Split (1)

train · ~16.8M rows (showing the first 4.97M)

text stringlengths 22 1.01M
## [Soln] 2013 Singapore MO (Junior Rd 2) Problem 1 2013 SMO (Junior-Rd2) 1. Let $a < b < c < d < e$ be real numbers. Among the 10 sums of the pairs of these numbers, the least three are 32, 36 and 37 while the largest two are 48 and 51. Find all possible values of $e$. The key objects are the pairwise sums of the 5 variables, and their relative sizes will be of importance. For a start, we could list out all 10 pairwise sums in a systematic way: \begin{aligned} &a+b && && &&\\ &a+c &&b+c && && \\ &a+d &&b+d &&c+d && \\ &a+e &&b+e &&c+e &&d+e \end{aligned} Because we have listed these sums out systematically, we can make some comments about their relative sizes. In the table above, for any 2 sums in the same column, the one above is smaller than the one below. Also, for any 2 sums in the same row, the one that is further left is smaller than the one that is further right. Hence, it is quite clear that $a+b$ is the smallest sum, $a+c$ is the 2nd smallest sum, $d+e$ is the largest sum while $c+e$ is the 2nd largest sum. (What is not obvious is what the 3rd smallest sum is: it could be either $a+d$ or $b+c$.) We can then write the information given in the question in equations: \begin{aligned} a + b &= 32 &&-(1) \\ a + c &= 36 &&-(2) \\ c+e &= 48 &&-(3) \\ d+e &= 51 &&-(4) \end{aligned} Some of the equations share a variable (e.g. equations 1 & 2), so by subtracting one from the other, we can tell how far apart some of the variables are (e.g. $c$ must be 4 more than $b$). By looking through all possible pairs of the equations, we get the following relationships: Letting the distance between $a$ and $b$ be denoted by $x$, and the distance between $d$ and $e$ be denoted by $y$, we can rewrite the 5 variables as $a, a+x, a+x+4, a+x+7, a+12.$ (It turns out that the variable $y$ wasn’t really needed at all!) Note that there are some restrictions on $x$ because we must have $a. In particular, $0 < x < 5$. (Why?) Let’s rewrite equations 1-4 with these variables to see if we can gain any further insight: \begin{aligned} &(1): &&a + (a + x) = 32, 2a + x = 32 \\ &(2): &&a + (a+x+4) = 36, 2a + x = 32 \\ &(3): &&(a+x+4) + (a+12) = 48, 2a + x = 32 \\ &(4): &&(a+x+7) + (a+12) = 51, 2a + x = 32 \end{aligned} All 4 equations ended up being the same. Maybe at this point it’ll be good to check if we can determine which is the 3rd smallest pairwise sum: \begin{aligned} a + d &= a + (a + x + 7) = 2a + x + 7, \\ b + c &= (a+x) + (a+x+4) = 2a + 2x + 4. \end{aligned} Which is bigger really depends on whether 7 or $x+4$ is bigger, i.e. whether 3 or $x$ is bigger. Case 1: $\boldsymbol{x>3}$ In this case, the 3rd smallest sum is $a + d$. Hence, $\begin{cases} 2a + x + 7 &= 37 \\ 2a + x &= 32 \end{cases}$ A contradiction! Hence, $x>3$ is not valid. Case 2: $\boldsymbol{x \leq 3}$ In this case, the 3rd smallest sum is $b + c$. Hence, $\begin{cases} 2a + 2x + 4 &= 37 \\ 2a + x &= 32 \end{cases}$ This means that $x = 1$, and $a = 15.5$. From here, we can derive the 5 numbers 15.5, 16.5, 20.5, 23.5, 27.5 It is easy to check that these 5 numbers satisfy the given conditions in the question. Hence, there is only 1 possible value for $e$, which is 27.5.
# Math Expressions Grade 3 Student Activity Book Unit 4 Lesson 1 Answer Key Make Place Value Drawings This handy Math Expressions Grade 3 Student Activity Book Pdf Answer Key Unit 4 Lesson 1 Make Place Value Drawings provides detailed solutions for the textbook questions. ## Math Expressions Grade 3 Student Activity Book Unit 4 Lesson 1 Make Place Value Drawings Answer Key Practice Place Value Drawings to 999 Write the number for each dot drawing. Question 1. 87 Explanation: The number for each doting given is 87. Question 2. 116 Explanation: The number for each doting given is 116. Write the number for each place value drawing. Question 3. 277 Explanation: Given, Hundred – 2 Tens – 7 Ones – 7 Therefore the number is 277. Question 4. 361 Explanation: Given, Hundred – 3 Tens – 6 Ones – 1 Therefore the number is 361. Question 5. 623 Explanation: Given, Hundred – 6 Tens – 2 Ones – 3 Therefore the number is 623. Question 6. 449 Explanation: Given, Hundred – 4 Tens – 4 Ones – 9 Therefore the number is 449 Make a place value drawing for each number. Question 7. 86 Eighty-six Explanation: The given number is Eighty-six. Question 8. 587 Five hundred and eighty-seven. Explanation: The given number is Five hundred and Eighty-seven. Practice with the Thousand Model Write the number for each place value drawing. Question 9. 1000 + 200 + 30 + 8 = 1238. Explanation: The number for the place value drawing is 1238. Question 10. 1000 + 00 + 90 + 3 1093. Explanation: The number for the place value drawing is 1093. Make a place value drawing for each number. Question 11. 2,368 Two thousand three hundred and sixty-eight. Explanation: Question 12. 5,017 Five thousand and seventeen. Explanation: Write Numbers for Word Names Write the number for the words. Question 13. eighty-two _________________ 82 Explanation: The number for the word given is 80 + 2 = 82 Question 14. ninety-nine _________________ 99 Explanation: The number for the word given is 90 + 9 = 99 Question 15. four hundred sixty-seven _________________ 467 Explanation: The number for the word given is 400 + 60 + 7 = 467. Question 16. nine hundred six _________________ 906. Explanation: The number for the word given is 900 + 6 = 906. Question 17. one thousand, fifteen _________________
# How to solve cube roots We will explore How to solve cube roots can help students understand and learn algebra. We will also look at some example problems and how to approach them. ## How can we solve cube roots In this blog post, we will take a look at How to solve cube roots. Solving problems is something that's a part of being human. We all need to solve problems in our lives; whether they be problems at work, at home, or with our relationships. And when you're able to solve problems, it can make you feel good about yourself and can even help you achieve other goals. There are lots of different ways to solve problems. You can talk to someone about your problem, try to find a solution on your own, or do both. If you want to be really good at solving problems, it's important to learn how to listen and ask questions, as well as how to use your imagination and think outside the box. And when you know how to solve problems well, you'll be able to get more done in less time. Square roots are useful for solving equations that contain square roots. They can be used to cancel out the square root and simplify an equation. These equations can then be solved by manipulating the variables. A square root is when you take a number and multiply it by itself. For example, if you want to take the square root of 16, you would get 4 because it takes four to make a square. If you want to take the square root of -16, you would get 2 because it takes two to make a square. Square roots are especially useful in order to solve trigonometry problems because you can use them to cancel out the square roots and simplify equations into simpler equations using just a few variables. This makes solving trigonometry problems much easier. In order to take a square root of an expression, begin by dividing both sides of the equation by the highest power of the denominator (that is, if the denominator is raised to a power of two then you divide both sides by 2). Then identify which side is negative and will yield a positive value when squared. If this side is negative, then multiply it by whichever positive value is larger (the smaller value will cancel out due to their relationship as opposites); otherwise, subtract this side from both sides: math>sqrt{(-x)^{2}} - sqrt{x} ight There’s no shortage of math apps on the market. If you prefer to learn math in a more creative way, there are plenty of free math games and apps out there to help you learn your numbers. One of the most popular free math apps is Count & Learn, which teaches basic math skills and basic addition, subtraction, multiplication and division. This app is a great choice for children because it includes engaging activities such as coloring and taking pictures while learning. Another great free app is MathwayPro, which has several different versions available. With this app, you can take notes in either English or Spanish, view diagrams or animations that explain each step of the process, and record your results with an interactive graph. You can also use this app to track your progress over time. Elimination is the process of removing something from your body. It's a great way to reduce or eliminate unwanted food or chemicals, as well as to reduce stress. There are two types of elimination: physical and dietary. Physical elimination involves flushing out toxins through your urinary system by peeing or by using a toilet (diarrhea). Dietary elimination involves removing food that you know is causing issues. The two most common reasons people experience constipation are dehydration and poor diet choices. Dehydration occurs when you don't drink enough water or when you're not drinking enough water in order to keep your electrolyte levels up. When this happens, your body can't produce enough water in the colon, which can lead to constipation. Poor diet choices include eating processed foods that contain lots of preservatives, high fructose corn syrup, and trans fats. These foods will make it harder for your body to absorb nutrients and minerals, which can cause constipation. I was skeptical to try this app if it will really work or not. My surprise that it really does, I've googled some equations with solutions and it's accurately the same. Great app! Would recommend it for all students and people who are struggling to figure out math. Oh, and I think it's a great bonus to have a solution explained to you so you can understand it as well. Hands down best app for solving mathematical problems! ### Khloe Baker I'm going to say this for my entire class. This app saved all of us for 2 years X'D the best part about this is that you can scan the questions right from your paper and it can also work as a quick equation solver in case your normal calculator can't solve them. In summary, a life-saver. ### Lena Sanders Basic mathematics problem solving Math home work 24 solver with fractions Solving algebra calculator Fraction solver calculator Quadratic function equation solver
# Examples of how to solve the power of a fraction product Perhaps the most convenient thing to do, before approaching each of the cases that can serve as examples to the correct way in which any operation that raises to an exponent the operation of sustained product between two fractions must be resolved, is to briefly revise the very definition of this operation, in order to understand each one of these examples in its precise context. ## Power of a product of fractions However, perhaps it is best to remember first that Mathematics defines fractions as all mathematical expressions, used to represent non-integer or non-exact quantities. Likewise, this discipline warns that fractions will always be composed of two elements: numerator, which will occupy the upper part of the expression, indicating how many parts of the whole have been taken; and denominator, which will serve to indicate into how many parts this whole is divided. As for the operation of power of a product of fractions, basically it can be said that it is a procedure consisting in elevating to a certain exponent a multiplication operation, where the factors are powers. In this sense, this product operation must be multiplied by itself, as many times as the corresponding exponent indicates, in order to comply with the condition of abbreviated multiplication of the potentiation operation. ## Steps to solve a fraction product power Likewise, Mathematics indicates that this type of procedure must be solved according to a specific method, consisting of a group of steps, which will be completed in the following order: 1. In the first place, it is necessary to begin by knowing the elements and characteristics that compose each one of the fractions that comprise the operation of the product. 2. Likewise, the value of the exponent to which this operation is elevated will be taken into account. 3. Each fraction will then be elevated to the exponent to which the product operation has been elevated. 4. Once this has been done, each rational-based potentiation operation obtained will be resolved separately, which will be done by elevating each element to the pertinent exponent. 5. Once obtained the fractions, product of the respective powers of rational base, it will be necessary to proceed to multiply both fractions, remembering that the numerator of one must be multiplied by the denominator of the second one, while it will be made equal with the denominators of each fraction. 6. Finally, we will try to simplify the fraction obtained. Likewise, the mathematical discipline indicates that the correct way to solve this operation can be expressed in the following way: ## Examples of how to solve the power of a product of fractions However, perhaps the most effective way to study this type of operations is through the exposition of some examples that allow us to see in a practical way how each of the steps indicated by Mathematics are fulfilled. Here are some exercises on how to find the power of the product of fractions: ## Example 1 Resolve the next operation: Once this approach has been made, each fraction will then be elevated to the exponent to which the operation of the product they both constitute has been elevated: In the second place, each one of the powers of rational base that have been generated will then be resolved: Once this result is obtained, the multiplication of fractions must be continued: Once the product operation is resolved, the fraction will be simplified: When the simplest possible form is reached, the operation is assumed to be resolved. ## Example 2 Resolve the next operation: In this case you have a product operation elevated to a negative exponent, to solve this operation you must elevate each multiplication factor to this exponent: At this point, prior to elevating each fraction to its respective exponent, it is best to make the exponent go from negative to positive, which will be achieved by inverting the elements of each operation: Having now a positive exponent, each one of the rational-based power operations will be solved: At this point the multiplication of fractions must be continued: The operation will then be simplified: What could have been done by suppressing the 27 found in both factors. Without being able to simplify the operation further, the operation will be considered resolved. ## Example 3 Resolve the next operation: When analyzing each one of the elements that constitute this operation, we will have to deal with a multiplication of fractions elevated to zero. We will begin, according to the procedure indicated by Mathematics, to elevate each of the multiplication factors to this exponent: When doing so, a solution must then be given to each rational base power; however, since each one of them is elevated to zero, one must remember the mathematical law that states that whenever a fraction is elevated to the exponent zero it will result in unity: In the same way, seeing in principle that the operation of the product was elevated to this exponent, this result could be interpreted without the need to elevate each factor of the multiplication to this number, in this way it could have simply been solved in this way: Image: pixabay.com Examples of how to solve the power of a fraction product Speed concept In the field of Physics, the concept of "Speed" is defined... Wilhelm Heinrich Walter Baade (24 March 1893 – 25 June 1960) (Schrodting... In the field of Universal Literature, the first history of modern science ... Back in the year 1412, in the very beginning of the 15TH century, there wa... ### Bibliography ► phoneia.com (October 25, 2019). Examples of how to solve the power of a fraction product. Bogotá: E-Cultura Group. Recovered from https://phoneia.com/en/education/examples-of-how-to-solve-the-power-of-a-fraction-product/
# How do you simplify 5(x + 3)(x + 2)- 3(x^2 + 2x + 1)? May 13, 2018 $2 {x}^{2} + 19 x + 27$ #### Explanation: Given: $5 \left(x + 3\right) \left(x + 2\right) - 3 \left({x}^{2} + 2 x + 1\right)$ $\textcolor{b l u e}{\text{Consider the } - 3 \left({x}^{2} + 2 x + 1\right)}$ Multiply everything inside the brackets by (-3) giving: $- 3 {x}^{2} - 6 x - 3$ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ $\textcolor{b l u e}{\text{Consider the } 5 \left(x + 3\right) \left(x + 2\right)}$ Multiply everything inside the first brackets by (+5): $5 \left(x + 3\right) = 5 x + 15$ giving: $5 \left(x + 3\right) \left(x + 2\right) \to \textcolor{red}{\left(5 x + 15\right)} \textcolor{g r e e n}{\left(x + 2\right)}$ Multiply everything in the right bracket by everything in the left. color(green)( color(red)(5x)(x+2)color(white)("ddd")color(red)(+15)(x+2) )larr" Notice the + followed the 15" $5 {x}^{2} \underbrace{+ 10 x \textcolor{w h i t e}{\text{ddd}} + 15 x} + 30$ $5 {x}^{2} \textcolor{w h i t e}{\text{dddd") +25xcolor(white)("dddd}} + 30$ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ $\textcolor{b l u e}{\text{Putting it all together}}$ $5 \left(x + 3\right) \left(x + 2\right) \textcolor{w h i t e}{\text{ddd}} - 3 \left({x}^{2} + 2 x + 1\right)$ $5 {x}^{2} + 25 x + 30 \textcolor{w h i t e}{\text{ddd}} - 3 {x}^{2} - 6 x - 3$ Collecting like terms $\left(5 {x}^{2} - 3 {x}^{2}\right) + \left(25 x - 6 x\right) + \left(30 - 3\right)$ $\textcolor{w h i t e}{\text{ddd")2x^2color(white)("ddddddd")+19xcolor(white)("ddddd}} + 27$ $2 {x}^{2} + 19 x + 27$
# What can equal 81? ## What can equal 81? The numbers which we multiply to get 81 are the factors of 81. Factors of 81 are 1, 3, 9, 27, and 81. 1 is a universal factor. ### What times what can equal 49? Pair Factors Of 49 1 x 49 = 49 therefore, we can say 1 and 49 are pair factors. 7 x 7 = 49 therefore, we can say 7 and 7 are pair factors. 49 x 1 = 49 therefore, we can say 49 and 1 are pair factors. What times what can equal 72? Well, Factors of 72 are the numbers, that when multiplied together in a pair of two return the result as 72. Therefore, 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, and 72 are the factors of 72. What are all the factors of 77? Factors of 77 • Factors of 77: 1, 7, 11 and 77. • Prime Factorization of 77: 77 = 7 × 11. ## Is 81 a perfect square? Informally: When you multiply an integer (a “whole” number, positive, negative or zero) times itself, the resulting product is called a square number, or a perfect square or simply “a square.” So, 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, and so on, are all square numbers. ### What can equal to 56? What are the Factors of 56 in Pairs? 4 x 14 = 56 or 56/4=14 and 56/14 = 4. Therefore, the factor pairs of number 56 are 1 & 56, 7 & 8, 2 & 28, and 4 & 14. What can equal to 54? The Different Factors of 54 are: 1, 2, 3, 6, 9, 18, 27, 54….As you notice, all these factor pairs would result in the number 54 when multiplied together. • x 54 = 54. • x 27 = 54. • x 18 = 54. What gives you 150? FAQs on Factors of 150 The factors of 150 are 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, and 150. ## What can equal 121? The factors of 121 are 1, 11, and 121. ### Is it 9 times 9 or 9 times 81? How about: 9 times 9 = 81 or 81 times 1 = 81 What is 9 times 9? 9 x 9 = 81 because if you count 9 times it would make up or be 81 What times 9 81? If you mean: 9 times 9 = 81 What is 9 times 9 a hundred times? 99=81 81100= 8100 What times what equals 81? 1 times 81 3 times 27 9 times 9 What times what eqauls 81? It is: 9*9 = 81 Can 9 go into 81? How many times what equals 81 equals what? Note that “what” and “what” in the above problem could be the same number or different numbers. Below is a list of all the different ways that what times what equals 81. 1 times 81 equals 81. 3 times 27 equals 81. 9 times 9 equals 81. 27 times 3 equals 81. 81 times 1 equals 81. Which is correct 81 divided by 9 or 9? 81 / 9 = 9. Instead of saying 81 divided by 9 equals 9, you could just use the division symbol, which is a slash, as we did above. Also note that all answers in our division calculations are rounded to three decimals if necessary. ## What are the combinations of two numbers that equal 81? For your convenience, we have made a list of all the combinations of two numbers multiplied by each other that will make 81: 1 x 81 = 81. 3 x 27 = 81. 9 x 9 = 81. 27 x 3 = 81. 81 x 1 = 81. Two Numbers Multiplied. Enter another number below to see which combinations of two numbers multiplied will equal that number.
# How Do You Find The Mode If No Number Is Repeated? ## How do you find the mode if there is no mode? In that case, you have to add those two numbers together and then divide by two to find the median. The mode of a data set refers to the number that occurs most often. If there is not a number that occurs more than any other, we say there is no mode for the data.. ## How do you find the mode if frequency is the same? There can be more than one mode in a list or set of numbers. Look at this list of numbers: 1, 1, 1, 3, 3, 3. In this list there are two modes, because both 1 and 3 are repeated same number of times. On the other hand, sometimes there is no mode at all. ## What is the mode if two numbers repeat? There are a few tricks to remember about mode: If there are two numbers that appear most often (and the same number of times) then the data has two modes. This is called bimodal. If there are more than 2 then the data would be called multimodal. ## Is it possible to have no mode? There may be no mode if no value appears more than any other. There may also be two modes (bimodal), three modes (trimodal), or four or more modes (multimodal). ## What is the mode if there is a tie? Calculating the Mode The mode is the number that appears the most often. A set of data can have more than one mode if there is a tie for the number that occurs most frequently. The number 4 is the mode since it appears the most frequently in Set S. ## What is the formula of mode in statistics? It is one of the three measures of central tendency, apart from mean and median. For example, mode of the set {3, 7, 8, 8, 9}, is 8….Mean Median Mode Comparison.MeanMedianModeMean = (2+2+3+4+5)/5 = 3.2Median = 3Mode = 22 more rows•Aug 31, 2019 ## How do you find the range of a data set? The Range is the difference between the lowest and highest values. Example: In {4, 6, 9, 3, 7} the lowest value is 3, and the highest is 9. So the range is 9 − 3 = 6. It is that simple! ## How do you find the mode in a frequency table? How To Obtain The Mean, Median And Mode From A Frequency Table? To find the mean: Multiply midpoints by frequencies, add the subtotals and divide by the total of the frequencies. To find the mode: Look for the largest frequency and the corresponding value is the modal value or modal class. ## What if there are no modes? It is possible for a set of data values to have more than one mode. If there are two data values that occur most frequently, we say that the set of data values is bimodal. If there is no data value or data values that occur most frequently, we say that the set of data values has no mode. ## What is the mode if all numbers appear once? The mode is an average that is calculated by finding the number in the list that occurs the most. If there are multiple numbers that occur more than others, those numbers are all modes; if all numbers do not occur more than others (in other words, if every number only occurs once), then there is no mode. ## How do you find mode? To find the mode, or modal value, it is best to put the numbers in order. Then count how many of each number. A number that appears most often is the mode. ## What happens when you have 3 modes? A set of numbers with two modes is bimodal, a set of numbers with three modes is trimodal, and any set of numbers with more than one mode is multimodal. ## How do you find the mean and mode? Mean: Add up all the numbers of the set. Divide by how many numbers there are. Mode: The number that occurs the most. ## What is the frequency of the mode? For data presented in a frequency table, the mode is the value associated with the greatest frequency (if there is a greatest frequency). In this case, the greatest frequency is 96 and the associated value is “1,” so the mode is “1.” More students received 1 parking ticket than any of the other possibilities. ## How do you interpret the mode in statistics? The mode is the value that occurs most frequently in a set of observations. Minitab also displays how many data points equal the mode. The mean and median require a calculation, but the mode is determined by counting the number of times each value occurs in a data set.
## Basic Continuous Distributions 26 Jan 2015 Curriculum Think Stats Statistics Book Distribution Exposition ### Continuous Distributions As a Data Scientist/Analyst, a huge part of the job is to characterize the data you’re dealing with. This abstraction step can help you apply functions to your data, since the overall structure has been smoothed out and simplified. Similarly, it can make communicating the shape of your data much easier, since by this characterization you can remove unhelpful noise in the data. The main method of data characterization is to use a well-known probability distribution to describe it. Doing so can add extra insight into your data, help you communicate the structure of your data, and give you a tailored toolset for working with the data. You’ll want to be sure your data set is a true fit, but if it is, then there is great benefit to abstracting your data in this way. In this post, we’ll look at 4 common continuous distributions. ### The Uniform Distribution The Uniform Distribution is the easiest distribution to understand. For all valid values of the distribution, the probibility is the same. If the minimum value is $$a$$ and the maximum value is $$b$$, then the CDF will look like this: $CDF(x) =$ \begin{eqnarray} &0& \text{ }\text{ }\text{ for }\text{ }\text{ } &x& < a \\ &\frac{x - a}{b - a}& \text{ }\text{ }\text{ for }\text{ }\text{ } &a& \le x < b \\ &1& \text{ }\text{ }\text{ for }\text{ }\text{ } &x& \ge b \end{eqnarray} The discrete version of this distribution is all too common in probability lectures. Just think back to all of those coin flips, die rolls, and card draws. As the number of outcomes grows beyond 2, 6, and 52, however, it can be considered continuous at a certain point. You might say that even 1,000,000 discrete outcomes would not a continuous distribution make, but considering that in data science we are usually measuring real-world values, and since our measurements are not infinitely accurate, there will necessarily be some level of discretization in nearly everything we do anyway. ### The Exponential Distribution The Exponential Distribution is a good model for events that occur and random, independent intervals at a constant rate for some span of time. In reality, a pure constant rate can be hard to find, but narrowing the time span or considering confounding effects in your analysis can help correct these issues. For example, incoming phone calls to a call center won’t come at a constant rate, but probably the incoming calls from 2pm to 4pm on a weekday come close. The CDF of an exponential distribution is $CDF(x) = 1 - e^{-\lambda x}$ and the $$\lambda$$ parameter defines the shape of the distribution. The mean of a exponential distribution is $$\frac{1}{\lambda}$$. Using the call center example from earlier, if phone calls arrive at an average of 1 call per 2 minutes, then the distribution graph (with minutes on the x axis) would match the exponential distribution where $$\lambda = 0.5$$. ### The Pareto Distribution The Pareto Distribution looks similar to the exponential distribution, but it has a heavier tail, meaning that as you move along the x-axis of the PDF, the probabilities don’t shrink as rapidly (meaning that the CDF doesn’t trend to 1 as rapidly). The reason for that is clear when looking at the CDF for this distribution. Unlike the exponential which approaches 1 at (gasp!) an exponential rate, the Pareto distribution rises at a fixed power rate. $CDF(x) = 1 - \left(\frac{x}{x_m}\right)^\alpha, \text{ where } x_m \text{ is the minimum value of the distribution}$ Just like $$\lambda$$ defined the shape of the exponential distribution, here, $$\alpha$$ determines the shape. Since the Pareto distribution is more heavily tailed than the exponential distribution, there are a range of examples that typically fall into this distribution. The most common are resources which follow the 80-20 rule, where 80% of the resources are owned by 20% of the people. This rule also extends to many other common real-world examples. The fact that the 80-20 rule also follows a power law distribution landed it its alternate name, the apropos Pareto principle. ### The Normal Distribution If you were familiar with any continuous distribution prior to reading this, then it was probably the normal distribution. Its PDF is the ubiquitous bell curve that we all know and love. So far we’ve been looking at CDFs, so we’ll stick with that here, too. The normal CDF has a sigmoid shape, which should be just a recognizable to a data scientist as the bell curve. Its shape is defined by its mean and standard deviation, $$\mu$$ and $$\sigma$$, respectively. #### The Central Limit Theorem The reason that the normal distribution is so ubiquitous is due to something called the Central Limit Theorem. It states that, under certain conditions, if you take a bunch of values independently from any distribution and sum those values up, then repeat this many times, the collection of resulting sums will be normally distributed. This theorem that applies to a wide range of distributions, including all of the ones we’ve looked at up to this point. For example, if you sum 100 values drawn from a uniform distribution, and then repeat the process 10,000 times, the resulting plot will resemble a normal distribution. import random #Generate a 2D list of uniformly distributed data points, 100 by 10000 data = [[random.random() for _ in range(100)] for _ in range(10000)] sums = [sum(row) for row in data] Running that python snippen and plotting sums in a histogram, we get the following: This fact is actually quite remarkable, and it’s the reason that the normal distribution shows up in so many different areas. Due to limited knowledge, we are often performing density estimation on unobservable, underlying PDFs. From Wikipedia: “In cases like electronic noise, examination grades, and so on, we can often regard a single measured value as the weighted average of a large number of small effects. Using generalisations of the central limit theorem, we can then see that this would often (though not always) produce a final distribution that is approximately normal.” So even though we may not be able to see the underlying distribution, if we can measure some value which is an accumulation of underlying factors, then that posterior measurement will probably come together as a normal distribution with which we can work.
Question 50 POINTS A pair of equations is shown below: y = 7x − 8 y = 5x − 2 Part A: Explain how you will solve the pair of equations by substitution or elimination. Show all the steps and write the solution. Part B: If the two equations are graphed, at what point will the lines representing the two equations intersect? Explain your answer x = 3 (3, 13) Step-by-step explanation: Given equations: • y = 7x − 8 • y = 5x − 2 ### Part A Solve by substitution: ⇒ y = y ⇒ 7x – 8 = 5x – 2 Subtract 5x from both sides: ⇒ 7x – 8 – 5x = 5x – 2 – 5x ⇒ 2x – 8 = – 2 ⇒ 2x – 8 + 8 = – 2 + 8 ⇒ 2x = 6 Divide both sides by 2: ⇒ 2x ÷ 2 = 6 ÷ 2 x = 3 ### Part B To find the y-value of the point of intersection, substitute the found value of x (from part A) into one of the equations: ⇒ y = 7(3) – 8 ⇒ y = 21 – 8 ⇒ y = 13 Therefore, the point at which the lines intersect is (3, 13)
## Engage NY Eureka Math 8th Grade Module 7 Lesson 7 Answer Key ### Eureka Math Grade 8 Module 7 Lesson 7 Example Answer Key Example 1. The number 0.253 is represented on the number line below. How can we locate the number 0.253 = $$\frac{2}{10}$$ + $$\frac{5}{100}$$ + $$\frac{3}{1000}$$ on the number line? We can see that 0.253 is a tad larger than 0.2 and is smaller than 0.3. So if we divide the line segment from 0 to 1 into tenths, 0.253 lies somewhere in the segment between 0.2 and 0.3. Now divide this segment into ten equal parts. (Those parts are hundredths of the original unit length.) We know that 0.253 is larger than 0.25 but smaller than 0.26, and so lies in the segment between these two values. Now divide this small segment into tenths again. (We are now dealing with thousandths of the original unit length.) We can now see where exactly to pin 0.253 on the number line. Writing 0.253 in its expanded decimal form of $$\frac{2}{10}$$ + $$\frac{5}{100}$$ + $$\frac{3}{1000}$$ illustrates this process: The first decimal digit of 0.253 is 0.2, or $$\frac{2}{10}$$, and this tells us within which tenth we are to place 0.253. The first two decimal digits of 0.253 are 0.25 which is equal to $$\frac{2}{10}$$ + $$\frac{5}{10^{2}}$$ , or $$\frac{25}{10^{2}}$$, and this tells us within which hundredth we are to place 0.253. The first three decimal digits of 0.253 are 0.253 which is equal to $$\frac{2}{10}$$ + $$\frac{5}{10^{2}}$$ + $$\frac{3}{10^{3}}$$, or $$\frac{253}{10^{3}}$$, and this tells us within which thousandth we are to place 0.253. And since the decimal terminates here, we are done. Have the students explain this process again in their own words, referring to the number line diagram as they do so. How do you think this process would change if we tried to locate an infinite decimal on the number line? The sequence for an infinite decimal would never end; it would go on infinitely. We need to introduce some notation. If a decimal has a repeating pattern, as for 0.3333â€¦ or 7.45454545â€¦, for instance, then a horizontal bar is used to indicate that a block of digits is being repeated. For example, 0.3333â€¦ is written as $$0 . \overline{3}$$ and 7.45454545â€¦ as $$7 . \overline{45}$$. It is conceivable that an infinite decimal could have no repeating pattern. Example 2. The number $$\frac{5}{6}$$, which is equal to 0.833333â€¦, or $$0 . \overline{83}$$ is partially represented on the number line below. Now, consider the equality $$\frac{5}{6}$$ = 0.833333″â€¦” = $$0 . \overline{83}$$. Notice that at the second step, the work feels as though it repeats, which coincides with the fact that the decimal digit of 3 repeats. What is the expanded form of the decimal 0.833333â€¦? 0.833333″â€¦” = $$\frac{8}{10} + \frac{3}{10^{2}} + \frac{3}{10^{3}} + \frac{3}{10^{4}} + \frac{3}{10^{5}} + \frac{3}{10^{6}} + \cdots$$ + ………… We see again that at the second step the work begins to repeat. Each step can be represented by increasing powers of 10 in the denominator: $$\frac{8}{10}, \frac{83}{10^{2}}, \frac{833}{10^{3}}, \frac{8333}{10^{4}}, \frac{83333}{10^{5}}, \frac{833333}{10^{6}}$$, and so on. When will it end? Explain. It will never end because the decimal is infinite. As we step through this process we are pinning the exact location of 0.8333â€¦ into smaller and smaller intervals, intervals with sizes shrinking to zero: an interval of a tenth, and then a hundredth, then a thousandth, and, if we kept going, an interval of size $$\frac{1}{10^{20}}$$, and later on to an interval of size $$\frac{1}{10^{100}}$$, and so on, supposedly forever! Okay. Letâ€™s now think deeply about 0.9999â€¦. Where do we find this number on the number line? Draw on the board a sequence of number-line segments akin to those in Example 2, and have students give instructions on how to pin down the location of 0.9999â€¦. In which tenth does it lie? In which hundredth? In which thousandth? And so on. Ask: At any stage of this process is 0.9999â€¦ right at the number 1 on the number line? No. We are always just to the left of 1. Right. And that makes sense as 0.9 is smaller than 1, and 0.99 is smaller than 1, as are 0.999 and 0.9999. At every stage of the process we are just shy of the number 1. ### Eureka Math Grade 8 Module 7 Lesson 7 Exercise Answer Key Opening Exercise a. Write the expanded form of the decimal 0.3765 using powers of 10. 0.3765 = $$\frac{3}{10} + \frac{7}{10^{2}} + \frac{6}{10^{3}} + \frac{5}{10^{4}}$$ b. Write the expanded form of the decimal 0.3333333â€¦ using powers of 10. 0.333333″â€¦” = $$\frac{3}{10} + \frac{3}{10^{2}} + \frac{3}{10^{3}} + \frac{3}{10^{4}} + \frac{3}{10^{5}} + \frac{3}{10^{6}} + \cdots$$ c. Have you ever wondered about the value of 0.99999â€¦? Some people say this infinite decimal has a value of 1. Some disagree. What do you think? Exercises 1â€“5 Exercise 1. a. Write the expanded form of the decimal 0.125 using powers of 10. 0.125 = $$\frac{1}{10} + \frac{2}{10^{2}} + \frac{5}{10^{3}}$$ b. Show on the number line the placement of the decimal 0.125. Exercise 2. a. Write the expanded form of the decimal 0.3875 using powers of 10. 0.3875 = $$\frac{3}{10} + \frac{8}{10^{2}} + \frac{7}{10^{3}} + \frac{5}{10^{4}}$$ b. Show on the number line the placement of the decimal 0.3875. Exercise 3. a. Write the expanded form of the decimal 0.777777â€¦ using powers of 10. 0.777777â€¦ = $$\frac{7}{10} + \frac{7}{10^{2}} + \frac{7}{10^{3}} + \frac{7}{10^{4}} + \frac{7}{10^{5}} + \frac{7}{10^{6}} + \cdots$$ b. Show the first few stages of placing the decimal 0.777777â€¦ on the number line. Exercise 4. a. Write the expanded form of the decimal $$0 . \overline{45}$$ using powers of 10. $$0 . \overline{45}$$ = $$\frac{4}{10} + \frac{5}{10^{2}} + \frac{4}{10^{3}} + \frac{5}{10^{4}} + \frac{4}{10^{5}} + \frac{5}{10^{6}} + \cdots$$ b. Show the first few stages of placing the decimal $$0 . \overline{45}$$ on the number line. Exercise 5. a. Order the following numbers from least to greatest: 2.121212, 2.1, 2.2, and $$2 . \overline{12}$$. 2.1, 2.121212, $$2 . \overline{12}$$, and 2.2 b. Explain how you knew which order to put the numbers in. Each number is the sum of the whole number 2 and a decimal. When you write each number in this manner, you get the following expansions. 2.121212 = 2 + $$\frac{1}{10} + \frac{2}{10^{2}} + \frac{1}{10^{3}} + \frac{2}{10^{4}} + \frac{1}{10^{5}} + \frac{2}{10^{6}}$$ 2.1 = 2 + $$\frac{1}{10}$$ 2.2 = 2 + $$\frac{2}{10}$$ $$2 . \overline{12}$$ = 2 + $$\frac{1}{10} + \frac{2}{10^{2}} + \frac{1}{10^{3}} + \frac{2}{10^{4}} + \frac{1}{10^{5}} + \frac{2}{10^{6}} + \frac{1}{10^{7}} + \frac{2}{10^{8}} + \cdots$$ In this form, it is clear that 2.1 is the least of the four numbers, followed by the finite decimal 2.121212, then the infinite decimal $$2 . \overline{12}$$, and finally 2.2. ### Eureka Math Grade 8 Module 7 Lesson 7 Problem Set Answer Key Question 1. a. Write the expanded form of the decimal 0.625 using powers of 10. 0.625 = $$\frac{6}{10} + \frac{2}{10^{2}} + \frac{5}{10^{3}}$$ b. Place the decimal 0.625 on the number line. Question 2. a. Write the expanded form of the decimal $$0 . \overline{370}$$ using powers of 10. $$0 . \overline{370}$$ = $$\frac{3}{10} + \frac{7}{10^{2}} + \frac{0}{10^{3}} + \frac{3}{10^{4}} + \frac{7}{10^{5}} + \frac{0}{10^{6}} + \cdots$$ b. Show the first few stages of placing the decimal 0.370370â€¦ on the number line. Question 3. Which is a more accurate representation of the fraction $$\frac{2}{3}$$: 0.6666 or $$0 . \overline{6}$$? Explain. Which would you prefer to compute with? The number $$\frac{2}{3}$$ is more accurately represented by the decimal 0.6 Ì… compared to 0.6666. The long division algorithm with $$\frac{2}{3}$$ shows that the digit 6 repeats. Then, the expanded form of the decimal $$0 . \overline{6}$$ is $$\frac{6}{10} + \frac{6}{10^{2}} + \frac{6}{10^{3}} + \frac{6}{10^{4}} + \frac{6}{10^{5}} + \frac{6}{10^{6}} + \cdots$$, and the expanded form of the decimal 0.6666 is $$\frac{6}{10} + \frac{6}{10^{2}} + \frac{6}{10^{3}} + \frac{6}{10^{4}}$$. For this reason, $$0 . \overline{6}$$ is precise, but 0.6666 is an approximation. For computations, I would prefer to use 0.6666. My answer would be less precise, but at least I would be able to compute with it. When attempting to compute with an infinite number, you would never finish writing it; thus, you could never compute with it. Question 4. Explain why we shorten infinite decimals to finite decimals to perform operations. Explain the effect of shortening an infinite decimal on our answers. We often shorten infinite decimals to finite decimals to perform operations because it would be impossible to represent an infinite decimal precisely since the sequence that describes infinite decimals has an infinite number of steps. Our answers are less precise; however, they are not that much less precise because with each additional digit we include in the sequence, we are adding a very small amount to the value of the number. The more decimals we include, the closer the value we add approaches zero. Therefore, it does not make that much of a difference with respect to our answer. Question 5. A classmate missed the discussion about why $$0 . \overline{9}$$ = 1. Convince your classmate that this equality is true. Answers will vary. Accept any reasonable explanation. One is provided below. Ask: Could there be any space between the locations of 0.9999â€¦ and 1 on the number line? We have that 0.9999â€¦ is larger than 0.9 and so is within one-tenth of 1 on the number line. We also have that 0.9999â€¦ is larger than 0.99 and so is within one-hundredth of 1 on the number line. And 0.9999â€¦ is larger than 0.999 and so is within one-thousandth of 1 on the number line, and so on. There can be no space between 0.9999â€¦ and 1 on the number line, as we can always argue that 0.9999â€¦ must be within any given distance from 1. Thus, 0.9999â€¦ and 1 must sit at the same location on the number line and so are the same number. Question 6. Explain why 0.3333 < 0.33333. 0.3333 = $$\frac{3}{10} + \frac{3}{10^{2}} + \frac{3}{10^{3}} + \frac{3}{10^{4}}$$, and 0.33333 = $$\frac{3}{10} + \frac{3}{10^{2}} + \frac{3}{10^{3}} + \frac{3}{10^{4}} + \frac{3}{10^{5}}$$. That means that 0.33333 is exactly $$\frac{3}{10^{5}}$$ larger than 0.3333. If we examined the numbers on the number line, 0.33333 is to the right of 0.3333, meaning that it is larger than 0.3333. ### Eureka Math Grade 8 Module 7 Lesson 7 Exit Ticket Answer Key Question 1. a. Write the expanded form of the decimal 0.829 using powers of 10. 0.829 = $$\frac{8}{10} + \frac{2}{10^{2}} + \frac{9}{10^{3}}$$ b. Show the placement of the decimal 0.829 on the number line. Question 2. a. Write the expanded form of the decimal 0.55555â€¦ using powers of 10. 0.55555″â€¦” = $$\frac{5}{10} + \frac{5}{10^{2}} + \frac{5}{10^{3}} + \frac{5}{10^{4}} + \frac{5}{10^{5}} + \frac{5}{10^{6}} + \cdots$$ b. Show the first few stages of placing the decimal 0.555555â€¦ on the number line. a. Write the expanded form of the decimal $$0 . \overline{573}$$ using powers of 10. $$0 . \overline{573}$$ = $$\frac{5}{10} + \frac{7}{10^{2}} + \frac{3}{10^{3}} + \frac{5}{10^{4}} + \frac{7}{10^{5}} + \frac{3}{10^{6}} + \cdots$$ b. Show the first few stages of placing the decimal $$0 . \overline{573}$$ on the number line.
# Expert Maths Tutoring in the UK How to Prove that a Matrix is Invertible How to Prove that a Matrix is Invertible with tutors mapped to your child’s learning needs. Inverse of 3×3 Matrix Before going to see how to find the inverse of a 3×3 matrix, let us recall the what the inverse mean. The inverse of a number is a number which when multiplied by the given number results in the multiplicative identity, 1. In the same way, the product of a matrix A and its inverse A-1 gives the identity matrix, I. i.e., AA-1 = A-1A = I. Let us see how to find the inverse of 3×3 matrix. Let us see the formula for finding the inverse of 3×3 matrix along with some other ways of finding it. Also, we will see some examples of finding the inverse of a 3×3 matrix. ## What is the Inverse of 3×3 Matrix? The inverse of a 3×3 matrix, say A, is a matrix of the same order denoted by A-1 where AA-1 = A-1A = I, where I is the identity matrix of order 3×3. i.e., I = $$\left[\begin{array}{rr}1 & 0 & 0 \\ 0&1&0 \\ 0 & 1&0 \end{array}\right]$$. For example, if A = $$\left[\begin{array}{rr}1 & 2 & -1 \\ 2&1&2 \\ -1 & 2&1 \end{array}\right]$$ then A-1 = $$\left[\begin{array}{rr}3 / 16 & 1 / 4 & -5 / 16 \\ 1 / 4 & 0 & 1 / 4 \\ -5 / 16 & 1 / 4 & 3 / 16 \\ \end{array}\right]$$. One can easily multiply these matrices and verify whether AA-1 = A-1A = I. We will see how to find the inverse of a 3×3 matrix in the upcoming section. ## Elements Used to Find Inverse of 3×3 Matrix Before going to know how to find the inverse of 3×3 matrix, let us see how to find the determinant and adjoint of 3×3 matrix. Let us use this same example (as in the previous section) in each explanation. ### Adjoint of a 3×3 Matrix The adjoint of a matrix A is obtained by finding the transpose of the cofactor matrix of A. To know how to find the adjoint of a matrix in detail click here. The cofactor of any element of a 3×3 matrix is the determinant of 2×2 matrix that is obtained by removing the row and the column containing the element. Also, we write alternate + and – signs while finding the cofactors. Here is an example. Let A = $$\left[\begin{array}{rr}1 & 2 & -1 \\ 2&1&2 \\ -1 & 2&1 \end{array}\right]$$. Then its cofactor matrix is: $$\left[\begin{array}{rr}\left\|\begin{array}{ll} 1 & 2 \\ 2 & 1 \end{array}\right\| & -\left\|\begin{array}{cc} 2 & 2 \\ -1 & 1 \end{array}\right\| & -\left\|\begin{array}{cc} 2 & 1 \\ -1 & 2 \end{array}\right\|\\ -\left\|\begin{array}{cc} 2 & -1 \\ 2 & 1 \end{array}\right\| & \left\|\begin{array}{cc} 1 & -1 \\ -1 & 1 \end{array}\right\|&-\left\|\begin{array}{cc} 1 & 2 \\ -1 & 2 \end{array}\right\| \\ \left\|\begin{array}{cc} 2 & -1 \\ 1 & 2 \end{array}\right\|& -\left\|\begin{array}{rr} 1 & -1 \\ 2 & 2 \end{array}\right\|&\left\|\begin{array}{ll} 1 & 2 \\ 2 & 1 \end{array}\right\| \end{array}\right]$$ Each 2×2 determinant is obtained by multiplying diagonals and subtracting the products (from left to right). So the cofactor matrix = $$\left[\begin{array}{ccc} 1-4 & -(2+2) & 4+1 \\ -(2+2) & 1-1 & -(2+2) \\ 4+1 & -(2+2) & 1-4 \end{array}\right]$$ = $$\left[\begin{array}{rrr} -3 & -4 & 5 \\ -4 & 0 & -4 \\ 5 & -4 & -3 \end{array}\right]$$ By transposing the cofactor matrix, we get the adjoint matrix. So adj A = $$\left[\begin{array}{ccc} -3 & -4 & 5 \\ -4 & 0 & -4 \\ 5 & -4 & -3 \end{array}\right]$$. (Of course, we have got both the cofactor matrix and adjoint matrix to be the same in this case. But it may not happen always). ### Determinant of a 3×3 Matrix To find the determinant of a 3×3 matrix, find the sum of the product of the elements of any of its row/column and their corresponding cofactors. Here is an example. A = $$\left[\begin{array}{rr}1 & 2 & -1 \\ 2&1&2 \\ -1 & 2&1 \end{array}\right]$$. Let us use the first row to find the determinant. det A = 1 (cofactor of 1) + 2 (cofactor of 2) + (-1) cofactor of (-1) = 1(-3) + 2(-4) + (-1)5 = -3 – 8 – 5 = -16 But here is a trick to find the determinant of any 3×3 A = $$\left[\begin{array}{ccc}a & b & c \\ p & q & r \\ x & y & z\end{array}\right]$$ matrix faster. Here, we just write the same matrix twice next to each other and then apply the trick. The inverse of a 3×3 matrix formula uses the determinant of the matrix. ## Inverse of 3×3 Matrix Formula The inverse of a 3×3 matrix A is calculated using the formula A-1 = (adj A)/(det A), where • det A = determinant of A det A is in the denominator in the formula of A-1. Thus, for A-1 to exist det A should not be 0. i.e., • A-1 exists when det A ≠ 0 (i.e., when A is nonsingular) • A-1 does not exist when det A = 0 (i.e., when A is singular) Thus, here are the steps to find the inverse of 3×3 matrix. The steps are explained through the same example A = $$\left[\begin{array}{rr}1 & 2 & -1 \\ 2&1&2 \\ -1 & 2&1 \end{array}\right]$$. Let us find A-1. • Step – 1: Find adj A. We have already seen that adj A = $$\left[\begin{array}{ccc} -3 & -4 & 5 \\ -4 & 0 & -4 \\ 5 & -4 & -3 \end{array}\right]$$. • Step – 2: Find det A. We have already seen that det A = -16 • Step – 3: Apply the inverse of 3×3 matrix formula A-1 = (adj A)/(det A). i.e., divide every element of adj A by det A. Then A-1 = $$\left[\begin{array}{ccc} -3/-16 & -4/-16 & 5/-16 \\ -4/-16 & 0/-16 & -4/-16 \\ 5/-16 & -4/-16 & -3/-16 \end{array}\right]$$ = $$\left[\begin{array}{rr}3 / 16 & 1 / 4 & -5 / 16 \\ 1 / 4 & 0 & 1 / 4 \\ -5 / 16 & 1 / 4 & 3 / 16 \\ \end{array}\right]$$. ## Finding Inverse of 3×3 Matrix Using Row Operations Like any other square matrix, we can use the elementary row operations to find the inverse of a 3×3 matrix as well. The process is explained below with an example. • We first write the given 3×3 matrix A and the identity matrix I of order 3×3 as an augmented matrix separated by a line where A is on the left side and I is on the right side. • Apply row operations so as to make the left side matrix to become the identity matrix I. • Then the matrix on the right side is A-1. We can an see an example for this in the upcoming sections. ## Solving System of 3×3 Equations Using Inverse We can solve the system of 3×3 equations using the inverse of a matrix. The steps for this are explained here with an example where we are going to solve the system of 3×3 equations x + 2y – z = 10, 2x + y + 2z = 5, and -x + 2y + z = 6. • Step – 1: Write the given system of equations as AX = B. $$\left[\begin{array}{rr}1 & 2 & -1 \\ 2&1&2 \\ -1 & 2&1 \end{array}\right]$$ $$\left[\begin{array}{rr}x \\y \\ z \end{array}\right]$$ = $$\left[\begin{array}{rr}10 \\ 5 \\ 6 \end{array}\right]$$ Here, A = $$\left[\begin{array}{rr}1 & 2 & -1 \\ 2&1&2 \\ -1 & 2&1 \end{array}\right]$$, X = $$\left[\begin{array}{rr}x \\y \\ z\end{array}\right]$$, and B = $$\left[\begin{array}{rr}10 \\ 5\\ 6 \end{array}\right]$$. • Step – 2: Find the inverse of the 3×3 matrix. i.e., find A-1. In one of the previous sections, we found that A-1 = $$\left[\begin{array}{rr}3 / 16 & 1 / 4 & -5 / 16 \\ 1 / 4 & 0 & 1 / 4 \\ -5 / 16 & 1 / 4 & 3 / 16 \\ \end{array}\right]$$. • Step – 3: Find the solution matrix X using the formula X = A-1B. X = $$\left[\begin{array}{rr}3 / 16 & 1 / 4 & -5 / 16 \\ 1 / 4 & 0 & 1 / 4 \\ -5 / 16 & 1 / 4 & 3 / 16 \\ \end{array}\right]$$ $$\left[\begin{array}{rr}10 \\ 5 \\ 6 \end{array}\right]$$ = $$\left[\begin{array}{rr}5/4 \\4 \\ -3/4 \end{array}\right]$$ Therefore, x = 5/4, y = 4, and z = -3/4 is the solution of the given system of equations. Important Notes on Inverse of 3×3 Matrix: • A matrix A is invertible (inverse of A exists) only when det A ≠ 0. • If A and A-1 are the inverses of each other, then AA-1 = A-1A = I. • The inverse of a 3×3 identity matrix is itself. i.e., I-1 = I. • The inverse of 3×3 matrix is used to solve a system of 3×3 equations in 3 variables. ☛ Related Topics: ## Inverse of 3×3 Matrix Examples 1. Example 1: Determine which of the following 3×3 matrices have an inverse. (a) A = $$\left[\begin{array}{rr}2 & 6 & 3 \\ 4 & -1 & 3 \\ 1 & 3 & 2 \end{array}\right]$$ (b) B = $$\left[\begin{array}{rr}-3 & 2 & 1\\ 4 & 8 & -1 \\ 6 & -4 & -2 \end{array}\right]$$. Solution: A 3×3 matrix A is invertible only if det A ≠ 0. So Let us find the determinant of each of the given matrices. (a) det A = 2 (-2 – 9) – 6 (8 – 3) + 3 (12 + 1) = 2(-11) -6(5) + 3(13) = -22 – 30 + 39 = -13 ≠ 0 Thus, A-1 exists. i.e., A is invertible. (b) det B = -3 (-16 – 4) – 2 ( -8 + 6) + 1 (-16 -48) = -3 (-20) – 2 (-2) + 1 (-64) = 60 + 4 – 64 = 0 Therefore, B-1 does NOT exist. i.e., B is NOT invertible. Answer: A has inverse but B does not. 2. Example 2: Find the inverse of A = $$\left[\begin{array}{rr}4 & 2 & -3 \\ 1 & -1 &2\\ 5 & 3 & 0 \end{array}\right]$$. Solution: We will find det A and adj A for the given matrix. Finding det A: det A = 4 (0 – 6) – 2 (0 – 10) – 3 (3 + 5) = -28 The co-factor matrix of A = $$\left[\begin{array}{rr}0-6 & -(0-10) & 3+5 \\ -(0+9) & 0+15 & -(12-10) \\ 4-3 & -(8+3) & -4-2 \end{array}\right]$$ = $$\left[\begin{array}{rr}-6 &10 & 8\\ -9 & 15 & -2 \\ 1 & -11 & -6 \end{array}\right]$$ Thus,adj A = $$\left[\begin{array}{rr}-6 &-9 & 1\\ 10 & 15 & -11 \\ 8 & -2 & -6 \end{array}\right]$$ Finding A-1: Substitute the values of adj A and det Ain the formula A-1 = (adj A) / (det A), A-1 = $$\left[\begin{array}{rr}-6/-28 &-9/-28 & 1/-28\\ 10/-28 & 15/-28 & -11/-28 \\ 8/-28 & -2/-28 & -6/-28 \end{array}\right]$$ = $$\left[\begin{array}{rr}3/14 &9/28 & -1/28\\ -5/14 & -15/28 & 11/28 \\ -2/7 & 1/14 & 3/14 \end{array}\right]$$ Answer: The inverse of the given 3×3 matrix is A-1 = $$\left[\begin{array}{rr}3/14 &9/28 & -1/28\\ -5/14 & -15/28 & 11/28 \\ -2/7 & 1/14 & 3/14 \end{array}\right]$$. 3. Example 3: Find the inverse of the 3×3 matrix A from Example 2 using elementary row operations. Verify whether the answer you get is the same as that of Example 2. Solution: Given A = $$\left[\begin{array}{rr}4 & 2 & -3 \\ 1 & -1 &2\\ 5 & 3 & 0 \end{array}\right]$$. Step 1: Write A and I as in a single matrix separated by a dotted line (as an augmented matrix). $$\left[\begin{array}{ccccccc} 4 & 2 & -3 & \| & 1 & 0 & 0 \\ 1 & -1 & 2 & \| & 0 & 1 & 0 \\ 5 & 3 & 0 & \| & 0 & 0 & 1 \end{array}\right]$$ Step 2: Apply elementary row operations to make the left side matrix converted to an identity matrix. Applying R$$_2$$ → 4R$$_2$$ – R$$_1$$ and R$$_3$$ → 4R$$_3$$ – 5R$$_1$$, $$\left[\begin{array}{ccccccc} 4 & 2 & -3 & \| & 1 & 0 & 0 \\ 0 & -6 & 11 & \| & -1 & 4 & 0 \\ 0 & 2 & 15 & \| & -5 & 0 & 4 \end{array}\right]$$ Applying R$$_1$$ → 3R$$_1$$ + R$$_2$$ and R$$_3$$ → 3R$$_3$$ + R$$_2$$, $$\left[\begin{array}{ccccccc} 12 & 0 & 2 & \|&2 & 4 & 0 \\ 0 & -6 & 11 & \| & -1 & 4 & 0 \\ 0 & 0 & 56 &\|& -16 & 4 & 12 \end{array}\right]$$ Applying R$$_1$$ → 28R$$_1$$ – R$$_3$$ and R$$_2$$ → 56R$$_2$$ -11R$$_3$$, $$\left[\begin{array}{cccccc} 336 & 0 & 0 &\|& 72 & 108 & -12 \\ 0 & -336 & 0 & \| & 120 & 180 & -132 \\ 0 & 0 & 56 &\|& -16 & 4 & 12 \end{array}\right]$$ Divide R$$_1$$ by 336, R$$_2$$ by -336, and R$$_3$$ by 56, $$\left[\begin{array}{cccccc} 1 & 0 & 0 & \| & 3 / 14 & 9 / 28 & -1 / 28 \\ 0 & 1 & 0 & \| & -5 / 14 & -15 / 28 & 11 / 28 \\ 0 & 0 &1 & \| & -2/7 & 1 / 14 & 3 / 14 \end{array}\right]$$ Step 3: The right side matrix is our inverse matrix. i.e., A-1 = $$\left[\begin{array}{rr}3/14 &9/28 & -1/28\\ -5/14 & -15/28 & 11/28 \\ -2/7 & 1/14 & 3/14 \end{array}\right]$$ ## FAQs on Inverse of 3×3 Matrix ### What is Meant by Inverse of a 3×3 Matrix? The inverse of a 3×3 matrix A is denoted by A-1. Here, AA-1 = A-1A = I, where I is the identity matrix of order 3×3. ### How to Find Inverse of a 3×3 Matrix? Here are the steps to find the inverse of a 3×3 matrix A: • Find det A. • Apply the formula A-1 = (adj A)/(det A). What is an Example of a 3×3 Matrix with No Inverse? A matrix cannot have inverse if its determinant is 0. A = $$\left[\begin{array}{rr}1 & 2 & 1 \\ 2&4&2 \\2 & 4 &5 \end{array}\right]$$ has no inverse as det A = 0 in this case. ### Are All 3×3 Matrices Invertible? No, all 3×3 matrices are not invertible as a matrix cannot have its inverse when its determinant is 0. For example, A = $$\left[\begin{array}{rr}0 & 0 & 0 \\ -1&3&2 \\5 & 7 &5 \end{array}\right]$$ is not invertible as det A = 0 in this case. ### What is the Inverse of 3×3 Matrix Formula? If A is a 3×3 matrix, its inverse formula is A-1 = (adj A)/(det A). Here, • det A = Determinant of the matrix A ### Does a 3×3 Matrix have an Inverse? A 3×3 matrix has inverse only if its determinant is not zero. If the determinant is zero, then the matrix has is not invertible (does not have inverse) and in that case, it is called a singular matrix. How to Find Inverse of a 3×3 Matrix Using Elementary Row Operations? For finding the inverse of a 3×3 matrix (A ) by elementary row operations, • Write A and I (identity matrix of same order) in a single matrix separating them by a vertical dotted line. • Apply elementary row operations so that the left side matrix becomes I. • The matrix that comes on the right side is A-1. visual curriculum You are watching: Expert Maths Tutoring in the UK. Info created by THVinhTuy selection and synthesis along with other related topics. Rate this post
Games Problems Go Pro! "Hi Professor Puzzler, in last week's post you said that if a number is irrational, it doesn't terminate or repeat. I wonder how you know that? Of all the irrational numbers, how do you know there isn't one that terminates or repeats?" This question comes from Trevor in Georgia. Hi Trevor, in answer to your question, I'll give you a short proof (with a bit of hand-waving over some details, just to keep the write-up from being too long!). What we'll do is, we'll start with numbers written in decimal notation, which we'll assume are irrational (which means, as you remember from the post about the square root of two, that it can't be written as a ratio of integers) and then we'll arrive at a contradiction, which will prove that they aren't irrational. So let's get started! ## Terminating Decimal Let n be an irrational number with a terminating decimal. This means that after a certain number of decimal places (let's call that k), the decimal ends. Since the decimal terminates after k decimal places, the number 10kn is an integer (every time you multiply by 10, you shift the decimal one place to the right, so multiplying n by 10k results in all the digits being to the left of the decimal). Now we can write n = (10kn)/(10k), which is ratio of two integers, and that contradicts our original assumption that n was irrational. Thus, if a number in decimal form terminates, it is not irrational. ## Repeating Decimal Let n be an irrational number with a repeating decimal. This means that after a certain number of decimal places (let's call that k), the decimal begins repeating every h digits, where h is some integer. For example, if the number is 3.125727272..., k = 3, h = 2. To simplify our math, let's multiply our number by 10k, which allows us to get the non-repeating part of the decimal over to the left of the decimal. We understand that when we're done with our manipulations, we have to divide by 10k to get our number back, but if we've proven that our result is rational, then dividing by 10k still gives us a rational number. Similarly, everything to the left of the decimal can then be subtracted, with the understanding that when we're done we must add it back in. But the same argument holds; if we've proved the result is rational, adding an integer to that rational number will result in another rational number. In other words, we can, without loss of generality, simplify our starting point to a decimal which looks like this: n = 0.a0a1...aha0a1...aha0a1...ah... Let m be the integer formed by the following digits: a0a1...ah Then n = m/(10h) + m/(102h) + m/(103h) + ... This is an infinite series with common ratio 1/(10h), and its sum is: S = [m/(10h)]/[1 - 1/(10h)] S = m/(10h - 1) Since m is an integer, and 10h - 1 is an integer, we've shown that our number can be expressed as a ratio of integers, which contradicts our assumption that it is irrational. From this we've shown that if a decimal repeats or terminates, it is not irrational.
