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Transformations - ESSENTIAL ALGEBRA 2 SKILLS - SAT Test Prep ## CHAPTER 11ESSENTIAL ALGEBRA 2 SKILLS ### Lesson 3: Transformations Functions with similar equations tend to have similar shapes. For instance, functions of the quadratic form have graphs that look like parabolas. You also should know how specific changes to the function equation produce specific changes to the graph. Learn how to recognize basic transformations of functions: shifts and reflections. To learn how changes in function equations produce changes in their graphs, study the graphs below until you understand how graphs change with changes to their equations. Horizontal Shifts The graph of is simply the graph of shifted k units to the left. Similarly, the graph of is the graph of y = f (x) shifted k units to the right. The graphs below show why. Vertical Shifts The graph of is simply the graph of shifted k units up. Similarly, the graph of is the graph of shifted k units downward. The graphs below show why. Reflections When the point (3, 4) is reflected over the y- axis, it becomes (–3, 4). That is, the x coordinate is negated. When it is reflected over the x- axis, it becomes (3, –4). That is, the y coordinate is negated. (Graph it and see.) Likewise, if the graph of is reflected over the x-axis, it becomes . Concept Review 3: Transformations 1. What equation describes the function after it has been shifted to the right five units? 2. What equation describes the function after it has been reflected over the x- axis? 3. How does the graph of compare with the graph of ? 4. What specific features do the graphs of and have in common? 5. What specific features do the graphs of and have in common? 6. The quadratic function h is given by , where a is a negative constant and c is a positive constant. Which of the following could be the graph of ? 7. The figure above is a graph showing the depth of water in a rectangular tank that is being drained at a constant rate over time. Which of the following represents the graph of the situation in which the tank starts with twice as much water, and the water drains out at twice the rate? SAT Practice 3: Transformations 1. The shaded region above, with area A, indicates the area between the x-axis and the portion of that lies above the x- axis. For which of the following functions will the area between the x- axis and the portion of the function that lies above the x-axis be greater than A? 2. The figure above shows the graph of the function , which has a minimum value at the point (1, –2). What is the maximum value of the function (A) 7 (B) 6 (C) 5 (D) 4 (E) It cannot be determined from the information given. 3. A point is reflected first over the line , then over the x-axis, and then over the y-axis. The resulting point has the coordinates (3, 4). What were the coordinates of the original point? (A) (3, 4) (B) (–3,-4) (C) (3, –4) (D) (–4, –3) (E) (4, 3) 4. In the figure above, point Q is the reflection of point P over the line , and point R is the reflection of point Q over the line . What is the length of line segment PR? (A) 10 (B) 11 (C) 12 (D) 13 (E) 14 5. If the functions f (x), g (x), and h (x) are defined by the equations , , and , then which of the following represents the graph of Concept Review 3 1. . Although the –5 seems to suggest a shift to the left (because when we subtract 5 from a number, we move five units to the left on the number line), this change actually shifts the graph to the right. To see why, look back at the first two examples in Lesson 3, and pay particular attention to how the changed equations produce the individual points on the graph and how these points compare with the points on the original graph. It also may help to pick a simple function, such as , graph it by hand (by choosing values for x, calculating the corresponding values for y, and plotting the ordered pairs), and then graph in the same way to see how the graphs compare. 2. . Since the point (x, –y) is the reflection of (x, y) over the x-axis, reflecting any function over the x- axis simply means multiplying y by –1. This “negates” every term in the function. 3. It is the original graph after it has been “flipped” vertically and “stretched” vertically. The graph of is a “vertically stretched” version of that also has been reflected over the x- axis. Every point on is four times farther from the x- axis as its corresponding point on and is also on the opposite side of the x- axis. 4. Overall shape and maximum and minimum values. The graph of is simply the graph of shifted to the left 15 units. It maintains the shape of the original graph and has the same maximum and minimum values. (That is, if the greatest value of y on is 10, then the greatest value of y on is also 10.) 5. Zeroes, vertical lines of symmetry, and x coordinates of maximum and minimum values. The graph of is simply the graph of that has been “vertically stretched” by a factor of 6. Imagine drawing the graph of on a rubber sheet and then attaching sticks across the top and bottom of the sheet and pulling the sheet until it”s six times as tall as it was originally. The stretched graph looks like . Although most of the points move because of this stretch, the ones on the x-axis do not. These points, called the zeroes because their ycoordinates are 0, remain the same, as do any vertical lines of symmetry and the x coordinates of any maximum or minimum points. 6. The graph of , since it is quadratic, looks like a parabola. If a is negative, the parabola is “open down.” (Remember that is the graph of the “open up” parabola after it has been “flipped” over the x- axis.) Also, notice that c is the “y-intercept” of the graph, since . Therefore, the graph is an upside-down parabola with a positive y-intercept. The only choice that qualifies is (B). 7. The point on the d-axis represents the starting depth of the water. If the tank begins with twice as much water, the starting point, or “d-intercept,” must be twice that of the original graph. Also, if the water drains out at twice the rate, the line must be twice as steep. Since twice as much water drains out at twice the rate, the tank should empty in the same amount of time it took the original tank to drain. The only graph that depicts this situation is (A). SAT Practice 3 1. D The transformation in (D) is a shift of the original function upward two units. This creates a triangular region above the x-axis with a greater height and base than those of the original graph, and therefore creates a greater overall area. The transformation in (A) will create a triangle with area 1/2 A, the transformations in (B) and (C) are horizontal shifts, and so will not change the area. The downward shift in (E) will reduce the height and base, and therefore the total area. 2. C The function h (x) is equivalent to the function g (x) after it has been reflected over the x-axis, vertically stretched by a factor of 3, and shifted downward one unit. After these transformations, the vertex of the parabola will be at (1, 5), so the maximum value is . 3. D Call the original point (a, b). Its reflection over the line is (b, a). (Draw a graph to see.) The reflection of this point over the x-axis is (b, -a), and the reflection of this point over the y-axis is (-b, -a). If the final point is (3, 4), then the original point was (–4, –3). 4. A A point and its reflection over a line are both equidistant to that line. Imagine that point Q has a y coordinate of 4 (any value between 1 and 6 will do). This implies that point Q is two units from the line , and therefore, point P also must be two units from the line and must have a y coordinate of 8. Point Q also must be three units from the line , so point R also must be three units from the line and must have a y coordinate of –2. Therefore, the length of PR is . 5. E You can determine the equation defining the function through substitution: , which describes an “open-down” parabola with vertex at (0,–1). Notice that this sequence of transformations takes the standard parabola and shifts it up one unit and then reflects the new graph over the x- axis.
# Cool Problems in Probabilistic Number Theory and Set Theory Cool Problems in Probabilistic Number Theory and Set Theory Here are a few off-the-beaten-path problems at the intersection of computer science (algorithms), probability, statistical science, set theory, and number theory. While they can easily be understood by beginners, finding a full solution to some of them is not easy, and some of the simple but deep questions below won’t be answered for a long time, if ever, even by the best mathematicians living today. In some sense, this is the opposite of classroom exercises, as there is no sure path that lead to full solutions. I offer partial solutions and directions, to help solve these problems. Mathematical tessellation artwork 1. A Special Number Here we try to construct an irrational number x that has 50% of zero’s, and 50% of one’s, in its binary representation (digits in base 2.) To this day, no one knows whether any classic mathematical constant (Pi, e, log 2, SQRT(2) and so on) has such a uniform distribution of 0 and 1 in base 2, or any other base. The number is constructed as follows. Let us denote as S(n) the set of strictly positive integers that are divisible by p(1) or p(2) or … or p(n) where the p(k)’s are prime numbers to be determined later. We formally define x as the limit, when n tends to infinity, of x(n), with The prime numbers p(1), p(2) and so on, are chosen such that x has 50% zero’s, and 50% one’s, in its base 2 representation. Which prime numbers we should choose is a relatively easy problem, achieved using a version of the greedy algorithm as follows: p(1) can not be 2 (why?) and the smallest value it can possibly be is p(1) = 3. Let’s pick p(1) = 3. Now, one third of the digits of x(1) are equal to 1. p(2) = 5 works. So let’s pick p(2) = 5. Now the proportion of 1 in the digits of x(2), is 7/15 still lower than 50%. p(3) = 7 does not work because x(3) would have more than 50% of its digits equal to 0. The smallest prime to choose next is p(3) = 17. With this choice, the proportion of 1 in the digits of x(3), is 0.498039216. The smallest next prime that works, is p(4) = 257. With this choice, the proportion of 1 in the digits of x(4), is 0.49999237. Now we have an algorithm to find the p(k)’s. It is easy to prove (see here) that the proportion of 1 in x(n), denoted as r(n), is equal to Note that this formula is correct only if the p(k)’s have no common divisors (why?), thus our focus is on prime numbers only. The fact that the above product, if taken over all prime numbers rather than those that we picked, is diverging, actually guarantees that we can grow our list of primes indefinitely, and at the limit as n tends to infinity, r(n) tends to 50%, as desired. Also note that for any fixed value of n, x(n) is a rational number and can be computed explicitly. Questions Can you find the first 10 values of p(n)? What about the first 50 values? (you will need high precision computing for that) Is the number x constructed here, rational or not? Compute the first 10 decimals (in base 10) of x. Compute the exact values of x(n) up to n = 5. Is the digit distribution of x truly random? The answer to the last question is no. The digits exhibit strong auto-correlations, by construction. It is also easy to prove that x is not a normal number, since sequences of digits such as “100001” never appear in its base 2 representation. 2. Another Special Number Again, the purpose here is to construct an irrational number x between 0 and 1, with the same proportion of digits equal either to 1 or 0, in base 2. I could call it the infinite mirror number, and it is built as follows, using some kind of mirror principle: The first digit is 0. First 2 digits: 0 \| 1 First 4 digits: 0, 1 \| 1, 0 First 8 digits: 0, 1, 1, 0 \| 1, 0, 0, 1 First 16 digits: 0, 1, 1, 0, 1, 0, 0, 1 \| 1, 0, 0, 1, 0, 1, 1, 0 And so on. Can you guess the pattern? Solution: at each iteration, concatenate two strings of digits: the one from the previous iteration, together with the one from the previous iteration with 1 replaced by 0 and zero replaced by 1. This leads to the following recursion, starting with x(1) = 0: That is, The number x, defined as the limit of x(n) as n tends to infinity, share a few properties with the number discussed in the first problem: x(n) is always a rational number, easy to compute exactly. The proportion of digits of x, equal to 0 in base 2, is 50% by construction. The number x is not normal: the sequence of digits “111” never appears in its binary expansion. Is x is a rational number? I could not find any period in the first few hundred digits of x in base 2; it makes me think that this number is probably irrational. Can you compute the first 10 decimals of x, in base 10? Finally, the digit 0 (in base 2) appears in the following positions: 1, 4, 6, 7, 10, 11, 13, 16, 18, 19, 21, 24, 25, 28, 30, 31, 34, 35, 37, 40, 41, 44, 46, 47, 49, 52, 54, 55, 58, 59, 61, 64, 66, 67, 69, 72, 73, 76, 78, 79, 81, 84, 86, 87, 90, 91, 93, 96, 97, 100, 102, 103, 106, 107, 109, 112, 114, 115, 117, 120, 121, 124, 126, 127, … Do you see any patterns in that sequence? Another number with digits uniformly distributed, although in base 10, is the Champernowne constant, defined as 0.12345678910111213… That number is even normal. Yet unlike Pi or SQRT(2), it fails many tests of randomness, see the section “failing the gap test” in chapter 4 in my book. Rather than using the concept of normal number, a better definition of a number with “random digits” is to use my concept of “good seed” (discussed in the same book) which takes into account the joint distribution of all the digits, not just some marginals. Note that for some numbers, the distribution of the digits does not even exist. See chapter 10 in the same book for an example. 3. Representing Sets by Numbers or Functions Here we try to characterize a set of strictly positive integer numbers by a real number, or a set of real numbers by a function. If S is a set of strictly positive integers, one might characterize S by the number f(S) = x, defined as follows:However, this approach has some drawbacks: for instance, f({1}) = f({2,3,4,5,6,…}) = 1/2. In short, all finite sets have a number representation that is also matched by an infinite set. How can we walk around this issue? If instead of using the representation using the base 2 system as above, we use the logistic map representation (see chapter 10 in my book), would this solve the issue? That is, in the logistic map numeration system, where digits are also equal to either 0 or 1, can two different sets be represented by the same number? Let’s turn now to sets consisting of real numbers. But first, let’s mention that there is enough real numbers in [0, 1] to characterize all sets consisting only of integer numbers: in short, in principle, there is a one-to-one mapping between sets of integers, and real numbers. Unfortunately, there isn’t enough real numbers to characterize all sets of real numbers, thus we can not uniquely characterize a set of real numbers, by a real number. However there is enough real-valued functions to characterize all of them uniquely. See this article for an introduction on the subject. The first idea that comes to my mind, to characterize a measurable bounded set S of real numbers, is to use the characteristic function of a uniform distribution on S. Two different sets will have two different characteristic functions. It is not difficult to generalize this concept to unbounded sets, but how can you generalize it to non measurable sets? Also, many characteristic functions won’t represent a set; for instance, in this framework, no set can have the characteristic function of a Gaussian distribution. By contrast, for sets of strictly positive integer numbers, any real number in [0, 1] clearly represents a specific set. How to go around this issue? Another interesting issue is to study how operations on sets (union, intersection, hyper-logarithm, and so on) look like when applied to the number or real-valued function that characterize them. The next tricky question of course, is how to represent sets of real-valued functions, by a mathematical object. There isn’t enough characteristic functions, by far, to uniquely characterize these sets. Operators, a branch of functional analysis, might fit the bill. These are functions whose argument is a function itself, such as the integral or derivative operators. Related Articles One Trillion Random Digits Curious Mathematical Problem Another Off-the-beaten-path Data Science Problem Two More Math Problems: Continued Fractions, Nested Square Roots, D… Difficult Probability Problem: Distribution of Digits in Rogue Systems Little Stochastic Geometry Problem: Random Circles Curious Mathematical Object: Hyperlogarithms 88 percent of all integers have a factor under 100 Fractional Exponentials – Dataset to Benchmark Statistical Tests Two Beautiful Mathematical Results – Part 2 Two Beautiful Mathematical Results Four Interesting Math Problems Number Theory: Nice Generalization of the Waring Conjecture Fascinating Chaotic Sequences with Cool Applications Representation of Numbers with Incredibly Fast Converging Fractions Yet Another Interesting Math Problem – The Collatz Conjecture Simple Proof of the Prime Number Theorem Factoring Massive Numbers: Machine Learning Approach Representation of Numbers as Infinite Products A Beautiful Probability Theorem Fascinating Facts and Conjectures about Primes and Other Special Nu… Three Original Math and Proba Challenges, with Tutorial For related articles from the same author, click here or visit www.VincentGranville.com. Follow me on on LinkedIn, or visit my old web page here. DSC Resources Invitation to Join Data Science Central Free Book: Applied Stochastic Processes Comprehensive Repository of Data Science and ML Resources Advanced Machine Learning with Basic Excel Difference between ML, Data Science, AI, Deep Learning, and Statistics Selected Business Analytics, Data Science and ML articles Hire a Data Scientist \| Search DSC \| Classifieds \| Find a Job Post a Blog \| Forum Questions Link: Cool Problems in Probabilistic Number Theory and Set Theory
Chapter Notes: Data Handling # Data Handling Class 7 Notes Maths Chapter 4 ## What is an Average? An average is a way to find out what the middle value or typical amount is when you have different numbers. It's like finding the "fair" amount. Example: Studying Hours Let's say you studied for different amounts of time each day this week. Here's how many hours you studied each day: • Monday: 2 hours • Tuesday: 3 hours • Wednesday: 1 hour • Thursday: 4 hours • Friday: 2 hours To find the average number of hours you studied each day, you follow these steps: 1. Add Up All the Hours: Add the number of hours you studied each day. $2+3+1+4+2=$12 hours 2. Count the Number of Days: There are 5 days in this example. 3. Divide the Total Hours by the Number of Days: Divide the total hours by the number of days to find the average. So, the average number of hours you studied each day is 2.4 hours. This brings us to the realization that the average is a number that represents the central tendency of a group of observations or data. It falls between the highest and lowest values in the data set, serving as a measure of central tendency. Different types of data require different representative values. Let’s look at some of the different representative values. Measures of Central Tendency ### Arithmetic Mean The most commonly used representative value for a group of data is known as the arithmetic mean or simply the mean. It lies between the highest and the lowest value of data. In other words, The average or arithmetic mean or mean of a given data is defined as : Mean = Sum of all observations / Number of observations Steps to calculate Mean: Step 1: Add all the observations together. Step 2: Divide the sum by the number of observations used. To better understand this concept, let's consider the following example: Example 1: We have two vessels containing 20 litres and 60 litres of milk respectively. If we were to divide the milk equally between both vessels, what amount would each vessel have? Sol: In the above case, the average or the arithmetic mean would be: = Total quantity of milk / Number of vessels = (20 + 60) / 2 litres = 40 litres. Thus, each vessel would have 40 litres of milk. Example 2: A batsman scored the following number of runs in six innings: 36, 35, 50, 46, 60, 55 Calculate the mean runs scored by him in an inning. Sol: Total runs = 36 + 35 + 50 + 46 + 60 + 55 = 282. In the above case, the average or the arithmetic mean would be: = Sum of all observations / Number of observations Therefore, in this case, mean = 282 / 6 = 47. Thus, the mean runs scored in an inning is 47. Question for Chapter Notes: Data Handling Try yourself:There are three boxes filled with marbles. The first box contains 30 marbles, the second box contains 50 marbles, and the third box contains 70 marbles. If we were to distribute the marbles equally among the three boxes, how many marbles would each box have? ### Range To understand how the observations are spread out, we can look at the difference between the highest and lowest values. To find this difference, we simply subtract the lowest value from the highest value. This difference is called the range of the observations. Steps to calculate Range: Step 1:Find out the highest and the lowest observation(or values from the given data) Step 2: Calculate their difference to get range Example 1: Find the range of the following data: 21, 16, 30, 15, 16, 18, 10, 24, 26, 20 Sol: Greatest number(or the highest value) = 30 Smallest number(or the lowest value) = 10 Range of the given data = (30 – 10) = 20 Example 2: The ages in years of 10 teachers of a school are: 32, 41, 28, 54, 35, 26, 23, 33, 38, 40 . (i) What is the age of the oldest teacher and that of the youngest teacher? (ii) What is the range of the ages of the teachers? Sol: (i) Age of the oldest teacher = 54 years Age of the youngest teacher = 23 years (ii) Range of the ages of the teachers = Highest Observation – Lowest Observation = (54 – 23) years = 31 years Question for Chapter Notes: Data Handling Try yourself:A batsman scored the following number of runs in six innings. Find the range of scores. 35, 30, 45, 65, 39, 20 ### Mode The mode of a set of observations is the observation that occurs most often. Note: A set of data can have more than 1 mode. Steps to calculate Mode: Step 1: Arrange the data in ascending order. Step 2: Tabulate the data in a frequency distribution table. Step 3: The most frequently occurring observation will be the mode. Example 1: Find the mode of data: 13, 16, 12, 14, 19, 12, 14, 13, 14 Sol: Let us arrange the data in ascending order i.e. 12,12,13,13,14,14,14,16,19 Here, 14 occurs the maximum number of times, hence it's the mode. Example 2: Find the mode of the given set of numbers: 1, 1, 2, 4, 3, 2, 1, 2, 2, 4 Sol: Arranging the numbers with the same values together, we get 1, 1, 1, 2, 2, 2, 2, 3, 4, 4 Mode of this data is 2 because it occurs more frequently than other observations Question for Chapter Notes: Data Handling Try yourself:The mode of the distribution 3,5, 7, 4, 2, 1, 4, 3, 4 is ### Mode of Large Data When we have many observations, it can be difficult to put them all together and count them. In such cases, we use a method called tabulation. Tally Chart depicting the frequency of people and their mode of transport Tabulation involves using tally marks and determining the frequency, which you might have learned in your previous class. This helps us organize the data in a more manageable way. Example: Following are the margins of victory in the football matches of a league. 1, 3, 2, 5, 1, 4, 6, 2, 5, 2, 2, 2, 4, 1, 2, 3, 1, 1, 2, 3, 2, 6, 4, 3, 2, 1, 1, 4, 2, 1, 5, 3, 3, 2, 3, 2, 4, 2, 1, 2 Find the mode of this data. Ans: Let us put the data in a tabular form: Looking at the table, we can see that the most common number of goals leading to victory is 2. This means that in most matches, teams have won with a margin of 2 goals. Hence the mode is 2. ### Median The median is the middle value when you arrange all the data in order. Half of the values are higher than the median, and the other half are lower than the median. Steps to calculate Median: Step 1: Arrange the data in ascending order. Step 2: The value that lies in the middle such that half of the observations lie above it and the other half below it will be the median. Example 1: The scores in mathematics test (out of 25) of 15 students is as follows: 19, 25, 23, 20, 9, 20, 15, 10, 5, 16, 25, 20, 24, 12, 20 Find the median of this data. Sol: The data is first arranged in ascending order. We get - 5, 9, 10, 12, 15, 16, 19, 20, 20, 20, 20, 23, 24, 25, 25. It can be seen from this arrangement that 20 (at the eighth place) is in the middle of the data. There are 7 observations below 20 and 7 observations above 20. Therefore, 20 is the median of the data. Example 2: The weights (in kg.) of 15 students in a class are: 38, 42, 35, 37, 45, 50, 32, 43, 43, 40, 36, 38, 43, 38, 47 Find the median of this data. Ans: Firstly, data will be arranged in ascending order in a tabular format. The middle term is 40, so 40 is the median. Question for Chapter Notes: Data Handling Try yourself:The median of the distribution 2, 3, 4, 7, 5, 1, 6 is ## Bar Graphs We learned about organizing information into frequency distribution tables and representing it visually through pictographs or bar graphs. By looking at these bar graphs, we can make observations and gather information from them. For instance, if the bar represents the frequency, we can identify the mode as the tallest bar on the graph. Bar Graph: The Subject Mathematics has the tallest bar ### Choosing a Scale A bar graph is a way of representing numbers using bars that have the same width. The lengths of the bars depend on the frequency and the scale chosen. For instance, in a bar graph where numbers are shown in units, each bar represents one observation. If the graph is displaying numbers in tens or hundreds, one unit length can represent 10 or 100 observations. Let's look at an example: Bar Graph Example: Two hundred students from 6th and 7th classes were asked to name their favourite colour to decide on the colour of their school building. The results are provided in the table below. Represent this data on a bar graph. Answer the following questions with the help of the bar graph: (i) Which is the most preferred colour and which is the least preferred? (ii) How many colours are there in all? What are they? Ans: Choose a suitable scale as follows: Start the scale at 0. The greatest value in the data is 55, so end the scale at a value greater than 55, such as 60. Use equal divisions along the axes, such as increments of 10. You know that all the bars would lie between 0 and 60. We choose the scale such that the length between 0 and 60 is neither too long nor too small. Here we take 1 unit for 10 students. We then draw and label the graph as shown. From the bar graph we conclude that: (i) Blue is the most preferred colour(Because the bar representing Blue is the tallest). Green is the least preferred colour(Because the bar representing Green is the shortest). (ii) There are five colours. They are Red, Green, Blue, Yellow and Orange. (These can be seen on the horizontal line) Important note: A scale: Large numbers of data cannot be represented directly, so the scaling factor is used to reduce or scale down large numbers. ## Drawing double-bar graphs Double bar graphs are used for comparing data between two different things. The difference between a bar graph and a double bar graph is that a bar graph displays one set of data, and a double bar graph compares two different sets of information or data. A Double Bar Graph Example: Sales of English and Hindi books in the years 1995, 1996, 1997,and 1998 are given below: Draw a double bar graph and answer the following questions: (a) In which year was the difference in the sale of the two language books least? (b) Can you say that the demand for English books rose faster? Justify. Ans: We can draw the double bar graph by taking the Years on the x-axis and the corresponding values of English and Hindi on the y-axis. Scale: 1 cm = 100 Books sold The above bar graph depicts the total production of motorbikes in two consecutive years. (a) In the year 1998, the difference in the sale of the two language books was least. (b) Yes, the demand for English books increased. From the graph, it is clear that the difference in the demand for English books from 1995 to 1998 is 620−350=270. Whereas for Hindi books the difference is 650−500=150. Hence, the demand for English books increased. The document Data Handling Class 7 Notes Maths Chapter 4 is a part of the Class 7 Course Mathematics (Maths) Class 7. All you need of Class 7 at this link: Class 7 ## Mathematics (Maths) Class 7 76 videos\|345 docs\|39 tests ## FAQs on Data Handling Class 7 Notes Maths Chapter 4 1. What is the purpose of using bar graphs in data handling? Ans. Bar graphs are used in data handling to visually represent data in a way that makes it easy to compare different categories or groups of data at a glance. 2. How can double-bar graphs be used to represent data effectively? Ans. Double-bar graphs are used to compare two different sets of data side by side, making it easier to identify trends, patterns, and differences between the two sets of data. 3. Can bar graphs be used to show averages of a data set? Ans. Yes, bar graphs can be used to show averages by including a line or shaded area on the graph to represent the average value of the data set. This can help viewers quickly see how individual data points compare to the average. 4. What are some common mistakes to avoid when creating bar graphs for data handling? Ans. Some common mistakes to avoid when creating bar graphs include using incorrect scaling, not labeling the axes correctly, and not providing a clear title or key to explain the data being represented. 5. How can bar graphs be used to identify outliers in a data set? Ans. Bar graphs can be used to identify outliers in a data set by visually displaying the range of data points and highlighting any values that fall significantly outside of the main group. This can help to pinpoint data points that may be unusual or require further investigation. ## Mathematics (Maths) Class 7 76 videos\|345 docs\|39 tests ### Up next Explore Courses for Class 7 exam ### Top Courses for Class 7 Signup to see your scores go up within 7 days! 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# What is cos(22.5Â°)? 1 teachers like this lesson Print Lesson ## Objective SWBAT derive and utilize the half angle formulas for sine and cosine. #### Big Idea The cosine of 22.5Â° has to be related to the cosine of 45Â°, right? ## Launch 15 minutes The goal for today is to derive a formula to find cos(1/2A) given cos(A); we will also find the formula for sin(1/2A) and tan(1/2A). Similar to the lesson where we found the double angle formulas, the focus will be on starting with a common misconception and the debunking it and trying to find the correct answer. To begin the lesson, I give students the notes worksheet and have them work on #1 in their table group. The focus is on MP3 (critique the reasoning of others) as students are going to respond to Gail’s argument that cos(22.5°) is half of cos(45°). I give students about 7 minutes to respond to her reasoning. If they agree I ask them to prove that it is true. If they disagree I ask them to give reasons why it is false. Most students will find out soon enough that Gail is incorrect. Their reasons for stating that this is false are usually varied, but here are some good ways to make it convincing. I will hand pick students with each one of these arguments. 1. The most common way they do this is to just plug both into their calculator and notice that cos(22.5°) is not half of cos(45°). Students may also try this for other angle measures. 2. Students think about what cosine represents (the x-coordinate on the unit circle) and reason that when the angle measure is divided in half, the x-coordinate is not half of what it was originally. In some cases (Quadrant II for example), the x-coordinate will actually increase when you take half of the original angle. 3. Students may think about the graph of y = cos(x) and y = cos(1/2x). The ½ in the second equation means that the graph is stretched horizontally. The amplitude will not change, so it wouldn’t make sense that the cosine value will be half of the original. If these reasons do not come out, I will try to lead the class there. Because this is such a common misconception, it is important to refute it in as many ways as possible to make it most convincing to all students. ## Explore 20 minutes After this initial investigation about Gail’s reasoning, students are hopefully convinced that cos(22.5°) is not equal to ½ of cos(45°). Now we want to focus on finding the correct answer. I believe that if students can pinpoint which existing formula will help them find cos(22.5°), then they can probably do this on their own. I give them about 10 minutes to work on questions #2 and #3 on the worksheet. Students are usually drawn to the double angle formula, because they know cos(45°) is double cos(22.5°). The tricky part is the substitution. In the formula cos(2x) = 2cos2(x) – 1, we must replace the (2x) with 45° and the (x) with 22.5°. Once students do this, they can algebraically solve for cos(22.5°). Here is our work from class. Once they square root both sides of the equation, students will usually wonder about the plus or minus. When students get to this point, it is a good idea to gather the call the class back together and discuss this as a class. I always point out that we choose plus or minus based on what sine the half angle should be, not the original angle. A common misconception is that students will choose the plus or minus based on the original angle. ## Share and Summarize 15 minutes After our initial work with finding cos(22.5°), I won’t derive the formulas for sin(1/2x) and tan(1/2x) in the interest of time. I think one derivation is sufficient for students to get the gist of this process. However, you could always assign it as a homework question. So for question #4 on the worksheet, I will simply give the students the correct formulas. When we move on to question #5, the discussion for choosing plus or minus will be more pressing. We don’t even know what the measure of angle A is, so I will ask students to several different angles in the given interval and they should notice that half of A will always be in quadrant II. Students will work on this one and I will select students to share their work. Sometimes I will choose a student who chose the wrong plus or minus sign to see if the class can catch their mistake. In this video, I give some ideas on how to wrap up this lesson.
