Once that has been done, imagine that the plane has been rotated into the x-y plane, which will make the problem much easier to solve!<br /><br />The two strategies for solution are: (Proof A) actually solve to find the circle. This is equivalent to finding the center of the circle (finding the equation of the circle is simple from there). But, you actually have to do some math to get this! If, while doing this, there is no possibility to obtain other values for the coordinates of the center of the circle, you have proved the assertion as well as obtained a method (and perhaps the formula) for doing so.<br />The other possibility (Proof B) is to assume that one center of such a circle exists, then show that another center of such a circle (i.e. the center of a different circle) could not exist for some reason. This is mathematically fun to do!<br /><br />So, (Proof B) runs as follows: let the three original non-collinear points be called A, B, and C. Let the proposed centers of the two different circles be designated as R1 and R2 (note that these are not radii, but actual points). Then, from the basic properties of a circle, any radius has the same length. Expressing each of the radii AR1, BR1, and CR1 by the Pythagorean theorem, using x(A) to be the x-coordinate of point A, and y(A) to be the y-coordinate of point A, etc.:<br /><br />((x(A) – x(R1))^2 + ( (y(A) – y(R1))^2 = <br />((x(B) – x(R1))^2 + ( (y(B) – y(R1))^2 = <br />((x(C) – x(R1))^2 + ( (y(C) – y(R1))^2 <br /><br />and also:<br /><br />((x(A) – x(R2))^2 + ( (y(A) – y(R2))^2 = <br />((x(B) – x(R2))^2 + ( (y(B) – y(R2))^2 = <br />((x(C) – x(R2))^2 + ( (y(C) – y(R2))^2 <br /><br />Now, these two sets of equalities look like three equations in 2 unknowns – i.e. overspecified! But this is not the case, because the equations are not equal to a constant, but to each other. So, each pair of two equations has infinitely many solutions – the line bisecting the segment joining the respective specified points – and the third equality serves to uniquely “fix” one point as the common intersection of the three such lines generated from each pair of points in the set {A, B, C} (there, now, you have the essential idea for calculating the center. But, leave that alone for now.).<br /><br />Now, these look like two formidable sets of equations to start drawing conclusions from! But, they are related, because the two proposed circle center points have an offset from each other, in possibly both their x and y coordinates. Therefore,<br />Let x(D) = x(R1) – x(R2) and<br />y(D) = y(R1) – y(R2)<br /><br />Then substitute these terms into the *second* set of three equations, to express them ONLY in terms of R1’s coordinates:<br /><br />(x(A) – x(R1) + x(D))^2 + ((y(A) – y(R1) + y(D))^2 =<br />(x(B) – x(R1) + x(D))^2 + ((y(B) – y(R1) + y(D))^2 =<br />(x(C) – x(R1) + x(D))^2 + ((y(C) – y(R1) + y(D))^2 <br /><br />Now, these equations are somewhat similar to the first set. Capitalize on the similarity, by considering each *(P) - *(R1) expression as a single term, but otherwise expanding the squares as binomial expansions (you’ll see why, in a minute):<br /><br />(x(A) – x(R1))^2 + 2x(D)(x(A) – x(R1)) + (x(D))^2 + (y(A) – y(R1))^2 + 2y(D)(y(A) – y(R1)) + (y(D))^2 =<br />(x(B) – x(R1))^2 + 2x(D)(x(B) – x(R1)) + (x(D))^2 + (y(B) – y(R1))^2 + 2y(D)(y(B) – y(R1)) + (y(D))^2 =<br />(x(C) – x(R1))^2 + 2x(D)(x(C) – x(R1)) + (x(D))^2 + (y(C) – y(R1))^2 + 2y(D)(y(C) – y(R1)) + (y(D))^2 <br /><br /><br />Now, compare the three equations you just derived. In common, they each have a term:<br /> ((x(P) – x(R1))^2 + ((y(P) – y(R1))^2 (where P is one of the three original points)<br />and also a term:<br />(x(D))^2 + ((y(D))^2<br /><br />We can drop (subtract out) both of these types of terms from the equalities, because, by our original conditions, the first set of terms just restates that the three points are all equidistant from point P1 (by the Pythagorean theorem), and the second set of terms are all a common constant (the square of the distance between the two proposed circle centers).