# Class 8 RD Sharma Solutions- Chapter 6 Algebraic Expressions And Identities – Exercise 6.2 ### (i) 3a2b, -4a2b, 9a2b Solution: 3a2b, -4a2b, 9a2b Now we have add the given expression = 3a2b + (-4a2b) + 9a2b = 3a2b – 4a2b + 9a2b = 8a2b ### (ii) 2/3a, 3/5a, -6/5a Solution: We have to add the given expression 2/3a + 3/5a + (-6/5a) 2/3a + 3/5a – 6/5a Now take LCM for 3 and 5 which will be 15 = (2×5)/(3×5)a + (3×3)/(5×3)a – (6×3)/(5×3)a = 10/15a + 9/15a – 18/15a = (10a+9a-18a)/15 = a/15 ### (iii) 4xy2 – 7x2y, 12x2y -6xy2, -3x2y + 5xy2 Solution: We have to add the given expression 4xy2 – 7x2y + 12x2y – 6xy2 – 3x2y + 5xy2 Now rearrange the expression: 12x2y – 3x2y – 7x2y – 6xy2 + 5xy2 + 4xy 3xy2 + 2x2y ### (iv) 3/2a – 5/4b + 2/5c, 2/3a – 7/2b + 7/2c, 5/3a + 5/2b – 5/4c Solution: 3/2a – 5/4b + 2/5c, 2/3a – 7/2b + 7/2c, 5/3a + 5/2b – 5/4c Now add the given expression 3/2a – 5/4b + 2/5c + 2/3a – 7/2b + 7/2c + 5/3a + 5/2b – 5/4c rearrange 3/2a + 2/3a + 5/3a – 5/4b – 7/2b + 5/2b + 2/5c + 7/2c – 5/4c Now take LCM of (2 and 3 is 6), (4 and 2 is 4), (5,2 and 4 is 20) (9a+4a+10a)/6 + (-5b-14b+10b)/4 + (8c+70c-25c)/20 23a/6 – 9b/4 + 53c/20 ### (v) 11/2xy + 12/5y + 13/7x, -11/2y – 12/5x – 137xy Solution: 11/2xy + 12/5y + 13/7x, -11/2y – 12/5x – 13/7xy Now add the given expression 11/2xy + 12/5y + 13/7x + -11/2y – 12/5x – 13/7xy Now rearrange 11/2xy – 13/7xy + 13/7x – 12/5x + 12/5y -11/2y Now take LCM for (2 and 7 is 14), (7 and 5 is 35), (5 and 2 is 10) (11xy-12xy)/14 + (65x-84x)/35 + (24y-55y)/10 51xy/14 – 19x/35 – 31y/10 ### (vi) 7/2x3 – 1/2x3 + 5/3, 3/2x3 + 7/4x2 – x + 1/3, 3/2x2 -5/2x -2 Solution: Now add the given expression 7/2x3 – 1/2x2 + 5/3 + 3/2x3 + 7/4x2 – x + 1/3 + 3/2x2 -5/2x – 2 Now rearrange =7/2x3 + 3/2x3 – 1/2x2 + 7/4x2 + 3/2x2 – x – 5/2x + 5/3 + 1/3 – 2 =10/2x3 + 11/4x2 – 7/2x + 0/6 =5x3 + 11/4x2 -7/2x ### (i) -5xy from 12xy Solution: Subtract the given expression = 12xy – (- 5xy) = 5xy + 12xy = 17xy ### (ii) 2a2 from -7a2 Solution: Subtract the given expression = (-7a2) – 2a2 = -7a2 – 2a2 = -9a2 ### (iii) 2a-b from 3a-5b Solution: Subtract the given expression =(3a – 5b) – (2a – b) = 3a – 5b – 2a + b = a – 4b ### (iv) 2x3 – 4x2 + 3x + 5 from 4x3 + x2 + x + 6 Solution: Subtract the given expression (4x3 + x2 + x + 6) – (2x3 – 4x2 + 3x + 5) 4x3 + x2 + x + 6 – 2x3 + 4x2 – 3x – 5 2x3 + 5x2 – 2x + 1 ### (v) 2/3y3 – 2/7y2 – 5 from 1/3y3 + 5/7y2 + y – 2 Solution: Subtract the given expression 1/3y3 + 5/7y2 + y – 2 – 2/3y3 + 2/7y2 + 5 On rearranging, 1/3y3 – 2/3y3 + 5/7y2 + 2/7y2 + y – 2 + 5 We will group similar expression: = -1/3y3 + 7/7y2 + y + 3 = -1/3y3 + y2 + y + 3 ### (vi) 3/2x – 5/4y – 7/2z from 2/3x + 3/2y – 4/3z Solution: Subtract the given expression 2/3x + 3/2y – 4/3z – (3/2x – 5/4y – 7/2z) On rearranging, 2/3x – 3/2x + 3/2y + 5/4y – 4/3z + 7/2z We will group similar expression: LCM of (3 and 2 is 6), (2 and 4 is 4), (3 and 2 is 6) =(4x-9x)/6 + (6y+5y)/4 + (-8z+21z)/6 = -5x/6 + 11y/4 + 13z/6 ### (vii) x2y – 4/5xy2 + 4/3xy from 2/3x2y + 3/2xy2 – 1/3xy Solution: Subtract the given expression 2/3x2y + 3/2xy2 – 1/3xy – (x2y – 4/5xy2 + 4/3xy) on rearrange 2/3x2y – x2y + 3/2xy2 + 4/5xy2 – 1/3xy – 4/3xy We will group similar expression: LCM of (3 and 1 is 3), (2 and 5 is 10), (3 and 3 is 3) -1/3x2y + 23/10xy2 – 5/3xy ### (viii) ab/7 – 35/3bc + 6/5ac from 3/5bc – 4/5ac Solution: Subtract the given expression 3/5bc – 4/5ac – (ab/7 – 35/3bc + 6/5ac) On rearrange 3/5bc + 35/3bc – 4/5ac – 6/5ac – ab/7 We will group similar expression: LCM of (5 and 3 is 15), (5 and 5 is 5) (9bc+175bc)/15 + (-4ac-6ac)/5 – ab/7 184bc/15 + -10ac/5 – ab/7 – ab/7 + 184bc/15 – 2ac ### (i) 6/5x2 – 4/5x3 + 5/6 + 3/2x from x3/3 – 5/2x2 + 3/5x + 1/4 Solution: Subtract the given expression 1/3x3 – 5/2x2 + 3/5x + 1/4 – (6/5x2 – 4/5x3 + 5/6 + 3/2x) On rearrange 1/3x3 + 4/5x3 – 5/2x2 – 6/5x2 + 3/5x – 3/2x + 1/4 – 5/6 By grouping similar expressions we get, LCM of (3 and 5 is 15), (2 and 5 is 10), (5 and 2 is 10), (4 and 6 is 24) 17/15x3 – 37/10x2 – 9/10x – 14/24 17/15x3 – 37/10x2 – 9/10x – 7/12 ### (ii) 5a2/2 + 3a3/2 + a/3 – 6/5 from 1/3a3 – 3/4a2 – 5/2 Solution: Subtract the given expression 1/3a3 – 3/4a2 – 5/2 – (5/2a2 + 3/2a3 + a/3 – 6/5) On rearrange 1/3a5 – 3/2a3 – 3/4a2 – 5/2a2 – a/3 – 5/2 + 6/5 By grouping similar expressions we get, LCM of (3 and 2 is 6), (4 and 2 is 4), (2 and 5 is 10) = (2a3 – 9a3)/6 – (3a2 + 10a2)/4 – a/3 + (-25+12)/10 = -7/6a3 – 13/4a2 – a/3 – 13/10 ### (iii) 7/4x3 + 3/5x2 + 1/2x + 9/2 from 7/2 – x/3 – x2/5 Solution: Subtract the given expression 7/2 – x/3 – 1/5x2 – (7/4x3 + 3/5x2 + 1/2x + 9/2) On rearranging, -7/4x3 – 1/5x2 – 3/5x2 – x/3 – x/2 + 7/2 – 9/2 By grouping similar expressions we get, LCM of (3 and 2 is 6) -7/4x3 – 4/5x2– (2x-3x)/6 + (7-9)/2 -7/4x3 – 4/5x2 – 5/6x – 1 ### (iv) y3/3 + 7/3y2 + 1/2y + 1/2 from 1/3 – 5/3y2 Solution: Subtract the given expression 1/3 – 5/3y2 – (1/3y2 + 7/3y2 + 1/2y + 1/2) On rearrange -1/3y3 – 5/3y2 – 7/3y2 – 1/2y + 1/3 – 1/2 By grouping similar expressions we get, LCM of (3 and 3 is 3), (3 and 2 is 6) -1/3y3 + (-5y2 – 7y2)/3 – 1/2y + (2-3)/6 -1/3y3 – 12/3y2 – 1/2y – 1/6 ### (v) 2/3ac – 5/7ab + 2/3bc from 3/2ab -7/4ac – 5/6bc Solution: Subtract the given expression 3/2ab – 7/4ac – 5/6bc – (2/3ac – 5/7ab + 2/3bc) On rearrange 3/2ab + 5/7ab – 7/4ac – 2/3ac – 5/6bc – 2/3bc By grouping similar expressions we get, LCM of (2 and 7 is 14), (4 and 3 is 12), (6 and 3 is 6) (21ab+10ab)/14 – (21ac-8ac)/12 – (5bc-4bc)/6 31/14ab – 29/12ac – 3/2bc ### Question 4. Subtract 3x – 4y – 7z from the sum of x – 3y + 2z and -4x + 9y – 11z. Solution: First we will find the sum: The sum of x – 3y + 2z and -4x + 9y – 11z is (x – 3y + 2z) + (-4x + 9y – 11z) On rearrange x – 4x – 3y + 9y + 2z – 11z = -3x + 6y – 9z Now Let’s subtract it from -3x + 6y – 9z (-3x + 6y – 9z) – (3x – 4y – 7z) On rearranging again = -3x – 3x + 6y + 4y – 9z + 7z = -6x + 10y – 2z ### Question 5. Subtract the sum of 3l – 4m – 7n2 and 2l + 3m – 4n2 from the sum of 9l + 2m – 3n2 and -3l + m + 4n2. Solution: Sum of 3l – 4m – 7n2 and 2l + 5m – 4n2 3l – 4m – 7n2 + 2l + 3m – 4n2 On rearrange 3l + 2l – 4m + 3m – 7n2 – 4n2 5l – m – 11n2 ……………………..eq. (1) Sum of 9l + 2m – 3n2 and -3l + m + 4n2 9l + 2m – 3n2 + (-3l + m + 4n2) On rearrange 9l – 3l + 2m + m – 3n2 + 4n2 6l + 3m + n2 ……………………….eq. (2) Let us subtract equ (i) from (ii), we get 6l + 3m + n2 – (5l – m – 11n2) On rearrange 6l – 5l + 3m + m + n2 + 11n2 l + 4m + 12n2 ### Question 6. Subtract the sum of 2x – x2 + 5 and -4x – 3 + 7x2 from 5. Solution: Sum of 2x – x2 + 5 and -4x – 3 + 7x2 is 2x – x2 + 5 + (-4x – 3 + 7x2) 2x – x2 + 5 – 4x – 3 + 7x2 On rearrange – x2 + 7x2 + 2x – 4x + 5 – 3 6x2 -2x + 2 …………eq (i) Let subtract eq (i) from 5 we will get, 5 – (6x2 -2x + 2) 5 – 6x2 + 2x – 2 3 + 2x – 6x2 ### (i) x2 – 3x + 5 – 1/2(3x2 – 5x + 7) Solution: x2 – 3x + 5 – 1/2(3x2 – 5x + 7) On rearrange x2 – 3/2x2 – 3x + 5/2x + 5 – 7/2 We will group similar expression: LCM of (1 and 2 is 2) = (2x2 – 3x2)/2 – (6x + 5x)/2 + (10-7)/2 = -1/2x2 – 1/2x + 3/2 ### (ii) [5 – 3x + 2y – (2x – y)] – (3x – 7y + 9) Solution: 5 – 3x + 2y – 2x + y – 3x + 7y – 9 On rearrange = – 3x – 2x – 3x + 2y + y + 7y + 5 – 9 We will group similar expression: = -8x + 10y – 4 ### (iii) 11/2x2y – 9/4xy2 + 1/4xy – 1/14y2x + 1/15yx2 + 1/2xy Solution: On rearrange 11/2x2y + 1/15x2y – 9/4xy2 – 1/14xy2 + 1/4xy + 1/2xy We will group similar expression: LCM of (2 and 15 is 30), (4 and 14 is 56), (4 and 2 is 4) = (165x2y + 2x2y)/30 + (-126xy2 – 4xy2)/56 + (xy + 2xy)/4 = 167/30x2y – 130/56xy2 + 3/4xy = 167/30x2y – 65/28xy2 + 3/4xy ### (iv) (1/3y2 – 4/7y + 11) – (1/7y – 3 + 2y2) – (2/7y – 2/3y2 + 2) Solution: On rearrange 1/3y2 – 2y2 – 2/3y2 – 4/7y – 1/7y – 2/7y + 11 + 3 – 2 We will group similar expression: LCM of (3, 1 and 3 is 3), (7, 7 and 7 is 7) = (y2 – 6y2 + 2y2)/3 – (4y – y – 2y)/7 + 12 = -3/3y2 – 7/7y + 12 = -y2 – y + 12 ### (v) -1/2a2b2c + 1/3ab2c – 1/4abc2 – 1/5cb2a2 + 1/6cb2a – 1/7c2ab + 1/8ca2b Solution: On rearrange -1/2a2b2c – 1/5a2b2c + 1/3ab2c + 1/6ab2c – 1/4abc2 – 1/7abc2 + 1/8a2bc We will group similar expression: LCM of (2 and 5 is 10), (3 and 6 is 6), (4 and 7 is 28) -7/10a2b2c + 1/2ab2c – 11/28abc2 + 1/8a2bc Previous Next
# Writing Equations ## Algebra: Writing Equations Read the following SAT test question and then select the correct answer. Approach all math questions the same way. Read the question carefully to avoid making careless mistakes. Identify the bottom line, the question you must solve, and note it on your test. Then assess your options and choose the most efficient method to attack the problem. When you have an answer, loop back to verify that the answer addresses the bottom line. First, 3 is subtracted from x and the square root of the difference is taken. Then, 5 is added to the result, giving a final result of 9. What is the value of x? Bottom line: x = ? Assess your options: You could try to plug in answer choices and see which one equals 9, but you may have to write and solve the equation multiple times. Instead, translate the two sentences into “math” and use algebra to find x. Attack the problem: Work through the words step by step. First, 3 is subtracted from x. Write: x – 3 The square root of the difference is taken. That means both numbers involved in the difference are under the radical. $\sqrt{x-3}$ Then 5 is added and the final result is 9. $\sqrt{x-3}\, +5=9$ Now that you have your equation written, all you have to do is solve for x: $\sqrt{x-3}\, +5=9$ (subtract 5 from each side) $\sqrt{x-3}\, =4$ (square each side to remove the radical) $x - 3= 16$ $x = 19$ Loop Back: You solved for your bottom line, so look down at the answer choices. (A) 3 (B) 4 (C) 5 (D) 16 (E) 19
# Section 9.3, Average and Instantaneous Rates of Change: The Derivative 1 ```Section 9.3, Average and Instantaneous Rates of Change: The Derivative 1 Average Rate of Change The average rate of change of a function y = f (x) from x = a to x = b is: f (b) − f (a) . b−a Note that this equals the slope of the line connecting the points (a, f (a)) and (b, f (b)). Example Find the average rate of change of f (x) = x2 − 2x + 4 on the interval [1, 3]. 7−3 f (3) − f (1) = =2 3−1 2 2 Derivative If y = f (x) is a function, then the derivative of f (x) at any value of x, denoted f 0 (x), is f (x + h) − f (x) , h→0 h f 0 (x) = lim if this limit exists. If f 0 (c) exists, we say that f is differentiable at c. If y = f (x), alternative notation for f 0 (x) includes y 0 , dy d dx , dx f (x), Dx y, and Dx f (x). The derivative has a variety of interpretations. First, f 0 (c) is the instantaneous rate of change of the function f at x = c. Secondly, f 0 (c) is the slope of the line tangent to the graph of y = f (x) at x = c. We will explore other uses of the derivative later this semester. Examples 1. Find the instantaneous rate of change of f (x) = 2x − 4 at x = −1. We need to find f 0 (−1). From the definition of the derivative, with x = −1, f 0 (−1) = lim h→0 f (−1 + h) − f (−1) 2(−1 + h) − 4 − (2(−1) − 4) 2h = lim = lim =2 h→0 h→0 h h h 2. Find the derivative of f (x) = x2 − x + 3. f (x + h) − f (x) (x + h)2 − (x + h) + 3 − (x2 − x + 3) = lim h→0 h→0 h h x2 + 2xh + h2 − x − h + 3 − x2 + x − 3 2xh + h2 − h = lim = lim h→0 h→0 h h = lim (2x + h − 1) = 2x − 1 f 0 (x) = lim h→0 3. Let f (x) = x2 − x + 3 again. Find the slope of the line tangent to the graph of y = f (x) at x = −3. This question is asking us to find f 0 (−3). Since we already calculated the derivative, we know that f 0 (x) = 2x − 1, so: f 0 (−3) = 2(−3) − 1 = −7 We will learn “shortcuts” to calculate derivatives later, but for this section, make sure that you use the definition of the derivative for your calculations. 3 Application: Velocity The velocity of an object is the rate at which its position is changing with respect to time. Example The height of a ball thrown upward at a speed of 30 ft/s from a height of 15 feet after t seconds is given by: S(t) = 15 + 30t − 16t2 Find the average velocity of the ball in the first 2 seconds after it is thrown. We are asked for the average rate of change of the position of the ball between times 0 and 2. Using our formula for average rate of change, S(2) − S(0) 11 − 15 = = −2 ft/s 2−0 2 ```
# Basic Algebra/Working with Numbers/Dividing Rational Numbers ## LessonEdit Dividing rational numbers. Dividing rational numbers covers a general area of equations. For an equation that is such has only to have a numerator and denominator that are both rational numbers. In turn, one will come out with a quotient that fits the terms applied to a "rational number". Given the fact that you already understand rational numbers, you will understand this unit. If, on the other hand, you have no clue what a rational number is, then you should do some research concerning this subject so that you can understand the explanation of dividing such numbers that follows this text. Anyway, dividing rational numbers, sometimes worded "quotients of rational expressions", is simply dividing a rational number by a rational number. For instance, look at the example problems, dividing rational numbers is very easy. If you have a fraction dividing another fraction then you simply flip the dividend and, by multiplying, one will come out with exactly the same number. The knowledge of expressing how this works is beyond the scope of this lesson. But, it works every time. You are still dividing, but you have switched your means of doing so. When you come to more complicated problems that have unknown variables the same method works. So if you have a fraction of 7 over 5 divided by 3 over 4, you will simply flip the 3 over 4 and multiply the fractions instead of dividing. This is a method that will be used again and again in math, so know it well. Look at the examples given and, although this is easy, make sure you know it. ## Example ProblemsEdit Example 1 $\frac{2}{7} \div \frac{14}{16}$ $\frac{2}{7} \times \frac{16}{14}$ (Change the division to multiplication and flip the fraction on the right.) $\frac{1}{7} \times \frac{16}{7}$ (Reduce fractions with any common factors on top and on bottom.) $\frac{1 \times 16}{7 \times 7}$ (Multiply the tops and bottoms together.) $\frac{16}{49}$ (Simplify) $\frac{14}{2} \div \frac{14}{2}$ $\frac{14 \times 2}{2 \times 14} =$ $\frac{28}{28} = 1$ $\frac{31}{2} = 15.5$ $\frac{77}{9} = 8.\overline{55}$ $\frac{12}{6} = 2$ $\frac{36}{8}= 4.5$ $\frac{55}{3}= 18.\overline{33}$ ## Practice GamesEdit put links here to games that reinforce these skills ## Practice ProblemsEdit (Note: put answer in parentheses after each problem you write) $\frac{59}{7} =$ (8.428571) $\frac{46}{5} =$ (9.2) $\frac{97}{4} =$ (24.25) $\frac{73}{4} =$ (18.25)
The fraction calculator will certainly add, subtract, multiply and also divide fountain with choose or uneven denominators. It will certainly also allow us to leveling fractions, transform fractions come decimals and also decimals come fractions. You are watching: 6 divided by 4 in fraction form First, merely input the worths a,b,c,d for the fountain $$\fracab$$ and also $$\fraccd$$, climate the mathematical operation you great to do (+, -, x, /). The calculator will instantly and also accurately carry out the operation and also give the answer in the most basic form. Girlfriend can additionally use the calculator to inspect your occupational that you’ve done manually. ### Like (Common) Denominators Add or subtract the numerators and keep the platform the same. Ex: $$\frac35 + \frac45$$ Since the denominator is 5 in both fractions, add 3 and also 4 to obtain 7. The denominator stays 5, so the price is 7/5. $$\frac76 – \frac56$$ Since the denominator is 6 in both fractions, subtract 5 native 7 to acquire 2. The fraction is then $$\frac26$$. But currently we deserve to simplify $$\frac26$$. Come simplify, look for a common factor. An alert that 2 divides evenly into both 2 and 6. Therefore, division both numerator and also denominator through 2 to gain $$\frac13$$. The fraction is currently simplified. ### Unlike Denominators To include and subtract unequal denominators, first calculate the usual denominator. The easiest method to carry out that is to main point the two denominators. This wont constantly give the lowest usual denominator, yet you can simplify after adding and subtracting. Ex: $$\frac25 + \frac47$$ A typical denominator is 5(7) = 35. Due to the fact that the denominator in the first fraction is multiply by 7, the numerator must likewise be multiplied by 7 to acquire $$\frac1435$$. Since the denominator in the second fraction is multiplied by 5, the numerator should be as well to gain $$\frac2035$$. Now include $$\frac1435+\frac2035=\frac3435$$ Subtraction is done the same way, simply subtract the 2 fractions after ~ rewriting the fractions with their common denominators. If you must simplify, mental to divide by the greatest usual factor.Adding and also Subtracting fractions Video ## Multiplying and also Dividing Fractions When multiply fractions, simply multiply across the molecule and across the denominators. Climate simplify. You can additionally simplify first before multiplying. Ex: $$\frac29\times\frac47$$ Multiply 2 and also 4 to acquire 8. Climate multiply 9 and also 7 to get 63. The an outcome is $$\frac863$$. Over there is no leveling needed because the greatest usual factor is 1. Now expect we great to divide $$\frac29 \div \frac47$$. When splitting fractions, take the an initial fraction and also multiply through the mutual of the second. The mutual is simply interchanging the numerator and also denominator. The department problem turns right into a multiplication problem. $$\frac29 \times \frac74$$ 2 × 7 = 14 and also 9 × 4 = 36. So the prize is $$\frac1436$$. But an alert this isn’t in most basic form. The greatest common factor is 2, so splitting both by 2 gives the simplified answer that $$\frac718$$.Multiplying and also Dividing fractions Video ## Converting fountain to Decimals The convert portion to decimal calculator will take any portion and change it come a decimal. The method to change a fraction to a decimal is fairly simple. Just divide the numerator by the denominator. Change $$\frac1425$$ to a decimal. Divide 14 by 25 to get 0.56. You can do this on a calculator or manually using lengthy division. Some fractions room not as easy to work by hand, an especially those that room non-terminating. Those room much easier to job-related on this calculator. But if you choose to resolve manually, the calculator provides a great tool to instantly examine your work.Converting fountain to decimal Video ## Converting decimal to Fractions Changing decimals to fractions is the train station of an altering fractions to decimals. The calculator will perform this promptly with precise results by merely entering the decimal value. To convert manually, take it the decimal and also convert to a entirety number, then divide by 10 raised to the variety of decimal locations moved to the appropriate to transform the number. From there you deserve to simplify the portion if needed. Ex: Convert 0.68 come a fraction. To readjust 0.68 to a totality number, move the decimal allude 2 places ot the ideal to get 68. Since we moved 2 decimal places, divide 68 by 10 elevated to the second power, i m sorry is 100. See more: How Long Is Yogurt Good Once Opened ? How Long Does Yogurt Last Once Opened That offers us $$\frac68100$$. Now we can simplify the fraction by looking for a typical factor. If friend don’t recognize the greatest common factor you deserve to start by separating by any common factor. An alert 68 and also 100 space both divisible through 2. This reduce the fraction to 34/50. Indigenous here, notice that both 34 and also 50 space divisible by 2. This reduces come $$\frac1725$$, i beg your pardon is the simplified answer. You can examine your hands-on calculations making use of this calculator or simply input the information for your certain problem for nearly instant, specific results!
mathleaks.com mathleaks.com Start chapters home Start History history History expand_more Community Community expand_more {{ filterOption.label }} {{ item.displayTitle }} {{ item.subject.displayTitle }} arrow_forward {{ searchError }} search {{ courseTrack.displayTitle }} {{ statistics.percent }}% Sign in to view progress {{ printedBook.courseTrack.name }} {{ printedBook.name }} # Manipulating Radical Functions ## Combining Radical Functions Radical functions can be combined, through addition, subtraction and multiplication. Depending on the original functions and which operations are used, the resulting function can have similar or new characteristics. fullscreen Exercise Consider the functions. Find h(x) if h(x)=f(x)g(x). Show Solution Solution When functions containing one or more radicals are multiplied, it is often helpful to rewrite the expressions using rational exponents. h(x)=f(x)g(x) Next, the properties of exponents can be used to simplify the product. Now that the function h(x) has been simplified completely, we can rewrite it with radical. ## Inverse Functions Inverse functions are two functions that undo each other. The functions f(x) and g(x) are inverses of each other if When a function is denoted f(x), its inverse is often referred to as ## Inverse Functions to Radical Functions Radicals can be eliminated by raising them to the same power as the index of the radical. Therefore, the inverse function to a radical function is typically a power function. The radical function has the inverse function This can be proven by showing Consider I. ### Rule Next, consider II. ### Rule It has been shown that and are inverses of each other. This is true for all values of n. fullscreen Exercise Determine if the two functions are inverses of each other. Show Solution Solution The two functions, f(x) and g(x), are inverses if Thus, if we can show that 1. f(g(x))=x, and 2. g(f(x))=x we can conclude that f and g are inverses. We'll begin with f(g(x)). f(g(x))=1+(x1) f(g(x))=1+x1 f(g(x))=x Now that we've satisfied 1, we'll attempt to satisfy 2. g(x)=(x1)3 g(f(x))=x Since both f and g are inverses of each other. fullscreen Exercise The graph of the function is shown. Use the graph to show that f(x) is invertible. Then, determine the inverse, and graph both functions. Show Solution Solution Graphically, it can be shown that a function is invertible by performing the horizontal line test. If we move an imaginary horizontal line across the graph of a function, and the line intersects the graph more than once anywhere, the function is not invertible. Here, we'll draw more than one line to be sure. The horizontal lines in the graph both intersect the function once. If we place a line anywhere else it still intersects the graph at the most once. Therefore, the function is invertible. Next, we can find the inverse, algebraically. To begin, replace f(x) with y in the given rule. Next step is to switch x and y in the function rule. Now we need to solve for y. The resulting equation will be the inverse of the given function. Thus, the inverse is We'll graph f and in the same coordinate plane. The domain of f(x) is restricted to x0. As a consequence, the domain of must be restricted to the same interval, even though is defined for all values of x. {{ 'mldesktop-placeholder-grade-tab' \| message }} {{ 'mldesktop-placeholder-grade' \| message }} {{ article.displayTitle }}!
# OpenStax_CNX You are here: Home » Content » Theorems on moment of inertia (Application) ### Recently Viewed This feature requires Javascript to be enabled. # Theorems on moment of inertia (Application) Module by: Sunil Kumar Singh. E-mail the author Summary: Solving problems is an essential part of the understanding process. Questions and their answers are presented here in the module text format as if it were an extension of the theoretical treatment of the topic. The idea is to provide a verbose explanation of the solution, detailing the application of theory. Solution presented here, therefore, is treated as the part of the understanding process – not merely a Q/A session. The emphasis is to enforce ideas and concepts, which can not be completely absorbed unless they are put to real time situation. ## Representative problems and their solutions We discuss problems, which highlight certain aspects of the application of theorem on moment of inertia. For this reason, questions are categorized in terms of the characterizing features pertaining to the questions : • Theorems on moment of inertia • Nature of MI • Part of a rigid body • Geometric shapes formed from wire ### Theorems on moment of inertia Example 1 Problem : A uniform circular disk of radius "R" lies in "xy" - plane such that its center is at the origin of coordinate system as shown in the figure. Its MI about the axis defined as x = R and y = 0 is equal to its MI about axis defined as y = d and z = 0. Then, find “d”. Solution : First, we need to visualize the two axes as defined. Then, we find disk's MI about them. Finally, we equate the MIs as given in the question to find "d". The coordinates x = R and y = 0 define the axis in "xz" - plane as "y" coordinate is zero. Further, the axis intersects x-axis at x = R = a constant. This axis (z'), therefore, is parallel to z-axis as shown in the figure. Note that z axis passes through center of mass. Thus by applying theorem of parallel axes, the MI of the disk about z' by the theorem of parallel axes is : I z’ = I z + M R 2 I z’ = I z + M R 2 I z’ = M R 2 2 + M R 2 = 3 M R 2 2 I z’ = M R 2 2 + M R 2 = 3 M R 2 2 The coordinates y = d and z = 0 define the axis in "xy" - plane as "z" coordinate is zero. Further, this axis intersects y-axis at y = d = a constant. This axis (x'), therefore, is parallel to x-axis as shown in the figure. The MI of the disk about x' by the theorem of parallel axes : I x’ = I x + M R 2 I x’ = I x + M R 2 I x’ = M R 2 4 + M d 2 I x’ = M R 2 4 + M d 2 According to the question, I x’ = I z’ I x’ = I z’ M R 2 4 + M d 2 = 3 M R 2 2 M R 2 4 + M d 2 = 3 M R 2 2 M d 2 = 3 M R 2 2 - M R 2 4 = 5 M R 2 4 M d 2 = 3 M R 2 2 - M R 2 4 = 5 M R 2 4 d = 5 R 2 d = 5 R 2 Example 2 Problem : Two identical uniform rods, each of mass "M" and length "L", are joined together to form a cross. Find the MI of the cross about the angle bisector "AA'" as shown in the figure. Solution : MI of each of the rod about perpendicular axis, which passes through COM (O) is : I rod = M L 2 12 I rod = M L 2 12 The MI of two rods joined together as a cross about perpendicular axis through the common point is : I cross = 2 I rod = 2 x M L 2 12 = M L 2 6 I cross = 2 I rod = 2 x M L 2 12 = M L 2 6 The cross can be treated as a planar object. As such, we can apply theorem of perpendicular axes to find MI about the bisector of the right angle. We consider another axis BB' perpendicular to AA' in the plane of cross. Then, according to theorem of perpendicular axes, I cross = I AA’ + I BB’ I cross = I AA’ + I BB’ The MIs about the two bisectors are equal by the symmetry of the cross about them. Hence, I AA’ = I BB’ I AA’ = I BB’ Thus, I cross = I AA’ + I BB’ = 2 I AA’ I cross = I AA’ + I BB’ = 2 I AA’ I AA’ = I cross 2 = M L 2 12 I AA’ = I cross 2 = M L 2 12 ### Nature of MI Example 3 Problem : The MI of a solid sphere is calculated about an axis parallel to one of its diameter, at a distance "x" from the origin. The values are plotted against distance. Which of the four plots, as shown below, correctly describes the variation of MI with respect to "x" for the solid sphere. Solution : It is evident that farther the axis from the origin, greater is MI about that axis. We can, thus, conclude that the plot between MI and the distance should be increasing one with increasing distance ("x"). Thus, options (a) and (d) are incorrect. Now, the MI of solid sphere about an axis parallel to one of diameters, at a distance "x" from the origin is given as : I = I O + M x 2 I = I O + M x 2 where I O I O is MI about one of the diameters and is a constant for the given solid sphere. Thus, the equation above takes the form of the equation of parabola, y = m x 2 + c y = m x 2 + c The correct plot, therefore, is an increasing parabola as shown in Figure (c). ### Part of a rigid body Example 4 Problem : Four holes of radius "L/4" are made in a thin square plate of mass "M" and side "L" in "xy" - plane, as shown in the figure. Find its MI about z – axis. Solution : MI is a scalar quantity. It means that MI has no directional attribute. Thus, we can conclude that MI of the complete plate is equal to the sum of MIs of remaining plate and that of the four circular disks taken out from the plate about z-axis. Hence, MI of remaining plate is obtained as : I (MI of remaining plate) = Iz (MI of complete plate) - 4 x Icz (MI of circular disks) I (MI of remaining plate) = Iz (MI of complete plate) - 4 x Icz (MI of circular disks) Now, according to theorem of perpendicular axes, the MI of complete square about z-axis is two times its MI about x-axis : I z = 2 x M L 2 12 = M L 2 6 I z = 2 x M L 2 12 = M L 2 6 Further, according to theorem of parallel axes, the MI of circular disk of mass “m” and radius “L/4” about the same z-axis is : I cz = m ( L 4 ) 2 + m ( 2 L 4 ) 2 I cz = m ( L 4 ) 2 + m ( 2 L 4 ) 2 I cz = m L 2 16 + 2 m L 2 16 = 3 m L 2 16 I cz = m L 2 16 + 2 m L 2 16 = 3 m L 2 16 The mass of the disk "m" is given by multiplying areal density with the area of the circular disk, m = M L 2 x π ( L 4 ) 2 = π M 16 m = M L 2 x π ( L 4 ) 2 = π M 16 Substituting for "m", we have : I cz = 3 m L 2 16 = 3 π M L 2 16 x 16 I cz = 3 m L 2 16 = 3 π M L 2 16 x 16 Thus, the required MI is : I = M L 2 6 - 4 x 3 π M L 2 16 x 16 = 16 x 16 M L 2 - 12 x 6 π M L 2 6 x 16 x 16 I = M L 2 6 - 4 x 3 π M L 2 16 x 16 = 16 x 16 M L 2 - 12 x 6 π M L 2 6 x 16 x 16 I = 32 M L 2 - 9 π M L 2 192 I = 32 M L 2 - 9 π M L 2 192 ### Geometric shapes formed from wire Example 5 Problem : A thin wire of length "L" and uniform linear mass density "λ" is bent into a circular loop. Find the MI of the loop about an axis AA' as shown in the figure. Solution : The MI of the circular loop about a tangential axis AA' is found out by the theorem of parallel axis : I AA’ = I XX’ + M R 2 I AA’ = I XX’ + M R 2 where IXX' is the MI about an axis which is one of the diameters of the ring and is parallel to tangential axis AA'. Now, MI about the diameter can be calculated applying theorem of perpendicular axes : I XX’ = I O 2 I XX’ = I O 2 where I O I O is MI about an axis perpendicular to the surface of loop and passing through center of mass (ZZ' axis). It is given by the expression : I O = M R 2 I O = M R 2 Combining two equations, we have : I XX’ = M R 2 2 I XX’ = M R 2 2 Putting this in the expression of I AA' I AA' , we have : I AA’ = M R 2 2 + M R 2 = 3 M R 2 2 I AA’ = M R 2 2 + M R 2 = 3 M R 2 2 Now, we are required to find mass and radius from the given data. Here, M = λ L M = λ L Since the wire is bent into a circle, the perimeter of the circle is equal to the length of the wire. Hence, 2 π R = L R = L 2 π 2 π R = L R = L 2 π Putting these values in the expression of MI about AA' axis is : I AA’ = 3 M R 2 2 = 3 λ L x L 2 2 ( 2 π ) 2 = 3 λ L 3 8 π 2 I AA’ = 3 M R 2 2 = 3 λ L x L 2 2 ( 2 π ) 2 = 3 λ L 3 8 π 2 Example 6 Problem : A uniform wire of mass "M" and length "L" is bent into a regular hexagonal loop. Find its MI about an axis passing through center of mass and perpendicular to its surface. Solution : We find MI of hexagonal loop in following three steps : 1. Treat each side as a wire of mass M/6 and length L/6 and find its MI about perpendicular axis through center of mass of the side. 2. Apply theorem of parallel axes to find MI of the side about perpendicular axis through center of mass of the loop. 3. Add MIs of individual sides to find the MI of the loop about perpendicular axis through center of mass. The MI of the side about perpendicular axis through center of mass of the side is : I AA’ = M 6 x ( L 6 ) 2 12 = M L 2 12 x 6 3 I AA’ = M 6 x ( L 6 ) 2 12 = M L 2 12 x 6 3 By the geometry of a side with respect to center, the traingle OPQ is a right angle triangle. We have perpendicular distance "R" between axes AA' and ZZ' as : tan 60 ° = R L 12 tan 60 ° = R L 12 R = L 12 x 3 = 3 12 L R = L 12 x 3 = 3 12 L Applying theorem of parallel axes, the MI about perpendicular axis through center of mass of the loop is : I O = I zz’ = I AA’ + M 6 x R 2 I O = M L 2 12 x 6 3 + M 6 x ( 3 L 12 ) 2 I O = M L 2 12 x 6 3 + 3 M L 2 12 2 x 6 I O = I zz’ = I AA’ + M 6 x R 2 I O = M L 2 12 x 6 3 + M 6 x ( 3 L 12 ) 2 I O = M L 2 12 x 6 3 + 3 M L 2 12 2 x 6 I O = M L 2 + 9 M L 2 12 x 6 3 = 10 M L 2 12 x 6 3 I O = M L 2 + 9 M L 2 12 x 6 3 = 10 M L 2 12 x 6 3 Now, adding MIs of each side, the required MI about perpendicular axis through center of mass of the loop is : I = 6 I O = 6 x 10 M L 2 12 x 6 3 = 5 M L 2 216 I = 6 I O = 6 x 10 M L 2 12 x 6 3 = 5 M L 2 216 ## Content actions PDF \| EPUB (?) ### What is an EPUB file? EPUB is an electronic book format that can be read on a variety of mobile devices. My Favorites (?) 'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'. \| A lens I own (?) #### Definition of a lens ##### Lenses A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust. ##### What is in a lens? Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content. ##### Who can create a lens? Any individual member, a community, or a respected organization. ##### What are tags? Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens. \| External bookmarks
# Video: GCSE Mathematics Foundation Tier Pack 5 β’ Paper 2 β’ Question 2 GCSE Mathematics Foundation Tier Pack 5 β’ Paper 2 β’ Question 2 01:37 ### Video Transcript Simplify four π¦ minus two π¦ plus π¦. To simplify an algebraic expression like the one we have here means that we need to collect like terms. Like terms are the terms which have the same letters and powers. For example, two π¦ and three π¦ squared are not like terms as although they have the same letters, they have different powers. The first term just has π¦, which is π¦ to the power of one, and the second term has π¦ squared. In our expression, all of the terms are actually like terms as they all have the same letter π¦ and they all have the same power, which although it isnβt written, is a power of one. To simplify then, we just need to look at how many π¦s we have and how many weβre adding or subtracting. The first part of the expression is four π¦ minus two π¦. And if you have four of something and then subtract two of the same thing, youβre left with two of them. So four π¦ minus two π¦ simplifies to two π¦. And therefore, the whole expression simplifies to two π¦ plus π¦. Now, remember π¦ is just a shorthand way of writing one π¦. We never write one π¦ in algebra. We just write it as π¦. So this means two π¦ plus one π¦. And as two plus one is equal to three, this simplifies to three π¦. The expression four π¦ minus two π¦ plus π¦ simplifies to three π¦.
# 1.5 - Additional Measures of Dispersion 1.5 - Additional Measures of Dispersion ## Overall Measures of Dispersion Sometimes it is also useful to have an overall measure of dispersion in the data. In this measure, it would be good to include all of the variables simultaneously, rather than one at a time. In the past, we looked at the individual variables and their variances to measure the individual variances. Here we are going to look at measures of dispersion of all variables together, particularly we are going to look at such measures that look at the total variation. The variance $$\sigma _{j}^{2}$$ measures the dispersion of an individual variable Xj. The following two are used to measure the dispersion of all variables together. • Total Variation • Generalized Variance To understand total variation we first must find the trace of a square matrix. A square matrix is a matrix that has an equal number of columns and rows. Important examples of square matrices include the variance-covariance and correlation matrices. Trace of an n x n Matrix The trace of an n x n matrix $$\mathbf{A}$$ is $$trace(\textbf{A}) = \sum_{i=1}^{n}a_{ii}$$ For instance, in a 10 x 10 matrix, the trace is the sum of the diagonal elements. Total Variation of a Random Vector, $$\mathbf{X}$$ The total variation, therefore, of a random vector $$\mathbf{X}$$ is simply the trace of the population variance-covariance matrix. $$trace (\Sigma) = \sigma^2_1 + \sigma^2_2 +\dots \sigma^2_p$$ Thus, the total variation is equal to the sum of the population variances. The total variation can be estimated by: $$trace(S) = s^2_1+s^2_2+\dots +s^2_p$$ The total variation is of interest for principal components analysis and factor analysis and we will look at these concepts later in this course. ## Example 1-6: Woman's Health Survey (Variance) Let us use the data from the USDA womenâ€™s health survey again to illustrate this. We have taken the variances for each of the variables from the software output and have placed them in the table below. Variable Variance Calcium 157829.4 Iron 35.8 Protein 934.9 Vitamin A 2668452.4 Vitamin C 5416.3 Total 2832668.8 The total variation for the nutrient intake data is determined by simply adding up all of the variances for each of the individual variables. The total variation equals 2,832,668.8. This is a very large number. Note! The problem with total variation is that it does not take into account correlations among the variables. ## Interpreting Correlation These plots show simulated data for pairs of variables with different levels of correlation. In each case, the variances for both variables are equal to 1, so that the total variation is 2. When the correlation r = 0, then we see a shotgun-blast pattern of points, widely dispersed over the entire range of the plot. Increasing the correlation to r = 0.7, we see an oval-shaped pattern. Note that the points are not as widely dispersed. Increasing the correlation to r = 0.9, we see that the points fall along a 45-degree line, and are even less dispersed. Thus, the dispersion of points decreases with increasing correlation. But, in all cases, the total variation is the same. The total variation does not take into account the correlation between the two variables. Fixing the variances, the scatter of the data will tend to decrease as $$\| r \| \rightarrow 1$$. ## The Determinant To take into account the correlations among pairs of variables an alternative measure of overall variance is suggested. This measure takes a large value when the various variables show very little correlation among themselves. In contrast, this measure takes a small value if the variables show a very strong correlation among themselves, either positive or negative. This particular measure of dispersion is the generalized variance. In order to define the generalized variance, we first define the determinant of the matrix. We will start simple with a 2 x 2 matrix and then we will move on to more general definitions for larger matrices. Let us consider the determinant of a 2 x 2 matrix $$\mathbf{B}$$ as shown below. Here we can see that it is the product of the two diagonal elements minus the product of the off-diagonal elements. $$\|\textbf{B}\| =\left\|\begin{array}{cc}b_{11} & b_{12}\\ b_{21} & b_{22}\end{array}\right\| = b_{11}b_{22}-b_{12}b_{21}$$ Here is an example of a simple matrix that has the elements 5, 1, 2, and 4. You will get the determinant 18. The product of the diagonal 5 x 4 subtracting the elements of the off-diagonal 1 x 2 yields an answer of 18: $$\left\|\begin{array}{cc}5 & 1\\2 & 4\end{array}\right\| = 5 \times 4 - 1\times 2= 20-2 =18$$ Determinant of a General $$p\ x\ p$$ Matrix $$\mathbf{B}$$ More generally the determinant of a general p x p matrix $$\mathbf{B}$$ is given by the expression shown below: $$\|\textbf{B}\| = \sum_{j=1}^{p}(-1)^{j+1}b_{1j}\|B_{1j}\|$$ The expression involves the sum over all of the first row of $$\mathbf{B}$$. Note that these elements are noted by $$b_{1j}$$. These are pre-multiplied by -1 raised to the $$[j + 1]^{th}$$ power, so basically we are going to have alternating plus and minus signs in our sum. The matrix $$B1_j$$ is obtained by deleting row 1 and column j from the matrix $$\mathbf{B}$$. By definition, the generalized variance of a random vector $$\mathbf{X}$$ is equal to $$\|\sum\|$$, the determinant of the variance/covariance matrix. The generalized variance can be estimated by calculating $$\|S\|$$, the determinant of the sample variance/covariance matrix. [1] Link ↥ Has Tooltip/Popover Toggleable Visibility
# Mathematics needs logic to be formulated ## Logic¶ The (propositional) logic is of fundamental importance for all sub-areas of mathematics. With the help of connecting words such as “and”, “or”, “exactly if” etc., several (partial) statements can be linked to form a compound statement. The following statement functions can be used in logic two (or more) statements too one form a new statement. The conjunction If you combine two statements and the word “and” creates the conjunction of the statements and , symbolically with designated. A conjunction of two statements is therefore only true if both (partial) statements are true. Examples: If you combine two statements and the word “or” creates the adjunction of the statements and , symbolically with designated. The adjunction is therefore true if one of the two statements is true (or both are true). Examples: • The adjunct of the true statement and the wrong statement is the true statement . • The true statement: "Either the earth is a cube or the sun is a star" is an adjunct of the false statement: "The earth is a cube" and the true statement: "The sun is a star". The implication If you combine two statements and the word “then” creates the implication of the statements and , symbolically with designated. The implication is true if both statements and are true or if the first statement is wrong. [1] Formally, one obtains an identical truth table if one uses the implication forms. [2] [3] Examples: Equivalence of two statements If you combine two statements and the word combination “then, and only then” creates the equivalence of the statements and , symbolically with designated. The equivalence of two partial statements is only true if either both partial statements are true or both are false. [4] Examples: • The true statement "In a right-angled triangle the rate of height applies" equivalently linked to the false statement "In a right-angled triangle all sides are of the same length" results in the false statement "In a right triangle all sides are the same length if and only if the rate of height applies" . • The equivalence combination of the false statement “The kilogram is a unit of length” with the true statement “A thousand meters equals one kilometer” is the false statement “The kilogram is a unit of length if and only then, if a thousand meters equals a kilometer”. Contravalence of two statements If you combine two statements and the word “either or” in the exclusive sense creates the contravalence of the statements and , with with designated. The contravalence of two partial statements is only true if exactly one of the two (partial) statements is true. Thus, as its name suggests, it is formally identical with the negation of equivalence. Example: • If you combine the true statement “The train is going to Munich” with the false statement “The train is going to Frankfurt”, the result is the true statement “The train is either going to Munich or to Frankfurt”. Rules for the combination of statements The following equivalences are defined between the statements and their links, some of which have a formal similarity to the rules for calculating with numbers: In addition, there are the following rules, which are sometimes useful for evidence proceedings and in computer science: In doing so, the link also called "tautology"; it is always true. [5] ### Variables, terms and proposition forms¶ A variable is a symbol for any element from a given basic set. In addition, there are no special rules or laws for calculating with variables. A term is a designation on the one hand for a single mathematical object (for example ), on the other hand also for a string of several constants, variables, brackets and arithmetic operators (for example ). [6] However, terms do not contain a relational sign, so they are neither true nor false. In addition to (at least) one variable and (at least) one term, a statement form always contains a relation symbol - for example or . However, in order to be able to assign a truth value to a statement form, all occurring variables must first be replaced by concrete elements from the basic set. Just like statements, several types of statements can be combined to create new types of statements through logical links. The dependence of a statement form on one or more variables will be in the form expressed. Statements can be divided into three types: Forms of assertion are treated as equations and inequalities, especially in algebra. 'For all' and 'There is' In addition to inserting specific objects for the occurring variables, forms of statements can also be made into statements in a second way: quantification. While an existence statement is true if the underlying statement form even just for a specific one is fulfilled, a universal statement can be made in the opposite case already through the proof of existence of a single "counterexample" be refuted as false. [7] [8] ### Direct and indirect evidence¶ The formal rules of logic can also be used in order to be able to draw conclusions about new laws using forms of statements that have already been proven to be true. Theorems obtained in this way (also called “theorems” or “propositions” for short) represent the basic structure of mathematical theory. In addition to already known theorems, so-called definitions are also used in order to be able to prove new theorems. When defining a term, it is clearly defined by defining essential, common features and is differentiated from other terms. Definitions are neither true nor false; rather, they serve as abbreviations for awkward formulations. The symbol is used as a definition symbol for mathematical terms , an abbreviation for "is equal by definition". For the actual "proofs", the following conclusive rules of logic are possible, among others: • Conclusion from a negation: The propositional expression is generally valid. A proposition can thus be proven by refuting the negation of the proposition. In the case of direct proofs, based on valid prerequisites and using admissible inference rules, after a finite number of steps, the assertion is inferred directly. In the case of indirect proofs, on the other hand, the negation of the assertion is added to the prerequisites. The complete induction Complete induction is a widely used technique for directly proving a statement. The logical conclusion is based on three steps: 1. With the "induction start" it is shown that a statement form for a (freely selectable) value is valid. 2. The "induction assumption" consists in the fact that the statement form for a particular is valid. 3. With the “induction conclusion”, a “proof in the proof”, it is shown that from the validity of the statement also the validity of the statement follows, in shorthand . Example: Remarks:
# 3 Digit Math Problems Three-digit numbers have three digits ranging from 100 to 999. These numbers are composed of hundreds, tens, and unit digits, which give them a unique value and place in the number system. Understanding and working with 3-digit numbers is an essential skill in mathematics. They play a crucial role in developing math skills as they provide real-life contexts for applying mathematical concepts. Solving word problems helps students understand how to use mathematical operations such as addition and subtraction in practical situations. It enhances critical thinking and problem-solving abilities and promotes a deeper understanding of mathematical concepts. ## Addition of 3 Digit Math Problems ### Problem 1: Adding two 3-digit numbers Example problem: 342 + 178 Solution steps and calculations: • Line up the numbers vertically, aligning the units, tens, and hundreds of digits. • Start by adding the units digits: 2 + 8 = 10. Write down 0 and carry over 1 to the tens column. • Add the tens digits: 4 + 7 + 1 (carried over) = 12. Write down 2 and carry over 1 to the hundreds column. • Add the hundreds of digits: 3 + 1 (carried over) = 4. • Therefore, 342 + 178 = 520. To add two 3-digit numbers, we start from the rightmost digit and move to the left, carrying over any excess values to the next column. By summing the corresponding digits in each column, we obtain the result. In the given example, the units column gives us 0, the tens column gives us 2, and the hundreds column gives us 4, resulting in 520. ### Problem 2: Adding a 2-digit and a 3-digit number Example problem: 97 + 534 Solution steps and calculations: • Line up the numbers vertically, aligning the units, tens, and hundreds of digits. • Add the units digits: 7 + 4 = 11. Write down 1 and carry over 1 to the tens column. • Add the tens digits: 9 + 3 + 1 (carried over) = 13. Write down 3 and carry over 1 to the hundreds column. • Add the hundreds of digits: 5 + 1 (carried over) = 6. • Therefore, 97 + 534 = 631. Adding a 2-digit and a 3-digit number is similar to adding two 3-digit numbers. We start from the rightmost digit and move to the left, carrying over any excess values. In the given example, the units column gives us 1, the tens column gives us 3, and the hundreds column gives us 6, resulting in 631. ## Subtraction 3 Digit Math Problems ### Problem 1: Subtracting a 3-digit number from another 3-digit number Example problem: 572 – 294 Solution steps and calculations: • Line up the numbers vertically, aligning the units, tens, and hundreds of digits. • Start by subtracting the units digits: 2 – 4. Since 4 is greater than 2, borrow 1 from the tens column, making it 12 – 4 = 8. • Subtract the tens digits: 7 – 9 (borrowed 1) = -2. Since we cannot have a negative digit, borrow 1 from the hundreds column, making it 17 – 9 = 8. • Subtract the hundreds of digits: 5 – 2 = 3. • Therefore, 572 – 294 = 278. To subtract a 3-digit number from another 3-digit number, we start from the rightmost digit and move to the left. If the digit being subtracted is greater than the corresponding digit in the minuend, we borrow from the next column. In the given example, the units column gives us 8, the tens column gives us -2, and the hundreds column gives us 3, resulting in a difference of 278. ### Problem 2: Subtracting a 2-digit number from a 3-digit number Example problem: 765 – 87 Solution steps and calculations: • Line up the numbers vertically, aligning the units, tens, and hundreds of digits. • Subtract the units digits: 5 – 7. Since 7 is greater than 5, borrow 1 from the tens column, making it 15 – 7 = 8. • Subtract the tens digits: 6 – 8 (borrowed 1) = -2. Since we cannot have a negative digit, borrow 1 from the hundreds column, making it 16 – 8 = 8. • Subtract the hundreds of digits: 7 – 0 = 7. • Therefore, 765 – 87 = 678. When subtracting a 2-digit number from a 3-digit number, the process is similar to subtracting two 3-digit numbers. We start from the rightmost digit and move to the left, borrowing from the next column if needed. In the given example, the units column gives us 8, the tens column gives us -2, and the hundreds column gives us 7, resulting in a difference of 678. ## Mixed Addition and Subtraction Word Problems ### Problem 1: Mixed addition and subtraction involving 3-digit numbers Example problem: 432 + 275 – 167 Solution steps and calculations: • Perform addition first: 432 + 275 = 707. • Then subtract 707 – 167 = 540. When faced with mixed addition and subtraction word problems involving 3-digit numbers, it is important to follow the order of operations (BIDMAS/BODMAS). In this case, addition is performed first, resulting in 707. Then, the subtraction is carried out, resulting in the final answer of 540. ### Problem 2: Mixed addition and subtraction with multiple 3-digit numbers Example problem: 846 – 219 + 374 Solution steps and calculations: • Perform the subtraction first: 846 – 219 = 627. • Then, perform the addition: 627 + 374 = 1001. In this example of mixed addition and subtraction with multiple 3-digit numbers, we start with subtraction to obtain 627. Then, we perform the addition operation, resulting in the final answer, 1001. ## Conclusion Practicing 3-digit addition and subtraction, especially through word problems, helps develop essential math skills such as critical thinking, problem-solving, and applying mathematical concepts in real-life scenarios. It enhances numerical fluency and provides a solid foundation for more advanced math topics. To strengthen math skills further, it is crucial to continue practicing and solving word problems involving 3-digit addition and subtraction. Regular practice will improve computational abilities and enhance logical reasoning and analytical thinking, empowering individuals to tackle more complex math problems in the future.
# Modern Algebra Lecture Notes: Rings and fields set 4 (Revision 2) Size: px Start display at page: Download "Modern Algebra Lecture Notes: Rings and fields set 4 (Revision 2)" ## Transcription 1 Modern Algebra Lecture Notes: Rings and fields set 4 (Revision 2) Kevin Broughan University of Waikato, Hamilton, New Zealand May 13, 2010 2 Remainder and Factor Theorem 15 Definition of factor If f (x) = g(x).h(x) in R[x] we say g(x) is a factor of f (x) in R[x]. Examples (1) The polynomial x + 2 is a factor of x 2 + 5x + 6 in Z[x]. (2) The polynomial x = (x + i)(x i) in Z[i], but has no factors in Z[x]. (3) Polys like f (x) = x 8 + 3x 6 6x 5 + 3x 4 12x x 2 6x + 9 are difficult to factor in any ring. In Z[x] we have f (x) = ( x ) ( x 3 + x 3 ) 2. How is this done? Theorem 15 Let F be a field, a F an element and f (x) F [x] a polynomial. Then f (x) = (x a)q(x) + f (a), i.e. f (a) is the remainder when we divide f (x) by x a. If f (a) = 0 then (x a) is a factor of f (x). 3 Proof of Theorem 15 By the division identity in F [x] there exist polynomials q(x), r(x) such that f (x) = (x a)q(x) + r(x) with r(x) = 0 or deg r(x) < deg x a = 1. Therefore r(x) = constant = r(a). But f (a) = (a a)q(a) + r(a) = r(a) so f (a) = r(a) and we can write f (x) = (x a)q(x) + f (a). If f (a) = 0 we have f (x) = (x a)q(x) + 0 = (x a)q(x) so x a is a factor of f (x). 4 Theorem 16: A poly over a field of degree n has at most n zeros in the field Proof The polynomial must be non-zero. If n is zero the polynomial has no zeros. Assume every polynomial of degree up to and including n has at most n zeros and let f (x) be a polynomial of degree n + 1. If f (x) has no zeros we are done. If a is a zero of f (x), by Theorem 15 we can write f (x) = (x a) k g(x), by Theorem 15, where k 1 and g(a) 0, deg f (x) = k + deg g(x), so the degree of g(x) is n. By the inductive hypothesis it has at most n zeros, a 1,..., a m say with m n. Every zero of g(x) is a zero of f (x). If b is a zero of f (x) other than a then 0 = f (b) = (b a) k g(b) so b is a zero of g(x). Hence the zeros of f (x) are {b, a 1,... a m} so are in number less than or equal to n + 1, completing the proof by induction. 5 Notes This is not true in general for polynomials over a ring: x 2 + x has 4 zeros in Z/6Z. If F K and K is also a field, a so-called extension field, then we have actually shown f (x) has at most deg f (x) zeros in K. Examples (1) x has no zeros in Q but a full set in Q(i) and in C. (2) x 2 2 has no zeros in Q but a full set in Q( 2). 6 Definition of a principal ideal domain A principal ideal domain or PID, is an integral domain R where every ideal has the form a for some a R. Examples: (1) Z: If A Z is an ideal let a > 0 be the smallest positive element of A. Then A = a. (2) If F is a field then F [x] is a principal ideal domain. This is Theorem 17 proved below. (3) The ring Z[ 5] is a principal ideal domain, but Z[ 5] is not. So being a PID is a big issue and quite subtle. 7 Proof of Theorem 17 Let A F [x] be an ideal. If A = {0} we have A = 0. If A {0} let g(x) be a polynomial in A of minimum degree. We can assume g(x) is monic. Then we claim A = g(x). To see this let f (x) A. Then, by the division identity there are polynomials q(x), r(x) such that f (x) = q(x).g(x) + r(x) where either r(x) = 0 or deg r(x) < deg g(x). If r(x) = 0 then f (x) = q(x).g(x) g(x). Otherwise r(x) = f (x) q(x).g(x) A. But deg r(x) < deg g(x) so this case is impossible. Therefore A = g(x). 8 Factoring polynomials Definitions If F is a field we say a polynomial f (x) F [x] is irreducible over F if it cannot be expressed as the product of two polynomials over F with strictly lower degrees than that of f (x). If R is an integral domain we say a polynomial f (x) R[x] is irreducible over R if whenever we write f (x) = g(x).h(x) we must have either g(x) or h(x) a unit in R[x]. Examples (1) x + 1 is irreducible over Q[x] and Z[x]. (2) A unit in R[x] is a constant polynomial f (x) = u where u is a unit in R. (3) 3x + 6 is irreducible over Q[x] but factors non-trivially as 3.(x + 2) in Z[x]. (4) x is irreducible in R[x] but factors as (x + i)(x i) in C[x]. (5) x 2 + x + 1 is irreducible in Q[x] but factors as (x + 2)(x + 2) in (Z/3Z)[x]. 9 The quadratic and cubic formulas Quadratic If f (x) = ax 2 + bx + c, a, b, c C and a 0 then in C[x] where α, β = b± b 2 4ac 2a f (x) = a(x α)(x β) Cubic If f (x) = x 3 + ax + b then f (x) = (x α)(x β)(x γ) where α = β = γ = a3 + 27b 2 9b 3 3 a 23 2/ a3 + 27b 2 9b ( 1 i ) ( 3 a 1 + i ) a3 + 27b 2 9b 2 2/ a3 + 27b 2 9b /3 ( 1 + i ) ( 3 a 1 i ) a3 + 27b 2 9b 2 2/ a3 + 27b 2 9b /3 10 Factors for low degree polynomials Degree 1 If a non-zero polynomial over a field has degree 0 it is irreducible. For degree 1, f (x) = ax + b, then f ( b/a) = 0 and f (x) is irreducible. Degree 2 or 3 If f (x) F [x] has a zero or root, f (a) = 0 then f (x) is reducible since f (x) = (x a)g(x). If f (x) has degree 2 or 3 then it is reducible if and only if it has a factor of degree 1 if and only if it has a zero/root. Degree 4 or more Over Q[x] (x 2 + 1)(x 2 + 2) = x 4 + 3x so the right hand side is reducible but has no root in Q. Polynomials over finite fields Finding roots is easy, just test each element of the field and see if the polynomial vanishes, e.g. f (x) = x 4 + 3x + 1 over Z/5Z has f (1) = 0 so x 1 = x + 4 divides exactly with quotient 4 + x + x 2 + x 3. This does not vanish at x = 0, 1, 2, 3, 4 modulo 5, therefore it is irreducible (over Z/5Z). 11 Gauss Lemma Definition Let f (x) = a nx n + + a 0 Z[x]. The content of f (x), denoted c(f (x)), is the gcd of the coefficients, i.e. gcd(a 0,..., a n). If c(f (x)) = 1 we say f (x) is primitive. Example (1) f (x) = 2x 2 4x + 12 = c(f (x)) = 2 f (x) = 2x 3 3x + 6 = c(f (x)) = 1. Statement of the Lemma Let f (x), g(x) Z[x] be primitive. Then f (x).g(x) is also primitive, i.e. the product of two primitive polynomials is primitive. Reduction Modulo p A useful operation on polynomials with integer coefficients is reduction modulo p where p is a fixed prime. This is a homomorphism of polynomial rings: f (x) Z[x] θ(f (x)) Z p[x] : a nx n + a 0 [a n]x n + + [a 0]. 12 Proof of Gauss Lemma Let f (x), g(x) be primitive and suppose f (x).g(x) is not primitive. Let the prime p divide the content, i.e. p divides every coefficient of f (x).g(x). Then in Z p[x], θ(f (x).g(x)) = 0. But this is an integral domain (because Z p is an integral domain) and, because θ is a homomorphism, 0 = θ(f (x)).θ(g(x)). Therefore θ(f (x)) = 0 or θ(g(x)) = 0. Suppose the former is true. But this means p divides each of the coefficients of f (x). But this is impossible, since f (x) is primitive. 13 Factoring over Q equals factoring over Z Example 6x 4 + 8x 3 + 3x 2 + 7x + 4 = (2x )(3x x ) = (3x + 4)(2x 3 + x + 1) Content fact If f (x) Z[x] then we can write f (x) = c(f (x)).f 1(x) where c(f 1(x)) = 1. Example f (x) = 12x 4 6x 3 + 3x + 18 = 3(4x 4 2x 3 + x + 6) = c(f (x)).f 1(x) where f 1(x) has content 1, i.e. is primitive. 14 Theorem 18 Statement If f (x) Z[x] and f (x) = g(x).h(x) factors in Q(x) then f (x) = g 1(x).h 1(x) in Z[x] and the degrees of deg g 1(x) = deg g(x), deg h 1(x) = deg h(x). Proof Make f (x) primitive by dividing both f (x) and g(x) by c(f (x)). Let a be the least common multiple of the denominators of the coefficients of g(x) and b the LCM of the denominators of h(x) so ag(x) Z[x] and bh(x) Z[x]. Let ag(x) = c(ag(x)).g 1(x) and bh(x) = c(bh(x)).h 1(x), so g 1(x), h 1(x) are primitive and in Z[x]. Then, since f (x) is primitive, c(abf (x) = ab. But abf (x) = ag(x)bh(x) = c(ag(x).g 1(x)c(bh(x)).h 1(x) = c(ag(x))c(bh(x)g 1(x).h 1(x). Since, by Gauss, g 1(x).h 1(x) is primitive the content of the RHS is c(ag(x)).c(bh(x)) which must be the content of the LHS, so they cancel leading to f (x) = g 1(x).h 1(x) the factorization over Z[x]. ### Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm. Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm. We begin by defining the ring of polynomials with coefficients in a ring R. After some preliminary results, we specialize More information ### PROBLEM SET 6: POLYNOMIALS PROBLEM SET 6: POLYNOMIALS 1. introduction In this problem set we will consider polynomials with coefficients in K, where K is the real numbers R, the complex numbers C, the rational numbers Q or any other More information ### it is easy to see that α = a 21. Polynomial rings Let us now turn out attention to determining the prime elements of a polynomial ring, where the coefficient ring is a field. We already know that such a polynomial ring is a UF. Therefore More information ### H/wk 13, Solutions to selected problems H/wk 13, Solutions to selected problems Ch. 4.1, Problem 5 (a) Find the number of roots of x x in Z 4, Z Z, any integral domain, Z 6. (b) Find a commutative ring in which x x has infinitely many roots. More information ### Introduction to Finite Fields (cont.) Chapter 6 Introduction to Finite Fields (cont.) 6.1 Recall Theorem. Z m is a field m is a prime number. Theorem (Subfield Isomorphic to Z p ). Every finite field has the order of a power of a prime number More information ### 7. Some irreducible polynomials 7. Some irreducible polynomials 7.1 Irreducibles over a finite field 7.2 Worked examples Linear factors x α of a polynomial P (x) with coefficients in a field k correspond precisely to roots α k [1] of More information ### POLYNOMIAL RINGS AND UNIQUE FACTORIZATION DOMAINS POLYNOMIAL RINGS AND UNIQUE FACTORIZATION DOMAINS RUSS WOODROOFE 1. Unique Factorization Domains Throughout the following, we think of R as sitting inside R[x] as the constant polynomials (of degree 0). More information ### 3 1. Note that all cubes solve it; therefore, there are no more Math 13 Problem set 5 Artin 11.4.7 Factor the following polynomials into irreducible factors in Q[x]: (a) x 3 3x (b) x 3 3x + (c) x 9 6x 6 + 9x 3 3 Solution: The first two polynomials are cubics, so if More information ### Unique Factorization Unique Factorization Waffle Mathcamp 2010 Throughout these notes, all rings will be assumed to be commutative. 1 Factorization in domains: definitions and examples In this class, we will study the phenomenon More information ### CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY January 10, 2010 CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY The set of polynomials over a field F is a ring, whose structure shares with the ring of integers many characteristics. More information ### MOP 2007 Black Group Integer Polynomials Yufei Zhao. Integer Polynomials. June 29, 2007 Yufei Zhao yufeiz@mit.edu Integer Polynomials June 9, 007 Yufei Zhao yufeiz@mit.edu We will use Z[x] to denote the ring of polynomials with integer coefficients. We begin by summarizing some of the common approaches used in dealing More information ### 3 Factorisation into irreducibles 3 Factorisation into irreducibles Consider the factorisation of a non-zero, non-invertible integer n as a product of primes: n = p 1 p t. If you insist that primes should be positive then, since n could More information ### JUST THE MATHS UNIT NUMBER 1.8. ALGEBRA 8 (Polynomials) A.J.Hobson JUST THE MATHS UNIT NUMBER 1.8 ALGEBRA 8 (Polynomials) by A.J.Hobson 1.8.1 The factor theorem 1.8.2 Application to quadratic and cubic expressions 1.8.3 Cubic equations 1.8.4 Long division of polynomials More information ### The Division Algorithm for Polynomials Handout Monday March 5, 2012 The Division Algorithm for Polynomials Handout Monday March 5, 0 Let F be a field (such as R, Q, C, or F p for some prime p. This will allow us to divide by any nonzero scalar. (For some of the following, More information ### Factoring Polynomials Factoring Polynomials Sue Geller June 19, 2006 Factoring polynomials over the rational numbers, real numbers, and complex numbers has long been a standard topic of high school algebra. With the advent More information ### calculating the result modulo 3, as follows: p(0) = 0 3 + 0 + 1 = 1 0, Homework #02, due 1/27/10 = 9.4.1, 9.4.2, 9.4.5, 9.4.6, 9.4.7. Additional problems recommended for study: (9.4.3), 9.4.4, 9.4.9, 9.4.11, 9.4.13, (9.4.14), 9.4.17 9.4.1 Determine whether the following polynomials More information ### Module MA3411: Abstract Algebra Galois Theory Appendix Michaelmas Term 2013 Module MA3411: Abstract Algebra Galois Theory Appendix Michaelmas Term 2013 D. R. Wilkins Copyright c David R. Wilkins 1997 2013 Contents A Cyclotomic Polynomials 79 A.1 Minimum Polynomials of Roots of More information ### Factorization in Polynomial Rings Factorization in Polynomial Rings These notes are a summary of some of the important points on divisibility in polynomial rings from 17 and 18 of Gallian s Contemporary Abstract Algebra. Most of the important More information ### SOLVING POLYNOMIAL EQUATIONS C SOLVING POLYNOMIAL EQUATIONS We will assume in this appendix that you know how to divide polynomials using long division and synthetic division. If you need to review those techniques, refer to an algebra More information ### PUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include 2 + 5. PUTNAM TRAINING POLYNOMIALS (Last updated: November 17, 2015) Remark. This is a list of exercises on polynomials. Miguel A. Lerma Exercises 1. Find a polynomial with integral coefficients whose zeros include More information ### minimal polyonomial Example Minimal Polynomials Definition Let α be an element in GF(p e ). We call the monic polynomial of smallest degree which has coefficients in GF(p) and α as a root, the minimal polyonomial of α. Example: We More information ### a 1 x + a 0 =0. (3) ax 2 + bx + c =0. (4) ROOTS OF POLYNOMIAL EQUATIONS In this unit we discuss polynomial equations. A polynomial in x of degree n, where n 0 is an integer, is an expression of the form P n (x) =a n x n + a n 1 x n 1 + + a 1 x More information ### Winter Camp 2011 Polynomials Alexander Remorov. Polynomials. Alexander Remorov alexanderrem@gmail.com Polynomials Alexander Remorov alexanderrem@gmail.com Warm-up Problem 1: Let f(x) be a quadratic polynomial. Prove that there exist quadratic polynomials g(x) and h(x) such that f(x)f(x + 1) = g(h(x)). More information ### (a) Write each of p and q as a polynomial in x with coefficients in Z[y, z]. deg(p) = 7 deg(q) = 9 Homework #01, due 1/20/10 = 9.1.2, 9.1.4, 9.1.6, 9.1.8, 9.2.3 Additional problems for study: 9.1.1, 9.1.3, 9.1.5, 9.1.13, 9.2.1, 9.2.2, 9.2.4, 9.2.5, 9.2.6, 9.3.2, 9.3.3 9.1.1 (This problem was not assigned More information ### FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z DANIEL BIRMAJER, JUAN B GIL, AND MICHAEL WEINER Abstract We consider polynomials with integer coefficients and discuss their factorization More information ### Quotient Rings and Field Extensions Chapter 5 Quotient Rings and Field Extensions In this chapter we describe a method for producing field extension of a given field. If F is a field, then a field extension is a field K that contains F. More information ### Math 345-60 Abstract Algebra I Questions for Section 23: Factoring Polynomials over a Field Math 345-60 Abstract Algebra I Questions for Section 23: Factoring Polynomials over a Field 1. Throughout this section, F is a field and F [x] is the ring of polynomials with coefficients in F. We will More information ### Factoring Cubic Polynomials Factoring Cubic Polynomials Robert G. Underwood 1. Introduction There are at least two ways in which using the famous Cardano formulas (1545) to factor cubic polynomials present more difficulties than More information ### 1 Homework 1. [p 0 q i+j +... + p i 1 q j+1 ] + [p i q j ] + [p i+1 q j 1 +... + p i+j q 0 ] 1 Homework 1 (1) Prove the ideal (3,x) is a maximal ideal in Z[x]. SOLUTION: Suppose we expand this ideal by including another generator polynomial, P / (3, x). Write P = n + x Q with n an integer not More information ### 6.2 Permutations continued 6.2 Permutations continued Theorem A permutation on a finite set A is either a cycle or can be expressed as a product (composition of disjoint cycles. Proof is by (strong induction on the number, r, of More information ### Factorization Algorithms for Polynomials over Finite Fields Degree Project Factorization Algorithms for Polynomials over Finite Fields Sajid Hanif, Muhammad Imran 2011-05-03 Subject: Mathematics Level: Master Course code: 4MA11E Abstract Integer factorization is More information ### 9. POLYNOMIALS. Example 1: The expression a(x) = x 3 4x 2 + 7x 11 is a polynomial in x. The coefficients of a(x) are the numbers 1, 4, 7, 11. 9. POLYNOMIALS 9.1. Definition of a Polynomial A polynomial is an expression of the form: a(x) = a n x n + a n-1 x n-1 +... + a 1 x + a 0. The symbol x is called an indeterminate and simply plays the role More information ### Zeros of a Polynomial Function Zeros of a Polynomial Function An important consequence of the Factor Theorem is that finding the zeros of a polynomial is really the same thing as factoring it into linear factors. In this section we More information ### Real Roots of Univariate Polynomials with Real Coefficients Real Roots of Univariate Polynomials with Real Coefficients mostly written by Christina Hewitt March 22, 2012 1 Introduction Polynomial equations are used throughout mathematics. When solving polynomials More information ### Cyclotomic Extensions Chapter 7 Cyclotomic Extensions A cyclotomic extension Q(ζ n ) of the rationals is formed by adjoining a primitive n th root of unity ζ n. In this chapter, we will find an integral basis and calculate More information ### Factoring of Prime Ideals in Extensions Chapter 4 Factoring of Prime Ideals in Extensions 4. Lifting of Prime Ideals Recall the basic AKLB setup: A is a Dedekind domain with fraction field K, L is a finite, separable extension of K of degree More information ### 15. Symmetric polynomials 15. Symmetric polynomials 15.1 The theorem 15.2 First examples 15.3 A variant: discriminants 1. The theorem Let S n be the group of permutations of {1,, n}, also called the symmetric group on n things. More information ### Chapter 13: Basic ring theory Chapter 3: Basic ring theory Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 42, Spring 24 M. Macauley (Clemson) Chapter 3: Basic ring More information ### Introduction to Algebraic Geometry. Bézout s Theorem and Inflection Points Introduction to Algebraic Geometry Bézout s Theorem and Inflection Points 1. The resultant. Let K be a field. Then the polynomial ring K[x] is a unique factorisation domain (UFD). Another example of a More information ### Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any. Algebra 2 - Chapter Prerequisites Vocabulary Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any. P1 p. 1 1. counting(natural) numbers - {1,2,3,4,...} More information ### 10 Splitting Fields. 2. The splitting field for x 3 2 over Q is Q( 3 2,ω), where ω is a primitive third root of 1 in C. Thus, since ω = 1+ 3 10 Splitting Fields We have seen how to construct a field K F such that K contains a root α of a given (irreducible) polynomial p(x) F [x], namely K = F [x]/(p(x)). We can extendthe procedure to build More information ### FACTORING IN QUADRATIC FIELDS. 1. Introduction. This is called a quadratic field and it has degree 2 over Q. Similarly, set FACTORING IN QUADRATIC FIELDS KEITH CONRAD For a squarefree integer d other than 1, let 1. Introduction K = Q[ d] = {x + y d : x, y Q}. This is called a quadratic field and it has degree 2 over Q. Similarly, More information ### A number field is a field of finite degree over Q. By the Primitive Element Theorem, any number Number Fields Introduction A number field is a field of finite degree over Q. By the Primitive Element Theorem, any number field K = Q(α) for some α K. The minimal polynomial Let K be a number field and More information ### Lagrange Interpolation is a method of fitting an equation to a set of points that functions well when there are few points given. Polynomials (Ch.1) Study Guide by BS, JL, AZ, CC, SH, HL Lagrange Interpolation is a method of fitting an equation to a set of points that functions well when there are few points given. Sasha s method More information ### Algebra 3: algorithms in algebra Algebra 3: algorithms in algebra Hans Sterk 2003-2004 ii Contents 1 Polynomials, Gröbner bases and Buchberger s algorithm 1 1.1 Introduction............................ 1 1.2 Polynomial rings and systems More information ### 1 Lecture: Integration of rational functions by decomposition Lecture: Integration of rational functions by decomposition into partial fractions Recognize and integrate basic rational functions, except when the denominator is a power of an irreducible quadratic. More information ### Galois Theory III. 3.1. Splitting fields. Galois Theory III. 3.1. Splitting fields. We know how to construct a field extension L of a given field K where a given irreducible polynomial P (X) K[X] has a root. We need a field extension of K where More information ### Non-unique factorization of polynomials over residue class rings of the integers Comm. Algebra 39(4) 2011, pp 1482 1490 Non-unique factorization of polynomials over residue class rings of the integers Christopher Frei and Sophie Frisch Abstract. We investigate non-unique factorization More information ### RESULTANT AND DISCRIMINANT OF POLYNOMIALS RESULTANT AND DISCRIMINANT OF POLYNOMIALS SVANTE JANSON Abstract. This is a collection of classical results about resultants and discriminants for polynomials, compiled mainly for my own use. All results More information ### Die ganzen zahlen hat Gott gemacht Die ganzen zahlen hat Gott gemacht Polynomials with integer values B.Sury A quote attributed to the famous mathematician L.Kronecker is Die Ganzen Zahlen hat Gott gemacht, alles andere ist Menschenwerk. More information ### 1.3 Algebraic Expressions 1.3 Algebraic Expressions A polynomial is an expression of the form: a n x n + a n 1 x n 1 +... + a 2 x 2 + a 1 x + a 0 The numbers a 1, a 2,..., a n are called coefficients. Each of the separate parts, More information ### COMMUTATIVE RINGS. Definition: A domain is a commutative ring R that satisfies the cancellation law for multiplication: COMMUTATIVE RINGS Definition: A commutative ring R is a set with two operations, addition and multiplication, such that: (i) R is an abelian group under addition; (ii) ab = ba for all a, b R (commutative More information ### The Method of Partial Fractions Math 121 Calculus II Spring 2015 Rational functions. as The Method of Partial Fractions Math 11 Calculus II Spring 015 Recall that a rational function is a quotient of two polynomials such f(x) g(x) = 3x5 + x 3 + 16x x 60. The method More information ### 11 Ideals. 11.1 Revisiting Z 11 Ideals The presentation here is somewhat different than the text. In particular, the sections do not match up. We have seen issues with the failure of unique factorization already, e.g., Z[ 5] = O Q( More information ### ALGEBRAIC NUMBER THEORY AND QUADRATIC RECIPROCITY ALGEBRAIC NUMBER THEORY AND QUADRATIC RECIPROCITY HENRY COHN, JOSHUA GREENE, JONATHAN HANKE 1. Introduction These notes are from a series of lectures given by Henry Cohn during MIT s Independent Activities More information ### ON GALOIS REALIZATIONS OF THE 2-COVERABLE SYMMETRIC AND ALTERNATING GROUPS ON GALOIS REALIZATIONS OF THE 2-COVERABLE SYMMETRIC AND ALTERNATING GROUPS DANIEL RABAYEV AND JACK SONN Abstract. Let f(x) be a monic polynomial in Z[x] with no rational roots but with roots in Q p for More information ### 1 = (a 0 + b 0 α) 2 + + (a m 1 + b m 1 α) 2. for certain elements a 0,..., a m 1, b 0,..., b m 1 of F. Multiplying out, we obtain Notes on real-closed fields These notes develop the algebraic background needed to understand the model theory of real-closed fields. To understand these notes, a standard graduate course in algebra is More information ### 1.5. Factorisation. Introduction. Prerequisites. Learning Outcomes. Learning Style Factorisation 1.5 Introduction In Block 4 we showed the way in which brackets were removed from algebraic expressions. Factorisation, which can be considered as the reverse of this process, is dealt with More information ### The Dirichlet Unit Theorem Chapter 6 The Dirichlet Unit Theorem As usual, we will be working in the ring B of algebraic integers of a number field L. Two factorizations of an element of B are regarded as essentially the same if More information ### 3.6 The Real Zeros of a Polynomial Function SECTION 3.6 The Real Zeros of a Polynomial Function 219 3.6 The Real Zeros of a Polynomial Function PREPARING FOR THIS SECTION Before getting started, review the following: Classification of Numbers (Appendix, More information ### Algebraic Structures II MAS 305 Algebraic Structures II Notes 12 Autumn 2006 Factorization in integral domains Lemma If a, b, c are elements of an integral domain R and ab = ac then either a = 0 R or b = c. Proof ab = ac a(b More information ### WARM UP EXERCSE. 2-1 Polynomials and Rational Functions WARM UP EXERCSE Roots, zeros, and x-intercepts. x 2! 25 x 2 + 25 x 3! 25x polynomial, f (a) = 0! (x - a)g(x) 1 2-1 Polynomials and Rational Functions Students will learn about: Polynomial functions Behavior More information ### Tim Kerins. Leaving Certificate Honours Maths - Algebra. Tim Kerins. the date Leaving Certificate Honours Maths - Algebra the date Chapter 1 Algebra This is an important portion of the course. As well as generally accounting for 2 3 questions in examination it is the basis for many More information ### Galois Theory. Richard Koch Galois Theory Richard Koch April 2, 2015 Contents 1 Preliminaries 4 1.1 The Extension Problem; Simple Groups.................... 4 1.2 An Isomorphism Lemma............................. 5 1.3 Jordan Holder................................... More information ### Prime Numbers and Irreducible Polynomials Prime Numbers and Irreducible Polynomials M. Ram Murty The similarity between prime numbers and irreducible polynomials has been a dominant theme in the development of number theory and algebraic geometry. More information ### EXERCISES FOR THE COURSE MATH 570, FALL 2010 EXERCISES FOR THE COURSE MATH 570, FALL 2010 EYAL Z. GOREN (1) Let G be a group and H Z(G) a subgroup such that G/H is cyclic. Prove that G is abelian. Conclude that every group of order p 2 (p a prime More information ### Field Fundamentals. Chapter 3. 3.1 Field Extensions. 3.1.1 Definitions. 3.1.2 Lemma Chapter 3 Field Fundamentals 3.1 Field Extensions If F is a field and F [X] is the set of all polynomials over F, that is, polynomials with coefficients in F, we know that F [X] is a Euclidean domain, More information ### THE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS THE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS KEITH CONRAD 1. Introduction The Fundamental Theorem of Algebra says every nonconstant polynomial with complex coefficients can be factored into linear More information ### p e i 1 [p e i i ) = i=1 Homework 1 Solutions - Sri Raga Velagapudi Algebra Section 1. Show that if n Z then for every integer a with gcd(a, n) = 1, there exists a unique x mod n such that ax = 1 mod n. By the definition of gcd, More information ### CM2202: Scientific Computing and Multimedia Applications General Maths: 2. Algebra - Factorisation CM2202: Scientific Computing and Multimedia Applications General Maths: 2. Algebra - Factorisation Prof. David Marshall School of Computer Science & Informatics Factorisation Factorisation is a way of More information ### Basics of Polynomial Theory 3 Basics of Polynomial Theory 3.1 Polynomial Equations In geodesy and geoinformatics, most observations are related to unknowns parameters through equations of algebraic (polynomial) type. In cases where More information ### Factoring polynomials over finite fields Factoring polynomials over finite fields Summary and et questions 12 octobre 2011 1 Finite fields Let p an odd prime and let F p = Z/pZ the (unique up to automorphism) field with p-elements. We want to More information ### Congruence properties of binary partition functions Congruence properties of binary partition functions Katherine Anders, Melissa Dennison, Jennifer Weber Lansing and Bruce Reznick Abstract. Let A be a finite subset of N containing 0, and let f(n) denote More information ### ALGEBRA HW 5 CLAY SHONKWILER ALGEBRA HW 5 CLAY SHONKWILER 510.5 Let F = Q(i). Prove that x 3 and x 3 3 are irreducible over F. Proof. If x 3 is reducible over F then, since it is a polynomial of degree 3, it must reduce into a product More information ### Zeros of Polynomial Functions Zeros of Polynomial Functions Objectives: 1.Use the Fundamental Theorem of Algebra to determine the number of zeros of polynomial functions 2.Find rational zeros of polynomial functions 3.Find conjugate More information ### 1.7. Partial Fractions. 1.7.1. Rational Functions and Partial Fractions. A rational function is a quotient of two polynomials: R(x) = P (x) Q(x). .7. PRTIL FRCTIONS 3.7. Partial Fractions.7.. Rational Functions and Partial Fractions. rational function is a quotient of two polynomials: R(x) = P (x) Q(x). Here we discuss how to integrate rational More information ### Zero: If P is a polynomial and if c is a number such that P (c) = 0 then c is a zero of P. MATH 11011 FINDING REAL ZEROS KSU OF A POLYNOMIAL Definitions: Polynomial: is a function of the form P (x) = a n x n + a n 1 x n 1 + + a x + a 1 x + a 0. The numbers a n, a n 1,..., a 1, a 0 are called More information ### Galois theory. a draft of Lecture Notes of H.M. Khudaverdian. Manchester, Autumn 2006 (version 16 XII 2006) Galois theory a draft of Lecture Notes of H.M. Khudaverdian. Manchester, Autumn 2006 (version 16 XII 2006) Contents 0.1................................... 2 0.2 Viète Theorem.......................... More information ### 4.3 Lagrange Approximation 206 CHAP. 4 INTERPOLATION AND POLYNOMIAL APPROXIMATION Lagrange Polynomial Approximation 4.3 Lagrange Approximation Interpolation means to estimate a missing function value by taking a weighted average More information ### First and raw version 0.1 23. september 2013 klokken 13:45 The discriminant First and raw version 0.1 23. september 2013 klokken 13:45 One of the most significant invariant of an algebraic number field is the discriminant. One is tempted to say, apart from the More information ### Mathematics. (www.tiwariacademy.com : Focus on free Education) (Chapter 5) (Complex Numbers and Quadratic Equations) (Class XI) ( : Focus on free Education) Miscellaneous Exercise on chapter 5 Question 1: Evaluate: Answer 1: 1 ( : Focus on free Education) Question 2: For any two complex numbers z1 and z2, prove that Re (z1z2) = More information ### Commutative Algebra Notes Introduction to Commutative Algebra Atiyah & Macdonald Commutative Algebra Notes Introduction to Commutative Algebra Atiyah & Macdonald Adam Boocher 1 Rings and Ideals 1.1 Rings and Ring Homomorphisms A commutative ring A with identity is a set with two binary More information ### Galois theory for dummies Galois theory for dummies Ruben Spaans May 21, 2009 1 Notes on notation To help avoid vertical figures, I use the notation E/F if E is an extension to the field F. This is the same notation as Wikipedia More information ### Theory of Matrices. Chapter 5 Chapter 5 Theory of Matrices As before, F is a field We use F[x] to represent the set of all polynomials of x with coefficients in F We use M m,n (F) and M m,n (F[x]) to denoted the set of m by n matrices More information ### FACTORING AFTER DEDEKIND FACTORING AFTER DEDEKIND KEITH CONRAD Let K be a number field and p be a prime number. When we factor (p) = po K into prime ideals, say (p) = p e 1 1 peg g, we refer to the data of the e i s, the exponents More information ### r + s = i + j (q + t)n; 2 rs = ij (qj + ti)n + qtn. Chapter 7 Introduction to finite fields This chapter provides an introduction to several kinds of abstract algebraic structures, particularly groups, fields, and polynomials. Our primary interest is in More information ### 26 Ideals and Quotient Rings Arkansas Tech University MATH 4033: Elementary Modern Algebra Dr. Marcel B. Finan 26 Ideals and Quotient Rings In this section we develop some theory of rings that parallels the theory of groups discussed More information ### 6 EXTENDING ALGEBRA. 6.0 Introduction. 6.1 The cubic equation. Objectives 6 EXTENDING ALGEBRA Chapter 6 Extending Algebra Objectives After studying this chapter you should understand techniques whereby equations of cubic degree and higher can be solved; be able to factorise More information ### The finite field with 2 elements The simplest finite field is The finite field with 2 elements The simplest finite field is GF (2) = F 2 = {0, 1} = Z/2 It has addition and multiplication + and defined to be 0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 0 0 0 = 0 0 1 = 0 More information ### Math 319 Problem Set #3 Solution 21 February 2002 Math 319 Problem Set #3 Solution 21 February 2002 1. ( 2.1, problem 15) Find integers a 1, a 2, a 3, a 4, a 5 such that every integer x satisfies at least one of the congruences x a 1 (mod 2), x a 2 (mod More information ### Lecture 6: Finite Fields (PART 3) PART 3: Polynomial Arithmetic. Theoretical Underpinnings of Modern Cryptography Lecture 6: Finite Fields (PART 3) PART 3: Polynomial Arithmetic Theoretical Underpinnings of Modern Cryptography Lecture Notes on Computer and Network Security by Avi Kak (kak@purdue.edu) January 29, 2015 More information ### Factoring Polynomials Factoring Polynomials Factoring Factoring is the process of writing a polynomial as the product of two or more polynomials. The factors of 6x 2 x 2 are 2x + 1 and 3x 2. In this section, we will be factoring More information ### FINITE FIELDS KEITH CONRAD FINITE FIELDS KEITH CONRAD This handout discusses finite fields: how to construct them, properties of elements in a finite field, and relations between different finite fields. We write Z/(p) and F p interchangeably More information ### SMT 2014 Algebra Test Solutions February 15, 2014 1. Alice and Bob are painting a house. If Alice and Bob do not take any breaks, they will finish painting the house in 20 hours. If, however, Bob stops painting once the house is half-finished, then the More information ### Application. Outline. 3-1 Polynomial Functions 3-2 Finding Rational Zeros of. Polynomial. 3-3 Approximating Real Zeros of. Polynomial and Rational Functions Outline 3-1 Polynomial Functions 3-2 Finding Rational Zeros of Polynomials 3-3 Approximating Real Zeros of Polynomials 3-4 Rational Functions Chapter 3 Group Activity: More information ### Basic Algorithms In Computer Algebra Basic Algorithms In Computer Algebra Kaiserslautern SS 2011 Prof. Dr. Wolfram Decker 2. Mai 2011 References Cohen, H.: A Course in Computational Algebraic Number Theory. Springer, 1993. Cox, D.; Little, More information ### 3.2 The Factor Theorem and The Remainder Theorem 3. The Factor Theorem and The Remainder Theorem 57 3. The Factor Theorem and The Remainder Theorem Suppose we wish to find the zeros of f(x) = x 3 + 4x 5x 4. Setting f(x) = 0 results in the polynomial More information ### 1.3 Polynomials and Factoring 1.3 Polynomials and Factoring Polynomials Constant: a number, such as 5 or 27 Variable: a letter or symbol that represents a value. Term: a constant, variable, or the product or a constant and variable. More information ### MA3D5 Galois theory. Miles Reid. Jan Mar 2004 printed Jan 2014 MA3D5 Galois theory Miles Reid Jan Mar 2004 printed Jan 2014 Contents 1 The theory of equations 3 1.1 Primitive question........................ 3 1.2 Quadratic equations....................... 3 1.3 The More information ### UNCORRECTED PAGE PROOFS number and and algebra TopIC 17 Polynomials 17.1 Overview Why learn this? Just as number is learned in stages, so too are graphs. You have been building your knowledge of graphs and functions over time. More information
# Estimating with Percents 2 teachers like this lesson Print Lesson ## Objective SWBAT estimate the percent of a number #### Big Idea Is there a difference between estimating a percent and finding a percent of a number? ## Launch 5 minutes • POD As students enter the room, they will have a seat, take out their Problem of the Day (POD) sheet and begin to work on the question on the SMARTboard. The POD allows students to use MP 3 continually based on the discussions we have about the problem each day. The POD focuses student thinking on converting percents, fractions, and decimals into other forms. As we begin to estimate percents, I don’t want students to get caught unable to estimate because the conversion is a problem for them. How do you change these percents to fractions and decimals: 20%, 30%, 40%, 50%, 60%, 70%, 80%, 90%. What are the decimal equivalents? How can you use 10% to find the fraction and decimal equivalents for the list of percents. Sometimes an exact answer is not needed when using percents. Name a situation when an exact answer may not be necessary. ## Explore 30 minutes To explore estimates we will do examples together that students can add to their notes. We will work through the examples as a whole class so that students can see how the problems are being solved and possibly see some alternate ways to solve the problem. After we work through the problems using estimation, we can go back through them finding exact answers. My goal is to have them see the effectiveness of a reasonable estimate along with how close the estimate is to the exact answer. We can then discuss the relevance of using one form over the other. ## Landing 5 minutes To end class, I want the exit ticket to help students make the connection between the last two activities we have done in class. The question will serve as a formative assessment to determine who understands the relationship between estimating percents and finding percents of numbers. Do they recognize the similarities of each process? EXIT TICKET How could you round 62%? What is 60% written as a fraction? How could you use this to estimate 62% of 500?