# How do you write an equation of a line going through (2,1) and (-2,-1)? Mar 18, 2016 $y = \frac{1}{2} x$ #### Explanation: Recall that the general equation for a line is: $\textcolor{t e a l}{\| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} y = m x + b \textcolor{w h i t e}{\frac{a}{a}} \|}}}$ where: $y =$y-coordinate $m =$slope $x =$x-coordinate $b =$y-intercept Determining the Equation of the Line $1$. Start by labelling the coordinates to either be coordinate $1$ or $2$. Coordinate $1$: $\left(\textcolor{red}{{x}_{1}} , \textcolor{t e a l}{{y}_{1}}\right) = \left(\textcolor{red}{2} , \textcolor{t e a l}{1}\right)$ Coordinate $2$: $\left(\textcolor{b l u e}{{x}_{2}} , \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{{y}_{2}}\right) = \left(\textcolor{b l u e}{- 2} , \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{- 1}\right)$ $2$. Find the slope between the two coordinates using the formula, $m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$. $m = \frac{\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{{y}_{2}} - \textcolor{t e a l}{{y}_{1}}}{\textcolor{b l u e}{{x}_{2}} - \textcolor{red}{{x}_{1}}}$ $m = \frac{\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{- 1} - \textcolor{t e a l}{1}}{\textcolor{b l u e}{- 2} - \textcolor{red}{2}}$ $m = \frac{- 2}{- 4}$ $\textcolor{v i o \le t}{m = \frac{1}{2}}$ $3$. Find the value of the y-intercept by substituting the slope and either coordinate $1$ or $2$ into $y = m x + b$. In this case, we will use coordinate $1$. $y = m x + b$ $\textcolor{t e a l}{1} = \textcolor{v i o \le t}{\frac{1}{2}} \left(\textcolor{red}{2}\right) + b$ $1 = 1 + b$ $b = 0$ $4$. Rewrite the equation. $\textcolor{g r e e n}{\| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} y = \frac{1}{2} x \textcolor{w h i t e}{\frac{a}{a}} \|}}}$
# The ratio of the coefficient of `x^15` to the term independent of `x` in `( x^2 + ( 2/x) )^15` txmedteach \| High School Teacher \| (Level 3) Associate Educator Posted on To solve this problem, we must get familiar with the binomial theorem. The theorem states the following: `(a+b)^n = ((n),(0)) a^n + ((n),(1))a^(n-1)b + ((n),(2))a^(n-2)b^2 + ... + ((n),(n))b^n` In our case, we allow `a = x^2` and `b = 2/x`. First, we need to find the conditions under which we have `x^15` and the conditions under which we have a term independent of `x`. Starting with the former case, we need to find a combination of exponents `m` and `n` so that the following holds true: `(x^2)^m(2/x)^n = C_15x^15` Here, `C_15` is simply a constant. Because of exponent rules, we know the following must be true: `2m - n = 15` We also know the following based on the binomial theorem: `m+n = 15` Solving this system of equations yields the following values for `m` and `n`: `m = 10` `n = 5` Therefore, our term containing `x^15` will have the following: `(x^2)^10 * (2/x)^5 = 32x^15` We can't stop here, though! If you look at the binomial theorem, there is another coefficient we need to worry about: that derived from the combination term. If you'll notice from the theorem, the top number in the combination will be the overall number of terms, 15, in our case. The bottom number will be the exponent of the second term, 5, in our case. Our overall term that generates `x^15` will now become: `((15),(5))(x^2)^10 (2/x)^5 = 3003x^2032/x^5 = 96096x^15` Therefore, our coefficient for `x^15` will be 96096. Now, we must find the term independent of `x`, which will be the constant in this problem. As above, we will use the following term: `(x^2)^m(2/x)^n` However, now, the exponents must cancel each other in the following way: `2m-n = 0` However, the other condition still holds true, where `m + n = 15` Solving this for `m` and `n`, we find: `m = 5` `n = 10` Finding the term including the binomial coefficient the way we did above: `((15),(10)) * (x^2)^5 * (2/x)^10 = 3003x^10 1024/x^10 = 3075072` Now, we may find the ratio of the coefficient for `x^15` and the constant term: `r = 96096/3075072 = 0.03125` Another way of finding this is to use the exact numbers involved: `r = ( ((15),(5))2^5)/( ((15),(10)) 2^10)` Now, we can use the fact that `((n),(k)) = ((n),(n-k))` to allow `((15),(5))` and `((15),(10))` to cancel. We can also use the rules of exponents to cancel `2^5` on the top and bottom, leaving us with the following exact result: `r = 1/2^5 = 1/32 = 0.03125` I hope this helps! Sources:
### Learning Objectives • State the constant, constant multiple, and power rules. • Apply the sum and difference rules to combine derivatives. • Compute the derivative of $e^x$ • Use the product rule for finding the derivative of a product of functions. • Use the quotient rule for finding the derivative of a quotient of functions. • Extend the power rule to functions with negative exponents. • Combine the differentiation rules to find the derivative of a polynomial or rational function. Finding derivatives of functions by using the definition of the derivative can be a lengthy and, for certain functions, a rather challenging process. For example, previously we found that $\frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}$ by using a process that involved multiplying an expression by a conjugate prior to evaluating a limit. The process that we could use to evaluate $\frac{d}{dx}(\sqrt[3]{x})$ using the definition, while similar, is more complicated. In this section, we develop rules for finding derivatives that allow us to bypass this process. We begin with the basics. # The Basic Rules The functions $f(x)=c$ and $g(x)=x^n$ where $n$ is a positive integer are the building blocks from which all polynomials and rational functions are constructed. To find derivatives of polynomials and rational functions efficiently without resorting to the limit definition of the derivative, we must first develop formulas for differentiating these basic functions. ## The Constant Rule We first apply the limit definition of the derivative to find the derivative of the constant function, $f(x)=c$. For this function, both $f(x)=c$ and $f(x+h)=c$, so we obtain the following result: $\begin{array}{ll}f^{\prime}(x) & =\underset{h\to 0}{\lim}\frac{f(x+h)-f(x)}{h} \\ & =\underset{h\to 0}{\lim}\frac{c-c}{h} \\ & =\underset{h\to 0}{\lim}\frac{0}{h} \\ & =\underset{h\to 0}{\lim}0=0 \end{array}$ The rule for differentiating constant functions is called the constant rule . It states that the derivative of a constant function is zero; that is, since a constant function is a horizontal line, the slope, or the rate of change, of a constant function is 0. We restate this rule in the following theorem. ### The Constant Rule Let $c$ be a constant. If $f(x)=c$, then $f^{\prime}(c)=0$. Alternatively, we may express this rule as $\frac{d}{dx}(c)=0$. ### Applying the Constant Rule Find the derivative of $f(x)=8$. #### Solution This is just a one-step application of the rule: $f^{\prime}(8)=0$. Find the derivative of $g(x)=-3$. #### Hint Use the preceding example as a guide. 0 # The Power Rule We have shown that $\frac{d}{dx}(x^2)=2x$ and $\frac{d}{dx}(x^{1/2})=\frac{1}{2}x^{-1/2}$. At this point, you might see a pattern beginning to develop for derivatives of the form $\frac{d}{dx}(x^n)$. We continue our examination of derivative formulas by differentiating power functions of the form $f(x)=x^n$ where $n$ is a positive integer. We develop formulas for derivatives of this type of function in stages, beginning with positive integer powers. Before stating and proving the general rule for derivatives of functions of this form, we take a look at a specific case, $\frac{d}{dx}(x^3)$. ### Differentiating $x^3$ Find $\frac{d}{dx}(x^3)$. #### Solution $\begin{array}{lllll}\frac{d}{dx}(x^3) & =\underset{h\to 0}{\lim}\frac{(x+h)^3-x^3}{h} & & & \\ & =\underset{h\to 0}{\lim}\frac{x^3+3x^2h+3xh^2+h^3-x^3}{h} & & & \begin{array}{l}\text{Notice that the first term in the expansion of} \\ (x+h)^3 \, \text{is} \, x^3 \, \text{and the second term is} \, 3x^2h. \, \text{All} \\ \text{other terms contain powers of} \, h \, \text{that are two or} \\ \text{greater.} \end{array} \\ & =\underset{h\to 0}{\lim}\frac{3x^2h+3xh^2+h^3}{h} & & & \begin{array}{l}\text{In this step the} \, x^3 \, \text{terms have been cancelled,} \\ \text{leaving only terms containing} \, h. \end{array} \\ & =\underset{h\to 0}{\lim}\frac{h(3x^2+3xh+h^2)}{h} & & & \text{Factor out the common factor of} \, h. \\ & =\underset{h\to 0}{\lim}(3x^2+3xh+h^2) & & & \begin{array}{l}\text{After cancelling the common factor of} \, h, \, \text{the} \\ \text{only term not containing} \, h \, \text{is} \, 3x^2. \end{array} \\ & =3x^2 & & & \text{Let} \, h \, \text{go to 0.} \end{array}$ Find $\frac{d}{dx}(x^4)$. #### Hint Use $(x+h)^4=x^4+4x^3h+6x^2h^2+4xh^3+h^4$ and follow the procedure outlined in the preceding example. #### Solution $4x^3$ As we shall see, the procedure for finding the derivative of the general form $f(x)=x^n$ is very similar. Although it is often unwise to draw general conclusions from specific examples, we note that when we differentiate $f(x)=x^3$, the power on $x$ becomes the coefficient of $x^2$ in the derivative and the power on $x$ in the derivative decreases by 1. The following theorem states that this power rule holds for all positive integer powers of $x$. We will eventually extend this result to negative integer powers. Later, we will see that this rule may also be extended first to rational powers of $x$ and then to arbitrary powers of $x$. Be aware, however, that this rule does not apply to functions in which a constant is raised to a variable power, such as $f(x)=3^x$. ### The Power Rule Let $n$ be a positive integer. If $f(x)=x^n$, then $f^{\prime}(x)=nx^{n-1}$. Alternatively, we may express this rule as $\frac{d}{dx}(x^n)=nx^{n-1}$. ## Proof For $f(x)=x^n$ where $n$ is a positive integer, we have $f^{\prime}(x)=\underset{h\to 0}{\lim}\frac{(x+h)^n-x^n}{h}$. Since $(x+h)^n=x^n+nx^{n-1}h+\binom{n}{2}x^{n-2}h^2+\binom{n}{3}x^{n-3}h^3+\cdots+nxh^{n-1}+h^n$, we see that $(x+h)^n-x^n=nx^{n-1}h+\binom{n}{2}x^{n-2}h^2+\binom{n}{3}x^{n-3}h^3+\cdots+nxh^{n-1}+h^n$. Next, divide both sides by $h$: $\large \frac{(x+h)^n-x^n}{h}=\frac{nx^{n-1}h+\binom{n}{2}x^{n-2}h^2+\binom{n}{3}x^{n-3}h^3+\cdots+nxh^{n-1}+h^n}{h}$. Thus, $\frac{(x+h)^n-x^n}{h}=nx^{n-1}+\binom{n}{2}x^{n-2}h+\binom{n}{3}x^{n-3}h^2+\cdots+nxh^{n-2}+h^{n-1}$. Finally, $\begin{array}{ll}f^{\prime}(x) & =\underset{h\to 0}{\lim}(nx^{n-1}+\binom{n}{2}x^{n-2}h+\binom{n}{3}x^{n-3}h^2+\cdots+nxh^{n-1}+h^n) \\ & =nx^{n-1} \end{array} _\blacksquare$ ### Applying the Power Rule Find the derivative of the function $f(x)=x^{10}$ by applying the power rule. #### Solution Using the power rule with $n=10$, we obtain $f^{\prime}(x)=10x^{10-1}=10x^9$. Find the derivative of $f(x)=x^7$. #### Hint Use the power rule with $n=7$. #### Solution $f^{\prime}(x)=7x^6$ # The Sum, Difference, and Constant Multiple Rules We find our next differentiation rules by looking at derivatives of sums, differences, and constant multiples of functions. Just as when we work with functions, there are rules that make it easier to find derivatives of functions that we add, subtract, or multiply by a constant. These rules are summarized in the following theorem. ### Sum, Difference, and Constant Multiple Rules Let $f(x)$ and $g(x)$ be differentiable functions and $k$ be a constant. Then each of the following equations holds. Sum Rule . The derivative of the sum of a function $f$ and a function $g$ is the same as the sum of the derivative of $f$ and the derivative of $g$. $\frac{d}{dx}(f(x)+g(x))=\frac{d}{dx}(f(x))+\frac{d}{dx}(g(x))$; that is, for $j(x)=f(x)+g(x), \, j^{\prime}(x)=f^{\prime}(x)+g^{\prime}(x)$. Difference Rule . The derivative of the difference of a function $f$ and a function $g$ is the same as the difference of the derivative of $f$ and the derivative of $g$. $\frac{d}{dx}(f(x)-g(x))=\frac{d}{dx}(f(x))-\frac{d}{dx}(g(x))$; that is, for $j(x)=f(x)-g(x), \, j^{\prime}(x)=f^{\prime}(x)-g^{\prime}(x)$. Constant Multiple Rule . The derivative of a constant $c$ multiplied by a function $f$ is the same as the constant multiplied by the derivative: $\frac{d}{dx}(kf(x))=k\frac{d}{dx}(f(x))$; that is, for $j(x)=kf(x), \, j^{\prime}(x)=kf^{\prime}(x)$. ## Proof We provide only the proof of the sum rule here. The rest follow in a similar manner. For differentiable functions $f(x)$ and $g(x)$, we set $j(x)=f(x)+g(x)$. Using the limit definition of the derivative we have $j^{\prime}(x)=\underset{h\to 0}{\lim}\frac{j(x+h)-j(x)}{h}$. By substituting $j(x+h)=f(x+h)+g(x+h)$ and $j(x)=f(x)+g(x)$, we obtain $j^{\prime}(x)=\underset{h\to 0}{\lim}\frac{(f(x+h)+g(x+h))-(f(x)+g(x))}{h}$. Rearranging and regrouping the terms, we have $j^{\prime}(x)=\underset{h\to 0}{\lim}(\frac{f(x+h)-f(x)}{h}+\frac{g(x+h)-g(x)}{h})$. We now apply the sum law for limits and the definition of the derivative to obtain $j^{\prime}(x)=\underset{h\to 0}{\lim}(\frac{f(x+h)-f(x)}{h})+\underset{h\to 0}{\lim}(\frac{g(x+h)-g(x)}{h})=f^{\prime}(x)+g^{\prime}(x) \, _\blacksquare$ ### Applying the Constant Multiple Rule Find the derivative of $g(x)=3x^2$ and compare it to the derivative of $f(x)=x^2$. #### Solution We use the power rule directly: $g^{\prime}(x)=\frac{d}{dx}(3x^2)=3\frac{d}{dx}(x^2)=3(2x)=6x$. Since $f(x)=x^2$ has derivative $f^{\prime}(x)=2x$, we see that the derivative of $g(x)$ is 3 times the derivative of $f(x)$. This relationship is illustrated in (Figure) . The derivative of $g(x)$ is 3 times the derivative of $f(x)$. ### Applying Basic Derivative Rules Find the derivative of $f(x)=2x^5+7$. #### Solution We begin by applying the rule for differentiating the sum of two functions, followed by the rules for differentiating constant multiples of functions and the rule for differentiating powers. To better understand the sequence in which the differentiation rules are applied, we use Leibniz notation throughout the solution: $\begin{array}{lllll}f^{\prime}(x) & =\frac{d}{dx}(2x^5+7) & & & \\ & =\frac{d}{dx}(2x^5)+\frac{d}{dx}(7) & & & \text{Apply the sum rule.} \\ & =2\frac{d}{dx}(x^5)+\frac{d}{dx}(7) & & & \text{Apply the constant multiple rule.} \\ & =2(5x^4)+0 & & & \text{Apply the power rule and the constant rule.} \\ & =10x^4. & & & \text{Simplify.} \end{array}$ Find the derivative of $f(x)=2x^3-6x^2+3$. #### Hint Use the preceding example as a guide. #### Solution $f^{\prime}(x)=6x^2-12x$. ### Finding the Equation of a Tangent Line Find the equation of the line tangent to the graph of $f(x)=x^2-4x+6$ at $x=1$. #### Solution To find the equation of the tangent line, we need a point and a slope. To find the point, compute $f(1)=1^2-4(1)+6=3$. This gives us the point $(1,3)$. Since the slope of the tangent line at 1 is $f^{\prime}(1)$, we must first find $f^{\prime}(x)$. Using the definition of a derivative, we have $f^{\prime}(x)=2x-4$ so the slope of the tangent line is $f^{\prime}(1)=-2$. Using the point-slope formula, we see that the equation of the tangent line is $y-3=-2(x-1)$. Putting the equation of the line in slope-intercept form, we obtain $y=-2x+5$. Find the equation of the line tangent to the graph of $f(x)=3x^2-11$ at $x=2$. Use the point-slope form. #### Hint Use the preceding example as a guide. #### Solution $y=12x-23$ # The Product Rule Now that we have examined the basic rules, we can begin looking at some of the more advanced rules. The first one examines the derivative of the product of two functions. Although it might be tempting to assume that the derivative of the product is the product of the derivatives, similar to the sum and difference rules, the product rule does not follow this pattern. To see why we cannot use this pattern, consider the function $f(x)=x^2$, whose derivative is $f^{\prime}(x)=2x$ and not $\frac{d}{dx}(x)\cdot \frac{d}{dx}(x)=1\cdot 1=1$. ### Product Rule Let $f(x)$ and $g(x)$ be differentiable functions. Then $\frac{d}{dx}(f(x)g(x))=\frac{d}{dx}(f(x))\cdot g(x)+\frac{d}{dx}(g(x))\cdot f(x)$. That is, if $j(x)=f(x)g(x)$ then $j^{\prime}(x)=f^{\prime}(x)g(x)+g^{\prime}(x)f(x)$. This means that the derivative of a product of two functions is the derivative of the first function times the second function plus the derivative of the second function times the first function. ## Proof We begin by assuming that $f(x)$ and $g(x)$ are differentiable functions. At a key point in this proof we need to use the fact that, since $g(x)$ is differentiable, it is also continuous. In particular, we use the fact that since $g(x)$ is continuous, $\underset{h\to 0}{\lim}g(x+h)=g(x)$. By applying the limit definition of the derivative to $j(x)=f(x)g(x)$, we obtain $j^{\prime}(x)=\underset{h\to 0}{\lim}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}$. By adding and subtracting $f(x)g(x+h)$ in the numerator, we have $j^{\prime}(x)=\underset{h\to 0}{\lim}\frac{f(x+h)g(x+h)-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x)}{h}$. After breaking apart this quotient and applying the sum law for limits, the derivative becomes $j^{\prime}(x)=\underset{h\to 0}{\lim}(\frac{f(x+h)g(x+h)-f(x)g(x+h)}{h})+\underset{h\to 0}{\lim}(\frac{f(x)g(x+h)-f(x)g(x)}{h})$. Rearranging, we obtain $j^{\prime}(x)=\underset{h\to 0}{\lim}(\frac{f(x+h)-f(x)}{h}\cdot g(x+h))+\underset{h\to 0}{\lim}(\frac{g(x+h)-g(x)}{h}\cdot f(x))$. By using the continuity of $g(x)$, the definition of the derivatives of $f(x)$ and $g(x)$, and applying the limit laws, we arrive at the product rule, $j^{\prime}(x)=f^{\prime}(x)g(x)+g^{\prime}(x)f(x) \, _\blacksquare$ ### Applying the Product Rule to Constant Functions For $j(x)=f(x)g(x)$, use the product rule to find $j^{\prime}(2)$ if $f(2)=3, \, f^{\prime}(2)=-4, \, g(2)=1$, and $g^{\prime}(2)=6$. #### Solution Since $j(x)=f(x)g(x), \, j^{\prime}(x)=f^{\prime}(x)g(x)+g^{\prime}(x)f(x)$, and hence $j^{\prime}(2)=f^{\prime}(2)g(2)+g^{\prime}(2)f(2)=(-4)(1)+(6)(3)=14$. ### Applying the Product Rule to Binomials For $j(x)=(x^2+2)(3x^3-5x)$, find $j^{\prime}(x)$ by applying the product rule. Check the result by first finding the product and then differentiating. #### Solution If we set $f(x)=x^2+2$ and $g(x)=3x^3-5x$, then $f^{\prime}(x)=2x$ and $g^{\prime}(x)=9x^2-5$. Thus, $j^{\prime}(x)=f^{\prime}(x)g(x)+g^{\prime}(x)f(x)=(2x)(3x^3-5x)+(9x^2-5)(x^2+2)$. Simplifying, we have $j^{\prime}(x)=15x^4+3x^2-10$. To check, we see that $j(x)=3x^5+x^3-10x$ and, consequently, $j^{\prime}(x)=15x^4+3x^2-10$. Use the product rule to obtain the derivative of $j(x)=2x^5(4x^2+x)$. #### Hint Set $f(x)=2x^5$ and $g(x)=4x^2+x$ and use the preceding example as a guide. #### Solution $j^{\prime}(x)=10x^4(4x^2+x)+(8x+1)(2x^5)=56x^6+12x^5$. # The Quotient Rule Having developed and practiced the product rule, we now consider differentiating quotients of functions. As we see in the following theorem, the derivative of the quotient is not the quotient of the derivatives; rather, it is the derivative of the function in the numerator times the function in the denominator minus the derivative of the function in the denominator times the function in the numerator, all divided by the square of the function in the denominator. In order to better grasp why we cannot simply take the quotient of the derivatives, keep in mind that $\frac{d}{dx}(x^2)=2x$, which is not the same as $\large \frac{\frac{d}{dx}(x^3)}{\frac{d}{dx}(x)} \normalsize =\frac{3x^2}{1}=3x^2$. ### The Quotient Rule Let $f(x)$ and $g(x)$ be differentiable functions. Then $\frac{d}{dx}\Big(\frac{f(x)}{g(x)}\Big)=\large\frac{\frac{d}{dx}(f(x))\cdot g(x)-\frac{d}{dx}(g(x))\cdot f(x)}{(g(x))^2}$. That is, if $j(x)=\frac{f(x)}{g(x)}$, then $j^{\prime}(x)=\large \frac{f^{\prime}(x)g(x)-g^{\prime}(x)f(x)}{(g(x))^2}$. The proof of the quotient rule is very similar to the proof of the product rule, so it is omitted here. Instead, we apply this new rule for finding derivatives in the next example. ### Applying the Quotient Rule Use the quotient rule to find the derivative of $k(x)=\frac{5x^2}{4x+3}$. #### Solution Let $f(x)=5x^2$ and $g(x)=4x+3$. Thus, $f^{\prime}(x)=10x$ and $g^{\prime}(x)=4$. Substituting into the quotient rule, we have $k^{\prime}(x)=\frac{f^{\prime}(x)g(x)-g^{\prime}(x)f(x)}{(g(x))^2}=\frac{10x(4x+3)-4(5x^2)}{(4x+3)^2}$. Simplifying, we obtain $k^{\prime}(x)=\frac{20x^2+30x}{(4x+3)^2}$. Find the derivative of $h(x)=\frac{3x+1}{4x-3}$. #### Hint Apply the quotient rule with $f(x)=3x+1$ and $g(x)=4x-3$. #### Solution $k^{\prime}(x)=-\frac{13}{(4x-3)^2}$. It is now possible to use the quotient rule to extend the power rule to find derivatives of functions of the form $x^k$ where $k$ is a negative integer. ### Extended Power Rule If $k$ is a negative integer, then $\frac{d}{dx}(x^k)=kx^{k-1}$. ## Proof If $k$ is a negative integer, we may set $n=-k$, so that $n$ is a positive integer with $k=-n$. Since for each positive integer $n, \, x^{-n}=\frac{1}{x^n}$, we may now apply the quotient rule by setting $f(x)=1$ and $g(x)=x^n$. In this case, $f^{\prime}(x)=0$ and $g^{\prime}(x)=nx^{n-1}$. Thus, $\frac{d}{dx}(x^{-n})=\frac{0(x^n)-1(nx^{n-1})}{(x^n)^2}$. Simplifying, we see that $\frac{d}{dx}(x^{-n})=\frac{-nx^{n-1}}{x^{2n}}=-nx^{(n-1)-2n}=-nx^{-n-1}$. Finally, observe that since $k=-n$, by substituting we have $\frac{d}{dx}(x^k)=kx^{k-1}. \, _\blacksquare$ ### Using the Extended Power Rule Find $\frac{d}{dx}(x^{-4})$. #### Solution By applying the extended power rule with $k=-4$, we obtain $\frac{d}{dx}(x^{-4})=-4x^{-4-1}=-4x^{-5}$. ### Using the Extended Power Rule and the Constant Multiple Rule Use the extended power rule and the constant multiple rule to find $f(x)=\frac{6}{x^2}$. #### Solution It may seem tempting to use the quotient rule to find this derivative, and it would certainly not be incorrect to do so. However, it is far easier to differentiate this function by first rewriting it as $f(x)=6x^{-2}$. $\begin{array}{lllll}f^{\prime}(x) & =\frac{d}{dx}(\frac{6}{x^2})=\frac{d}{dx}(6x^{-2}) & & & \text{Rewrite} \, \frac{6}{x^2} \, \text{as} \, 6x^{-2}. \\ & =6\frac{d}{dx}(x^{-2}) & & & \text{Apply the constant multiple rule.