<br /><br />Then, we are left with only the middle terms of the binomial expansions:<br /><br />2x(D)(x(A) – x(R1)) + 2y(D)(y(A) – y(R1)) =<br />2x(D)(x(B) – x(R1)) + 2y(D)(y(B) – y(R1)) =<br />2x(D)(x(C) – x(R1)) + 2y(D)(y(C) – y(R1))<br /><br />These are really simplifying nicely, but we’re not done yet; take out the remaining common terms, and divide through by 2, to obtain:<br /><br />x(D)x(A) + y(D)y(A) = x(D)x(B) + y(D)y(B) = x(D)x(C) + y(D)y(C)<br /><br />Do you realize what you have here? It’s a set of equations, where x(D) and y(D) are the constants “a” and “b” (assuming x(D) and y(D) are non-zero values), all saying ax + by = some common value -- this is the equation for a single line. More importantly, since coordinates for points A, B, and C all satisfy the equation, all three points must lie on that line! But, we stated above, A, B, and C are *non-*collinear! Therefore, the only way that the equation can be true is if x(D) and y(D) are both zero, meaning that the “second” circle center *must* be the same as the “first” circle center, -- in other words, there can only be one, unique, circle that contains the 3 non-collinear points A, B, and C.<br /><br />It’s perhaps instructive to consider what would be the meaning of three *collinear* points on a circle. In this case, the circle would have to be of infinite radius, the center to one side or the other of the collinearity line – and in this case, the center could be offset “laterally” (i.e. in a direction parallel to the collinearity line) to any amount one desires, just as the equation above implies.</div>Sun, 01 Dec 2013 17:07:18 -06002013-12-01T17:07:18-06:00246616http://www.wyzant.com/resources/blogs/246616/how_to_find_a_distance_between_two_parallel_lines_2Kirill Z.http://www.wyzant.com/resources/users/view/81465830How to find a distance between two parallel lines?-2<div>This is another way to find a distance between two parallel lines. This derivation was suggested to me by Andre and I highly recommend him and his answers to any student, who wants to learn math ans physics. This derivation requires the knowledge of trigonometry and some simple trigonometric identities, so this may be suitable for more advanced students.</div>
<div> </div>
<div>Once again, we have two lines.</div>
<div> </div>
<div>y=mx+b<sub>1</sub> (1)--equation for the first line.<br /><br /></div>
<div>y=mx+b<sub>2</sub> (2)--equation for the second line.</div>
<div> </div>
<div>Now recall that the slope of the line is the tangent of an angle this line forms with the x-axis. Indeed, m=(y<sub>2</sub>-y<sub>1</sub>)/(x<sub>2</sub>-x<sub>1</sub>), where x<sub>1</sub>, x<sub>2</sub>, y<sub>1</sub>, y<sub>2 </sub>are the x- and y-coordinates of any two distinct points on the line. If one draws the picture, it will be immediately obvious that m is the tangent of the angle between the line and the x-axis.</div>
<div> </div>
<div>The difference b<sub>2</sub>-b<sub>1</sub> gives the relative displacement along the y-axis of two lines. Since b<sub>2</sub> is the y-intercept of the second line and b<sub>1</sub> is the y-intercept of the first line, the vertical displacement of one line with respect to another is given by |b<sub>2</sub>-b<sub>1</sub>|. The perpendicular line segment between two lines and this displacement form the same angle between each other as the lines and the x-axis have between each other. So its tangent is m. Now recall the identity:</div>