# Dll grade 6 quarter 1 week 6 melc based dll grade 6 quarter 1 week 6 melc based Grade 6 DLL (Detailed Lesson Plan) for Quarter 1, Week 6 (MELC Based) Subject: [Insert Subject Here] I. Objectives At the end of the lesson, students should be able to: 1. [Outcome 1] Understand and demonstrate [specific skill/concept]. 2. [Outcome 2] Apply the concept of [lesson-specific topic] in real-world scenarios. 3. [Outcome 3] Analyze [specific data/concept/text] and draw pertinent conclusions. II. Subject Matter • Topic: [Insert Topic Here] • Lesson Reference: [Insert book/reference material details] • Materials: [List all required materials, including visual aids, handouts, and other resources] III. Procedure A. Review of Previous Lessons • Briefly discuss the key points from the previous lesson. • Ask the students some review questions to refresh their memory. B. Motivation • Show an engaging video clip or image related to the lesson. • Ask students to share their thoughts and predictions based on the visual. C. Lesson Proper 1. Presentation of the New Material • Introduce the lesson topic and objectives. • Explain the relevance of the lesson to daily life. • Discuss key concepts and terms. Solution By Steps: Step 1: Introduction to the Topic • Define and explain key terms and concepts. • Use real-life examples to illustrate the topic. Example: If the topic is fractions, define what a fraction is, and give examples of where we see fractions in real life, such as dividing a pizza. Step 2: Demonstration/Modeling • Conduct a demonstration or model the concept. • Use visual aids such as charts, and diagrams for better understanding. Example: Solve a sample fraction problem on the board and explain each step. $$\text{For example: Solve } \frac{1}{4} + \frac{3}{4} = \frac{1+3}{4} = \frac{4}{4} = 1$$ Step 3: Guided Practice • Provide exercises for students to work on with guidance. • Circulate around the room to provide assistance. Example: Give students similar fraction problems to solve in pairs. $$\text{Examples: } \frac{2}{5}+\frac{2}{5}, \frac{3}{8}+\frac{1}{8}$$ 2. Group Activity/Experimentation • Organize students into groups. • Conduct an activity or experiment related to the lesson topic. • Allow students to discuss and collaborate. 3. Analysis and Generalization • Have students report their findings and observations. • Facilitate a discussion to synthesize the information learned. D. Application • Provide real-life scenarios or problems for students to solve individually or in groups. • Encourage students to apply the concepts they have learned to these new situations. Example: Present a scenario where students need to use fractions to solve a problem, such as sharing a certain amount of supplies among a group. $$\text{If you have 6 students and 3 markers, how many markers does each student get? } \frac{3}{6} = \frac{1}{2}$$ E. Assessment • Give a quiz or worksheet to check students’ understanding of the lesson. • Use formative assessment techniques like exit cards or quick writes. F. Assignment • Assign homework that reinforces the day’s lesson. • Provide instructions for any research or project work related to the next lesson. IV. Reflection • Reflect on what worked well and what needs improvement. • Take note of students who may need additional support or challenge in future lessons. V. Remarks • Any important reminders or announcements for the next class. VI. Agreement • Confirm the next lesson’s topic and any preparatory work students need to do. This template ensures that the lesson is structured, thorough, and engaging for Grade 6 students, aligning with the MELC (Most Essential Learning Competencies) based curriculum requirements. The detailed steps and examples aid in making the lesson interactive and educational.
How Things Work - Chapter 3 Demonstrations Section 3.1 Spring Scales Demonstration 3.1.1: "Measuring" Mass by Shaking Description: A massive object is found inside two identical containers by shaking them to "measure" the masses of those containers. Purpose: To illustrate that you can quantify "how much" there is of an object by measuring its mass Supplies: 2 identical low-mass containers, each hanging from a cord. 1 massive object, to hide in one of the containers. Procedure: Have a student close his/her eyes while you hide the massive object in one of the containers. Then have the student find the object without lifting either container (weighing is forbidden). Of course, the student will shake the two containers and easily discover which one is the more massive and thus which one contains the object. Explanation: Shaking the container measures its mass. The more massive it is, the less it will accelerate in response to a given force. Demonstration 3.1.2: "Measuring" Weight by Lifting Description: A heavy object is found inside two identical containers by lifting them to "measure" the weights of those containers. Purpose: To illustrate that you can quantify "how much" there is of an object by measuring its weight Supplies: 2 identical low-weight containers, each hanging from a cord. 1 heavy object, to hide in one of the containers. Procedure: Have a student close his/her eyes while you hide the heavy object in one of the containers. Then have the student find the object without shaking either container (measuring mass is forbidden). Of course, the student will slowly lift the two containers and easily discover which one is heavier and thus which one contains the object. Explanation: Lifting the container measures its weight. The heavier it is, the more upward force the student will have to exert on it to support its weight so that it doesn't accelerate. Demonstration 3.1.3: An Object Hanging in Equilibrium Description: The heavy object from the previous demonstrations simply hangs at equilibrium, its weight supported perfectly by the upward pull of a cord. Purpose: To illustrate equilibrium. Supplies: 1 heavy object, hanging from a cord. Procedure: Hang the heavy object and let everything settle so that nothing is accelerating. Ask whether the object has lost its weight. Ask why the object isn't falling. Ask how hard the cord must be pulling up on the object and have someone explain that answer. Ask what question you might pose to the cord that would allow you to determine the object's weight. Discuss also whether the object must be motionless for this procedure to work, or whether they must merely be at constant velocity. If you're ambitious, you could carry the hanging object steadily across the room (at constant velocity) to illustrate that equilibrium means zero acceleration, not (necessarily) zero velocity. Explanation: The cord is supporting the object's weight: it is exerting an upward force on the object that's equal in amount to the object's weight, but in the upward direction. The object is thus in equilibrium, which is why it doesn't fall. If you could get the cord to tell you how hard it is pulling upward on the object, you'd know the weight of the object. Demonstration 3.1.4: A Spring's Behavior At and Near Equilibrium Description: A coil spring, supported from above, is shown to exert a restoring force that's proportional to how far it's distorted away from its equilibrium shape or position. Purpose: To illustrate equilibrium and to show that a spring's restoring forces are proportional to its distortion (Hooke's law). Supplies: 1 coil spring 1 support for the coil spring (supporting it from above) 1 ruler or other measuring device (it should be fixed in place next to the coil spring so that the students can see how the spring's length changes) 3 or more identical reference weights Procedure: Suspend the coil spring from the support and align the ruler next to it so that the ruler's zero is next to the spring's free end. Point out that the end of the spring is motionless and not accelerating, so that it must be experiencing a net force of zero—it must be in equilibrium. Now hang first one, then two, then all three reference weights from the spring. Note that the spring adopts a new equilibrium height after each addition. Since the weights' weights are being exerted downward on the spring, the spring must be pulling upward on the weights with a force that's equal in magnitude to their weights. Point out that the spring has had to stretch in order to exert this upward force on the weights and that the extent of this stretch is proportional to the upward force the spring is exerting on the weights. Determine the distance stretch per unit of force exerted, thereby calibrating the spring. Explanation: The spring obeys Hooke's law, exerting an upward force on the objects that's proportional to how far the spring has been stretched downward. Since two objects weigh twice as much as a single object, the spring must stretch twice as far. For three objects, it must stretch three times as far. Demonstration 3.1.5: A Ruler is a Spring Description: A flexible ruler that extends from the edge of a table acts as a spring when objects are placed on it. Purpose: To show that almost everything acts as a spring when deformed slightly. Supplies: 1 flexible ruler (a plastic ruler or a wooden meter stick) 3 identical objects 1 heavy book (or any other anchor for the ruler) Procedure: Extend the ruler from the edge of a table, using the book to anchor its supported end to the table. Note that the free end of the ruler adopts an equilibrium height. Now add first one, then two, then three objects to the end of the ruler. Point out that the ruler deforms downward with each additional object and that the amount of its deformation is proportional to the weight that it's supporting. Explanation: The ruler is acting as a spring, deforming downward by an amount that's proportional to the restoring force it's experiencing. This restoring force is supporting the weight of the objects on its end. Since the objects have equal weights, the ruler's restoring force and its deformation are both proportional to the number of objects the ruler is supporting. Demonstration 3.1.6: Weigh An Object Hanging in Equilibrium Using the Spring Description:The heavy object from the previous demonstrations is hung from the spring at equilibrium, and its weight is measured by observing the spring's distortion. Purpose: To show how a spring scale measures weight. Supplies: 1 heavy object 1 coil spring 1 support for the coil spring (supporting it from above) 1 ruler or other measuring device (it should be fixed in place next to the coil spring so that the students can see how the spring's length changes) Procedure: Let the spring settle at equilibrium. Note the current position of the spring's unsupported end. Now hang the heavy object from the spring and again let the system settle at equilibrium. Note the new position of the spring's unsupported end. Using the calibration of the spring from the previous demonstrations, determine how much the object weighs. Ask why it's important to not touch the object during weighing. Ask why it's important that the object not be accelerating during weighing. Explanation: The spring is now supporting the weight of the object, so it is exerting an upward force on the object equal in magnitude to the object's weight. The spring is using its restoring force to exert that upward force, so it is distorted by an amount that is proportional to the object's weight. Having calibrated the spring (its restoring force per unit of distortion), it's now possible to measure the object's weight. Demonstration 3.1.7: A Hanging Spring Scale Description: As you hang an object from a hanging spring scale, it's spring stretches and its needle indicates the object's weight. Purpose: To show that a hanging scale is a spring scale. Supplies: 1 hanging spring scale 1 weight to hang from it Procedure: Hang the weight from the spring scale and watch the spring stretch and the needle move. Show that the needle is actually reporting the spring's stretch, calibrated so that it appears in units of weight. Show the importance of waiting until nothing is accelerating. Explanation: The hanging scale contains a spring that deforms as you add weights to the scale. This deformation allows the spring to exert the upward support force that keeps the weights from falling. The scale's hanger descends until the spring's restoring force is just enough to provide an upward support force that's equal in magnitude to the weights of the objects the scale is supporting. Demonstration 3.1.8: A Pan Spring Scale Description: As you place an object on a pan spring scale, it's spring stretches and its needle indicates the object's weight. Purpose: To show that a pan scale is a spring scale. Supplies: 1 pan spring scale 1 weight to place on it Procedure: Place the weight on the spring scale and watch the spring stretch and the needle move. Show that the needle is actually reporting the spring's stretch, calibrated so that it appears in units of weight. Show the importance of waiting until nothing is accelerating. Explanation: The pan scale contains a spring that deforms as you add weights to the scale. This deformation allows the spring to exert the upward support force that keeps the weights from falling. The scale's pan descends until the spring's restoring force is just enough to provide an upward support force that's equal in magnitude to the weights of the objects the scale is supporting. Demonstration 3.1.9: A Bathroom Scale Description: As you step on a bathroom scale, its surface descends slightly as the spring inside it deforms. The scale measures this deformation in order to determine your weight. Purpose: To show that even a bathroom scale is really a spring scale. Supplies: 1 bathroom scale Procedure: Step on a bathroom scale and watch the scale's surface descend. Show that the more weight you place on this scale, the farther downward the scale's surface goes. Pick up the scale and squeeze it to show the direct relationship between how far inward you push its surface and the weight that it reports on its dial. Explanation: The bathroom scale contains a spring that deforms as you step on the scale. This deformation allows the spring to exert the upward support force that keeps you from falling into the scale's surface. The scale's surface descends until the spring's restoring force is just enough to provide its surface with an upward support force that's equal in magnitude to your weight. Demonstration 3.1.10: Jumping on a Bathroom Scale Description: As you jump on a bathroom scale, it reads heavy and then light. Purpose: To show that a bathroom scale reports the force it exerts on you, not necessarily your weight. Supplies: 1 bathroom scale Procedure: Step on a bathroom scale and, after letting it settle to equilibrium, observe the scale's reading -- your weight. Now jump upward and watch as the scale reads more than your weight, then less than your weight. Ask what the scale is reporting. Ask if your weight changed during that experiment. Ask why it's so important that you not jump during weighing. Explanation: The bathroom scale reports the force that it exerts on you. If you're not accelerating, then that force is equal in magnitude to your weight. But if you're jumping upward and accelerating upward, then the upward force that the scale exerts on you is greater in magnitude than your weight. And as you subsequently fall, the scale exerts less force on you than the magnitude of your weight. Demonstration 3.1.11: An Accelerating Spring Scale Description: An object hangs from a spring scale as both bounce slowly up and down on the end of a very long spring. The spring scale reads alternately more or less than the object's actual weight. Purpose: To show that when a scale and the object it's supporting accelerate, the force that the scale exerts on the object isn't equal in magnitude to the object's weight. Supplies: 1 spring scale 1 object 1 very long spring or elastic cord Procedure: Suspend the object from the spring scale and then suspend the spring scale from the very long spring. Allow the scale to hang motionless from the spring and note that the scale reads the true weight of the object. Then make the scale and object bounce gently up and down. The scale will alternately read more or less than the object's weight. Explanation: Whenever the scale and object are below their equilibrium position and the very long spring is stretched downward, the scale and object are accelerating upward. The scale must therefore pull upward extra hard on the object and the scale reads more than the object's weight. Whenever the scale and object are above their equilibrium position, they are accelerating downward and the scale reads less than the object's weight. Follow-up: Allow the scale and object to bounce so high that they enter free fall. At that point, the scale will read zero! Be careful that the object doesn't fall off the spring scale. Demonstration 3.1.12: A Hanging Pan Centers Itself Automatically Description: A hanging pan automatically adjusts its angle so that its center of gravity is directly below its support. Purpose: To show that a hanging object will tip until its center of gravity is directly below the point from which it's being supported. Supplies: 1 hanging pan (or any basket that's supported by strings that merge to a single point of support) 1 support for the pan 3 objects to put in the pan Procedure: Suspend the hanging pan from the support and allow it to settle. Its center of gravity should then be directly below the point from which it's supported. Now begin adding the objects to the pan. Show that the pan tips until its new center of gravity is directly below the support point. Explanation: When the pan's center of gravity is directly below its support point, the pan is in equilibrium—it experiences no net force and no net torque. But whenever the pan's center of gravity isn't below the support point, it experiences a torque about its support point that restores it to its equilibrium position. This torque is a restoring torque because, just as restoring force of a spring returns it to its equilibrium position, this restoring torque always returns the pan to its equilibrium orientation. Demonstration 3.1.13: The Difficulty in Weighing an Astronaut Description: You jump off a stool while holding a loaded spring scale. The scale reads zero while you are falling. Purpose: To show that a spring scale only reads the weight of the objects its holding when the objects aren't accelerating. Supplies: 1 spring scale 1 object 1 short stool Procedure: Hang the object from the spring scale and stand motionless on the stool. The scale will read the weight of the object. Now jump carefully from the stool and allow the scale and the object to fall with you. Pay attention to your landing so that you don't hurt yourself. During the time that you, the scale, and the object are in free fall, the scale will read zero. Explanation: When the object is falling, the only force acting on it is gravity. Since the scale doesn't exert any upward force on the object, the scale's spring doesn't distort and the scale reports that it's exerting zero force on the object. Section 3.2 Bouncing Balls Demonstration 3.2.1: How Different Balls Bounce Description: Several different balls rebound to different heights after being dropped. Purpose: To show that different balls have different coefficients of restitution and thus return different fractions of the collision energy as rebound energy. Supplies: 3 different balls (or more) 1 set of happy and unhappy balls (optional—available from a scientific supply company) Procedure: Drop the different balls one at a time from a set height. Show that, while none of them return to their original heights, some bounce higher than others. Discuss how energy changes form from gravitational potential energy, to kinetic energy, to elastic potential energy (in the ball), to kinetic energy, and back to gravitational potential energy as the ball bounces. If you can obtain a happy/unhappy ball pair, show how well the happy ball bounces (essentially a superball) and how incredibly poorly the unhappy ball bounces (a remarkably dead ball—it barely bounces at all). Explanation: As they deform during a collision, different balls have different efficiencies at storing energy. Hard balls that store energy via compression tend to bounce much better than soft balls that store energy via bending surfaces. Demonstration 3.2.2: Bouncing from a Trampoline Description: A relatively dead ball bounces nicely from a tightly inflated plastic bag. Purpose: To show that the surface from which a ball bounces can contribute to the rebound. Supplies: 1 non-lively ball (an "unhappy" ball is ideal and so is a beanbag) 1 air-filled plastic bag Procedure: Show that the ball doesn't bounce well from a solid surface. Then show that the ball bounces reasonably well from the surface of the plastic bag. Explanation: During the collision between ball and surface, the one that deforms the most receives the majority of the collision energy. Since a non-lively ball doesn't return much of the collision energy, its rebound depends critically on how lively the surface it strikes is and on what fraction of the collision energy goes into that surface. Since the plastic bag deforms easily and stores energy well, it allows even an "unhappy" ball to bounce well. While the ball returns no rebound energy, the surface does. Demonstration 3.2.3: Bouncing from Rising Surface or Bat Description: A baseball bounces especially high from a rising board or bat. Purpose: To show that the movement of a surface from which a ball bounces can contribute to the rebound. Supplies: 1 baseball 1 thick board or baseball bat Procedure: Let baseball fall and bounce from the non-moving board (or bat) and watch how high it rebounds. Now let the baseball bounce from the board as that board is rising rapidly upward. The ball will rebound much higher. Discuss the approach speed between the ball and board in the two cases. Discuss the separation speeds in those cases. Ask where the extra energy comes from to propel the ball so high. Can you see that energy flow into the ball by way of work? Explanation: When the ball rebounds from the stationary board, the ball's approach speed is simply its own speed. But when the board is rising to meet the falling ball, the approach speed is much greater. The separation speed is a constant fraction of the approach speed, so when the board and ball are both moving toward one another before the impact, their separation speed after the impact is that much greater. And because the ball is separating from an upward moving board, it has to move upward especially fast. In short, everything about the impact between the rising board and falling ball acts to toss the rebounding ball upward especially fast. Demonstration 3.2.4: Ball Bouncing from a Moving Baseball Bat Description: A human "animation" of a ball bouncing from a moving bat. Purpose: To show the importance of inertial reference frame in following a collision between moving objects and to show how a baseball can rebound from a moving bat with a greater speed than it had before it hit the bat. Supplies: 1 baseball 1 baseball bat Procedure: Walk the students carefully through a collision between a ball moving at 100 km/h toward home plate and a bat moving at 100 km/h toward the pitcher. Start in the spectators' reference frame with the baseball and bat moving toward one another. Point out that the closing speed between the two is 200 km/h. Then shift to the bat's frame of reference, in which the bat is stationary, the pitcher (and the whole stadium) is heading toward home plate at 100 km/h, and the ball is heading toward home plate at 200 km/h. Allow the ball to bounce from the stationary bat (assume infinite mass for the bat) and rebound at 110 km/h (assume a coefficient of restitution of 0.55). Now the ball is heading toward the pitcher at 110 km/h from the bat's frame of reference. Finally, shift back to the spectators' frame of reference. The pitcher is stationary again and the bat is now heading toward pitcher at 100 km/h, so the ball must be heading toward the pitcher at a speed of 210 km/h (110 km/h + 100 km/h). The ball is now moving faster than anything else in the stadium! Explanation: The collision appears simple only in the inertial frame of reference of the bat. In the spectators' frame of reference, the ball and bat are both moving when they collide and the result is somewhat counterintuitive. Demonstration 3.2.5: A Tennis Ball Bouncing from a Basketball Description: A tennis ball sits atop a basketball as the two are dropped from a modest height. When they hit the floor, the tennis ball leaps into the air and rises well above its original height. Purpose: To show that bouncing from a moving surface can lead to counterintuitive results and to show the importance of picking a good inertial frame of reference from which to observe a bounce. Supplies: 1 tennis ball (or another small, elastic ball) 1 basketball (or another massive, elastic ball) Procedure: Balance the tennis ball atop the basketball and hold them a few feet above the floor or a firm table. Now drop the pair. The tennis ball should rebound to a height much greater than its original height. Explanation: The tennis ball effectively completes its bounce from the basketball after the basketball has already bounced off the floor. As a result, the tennis ball is bouncing from a rising surface. Like a ball that's been hit by a rising baseball bat, the tennis ball rebounds with a speed that's larger than its speed before it hit the basketball. If both balls were ideally elastic (coefficients of restitution of 1.0) and if the basketball were infinitely massive, the tennis ball would rebound to 9 times its original height. Demonstration 3.2.6: A Bouncy Ball Transfers More Momentum Description: A bouncy object swings into a block that's balanced on its end and the block falls over. A less bouncy object of identical mass swings into the block but this time the block doesn't fall over. Purpose: To show that a bouncy object transfers momentum during a collision both as the object slows to a stop and as it rebounds backward. An object that doesn't bounce transfers momentum only as it slows to a stop and thus transfers less momentum. Supplies: 1 hard block that can be balanced on its end 1 bouncy ball (such as a happy ball) 1 non-bouncy ball (such as an unhappy ball) string 1 support for balls Procedure: Use the string to suspend the two balls from the support. Place the block on end in front of the bouncy ball, pull the ball back, and let it swing into the block. Determine how far back you must pull the ball in order to knock the block over. Now try the same experiment with the non-bouncy ball. You should have to pull it back much farther in order to knock over the block. Alternative Procedure: Use a metal rod as a battering ram—suspend it on several strings so that it swings forward smoothly and strikes the block. Now put Silly Putty on the block to create a bouncy surface for the metal rod to hit. Determine how far back you must pull the battering ram in order to knock over the block. Now try the same experiment but replace the Silly Putty with a soft, non-elastic putty. You will have to pull the battering ram much farther back in order to knock over the block when the battering ram hits the non-elastic putty. Explanation: When an object slows to a stop during a collision, it transfers all of its forward momentum to the surface it hits. If that object rebounds, it will then transfer additional forward momentum to the surface so that the object leaves with backward momentum. Demonstration 3.2.7: Sending a Croquet Ball Description: Two croquet balls are touching one another when one of the balls is hit sharply with a mallet. The struck ball immediately hits the second ball head on and comes to a stop. The second ball continues on with the momentum and energy of the first ball. Purpose: To illustrate the transfers of energy and momentum that occur during a collision. Supplies: 2 croquet balls 1 croquet mallet Procedure: Place the two croquet balls side by side so that they almost touch. Now strike the outside surface of one ball firmly and briefly so that the ball travels directly toward the second ball. When the first ball hits the second ball, the first ball will come to a stop and the second ball will take over its motion. Explanation: Croquet balls are highly elastic, so that when they collide, they exchange both momentum and energy. The first ball pushes on the second ball both as they approach one another and as they rebound. This relatively elastic collision allows the first ball to transfer virtually all of its momentum and most of its energy to the second ball and the second ball takes over the first ball's motion. Follow-up: Try a similar experiment with pool or billiard balls and a cue stick. Also, experiment with an "executive toy," a toy (discussed below) with several steel balls that are suspended from a wooden frame so that they can swing into one another. If you pull one of the balls back and let it swing into the others, only the last ball in the chain will swing out. You can do this same experiment with a row of identical coins on a tabletop. Demonstration 3.2.8: The Executive Toy Description: A set of metal balls suspended from a frame exchange energy and momentum with one another. Purpose: To illustrate the transfers of energy and momentum that occur during collisions. Supplies: 1 executive toy Procedure: Pull one ball out and let go of it. It will swing back toward its equilibrium position and collide with the other balls when it arrives there. Its momentum and energy will be conveyed from ball to ball until the far ball pops out of the line and continues the motion of the original ball. That last ball will swing up and out until gravity and its support strings reverse its motion and the process repeats in reverse. Explanation: The metal balls are highly elastic and have equal masses. Like billiard balls, they exchange energy and momentum beautifully and almost completely. In this case, the first ball transfers essentially all of its forward momentum and all of its kinetic energy to the second ball during the impact, coming to a stop. The second ball does the same to the third ball, and so on until the last ball continues the first ball's motion. Demonstration 3.2.9: Hardball and Less-Hardball Hammers Description: Two identical looking hammers are used to pound a nail into a board. The hardball hammer works fine, but the less-hardball hammer merely bounces off the nail. Purpose: To illustrate the enormous peak impact force that can arise when two hard objects collide. Supplies: 1 hardball on with a mallet handle (you can buy a mallet handle and install it in a hardball with a hole drilled in it. Glue and friction will hold it in place, although you should always be careful not to let the ball fling off its handle). 1 less-hardball on with a mallet handle. These balls look and feel almost like hardballs, but they are slightly softer. 1 board. Procedure: Have a student pound the nail in with the hardball hammer. It should be relatively easy to make progress. Now have the student try the less-hardball hammer. The hammer will just bounce off the nailhead and the nail won't make any progress into the board. Explanation: Although the two ball-mallets look and feel nearly identical, the true hardball is much harder than the other ball. The true hardball therefore conveys all of its downward momentum to the nail in a much shorter period of time and with a much larger force--an impact force. Since the nail is being held in place by static friction, it won't make any progress into the board until the force exerted on it exceeds some large peak value. The hardball's impact force easily exceeds that threshold, but the other ball can't exert enough impact force to start the nail moving. Instead, the less-hard ball transfers its downward momentum too slow to the nail and subsequently bounces back up from the nail without driving the nail into the board. Demonstration 3.2.10: A Baseball Bat's Center of Percussion Description: A baseball bat, hanging by its handle from a support, is struck at various places with a rubber mallet. Only when the bat is struck at its center of percussion does the handle remain in place. Purpose: To show that there is a special point on the bat, its center of percussion, at which you can hit the ball without causing the bat's handle to accelerate. Supplies: 1 baseball bat 1 rubber mallet 1 support string putty Procedure: Attach the string to the bat's handle and hang the bat from the support. With the bat hanging motionless below the support, strike the bat firmly at various points on its business end. Only when you strike the bat on its center of percussion will the handle remain in place (although the bat's body will accelerate away from the impact and the bat will begin to rotate). If you hit the bat almost at its end, the handle will jerk toward the mallet. If you hit the bat near its middle, the handle will jerk away from the mallet. You can show this jerking motion by sticking the putty to the bat's handle. The bat will fling the putty in the direction of its jerk. When you hit the bat exactly at its center of percussion, the putty may still come off the bat because of vibrations, but it will drop more or less straight down. Explanation: When you hit the bat, the bat's center of mass will accelerate away from the mallet but the bat will also begin to rotate about that center of mass. If you hit the bat at its center of percussion, these two motions will cancel at the handle and the handle itself won't accelerate. Demonstration 3.2.11: A Baseball Bat's Vibrational Node Description: A wooden baseball bat, hanging by a string from a support, is struck at several places with a rubber mallet. Only when it's struck at its vibrational node does the bat emit a clear "crack" sound. When struck at other places, the bat emits a buzzing sound. Purpose: To show that a bat can vibrate and that you can avoid making it vibrate only by hitting it at its vibrational node. Supplies: 1 wooden baseball bat 1 rubber mallet 1 support string Procedure: Attach the string to the handle of the bat and hang it from the support. Strike the bat a sharp blow with the mallet and listen to the buzzing sound it emits. When you hit the bat's vibrational node, there should be a significant change in the sound, with it emitting the sharp "crack" sound we associate with a solid impact. The bat's vibrational node should roughly coincide with its center of percussion. Explanation: The bat vibrates in much the same way that a xylophone plate vibrates—the middle of the bat moves in the opposite direction from its two ends. That motion is its fundamental vibrational mode. This mode of vibration leaves two points on the bat motionless and these two vibrational nodes are located along its handle and part way along the business end of the bat. When you strike the bat at one of these nodes, you don't excite its fundamental vibrational mode and it thus emits very little sound. Any sound that the bat emits is at much higher frequencies because it involves higher order vibrational modes. Follow-up: Why is a xylophone plate supported at two points that are each about half way between the middle and end of the plate? Another Follow-up: Hold an 18" C-Thru plastic ruler horizontally by its middle and flap it up and down rapidly. You'll see that the middle and ends are anti-nodes and that it has a node part way toward each end. Demonstration 3.2.12: Rotation and Translation During a Bounce Description: A spinning basketball that's dropped on the floor leaps forward or backward after it hits. Purpose: To show that friction between a ball and the surface it hits can cause the ball's rotational and translational motions to interact with one another. Supplies: 1 basketball (or another large ball) Procedure: Spin the basketball as you drop it and watch what happens when it hits the floor. It will leap forward or backward, depending on its direction of spin. Point out that the support force from the floor is directly upward, so that this effect must be due to friction between the ball's surface and that of the floor. Explanation: The frictional force on the ball causes it to accelerate, increasing its translational motion. This same frictional force also produces a torque that slows the ball's rotational motion. Demonstration 3.2.13: Balls Rolling Down a Ramp or the Importance of Properly Inflated Tires Description: Two identical-looking balls roll down a ramp and the one that bounces poorly takes longer. Purpose: To show that balls that bounce poorly also tend to waste energy as they roll. Supplies: 1 happy ball 1 unhappy ball 1 hard-surfaced ramp Procedure: Show the students that the two ball bounce differently by dropping them onto a hard surface. The unhappy ball will barely bounce at all while the happy ball will bounce nicely. Then ask the students to predict which of the two balls, if any, will reach the bottom of the ramp first. In all likelihood, they won’t have any firm answer. Simultaneously roll the two balls down the ramp from rest. The happy ball should reach the bottom well ahead of the unhappy ball. Explanation: As the balls roll, their surface dent and undent continuously. The happy ball wastes little energy in this process while the unhappy ball wastes a considerable amount. This demonstration explains why poorly inflated tires lower a car’s gas mileage: they waste energy while denting and undenting, becoming hot and squandering gasoline. Section 3.3 Carousels and Roller Coasters Demonstration 3.3.1: The Experience of Weight Description: You sit in a chair and describe how you experience your weight. Purpose: To show that we experience weight in terms of the internal stresses involved in supporting the parts of our bodies. Supplies: 1 chair Explanation: We don't feel weight directly; we feel the forces that develop inside us when we try not to accelerate in response to our weight. Demonstration 3.3.2: The Experience of Acceleration Description: You sit in a chair on a wheeled cart and describe how you experience acceleration. Purpose: To show that we experience acceleration in terms of the internal stresses involved in making the parts of our bodies accelerate together. Supplies: 1 chair 1 wheeled cart on which the chair will fit (or simply use an office chair that has its own wheel). Procedure: Sit in the chair and have a student make you accelerate forward (briefly). Describe what you feel while you are accelerating. You feel your neck pushing your head forward to keep it accelerating along with the rest of you. You feel your shoulders pushing your neck forward. And all the way to your back, which is being pushed forward by the chair. Explanation: We don't feel acceleration directly; we feel the forces that develop inside us when we try not to come apart as we accelerate. Demonstration 3.3.3: Sweeping a Bowling Ball in a Circle Description: A student tries to keep a bowling ball moving in a circle, using only a broom to guide it. She quickly discovers that she has to push it always toward the center of the circle. Purpose: To show that uniform circular motion requires a centripetal acceleration caused by a centripetal force. Supplies: 1 large ball (a bowling ball is ideal) 1 broom 1 marker for the center of the circle Procedure: Mark the center of your circle on the floor. Have a student use the broom to sweep the ball in a circular path around the marker at a steady pace. After getting the ball moving, she'll find that she has to keep pushing the ball toward the center of the circle. That requirement is somewhat unexpected and counter-intuitive, so the student will have to find it by trial and error. Explanation: To keep moving steadily in a circle, an object needs a central force. The broom makes this need for a central force obvious because it doesn't exert much friction and can't mask the direction of the force it's exerting on the ball. Demonstration 3.3.4: Swinging a Ball on a String Description: A rubber ball attached to a string circles your head. The only force on the ball (ignoring gravity) is the inward pull of the string. Purpose: To show that a centripetal force can cause uniform circular motion. Supplies: 1 rubber ball attached to a string Procedure: Swing the ball around your head on the string. Point out that the only horizontal force on the ball is the inward pull of the string. Point out that the ball is always accelerating toward your hand (the center of the circle) because of the inward pull of the string (a centripetal force). Ask the students what will happen if you let go of the string. They should recognize that it will immediately begin to travel in a straight line, continuing forward in the direction it was traveling at the moment you let go of the string. Show them that this is the case. Explanation: Uniform circular motion involves a centripetal acceleration. For an object to travel in a circle at a uniform speed, it must be experiencing a centripetal force. Demonstration 3.3.5: A Book-in-Hand Loop-the-Loop Description: A book held in your open palm remains in your palm as you move it in a vertical circle, even though the book is beneath your palm as you pass through the top of that circle. Purpose: To show that an object traveling in a circle is experiencing a centripetal acceleration and that that acceleration can exceed the acceleration due to gravity. Supplies: 1 book (a paperback novel works just fine) Procedure: Place the book on top of your open palm. Now move your palm quickly in a large vertical circle so that your palm is always facing the center of that circle. When you finish, your arm will have become twisted one full turn. If you do this motion quickly enough, the book will follow your palm and will remain pressed against it even as your hand travels over the top of the circle and the book is below your open palm. You can then circle backward to untwist your arm. Explanation: As long as you make the book accelerate toward the center of the circle faster than the acceleration due to gravity, your palm will have to provide at least part of the centripetal force and the book will remain pressed against your palm. Demonstration 3.3.6: Swinging a Wine Glass on a Pizza Pan Description: A full wineglass remains in place on a pizza pan as you swing that pizza pan in a vertical circle at the end of a rope. Purpose: To show that when an object on a platter is made to travel rapidly in a circle, the object experiences a large centripetal acceleration and needs a large centripetal force from the platter. If that centripetal acceleration exceeds the acceleration due to gravity, then the platter will have to push inward on the object and the object will push back on the platter. Even a liquid (wine) in the object will push against the platter and remain in the object. Supplies: 1 wineglass (or a brandy snifter because a shorter stem makes the starting and stopping easier). 1 pizza platter rope red wine (or red disappearing ink: about 1/4 tsp. of phenolphthalein in 1 liter water, with just enough sodium hydroxide—about 1/16 tsp.—to turn it pink. When exposed to air, carbon dioxide gradually deactivates the sodium hydroxide and renders the mixture colorless.) Procedure: Attach three pieces of rope to the edge of the pizza platter at three evenly spaced locations and join those ropes together about 0.5 m above the platter. Attach a single rope about 1 m long to the three joined ropes. You should be able to hold that single rope and swing the platter in a vertical circle, and the platter's surface should always face the center of the circle. After some practice with non-fragile objects in the platter, particular with a cup of water, try the wineglass. Starting and stopping are much harder than keeping the wineglass going. You must always let the platter swing freely during the starting and stopping—it will lag behind your hand briefly as you start and it will swing past your hand briefly as you stop. If you let it swing properly, the wineglass will remain in place on the platter and everything will go well. Make sure that you start aggressively enough that you go over the top of the circle the first time at full speed. If you go slowly over the top of the circle, you'll have a disaster. Explanation: Traveling in a circle requires a centripetal force. The platter exerts that centripetal force on the wineglass and the wineglass exerts that centripetal force on the wine. Since the wineglass pushes inward on the wine, the wine pushes outward on the wineglass and the two remain pressed against one another, even as they pass upside-down over the top of the circle. Demonstration 3.3.7: Ball on Loop-the-Loop Track Description: A car or ball rolls down the hill of a track and then around a circular loop-the-loop. It remains pressed against the track, even at the top of the loop-the-loop. Purpose: To show that an object traveling in a circle is undergoing centripetal acceleration and requires a centripetal force. Supplies: 1 car or ball 1 toy track with a hill and a loop-the-loop Procedure: Assemble the track so that you have a hill and a loop-the-loop. Make sure that the hill is high enough (at least 5/2 as tall as the loop-the-loop) that the car or ball will move fast enough to remain on the loop-the-loop. Now roll the car or ball down the hill and let it go around the loop-the-loop. If it's traveling fast enough, it will remain pressed against the track, even at the top of the loop-the-loop. Now repeat this experiment from lower points on the hill and show that without sufficient speed, the centripetal acceleration will be less than the acceleration due to gravity and the car or ball will begin to fall rather than follow the track. Explanation: As long as the centripetal acceleration of the car or ball exceeds the acceleration due to gravity, the track will have to provide at least part of the centripetal force on the car or ball and the two will push against one another. Even at the top of the track, the car or ball will remain pressed against the track. Demonstration 3.3.8: Spin Drying Description: A wet towel is swung rapidly in a circle, causing the water to leave it and travel in a straight line. Purpose: To show how a spin dryer works. Supplies: 1 wet towel Procedure: Hold one end of the towel in your hand and swing it rapidly around in a circle. Water will spray off the other end of the towel and travel in a straight line (though it will also fall). Swing it both in a horizontal plane and a vertical plane to show that this effect is essentially independent of gravity. Explanation: For the far end of the towel to travel in a circle, it must experience a centripetal force. This force is provided by your hand and by the tension in the towel. The water, which is not very well attached to the towel, can break free of the towel and travel in a straight line. Demonstration 3.3.9: Break a String with Forces Due to Acceleration Description: A small weigh hangs easily from a piece of thin string until you try to accelerate the weight too quickly. The string breaks. Purpose: To demonstrate that a force is needed to cause acceleration. Supplies: 1 piece of thin string. 1 weight that can be held relatively easily by the string. Procedure: Hang the weight from the string to show that the string is strong enough to support it. Now show that if you try to accelerate the weight rapidly in any direction, the force required may be more than the string can tolerate. You should be able to break the string by accelerating the weight in any direction, even downward. You can also drop the weight and let it pull taut and break the string to show that an object moving downward can also be accelerating upward. Explanation: The force an object exerts on a string is not limited to its weight. If you are causing that object to accelerate, it will pull back on you with a force equal to its mass times its acceleration (in addition to any forces needed to support its weight). If you can’t tolerate that much force, you’re in trouble.
# What is The KenKen Game? In this article, I will explain about a new puzzle game called as KenKen game. I will explain rules of the game and how we can play this game. KenKen game s mathematical and logical abilities. KenKen can be played online through various websites. # KenKen Game ## Background KenKen is a mathematical puzzle game. In 2004, Tetsuya Miyamoto invented the game. Tetsuya Miyamoto is a Japanese teacher. This game is also known with various names such as MathDocu, KenDocu and CaluDoku. This game is becoming popular now-a-days. Over the world, this game is published in nearly about 40 newspapers. KenKen increases players mathematical and logical abilities. This game can be played in various grids like 3X3, 4X4, 5X5, 6X6, 7X7, 8X8 and 9X9. The 5X5 grid means, the box have 5 rows and 5 columns. ## Rules:- 1. Each row and column contains exactly one number i.e. no number is repeated in row or column. 2. If you want to play in 5X5 grid, then you can fill number from 1 to 5 only. 3. Each bold shaded group of cells called cage.(refer Typical KenKen game) There is a target number written in upper left corner in one cell of the cage with a mathematical operators (+, -, *, % (division)). You have to use that mathematical operator on numbers to get that target number. 4. You can repeat the numbers in the cage but you cannot repeat that numbers in that row or column. ## How to Play In attachment, consider first cage (bold lined) which have the number 11 and mathematical operator is '+'. This means that addition of two numbers should be 11. The possible number which we can fill are 9,2 or 2,9 or 3,8 or 8, 3 or 7,4 or 4,7 or 6,5 or 5,6. Now consider second cage which have the number 2 with mathematical operator '%'. This means that division of two numbers should be 2. The possible numbers are 2,1 or 1,2. KenKen can be played online. You can get the link on good link to play KenKen game. Enjoy the playing game and increase your mathematical ability. ## Attachments Author: Medha Srikanth02 Jun 2011 Member Level: Gold Points : 0 O really.. I have never heard of this earlier. Good post. Regards Medha Srikanth Author: Tushar03 Jun 2011 Member Level: Silver Points : 0 Author: Kiran11 Jun 2011 Member Level: Gold Points : 1 Hi Tushar,
Â Â Â Â Â Â Â Â Â Â # Graphing Linear Relations And Functions A Set of Ordered Pair is known as relation. Suppose we have (2, 5), (-3, 4), (12, -5), (5, 7), (0, 5), (6, 9), (1, 6); The value lies on x – axis and it is known as Domain of given set and the value of y – axis is known as range of a set. So in these values 2, -3, 12, 5, 0, 6, 1 are the domain of a set. 5, 4, -5, 7, 5, 9, 6 is range of a set. Function is set of order pair in which one element of domain is exactly mapped with one element of range. Now we see how we plot the graphs of linear Functions. Step 1: First we take some function. Step 2: Then we know that the values present in the domain are plotted towards the x – axis. Step 3: And range value is plotted towards the y – axis. Step 4: At last we join all the points in a graph. Suppose we have (2, 3), (3, 4), (5, 6); In these values 2, 3, 5 is the domain of the function. 3, 4, 6 is The Range of the function. So when we plot graph then the domain values are plotted towards the x – axis and range value towards the y – axis. For Graphing linear Relations we follow the steps which are given below: Step 1: In the relation the domain values are plotted towards the x – axis. Step 2: The range value is plotted towards the y – axis. Step 3: At last we join the points with a line and we get a graph of relation. With the help of these steps we can easily plot the graph of relation and function. This is the process of graphing linear functions. ## Linear Inequalities Linear inequalities can be defined as expressions which include Combination of variables and constants as well as operators. Since it is an inequality it must contain either a greater than sign or a less than sign. These signs are represented as ‘<’ and ‘>’. It can also include less than equal and greater than equal, these are also included in inequalities. An inequa...Read More ## Linear Equations with Two Variables Linear Equations with two variables are expressions in mathematics that consist of two unknown quantities in same equation and we need at least one more such equation to solve for unknowns. On substituting the values of variables in the equation we get the value of complete expression. These lines can be plotted on Cartesian plane by finding out the c...Read More ## Slope in Linear Function Slope can be defined as a number that explains inclination of line which is going up or down. Slope is normally denoted by letter ‘m’. If line is going up that is if we are going from left to the right and during this movement, we go up then it will be said positive slope. If slope of line is going down that is we are going from right to left and coming down, it will be called as ...Read More ## Linear Equations with One Variable A system of Linear Equations can be defined as two or more linear equations which can be solved simultaneously. While solving linear equations Algebra 2 there are three possibilities of solutions: 1. one solution 2. no solution 3. infinite solution Let’s have small introduction about solutions. ## Special Functions Special Functions are mathematical functions that have one or more than one names. Special functions have many different names which are acceptable in mathematics. There is no specified definition of special functions; we can denote the special functions in many ways. There is a list of functions that are accepted as special functions. Elementary functions are also ...Read More ## Writing Linear Equations Linear equation is an algebraic function which consists of Cartesian coordinates x and y on plane. Linear equation with x and y coordinates gives a straight line when drawn on graph. We can draw the line on graph by placing the value of x and y coordinates on graph. Let us see standard form of writing a linear equation: Steps for writing Linear Equations are:
# Add in Columns: Regrouping Twice This is a complete lesson with instruction and exercises about regrouping twice in addition: both in tens and in hundreds (carrying two times), meant for 2nd or 3rd grade. First, the lesson shows with base ten blocks what happens when we need to regroup two times. Then, students do exercises using visual models while recording the sums horizontally. Then they write the addends under each other while still using the visual model. Lastly, we go to the abstract only, without a visual model. The lesson also has word problems and a puzzle problem in the end. 145 + 79 10 ones form a new ten. 10 tens form a new hundred.The total is 224. Can you see that in the picture? (You can also use manipulatives to do this problem.) 166 + 138 10 ones form a new ten. 10 tens form a new hundred.The total is 304. Can you see that in the picture? (You can also use manipulatives to do this problem.) You have to regroup the ones and the tens. You have to regroup two times. 1. Circle ten 1-dots to make a new ten, AND circle ten 10-sticks to make a new hundred. Alternatively, you can do these exercises using base-ten blocks or similar manipulatives. a. + __________ + ________ = __________ b. + __________ + ________ = __________ c. + __________ + ________ = __________ d. + __________ + ________ = __________ hundreds 1 tens 1 ones 1 8 7 + 1 3 8 3 2 5 Add in the ones column: 7 + 8 = 15. There are more than 10 ones, so regroup them as 1 ten 5 ones, writing “1” in the tens column. Add in the tens column: 1 + 8 + 3 = 12. There are 10 tens, so regroup them as 1 hundred, writing “1” in the hundreds column. 2. Write the numbers in the grid, and add. Regroup. You can circle 10 ten-sticks AND Alternatively, you can do these exercises using base-ten blocks or similar manipulatives. a. + 96 + 145 b. + 185 + 157 c. + 359 + 63 d. + 267 + 385 and 384 using the picture. Explain how she did it. 256 + 384 = 6 hundreds and 4 tens = 640 4. Add. Regroup two times if necessary. a. 3 0 6 + 4 6 1 b. 2 9 9 + 2 2 5 c. 4 8 8 + 3 2 2 d. 1 1 5 + 5 3 6 e. 7 0 4 + 1 5 6 f. 2 6 0 + 3 4 1 g. 2 4 8 + 3 7 6 h. 1 7 3 + 6 4 6 i. 4 0 4 1 9 9 + 1 5 6 j. 7 0 1 1 2 9 + 1 0 1 k. 3 3 5 2 1 9 + 2 7 8 l. 1 0 3 2 8 0 + 5 4 7 5. Matt solved 650 + 331 in an interesting way. Can you follow his thinking? Fill in. First I check the hundreds: 600 + 300 makes _________. Then I add the _____________, and I get 50 + ______ = ______ . Lastly in the ones, there's just 0 and 1, which is 1. Okay, so I have these parts: 900, 80, and ____, so that makes _______________. 6. Solve the word problems. a. From Flowertown to Princetown is 148 miles. You travel from Flowertown to Princetown and back to Flowertown. How many miles is that? b. The school bought pencils for \$128, pens for \$219, and notebooks for \$549. Find the total cost of the items. c. Find how many feet it is if you walk all of the way around this triangle. What numbers are missing from the addition problems? 3 + 1 9 3 9 1 2 + 3 6 5 1 7 1 6 9 + 5 7 4 8 8 + 7 9 0 0 This lesson is taken from Maria Miller's book Math Mammoth Add & Subtract 3, and posted at www.HomeschoolMath.net with permission from the author. Copyright © Maria Miller. #### Math Mammoth Add & Subtract 3 A self-teaching worktext for 2nd-3rd grade that covers mental addition & subtraction with three-digit numbers, regrouping with three-digit numbers (carrying and borrowing), rounding, estimating, and word problems.
# Equivalent fraction for 4/6 2/3 ## How do you write an equivalent fraction? Summary: • You can make equivalent fractions by multiplying or dividing both top and bottom by the same amount. • You only multiply or divide, never add or subtract, to get an equivalent fraction. • Only divide when the top and bottom stay as whole numbers. ## What fraction is 4 out of 6? We do this by first finding the greatest common factor of 4 and 6, which is 2. Then, we divide both 4 and 6 by the greatest common factor to get the following simplified fraction: 2/3. Therefore, this equation is true: 4/6 = 2/3. ## How to make an equivalent fraction? It is possible by these methods: • Method 1: Make the Denominators the same • Method 2: Cross Multiply • Method 3: Convert to decimals ## What is the simplest form of 4/6? The simplest form of 4 6 is 2 3. Steps to simplifying fractions Find the GCD (or HCF) of numerator and denominator GCD of 4 and 6 is 2 ## Are the fractions 4 6 and 6 9 equivalent? They are the same. ## Are 2 3 and 4/6 equivalent Yes or no? So notice, 4/6 is the exact same fraction of the whole as 2/3. These are equivalent fractions. We could say that 2/3 is equal to 4/6. ## How do you write 4/5 as an equivalent fraction? Equivalent fraction of 4/5 Thus 8/10, 12/15 and 16/20, 20/25 are equivalent fraction to 4/5. ## How do you find equivalent fractions? To find the equivalent fractions for any given fraction, multiply the numerator and the denominator by the same number. For example, to find an equivalent fraction of 3/4, multiply the numerator 3 and the denominator 4 by the same number, say, 2. Thus, 6/8 is an equivalent fraction of 3/4. ## What fraction is 4 6 equivalent to? 2/3To do so, evaluate the greatest common factor of the numerator and the denominator. Here, the GCF of 4 and 6 is 2, so 4/6 is an equivalent fraction of 2/3 , and the latter is the simplest form of this ratio. ## What is equivalent calculator? Equivalent Expression Calculator is a free online tool that displays the equivalent expressions for the given algebraic expression. BYJU’S online equivalent expression calculator tool makes the calculations and simplification faster and it displays the equivalent expression in a fraction of seconds. ## Are the numbers and 4/5 equivalent? They are all equal, they just look different! ## What fractions are equal to 4 7? Decimal and Fraction Conversion ChartFractionEquivalent Fractions4/78/1432/565/710/1440/566/712/1448/561/82/168/6423 more rows ## What is 4/5 as a decimal? 0.8Answer: 4/5 as a decimal is 0.8. 1 2 and 6 12 . ## What is 3/5 equivalent to as a fraction? Answer: The fractions equivalent to 3/5 are 6/10, 9/15, 12/20, etc. Equivalent fractions have the same value in the reduced form. ## What is 1/4 equal to as a fraction? 2/8Answer: The fractions equivalent to 1/4 are 2/8, 3/12, 4/16, etc. Equivalent fractions have the same value in their reduced form. ## How do you find equivalent equations? Adding or subtracting the same number or expression to both sides of an equation produces an equivalent equation. Multiplying or dividing both sides of an equation by the same non-zero number produces an equivalent equation….For example, these three equations are equivalent to each other:3 + 2 = 5.4 + 1 = 5.5 + 0 = 5. ## What is the equivalent fraction of 2 by 3? Answer: The fractions equivalent to 2/3 are 4/6, 6/9, 8/12, etc. Equivalent fractions have the same value in the reduced form. Explanation: Equivalent fractions can be written by multiplying or dividing both the numerator and the denominator by the same number. ## How do you find equivalent fractions 5th grade? 14:2615:58Equivalent Fractions – 5th Grade Math – YouTubeYouTubeStart of suggested clipEnd of suggested clipIf I multiply by the top by 4 in the bottom by 4 on the top I’ll get 8 and on the bottom I will getMoreIf I multiply by the top by 4 in the bottom by 4 on the top I’ll get 8 and on the bottom I will get 12. So I look for 8. 12. ## How do you find equivalent fractions 4th grade? 5:139:134th Grade: Equivalent Fractions – YouTubeYouTubeStart of suggested clipEnd of suggested clip3. So 7 21 is equal to 1/3 so there are two ways to find equivalent fractions you can eitherMore3. So 7 21 is equal to 1/3 so there are two ways to find equivalent fractions you can either multiply both the numerator and the denominator. By the same number or you can divide the numerator. ## What are Equivalent Fractions? Equivalent fractions are fractions with different numbers representing the same part of a whole. They have different numerators and denominators, but their fractional values are the same. ## How to make a fraction equivalent? Multiply both the numerator and denominator of a fraction by the same whole number. As long as you multiply both top and bottom of the fraction by the same number, you won’t change the value of the fraction , and you’ll create an equivalent fraction. ## What is half of a fraction? For example, think about the fraction 1/2. It means half of something. You can also say that 6/12 is half, and that 50/100 is half. They represent the same part of the whole. These equivalent fractions contain different numbers but they mean the same thing: 1/2 = 6/12 = 50/100 ## How to find equivalent fractions? To find equivalent fractions, you just need to multiply the numerator and denominator of that reduced fraction ( 32) by the same natural number, ie, multiply by 2, 3, 4, 5, 6 ## Is 6 4 a fraction? Important: 6 4 looks like a fraction, but it is actually an improper fraction. ## Can you convert fractions to decimals? This Equivalent Fractions Table/Chart contains common practical fractions. You can easily convert from fraction to decimal, as well as, from fractions of inches to millimeters. ## What is fraction in math? In mathematics, a fraction is a number that represents a part of a whole. It consists of a numerator and a denominator. The numerator represents the number of equal parts of a whole, while the denominator is the total number of parts that make up said whole. For example, in the fraction. 3. ## How to convert decimals to fractions? It does however require the understanding that each decimal place to the right of the decimal point represents a power of 10; the first decimal place being 10 1, the second 10 2, the third 10 3, and so on. Simply determine what power of 10 the decimal extends to , use that power of 10 as the denominator, enter each number to the right of the decimal point as the numerator, and simplify. For example, looking at the number 0.1234, the number 4 is in the fourth decimal place which constitutes 10 4, or 10,000. This would make the fraction#N#1234#N##N#10000#N#, which simplifies to#N#617#N##N#5000#N#, since the greatest common factor between the numerator and denominator is 2. ## How to multiply fractions? Just multiply the numerators and denominators of each fraction in the problem by the product of the denominators of all the other fractions (not including its own respective denominator) in the problem. ## How to find common denominator of fractions? One method for finding a common denominator involves multiplying the numerators and denominators of all of the fractions involved by the product of the denominators of each fraction. Multiplying all of the denominators ensures that the new denominator is certain to be a multiple of each individual denominator. The numerators also need to be multiplied by the appropriate factors to preserve the value of the fraction as a whole. This is arguably the simplest way to ensure that the fractions have a common denominator. However, in most cases, the solutions to these equations will not appear in simplified form (the provided calculator computes the simplification automatically). Below is an example using this method. ## How to divide fractions? In order to divide fractions, the fraction in the numerator is multiplied by the reciprocal of the fraction in the denominator. The reciprocal of a number a is simply. . When a is a fraction, this essentially involves exchanging the position of the numerator and the denominator. The reciprocal of the fraction. ## What is fraction used for in engineering? In engineering, fractions are widely used to describe the size of components such as pipes and bolts. The most common fractional and decimal equivalents are listed below. ## Is it easier to work with simplified fractions? It is often easier to work with simplified fractions. As such, fraction solutions are commonly expressed in their simplified forms.#N#220#N##N#440#N#for example, is more cumbersome than#N#1#N##N#2#N#. The calculator provided returns fraction inputs in both improper fraction form, as well as mixed number form. In both cases, fractions are presented in their lowest forms by dividing both numerator and denominator by their greatest common factor. ## What are Equivalent Fractions? Two or more fractions representing the same part of a whole are called equivalent fractions. ## How to get fraction equivalent to given fraction? To get a fraction equivalent to a given fraction, we multiply or divide the numerator and the denominator of the given fraction by the same non-zero number. ## What are the denominators of 3 9 and 6 18? The denominators of the fractions, 3 9 and 6 18 are 9 and 18. The least common multiple LCM of the denominators 9 and 18 is 18. Now make the denominators of both fractions 18 by multiplying them with suitable numbers.#N#So, 3 9 = 3 × 2 9 × 2 = 6 18#N#And, 6 18 = 6 × 1 18 × 1 = 6 18#N#Note that both the fractions are equivalent to the same fraction 6 18 Thus, the given fractions are equivalent.#N#Note: If the fractions are not equivalent fractions, we can check the greater or smaller fraction by looking at the numerator of both the resultant fractions. Hence, this method can also be used for compairing fractions. ## What is a group of fractions with different denominators called? The group of two or more fractions with different denominators is called Unlike Fractions. Examples: 1 2, 2 4, 3 5, 4 7, 5 8, etc., are all unlike fractions. 6. Unit Fractions: A unit fraction is any fraction with 1 as its numerator and a whole number for the denominator. Examples: 1 2, 1 4, 1 5, 1 7, 1 8, etc., are all unit fractions. ## What is a combination of a whole number and a proper fraction called? A combination of a whole number and a proper fraction is called a mixed fraction.#N#Exam ples: 12 4, 3 5 10, 51 4, 6 3 17, etc., are all mixed fractions. ## Is 4 7 a fraction? Here, 4 7 is a fraction and four – sevenths is a fractional number. ## Can you simplify fractions to find equivalent fractions? We should simplify the given fractions to find whether they are equivalent fractions or not . Simplification to get equivalent fractions can be done until both the numerator and denominator should still be non-zero integers.