} \\ & =6(-2x^{-3}) & & & \text{Use the extended power rule to differentiate} \, x^{-2}. \\ & =-12x^{-3} & & & \text{Simplify.} \end{array}$ Find the derivative of $g(x)=\frac{1}{x^7}$ using the extended power rule. #### Hint Rewrite $g(x)=\frac{1}{x^7}=x^{-7}$. Use the extended power rule with $k=-7$. #### Solution $g^{\prime}(x)=-7x^{-8}$. # Combining Differentiation Rules As we have seen throughout the examples in this section, it seldom happens that we are called on to apply just one differentiation rule to find the derivative of a given function. At this point, by combining the differentiation rules, we may find the derivatives of any polynomial or rational function. Later on we will encounter more complex combinations of differentiation rules. A good rule of thumb to use when applying several rules is to apply the rules in reverse of the order in which we would evaluate the function. ### Combining Differentiation Rules For $k(x)=3h(x)+x^2g(x)$, find $k^{\prime}(x)$. #### Solution Finding this derivative requires the sum rule, the constant multiple rule, and the product rule. $\begin{array}{lllll}k^{\prime}(x) & =\frac{d}{dx}(3h(x)+x^2g(x))=\frac{d}{dx}(3h(x))+\frac{d}{dx}(x^2g(x)) & & & \text{Apply the sum rule.} \\ & =3\frac{d}{dx}(h(x))+(\frac{d}{dx}(x^2)g(x)+\frac{d}{dx}(g(x))x^2) & & & \begin{array}{l}\text{Apply the constant multiple rule to} \\ \text{differentiate} \, 3h(x) \, \text{and the product} \\ \text{rule to differentiate} \, x^2g(x). \end{array} \\ & =3h^{\prime}(x)+2xg(x)+g^{\prime}(x)x^2 & & & \end{array}$ ### Extending the Product Rule For $k(x)=f(x)g(x)h(x)$, express $k^{\prime}(x)$ in terms of $f(x), \, g(x), \, h(x)$, and their derivatives. #### Solution We can think of the function $k(x)$ as the product of the function $f(x)g(x)$ and the function $h(x)$. That is, $k(x)=(f(x)g(x))\cdot h(x)$. Thus, $\begin{array}{lllll}k^{\prime}(x) & =\frac{d}{dx}(f(x)g(x))\cdot h(x)+\frac{d}{dx}(h(x))\cdot (f(x)g(x)) & & & \begin{array}{l}\text{Apply the product rule to the product} \\ \text{of} \, f(x)g(x) \, \text{and} \, h(x). \end{array} \\ & =(f^{\prime}(x)g(x)+g^{\prime}(x)f(x))h(x)+h^{\prime}(x)f(x)g(x) & & & \text{Apply the product rule to} \, f(x)g(x). \\ & =f^{\prime}(x)g(x)h(x)+f(x)g^{\prime}(x)h(x)+f(x)g(x)h^{\prime}(x). & & & \text{Simplify.} \end{array}$ ### Combining the Quotient Rule and the Product Rule For $h(x)=\large \frac{2x^3k(x)}{3x+2}$, find $h^{\prime}(x)$. #### Solution This procedure is typical for finding the derivative of a rational function. $\begin{array}{lllll}h^{\prime}(x) & =\large \frac{\frac{d}{dx}(2x^3k(x))\cdot (3x+2)-\frac{d}{dx}(3x+2)\cdot (2x^3k(x))}{(3x+2)^2} & & & \text{Apply the quotient rule.} \\ & =\large \frac{(6x^2k(x)+k^{\prime}(x)\cdot 2x^3)(3x+2)-3(2x^3k(x))}{(3x+2)^2} & & & \begin{array}{l}\text{Apply the product rule to find} \\ \frac{d}{dx}(2x^3k(x)). \, \text{Use} \, \frac{d}{dx}(3x+2)=3. \end{array} \\ & =\large \frac{-6x^3k(x)+18x^3k(x)+12x^2k(x)+6x^4k^{\prime}(x)+4x^3k^{\prime}(x)}{(3x+2)^2} & & & \text{Simplify.} \end{array}$ Find $\frac{d}{dx}(3f(x)-2g(x))$. #### Hint Apply the difference rule and the constant multiple rule. #### Solution $3f^{\prime}(x)-2g^{\prime}(x)$. ### Determining Where a Function Has a Horizontal Tangent Determine the values of $x$ for which $f(x)=x^3-7x^2+8x+1$ has a horizontal tangent line. #### Solution To find the values of $x$ for which $f(x)$ has a horizontal tangent line, we must solve $f^{\prime}(x)=0$. Since $f^{\prime}(x)=3x^2-14x+8=(3x-2)(x-4)$, we must solve $(3x-2)(x-4)=0$. Thus we see that the function has horizontal tangent lines at $x=\frac{2}{3}$ and $x=4$ as shown in the following graph. ### Finding a Velocity The position of an object on a coordinate axis at time $t$ is given by $s(t)=\frac{t}{t^2+1}$. What is the initial velocity of the object? #### Solution Since the initial velocity is $v(0)=s^{\prime}(0)$, begin by finding $s^{\prime}(t)$ by applying the quotient rule: $s^{\prime}(t)=\frac{1(t^2+1)-2t(t)}{(t^2+1)^2}=\frac{1-t^2}{(t^2+1)^2}$. After evaluating, we see that $v(0)=1$. Find the value(s) of $x$ for which the line tangent to the graph of $f(x)=4x^2-3x+2$ is parallel to the line $y=2x+3$. #### Hint Solve the equation $f^{\prime}(x)=2$. #### Solution $\frac{5}{8}$ # Derivative of the Exponential Function Just as when we found the derivatives of other functions, we can find the derivatives of exponential and logarithmic functions using formulas. As we develop these formulas, we need to make certain basic assumptions. The proofs that these assumptions hold are beyond the scope of this course. First of all, we begin with the assumption that the function $B(x)=b^x, \, b < 0$, is defined for every real number and is continuous. In previous courses, the values of exponential functions for all rational numbers were defined—beginning with the definition of $b^n$, where $n$ is a positive integer—as the product of $b$ multiplied by itself $n$ times. Later, we defined $b^0=1, \, b^{-n}=\frac{1}{b^n}$ for a positive integer $n$, and $b^{s/t}=(\sqrt[t]{b})^s$ for positive integers $s$ and $t$. These definitions leave open the question of the value of $b^r$ where $r$ is an arbitrary real number. By assuming the continuity of $B(x)=b^x, \, b < 0$, we may interpret $b^r$ as $\underset{x\to r}{\lim}b^x$ where the values of $x$ as we take the limit are rational. For example, we may view ${4}^{\pi}$ as the number satisfying $\begin{array}{l}4^3 < 4^{\pi} < 4^4, \, 4^{3.1} < 4^{\pi} < 4^{3.2}, \, 4^{3.14} < 4^{\pi} < 4^{3.15},\\ 4^{3.141} < 4^{\pi} < 4^{3.142}, \, 4^{3.1415} < 4^{\pi} < 4^{3.1416}, \, \cdots \end{array}$ As we see in the following table, $4^{\pi}\approx 77.88$. Approximating a Value of $4^{\pi}$ $x$ $4^x$ $x$ $4^x$ $4^3$ 64 $4^{3.141593}$ 77.8802710486 $4^{3.1}$ 73.5166947198 $4^{3.1416}$ 77.8810268071 $4^{3.14}$ 77.7084726013 $4^{3.142}$ 77.9242251944 $4^{3.141}$ 77.8162741237 $4^{3.15}$ 78.7932424541 $4^{3.1415}$ 77.8702309526 $4^{3.2}$ 84.4485062895 $4^{3.14159}$ 77.8799471543 $4^4$ 256 We also assume that for $B(x)=b^x, \, b < 0$, the value $B^{\prime}(0)$ of the derivative exists. In this section, we show that by making this one additional assumption, it is possible to prove that the function $B(x)$ is differentiable everywhere. We make one final assumption: that there is a unique value of $b < 0$ for which $B^{\prime}(0)=1$. We define $e$ to be this unique value, as we did in Introduction to Functions and Graphs . (Figure) provides graphs of the functions $y=2^x, \, y=3^x, \, y=2.7^x$, and $y=2.8^x$. A visual estimate of the slopes of the tangent lines to these functions at 0 provides evidence that the value of $e$ lies somewhere between 2.7 and 2.8. The function $E(x)=e^x$ is called the natural exponential function . Its inverse, $L(x)=\log_e x=\ln x$ is called the natural logarithmic function . For a better estimate of $e$, we may construct a table of estimates of $B^{\prime}(0)$ for functions of the form $B(x)=b^x$. Before doing this, recall that $B^{\prime}(0)=\underset{x\to 0}{\lim}\frac{b^x-b^0}{x-0}=\underset{x\to 0}{\lim}\frac{b^x-1}{x} \approx \frac{b^x-1}{x}$ for values of $x$ very close to zero. For our estimates, we choose $x=0.00001$ and $x=-0.00001$ to obtain the estimate $\frac{b^{-0.00001}-1}{-0.00001} < B^{\prime}(0) < \frac{b^{0.00001}-1}{0.00001}$. See the following table. < table id="fs-id1169738019199" summary="This table has six rows and four columns. The first row is a header row and it labels each column. The first column header is b, the second column header is (b-0.00001 – 1)/-0.00001 < B’(0) < (b0.00001 – 1)/0.00001, the third column header is b, and the fourth column header is (b-0.00001 – 1)/-0.00001 < B’(0) < (b0.00001 – 1)/0.00001. Under the first column are the values 2, 2.7, 2.71, 2.718, and 2.7182. Under the second column are the values 0.693145 < B’(0) < 0.69315, 0.993247 < B’(0) < 0.993257, 0.996944 < B’(0) < 0.996954, 0.999891 < B’(0) < 0.999901, and 0.999965 < B’(0) < 0.999975. Under the third column are the values 2.7183, 2.719, 2.72, 2.8, and 3. Under the fourth column are the values 1.000002 < B’(0) < 1.000012, 1.000259 < B’(0) < 1.000269, 1.000627 < B’(0) < 1.000637, 1.029614 < B’(0) < 1.029625, and 1.098606 < B’(00 Estimating a Value of $e$$b$$\frac{b^{-0.00001}-1}{-0.00001} < B^{\prime}(0) < \frac{b^{0.00001}-1}{0.00001}$$b$$\frac{b^{-0.00001}-1}{-0.00001} < B^{\prime}(0) < \frac{b^{0.00001}-1}{0.00001}$2$0.693145 < B^{\prime}(0) < 0.69315$2.7183$1.000002 < B^{\prime}(0) < 1.000012$2.7$0.993247 < B^{\prime}(0) < 0.993257$2.719$1.000259 < B^{\prime}(0) < 1.000269$2.71$0.996944 < B^{\prime}(0) < 0.996954$2.72$1.000627 < B^{\prime}(0) < 1.000637$2.718$0.999891 < B^{\prime}(0) < 0.999901$2.8$1.029614 < B^{\prime}(0) < 1.029625$2.7182$0.999965 < B^{\prime}(0) < 0.999975$3$1.098606 < B^{\prime}(0) < 1.098618$ The evidence from the table suggests that $2.7182 < e < 2.7183$. The graph of $E(x)=e^x$ together with the line $y=x+1$ are shown in (Figure) . This line is tangent to the graph of $E(x)=e^x$ at $x=0$. Now that we have laid out our basic assumptions, we begin our investigation by exploring the derivative of $B(x)=b^x, \, b < 0$. Recall that we have assumed that $B^{\prime}(0)$ exists. By applying the limit definition to the derivative we conclude that $B^{\prime}(0)=\underset{h\to 0}{\lim}\frac{b^{0+h}-b^0}{h}=\underset{h\to 0}{\lim}\frac{b^h-1}{h}$. Turning to $B^{\prime}(x)$, we obtain the following. $\begin{array}{lllll} B^{\prime}(x) & =\underset{h\to 0}{\lim}\frac{b^{x+h}-b^x}{h} & & & \text{Apply the limit definition of the derivative.} \\ & =\underset{h\to 0}{\lim}\frac{b^xb^h-b^x}{h} & & & \text{Note that} \, b^{x+h}=b^x b^h. \\ & =\underset{h\to 0}{\lim}\frac{b^x(b^h-1)}{h} & & & \text{Factor out} \, b^x. \\ & =b^x\underset{h\to 0}{\lim}\frac{b^h-1}{h} & & & \text{Apply a property of limits.} \\ & =b^x B^{\prime}(0) & & & \text{Use} \, B^{\prime}(0)=\underset{h\to 0}{\lim}\frac{b^{0+h}-b^0}{h}=\underset{h\to 0}{\lim}\frac{b^h-1}{h}. \end{array}$ We see that on the basis of the assumption that $B(x)=b^x$ is differentiable at $0, \, B(x)$ is not only differentiable everywhere, but its derivative is $B^{\prime}(x)=b^x B^{\prime}(0)$. For $E(x)=e^x, \, E^{\prime}(0)=1$. Thus, we have $E^{\prime}(x)=e^x$. (The value of $B^{\prime}(0)$ for an arbitrary function of the form $B(x)=b^x, \, b < 0$, will be derived later.) ### Derivative of the Natural Exponential Function Let $f(x)=e^x$ be the natural exponential function. Then $f^{\prime}(x)=e^x$. ### Combining Differentiation Rules Find the derivative of $y=\frac{e^{x}}{x}$. #### Solution Use the derivative of the natural exponential function and the quotient rule. $\begin{array}{lllll} y^{\prime} & =\large \frac{(e^{x}) \cdot x - 1 \cdot e^{x}}{x^2} & & & \text{Apply the quotient rule.} \\ & = \large \frac{e^{x}(x-1)}{x^2} & & & \text{Simplify.} \end{array}$ Find the derivative of $h(x)=(x^2+1)e^{x}$. #### Hint Don’t forget to use the product rule. #### Solution $h^{\prime}(x)=2xe^{x}+(x^2+1)e^{x}$ ### Student Project — Formula One Grandstands Formula One car races can be very exciting to watch and attract a lot of spectators. Formula One track designers have to ensure sufficient grandstand space is available around the track to accommodate these viewers. However, car racing can be dangerous, and safety considerations are paramount. The grandstands must be placed where spectators will not be in danger should a driver lose control of a car ( (Figure) ). ********** Safety is especially a concern on turns. If a driver does not slow down enough before entering the turn, the car may slide off the racetrack. Normally, this just results in a wider turn, which slows the driver down. But if the driver loses control completely, the car may fly off the track entirely, on a path tangent to the curve of the racetrack. Suppose you are designing a new Formula One track. One section of the track can be modeled by the function $f(x)=x^3+3x^2+x$ ( (Figure) ). The current plan calls for grandstands to be built along the first straightaway and around a portion of the first curve. The plans call for the front corner of the grandstand to be located at the point $(-1.9,2.8)$. We want to determine whether this location puts the spectators in danger if a driver loses control of the car. 1. Physicists have determined that drivers are most likely to lose control of their cars as they are coming into a turn, at the point where the slope of the tangent line is 1. Find the $(x,y)$ coordinates of this point near the turn. 2. Find the equation of the tangent line to the curve at this point. 3. To determine whether the spectators are in danger in this scenario, find the $x$-coordinate of the point where the tangent line crosses the line $y=2.8$. Is this point safely to the right of the grandstand? Or are the spectators in danger? 4. What if a driver loses control earlier than the physicists project? Suppose a driver loses control at the point $(-2.5,0.625)$. What is the slope of the tangent line at this point? 5. If a driver loses control as described in part 4, are the spectators safe? 6. Should you proceed with the current design for the grandstand, or should the grandstands be moved? ### Key Concepts • The derivative of a constant function is zero. • The derivative of a power function is a function in which the power on $x$ becomes the coefficient of the term and the power on $x$ in the derivative decreases by 1. • The derivative of a constant $c$ multiplied by a function $f$ is the same as the constant multiplied by the derivative. • The derivative of the sum of a function $f$ and a function $g$ is the same as the sum of the derivative of $f$ and the derivative of $g$ . • The derivative of the difference of a function $f$ and a function $g$ is the same as the difference of the derivative of $f$ and the derivative of $g$ . • The derivative of a product of two functions is the derivative of the first function times the second function plus the derivative of the second function times the first function. • The derivative of the quotient of two functions is the derivative of the first function times the second function minus the derivative of the second function times the first function, all divided by the square of the second function. • We used the limit definition of the derivative to develop formulas that allow us to find derivatives without resorting to the definition of the derivative. These formulas can be used singly or in combination with each other. For the following exercises, find $f^{\prime}(x)$ for each function. 1. $f(x)=x^7+10$ 2. $f(x)=5x^3-x+1$ #### Solution $f^{\prime}(x)=15x^2-1$ 3. $f(x)=4x^2-7x$ 4. $f(x)=8x^4+9x^2-1$ #### Solution $f^{\prime}(x)=32x^3+18x$ 5. $f(x)=x^4+\frac{2}{x}$ 6. $f(x)=3x(18x^4+\frac{13}{x+1})$ #### Solution $f^{\prime}(x)=270x^4+\frac{39}{(x+1)^2}$ 7. $f(x)=(x+2)(2x^2-3)$ 8. $f(x)=x^2(\frac{2}{x^2}+\frac{5}{x^3})$ #### Solution $f^{\prime}(x)=\frac{-5}{x^2}$ 9. $f(x)=\frac{x^3+2x^2-4}{3}$ 10. $f(x)=\frac{4x^3-2x+1}{x^2}$ #### Solution $f^{\prime}(x)=\frac{4x^4+2x^2-2x}{x^4}$ 11. $f(x)=\frac{x^2+4}{x^2-4}$ 12. $f(x)=\frac{x+9}{x^2-7x+1}$ #### Solution $f^{\prime}(x)=\frac{-x^2-18x+64}{(x^2-7x+1)^2}$ 13. $f(x)=\frac{e^{x}}{x^3+1}$ 14. $f(x)=x^2 e^x$ #### Solution $f^{\prime}(x) = 2xe^x+x^2 e^x$ 15. $f(x)=x^2 e^x$ 16. $f(x)=\frac{5e^x}{2x}$ #### Solution $f^{\prime}(x) = \frac{5}{2} (\frac{e^x}{x} - \frac{e^x}{x^2}})$ 17. $f(x)=\frac{2x^{\pi} - 5}{x^5}$ 18. $f(x)=2\sqrt[3]{x}e^x$ Hint: $\sqrt[3]{x} = x^{\frac{1}{3}}$. #### Solution $f^{\prime}(x) = 2 (\frac{1}{3}x^{\frac{-2}{3}}e^x + x^{\frac{1}{3}} e^x)$ 19. $f(x)=e^{\pi}x^{2021}$ 20. $f(x)=\sqrt[4]{e}x^{\frac{\pi}{2}}$ #### Solution $f^{\prime}(x) = \frac{\sqrt[4]{e}{\pi}}{2} x^{\frac{\pi -2}{2}}$ 21. $f(x)=\frac{\sqrt[5]{x}^2}{7e^x}$ 22. $f(x)=\frac{3e^x -2x^5}{\sqrt[3]{x}^5}$. Do not simplify final answer. #### Solution $f^{\prime}(x) = \frac{(3e^x-10x^4)(x^{\frac{5}{3}})-(3e^x-2x^5)(\frac{5}{3}x^{\frac{2}{3}})}{(x^{\frac{5}{3}})^2}$ 23. $f(x)= \frac{\sqrt[3]{x}^7 + 3e^x}{\frac{1}{x^2}}$. Do not simplify final answer. 24. $f(x)=\frac{(7\sqrt{x} - 3x^{\pi})(2x^3-6)}{2e^x - \frac{5}{x^3}}.$ Do not simplify final answer. #### Solution $f^{\prime}(x) = \frac{[(\frac{7}{2}x^{-\frac{1}{2}}-3\pi x^{\pi-1})(2x^3-6)+(7\sqrt{x}-3x^{\pi})(6x^2)](2e^x-\frac{5}{x^3}) - (7\sqrt{x} - 3x^{\pi})(2x^3-6)(2e^x +\frac{15}{x^4})}{(2e^x-\frac{5}{x^3})^2}$ For the following exercises, find the equation of the tangent line $T(x)$ to the graph of the given function at the indicated point. Use a graphing calculator or app to graph the function and the tangent line. 25. [T] $y=3x^2+4x+1$ at $(0,1)$ 26. [T] $y=2\sqrt{x}+1$ at $(4,5)$ #### Solution $T(x)=\frac{1}{2}x+3$ 27. [T] $y=\frac{2x}{x-1}$ at $(-1,1)$ 28. [T] $y=\frac{2}{x}-\frac{3}{x^2}$ at $(1,-1)$ #### Solution $T(x)=4x-5$ For the following exercises, assume that $f(x)$ and $g(x)$ are both differentiable functions for all $x$. Find the derivative of each of the functions $h(x)$. 29. $h(x)=4f(x)+\frac{g(x)}{7}$ 30. $h(x)=x^3f(x)$ #### Solution $h^{\prime}(x)=3x^2f(x)+x^3f^{\prime}(x)$ 31. $h(x)=\frac{f(x)g(x)}{2}$ 32. $h(x)=\frac{3f(x)}{g(x)+2}$ #### Solution $h^{\prime}(x)=\frac{3f^{\prime}(x)(g(x)+2)-3f(x)g^{\prime}(x)}{(g(x)+2)^2}$ For the following exercises, assume that $f(x)$ and $g(x)$ are both differentiable functions with values as given in the following table. Use the following table to calculate the following derivatives. $x$ 1 2 3 4 $f(x)$ 3 5 -2 0 $g(x)$ 2 3 -4 6 $f^{\prime}(x)$ -1 7 8 -3 $g^{\prime}(x)$ 4 1 2 9 33. Find $h^{\prime}(1)$ if $h(x)=xf(x)+4g(x)$. 34. Find $h^{\prime}(2)$ if $h(x)=\frac{f(x)}{g(x)}$. #### Solution $\frac{16}{9}$ 35. Find $h^{\prime}(3)$ if $h(x)=2x+f(x)g(x)$. 36. Find $h^{\prime}(4)$ if $h(x)=\frac{1}{x}+\frac{g(x)}{f(x)}$. #### Solution Undefined For the following exercises, use the following figure to find the indicated derivatives, if they exist. 37. Let $h(x)=f(x)+g(x)$. Find 1. $h^{\prime}(1)$ 2. $h^{\prime}(3)$ 3. $h^{\prime}(4)$ 38. Let $h(x)=f(x)g(x)$. Find 1. $h^{\prime}(1)$ 2. $h^{\prime}(3)$ 3. $h^{\prime}(4)$ #### Solution a. 2 b. does not exist c. 2.5 39. Let $h(x)=\frac{f(x)}{g(x)}$. Find 1. $h^{\prime}(1)$ 2. $h^{\prime}(3)$ 3. $h^{\prime}(4)$ For the following exercises, 1. Evaluate $f^{\prime}(a)$, and 2. Graph the function $f(x)$ and the tangent line at $x=a$. 