<div> </div>
<div>tan<sup>2</sup>(x)+1=1/cos<sup>2</sup>(x) (3)</div>
<div> </div>
<div>From (3) it follows that cos(x)=1/√(1+tan<sup>2</sup>(x))</div>
<div> </div>
<div>In our case the cosine of the angle α between the vertical line, which length is |b<sub>2</sub>-b<sub>1</sub>|, and the perpendicular line, which length d we are trying to determine, is 1/√(1+m<sup>2</sup>). The distance d is them given by:</div>
<div> </div>
<h2><strong>d=|b<sub>2</sub>-b<sub>1</sub>|cos(α)=|b<sub>2</sub>-b<sub>1</sub>|/√(1+m<sup>2</sup>) (4)</strong></h2>
<div>This once again proves the formula derived earlier using coordinate approach.</div>Wed, 27 Nov 2013 21:36:10 -06002013-11-27T21:36:10-06:00246582http://www.wyzant.com/resources/blogs/246582/how_to_find_a_distance_between_two_parallel_linesKirill Z.http://www.wyzant.com/resources/users/view/81465830How to find a distance between two parallel lines?<div>Suppose, one have two parallel lines given by the equations:</div>
<div> </div>
<div><em>y=mx+b<sub>1</sub></em> and <em>y=mx+b<sub>2</sub></em>. Remember, if the lines are parallel, their slopes must be the same, so <em>m</em> is the same for two lines, hence no subscript for <em>m</em>. How would one approach the problem of finding the distance between those lines?</div>
<div> </div>
<div>First, if one draws a picture, he or she shall immediately realize that if a point is A chosen on one of the lines, with coordinates (x<sub>1</sub>, y<sub>1</sub>), and a perpendicular line is drawn from that point to the second line, the length of the segment of this new line between two parallel lines give us the sought distance. Let us denote the point of intersection of our perpendicular line with the second line as B(x<sub>2</sub>,y<sub>2</sub>).</div>
<div> </div>
<div>What do we know of point A and B?</div>
<div> </div>
<div>First, since A lies on the first parallel line, its coordinates must satisfy the equation for the first line, that is,</div>
<div> </div>
<div>y<sub>1</sub>=mx<sub>1</sub>+b<sub>1 </sub> (1)</div>
<div> </div>
<div>Same is true for B and the second line. So we can write:</div>
<div> </div>
<div>y<sub>2</sub>=mx<sub>2</sub>+b<sub>2 </sub> (2)</div>
<div> </div>
<div>We also know that line AB is perpendicular to both parallel lines. It is a known fact that the slope of the line, which is perpendicular to the given line, is inverse reciprocal of the slope of that line. So the line AB MUST have the slope equal to -1/m. The slope of the line connecting two points, A and B, is given by:</div>
<div> </div>
<div>slope=(y<sub>2</sub>-y<sub>1</sub>)/(x<sub>2</sub>-x<sub>1</sub>). In this case it is -1/m. So we obtain:</div>
<div> </div>
<div>(y<sub>2</sub>-y<sub>1</sub>)/(x<sub>2</sub>-x<sub>1</sub>)=-1/m (3)</div>
<div> </div>
<div>The distance between two points, A and B, is given by:</div>
<div> </div>
<div>d=√[(y<sub>2</sub>-y<sub>1</sub>)<sup>2</sup>+(x<sub>2</sub>-x<sub>1</sub>)<sup>2</sup>] (4)</div>
<div> </div>
<div>From equations (1) and (2) we obtain:</div>
<div> </div>
<div>(y2-y1)=m(x<sub>2</sub>-x<sub>1</sub>)+(b<sub>2</sub>-b<sub>1</sub>) (5)</div>
<div> </div>
<div>Plug in (y<sub>2</sub>-y<sub>1</sub>) from eq. (5) into eq. (3) and solve for (x<sub>2</sub>-x<sub>1</sub>). We obtain upon solving:</div>
<div> </div>
<div>(x<sub>2</sub>-x<sub>1</sub>)=m(b<sub>2</sub>-b<sub>1</sub>)/(m<sup>2</sup>+1); (6)</div>
<div> </div>
<div> </div>