Upon completion you will be able to: Understand and use basic graph terminology and concepts # Upon completion you will be able to: Understand and use basic graph terminology and concepts ## Upon completion you will be able to: Understand and use basic graph terminology and concepts - - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - - ##### Presentation Transcript 1. Graphs • Upon completion you will be able to: • Understand and use basic graph terminology and concepts • Define and discuss basic graph and network structures • Design and implement graph and network applications • Design and implement applications using the graph ADT • Define and discuss Dijkstra's shortest path algorithm Data Structures: A Pseudocode Approach with C, Second Edition 2. 11-1 Basic Concepts • A graph is a collection of nodes, called vertices, and a collection of segments, called lines or edges, connecting pairs of vertices. In Basic Concepts we develop the terminology and structure for graphs. In addition to the graph definitions, discussion points include: • Directed and Undirected Graphs • Cycles and Loops • Connected and Disjoint Graphs Data Structures: A Pseudocode Approach with C, Second Edition 3. A directed graph or digraph for short, is a graph in which line has a direction (arrow head) to its successor. The lines in a directed graphs are known as arcs. • An undirected graph is a graph in which there is no direction on the lines, known as edges. Data Structures: A Pseudocode Approach with C, Second Edition 4. Two vertices in a graph are said to be adjacent vertices if there exists an edge that directly connects them. • A path is a sequence of vertices in which each vertex is adjacent to the next one. It does not make any difference whether or not the graph is directed, it may still have paths. In an undirected graph, you may travel in either directions. Simple path is a path such that all its vertices and edges are distinct. • A cycle is a path consisting of at least three vertices that starts and ends with the same vertex. • A loop is a special case of a cycle in which a single arc begins and ends with the same vertex. In a loop, the end points of the edge are the same. Data Structures: A Pseudocode Approach with C, Second Edition 5. Two vertices are said to be connectedif there is a path between them. A graph is said to be connected if, suppressing direction, there is a path from any vertex to any other vertex. • A directed graph is strongly connected if there is a path from each vertex to every other vertex in the digraph. • A directed graph is weakly connected when there are at least two vertices that are not connected. • A graph is disjoint if it is not connected. • The degree of a vertex is the number of lines incident to it. • The outdegree of a vertex in a digraph is the number of arcs leaving the vertex. • The indegree is the number of arcs entering the vertex. Data Structures: A Pseudocode Approach with C, Second Edition 6. 11-2 Operations • We define and discuss the six primitive graph operations required to maintain a graph. • Insert Vertex • Delete Vertex • Add Edge • Delete Edge • Find Vertex • Traverse Graph Data Structures: A Pseudocode Approach with C, Second Edition 7. We begin by pushing the 1st vertex, A, into the stack. • We then loop, popping the stack and after processing the vertex, pushing all of the adjacent vertices into the stack. To process X at step 2, we pop Xfrom the stack, process it and then push G and H into the stack giving the stack contents for step 3 to process H and G. • When the stack is empty, the traversal is complete. Data Structures: A Pseudocode Approach with C, Second Edition 8. We begin by enqueuing vertex, A, in the queue. • We then loop, dequeuing the queue and processing the vertex from the front of the queue. After processing the vertex, we place all of its adjacent vertices into thequeue. We are then ready for the next step. • When the queue is empty, the traversal is complete. Data Structures: A Pseudocode Approach with C, Second Edition 9. 11-3 Graph Storage Structures • To represent a graph, we need to store two sets. The first set represents the vertices of the graph, and the second set represents the edges or arcs. The two most common structures used to store these sets are arrays and linked lists. • Adjacency Matrix • Adjacency List Data Structures: A Pseudocode Approach with C, Second Edition 10. 11-6 Networks • A network is a graph whose lines are weighted. It is also known as a weighted graph. Included in this section are two graph applications that process networks. • Minimum Spanning Tree • Shortest Path Algorithm Data Structures: A Pseudocode Approach with C, Second Edition 11. A spanning tree is a tree that contains all of the vertices in a graph. A minimum spanning tree is a spanning tree in which the total weight of the lines are guaranteed to be the minimum of all possible trees in the graph. If the weights are unique, then there will be only one minimum spanning tree. Minimum Spanning Tree Data Structures: A Pseudocode Approach with C, Second Edition 12. Algorithm of MST (Prim Algoritm) 1. Insert the first vertex (pick any vertex) into the tree 2. From every vertices already in the tree, examine the edges that connected to all adjacent vertices not in the tree. Select the edge with the minimum weight to a vertex not currently in the tree. Insert that minimum-weight edge and the vertex into the tree. 3. Repeat step 2 until all vertices are in the tree. Data Structures: A Pseudocode Approach with C, Second Edition 13. (continued) Data Structures: A Pseudocode Approach with C, Second Edition
In 4th grade students use strategies to estimate numbers. One strategy for estimation is rounding. This is built off of the foundation they received in 3rd grade. Below is how students learned how to use number lines to estimate numbers in 3rd grade. Five or more raise the score, four or less give it a rest! This saying and others similar to it are often used to help students remember how to round numbers. When these students are asked why they rounded 48 to 50, they explain that the 8 in the number 48 is five or more, so they raised the score. They are not able to explain that because 48 is only 2 away from 50, but is 8 away from 40, 48 should be rounded to 50. Open number lines can help students build conceptual understanding of how to round numbers. How to get students to use open number lines when rounding: To build understanding of using open number lines to round, engage students with a human number line. Challenge students to round the number 76 to the nearest 10. Write the number 76 on an index card. Use a string (that 2 students hold) and have a student place the index card on the string with a clothes pin. Pose the question “What are the two closest tens to 76?” Students place the numbers 70 and 80 on either side of 76. Finding the mid-point between 70 and 80 will also help students know exactly where 76 should be on the number line. Ask students “Which ten is 76 closest to?” They will be able to see that 76 is closer to 80. They should also be able to explain that 76 is 4 away from 80. After practice with a human number line, some students will be able to sketch an open number when rounding. Continue to expose students to this strategy when sequencing students while sharing their work. If you want to check out more about 3rd grade open number lines click the following link http://smathsmarts.com/rounding-number-lines/ Using an open number line to find benchmark numbers can help when rounding. If you have an example like the one below and were asked who is closer to the bakery? or who is closer to the pizza? This would be a easy visual and a way to prove your solution. Using that same idea we can use benchmark numbers to help us round on a number line. Below is an example of finding benchmark numbers for 2,356. If you are rounding to the nearest thousand you will get a much different solution than if you are rounding to the hundreds. We would use each of these strategies for different purposes. If you wanted just a ball park number you might want to round to the thousands place but if you wanted a more precise estimation, rounding to the hundreds place would be more appropriate. Either way, using a number line to estimate is a great tool. To learn more about rounding to estimate using benchmark number click on the learnzillion lesson below. https://learnzillion.com/lesson_plans/6250
## Solve out the equations using simpler method, Algebra Assignment Help: Example Solve out each of the following equations. 7x = 9 Solution Okay, although we say above that if we contained a logarithm in front the left side we could obtain the x out of the exponent. That's simple enough to do. We'll only put a logarithm in front of the left side. But, if we put a logarithm there we also have to put a logarithm in front of the right side. Commonly this is referred to as taking the logarithm of both sides. We can employ any logarithm that we'd like to thus let's try the natural logarithm. ln 7x = ln 9 x ln 7 = ln 9 Now, we have to solve for x. it is easier than it looks. If we had 7 x = 9 then we could all solve out for x simply by dividing both of the sides by 7. It works in accurately the same manner here. Both ln7 & ln9 are only numbers. Admittedly, it would take a calculator to find out just what those numbers are, however they are numbers & so we can do the similar thing here. x ln 7/ ln 7 = ln 9/ ln 7 x = ln 9/ ln 7 Now, technically i.e. the exact answer. Though, in this case usually it's best to get a decimal answer so let's go one step further. x = ln 9 / ln 7= 2.19722458 /1.94591015= 1.12915007 Note down that the answers to these are decimal answers more frequently than not. Also, be careful to not make the following mistake here. 1.12915007 = ln 9/ ln 7 ≠ ln ( 9 /7)=0.2513144283 The two are apparently different numbers. At last, let's also utilizes the common logarithm to ensure that we get the same answer. log 7x = log 9 x log 7 =log 9 x =log 9/ log 7 = 0.954242509 /0.845098040= 1.12915007 Thus, sure enough the similar answer. We can use either logarithm, even though there are times while it is more convenient to employ one over the other. #### Percents, How do I figure out what is 20% of 100? How do I figure out what is 20% of 100? #### Write phrase as an algebraic expressio, carly spends \$5 for n notebooks.w carly spends \$5 for n notebooks.w #### Writing Equivalent Ratios, John can drive his car 15 miles on 2 gallons. Ho... John can drive his car 15 miles on 2 gallons. How many gallons of gas will john need to drive 195 miles? #### I need help, Do you accept screen shot because it is a graph. Do you accept screen shot because it is a graph. #### Quiz #5., Working together Jack and Bob can clean a place in 30 minutes. On... Working together Jack and Bob can clean a place in 30 minutes. On his own, Jack can clean this place in 50 minutes. How long does it take Bob to clean the same place on his own? 5/4:3/7 #### Permutations, Subject identification numbers in a certain scientific resear... Subject identification numbers in a certain scientific research project consist of three digits followed by three letters and then three more digits. Assume repetitions are not all #### Slopes, why is the slope of two perpendicular line a negative reciprocal why is the slope of two perpendicular line a negative reciprocal 7 ?(2+N)2 N=0 #### Basic fact, What is a Math Basic Fact? What is a Math Basic Fact? Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!
Pages Sequence - Part 8 In the last post, we learn how to determine a trigonometric formula for a sequence in the case when the characteristic equation has complex roots. Today we will solve more exercises for this case. Let us recall the method to determine a trigonometric formula for a sequence. Suppose we want to determine a general formula for a real number sequence $\{f_n\}$ that satisfies the following recurrence equation $$a_k ~f_n + a_{k−1} ~f_{n−1} + a_{k−2} ~f_{n−2}+ \dots + a_0 ~f_{n−k}=0.$$ The coefficients $a_0, a_1, \dots, a_k$ are real numbers but the characteristic equation $$a_k ~x^k + a_{k−1} ~x^{k−1} + \dots + a_1 ~x + a_0=0$$ has complex roots. We classify the roots into two groups: • Real roots: suppose that the characteristic equation has $t$ real roots $x_1$, $x_2$, ..., $x_t$, where $x_1$ is a root of multiplicity $u_1$, $x_2$ is a root of multiplicity $u_2$, etc... • Complex roots: suppose that the characteristic equation has $s$ pairs of complex roots $z_1$, $\overline{z_1}$, $z_2$, $\overline{z_2}$, ..., $z_s$, $\overline{z_s}$, where $z_1$, $\overline{z_1}$ is a root pair of multiplicity $v_1$, $z_2$, $\overline{z_2}$ is a root pair of multiplicity $v_2$, etc... Write these complex roots in trigonometric form as follows $$z_1, \overline{z_1} = r_1 (\cos{\phi_1} \pm i ~ \sin{\phi_1}); ~\dots; ~z_s, \overline{z_s} = r_s (\cos{\phi_s} \pm i ~ \sin{\phi_s}).$$ Then the sequence can be written in trigonometric form as follows $$f_n = p_1(n) ~x_1^{n} + \dots + p_t(n) ~x_t^{n}$$ $$+ r_1^n (g_1(n) ~\cos{n \phi_1} + h_1(n) ~ \sin{n \phi_1}) + \dots + r_s^{n} (g_s(n) ~ \cos{n \phi_s} + h_s(n) ~ \sin{n \phi_s}),$$ where • $p_1(n)$, ..., $p_t(n)$ are polynomials of real coefficients whose degrees are less than $u_1$, ..., $u_t$, respectively; • $g_1(n)$, $h_1(n)$, ..., $g_s(n)$, $h_s(n)$ are polynomials of real coefficients whose degrees are less than $v_1$, ..., $v_s$. Let us look at some examples. • If the characteristic equation is factored as $(x - z_1)(x - \overline{z_1})=0$ with $$z_1, \overline{z_1} = r_1 (\cos{\phi_1} \pm i ~ \sin{\phi_1})$$ then $$f_n = (g_1 ~\cos{n \phi_1} + h_1 ~ \sin{n \phi_1}) ~r_1^n,$$ where $g_1$, $h_1$ are two real numbers. • If the characteristic equation is factored as $(x - x_1)(x - z_1)(x - \overline{z_1})=0$ then $$f_n = p_1 ~x_1^n + (g_1 ~\cos{n \phi_1} + h_1 ~ \sin{n \phi_1}) ~r_1^n,$$ where $p_1$, $g_1$, $h_1$ are real numbers. • If the characteristic equation is factored as $(x - x_1)^2 (x - z_1)(x - \overline{z_1})=0$ then $$f_n = (p_{11} + p_{12} n) ~x_1^n + (g_1 ~\cos{n \phi_1} + h_1 ~ \sin{n \phi_1}) ~r_1^n.$$ • If the characteristic equation is factored as $(x - x_1)^2 (x - z_1)^2(x - \overline{z_1})^2=0$ then $$f_n = (p_{11} + p_{12} n) ~x_1^n + [(g_{11} + g_{12} n) ~\cos{n \phi_1} + (h_{11}+ h_{12} n) ~ \sin{n \phi_1}] ~r_1^n .$$ • If the characteristic equation is factored as $(x - x_1) (x - z_1)^2(x - \overline{z_1})^2=0$ then $$f_n =p_1 ~x_1^n + [(g_{11} + g_{12} n) ~\cos{n \phi_1} + (h_{11}+ h_{12} n) ~ \sin{n \phi_1}] ~r_1^n .$$ • If the characteristic equation is factored as $(x - z_1)(x - \overline{z_1})(x - z_2)(x - \overline{z_2})=0$ then $$f_n = (g_1 ~\cos{n \phi_1} + h_1 ~ \sin{n \phi_1}) ~r_1^n + (g_2 ~\cos{n \phi_2} + h_2 ~ \sin{n \phi_2}) ~r_2^n.$$ Let us do some exercises. Problem 1: Determine a general formula for the sequence $$f_0=1, ~f_1=4, ~f_n= 2 f_{n−1} − 4 f_{n−2}.$$ Solution: From the recurrence equation $f_n= 2 f_{n−1} − 4 f_{n−2}$ we have the following characteristic equation $$x^2 − 2 x + 4 =0.$$ This quadratic equation has a pair of complex roots $1 \pm i~ \sqrt{3}$. We will express the roots in trigonometric form. First we calculate their absolute value $$\| 1 \pm i~ \sqrt{3} \| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2.$$ Thus, $$1 \pm i~ \sqrt{3} = 2 ~\left( \frac{1}{2} \pm i ~\frac{\sqrt{3}}{2} \right) = 2 (\cos{\frac{\pi}{3}} \pm i ~ \sin{\frac{\pi}{3}}).$$ The sequence has the following form $$f_n = (\alpha ~ \cos{\frac{n \pi}{3}} + \beta ~ \sin{\frac{n \pi}{3}} ) ~ 2^n.$$ With $n=0,1$, we have $$f_0= \alpha = 1,$$ $$f_1= (\alpha ~\frac{1}{2} + \beta ~\frac{\sqrt{3}}{2}) 2 = 4.$$ Solving these equations we obtain $\alpha = 1$ and $\beta = \sqrt{3}$. Therefore, $$f_n = (\cos{\frac{n \pi}{3}} + \sqrt{3} ~ \sin{\frac{n \pi}{3}} ) ~ 2^n.$$ Problem 2: Determine a general formula for the sequence $$f_0=2, ~f_1=4, ~f_n = f_{n−1} − f_{n−2}.$$ Solution: From the recurrence equation $f_n= f_{n−1} − f_{n−2}$ we have the following characteristic equation $$x^2 − x + 1 =0.$$ This quadratic equation has a pair of complex roots $$\frac{1 \pm i ~\sqrt{3}}{2}.$$ We will express the roots in trigonometric form. First we calculate their absolute value $$\left\| \frac{1 \pm i ~\sqrt{3}}{2} \right\| = \sqrt{\left( \frac{1}{2}\right)^2 + \left( \frac{\sqrt{3}}{2}\right)^2} = 1.$$ Thus, $$\frac{1 \pm i ~\sqrt{3}}{2} = \cos{\frac{\pi}{3}} \pm i ~ \sin{\frac{\pi}{3}}.$$ The sequence has the following form $$f_n = \alpha ~ \cos{\frac{n \pi}{3}} + \beta ~ \sin{\frac{n \pi}{3}}.$$ With $n=0,1$, we have $$f_0= \alpha = 2,$$ $$f_1= \alpha ~\frac{1}{2} + \beta ~\frac{\sqrt{3}}{2} = 4.$$ Solving these equations we obtain $\alpha = 2$ and $\beta = 2 \sqrt{3}$. Therefore, $$f_n = 2 ~ \cos{\frac{n \pi}{3}} + 2 \sqrt{3} ~ \sin{\frac{n \pi}{3}} .$$ Problem 3: Determine a general formula for the sequence $$f_0=5, ~f_1=6, ~f_n = 3 f_{n−1} − 3 f_{n−2}.$$ Solution: From the recurrence equation $f_n = 3 f_{n−1} − 3 f_{n−2}$ we have the following characteristic equation $$x^2 − 3 x + 3 =0.$$ This quadratic equation has a pair of complex roots $$\frac{3 \pm i ~\sqrt{3}}{2}.$$ We will express the roots in trigonometric form. First we calculate their absolute value $$\left\| \frac{3 \pm i ~\sqrt{3}}{2} \right\| = \sqrt{\left( \frac{3}{2}\right)^2 + \left( \frac{\sqrt{3}}{2}\right)^2} = \sqrt{3}.$$ Thus, $$\frac{3 \pm i ~\sqrt{3}}{2} = \sqrt{3} \left( \frac{\sqrt{3}}{2} \pm i ~ \frac{1}{2}\right) = \sqrt{3} (\cos{\frac{\pi}{6}} \pm i ~ \sin{\frac{\pi}{6}}).$$ The sequence has the following form $$f_n = (\alpha ~ \cos{\frac{n \pi}{6}} + \beta ~ \sin{\frac{n \pi}{6}}) ~(\sqrt{3})^n.$$ With $n=0,1$, we have $$f_0= \alpha = 5,$$ $$f_1= (\alpha ~\frac{\sqrt{3}}{2} + \beta ~\frac{1}{2}) ~\sqrt{3} = 6.$$ Solving these equations we obtain $\alpha = 5$ and $\beta = - \sqrt{3}$. Therefore, $$f_n = (5 ~ \cos{\frac{n \pi}{6}} - \sqrt{3} ~ \sin{\frac{n \pi}{6}}) ~(\sqrt{3})^n .$$ Problem 4: Determine a general formula for the sequence $$f_0=2, ~f_1=1, ~f_2=10, ~f_n= 4 f_{n−1} − 24 f_{n−3}.$$ Solution: From the recurrence equation $f_n= 4 f_{n−1} − 24 f_{n−3}$ we have the following characteristic equation $$x^3 − 4 x^2 + 24 =0.$$ We can factor it as $$x^3 − 4 x^2 + 24 = (x+2)(x^2 - 6x + 12) = 0.$$ Thus, the characteristic equation has one real root (-2) and two complex roots $3 \pm i~ \sqrt{3}$. We will express the complex roots in trigonometric form. First we calculate their absolute value $$\left\| 3 \pm i~ \sqrt{3} \right\| = \sqrt{3^2 + (\sqrt{3})^2} = \sqrt{12} = 2 \sqrt{3}.$$ Thus, $$3 \pm i~ \sqrt{3} = 2 \sqrt{3} ~\left( \frac{\sqrt{3}}{2} \pm i ~\frac{1}{2} \right) = 2 \sqrt{3} (\cos{\frac{\pi}{6}} \pm i ~ \sin{\frac{\pi}{6}}).$$ The sequence has the following form $$f_n = \alpha ~(-2)^n + (\beta ~ \cos{\frac{n \pi}{6}} + \gamma ~ \sin{\frac{n \pi}{6}} ) ~(2 \sqrt{3})^n.$$ With $n=0,1,2$, we have $$f_0= \alpha + \beta= 2,$$ $$f_1= -2 \alpha + (\beta ~\frac{\sqrt{3}}{2} + \gamma ~\frac{1}{2}) 2 \sqrt{3} = 1,$$ $$f_2 = 4 \alpha +(\beta ~\frac{\sqrt{1}}{2} + \gamma ~\frac{\sqrt{3}}{2}) 12 =10.$$ Solving these equations we obtain $\alpha = 1$, $\beta = 1$ and $\gamma=0$. Therefore, $$f_n = (-2)^n + (2 \sqrt{3})^n ~ \cos{\frac{n \pi}{6}} .$$ Problem 5: Determine a recurrence condition for the sequence $$f_n = (5 ~\cos{\frac{n \pi}{4}} + 3 ~\sin{\frac{n \pi}{4}}) (\sqrt{2})^n.$$ Solution: We need to find a characteristic equation that has two complex roots $$z_{1}, \overline{z_1} = \sqrt{2}(\cos{\frac{\pi}{4}} \pm i ~\sin{\frac{\pi}{4}}).$$ We have $$z_1 \overline{z_1} = (\sqrt{2})^2 = 2,$$ $$z_1 + \overline{z_1} = 2 \sqrt{2} \cos{\frac{\pi}{4}} = 2 \sqrt{2} \frac{\sqrt{2}}{2} = 2,$$ therefore, by Vieta's formula, $z_1$ and $\overline{z_1}$ are two roots of the following quadratic equation $$x^2 - 2 x + 2 =0.$$ Thus, we obtain the recurrence equation $$f_n= 2 f_{n−1} − 2 f_{n−2}.$$ Combining with the initial condition, we have $$f_0 = 5, ~~f_1 = 8, ~~f_n = 2 f_{n-1} - 2 f_{n-2}.$$ Problem 6: Determine a recurrence condition for the sequence $$f_n = \cos{\frac{n \pi}{4}} + \sin{\frac{n \pi}{4}}.$$ Solution: We need to find a characteristic equation that has two complex roots $$z_{1}, \overline{z_1} = \cos{\frac{\pi}{4}} \pm i ~\sin{\frac{\pi}{4}}.$$ We have $$z_1 \overline{z_1} = 1,$$ $$z_1 + \overline{z_1} = 2 \cos{\frac{\pi}{4}} = 2 \frac{\sqrt{2}}{2} = \sqrt{2},$$ therefore, by Vieta's formula, $z_1$ and $\overline{z_1}$ are two roots of the following quadratic equation $$x^2 - \sqrt{2} x + 1 =0.$$ Thus, we obtain the recurrence equation $$f_n= \sqrt{2} f_{n−1} − f_{n−2}.$$ Combining with the initial condition, we have $$f_0 = 1, ~~f_1 = \sqrt{2}, ~~f_n = \sqrt{2} f_{n-1} - f_{n-2}.$$ Problem 7: Determine a recurrence condition for the sequence $$f_n = 2n+1 + (3 ~\cos{\frac{n \pi}{6}} - \sqrt{3} ~\sin{\frac{n \pi}{6}}) (\sqrt{3})^n.$$ Solution: We have $$f_n = (2n+1)~ 1^n + (3 ~\cos{\frac{n \pi}{6}} - \sqrt{3} ~\sin{\frac{n \pi}{6}}) (\sqrt{3})^n.$$ We need to find a characteristic equation that has one real root $x_1 = 1$ of multiplicity 2 and two complex roots $$z_{1}, \overline{z_1} = \sqrt{3} (\cos{\frac{\pi}{6}} \pm i ~\sin{\frac{\pi}{6}}).$$ We have $$z_1 \overline{z_1} = (\sqrt{3})^2 = 3,$$ $$z_1 + \overline{z_1} = 2 \sqrt{3}~\cos{\frac{\pi}{6}} = 2 \sqrt{3} \frac{\sqrt{3}}{2} = 3,$$ therefore, by Vieta's formula, $z_1$ and $\overline{z_1}$ are two roots of the following quadratic equation $$x^2 - 3 x + 3 =0.$$ Thus, we obtain the characteristic equation $$(x - 1)^2 (x^2 - 3x + 3) = x^4 - 5x^3 + 10 x^2 - 9 x + 3 =0.$$ The corresponding recurrence equation is $$f_n= 5 f_{n−1} − 10 f_{n−2} + 9 f_{n-3} - 3 f_{n-4}.$$ Combining with the initial condition, we have $$f_0 = 4, ~~f_1 = 6, ~~f_2 = 5, ~~f_3 = -2, ~~f_n= 5 f_{n−1} − 10 f_{n−2} + 9 f_{n-3} - 3 f_{n-4}.$$ Problem 8: Determine a recurrence condition for the sequence $$f_n = (2n ~\cos{\frac{n \pi}{3}} - 2 \sqrt{3} ~\sin{\frac{n \pi}{3}}) 3^n.$$ Solution: We need to find a characteristic equation that has two complex roots of multiplicity 2 $$z_{1}, \overline{z_1} = 3 (\cos{\frac{\pi}{3}} \pm i ~\sin{\frac{\pi}{3}}).$$ We have $$z_1 \overline{z_1} = 3^2 = 9,$$ $$z_1 + \overline{z_1} = 6~\cos{\frac{\pi}{3}} = 3,$$ therefore, by Vieta's formula, $z_1$ and $\overline{z_1}$ are two roots of the following quadratic equation $$x^2 - 3 x + 9 =0.$$ Thus, we obtain the characteristic equation $$(x^2 - 3x + 9)^2 = x^4 - 6 x^3 + 27 x^2 - 54 x + 81 =0.$$ The corresponding recurrence equation is $$f_n= 6 f_{n−1} − 27 f_{n−2} + 54 f_{n-3} - 81 f_{n-4}.$$ Combining with the initial condition, we have $$f_0 = 0, ~~f_1 = -6, ~~f_2 = -45, ~~f_3 = -162, ~~f_n= 6 f_{n−1} − 27 f_{n−2} + 54 f_{n-3} - 81 f_{n-4}.$$ Let us stop here for now. The next post will be the last post of our series on sequence. Hope to see you again then. Homework. 1. Prove that for any values of $f_0$ and $f_1$, the following sequence is periodic $$f_n = 2 \cos{\frac{\pi}{2013}} f_{n-1} - f_{n-2}.$$ (A sequence is called periodic if there is a period $K \neq 0$ such that $f_{n+K} = f_n$ for all $n$.) 2. Given the following sequence $$f_0 = 1, ~~f_1 = 2 \cos{\frac{\pi}{2013}}, ~~ f_n = 4 \cos{\frac{\pi}{2013}} f_{n-1} - 4 f_{n-2}$$ Prove that $f_{2013}$ is an integer.
# Thread: this is as far as I got, not sure of the next step, group like terms but how? 1. ## this is as far as I got, not sure of the next step, group like terms but how? -3(2x -2) +5x = -4x + 2(x-1) -6 + 6 + 5x = -4x - 2x - 2 2. ## Re: this is as far as I got, not sure of the next step, group like terms but how? Your first term on the left side should include x. Your second term on the right side should be positive. This gives you: $\displaystyle -6x+6+5x=-4x+2x-2$ Grouping like terms, we may write: $\displaystyle (5x-6x)+6=(2x-4x)-2$ Combining the like terms, we find: $\displaystyle -x+6=-2x-2$ Can you proceed from here? 3. ## Re: this is as far as I got, not sure of the next step, group like terms but how? is it ? -2x -x = 6 -2 x = 4 ? 4. ## Re: this is as far as I got, not sure of the next step, group like terms but how? No. If you are unsure about your solution, substitute it into the original equation to see if it satisfies the equation. For $\displaystyle x=4$, we get: $\displaystyle -3(2(4)-2)+5(4)=-4(4)+2(4-1)$ $\displaystyle -3(8-2)+20=-16+2(3)$ $\displaystyle -3(6)+20=-16+6$ $\displaystyle -18+20=-10$ $\displaystyle 2=-10$ This is not true, so the solution is wrong. We have: $\displaystyle -x+6=-2x-2$ We may first add $\displaystyle 2x$ to both sides: $\displaystyle -x+2x+6=-2x+2x-2$ Combining like terms, we find: $\displaystyle x+6=-2$ Now subtract 6 from both sides, and you get: $\displaystyle x+6-6=-2-6$ $\displaystyle x=-8$ Now, checking the solution: $\displaystyle -3(2(-8)-2)+5(-8)=-4(-8)+2(-8-1)$ $\displaystyle -3(-16-2)-40=32+2(-9)$ $\displaystyle -3(-18)-40=32-18$ $\displaystyle 54-40=14$ $\displaystyle 14=14$ This is true so we know the solution is correct. 5. ## Re: this is as far as I got, not sure of the next step, group like terms but how? wow thanks my issue is that I can't seem to get each step in the correct order I feel so lost but your technique is very very helpful. Again thanks
## Elementary Linear Algebra 7th Edition (a) ${u} \cdot {v}=-6$. (b) ${v} \cdot {v}=13$. (c) $\\|u\\|^{2}=25$. (d) $({u} \cdot {v})v=-6\left[\begin{array}{l}{2}\\ {-3}\end{array}\right]$. (e) ${u} \cdot {5v}=-30$. Given ${u}=\left[\begin{array}{l}{3} \\ {4}\end{array}\right]$ and ${v}=\left[\begin{array}{l}{2}\\ {-3}\end{array}\right]$, then we have (a) ${u} \cdot {v}={u}^{T} {v}=[3 \ \ 4]\left[\begin{array}{l}{2} \\ {-3}\end{array}\right]=[(3)(2)+(-3)(4)]=-6$. (b) ${v} \cdot {v}={v}^{T} {v}=[2 \ \ -3]\left[\begin{array}{l}{2} \\ {-3}\end{array}\right]=[(2)(2)+(-3)(-3)]=13$. (c) $\\|u\\|^{2}={u} \cdot {u}={u}^{T} {u}=[3 \ \ 4]\left[\begin{array}{l}{3} \\ {4}\end{array}\right]=[(3)(3)+(4)(4)]=25$. (d) $({u} \cdot {v})v=({u}^{T} {v}) v=[3 \ \ 4]\left[\begin{array}{l}{2} \\ {-3}\end{array}\right] \left[\begin{array}{l}{2}\\ {-3}\end{array}\right]=-6\left[\begin{array}{l}{2}\\ {-3}\end{array}\right]$. (e) ${u} \cdot {5v}={u}^{T} {5v}=[3 \ \ 4]\left[\begin{array}{l}{10} \\ {-15}\end{array}\right]=[(3)(10)+(4)(-15)]=-30$.
# Texas Go Math Grade 4 Lesson 10.2 Answer Key Divide Using Partial Quotients Refer to our Texas Go Math Grade 4 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 4 Lesson 10.2 Answer Key Divide Using Partial Quotients. ## Texas Go Math Grade 4 Lesson 10.2 Answer Key Divide Using Partial Quotients Essential Question How can you use partial quotients to divide by 1-digit divisors? Unlock the Problem At camp, there are 5 players on each lacrosse team. If there are 125 people on lacrosse teams, how many teams are there? • Underline what you are asked to find. • Circle what you need to use. • What operation can you use to find the number of teams? One Way Use partial quotients. In the partial quotient method of dividing, multiples of the divisor are subtracted from the dividend and then the partial quotients are added together. Divide. 125 ÷ 5 Write. STEP 1 Start by subtracting a greater multiple, such as 10 times the divisor. For example, you know that you can make at least 10 teams of 5 players. Continue subtracting until the remaining number is less than the multiple, 50. On adding partial quotients : 20 + 5 = 25 Remainder : 0 STEP 2 Subtract smaller multiples, such as 5, 2, or 1 times the divisor until the remaining number is less than the divisor. In other words, keep going until you no longer have enough players to make a team. Then add the partial quotients to find the quotient. So, there are ____________ lacrosse teams. So, there are 25 lacrosse teams. Math Talk Mathematical Processes Explain how you found the total number of teams after finding the partial quotients. We can find the total number of teams by adding the partial quotients. Another Way Use area models to record the partial quotients. Jawd and Mi also found the number of teams using partial quotients. They recorded the partial quotients using area models. they each still had 25 as the quotient. Jarod : 10+10+5 = 25 Mi : 20+5 = 25 Math Talk Mathematical Processes Explain why you might prefer to use one method rather than the other. Answer: We choose the partial quotient division method to complete the given problem in fewer steps. Share and Show Question 1. Lacrosse is played on a field 330 ft long. How many yards long is a lacrosse held? (3 feet = 1 yard) Divide. Use partial quotients. So, the lacrosse field is _____ yards long. 100 x 1 = 100 10 x 1 = 10 Total : 100 + 10 = 110 Divide. Use partial quotients. Question 2. On adding partial quotients = 70+5 = 75 Remainder = 0 Divide. Use area models to record the partial quotients. Question 3. 428 ÷ 4 = __________ On adding partial quotients = 100+7 = 107 Remainder = 0 Math Talk Mathematical Processes Explain how you could Problems 2 and 3 a different way. We can solve the above problems in long division way : 2. Quotient : 75 Remainder : 0 3. 428 ÷ 4 Quotient : 107 Remainder : 0 Problem Solving Use the table for 4-5. Question 4. Multi-Step Rob filled 9 plastic boxes with basketball cards and hockey cards. There were the same number of cards in each box. How many cards did he put iii each box? Number of basket ball cards = 189 Number of hockey cards = 63 Total number = 189 + 63 =252 Now, Total number of plastic cards = 9 So, 252/9 = 28 On adding partial quotients = 20+8 = 28 Therefore, there are 28 cards in each box. Remainder = 0 Question 5. H.O.T. Multi-Step Rob filled 3 fewer plastic boxes with football cards than the lime boxes he filled with basketball cards and hockey cards. How many boxes did he fill? How many football cards were in each box? Question 6. H.O.T. Communicate What is the least number you can divide by 5 to get a three-digit quotient? Explain how you found ‘our answer. Question 7. On the American frontier, salt was 4 times as expensive as beef Suppose a family on the frontier buys 112 pounds of beef. How much salt could they purchase with the same amount of money? (A) 28 pounds (B) 13 pounds (C)22 pounds (D) 33 pounds Given that, salt was 4 times expensive as beef. So, According to given condition 112 pounds of beef = Therefore, 28 pounds of salt they purchase with the same amount of money. Question 8. Multi-Step A family that owns an orchard collects 153 pounds of apples in one day and 126 pounds the next day. If the harvest is divided equally among 9 containers, how many pounds of apples are in each container? (A) 19 pounds (B) 37 pounds (C) 31 pounds (D) 27 pounds The number of apples collected in one day = 153 pounds The number of apples collected in next day = 126 pounds Total = 153 + 126 = 279 Number of containers = 9 Now, 279/9 = Therefore, 31 pounds of apples are in each container Question 9. Apply Zachary’s family is planning a trip to Washington D.C. They will drive 837 miles from their home in St. Louis, Missouri. Zachary’s dad plans to make the trip in 3 days. How many miles should they plan to drive in each day if they want to drive the same distance every day? (A) 314miles (B) 239 miles (C) 409miles (D) 279 miles The total distance Zachary’s family travels = 837 miles Total number of days = 3 According to given condition, 837/3 Therefore, 31 miles they should plan to drive in each day if they want to drive the same distance every day. TEXAS Test Prep Question 10. There are 80 fourth grade students and 46 fifth grade students who signed up to learn how to play lacrosse. If there are 6 students in each group, how many groups are there? (A) 120 (B) 20 (C) 21 (D) 12 Number of forth grade students = 80 Number of fifth grade students = 46 Total = 126 Given, If there are 6 students un each group. Now, 126/6 = Therefore, there are 21 groups. ### Texas Go Math Grade 4 Lesson 10.2 Homework and Practice Answer Key Divide. Use partial quotients. Question 1. On adding partial quotients = 90+0 = 90 Remainder = 0 Question 2. On adding partial quotients = 40+1 = 41 Remainder = 0 Question 3. On adding partial quotients = 20+1 = 21 Remainder = 0 Question 4. On adding partial quotients = 40+6 = 46 Remainder = 0 Question 5. On adding partial quotients = 20+5 = 25 Remainder = 0 Question 6. On adding partial quotients = 30+2 = 32 Remainder = 0 Divide. Use area models to record the partial quotients. Question 7. 104 ÷ 8 = _________ On adding partial quotients = 10+3 = 13 Remainder = 0 Question 8. 210 ÷ 7 = _________ On adding partial quotients = 30+0 = 30 Remainder = 0 Question 9. 365 ÷ 5 = _________ On adding partial quotients = 70+3 = 73 Remainder = 0 Problem Solving Question 10. Valerie drove 320 miles in 5 hours. If she drove the same amount of miles each hour, how many miles did she drive in one hour? Total distance = 320 miles Total time = 5 hours According to given condition, if she drove the same amount of miles each hour Distance travels in 1 hour = 320/5 = Therefore, she drives 64 miles in each hour. Question 11. Keisha has 128 photos to put in a photo album. Each page holds 8 photos. How many pages will Keisha fill? Total number of photos = 128 Number of photos each photo hold = 8 Now, Therefore, Keisha can fill 16 pages. Lesson Check Question 12. Mr. Carson spent $144 on 8 bushes for his ‘yard. If each hush costs the same amount, how much did each bush cost? (A)$22 (B) $18 (C)$23 (D) $20 Answer: Number of bushes = 8 Total cost of 8 bushes =$144 If each hush costs the same amount, Now, 144/8 = Therefore, each bush cost $18. Question 13. A furniture company moved 126 couches from their warehouse to the showroom. The truck made 9 trips to deliver the couches. If the truck carried the same number of couches in each trip, how many couches did the truck carry in one trip? (A) 13 (B) 18 (C) 15 (D) 14 Answer: The number of couches company moves to showroom = 126 The number of trips the truck made = 9 If the truck carried the same number of couches in each trip, Now Therefore, The truck carries 14 couches in one trip. Question 14. Leticia has$138 to spend on gifts for 6 friends. Each gift will cost the same amount of money. How much will she spend on each gift? (A) $23 (B)$21 (C) $24 (D)$22 Total number of Leticia friends = 6 The amount she spent = $138 Also given, Each gift will cost the same amount of money. So, Therefore, Leticia spent$23 for each gift. Question 15. A custom order baseball cap company can make 320 caps in 5 days. If they make the same number of caps each day, how many caps can they make in one day? (A) 50 (B) 80 (C) 64 (D) 70 The number of caps Baseball cap company make = 320 Number of days = 5 If they make the same number of caps each day, Now, Therefore, they can make 64 caps in one day. Question 16. Multi-Step Marshall can buy 5 tee shirts for $60. If each shirt costs the same amount, what is the cost of 4 tee shirts? (A)$12 (B) $48 (C)$15 (D) $60 Answer: Number of tee shirts = 5 Total amount for 5 tee shirts =$60 If each shirt costs the same amount, Now, The cost of 1 each shirt = $12 Now, the cost of 4 tee shirts =$12 x 4 = 48 Therefore, the cost 4 tee shirts = $48. Question 17. Multi-Step The Jones family paid$36 for 2 burger meals and 4 chicken strip meals. Each of the meals costs the same amount of money. What would the Jones family have paid if they bought only the 4 chicken strip meals? (A) $24 (B)$18 (C) $12 (D)$6 The amount Jones family paid for 2 burger and 4 chicken strips = $36 Given, Each of the meals costs the same amount of money.$36/6 = $6 So, each meal cost$6 each. 4 x $6 =$24.
# A sphere of diameter 12 cm, is dropped in a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water, the water level in the cylindrical vessel rises by 3 5/9 cm. Find the diameter of the cylindrical vessel. - Mathematics A sphere of diameter 12 cm, is dropped in a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water, the water level in the cylindrical vessel rises by 3 5/9 cm. Find the diameter of the cylindrical vessel. #### Solution Radius of sphere = r = 6 cm Volume of sphere = 4/3pir^3=4/3pixx(6)^3 = 288pi ""cm^3 Let R be the radius of cylindrical vessel. Reise in the water level of cylindrical vessel = h=3 5/9 "cm" = 32/9 "cm" Increase in volume of cylindrical vessel = piR^2h=piR^2xx32/9=32/9piR^2 Now, volume of water displaced by the sphere is equal to volume of sphere. :.32/9piR^2=288pi :. R^2=(288xx9)/32=81 ∴ R = 9 cm ∴ Diameter of the cylindrical vessel 2xR = 2x9 =18 cm Concept: Concept of Surface Area, Volume, and Capacity Is there an error in this question or solution?
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ # Appendix B: Answers for the Progress Checks $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$ Section 1.1 Progress Check 1.2 1. This proposition is false. A counterexample is $$a = 2$$ and $$b = 1$$. For these values, $$(a + b)^2 = 9$$ and $$a^2 + b^2 = 5$$. 2. This proposition is true, as we can see by using $$x = 3$$ and $$y = 7$$. We could also use $$x = -2$$ and $$y = 9$$. There are many other possible choices for $$x$$ and $$y$$. 3. This proposition appears to be true. Anytime we use an example where $$x$$ is an even integer, the number $$x^2$$ is an even integer. However, we cannot claim that this is true based on examples since we cannot list all of the examples where $$x$$ is an even integer. 4. This proposition appears to be true. Anytime we use an example where $$x$$ and $$y$$ are both integers, the number $$x \cdot y$$ is an odd integer. However, we cannot claim that this is true based on examples since we cannot list all of the examples where both x and y are odd integers. Progress Check 1.4 1. (a) This does not mean the conditional statement is false since when $$x = -3$$, the hypothesis is false, and the only time a conditional statement is false is when the hypothesis is true and the conclusion is false. (b) This does not mean the conditional statement is true since we have not checked all positive real numbers, only the one where $$x = 4$$. (c) All examples should indicate that the conditional statement is true. 2. The number ($$n2 - n + 41$$) will be a prime number for all examples of $$n$$ that are less than 41. However, when $$n = 41$$, we get $\begin{array} {rcl} {n^2 - n + 41} &= & {41^2 - 41 + 41} \\ {n^2 - n + 41} &= & {41^2} \end{array}$ So in the case where $$n = 41$$, the hypothesis is true (41 is a positive integer) and the conclusion is false ($$41^2$$ is not prime). Therefore, 41 is a counterexample that shows the conditional statement is false. There are other counterexamples (such as $$n = 42$$, $$n = 45$$, and $$n = 50$$), but only one counterexample is needed to prove that the statement is false. Progress Check 1.5 1. We can conclude that this function is continuous at 0. 4. We can conclude that this function is not differentiable at 0. Progress Check 1.7 1. The set of rational numbers is closed under addition since $$\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{ad + bc}{bd}$$. 2. The set of integers is not closed under division. For example, $$\dfrac{2}{3}$$ is not an integer. 3. The set of rational numbers is closed under subtraction since $$\dfrac{a}{b} - \dfrac{c}{d} = \dfrac{ad - bc}{bd}$$. Section 1.2 Progress Check 1.10 All examples should indicate the proposition is true. Following is a proof. Proof. We assume that $$m$$ is an odd integer and will prove that ($$3m^2 + 4m + 6$$). Since $$m$$ is an odd integer, there exists an integer $$k$$ such that $$m D= 2k + 1$$. Substituting this into the expression ($$3m^2 + 4m + 6$$) and using algebra, we obtain $$\begin{array} {rcl} {3m^2 + 4m + 6} &= & {3(2k + 1)^2 + 4(2k + 1) + 6} \\ {} &= & {(12k^2 + 12k + 3) + (8k + 4) + 6} \\ {} &= & {12k^2 + 20k + 13} \\ {} &= & {12k^2 + 20k + 12 + 1} \\ {} &= & {2(6k^2 + 10k + 6) + 1} \end{array}$$ Progress Check 1.11 Proof. We let $$m$$ be a real number and assume that $$m$$, $$m + 1$$, and $$m + 2$$ are the lengths of the three sides of a right triangle. We will use the Pythagorean Theorem to prove that $$m = 3$$. Since the hypotenuse is the longest of the three sides, the Pythagorean Theorem implies that $$m^2 + (m + 1)^2 = (m + 2)^2$$. We will now use algebra to rewrite both sides of this equation as follows: $$\begin{array} {rcl} {m^2 + (m^2 + 2m + 1)} &= & {m^2 + 4m + 4} \\ {2m^2 + 2m + 1} &= & {m^2 + 4m + 4} \end{array}$$ The last equation is a quadratic equation. To solve for $$m$$, we rewrite the equation in standard form and then factor the left side. This gives $$\begin{array} {rcl} {m^2 - 2m - 3} &= & {0} \\ {(m - 3) (m + 1)} &= & {0} \end{array}$$ The two solutions of this equation are $$m = 3$$ and $$m = -1$$. However, since $$m$$ is the length of a side of a right triangle, $$m$$ must be positive and we conclude that $$m = 3$$. This proves that if $$m$$, $$m + 1$$, and $$m + 2$$ are the lengths of the three sides of a right triangle, then $$m = 3$$. Section 2.1 Progress Check 2.1 1. Whenever a quadrilateral is a square, it is a rectangle, or a quadrilateral is a rectangle whenever it is a square. 2. A quadrilateral is a square only if it is a rectangle. 3. Being a rectangle is necessary for a quadrilateral to be a square. 4. Being a square is sufficient for a quadrilateral to be a rectangle. Progress Check 2.2 $$P$$ $$Q$$ $$P \wedge \urcorner Q$$ $$\urcorner (P \wedge Q)$$ $$\urcorner P \wedge \urcorner Q$$ $$\urcorner P \vee \urcorner Q$$ T T F F F F T F T T F T F T F T F T F F F T T T Statements (2) and (4) have the same truth table. Progress Check 2.4 $$P$$ $$\urcorner P$$ $$P \vee \urcorner P$$ $$P \wedge \urcorner P$$ T F T F F T T F $$P$$ $$Q$$ $$P \vee Q$$ $$P \to (P \vee Q)$$ T T T T T F T T F T T T F F F T Section 2.2 Progress Check 2.7 1. Starting with the suggested equivalency, we obtain $\begin{array} {rcl} {(P \wedge \urcorner Q) \to R} &\equiv & {\urcorner (P \wedge \urcorner Q) \vee R} \\ {} &\equiv & {(\urcorner P \vee \urcorner (\urcorner Q)) \vee R} \\ {} &\equiv & {\urcorner P \vee (Q \vee R)} \\ {} &\equiv & {P \to (Q \vee R)} \end{array}$ 2. For this, let $$P$$ be,“3 is a factor of $$a \cdot b$$,” let $$Q$$ be, “is a factor of $$a$$,” and let $$R$$ be, “3 is a factor of $$b$$.” Then the stated proposition is written in the form $$P \to (Q \vee R)$$. Since this is logically equivalent to $$(P \wedge \urcorner Q) \to R$$, if we prove that if 3 is a factor of $$a \cdot b$$ and 3 is not a factor of $$a$$, then 3 is a factor of $$b$$, then we have proven the original proposition. Section 2.3 Progress Check 2.9 1. $$10 \in A$$, $$22 \in A$$, $$13 \notin A$$, $$0 \in A$$, $$-12 \notin A$$ 2. $$A = B, A \subseteq B, B \subseteq A, A \subseteq C, A \subseteq D, B \subseteq C, B \subseteq D$$ Progress Check 2.11 1. (a) Two values of $$x$$ for which $$P(x)$$ is false are $$x = 3$$ and $$x = -4$$. (b) The set of all $$x$$ for which $$P(x)$$ is true is the set {-2, -1, 0, 1, 2}. 2. (a) Two examples for which $$R(x, y, z)$$ is false are: $$x = 1, y = 1, z = 1$$ and $$x = 3$$, $$y = -1$$, $$z = 5$$. (b) Two examples for which $$R(x, y, z)$$ is true are: $$x = 3, y = 4, z = 5$$ and $$x = 5, y = 12, z = 13$$. Progress Check 2.13 1. The truth set is the set of all real numbers whose square is less than or equal to 9. The truth set is $$\{x \in \mathbb{R}\ \|\ x^2 \le 9\} = \{x \in \mathbb{R}\ \|\ -3 \le x \le 3\}$$. 2. The truth set is the set of all integers whose square is less than or equal to 9. The truth set is {-3, -2, -1, 0, 1, 2, 3}. 3. The truth sets in Parts (1) and (2) equal are not equal. One purpose of this progress check is to show that the truth set of a predicate depends on the predicate and on the universal set. Progress Check 2.15 $$A = \{4n - 3\ \|\ n \in \mathbb{N}\} = \{x \in \mathbb{N}\ \|\ x = 4n - 3 \text{ for some natural number $n$$}\}.$ $$B = \{-2n\ \|\ \text{$n$$ is a nonnegative integer\}.$ $$C = \{(\sqrt{2})^{2m - 1}\ \|\ m \in \mathbb{N}\} = \{(\sqrt{2})^n\ \|\ \text{$n$$ is an odd natural number}\}.$ $$D = \{3^n\ \|\ \text{$n$$ is a nonnegative integer}\}.$ Section 2.4 Progress Check 2.18 1. $$\bullet$$ For each real number $$a$$, $$a + 0 = a$$. $$\bullet$$ $$(\exists a \in \mathbb{R}) (a + 0 \ne a).$$ $$\bullet$$ There exists a real number $$a$$ such that $$a + 0 \ne a$$. 2. $$\bullet$$ For each real number $$x$$, sin (2$$x$$) = 2(sin $$x$$)(cos $$x$$). $$\bullet$$ $$(\exists x \in \mathbb{R})$$ (sin (2$$x$$) $$\ne$$ 2 (sin $$x$$) (cos $$x$$)). $$\bullet$$ There exists a real number $$x$$ such that sin (2$$x$$) $$\ne$$ 2 (sin $$x$$) (cos $$x$$). 3. $$\bullet$$ For each real number $$x$$, $$\text{tan}^2 x + 1 = \text{sec}^2 x$$. $$\bullet$$ $$(\exists x \in \mathbb{R}) (\text{tan}^2 x + 1 \ne \text{sec}^2 x)$$. $$\bullet$$ There exists a real number $$x$$ such that $$\text{tan}^2 x + 1 \ne \text{sec}^2 x$$. 4. $$\bullet$$ There exists a rational number $$x$$ such that $$x^2 - 3x - 7 = 0$$. $$\bullet$$ $$(\forall x \in \mathbb{Q})(x^2 - 3x - 7 \ne 0).$$ $$\bullet$$ For each rational number $$x$$, $$x^2 - 3x - 7 \ne 0$$. 5. $$\bullet$$ There exists a real number $$x$$ such that $$x^2 + 1 = 0$$. $$\bullet$$ $$(\forall x \in \mathbb{R})(x^2 + 1 \ne 0).$$ $$\bullet$$ For each real number $$x$$, $$x^2 + 1 \ne 0$$. Progress Check 2.19 1. A counterexample is $$n = 4$$ since $$4^2 + 4 + 1 = 21$$, and 21 is not prime. 2. A counterexample is $$x = \dfrac{1}{4}$$ since $$\dfrac{1}{4}$$ is positive and $$2(\dfrac{1}{4})^2 = \dfrac{1}{8}$$ and $$\dfrac{1}{8} \le \dfrac{1}{4}$$. Progress Check 2.20 1. An integer $$n$$ is a multiple of 3 provided that $$\exists k \in \mathbb{Z})(n = 3k)$$. 4. An integer $$n$$ is not a multiple of 3 provided that $$\forall k \in \mathbb{Z})(n \ne 3k)$$. 5. An integer $$n$$ is not a multiple of 3 provided that for every integer $$k$$, $$n \ne 3k$$. Progress Check 2.21 • $$(\exists x \in \mathbb{Z})(\exists y \in \mathbb{Z}) (x + y \ne 0)$$. • There exist integers $$x$$ and $$y$$ such that $$x + y \ne 0$$. Section 3.1 Progress Check 3.2 1. For each example in Part (1), the integer $$a$$ divides the sum $$b + c$$. 2. Conjecture: For all integers $$a$$, $$b$$, and $$c$$ with $$a \ne 0$$, if $$a$$ divides $$b$$ and $$a$$ divides $$c$$, then $$a$$ divides $$b + c$$. 3. A Know-show table for a proof of the conjecture in Part (3). Step Know Reason $$P$$ $$a\ \|\ b$$ and $$a\ \|\ c$$ Hypothesis $$P$$1 $$(\exists s \in \mathbb{Z})(b = a \cdot s)$$ $$(\exists t \in \mathbb{Z})(c = a \cdot t)$$ Definition of "divides" $$P$$2 $$b + c = as + at$$ Substituting for $$b$$ and $$c$$ $$P$$3 $$b + c = a(s + t)$$ Distributive property $$Q$$1 $$s + t$$ is an integer $$\mathbb{Z}$$ is closed under addition $$Q$$ $$a\ \|\ (b + c)$$ Definition of "divides" Step Show Reason Progress Check 3.3 A counterexample for this statement will be values of a and b for which 5 divides $$a$$ or 5 divides $$b$$, and 5 does not divide $$5a + b$$. One counterexample for the statement is $$a = 5$$ and $$b = 1$$. For these values, the hypothesis is true since 5 divides a and the conclusion is false since $$5a + b = 26$$ and 5 does not divide 26. Progress Check 3.4 1. Some integers that are congruent to 5 modulo 8 are -11, -3, 5, 13, and 21. 2. $$\{x \in \mathbb{Z}\ \|\ x \equiv 5\text{ (mod 8)\} = \{..., -19, -11, -3, 5, 13, 21, 29, ...\}$$. 3. For example, -3 + 5 = 2, -11 + 29 = 18, 13 + 21 = 34. 4. If we subtract 2 from any of the sums obtained in Part (3), the result will be a multiple of 8. This means that the sum is congruent to 2 modulo 8. For example, $$2 - 2 = 0$$, $$18 - 2 = 16$$, $$34 - 2 = 32$$. Progress Check 3.6 1. To prove that 8 divides $$(a + b - 2)$$, we can prove that there exists an integer $$q$$ such that ($$a + b - 2 = 8q$$). 2. Since 8 divides ($$a - 5$$) and ($$b - 5$$), there exist integers $$k$$ and $$m$$ such that $$a - 5 - 8k$$ and $$b - 5 = 8m$$. 3. $$a = 5 + 8k$$ and $$b = 5 + 8m$$. 4. $$a + b - 2 = (5 + 8k) + 5 + 8m) - 2 = 8 + 8k + 8m = 8(1 + k + m)$$. 5. Proof. Let a and b be integers and assume that $$a \equiv 5$$ (mod 8) and $$b \equiv 5$$ (mod 8). We will prove that $$(a + b) \equiv 2$$ (mod 8\). Since 8 divides $$(a - 5)$$ and $$(b - 5)$$, there exist integers $$k$$ and $$m$$ such that $$a - 5 = 8k$$ and $$b - 5 = 8m$$. We then see that $\begin{array} {a + b - 2} &= & {(5 + 8k) + (5 + 8m) - 2] \\ {} &= & {8 + 8k + 8m} \\ {} &= & {8(1 + k + m)} \end{array}$ By the closure properties of the integers, $$(1 + k + m)$$ is an integer and so the last equation proves that 8 divides ($$a + b - 2$$) and hence, $$(a + b) \equiv 2$$ (mod 8). This proves that if $$a \equiv 5$$ (mod 8) and $$b \equiv 5$$ (mod 8), then $$(a + b) \equiv 2$$ (mod 8). Section 3.2 Progress Check 3.8 1. For all real numbers $$a$$ and $$b$$, if $$ab = 0$$, then $$a = 0$$ or $$b = 0$$. 2. For all real numbers $$a$$ and $$b$$, if $$ab = 0$$ and $$a \ne 0$$, $$b = 0$$ 3. This gives $\dfrac{1}{a}(ab) = \dfrac{1}{a} \cdot 0.$ We now use the associative property on the left side of this equation and simplify both sides of the equation to obtain $\begin{array} {rcl} {(\dfrac{1}{a} \cdot a) b} &= & {0} \\ {1 \cdot b} &= & {0} \\ {b} &= & {0} \end{array}$ Therefore, $$b = 0$$ and this completes the proof of a statement that is logically equivalent to the contrapositive. Hence, we have proved the proposition. Section 3.3 Progress Check 3.15 1. There exists a real number $$x$$ such that $$x$$ is irrational and $$\sqrt[3]{x}$$ is rational. 2. There exists a real number $$x$$ such that $$x + \sqrt{2}$$ is rational and $$(-x + \sqrt{2})$$ is rational. 3. There exist integers $$a$$ and $$b$$ such that 5 divides $$ab$$ and 5 does not divide $$a$$ and 5 does not divide $$b$$. 4. There exist real numbers $$a$$ and $$b$$ such that $$a > 0$$ and $$b > 0$$ and $$\dfrac{2}{a} + \dfrac{2}{b} = \dfrac{4}{a + b}$$. Progress Check 3.16 1. Some integers that are congruent to 2 modulo 4 are -6. -2, 2, 6, 10, and some integers that are congruent to 3 modulo 6 are: -9, -3, 3, 9, 15. There are no integers that are in both of the lists. 2. For this proposition, it is reasonable to try a proof by contradiction since the conclusion is stated as a negation. 3. Proof. We will use a proof by contradiction. Let $$n \in \mathbb{Z}$$ and assume that $$n \equiv 2$$ (mod 4) and that $$n \equiv 3$$ (mod 6). Since $$n \equiv 2$$ (mod 4), we know that 4 divides $$n - 2$$. Hence, there exists an integer $$k$$ such that $n - 2 = 4k.$ We can also use the assumption that $$n \equiv 3$$ (mod 6) to conclude that 6 divides $$n - 3$$ and that there exists an integer $$m$$ such that $n - 3 = 6m.$ If we now solve equations (B.5) and (B.6) for n and set the two expressions equal to each other, we obtain $4k + 2 = 6m + 3.$ However, this equation can be rewritten as $2(2k + 1) = 2(3m + 1) + 1.$ Since $$2k + 1$$ is an integer and $$3m + 1$$ is an integer, this last equation is a contradiction since the left side is an even integer and the right side is an odd integer. Hence, we have proven that if $$n \equiv 2$$ (mod 4), then $$n \equiv 3$$ (mod 6). Progress Check 3.18 1. $$x^2 + y^2 = (2m + 1)^2 + (2n + 1)^2 = 2(2m^2 + 2m + 2n^2 + 2n + 1).$$ 2. Using algebra to rewrite the last equation, we obtain $4m^2 + 4m + 4n^2 + 4n + 2 = 4k^2.$ If we divide both sides of this equation by 2, we see that $$2m^2 + 2m + 2n^2 + 2n + 1 = 2k^2$$ or $2(m^2 + m + n^2 + n) + 1 = 2k^2.$ However, the left side of the last equation is an odd integer and the right side is an even integer. This is a contradiction, and so we have proved that for all integers $$x$$ and $$y$$, if $$x$$ and $$y$$ are odd integers, then there does not exist an integer $$z$$ such that $$x^2 + y^2 = z^2$$. Section 3.4 Progress Check 3.21 Proposition. For each integer $$n$$, $$n^2 - 5n + 7$$ is an odd integer. Proof. Let $$n$$ be an integer. We will prove that $$n^2 - 5n + 7$$ is an odd integer by examining the case where $$n$$ is even and the case where $$n$$ is odd. In the case where $$n$$ is even, there exists an integer m such that $$n = 2m$$. So in this case, $$\begin{array} {rcl} {n^2 - 5n + 7} &= & {(2m^2) - 5(2m) + 7} \\ {} &= & {4m^2 - 10m + 6 + 1} \\ {} &= & {2(2m^2 - 5m + 3) + 1.} \end{array}$$ Since ($$2m^2 - 5m + 3$$) is an integer, the last equation shows that if $$n$$ is even, then $$n^2 - 5n + 7$$ is odd. In the case where $$n$$ is odd, there exists an integer $$m$$ such that $$n = 2m + 1$$. So in this case, $$\begin{array} {rcl} {n^2 - 5n + 7} &= & {(2m + 1)^2 - 5(2m + 1) + 7} \\ {} &= & {4m^2 - 14m + 3} \\ {} &= & {2(2m^2 - 7m + 1) + 1.} \end{array}$$ Since ($$2m^2 - 7m + 1$$) is an integer, the last equation shows that if $$n$$ is odd, then $$n^2 - 5n + 7$$ is odd. Hence, by using these two cases, we have shown that for each integer $$n$$, $$n^2 - 5n + 7$$ is an odd integer. Progress Check 3.24 1. \|4.3\| = 4.3 and \|-$$\pi$$\| = $$\pi$$ 2. (a) $$t = 12$$ or $$t = -12$$ (b) $$t + 3 = 5$$ or $$t + 3 = -5$$. So $$t = 2$$ or $$t = -8$$. (c) $$t - 4 = \dfrac{1}{5}$$ or $$t - 4 = -\dfrac{1}{5}$$. So $$t = \dfrac{21}{5}$$ or $$t = \dfrac{19}{5}$$. (d) $$3t - 4 = 8$$ or $$3t - 4 = -8$$. So $$t = 4$$ or $$t = -\dfrac{4}{3}$$. Section 3.5 Progress Check 3.26 1. (a) The possible remainders are 0, 1, 2, and 3. (b) The possible remainders are 0, 1, 2, 3, 4, 5, 6, 7, and 8. (a) $$17 = 5 \cdot 3 + 2$$ (b) $$-17 = (-6) \cdot 3 + 1$$ (c) $$73 = 10 \cdot 7 + 3$$ (d) $$-73 = (-11) \cdot 7 + 4$$ (e) $$436 = 16 \cdot 27 + 4$$ (f) $$539 = 4 \cdot 110 + 99$$ Progress Check 3.29 Proof. Let $$n$$ be a natural number and let $$a, b, c$$ and $$d$$ be integers.We assume that $$a \equiv b$$ (mod $$n$$) and $$c \equiv d$$ (mod $$n$$) and will prove that $$(a + c) \equiv (b + d)$$ (mod $$n$$). Since $$a \equiv b$$ (mod $$n$$) and $$c \equiv d$$ (mod $$n$$), $$n$$ divides $$a - b$$ and $$c - d$$ and so there exist integers $$k$$ and $$q$$ such that $$a - b = nk$$ and $$c - d = nq$$. We can then write $$a = b + nk$$ and $$c = d + nq$$ and obtain $$\begin{array} {rcl} {a + c} &= & {(b + nk) + (d + nq)} \\ {} &= & {(b + d) + n(k + q)} \end{array}.$$ By subtracting $$(b + d)$$ from both sides of the last equation, we see that $$(a + c) - (b + d) = n(k + q).$$ Since $$(k + q)$$ is an integer, this proves that $$n$$ divides $$(a + c) - (b + d)$$, and hence, we can conclude that $$(a + c) \equiv (b + d)$$ (mod $$n$$). Progress Check 3.34 Case 2. ($$a \equiv 2$$ (mod 5)). In this case, we use Theorem 3.28 to conclude that $$a^2 \equiv 2^2$$ (mod 5) or $$a^2 \equiv 4$$ (mod 5). This proves that if $$a \equiv 2$$ (mod 5), then $$a^2 \equiv 4$$ (mod 5). Case 3. ($$a \equiv 3$$ (mod 5)). In this case, we use Theorem 3.28 to conclude that $$a^2 \equiv 3^2$$ (mod 5) or $$a^2 \equiv 9$$ (mod 5). We also know that $$9 \equiv 4$$ (mod 5). So we have $$a^2 \equiv 9$$ (mod 5) and $$9 \equiv 4$$ (mod 5), and we can now use the transitive property of congruence (Theorem 3.30) to conclude that $$a^2 \equiv 4$$ (mod 5). This proves that if $$a \equiv 3$$ (mod 5), then $$a^2 \equiv 4$$ (mod 5). Section 4.1 Progress Check 4.1 1. It is not possible to tell if $$1 \in T$$ and $$5 \in T$$. 2. True. 3. True. The contrapositive is, "If $$2 \in T$$, then $$5 \in T$$," which is true. 4. True. 5. False. If $$k \in T$$, then $$k + 1 \in T$$. 6. True, since "k \notin t\) OR $$k + 1 \in T$$" is logically equivalent to "If $$k \in T$$, then $$k + 1 \in T$$." 7. It is not possible to tell if this is true. It is the converse of the conditional statement, “For each integer $$k$$, if $$k \in T$$, then $$k + 1 \in T$$." 8. True. This is the contrapositive of the conditional statement, “For each integer $$k$$, if $$k \in T$$, then $$k + 1 \in T$$." Progress Check 4.3 Proof. Let $$P(n)$$ be the predicate, "$$1 + 2 + 3 + \cdot\cdot\cdot + n = \dfrac{n(n + 1)}{2}$$." For basis step, notice that the equation $$1 = \dfrac{1(1 + 1)}{2}$$ shows that $$P(1)$$ is true. Now let $$k$$ be a natural number an assume that $$P(k)$$ is true. That is, assume that $1 + 2 + 3 + \cdot\cdot\cdot + k = \dfrac{k(k + 1)}{2}.$ We now need to prove that $$P(k + 1)$$ is true or that $1 + 2 + 3 + \cdot\cdot\cdot + k + (k + 1) = \dfrac{(k + 1)(k + 2)}{2}.$ By adding $$(k + 1)$$ to both sides of equation (B.11), we see that $$\begin{array} {rcl} {1 + 2 + 3 + \cdot\cdot\cdot + k + (k + 1)} &= & {\dfrac{k(k + 1)}{2} + (k + 1)} \\ {} &= & {\dfrac{k(k + 1) + 2(k + 1)}{2}} \\ {} &= & {\dfrac{k^2 + 3k + 2}{2}} \\ {} &= & {\dfrac{(k + 1)(k + 2)}{2}.} \end{array}$$ By comparing the last equation to equation (2), we see that we have proved that if $$P(k)$$ is true, then $$P(k + 1)$$ is true, and the inductive step has been established. Hence, by the Principle of Mathematical Induction, we have proved that for each integer $$n$$, $$1 + 2 + 3 + \cdot\cdot\cdot + n = \dfrac{n(n + 1)}{2}$$. Progress Check 4.5 For the inductive step, let $$k$$ be a natural number and assume that $$P(k)$$ is true. That is, assume that $$5^k \equiv 1$$ (mod 4). 1. To prove that $$P(k + 1)$$ is true, we must prove $$5^{k + 1} \equiv 1$$ (mod 4). 2. Since $$5^{k + 1} = 5 \cdot 5^{k}$$, we multiply both sides of the congruence $$5^{k} \equiv 1$$ (mod 4) by 5 and obtain $5 \cdot 5^k \equiv 5 \cdot 1 \text{ (mod 4) or}\ \ \ \ \ \ \ \ 5^{k + 1} \equiv 5 \text{ (mod 4).}$ 3. Since $$5^{k + 1} \equiv 5$$ (mod 4) and we know that $$5 \equiv 1$$ (mod 4), we can use the transitive property of congruence to obtain $$5^{k + 1} \equiv 1$$ (mod 4). This proves that if $$P(k)$$ is true, then $$P(k + 1)$$ is true, and hence, by the Principle of Mathematical Induction, we have proved that for each natural number $$n$$, $$5^n \equiv 1$$ (mod 4). Section 4.2 Progress Check 4.8 1. For each natural number $$n$$, if $$n \ge 3$$, then $$3^n > 1 + 2^n$$. 2. For each natural number $$n$$, if $$n \ge 6$$, then $$2^n > (n + 1)^2$$. 3. For each natural number $$n$$, if $$n \ge 6$$, then $$(1 + \dfrac{1}{n})^n > 2.5$$. Progress Check 4.10 Construct the following table and use it to answer the first two questions. The table shows that $$P(3)$$, $$P(5)$$, and $$P(6)$$ are true. We can also see that $$P(2)$$, $$P(4)$$, and $$P(7)$$ are false. It also appears that if $$n \in \mathbb{N}$$ and $$n \ge 8$$, then $$P(n)$$ is true. $$x$$ 0 1 2 3 4 0 1 2 0 1 1 $$y$$ 0 0 0 0 0 1 1 1 2 2 3 $$3x+ 5y$$ 0 3 6 9 12 5 8 11 10 13 18 The following proposition provides answers for Problems (3) and (4). Proposition 4.11. For all natural numbers $$n$$ with $$n \ge 8$$, there exist non-negative integers $$x$$ and $$y$$ such that $$n = 3x + 5y$$. Proof. (by mathematical induction) Let $$\mathbb{Z}^{\ast} = \{x \in \mathbb{Z}\ \|\ x \ge 0\}$$, and for each natural number $$n$$, let $$P(n)$$ be, "there exist $$x, y \in \mathbb{Z}^{\ast}$$ such that $$n = 3x + 5y$$." Basis Step: Using the table above, we see that $$P(8)$$, $$P(9)$$, and $$P(10)$$ are true. Inductive Step: Let $$k \in \mathbb{N}$$ with $$k \ge 13$$. assume that $$P(8)$$, $$P(9)$$, ..., $$P(k)$$ are true. Now, notice that $$k + 1 = 3 + (k - 2).$$ Since $$k \ge 10$$, we can conclude that $$k - 2 \ge 8$$ and hence $$P(k - 2)$$ is true. Therefore, there exist non-negative integers $$u$$ and $$v$$ such that $$k - 2 = (3u + 5v)$$. Using this equation, we see that $$\begin{array} {k + 1} &= & {3 + (3u + 5v)} \\ {} &= & {3(1 + u) + 5v}. \end{array}$$ Hence, we can conclude that $$P(k + 1)$$ is true. This proves that if $$P(8)$$, $$P(9)$$, ..., $$P(k)$$ are true, then $$P(k + 1)$$ is true. Hence, by the Second Principle of Mathematical Induction, for all natural numbers $$n$$ with $$n \ge 8$$, there exist nonnegative integers $$x$$ and $$y$$ such that $$n = 3x + 5y$$. Section 4.3 Progress Check 4.12 Proof. We will use a proof by induction. For each natural number $$n$$, we let $$P(n)$$ be, $$f_{3n}$$ is an even natural number. Since $$f_3 = 2$$, we see that $$P(1)$$ is true and this proves the basis step. For the inductive step, we let $$k$$ be a natural number and assume that $$P(k)$$ is true. That is, assume that $$f_{3k}$$ is an even natural number. This means that there exists an integer m such that $f_{3k} = 2m.$ We need to prove that $$P(k + 1)$$ is true or that $$f_{3(k + 1)}$$ is even. Notice that $$3(k + 1) = 3k + 3$$ and, hence, $$f_{3(k + 1) = f_{3k + 3}$$. We can now use the recursion formula for the Fibonacci numbers to conclude that $$f_{3k + 3} = f_{3k + 2} + f_{3k + 1}$$. Using the recursion formula again, we get $$f_{3k + 2} = f_{3k + 1} + f_{3k}$$. Putting this all together, we see that $\begin{array} {rcl} {f_{3(k + 1)}} &= & {f_{3k + 3}} \\ {} &= & {f_{3k + 2} +f_{3k + 1}} \\ {} &= & {(f_{3k + 1} + f_{3k}) + f_{3k + 1}} \\ {} &= & {2f_{3k + 1} + f_{3k}.} \end{array}$ We now substitute the expression for $$f_{3k}$$ in equation (B.14) into equation (B.15). This gives $\begin{array} {rcl} {f_{3(k + 1)}} &= & {2f_{3k + 1} + 2m} \\ {f_{3(k + 1)}} &= & {2(f_{3k + 1} + m)} \end{array}$ This preceding equation shows that $$f_{3(k + 1)}$$ is even. Hence it has been proved that if $$P(k)$$ is true, then $$P(k + 1)$$ is true and the inductive step has been established. By the Principle of Mathematical Induction, this proves that for each natural number $$n$$, the Fibonacci number $$f_{3n}$$ is an even natural number. Section 5.1 Progress Check 5.3 Progress Check 5.4 1. Using the standard Venn diagram for three sets shown above: (a) For the set $$(A \cap B) \cap C$$, region 5 is shaded. (b) For the set $$(A \cap B) \cup C$$, the regions 2, 4, 5, 6, 7 are shaded. (c) For the set $$(A^{c} \cup B)$$, the regions 2, 3, 5, 6, 7, 8 are shaded. (d) For the set $$(A^{c} \cap (B \cup C))$$, the regions 3, 6, 7 are shaded. Section 5.2 Progress Check 5.8 $$A = \{x \in \mathbb{Z}\ \|\ x \text{ is multiple of 9}\}$$ and $$B = \{x \in \mathbb{Z}\ \|\ x \text{ is a multiple of 3}\}.