40. [T] $f(x)=2x^3+3x-x^2, \, a=2$ #### Solution a. 23 b. $y=23x-28$ 41. [T] $f(x)=\frac{1}{x}-x^2, \, a=1$ 42. [T] $f(x)=x^2-x^{12}+3x+2, \, a=0$ #### Solution a. 3 b. $y=3x+2$ 43. [T] $f(x)=\frac{1}{x}-x^{2/3}, \, a=-1$ 44. Find the equation of the tangent line to the graph of $f(x)=2x^3+4x^2-5x-3$ at $x=-1$. #### Solution $y=-7x-3$ 45. Find the equation of the tangent line to the graph of $f(x)=x^2+\frac{4}{x}-10$ at $x=8$. 46. Find the equation of the tangent line to the graph of $f(x)=(3x-x^2)(3-x-x^2)$ at $x=1$. #### Solution $y=-5x+7$ 47. Find the point on the graph of $f(x)=x^3$ such that the tangent line at that point has an $x$ intercept of 6. 48. Find the equation of the line passing through the point $P(3,3)$ and tangent to the graph of $f(x)=\frac{6}{x-1}$. #### Solution $y=-\frac{3}{2}x+\frac{15}{2}$ 49. Determine all points on the graph of $f(x)=x^3+x^2-x-1$ for which the slope of the tangent line is 1. horizontal 2. -1. 50. Find a quadratic polynomial such that $f(1)=5, \, f^{\prime}(1)=3$, and $f''(1)=-6$. #### Solution $y=-3x^2+9x-1$ 51. A car driving along a freeway with traffic has traveled $s(t)=t^3-6t^2+9t$ meters in $t$ seconds. 1. Determine the time in seconds when the velocity of the car is 0. 2. Determine the acceleration of the car when the velocity is 0. 52. [T] A herring swimming along a straight line has traveled $s(t)=\frac{t^2}{t^2+2}$ feet in $t$ seconds. Determine the velocity of the herring when it has traveled 3 seconds. #### Solution $\frac{12}{121}$ or 0.0992 ft/s 53. The population in millions of arctic flounder in the Atlantic Ocean is modeled by the function $P(t)=\frac{8t+3}{0.2t^2+1}$, where $t$ is measured in years. 1. Determine the initial flounder population. 2. Determine $P^{\prime}(10)$ and briefly interpret the result. 54. [T] The concentration of antibiotic in the bloodstream $t$ hours after being injected is given by the function $C(t)=\frac{2t^2+t}{t^3+50}$, where $C$ is measured in milligrams per liter of blood. 1. Find the rate of change of $C(t)$. 2. Determine the rate of change for $t=8, \, 12, \, 24$, and $36$. 3. Briefly describe what seems to be occurring as the number of hours increases. #### Solution a. $\frac{-2t^4-2t^3+200t+50}{(t^3+50)^2}$ b. -0.02395 mg/L-hr, -0.01344 mg/L-hr, -0.003566 mg/L-hr, -0.001579 mg/L-hr c. The rate at which the concentration of drug in the bloodstream decreases is slowing to 0 as time increases. 55. A book publisher has a cost function given by $C(x)=\frac{x^3+2x+3}{x^2}$, where $x$ is the number of copies of a book in thousands and $C$ is the cost, per book, measured in dollars. Evaluate $C^{\prime}(2)$ and explain its meaning. 56. [T] According to Newton’s law of universal gravitation, the force $F$ between two bodies of constant mass $m_1$ and $m_2$ is given by the formula $F=\frac{G m_1 m_2}{d^2}$, where $G$ is the gravitational constant and $d$ is the distance between the bodies. 1. Suppose that $G, \, m_1$, and $m_2$ are constants. Find the rate of change of force $F$ with respect to distance $d$. 2. Find the rate of change of force $F$ with gravitational constant $G=6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2$, on two bodies 10 meters apart, each with a mass of 1000 kilograms. #### Solution a. $F^{\prime}(d)=\frac{-2Gm_1 m_2}{d^3}$ b. $-1.33 \times 10^{-7}$ N/m ## Glossary constant multiple rule the derivative of a constant $c$ multiplied by a function $f$ is the same as the constant multiplied by the derivative: $\frac{d}{dx}(cf(x))=cf^{\prime}(x)$ constant rule the derivative of a constant function is zero: $\frac{d}{dx}(c)=0$, where $c$ is a constant difference rule the derivative of the difference of a function $f$ and a function $g$ is the same as the difference of the derivative of $f$ and the derivative of $g$: $\frac{d}{dx}(f(x)-g(x))=f^{\prime}(x)-g^{\prime}(x)$ power rule the derivative of a power function is a function in which the power on $x$ becomes the coefficient of the term and the power on $x$ in the derivative decreases by 1: If $n$ is an integer, then $\frac{d}{dx}(x^n)=nx^{n-1}$ product rule the derivative of a product of two functions is the derivative of the first function times the second function plus the derivative of the second function times the first function: $\frac{d}{dx}(f(x)g(x))=f^{\prime}(x)g(x)+g^{\prime}(x)f(x)$ quotient rule the derivative of the quotient of two functions is the derivative of the first function times the second function minus the derivative of the second function times the first function, all divided by the square of the second function: $\frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{f^{\prime}(x)g(x)-g^{\prime}(x)f(x)}{(g(x))^2}$ sum rule the derivative of the sum of a function $f$ and a function $g$ is the same as the sum of the derivative of $f$ and the derivative of $g$: $\frac{d}{dx}(f(x)+g(x))=f^{\prime}(x)+g^{\prime}(x)$
# Partial Derivatives of Multivariable Functions Examples and exercises on the calculations of partial derivatives are presented. While ordinary derivatives deal with functions of a single variable, partial derivatives are a type of derivative that generalize the concept of ordinary derivatives to multivariable functions. Formally, the partial derivative of a function $$f(x_1, x_2, ..., x_n)$$ with respect to one of its variables, say $$x_i$$, is denoted by $$\dfrac{\partial f}{\partial x_i}$$. It represents the rate of change of the function $$f$$ with respect to the variable $$x_i$$, while holding all other variables constant. Mathematically, the partial derivative of $$f$$ with respect to $$x_i$$ is defined as: $\dfrac{\partial f}{\partial x_i} = \lim_{h \to 0} \dfrac{f(x_1, x_2, ..., x_i + h, ..., x_n) - f(x_1, x_2, ..., x_i, ..., x_n)}{h}$ In words, this definition says that the partial derivative of $$f$$ with respect to $$x_i$$ is the limit of the difference quotient as $$h$$ approaches zero, where $$h$$ represents a small change in the variable $$x_i$$, while keeping all other variables constant. Partial derivatives allow us to analyze how a function changes with respect to one of its variables while keeping the others fixed. When calculating the partial derivative of a function $$f(x, y)$$ with respect to $$x$$, denoted as $$\dfrac{\partial f}{\partial x}$$, we treat $$y$$ as a constant. Similarly, when calculating the partial derivative of $$f$$ with respect to $$y$$, denoted as $$\dfrac{\partial f}{\partial y}$$, we treat $$x$$ as a constant. We only consider how $$f$$ changes with respect to variations in $$y$$, while keeping $$x$$ constant. This is the fundamental concept behind partial derivatives, allowing us to analyze how a function changes with respect to one variable while holding others constant. Partial derivatives are used extensively in calculus, differential equations, optimization, and various fields of science and engineering, including physics, economics, and engineering. They play a crucial role in the study of multivariable calculus and the analysis of systems with multiple independent variables. A partial derivative calculator is included and may be used to check calculkations. ## Examples with Solutions ### Example 1 Calculate the partial derivatives of $$f$$ with respect to $$x$$ denoted by $$\dfrac{\partial f}{\partial x}$$ and the partial derivatives of $$f$$ with respect to $$y$$ denoted by $$\dfrac{\partial f}{\partial y}$$ where $$f$$ is given by $f(x, y) = 3x^2 + 4xy - y^2$. ### Solution to Example 1 #### 1. Derivative of $$3x^2 + 4xy - y^2$$ with respect to $$x$$: We calculate the partial derivative of $$f$$ with respect to $$x$$, denoted as $$\dfrac{\partial f}{\partial x}$$. $\dfrac{\partial f}{\partial x} = \dfrac{\partial}{\partial x} (3x^2 + 4xy - y^2)$ Using the rule of the sum, we write $\dfrac{\partial f}{\partial x} = \dfrac{\partial}{\partial x} (3x^2) + \dfrac{\partial}{\partial x} (4xy) - \dfrac{\partial}{\partial x} (y^2)$ NOTE that when calculating the partial derivative of a function $$f(x, y)$$ with respect to $$x$$, we treat $$y$$ as a constant. Using the power rule for differentiation, we have: $\dfrac{\partial}{\partial x} (3x^2) = 6x$ $\dfrac{\partial}{\partial x} (4xy) = 4y$ $\dfrac{\partial}{\partial x} (y^2) = 0$ Hence $\dfrac{\partial f}{\partial x} = 6x + 4y - 0$ $\dfrac{\partial f}{\partial x} = 6x + 4y$. #### 2. Derivative of $$3x^2 + 4xy - y^2$$ with respect to $$y$$: We now calculate the partial derivative of $$f$$ with respect to $$y$$, denoted as $$\dfrac{\partial f}{\partial y}$$. $\dfrac{\partial f}{\partial x} = \dfrac{\partial}{\partial y} (3x^2 + 4xy - y^2)$ Using the rule of the sum, we write $\dfrac{\partial f}{\partial x} = \dfrac{\partial}{\partial y} (3x^2) + \dfrac{\partial}{\partial y} (4xy) - \dfrac{\partial}{\partial y} (y^2)$ NOTE that when calculating the partial derivative of a function $$f(x, y)$$ with respect to $$y$$, we treat $$x$$ as a constant. Using different rules for differentiation, we have: $\dfrac{\partial}{\partial y} (3x^2) = 0$ $\dfrac{\partial}{\partial y} (4xy)= 4x$ $\dfrac{\partial}{\partial y} (y^2) = 2y$ Hence $\dfrac{\partial f}{\partial y} = 0 + 4x - 2y$ $\dfrac{\partial f}{\partial y} = 4x - 2y$. ### Example 2 Calculate the partial derivatives of $$g$$ with respect to $$x$$, $$y$$ and $$z$$, where $$g$$ is given by $g(x, y, z) = e^{xy} \cos(z)$. ### Solution to Example 2 To calculate the partial derivatives of $$g(x, y, z) = e^{xy} \cos(z)$$ with respect to $$x$$, $$y$$, and $$z$$, we treat each variable as independent and differentiate each term of $$g$$ with respect to the respective variable while holding the other variables constant. Let's calculate each partial derivative step by step: #### 1. Partial derivative with respect to $$x$$: $\dfrac{\partial g}{\partial x} = \dfrac{\partial}{\partial x} \left( e^{xy} \cos(z) \right)$ Using the product rule for differentiation: $\dfrac{\partial}{\partial x} \left( e^{xy} \cos(z) \right) = \dfrac{\partial}{\partial x} \left( e^{xy} \right) \cos(z) + e^{xy} \dfrac{\partial}{\partial x} \left( \cos(z) \right)$ Now, we calculate each term separately: $\dfrac{\partial}{\partial x} \left( e^{xy} \right) = y e^{xy}$ $\dfrac{\partial}{\partial x} \left( \cos(z) \right) = 0$ So, putting it all together: $\dfrac{\partial g}{\partial x} = y e^{xy} \cos(z)$ #### 2. Partial derivative with respect to $$y$$: $\dfrac{\partial g}{\partial y} = \dfrac{\partial}{\partial y} \left( e^{xy} \cos(z) \right)$ Using the product rule for differentiation: $\dfrac{\partial}{\partial y} \left( e^{xy} \cos(z) \right) = \dfrac{\partial}{\partial y} \left( e^{xy} \right) \cos(z) + e^{xy} \dfrac{\partial}{\partial y} \left( \cos(z) \right)$ Now, we calculate each term separately: $\dfrac{\partial}{\partial y} \left( e^{xy} \right) = x e^{xy}$ Since $$\cos(z)$$ does not depend on $$y$$, its derivative with respect to $$y$$ is zero: $\dfrac{\partial}{\partial y} \left( \cos(z) \right) = 0$ So, putting it all together: $\dfrac{\partial g}{\partial y} = x e^{xy} \cos(z)$ #### 3. Partial derivative with respect to $$z$$: $\dfrac{\partial g}{\partial z} = \dfrac{\partial}{\partial z} \left( e^{xy} \cos(z) \right)$ Using the product rule for differentiation: $\dfrac{\partial}{\partial z} \left( e^{xy} \cos(z) \right) = \dfrac{\partial}{\partial z} \left( e^{xy} \right) \cos(z) + e^{xy} \dfrac{\partial}{\partial z} \left( \cos(z) \right)$ Now, we calculate each term separately: a. Derivative of $$e^{xy}$$ with respect to $$z$$: Since $$e^{xy}$$ does not depend on $$z$$, its derivative with respect to $$z$$ is zero: $\dfrac{\partial}{\partial z} \left( e^{xy} \right) = 0$ b. Derivative of $$\cos(z)$$ with respect to $$z$$: $\dfrac{\partial}{\partial z} \left( \cos(z) \right) = -\sin(z)$ So, putting it all together: $\dfrac{\partial g}{\partial z} = - e^{xy} \sin(z)$ Therefore, the partial derivatives of $$g$$ with respect to $$x$$, $$y$$, and $$z$$ are: $\dfrac{\partial g}{\partial x} = y e^{xy} \cos(z)$ $\dfrac{\partial g}{\partial y} = x e^{xy} \cos(z)$ $\dfrac{\partial g}{\partial z} = - e^{xy} \sin(z)$ ## Exercises with Solutions Find the partial derivatives of the functions 1. $$g(u,v) = u^2 \; v^2 + e^{u^2+v^2}$$ 2. $$f(x,y,z) = \sin (xy )\;\ln (xyz )$$ 3. $$h(x,y,z) = \dfrac{z}{x \;y \;z +1}$$ ### Solutions to the Above Exercises 1. $$\dfrac{\partial g}{\partial u} = 2 \;u \;v^2 + 2u \; e^{u^2+v^2}$$ $$\dfrac{\partial g}{\partial v} = 2 \;v \;u^2 + 2v \; e^{u^2+v^2}$$ 2. $$\dfrac{\partial f}{\partial x} = y \;\cos(xy)\;\ln (xyz )+\dfrac{\sin (xy )}{x}$$ $$\dfrac{\partial f}{\partial y} = x\;\cos (xy )\;\ln (xyz )+\dfrac{\sin (xy )}{y}$$ $$\dfrac{\partial f}{\partial z} = \dfrac{\sin (xy)}{z}$$ 3. $$\dfrac{\partial h}{\partial x} = -\dfrac{y z^2 }{\left(zxy+1\right)^2}$$ $$\dfrac{\partial h}{\partial y} = -\dfrac{x z^2}{\left(zxy+1\right)^2}$$ $$\dfrac{\partial h}{\partial z} = \dfrac{1}{\left(zxy+1\right)^2}$$
Stat Trek Teach yourself statistics Teach yourself statistics # Combinations and Permutations Calculator Find the number of combinations and permutations when you choose a subset of r elements from a set of n elements. For help in using the calculator, read the Frequently-Asked Questions or review the Sample Problem. • Enter the number of elements in the subset (r). • Enter the number of elements in the set (n). • Click the Calculate button to display results. Subset size (r) Set size (n) Combinations Permutations Instructions: To find the answer to a frequently-asked question, simply click on the question. ### What is a permutation? A permutation is an arrangement of all or part of a set of objects, with regard to the order of the arrangement. For example, suppose we have a set of three letters: A, B, and C. We might ask how many ways we can arrange 2 letters from that set. Each possible arrangement would be an example of a permutation. The complete list of possible permutations would be: AB, AC, BA, BC, CA, and CB. When statisticians refer to permutations, they use a specific terminology. They describe permutations as n distinct objects taken r at a time. Translation: n refers to the number of objects from which the permutation is formed; and r refers to the number of objects used to form the permutation. Consider the example from the previous paragraph. The permutations were formed from 3 letters (A, B, and C), so n = 3; and each permutation consisted of 2 letters, so r = 2. Note that AB and BA are considered to be two permutations, because the order in which objects are selected matters. This is the key distinction between a combination and a permutation. A combination focuses on the selection of objects without regard to the order in which they are selected. A permutation, in contrast, focuses on the arrangement of objects with regard to the order in which they are arranged. Bottom line: AB and BA represent two permutations, but only one combination. For an example that counts permutations, see Sample Problem 1. ### What is a combination? A combination is a selection of all or part of a set of objects, without regard to the order in which objects are selected. For example, suppose we have a set of three letters: A, B, and C. We might ask how many ways we can select 2 letters from that set. Each possible selection would be an example of a combination. The complete list of possible selections would be: AB, AC, and BC. When statisticians refer to combinations, they use a specific terminology. They describe combinations as n distinct objects taken r at a time. Translation: n refers to the number of objects from which the combination is formed; and r refers to the number of objects used to form the combination. Consider the example from the previous paragraph. The combinations were formed from 3 letters (A, B, and C), so n = 3; and each combination consisted of 2 letters, so r = 2. Note that AB and BA are considered to be one combination, because the order in which objects are selected does not matter. This is the key distinction between a combination and a permutation. A combination focuses on the selection of objects without regard to the order in which they are selected. A permutation, in contrast, focuses on the arrangement of objects with regard to the order in which they are arranged. Bottom line: AB and BA represent two permutations, but only one combination. For an example that counts the number of combinations, see Sample Problem 2. ### How do you count the number of combinations? A combination is a selection of all or part of a set of objects, without regard to the order in which they were selected. This means that XYZ is considered the same combination as ZYX. The number of combinations of r objects that can be selected from a set of n objects is denoted by nCr. And the formula for computing that number is: nCr = n(n - 1)(n - 2) ... (n - r + 1)/r! = n! / r!(n - r)! Note: In the formula above, n! refers to n factorial, where n! is equal to n(n-1)(n-2) ... (3)(2)(1). ### How do you count the number of permutations? A permutation is a selection of all or part of a set of objects, with regard to the order in which they were selected. This means that XYZ is considered a different permutation than ZYX. The number of permutations of r objects that can be selected from a set of n objects is denoted by nPr. And the formula for computing that number is: nPr = n(n - 1)(n - 2) ... (n - r + 1) = n! / (n - r)! Note: In the formula above, n! refers to n factorial, where n! is equal to n(n-1)(n-2) ... (3)(2)(1). ### What is the difference between a combination and a permutation? The distinction between a combination and a permutation has to do with the sequence or order in which objects appear. A combination focuses on the selection of objects without regard to the order in which they are selected. A permutation, in contrast, focuses on the arrangement of objects with regard to the order in which they are arranged. For example, consider the letters A and B. Using those letters, we can create two 2-letter permutations - AB and BA. Because order is important to a permutation, AB and BA are considered different permutations. However, AB and BA represent only one combination, because order is not important to a combination. ### What is E Notation? The Combinations and Permutations Calculator uses E notation to express very large numbers. E notation is a way to write numbers that are too large or too small to be concisely written in a decimal format. With E notation, the letter E represents "times ten raised to the power of". Here is an example of a number written using E notation: 3.02E+12 = 3.02 * 1012 = 3,020,000,000,000 ### How accurate is this calculator? When the number of combinations or permutations is displayed as an ordinary integer, the result is exact. When the number of combinations or permutations is displayed using E notation, the result is not exact; but it is a very good approximation, accurate to 16 decimals. ## Sample Problem 1. How many 3-digit numbers can be formed from the digits 1, 2, 3, 4, 5, 6, and 7, if each digit can be used only once? Solution: The solution to this problem involves counting the number of permutations of 7 distinct objects, taken 3 at a time. The number of permutations of n distinct objects, taken r at a time is: nPr = n! / (n - r)! 7P3 = 7! / (7 - 3)! = 7! / 4! = (7)(6)(5) = 210 Thus, 210 different 3-digit numbers can be formed from the digits 1, 2, 3, 4, 5, 6, and 7. To solve this problem using the Combination and Permutation Calculator, do the following: • Enter "3" for "Subset size". • Enter "7" for "Set size". • Click the "Calculate" button. The answer, 210, is displayed in the "Permutations" textbox, as shown below. 1. The Atlanta Braves are having a walk-on tryout camp for baseball players. Thirty players show up at camp, but the coaches can choose only four. How many ways can four players be chosen from the 30 that have shown up? Solution: The solution to this problem involves counting the number of combinations of 30 players, taken 4 at a time. The number of combinations of n distinct objects, taken r at a time is: nCr = n! / r! (n - r)! 30C4 = 30! / 4!(30 - 4)! = 30! / 4! 26! = 27,405 Thus, 27,405 different groupings of 4 players are possible. To solve this problem using the Combination and Permutation Calculator, do the following: • Enter "4" for "Subset size". • Enter "30" for "Set size". • Click the "Calculate" button. The answer, 27,405, is displayed in the "Combinations" textbox, as shown below. ## Problem? Oops! Something went wrong.