<div>Substituting (6) into (3) immediately yields the following:</div>
<div> </div>
<div>(y<sub>2</sub>-y<sub>1</sub>)=-(b<sub>2</sub>-b<sub>1</sub>)/(m<sup>2</sup>+1) (7)</div>
<div> </div>
<div>Now final step is to plug eq. (6) and (7) into eq. (4). After some transformation we obtain:</div>
<div> </div>
<div>d=√[(b<sub>2</sub>-b<sub>1</sub>)<sup>2</sup>/(m<sup>2</sup>+1)<sup>2</sup>+m<sup>2</sup>(b<sub>2</sub>-b<sub>1</sub>)<sup>2</sup>/(m<sup>2</sup>+1)<sup>2</sup>]=[|b<sub>2</sub>-b<sub>1</sub>|/(m<sup>2</sup>+1)]√(m<sup>2</sup>+1)</div>
<div> </div>
<div>Final answer is as follows:</div>
<div> </div>
<h2><strong>d=|b<sub>2</sub>-b<sub>1</sub>|/√(m<sup>2</sup>+1) (8)</strong></h2>Wed, 27 Nov 2013 17:24:47 -06002013-11-27T17:24:47-06:00245878http://www.wyzant.com/resources/blogs/245878/christmas_presentAndrew L.http://www.wyzant.com/resources/users/view/75937410Christmas Present!!<div>Hi All!</div>
<div> </div>
<div>In the spirit of giving, starting on 11/29/2013, I will be offering a few brainteasers/ trivia questions where the first 3 people to email me the correct answer will receive a free, one hour, tutoring session in any subject that I offer tutoring for (via the online platform)! That's right free! Get your thinking hats on everyone!</div>
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<h1><span class="greenText"><strong>Merry <span class="orange">Christmas</span>!! </strong></span></h1>
<div><span class="greenText"><strong><a href="http://www.wyzant.com/Tutors/AndrewLane">Andrew L. Profile</a></strong></span></div>
<div> </div>Thu, 21 Nov 2013 01:35:13 -06002013-11-21T01:35:13-06:00243335http://www.wyzant.com/resources/blogs/243335/math_equations_formulas_and_vocabularyKayleigh T.http://www.wyzant.com/resources/users/view/84291860Math Equations, Formulas and Vocabulary<div>
<ul>
<li>Area, Volume and Circumference equations:</li>
<li>Area of a Square</li>
</ul>
</div>
<div>A=S<sup>2</sup></div>
<div>
<ul>
<li>Area of a Triangle</li>
</ul>
</div>
<div>A=1/2bh</div>
<div>
<ul>
<li>Area of a Rectangle</li>
</ul>
</div>
<div>A=LW</div>
<div>
<ul>
<li>Right Triangle/Pythagorean Theorem</li>
</ul>
</div>
<div>a<sup>2</sup>+b<sup>2</sup>=c<sup>2</sup></div>
<div>
<ul>
<li>Area of Parallelogram</li>
</ul>
</div>
<div>A=bh</div>
<div>
<ul>
<li>Area of a Trapezoid</li>
</ul>
</div>
<div>A=1/2h(a+b)</div>
<div>
<ul>
<li>Area of a Circle</li>
</ul>
</div>
<div>A=πr<sup>2</sup></div>
<div>
<ul>
<li>Circumference of a Circle</li>
</ul>
</div>
<div>c=πd or c=2πr</div>
<div>
<ul>
<li>Volume of a Sphere</li>
</ul>
</div>
<div>V=4/3πr<sup>3</sup></div>
<div>
<ul>
<li>Surface Area of a Sphere</li>
</ul>
</div>
<div>SA=4πr<sup>2</sup></div>
<div>
<ul>
<li>Volume of a Cube</li>
</ul>
</div>
<div>V=s<sup>3</sup></div>
<div>
<ul>
<li>Volume of a Rectangular Solid</li>
</ul>
</div>
<div>V=lwh</div>
<div>
<ul>
<li>Slope of a line Equations</li>
<li>Slope-intercept form</li>
</ul>
</div>
<div>y=mx+b</div>
<div>m is the slope</div>
<div>b is the y-intercept</div>
<div>y is a y coordinate on the graph (that coincides with the line)</div>
<div>x is an x coordinate on the graph (that coincides with the line)</div>
<div>
<ul>
<li>Horizontal line</li>
</ul>
y=b</div>
<div>
<ul>
<li>Vertical line</li>
</ul>
x=a</div>
<div>
<ul>
<li>Finding slope of line containing points (x<sub>1</sub>,y<sub>1</sub>) to (x<sub>2</sub>,y<sub>2</sub>)</li>
</ul>
m=<span style="text-decoration: underline;">y<sub>2</sub>-y<sub>1</sub></span></div>
<div> x<sub>2</sub>-x<sub>1</sub></div>
<div>
<ul>
<li>Distance from one point to another</li>
</ul>
(x<sub>1</sub>,y<sub>1</sub>) to (x<sub>2</sub>,y<sub>2</sub>)<span style="font-size: 9px;"><br /></span></div>