$$ 1. The set $$A$$ is a subset of $$B$$. To prove this, we let $$x \in A$$. Then there exists an integer $$m$$ such that $$x = 9m$$, which can be written as $x = 3(3m).$ Since $$3m \in \mathbb{Z}$$, the last equation proves that $$x$$ is a multiple of 3 and so $$x \in B$$. Therefore, $$A \subseteq B$$. 2. The set $$A$$ is not equal to the set $$B$$. We note that $$3 \in B$$ but $$3 \notin A$$. Therefore, $$B \not\subseteq A$$ and, hence, $$A \ne B$$. Progress Check 5.9 Step Know Reason $$P$$ $$A \subseteq B$$ Hypothesis $$P$$1 Let $$x \in B^{c}$$. Choose an arbitrary element of $$B^{c}$$. $$P$$2 If $$x \in A$$, then $$x \in B$$. Definition of "subset" $$P$$3 If $$x \notin B$$, then $$x \notin A$$. Contrapositive $$P$$4 If $$x \in B^{c}$$, then $$x \in A^{c}$$. Step $$P$$3 and definition of “complement” $$Q$$2 The element $$x$$ is in $$A^{c}$$. Step $$P$$1 and $$P$$4 $$Q$$1 Every element of $$B^{c}$$ is an element of $$A^{c}$$. The choose-an-element method with Steps $$P$$1 and $$Q$$2. $$Q$$ $$B^{c} \subseteq A^{c}$$ Definition of "subset" Progress Check 5.12 Proof. Let $$A$$ and $$B$$ be subsets of some universal set. We will prove that $$A - B = A \cap B^{c}$$ by proving that each set is a subset of the other set. We will first prove that $$A - B \subseteq A \cap B^{c}$$. Let $$x \in A - B$$. We then know that $$x \in A$$ and $$x \notin B$$. However, $$x \notin B$$ implies that $$x \in B^c$$. Hence, $$x \in A$$ and $$x \in B^{c}$$, which means that $$x \in A \cap B^{c}$$. This proves that $$A - B \subseteq A \cap B^{c}$$. To prove that $$A \cap B^{c} \subseteq A - B$$, we let $$y \in A \cap B^{c}$$. This means that $$y \in A$$ and $$y \in B^{c}$$, and hence, $$y \in A$$ and $$y \notin B$$. Therefore, $$y \in A - B$$ and this proves that $$A \cap B^{c} \subseteq A - B$$. Since we have proved that each set is a subset of the other set, we have proved that $$A - B = A \cap B^{c}$$. Progress Check 5.15 Proof. Let $$A = \{x \in \mathbb{Z}\ \|\ x \equiv 3 \text{ (mod 12)}\}$$ and $$B = \{y \in \mathbb{Z}\ \|\ y \equiv 2 \text{ (mod 8)}\}$$. We will use a proof by contradiction to prove that $$A \cap B = \emptyset$$. So we assume that $$A \cap B \ne \emptyset$$ and let $$x \in A \cap B$$. We can then conclude that $$x \equiv 3 \text{ (mod 12)}$$ and that $$x \equiv 2$$ (mod 8). This means that there exist integers $$m$$ and $$n$$ such that $$x = 3 + 12m$$ and $$x = 2 + 8n$$. By equating these two expressions for $$x$$, we obtain $$3 + 12m = 2 + 8n$$, and this equation can be rewritten as $$1 = 8n - 12m$$. This is a contradiction since 1 is an odd integer and $$8n - 12m$$ is an even integer. We have therefore proved that $$A \cap B = \emptyset$$. Section 5.3 Progress Check 5.19 1. In our standard configuration for a Venn diagram with three sets, regions 1, 2, 4, 5, and 6 are the shaded regions for both $$A \cup (B \cap C)$$ and $$(A \cup B) \cap (A \cup C)$$. 2. Based on the Venn diagrams in Part (1), it appears that $$A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$$. Progress Check 5.21 1. Using our standard configuration for a Venn diagram with three sets, regions 1, 2, and 3 are the regions that are shaded for both $$(A \cup B) - C$$ and $$(A - C) \cup (B- C)$$. 2. $$\begin{array} {rclcr} {(A \cup B) - C} &= & {(A \cup B) \cap C^{c}} & & {\text{(Theorem 5.20)}} \\ {} &= & {C^{c} \cap (A \cup B)} & & {\text{(Commutative Property)}} \\ {} &= & {(C^{c} \cap A) \cup (C^{c} \cap B)} & & {\text{(Distributive Property)}} \\ {} &= & {(A \cap C^{c}) \cup (B \cap C^{c})} & & {\text{(Commutative Property)}} \\ {} &= & {(A - C) \cup (B - C)} & & {\text{(Theorem 5.20)}} \end{array}$$ Section 5.4 Progress Check 5.23 1. Let $$A = \{1, 2, 3\}$$, $$T = \{1, 2\}$$, $$B = \{a, b\}$$, and $$C = \{a, c\}$$. (a) $$A \times B = \{(1, a), (1, b), (2, a), (2, b), (3, a), (3, b)\}$$ (b) $$T \times B = \{(1, a), (1, b), (2, a), (2, b)\}$$ (c) $$A \times C = \{(1, a), (1, c), (2, a), (2, c), (3, a), (3, c)\}$$ (d) $$A \times (B \cap C) = \{(1, a), (2, a), (3, a)\}$$ (e) $$(A \times B) \cap (A \times C) = \{(1, a), (2, a), (3, a)\}$$ (f) $$A \times (B \cup C) = \{(1, a), (1, b), (1, c), (2, a), (2, b), (2, c), (3, a), (3, b), (3, c)\}$$ (g) $$(A \times B) \cup (A \times C) = \{(1, a), (1, b), (1, c), (2, a), (2, b), (2, c), (3, a), (3, b), (3, c)\}$$ (h) $$A \times (B - C) = \{(1, b), (2, b), (3, b)\}$$ (i) $$(A \times B) - (A \times C) = \{(1, b), (2, b), (3, b)\}$$ (j) $$B \times A = \{(a, 1), (b, 1), (a, 2), (b, 2), (a, 3), (b, 3)\}$$ 2. $$\begin{array} {lcl} {T \times B \subseteq A \times B} & & {A \times (B \cup C) = (A \times B) \cup (A \times C)} \\ {A \times (B \cap C) = (A \times B) \cap (A \times C)} & & {A \times (B - C) = (A \times B) - (A \times C)} \end{array}$$ Progress Check 5.24 1. (a) $$A \times B = \{(x, y) \in \mathbb{R} \times \mathbb{R}\ \|\ 0 \le x \le 2 \text{ and } 2 \le y < 4\}$$ (b) $$T \times B = \{(x, y) \in \mathbb{R} \times \mathbb{R}\ \|\ 1 < x < 2 \text{ and } 2 \le y < 4\}$$ (c) $$A \times C = \{(x, y) \in \mathbb{R} \times \mathbb{R}\ \|\ 0 \le x \le 2 \text{ and } 3 < y \le 6\}$$ (d) $$A \times (B \cap C) = \{(x, y) \in \mathbb{R} \times \mathbb{R}\ \|\ 0 \le x \le 2 \text{ and } 3 < y < 4\}$$ (e) $$(A \times B) \cap (A \times C) = \{(x, y) \in \mathbb{R} \times \mathbb{R}\ \|\ 0 \le x \le 2 \text{ and } 3 < y < 4\}$$ (f) $$A \times (B \cup C) = \{(x, y) \in \mathbb{R} \times \mathbb{R}\ \|\ 0 \le x \le 2 \text{ and } 2 \le y \le 5\}$$ (g) $$(A \times B) \cup (A \times C) = \{(x, y) \in \mathbb{R} \times \mathbb{R}\ \|\ 0 \le x \le 2 \text{ and } 2 \le y \le 5\}$$ (h) $$A \times (B - C) = \{(x, y) \in \mathbb{R} \times \mathbb{R}\ \|\ 0 \le x \le 2 \text{ and } 2 \le y \le 3\}$$ (i) $$(A \times B) - (A \times C) = \{(x, y) \in \mathbb{R} \times \mathbb{R}\ \|\ 0 \le x \le 2 \text{ and } 2 \le y \le 3\}$$ (j) $$B \times A = \{(x, y) \in \mathbb{R} \times \mathbb{R}\ \|\ 2 \le x <4 \text{ and } 0 \le y \le 2\}$$ 2. $$T \times B \subseteq A \times B$$ $$A \times (B \cap C) = (A \times C) \cap (A \times C)$$ $$A \times (B \cup C) = (A \times C) \cup (A \times C)$$ $$A \times (B - C) = (A \times C) - (A \times C)$$ Section 5.5 Progress Check 5.26 1. $$\bigcup_{j = 1}^{6} A_j = \{1, 2, 3, 4, 5, 6, 9, 16, 25, 36\}$$ 2. $$\bigcap_{j = 1}^{6} A_j = \{1\}$$ 3. $$\bigcup_{j = 3}^{6} A_j = \{3, 4, 5, 6, 9, 16, 25, 36\}$$ 4. $$\bigcap_{j = 3}^{6} A_j = \{1\}$$ 5. $$\bigcup_{j = 1}^{\infty} A_j = \mathbb{N}$$ 6. $$\bigcap_{j = 1}^{\infty} A_j = \{1\}$$ Progress Check 5.27 1. $$A_{1} = \{7, 14\}$$, $$A_{2} = \{10, 12\}$$, $$A_{3} = \{10, 12\}$$, $$A_{4} = \{8, 14\}$$. 2. The statement is false. For example, $$2 \ne 3$$ and $$A_2 = A_3$$. 3. The statement is false. For example, $$1 \ne -1$$ and $$B_1 = B_{-1}$$. Progress Check 5.29 1. Since $$\bigcup_{\alpha \in \mathbb{R}^{+}} A_{\alpha} = (-1, \infty)$$, $$(\bigcup_{\alpha \in \mathbb{R}^{+}} A_{\alpha})^{c} = (-\infty, 1]$$. 2. $$\bigcap_{\alpha \in \mathbb{R}^{+}} A_{\alpha}^{c} = (-\infty, -1]$$. 3. Since $$\bigcap_{\alpha \in \mathbb{R}^{+}} A_{\alpha} = (-1, 0]$$, $$(\bigcap_{\alpha \in \mathbb{R}^{+}} A_{\alpha})^{c} = (-\infty, 1] \cup (0, \infty)$$. 4. $$\bigcup_{\alpha \in \mathbb{R}^{+}} A_{\alpha}^{c} = (-\infty, -1] \cup (0, \infty)$$. Progress Check 5.32 All three families of sets ($$\mathcal{A}$$, $$\mathcal{B}$$, and $$\mathcal{C}$$ are disjoint families of sets. One the family $$\mathcal{A}$$ is a pairwise disjoint family of sets. Section 6.1 Progress Check 6.1 1. $$f(-3) = 24$$ $$f(\sqrt{8}) = 8 - 5\sqrt{8}$$ 2. $$g(2) = -6, g(-2) = 14$$ 3. {-1, 6} 4. {-1, 6} 5. $$\{\dfrac{5 + \sqrt{33}}{2}, \dfrac{5 - \sqrt{33}}{2}\}$$ 6. $$\emptyset$$ Progress Check 6.2 1. (a) The domain of the function $$f$$ is the set of all people. (b) A codomain for the function $$f$$ is the set of all days in a leap year. (c) This means that the range of the function $$f$$ is equal to its codomain. 2. (a) The domain of the function $$s$$ is the set of natural numbers. (b) A codomain for the function $$s$$ is the set of natural numbers. (c) This means that the range of $$s$$ is not equal to the set o natural numbers. Progress Check 6.3 1. $$f(-1) \thickapprox -3$$ and $$f(2) \thickapprox -2.5$$. 2. Values of $$x$$ for which $$f(x) = 2$$ are approximately -2.8, -1.9, 0.3, 1.2, and 3.5. 3. The range of $$f$$ appears to be the closed interval [-3.2, 3.2] or $$\{y \in \mathbb{R}\ \|\ -3.2 \le y \le 3.2\}$$. Progress Check 6.4 Only the arrow diagram in Figure (a) can be used to represent a function from $$A$$ to $$B$$. The range of this function is the set $$\{a, b\}$$. Section 6.2 Progress Check 6.5 1. $$f(0) = 0$$, $$f(1) = 1$$, $$f(2) = 1$$, $$f(3) = 1$$, $$f(4) = 1$$. 2. $$g(0) = 0$$, $$g(1) = 1$$, $$g(2) = 2$$, $$g(3) = 3$$, $$g(4) = 4$$. Progress Check 6.6 $$I_{\mathbb{Z}_5} \ne f$$ and $$I_{\mathbb{Z}_5} = g$$ Progress Check 6.7 1. 3.5 2. 4.02 3. $$\dfrac{\pi + \sqrt{2}{4}$$ 4. The process of finding the average of a finite set of real numbers can be thought of as a function from $$\mathcal{F}(\mathbb{R})$$ to $$\mathbb{R}$$. So the domain is $$\mathcal{F}(\mathbb{R})$$, the codomain is $$\mathbb{R}$$, and we can define a function avg: $$\mathcal{F}(\mathbb{R}) \to \mathbb{R}$$ as follows: If $$A \in \mathcal{F}(\mathbb{R})$$ and $$A = \{a_1, a_2, ..., a_n\}$$, then ave$$(A) = \dfrac{a_1 + a_2 + \cdot\cdot\cdot a_n}{n}$$. Progress Check 6.8 1. The sixth terms is $$\dfrac{1}{18}$$ and the tenth term is $$\dfrac{1}{30}$$. 2. The sixth terms is $$\dfrac{1}{36}$$ and the tenth term is $$\dfrac{1}{100}$$. 3. The sixth terms is 1 and the tenth term is 1. Progress Check 6.9 1. $$g(0, 3) = -3$$; $$g(3, -2) = 11$$; $$g(-3, -2) = 11$$; $$g(7, -1) = 50$$. 2. $$\{(m, n) \in \mathbb{Z} \times \mathbb{Z}\ \|\ n = m^2\}$$ 3. $$\{(m, n) \in \mathbb{Z} \times \mathbb{Z}\ \|\ n = m^2 - 5\}$$ Section 6.3 Progress Check 6.10 The functions $$k$$, $$F$$, and $$s$$ are injections. The functions $$f$$ and $$h$$ are not injections. Progress Check 6.11 The functions $$f$$ and $$s$$ are surjections. The functions $$k$$ and $$F$$ are not surjections. Progress Check 6.15 The function $$f$$ is an injection but not a surjection. To see that it is an injection, let $$a, b \in \mathbb{R}$$ and assume that $$f(a) = f(b)$$. This implies that $$e^{-a} = e^{-b}$$. Now use the natural logarithm function to prove that $$a = b$$. Since $$e^{-x} > 0$$ for each real number $$x$$, there is no $$x \in \mathbb{R}$$ such that $$f(x) = -1$$. So $$f$$ is not a surjection. The function $$g$$ is an injection and is a surjection. The proof that g is an injection is basically the same as the proof that $$f$$ is an injection. To prove that g is a surjection, let $$b \in R_{+}$$. To construct the real number a such that g.a/ D b, solve the equation $$e^{-a} = b$$ for $$a$$. The solution is $$a = -\text{ln}b$$. It can then be verified that $$g(a) = b$$. Progress Check 6.16 1. There are several ordered pairs $$(a, b) \in \mathbb{R} \times \mathbb{R}$$ such that $$g(a, b) = 2$$. For example, $$g(0, 2) = 2$$, $$g(-1, 4) = 2$$, and $$g(2, -2) = 2$$. 2. For each $$z \in \mathbb{R}$$, $$g(0, z) = z$$. 3. Part (1) implies that the function $$g$$ is not an injection. Part (2) implies that the function $$g$$ is a surjection since for each $$z \in \mathbb{R}$$, ($$0, z$$) is in the domain of $$g$$ and $$g(0, z) = z$$. Section 6.4 Progress Check 6.17 The arrow diagram for $$g \circ f: A \to B$$ should show the following: $$\begin{array} {rclcrcl} {(g \circ f)(a)} &= & {g(f(a))} & & {(g \circ f)(b)} &= & {g(f(b))} \\ {} &= & {g(2) = 1} & & {} &= & {g(3) = 2} \\ {(g \circ f)(c)} &= & {g(f(c))} & & {(g \circ f)(d)} &= & {g(f(d))} \\ {} &= & {g(1) = 3} & & {} &= & {g(2) = 1} \end{array}$$ The arrow diagram for $$g \circ g: B \to B$$ should show the following: $$\begin{array} {rclcrcl} {(g \circ f)(1)} &= & {g(g(1))} & & {(g \circ g)(2)} &= & {g(g(2))} \\ {} &= & {g(3) = 2} & & {} &= & {g(1) = 3} \\ {(g \circ g)(3)} &= & {g(g(3))} & & {} & & {g(f(d))} \\ {} &= & {g(2) = 1} & & {} & & {} \end{array}$$ Progress Check 6.18 1. $$F = g \circ f$$, where $$f: \mathbb{R} \to \mathbb{R}$$ by $$f(x) = x^2 + 3$$, and $$g: \mathbb{R} \to \mathbb{R}$$ by $$g(x) = x^3$$. 2. $$G = h \circ f$$, where $$f: \mathbb{R} \to \mathbb{R}^{+}$$ by $$f(x) = x^2 + 3$$, and $$h: \mathbb{R}^{+} \to \mathbb{R}$$ by $$h(x) = \text{In}x$$. 3. $$f = g \circ k$$, where $$k: \mathbb{R} \to \mathbb{R}$$ by $$k(x) = x^2 - 3$$, and $$g: \mathbb{R} \to \mathbb{R}$$ by $$g(x) = \|x\|$$. 4. $$g = h \circ f$$, where $$f: \mathbb{R} \to \mathbb{R}$$ by $$f(x) = \dfrac{2x - 3}{x^2 + 1}$$, and $$h: \mathbb{R} \to \mathbb{R}$$ by $$h(x) = \text{cos} x$$. Progress Check 6.19 For the examples that are constructed: 1. $$g \circ f$$ should be an injection. 2. $$g \circ f$$ should be a surjection. 3. $$g \circ f$$ should be a bijection. Section 6.5 Progress Check 6.23 Neither set can be used to define a function. 1. The set $$F$$ does not satisfy the first condition of Theorem 6.22. 2. The set $$G$$ does not satisfy the second condition of Theorem 6.22. Progress Check 6.24 2. $$f^{-1} = \{(r, a), (p, b), (q, c)\}$$ $$h^{-1} = \{(p, a), (q, b), (r, c), (q, d)\}$$ $$g^{-1} = \{(p, a), (q, b), (p, c)\}$$ 3. (a) $$f^{-1}$$ is a function from $$C$$ to $$A$$. (b) $$g^{-1}$$ is not a function from $$C$$ to $$A$$ since $$(p, a) \in g^{-1}$$ and $$(p, c) \in g^{-1}$$. (c) $$h^{-1}$$ is not a function from $$C$$ to $$B$$ since $$(q, b) \in h^{-1}$$ and $$(q, d) \in h^{-1}$$. 5. In order for the inverse of a function $$F: S \to T$$ to be a function from $$T$$ to $$S$$, the function $$F$$ must be a bijection. Section 6.6 Progress Check 6.30 1. $$f(A) = \{s, t\}$$ 2. $$f(B) = \{f(x) \ \|\ x \in b\} = \{x\}$$ 3. $$f^{-1}(C) = \{x \in S\ \|\ f(x) \in C\} = \{a, b, c, d\}$$ 4. $$f^{-1}(D) = \{x \in S\ \|\ f(x) \in D\} = \{a, d\}$$ Progress Check 6.32 1. $$f(0) = 2$$ $$f(2) = 6$$ $$f(4) = 2$$ $$f(6) = 6$$ $$f(1) = 3$$ $$f(3) = 3$$ $$f(5) = 3$$ $$f(7) = 3$$ 2. (a) $$f(A) = \{2, 3, 6\}$$ $$f^{-1}(C) = \{0, 1, 3, 4, 5, 7\}$$ $$f(B) = \{2, 3, 6\}$$ $$f^{-1}(D) = \{1, 3, 5, 7\}$$ 3. (a) $$f(A) \cap f(B) = \{2\}$$ $$f(A) \cap f(B) = \{2, 3, 6\}$$ So in this case, $$f(A \cap B) \subseteq f(A) \cap f(B)$$. (b) $$f(A) \cup f(B) = \{2, 3, 6\}$$ $$f(A \cup B) = \{2, 3, 6\}$$ So in this case, $$f(A \cap B) \subseteq f(A \cup B)$$. (c) $$f^{-1}(C) \cap f^{-1}(D) = f^{-1}(C \cap D) = \{1, 3, 5, 7\}$$. So in this case, $$f^{-1}(C \cap D) = f^{-1}(C) \cap f^{-1}(D)$$. (d) $$f^{-1}(C) \cup f^{-1}(D) = f^{-1}(C \cup D) = \{0, 1, 3, 4, 5, 7\}$$. So in this case, $$f^{-1}(C \cup D) = f^{-1}(C) \cup f^{-1}(D)$$. 4. $$f(A) = \{2, 3, 6\}$$. Hence, $$f^{-1}(f(A)) = \{0, 1, 2, 3, 4, 5, 6, 7\}$$. So in this case, $$A \subseteq f^{-1}(f(A))$$. Section 7.1 Progress Check 7.2 1. (a) $$T$$ is a relation on $$\mathbb{R}$$ since $$S$$ is a subset of $$\mathbb{R} \times \mathbb{R}$$. (b) Solve the equation $$x^2 + 4^2 = 64$$. This gives $$x = \pm \sqrt 48$$. Solve the equation $$x^2 + 9^2 = 64$$. There are no real number solutions. So there does not exist an $$x \in \mathbb{R}$$ such that $$(x, 9) \in S$$. (c) dom$$(T) = \{x \in \mathbb{R}\ \|\ -8 \le x \le 8\}$$ range$$(T) = \{y \in \mathbb{R}\ \|\ -8 \le y \le 8\}$$ (d) The graph is a circle of radius 8 whose center is at the origin. 2. (a) $$R$$ is a relation on $$A$$ since $$R$$ is a subset of $$A \times A$$. (b) If we assume that each state except Hawaii has a land border in common with itself, then the domain and range of $$R$$ are the set of all states except Hawaii. If we do not make this assumption, then the domain and range are the set of all states except Hawaii and Alaska. (c) The first statement is true. If x has a land border with y, then y has a land border with x. The second statement is false. Following is a counterexample: (Michigan, Indiana) $$\in R$$, (Indiana, Illinois) $$\in R$$, but (Michigan, Illinois) $$\notin R$$. Progress Check 7.3 1. The domain of the divides relation is the set of all nonzero integers. The range of the divides relation is the set of all integers. 2. (a) This statement is true since for each $$a \in \mathbb{Z}$$, $$a = a\cdot 1$$. (b) This statement is false: For example, 2 divides 4 but 4 does not divide 2. (c) This statement is true by Theorem 3.1 on page 88. Progress Check 7.4 1. Each element in the set $$F$$ is an ordered pair of the form $$(x, y)$$ where $$y = x^2$$. 2. (a) $$A = \{-2, 2\}$$ (b) $$B = \{-\sqrt{10}, \sqrt{10}\}$$ (c) $$C = \{25\}$$ (d) $$D = \{9\}$$ 3. The graph of $$y = x^2$$ is a parabola with vertex at the origin that is concave up. Progress Check 7.5 The directed graph for Part (a) is on the left and the directed graph for Part (b) is on the right. Section 7.2 Progress Check 7.7 The relation $$R$$: • Is not reflexive since $$(c, c) \notin R$$ and $$(d, d) \notin R$$. • Is symmetric. • Is not transitive. For example, $$(c, a) \in R$$, $$(a, c) \in R$$, but $$(c, c) \notin R$$. Progress Check 7.9 • Proof that the relation $$\sim$$ is symmetric: Let $$a, b \in \mathbb{Q}$$ and assume that $$a \sim b$$. This means that $$a - b \in \mathbb{Z}$$. Therefore, $$-(a - b) \in \mathbb{Z}$$ and this means that $$b - a \in \mathbb{Z}$$, and hence, $$b \sim a$$. • Proof that the relation $$\sim$$ is transitive: Let $$a, b, c \in \mathbb{Q}$$and assume that $$a \sim b$$ and $$b \sim c$$. This means that $$a - b \in \mathbb{Z}$$ and that $$b - c \in \mathbb{Z}$$. Therefore, $$((a - b) + (b - c)) \in \mathbb{Z}$$ and this means that $$a - c \in \mathbb{Z}$$, and hence, $$a \sim c$$. Progress Check 7.11 The relation $$\thickapprox$$ is reflexive on $$\mathcal{P}(U)$$ since for all $$A \in \mathcal{P}(U)$$, card($$A$$ )= card($$A$$). The relation $$\thickapprox$$ is symmetric since for all $$A, B \in \mathcal{P}(U)$$, if card($$A$$) = card($$B$$), then using the fact that equality on $$\mathbb{Z}$$ is symmetric, we conclude that card($$B$$) = card($$A$$). That is, if $$A$$ has the same number of elements as $$B$$, then $$B$$ has the same number of elements as $$A$$. The relation $$\thickapprox$$ is transitive since for all $$A, B, C \in \mathcal{P}(U)$$, if card($$A$$) = card($$B$$) and card($$B$$) = card($$C$$), then using the fact that equality on $$\mathbb{Z}$$ is transitive, we conclude that card($$A$$) = card($$C$$). That is, if $$A$$ and $$B$$ have the same number of elements and $$B$$ and $$C$$ have the same number of elements, then $$A$$ and $$C$$ have the same number of elements. Therefore, the relation $$\thickapprox$$ is an equivalence relation on $$\mathcal{P}(U)$$. Section 7.3 Progress Check 7.12 The distinct equivalence classes for the relation $$R$$ are: $$\{a, b, e\}$$ and $$\{c, d\}$$. Progress Check 7.13 The distinct congruence classes for congruence modulo 4 are [0] = {..., -12, -8, -4, 0, 4, 8, 12, ...} [1] = {..., -11, -7, -3, 1, 5, 9, 13, ...} [2] = {..., -10, -6, -2, 2, 6, 10, 14, ...} [1] = {..., -9, -5, -1, 3, 7, 11, 15, ...} Progress Check 7.15 1. [5] = [-5] = {-5, 5} [$$\pi$$] = [-$$\pi$$] = {-$$\pi$$, $$\pi$$} [10] = [-10] = {-10, 10} 2. [0] = {0} 3. [$$a$$] = {-$$a$$, $$a$$} Section 7.4 Progress Check 7.2 1. For all $$a, b \in \mathbb{Z}$$, if $$a \ne 0$$ and $$b \ne 0$$, then $$ab \ne 0$$. 2. The statement in (a) is true and the statement in (b) is false. For example, in $$\mathbb{Z}_{6}$$, $$[2] \odot [3] = [0]$$. Section 8.1 Progress Check 8.2 1. The remainder is 8. 2. gcd(12, 8) = 4 3. $$12 = 8 \cdot 1 + 4$$ and gcd($$r$$, $$r_2$$) = gcd(8, 4) = 4. Progress Check 8.4 1. Original Pair Equation from Division Algorithm New Pair (180, 126) $$180 = 126 \cdot 1 + 54$$ (126, 54) (126, 54) $$126 = 54 \cdot 2 + 18$$ (54, 18) (54, 18) $$54 = 18 \cdot 3 + 0$$ Consequently, gcd(180, 126) = 18. 2. Original Pair Equation from Division Algorithm New Pair (4208, 288) $$4208 = 288 \cdot 14 + 176$$ (288, 176) (288, 176) $$288 = 126 \cdot 1+ 112$$ (176, 112) (176, 112) $$176 = 112 \cdot 1+ 64$$ (112, 64) (112, 64) $$112 = 64 \cdot 1+ 48$$ (64, 48) (64, 48) $$64 = 48 \cdot 1+ 16$$ (48, 16) (48, 16) $$48 = 16 \cdot 3+ 0$$ Consequently, gcd(4208. 288) = 16. Progress Check 8.7 1. From Progress Check 8.4, gcd(180, 126) = 18. $\begin{array}{rcl} {18} &= & {126 - 54 \cdot 2} \\ {} &= & {126 - (180 - 126) \cdot 2} \\ {} &= & {126 \cdot 3 + 180 \cdot (-2)} \end{array}$ So gcd(180, 126) = 18, and $$18 = 126 \cdot 3 + 180 \cdot (-2)$$. 2. From Progress Check 8.4, gcd(4208. 288) = 16. $\begin{array}{rcl} {16} &= & {64 - 48} \\ {} &= & {64 - (112 - 64) = 64 \cdot 2 - 112} \\ {} &= & {(176 - 112) \cdot 2 - 112 = 176 \cdot 2 - 112 \cdot 3} {} &= & {176 \cdot 2 - (288 - 176) \cdot 3 = 176 \cdot 5 - 288 \cdot 3} \\ {} &= & {(4208 - 288 \cdot 14) \cdot 5 - 288 \cdot 3} \\ {} &= & {4208 \cdot 5 + 288 \cdot (-73)} \end{array}$ So gcd(4208. 288) = 16, and $$16 = 4208 \cdot 5 + 288 \cdot (-73)$$. Section 8.2 Progress Check 8.10 1. If $$a, p \in \mathbb{Z}$$, $$p$$ is prime, and $$p$$ divides $$a$$, then gcd($$a$$, $$p$$) = $$p$$. 2. If $$a, p \in \mathbb{Z}$$, $$p$$ is prime, and $$p$$ does not divide $$a$$, then gcd($$a$$, $$p$$) = 1. 3. Three examples are gcd(4, 9) = 1, gcd(15, 16) = 1, gcd(8, 25) = 1. Progress Check 8.13 Theorem 8.12. Let $$a$$, $$b$$, and $$c$$ be integers. If $$a$$ and $$b$$ are relatively prime and $$a\ \|\ (bc)$$. We will prove that $$a$$ divides $$c$$. Proof. Let $$a$$, $$b$$, and $$c$$ be integers. Assume that $$a$$ and $$b$$ are relatively prime and $$a\ \|\ (bc)$$. We will prove that $$a$$ divides $$c$$. Since $$a$$ divides $$bc$$, there exists an integer $$k$$ such that $bc = ak.$ In addition, we are assuming that $$a$$ and $$b$$ are relatively prime and hence gcd($$a$$, $$b$$) = 1. So by Theorem 8.9, there exist integers $$m$$ and $$n$$ such that $am + bn = 1.$ We now multiply both sides of equation (B.21) by $$c$$. This gives $\begin{array} {rcl} {(am + bn)c} &= & {1 \cdot c} \\ {acm + bcn} &= & {c} \end{array}$ We can now use equation (B.20) to substitute $$bc = ak$$ in equation (B.22) and obtain $$acm + akn = c.$$ If we now factor the left side of this last equation, we see that $$a(cm + kn) = c$$. Since ($$cm + kn$$) is an integer, this proves that $$a$$ divides $$c$$. Hence, we have proven that if a and b are relatively prime and $$a\ \|\ (bc)$$, then $$a\ \|\ c$$. Section 8.3 Progress Check 8.20 2. $$x = 2 + 3k$$ and $$y = 0 - 2k$$, where $$k$$ can be any integer. Again, this does not prove that these are the only solutions. Progress Check 8.21 One of the Diophantine equations in Preview Activity 2 was $$3x + 5y = 11$$. We were able to write the solutions of this Diophantine equation in the form $$x = 2 + 5k$$ and $$y = 1 - 3k$$, where $$k$$ is an integer. Notice that $$x = 2$$ and $$y = 1$$ is a solution of this equation. If we consider this equation to be in the form $$ax + by = c$$, then we see that $$a = 3$$, $$b = 5$$, and $$c = 11$$. Solutions for this equation can be written in the form $$x = 2 + bk$$ and $$y = 1 - ak$$, where $$k$$ is an integer. The other equation was $$4x + 6y = 16$$. So in this case, $$a = 4$$, $$b = 6$$, and $$c = 16$$. Also notice that $$d = \text{gcd}(4, 6) = 2$$. We note that $$x = 4$$ and $$y = 0$$ is one solution of this Diophantine equation and solutions can be written in the form $$x = 4 + 3k$$ and $$y = 0 - 2k$$, where $$k$$ is an integer. Using the values of $$a$$, $$b$$, and $$d$$ given above, we see that the solutions can be written in the form $$x = 2 + \dfrac{b}{d}k$$ and $$y = 0 - \dfrac{a}{d}$$, where $$k$$ is an integer. Progress Check 8.24 1. Since 21 does not divide 40, Theorem 8.22 tells us that the Diophantine equation $$63x + 336y = 40$$ has no solutions. Remember that this means there is no ordered pair of integers ($$x$$, $$y$$) such that $$63x + 336y = 40$$. However, if we allow $$x$$ and $$y$$ to be real numbers, then there are real number solutions. In fact, we can graph the straight line whose equation is $$63x + 336y = 40$$ in the Cartesian plane. From the fact that there is no pair of integers $$x$$, $$y$$ such that $$63x + 336y = 40$$, we can conclude that there is no point on the graph of this line in which both coordinates are integers. 2. To write formulas that will generate all the solutions, we first need to find one solution for $$144x + 225y = 27$$. This can sometimes be done by trial and error, but there is a systematic way to find a solution. The first step is to use the Euclidean Algorithm in reverse to write gcd(144, 225) as a linear combination of 144 and 225. See Section 8.1 to review how to do this. The result from using the Euclidean Algorithm in reverse for this situation is $144 \cdot 11 + 225 \cdot (-7) = 9.$ If we multiply both sides of this equation by 3, we obtain $144 \cdot 33 + 225 \cdot (-21) = 27.$ This means that $$x_0 = 33$$, $$y_0 = -21$$ is a solution of the linear Diophantine equation $$144x + 225y = 27$$. We can now use Theorem 8.22 to conclude that all solutions of this Diophantine equation can be written in the form $x = 33 + \dfrac{225}{9}k\ \ \ \ \ \ \ \ \ y = -21 - \dfrac{144}{9}k,$ where $$k \in \mathbb{Z}$$. We check this general solution as follows: Let $$k \in \mathbb{Z}$$. Then $\begin{array} {rcl} {144x + 225y} &= & {144(33 + 25k) + 225(-21 - 16k)} \\ {} &= & {(4752 + 3600k) + (-4725 - 3600k)} \\ {} &= & {27.} \end{array}$ Section 9.1 Progress Check 9.2 1. We first prove that $$f: A \to B$$ is an injection. So let $$x, y \in A$$ and assume that $$f(x) = f(y)$$. Then $$x + 350 = y + 350$$ and we can conclude that $$x = y$$. Hence, $$f$$ is an injection. To prove that $$f$$ is a surjection, let $$b \in B$$. Then $$351 \le b \le 450$$ and hence, $$1 \le b - 350 \le 100$$ and so $$b - 350 \in A$$. In addtion, $$f(b - 350) = (b - 350) + 350 = b$$. This proves that $$f$$ is a surjection, Hence, the function $$f$$ is a bijection, and so, $$A \thickapprox B$$. 2. If $$x$$ and $$t$$ are even integers and $$F(x) = F(t)$$, then $$x + 1 = t + 1$$ and, hence, $$x = t$$. Therefore, $$F$$ is an injection. To prove that $$F$$ is a surjection, let $$y \in D$$. This means that $$y$$ is an odd integer and, hence, $$y - 1$$ is an even integer. In addition, $F(y - 1) = (y - 1 + 1 = y.$ Therefore, $$F$$ is a surjection and hence, $$F$$ is a bijection. We conclude that $$E \thickapprox D$$. 3. Let $$x, t \in (0,1)$$ and assume that $$f(x) = f(t)$$. Then $$bx = bt$$ and, thence, $$x = t$$. Therefore, $$f$$ is an injection. To prove that $$f$$ is a surjection, let $$y \in (0, b)$$. Since $$0 < y < b$$, we conclude that $$0 < \dfrac{y}{b} < 1$$ and that $f(\dfrac{y}{b}) - b(\dfrac{y}{b}\) = y.$ Therefore, $$f$$ is a surjection and hence $$f$$ is a bijection. Thus, $$(0, 1) \thickapprox (0, b)$$. Section 9.2 Progress Check 9.11 1. The set of natural numbers $$\mathbb{N}$$ is a subset of $$\mathbb{Z}$$, $$\mathbb{Q}$$, and $$\mathbb{R}$$. Since $$\mathbb{N}$$ is an infinite set, we can use Part (2) of Theorem 9.10 to conclude that $$\mathbb{Z}$$, $$\mathbb{Q}$$, and $$\mathbb{R}$$ are infinite sets. 2. Use Part (1) of Theorem 9.10. 3. Prove that $$\mathbb{E}^{+} \thickapprox \mathbb{N}$$ and use Part (1) of Theorem 9.10. Progress Check 9.12 1. Use the definition of a countably infinite set. 2. Since $$\mathbb{E}^{+} \thickapprox \mathbb{N}$$, we can conclude that card$$(E^{+}) = \aleph_{0}$$. 3. One function that can be used is $$f: S \to \mathbb{N}$$ defined by $$f(m) = \sqrt{m}$$ for all $$m \in S$$. Progress Check 9.23 Player Two has a winning strategy. On the $$k$$th turn, whatever symbol Player One puts in the $$k$$th position of the $$k$$th row, Player Two must put the other symbol in the $$k$$th position of his or her row. This guarantees that the row of symbols produced by Player Two will be different that any of the rows produced by Player One. This is the same idea used in Cantor’s Diagonal Argument. Once we have a “list” of real numbers in normalized form, we create a real number that is not in the list by making sure that its $$k$$th decimal place is different than the $$k$$th decimal place for the $$k$$th number in the list. The one complication is that we must make sure that our new real number does not have a decimal expression that ends in all 9’s. This was done by using only 3’s and 5’s. Progress Check 9.25 1. Proof. In order to find a bijection $$f: (0, 1) \to (a, b)$$, we will use the linear function through the points $$(0, a)$$ and $$(1, b)$$. The slope is $$(b - a)$$ and the $$y$$-intercept is $$(0, a)$$. So define $$f: (0, 1) \to (a, b)$$ by $f(x) = (b - a) x + a, \text{for each } x \in (0, 1).$ Now, if $$x, t \in (0, 1)$$ and $$f(x) = f(t)$$, then $(b - a)x + a = (b - a)t + a.$ This implies that $$(b - a) x = (b - a) t$$, and since $$b - a \ne 0$$, e can conclude that $$x = t$$. Therefore, $$f$$ is an injection. To prove that $$f$$ is a surjection, we let $$y \in (a, b)$$. If $$x = \dfrac{y - a}{b - a}$$, then $\begin{array} {rcl} {f(x)} &= & {f(\dfrac{y - a}{b - a})} \\ {} &= & {(b - a) (\dfrac{y - a}{b - a}) + a} \\ {} &= & {(y - a) + a} \\ {} &= & {y} \begin{array}$ This proves that $$f$$ is a surjection. Hence, $$f$$ is a bijection and $$(0, 1) \thickapprox (a, b)$$. Therefore, $$(a, b)$$ is uncountable and has cardinality $$c$$. 2. Now, if $$a, b, c, d$$ are real number with $$a < b$$ and $$c < d$$, then we know that $(a, b) \thickapprox (0, 1) \text{ and } (c, d) \thickapprox (0, 1).$ Since $$\thickapprox$$ is an equivalence relation, we can conclude that $$(a, b) \thickapprox (c, d)$$.
## Formulas An equation is a mathematical statement indicating that two expressions are equal. The expressions can be numerical or algebraic. The equation is not inherently true or false, but only a proposition. The values that make the equation true, the solutions, are found using the properties of real numbers and other results. For example, the equation $2x+1=7$ has the unique solution $x=3$ because when we substitute 3 for $x$ in the equation, we obtain the true statement $2\left(3\right)+1=7$. A formula is an equation expressing a relationship between constant and variable quantities. Very often, the equation is a means of finding the value of one quantity (often a single variable) in terms of another or other quantities. One of the most common examples is the formula for finding the area $A$ of a circle in terms of the radius $r$ of the circle: $A=\pi {r}^{2}$. For any value of $r$, the area $A$ can be found by evaluating the expression $\pi {r}^{2}$. ### Example 11: Using a Formula A right circular cylinder with radius $r$ and height $h$ has the surface area $S$ (in square units) given by the formula $S=2\pi r\left(r+h\right)$. Find the surface area of a cylinder with radius 6 in. and height 9 in. Leave the answer in terms of $\pi$. Figure 3. Right circular cylinder ### Solution Evaluate the expression $2\pi r\left(r+h\right)$ for $r=6$ and $h=9$. $\begin{array}\text{ }S\hfill&=2\pi r\left(r+h\right) \\ \hfill& =2\pi\left(6\right)[\left(6\right)+\left(9\right)] \\ \hfill& =2\pi\left(6\right)\left(15\right) \\ \hfill& =180\pi\end{array}$ The surface area is $180\pi$ square inches. ### Try It 11 Figure 4 A photograph with length L and width W is placed in a matte of width 8 centimeters (cm). The area of the matte (in square centimeters, or cm2) is found to be $A=\left(L+16\right)\left(W+16\right)-L\cdot W$. Find the area of a matte for a photograph with length 32 cm and width 24 cm. Solution
Question of Solved Question # Question Three equal cubes are placed adjacently in a row. Find the ratio of the total surface area of the new cuboids to that of the sum of the surface areas of three cubes. Represent Root 9 point 3 on the number line Solution: Explanation: Step 1: Draw a line segment AB of length 9.3 units. Step 2: Now, Extend the line by 1 unit more such that BC=1 unit . Step 3: Find the midpoint of AC. Step 4: Draw a line BD perpendicular to AB and let it intersect the semicircle at point D. Step 5: Draw an arc DE such that BE=BD. Hence, Number line of √ 9.3 is attached below. Which one of the following statement is true A: Only one line can pass through a single point. B: There are an infinite number of lines which pass through two distinct points. C: Two distinct lines cannot have more than one point in common. D: If two circles are equal, then their radii are not equal. Solution: Explanation: From one point there is an uncountable number of lines that can pass through. Hence, the statement “ Only one line can pass through a single point” is false. We can draw only one unique line passing through two distinct points. Hence, the statement “There are an infinite number of lines which pass through two distinct points” is false. Given two distinct points, there is a unique line that passes through them. Hence, the statement “Two distinct lines cannot have more than one point in common” is true. If circles are equal, which means the circles are congruent. This means that circumferences are equal and so the radii of two circles are also equal. Hence, the statement “If two circles are equal, then their radii are not equal” is false. The correct option is (C) Two distinct lines cannot have more than one point in common. The class mark of the class 90-120 is A: 90 B: 105 C: 115 D: 120 Solution: Explanation: To find the class mark of a class interval, we find the sum of the upper limit and lower limit of a class and divide it by 2 Thus, Class -mark=Upper limit + Lower limit/2 Here, the lower limit of 90-120=90 And the upper limit of 90-120=120 So, Class -mark=120+90/2 =210/2 =105 Hence, the class mark of the class 90-120 is 105 That is, option (B) is correct. Option (B) 105 is correct. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD Show that i)ΔAPBΔCQD ii) AP = CQ Solution: Find the roots of the following equation A: - 1, - 2 B: - 1, - 3 C: 1,3 D: 1,2 Solution: Explanation:-
Courses Courses for Kids Free study material Offline Centres More Store State whether the statement given below is true or false:$\dfrac{{\sqrt 3 }}{2}$ and $\dfrac{{2\sqrt 3 }}{3}$ are reciprocal of each other. Last updated date: 13th Jun 2024 Total views: 412.2k Views today: 8.12k Verified 412.2k+ views Hint: We will first find the reciprocal of the first number and then rationalize it to compare to the given second number. If they are the same, then we know that the statement is true. The first number given to us is $\dfrac{{\sqrt 3 }}{2}$. We know that the reciprocal of a number $x$ is $e$, then $x \times e = 1 = e \times x$. So, we get:- $e = \dfrac{1}{x}$. Using this concept by taking $x = \dfrac{{\sqrt 3 }}{2}$, then we have:- Reciprocal of $x$ is $\dfrac{1}{x} = \dfrac{1}{{\dfrac{{\sqrt 3 }}{2}}}$ We know that $\dfrac{1}{{\dfrac{a}{b}}} = \dfrac{1}{a} \times b = \dfrac{b}{a}$ So, we get:- $\dfrac{1}{x} = \dfrac{1}{{\dfrac{{\sqrt 3 }}{2}}} = \dfrac{2}{{\sqrt 3 }}$ We know that multiplying a number by 1 doesn’t change its value because $a \times 1$ = $1 \times a$ = $a$. So, $\dfrac{2}{{\sqrt 3 }} = \dfrac{2}{{\sqrt 3 }} \times 1$ We can rewrite it as:- $\dfrac{2}{{\sqrt 3 }} \times 1 = \dfrac{2}{{\sqrt 3 }} \times \dfrac{{\sqrt 3 }}{{\sqrt 3 }}$. This is done because we know that $\dfrac{a}{a}$ = 1 for all a. This implies that:- $\dfrac{1}{x} = \dfrac{{2 \times \sqrt 3 }}{{{{\left( {\sqrt 3 } \right)}^2}}}$ We know that square root and square cancel each other. So, we get:- $\dfrac{1}{x} = \dfrac{{2\sqrt 3 }}{3}$. We clearly see that this is equal to the second number given to us. Hence, the statement is true. Note: We can think of a whole number as being $\dfrac{{number}}{1}$, so the reciprocal is just like "flipping it over" and by that we receive $\dfrac{1}{{number}}$. We need to remember the fact that sometimes there might be a possibility that the second number is not rationalized and the denominator already contains a number with a square root. So, then we would not have to rationalize the reciprocal we got from the first number. We must read the question properly as there might be cases that we are already given no in the question. For example:- 1 is not reciprocal of 1. Then its answer will be false because reciprocal of 1 is always 1. Fun Fact:- We cannot find a reciprocal of 0 because denominator a number cannot be 0. So, its reciprocal is not defined but for a bit ease, we say it to be $\infty$.
# 2013 AMC 10B Problems/Problem 16 ## Problem In triangle $ABC$, medians $AD$ and $CE$ intersect at $P$, $PE=1.5$, $PD=2$, and $DE=2.5$. What is the area of $AEDC$? $\qquad\textbf{(A) }13\qquad\textbf{(B) }13.5\qquad\textbf{(C) }14\qquad\textbf{(D) }14.5\qquad\textbf{(E) }15$ $[asy] pair A,B,C,D,E,P; A=(0,0); B=(80,0); C=(20,40); D=(50,20); E=(40,0); P=(33.3,13.3); draw(A--B); draw(B--C); draw(A--C); draw(C--E); draw(A--D); draw(D--E); dot(A); dot(B); dot(C); dot(D); dot(E); dot(P); label("A",A,NNW); label("B",B,NNE); label("C",C,ENE); label("D",D,ESE); label("E",E,SSE); label("P",P,SSE); [/asy]$ ## Solution 1 Let us use mass points: Assign $B$ mass $1$. Thus, because $E$ is the midpoint of $AB$, $A$ also has a mass of $1$. Similarly, $C$ has a mass of $1$. $D$ and $E$ each have a mass of $2$ because they are between $B$ and $C$ and $A$ and $B$ respectively. Note that the mass of $D$ is twice the mass of $A$, so AP must be twice as long as $PD$. PD has length $2$, so $AP$ has length $4$ and $AD$ has length $6$. Similarly, $CP$ is twice $PE$ and $PE=1.5$, so $CP=3$ and $CE=4.5$. Now note that triangle $PED$ is a $3-4-5$ right triangle with the right angle $DPE$. Since the diagonals of quadrilaterals $AEDC$, $AD$ and $CE$, are perpendicular, the area of $AEDC$ is $\frac{6 \times 4.5}{2}=\boxed{\textbf{(B)} 13.5}$ ## Solution 2 Note that triangle $DPE$ is a right triangle, and that the four angles (angles $APC, CPD, DPE,$ and $EPA$) that have point $P$ are all right angles. Using the fact that the centroid ($P$) divides each median in a $2:1$ ratio, $AP=4$ and $CP=3$. Quadrilateral $AEDC$ is now just four right triangles. The area is $\frac{4\cdot 1.5+4\cdot 3+3\cdot 2+2\cdot 1.5}{2}=\boxed{\textbf{(B)} 13.5}$ ## Solution 3 From the solution above, we can find that the lengths of the diagonals are $6$ and $4.5$. Now, since the diagonals of AEDC are perpendicular, we use the area formula to find that the total area is $\frac{6\times4.5}{2} = \frac{27}{2} = 13.5 = \boxed{\textbf{(B)} 13.5}$ ## Solution 4 From the solutions above, we know that the sides CP and AP are 3 and 4 respectively because of the properties of medians that divide cevians into 1:2 ratios. We can then proceed to use the heron's formula on the middle triangle EPD and get the area of EPD as 3/2, (its simple computation really, nothing large). Then we can find the areas of the remaining triangles based on side and ratio length of the bases. ## Solution 5 We know that $[AEDC]=\frac{3}{4}[ABC]$, and $[ABC]=3[PAC]$ using median properties. So Now we try to find $[PAC]$. Since $\triangle PAC\sim \triangle PDE$, then the side lengths of $\triangle PAC$ are twice as long as $\triangle PDE$ since $D$ and $E$ are midpoints. Therefore, $\frac{[PAC]}{[PDE]}=2^2=4$. It suffices to compute $[PDE]$. Notice that $(1.5, 2, 2.5)$ is a Pythagorean Triple, so $[PDE]=\frac{1.5\times 2}{2}=1.5$. This implies $[PAC]=1.5\cdot 4=6$, and then $[ABC]=3\cdot 6=18$. Finally, $[AEDC]=\frac{3}{4}\times 18=\boxed{13.5}$. ~CoolJupiter ## Solution 6 As from Solution 4, we find the area of $\triangle DPE$ to be $\frac{3}{2}$. Because $DE = \frac{5}{2}$, the altitude perpendicular to $DE = \frac{6}{5}$. Also, because $DE \|\| AC$, $\triangle ABC$ is similar to $\triangle{DBE}$ with side length ratio $2:1$, so $AC=5$ and the altitude perpendicular to $AC = \frac{12}{5}$. The altitude of trapezoid $ACDE$ is then $\frac{18}{5}$ and the bases are $\frac{5}{2}$ and $5$. So, we use the formula for the area of a trapezoid to find the area of $ACDE = \boxed{13.5}$
# Law of sines In trigonometry, the law of sines, sine law, sine formula, or sine rule is an equation relating the lengths of the sides of a triangle (any shape) to the sines of its angles. According to the law, ${\displaystyle {\frac {a}{\sin A}}\,=\,{\frac {b}{\sin B}}\,=\,{\frac {c}{\sin C}}\,=\,d,}$ where a, b, and c are the lengths of the sides of a triangle, and A, B, and C are the opposite angles (see the figure to the right), while d is the diameter of the triangle's circumcircle. When the last part of the equation is not used, the law is sometimes stated using the reciprocals; ${\displaystyle {\frac {\sin A}{a}}\,=\,{\frac {\sin B}{b}}\,=\,{\frac {\sin C}{c}}.}$ The law of sines can be used to compute the remaining sides of a triangle when two angles and a side are known—a technique known as triangulation. It can also be used when two sides and one of the non-enclosed angles are known. In some such cases, the triangle is not uniquely determined by this data (called the ambiguous case) and the technique gives two possible values for the enclosed angle. The law of sines is one of two trigonometric equations commonly applied to find lengths and angles in scalene triangles, with the other being the law of cosines. The law of sines can be generalized to higher dimensions on surfaces with constant curvature.[1] ## Proof The area T of any triangle can be written as one half of its base times its height. Selecting one side of the triangle as the base, the height of the triangle relative to that base is computed as the length of another side times the sine of the angle between the chosen side and the base. Thus, depending on the selection of the base the area of the triangle can be written as any of: ${\displaystyle T={\frac {1}{2}}b(c\sin A)={\frac {1}{2}}c(a\sin B)={\frac {1}{2}}a(b\sin C)\,.}$ Multiplying these by 2/abc gives ${\displaystyle {\frac {2T}{abc}}={\frac {\sin A}{a}}={\frac {\sin B}{b}}={\frac {\sin C}{c}}\,.}$ ## The ambiguous case of triangle solution When using the law of sines to find a side of a triangle, an ambiguous case occurs when two separate triangles can be constructed from the data provided (i.e., there are two different possible solutions to the triangle). In the case shown below they are triangles ABC and AB′C′. Given a general triangle, the following conditions would need to be fulfilled for the case to be ambiguous: • The only information known about the triangle is the angle A and the sides a and c. • The angle A is acute (i.e., A < 90°). • The side a is shorter than the side c (i.e., a < c). • The side a is longer than the altitude h from angle B, where h = c sin A (i.e., a > h). If all the above conditions are true, then each of angles C and C′ produces a valid triangle, meaning that both of the following are true: ${\displaystyle C'=\arcsin {\frac {c\sin A}{a}}{\text{ or }}C=\pi -\arcsin {\frac {c\sin A}{a}}.}$ From there we can find the corresponding B and b or B′ and b′ if required, where b is the side bounded by angles A and C and b′ bounded by A and C′. Without further information it is impossible to decide which is the triangle being asked for. ## Examples The following are examples of how to solve a problem using the law of sines. ### Example 1 Given: side a = 20, side c = 24, and angle C = 40°. Angle A is desired. Using the law of sines, we conclude that ${\displaystyle {\frac {\sin A}{20}}={\frac {\sin 40^{\circ }}{24}}.}$ ${\displaystyle A=\arcsin \left({\frac {20\sin 40^{\circ }}{24}}\right)\approx 32.39^{\circ }.}$ Note that the potential solution A = 147.61° is excluded because that would necessarily give A + B + C > 180°. ### Example 2 If the lengths of two sides of the triangle a and b are equal to x, the third side has length c, and the angles opposite the sides of lengths a, b, and c are A, B, and C respectively then {\displaystyle {\begin{aligned}&A=B={\frac {180^{\circ }-C}{2}}=90^{\circ }-{\frac {C}{2}}\\[6pt]&\sin A=\sin B=\sin \left(90^{\circ }-{\frac {C}{2}}\right)=\cos \left({\frac {C}{2}}\right)\\[6pt]&{\frac {c}{\sin C}}={\frac {a}{\sin A}}={\frac {x}{\cos \left({\frac {C}{2}}\right)}}\\[6pt]&{\frac {c\cos \left({\frac {C}{2}}\right)}{\sin C}}=x\end{aligned}}} ## Relation to the circumcircle In the identity ${\displaystyle {\frac {a}{\sin A}}={\frac {b}{\sin B}}={\frac {c}{\sin C}},}$ the common value of the three fractions is actually the diameter of the triangle's circumcircle which dates back to Ptolemy.[2][3] ### Proof As shown in the figure, let there be a circle with inscribed${\displaystyle \bigtriangleup ABC}$ and another inscribed ${\displaystyle \bigtriangleup ADB}$ that passes through the circle's center O. The ${\displaystyle \angle AOD}$ has a central angle of ${\displaystyle 180^{\circ }}$ and thus ${\displaystyle \angle ABD=90^{\circ }}$. Since ${\displaystyle \bigtriangleup ABD}$ is a right triangle, ${\displaystyle \sin D={\frac {\text{opposite}}{\text{hypotenuse}}}={\frac {c}{2R}},}$ where ${\displaystyle R}$ is the radius of the circumscribing circle of the triangle.[3] Angles ${\displaystyle \angle C}$ and ${\displaystyle \angle D}$ have the same central angle thus they are the same: ${\displaystyle \angle C=\angle D}$. Therefore, ${\displaystyle \sin D=\sin C={\frac {c}{2R}}.}$ Rearranging yields ${\displaystyle 2R={\frac {c}{\sin C}}.}$ Repeating the process of creating ${\displaystyle \bigtriangleup ADB}$ with other points gives ${\displaystyle {\frac {a}{\sin A}}={\frac {b}{\sin B}}={\frac {c}{\sin C}}=2R.}$ ### Relationship to the area of the triangle The area of a triangle is given by ${\displaystyle T={\frac {1}{2}}ab\sin \theta }$, where ${\displaystyle \theta }$ is the angle enclosed by the sides of lengths a and b. Substituting the sine law into this equation gives ${\displaystyle T={\frac {1}{2}}ab\cdot {\frac {c}{2R}}.}$ Taking ${\displaystyle R}$ as the circumscribing radius,[4] ${\displaystyle T={\frac {abc}{4R}}.}$ It can also be shown that this equality implies {\displaystyle {\begin{aligned}{\frac {abc}{2T}}&={\frac {abc}{2{\sqrt {s(s-a)(s-b)(s-c)}}}}\\[6pt]&={\frac {2abc}{\sqrt {{(a^{2}+b^{2}+c^{2})}^{2}-2(a^{4}+b^{4}+c^{4})}}},\end{aligned}}} where T is the area of the triangle and s is the semiperimeter ${\displaystyle s={\frac {a+b+c}{2}}.}$ The second equality above readily simplifies to Heron's formula for the area. The sine rule can also be used in deriving the following formula for the triangle’s area: Denoting the semi-sum of the angles' sines as ${\displaystyle S={\frac {\sin A+\sin B+\sin C}{2}}}$, we have[5] ${\displaystyle T=D^{2}{\sqrt {S(S-\sin A)(S-\sin B)(S-\sin C)}}}$ where ${\displaystyle D}$ is the diameter of the circumcircle: ${\displaystyle D=2R={\frac {a}{\sin A}}={\frac {b}{\sin B}}={\frac {c}{\sin C}}}$. ## Curvature The law of sines takes on a similar form in the presence of curvature. ### Spherical case In the spherical case, the formula is: ${\displaystyle {\frac {\sin A}{\sin \alpha }}={\frac {\sin B}{\sin \beta }}={\frac {\sin C}{\sin \gamma }}.}$ Here, α, β, and γ are the angles at the center of the sphere subtended by the three arcs of the spherical surface triangle a, b, and c, respectively. A, B, and C are the surface angles opposite their respective arcs. ### Vector proof Consider a unit sphere with three unit vectors OA, OB and OC drawn from the origin to the vertices of the triangle. Thus the angles α, β, and γ are the angles a, b, and c, respectively. The arc BC subtends an angle of magnitude a at the centre. Introduce a Cartesian basis with OA along the z-axis and OB in the xz-plane making an angle c with the z-axis. The vector OC projects to ON in the xy-plane and the angle between ON and the x-axis is A. Therefore, the three vectors have components: ${\displaystyle \mathbf {OA} ={\begin{pmatrix}0\\0\\1\end{pmatrix}},\quad \mathbf {OB} ={\begin{pmatrix}\sin c\\0\\\cos c\end{pmatrix}},\quad \mathbf {OC} ={\begin{pmatrix}\sin b\cos A\\\sin b\sin A\\\cos b\end{pmatrix}}.}$ The scalar triple product, OA · (OB × OC) is the volume of the parallelepiped formed by the position vectors of the vertices of the spherical triangle OA, OB and OC. This volume is invariant to the specific coordinate system used to represent OA, OB and OC. The value of the scalar triple product OA · (OB × OC) is the 3 × 3 determinant with OA, OB and OC as its rows. With the z-axis along OA the square of this determinant is {\displaystyle {\begin{aligned}{\bigl (}\mathbf {OA} \cdot (\mathbf {OB} \times \mathbf {OC} ){\bigr )}^{2}&={\bigl (}\det(\mathbf {OA} ,\mathbf {OB} ,\mathbf {OC} ){\bigr )}^{2}\\[4pt]&=\left({\begin{vmatrix}0&0&1\\\sin c&0&\cos c\\\sin b\cos A&\sin b\sin A&\cos b\end{vmatrix}}\right)^{2}=\left(\sin b\sin c\sin A\right)^{2}.\end{aligned}}} Repeating this calculation with the z-axis along OB gives (sin c sin a sin B)2, while with the z-axis along OC it is (sin a sin b sin C)2. Equating these expressions and dividing throughout by (sin a sin b sin c)2 gives ${\displaystyle {\frac {\sin ^{2}A}{\sin ^{2}a}}={\frac {\sin ^{2}B}{\sin ^{2}b}}={\frac {\sin ^{2}C}{\sin ^{2}c}}={\frac {V^{2}}{\sin ^{2}a\sin ^{2}b\sin ^{2}c}},}$ where V is the volume of the parallelepiped formed by the position vector of the vertices of the spherical triangle. It is easy to see how for small spherical triangles, when the radius of the sphere is much greater than the sides of the triangle, this formula becomes the planar formula at the limit, since ${\displaystyle \lim _{\alpha \rightarrow 0}{\frac {\sin \alpha }{\alpha }}=1}$ and the same for sin β and sin γ. ### Geometric proof Consider a unit circle with: ${\displaystyle OA=OB=OC=1}$ Construct point ${\displaystyle D}$ and point ${\displaystyle E}$ such that ${\displaystyle \angle ADO=\angle AEO=90^{\circ }}$ Construct point ${\displaystyle A'}$ such that ${\displaystyle \angle A'DO=\angle A'EO=90^{\circ }}$ It can therefore be seen that ${\displaystyle \angle ADA'=B}$ and ${\displaystyle \angle AEA'=C}$ Notice that ${\displaystyle A'}$ is the projection of ${\displaystyle A}$ on plane ${\displaystyle OBC}$. Therefore ${\displaystyle \angle AA'D=\angle AA'E=90^{\circ }}$ By basic trigonometry, we have: ${\displaystyle AD=\sin c}$ ${\displaystyle AE=\sin b}$ But ${\displaystyle AA'=AD\sin B=AE\sin C}$ Combining them we have: ${\displaystyle \sin c\sin B=\sin b\sin C}$ ${\displaystyle {\frac {\sin B}{\sin b}}={\frac {\sin C}{\sin c}}}$ By applying similar reasoning, we obtain the spherical law of sine: ${\displaystyle {\frac {\sin A}{\sin a}}={\frac {\sin B}{\sin b}}={\frac {\sin C}{\sin c}}}$ ### Other proofs A purely algebraic proof can be constructed from the spherical law of cosines.. From the identity ${\displaystyle \sin ^{2}A=1-\cos ^{2}A}$ and the explicit expression for ${\displaystyle \cos A}$ from the spherical law of cosines {\displaystyle {\begin{aligned}\sin ^{2}\!A&=1-\left({\frac {\cos a-\cos b\,\cos c}{\sin b\,\sin c}}\right)^{2}\\&={\frac {(1-\cos ^{2}\!b)(1-\cos ^{2}\!c)-(\cos a-\cos b\,\cos c)^{2}}{\sin ^{2}\!b\,\sin ^{2}\!c}}\\{\frac {\sin A}{\sin a}}&={\frac {[1-\cos ^{2}\!a-\cos ^{2}\!b-\cos ^{2}\!c+2\cos a\cos b\cos c]^{1/2}}{\sin a\sin b\sin c}}.\end{aligned}}} Since the right hand side is invariant under a cyclic permutation of ${\displaystyle a,\;b,\;c}$ the spherical sine rule follows immediately. The figure used in the Geometric proof above is used by and also provided in Banerjee[6] (see Figure 3 in this paper) to derive the sine law using elementary linear algebra and projection matrices. ### Hyperbolic case In hyperbolic geometry when the curvature is −1, the law of sines becomes ${\displaystyle {\frac {\sin A}{\sinh a}}={\frac {\sin B}{\sinh b}}={\frac {\sin C}{\sinh c}}\,.}$ In the special case when B is a right angle, one gets ${\displaystyle \sin C={\frac {\sinh c}{\sinh b}}}$ which is the analog of the formula in Euclidean geometry expressing the sine of an angle as the opposite side divided by the hypotenuse. ### Unified formulation Define a generalized sine function, depending also on a real parameter K: ${\displaystyle \sin _{K}x=x-{\frac {Kx^{3}}{3!}}+{\frac {K^{2}x^{5}}{5!}}-{\frac {K^{3}x^{7}}{7!}}+\cdots .}$ The law of sines in constant curvature K reads as[1] ${\displaystyle {\frac {\sin A}{\sin _{K}a}}={\frac {\sin B}{\sin _{K}b}}={\frac {\sin C}{\sin _{K}c}}\,.}$ By substituting K = 0, K = 1, and K = −1, one obtains respectively the Euclidean, spherical, and hyperbolic cases of the law of sines described above. Let pK(r) indicate the circumference of a circle of radius r in a space of constant curvature K. Then pK(r) = 2π sinK r. Therefore the law of sines can also be expressed as: ${\displaystyle {\frac {\sin A}{p_{K}(a)}}={\frac {\sin B}{p_{K}(b)}}={\frac {\sin C}{p_{K}(c)}}\,.}$ This formulation was discovered by János Bolyai.[7] ## Higher dimensions For an n-dimensional simplex (i.e., triangle (n = 2), tetrahedron (n = 3), pentatope (n = 4), etc.) in n-dimensional Euclidean space, the absolute value of the polar sine of the normal vectors of the faces that meet at a vertex, divided by the hyperarea of the face opposite the vertex is independent of the choice of the vertex. Writing V for the hypervolume of the n-dimensional simplex and P for the product of the hyperareas of its (n−1)-dimensional faces, the common ratio is ${\displaystyle {\frac {(nV)^{n-1}}{(n-1)!P}}.}$ For example, a tetrahedron has four triangular faces. The absolute value of the polar sine of the normal vectors to the three faces that share a vertex, divided by the area of the fourth face will not depend upon the choice of the vertex: {\displaystyle {\begin{aligned}&{\frac {{\bigl \|}\operatorname {psin} (\mathbf {n_{2}} ,\mathbf {n_{3}} ,\mathbf {n_{4}} ){\bigr \|}}{\mathrm {Area} _{1}}}={\frac {{\bigl \|}\operatorname {psin} (\mathbf {n_{1}} ,\mathbf {n_{3}} ,\mathbf {n_{4}} ){\bigr \|}}{\mathrm {Area} _{2}}}={\frac {{\bigl \|}\operatorname {psin} (\mathbf {n_{1}} ,\mathbf {n_{2}} ,\mathbf {n_{4}} ){\bigr \|}}{\mathrm {Area} _{3}}}={\frac {{\bigl \|}\operatorname {psin} (\mathbf {n_{1}} ,\mathbf {n_{2}} ,\mathbf {n_{3}} ){\bigr \|}}{\mathrm {Area} _{4}}}\\[4pt]={}&{\frac {(3\operatorname {Volume} _{\mathrm {tetrahedron} })^{2}}{2!~\mathrm {Area} _{1}\mathrm {Area} _{2}\mathrm {Area} _{3}\mathrm {Area} _{4}}}\,.\end{aligned}}} ## History According to Ubiratàn D'Ambrosio and Helaine Selin, the spherical law of sines was discovered in the 10th century. It is variously attributed to Abu-Mahmud Khojandi, Abu al-Wafa' Buzjani, Nasir al-Din al-Tusi and Abu Nasr Mansur.[8] Ibn Muʿādh al-Jayyānī's The book of unknown arcs of a sphere in the 11th century contains the general law of sines.[9] The plane law of sines was later stated in the 13th century by Nasīr al-Dīn al-Tūsī. In his On the Sector Figure, he stated the law of sines for plane and spherical triangles, and provided proofs for this law.[10] According to Glen Van Brummelen, "The Law of Sines is really Regiomontanus's foundation for his solutions of right-angled triangles in Book IV, and these solutions are in turn the bases for his solutions of general triangles."[11] Regiomontanus was a 15th-century German mathematician. ## References 1. "Generalized law of sines". mathworld. 2. Coxeter, H. S. M. and Greitzer, S. L. Geometry Revisited. Washington, DC: Math. Assoc. Amer., pp. 1–3, 1967 3. "Law of Sines". www.pballew.net. Retrieved 2018-09-18. 4. Mr. T's Math Videos (2015-06-10), Area of a Triangle and Radius of its Circumscribed Circle, retrieved 2018-09-18 5. Mitchell, Douglas W., "A Heron-type area formula in terms of sines," Mathematical Gazette 93, March 2009, 108–109. 6. Banerjee, Sudipto (2004), "Revisiting Spherical Trigonometry with Orthogonal Projectors", The College Mathematics Journal, Mathematical Association of America, 35: 375–381Text online 7. Katok, Svetlana (1992). Fuchsian groups. Chicago: University of Chicago Press. p. 22. ISBN 0-226-42583-5. 8. Sesiano just lists al-Wafa as a contributor. Sesiano, Jacques (2000) "Islamic mathematics" pp. 137–157, in Selin, Helaine; D'Ambrosio, Ubiratan (2000), Mathematics Across Cultures: The History of Non-western Mathematics, Springer, ISBN 1-4020-0260-2 9. Berggren, J. Lennart (2007). "Mathematics in Medieval Islam". The Mathematics of Egypt, Mesopotamia, China, India, and Islam: A Sourcebook. Princeton University Press. p. 518. ISBN 978-0-691-11485-9. 10. Glen Van Brummelen (2009). "The mathematics of the heavens and the earth: the early history of trigonometry". Princeton University Press. p.259. ISBN 0-691-12973-8
# Probability, Part 6, The Problem with Probability In my last few posts, I’ve talked about probability and how to calculated a basic probability: Probability = $\frac{{\mathrm{Number}}\hspace{0.33em}{\mathrm{of}}\hspace{0.33em}{\mathrm{favorable}}\hspace{0.33em}{\mathrm{outcomes}}}{{\mathrm{Total}}\hspace{0.33em}{\mathrm{number}}\hspace{0.33em}{\mathrm{of}}\hspace{0.33em}{\mathrm{possible}}\hspace{0.33em}{\mathrm{outcome}}{s}}$ This formula is simple if you know the number of favourable outcomes and the number of possible outcomes. This works well if asking questions like what is the probability of rolling a 7 with a pair of dice. To calculate the number of total outcomes, there are 6 possible ways a single die can be thrown, and for each of these, the other die can have 6 possible value. So the total will be 6 × 6 or 36. This illustrates the multiplication rule for counting things: If there are m ways for one thing to occur and n ways for a second thing to occur, then there are m × n ways to do both. Manually counting the ways to get a 7 where the first number is from die 1 and the second from die 2 gives: 1 + 6, 2 + 5, 3 + 4, 4 + 3, 5 + 2, and 6 + 1. Six ways So the probability of rolling a 7 is 6/36 or 1/6. Now what if I asked what is the probability of getting four aces in a 5-card poker hand? How do you even begin to count the number of possible poker hands? There are two ways to count large possibilities like this: combinations and permutations. A combination is the number of ways where a collection of objects can be arranged where you are not concerned with order. For example, in the card example, a hand of 2, 3, 4, 5, and 6 of hearts would be the same as a 6, 5, 4, 3, an 2 of hearts and you would only want to count these two possibilities as one along with any other arrangement of these five cards. A permutation is where order does count and these two card combinations would be counted as two permutations. In our card example, order doesn’t count, so we want the number of combinations of taking 52 cards, 5 at a time. Fortunately, there is a formula and notation used to simplify this. Before I present this, there is another math operation that needs to be explained: factorials. You may have a calculator with a “!” symbol or “x!” on one of the keys. This is a factorial operation. A factorial is successively multiplying an integer by one less for each factor. For example, 5! = 5 × 4 × 3 × 2 ×1 = 120. Factorials get large very quickly. For example, 30! is 265252859812191058636308480000000. To make the formulas using factorials consistent, a special definition 0! = 1 is made. So the notation for the number of r combinations of n objects is ${}_{n}{C}_{r}$ or more commonly C(nr). So in our case, we want to calculate C(52,5), this is the number of possible 5-card combinations out of a deck of 52 cards. The general formula for combinations is ${C}{(}{n}{,}\hspace{0.33em}{r}{)}\hspace{0.33em}{=}\hspace{0.33em}\frac{n!}{{r}{!(}{n}\hspace{0.33em}{-}\hspace{0.33em}{r}{)!}}$ In our poker hand example, the number of possible poker hands is ${C}{(}{52}{,}\hspace{0.33em}\hspace{0.33em}{5}{)}\hspace{0.33em}{=}\hspace{0.33em}\frac{52!}{{5}{!(}{52}\hspace{0.33em}{-}\hspace{0.33em}{5}{)!}}\hspace{0.33em}{=}\hspace{0.33em}\frac{52!}{5!(47)!}$ Now before you go off and calculate this, remember how large factorials can get? Many calculators cannot keep the number of digits necessary to accurately store very large numbers and the accuracy of the calculation will be poor. So when dealing with combination and permutation formulas, it is always best to simplify before calculating the answer. See if you can see where we can simplify the expression on the right side. I will continue this example in my next post.
ziphumulegn 2022-07-04 Subtraction of two fractions. I'm learning math by reading books only and the book I'm currently studying is Smart Math by Marcia Learner in which I practice basic operation on fractions. I can't solve the two following problems. Let me show you how I did it but my answers don't match the answers the book presents. 1. $-7/8-\left(-1/3\right)=?$ $-7/8X3/3=21/24$ and $-1/3X8/8=8/24$ $-21/24+8/24=13/24$ My answer is $13/24$ but, according to the book the answer is $-1/2$ 2. $12/13-\left(-3/7\right)=?$ $84/91-\left(-39/91\right)$ $84/91+39/91=123/91$ Book's answer is $105/14$ Please tell me where I've done the mistakes. I correctly did the other problems of similar kind but couldn't understand where I've done wrong. amanhantmk When adding or subtracting fractions, the denominators of the terms need to be the same. First let's simplify the number of signs: $-\frac{7}{8}-\left(-\frac{1}{3}\right)=-\frac{7}{8}+\frac{1}{3}=\frac{1}{3}-\frac{7}{8}.$ Then we'll make the denominators the same by multiplying top and bottom by the same number on each term separately: $\frac{1}{3}-\frac{7}{8}=\frac{1\cdot 8}{3\cdot 8}-\frac{7\cdot 3}{8\cdot 3}=\frac{8}{24}-\frac{21}{24}.$ Now we can subtract the numerators straight away: $\frac{8}{24}-\frac{21}{24}=\frac{8-21}{24}=-\frac{13}{24}.$ So you almost got it except for the sign. Do the same thing with the second problem. Do you have a similar question?