# 2018 AMC 8 Problems/Problem 19 (diff) ← Older revision \| Latest revision (diff) \| Newer revision → (diff) ## Problem In a sign pyramid a cell gets a "+" if the two cells below it have the same sign, and it gets a "-" if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a "+" at the top of the pyramid? $[asy] unitsize(2cm); path box = (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--cycle; draw(box); label("+",(0,0)); draw(shift(1,0)box); label("-",(1,0)); draw(shift(2,0)box); label("+",(2,0)); draw(shift(3,0)box); label("-",(3,0)); draw(shift(0.5,0.4)box); label("-",(0.5,0.4)); draw(shift(1.5,0.4)box); label("-",(1.5,0.4)); draw(shift(2.5,0.4)box); label("-",(2.5,0.4)); draw(shift(1,0.8)box); label("+",(1,0.8)); draw(shift(2,0.8)box); label("+",(2,0.8)); draw(shift(1.5,1.2)box); label("+",(1.5,1.2)); [/asy]$ $\textbf{(A) } 2 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16$ ## Solution 1 You could just make out all of the patterns that make the top positive. In this case, you would have the following patterns: +−−+, −++−, −−−−, ++++, −+−+, +−+−, ++−−, −−++. There are 8 patterns and so the answer is $\boxed{\textbf{(C) } 8}$. -NinjaBoi2000 ## Solution 2 The top box is fixed by the problem. Choose the left 3 bottom-row boxes freely. There are $2^3=8$ ways. Then the left 2 boxes on the row above are determined. Then the left 1 box on the row above that is determined Then the right 1 box on that row is determined. Then the right 1 box on the row below is determined. Then the right 1 box on the bottom row is determined, completing the diagram. So the answer is $\boxed{\textbf{(C) } 8}$. ~BraveCobra22aops ## Solution 3 Let the plus sign represent 1 and the negative sign represent -1. The four numbers on the bottom are $a$, $b$, $c$, and $d$, which are either 1 or -1. $[asy] unitsize(2cm); path box = (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--cycle; draw(box); label("a",(0,0)); draw(shift(1,0)box); label("b",(1,0)); draw(shift(2,0)box); label("c",(2,0)); draw(shift(3,0)box); label("d",(3,0)); draw(shift(0.5,0.4)box); label("ab",(0.5,0.4)); draw(shift(1.5,0.4)box); label("bc",(1.5,0.4)); draw(shift(2.5,0.4)box); label("cd",(2.5,0.4)); draw(shift(1,0.8)box); label("ab^2c",(1,0.8)); draw(shift(2,0.8)box); label("bc^2d",(2,0.8)); draw(shift(1.5,1.2)box); label("ab^3c^3d",(1.5,1.2)); [/asy]$ Which means $ab^3c^3d$ = 1. Since $b$ and $c$ are either 1 or -1, $b^3 = b$ and $c^3 = c$. This shows that $abcd$ = 1. Therefore either $a$, $b$, $c$, and $d$ are all positive or negative, or 2 are positive and 2 are negative. There are 2 ways where $a$, $b$, $c$, and $d$ are 1 (1, 1, 1, 1) and (-1, -1, -1, -1) There are 6 ways where 2 variables are positive and 2 are negative: (1, 1, -1, -1), (1, -1, 1, -1), (-1, 1, 1, -1), (-1, -1, 1, 1), (-1, 1, -1, 1), and (-1, -1, 1, 1). So the answer is $\boxed{\textbf{(C) } 8}$. ~atharvd ~savannahsolver
# Reciprocals – Definition & Examples ## What is a Reciprocal? In mathematics, the reciprocal, also known as multiplicative inverse, is the inverse of a number x. denoted as 1/x or x-1. This means that the product of a number x and its reciprocal yields 1. The inverse of a fraction a/b is denoted as (a/b)-1, which is b/a. This article discusses the steps on how to find the reciprocal of a number, mixed numbers, fractions and decimals. ## How to Find Reciprocals? The reciprocal of a number is simply the number that has been flipped or inverted upside-down. This entails transposing a number such that the numerator and denominator are placed at the bottom and top respectively. To find the reciprocal of a whole number, just convert it into a fraction in which the original number is the denominator and the numerator is 1. Example 1 The reciprocal of 2/3 is 3/2. The product of 2/3 and its reciprocal 3/2 is 1. 2/3 x 3/2 = 1 Example 2 The reciprocal of a whole number 7 is 1/7 because 7 x 1/7 = 1. ### How to Find the Reciprocal of a Mixed Number? In order to find the reciprocal of a mixed fraction, convert it into improper fraction first and then apply the same rule we learnt above. Example 3 Find the reciprocal of 4 1/2. Solution • Convert a mixed fraction into an improper fraction as calculated below. 4 1/2 = {(4 x 2) + 1}/ 2 = 9/2 • Now flip the numerator and denominator of 9/2. • Therefore, the solution for the reciprocal of 4 1/2 is 2/9. ### How to Find the Reciprocal of Decimal Numbers? Like other numbers, decimal numbers too have reciprocals. Calculating the reciprocal of a decimal number can be done in the following ways: • Convert the decimal into an equivalent fraction, for example, 0.25 = 1/4, and therefore, the reciprocal is 4/1 = 4. • You can also use a calculate to divide 1 by the fraction. For example, the reciprocal of 0.25 = 1/0.25 = 4 It can be noted that dividing 1 by a fraction is the same as multiplying the reciprocal of the number by 1. For example, 5 ÷ 1/4 = 5 x 4/1 = 20 Example 4 Solve the following problems: a. Find the reciprocal of 5 Solution 5 = 5/1 So, the reciprocal of 3 = 1/5 b. Find the reciprocal of 1/4 Solution To find the reciprocal of 1/4, invert the numerator and denominator. The reciprocal of 1/4 = 4 c. Determine the reciprocal of 10/3 Solution Step 1: To find the reciprocal of 10/3, flip the numerator and denominator. The reciprocal = 3/10. Example 5 If 4/7 of a number x is 84. What is the value of x? Solution 4/7 of a number x = 84 Write the mathematical equation: (4/7) x = 84 Multiply both sides by the reciprocal of 4/7 Number x = 84 × 7/4 = 21 × 7 = 147 And thus, the number x is 147. Example 6 A half of the students in a college are boys, 3/5 of these boys take science courses and the rest take humanities. What fraction of the boys take humanities? Solution Fraction of boys in the college = 1/2 Fraction of boys who take sciences = 3/5 of 1/2 = 3/5 × 1/2 = 3 × 1/5 × 2 = 3/10 Therefore, 3/10 of the boys take humanities. Example 7 Pedro has written three-fifth of his 75 paged research work. How many pages are left to complete writing his research? Solution Number of pages written = 3/5 of 75 = 3/5 × 7 = 45 pages. Number of pages left= 75 – 45. = 30 pages. Example 8 A herd of cows in a farm produces 99 liters of milk daily. If each cow produces one-third of total milk produced in a day. How many cows are in the farm if 7700 liters of milk is produced weekly. Solution A herd of cows produces 99 liters of milk daily. One cow produces 1/3 of total milk daily = 1/3 of 99 Therefore, one cow produces 11 liters. Total number of animals in the farm= (7700/7) ÷ 11 = 100 cows
Ex 5.2 Chapter 5 Class 10 Arithmetic Progressions Serial order wise ### Transcript Ex 5.2, 20 Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the nth week, her weekly savings become Rs 20.75, find n. Saving made first week = Rs 5 Saving made in second week = Rs 5 + 1.75 = Rs 6.75 Saving made in third week = 6. 75 + 1.75 = Rs 8.50 So, the series is 5, 6.75, 8.50 …… Since difference is same, it is an AP First Term = a = 5 Common difference = d = 1.75 Given, In nth week, her savings become Rs 20.75 So, an = 20.75 We need to find n We know that an = a + (n – 1) d Putting a = 5 , d = 1.75, an = 20.75 in formula 20.75 = 5 + (n – 1) × 1.75 20.75 – 5 = (n – 1) × 1.75 15.75 = (n – 1) × 1.75 15.75/1.75 = n – 1 1575/175 = n – 1 9 = n – 1 n – 1 = 9 n = 9 + 1 n = 10 Hence, in 10th week, her weekly savings become Rs 20.75
# Important CAT Remainder Theorem Questions (with Notes) PDF 0 711 The Remainder theorem is a part of the Number Systems topic in the CAT Quant Section. You can check out these CAT Remainder Questions from Previous years. In this article, we will look into some important CAT Remainder theorem Questions (with Notes) PDF. These are a good source for practice; If you want to practice these questions, you can download these CAT Remainder theorem Questions PDF below, which is completely Free. • CAT Remainder theorem Questions – Tip 1: If you’re starting the prep, firstly understand the CAT Number System Syllabus; Based on our analysis of the previous years CAT number system questions,Â only a few questions were asked from CAT Remainder theorem questions. • CAT Remainder theorem Questions – Tip 2: If you’re not very strong in this topic, don’t spend too much time on these theorems. Only after you have completed all other topics, you can go to CAT remainder theorem questions and concepts. • You can practice these CAT remainder theorem questions in PDF with video solutions. Learn all the major formulae from these concepts and the Important Number System for CAT tricks Formulas PDF here. Question 1:Â A positive whole number M less than 100 is represented in base 2 notation, base 3 notation, and base 5 notation. It is found that in all three cases the last digit is 1, while in exactly two out of the three cases the leading digit is 1. Then M equals a)Â 31 b)Â 63 c)Â 75 d)Â 91 Solution: SinceÂ Â in all three cases the last digit is 1, the number should give remainder 1 when divided individually by 2,3,5 . So the no. may be 31 or 91 . Now 31 in base 2,3 and 5 give first digit as 1 in all the 3 cases while 91 givesÂ exactly two out of the three cases the leading digit as 1. Hence option D. Question 2:Â The remainder, when $(15^{23} + 23^{23})$ is divided by 19, is a)Â 4 b)Â 15 c)Â 0 d)Â 18 Solution: The remainder when $15^{23}$ is divided by 19 equals $(-4)^{23}$ The remainder when $23^{23}$ is divided by 19 equals $4^{23}$ So, the sum of the two equals$(-4)^{23}+(4)^{23}=0$ Question 3:Â If x = $(16^3 + 17^3+ 18^3+ 19^3 )$, then x divided by 70 leaves a remainder of a)Â 0 b)Â 1 c)Â 69 d)Â 35 Solution: We know that x = $16^3 + 17^3 + 18^3 + 19^3 = (16^3 + 19^3) + (17^3 + 18^3)$ = $(16 + 19)(16^2 – 16 * 19 + 19^2) + (17 + 18)(17^2 – 17 * 18 + 18^2)$ = 35 Ã— odd + 35 Ã— odd = 35 Ã— even = 35 Ã— (2k) =>Â x = 70k => Remainder when divided by 70 is 0. Question 4:Â Let $n!=123* …n$ for integer $n \geq 1$. If $p = 1!+(22!)+(33!)+… +(1010!)$, then $p+2$ when divided by 11! leaves a remainder of a)Â 10 b)Â 0 c)Â 7 d)Â 1 Solution: According to given condiiton we have p = (1 Ã— 1!) + (2 Ã— 2!) + (3 Ã— 3!) + (4 Ã— 4!) + â€¦ + (10 Ã— 10!) . SoÂ n Ã— n! = [(n + 1) – 1] Ã— n! = (n + 1)! – n!. SoÂ equation becomesÂ p = 2! – 1! + 3! – 2! + 4! – 3! + 5! – 4! +â€¦ + 11! – 10!. Â SoÂ p = 11! – 1! = 11! – 1.Â Â p + 2 = 11! + 1 .So when it is Â divided by 11! gives a remainder of 1. Hence, option 4. Question 5:Â Let N = 1421 * 1423 * 1425. What is the remainder when N is divided by 12? a)Â 0 b)Â 9 c)Â 3 d)Â 6 Solution: The numbers 1421, 1423 and 1425 when divided by 12 give remainder 5, 7 and 9 respectively. 579 mod 12 = 11 * 9 mod 12 = 99 mod 12 = 3 Question 6:Â The integers 34041 and 32506 when divided by a three-digit integer n leave the same remainder. What is n? a)Â 289 b)Â 367 c)Â 453 d)Â 307 Solution: The difference of the numbers = 34041 – 32506 = 1535 The number that divides both these numbers must be a factor of 1535. 307 is the only 3 digit integer that divides 1535. Question 7:Â Convert the number 1982 from base 10 to base 12. The result is: a)Â 1182 b)Â 1912 c)Â 1192 d)Â 1292 Solution: Quotient of 1982/12 = 165, remainder = 2 Quotient of 165/12 = 13, remainder = 9 Quotient of 13/12 = 1, remainder = 1 Remainder of 1/12 = 1 So, the required number in base 12 = 1192 Question 8:Â When $2^{256}$ is divided by 17, the remainder would be a)Â 1 b)Â 16 c)Â 14 d)Â None of these Solution: $2^4 = 16 = -1$ (mod $17$) So, $2^{256} = (-1)^{64}$(mod $17$) $= 1$ (mod $17$) Hence, the answer is 1. Option a). Question 9:Â After the division of a number successively by 3, 4 and 7, the remainders obtained are 2, 1 and 4 respectively. What will be the remainder if 84 divides the same number? a)Â 80 b)Â 75 c)Â 41 d)Â 53 Solution: Since after division of a number successively by 3, 4 and 7, the remainders obtained are 2, 1 and 4 respectively, the number is of form ((((44)+1)3)+2)k = 53K Let k = 1; the number becomes 53 If it is divided by 84, the remainder is 53. Option d) is the correct answer. Alternative Solution. Consider only for 3 and 4 and the remainders are 2 and 1 respectively. So 5 is the first number to satisfy both the conditions. The number will be of the form 12k+5. Put different integral values of k to find whether it will leave remainder 5 when divided by 7. So the first number to satisfy such condition is 48×4+5= 53 Question 10:Â The remainder when $2^{60}$ is divided by 5 equals a)Â 0 b)Â 1 c)Â 2 d)Â None of these Solution: $2^{60}$ or $4^{30}$ when divided by 5 So according to remainder theorem remainder will be $(-1)^{30}$ = 1. Question 11:Â Find the minimum integral value of n such that the division $\frac{55n}{124}$ leaves no remainder. a)Â 124 b)Â 123 c)Â 31 d)Â 62 Solution: As 55 and 124 don’t have any common factor, and n has to be aÂ minimumÂ integer, Hence, it should be 124 only. So that given equation won’t have a remainder. Question 12:Â A certain number, when divided by 899, leaves a remainder 63. Find the remainder when the same number is divided by 29. a)Â 5 b)Â 4 c)Â 1 d)Â Cannot be determined Solution: Let’s say N is our number N = (899K + 63) or N = ($29 \times 31$K) + 63 So when it is divided by 29, remainder will be $\frac{63}{29}$ = 5 Question 13:Â A is the set of positive integers such that when divided by 2, 3, 4, 5, 6 leaves the remainders 1, 2, 3, 4, 5 respectively. How many integers between 0 and 100 belong to set A? a)Â 0 b)Â 1 c)Â 2 d)Â None of these Solution: Let the number ‘n’ belong to the set A. Hence, the remainder when n is divided by 2 is 1 The remainder when n is divided by 3 is 2 The remainder when n is divided by 4Â is 3 The remainder when n is divided by 5Â is 4 and The remainder when n is divided by 6Â is 5 So, when (n+1) is divisible by 2,3,4,5 and 6. Hence, (n+1) is of the form 60k for some natural number k. And n is of the form 60k-1 Between numbers 0 and 100, only 59 is of the form above and hence the correct answer is 1 Question 14:Â A number is formed by writing first 54 natural numbers next to each other as 12345678910111213 … Find the remainder when this number is divided by 8. a)Â 1 b)Â 7 c)Â 2 d)Â 0 Solution: For a number to be divisible by 8, last 3 digits must be divisible by 8. Last 3 digits of this number are 354. 354 mod 8 = 2 Hence, 2 is the remainder. Question 15:Â The remainder when $7^{84}$ is divided by $342$Â is : a)Â 0 b)Â 1 c)Â 49 d)Â 341 $7^3$ = 343 $7^{84}$ = $(7^3)^{28}$ = $343^{28}$ $343^{28}$ mod 342 = $1^{28}$ mod 342 = 1
# RD Sharma Solutions for Class 6 Chapter 7: Decimals Exercise 7.4 Learning Mathematics is made fun for the students with the help of RD Sharma Solutions Class 6. Exercise 7.4 covers major concepts like steps followed to convert the given fractions as decimals and vice versa. The fractions can be expressed as decimals using 10, 100 and 1000 as the denominator. Students gain a better hold on the concepts covered in Chapter 7 using RD Sharma Solutions for Class 6 Maths Chapter 7 Decimals Exercise 7.4 PDF which are provided here. ## RD Sharma Solutions for Class 6 Chapter 7: Decimals Exercise 7.4 Download PDF ### Access RD Sharma Solutions for Class 6 Chapter 7: Decimals Exercise 7.4 1. Express the following fractions as decimals: (i) 23/10 (ii) 139/100 (iii) 4375/1000 (iv) 12 1/2 (v) 75 1/4 (vi) 25 1/8 (vii) 18 3/24 (viii) 39 7/35 (ix) 15 1/25 (x) 111/250 Solution: (i) 23/10 It can be written as = 20 + 3/10 We get = 20/10 + 3/10 = 2 + 3/10 So we get = 2.3 (ii) 139/100 It can be written as = 100 + 30 + 9/100 We get = 100/100 + 30/100 + 9/100 = 1 + 3/10 + 9/100 So we get = 1.39 (iii) 4375/1000 It can be written as = 4000 + 300 + 70 + 5/1000 We get = 4000/1000 + 300/1000 + 70/1000 + 5/1000 = 4 + 3/10 + 7/100 + 5/1000 So we get = 4.375 (iv) 12 1/2 It can be written as = 12 + 1/2 Multiplying and dividing by 5 to get denominator as 10 = 12 + [(1/2) Ã— (5/5)] On further calculation = 12 + 5/10 So we get = 12.5 (v) 75 1/4 It can be written as = 75 + 1/4 Multiplying and dividing by 25 to get 100 as denominator = 75 + [(1/4) Ã— (25/25)] On further calculation = 75 + 25/100 = 75.25 (vi) 25 1/8 It can be written as = 25 + 1/8 Multiplying and dividing by 125 to get 1000 as denominator = 25 + [(1/8) Ã— (125/125)] On further calculation = 25 + 125/1000 = 25.125 (vii) 18 3/24 It can be written as = 18 + 3/24 We get = 18 + 1/8 Multiplying and dividing by 125 to get 1000 as denominator = 18 + [(1/8) Ã— (125/125)] On further calculation = 18 + 125/1000 = 18.125 (viii) 39 7/35 It can be written as = 39 + 7/35 We get = 39 + 1/5 Multiplying and dividing by 2 to get 10 as denominator = 39 + [(1/5) Ã— (2/2)] On further calculation = 39 + 2/10 = 39.2 (ix) 15 1/25 It can be written as = 15 + 1/25 Multiplying and dividing by 4 to get 100 as denominator = 15 + [(1/25) Ã— (4/4)] On further calculation = 15 + 4/100 = 15.04 (x) 111/250 It can be written as = 111 Ã— [(1/250) Ã— (4/4)] On further calculation = 444/1000 By division = 0.444 2. Express the following decimals as fractions in the lowest form: (i) 0.5 (ii) 2.5 (iii) 0.60 (iv) 0.18 (v) 5.25 (vi) 7.125 (vii) 15.004 (viii) 20.375 (ix) 600.75 (x) 59.48 Solution: (i) 0.5 It can be written as = 5/10 By division = 1/2 (ii) 2.5 It can be written as = 25/10 By division = 5/2 (iii) 0.60 It can be written as = 60/100 By division = 3/5 (iv) 0.18 It can be written as = 18/100 By division = 9/50 (v) 5.25 It can be written as = 525/100 By division = 21/4 (vi) 7.125 It can be written as = 7125/1000 By division = 57/8 (vii) 15.004 It can be written as = 15004/1000 By division = 3751/250 (viii) 20.375 It can be written as = 20375/1000 By division = 163/8 (ix) 600.75 It can be written as = 60075/100 By division = 2403/4 (x) 59.48 It can be written as = 5948/100 By division = 1487/25
### 4.OA: Operations and Algebraic Thinking #### 1.1: Use the four operations with whole numbers to solve problems 4.OA.1: Interpret a multiplication equation as a comparison, e.g., interpret 35 = 5 × 7 as a statement that 35 is 5 times as many as 7 and 7 times as many as 5. Represent verbal statements of multiplicative comparisons as multiplication equations. 4.OA.2: Multiply or divide to solve word problems involving multiplicative comparison, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem, distinguishing multiplicative comparison from additive comparison. 4.OA.3: Solve multistep (two or more operational steps) word problems posed with whole numbers and having whole-number answers using the four operations, including problems in which remainders must be interpreted. Represent these problems using equations with a letter standing for the unknown quantity. Assess the reasonableness of answers using mental computation and estimation strategies including rounding. #### 1.2: Gain familiarity with factors and multiples 4.OA.4: Find all factor pairs for a whole number in the range 1–100. Recognize that a whole number is a multiple of each of its factors. Determine whether a given whole number in the range 1–100 is a multiple of a given one-digit number. Determine whether a given whole number in the range 1–100 is prime or composite. #### 1.3: Generate and analyze patterns 4.OA.5: Generate a number or shape pattern that follows a given rule. Identify apparent features of the pattern that were not explicit in the rule itself. ### 4.NBT: Number and Operations in Base Ten #### 2.1: Generalize place value understanding for multi-digit whole numbers 4.NBT.1: Recognize that in a multi-digit whole number, a digit in one place represents ten times what it represents in the place to its right. 4.NBT.2: Read and write multi-digit whole numbers using base-ten numerals, number names, and expanded form. Compare two multi-digit numbers based on meanings of the digits in each place, using >, =, and < symbols to record the results of comparisons. 4.NBT.3: Use place value understanding to round multi-digit whole numbers to any place. #### 2.2: Use place value understanding and properties of operations to perform multi-digit arithmetic 4.NBT.4: Fluently add and subtract (including subtracting across zeros) multi-digit whole numbers using the standard algorithm. 4.NBT.5: Multiply a whole number of up to four digits by a one-digit whole number, and multiply two two-digit numbers, using strategies based on place value and the properties of operations. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models. 4.NBT.6: Find whole-number quotients and remainders with up to four-digit dividends and one-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models. ### 4.NF: Number and Operations—Fractions #### 3.1: Extend understanding of fraction equivalence and ordering 4.NF.1: Recognizing that the value of “n” cannot be 0, explain why a fraction a/b is equivalent to a fraction (n × a)/ (n × b) by using visual fraction models, with attention to how the number and size of the parts differ even though the two fractions themselves are the same size. Use this principle to recognize and generate equivalent fractions. 4.NF.2: Compare two fractions with different numerators and different denominators, e.g., by creating common denominators or numerators, or by comparing to a benchmark fraction such as 1/2. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with symbols >, =, or <, and justify the conclusions, e.g., by using a visual fraction model. #### 3.2: Build fractions from unit fractions by applying and extending previous understandings of operations on whole numbers 4.NF.3: Understand a fraction a/b with a > 1 as a sum of fractions 1/b. 4.NF.3a: Understand addition and subtraction of fractions as joining and separating parts referring to the same whole. 4.NF.3b: Decompose a fraction into a sum of fractions with the same denominator in more than one way, recording each decomposition by an equation. Justify decompositions, e.g., by using a visual fraction model (including, but not limited to: concrete models, illustrations, tape diagram, number line, area model, etc.). 4.NF.3c: Add and subtract mixed numbers with like denominators, e.g., by replacing each mixed number with an equivalent fraction, and/or by using properties of operations and the relationship between addition and subtraction. 4.NF.3d: Solve word problems involving addition and subtraction of fractions referring to the same whole and having like denominators, e.g., by using visual fraction models and equations to represent the problem. 4.NF.4: Apply and extend previous understandings of multiplication to multiply a fraction by a whole number. 4.NF.4a: Understand a fraction a/b as a multiple of 1/b. 4.NF.4b: Understand a multiple of a/b as a multiple of 1/b, and use this understanding to multiply a fraction by a whole number. #### 3.3: Understand decimal notation for fractions, and compare decimal fractions 4.NF.6: Use decimal notation for fractions with denominators 10 or 100. 4.NF.7: Compare two decimals to hundredths by reasoning about their size. Recognize that comparisons are valid only when the two decimals refer to the same whole. Record the results of comparisons with the symbols >, =, or <, and justify the conclusions, e.g., by using a visual model. ### 4.MD: Measurement and Data #### 4.1: Solve problems involving measurement and conversion of measurements from a larger unit to a smaller unit 4.MD.1: Know relative sizes of measurement units within one system of units including km, m, cm, mm; kg, g, mg; lb, oz.; l, ml; hr, min, sec. Within a single system of measurement, express measurements in a larger unit in terms of a smaller unit. Record measurement equivalents in a two-column table. 4.MD.2: Use the four operations to solve word problems involving intervals of time, money, distances, liquid volumes and masses of objects including problems involving simple fractions or decimals, and problems that require expressing measurements given in a larger unit in terms of a smaller unit. Represent measurement quantities using diagrams such as number line diagrams that feature a measurement scale. 4.MD.3: Apply the area and perimeter formulas for rectangles in real world and mathematical problems. #### 4.3: Geometric measurement: understand concepts of angle and measure angles 4.MD.7: Recognize angle measure as additive. When an angle is decomposed into nonoverlapping parts, the angle measure of the whole is the sum of the angle measures of the parts. Solve addition and subtraction problems to find unknown angles on a diagram in real world and mathematical problems, e.g., by using an equation with a symbol for the unknown angle measure. ### 4.G: Geometry #### 5.1: Draw and identify lines and angles, and classify shapes by properties of their lines and angles 4.G.1: Draw points, lines, line segments, rays, angles (right, acute, obtuse), and perpendicular and parallel lines. Identify these in two-dimensional figures. 4.G.2: Classify two-dimensional figures based on the presence or absence of parallel or perpendicular lines, or the presence or absence of angles of a specified size. Recognize right triangles as a category, and identify right triangles. 4.G.3: Recognize a line of symmetry for a two-dimensional figure as a line across the figure such that the figure can be folded along the line into matching parts. Identify line-symmetric figures and draw lines of symmetry. Correlation last revised: 9/15/2020 This correlation lists the recommended Gizmos for this state's curriculum standards. Click any Gizmo title below for more information.