<div>d=√(x<sub>2</sub>-x<sub>1</sub>)<sup>2</sup>+(y<sub>2</sub>-y<sub>1</sub>)<sup>2</sup></div>
<div>
<ul>
<li>Midpoint of a Line Segment</li>
</ul>
( <sup><span style="text-decoration: underline;">x1+x2 </span> , <span style="text-decoration: underline;">y1+y2</span> </sup>)</div>
<div> <sup>2 2</sup></div>
<div>
<ul>
<li>Quadratic Formula</li>
</ul>
x=<span style="text-decoration: underline;">-b±√b<sup>2</sup>-4ac</span></div>
<div> 2a</div>
<div>
<ul>
<li>Math Vocabulary</li>
</ul>
<ol>
<li>Equilateral Triangle- Three sides of equal length and 3 angles 60° each</li>
<li>Scalene Triangle- Three unequal sides and 3 unequal angles</li>
<li>Isosceles Triangle- Two sides of equal length and two base angles are equal</li>
<li>Commutative Property- a+b=b+a; ab=ba</li>
<li>Associative Property- a=(b+c)=(a+b)+c; a(b+c)=ab+ac</li>
<li>Distributive Property- a9b+c)=ab+ac</li>
<li>Additive Identity- a+0=0+a=a</li>
<li>Multiplicative Identity- a•1=1•a=a</li>
<li>Additive Inverse- -a+a=a+(-a)=0</li>
<li>Multiplicative Inverse- a•1/a=1, a≠0</li>
</ol></div>Wed, 06 Nov 2013 00:09:02 -06002013-11-06T00:09:02-06:00237734http://www.wyzant.com/resources/blogs/237734/maybe_i_m_just_no_good_at_mathJenna G.http://www.wyzant.com/resources/users/view/78162900Maybe I'm Just No Good At Math.<div>I've heard this sentiment over and over--sometimes from students, and sometimes, I'll admit, in my own head.<br /><br />Last night, I was working on my own math homework, and there was one problem I just couldn't get my head around. I read the book, looked back at my class notes, and even sat down with a tutor for a while, and still, when I tried a new problem of the same type on my own, it just didn't work!<br /><br />"Maybe I'm not as good at math as I thought," I told myself. "Am I REALLY smart enough for bioengineering?"<br /><br />It was hard, but I told myself "YES!" And I kept working. I laid the assigned problems aside and started doing other problems of the same type from the book. I checked my work every time. Each problem took at least ten minutes to solve, and the first three were ALL wrong! I kept going. I got one right, and it made sense! I did another, and it was half right, but there was still a problem. I did another, and it was right!<br /><br />Eventually I had a page of right answers, and I went back to the same homework problem I'd tried four times and suddenly it worked! I don't know why this one method was so difficult for me to understand, but at this point, it doesn't matter. My homework is done, and I'll be able to do it right on the test.<br /><br />This is what "good at math" means. We all want to be good without trying, but math is a vast set of skills. Sometimes we're lucky, and we're able to pick up some or even most things without really grappling with them. Sometimes, we have to trudge through a skill an inch at a time. Many students think of math classes as something you just barely survive, where there will come a day when one just gets you, and you're done. When you're trying to survive math, a tough problem like this looks like the end. This is it. You've understood everything you will ever understand, and everything harder is just Too Hard. The student who is Good At Math gets those hard problems too, but she sees them as places she has to put in some hard work. Even if it takes you twice as long as everyone else to get this concept, the next one might be a breeze. And even if EVERY concept is a tough uphill climb, at the top of the hill, you've beaten it.<br /><br /></div>Thu, 19 Sep 2013 13:56:24 -05002013-09-19T13:56:24-05:00