# Question: What’S The Perfect Square Of 48? ## How do you simplify radical 45? Since radical 45 is equal to radical 9 times radical 5, and because radical 9 is equal to 3 (since 9 is a perfect square), we can simplify radical 45 to 3 times radical 5 (see the diagram below for a more detailed look on how to simplify square roots).. ## How do you simplify radicals? Step 1: Find the prime factorization of the number inside the radical. Step 2: Determine the index of the radical. In this case, the index is two because it is a square root, which means we need two of a kind. Step 3: Move each group of numbers or variables from inside the radical to outside the radical. ## Is 48 a perfect square? Is 48 a perfect square number? A number is a perfect square (or a square number) if its square root is an integer; that is to say, it is the product of an integer with itself. Here, the square root of 48 is about 6.928. ## What is the square of 48? So the square of 48 is, by definition, a perfect square. For example: 9 is the square of 3, we say “9 is a perfect square” (“it is the square of 3”). We do not need to say (and so we don’t) that “9 is the perfect square of 3”. ## What is the perfect square of 47? 2,209Table of Squares and Square RootsNUMBERSQUARESQUARE ROOT441,9366.633452,0256.708462,1166.782472,2096.85696 more rows ## What are the prime numbers of 48? Answer and Explanation: The prime factorization of 48 is 2 × 2 × 2 × 2 × 3. We can find the prime factorization of 48 using a factor tree. ## What are the factors of 48? 48 is a composite number. 48 = 1 x 48, 2 x 24, 3 x 16, 4 x 12, or 6 x 8. Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48. Prime factorization: 48 = 2 x 2 x 2 x 2 x 3, which can also be written 48 = 2⁴ x 3. ## What is the product of prime factors of 48? The number 48 expressed as a product of its prime factors is 2 x 2 x 2 x 2 x 3. ## Is 47 a prime number Yes or no? When a number has more than two factors it is called a composite number. Here are the first few prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, etc. ## Is 50 a perfect square? Explanation: 50 is not the perfect square of an integer or rational number. … It is a square of an irrational, algebraic, real number, namely 5√2 , therefore you could call it a perfect square in the context of the algebraic numbers. ## Is 2 a perfect square? Answer: YES, 2 is in the list of numbers that are never perfect squares. The number 2 is NOT a perfect square and we can stop here as there is not need to complete the rest of the steps. ## What are the common factors of 18 and 48? We found the factors and prime factorization of 18 and 48. The biggest common factor number is the GCF number. So the greatest common factor 18 and 48 is 6. ## What are the factors of 48 and 64? The gcf of 48 and 64 can be obtained like this:The factors of 48 are 48, 24, 16, 12, 8, 6, 4, 3, 2, 1.The factors of 64 are 64, 32, 16, 8, 4, 2, 1.The common factors of 48 and 64 are 16, 8, 4, 2, 1, intersecting the two sets above.In the intersection factors of 48 ∩ factors of 64 the greatest element is 16.More items…
# Why Do Signs Change in Algebra? (Article 58) In article 58 of Algebra I, we read the following rule: 1. Write the quantity to be subtracted under that from which it is to be taken, placing similar terms under each other. 2. Conceive the signs of all the terms of the subtrahend to be changed, and then reduce by the rule for Addition. Step 2 of this rule leads to some confusion. Why should we “conceive the signs of all the terms of the subtrahend to be changed”? The reason is because operations are being performed within the subtrahend before it is subtracted from the minuend in the problem. These operations have an effect on the outcome of the final subtraction. Let us look at this example: From 3ax – 2y, take 2ax + 3y. Here, we are subtracting one quantity from another. The first quantity is 3ax – 2y. The second quantity is 2ax + 3y. It is helpful to set these in parentheses so that we remember that they are separate quantities. Thus, from (3ax – 2y), we will take (2ax + 3y). We are not subtracting the quantities 2ax and 3y from the quantity (3ax – 2y). We are subtracting the quantity (2ax + 3y) from the quantity (3ax – 2y). Why does this matter? Subtracting the two quantities from the minuend will produce a different result than subtracting the result of their operation as a quantity. Let a = 2, x = 3 and y = 4. 1. (3ax – 2y) – (2ax + 3y) 2. (18 – 8) – (12 + 12) 3. (10) – (24) = -14 Note the value of the subtrahend in line 3 and the answer, in bold. Now, let us take the quantities of the subtrahend out from the parentheses and see how this operation is performed incorrectly: 1. (3ax – 2y) – 2ax + 3y 2. (18 – 8) – 12 + 12 3. (10) – 12 + 12 = 10 In the first operation, we subtracted 12 from the minuend and then subtracted 12 more, leaving a difference of -14. In the second operation, we subtracted 12 from the minuend, but then added 12 back! This is not what the problem asked us to do. When we keep the quantities in parentheses, the solution is easy. However, when we take the quantities of the subtrahend out from the parentheses, we must change the signs to make sure that the same result is obtained. 1. (3ax – 2y) – 2ax + 3y 2. (18 – 8) – 12 + 12 3. (10) – 12 – 12 = -14 Here, we show that after subtracting 12, we are subtracting 12 again. When the quantities of the subtrahend are removed from the parentheses and subtracted individually, the signs must be changed to make sure the same effect is had on the minuend. Now, let us read the rule again and understand: 1. Write the quantity to be subtracted under that from which it is to be taken, placing similar terms under each other. 2. Conceive the signs of all the terms of the subtrahend to be changed, and then reduce by the rule for Addition.
Solutions for Session 7, Part C See solutions for Problems: C1 \| C2 \| C3 \| C4 Problem C1 a. If the numerator is larger than the denominator, the fraction is greater than 1. b. If twice the numerator is larger than the denominator, the fraction is greater than 1/2. If twice the numerator is smaller than the denominator, the fraction is less than 1/2. Problem C2 Here are the fractions within each range: • 0 to 1/2: 17/35, 2/9 • 1/2 to 1: 4/7, 14/15 • Greater than 1: 25/23 Problem C3 a. These fractions have the same denominator. The order is 4/17, 7/17, 12/17. b. These fractions have the same numerator. The order is 3/8, 3/7, 3/4. c. These fractions all have numerators that are one less than their denominators. The order is 3/4, 5/6, 7/8. d. These fractions all have numerators that are five less than their denominators. The order is 1/6, 8/13, 12/17. e. These fractions all have numerators that are one less than their denominators. The order is 2/3, 5/6, 10/11. Problem C4 First divide the list into fractions larger and smaller than 1/2: • Smaller than 1/2: 2/5, 1/3, 1/4 • Larger than 1/2: 5/8, 3/4, 2/3, 4/7 For those smaller than 1/2, 2/5 is larger than 1/4 (numerators three less than denominators), 1/3 is larger than 1/4 (same numerator), and 2/5 is larger than 1/3 (compare using equivalent fractions 6/15 and 5/15). The order is 1/4, 1/3, 2/5. For those larger than 1/2, convert 3/4 and 2/3 to fractions where the numerator is three less than the denominator. (Or you may be able to visualize that 3/4 is greater than 2/3.) The list becomes 5/8, 9/12, 6/9, 4/7. This makes it easy to put the list in order: 4/7, 5/8, 6/9, 9/12. The complete list is 1/4, 1/3, 2/5, 4/7, 5/8, 2/3, 3/4.
## What is a Mathematical Equation? A mathematical equation is a statement that consists of two or more expressions connected by the equals sign (=). It represents the relationship between two sides, where the left-hand side (LHS) and the right-hand side (RHS) are equal to each other. The purpose of an equation is to solve for the value of an unknown variable, which is represented by one of the expressions in the equation. It is important to note that an equation needs to have an equals sign; if it’s missing, the numbers or values are considered to be an “expression” rather than an equation. It’s important to note that an equation requires an equals sign to be considered as such. If the equals sign is absent, the numbers or values are instead considered to be an “expression.” And this concept is well explained in Geneo App by the help of interactive examples and simulations. Learners can easily visualize the theorem and see how it is applied in real-life examples, making the learning experience more relatable and meaningful. Each math equation usually has several components, which might include the following: • Coefficients • Variables • Terms • Constants • Expressions ## Types of Math Equations There are different forms of mathematical equations, some of which are as follows: • Linear equation • Rational equation • Cubic equation Linear equations are those in which the highest power of a variable is 1, while cubic equations involve one variable raised to the exponent 3. Similarly, a quadratic equation will always have one variable raised to the exponent 2. These equations are commonly used in various aspects of daily life, and they have a wide range of applications in both academic and non-academic settings. They are used in various fields like physics, chemistry, engineering, economics, and even in everyday problems. They can help us in understanding the relationship between different variables and help us to make predictions and take decisions based on them. Let’s explore different scenarios, the various ways in which mathematical equations are used in real-world scenarios with the help of Geneo Learning App, in order to deepen your understanding of their applications and significance. ## The application of the Pythagorean theorem in architectural design. This is one of the most popular applications of mathematical equations in real life. The Pythagorean theorem, named after the Greek mathematician Pythagoras, is commonly used to find the length of an unknown side or the steepness of a slope and the angle of a triangle. It is used extensively in the field of architectural design, and it’s an important part of the Geneo Learning App curriculum. This theorem can be applied in various scenarios like in the building of bridges, houses, and ramps. The Geneo App is designed to make learning about the Pythagorean theorem easy and intuitive. The app provides interactive and engaging visual aids that help students understand the concepts behind the theorem. ## For Measuring the pH of a Substance or Chemical Measuring the pH of a substance or chemical is an essential step in understanding the acidity level of the substance. The process of measuring pH is done by using small strips of paper, often called pH paper, which change color when dipped in a chemical substance. The change in color on the pH paper indicates the acidity level of the substance. This process makes use of logarithms, which are a mathematical concept that allows us to determine the power to which a number is raised to get a new value. Logarithms are employed to calculate the acidity level of the substance, and the result is known as the pH scale. ### Uses of Mathematical Equation In the Field of Cryptology Cryptology, the study of codes and their solutions, is another area where mathematical equations are frequently used in daily life. Researchers and experts in the field use mathematical equations to encrypt confidential data on the Internet and protect it from unauthorized access. One example of a math equation used in cryptology is the Fibonacci series, a sequence of numbers in which each number is the sum of the two preceding ones. This concept is not only used in cryptology but also in nature, often found in patterns in trees and flowers. ### The Vital Role of Mathematical Equations in Research and Development In the field of research and development, mathematical equations play a crucial role in helping scientists and researchers understand and analyze complex phenomena. It is used to model different systems and processes, as well as in the development of new technologies. One particular area where mathematical equations are heavily used is in the study of bacterial growth and decay. Researchers use mathematical equations to model how bacteria react to different combinations and situations, allowing them to make predictions and understand the underlying mechanisms. ### In the Field of Baking or Cooking Fractions play a crucial role in the kitchen, often used by chefs to measure portions, especially when preparing recipes without specific serving sizes. By using fractions, chefs are able to accurately divide ingredients and make precise measurements, resulting in consistent and precise portions. This is especially important when cooking for a large group, where precise measurements are essential in avoiding food waste. ### Unlock the full potential of your education with the Geneo Learning App With mathematical equations present in many aspects of our lives, it’s important to have a solid understanding of them. The Geneo Learning App provides curated, NCERT-aligned content for students in grades 5-10, allowing them to follow the same curriculum as taught in school, promoting better overall learning and building a strong conceptual foundation. With self-paced learning and AI-guided curriculum tailored to each student, it ensures maximum engagement and retention. Unleash the power of personalized learning with the Geneo App! The Geneo Learning App offers a comprehensive and convenient way to supplement traditional education. With self-paced learning and AI-guided curriculum tailored to each student, it ensures maximum engagement and retention. In addition to this, it provides mentor support and Live teacher-led online classes, assignments, and assessments to support and fortify students’ conceptual understanding, making Geneo App a perfect supplement to traditional classroom learning.
Mathematic # Concept of Locus Locus is a curve or other figure formed by all the points satisfying a particular equation of the relation between coordinates, or by a point, line, or surface moving according to mathematically defined conditions. It is the set of all points (usually forming a curve or surface) satisfying some condition. For example, the locus of points in the plane equidistant from a given point is a circle, and the set of points in three-space equidistant from a given point is a sphere. Example: A Circle is “the locus of points on a plane that are a certain distance from a central point”. Consider a point that is moving in a plane. If we join each of the positions through which the moving point passes, we will get the path followed by the point. If the movement of the point is random, its path will be irregular, and we shall not be able to predict, for example, how its path will look when the point stops. Therefore, the path of a randomly moving point cannot be determined. On the other hand, if the point’s movement obeys a given set of rules, the resulting path will be predictable and can be a circle, a straight line, an ellipse, and so on. The word locus describes the position of points which obey a certain rule Three important loci are: • The circle – the locus of points which are equidistant from a fixed point, the centre. • The perpendicular bisector – the locus of points which are equidistant from two fixed points A and B. • The angle bisector – the locus of points which are equidistant from two fixed lines. So, you can look upon a circle, a straight line, and so on, as a set of points positioned according to a given rule or rules. Definition of locus: A path consisting of a set of points whose location is governed by some definite rule of rules is called locus. Locus in Latin means “position” or “location”. Examples: Let a point P move in such a way that it is always at a constant distance, r, from ma fixed point O. The locus of a point P is a circle which has its center at O and has a radius equal to r units. Strategy for Solving Locus Problems: 1. Draw a diagram using the given information. 2. Read carefully to determine what conditions need to be satisfied. 3. Start by finding ONE point that satisfies the needed conditions and plot it on your diagram. Then find several additional points that satisfy the conditions and plot them as well. Continue plotting points until you can see the pattern, or path, that is developing. 4. Draw a dashed line, or circle, through the plotted points to show the locus. 5. If you are asked to describe your locus in writing, be very specific so you will give a clear description of your work. 6. If TWO, or more, conditions exist in the problem (a compound locus), repeat steps 2 through 5 for the additional conditions on the same diagram. You will know you have a compound locus when you see the words “AND” or “AND ALSO” separating the conditions. 7. For a compound locus, count the number of locations where the two loci intersect (where the dashed drawings cross). Clearly label these locations on the diagram, or describe them in words. Information Source:
# How do you solve x=sqrt(x+6) and find any extraneous solutions? Apr 30, 2018 $x = 3 , x = - 2 \text{ is extraneous}$ #### Explanation: $\textcolor{b l u e}{\text{square both sides}}$ $\text{note that } \sqrt{a} \times \sqrt{a} = {\left(\sqrt{a}\right)}^{2} = a$ ${x}^{2} = {\left(\sqrt{x + 6}\right)}^{2}$ $\Rightarrow {x}^{2} = x + 6$ "express in "color(blue)"standard form ";ax^2+bx+c=0 $\Rightarrow {x}^{2} - x - 6 = 0$ $\text{the factors of - 6 which sum to - 1 are - 3 and + 2}$ $\Rightarrow \left(x - 3\right) \left(x + 2\right) = 0$ $\text{equate each factor to zero and solve for x}$ $x + 2 = 0 \Rightarrow x = - 2$ $x - 3 = 0 \Rightarrow x = 3$ $\textcolor{b l u e}{\text{As a check}}$ Substitute these values into the equation and if both sides equate then they are the solutions. $x = - 2 \to \text{ left "=-2" right } = \sqrt{4} = 2$ $- 2 \ne 2 \Rightarrow x = - 2 \text{ is extraneous}$ $x = 3 \to \text{left "=3" right } = \sqrt{9} = 3$ $\Rightarrow x = 3 \text{ is the solution}$
# The ninth grade class has 17 more girls than boys. There are 431 in all. How many boys are there? How many girls are there? Jun 11, 2016 224 girls #### Explanation: Method 1: You can use an equation. The number of boys is less than the number of girls, so call that $x$ Let $x$ be the number of boys, then the number of girls is $x + 17$ Altogether there are 431. $x + x + 17 = 431$ $2 x = 431 - 17$ $2 x = 414$ $x = 207 \text{ this is the number of boys}$ There are 207 + 172 = 224 girls. Method 2: Consider that the number of boys and girls are the same, apart from the 17. If we get the difference of 17 out of the way first, then the total can be divided by 2. $431 - 17 = 414.$ 414 ÷ 2 = 207 Boys: 207 Girls: 207 + 17 = 224 (the initial difference).
# Simplifying Expressions Math Activities Instructor: Clio Stearns Clio has taught education courses at the college level and has a Ph.D. in curriculum and instruction. Teaching students to simplify math expressions is an important part of helping students learn how to work and think algebraically. This lesson offers activities that will keep your students engaged while teaching them how to simplify expressions. ## Teaching Simplifying Expressions As a high school algebra teacher, a great deal of the work you do hinges on your students' ability to simplify expressions. Simplifying expressions involves knowing the order of operations and how to use it, understanding what it means to combine like terms, and knowing how to work with variables and coefficients. Once students know how to simplify expressions, they are well positioned to move on to more complex work in algebra. Teaching students to simplify expressions can also be great fun, especially if you work with activities as part of your teaching. The activities in this lesson appeal to different learning styles while helping your students simplify expressions. ## Visual Activities This section provides activities that will appeal to visual learners, who often benefit from images and graphic organizers as they work and learn. ### Graphic Guide This is an activity students can do independently or with partners. Give each student one or two algebraic expressions to work with, ensuring that they have expressions that are well suited to their mathematical abilities. Then, ask students to make a step-by-step, illustrated guide showing how to simplify the expression. They should break down their simplification according to order of operations, showing what they do using colors and icons to illustrate their processes. Finally, let students share their guides and simplification with their classmates. ### Sorting Expressions This activity will give good practice at combining like terms in algebraic expressions. Have students work in partnerships and give each pair a set of ten to 15 expressions to work with. Ask them to rewrite the expressions the same way, but use the same colors for terms that can then be combined. For instance, in 2x + 3x = 17, they would rewrite 2x and 3x both in red. More complex expressions will obviously involve more colors. Then, once students have sorted their expressions by colors, they will be able to simplify them more easily. ## Games Here, you will find activities designed to make it fun and active for students to work on simplifying expressions. Start this game by writing a series of long form expressions and their corresponding simplified forms on slips of paper, resulting in enough papers for each student in your class. Then, tape one slip to each student's back. Students should mill around the classroom, guiding each other to find their corresponding pair. For example, students would guide 2x + 3x toward 5x, etc. After a while, bring students together to share what they came up with. ### Expressions Ball Toss Let students work in small groups for this activity. Each group should start with a striped beach ball, with one expression written in permanent marker on each stripe. To unlock this lesson you must be a Study.com Member. ### Register to view this lesson Are you a student or a teacher? #### See for yourself why 30 million people use Study.com ##### Become a Study.com member and start learning now. Back What teachers are saying about Study.com ### Earning College Credit Did you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.
# How do you factor x^2+3x+12=0? Mar 5, 2018 There are no Real factors of ${x}^{2} + 3 x + 12$ #### Explanation: If you are willing to include Complex values in your solution then you can use the quadratic formula to get $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{- 3 \pm \sqrt{{3}^{2} - \left(4 \cdot 1 \cdot 12\right)}}{2 \cdot 1} = \frac{- 3 \pm \sqrt{39} i}{2}$ for the factoring: $\left(x + \frac{3 + \sqrt{39} i}{2}\right) \left(x + \frac{3 - \sqrt{39} i}{2}\right) = 0$ Mar 5, 2018 $\left[x + \frac{3}{2} + \frac{\sqrt{39} \textcolor{w h i t e}{. .} i}{2}\right] \textcolor{w h i t e}{. .} \left[x + \frac{3}{2} - \frac{\sqrt{39} \textcolor{w h i t e}{. .} i}{2}\right] = 0$ #### Explanation: First of all there are no whole number factors of 12 that can be use to sum to 3 from the $3 x$. Consequently we are looking at fractional factors. Lets check to see if the plot actually crosses the x-axis. If it does then the determinate part of $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ will be positive. Determinate $\to {b}^{2} - 4 a c \textcolor{w h i t e}{\text{d") ->color(white)("d}} {3}^{2} - 4 \left(1\right) \left(12\right) = 9 - 48 < 0$ Thus the roots do not belong to the set of real numbers ( $\mathbb{R}$ ) They belong to the set of numbers called complex ( $\mathbb{C}$ ) ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ $x = \frac{- 3 \pm \sqrt{- 39}}{2} = - \frac{3}{2} \pm \frac{\left(\sqrt{39}\right) \textcolor{w h i t e}{. .} i}{2}$ $x = \left(- \frac{3}{2} - \frac{\left(\sqrt{39}\right) \textcolor{w h i t e}{. .} i}{2}\right)$ $x = \left(- \frac{3}{2} + \frac{\left(\sqrt{39}\right) \textcolor{w h i t e}{. .} i}{2}\right)$ $\left[x + \frac{3}{2} + \frac{\left(\sqrt{39}\right) \textcolor{w h i t e}{. .} i}{2}\right] \textcolor{w h i t e}{. .} \left[x + \frac{3}{2} - \frac{\left(\sqrt{39}\right) \textcolor{w h i t e}{. .} i}{2}\right] = 0$
## John and Erica each have a marble collection. • The number of marbles in John’s collection can be represented by x. • The number of marbles Question John and Erica each have a marble collection. • The number of marbles in John’s collection can be represented by x. • The number of marbles in Erica’s collection is 10 more than 4 times the number of marbles in John’s collection. • The total number of marbles in both collections is 270. Find the number of marbles in John’s collections. in progress 0 3 years 2021-09-03T10:36:20+00:00 1 Answers 1 views 0 1. Answer: John has 52 marbles in his collection Step-by-step explanation: Step 1 LET The number of marbles in John’s collection can be represented by x. The and the number of marbles in Erica collection can be represented as =4x+ 10 The total number of marbles in both collections is 270 such that the expression to solve the equation will be JOHN + ERICA = 270marbles X + 4X+ 10= 270 Step 2— solving X + 4X+ 10= 270 5X+ 10= 270 5X= 270-10 5X= 260 X= 260/5 =52 John has 52 marbles in his collection Erica has 4x + 10 = 4x 52 + 10=218 marbles
# Definition of a factorial The factorial of a non-negative integer number n, denoted as n! is the product of all the integers ranging from 1 to n. That is: n! = n.(n-1).(n-2)...3.2.1 Examples: 5! = 5.4.3.2.1 = 120 9! = 9.8.7.6.5.4.3.2.1 = 362880 Note that 9! = 9.8.7.6.5! It means we can shorten the expression of a factorial as much as desired for simplifying purposes A special rule of factorials state that 0! = 1 ## Simplifying factorial expressions The above mentioned properties of the factorials can be used to simplify expressions where several factorials can be found. Let's simplify the expression: To better simplify the expression, the largest factorial should be expanded until it matches with the largest factorial in the opposite side of the fraction. In the example provided, the 9! will be expanded up to 7! to simplify them both. The remaining 4! must be expanded completely When the argument of a factorial is a sum, a subtraction, or any other operation, they must be performed before the factorial, since the factorial of a sum is in general, different from the sum of factorials. The same applies to the subtraction and other operations. Example: (1+3)! = 4! = 4.3.2.1 = 24 (5-5)! = 0! = 1
# Australian Bureau of Statistics CensusAtSchool #search{vertical-align:-3.5px; } Education Services Education Services homepage Concepts and definitions Click on the triangles to open a section. The table below is a list of the concepts covered in each section. Statistics Variables Sampling Frequency and distribution The frequency (f) of a particular observation is the number of times the observation occurs in that data. Cumulative frequency Cumulative frequency is the total of a frequency and all frequencies below it in a frequency distribution. It is the running total of frequencies. Relative frequency Relative frequency is another term for proportion. It is the number of times a particular observations occurs divided by the total number of observations. Distribution The distribution of a variable is the pattern of values of the observations. Graphs and displays Graph A graph is a diagram representing a system of connections or interrelations among two or more variables by a number of distinctive dots, lines, bars, etc. Chart A chart is a visual representation of data. Bar, line, pie and other types of charts are examples of charts. Box and whisker plots (often called ‘box plots’) can be used to show the interquartile range. Figure 1 shows a box and whisker plot of student ages. Notice that a scale is drawn underneath. Box plots can be drawn horizontally or vertically. Frequency distribution tables can be used for nominal and numeric variables. Example: Twenty people were asked how many cars were registered to their households. The results were recorded as follows: 1, 2, 1, 0, 3, 4, 0, 1, 1, 1, 2, 2, 3, 2, 3, 2, 1, 4, 0, 0. This data can be presented in a frequency distribution table – see Figure 2. Stem and leaf plots are a convenient way to organise data. Each observation value is considered to consist of two parts - a stem and a leaf. • the stem is the first digit or digits • the leaf is the final digit Example: The number of books ten students read in one year were as follows: 12, 23, 19, 6, 10, 7, 15, 25, 21, 12. In ascending order, these are: 6, 7, 10, 12, 12, 15, 19, 21, 23, 25. Figure 3 is a stem and leaf plot of this data. In the stem and leaf plot (fig 3): • the stem '0' represents the class interval 0-9 • the stem '1' represents the class interval 10-19 • the stem '2' represents the class interval 20-29. If there are a large number of observations for each stem, the stem can be split in two. For example the interval 0-9 could be split into intervals 0-4 and 5-9. The stem would then be written as 0(0) and 0(5). Time series A time series is a collection of observations of well-defined data items obtained through repeated measurements over time. For example, measuring the value of retail sales each month of the year would comprise a time series. Trend The ABS defines a trend as the long term movement in a time series without calendar related and irregular effects, and is a reflection of the underlying change in that measure. It is the result of influences such as population growth, price inflation and general economic changes. Fig 1 Box and whisker plot Number of cars (x) Tally Frequency (f) 0 l l l l 4 1 l l l l l 6 2 l l l l 5 3 l l l 3 4 l l 2 Fig 2 Frequency distribution table Stem Leaf 0 1 2 6 7 0 2 2 5 9 1 3 5 Fig 3 Stem and leaf plot Summary statistics Mean The mean of a numeric variable is calculated by adding together the values of all observations in a dataset and then dividing by the number of observations in the set. It is often referred to as the average. Thus: Mean = sum of all the observations ÷ number of observations For example, find the mean of these numbers 5, 3, 4, 5, 7, 6. Mean = (5 + 3 + 4 + 5 + 7 + 6) ÷ 6 = 30 ÷ 6 = 5 Notice that the value of every member of the dataset is used to calculate the mean. Median The median is the middle value of a set of odd numbered data, or the mean of the middle two in an even numbered set after the data have been placed in ascending order. For example, dataset A contains 3, 7, 1, 9, 2, 5, 9. Rearranged in ascending order it becomes: 1, 2, 3, 5, 7, 9, 9. The middle number is 5 so, the median is 5. Dataset B contains 1, 3, 4, 5, 10, 12, 13, and also has a median of 5 although the values of the data vary considerably. The position of the median can also be found by using the formula (n + 1) ÷ 2 , where n is the number of values in a set of ordered data. For dataset A: n = 7 So the position of the median = (7 + 1) ÷ 2 = 8 ÷ 2 = 4 The median is the fourth number which has a value of 5. The above example is for an odd number of observation, i.e. n = 7. However, an extra step is necessary when the number of observations is even. For example, if n = 8 then the position of the median = (8 + 1) ÷ 2 = 9 ÷ 2 = 4.5 This means that the position of the median lies between the fourth and fifth observations. To find the value of the median, add together the fourth and fifth observations and divide by two. For example, if the dataset is: 1, 1, 4, 4, 8, 9, 9, 10 then the median is, (4 + 8) / 2 = 6. The median value is decided by its location in the ordered dataset and not because of its actual value. Notice that the values of the other members of the dataset are not taken into consideration, only their position. There are as many values above the median as there are below. The median is usually calculated for numeric variables but may also be calculated for an ordinal nominal variable. Mode The mode is the most frequently observed value in a dataset. Mode is the only measure you can use when the data is categorical and has no order – for example, place of birth, favourite colour and hair colour. As the dataset is not numbers, you cannot add and divide, so you cannot find a mean. The dataset cannot be sorted from smallest to largest so you cannot find the middle value and median. The mode does not necessarily give an indication of a dataset’s centre. A set of data can have more than one mode (see Figure 1). For example, a group friends in Year 10 have the following hair colours: red, brown, blonde, black, blonde, black, brown, brown, black, blonde, brown, brown, black. HAIR COLOUR FREQUENCY Red 1 1 Brown 5 5 Black 4 4 Blonde 3 3 The most common hair colour is brown so the mode is brown. Range The range is the actual spread of data including any outliers. It is the difference between the highest and lowest observation. Range = maximum value – minimum value For the following dataset of students' ages: 17, 15, 14, 16, 14, 15, 16, 12, 17, 13, 12, 17, 13, 16, 15 Maximum value Minimum value = 17 = 12 Range = maximum value – minimum value = 17 – 12 = 5 The range of the student's ages is 5 years. Quartiles Quartiles divide data into four equal groups. Using the example of 15 students above, we have the following ordered dataset: 12, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 16, 17, 17, 17. We can divide this set into four equal sized groups with each group containing one quarter of the data: • The first quartile (Q1) is the value that 25% of the data is below. • The second quartile (Q2) is the value that 50% of the data is below. This is the same as the median. • The third quartile (Q3) is the value that 75% of the data is below. In the example: Q1 = 13 Q2 = 15 Q3 = 16 Interquartile range The interquartile range refers to the middle 50% of data. Another way to put it is the interquartile range is the difference between the upper (75%) and lower (25%). The interquartile range is an indicator of the spread of the data. It eliminates the influence of outliers since the highest and lowest quarters are removed. The interquartile range is found by subtracting Q1 from Q3. Five number summary (quartiles) This is a useful way to summarise data. It consists of: • the lowest value • the highest value • the first quartile (Q1) • the third quartile (Q3) • the second quartile (Q2). The range can be found from the difference between the highest and lowest value. The median is the second quartile (Q2) and the interquartile range is the difference between the third and first quartiles (Q3 – Q1). Standard deviation Standard deviation (s) is the measure of spread most commonly used when the mean is the measure of centre. Standard deviation is most useful for symmetric distributions with no outliers. The standard deviation for a discrete variable made up of n observations is the positive square root of the variance as show in Figure 3. Fig 1 Unimodal, bimodal and multimodal Fig 2 Quartiles Fig 3 Standard deviation formula
# Algebra: Substitution with Squares and Square Roots (1) In this worksheet, students substitute numbers into expressions containing squares and square roots. Key stage: KS 3 Curriculum topic: Algebra Curriculum subtopic: Substitute Numerical Values for Formulae/Expressions Difficulty level: ### QUESTION 1 of 10 When we substitute a number for a letter, we simply replace the letter with the number. However, we must remember that 4b means 4 × b and 4b2 means 4 × b × b. Example Work out the value of the following expression when a = 4 and b = 9 3a2 - 7 - 2√b 3a2 - 7 - 2b = 3 × a × a - 7 - 2 × b = 3 × 4 × 4 - 7 - 2 × √9 = 48 - 7 - 2 × 3 = 48 - 7 - 6 = 35 Work out the value of the following expression when a = 4 3a2 Work out the value of the following expression when a = -4 3a2 Work out the value of the following expression when a = 4 √a Work out the value of the following expression when a = 4 5√a Work out the value of the following expression when a = 16 4√a Work out the value of the following expression when a = 3 4a2 - 3 Work out the value of the following expression when a = 3 4a2 + 3a Work out the value of the following expression when a = 4 2√a - 3 Work out the value of the following expression when a = 25 2√a - a Work out the value of the following expression when a = 4 3√a + a2 • Question 1 Work out the value of the following expression when a = 4 3a2 48 • Question 2 Work out the value of the following expression when a = -4 3a2 48 • Question 3 Work out the value of the following expression when a = 4 √a 2 • Question 4 Work out the value of the following expression when a = 4 5√a 10 • Question 5 Work out the value of the following expression when a = 16 4√a 16 • Question 6 Work out the value of the following expression when a = 3 4a2 - 3 33 • Question 7 Work out the value of the following expression when a = 3 4a2 + 3a 45 • Question 8 Work out the value of the following expression when a = 4 2√a - 3 1 • Question 9 Work out the value of the following expression when a = 25 2√a - a -15 • Question 10 Work out the value of the following expression when a = 4 3√a + a2 22 ---- OR ---- Sign up for a £1 trial so you can track and measure your child's progress on this activity. ### What is EdPlace? We're your National Curriculum aligned online education content provider helping each child succeed in English, maths and science from year 1 to GCSE. With an EdPlace account you’ll be able to track and measure progress, helping each child achieve their best. We build confidence and attainment by personalising each child’s learning at a level that suits them. Get started Start your £1 trial today. Subscribe from £10/month. • Tuition Partner
## When is a triangle like a circle? For a circle with radius $r$, its area is $\pi r^2$ and its circumference is $2\pi r$. If you take the derivative of the area formula with respect to $r$, you get the circumference formula! Let’s define the term “differential radius.” The differential radius $r$ of a shape with area $A$ and perimeter $P$ (both functions of $r$) has the property that $dA/dr = P$. (Note that $A$ always scales with $r^2$ and $P$ always scales with $r$.) For example, consider a square with side length $s$. Its differential radius is $r = s/2$. The square’s area is $s^2$, or $4r^2$, and its perimeter is $4s$, or $8r$. Sure enough, $dA/dr = d(4r^2)/dr = 8r = P$. What is the differential radius of an equilateral triangle with side length s? Extra credit: What is the differential radius of a rectangle with sides of length $a$ and $b$? My solution: [Show Solution] ## Chessboard race This week’s Fiddler is a puzzle with a surprise connection to physics! A tiny ant is racing across a 2-by-2 chessboard, as shown below, where each smaller square has a side length of 1 cm. The ant starts at the bottom-left corner of the bottom-left black square and is trying to reach the top-right corner of the top-right black square. The ant moves faster on the white squares than on the black squares. Speed on the white squares is 1 cm per minute, while speed on the black squares is 0.9 cm per minute. What’s the least amount of time it will take the ant to reach the finish? Extra Credit: Instead, the board is now 8-by-8, as shown below. My solution: [Show Solution] ## Pancake race This week’s Fiddler is a logic puzzle about getting home as fast as possible. Alice, Bob, and Carey start together and each walk home at a different constant speed. Once all three get home, they can have pancakes! Alice can walk home in 10 minutes, Bob can do it in 20, and Carey in 30. Fortunately, any of them can carry any of the others on their back without reducing their own walking speed. Assume that they can pick someone up, set someone down, and change direction instantaneously. What is the fastest they can get to eat pancakes? Extra Credit There is now a fourth: Dee. Dee is the slowest, needing 60 minutes to walk home. As before, anyone can carry anyone else, and they won’t get pancakes until everyone gets home. What is the fastest this can happen? My solution: [Show Solution] ## The weaving loom problem This week’s Fiddler is a classic problem. A weaving loom consists of equally spaced hooks along the x and y axes. A string connects the farthest hook on the x-axis to the nearest hook on the y-axis, and continues back and forth between the axes, always taking up the next available hook. This leads to a picture that looks like this: As the number of hooks goes to infinity, what does the shape trace out? Extra credit: If four looms are rotated and superimposed as shown below, what is the area of the white region in the middle? My solution: [Show Solution] ## Ellipse packing You’ve heard of circle packing… Well this week’s Riddler Classic is about ellipse packing! This week, try packing three ellipses with a major axis of length 2 and a minor axis of length 1 into a larger ellipse with the same aspect ratio. What is the smallest such larger ellipse you can find? Specifically, how long is its major axis? Extra credit: Instead of three smaller ellipses, what about other numbers of ellipses? My solution: [Show Solution] ## Randomly cutting a sandwich This week’s Riddler Classic is geometry puzzle about randomly slicing a square sandwich. I have made a square sandwich, and now it’s time to slice it. But rather than making a standard horizontal or diagonal cut, I instead pick two random points along the perimeter of the sandwich and make a straight cut from one point to the other. (These points can be on the same side.) What is the probability that the smaller resulting piece has an area that is at least one-quarter of the whole area? My solution: [Show Solution] ## Perfect pizza sharing This week’s Riddler Classic is about how to cut a pizza to achieve precise area ratios between the slices. Dean made a pizza to share with his three friends. Among the four of them, they each wanted a different amount of pizza. In particular, the ratio of their appetites was 1:2:3:4. Therefore, Dean wants to make two complete, straight cuts (i.e., chords) across the pizza, resulting in four pieces whose areas have a 1:2:3:4 ratio. Where should Dean make the two slices? Extra credit: Suppose Dean splits the pizza with more friends. If six people are sharing the pizza and Dean cuts along three chords that intersect at a single point, how close to a 1:2:3:4:5:6 ratio among the areas can he achieve? What if there are eight people sharing the pizza? My solution: [Show Solution] To jump straight to the results: [Show Solution] ## How high should you climb up the tower? This week’s Riddler classic is a neat geometry problem. Two people climb two of the tallest towers on an planet, which happen to be in neighboring cities. You both travel 100 meters up each tower on a clear day. Due to the curvature of the planet, they can barely make each other out. The first person returns to the ground floor of their tower. How high up their tower must the second person be you can barely make each other out again? My solution: [Show Solution] ## Desert escape This week’s Riddler classic is about geometry and probability, and desert escape! Here is the (paraphrased) problem: There are $n$ travelers who are trapped on a thin and narrow oasis. They each independently pick a random location in the oasis from which to start and a random direction in which to travel. What is the probability that none of their paths will intersect, in terms of $n$? My solution: [Show Solution] ## Cone crawling This week’s Riddler Classic is a geometry problem about traversing the surface of a cone The circular base of the cone has a radius of 2 meters and a slant height of 4 meters. We start on the base, a distance of 1 meter away from the center. The goal is to reach the point half-way up the cone, 90 degrees around the cone’s central axis from the start, as shown. What is the shortest path? Here is my solution: [Show Solution]
# Probability Without Replacement – Explanation & Examples Did you ever take candy from a box, gave it to your older kid (who ate the candy), then took a second candy, and gave it to your younger kid? If your answer is yes, you have already come across a nice story involving dependent probability, generally termed as Probability Without Replacement. But if your older kid had returned the candy and you had put it back in the box, the second event would have been independent. The box of sweets (sample space) would have remained the same for the second event — probability with replacement. Probability without replacement involves dependent events where the preceding event has an effect on the probability of the next event. Probability Without Replacement may initially sound tricky but trust me; it is one of the most specific mathematics topics. This lesson will clear your concept about dependent probability, and we will learn step by step how to calculate probability without replacement. This lesson will also help you understand how to use a probability tree diagram involving dependent probability — probability without replacement. ## What does probability without replacement mean? Probability without replacement means once we draw an item, then we do not replace it back to the sample space before drawing a second item. In other words, an item cannot be drawn more than once. For example, if we draw a candy from a box of 9 candies, and then we draw a second candy without replacing the first candy. Of course, the sample space would no longer remain 9 for the second event because we have not replaced the first candy. Thus, the sample space would be 8 for the second event. In other words, the sample space has been changed for the second event. We use probability without replacement to solve the problems where the sample space changes for different events and the occurrence of the next event depends upon what happens in the preceding event. ## How to calculate probability without replacement? Now that we briefly understand dependent probability – before diving deep into calculating probability without replacement – let’s first try to visualize what dependent events are and how the preceding event can affect the next event. For example, a box contains $4$ blue candies and $5$ orange candies, as shown below. Let us consider that event A is getting a blue candy. As there are $4$ blue candies, so the number of ways it can happen = $4$. Also, the total number of candies is $9$, so the total number of outcomes = $9$. Let’s suppose $P(A)$ represents the probability of getting a blue candy. Thus, we can calculate the probability of getting a blue candy such as: P(A) = number of ways it can happen $\displaystyle /$ total number of outcomes $P(A) = \frac{4}{9}$ Thus, the probability of getting a blue candy is $4$ in $9$, as shown below. Now going ahead, we need to understand that the probabilities change if we take any candy out. Here, let us check the two scenarios such as: Scenario 1: If we draw a orange candy before, then the probability of a blue candy next is $4$ in $8$, as shown below. Scenario 2: And if we draw a blue candy before, then the probability of a blue candy next is $3$ in $8$, as shown below. Here is the complete snapshot of what we learned. The probabilities change because we are removing candies from the box. The next event tends to depend on what occurs in the preceding event, also known as a dependent. Thus, when we remove the candies each time without replacement, then the probabilities change, and the events are dependent. A tree diagram is a pleasing way to visualize the concept involving probability without replacement. Taking the same example, there is a $\frac{5}{9}$ chance of drawing a orange candy from the box, and a $\frac{4}{9}$ chance for blue candy such as: Going ahead, let’s check what occurs when we draw a second candy. If a blue candy is chosen first, there is a $\frac{4}{8}$ probability of getting a blue candy and a $\frac{4}{8}$ probability of getting a blue candy. The reason is that after $1$ orange candy is taken, we have $8$ candies left, of which $4$ are orange, and $4$ are blue. If a blue candy is chosen first, there is a $\frac{5}{8}$ probability of getting a orange candy and a $\frac{3}{8}$ probability of getting a blue candy. Again, after $1$ blue candy is taken, we have $8$ candies left, of which $3$ are orange, and $5$ are blue. Now, we can determine a variety of probabilities without replacement with the help of the above tree diagram. For example, let’s say we need to determine the probability of drawing $2$ blue candies. We can visualize from the tree diagram that it is a $\frac{4}{9}$ probability followed by a $\frac{3}{8}$ probability, as shown in the tree diagram below. From the tree diagram, we can easily figure out that we have just multiplied the probabilities such as: $\frac{4}{9}\times \frac{3}{8}=\frac{12}{72}=\frac{1}{6}$ Thus, the probability of drawing 2 blue candies is $\frac{1}{6}$. ### Probability without replacement formula In our example, event $A$ is getting a blue candy, and $P(A)$ represents the probability of getting a blue candy with a probability of $\frac{4}{9}$: $P(A) = \frac{4}{9}$ Also, event $B$ is getting a blue candy second, but for that, we have two scenarios such as: • If we chose a blue candy first, the probability is now $\frac{3}{8}$. • If we chose a orange candy first, the probability is now $\frac{4}{8}$. It is up to us which one we chose. And we can use the notation $P(B\mid A)$ read as ‘the probability of $B$ given $A$‘. Please note that the symbol ‘\$\|
All the solutions provided in McGraw Hill My Math Grade 5 Answer Key PDF Chapter 5 Add and Subtract Decimals will give you a clear idea of the concepts. Essential Question How can I use place value and properties to add and subtract decimals? 1. First, line up the decimal points vertically. 2. Fill in any 0’s wherever it is necessary. 3. Add or subtract the numbers as if the given numbers are whole numbers. 4. Place the decimal point in the sum or difference with the numbers being added or subtracted. Name the place-value position of each underlined digit. Question 1. 52 ______________ 5 tens. Explanation: The place-value position of the underlined digit is 5 tens. Question 2. 138 ______________ 8 ones. Explanation: The place-value position of the underlined digit is 8 ones. Question 3. 4.3 ______________ 3 ones. Explanation: The place-value position of the underlined digit is 5 ones. Question 4. 901 ______________ 9 Hundred. Explanation: The place-value position of the underlined digit is 9 hundreds. Question 5. 1.216 _______________ 1 Tens. Explanation: The place-value position of the underlined digit is 1 Tens. Question 6. 2,785 _______________ 2 Thousands. Explanation: The place-value position of the underlined digit is 2 Thousands. Round each number to the underlined place. Question 7. 19 = ____________ 1 Tens. Explanation: The place-value position of the underlined digit is 1 tens. Question 8. 681 = ______________ 8 Thousands. Explanation: The place-value position of the underlined digit is 5 Thousands. Question 9. 735 = _______________ 7 Thousands. Explanation: The place-value position of the underlined digit is 5 ones. Question 10. 3,705 = _______________ 3 Thousands Explanation: The place-value position of the underlined digit is 3 Thousands. Question 11. 106,950 = _________________ 6 Thousands. Explanation: The place-value position of the underlined digit is 6 Thousands. Question 12. 5,750 = ________________ 7 Hundreds. Explanation: The place-value position of the underlined digit is 7 Hundreds. Question 13. 38 + 16 = ______________ 54 Explanation: The addition of the given two numbers is 54. Question 14. 151 + 218 = ______________ 369 Explanation: The addition of the given two numbers is 369. Question 15. 260 + 398 = _____________ 658 Explanation: The addition of the given two numbers is 658. Question 16. 235 + 68 = ______________ 303 Explanation: The addition of the given two numbers is 303. Question 17. The Pham family and the Weber family have many pets. How many more pets does the Pham family have than the Weber family? The Pham family has 4 more pets than the Weber family. Explanation: Given, The Pham family has 3 dogs, 1 cat, and 6 fish Total no of pets Pham family has = 3 + 1 + 6 = 10 The weber family has 2 dogs, 3 gerbils, and 1 turtle Total no of pets weber family has = 2 + 3 + 1 = 6 10 – 6 = 4 Therefore the Pham family has 4 more pets than the Weber family. How Did I Do? My Math Words Review Vocabulary greater than (>) less than (<) equal to (=) Making Connections Compare the numbers in each row. Use the review vocabulary to compare the two numbers in each row using >, <, or =. Describe how you used place value to complete the chart. Given, The number is 57.2 and 57.02 Now compare the given numbers Check the place values 2 ones > 0 ones 57.2 > 57.02 Numbers are 12.01 and 12.1 Now compare the numbers 0 ones < 1 ones Numbers are 12.6 and 12.60 6 ones is equal to 6 ones 6 = 6 12.6 = 12.60 Numbers are 24.56 and 24.5 Now compare the numbers 6 ones > 0 ones 24.56 > 24.5 Given Numbers are 6.99 and 6.89 9 > 8 6.99 > 6.89 Explanation: My Vocabulary Cards Ideas for Use • Write a tally mark on each card every time you read the word in this chapter or use it in your writing. Challenge yourself to use at least 10 tally marks for each card. • Use the blank cards to write your own vocabulary cards. The order in which numbers are added does not change the sum. Commutative means “something that involves substitution.” How does this definition relate to the Commutative Property? Explanation: The related to the Commutative Property is called Commutative Property of addition. The way in which numbers are grouped does not change the sum. Compare the example on this card with the example on the Commutative Property card. What differences do you notice? It is called associative property of addition. Explanation: The way in which numbers are grouped does not change the sum is called associative property of addition. The commutative property means that a+b is the same as b+a. The associative property means that (a+b)+c is the same as (a+c)+b. Operations that undo each other. What are the inverse operations for addition and division? Inverse operations are opposite operations. The division is the inverse of multiplication and subtraction is the inverse of addition and The sum of any number and 0 equals the number. Identity property My Foldable
# 2TH/vwo lesson 3 - Equations/Graphs First let's look at the steps we need to plot a linear graph 1 / 18 Slide 1: Tekstslide Middelbare schoolvwoLeerjaar 1 In deze les zitten 18 slides, met interactieve quizzen en tekstslides. Lesduur is: 50 min ## Onderdelen in deze les First let's look at the steps we need to plot a linear graph Formula Table Graph #### Slide 2 -Tekstslide Sometimes we only have a graph and not the formula. What do we do then? Graph Table Formula #### Slide 4 -Tekstslide Steps 1. Find some values on the graph that can be read off easily 2. Make a table 4. Make a linear formula #### Slide 5 -Tekstslide Remember this? • Make a table with values of t on the top row and corresponding values of N on the bottom row t 0 1 2 3 4 N 2 2.5 3 3.5 4 • Use the table to find the y-intercept (value of N when t=0) so here y-intercept = 2 • Use the table to calculate the gradient increase in N /increase in t = 1/2 or 0.5 • Fill these values into the standard formula y=ax + b where a is the gradient and b is the y-intercept • y = 0.5x + 2 #### Slide 6 -Tekstslide What is the difference between a formula and an equation? Formula - is a rule about how you calculate for all values Equation - When a formula has a specific answer. You can solve the equation to find which value of the variable gives the correct answer. #### Slide 7 -Tekstslide How do you find a value that is not easy to read off a graph? Use the steps above to find the linear formula Write down the corresponding equation Solve the equation #### Slide 8 -Tekstslide The graph corresponds to the growth of a seedling. After how many days is the seedling 50cm high? Graph Table time in days (t) 0 2 4 6 height in cm (H) 5 10 15 20 y-intercept = 5 formula: H = 5 + 2.5t but H=50 so equation 50 = 5 + 2.5t Solving 45 = 2.5t 45/2.5 = 18 = t so after 18 days the seedling is 50cm high #### Slide 9 -Tekstslide What is a formula? #### Slide 10 -Open vraag What is an Equation? #### Slide 11 -Open vraag How many coins weigh 875 grams? A 75 B 7.5 C 100 D 65 #### Slide 12 -Quizvraag Here is the graph for the formula b = 10 + 5t Solving the equation 21.50 = 10 + 5t gives t=? A 3.2 B 2.3 C 11.5 D 5 #### Slide 13 -Quizvraag L= 3m +12 L = 4m + 12 L = 1/3m + 12 What are the values of m when L=48? A m=3, m=4, m=1/3 B m=16, m=12, m= 144 C m=1/3, m=4, m=3 D m=12, m=9, m=108 #### Slide 14 -Quizvraag More than 1 mistake ex 15,17,18 ex 38,40, T3 ex B4,B5 Only 1 mistake Ex 38,40, T3 ex B4,B5 All correct ex T3,B4,B5 #### Slide 17 -Tekstslide What do you remember from today's lesson? Graphs
Subset In mathematics, a set A is a subset of a set B, or equivalently B is a superset of A, if A is "contained" inside B, that is, all elements of A are also elements of B. A and B may coincide. The relationship of one set being a subset of another is called inclusion or sometimes containment. A is a subset of B may also be expressed as B includes A; or A is included in B. The subset relation defines a partial order on sets. The algebra of subsets forms a Boolean algebra in which the subset relation is called inclusion. ## Definitions If A and B are sets and every element of A is also an element of B, then: • A is a subset of B, denoted by ${\displaystyle A\subseteq B,}$ or equivalently • B is a superset of A, denoted by ${\displaystyle B\supseteq A.}$ If A is a subset of B, but A is not equal to B (i.e. there exists at least one element of B which is not an element of A), then • A is also a proper (or strict) subset of B; this is written as ${\displaystyle A\subsetneq B.}$ or equivalently • B is a proper superset of A; this is written as ${\displaystyle B\supsetneq A.}$ For any set S, the inclusion relation ? is a partial order on the set ${\displaystyle {\mathcal {P}}(S)}$ of all subsets of S (the power set of S) defined by ${\displaystyle A\leq B\iff A\subseteq B}$. We may also partially order ${\displaystyle {\mathcal {P}}(S)}$ by reverse set inclusion by defining ${\displaystyle A\leq B\iff B\subseteq A.}$ When quantified, A ? B is represented as: ?x{x?A -> x?B}.[1] ## Properties • A set A is a subset of B if and only if their intersection is equal to A. Formally: ${\displaystyle A\subseteq B\Leftrightarrow A\cap B=A.}$ • A set A is a subset of B if and only if their union is equal to B. Formally: ${\displaystyle A\subseteq B\Leftrightarrow A\cup B=B.}$ • A finite set A is a subset of B if and only if the cardinality of their intersection is equal to the cardinality of A. Formally: ${\displaystyle A\subseteq B\Leftrightarrow \|A\cap B\|=\|A\|.}$ ## ? and ? symbols Euler diagram showing A is a proper subset of B, A?B, and conversely B is a proper superset of A. Some authors use the symbols ? and ? to indicate subset and superset respectively; that is, with the same meaning and instead of the symbols, ? and ?.[2] So for example, for these authors, it is true of every set A that . Other authors prefer to use the symbols ? and ? to indicate proper subset and proper superset respectively; that is, with the same meaning and instead of the symbols, ? and ?.[3] This usage makes ? and ? analogous to the inequality symbols x y then x may or may not equal y, but if , then x definitely does not equal y, and is less than y. Similarly, using the convention that ? is proper subset, if , then A may or may not equal B, but if , then A definitely does not equal B. ## Examples The regular polygons form a subset of the polygons • The set A = {1, 2} is a proper subset of B = {1, 2, 3}, thus both expressions A ? B and A ? B are true. • The set D = {1, 2, 3} is a subset (but not a proper subset) of E = {1, 2, 3}, thus D ? E is true, and D ? E is not true (false). • Any set is a subset of itself, but not a proper subset. (X ? X is true, and X ? X is false for any set X.) • The empty set { }, denoted by ?, is also a subset of any given set X. It is also always a proper subset of any set except itself. • The set {x: x is a prime number greater than 10} is a proper subset of {x: x is an odd number greater than 10} • The set of natural numbers is a proper subset of the set of rational numbers; likewise, the set of points in a line segment is a proper subset of the set of points in a line. These are two examples in which both the subset and the whole set are infinite, and the subset has the same cardinality (the concept that corresponds to size, that is, the number of elements, of a finite set) as the whole; such cases can run counter to one's initial intuition. • The set of rational numbers is a proper subset of the set of real numbers. In this example, both sets are infinite but the latter set has a larger cardinality (or power) than the former set. Another example in an Euler diagram: ## Other properties of inclusion A ? B and B ? C imply A ? C Inclusion is the canonical partial order in the sense that every partially ordered set (X, ${\displaystyle \preceq }$) is isomorphic to some collection of sets ordered by inclusion. The ordinal numbers are a simple example--if each ordinal n is identified with the set [n] of all ordinals less than or equal to n, then a b if and only if [a] ? [b]. For the power set ${\displaystyle {\mathcal {P}}(S)}$ of a set S, the inclusion partial order is (up to an order isomorphism) the Cartesian product of k = \|S\| (the cardinality of S) copies of the partial order on {0,1} for which 0 < 1. This can be illustrated by enumerating S = {s1, s2, ..., sk} and associating with each subset T ? S (which is to say with each element of 2S) the k-tuple from {0,1}k of which the ith coordinate is 1 if and only if si is a member of T. ## References 1. ^ Rosen, Kenneth H. (2012). Discrete Mathematics and Its Applications (7th ed.). New York: McGraw-Hill. p. 119. ISBN 978-0-07-338309-5. 2. ^ Rudin, Walter (1987), Real and complex analysis (3rd ed.), New York: McGraw-Hill, p. 6, ISBN 978-0-07-054234-1, MR 0924157 3. ^ Subsets and Proper Subsets (PDF), retrieved Connect with defaultLogic What We've Done Led Digital Marketing Efforts of Top 500 e-Retailers. Worked with Top Brands at Leading Agencies. Successfully Managed Over \$50 million in Digital Ad Spend. Developed Strategies and Processes that Enabled Brands to Grow During an Economic Downturn.
# NCERT Solutions Ch- 4 Quadratic Equations ## Ex 4.3 #### Question 2. Find the roots of the quadratic equations by applying the quadratic formula. (i) 2x2 – 7x + 3 = 0 (ii) 2x2 – x + 4 = 0 (iii) 4x2 – 4√3x + 3 = 0 (iv) 2x2 – x + 4 = 0 Solution: (i) 2x2 – 7x + 3 = 0 On comparing the given equation with ax2 + bx + c = 0, we get, a = 2, b = -7 and c = 3 ⇒ x = (7±√(49 – 24))/4 ⇒ x = (7±√25)/4 ⇒ x = (7±5)/4 ⇒ x = (7+5)/4 or x = (7-5)/4 ⇒ x = 12/4 or 2/4 ∴ x = 3 or 1/2 (ii) 2x2 + x – 4 = 0 On comparing the given equation with ax2 + bx + c = 0, we get, a = 2, b = 1 and c = -4 ⇒x = -1±√1+32/4 ⇒x = -1±√33/4 ∴ x = -1+√33/4 or x = -1-√33/4 (iii) 4x2 + 4√3x + 3 = 0 On comparing the given equation with ax2 + bx + c = 0, we get a = 4, b = 4√3 and c = 3 ⇒ x = -4√3±√48-48/8 ⇒ x = -4√3±0/8 ∴ x = -√3/2 or x = -√3/2 (iv) 2x2 + x + 4 = 0 On comparing the given equation with ax2 + bx + c = 0, we get, a = 2, b = 1 and c = 4 ⇒ x = -1±√1-32/4 ⇒ x = -1±√-31/4 As we know, the square of a number can never be negative. Therefore, there is no real solution for the given equation. 3. Find the roots of the following equations: (i) x-1/x = 3, x ≠ 0 (ii) 1/x+4 – 1/x-7 = 11/30, x = -4, 7 Solutions: (i) x-1/x = 3 ⇒ x2 – 3x -1 = 0 On comparing the given equation with ax2 + bx + c = 0, we get a = 1, b = -3 and c = -1 ⇒ x = 3±√9+4/2 ⇒ x = 3±√13/2 ∴ x = 3+√13/2 or x = 3-√13/2 (ii) 1/x+4 – 1/x-7 = 11/30 ⇒ x-7-x-4/(x+4)(x-7) = 11/30 ⇒ -11/(x+4)(x-7) = 11/30 ⇒ (x+4)(x-7) = -30 ⇒ x2 – 3x – 28 = 30 ⇒ x2 – 3x + 2 = 0 We can solve this equation by factorization method now, ⇒ x2 – 2x – x + 2 = 0 ⇒ x(x – 2) – 1(x – 2) = 0 ⇒ (x – 2)(x – 1) = 0 ⇒ x = 1 or 2 4. The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is 1/3. Find his present age. Solutions: Let us say, present age of Rahman is x years. Three years ago, Rehman’s age was (x – 3) years. Five years after, his age will be (x + 5) years. Given, the sum of the reciprocals of Rehman’s ages 3 years ago and after 5 years is equal to 1/3. ∴ 1/x-3 + 1/x-5 = 1/3 (x+5+x-3)/(x-3)(x+5) = 1/3 (2x+2)/(x-3)(x+5) = 1/3 ⇒ 3(2x + 2) = (x-3)(x+5) ⇒ 6x + 6 = x2 + 2x – 15 ⇒ x2 – 4x – 21 = 0 ⇒ x2 – 7x + 3x – 21 = 0 ⇒ x(x – 7) + 3(x – 7) = 0 ⇒ (x – 7)(x + 3) = 0 ⇒ x = 7, -3 As we know, age cannot be negative. Therefore, Rahman’s present age is 7 years. 5. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects. Solutions: Let us say, the marks of Shefali in Maths be x. Then, the marks in English will be 30 – x. As per the given question, (x + 2)(30 – x – 3) = 210 (x + 2)(27 – x) = 210 ⇒ -x2 + 25x + 54 = 210 ⇒ x2 – 25x + 156 = 0 ⇒ x2 – 12x – 13x + 156 = 0 ⇒ x(x – 12) -13(x – 12) = 0 ⇒ (x – 12)(x – 13) = 0 ⇒ x = 12, 13 Therefore, if the marks in Maths are 12, then marks in English will be 30 – 12 = 18 and the marks in Maths are 13, then marks in English will be 30 – 13 = 17. 6. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field. Solutions: Let us say, the shorter side of the rectangle be x m. Then, larger side of the rectangle = (x + 30) m As given, the length of the diagonal is = x + 30 m Therefore, ⇒ x2 + (x + 30)2 = (x + 60)2 ⇒ x2 + x2 + 900 + 60x = x2 + 3600 + 120x ⇒ x2 – 60x – 2700 = 0 ⇒ x2 – 90x + 30x – 2700 = 0 ⇒ x(– 90) + 30(x -90) = 0 ⇒ (– 90)(x + 30) = 0 ⇒ = 90, -30 However, side of the field cannot be negative. Therefore, the length of the shorter side will be 90 m. and the length of the larger side will be (90 + 30) m = 120 m. 7. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers. Solution: Let us say, the larger and smaller number be x and y respectively. As per the question given, x– y2 = 180 and y2 = 8x ⇒ x– 8x = 180 ⇒ x– 8x – 180 = 0 ⇒ x– 18x + 10x – 180 = 0 ⇒ x(x – 18) +10(x – 18) = 0 ⇒ (x – 18)(x + 10) = 0 ⇒ x = 18, -10 However, the larger number cannot considered as negative number, as 8 times of the larger number will be negative and hence, the square of the smaller number will be negative which is not possible. Therefore, the larger number will be 18 only. x = 18 ∴ y2 = 8x = 8 × 18 = 144 ⇒ y = ±√144 = ±12 ∴ Smaller number = ±12 Therefore, the numbers are 18 and 12 or 18 and -12. 8. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train. Solution: Let us say, the speed of the train be x km/hr. Time taken to cover 360 km = 360/x hr. As per the question given, ⇒ (x + 5)(360-1/x) = 360 ⇒ 360 – x + 1800-5/x = 360 ⇒ x+ 5x + 10x – 1800 = 0 ⇒ x(x + 45) -40(x + 45) = 0 ⇒ (x + 45)(x – 40) = 0 ⇒ x = 40, -45 As we know, the value of speed cannot be negative. Therefore, the speed of train is 40 km/h. 9. Two water taps together can fill a tank in 9 3/8 hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank. Solution: Let the time taken by the smaller pipe to fill the tank = x hr. Time taken by the larger pipe = (x – 10) hr Part of tank filled by smaller pipe in 1 hour = 1/x Part of tank filled by larger pipe in 1 hour = 1/(– 10) As given, the tank can be filled in 9 3/8 = 75/8 hours by both the pipes together. Therefore, 1/x + 1/x-10 = 8/75 x-10+x/x(x-10) = 8/75 ⇒ 2x-10/x(x-10) = 8/75 ⇒ 75(2x – 10) = 8x2 – 80x ⇒ 150x – 750 = 8x2 – 80x ⇒ 8x2 – 230x +750 = 0 ⇒ 8x2 – 200x – 30x + 750 = 0 ⇒ 8x(x – 25) -30(x – 25) = 0 ⇒ (x – 25)(8x -30) = 0 ⇒ x = 25, 30/8 Time taken by the smaller pipe cannot be 30/8 = 3.75 hours, as the time taken by the larger pipe will become negative, which is logically not possible. Therefore, time taken individually by the smaller pipe and the larger pipe will be 25 and 25 – 10 =15 hours respectively. 10. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speeds of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains. Solution: Let us say, the average speed of passenger train = x km/h. Average speed of express train = (x + 11) km/h Given, time taken by the express train to cover 132 km is 1 hour less than the passenger train to cover the same distance. Therefore, (132/x) – (132/(x+11)) = 1 132(x+11-x)/(x(x+11)) = 1 132 × 11 /(x(x+11)) = 1 ⇒ 132 × 11 = x(x + 11) ⇒ x2 + 11x – 1452 = 0 ⇒ x2 + 44x -33x -1452 = 0 ⇒ x(x + 44) -33(x + 44) = 0 ⇒ (x + 44)(x – 33) = 0 ⇒ x = – 44, 33 As we know, Speed cannot be negative. Therefore, the speed of the passenger train will be 33 km/h and thus, the speed of the express train will be 33 + 11 = 44 km/h. 11. Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares. Solutions: Let the sides of the two squares be x m and y m. Therefore, their perimeter will be 4x and 4y respectively And area of the squares will be x2 and y2 respectively. Given, 4x – 4y = 24 x – y = 6 x = y + 6 Also, x+ y2 = 468 ⇒ (6 + y2) + y2 = 468 ⇒ 36 + y2 + 12y + y2 = 468 ⇒ 2y2 + 12y + 432 = 0 ⇒ y2 + 6y – 216 = 0 ⇒ y2 + 18y – 12y – 216 = 0 ⇒ y(+18) -12(y + 18) = 0 ⇒ (y + 18)(y – 12) = 0 ⇒ y = -18, 12 As we know, the side of a square cannot be negative. Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m. ## Ex 4.4 1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them; (i) 2x2 – 3x + 5 = 0 (ii) 3x2 – 4√3x + 4 = 0 (iii) 2x2 – 6x + 3 = 0 Solutions: (i) Given, 2x2 – 3x + 5 = 0 Comparing the equation with ax2 + bx c = 0, we get a = 2, b = -3 and c = 5 We know, Discriminant = b2 – 4ac ( – 3)2 – 4 (2) (5) = 9 – 40 = – 31 As you can see, b2 – 4ac < 0 Therefore, no real root is possible for the given equation, 2x2 – 3x + 5 = 0. (ii) 3x2 – 4√3x + 4 = 0 Comparing the equation with ax2 + bx c = 0, we get a = 3, b = -4√3 and c = 4 We know, Discriminant = b2 – 4ac = (-4√3)– 4(3)(4) = 48 – 48 = 0 As b2 – 4ac = 0, Real roots exist for the given equation and they are equal to each other. Hence the roots will be –b/2a and –b/2a. b/2= -(-4√3)/2×3 = 4√3/6 = 2√3/3 = 2/√3 Therefore, the roots are 2/√3 and 2/√3. (iii) 2x2 – 6x + 3 = 0 Comparing the equation with ax2 + bx c = 0, we get a = 2, b = -6, c = 3 As we know, Discriminant = b2 – 4ac = (-6)2 – 4 (2) (3) = 36 – 24 = 12 As b2 – 4ac > 0, Therefore, there are distinct real roots exist for this equation, 2x2 – 6x + 3 = 0. =( -(-6) ± √(-62-4(2)(3)) )/ 2(2) = (6±2√3 )/4 = (3±√3)/2 Therefore the roots for the given equation are (3+√3)/2 and (3-√3)/2 2. Find the values of k for each of the following quadratic equations, so that they have two equal roots. (i) 2x2 + kx + 3 = 0 (ii) kx (x – 2) + 6 = 0 Solutions: (i) 2x2 + kx + 3 = 0 Comparing the given equation with ax2 + bx c = 0, we get, a = 2, b = k and c = 3 As we know, Discriminant = b2 – 4ac = (k)2 – 4(2) (3) k2 – 24 For equal roots, we know, Discriminant = 0 k2 – 24 = 0 k2 = 24 k = ±√24 = ±2√6 (ii) kx(x – 2) + 6 = 0 or kx2 – 2kx + 6 = 0 Comparing the given equation with ax2 + bx c = 0, we get a = kb = – 2k and c = 6 We know, Discriminant = b2 – 4ac = ( – 2k)2 – 4 (k) (6) = 4k2 – 24k For equal roots, we know, b2 – 4ac = 0 4k2 – 24k = 0 4k (k – 6) = 0 Either 4k = 0 or k = 6 = 0 k = 0 or k = 6 However, if k = 0, then the equation will not have the terms ‘x2‘ and ‘x‘. Therefore, if this equation has two equal roots, k should be 6 only. 3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find its length and breadth. Solutions: Let the breadth of mango grove be l. Length of mango grove will be 2l. Area of mango grove = (2l) (l)= 2l2 2l= 800 l= 800/2 = 400 l– 400 =0 Comparing the given equation with ax2 + bx c = 0, we get a = 1, b = 0, c = 400 As we know, Discriminant = b2 – 4ac => (0)2 – 4 × (1) × ( – 400) = 1600 Here, b2 – 4ac > 0 Thus, the equation will have real roots. And hence, the desired rectangular mango grove can be designed. = ±20 As we know, the value of length cannot be negative. Therefore, breadth of mango grove = 20 m Length of mango grove = 2 × 20 = 40 m 4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48. Solution: Let’s say, the age of one friend be x years. Then, the age of the other friend will be (20 – x) years. Four years ago, Age of First friend = (x – 4) years Age of Second friend = (20 – x – 4) = (16 – x) years As per the given question, we can write, (x – 4) (16 – x) = 48 16x – x2 – 64 + 4x = 48 – x2 + 20x – 112 = 0 x2 – 20x + 112 = 0 Comparing the equation with ax2 + bx c = 0, we get a = 1b = -20 and c = 112 Discriminant = b2 – 4ac => (-20)2 – 4 × 112 => 400 – 448 = -48 b2 – 4ac < 0 Therefore, there will be no real solution possible for the equations. Hence, condition doesn’t exist. 5. Is it possible to design a rectangular park of perimeter 80 and area 400 m2? If so find its length and breadth. Solution: Let the length and breadth of the park be and b. Perimeter of the rectangular park = 2 (l + b) = 80 So, l + b = 40 Or, b = 40 – l Area of the rectangular park = l×b = l(40 – l) = 40– l= 400 l2 40l + 400 = 0, which is a quadratic equation. Comparing the equation with ax2 + bx c = 0, we get a = 1, b = -40, c = 400 Since, Discriminant = b2 – 4ac =>(-40)2 – 4 × 400 => 1600 – 1600 = 0 Thus, b2 – 4ac = 0 Therefore, this equation has equal real roots. Hence, the situation is possible. Root of the equation, l = –b/2a l = (40)/2(1) = 40/2 = 20 Therefore, length of rectangular park, = 20 m And breadth of the park, = 40 – = 40 – 20 = 20 m.