Presentation is loading. Please wait. # 14.4 The Normal Distribution ## Presentation on theme: "14.4 The Normal Distribution"— Presentation transcript: 14.4 The Normal Distribution Probability and Statistics Normal Distribution or Normal Curve Most common distribution in real-life and in statistics Curve Shaped AKA: Bell-Curve Properties of a Normal Distribution A normal curve is bell shaped. The highest point on the curve is at the mean of the distribution. (also median and mode) The mean, median, and mode of the distribution are the same. The total area under the curve is 1. The Rule The Rule 68% of the data is within 1 standard deviation of the mean (highest point). This means there is 34% on each side. 95% of the data is within 2 standard deviations of the mean. This means that there is 47.5% on each side. 99.7% of the data is within 3 standard deviations of the mean. This means that there is 49.85% on each side. The Rule 34% 34% 2.35% 2.35% 13.5% 13.5% 68 % 95 % 99.7 % Nearly all (99.7%) of the values lie within 3 standard deviations of the mean (or between the mean minus 3 times the standard deviation and the mean plus 3 times the standard deviation). EXAMPLE: Suppose a distribution of 1,000 scores represent scores on a standardized test. The mean of the distribution is 450 and the standard deviation is 25. How many scores do we expect to fall between 425 and 475? How many scores do we expect to fall above 500? Part 1 If the mean is 450, then 425 and 475 represent one standard deviation below and one standard deviation above the mean. This falls with 68% of the data. So… 68% of 1000 is: .68 x 1000 = 680 So 680 scores are in the range of Part 2 500 represents 2 standard deviations above the mean. 95% of our data falls within that range. We want to find the values above 500. This would be represented by (1/2) of 5% or 2.5% of the data. So, 2.5% of 1000 scores = .025 x 1000 = scores would be above 500. Assignment: pg. 831 #6-16 even 14.4. Notes Part 2 What would be the z-score? Z-Scores Represents the number of standard deviations a data value is from the mean. Positive above mean, negative below mean In the last example, we said that 500 was 2 standard deviations above the mean (the mean was 450 and standard deviation 25)… What would be the z-score? Z = 2 Z-Scores The z-score table can be used for both positive and negative z-scores. The table tells you “A” - represents the area underneath the curve. (Remember that the area underneath the entire curve is %). Area represents the percentage of scores within the specified range. For example, a z-score of 1 (1 standard deviation above the mean) shows an area of 0.34 34% of the data is 1 standard deviation above the mean… Finding Areas under a Normal Curve Using your z-score table, find the percentage of data (area under the curve) that lie in the following regions for a standard normal distribution: draw a diagram for each! A). Between z = 0 and z = 1.3 B). Between z = 1.5 and z = 2.1 C). Between z = 0 and z = A). Between z = 0 and z = 1.3 By the z-score table, I look at z = 1.3 The area of the region is .403 This means 40.3% of the data will fall within this range. B). Between z = 1.5 and z = 2.1 Z= % (represents between z=0 and z=1.5) Z= % (represents between z=0 and z=2.1) So between those scores (1.5 – 2.1) we need to subtract. = 4.9% SSS (same-side-subtract) C). z = 0 and z = -1.83 Since the data is evenly distributed on both sides of the mean (z=0), just look up 1.83 on your chart. z = 1.83 same as z = -1.83, just on opposite sides of the mean… Z= % Assignment 14.4 Worksheet Converting Raw Scores to Z-Scores Value Mean Standard Deviation Examples: Suppose the mean of a normal distribution is 20 and its standard deviation is 3. draw a diagram to represent each situation, and answer the question. a). Find the z-score that corresponds to a raw score of 25. b). Find the z-score that corresponds to a raw score of 16. c). Find the raw score that corresponds to a z score of 2.1 Suppose the mean of a normal distribution is 20 and its standard deviation is 3. a). Find the z-score that corresponds to a raw score of 25. Suppose the mean of a normal distribution is 20 and its standard deviation is 3. b). Find the z-score that corresponds to a raw score of 16. Suppose the mean of a normal distribution is 20 and its standard deviation is 3. c). Find the raw score that corresponds to a z score of 2.1 26.3 = x Suppose the mean of a normal distribution is 25 and its standard deviation is 5. Find the percentage of values that fall between: *draw diagram for each.. 25 and 29 22 and 25 Under 17 Percentile If the mean salary for teachers is \$55,000 with a standard deviation of \$5000, to what percentile does your salary of \$49,000 correspond? Your friend’s salary of \$63,000? The heights of 5th graders have a mean of 4’8’’ and standard deviation of 2’’. If there are 100 5th graders, how many will be: Over 5’ tall? Under 4’5’’ tall? ASSIGNMENT: Pg. 831 #18-34 even, #44-54 even, 65, 66, 79, 80 Download ppt "14.4 The Normal Distribution" Similar presentations Ads by Google
Math Practice Online > free > lessons > Florida > 9th grade > Probability 2 These sample problems below for Probability 2 were generated by the MathScore.com engine. ## Sample Problems For Probability 2 ### Complexity=5 Find the probability that if you randomly select one piece from each of the three shapes, you will get 3 highlighted pieces and the odds of getting at least one unhighlighted piece. 1. Probability: Odds: 2. Probability: Odds: ### Complexity=6 Find the probability that if you randomly select one piece from each of the three shapes, you will get 3 highlighted pieces and the odds of getting at least one unhighlighted piece. 1. Probability: Odds: 2. Probability: Odds: ### Complexity=7 Find the probability that if you randomly select one piece from each of the three shapes, you will get 3 highlighted pieces and the odds of getting at least one unhighlighted piece. 1. Probability: Odds: 2. Probability: Odds: ### Complexity=8 Find the probability that if you randomly select one piece from each of the three shapes, you will get 3 highlighted pieces and the odds of getting at least one unhighlighted piece. 1. Probability: Odds: 2. Probability: Odds: ### Complexity=9 Find the probability that if you randomly select one piece from each of the three shapes, you will get 3 highlighted pieces and the odds of getting at least one unhighlighted piece. 1. Probability: Odds: 2. Probability: Odds: ### Complexity=10 Find the probability that if you randomly select one piece from each of the three shapes, you will get 3 highlighted pieces and the odds of getting at least one unhighlighted piece. 1. Probability: Odds: 2. Probability: Odds: ### Complexity=5 Find the probability that if you randomly select one piece from each of the three shapes, you will get 3 highlighted pieces and the odds of getting at least one unhighlighted piece. 1 Probability: Odds: Solution Probability that all 3 pieces are highlighted: P(1st piece highlighted) = 1/4. P(2nd piece highlighted) = 4/5. P(3rd piece highlighted) = 1/3. P(all highlighted) = 1/4 × 4/5 × 1/3 = 1/15. Odds of choosing at least one unhighlighted piece: P(at least one piece unhighlighted) = 1 - P(all highlighted) = 1 - 1/15 = 14/15. Odds(at least one piece unhighlighted) = 14:(15 - 14) = 14:1. 2 Probability: Odds: Solution Probability that all 3 pieces are highlighted: P(1st piece highlighted) = 2/5. P(2nd piece highlighted) = 4/5. P(3rd piece highlighted) = 4/5. P(all highlighted) = 2/5 × 4/5 × 4/5 = 32/125. Odds of choosing at least one unhighlighted piece: P(at least one piece unhighlighted) = 1 - P(all highlighted) = 1 - 32/125 = 93/125. Odds(at least one piece unhighlighted) = 93:(125 - 93) = 93:32. ### Complexity=6 Find the probability that if you randomly select one piece from each of the three shapes, you will get 3 highlighted pieces and the odds of getting at least one unhighlighted piece. 1 Probability: Odds: Solution Probability that all 3 pieces are highlighted: P(1st piece highlighted) = 3/5. P(2nd piece highlighted) = 1/3. P(3rd piece highlighted) = 2/4 = 1/2. P(all highlighted) = 3/5 × 1/3 × 1/2 = 1/10. Odds of choosing at least one unhighlighted piece: P(at least one piece unhighlighted) = 1 - P(all highlighted) = 1 - 1/10 = 9/10. Odds(at least one piece unhighlighted) = 9:(10 - 9) = 9:1. 2 Probability: Odds: Solution Probability that all 3 pieces are highlighted: P(1st piece highlighted) = 2/5. P(2nd piece highlighted) = 2/6 = 1/3. P(3rd piece highlighted) = 2/6 = 1/3. P(all highlighted) = 2/5 × 1/3 × 1/3 = 2/45. Odds of choosing at least one unhighlighted piece: P(at least one piece unhighlighted) = 1 - P(all highlighted) = 1 - 2/45 = 43/45. Odds(at least one piece unhighlighted) = 43:(45 - 43) = 43:2. ### Complexity=7 Find the probability that if you randomly select one piece from each of the three shapes, you will get 3 highlighted pieces and the odds of getting at least one unhighlighted piece. 1 Probability: Odds: Solution Probability that all 3 pieces are highlighted: P(1st piece highlighted) = 5/7. P(2nd piece highlighted) = 5/6. P(3rd piece highlighted) = 1/3. P(all highlighted) = 5/7 × 5/6 × 1/3 = 25/126. Odds of choosing at least one unhighlighted piece: P(at least one piece unhighlighted) = 1 - P(all highlighted) = 1 - 25/126 = 101/126. Odds(at least one piece unhighlighted) = 101:(126 - 101) = 101:25. 2 Probability: Odds: Solution Probability that all 3 pieces are highlighted: P(1st piece highlighted) = 2/3. P(2nd piece highlighted) = 2/6 = 1/3. P(3rd piece highlighted) = 3/5. P(all highlighted) = 2/3 × 1/3 × 3/5 = 2/15. Odds of choosing at least one unhighlighted piece: P(at least one piece unhighlighted) = 1 - P(all highlighted) = 1 - 2/15 = 13/15. Odds(at least one piece unhighlighted) = 13:(15 - 13) = 13:2. ### Complexity=8 Find the probability that if you randomly select one piece from each of the three shapes, you will get 3 highlighted pieces and the odds of getting at least one unhighlighted piece. 1 Probability: Odds: Solution Probability that all 3 pieces are highlighted: P(1st piece highlighted) = 2/7. P(2nd piece highlighted) = 2/6 = 1/3. P(3rd piece highlighted) = 1/8. P(all highlighted) = 2/7 × 1/3 × 1/8 = 1/84. Odds of choosing at least one unhighlighted piece: P(at least one piece unhighlighted) = 1 - P(all highlighted) = 1 - 1/84 = 83/84. Odds(at least one piece unhighlighted) = 83:(84 - 83) = 83:1. 2 Probability: Odds: Solution Probability that all 3 pieces are highlighted: P(1st piece highlighted) = 4/5. P(2nd piece highlighted) = 1/5. P(3rd piece highlighted) = 1/7. P(all highlighted) = 4/5 × 1/5 × 1/7 = 4/175. Odds of choosing at least one unhighlighted piece: P(at least one piece unhighlighted) = 1 - P(all highlighted) = 1 - 4/175 = 171/175. Odds(at least one piece unhighlighted) = 171:(175 - 171) = 171:4. ### Complexity=9 Find the probability that if you randomly select one piece from each of the three shapes, you will get 3 highlighted pieces and the odds of getting at least one unhighlighted piece. 1 Probability: Odds: Solution Probability that all 3 pieces are highlighted: P(1st piece highlighted) = 3/7. P(2nd piece highlighted) = 1/4. P(3rd piece highlighted) = 1/6. P(all highlighted) = 3/7 × 1/4 × 1/6 = 1/56. Odds of choosing at least one unhighlighted piece: P(at least one piece unhighlighted) = 1 - P(all highlighted) = 1 - 1/56 = 55/56. Odds(at least one piece unhighlighted) = 55:(56 - 55) = 55:1. 2 Probability: Odds: Solution Probability that all 3 pieces are highlighted: P(1st piece highlighted) = 2/3. P(2nd piece highlighted) = 1/7. P(3rd piece highlighted) = 5/8. P(all highlighted) = 2/3 × 1/7 × 5/8 = 5/84. Odds of choosing at least one unhighlighted piece: P(at least one piece unhighlighted) = 1 - P(all highlighted) = 1 - 5/84 = 79/84. Odds(at least one piece unhighlighted) = 79:(84 - 79) = 79:5. ### Complexity=10 Find the probability that if you randomly select one piece from each of the three shapes, you will get 3 highlighted pieces and the odds of getting at least one unhighlighted piece. 1 Probability: Odds: Solution Probability that all 3 pieces are highlighted: P(1st piece highlighted) = 2/5. P(2nd piece highlighted) = 1/6. P(3rd piece highlighted) = 3/6 = 1/2. P(all highlighted) = 2/5 × 1/6 × 1/2 = 1/30. Odds of choosing at least one unhighlighted piece: P(at least one piece unhighlighted) = 1 - P(all highlighted) = 1 - 1/30 = 29/30. Odds(at least one piece unhighlighted) = 29:(30 - 29) = 29:1. 2 Probability: Odds: Solution Probability that all 3 pieces are highlighted: P(1st piece highlighted) = 3/7. P(2nd piece highlighted) = 7/9. P(3rd piece highlighted) = 2/3. P(all highlighted) = 3/7 × 7/9 × 2/3 = 2/9. Odds of choosing at least one unhighlighted piece: P(at least one piece unhighlighted) = 1 - P(all highlighted) = 1 - 2/9 = 7/9. Odds(at least one piece unhighlighted) = 7:(9 - 7) = 7:2. MathScore.com Copyright Accurate Learning Systems Corporation 2008.
# Montessori Number Rods: Explanation and Presentation Montessori teaches math in a hands-on fashion and in order of concrete to abstract. The Montessori Number Rods are one of many examples of this type of sensorial learning. In this article, you will learn about the Number Rods, the reason behind the red and blue design, and how to present this Montessori material. By the end of this post, you should feel comfortable demonstrating the use of this math material in a classroom setting or in your home, if you are homeschooling the Montessori way. (This post contains affiliate links. Purchasing from these links costs you nothing extra, but helps with our website upkeep.) ## What are the Montessori Number Rods? The Number Rods are a Montessori math activity presented around the age of 4. In fact, they are one of the very first Montessori math materials a child will work with while attending a Montessori primary program. This material consists of 10 wooden rods whose length increases by increments of 10cm rod-by-rod and whose width remains consistent at 2.5cm x 2.5 cm. The dimensions of this material are the same as the dimensions of the Long Red Rods, a Montessori sensorial material. The single difference between the Long Red Rods and the Number Rods is the latter is painted red and blue, in alternating increments of 10cm, to signify separate units. As children work with this material, they gain the understanding that the length of each rod corresponds to a greater number of units. This teaches children helps children associate quantities with numbers (1:1 correspondence), introduces children to the decimal system, and shows them that single units can be combined to produce various quantities. ### Aims of Montessori Number Rods activity • to show that numbers can represent quantities • to show that units can be combine to create quantities • counting to ten ### Indirect aims • a sensorial representation of the decimal system • to help a child grasp the concept of numbers ### Control of error • the graduated length of the rods (a stair) ### Point of interest • the weight of each rod as it's carried to the mat in relation to its length • the pattern of the rods as they build the stair ## The Montessori Number Rods materials, presentation, and extensions ### Presentation • Inviting the child to the shelf, demonstrate carrying the Number Rods to the work mat, one-by-one, starting with the smallest rod. The shortest 4 rods should be carries with one hand and the remaining rods should be carried carefully in front of the body with 2 hands. • Randomly place each rod carefully on the mat, allowing a little space between each one. • Invite the child to help you arrange the Number Rods in order of longest to shortest, like a stair. • Place the longest rod toward the top of the mat, red segment to the left. • Find the next longest rod and line it up just below, and flush with, the first rod. The red segment should again be to the left. • Continue with the subsequent rods, always keeping the red segment on the left side and being sure to place the rods flush with each other, for a proper stair. Be sure to move slowly and be precise with your actions so the child can properly observe. • Begin the 3-period lesson. (below) ### 3period lesson • Take the 1-rod (the single unit, smallest rod) and place it apart from the stair and directly in front of the child. • Point to the rod with and tell the child, “One. This is one.” • Invite your child to touch the rod with 2 finger and say, “One”. • Take the 2-rod from the stair and place it directly above the 1-rod and say, “Two. This is two.” • Place your pointing fingers on the first segment (red) or the 2-rod and say, “One”. Then place them on the second segment (blue) and say, “Two”. • Ask the child to touch each segment, as you have done, and count. • Do these steps again with the 3-rod. • Randomly pick a number 1-3 and have the child point to the corresponding rod. Repeat this until you feel certain the child understands. • Point to random Number Rods 1-3 and ask the child, “What is this?”. Repeat this until you feel certain the child understands. • Place the 3 rods back in the stair. • Show the child how to carry the rods back to the self, starting with the longest one. *The 3-period lesson should be done just a few rods at a time and on separate days. When larger number rods are introduced in the lesson, first review the smaller number rods, just to verify the child is understanding. ## Montessori Number Rod Extensions ### Building a maze When the child can count to 55, count the total segments of all the Number Rods after they have constructed a maze. ### Discrimination of quantities Select 2 random Number Rods from the stair and place them in front of the child. Ask the child which rod is longer and how many more units the longer rod has than the shorter rod.
# How do you write the partial fraction decomposition of the rational expression (3x^2 -7x+1) / (x-1)^3? Nov 21, 2016 The answer is $= \frac{- 3}{x - 1} ^ 3 - \frac{1}{x - 1} ^ 2 + \frac{3}{x - 1}$ #### Explanation: The decomposition in partial fractions is $\frac{3 {x}^{2} - 7 x + 1}{x - 1} ^ 3 = \frac{A}{x - 1} ^ 3 + \frac{B}{x - 1} ^ 2 + \frac{C}{x - 1}$ $= \frac{A + B \left(x - 1\right) + C {\left(x - 1\right)}^{2}}{x - 1} ^ 3$ Therefore, $\left(3 {x}^{2} - 7 x + 1\right) = A + B \left(x - 1\right) + C {\left(x - 1\right)}^{2}$ Let $x = 1$, $\implies$, $- 3 = A$ $A + B + C = 1$ Coefficients of ${x}^{2}$ $3 = C$ Coefficients of $x$ $- 7 = - B - 2 C$, $\implies$, $B = - 1$ so, $\frac{3 {x}^{2} - 7 x + 1}{x - 1} ^ 3 = \frac{- 3}{x - 1} ^ 3 - \frac{1}{x - 1} ^ 2 + \frac{3}{x - 1}$
# O is the centre of a circle of radius 5 cm. At a distance of 13 cm from O, a point P is taken. Question: O is the centre of a circle of radius 5 cm. At a distance of 13 cm from O, a point P is taken. From this point, two tangents PQ and PR are drawn to the circle. Then, the area of quadrilateral PQOR is (a) 60 cm2 (b) 32.5 cm2 (c) 65 cm2 (d) 30 cm2 Solution: (a) 60 cm2 Given, $O Q=O R=5 \mathrm{~cm}, O P=13 \mathrm{~cm} .$ $\angle O Q P=\angle O R P=90^{\circ}$ (T angents drawn from an external point are perpendicular to the radius at the point of contact) From right - angled $\triangle P O Q$ : $P Q^{2}=\left(O P^{2}-O Q^{2}\right)$ $\Rightarrow P Q^{2}=13^{2}-5^{2}$ $\Rightarrow P Q^{2}=169-25$ $\Rightarrow P Q^{2}=144$ $\Rightarrow P Q=\sqrt{144}$ $\Rightarrow P Q=12 \mathrm{~cm}$ $\therefore a r(\triangle O Q P)=\frac{1}{2} \times P Q \times O Q$ $\Rightarrow a r(\triangle O Q P)=\left(\frac{1}{2} \times 12 \times 5\right) \mathrm{cm}^{2}$ $\Rightarrow \operatorname{ar}(\triangle O Q P)=30 \mathrm{~cm}^{2}$ Similarly, $a r(\triangle O R P)=30 \mathrm{~cm}^{2}$ $\therefore \operatorname{ar}($ quad. PQOR $)=(30+30) \mathrm{cm}^{2}=60 \mathrm{~cm}^{2}$
# What is 1.1970 as a fraction? Looking to convert 1.1970 to a fraction? If so, you're in the right place! In this step-by-step guide, we'll show you exactly what the fractional form of 1.1970 is and show you exactly how to calculate it so you can convert any decimal number to a fraction. Let's go! Want to quickly learn or show students how to convert 1.1970 to a fraction? Play this very quick and fun video now! There are many reasons why you might want to convert a decimal to a fraction. The most common reason is because your teacher told you to! But beyond that, the decimal form of a fraction makes it very easy to compare two fractions quickly without having to think about it. A quick fraction recap here before we begin. The number above the fraction line is the numerator, and the number below the fraction line is the denominator. If you've done much work with fractions you probably already know that, but it never hurts to double check! Okay, so the first thing to do here is show you that any number can be a fraction if you use a 1 as the denominator. Take a look: 1.1970 / 1 What we really want to do though, is get rid of the decimal places completely so that the numerator in our fraction is a whole number. To do this, we have to count the numbers after the decimal point, which in this case is 1970. To get a whole fraction we need to multiply both the numerator and the denominator by 10 if there is one number after the decimal point, 100 if there are two numbers, 1,000 if it's three numbers and 10,000 if it's...well, you get the idea! In our case 1970 is 4 digits long so we need to multiply the numerator and denominator by 10000. Now we just need to do that multiplication to get our whole fraction: 1.1970 x 10000 / 1 x 10000 = 11970 / 10000 The next step is to simplify this fraction and, to do that, we need to find the greatest common factor (GCF). This is sometimes also known as: • Greatest Common Divisor (GCD) • Highest Common Factor (HCF) • Greatest Common Denominator (GCD) The GCF can be a bit complicated to work out by hand but you can use our handy GCF calculator to figure it out. In the case of 11970 and 10000, the greatest common divisor is 10. This means that to simplify the fraction we can divide by the numerator and the denominator by 10 and we get: 11970/10 / 10000/10 = 1197 / 1000 And there you have it! In just a few short steps we have figured out what 1.1970 is as a fraction. The complete answer for your enjoyment is below: 1 197/1000 Note: because 11970 is greater than 10000 we have simplified this fraction even further to a mixed fraction. Hopefully this tutorial has helped you to understand how to convert a decimal number into a fraction. You can now go forth and convert decimals to fractions as much as your little heart desires! If you found this content useful in your research, please do us a great favor and use the tool below to make sure you properly reference us wherever you use it. We really appreciate your support! • "What is 1.1970 as a fraction?". VisualFractions.com. Accessed on January 19, 2022. http://visualfractions.com/calculator/decimal-as-fraction/what-is-1-1970-as-a-fraction/. • "What is 1.1970 as a fraction?". VisualFractions.com, http://visualfractions.com/calculator/decimal-as-fraction/what-is-1-1970-as-a-fraction/. Accessed 19 January, 2022. • What is 1.1970 as a fraction?. VisualFractions.com. Retrieved from http://visualfractions.com/calculator/decimal-as-fraction/what-is-1-1970-as-a-fraction/. ### Preset List of Decimals to Fractions Below are links to some preset decimal calculations that are commonly searched for: ### Random Decimal to Fraction Calculations If you really love decimal to fraction conversions and just can't get enough, here are some random calculations for you:
Three Terms Inversely Proportional If $3$ workers assemble $5$ computers parts in $3$ hours. How many computer parts it will take $9$ workers to assemble in $5$ hours. I always get stuck with three term proportionality problems. What is the general formula to solve this? $$\frac{9\cdot5\cdot5}{3\cdot3} = 25$$ • The proportion must be $\dfrac 5 {3 \cdot 3 } = \dfrac x {9 \cdot 5}$ from which ... Dec 4, 2016 at 16:10 • If the $A$ computers are produced with total work $B$ (the product of the workers for the time they are working) then $x$ computers will be produced by total work $D$, i.e. : $\dfrac A B = \dfrac x D$ and then solve for $x$. Dec 4, 2016 at 16:18 You can do this in two steps. In each step, change just one of the variables to the desired final value. If $3$ workers assemble $5$ computers parts in $3$ hours. How many computer parts it will take $9$ workers to assemble in $5$ hours. We need to increase the number of workers from $3$ to $9$ and the number of hours from $3$ to $5.$ Choose one thing to do first. Suppose you choose to increase the number of workers first from $3$ to $9.$ Then $9$ workers can assemble $\frac{9}{3}\times5=15$ parts in $3$ hours. Now increase the number of hours. We know $9$ workers can assemble $15$ parts in $3$ hours, so in $5$ hours, the same $9$ workers can assemble $\frac{5}{3}\times15=25$ parts. To do the whole thing in one equation, just apply the second ratio without first simplifying the first multiplication. So instead of $\frac{9}{3}\times5=15$ and then $\frac{5}{3}\times15=25,$ you have $$\frac{5}{3}\times\frac{9}{3}\times5= \frac{5\times9\times5}{3\times3}=25.$$ You need to figure out how many computer parts one person can do in an hour. Since it takes 3 worker to build 5 in 3 hours, 3 workers can build $\frac{5}{3}$ of a computer part in an hour so one worker can build $\frac{5}{9}$ Since you have 9 workers and 5 hours you multiply that by 9 and 5 There are 3 quantities. They are interdependent. They are related by proportionality and inverse proportionality. One has to express the wanted quantity by the other quantities. The requested quantity is the number of computer parts. The number of computer parts (p) per number of workers (w) and number of hours (h) is: $\frac{p}{w\cdot h}=\frac{5}{3\cdot 3}=\frac{5}{9}$. Solving for the wanted $p$ yields $p=\frac{5}{9}\cdot w\cdot h$. The result for your example is: $p=\frac{5}{9}\cdot 9\cdot 5=25$.
# What is the exact value of the square root of 32 over 5 the square root of 14? Oct 6, 2015 $\frac{4 \sqrt{7}}{35}$ #### Explanation: $\frac{\sqrt{32}}{5 \sqrt{14}}$ Simplify $\sqrt{32}$. $\frac{\sqrt{2 \times 2 \times 2 \times 2 \times 2}}{5 \sqrt{14}} =$ $\frac{\sqrt{{2}^{2} \times {2}^{2} \times 2}}{5 \sqrt{14}} =$ Apply square root rule $\sqrt{{a}^{2}} = a$. $\frac{2 \times 2 \sqrt{2}}{5 \sqrt{14}} =$ $\frac{4 \sqrt{2}}{5 \sqrt{14}}$ Rationalize the denominator. $\frac{4 \sqrt{2}}{5 \sqrt{14}} \times \frac{\sqrt{14}}{\sqrt{14}} =$ $\frac{4 \sqrt{2} \sqrt{14}}{5 \times 14} =$ $\frac{4 \sqrt{28}}{70} =$ Simplify $\left(4 \sqrt{28}\right)$. $\frac{4 \sqrt{2 \times 2 \times 7}}{70} =$ $\frac{4 \sqrt{{2}^{2} \times 7}}{70} =$ $\frac{4 \times 2 \sqrt{7}}{70} =$ $\frac{8 \sqrt{7}}{70}$ Simplify. $\frac{4 \sqrt{7}}{35}$
# Solution to Hannah’s Sweets Viral Probability Problem A probability problem which is GCSE takers find it “unfair” went viral on Twitter after students tweets about this “Hanna’s Sweets” problem. This post will enlighten the students and everyone who wanted to know the solution. Mathematics problems seldom made its way to the trending topic in social media. Recently, grade 5 olympiad problem about Cheryl’s birthday hit the internet like a storm. Now, my favorite topic, probability. The problem says, Hanna’s sweets viral probability problem Let us show elementary solution to the problem so that everyone would be able to understand it. Since there are n sweets and 6 of them are orange, the probability of obtaining the first sweet is $\dfrac{6}{n}$. (6 possibilities out of n total number of sweets). Now, since Hanna has taken 1 orange sweet, there are only 5 orange sweets remaining. Also, the total number of sweets from n must become n-1 since 1 sweet was removed. In her second time taking an orange sweet, the probability must be $\dfrac{5}{n-1}$. (5 possibilities out of n-1 total sweets). The total probability in obtaining 2 consecutive orange sweets can be found by multiplying the individual probabilities solved. Thus, $P=\dfrac{6}{n}\cdot \dfrac{5}{n-1}$ $P=\dfrac{30}{n(n-1)}$ $P=\dfrac{30}{n^2-n}$ Going back to the problem, we are given by the total probability for this event which is 1/3. This means that the expression at the top in terms of n is equal to 1/3. $\dfrac{30}{n^2-n}=\dfrac{1}{3}$ By cross multiplication, $30\cdot 3=n^2-n$ $90=n^2-n$ $\boxed{n^2-n-90=0}$ Another solution(the practical approach): Let’s think of taking 2 oranges from n total number of sweets. The number of ways to choose 2 sweets from n sweets can be solved as follows, $_nC_2=\dfrac{n!}{(n-2)!2!}$ $_nC_2=\dfrac{n(n-1)(n-2)!}{(n-2)!2!}$ $_nC_2=\dfrac{n(n-1)}{2}$ The total ways to choosing 2 orange sweets from 6 orange sweets can be solved as follows, $_6C_2=\dfrac{6!}{(6-2)!2!}$ $_6C_2=\dfrac{6!}{4!2!)}$ $_6C_2=\dfrac{6\cdot 5}{2}$ $_6C_2=15$ The desired probability is, $P=\dfrac{15}{\dfrac{n(n-1)}{2}}$ $P=15\cdot\dfrac{2}{n(n-1)}$ $P=\dfrac{30}{n(n-1)}$ Since this probability is equal to 1/3 we have, $\dfrac{1}{3}=\dfrac{30}{n(n-1)}$ By cross multiplication, $n(n-1)=90$ $n^2-n=90$ $\boxed{n^2-n-90=0}$ Now, if you want to find the number of yellow sweets. Solve for n using factoring. Subtract the number of orange sweets(6) from the total number of sweets(n). To give you an idea, there are 10 total number of sweets where 4 of them are yellow and 6 of them are orange.
# What is 201/141 as a decimal? ## Solution and how to convert 201 / 141 into a decimal 201 / 141 = 1.426 Fraction conversions explained: • 201 divided by 141 • Numerator: 201 • Denominator: 141 • Decimal: 1.426 • Percentage: 1.426% 201/141 converted into 1.426 begins with understanding long division and which variation brings more clarity to a situation. Both represent numbers between integers, in some cases defining portions of whole numbers In certain scenarios, fractions would make more sense. Ex: baking, meal prep, time discussion, etc. While decimals bring clarity to others including test grades, sale prices, and contract numbers. Now, let's solve for how we convert 201/141 into a decimal. 201 / 141 as a percentage 201 / 141 as a fraction 201 / 141 as a decimal 1.426% - Convert percentages 201 / 141 201 / 141 = 1.426 ## 201/141 is 201 divided by 141 Converting fractions to decimals is as simple as long division. 201 is being divided by 141. For some, this could be mental math. For others, we should set the equation. The two parts of fractions are numerators and denominators. The numerator is the top number and the denominator is the bottom. And the line between is our division property. We use this as our equation: numerator(201) / denominator (141) to determine how many whole numbers we have. Then we will continue this process until the number is fully represented as a decimal. This is how we look at our fraction as an equation: ### Numerator: 201 • Numerators are the parts to the equation, represented above the fraction bar or vinculum. Overall, 201 is a big number which means you'll have a significant number of parts to your equation. 201 is an odd number so it might be harder to convert without a calculator. Large numerators make converting fractions more complex. Let's take a look below the vinculum at 141. ### Denominator: 141 • Denominators differ from numerators because they represent the total number of parts which can be found below the vinculum. 141 is a large number which means you should probably use a calculator. But 141 is an odd number. Having an odd denominator like 141 could sometimes be more difficult. Have no fear, large two-digit denominators are all bark no bite. Let's start converting! ## How to convert 201/141 to 1.426 ### Step 1: Set your long division bracket: denominator / numerator $$\require{enclose} 141 \enclose{longdiv}{ 201 }$$ We will be using the left-to-right method of calculation. This is the same method we all learned in school when dividing any number against itself and we will use the same process for number conversion as well. ### Step 2: Solve for how many whole groups you can divide 141 into 201 $$\require{enclose} 00.1 \\ 141 \enclose{longdiv}{ 201.0 }$$ Since we've extended our equation we can now divide our numbers, 141 into 2010 (remember, we inserted a decimal point into our equation so we we're not accidentally increasing our solution) Multiply this number by 141, the denominator to get the first part of your answer! ### Step 3: Subtract the remainder $$\require{enclose} 00.1 \\ 141 \enclose{longdiv}{ 201.0 } \\ \underline{ 141 \phantom{00} } \\ 1869 \phantom{0}$$ If you don't have a remainder, congrats! You've solved the problem and converted 201/141 into 1.426 If you still have a remainder, continue to the next step. ### Step 4: Repeat step 3 until you have no remainder Remember, sometimes you won't get a remainder of zero and that's okay. Round to the nearest digit and complete the conversion. There you have it! Converting 201/141 fraction into a decimal is long division just as you learned in school. ### Why should you convert between fractions, decimals, and percentages? Converting between fractions and decimals depend on the life situation you need to represent numbers. Remember, fractions and decimals are both representations of whole numbers to determine more specific parts of a number. Same goes for percentages. Though we sometimes overlook the importance of when and how they are used and think they are reserved for passing a math quiz. But 201/141 and 1.426 bring clarity and value to numbers in every day life. Without them, we’re stuck rounding and guessing. Here are real life examples: ### When you should convert 201/141 into a decimal Pay & Salary - Anything to do with finance or salary will leverage decimal format. If you look at your pay check, you will see your labor is worth $20.142 per hour and not$20 and 201/141. ### When to convert 1.426 to 201/141 as a fraction Distance - Any type of travel, running, walking will leverage fractions. Distance is usually measured by the quarter mile and car travel is usually spoken the same. ### Practice Decimal Conversion with your Classroom • If 201/141 = 1.426 what would it be as a percentage? • What is 1 + 201/141 in decimal form? • What is 1 - 201/141 in decimal form? • If we switched the numerator and denominator, what would be our new fraction? • What is 1.426 + 1/2? ### Convert more fractions to decimals From 201 Numerator From 141 Denominator What is 201/131 as a decimal? What is 191/141 as a decimal? What is 201/132 as a decimal? What is 192/141 as a decimal? What is 201/133 as a decimal? What is 193/141 as a decimal? What is 201/134 as a decimal? What is 194/141 as a decimal? What is 201/135 as a decimal? What is 195/141 as a decimal? What is 201/136 as a decimal? What is 196/141 as a decimal? What is 201/137 as a decimal? What is 197/141 as a decimal? What is 201/138 as a decimal? What is 198/141 as a decimal? What is 201/139 as a decimal? What is 199/141 as a decimal? What is 201/140 as a decimal? What is 200/141 as a decimal? What is 201/141 as a decimal? What is 201/141 as a decimal? What is 201/142 as a decimal? What is 202/141 as a decimal? What is 201/143 as a decimal? What is 203/141 as a decimal? What is 201/144 as a decimal? What is 204/141 as a decimal? What is 201/145 as a decimal? What is 205/141 as a decimal? What is 201/146 as a decimal? What is 206/141 as a decimal? What is 201/147 as a decimal? What is 207/141 as a decimal? What is 201/148 as a decimal? What is 208/141 as a decimal? What is 201/149 as a decimal? What is 209/141 as a decimal? What is 201/150 as a decimal? What is 210/141 as a decimal? What is 201/151 as a decimal? What is 211/141 as a decimal? ### Convert similar fractions to percentages From 201 Numerator From 141 Denominator 202/141 as a percentage 201/142 as a percentage 203/141 as a percentage 201/143 as a percentage 204/141 as a percentage 201/144 as a percentage 205/141 as a percentage 201/145 as a percentage 206/141 as a percentage 201/146 as a percentage 207/141 as a percentage 201/147 as a percentage 208/141 as a percentage 201/148 as a percentage 209/141 as a percentage 201/149 as a percentage 210/141 as a percentage 201/150 as a percentage 211/141 as a percentage 201/151 as a percentage
• A A # Difference Between Expression and Equation Expression vs Equation Expression and Equation are terms that are very often encountered in mathematics. However, if you were to ask the difference between an expression and an equation to even those who are students of Math, chances are you may not get a satisfactory answer. Both are however important in understanding different concepts in Mathematics. Both make use of numbers and variables, however, the difference lies in their arrangement. This article will highlight the differences between the expression and equation and make it easier for you to pick up an equation from an expression. While an equation is a sentence, an expression is a phrase. For example, ‘Ten is five less than a number’ is an equation that can be represented by a formula. 10= x-5. On the other hand, a number less than five is a phrase, and hence an expression. If you are given an expression A+2A, you cannot make out anything unless you know the value of the variable A. So, while A+2A is just an expression, A+2A=3A becomes and equation. An equation is a combination of two expressions usually separated by an equals sign, which means that both expressions must equal each other. For example x-4=5 means x can have only one value that is 9. An expression can be evaluated, whereas an equation can be solved. An expression is basically an incomplete mathematical equation. It cannot have an answer or solution. If we compare with English language, an equation is like a complete sentence, while an expression is just like a phrase. If you have any difficulty in identifying an equation or an expression, looking for the equality sign will remove all your doubts. Knowing that equations involve relationships, it is easy to identify a mathematical equation. Also, when you see an equation, you have to solve it to arrive at an answer, whereas you only evaluate an expression. Summary • Equations and expressions are often encountered when understanding mathematical concepts. • If compared with language, expressions are like phrases while equations are complete sentences. • Expressions have no relations whereas equations reveal relationships. • You need to solve equations while expressions can only be evaluated. • Equations have an equality sign while expressions do not have any equals sign. Related posts:
# Relationship between chord length and radius ### Arcs and Chords The relation is known in literature (Weinberg,. A.M., Wigner, E.P. average chord length of a sphere in an isotropic flux equals r, the radius of the sphere, he. Chord Length Formula, length of chord of circle, Formula to find the length of a chord when radius is known, length of common chord of two circles formula. The circumference is always the same distance from the centre - the radius. Arc length. A chord separates the circumference of a circle into two sections - the. Chords of a Circle Theorems In these lessons, we will learn theorems that involve chords of a circle. Perpendicular bisector of a chord passes through the center of a circle. Congruent chords are equidistant from the center of a circle. If two chords in a circle are congruent, then their intercepted arcs are congruent. If two chords in a circle are congruent, then they determine two central angles that are congruent. The following diagrams give a summary of some Chord Theorems: Perpendicular Bisector and Congruent Chords. Scroll down the page for examples, explanations, and solutions. A chord is a straight line joining 2 points on the circumference of a circle. ### Given the arc length and chord length, what is the radius? - Math Central A radius or diameter that is perpendicular to a chord divides the chord into two equal parts and vice versa. The perpendicular bisector of a chord passes through the center of a circle. In the above circle, OA is the perpendicular bisector of the chord PQ and it passes through the center of the circle. OB is the perpendicular bisector of the chord RS and it passes through the center of the circle. We can use this property to find the center of any given circle. Determine the center of the following circle. Draw 2 non-parallel chords Step 2: Construct perpendicular bisectors for both the chords. ## Chord Length Formula The center of the circle is the point of intersection of the perpendicular bisectors. Circles, Radius Chord relationships, distance from the center to a chord This video shows how to define a chord; how to describe the effect of a perpendicular bisector of a chord and the distance from the center of the circle. The equation of circle with radius r and center at the start of Cartesian coordinate: The equation of circle with radius r and center at point with coordinates a, b in the Cartesian coordinate: Parametric equations of circle with a radius r and center at point with coordinates a, b in the Cartesian coordinate: Tangent is a coplanar straight line that touches the circle at a single point. Tangent is always perpendicular to the radius of the circle drawn at the point of contact. The shortest distance from the center of circle to tangent is radius of the circle. If two tangents from touch points B and C on the same circle are not parallel, they intersect at point A and the point of contact between the segment and the point of intersection of a tangent is the same segment on a different tangent: Secant of a circle is a straight line, connecting two points of circle. • The Relationship Between Chords, Diameters, and Minor Arcs of a Circle • What is the relationship between radius and chord in a circle? • Arcs and Chords Properties of circle secant 1. If a point outside the circle Q obtained two secant, crossing the circle at two points A and B for a first secant and C and D for another secant, the products of two intersecting segments are equal: If the point comes out of Q circle secant, crossing the circle at two points A and B, and the tangent point of contact C, then product segments the secant lengths equal to the square of the tangent: Chord of a circle is a segment that connects two points of circle. Chord is a segment of tangent. Length of chord 1.
# Multiplication ## Elementary multiplication Multiplication is the third elementary operation of arithmetic discussed in this book. Multiplying natural numbers is basically counting the values of repeated copies of a number together. Counting 3 copies of five equals 15 together: 3 × 5 = 15. In other words: multiplication represents repeated addition: 2 × 5 = 5 + 5 3 × 5 = 5 + 5 + 5 4 × 5 = 5 + 5 + 5 + 5 Verbally 3 × 5 = 15 is expressed as: "three times five equals fifteen" or "five multiplied by three is fifteen". The multiplication sign is generally a cross, "×", but in mathematics or in computer languages also other symbols are often used, such as a (centered) dot (3 · 5 = 15) or an asterisk (3 * 5 = 15). In algebra, the multiplication sign is generally not used at all, when dealing with variables: 3a = 3 × a. Numbers multiplied in a multiplication are called factors and the result is called the product. In 3 × 5 = 15, 3 and 5 are the "factors" and 15 is the "product". Figure 1 shows a rectangular area of 15 unit squares as a row of 5 plus a row of 5 plus a row of 5, stacked on each other: 3 × 5 = 5 + 5 + 5 = 15. This same rectangle turned on its other side will show a stack of 5 rows of 3 squares each, or 3 + 3 + 3 + 3 + 3. So apparently: 3 × 5 = 5 × 3. Conclusion: next to addition also multiplication is commutative, meaning that changing the order of the factors does not change the product: 4 × 3 = 3 × 4 3 + 3 + 3 + 3 = 4 + 4 + 4. Multiplication by zero is always zero, and if the result of a multiplication is zero, then at least one of the factors is zero: 2 × 0 = 0. 0 × 2017 = 0 0 × 0 = 0 If a × b = 0, then a or b or both must be zero. Multiplication under ten is typically learned by memorizing the times table or multiplication table (fig.2). ## Multiplying factors > 10 Using the fact that multiplication is commutative (order of factors is arbitrary): 2 × 300 = 2 × 3 × 100 = 6 × 100 = 600. 3 × 40 = 3 × 4 × 10 = 12 × 10 = 120. 30 × 40 = 3 × 4 × 10 × 10 = 12 × 100 = 1200. 10 × 100 (= 10 × 10 × 10) = 1000. 70 × 800 = 7 × 8 × 10 × 100 = 56 × 1000 = 56 000. 30 × 6000 = 3 × 6 × 10 × 1000 = 18 × 10 000 = 180 000. ### Distributive property How to calculate 2 × 13? 700 + 500 = (7+5) × 100 = 12 × 100 = 1200. Both 7 and 5 are in the ones position, so they can be added, 7 + 5 = 12, and in the hundreds position this makes 1200. We actually applied distribution or the distributive law or the distributive property: (7 + 5) × 100 = 700 + 500. If we apply this to 2 × 13: 2 × 13 = 2 × (10 + 3) = (2 × 10) + (2 × 3) = 20 + 6 = 26. Some examples: 12 × 10 = (10 + 2) × 10 = 100 + 20 = 120. 3 × 26 = 3 × (20 + 6) = (3 × 20) + (3 × 6) = 60 + 18 = 78. 6 × 78 = 6 × (70 + 8) = (6 × 70) + (6 × 8) = 420 + 48 = 468. 7 × 36 = 7 × (30 + 6) = (7 × 30) + (7 × 6) = 210 + 42 = 252. 4 × 370 = 4 × (300 + 70) = (4 × 300) + (4 × 70) = 1200 + 280 = 1480. 4 × 370 = 4 × 37 × 10 = ((4 × 30) + (4 × 7)) × 10 = (120 + 28) × 10 = 1480. 3 × 2951 = 3 × (2000 + 900 + 50 + 1) = (3 × 2000) + (3 × 900) + (3 × 50) + (3 × 1) = 6000 + 2700 + 150 + 3 = 8853. 17 × 36 = (10 + 7) × (30 + 6) = ((30 + 6) × 10) + ((30 + 6) × 7) = 360 + (7 × 30) + (7 × 6) = 360 + 210 + 42 = 612. ## Multiplication algorithm The standard multiplication algorithm, also called long multiplication or column multiplication, puts calculation steps applied in the examples above in an orderly, systematic method that always works for any multiplication of two natural numbers: In the above examples long multiplication is used to calculate 7 × 36 = 252 and 17 × 36 = 612. The recipe is as follows: We work from right to left. We start with 7 × 6 = 42. We write the 2 in the ones place of the result row and carry the 4 to the tens column. Then we calculate 7 × 3 = 21 plus the carried 4 makes 25 and we write it in the result row. Done! So, we do not just multiply digits per same column, we multiply the 7 in the lower row by 6 in the ones column, top row and multiply the same 7 by 3 (= 30) in the tens column, top row. Finally we add both results. To calculate 17 × 36 = 612, we first calculate 7 × 36 as shown, and than, working from right to left, we do the same with the next digit in 17, being 1 (= 10). So, we calculate 1 × 6 and 1 × 3 and write the results in the second result row, but not before we put a 0 most right, because the 1 in the tens position is actually a 10 (10 × 6 = 1 × 6 × 10 and 10 × 30 = 1 × 3 × 10 × 10). Finally we need to add 252 and 360 to find the result of 17 × 36. This still may look a little complicated, therefore a step by step example: In this example 1234 × 333 = 410 922 is calculated using long multiplication. The steps in words: 1. Step 1: The 3 in the ones position times 1234. 1. 3 × 4 = 12 (carry the 1 to the tens position). 2. 3 × 3 = 9 plus the carried 1 equals 10 (carry the 1 to the hundreds position). 3. 3 × 2 = 6 plus the carried 1 equals 7. 4. 3 × 1 = 3. 2. Step 2: The 3 in the tens position times 1234. 1. First write a 0. 2. Then the same digit calculations as in step 1. 3. Step 3: The 3 in the hundreds position times 1234. 1. First write 00 (a zero in the ones, a zero in the tens position). 2. Then the same digit calculations as in step 1. 4. Step 4: Add the results of the steps before: 3702 + 37020 + 370200 = 410 922. Now you try to execute this algorithm with 1234 and 333 swapped, that is, 333 in the top row and 1234 in the row below. You know you must get the same result. Two more examples: Suppose you and four of your friends are going to the cinema. One ticket costs \$28. How much do the tickets cost together?
# Applied Combinatorics ## Section10.6Probability Spaces with Infinitely Many Outcomes To this point, we have focused entirely on probability spaces $$(S,P)$$ with $$S$$ a finite set. More generally, probability spaces are defined where $$S$$ is an infinite set. When $$S$$ is countably infinite, we can still define $$P$$ on the members of $$S\text{,}$$ and now $$\sum_{x\in S} P(x)$$ is an infinite sum which converges absolutely (since all terms are non-negative) to $$1\text{.}$$ When $$S$$ is uncountable, $$P$$ is not defined on $$S\text{.}$$ Instead, the probability function is defined on a family of subsets of $$S\text{.}$$ Given our emphasis on finite sets and combinatorics, we will discuss the first case briefly and refer students to texts that focus on general concepts from probability and statistics for the second. ### Example10.27. Consider the following game. Nancy rolls a single die. She wins if she rolls a six. If she rolls any other number, she then rolls again and again until the first time that one of the following two situations occurs: (1) she rolls a six, which now this results in a loss or (2) she rolls the same number as she got on her first roll, which results in a win. As an example, here are some sequences of rolls that this game might take: 1. $$(4,2,3,5,1,1,1,4)\text{.}$$ Nancy wins! 2. $$(6)\text{.}$$ Nancy wins! 3. $$(5,2, 3,2,1,6)\text{.}$$ Nancy loses. Ouch. So what is the probability that Nancy will win this game? Nancy can win with a six on the first roll. That has probability $$1/6\text{.}$$ Then she might win on round $$n$$ where $$n\ge2\text{.}$$ To accomplish this, she has a $$5/6$$ chance of rolling a number other than six on the first roll; a $$4/6$$ chance of rolling something that avoids a win/loss decision on each of the rolls, $$2$$ through $$n-1$$ and then a $$1/6$$ chance of rolling the matching number on round $$n\text{.}$$ So the probability of a win is given by: \begin{equation} \frac{1}{6}+\sum_{n\ge 2}\frac{5}{6}\left(\frac{4}{6}\right)^{n-2}\frac{1}{6} = \frac{7}{12}. \end{equation} ### Example10.28. You might think that something slightly more general is lurking in the background of the preceding example—and it is. Suppose we have two disjoint events $$A$$ and $$B$$ in a probability space $$(S,P)$$ and that $$P(A)+P(B)\lt 1\text{.}$$ Then suppose we make repeated samples from this space with each sample independent of all previous ones. Call it a win if event $$A$$ holds and a loss if event $$B$$ holds. Otherwise, it’s a tie and we sample again. Now the probability of a win is: \begin{equation} P(A)+P(A)\sum_{n\ge 1}(1-P(A)-P(B))^n=\frac{P(A)}{P(A)+P(B)}. \end{equation}
# Arithmetic Reasoning • 13 Questions • 54 Problems • 36 Flash Cards ### Fundamentals ##### Number Properties ###### Integers An integer is any whole number, including zero. An integer can be either positive or negative. Examples include -77, -1, 0, 55, 119. ###### Rational Numbers A rational number (or fraction) is represented as a ratio between two integers, a and b, and has the form $${a \over b}$$ where a is the numerator and b is the denominator. An improper fraction ($${5 \over 3}$$) has a numerator with a greater absolute value than the denominator and can be converted into a mixed number ($$1 {2 \over 3}$$) which has a whole number part and a fractional part. ###### Absolute Value The absolute value is the positive magnitude of a particular number or variable and is indicated by two vertical lines: $$\left\|-5\right\| = 5$$. In the case of a variable absolute value ($$\left\|a\right\| = 5$$) the value of a can be either positive or negative (a = -5 or a = 5). ###### Factors & Multiples A factor is a positive integer that divides evenly into a given number. The factors of 8 are 1, 2, 4, and 8. A multiple is a number that is the product of that number and an integer. The multiples of 8 are 0, 8, 16, 24, ... ###### Greatest Common Factor The greatest common factor (GCF) is the greatest factor that divides two integers. ###### Least Common Multiple The least common multiple (LCM) is the smallest positive integer that is a multiple of two or more integers. ###### Prime Number A prime number is an integer greater than 1 that has no factors other than 1 and itself. Examples of prime numbers include 2, 3, 5, 7, and 11. ##### Operations on Fractions ###### Simplifying Fractions Fractions are generally presented with the numerator and denominator as small as is possible meaning there is no number, except one, that can be divided evenly into both the numerator and the denominator. To reduce a fraction to lowest terms, divide the numerator and denominator by their greatest common factor (GCF). Fractions must share a common denominator in order to be added or subtracted. The common denominator is the least common multiple of all the denominators. ###### Multiplying & Dividing Fractions To multiply fractions, multiply the numerators together and then multiply the denominators together. To divide fractions, invert the second fraction (get the reciprocal) and multiply it by the first. ##### Operations on Exponents ###### Defining Exponents An exponent (cbe) consists of coefficient (c) and a base (b) raised to a power (e). The exponent indicates the number of times that the base is multiplied by itself. A base with an exponent of 1 equals the base (b1 = b) and a base with an exponent of 0 equals 1 ( (b0 = 1). To add or subtract terms with exponents, both the base and the exponent must be the same. If the base and the exponent are the same, add or subtract the coefficients and retain the base and exponent. For example, 3x2 + 2x2 = 5x2 and 3x2 - 2x2 = x2 but x2 + x4 and x4 - x2 cannot be combined. ###### Multiplying & Dividing Exponents To multiply terms with the same base, multiply the coefficients and add the exponents. To divide terms with the same base, divide the coefficients and subtract the exponents. For example, 3x2 x 2x2 = 6x4 and $${8x^5 \over 4x^2}$$ = 2x(5-2) = 2x3. ###### Exponent to a Power To raise a term with an exponent to another exponent, retain the base and multiply the exponents: (x2)3 = x(2x3) = x6 ###### Negative Exponent A negative exponent indicates the number of times that the base is divided by itself. To convert a negative exponent to a positive exponent, calculate the positive exponent then take the reciprocal: $$b^{-e} = { 1 \over b^e }$$. For example, $$3^{-2} = {1 \over 3^2} = {1 \over 9}$$ Radicals (or roots) are the opposite operation of applying exponents. With exponents, you're multiplying a base by itself some number of times while with roots you're dividing the base by itself some number of times. A radical term looks like $$\sqrt[d]{r}$$ and consists of a radicand (r) and a degree (d). The degree is the number of times the radicand is divided by itself. If no degree is specified, the degree defaults to 2 (a square root). The radicand of a simplified radical has no perfect square factors. A perfect square is the product of a number multiplied by itself (squared). To simplify a radical, factor out the perfect squares by recognizing that $$\sqrt{a^2} = a$$. For example, $$\sqrt{64} = \sqrt{16 \times 4} = \sqrt{4^2 \times 2^2} = 4 \times 2 = 8$$. To add or subtract radicals, the degree and radicand must be the same. For example, $$2\sqrt{3} + 3\sqrt{3} = 5\sqrt{3}$$ but $$2\sqrt{2} + 2\sqrt{3}$$ cannot be added because they have different radicands. To multiply or divide radicals, multiply or divide the coefficients and radicands separately: $$x\sqrt{a} \times y\sqrt{b} = xy\sqrt{ab}$$ and $${x\sqrt{a} \over y\sqrt{b}} = {x \over y}\sqrt{a \over b}$$ ###### Square Root of a Fraction To take the square root of a fraction, break the fraction into two separate roots then calculate the square root of the numerator and denominator separately. For example, $$\sqrt{9 \over 16}$$ = $${\sqrt{9}} \over {\sqrt{16}}$$ = $${3 \over 4}$$ ##### Miscellaneous ###### Scientific Notation Scientific notation is a method of writing very small or very large numbers. The first part will be a number between one and ten (typically a decimal) and the second part will be a power of 10. For example, 98,760 in scientific notation is 9.876 x 104 with the 4 indicating the number of places the decimal point was moved to the left. A power of 10 with a negative exponent indicates that the decimal point was moved to the right. For example, 0.0123 in scientific notation is 1.23 x 10-2. ###### Factorials A factorial has the form n! and is the product of the integer (n) and all the positive integers below it. For example, 5! = 5 x 4 x 3 x 2 x 1 = 120. ### Applications ##### Order of Operations ###### PEMDAS Arithmetic operations must be performed in the following specific order: 1. Parentheses 2. Exponents 3. Multiplication and Division (from L to R) 4. Addition and Subtraction (from L to R) The acronym PEMDAS can help remind you of the order. ###### Distributive Property - Multiplication The distributive property for multiplication helps in solving expressions like a(b + c). It specifies that the result of multiplying one number by the sum or difference of two numbers can be obtained by multiplying each number individually and then totaling the results: a(b + c) = ab + ac. For example, 4(10-5) = (4 x 10) - (4 x 5) = 40 - 20 = 20. ###### Distributive Property - Division The distributive property for division helps in solving expressions like $${b + c \over a}$$. It specifies that the result of dividing a fraction with multiple terms in the numerator and one term in the denominator can be obtained by dividing each term individually and then totaling the results: $${b + c \over a} = {b \over a} + {c \over a}$$. For example, $${a^3 + 6a^2 \over a^2} = {a^3 \over a^2} + {6a^2 \over a^2} = a + 6$$. ###### Commutative Property The commutative property states that, when adding or multiplying numbers, the order in which they're added or multiplied does not matter. For example, 3 + 4 and 4 + 3 give the same result, as do 3 x 4 and 4 x 3. ##### Ratios ###### Ratios Ratios relate one quantity to another and are presented using a colon or as a fraction. For example, 2:3 or $${2 \over 3}$$ would be the ratio of red to green marbles if a jar contained two red marbles for every three green marbles. ###### Proportions A proportion is a statement that two ratios are equal: a:b = c:d, $${a \over b} = {c \over d}$$. To solve proportions with a variable term, cross-multiply: $${a \over 8} = {3 \over 6}$$, 6a = 24, a = 4. ###### Rates A rate is a ratio that compares two related quantities. Common rates are speed = $${distance \over time}$$, flow = $${amount \over time}$$, and defect = $${errors \over units}$$. ###### Percentages Percentages are ratios of an amount compared to 100. The percent change of an old to new value is equal to 100% x $${ new - old \over old }$$. ##### Statistics ###### Averages The average (or mean) of a group of terms is the sum of the terms divided by the number of terms. Average = $${a_1 + a_2 + ... + a_n \over n}$$ ###### Sequence A sequence is a group of ordered numbers. An arithmetic sequence is a sequence in which each successive number is equal to the number before it plus some constant number. ###### Probability