# Lesson 1Under ConstructionDevelop Understanding ## Jump Start Examine the diagram. (Note: points and are at the center of their respective circles.) ### 1. Write as many equality statements as you can to represent equal lengths in the diagram. ### 2. Write as many congruence statements as you can to represent congruent segments in the diagram. ### 3. Be prepared to state how you know which segments are congruent. Are there any segments you are wondering about, but aren’t sure if they are really congruent? ## Learning Focus Construct a rhombus, a perpendicular bisector, and a square using only a compass and a straightedge (unmarked ruler) as tools. How do I use geometric objects, such as circles and lines, to construct geometric figures like rhombuses and squares, rather than using measurement tools, such as rulers and protractors, to draw such figures? ## Open Up the Math: Launch, Explore, Discuss In ancient times, one of the only tools builders and surveyors had for laying out a plot of land or the foundation of a building was a piece of rope. There are two geometric figures you can create with a piece of rope: you can pull it tight to create a line segment, or you can fix one end, and—while extending the rope to its full length—trace out a circle with the other end. Geometric constructions have traditionally mimicked these two processes using an unmarked straightedge to create a line segment and a compass to trace out a circle (or sometimes a portion of a circle called an arc). Using only these two tools, you can construct all kinds of geometric shapes. Suppose you want to construct a rhombus using only a compass and straightedge. You might begin by drawing a line segment to define the length of a side, and drawing another ray from one of the endpoints of the line segment to define an angle, as in the sketch. Now the hard work begins. We can’t just keep drawing line segments because we have to be sure that all four sides of the rhombus are the same length. We have to stop drawing and start constructing. Construct a Rhombus Knowing what you know about circles and line segments, how might you locate point on the ray in the diagram given, so the distance from to is the same as the distance from to ? ### 1. Describe how you will locate point and how you know , then construct point on the diagram given. Now that we have three of the four vertices of the rhombus, we need to locate point , the fourth vertex. ### 2. Describe how you will locate point and how you will know , then construct point on the diagram. Pause and Reflect Construct a Perpendicular Bisector of a Segment and a Square (a rhombus with right angles) The only difference between constructing a rhombus and constructing a square is that a square contains right angles. Therefore, we need a way to construct perpendicular lines using only a compass and a straightedge. We will begin by inventing a way to construct a perpendicular bisector of a line segment. ### 3. Given , fold and crease the paper so that point is reflected onto point . Based on the definition of reflection, what do you know about this “crease line”? You have “constructed” a perpendicular bisector of by using a paper-folding strategy. Is there a way to construct this line using a compass and a straightedge? ### 4. Experiment with the compass to see if you can develop a strategy to locate points on the “crease line.” When you have located at least two points on the “crease line,” use the straightedge to finish your construction of the perpendicular bisector. Describe your strategy for locating points on the perpendicular bisector of . Now that you have created a line perpendicular to , we will use the right angle formed to construct a square. ### 5. Label the midpoint of on the diagram as point . Using as one side of the square, and the right angle formed by and the perpendicular line drawn through point as the beginning of a square, finish constructing this square on the diagram. (Hint: Remember that a square is also a rhombus, and you have already constructed a rhombus in the first part of this task.) Draw a line and select two arbitrary points on the line. Treating the line segment between the two points you selected as one side of a square, use the construction strategies you invented in today’s task to construct the square. Demonstrate your thinking by showing all of the circles, or portions of circles, you use in your construction. ## Takeaways I used circles and lines as a construction tool today. • Circles are useful construction tools because • Congruent circles are useful construction tools because • The congruent circles in the construction of the perpendicular bisector helped me to notice that ## Lesson Summary In this lesson, we learned about constructions: creating geometric figures precisely, using only a compass and a straightedge. Using only these tools, we constructed a rhombus with a given side and angle, constructed the perpendicular bisector of a side, and constructed a square with a given right angle and line segment for a side. We learned the value of the definition of a circle—the set of all points in a plane equidistant from a fixed center point—since circles allow us to construct congruent line segments. ## Retrieval ### 1. Figure is a rhombus. 1. Use a straightedge to draw the two diagonals. 2. Using a compass, construct a circle with center at point and a radius of length . Then construct a circle at point with radius length . 3. What do you notice about the intersections of the two circles? ### 2. Solve the system of equations. Choose an appropriate method: graphing, substitution, or elimination.
# Difference between revisions of "2013 AMC 12B Problems/Problem 14" The following problem is from both the 2013 AMC 12B #14 and 2013 AMC 10B #21, so both problems redirect to this page. ## Problem Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is $N$. What is the smallest possible value of $N$ ? $\textbf{(A)}\ 55 \qquad \textbf{(B)}\ 89 \qquad \textbf{(C)}\ 104 \qquad \textbf{(D)}\ 144 \qquad \textbf{(E)}\ 273$ ## Solution 1 Let the first two terms of the first sequence be $x_{1}$ and $x_{2}$ and the first two of the second sequence be $y_{1}$ and $y_{2}$. Computing the seventh term, we see that $5x_{1} + 8x_{2} = 5y_{1} + 8y_{2}$. Note that this means that $x_{1}$ and $y_{1}$ must have the same value modulo $8$. To minimize, let one of them be $0$; WLOG assume that $x_{1} = 0$. Thus, the smallest possible value of $y_{1}$ is $8$; and since the sequences are non-decreasing we get $y_{2} \ge 8$. To minimize, let $y_{2} = 8$. Thus, $5y_{1} + 8y_{2} = 40 + 64 = \boxed{\textbf{(C) }104}$. ## Solution 2 WLOG, let $a_i$, $b_i$ be the sequences with $a_1. Then $$N=5a_1+8a_2=5b_1+8b_2$$ or $$5a_1+8a_2=5(a_1+c)+8(a_2-d)$$ for some natural numbers $c$, $d$. Thus $5c=8d$. To minimize $c$ and $d$, we have $(c,d)=(8,5)$, or $$5a_1+8a_2=5(a_1+8)+8(a_2-5).$$ To minimize $a_1$ and $b_1$, we have $(a_1,b_1)=(0,0+c)=(0,8)$. Using the same method, since $b_2\ge b_1$, we have $b_2\ge8$. Thus the minimum $N=5b_1+8b_2=104\boxed{\mathrm{(C)}}$ ~ Nafer ~IceMatrix ## See also 2013 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 13 Followed byProblem 15 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions 2013 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 20 Followed byProblem 22 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. Invalid username Login to AoPS
# A* grade maths GCSE borders coursework investigation ## Introduction Borders Coursework The diagrams below show a square that has been surrounded by more squares to make a cross shape. The cross shape has then been surrounded by more squares to make an even bigger cross shape. The first diagram consists of 1 square, the second diagram consists of 5 squares while the third diagram has 13 squares in total. The next cross shape is always made by surrounding the previous cross shape with a certain number of squares. I am going to investigate, using algebra, how many squares would be needed to make any cross shape built up in this way. Once I have found the formula for the first part of my investigation, I will extend my investigation to three dimensions and find out the formula for the number of squares added each time. When I find out what the formulas are in each case, I will check them and make sure they work. ## Part 1 I am going to draw some diagrams to help me find out how many squares are in each diagram. I will then make a table so I can see clearly see the number of squares in each diagram. I have surrounded each diagram with a different colour of squares because it makes it clearer which layer is the new one each time. The sequence is a quadratic sequence because the second difference is constant. I am going to find out the formula using a quadratic equation. In general, a quadratic equation has the form: y = ax2 + bx + c where x is the number. I am now going to work out what the formula is using the quadratic equation. 1) 1 = a + b + c - this is 12 = 1, x a which equals a, + 1 x b which equals b + c 2) 5= 4a + 2b + c - this is 22 = 4, x a which equals 4a, + 2 x b which equals 2b + c 3) 13 = 9a + 3b + c - this is 32 = 9, x a which equals 9a, + 3 x b which equals 3b + c I am going to solve these equations by means of using simultaneous equations. I only need to use 3 equations because there are only 3 unknowns: a, b and c. If 2a = 4, a = 2 So a = 2, using 5a + b = 8 - 5 x 2 = 10, + b = 8, this means b = -2 Then using a + b + c = 1 2 – 2 + c = 1 so c = 1 2 – 2 + 1 In general if the quadratic equation has the form y = ax2 + bx + c Then the formula to find out the numbers in the sequence is y=2x2 – 2x + 1 The formula for the 2 dimensions is 2x2 – 2x + 1 This can also be written as x2(x-1)2 To check the formula I will work out the 1st and 2nd terms in the sequence. Both these result are right so my formula is correct. I am also going to find out what the 7th and 8th terms of the sequence are and see if they fit in with the number pattern of the sequence. ## Three dimensions Three dimensions I am now going to extend my investigation into three dimensions. So diagram 1 will consist of 1 cube, diagram 2 will consist of 2 x diagram 1 ( above and below) and 5 cubes in the middle while diagram 3 will consist of 2 x diagram 2, 2 x diagram 1 and 13 cubes. The diagrams below show three dimensions, There is a common difference, the first and second differences are not the same but the 3rd difference is. I think the formula for the sequence contains x3 because I am making it 3D. As before the number in front of x3 is half the constant difference, so half of 8 is 4. Therefore the formula contains 4x3 Because the equation to find out the 2D sequence was y = ax2 + bx + c I think the equation to find out the 3D sequence will be cubic because it is 3D. The general equation to find out the formula is y= ax3 + bx2 + cx + d Unlike in the first part of my investigation where I only had to use 3 equations I now have to use 4 equations to work out the formula because there are 4 unknowns: a, b, c and d. Using the table – 1) a + b + c + d = 1 - this is 13=1, x a which = a, + 12 = 1, x b = b + 1 x c = c + d 2) 8a + 4b + 2c + d = 7 - this is 23=8, x a which = 8a, + 22 = 4, x b = 4b + 2 x c = 2c + d 3) 27a + 9b + 3c + d = 25 – this is 33=27, x a which = 27a, + 32 = 9, x b = 9b + 3 x c =3c + d 4) 64 a + 16b + 4c + d = 63 – this is 43=64, x a which = 64a, + 42 = 16, x b = 16b = 4 x c = 4c + d Like the equation in the first part of my investigation, I am going to solve these by using simultaneous equations. I have now found both, the formula for 2 dimensions and the formula for 3 dimensions. Two dimensions was relatively easy as the second difference was constant. In the 3D sequence it wasn’t until the third difference until it was constant. This gave me the idea, as in the first part of my investigation I worked out that the formula contained x2 because the 2nd difference was constant, I thought that the formula for 3D will have x3 in it because the 3rd difference is constant. The number in front of x2 in the first part of my investigation was half the constant difference which was 4 so I thought the number in front of x3 will also be half the constant difference which was 4. I also noticed, that the result of simultaneous equations a, b and c matched the pattern of the second difference of the number pattern. 5 24 0 3 2 2 14
# x Intercept Definition ## Online Tutoring Is The Easiest, Most Cost-Effective Way For Students To Get The Help They Need Whenever They Need It. The point on a graph where the curve of a function crosses the x – axis is known as its x – intercept. The value of y – coordinate is zero at the point where the graph intersects the x – axis. To find the x – intercept of an equation, we put the value of y as zero in the equation and solve the equation to find the corresponding value of x. Example 1: Find the x – intercept for the following equation of a line: 3x – 5y = 12. Solution: In order to find the x – intercept we need to put the value of y = 0 in the equation and solve the equation to find the corresponding value of x. 3x – 5 (0) = 12 3x – 0 = 12 3x = 12 x = 4 Therefore the x – intercept for the above equation of line is (4, 0) Example 2: Find the x – intercept for the function y = x2 – x – 6 Solution: For finding the x – intercept of the above function, the value of y is taken as 0. As a result we get an equation with terms having variables x and x2 on one side and zero on the other side. We need to solve this equation to get the corresponding value of x. x2 – x – 6 = 0 x2 – 3x + 2x – 6 = 0 x( x- 3) + 2(x-3) = 0 (x+2) (x-3) = 0 x +2 = 0 or x – 3 = 0 x = - 2 x = 3 Thus the x – intercepts for the above function are (-2, 0) and (3, 0).
If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. ## Algebra II (2018 edition) ### Unit 4: Lesson 14 Introduction to symmetry of functions # Function symmetry introduction Learn what even and odd functions are, and how to recognize them in graphs. #### What you will learn in this lesson A shape has reflective symmetry if it remains unchanged after a reflection across a line. For example, the pentagon above has reflective symmetry. Notice how line l is a line of symmetry, and that the shape is a mirror image of itself across this line. This idea of reflective symmetry can be applied to the shapes of graphs. Let's take a look. ## Even functions A function is said to be an even function if its graph is symmetric with respect to the y-axis. For example, the function f graphed below is an even function. Verify this for yourself by dragging the point on the x-axis from right to left. Notice that the graph remains unchanged after a reflection across the y-axis! 1) Which of the graphs represent even functions? ### An algebraic definition Algebraically, a function f is even if f, left parenthesis, minus, x, right parenthesis, equals, f, left parenthesis, x, right parenthesis for all possible x values. For example, for the even function below, notice how the y-axis symmetry ensures that f, left parenthesis, x, right parenthesis, equals, f, left parenthesis, minus, x, right parenthesis for all x. ## Odd functions A function is said to be an odd function if its graph is symmetric with respect to the origin. Visually, this means that you can rotate the figure 180, degrees about the origin, and it remains unchanged. Another way to visualize origin symmetry is to imagine a reflection about the x-axis, followed by a reflection across the y-axis. If this leaves the graph of the function unchanged, the graph is symmetric with respect to the origin. For example, the function g graphed below is an odd function. Verify this for yourself by dragging the point on the y-axis from top to bottom (to reflect the function over the x-axis), and the point on the x-axis from right to left (to reflect the function over the y-axis). Notice that this is the original function! Which of the graphs represent odd functions? ### An algebraic definition Algebraically, a function f is odd if f, left parenthesis, minus, x, right parenthesis, equals, minus, f, left parenthesis, x, right parenthesis for all possible x values. For example, for the odd function below, notice how the function's symmetry ensures that f, left parenthesis, minus, x, right parenthesis is always the opposite of f, left parenthesis, x, right parenthesis. ## Reflection question Can a function be neither even nor odd? ## Want to join the conversation? • What is the use of describing a function as "even" or "odd"? • Even and odd functions have properties that can be useful in different contexts. The most basic one is that for an even function, if you know f(x), you know f(-x). Similarly for odd functions, if you know g(x), you know -g(x). Put more plainly, the functions have a symmetry that allows you to find any negative value if you know the positive value, or vice versa. • Can an equation be both even and odd? • The only function which is both even and odd is f(x) = 0, defined for all real numbers. This is just a line which sits on the x-axis. If you count equations which are not a function in terms of y, then x=0 would also be both even and odd, and is just a line on the y-axis. • How can you prove definitively that a function is even or odd (or neither) just by its equation? Is there even a way? • Mona's explanation works very well for polynomials. Two things to keep in mind: 1) Odd functions cannot have a constant term because then the symmetry wouldn't be based on the origin. 2) Functions that are not polynomials or that don't have exponents can still be even or odd. For example, f(x)=cos(x) is an even function. • How can a function be neither even or odd? • Even and odd describe 2 types of symmetry that a function might exhibit. 1) Functions do not have to be symmetrical. So, they would not be even or odd. 2) If a function is even, it has symmetry around the y-axis. What is a function has symmetry around y=5? It would not be even, because the symmetry is not around the Y-axis. 3) Similarly, odd functions have symmetry around the origin. Functions might have symmetry based on some point other than the origin. So, they would not be odd. Hope this helps. • Let's say the parent function y=x^2 gets translated to the left by 4. So now the equation is y=(x+4)^2. Is it still an even function? It is confusing because now the graph is not symmetric over the y-axis. So does this mean it is an odd function now? Or is it neither? • Even function are strictly symmetrical about the y axis, so it's neither. • How about symmetry with respect to x-axis only? Is it a thing? Why did we define an even function to be symmetric with respect to y-axis and not the x one? • Remember the vertical line test? A curve cannot be a function when a vertical line interesects it more than once. And a curve that is symmetrical around the x-axis will always fail the vertical line test (unless that function is f(x) = 0). So, a function can never be symmetrical around the x-axis. Just remember: symmetry around x-axisfunction To answer your second question, "even" and "odd" functions are named for the exponent in this power function: f(x) = xⁿ - if n is an even integer, then f(x) is an "even" function - if n is an odd integer, then f(x) is an "odd" function Hope this helps! • What is the name for a function that is neither even nor odd? • There is no such terminology, it's just that a function that does not exhibit both the symmetry i.e. even or odd. • I know that a function can be neither even or odd. The only way this is possible is if it's a line right? • No, there are other ways that it can happen. You can have a functions that has multiple curves and the curves are not symmetrical according to the rules for even or odd symmetry. • How do you know whether a function is even or odd if the functions consists of sines and cosines? • Use the fact that sine is odd and cosine is even and observe the function's behavior when you plug in -𝑥. Just see if the function satisfies 𝑓(-𝑥) = 𝑓(𝑥) or 𝑓(-𝑥) = -𝑓(𝑥).
Courses Courses for Kids Free study material Offline Centres More Store # How do you find the derivative of $f(x)=\sqrt{\sin x\left( 2x \right)}?$ Last updated date: 13th Jun 2024 Total views: 374.1k Views today: 9.74k Verified 374.1k+ views Hint: As derivative of a function of real variable measures the sensitivity to change of the function value with respect to change in its argument. Derivatives are a fundamental tool of calculus. Use chain rule to find the derivative of $f(x)=\sqrt{\sin \left( 2x \right)}$ Chain Rule:- $fg'(x).g'(x)$ Example: $\sin (5x)$ a competitive function. $f(x)=\sin x\Rightarrow f'(x)=\cos x$ $g(x)=\sin x\Rightarrow g'(x)=5$ So, the derivative will be equal to $\cos \left( 5x \right)5$ Complete step by step solution: You know that, given function is $f(x)=\sqrt{\sin \left( 2x \right)}$ Firstly, let $y=\sqrt{\sin \left( 2x \right)}$ And let $u=\sin \left( 2x \right)$ This mean $y={{u}^{\dfrac{1}{2}}}$ Therefore, $\dfrac{dy}{dx}=\dfrac{dy}{du}.\dfrac{du}{dx}$ $\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2}{{u}^{\dfrac{-1}{2}}}.2\cos \left( 2x \right)$ Which implies, as ${{u}^{\dfrac{-1}{2}}}=-\sqrt{u}$ therefore as you transfer to denominator it gets reprobated into $\sqrt{4}$ $\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2}.\dfrac{2\cos 2x}{\sqrt{u}}$ And $'2'$ of denominator and $2$ of $'2\cos 2x'$ gets canceled. Therefore, simplified equation will be $\dfrac{dy}{dx}=\dfrac{\cos 2x}{\sqrt{u}}$ Replace $u=\sin 2x$ to get, $\Rightarrow \dfrac{dy}{dx}=\dfrac{\cos 2x}{\sqrt{\sin 2x}}$ $\dfrac{d}{dx}\left[ f\left( f(x) \right) \right]=f'(g(x)g'(x))$ A function is composite if you can write it as $f(g(x))$. In other words, it is a function within a function, or a function of function. For example: $\cos \left( {{x}^{2}} \right)$ is composite because if we let $f(x)=\cos (x')$ and $g(x)={{x}^{2}}$ then $\cos \left( {{x}^{2}} \right)=f\left( g\left( x \right) \right)$ $'g'$ is a function within $'f'$, so you call $'g'$ inner function and $'f'$ outer function. On the outer hand, $\cos (x).{{x}^{2}}$ is not a composite function. It is a product of $f(x)=\cos (x)$ and $g(x)={{x}^{2}}$ but neither function is within the other one. Usually, The only way to differentiate a composite function is to recognize that a function is composite and that the chain rule must be applied, You will not be able to differentiate correctly. Note: Apply chain rule in given function. In some cases if you recognize composite functions you may get the inner and outer functions wrong. This will give you a derivative. For example, in the composite function ${{\cos }^{2}}(x)$ the outer function is ${{x}^{2}}$ and the inner function is $\cos (x)$ the outer function is ${{x}^{2}}$ and the inner function is $\cos (x)$ Sometimes you may get confused by this type of question and think $\cos (x)$ is the outer function.
# Relation Between Arithmetic Geometric and Harmonic mean Here you will learn formula for arithmetic geometric and harmonic mean and relation between arithmetic geometric and harmonic mean. Let’s begin – ## Arithmetic Mean Formula If three terms are in A.P. then the middle term is called the A.M. between the other two, so if a, b, c are in A.P., b is A.M. of a & c. So A.M. of a and c = $${a+b}\over 2$$ = b #### n-Arithmetic Means between two numbers : If a, b be any two given numbers & a, $$A_1$$, $$A_2$$……$$A_n$$, b are in AP, then $$A_1$$, $$A_2$$……$$A_n$$ are the ‘n’ A.M’s between a & b then. $$A_1$$ = a + d, $$A_2$$ = a + 2d ,……., $$A_n$$ = a + nd or b – d, where d = $${b-a}\over {n+1}$$ $$\implies$$ $$A_1$$ = a + $${b-a}\over {n+1}$$, $$A_2$$= a + $$2({b-a})\over {n+1}$$,….. Note : Sum of n A.M’s inserted between a & b is equal to n times the single A.M. between a & b. i.e. $${\sum_{r=1}^{n}A_r}$$ = nA where A is the single A.M. between a & b. ## Geometric Mean – Formula for Geometric Mean If a, b, c are in G.P., then b is the G.M. between a & c, $$b^2$$ = ac. So G.M. of a and c = $$\sqrt{ac}$$ = b #### n-Geometric Means between two numbers : If a, b be any two given positive numbers & a, $$G_1$$, $$G_2$$…… $$G_n$$, b are in G.P. Then $$G_1$$, $$G_2$$……$$G_n$$ are the ‘n’ G.M’s between a & b, where b = $$ar^{n+1}$$ => r = $$(b/a)^{1\over {n+1}}$$ $$G_1$$ = a$$(b/a)^{1\over {n+1}}$$, $$G_2$$= a$$(b/a)^{2\over {n+1}}$$,……. $$G_n$$= a$$(b/a)^{n\over {n+1}}$$ Note : The product of n G.Ms between a & b is equal to $$n^{th}$$ power of the single G.M. between a & b i.e. $$\prod_{r=1}^{\infty} G_{r}$$ = $$(G)^n$$ where G is the single G.M. between a & b ## Harmonic Mean – Formula for harmonic mean If a, b, c are in H.P., then b is H.M. between a & c. So H.M. of a and c = $$2ac\over{a+c}$$ = b #### Insertion of ‘n’ HM’s between a and b : a, $$H_1$$, $$H_2$$, $$H_3$$,……,$$H_n$$, b $$\rightarrow$$ H.P $$1\over a$$, $$1\over{H_1}$$, $$1\over{H_2}$$, $$1\over{H_3}$$,……..,$$1\over{H_n}$$, $$1\over b$$ $$\rightarrow$$ A.P. $$1\over b$$ = $$1\over a$$ + (n + 1)D => D = $${{1\over a}-{1\over b}}\over {n+1}$$ $$1\over{H_n}$$ = $$1\over a$$ + n($${{1\over a}-{1\over b}}\over {n+1}$$) ## Relation Between Arithmetic Geometric and Harmonic mean (i) If A, G, H, are respectively A.M., G.M., H.M. between two positive number a & b then (a) $$G^2$$ = AH (A, G, H constitute a GP) (b) $$A \ge G \ge H$$ (c) A = G = H $$\Leftrightarrow$$ a = b (ii) Let $$a_1$$ + $$a_2$$ + $$a_3$$ + ……… + $$a_n$$ be n positive real numbers, then we define their arithmetic mean(A), geometric mean(G) and harmonic mean(H) as A = $${a_1 + a_2 + a_3 + ……… + a_n}\over n$$ G = $$(a_1 + a_2 + a_3 + ……… + a_n)^{1\over n}$$ and H = $$n\over {1\over {a_1}} + {1\over {a_2}} +…..{1\over {a_n}}$$ It can be shown that $$A \ge G \ge H$$. Moreover equality holds at either place if and only if $$a_1$$ = $$a_2$$ =…..= $$a_n$$.
# How do you solve x^2 + 7x - 44 = 0? Apr 8, 2016 $x = - 11 , 4$ #### Explanation: Begin with writing the equation into brackets. $\left(x + a\right) \left(x + b\right) = 0$ You want to find $a$ and $b$ such that $a + b = 7$ and $a \cdot b = - 44$. Through a process of trial and error, this gives $a = - 4$ and $b = 11$. $\left(x - 4\right) \left(x + 11\right) = 0$ Anything multiplied by $0$ is $0$, meaning at least one of the brackets has to be equal to $0$. This actually gives two answers for $x$. Set each one equal to zero and then solve. $x - 4 = 0$ $x = 4$ or $x + 11 = 0$ $x = - 11$
# File ```TEKS Focus:  (6)(B) Prove two triangles are congruent by applying the Side-Angle-Side, AngleSide-Angle, Side-Side-Side, AngleAngle-Side, and Hypotenuse-Leg congruence conditions.  (1)(G) Display, explain, or justify mathematical ideas and arguments using precise mathematical language in written or oral communication.  Geometric figures are congruent if they are the same size and shape. Corresponding angles and corresponding sides are in the same position in polygons with an equal number of sides. Two polygons are congruent polygons if and only if their corresponding sides are congruent. Thus triangles that are the same size and shape are congruent. CPCTC is an abbreviation for the phrase “Corresponding Parts of Congruent Triangles are Congruent.” It can be used as a justification in a proof after you have proven two triangles congruent. To name a polygon, write the vertices in consecutive order. For example, you can name polygon PQRS as QRSP or SRQP, but not as PRQS. In a congruence statement, the order of the vertices indicates the corresponding parts. P Q S R Some hikers come to a river in the woods. They want to cross the river but decide to find out how wide it is first. So they set up congruent right triangles. The figure shows the river and the triangles. Find the width of the river, GH. Explain. 5 meters, by using CPCTC Example: 2 Use the diagram to prove the following. STATEMENT REASON 1. BA  DA 1. Given 2. Vertical Angle Theorem 3. CA  EA 3. Given 4. BAC  DAE 4. SAS 5. C E 5. CPCTC Example 3: STATEMENT REASON 1. AB  AC 1. Given 2. M is the midpoint of BC 2. Given 2. BM  MC 2. Definition of Midpoint 3. AM  AM 3. Reflexive Prop. of Congruence 4. AMB  AMC 4. SSS 5. AMB  AMC 5. CPCTC Example 4: STATEMENT REASON 1. KN and LM bisect each other 1. Given 2. KH  HN and LH  HM 2. Definition of Segment Bisector 3. KHL   NHM 3. Vertical Angle Theorem 4. KHL  NHM 4. SAS 5. KLH  NMH 5. CPCTC 6. KL \|\| MN 6. Converse of Alternate Interior Angles Theorem ```
# M4ML – Linear Algebra – 2.2 – Part 2: Cosine & Dot Product [MUSIC] Let’s take the cosine rule from algebra, which you’ll remember probably vaguely from school. And that said if we had a triangle with sides a, b, and c, then what the cosine rule said, Was that c squared= a squared + b squared- 2ab times the cos of the angle between a and b, cos that angle theta there. Now, we can translate that into our vector notation. If we call this vector r here, and we call this vector s here. Then this vector will be minus s plus r, so that vector would be r-s, minus s + r. So we could say that c squared was the modulus of r-s squared. And that would be equal to the modulus, the size, of r squared plus the size of s squared -2 mod r mod s cos theta. Now, here’s the cool bit. We can multiply this out using our dot product, because we know that the size of (r- s) squared is equal to (r- s) dotted with itself. Now, that’s just that. And we can multiply that out, and then we’ll compare it to this right hand side here. So (r- s).(r- s), well that’s going to be, if, we need to figure out how to multiply that out, that’s going to be equal to r.r, and then take the next one, -s.r -s.r again. If you take that -s and that r, -s.r, again, and then -s.-s. So that is, we’ve got the modulus of r squared here and we dot r with itself, -2s.r and then -s.-s. Well, that’s going to be the size of -s squared, which is just the size of s squared. And then we can compare that to the right-hand side. And when we do that comparison, compare that to the right-hand side, the -r squareds are going to cancel, the r squareds, even. The s squareds are going to cancel, and so we get a result which is that -2s.r=-2 modulus of r modulus of s cos theta. And then we could lose the minus sign, right? Minus signs will cancel out. Just multiply through by minus one and then the 2s we can cancel out again. So we can say that the dot product r.s, just to put it in a more familiar form, =mod r mod s cos theta. So what we found here is that the dot product really does something quite profound. It takes the size of the two vectors, if these were both unit length vectors, those would be one, and multiplies by cos of the angle between them. It tells us something about the extent to which the two vectors go in the same direction. because if theta was zero, then cos theta would be one, and r.s would just be the size of the two vectors multiplied together. If the two vectors, on the other hand, were at 90 degrees to each other, if they were r was like this and s was like this, and the angle between them, theta, was equal to 90 degrees, cos theta, cos 90, is 0. And then r.s is going to be, we could immediately see, r.s is going to be some size of r, some size of s times 0. So if the two vectors are pointing at 90 degrees to each other, if they’re what’s called orthogonal to each other, then the dot product’s going to give me zero. If they’re both pointing in the same direction, say s was like that, and the angle between them is nought, cos 0=1. And then r.s, is equal to mod r times mod s, just the multiplication of the two sizes together. Fun one, last fun one here, is that r and s are in the opposite directions. So let’s say s was now going this way and the angle between them was 180 degrees. Cos 180 degrees=-1, so then r.s would be equal to minus the size of r times the size of s. And so what the dot product here really does with this cos, it tells us when we get the minus sign out, that they’re going in opposite directions. So there’s some property here in the dot product we’ve derived by looking at the cosine rule that we’ve derived here. When the dot product’s 0, they’re at 90 degrees to each other, orthogonal. When they’re going the same way, we get a positive answer. When they’re going more or less in opposite directions, we get a negative answer for the dot product. [MUSIC]
# What is the slope of any line perpendicular to the line passing through (-7,3) and (-14,14)? May 13, 2016 7/11 #### Explanation: The slope of any line perpendicular to another is the inverse of the slope of the reference line. The general line equation is y = mx + b, so the set of lines perpendicular to this will be y = -(1/m)x + c. y = mx + b Calculate the slope, m, from the given point values, solve for b by using one of the point values, and check your solution using the other point values. A line can be thought of as the ratio of the change between horizontal (x) and vertical (y) positions. Thus, for any two points defined by Cartesian (planar) coordinates such as those given in this problem, you simply set up the two changes (differences) and then make the ratio to obtain the slope, m. Vertical difference “y” = y2 – y1 = 14 – 3 = 11 Horizontal difference “x” = x2 – x1 = -14 – -7 = -7 Ratio = “rise over run”, or vertical over horizontal = 11/-7 = -11/7 for the slope, m. A line has the general form of y = mx + b, or vertical position is the product of the slope and horizontal position, x, plus the point where the line crosses (intercepts) the x-axis (the line where z is always zero.) So, once you have calculated the slope you can put any of the two points known into the equation, leaving us with only the intercept 'b' unknown. 3 = (-11/7)(-7) + b ; 3 = 11 + b ; -8 = b Thus the final equation is y = -(11/7)x - 8 We then check this by substituting the other known point into the equation: 14 = (-11/7)(-14) - 8 ; 14 = 22 - 8 ; 14 = 14 CORRECT! SO, if our original equation is y = -(11/7)x – 8, the set of lines perpendicular to it will have a slope of 7/11.
PDF search ## Pre-Algebra Math 1. ### Math 2. #### Integers Adding Integers (A) Answers Use an integer strategy to find each answer (-11) + (-5) = 12 + 2 = 10 + (-13) = = (-16) = 14 = (-3) (-8) + (-5) = PDF ## [PDF] Rules for Integers Adding Integers Rule: If the signs are the same, add and keep the same sign (+) + (+) = add the numbers and the answer is positive PDF ## [PDF] Rules for Adding Integers - Hewlett-Woodmere Rules for Adding Integers: ❖ If the integers have the same sign you ADD the numbers (based on absolute value) together and keep the sign PDF ## [PDF] 521 Adding Integers - OpenTextBookStore See if you can find a rule for adding numbers without using the number line Notice that when you add a positive integer and a negative integer, you move PDF ## [PDF] Adding and Subtracting Integers Adding and Subtracting Integers Find each sum 1) 3 + (−7) 2) (−10) + 12 3) (−5) + 8 4) (−4) + 6 5) (−8) + 12 6) 6 + (−11) 7) (−7) + 2 PDF ## [PDF] Adding Integers with the Same Sign Adding Integers with the Same Sign Practice and Problem Solving: A/B Find each sum White counters are positive Black counters are negative PDF ## [PDF] Understanding Integer Addition and Subtraction Concepts Using students explain and interpret rules for adding, subtracting, multiplying and dividing with negative numbers” (p 46) The set of integers is a subset of the PDF ## [PDF] Adding and Subtracting Integers - Nasco Students will learn how to add and subtract integers Materials: Two-Color Counters Desk-topper Number Strips Steps: 1 Place students into small groups PDF ## [PDF] Addition and Subtraction of Integers - Alamo Colleges Addition and Subtraction of Integers Integers are the negative numbers, zero, and positive numbers Addition of integers An integer can be represented or PDF ## [PDF] Inquiry Based Lesson – Adding Integers using Integer Chips Finally, students will be asked to generalize how to add any two integers 3 Key Words: Integers, modeling, zero pairs, addition, integer chips 4 Background PDF When you add two opposites, their sum is zero So, opposites are called additive inverses when you're adding them together Therefore, the sum of an integer and ppt ## [PPT] Add and Subtract Integers You already learned that subtracting a negative integer gives the same solution as adding its opposite Vocabulary Define integers: a positive or negative ppt Adding Subtracting Integers Adding Integers Same Sign Add and Keep; Different Signs Subtract; Keep the sign of the larger number; Then you'll be exact!! pptx ## [PPT] 9 + 3 = -6 Let's try a few examples using the rules for adding integers A positive integer paired with a negative integer form a zero pair The value of a zero pair is 0 Add -2 + (-4) = -6 Add 5 + 6 = pptx ## [DOC] ADDING and SUBTRACTING INTEGERS - Bath County Schools ADDING and SUBTRACTING INTEGERS INTEGER – all positive and negative whole numbers ABSOLUTE VALUE – the distance a number is away from zero docx Integer Rules Adding Integers The rule for subtracting integers is when subtracting add the Additive Inverse; Keep the first number, Change the sign to doc ## [PPT] Add and Subtract Integers - This is the teachershenricok12vaus Graph integers on a number line 3 Add and subtract integers When subtracting, change the subtraction to adding the opposite (keep-change-change) and ppt ## [PPT] Subtracting integers 3 + 6 = = 9 We can use these to add integers ─ + © Boardworks Ltd 2004 5 of 42 Adding Integers using Counters Let's ppt ## [PPT] ILLUSTRATING INTEGERS Red tiles will represent negative integers, and yellow tiles will INTEGERS ADDING INTEGERS We can model integer addition with the colored tiles ppt ## [DOC] Math 7 Unit 2: Integers - Understanding By Design Unit Template Illustrate a number line, the results of adding or subtracting negative and positive integers Add two integers using colored integer tiles record symbolically doc When adding integers with different signs Introduction to integers ppt Subtraction of integers ppt Operations on integers ppt Integers Free PDF Document 1 PDF search
### Different Ways to Investigate Conic Sections by Jennifer Roth I would like to discuss why I chose to do this essay before I get started on the topic. I have to admit that I need to learn much more about this topic before I feel comfortable teaching it, so this is really an investigation so I can learn more about the topic. I think it is also an exciting idea for and entire class to have access to the software that we have available to us. There would be so many materials available to use in teaching these topics to our students. The following unit plan includes using the different software that we have available to us. Conic sections are the curves formed when a plane intersects the surface of a right cylindrical cone. The general equation for a Conic Section is . These curves are the circle, ellipse, hyperbola, and the parabola. For a circle A=C; for a parabola either A or C is zero; for an ellipse A and C have the same sign, and A is not equal to C; for an hyberbola A and C have opposite signs. I am going to look at each of these 4 conic sections separately. First, we will review what we know about circles. ### Circles The equation of a circle can be derived from the distance formula. If we have a circle with center (h,k) and radius r as follows by definition, a point (x,y) is on this circle if and only if the distance from (x,y) to (h,k) is r. Thus we have or . The center is the point (h,k) and the circle has radius r. So, how is a circle formed from a plane intersecting a right circular cone. The plane has to be parallel to the xy-plane, and for it to be a circle z is not equal to 0. In this case, the intersection is a point. So if the above equation is expanded, you get in which h, k, and r are constants. Therefore, the general form for the equation of a circle is When the equation of a circle is given in general form, it can be rewritten to find its center and radius by completing the square. Now I will start with the basic equation for a circle like . This is the unit circle, and has many applications is trigonometry. When graphed it looks like the following. So, what do different values for D, E, and F do to the graph of the circle. If E = 1, and F = 0 and we let D = -3, -2, -1, 1, 2, 3, you get the following picture It appears that the circle grows, but maintains the x-intercept values. 2 of the above equations are as follows So, if we set x = 0 in both of these we get And, therefore, our x-intercepts are the same. If we complete the square on the above 2 equations we get the following equations The centers are at (-1/2, -1/2) and (1/2,-1/2) and radius (sqrt(2))/2. We can go through a similar process to see what effect changes to the E and F values have on the graph. For a more geometric approach to circles, Click Here to view a 10-day unit on the geometry of circles. ### Parabolas The graph of any quadratic function is a parabola. Quadratic functions are of the form where a, b, and c are constrants and a is not zero. So all parabola's are similar in shape to the graph of (As seen in the following picture). Here, the vertex is the point (0,0) and the axis of symetry is the y-axis. By completing the square, the equation of the parabola can be rewritten as where the vertex of the parabola is (h,k) and the axis of symetry is the line x=h. Let's look at a few graphs and do some algebra and see what we can figure out about different values for a, b, and c using Algebra Expressor. In the following picture I let b = 0, c = 0, while setting a = -3, -1, -1/2, 1/2, 1, and 3. Therefore, you can see the parabola opens upward if a > 0 and downward if a < 0. There also appears to be another pattern. It looks like if the absolut value of a is greater than 1 then the the curve is wider and if it is smaller it is more narrow. Does this match what we know about parabolas? In the next picture I let a = 1, c = 0, while setting b = 1 The equation of the above graph is and if we complete the square, we get the equation Therefore, our vertex is ( -1/2, -1/4) and the line of symetry is x = -1/2. Is this reinforced by the graph? So, what is b doing to the graph. It looks like setting b = 1 moved the graph of in the negative x direction by 1/2 and the negative y direction by 1/4. What does other values for b do? In the following picture a = 1, c = 0, and b = -3, -1, -1/2, 1/2, 1, and 3. Therefore this is the picture of the graph of the following equations: How is each equation moving the graph of the original equation in the form . What if we were to complete the square for these equations? And, what if we had a different value for a? In the following picture a = -1, c = 0, and b = -3, -1, -1/2, 1/2, 1, and 3. Now, compare this with the previous graph. This is fairly reinforcing. If we just look at the equation in which a = -1, b = -3 and c = 0, and we get the following picture If we complete the square of , we get the equation . So, what changes have the values of a and b made to the graph . The negative appears to have inverted the graph. The gragh also appears to be moved over to the left 3/2 and up 9/4. Where do these two graphs intersect? If we set them equal to each other, they intersect at the points (0,0) and (-3/2, 9/4). Maybe one more will help us make some decions about what is going on. In the following picture a = 3, c = 0, and b = -3, -1, -1/2, 1/2, 1, and 3. If we just look at the equation in which a = 3, b = -1, and c = 0, and we get the following picture. If we complete the square of , we get the equation . So, what changes have the values of a and b made to the graph .The graph appears to be moved over to the right 1/6 and down 1/12. What is different about where these two graphs intersect? The 3 is what is making the graph of more narrow than that of . Now let's look at different values for all a, b, and c for . If we set a = 1, b = 1 and let c = -3, -1, -1/2, 0, 1/2, 1, 3 , we get the following picture. It is fairly easy to see what is going on with the c values. There are also parabolas of the form Such a parabola is expressed in the following picture The equation of this parabola is . Similar investigations can be done with a parabola whose directrix is parallel to the y-axis. The investigations above involved parabolas whose directrix is parallel to the x-axis How can we use a spreadsheet to investigate parabolas. The following is a graph created in Excel Using the following columns In which the B-column contains the formula: A1A1-10A1+25 A more geometric definition of the parabola is the set of all points in the plane equally distant from a fixed line and a fixed point not on the line. The fixed line is called the directrix. The fixed point is called the focus. The following is part of a parabola created using GSP. I started with line AB and a point C on that line. I constructed a perpendicular line k to line AB through C. Then I pick an arbitrary point not on the line which will be the focus, point D. Find the midpoint on the segment CD and construct the perpendicular through this point, F. Where this line intersects k is the point, G, of tangency, and the perpendicular, m, through F is the line tangent to the parabola. Click here to view the animation in GSP. As C moves along AB G traces the parabola. ### The Ellipse The standard form of the equation of the ellipse with center at (h,k) and major axis of length 2a units, where , is as follows: , when the major axis is parallel to the x-axis or , when the major axis is parallel to the y-axis An example of an ellipse graphed using algebra expressor is as follows The equation of this ellipse is . Another example is as follows A more geometric definition of the ellipse is the set of all points in the plane, the sum of whose distances from two fixed points is constant. Each fixed point is called a focus. The following is an ellipse created using GSP. The construcstion is as follows: Construct any point C on the circle. Construct a segment through point C and the center of the circle, A. Construct another point D within the circle. Construct segment CD. Find the midpoint, F, of this segment. Construct the line perpendicular to CD, through F. Find the point of intersection of this line with the line through CA. Trace that point, H, as the point C travels along the circumference of the circle. The following is the picture with the appropriate labels. D and A are the foci, and V1-V4 are the vertices. H is the point that defines the ellipse. Click here to view the GSP file that demonstrates this. As D is moved within the circle, you will get a different ellipse as seen in the following picture. ### The Hyberbola The standard form of the equation of a hyperbola with center at (h,k) and tansverse axis of lenght 2a units, where , is as follows. , when the transverse axis is parallel to the x-axis, or , when the tansverse axis is parallel to the y-axis. An example of a hyberbola graphed using Algebra Expressor is as follows: The equation of this hyberbola is The equation of this hyberbola is . A more geometric definition of a hyperbola is the set of all points in the plane, the difference of whose distances from two fixed points is a positive constant. How can we use the GSP construction above to create a hyperbola. The above hyberbola was constructed by moving the point in the ellipse construction above outside the circle. The following is the picture with the appropriate labels. As D is moved around on the outside of the circle, you will get a different hyberbola. Click here to view the GSP file that demonstrates this. The following is an example of such a hyberbola. And that concludes my investigation of conics. Of course, you can see that there are several different ways you can explore conics using different computer software.
# 10th cbse maths solution for exercise 4.2 part 3 This page 10th cbse maths solution for exercise 4.2 part 3 is going to provide you solution for every problems that you find in the exercise no 4.2 ## 10th CBSE maths solution for Exercise 4.2 part 3 (5) The altitude of a right triangle is 7cm less than its base. If the hypotenuse is 13 cm, find the other two sides. Solution: Let 'x' be the base of the triangle Altitude = x - 7 (Hypotenuse side)² = (Base)² + (Height)² 13² = x² + (x - 7)² 169 = x² + x² - 2 (x) (7) + 7² 169 = x² + x² - 14 x + 49 0 = 2 x²- 14 x + 49 - 169 2 x²- 14 x - 120 = 0 dividing the whole equation by 2 => x²- 7 x - 60 = 0 x²- 12 x + 5 x - 60 = 0 x (x - 12) + 5 (x - 12) = 0 (x + 5) (x - 12) = 0 x + 5 = 0 x - 12 = 0 x = -5 x = 12 Base = 12 cm Height = 12 - 7 = 5 cm (6) A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article(in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs.90,find the number of articles produced and the cost of each article. Solution: Let 'x' be the number of articles produced on that day cost of production of each article = 2 x + 3 Total cost of production = number of article x cost of one article 90 = x (2 x + 3) 90 = 2 x² + 3 x 2 x² + 3 x - 90 = 0 2 x² - 12 x + 15 x - 90 = 0 2 x (x - 6) + 15 (x - 6) = 0 (2 x + 15) (x - 6) = 0 2 x + 15 = 0 x - 6 = 0 2 x = -15 x = 6 x = -15/2 Therefore the number of articles produced on that particular day = 6 Related pages
## Bakuro Binary Number Puzzles Bakuro puzzles are a fun way to learn about binary numbers and how to calculate their values. Bakuro puzzles are like Sudoku puzzles with binary numbers to determine what value to place in empty puzzle squares. It’s a fun way to learn about binary numbers and solve complex puzzles. Bakuro puzzles are Kakuro puzzles that use binary numbers. Paul Curzon at TeachingLondonComputing.org, with support from Google, has created a set of Bakuro puzzles plus blank forms to create your own. This article describes the basics of Bakuro with links at the bottom to explore more, including Kakuro puzzles. ### What are Binary Numbers? If you don’t know, in their simplest form, binary numbers use a 1 or 0 to indicate a number value based on their location in a string of binary numbers. Each position in the binary number string — counting from right to left — has a number value of 1, 2, 4, 8, 16, and so on. The position of each 1 in the string is used to add up the value of the binary number. Here’s an example with a 5-digit binary number: 5 4 3 2 1 <== position of value, counted from right to left 16 8 4 2 1 <== value of each position, as a power of 2 These binary numbers and their values are added up based on the position of the 1 within the 5-digit binary number: 00001 = 1 00011 = 2 + 1 = 3 00111 = 4 + 2 + 1 = 7 01111 = 8 + 4 + 2 + 1 = 15 Do you see how these binary numbers equal one less than the value of the next position in the binary number? 00111 equals 7 while 01000 equals 8. And 00011 equals 3 while 00100 equals 4. Binary numbers are interesting. To convert numbers into binary numbers, start with the nearest power of 2 represented in the binary number. For example, to convert the number 6 into binary, the nearest power of 2 is 4 and leaves you with a remainder of 2 (6 – 4). The remainder of 2 can be converted into binary with a 1 placed in the second position, counting from the right of the binary number, as shown here: 0 0 1 1 0 = 4 + 2 = 6 16 8 4 2 1 <== value of each position, as a power of 2 5 4 3 2 1 <== position of value, counted from right to left ### How to Solve a Basic Bakuro Binary Numbers Puzzle With this basic understanding of how binary numbers are calculated and how they work, you can start to solve basic Bakuro puzzles like this one: In a Bakuro puzzle, empty cells in the grid must be filled only with powers of 2: 1, 2, 4, and 8 in this case with 4-digit binary numbers. The numbers or blocks must equal the number given in the binary number clue at the top and left of the puzzle. In the puzzle above, for example, the first column cells must equal the binary number 1001 which converts to 9. The two 1 digits are in the 8 and 1 position, as explained above. Here’s how to work this simple Bakuro puzzle to a solution. Because you’re smart, you’ll realize the value of each position of a 1 digit tells you what numbers to enter in the cell (but not in what order: you have to look at the clue on the left of each row to learn the correct order). In the simple Bakuro example below, your top left most empty column cell must include an 8 or a 1 because the 1 digits are in the 8 and 1 positions. The 1001 binary number adds up to 8 + 1 or 9. Now look at the far left column and the 0011 and 1100 numbers. The top row binary number converts to 3 (0011) and the bottom row converts to 12 (1100). For 0011, the 1 digits are in the 1 and 2 value positions. For 1100, the 1 digits are in the 4 and 8 value positions. In the top row, we need to figure out how to divide up the left column 1001 binary number, which converts to 8 and 1. The top left row 0011 binary number tells us the row values are either a 1 or a 2 because the 1 digits are in the columns with value of 1 and 2 (1+2=3). We can put a 1 in the top left empty cell with some confidence. The bottom row 1100 binary number tells us that row uses an 8 or 4 because the 1 digits are in the binary number columns with a value of 4 and 8 (8+4=12). Therefore, we can say the top left empty cell should be filled with 1 and the bottom left empty cell should be filled with an 8. This equals 1001 or 9 at the top of the left column. This leaves us with the right empty column to fill. With a 1 in the top left solution cell and an 8 in the bottom left solution cell, we need the top row to equal 3 and bottom row equal 12. The 0011 binary number for the top row has the 1 digits in the 2 and 1 value positions. Therefore, the top right empty cell must be a 2. The top left empty cell already has a 1. The 1100 binary number for the bottom row works the same way. The 1 digits are in the 8 and 4 value positions. The bottom right empty cell, therefore, must be a 4. The bottom left empty cell already has an 8. We can double check this solution by calculating the value of the right column. The binary number is 0110 with the 1 digits in the 4 and 2 value positions. Our right column solution of 2 and 4 equals 6 so we’ve solved this basic puzzle. Congratulations! ### Beyond Basic Bakuro Puzzles Bakuro puzzles can be as addictive as Sudoku puzzles. However, you get far more geek points because Sudoku uses mere numbers. Anyone can play with numbers. Bakuro requires you to know binary numbers and how to calculate them. This simple Bakuro puzzle is from the Teaching London Computing website, a resource for teachers, parents, and students. Here’s another puzzle if you want to print and solve it: All puzzles from Paul Curson at TeachingLondonComputing.org/bakuro with help from Google. Below are links to more Bakuro puzzles you can find online. These are from Computing at Schools, an organization in the United Kingdom that supports computer science teachers. I’ve also linked to Kakuro puzzles if you want to relax and work with mere numbers. ## Author • Tim is an award-winning writer and technologist who enjoys teaching tech to non-technical people. He has many years experience with web sites and applications in business, technical, and creative roles. He and his wife have two kids, now teenagers, who are mad about video games. ### Also In The April 2019 Issue Use SketchUp to create this fascinating mathematical pattern that appears everywhere in nature. Learn about the STEAM star’s amazing journey onto Mythbusters Junior and beyond. What activities are best for teaching STEM to young kids? What’s the best way to choose a classroom lunch? Or the best way to elect a leader? The answer isn’t so simple. How a teapot became the most important object in computer graphics. Bring your coding skills and your desserts to new levels in this simple Python coding activity. Bakuro puzzles are a fun way to learn about binary numbers and how to calculate their values. Learn about the shiny new technology that allows us to be connected like never before. Squares, checkerboards, and hollow boxes… what pattens can you imagine in Python? A fun, DIY electronics project that’ll keep you from bumping around in the dark! It’s time for a throw back to old school programming. Dive into the nuts and bolts of coding instructions! Use your favourite block language to animate this fascinatingly odd game. Can we make a computer using only three simple rules? How science and tech led to an exciting discovery in one of the most dangerous areas of space. How did video games become popular before the internet? It’s all about shareware, floppy disks, and human cleverness! Links from the bottom of all the April 2019 articles, collected in one place for you to print, share, or bookmark. Interesting stories about science and technology for April 2019. Interested but not ready to subscribe? Sign-up for our free monthly email newsletter with curated site content and a new issue email announcement that we send every two months. No, thanks!
## Steps to Turning a Quadratic Equation into Standard Form : Equation : 4x^2 - 2x = x^2 + 5 • To solve this equation we must put it into standard form or ax^2+bx+c = 0. • To do that, we must first move x^2 to the left of the equal sign. Notice that 4x^2 and x^2 both have "^2" or are to the power of 2. Because of this, we can combine both numbers. • Because x^2 is a positive number, we're going to subtract it from 4x^2. Always do the opposite of what the number is. If x^2 was a negative number, we would be adding it. • 4x^2 -x^2 = 3x^2. When subtracting or adding numbers with powers, never subtract or add the powers. The ^2 stays the same. • Now our equation is 3x^2 - 2x = 5. • Since no other number to the left of the equal sign is similar to 5, we are just going to move it over. 5 is positive, so we'll make the 5 we move over negative. Our equation is now 3x^2 - 2x - 5 = 0. ## Steps to Solving/Factoring a Quadratic Formula : • Now we must factor the equation. • First, multiply 3x and -5, and get -15x-. Now, find two numbers that when multiplied equal -15, and when added equal -2. Those two numbers are -5, and 3. • Now, replace "-2x" with "-5x" and "3x". Our equation is now 3x^2 + 3x - 5x - 5 = 0. • Divide the equation in half, and find the greatest common factor or GCF from each half. 3x^2+3x and -5x-5. • The greatest common factor from 3x^2 + 3x is 3x, and the greatest common factor from -5x -5 is -5. • Now, divide the GCF from the first half by 3x. That leaves us with 1. Next, divide the GCF of the second half by -5. That leaves us 1. • I'm going to put the GCF outside of my parenthesis and put "x+1" inside my parenthesis. • 3x( x + 1 ) -5( x + 1) • Now, I'll take the GCFs and put them in their own parenthesis, and put x+1 in another parenthesis. • ( 3x - 5 ) ( x + 1) • In my second parenthesis, I am left with x + 1, or x + 1 = 0. I am first going to subtract 1 (remember to subtract if the number is positive, or add if the number is negative) from 1 and 0. That leaves me with x + 0 = -1, so the answer is x = -1. • In the first parenthesis, I have 3x - 5, or 3x - 5 = 0. First, add 5 to -5 and 0. That leaves me with 3x = 5. Divide 3x and 5 by 3, which leaves us with x = 5/3. • The answer to 4x^2 - 2x = x^2 + 5 is {-1, 5/3} ## Questions Relating to Quadratic Equations : 1. What jobs use quadratic equations? Quadratic equations are obviously used in engineering, and jobs that fall in math and science categories, but are also used in jobs relating to insurance, because many insurance plans are based on these equations, as well as agriculture, because they may need this equation to find areas of field or calculate lengths and widths. 2. What is the first step in solving this equation : x^2 - 4x + 4 ? The first step in solving this equation is to factor, or find two numbers that equal 4 when multiplied, and equal -4 when added. 3. How are quadratic equations used in everyday life? Quadratic equations are used to find speed, figure out profit, find areas, finding velocity, etc. 4. What makes an equation quadratic? An equation is quadratic when the first number and variable are squared, or to the second power. For example, 5x^2. 1. What are the characteristics of a quadratic equation?
× # Expected Area of a Triangle Paradox There have been several recent problems about expected values and unit circles. I came up with another problem like this, but I noticed something interesting. Here's the problem. It is a well known fact that the triangle with vertices on opposite ends of the diameter of a circle and a third point on the circle is a right triangle. In a circle of radius $$1,$$ find the expected area of such a triangle. Seems pretty simple, right? Actually, it has two different answers, depending on how you interpret the problem. Consider the following. It is a well known fact that the triangle with vertices on opposite ends of the diameter of a circle and a third point on the circle is a right triangle. In a circle of radius $$1,$$ a diameter and a random point not on the diameter are randomly picked and connected to form a triangle. Find the expected area of this triangle. And It is a well known fact that the triangle with vertices on opposite ends of the diameter of a circle and a third point on the circle is a right triangle. In a circle of radius $$1,$$ an $$x\text{-value}$$ is randomly selected from the diameter and its corresponding $$y\text{-value}$$ is plotted on the circle. Find the expected area of this triangle. Assume WLOG that the diameters of these circles are on the $$x\text{-axis}$$ These seem like the exact same problem. But there's two different ways to solve them, given the wording of the problem. $$\textbf{Solving the first problem}$$ The points on the diameter are $$(-1,0)$$ and $$(1,0).$$ These are the base of the triangle. We just have to find the average height of the circle. Any point on the unit circle can be represented by $$(\cos\theta,\sin\theta),$$ where $$\theta$$ is the angle formed by drawing the ray starting at the origin and passing through $$(1,0)$$ and the ray starting at the origin and passing through the point. The height of the triangle is going to be $$\sin\theta.$$ The angle will be anywhere in the range $$[0,2\pi).$$ We can just find the average of $$\|\sin\theta\|$$ over the range $$[0,2\pi)$$ to find the expected height. This will be the same as the average value of $$\sin\theta$$ over $$[0,\pi).$$ \begin{align} \dfrac{1}{\pi}\int_0^\pi\sin\theta\text{ }d\theta&=\dfrac{-1}{\pi}\left.\cos\theta\right\|^\pi_0\\ &=\dfrac{-1}{\pi}(\cos\pi-\cos0)\\ &=\dfrac{-1}{\pi}(-1-1)\\ &=\dfrac{2}{\pi} \end{align} The height of this triangle has an average value of $$\frac{2}{\pi}$$ and the base is a constant $$2,$$ so the expected area of the triangle is $$\frac{1}{2}\times2\times\frac{2}{\pi}=\boxed{\frac{2}{\pi}}$$ $$\textbf{Solving the second problem}$$ In this problem, you are randomly picking an $$x\text{-value}$$ in the range $$(-1,1)$$ and plotting it on the circle, which we can again remove the parts of the circle in the third and fourth quadrants because of symmetry. We can model the semicircle by $$y=\sqrt{1-x^2}.$$ The average value of this function is the area of the semicircle divided by $$2,$$ which is $$\frac{\frac{\pi}{2}}{2}=\frac{\pi}{4}.$$ The base of the triangle is $$2$$ and the average height is $$\frac{\pi}{4},$$ so the expected area of the triangle is $$\frac{1}{2}\times2\times\frac{\pi}{4}=\boxed{\frac{\pi}{4}}.$$ Wait, we've gotten two different answers to the same problem. Why is that? Consider the length of the arc we are picking a point from. In the first problem, we are picking any point from an arc of length $$\pi.$$ In the second problem, the arc has length $$2.$$ But these points are still going to result in the same triangles. The variable arc in the first problem $$(A_1)$$ and the variable arc in the second problem $$(A_2)$$ are different; $$\frac{A_1}{A_2}=\frac{\pi}{2}.$$ So we can then expect to see a difference in area. See if you can try to explain the difference mathematically! Note by Trevor B. 2 years, 5 months ago Sort by: The difference in the two cases is that the probability distributions are different. Hence, the expected value is different in each case. In the first case, you are picking the third point with uniform distribution on the circular arc, whereas in second case, the uniform distribution in on the x-axis (diameter). In the first case, the distribution over x-axis would be a bowl shaped distribution, i.e. it would produce higher probability for smaller-area triangles, whereas in second case, as stated earlier, the distribution is uniform over x-axis. Hence, you get lower expected area in first case. · 2 years, 5 months ago In the first case $$\theta$$ is uniformly distributed over [0,$$\pi$$] so by transform of variable distribution function of $$x= \cos \theta$$ will be= $$\frac {1}{\pi \sqrt{1-x^2}}$$ over [-1,1] In the second case x is distributed uniformly over [-1,1] . · 2 years, 5 months ago Can i ask you one thing....how much time did it take to write such a long note with latex · 2 years, 5 months ago This took maybe about $$15$$ minutes. LaTeX isn't that hard if you've used it a lot. Do you have anything to say about the paradox? · 2 years, 5 months ago
# Class 11 NCERT Solutions- Chapter 15 Statistics – Miscellaneous Exercise On Chapter 15 • Last Updated : 03 Mar, 2021 ### Question 1. The mean and variance of eight observations are 9 and 9.25, respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations. Solution: Given, We are provided with six of the observations 6, 7, 10, 12, 12 and 13. Let us assume the missing observations to be a and b. Now, Mean = 9 9 = (6 + 7 + 10 + 12 + 12 + 13 + a + b)/8 But, Solving for a + b, we get, a + b = 12 Also, Variance Equating We have, 9.25 = 1/8(62 + 72 + 102 + 122 + 122 + 132 + a2 + b2) – 92 => 9.25 + 81 = 1/8(36 + 49 + 100 + 144 + 144 + 169 + a2 + b2) => 90.25 * 8 = 642 + a2 + b2 => a2 + b2 = 80 We have, b = 12 – a On substituting the value, a2 + (12 – a)2= 80 => 2a2 – 24a + 64 = 0 On dividing by 2, we get a2 – 12a + 32 = 0 Therefore, a = 4, 8 Now, for a = 4, b = 8 And, for a = 8, b = 4 ### Question 2. The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12, 14. Find the remaining two observations. Solution: Given, We are provided with five of the observations 2, 4, 10, 12, 14. Let us assume the missing observations to be a and b. Now, Mean But, 8 = (2 + 4 + 10 + 12 + 14 + a + b)/(7) Solving for a + b, we get, a + b = 14 Also, Variance = Equating We have, 16 = 1/7(22 + 42 + 102 + 122 + 142 + a2 + b2) – 64 => 16 + 64 = 1/7(4 + 16 + 100 + 144 + 196 + a2 + b2) => 560 = 460 + a2 + b2 => a2+b2 = 100 We have, b = 14 – a On substituting the value, we get a2 + (14 – a)2= 100 => 2a2 – 28a + 96 = 0 On dividing by 2, we get a2 – 14a + 48 = 0 => (a – 8)(a – 6) = 0 Therefore, a = 6, 8 Now, for a = 6, b = 8 And, for a = 8, b = 6 ### Question 3. The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations. Solution: Mean of six observations = 8 Standard deviation of six observations = 4 Let the six observations be x1, x2, x3, x4, x5, x Therefore, Mean of observations, = (x1 + x2 + x3 + x4 + x5 + x6)/6 = 8 If each observation is multiplied by 3 and the resultant observations are yi then, yi = 3xi xi = (1/3)yi, where i = 1….6 So, new mean = (y1 + y2 + y3 + y4 + y5 + y6)/6 = 3(x1 + x2 + x3 + x4 + x5 + x6)/6 = 3 * 8 = 24 Standard Deviation Substituting values, we get, Therefore, Variance of new observation = 1/6 x 864 = 144 Standard Deviation = √144 = 12 ### Question 4. Given that x̅ is the mean and σ2 is the variance of n observations x1, x2, …,xn . Prove that the mean and variance of the observations ax1, ax2, ax3, …., axn are ax̅ and a2σ2, respectively, (a ≠ 0). Solution: Let us assume the observations to be x1, …xn Mean of n observations = Variance of n observations = We know, y Now, Mean of the observations, ax1, ax2, …..axn By substituting values, we get, Variance = ### Question 5. The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases: (i) If wrong item is omitted. (ii) If it is replaced by 12 Solution: (i) On omission of wrong item n = 20 Incorrect mean = 20 Incorrect SD = 2 Now, Incorrect sum of observations = 200 Correct sum of observations = 200 – 8 = 192 Therefore, Correct mean = Correct sum/19 = 192/19 = 10.1 Standard Deviation 4 = 1/20 Incorrect – 100 Incorrect = 2080 Therefore, Correct = Incorrect – (8)2 = 2080 – 64 = 2016 Calculating correct standard deviation, Correct SD = = √(1061.1 – 102.1) = 2.02 (ii) If it is replaced by 12, Incorrect sum of observations, n = 200 Correct sum of observations n = 200 – 8 + 12 n = 204 Correct Mean = Correct Sum / 20 = 204/20 = 10.2 Standard Deviation 4 = 1/20 Incorrect – 100 Incorrect = 2080 Therefore, Correct = Incorrect – (8)2 + (12) = 2080 – 64 + 144 = 2160 Calculating correct standard deviation, = √(108 – 104.04) = 1.98 ### Which of the three subjects shows the highest variability in marks and which shows the lowest? Solution: Given values, Mean of Mathematics = 42 Standard deviation of Mathematics = 12 Mean of Physics = 32 Standard deviation of physics = 15 Mean of Chemistry = 40.9 Standard deviation of chemistry = 20 Now, Coefficient of Variation (C.V) = CV for Mathematics = 12/42 x 100 = 28.57 CV for Physics = 15/32 x 100 = 46.87 CV for Chemistry = 20/40.9 x 100 = 48.89 The Highest Variability of the subject is of Chemistry. ### Question 7. The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted. Solution: Given: n = 100 Incorrect mean, (x̅) = 20 Incorrect standard deviation (σ) = 3 Therefore, On solving, we get Incorrect sum of observations = 2000 Now, Correct sum of observations = 2000 – 21- 21 – 18 = 1940 Correct mean = Correct Sum / 97 = 1940/97 = 20 Also, Standard Deviation Incorrect = 100(9 + 400) Incorrect = 40900 Correct = Incorrect – (21)2 – (21)2 – (18) = 40900 – 441 – 441 – 324 = 40900 – 1206 = 39694 Therefore, Correct S.D = = √(409.216 – 400) = 3.036 My Personal Notes arrow_drop_up
# SICP 1.2 - Procedures and the Processes They Generate, Part 2 Covering SICP Exercises 1.11 through 1.13 ### Exercise 1.11 ✅ Exercise 1.11: A function f is defined by the rule that $f(n)=n$ if $n<3$ and $f(n)=f(n−1)+2f(n−2)+3f(n−3)$ if $n≥3$. Write a procedure that computes f by means of a recursive process. Write a procedure that computes f by means of an iterative process. #### Recursive Process ✅ (define (fr n) (if (< n 3) n (+ (fr (- n 1)) (* 2 (fr (- n 2))) (* 3 (fr (- n 3)))))) #### Iterative Process ✅ ##### Discussion Let's start with (fi 3) (the "i" standing for iterative). For (fi n) where n = 3: (fi (- n 1)) = (fi 2) = 2 (fi (- n 2)) = (fi 1) = 1 (fi (- n 3)) = (fi 0) = 0 To figure out (fi 3), we want sum up (fi 2) (or 2), 2 multiplied by (fi 1) (or 2), and 3 times (fi 0) (or 0). We get a total of 4 from doing this. Suppose we want to figure out (fi 4). To figure out (fi 4), we'll need the values of (fi 3), (fi 2), and (fi 1). When n = 3 for (fi n), those values are: for (fi 3), the sum of (fi 2), 2 times (fi 1), and 3 times (fi 0). This is 4. for (fi 2), the value of (fi (- n 1), which is 2. for (fi 1), the value of (fi (- n 2), which is 1. So within (fi 3), we can generate all the values we need for calculating (fi 4). We could think of there being three key values. For a given n, there is the value of fi one step back (fi (- n 1), and the value of fi two step backs, and the value of fi three steps back ((fi (- n 2)). We might call these 1ago, 2ago, and 3ago. For (fi 3), these values are 2, 1, 0. For (fi 4), these values are 4, 2, 1. For (fi 5), these values are 11, 4, 2. For (fi 6), these values are 25, 11, 4. For (fi 7), these values are 50, 25, 11. Notice that a new value is generated using the existing values in the first slot (for 1ago), and then that value recurs as we increase n by 1, first as 2ago, then as 3ago. In order to have an iterative procedure that can solve Exercise 1.11 for arbitrary values, we will want to do a recursive call to that iterative procedure with appropriate transformations of these values. So, suppose we start at a calculation of (fi 3) as our initial starting place with initial values of 2, 1, 0 for 1ago, 2ago, 3ago. How will we transform these values into a suitable form for calculating (fi 4)? As our new 1ago argument, we'll want to provide (+ 1ago (* 2 2ago)(3 3ago)). This is 4. As our new 2ago argument, we'll want to provide our current 1ago argument, which is (fi 2) or 2. As our new 3ago argument, we'll want to provide our current 2ago argument, which is (fi 1) or 1. With these values provided, our procedure will be able to calculate (fi 4) by engaging in the appropriate transformations of them per the rule specified in the exercise. ##### First Attempt (wrong) ❌ I went awry partially cuz I was doing unnecessary operations because I hadn't thought in sufficient detail about which operations were necessary. He's some mistaken code that that I wrote: ;bad wrong code; not the answer! (define (fi n) (if (< n 3) n (fi-helper 2 1 0 3 n))) (define (fi-helper 1ago 2ago 3ago current max) (if (= current max) (+ 1ago (* 2 2ago) (* 3 3ago)) (fi-helper (+ 1ago 2ago 3ago) ;WRONG! (* 2 1ago) ;WRONG! (* 3 2ago) ;WRONG! (+ current 1) max))) Multiple problems here. One is that, for some reason, I'm doing a different operation on the numbers in my recursive call to fi-helper (specifically, (+ 1ago 2ago 3ago)) than I do as the consequent for the if (specifically, (+ 1ago (* 2 2ago) (* 3 3ago)). It was only after going through and writing all the stuff above that I realized clearly that these should be the same operation, and specifically that they should be (+ 1ago (* 2 2ago) (* 3 3ago)). The second issue is that I am unnecessarily multiplying the 1ago and 2ago values by 2 and 3, respectively, in providing them as arguments for my recursive call. Going through the above analysis helped me realize that these values should be left untouched. ##### Working Solution ✅ Below is a fixed and functioning fi program. Note that I also keep track of which n value we're currently calculating (using current) and the final n we need to reach (with max). (define (fi n) (if (< n 3) n (fi-helper 2 1 0 3 n))) (define (fi-helper 1ago 2ago 3ago current max) (if (= current max) (+ 1ago (* 2 2ago) (* 3 3ago)) (fi-helper (+ 1ago (* 2 2ago) (* 3 3ago)) 1ago 2ago (+ current 1) max))) ##### Looking at Other Solutions I compared my solution to other people's. Here's one that's similarly structured: define (f n) (if (< n 3) n (f-iter 2 1 0 (- n 2)))) (define (f-iter n1 n2 n3 left) (if (= left 0) n1 (f-iter (+ n1 (* 2 n2) (* 3 n3)) n1 n2 (- left 1)))) Note, though, that instead of using two separate current and max values and increasing the current value until it is equal to the max value, this guy just has 1 value, left, that increments down to 0. That's less complicated, so it's better. Other solutions nested the helper procedure inside the main procedure, like Anne's and Sebastien's. I'm not sure how I feel about that style yet. ### Exercise 1.12 ✅ The following pattern of numbers is called Pascal’s triangle. The numbers at the edge of the triangle are all 1, and each number inside the triangle is the sum of the two numbers above it. Write a procedure that computes elements of Pascal’s triangle by means of a recursive process. #### Discussion I found the book's specification of the problem really vague. I found a Reddit thread by a guy who said he had a Bachelor's in math and was struggling about what it was asking for and gave a bunch of possibilities, and the reply he got was basically that you're supposed to figure it out and that's part of the exercise. I watched Brian Harvey talking about a Pascal's triangle procedure a bit but didn't really look at the body of his procedure and stopped the video when it came on screen (cuz I wanted to figure things out for myself, just needed a bit of a hint to get started). Harvey said that the pascal procedure he was demonstrating takes a row and a column. If column = 0, then the you're on the left edge of the triangle and the answer is 1. That makes sense since the 0 column will always be the first column in any given row. He also says that if row = column, then you're on the right edge of the triangle and the answer is 1. I think that the "tip" of the triangle - the 1 at the summit - would be defined by such a procedure as being located at (0, 0). So anyways from this description, I got the sense that basically Harvey's procedure would take a column and row value and give you the index at that location. I decided to try implementing this approach in my ownprocedure. So let's assume the format is (column, row). For the row that appears below the summit... ...the left 1 would be at (0, 1) and the right 1 would be at (1, 1). And then the next row down would have the range of values from(0, 2) through (2, 2), and the row after that would have the range of possible values from (0, 3) through (3, 3), and so on. So let's say we want to figure out a particular value at a point. Let's say we want (1, 2): While, for some values, there will be lots of recursive calls, at any particular step, we're only going to have to worry about 2 values. So to figure out that, we've got to look at the values in row 1. So we are starting at (1, 2) and we're going to want to look at (0, 1) (the left 1) and (1, 1) (the right 1). So we started at an initial index we can call (c, r). To calculate the value at that index, we look to two places: the value at the place where column and row are both reduced by 1, and the value at the place where just the row has been reduced by 1. And then we want to add these two values. Another way of saying this is to calculate some given (c, r) value we want to figure out the value at the index where c = (- c 1) and r = (- r 1), and also figure out the value at the index where c is the same and r = (- r 1). There are "edges" that to our triangle that are relevant to our calculaions. Suppose we tried to get the value of (0, 1) in our hypothetical Pascal's triangle procedure, which would be the 1 in the lower left on the following diagram: If we follow the method I laid out and try to get the value of the index where c = (- c 1) and r = (- r 1) for a (c,r) of (0, 1), then we wind up trying to calculate the value of (-1, 0), which represents the location of the 0 in the upper left of the diagram. I think whenever we have a negative value in our index values, we want to return 0 to have everything come out okay. One thing that came out in testing was that we also want to return 0 whenever column exceeds row. A negative column indicates we've gone "off" the left side of our triangle in looking for something to add together to get the value underneath, but a column exceeding the row indicates we've gone "off" the right side in the same way, so we want to return 0 for that. In the diagram above, the 0 depicted on the right side would be at index (2, 1), but the only valid indices in that row are where the 1 values are - these would be indices (0,1) and (1,1) respectively. If we don't handle this kind of case, then we'll have problems. So, boiling down the above stuff to a simple statement in English: 1. If either column or row are less than 0, or if column is greater than row, we're looking "past" the "edge" of our triangle, and so we want to return 0 as the value of such a location. 2. In the summit or peak, at location (0,0), we want to return 1. 3. For all other indices at some (c, r), we determine their value by summing the value of the indices directly "above" the current value. The first of these indices is calculated by subtracting 1 from both the current c and r, and the second of these values is calculated by using the current c and subtracting 1 from the current r. #### Solution In code, this is: (define (pascal column row) (cond ((or(< column 0) (< row 0) (> column row)) 0) ((and (= row 0)(= column 0)) 1) (else (+ (pascal (- column 1)(- row 1)) (pascal column (- row 1)))))) #### Looking at Other Solutions I looked at other people's solutions and noticed some differences. I was curious if I could come up with a solution that was more precise, so I investigated what the differences were. Here's aelanteno's: (define (pascal row column) (cond ((= column 1) 1) ((= column row) 1) (else (+ (pascal (- row 1) (- column 1)) (pascal (- row 1) column))))) This solution starts out the counting at 1 and has the arguments in a different order. The ((= column 1) 1) defines the left boundary of the triangle. Whenever the column is 1, indicating that you're at the left most side of the triangle, then the value returned should be 1. ((= column row) 1) defines the right boundary of the triangle - column and row being equal indicates you're at the rightmost side of the triangle in whatever row you are in and thus that the value should be 1. The recursive case is essentially identical to mine except for the order. I tried "translating" this procedure to something closer to mine by changing the order of the arguments and having it start at 0, so I could compare the solutions "side by side" more and examine the differences. Here's my "translated" version: (define (pascal column row) (cond ((= column 0) 1) ((= column row) 1) (else (+ (pascal (- column 1) (- row 1)) (pascal column (- row 1)))))) And here is my current solution: (define (pascal column row) (cond ((or (< column 0) (< row 0) (> column row)) 0) ((and (= row 0)(= column 0)) 1) (else (+ (pascal (- column 1)(- row 1)) (pascal column (- row 1)))))) So we can see that the recursive case is identical, and the only real differences are in the non-recursive cases. So I think a short summary of how I was thinking of the problem is helpful. My first cond clause, the big or, is all about defining no-go zones "outside" the triangle for which the procedure should return 0. I think ruling out negative numbers for column and row makes intuitive sense, and as I said earlier, if column is greater than row then that indicates we've gone off the edge. After that, in the second line, I explicitly defined the summit of the triangle. with ((and (= row 0)(= column 0)) 1). So my conception of the problem was this: we have the summit. That's the really significant/important base case. Then we have some no-go zones outside the triangle's boundaries where we want to return 0. And for everything else, we'll return a value based on the recursive definition of pascal in else, with the number values ultimately "flowing down" from recursive calculations that ultimately originate in that summit with the value of 1. A different way of thinking about the problem, which leads to a simpler implementation in code that seems more common than how I went about it, is to define the entire left and right boundaries of the triangles as 1 explicitly by returning a 1 when the indices of column or column+row indicate that you're at the boundary of the triangle. So the triangle's entire boundaries become the base case. These base cases include the summit. And you don't have to worry about going "outside" the triangle because whenever you hit an edge you just return 1, full stop. So one thing worth noting after this comparison is that while in my head I was thinking of the summit of the triangle as the "important" base case, what I effectively did was define the coordinates just outside the triangle's edge as another base case - one for which I would return 0. This led to a bit of extra complexity. OTOH, my program is more robust at handling out-of-range inputs without looping, since while I had the coordinates just beyond the "edge" of the triangle in mind as the important case, I actually handled a whole range of other cases in how I implemented my solution. ### Exercise 1.13 ❌🤯 (Needed this one explained to me in detail and at length - thanks Max. While I'm not giving myself "credit" for having figured this out without help, I persisted quite a lot and figured a bunch of stuff out 🙂). For a discussion of mathematical induction in the context of a simpler example, see this earlier post. The definition of Fibonacci numbers: #### Starting with the Hint The problem starts by asking you to prove that $Fib(n)$ is the closest integer to $\frac{\psi^n}{\sqrt5}$ but then gives you a big "hint". We start with the hint in our analysis. So we start with using mathematical induction to try to prove that $Fib(n)= \frac{\phi^n - \psi^n}{\sqrt{5}}$. ##### Some Initial Facts $Fib(n) = Fib(n - 1) + Fib (n - 1)$ And we also know $\psi = \frac{1 - \sqrt{5}}{2}$ $\phi = \frac{1 + \sqrt{5}}{2}$ Some other stuff we can figure out: $\phi^2 = \phi + 1$ This is shown from: $\phi^2 = (\frac{1 + \sqrt{5}}{2})^2$ $= \frac{1 + 2\sqrt{5} + 5}{4}$ $= \frac{6 + 2\sqrt{5}}{4}$ $= \frac{3 + \sqrt{5}}{2}$ $= \frac{1 + \sqrt{5}}{2} + 1$ $= \phi + 1$ And we can by similar means show that $\psi^2 = \psi + 1$ So those are some things we can know up front. ##### Base Cases With Max's help I moved onto evaluating some base cases. $Fib(1) = 1$ So we want to show that $\frac{\phi^1 - \psi^1}{\sqrt{5}} = 1$ Step 1: $\frac{\phi^1 - \psi^1}{\sqrt{5}}$ Step 2: $\frac{1}{\sqrt{5}}(\frac{1 + \sqrt{5}}{2} - \frac{1 - \sqrt{5}}{2})$ Step 3: $\frac{1}{\sqrt{5}}{(\frac{2\sqrt{5}}{2})}$ Step 4: 1 So $Fib(2) = 1$ per our definition of $Fib(n)$. So we want to show that: $\frac{\phi^2 - \psi^2}{\sqrt{5}} = 1$ Step 1: $\frac{\phi^2 - \psi^2}{\sqrt{5}}$ Step 2: $\frac{(\phi - \psi)({\phi + \psi})}{\sqrt{5}}$ Step 3: $\frac{1}{\sqrt{5}}{(\phi - \psi)({\phi + \psi})}$ Step 4: $\frac{1}{\sqrt{5}}{( \frac{1 + \sqrt{5}}{2} - \frac{1 - \sqrt{5}}{2})({\frac{1 + \sqrt{5}}{2} + \frac{1 - \sqrt{5}}{2}})}$ Step 5: $\frac{1}{\sqrt{5}}{(\frac{2\sqrt{5}}{2})({\frac{2}{2}})}$ Step 6: 1 ##### Inductive Hypothesis This is where I started to have lots of difficulty. Suppose that $n$ is some $k$. Assuming that $Fib(k)= \frac{\phi^k \,- \,\psi^k}{\sqrt{5}}$ where $k = n - 1$ and $k = n - 2$, show that it works when $k = n$. In other words, show that the following is true: $\frac{\phi^{k-1} \,- \,\psi^{k - 1}}{\sqrt{5}} + \frac{\phi^{k-2} \,-\, \psi^{k - 2}}{\sqrt{5}} = \frac{\phi^{k} \,-\, \psi^{k}}{\sqrt{5}}$ ##### Inductive Step $\frac{\phi^{k-1} \,-\, \psi^{k - 1}}{\sqrt{5}} + \frac{\phi^{k-2} \,-\, \psi^{k - 2}}{\sqrt{5}}$ $= \frac{\phi^{k-1} \,-\, \psi^{k - 1} \, + \, \phi^{k-2} \,-\, \psi^{k - 2}}{\sqrt{5}}$ $= \frac{\phi^{k-1} \, + \, \phi^{k-2} \,-\, \psi^{k - 1} \,-\, \psi^{k - 2}}{\sqrt{5}}$ $= \frac{\phi^{k-2}\phi^1 \, + \, \phi^{k-2} \,-\, \psi^{k - 2}\psi^1 \,-\, \psi^{k - 2}}{\sqrt{5}}$ (note: the step above follows from the fact that $a^m = a^{m + 1 - 1} = a^{m - 1} \times a^1$) $= \frac{\phi^{k-2}(\phi^1 + 1) \,-\, \psi^{k - 2}(\psi^1+\, 1)}{\sqrt{5}}$ $= \frac{\phi^{k-2}(\phi^1 + 1) \,-\, \psi^{k - 2}(\psi^1+\, 1)}{\sqrt{5}}$ Next, per some stuff we partially proved in "Some Initial Facts", $= \frac{\phi^{k-2}(\phi^2) \,-\, \psi^{k - 2}(\psi^2)}{\sqrt{5}}$ $= \frac{\phi^{k-2+2} \,-\, \psi^{k - 2 + 2}}{\sqrt{5}}$ $= \frac{\phi^{k} \,-\, \psi^{k}}{\sqrt{5}}$ #### Doubling Back Ok, now that we've shown that $\frac{\phi^{k-1} \,- \,\psi^{k - 1}}{\sqrt{5}} + \frac{\phi^{k-2} \,-\, \psi^{k - 2}}{\sqrt{5}} = \frac{\phi^{k} \,-\, \psi^{k}}{\sqrt{5}}$, we can turn back to the initial question, which asked us to prove that $Fib(n)$ is the closest integer to $\frac{\phi^n}{\sqrt{5}}$. So for every n we can give to $Fib(n)$, we get an integer out, cuz that's how $Fib(n)$ is defined - it adds up integers. Now, if we want to say that $Fib(n)$ is the closest integer to $\frac{\phi^n}{\sqrt{5}}$, we have to show that the absolute value of the difference between $Fib(n)$ and $\frac{\phi^n}{\sqrt{5}}$ is less than or equal to 1/2. At this point I'm going to link a solid explanation I found for Exercise 1.13 which appears in a Google Group. The figure below comes from that explanation: So again, $fib(n)$ is always an integer. If $fib(n)$ for some given $n$ were to be greater than $\frac{1}{2}$ away from $\frac{\phi^n}{\sqrt{5}}$ for some given $n$, then that would mean that there was some other integer, besides $fib(n)$, which was the closest integer. Consider an example: $fib(10)$. That's 55. And $\frac{\phi^{10}}{\sqrt{5}}\approx 55$ (it's a tiny bit over 55). But if $\frac{\phi^{10}}{\sqrt{5}}\approx 55.6$, that would mean another integer was closer to $\frac{\phi^{10}}{\sqrt{5}}$ than $fib(10)$ (namely, 56). So that's all by way of saying that we want to prove that $\left\| fib(n) - \frac{\phi^n}{\sqrt{5}} \right\| \leq \frac{1}{2}$, which will show that $Fib(n)$ is the closest integer to $\frac{\phi^n}{\sqrt{5}}$. So let's prove that $\left\| fib(n) - \frac{\phi^n}{\sqrt{5}} \right\| \leq \frac{1}{2}$ This is where that whole business we went through in the previous section "Starting with the Hint" comes in handy. Because we know already know that $Fib(n)= \frac{\phi^n - \psi^n}{\sqrt{5}}$. So we can substitute the left side: $\left\| \frac{\phi^n - \psi^n}{\sqrt{5}} - \frac{\phi^n}{\sqrt{5}} \right\| \leq \frac{1}{2}$ And then we're left with: $\left\| -\frac{\psi^n}{\sqrt{5}} \right\| \leq \frac{1}{2}$ and since we don't have to worry about the negative inside absolute value bars: $\left\|\frac{\psi^n}{\sqrt{5}} \right\| \leq \frac{1}{2}$ The $\psi^n$ part is the part whose sign varies depending on the $n$. We don't have to worry about negative signs hiding in the square root of 5, so we can do: $\frac {\left\|\psi^n \right\| }{\sqrt{5}} \leq \frac{1}{2}$ ${\left\|\psi^n \right\| } \leq \frac{\sqrt{5}}{2}$ The highest value ${\left\|\psi^n \right\| }$ achieves is 1, when n = 0. Past that, for each increase in $n$, ${\left\|\psi^n \right\| }$ becomes smaller. $\left\|\psi^1 \right\| = \frac{1 - \sqrt{5}}{2} \approx 0.61$, and from there on we're just multiplying more and more sub-1 values together as we increase $n$. So ${\left\|\psi^n \right\| }$ is always $\leq \frac{\sqrt{5}}{2}$ #### Why Do We Care Again? On the Critical Fallibilism Discourse forum, there was some discussion of an unanswered question from a previous post starting around here. I had asked why the book was talking about the golden ratio and the inefficiency of the recursive fibonacci procedure at an earlier point in the book (before Exercise 1.13) and Max said: So the point was just to rigorously show the complexity of the fib procedure as we had implemented it. ## Following Up ### More on Exercise 1.10 I realized that I forgot to check my answers against other people's in my last post. Turns out I was right, but there was one point that I thought warranted further clarification. For the part of the problem that asked for a concise mathematical definition of (define (h n) (A 2 n)), I saw a couple of answers along the lines of aelanteno's here that used notation that looks more standard, using standard exponentiation and a composition of functions instead of the Rudy Rucker notation for tetration: Basically I think the analysis is like: (g n) = $g(n)$ = $2^n$ (h n) = $h(n)$ =(A 2 n) $g(h(n))$ I liked the Rudy Rucker notation where you do e.g. ${^{3}2}$ for $2^{2^{2}}$ because it matches my intuitions better. I basically want to include the "bottom" 2 as part of the "tower" of stuff that is getting repeatedly exponentiated (or maybe the word is tetrated). If I used regular exponents, I'd have to use something like $2^{n-1}$
# 1984 AIME Problems/Problem 3 ## Problem A point $P$ is chosen in the interior of $\triangle ABC$ such that when lines are drawn through $P$ parallel to the sides of $\triangle ABC$, the resulting smaller triangles $t_{1}$, $t_{2}$, and $t_{3}$ in the figure, have areas $4$, $9$, and $49$, respectively. Find the area of $\triangle ABC$. $[asy] size(200); pathpen=black;pointpen=black; pair A=(0,0),B=(12,0),C=(4,5); D(A--B--C--cycle); D(A+(B-A)3/4--A+(C-A)3/4); D(B+(C-B)5/6--B+(A-B)5/6);D(C+(B-C)5/12--C+(A-C)5/12); MP("A",C,N);MP("B",A,SW);MP("C",B,SE); /* sorry mixed up points according to resources diagram. / MP("t_3",(A+B+(B-A)3/4+(A-B)5/6)/2+(-1,0.8),N); MP("t_2",(B+C+(B-C)5/12+(C-B)5/6)/2+(-0.3,0.1),WSW); MP("t_1",(A+C+(C-A)3/4+(A-C)*5/12)/2+(0,0.15),ESE); [/asy]$ ## Solution By the transversals that go through $P$, all four triangles are similar to each other by the $AA$ postulate. Also, note that the length of any one side of the larger triangle is equation to the sum of the sides of each of the corresponding sides on the smaller triangles. We use the identity $K = absinC$ to show that the areas are proportional (the sides are proportional and the angles are equal) Hence, we can write the lengths of corresponding sides of the triangle as $2x,\ 3x,\ 7x$. Thus, the corresponding side on the large triangle is $12x$, and the area of the triangle is $12^2 = \boxed{144}$.
# Radians to Revolutions – Formulas and Examples Revolutions are a way of measuring complete turns in circles. Revolutions are used in various situations where an object makes a large number of turns and it is more convenient to measure the number of turns the object makes per minute or per second. For example, revolutions per minute (RPM) are used to measure the rotation of engines, tires, or other objects. One revolution is equivalent to 2π radians. Therefore, to convert from radians to revolutions, we divide the radians by 2π. Here, we will use the radians to revolutions transformation formula to solve some practice problems. ##### TRIGONOMETRY Relevant for Learning to transform from radians to revolutions with examples. See examples ##### TRIGONOMETRY Relevant for Learning to transform from radians to revolutions with examples. See examples ## How to convert from radians to revolutions? We can convert from radians to revolutions by dividing the number of radians by 2π and we will get the number of turns that is equal to the given radians. This means that we have the following formula: where y represents the given radians and x is the response in revolutions. The formula for the transformation from radians to revolutions is derived by considering that we have 2π radians when we make a complete revolution of a circle. Radians are related to the radius of a circle as we can see in the following animation. Also, we know that one revolution equals one complete turn. Therefore, we have the relationship 1 rev = 2π rad. Therefore, to get the number of revolutions in a given number of radians, we simply divide the number of radians by 2π. Each of the following examples is solved using the transformation formula from radians to revolutions seen above. Try to solve the problems yourself before looking at the answer. ### EXAMPLE 1 How many revolutions are equal to 4π radians? Using the formula given above with the given value, we have: $latex \frac{y \text{ rad}}{2\pi}=x \text{ rev}$ $latex \frac{4\pi \text{ rad}}{2\pi}=2 \text{ rev}$ Therefore, 4π radians is equal to 2 revolutions. ### EXAMPLE 2 If we have 12π radians, what is its equivalent in revolutions? We substitute the given value in the transformation formula to obtain: $latex \frac{y \text{ rad}}{2\pi}=x \text{ rev}$ $latex \frac{12\pi \text{ rad}}{2\pi}=6 \text{ rev}$ Therefore,12π radians is equivalent to 6 revolutions. ### EXAMPLE 3 How many revolutions are equal to 7π radians? We have $latex y=7\pi$. Using this value in the formula given above, we have: $latex \frac{y \text{ rad}}{2\pi}=x \text{ rev}$ $latex \frac{7\pi \text{ rad}}{2\pi}=3.5 \text{ rev}$ Therefore, 7π radians is equal to 3.5 revolutions. ### EXAMPLE 4 If we have 12 radians, how many revolutions do we have? In this case, we have $latex y = 12$. By substituting this value in the formula, we get: $latex \frac{y \text{ rad}}{2\pi}=x \text{ rev}$ $latex \frac{12 \text{ rad}}{2\pi}=1.91 \text{ rev}$ Therefore, 12 radians is equivalent to 1.91 revolutions. ### EXAMPLE 5 How many revolutions are equivalent to 24 radians? We substitute the value $latex y = 24$ in the formula to obtain: $latex \frac{y \text{ rad}}{2\pi}=x \text{ rev}$ $latex \frac{24 \text{ rad}}{2\pi}=3.82 \text{ rev}$ Therefore, 24 radians is equal to 3.82 revolutions. ## Radians to revolutions – Practice problems Solve the following practice problems using the radians to revolutions transformation formula. Select an answer and check it to see if you got the correct answer.
# How many ways can a committtee of $2$ women and $3$ men be selected from a group of $5$ women and $7$ men if two of the men refuse to serve together? I am trying to learn permutations and combinations. Please help me to solve the second part of the question. From a group of $5$ women and $7$ men, how many different committees of $2$ women and $3$ men can be formed? What if $2$ of the men are feuding and refuse to serve on the committee together? I gather that you understand that the number of committees of two women and three men that can be selected from five women and seven men is $$\binom{5}{2}\binom{7}{3}$$ Below is a hint for the second part of the question. Hint: Consider three groups of people: the five women, the five men who are not feuding, and the two men who are feuding. There are two ways to form the committee so that the two men who are feuding are not both selected: 1. Neither of the feuding men is selected. Choose two of the five women and three of the five men who are not feuding. 2. Exactly one of the feuding men is selected. Choose two of the five women, two of the five men who are not feuding, and one of the two men who are feuding. A committee with 2 women and 3 men is a group of 2 women together with a group of 3 men. How many groups of 2 women are there? You have 5 ways to pick the first woman, 4 ways to pick the second, and then you have to divide by 2, because picking Alice and then Beatrice is the same as picking Beatrice and then Alice. Order does not matter. Thus you have $$\frac{5\cdot4}{2} = {5\choose2}$$ Similarly for the group of 3 men. You have $7\cdot6\cdot5$ ways of picking them, but then you have to discount the different orders that pick the same groups. Picking the men in these orders form always the same group: $$ABC\\ ACB\\ BAC\\ BCA\\ CAB\\ CBA$$ So 6 different orders produce the same group, so every 6 different orders give the same group. Why 6? Because 6, $3!$, is the number of ways you can order 3 elements. Thus you have $$\frac{7\cdot6\cdot5}{3\cdot2} = {7\choose3}$$ You now multiply both together and get $${5\choose2}\cdot{7\choose3}$$ diferent committees. In the general case, if you have a group of $n$ people and you want to form a group with $k$ of them, you get $${n\choose k }$$ total groups, if there are no further restrictions. Notice that this formula matches our results for the group of 2 women and for the group of 3 men. For the second part, let us call Bob and John to the two guys refusing to work together. The number of valid committees is the number of committees with none of them plus the number of committees with Bob and 2 guys other than John plus the number of committees with John and two guys other than Bob. Can you try and calculate that?
# Factoring View: Sorted by: ### Find the Absolute Value of a Number The absolute value of a number is the positive value of that number. It tells you how far away from 0 a number is on the number line. The symbol for absolute value is a set of vertical bars. ### How to Find the Factors of a Number The factors of a number are all those numbers that can divide evenly into the number with no remainder. The greatest factor of a number is the number itself, so you can always list all the factors of any ### How to Find the Least Common Multiple The least common multiple (LCM) of a set of numbers is the lowest positive number that is a multiple of every number in that set. For example, the LCM of the numbers 2, 3, and 5 is 30 because ### How to Find the Greatest Common Factor The greatest common factor (GCF) of a set of numbers is the largest number that is a factor of all those numbers. For example, the GCF of the numbers 4 and 6 is 2 because 2 is the greatest number that’s ### 6 Mathematical Ways to Say the Same Thing In math, factors and multiples are two important connected concepts. Multiplication and division are inverse operations. You may have noticed that, in math, you tend to run into the same ideas over and ### How to Generate Multiples Even though multiples tend to be larger numbers than factors, most students find them easier to work with. Finding all the factors is possible because factors of a number are always less than or equal ### How to Conduct Divisibility Tests When one number is divisible by a second number, you can divide the first number by the second without having anything left over. For example, 16 is divisible by 8 because 16 / 8 ### Factors and Multiples The concept of divisibility — for example, 12 is divisible by 3 because 12 / 3 = 4, with no remainder — can also be described using the words factor and ### How to Identify Prime (and Composite) Numbers Every counting number greater than 1 is either a prime number or a composite number. A prime number has exactly two factors — 1 and the number itself. For example, the number 5 is prime because its only ### How to Generate a Number’s Factors When one number is divisible by a second number, that second number is a factor of the first. For example, 10 is divisible by 2, so 2 is a factor of 10. ### How to Decompose a Number into Its Prime Factors Every number is the product of a unique set of prime factors, a group of prime numbers (including repeats) that, when multiplied together, equals that number. You can find those prime factors for a given ### How to Find the Greatest Common Factor The greatest common factor (GCF) of a set of numbers is the largest number that’s a factor of every number in that set. Finding the GCF is helpful when you want to reduce a fraction to its lowest terms ### How to Generate the Multiples of a Number Generating the multiples of a number is easier than generating the factors: Just multiply the number by 1, 2, 3, and so forth. But unlike the factors of a number — which are always less than the number ### How to Find the Least Common Multiple The least common multiple (LCM) of a set of numbers is the smallest number that’s a multiple of every number in that set. For small numbers, you can simply list the first several multiples of each number
Question 10 # The cost of fencing a rectangular plot is ₹ 200 per ft along one side, and ₹ 100 per ft along the three other sides. If the area of the rectangular plot is 60000 sq. ft, then the lowest possible cost of fencing all four sides, in INR, is Solution Let us draw the rectangle. Now, definitely, three sides should be fenced at Rs 100/ft, and one side should be fenced at Rs 200/ft. In this question, we are going to assume that the L is greater than B. Hence, the one side painted at Rs 200/ft should be B to minimise costs. Hence, the total cost = 200B + 100B + 100L + 100L = 300B + 200L Now, L x B = 60000 B = 60000/L Hence, total cost = 300B + 200L = 18000000/L + 200L To minimise this cost, we can use AM>=GM, $$\frac{\frac{18000000}{L}+200L}{2}\ge\sqrt{\ \frac{18000000}{L}\times\ 200L}$$ $$\frac{18000000}{L}+200L\ge2\sqrt{\ 18000000\times\ 200}$$ $$\frac{18000000}{L}+200L\ge2\times\ 60000$$ Hence, minimum cost = Rs 120000.
💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here! # A drug response curve describes the level of medication in the bloodstream after a drug is administered. A surge function $S(t) = At^p e^{-kt}$ is often used to model the response curve, reflecting an initial surge in the drug level and then a more gradual decline. If, for a particular drug, $A = 0.01$, $p = 4$, $k = 0.07$, and $t$ is measured in minutes, estimate the times corresponding to the inflection points and explain their significance. If you have graphing device, use it to graph the drug response curve. ## $S(t)=A t^{p} e^{-k t}$ with $A=0.01, p=4,$ and $k=0.07 .$ We will find thezeros of $f^{\prime \prime}$ for $f(t)=t^{p} e^{-k t}$\begin{aligned}f^{\prime}(t) &=t^{p}\left(-k e^{-k t}\right)+e^{-k t}\left(p t^{p-1}\right)=e^{-k t}\left(-k t^{p}+p t^{p-1}\right) \\f^{\prime \prime}(t) &=e^{-k t}\left(-k p t^{p-1}+p(p-1) t^{p-2}\right)+\left(-k t^{p}+p t^{p-1}\right)\left(-k e^{-k t}\right) \\&=t^{p-2} e^{-k t}\left[-k p t+p(p-1)+k^{2} t^{2}-k p t\right] \\&=t^{p-2} e^{-k t}\left(k^{2} t^{2}-2 k p t+p^{2}-p\right)\end{aligned} Derivatives Differentiation Volume ### Discussion You must be signed in to discuss. ##### Catherine R. Missouri State University ##### Heather Z. Oregon State University ##### Kristen K. University of Michigan - Ann Arbor Lectures Join Bootcamp ### Video Transcript called the search function and it's equal to uh some number eight times T. To the p power times E. To the negative K. Times T. Power S. Is the level of medication and bloodstream T. Is the number of minutes that have elapsed for this particular Search function A. is .01 times T. To the P. Power where P. Is given to be four. And then that multiplies E. To the negative K. T. K. Is given to be point oh seven. So that's E. To the negative zero. Actually .07. Each of the negative .07. Each of negative Katie each of negative .07 R. T. So we are going to graft dysfunction using dez mose. And then we're going to look for the inflection points and kind of describe what's going on with this search function at those uh inflection points. Okay so using Dez most I graphed the search function. Dez most uh prefers to work with the variable X. So instead of the variable T. Which was representing minutes, we're just using the variable X. And since the number of minutes that have elapsed since the injection uh should be greater than or equal to zero. Uh graphing dysfunction for uh time less than zero really doesn't mean anything. So to uh make the graph, you know look a little less confusing. I restricted uh dysfunction to being graft off when the number of minutes X. Is greater than or equal to zero. Uh The goal here is to estimate the time which would be the horizontal X. Axis, estimate the time at which inflection points occur. Now, an inflection point occurs when the second derivative of the function is zero. Uh Basically it's signals an inflection point signals when the first derivative uh will change from increasing to decreasing if the second derivative is zero, that means the first derivative reaches a maximum point meaning the first derivative or the steepness to slope. If you will was increasing, reached a maximum point now will start decreasing or vice versa. Uh An inflection point being where the second derivative is zero. Inflection point could also be where the first derivative reaches a minimum, the first derivative is decreasing, reaches a minimum and then starts increasing. Now if we're looking for the first derivative to be at a maximum point or minimum point. That means, for example, if we say the first derivative is reaching a maximum, that means to first derivative was increasing, reaches a maximum point will start decreasing. Well, the first derivative lets you know, you know just how steep the function itself is increasing or decreasing. So if we're looking for where the first derivative is going to change from increasing or decreasing, we want to look for a change in the steepness of the curve. Let me go back to our uh whiteboard here and give you an idea of what um what an inflection point would look like. Just so you have an idea of what we're looking for. If I have a function it's going up steeply. Okay, But then at this point it's still gonna be increasing but at a at a slower rate. Okay, this would be an inflection point. The function is still increasing, but the rate at which it's increasing uh as indicated by the first derivative um is changing here to function is increasing at a quicker and quicker rate. So to slope or the first derivative is increasing. Then at this inflection point, uh the second derivative zero. The first derivative reaches a maximum. Uh The function itself still increases but now at a slower rate because F prime of X now is starting to decrease. So this would be the graph of the function F and uh f prime of X would change from the slope. Would at this inflection point changes from uh increasing and increasing to where the slope or the derivative is now uh decreasing. Okay, so at prime of X changes from positive two negative. The second derivative at this inflection point would be zero. So this is what happens at an inflection point. At an inflection point. Uh function will still be increasing. And of course this can also be shown for a function that is decreasing through an inflection point. Let me give you no idea what that looks looks like. Um If I uh three very set. If I have a function that's decreasing reaches an inflection point still decreases. But now at a slower rate. This would also be an inflection point Um at an inflection point the second derivative is zero and the first derivative changes signs. So here the first derivative is changing signs at this point here the first derivative is positive. Uh Here the first derivative will be negative at this inflection point, The second derivative will be zero here, the first derivative is negative and then at the inflection point, first derivative becomes positive even though the function is still decreasing through that inflection point. So we're looking for a bend or you change in the bend of the curve. That's where these, that's where these inflection points are. C Okay, we're going up but then we change how steeply you can think of it as common cavity kind of concave duck. And then we start con caving down here, we're concave down and then we start uh at the inflection point becoming concave uh So we can estimate um Here you can see that the function is increasing. And then approximately around here there might be an inflection point right around here at approximately uh t equals 25 Maybe 30. Um so when T is 30 minutes, uh the function is still increasing fastest inflection point. Um but not as deeply. Okay, here it's climbing up steeply reaches an inflection point and then starts, you know kind of rounding off. Think of this as concave up Until time is 30 minutes and then concave down. So it's a little hard to see here if we zoomed in and be a little bit easier but we have a point of inflection here and then we have another point of inflection here here the function is coming down. it's the inflection point. The function keeps decreasing but now at a slower rate. Okay here is kind of concave down if I exaggerate it. And then at this inflection point here at approximately 100 minutes uh to function changes come cavity and starts becoming concave, let's go back to our white board and write down those two times. So the two values of T. Of course on our graft the variables X 22 values of T. That are inflection points would be, It looks like when T. 35 minutes We have an inflection point and then we have another inflection point uh when X in the graph or are variable t. 100 minutes. So let's go back to our graph and uh kind of describe what's going on with our search function at these inflection points. So uh when uh t or in this case ex uh number the number of minutes. The time is 30 minutes. We have an inflection point. So for 30 minutes we are at this point on the graph. So the surge is increasing uh rapidly. Um until we reach a time of 30 minutes and then the surge continues increasing but at a slower rate that's what happens. Uh the rate at which it's increasing uh changes. Okay, the surge is increasing and increasing the rate at which the surge, the rate at which the function is increasing is itself increasing until the inflection point. Then the function the surge keeps increasing after 30 minutes. But at a slower rate. Now, when the time was 100 minutes, we estimated there was another inflection point here, at 100 minutes. Uh here the surge is decreasing rapidly until we reach 100 minutes. That's another point of inflection and then the search is still decreasing. But at a slower rate, that's what's happening at those two inflection points. Temple University Derivatives Differentiation Volume Lectures Join Bootcamp
### Round and Round the Circle What happens if you join every second point on this circle? How about every third point? Try with different steps and see if you can predict what will happen. ### Making Cuboids Let's say you can only use two different lengths - 2 units and 4 units. Using just these 2 lengths as the edges how many different cuboids can you make? How many DIFFERENT quadrilaterals can be made by joining the dots on the 8-point circle? # Always, Sometimes or Never? Number ## Always, Sometimes or Never? Number Are the following statements always true, sometimes true or never true? How do you know? What about these more complex statements? You could cut out the cards from this sheet (wordpdf) and arrange them in this grid. Can you find examples or counter-examples for each one? For the “sometimes” cards can you explain when they are true? Or rewrite them so that they are always true or never true? ### Why do this problem? These tasks are a great opportunity for learners to use reasoning to decipher mathematical statements. We often make mathematical claims that are only true in certain contexts and it is important for learners to be able to look critically at statements and understand in what situations they apply. The examples here only refer to one key topic but similar statements could be created for any area of maths - some similar problems about shape can be found here. ### Possible approach You may want to start with one statement and have a class discussion about whether it is true. Ask learners to think of some examples to illustrate the statement and decide whether it is always, sometimes or never true. If they decide it is sometimes true, they could think about what conditions make it true. Groups of learners could be given the set of statements (wordpdf) to sort into the grid (wordpdf). Taking each card in turn they could decide if it is always, sometimes or never true. Then they could justify their reasoning. If they think it is always true or never true, they could explain why they think this is. If they think it is sometimes true they could start by coming up with cases for each and trying to generalise. For learners who have had more experience of reasoning it might be good to ask them to try and write their ideas down in a clear way, perhaps for just one or two of the statements to start with. It would be worth sharing ideas as a class at the end. You could pick up on a statement that has been problematic or where there does not seem to be a consensus and support a whole class discussion. ### Key questions Can you think of an example when it isn’t true? How do you know that it is always true? Is it possible to check all examples? Is there another way of knowing? ### Possible extensions Learners could be asked to come up with their own statements for things that are always, sometimes and never true within a topic area. Again they should try to justify their reasons and specify the conditions necessary. ### Possible support When discussing as a class, suggest types of numbers to try or specific shapes to consider. Learners often need to start with concrete examples to develop their understanding of a particular concept before they can before they can develop their reasoning within that area. Concrete resources can be useful for developing an understanding of the structure of numbers, and can be used by all learners to support their arguments. The similar problems found here and here might be a good starting point for pupils who need more support.
# Video: AQA GCSE Mathematics Higher Tier Pack 4 • Paper 1 • Question 10 Which of 3/5 and 2/6 is closer in value to 1/2? Show all your working. 03:33 ### Video Transcript Which of three-fifths and two-sixths is closer in value to a half? Show all your working. So what we’re gonna do with this question is take a look at the three fractions we have, which is three-fifths, a half, and two-sixths. And to help us look at them and compare them, what we’re gonna want to do is convert them to an equivalent fraction that has a common denominator. So to help us do that and work out what is the lowest common multiple which is gonna be our common denominator, I’m just gonna write out a few multiples of each of the numbers that we have as the denominators. So first of all, I’ve written out the first few multiples of six. We got six, 12, 18, 24, 30, 36. And then, I’ve written out the first three multiples of five. So I got five, 10, 15, 20, 25, and 30. And I’ve stopped at 30. And the reason I’ve stopped at 30 is because 30 is a common multiple for both six and five. And 30 is also a multiple of two because 30 divided by two gives us 15. So great, this is gonna be our common denominator. So now, what we’re gonna do is convert each of our fractions into fractions where 30 is the denominator. So in our first fraction, to get from five as the denominator to 30 as this denominator, we have to multiply by six. Therefore, we’ll have to do the same to the numerators because whatever you do to the bottom, you must do to the top. So we’re gonna multiply three by six which is gonna give us 18. So three-fifths is equal to 18 over 30. Then, we’ll do the same process with our second fraction. So we can see that we’d have to multiply two by 15 to get to 30. So we’re gonna do the same to the top, so the numerator. So we do one multiplied by 15 which gives us 15. So we can say that a half is equal to 15 over 30. And then, finally, for our bottom fraction, we can say that we’d have to multiply six by five to get to 30. So therefore, again, we do the same to the top or the numerator. So we get two multiplied by five which gives us 10. So we can say that two-sixths is equal to 10 over 30. So now, what we want to do is actually compare three-fifths and two-sixths with a half because what we want to see is which one of them is closer in value to a half. So to do that, what I’m gonna do is actually find the difference between them. So first of all, we’re gonna find the difference between three-fifths and a half. So we’re gonna do 18 over 30 minus 15 over 30. So this is gonna give us a difference of three over 30. And then to see the difference between two-sixths and a half, I’m gonna do 15 over 30 minus 10 over 30 which is gonna give us a difference of five over 30. Okay, so now, we need to decide which is closer in value to a half: three-fifths or two-sixths? Well, if we compare our differences, we can see that three over 30 is less than five over 30. So therefore, we can say that three-fifths is closer to a half than two-sixths. And that’s because the difference between three-fifths and a half is less than the difference between two-sixths and a half.
Courses Courses for Kids Free study material Offline Centres More Store # Addition Facts for Class 1 Reviewed by: Last updated date: 10th Aug 2024 Total views: 175.5k Views today: 1.75k ## Introduction One of the arithmetic operations in mathematics is addition. You must have added at least once before. For example, your parent may offer you an apple in the morning and another apple in the evening. You have eaten a total of 2 apples in one day. How can one get the total? The only way to do this is by adding. As a result, you must add 1 to 1 to obtain the amount. The addition is not just a process of adding numbers. It is also a foundation for numerous math operations, ranging from simple concepts like “two plus three equals five” to more complex topics like algebraic expressions. One addition fact is that addition can be defined as the sum of two or more addends. Example: 2 + 3 = 5 In addition, the place value of the numbers is given more importance. Here, the place value is expressed as ones and tens. To add numbers, we need to arrange them column-wise according to their place values. Each digit is individually added. The answer is known as their total, and they are known as addends. There are two methods for adding two digits. The first two are the "without regrouping" approach and the "regrouping" method, both referred to as carrying forward. The sums from 1 + 1 to 9 + 9 are addition facts. Children must learn these 81 sums since they are the foundation for the rest of elementary mathematics. The full list of addition facts is provided below: • The addition of 1-digit numbers can be done horizontally. Example : 2+3 =5 • The addition of zero to any number does not change the sum. Example : 5 + 0= 5 • If we add one to any number, it gives the after that number as a sum. ## Solved Numericals: Example 1: Find out the missing numbers - 1. 1 + 5 = _ 2. 4 + _ = 7 3. _ + 3 = 3 Ans: a. 6 b. 3 c. 0 Example 2: Solve the following sums and match them with the correct answer. Column A Column B 1. 4 + 5 4 2. 6 + 2 7 3. 3 + 1 9 4. 5 + 2 8 Ans: a . 4 + 5 = 9 b . 6 + 2 = 8 c . 3 + 1 = 4 d . 5 + 2 = 7 Example 3: Find the sum of the following numbers - 1. 2 + 7 = 2. 1 + 4 = 3. 6 + 3 = 4. 1 + 0 = Ans: a. 9 b. 5 c. 9 d. 1 Example 4: Fill in the missing numbers - 1. 6 + _ = 7 2. _ + 3 = 9 3. 7 + _ = 0 4. 1 + _ = 8 Ans: a. 1 b. 6 c. 0 d. 7 ## Conclusion: This article taught us about adding 1 - digit numbers with solved numerals. Adding numbers is a fundamental mathematical process combining two or more numerical values. This operation is used in our daily lives; some simple examples include calculating money, calculating time, counting students in a class, and so on. ## FAQs on Addition Facts for Class 1 1. What is the significance of addition? Adding numbers is a fundamental mathematical concept required for even the most basic problems in our daily lives. One of the most common uses is when working with money, such as adding bills and receipts. 2. What is the difference between addition with and without regrouping? When the sum of the digits in each place value column is less than or equal to 9, it is considered to be added without regrouping. When the sum of the digits in at least one of the place value columns is greater than 9, this is considered an addition with regrouping. 3. Why is it important to add math? Additional mathematics is an extremely beneficial learning experience for students who have mastered lower secondary school mathematics. It helps you develop deeper mathematical competency and opens up many opportunities for further education.
# UNDERSTANDING SETS AND SUBSETS OF REAL NUMBERS By understanding which sets are subsets of types of numbers, we can verify whether statements about the relationships between sets are true or false. The picture shown below clearly illustrates the subsets of real numbers. From the above picture, we can list out the following important subsets of real numbers. (i) Rational numbers (ii) Irrational numbers (iii) Whole numbers (iv) Integers ## Examples Example 1 : Tell whether the given statement is true or false. Explain your choice. "All irrational numbers are real numbers" True. Every irrational number is included in the set of real numbers. The irrational numbers are a subset of the real numbers. Example 2 : Tell whether the given statement is true or false. Explain your choice. "No rational numbers are whole numbers" False. A whole number can be written as a fraction with a denominator of 1, so every whole number is included in the set of rational numbers. The whole numbers are a subset of the rational numbers. Example 3 : Tell whether the given statement is true or false. Explain your choice. "All rational numbers are integers" False. Every integer is a rational number, but not every rational number is an integer. For example, rational numbers such as 3/5 and -5/2 are not integers. Example 4 : Tell whether the given statement is true or false. Explain your choice. "Some irrational numbers are integers" False. Real numbers are either rational or irrational numbers. Integers are rational numbers, so no integers are irrational numbers. Example 5 : Tell whether the given statement is true or false. Explain your choice. "All whole numbers are rational numbers" True. Whole numbers are a subset of the set of rational numbers and can be written as a ratio of the whole number to 1. Example 6 : Tell whether the given statement is true or false. Explain your choice. "No irrational numbers are whole numbers" True. All whole numbers are rational numbers. Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here. You can also visit our following web pages on different stuff in math. WORD PROBLEMS Word problems on simple equations Word problems on linear equations Algebra word problems Word problems on trains Area and perimeter word problems Word problems on direct variation and inverse variation Word problems on unit price Word problems on unit rate Word problems on comparing rates Converting customary units word problems Converting metric units word problems Word problems on simple interest Word problems on compound interest Word problems on types of angles Complementary and supplementary angles word problems Double facts word problems Trigonometry word problems Percentage word problems Profit and loss word problems Markup and markdown word problems Decimal word problems Word problems on fractions Word problems on mixed fractrions One step equation word problems Linear inequalities word problems Ratio and proportion word problems Time and work word problems Word problems on sets and venn diagrams Word problems on ages Pythagorean theorem word problems Percent of a number word problems Word problems on constant speed Word problems on average speed Word problems on sum of the angles of a triangle is 180 degree OTHER TOPICS Profit and loss shortcuts Percentage shortcuts Times table shortcuts Time, speed and distance shortcuts Ratio and proportion shortcuts Domain and range of rational functions Domain and range of rational functions with holes Graphing rational functions Graphing rational functions with holes Converting repeating decimals in to fractions Decimal representation of rational numbers Finding square root using long division L.C.M method to solve time and work problems Translating the word problems in to algebraic expressions Remainder when 2 power 256 is divided by 17 Remainder when 17 power 23 is divided by 16 Sum of all three digit numbers divisible by 6 Sum of all three digit numbers divisible by 7 Sum of all three digit numbers divisible by 8 Sum of all three digit numbers formed using 1, 3, 4 Sum of all three four digit numbers formed with non zero digits Sum of all three four digit numbers formed using 0, 1, 2, 3 Sum of all three four digit numbers formed using 1, 2, 5, 6
# What Is 12 Out Of 16? Are you curious to know what is code Adam? You have come to the right place as I am going to tell you everything about code Adam in a very simple explanation. Without further discussion let’s begin to know what is code Adam? Ratios are mathematical expressions that help us understand the relationship between different quantities. When you encounter a ratio like “12 out of 16,” it represents a comparison between a part (12) and a whole (16). In this blog, we’ll explore the meaning of this ratio, simplify it, and discuss how ratios are applied in various real-life situations. ## What Is 12 Out Of 16? The ratio “12 out of 16” signifies a comparison between two quantities, where 12 represents the part, and 16 represents the whole. Here are several ways to interpret this ratio: 1. Fraction: “12 out of 16” can be expressed as a fraction, where 12 is the numerator (the top number) and 16 is the denominator (the bottom number). In this case, it represents the fraction 12/16. 2. Percentage: Ratios can also be expressed as percentages. To convert the ratio to a percentage, divide the part (12) by the whole (16) and multiply by 100: Percentage = (12 / 16) * 100 = 75% Therefore, “12 out of 16” is equivalent to 75%. 1. Proportion: Ratios can represent proportions or relationships between quantities. In this context, “12 out of 16” signifies that 12 is a portion of the whole 16. 2. Simplified Form: Ratios can often be simplified to their simplest form by dividing both the part and the whole by their greatest common divisor. For “12 out of 16,” the greatest common divisor is 4: Simplified Ratio = (12 ÷ 4) out of (16 ÷ 4) = 3 out of 4 So, “12 out of 16” simplifies to “3 out of 4.” ## Real-Life Applications Ratios are prevalent in various real-life situations, including: 1. Cooking: Recipes often involve ratios, specifying the proportion of ingredients required to create a dish. For example, a cake recipe might call for 1 cup of flour out of 2 cups of flour. 2. Finance: Financial ratios are used to analyze the performance and health of businesses. The debt-to-assets ratio, for instance, compares a company’s debt (part) to its total assets (whole). 3. Education: Student-to-teacher ratios are used in education to determine class sizes and teacher-to-student relationships. 4. Health: In healthcare, ratios like the body mass index (BMI) compare a person’s weight to their height to assess their health status. 5. Sports: Sports statistics often involve ratios. Batting averages in baseball, for example, compare the number of hits (part) to the number of at-bats (whole). ## Conclusion Ratios are versatile mathematical tools used to express relationships between quantities. When you encounter a ratio like “12 out of 16,” it indicates a comparison between a part and a whole, and it can be expressed as a fraction, a percentage, or in simplified form. Understanding ratios is essential in a wide range of real-life contexts, from cooking and finance to education and health, where they help us make informed decisions and assessments based on quantitative relationships. ## FAQ ### What Is 12 Out Of 16 As A Percent? 12/16 as a percent is 75% For instance, in the fraction 12/16, we could say that the value is 12 portions, out of a possible 16 portions to make up the whole. ### What Is A 13 Out Of 16? 13/16 as a percent is 81.25% ### What Is A 14 Out Of 16 Grade? A score of 14 out of 16 on a test, assignment or class is a 87.5% percentage grade. 2 questions were wrong or points missed. A 87% is a B+ letter grade. A letter grade B+ means good or above average performance. ### Is A 12 Out Of 16 Passing? A score of 12 out of 16 on a test, assignment or class is a 75% percentage grade. 4 questions were wrong or points missed. A 75% is a C letter grade. A letter grade C means satisfactory or average performance. I Have Covered All The Following Queries And Topics In The Above Article What Is A 12 Out Of 16 What Is 12 Out Of 16 What Percent Is 12 Out Of 16 What Percentage Is 12 Out Of 16 What Grade Is A 12 Out Of 16 What Is 12 Out Of 16 As A Percentage What Grade Is 12 Out Of 16 What Is A 12 Out Of 16 Letter Grade What Is 12 Out Of 16 As A Grade What Is A 12 Out Of 16 Grade What Is The Percentage Of 12 Out Of 16 What Letter Grade Is A 12 Out Of 16 What Percent Is A 12 Out Of 16 12 Out Of 16 Is What Percent What Is A 12 Out Of 16 As A Percentage What Is 12 Out Of 16 In Percent What Is A 12 Out Of 16 In Percentage What Is 16 Out Of 12 Percent Shotsd What Is A 12 Out Of 16 As A Grade What Is 16 Out Of 12 Percent What Is 12 Out Of 16
## Intermediate Algebra (12th Edition) $(-7,-6)$ $\bf{\text{Solution Outline:}}$ To solve the given inequality, $-15 \lt 3x+6 \lt -12 ,$ use the properties of inequality. For the interval notation, use a parenthesis for the symbols $\lt$ or $\gt.$ Use a bracket for the symbols $\le$ or $\ge.$ For graphing inequalities, use a hollowed dot for the symbols $\lt$ or $\gt.$ Use a solid dot for the symbols $\le$ or $\ge.$ $\bf{\text{Solution Details:}}$ Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} -15-6 \lt 3x+6-6 \lt -12-6 \\\\ -21 \lt 3x \lt -18 \\\\ -\dfrac{21}{3} \lt \dfrac{3x}{3} \lt -\dfrac{18}{3} \\\\ -7 \lt x \lt -6 .\end{array} In interval notation, the solution set is $(-7,-6) .$ The red graph is the graph of the solution set.
# 30 SAT Math Formulas You Need to Know ## Is your SAT score enough to get you into your dream school? Our free chancing engine takes into consideration your SAT score, in addition to other profile factors, such as GPA and extracurriculars. Create a free account to discover your chances at hundreds of different schools. Have you ever been stumped on a math test question, only to realize how simple it was when you later reviewed your test? That’s exactly what happens to many students who take the SAT. The SAT covers a wide range of math—from as early as elementary school all the way to senior year of high school. While you probably learned these formulas at some point, it may have been a while since you’ve had to use them. This is part of what makes the SAT challenging: since it draws on many different types of math, you need to get out of the high school math mindset (where you only remember what you’ve learned the last month or so to ace the test) and review the math you’ve learned over the years. If you don’t do any studying for the SAT, you’ll take longer to recall certain helpful formulas and concepts. While there’s always more than one way to reach the right answer, being able to quickly remember these math facts will help you answer questions more efficiently and minimize careless mistakes. We’ve categorized these formulas to help you focus your preparation, and we’ve provided a quick review of what each concept is. ## Arithmetic and Algebra ### 1. Slope-Intercept Form of a Line $$y=mx+b$$ The $$m$$ in the equation represents the slope of the equation, and the $$b$$ represents the y-value of the y-intercept. For example, if we have the equation $$y=2x+4$$, then the slope is $$2$$ and the y-intercept is $$(0,4)$$. ### 2. Vertex Form of a Parabola/Quadratic $$y=a(x-h)^2+k$$ You may be more familiar with a quadratic in its factored form, or in the form $$y=ax^2+bx+c$$. However, you should be able to recognize vertex form and convert quadratics to this form for the SAT. The values of $$h$$ and $$k$$ give you the coordinates of the vertex, $$(h,k)$$. ### 3. Distance Formula $$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$ The distance formula is derived from the Pythagorean Theorem (covered later in this post) and it’s useful for quickly finding the distance between two points. Take the values of the coordinates and plug them into this formula to find the distance, and be sure to apply the squares and the square root at the right step. $$x=\frac{-b\:\pm\:\sqrt{b^2-4ac}}{2a}$$ The quadratic formula helps you find the roots of a quadratic equation (parabola) if you can’t easily factor it. You need the quadratic to be in the form $$y=ax^2+bx+c$$, and then you simply plug the coefficients and constants into the formula. Note that you will get two answers because there is the plus and minus sign in the numerator. ### 5. Exponent Rule (Multiplication) $$a^na^m=a^{n+m}$$ Knowing how to manipulate exponents in a variety of ways will help you tremendously on the SAT, especially the no-calculator portion. In this case, if you have the same base number raised to different powers being multiplied together, you can add the exponents together. ### 6. Exponent Rule (Division) $$\frac{a^m}{a^n}=a^{m-n}$$ Similar to the multiplication rule above, if you have the same base number raised to different powers being divided, you can subtract the exponents. You can also rewrite the expression on the right to mirror the one on the left. ### 7. Exponent Rule (Power Raised to a Power) $$(a^n)^m=a^{n\:\cdot\:m}$$ To continue the exponent rules, raising a power to another power is the same as multiplying the exponents together. This is not an exhaustive list of ways to manipulate exponents, so if you find that you don’t remember these at all, be sure to brush up! ### 8. Binomial Product 1—Difference of Squares $$(x-y)(x+y)=x^2-y^2$$ The best times to recognize the binomial products and quickly factor them is on the no-calculator section. You don’t have to FOIL or use any other method—you can quickly convert from the factored form to the expanded form on sight. The difference of squares is used often by SAT test makers in a variety of contexts. ### 9. Binomial Product 2—Perfect Squares Trinomial (Positive) $$(x+y)^2=x^2+2xy+y^2$$ Students sometimes forget the perfect squares trinomial once they’ve left their algebra class, but this is also a good one to recognize. It saves you time because you can quickly convert from one form to another, but it’s a little more difficult to catch than the expanded difference of squares. A good way to know if you’re dealing with one is to look at the first and last values—are they perfect squares? ### 10. Binomial Product 2—Perfect Squares Trinomial (Negative) $$(x-y)^2=x^2-2xy+y^2$$ This is similar to the above trinomial, except that the quantity involves subtraction rather than addition. While the factored form doesn’t include coefficients, the binomial products on the SAT often do. Practice recognizing these patterns by inputting coefficients in front of $$x$$ and constants for $$y$$ on the left-hand side. Then multiply out the expression to see how the pattern works with different combinations. ### 11. Complex Conjugate $$(a+bi)(a-bi)=a^2+b^2$$ On most SAT Math tests, there will be at least one question that involves manipulating imaginary numbers. The complex conjugate allows you to get rid of the imaginary part of a complex number and leaves with you a real number (notice how it resembles the difference of squares!). When given a complex number in the form $$a+bi$$, the conjugate is $$a-bi$$. ### 12. Exponential Growth and Decay $$y=a(1\:\pm\:r)^x$$ This is technically two different equations, one where there is a plus in the equation and one where there is a minus. Knowing the general format of exponential equations will help you on several SAT questions, as you may need to interpret or manipulate these equations. The value $$a$$ is the initial value, $$r$$ is the rate of growth when it’s positive and the rate of decay when it’s negative. As part of our free guidance platform, our Admissions Assessment tells you what schools you need to improve your SAT score for and by how much. Sign up to get started today. ## Ratios, Percentages, and Statistics ### 13. Simple interest $$A=Prt$$ This one appears less often than compound interest on the SAT, but it still shows up, so it’s worth knowing. $$P$$ represents the principal amount, $$r$$ is the interest rate expressed as a decimal, and $$t$$ is for time, usually in years. ### 14. Compound interest $$A = P(1 + \frac{r}{n})^{nt}$$ The good news is that $$P$$, $$r$$, and $$t$$ mean the same thing in this equation as they do in simple interest. The $$n$$ represents the number of times that the interest is compounded during $$1\:t$$. For example, if the interest is compounded quarterly over the course of a year, then $$n=4$$. ### 15. Average/Mean In math, the words average and mean are the same thing: the number you get when you take the sum of a set and divide it by the number of values in the set. You could also think of it as the sum divided by the count. You should know how to calculate an average and interpret it. Be sure to understand the difference between mean and median. ### 16. Random Sampling This isn’t technically a formula, but many of the statistics-based problems on the SAT focus more on interpreting concepts in context rather than performing mathematical operations. Random sampling is when you select participants for a study at random within your population. It ensures that your study is representative of the population. ### 17. Random Assignment Random assignment is when the participants in a study are assigned a treatment or trial at random. It reduces bias in your study, and means that you can attribute causation in regards to the treatment. On the SAT, you’re often asked about what will reduce bias, or how much you can generalize results to the rest of the population. In these instances, you need to identify random sampling and random assignment. ### 18. Standard deviation You won’t need to calculate standard deviation for the SAT, but you will be tested on it conceptually, as with with random sampling and random assignment. Standard deviation is the measure of spread in the data set. A higher standard deviation means greater spread, and lower standard deviations mean smaller spread. You’ll need to know how changes in the data set might affect the standard deviation by making it greater or smaller. ## Geometry and Trigonometry ### 19. Area of an Equilateral Triangle $$A=\frac{\sqrt{3}s^2}{{4}}$$ The regular area of a triangle formula is provided on the SAT reference sheet, but it requires that you know the height of a triangle. Sometimes you aren’t given the height and you’ll need to calculate it, but you can quickly find the area of an equilateral triangle by plugging the length of one of its sides into the formula above. No need to calculate the height! ### 20. Equation of a Circle $$(x-h)^2+(y-k)^2=r^2$$ There is usually one question involving the equation of a circle. In this equation, $$(h,k)$$ is the coordinate for the center of the circle, and $$r$$ is the radius of the circle. ### 21. Sine Ratio Some students get nervous when they hear that trig is on the SAT, but it most often appears in the form of trig ratios. Remember that for a given angle in a right triangle, the value of sine is the length of the opposite side divided by the length of the hypotenuse, or opposite/hypotenuse. ### 22. Cosine Ratio Just like with sine, remember what the cosine ratio is: the length of the adjacent side divided by the length of the hypotenuse, or adjacent/hypotenuse. ### 23. Tangent Ratio Last but not least, the tangent ratio is the length of the opposite side divided by the length of the adjacent side, or opposite/adjacent. Some students find the mnemonic SOH CAH TOA helpful for remembering trig ratios. While the most common form of trig are the basic ratios, you may encounter things like the unit circle or more advanced math. If you need to convert degrees to radians, multiply the degrees by $$\frac{\pi}{180}$$. If you need to convert radians to degrees, multiply the radians by $$\frac{180}{\pi}$$. ### 25. Pythagorean Theorem $$a^2+b^2=c^2$$ The Pythagorean Theorem applies to right triangles, and allows you to solve for one of the side lengths given any other side length. $$a$$ and $$b$$ are the legs of the triangle, and $$c$$ is the hypotenuse. ### 26. Regular Polygon Interior Angle $$\frac{(n-2)180}{n}$$ The SAT will probably involve one question with a regular polygon that isn’t a triangle or square. Regular polygons have unique and consistent properties based on their number of sides, and knowing these properties can help you solve these problems. This equation tells you what the degree measure at each angle is based on the number of sides $$n$$. ### 27. 3-4-5 triangle The SAT provides you with two special right triangles you may already be familiar with on your reference sheet—the 30-60-90 and 45-45-90 triangles. However, the 3-4-5 is a special right triangle with sides that are straightforward integers. This triangle is often incorporated into SAT problems, especially the no-calculator portion, so be on the lookout for it! It can save you having to use the Pythagorean theorem. ### 28. 5-12-13 triangle Another special right triangle with whole-number sides, the 5-12-13 triangle is less well-known and shows up less often than 3-4-5. Still, it helps to be able to quickly solve the remaining sides without the Pythagorean theorem, so check for these numbers or their multiples in triangle problems. ### 29. Length of Arc in a Circle $$length\:of\:arc = \frac{central\:angle}{360}\pi d$$ Although geometry questions don’t make up a huge portion of the SAT, you may still find a question either about arcs or sectors in a circle. An arc is the length between two points on a circle, usually measured by extending two radii from the center of the circle with an angle formed between them. You can use the degree measure of the arc as a fraction of $$360$$ and multiply it by the equation for the circumference to find the length of the arc. ### 30. Area of Sector in a Circle $$area\:of\:sector=\frac{central\:angle}{360}\pi r^2$$ Like an arc, the sector is the area in between two radii extending from the circle, sort of like a slice of pie. Again, multiply the degree measure as a fraction of $$360$$ and multiply it by the equation for the area of a circle to find the area of the sector. ## Wrapping it Up Before you go, we’re going to offer you a bonus tip: you may want to memorize the perfect squares and perfect cubes. This can help you with quadratic equations that often involve squares, and cubes are often used in solving problems with exponents. Memorizing these will cut down on your need to do math with scratch paper or calculator. The best way to be able to remember formulas is to practice using them. Unlike your high school math test, where you know what topics will be covered, the SAT will simply present you with a questionit’s up to you to determine what formulas apply. When you practice using formulas with a variety of problems, you’ll be able to quickly identify which formula to use. Preparing for the SAT? Download our free guide with our top 8 tips for mastering the SAT. Check out some of our other posts on math prep: ### Want more SAT tips sent to you? Sign up below and we'll send you expert SAT tips and guides. Gianna Cifredo Blogger at CollegeVine Short bio Gianna Cifredo is a graduate of the University of Central Florida, where she majored in Philosophy. She has six years of higher education and test prep experience, and now works as a freelance writer specializing in education. She currently lives in Orlando, Florida and is a proud cat mom. Other articles by Gianna
Construct An Angle Bisector In these lessons, we will learn • how to construct an angle bisector of a given angle. • how to use an angle bisector to construct some angles for example, 90 degrees, 45 degrees, 60 degrees, 30 degrees, 120 degrees, 135 degrees, 15 degrees. Related Pages Types Of Angles Loci More Geometry Lessons An angle bisector is a straight line that divides the angle into two equal parts. Example: Construct an angle bisector for the following angle: Solution: Step 1: Put the sharp end of your compasses at point B and make one arc on the line BC (point S) and another arc on line AB (point T). Step 2: Put the sharp end of the compasses at S and make an arc within the lines AB and BC. Do the same at T and make sure that the second arc intersects the first arc. Step 3: Draw a line from point B to the points of intersection of the 2 arcs. This line bisects ∠ABC. The steps to construct an angle bisector can be summarized as follows: 1. From the vertex, draw an arc across both rays of the angle. 2. From each arc intersection draw another pair of arcs that intersect each other. 3. Draw a line from the vertex to the intersection point to form the angle bisector. How to draw an angle bisector using a compass and a straight edge? 1. Place compass point on the vertex, and draw an arc across each ray. 2. Place the compass on each arc intersection and draw a further pair of arcs which intersect each other. 3. Use a straight edge to connect the intersection point to the vertex. How to bisect a given angle using only a compass and straightedge? Use Angle Bisector To Construct Angles We can use the angle bisector to construct some other angles from existing angles. For examples: A 30˚ angle can be obtained by bisecting a 60˚ angle. A 15˚ angle can be obtained by bisecting a 30˚ angle. A 45˚ angle can be obtained by bisecting a 90˚ angle. A 22.5˚ angle can be obtained by bisecting a 45˚ angle. Example: The figure shows a point A on a straight line. Construct an angle of 45˚ at point A. Solution: Construct a 90˚ angle, and then construct an angle bisector to obtain a 45˚ angle. How to construct 30, 45, 60, 90, and 120 degree angles with a compass by constructing angle bisectors? How to construct a 75° angle by constructing a 60° angle and a 15° angle? Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
## Order of Operations The “Order of Operations” is a convention that ensures everyone interprets algebraic notation the same way. It is used in conjunction with the properties of each operation (addition, subtraction, multiplication, division, and exponentiation). I summarize the order of operations for myself as “do the most powerful operations first”. The answer shown below is wrong. Try working your way through the problem backwards, from the answer up. As you find each mistake, try to identify the thinking behind it… what perspective led to the mistake? What should have been done instead? $3+2(15 \div 3-1)^2$ $=5(15 \div 3-1)^2$ $=(75 \div 15-5)^2$ $=(5-5)^2$ $=0^2$ $=0$ . .. Hint 1: The correct answer to the original problem is: $35$ Hint 2: What if the problem had been written $3+2\cdot (15 \div 3-1)^2$ Hint 3: Two mistakes have been made in the work above ## Collecting Like Terms #1 “Collecting Like Terms” is a phrase that is used often in mathematics, yet it is a process that can feel a bit arbitrary at first as it relies on several algebraic principles. The answer shown below is wrong. Try working your way through the problem backwards, from the answer up. As you find each mistake, try to identify the thinking behind it… what perspective led to the mistake? What should have been done instead? $3x+5-4x+2$ $=8x-4x+2$ $=4x-2$ . .. Hint 1: The correct answer to the original problem is: $-x+7$ Hint 2: What if the problem had been written $3x+4+~^-4x+2$ or perhaps $3\cdot x+4+(-4)\cdot x+2$ Hint 3: Two mistakes have been made in the work above ## Sums and Products with Exponents Powers and roots can be distributed over products and quotients. They may not be distributed over sums or differences, no matter how tempting it may be. Sums or differences raised to a power must be used as a factor the indicated number of times, then multiplied using the distributive property. An exponent applies only to the factor immediately below it unless parentheses have been used to indicate otherwise. “Like terms” have the same variables, to the same powers… and only “like” terms may be combined by adding their coefficients. The answer shown below is wrong. Try working your way through the problem backwards, from the answer up. As you find each mistake, try to identify the thinking behind it… what perspective led to the mistake? What should have been done instead? $(f-g^2)^2-fg^2-(-fg)^2-f^2$ $=f^2-g^4-fg^2-(-fg)^2-f^2$ $=f^2-g^4-f^2g^2-(-fg)^2-f^2$ $=f^2-g^4-f^2g^2+fg^2-f^2$ $=-g^4$ . .. Hint 1: The correct answer to the problem on the first line above is: $g^4-3fg^2-f^2g^2$ Hint 2: The work shown above contains sign errors, simplification errors, and exponentiation errors. Re-read the text at the top of the posting if you have not found them all… ## Distributes Over… The “distributive property” of multiplication and division. That’s not the proper full name, but it’s what many people say… so, when do you distribute, and when don’t you? That is the question. The answer shown below is wrong. Try working your way through the problem backwards, from the answer up. As you find each mistake, try to identify the thinking behind it… what perspective led to the mistake? What should have been done instead? $\dfrac{(4)(3)(2k+6)}{6}$ $=\dfrac{(12)(8k+24)}{6}$ $=\dfrac{(12)(8k+24)}{(2)(3)}$ $=\dfrac{(6)(4k+12)}{3}$ $=(2)(4k+4)$ $=8k+4$ . .. Hint 1: The correct answer to the original problem is: $4k+12$ Hint 2: The complete description of the distributive property of multiplication is: “the distributive property of multiplication over…”? Hint 3: Several mistakes have been made in the work above – don’t just seek the correct answer – look for mistakes between every pair of lines. ## Negative Exponents Negative exponents can be another source of confusion. No matter where you find a negative exponent, you can turn it into a positive exponent by taking the reciprocal of the expression it applies to. The answer shown below is wrong. Try working your way through the problem backwards, from the answer up. As you find each mistake, try to identify the thinking behind it… what perspective led to the mistake? What should have been done instead? $\dfrac{c^{-2}d^3}{c^3d^{-4}}$ $=\dfrac{d^3d^4}{c^{-2}c^3}$ $=\dfrac{d^7}{c^{-6}}$ $=d$ . .. Hint 1: The correct answer to the original problem is: $\dfrac{d^7}{c^5}$ Hint 2: Several mistakes have been made in the work above – don’t just seek the correct answer – look for the mistakes between every pair of lines. ## Laws of Exponents The “laws of exponents” are a frequent source of errors. The rules that apply when simplifying expressions that involve exponents can be figured out quickly on your own if you (in your mind’s eye) expand integral exponents into repeated multiplication. I encourage students to master being able to explain why each of these rules is as it is instead of memorizing them, as memorized versions are more likely to get jumbled together in your thinking when working problems. The answer shown below is wrong. Try working your way through the problem backwards, from the answer up. As you find each mistake, try to identify the thinking behind it… what perspective led to the mistake? What should have been done instead? $\dfrac{(ab^2)^3(5a^2)}{a^2b}$ $=\dfrac{a^5b^525a^2}{a^2b}$ $=\dfrac{25a^{10}b^5}{a^2b}$ $=25a^5b^5$. . .. Hint 1: The correct answer to the original problem is: $5a^3 b^5$ Hint 2: Several mistakes have been made in the work above – don’t just seek the correct answer – look for the mistakes between every pair of lines. ## Negative Signs and Fractions Negative signs bother many students, particularly when they are followed by fractions. In such situations, it is important to remember that the vinculum (the horizontal line between numerator and denominator) serves as a grouping symbol – like parentheses would. I recommend that students always put a numerator with more than one term in parentheses before bringing a leading negative sign into the numerator. The answer shown below is wrong. Try working your way through the problem backwards, from the answer up. As you find each mistake, try to identify the thinking behind it… what perspective led to the mistake? What should have been done instead? $\dfrac{2-w}{3}-\dfrac{2w-5}{2}$ $=\dfrac{2-w}{3}\cdot\dfrac{2}{2}-\dfrac{2w-5}{2}$ $=\dfrac{2(2-w)}{6}+\dfrac{-2w-5}{6}$ $=\dfrac{4-w-2w-5}{6}$ $=\dfrac{-1-3w}{6}$ . .. Hint 1: the correct answer to the original problem is: $\dfrac{-8w+19}{6}$ Hint 2: several mistakes have been made in the work above – they are not all sign or distribution errors ## Negative Signs and Parentheses The distributive property is a source of many mistakes in algebra. When a negative sign, multiplication, or division is next to a set of parentheses, everything inside the parentheses will be affected. Negative signs in particular bother many students. The most reliable approach is probably to rewrite subtraction as the addition of a negative, then distribute the negative sign to all terms inside the parentheses. The answer shown below is wrong. Try working your way through the problem backwards, from the answer up. As you find each mistake, try to identify the thinking behind it… what perspective led to the mistake? What should have been done instead? $(1-d+e)-(2d-3e+4)-(-4e+3d-2)$ $=(1-d+e)-(-2d+3e-4)+(4e-3d-2)$\$ $=1-d+e+2d-3e+4+4e-3d-2$ $=-d+2d-3d+e-3e+4e+1+4-2$ $=-6d+2e+3$ . .. Hint 1: the correct answer to the original problem is: $8e-6d-1$ Hint 2: a number of mistakes have been made in the work above – they are not all sign errors. Identify all of the errors made at each step and the likely reasons they were made before solving the problem correctly yourself. ## Simplifying Algebraic Fractions #2 The rules of fractions are no different for expressions that involve variables than they are for expressions involving only constants. Many students make errors when combining and simplifying fractions, sometimes because they are so used to working with equations that they have forgotten that some steps only work with equations. If you wish to simplify a fraction, you may only do so if the term to be cancelled is a factor of both the numerator and the denominator. The answer shown below is wrong. Try working your way through the problem backwards, from the answer up. As you find each mistake, try to identify the thinking behind it… what perspective led to the mistake? What should have been done instead? $\dfrac{3p}{5}+\dfrac{2+5p}{7}$ $=5\cdot\dfrac{3p}{5}+\dfrac{2+5p}{7}$ $=3p+\dfrac{2+5p}{7}$ $=3p+7\cdot \dfrac{2+5p}{7}$ $=3p+2+5p$ $=8p+2$ . .. Hint 1: The correct answer to the problem on the first line is $\dfrac{46p+10}{35}$ Hint 2: The same type of error is made twice in the work above, but even if it were a valid approach to simplifying this expression, another type of error was also made. ## Simplifying Algebraic Fractions #1 The rules of fractions are no different for expressions that involve variables than they are for expressions involving only constants. Many students make errors when simplifying fractions, most often by “cancelling out” terms that are not factors of both the numerator and the denominator. If you wish to simplify a fraction, you may only do so if the term to be cancelled is a factor of both the numerator and the denominator. The answer shown below is wrong. Try working your way through the problem backwards, from the answer up. As you find each mistake, try to identify the thinking behind it… what perspective led to the mistake? What should have been done instead? $\dfrac{2m+6}{2m}+\dfrac{3-9m}{3m}$ $= 6+\dfrac{3-9m}{3m}$ $=6+\dfrac{3(3m)}{3m}$ $=6+3$ $=9$ . .. .. Hint 1: The correct answer to the problem in the first line above is $\dfrac{-2m+4}{m}$ Hint 2: Two mistakes have been made in the work shown above.

Subsets and Splits

No community queries yet

The top public SQL queries from the community will appear here once available.