Probability is the numerical likelihood that a specific outcome will occur. Probability = $${ \text{outcomes of interest} \over \text{possible outcomes}}$$. To find the probability that two events will occur, find the probability of each and multiply them together. ##### Word Problems ###### Practice Many of the arithmetic reasoning problems on the ASVAB will be in the form of word problems that will test not only the concepts in this study guide but those in Math Knowledge as well. Practice these word problems to get comfortable with translating the text into math equations and then solving those equations. # Electronics Information • 104 Questions • 9 Problems • 77 Flash Cards ### Electricity ##### Electron Flow ###### Electrons All electricity is the movement of electrons which are subatomic particles that orbit the nucleus of an atom. Electrons occupy various energy levels called shells and how well an element enables the flow of electrons depends on how many electrons occupy its outer (valence) electron shell. ###### Conductors Conductors are elements that allow electrons to flow freely. Their valence shell is less than half full of electrons that are able to move easily from one atom to another. ###### Insulators Insulators have valence shells that are more than half full of electrons and, as such, are tightly bound to the nucleus and difficult to move from one atom to another. ###### Semiconductors Semiconductors have valence shells that are exacly half full and can conduct electricity under some conditions but not others. This property makes them useful for the control of electrical current. ###### Current Current is the rate of flow of electrons per unit time and is measured in amperes (A). A coulomb (C) is the quantity of electricity conveyed in one second by a current of one ampere. ###### Voltage Voltage (V) is the electrical potential difference between two points. Electrons will flow as current from areas of high potential (concentration of electrons) to areas of low potential. Voltage and current are directly proportional in that the higher the voltage applied to a conductor the higher the current that will result. ###### Resistance Resistance is opposition to the flow of current and is measured in ohms (Ω). One ohm is defined as the amount of resistance that will allow one ampere of current to flow if one volt of voltage is applied. As resistance increases, current decreases as resistance and current are inversely proportional. ###### Conductive Materials All conductors have resistance and the amount of resistance varies with the element. But, resistance isn't the only consideration when choosing a conductor as the most highly conductive elements like silver and gold are also more expensive and more brittle than slightly less conductive elements like copper. A balance needs to be struck between the electrical qualities of a material and its cost and durability. ###### Power Electrical power is measured in watts (W) and is calculated by multiplying the voltage (V) applied to a circuit by the resulting current (I) that flows in the circuit: P = IV. In addition to measuring production capacity, power also measures the rate of energy consumption and many loads are rated for their consumption capacity. For example, a 60W lightbulb utilizes 60W of energy to produce the equivalent of 60W of heat and light energy. ##### Circuits A load is a source of resistance that converts electrical energy into another form of energy. The components of a microwave, for example, are loads that work together to convert household electricity into radation that can be used to quickly cook food. ###### Open & Closed Circuits A closed circuit is a complete loop or path that electricity follows. It consists of a source of voltage, a load, and connective conductors. If the circuit is interrupted, if a wire is disconnected or cut for example, it becomes an open circuit and no electricity will flow. ###### Ohm's Law Ohm's law specifies the relationship between voltage (V), current (I), and resistance (R) in an electrical circuit: V = IR. ###### Series Circuit A series circuit has only one path for current to flow. In a series circuit, current (I) is the same throughout the circuit and is equal to the total voltage (V) applied to the circuit divided by the total resistance (R) of the loads in the circuit. The sum of the voltage drops across each resistor in the circuit will equal the total voltage applied to the circuit. ###### Parallel Circuit In a parallel circuit, each load occupies a separate parallel path in the circuit and the input voltage is fully applied to each path. Unlike a series circuit where current (I) is the same at all points in the circuit, in a parallel circuit, voltage (V) is the same across each parallel branch of the circuit but current differs in each branch depending on the load (resistance) present. ###### Series-Parallel Circuits Circuits are not limited to only series or only parallel configurations. Most circuits contain a mix of series and parallel segments. A good example is a household circuit breaker. Electrical outlets in each section of the house are wired in parallel with the circuit breaker for that section wired in series making it easy to cut off electricity to the parallel parts of the circuit when needed. ###### Battery Configurations Batteries can be connected together in various combinations to increase their total voltage and/or total current. Connecting batteries in series combines their voltage while keeping their current the same, connecting batteries in parallel combines their current while keeping their voltage the same, and using a series-parallel configuration, half the batteries can be connected in series and half in parallel to combine both voltage and current. ### Electrical Systems ##### Types of Current ###### Direct Current (DC) Direct current flows in only one direction in a circuit, from the negative terminal of the voltage source to the positive. A common source of direct current (DC) is a battery. ###### Alternating Current (AC) In contrast to the constant one-way flow of direct current, alternating current changes direction many times each second. Electricity is delivered from power stations to customers as AC because it provides a more efficient way to transport electricity over long distances. ##### Electronic Components ###### Resistors Resistors are used to limit voltage and/or current in a circuit and can have a fixed or variable resistance. Variable resistors (often called potentiometers or rheostats) are used when dynamic control over the voltage/current in a circuit is needed, for example, in a light dimmer or volume control. ###### Fuses Fuses are thin wires that melt when the current in a circuit exceeds a preset amount. They help prevent short circuits from damaging circuit components when an unusually large current is applied to the circuit, either through component failure or spikes in applied voltage. ###### Circuit Breakers Like fuses, circuit breakers stop current flow once it reaches a certain amount. They have the advantage of being reusable (fuses must be replaced when "blown") but respond more slowly to current surges and are more expensive than fuses. ###### Capacitors Capacitors store electricity and are used in circuits as temporary batteries. Capacitors are charged by DC current (AC current passes through a capacitor) and that stored charge can later be dissipated into the circuit as needed. Capacitors are often used to maintain power within a system when it is disconnected from its primary power source or to smooth out or filter voltage within a circuit. ###### Diodes A diode allows current to pass easily in one direction and blocks current in the other direction. Diodes are commonly used for rectification which is the conversion of alternating current (AC) into direct current (DC). Because a diode only allows current flow in one direction, it will pass either the upper or lower half of AC waves (half-wave rectification) creating pulsating DC. Multiple diodes can be connected together to utilize both halves of the AC signal in full-wave rectification. ###### Transistors The transistor is the foundation of modern electronic devices. It is made entirely from semiconductor material (making it a solid state device) and can serve many different functions in a circuit including acting as a switch, amplifier, or current regulator. A transistor works by allowing a small amount of current applied at the base to control general current flow from collector to emitter through the transistor. ###### Integrated Circuits Circuits containing transistors are packaged into integrated circuit chips that allow encapsulating complex circuit designs (CPU, memory, I/O) for easier integration into electronic devices and machines. ###### Thermocouples A thermocouple is a temperature sensor that consists of two wires made from different conductors. The junction of these two wires produces a voltage based on the temperature difference between them. ##### Magnetism ###### Magnetic Fields A moving electric current produces a magnetic field proportional to the amount of current flow. This magnetic field can be made stronger by winding the wire into a coil and further enhanced if done around an iron containing (ferrous) core. ###### Inductors An inductor is coiled wire that stores electric energy in the form of magnetic energy and resists changes in the electric current flowing through it. If current is increasing, the inductor produces a voltage that slows the increase and, if current is decreasing, the magnetic energy in the coil opposes the decrease to keep the current flowing longer. In contrast to capacitors, inductors allow DC to pass easily but resist the flow of AC. ###### Transformers A transformer utilizes an inductor to increase or decrease the voltage in a circuit. AC flowing in a coil wrapped around an iron core magnetizes the core causing it to produce a magnetic field. This magnetic field generates a voltage in a nearby coil of wire and, depending on the number of turns in the wire of the primary (source) and secondary coils and their proximity, voltage is induced in the secondary coil. # General Science • 264 Questions • 315 Flash Cards ### Nutrition & Health ##### Macronutrients ###### Proteins Found in both animal sources (meat, fish, eggs, cheese) and vegetables (beans, nuts, some grains), proteins are important for the body's maintenance, growth, and repair. ###### Carbohydrates Carbohydrates are major sources of energy for the body and are found in sugars (fruit, cane sugar, beets) and starches (bread, rice, potatoes, pasta). ###### Fats Like carbohydrates, fats provide energy to the body. The difference is energy from fats tends to be longer burning as opposed to the quick fuel provided by carbohydrates. Fats come in three types, saturated (meats, shellfish, eggs, milk), monounsaturated (olives, almonds, avocados), and polyunsaturated (vegetable oils). Saturated fats can raise LDL ("bad") cholesterol while unsaturated fats can decrease it. ##### Micronutrients ###### Minerals Small quantities of certain minerals like iron, calcium, magnesium, and salt are important for nutrition and health. ###### Vitamins Vitamins are necessary for a wide variety of bodily processes. Some vitamins like Vitamins A and C come from diet but others, like Vitamin D, are generated in response to sunlight. ###### Fiber Fiber provides bulk to help the large intestine carry away waste. Good sources of fiber are leafy vegetables, beans, potatoes, fruits, and whole grains. ##### Nutritional Deficiencies A diet lacking healthy amounts of necessary nutrients can lead to a variety of health conditions and diseases. Examples include anemia which is caused by a lack of iron and scurvy which is caused by a lack of Vitamin C. ###### Health Benefits of Vitamins & Minerals Vitamin / Mineral Sources Health Benefits Calcium Dairy products (milk, yogurt, cheese), spinach. Aids bone growth and repair, muscle function. Iron Red meat, beans, whole grains. Allows red blood cells to transfer oxygen to body tissues. Magnesium Nuts, whole grains, green leafy vegetables. Muscle, nerve, and enzyme function. Potassium Bananas, nuts, seeds. Helps balance fluid levels in the body. Vitamin A Liver, milk, eggs, carrots. Vision, immune system, cell growth. Vitamin C Green and red peppers, citrus fruits, broccoli. Collagen formation, immune system function, antioxidant (helps protect cells from damage). Vitamin D Exposure to sunlight. Helps calcium strengthen bones, muscle, nerve, and immune system function. ### Human Body Systems ##### Skeleton & Muscles ###### Exoskeleton An exoskeleton (external skeleton) is common in arthropods like insects, spiders, and crustaceans. ###### Endoskeleton An endoskeleton (internal skeleton) is a charateristic of vertebrate animals, including humans. ###### Bones & Cartilage Hard bones provide primary support for the endoskeleton while more flexible cartilage is found at the end of all bones, at the joints, and in the nose and ears. In addition to providing support and protecting bodily organs, bones also produce blood cells and store minerals like calcium. ###### Tendons & Ligaments Tough fibrous cords of connective tissue called tendons connect muscles to the skeleton while another type of connective tissue called ligaments connect bones to other bones at joints (elbow, knee, fingers, spinal column). ##### Respiratory System ###### Respiration The respiratory system manages respiration which is the process by which blood cells absorb oxygen and eliminate carbon dioxide. ###### Filtering Air After air enters through the nose, it passes through the nasal cavity which filters, moistens, and warms it. Further filtering takes place in the pharynx, which also helps protect against infection, and then in the trachea which is just past the epiglottis, responsible for preventing food from entering the airway. ###### Lungs The trachea branches into the left and right bronchi which each lead to a lung where the bronchi subdivide into smaller tubes called bronchioles. Each bronchiole ends in a small sac called an alveolus which allows oxygen from the air to enter the bloodstream via tiny blood vessels called capillaries. ###### Breathing The diaphragm is a system of muscles that allows breathing. During inhalation, the diaphragm expands and air rushes in to fill the space created. Then, during exhalation, the diaphragm contracts and forces the air back out. ##### Circulatory System ###### Circulation Like the respiratory system, the circulatory system serves to transport oxygen throughout the body while removing carbon dioxide. In addition, the circulatory system transports nutrients from the digestive system. ###### Heart The heart is the organ that drives the circulatory system. In humans, it consists of four chambers with two that collect blood called atria and two that pump blood called ventricles. The heart's valves prevent blood pumped out of the ventricles from flowing back into the heart. ###### Pulmonary Artery & Vein The two largest veins in the body, the venae cavae, pass blood to the right ventricle which pumps the blood to the lungs through the pulmonary artery. Blood picks up oxygen in the lungs and returns it to the left atrium via the pulmonary vein. ###### Blood Flow To provide oxygen to the body, blood flows through the heart in a path formed by the right atrium → right ventricle → lungs → left atrium → left ventricle → body. When blood enters the right side of the heart it is deoxygenated. It enters the left side of the heart oxygenated after traveling to the lungs. ###### Arteries The aorta is the body's largest artery and receives blood from the pulmonary vein via the left ventricle. From there, blood is circulated through the rest of the body through smaller arteries called arterioles that branch out from the heart. Finally, blood is delivered to bodily tissues through capillaries. ###### Veins Veins carry blood back to the heart from the body. While arteries are thick-walled because they carry oxygenated blood at high pressure, veins are comparatively thin-walled as they carry low-pressure deoxygenated blood. Like the heart, veins contain valves to prevent blood backflow. ###### Capillaries Capillaries are small thin-walled vessels that permit the exchange of oxygen, carbon dioxide, nutrients, and waste between blood and the body's cells. This process of exchange is called diffusion. ###### Blood Cells Blood is created in bone marrow and is made up of cells suspended in liquid plasma. Red blood cells carry oxygen, white blood cells fight infection, and platelets are cell fragments that allow blood to clot. ###### Blood Types Blood is categorized into four different types (A, B, AB, and O) based on the type of antigens found on the outside of the red blood cells. Additionally, each type can be negative or positive based on whether or not the cells have an antigen called the Rh factor. ###### Blood Transfer Blood transfer is limited by the type and Rh factor of the blood. Someone who has Rh-factor negative blood cannot receive blood with a positive type but a person with Rh-factor positive type blood can receive Rh-negative blood. Type O negative blood is the universal donor because it can be given to a person with any blood type. Type AB positive is the universal recipient meaning someone with this blood type can receive any other type of blood. ##### Digestive System ###### Mouth & Throat Digestion begins in the mouth where the teeth and tongue break down food mechanically through chewing and saliva, via the enzyme salivary amylase, starts to break starches down chemically. From the mouth, food travels down the esophagus where contractions push the food into the stomach. ###### Stomach Food is mixed with gastric acid and pepsin in the stomach to help break down protein. ###### Small Intestine The small intestine is where most digestion takes place. As food travels along the small intestine it gets broken down completely by enzymes secreted from the walls. These enzymes are produced in the small intestine as well as in the pancreas and liver. After the enzymes break down the food, the resulting substances are then absorbed into the blood via capillaries in the small intestine walls. ###### Pancreas The acids produced by the pancreas contain several enzymes that aid in digestion. Lipase converts fat to glycerol and fatty acids. Pancreatic amylase breaks down complex carbohydrates into simple sugars. Trypsin converts polypeptides (the building blocks of protein) into amino acids. ###### Liver The liver produces bile which emulsifies (separates) fat. ###### Large Intestine The large intestine (colon) follows the small intestine and processes the physical waste produced by digestion, absorbing water and minerials that remain back into the body. Solid waste is then stored in the rectum while liquid waste is stored in the bladder. Chemical waste like excess water, minerals, and salt are filtered from the blood by the kidneys and secreted into the urine. Urine is transported from the kidneys to the bladder through ureters. ##### Nervous System ###### Central & Peripheral Nervous Systems The nervous system consists of the brain and spinal cord (central nervous system) and the peripheral nervous system which is the network of nerve cells (neurons) that collect and distribute signals from the central nervous system throughout the body. ###### Cerebrum The cerebrum is the major part of the brain and is responsible for the main senses (thinking, hearing, seeing). ###### Cerebellum The cerebellum is a large cluster of nerves at the base of the brain that's responsible for balance, movement, and muscle coordination. ###### Medulla Part of the brainstem, the medulla is the connection between the brain and the spinal cord. It controls involuntary actions like breathing, swallowing, and heartbeat. ###### Spinal Cord The spinal cord connects the brain to the body's network of nerves. It carries impulses between all organs and the brain and controls simple reflexes. ###### Somatic Nervous System Part of the peripheral nervous system, the somatic nervous system is made up of nerve fibers that send sensory information to the central nervous system and control voluntary actions. ###### Autonomic Nervous System Part of the peripheral nervous system, the autonomic nervous system regulates involuntary activity in the heart, stomach, and intestines. ##### Reproductive System ###### Ovulation Approximately every 28 days during female ovulation an egg (ovum) is released from one of the ovaries and travels through the oviduct (fallopian tube) and into the uterus. At the same time, the endometrial lining of the uterus becomes prepared for implantation. ###### Reproduction During intercourse, the penis ejaculates sperm, produced in the testes, into the vagina. Some of the sperm makes their way to the uterus where, if they encounter an egg to fertilize, unite with the ovum to form a fertilized egg or zygote. The zygote then may implant in the uterus and eventually develop into a fetus. ###### Menstruation If the ovum fails to become fertilized, the lining of the uterus sloughs off during menstruation. From puberty to menopause, this cycle of menstruation repeats monthly (except during pregnancy). ### Cellular Biology ##### Genetics ###### Heredity Heredity is the passing on of physical or mental characteristics genetically from one generation to another. Heredity is made possible via large strings of chromosomes which carry information encoded in genes. ###### Meiosis Reproductive (haploid) cells known as gametes have half as many (23) pairs of chromosomes as normal (diploid) cells. When the male gamete (sperm) combines with the female gamete (ovum) through meiosis to form a zygote, each gamete supplies half the chromosomes needed to form the normal diploid cells. ###### DNA Deoxyribonucleic acid (DNA) is the molecule that contains genetic information. DNA is encoded through a combination of nucleotides that bind together in a specific double helix pattern. ###### Genes The gene is the base unit of inheritance and is contained within DNA. A gene may come in several varieties (alleles) and there are a pair of alleles for every gene. If the alleles are alike, a person is homozygous for that gene. If the alleles are different, heterozygous. ###### Traits The traits represented by genes are inherited independently of each other (one from the male and one from the female gamete) and a trait can be dominant or recessive. A dominant trait will be expressed when paired with a recessive trait while two copies of a recessive trait (one from each parent) must be present for the recessive trait to be expressed. ###### Genetic Type A person's genotype is their genetic makeup and includes both dominant and recessive alleles. Phenotype is how the genes express themselves in physical characteristics. ##### Cell Structures ###### Nucleus Cells are classified into one of two groups based on whether or not they have a nucleus. Eukaryotic cells have a nucleus, prokaryotic cells do not have a nucleus and therefore have a less complex structure than eukaryotic cells. ###### Cytoplasm The nucleus of a eukaryotic cell contains the genetic material of the cell and is surrounded by cytoplasm which contains many organelles. These include: Organelle Function ribosomes produce proteins mitochondria produce energy endoplasmic reticulum helps synthesize proteins and fats Golgi apparatus prepare proteins for use lysosomes help the cell manage waste centrosomes guide cell reproduction ###### Cell Membrane Animal cells are surrounded by a semipermeable membrane which allows for the transfer of water and oxygen to and from the cell. In plant cells, the cell membrane is surrounded by a somewhat rigid cell wall which provides the cell structure and support. ###### Cell Energy Some plant cells produce their own energy through photosynthesis which is the process by which sunlight, carbon dioxide, and water react to make sugar and oxygen. Animal cells cannot produce their own energy and, instead, generate energy when mitochondria consume outside sugar and oxygen through aerobic respiration. ###### Fermentation If no oxygen is present, cellular respiration is anaerobic and will result in fermentation where either lactic acid or alcohol is used instead of oxygen. ###### Cell Division Cell division is the process by which cells replicate genetic material in the nucleus. Cell division consists of several phases: Phase Major Process interphase chromosomes replicate into chromatids and the cell grows prophase chromatids pair up metaphase paired chromatids move to opposite sides of the cell anaphase cell elongates and nucleus begins to separate telophase separation of nucleus is complete resulting in two new nuclei cytokinesis cytoplasm and cell membranes complete their separation resulting in two separate cells ### Ecology ##### Classifications ###### Biosphere The biosphere is the global ecological system integrating all living beings and their relationships. This includes their interactions with the lithosphere (the rigid outer part of the earth, consisting of the crust and upper mantle), hydrosphere (all surface water), and atmosphere (the envelope of gases surrounding the planet). ###### Biome A biome is a large naturally occurring community of flora (plants) and fauna (animals) occupying a major habitat (home or environment). ###### Ecosystem An ecosystem is a biological community of interacting organisms and their physical environment. This includes both the biotic (living) and abiotic (non-living). ###### Species Groups A population is a group of organisms of the same species who live in the same area at the same time. A community is a group of populations living and interacting with each other in an area. ##### Food Chain ###### Producers Producers (autotrophs) serve as a food source for other organisms. Typical producers are plants that can make their own food through photosynthesis and certain bacteria that are capable of converting inorganic substances into food through chemosynthesis ###### Decomposers Decomposers (saprotrophs) are organisms such as bacteria and fungi that break down the organic matter in the dead bodies of plants and animals into simple nutrients. ###### Scavengers Like decomposers, scavengers also break down the dead bodies of plants and animals into simple nutrients. The difference is that scavengers operate on much larger refuse and dead animals (carrion). Decomposers then consume the much smaller particles left over by the scavengers. ###### Consumers Most animals consume other organisms to survive. Consumers (heterotrophs) are divided into three types, primary, secondary, and tertiary, based on their place in the food chain. ###### Primary Consumers Primary consumers (herbivores) subsist on producers like plants and fungus. Examples are grasshoppers, cows, and plankton. ###### Secondary Consumers Secondary consumers (carnivores) subsist mainly on primary consumers. Omnivores are secondary consumers that also eat producers. Examples are rats, fish, and chickens. ###### Tertiary Consumers Tertiary consumers eat primary consumers and secondary consumers and are typically carnivorous predators. Tertiary consumers may also be omnivores. Examples include wolves, sharks, and human beings. ##### Taxonomy ###### Domain The broadest classification of life splits all organisms into three groups called domains. The three domains of life are bacteria, archaea and eukaryota. ###### Kingdom Below domain, life is classified into six kingdoms: plants, animals, archaebacteria, eubacteria, and fungi. The last kingdom, protists, include all microscopic organisms that are not bacteria, animals, plants or fungi. (Archaebacteria and eubacteria are sometimes combined into a single kingdom, monera.) ###### Classification of Life Classifications of life are too numerous to enumerate, here's an overview of the classifications from broadest to narrowist: Classification Contains Related Kingdom phyla Phylum classes Class orders Order families Family genera Genus species Species organisms ###### Species The narrowest classification of life, species, contains organisms that are so similar that they can only reproduce with others of the same species. ### Earth Science ##### Structure of the Earth ###### Crust The crust is the Earth's outermost layer and is divided into oceanic and continental types. Oceanic crust is 3 miles (5 km) to 6 miles (10 km) thick and is composed primarily of denser rock. Continental crust is 20 to 30 miles (30 to 50 km) thick and composed primarily of less dense rock. The crust makes up approximately one percent of the Earth's total volume. ###### Mantle Mantle makes up 84% of the Earth's volume and has an average thickness of approximately 1,800 miles (2,900 km). It is dense, hot, and primarily solid although in places it behaves more like a viscous fluid as the plates of the upper mantle and crust gradually "float" along its circumference. ###### Core The Earth's core is divided into the liquid outer core (1,430 miles or 2,300 km radius) and the solid inner core (745 miles or 1,200 km radius). ###### Plate Tectonics The crust and the rigid lithosphere (upper mantle) is made up of approximately thirty separate plates. These plates more very slowly on the slightly more liquid mantle (asthenosphere) beneath them. This movement has resulted in continental drift which is the gradual movement of land masses across Earth's surface. Continental drift is a very slow process, occurring over hundreds of millions of years. ###### Types of Rock The Earth's rocks fall into three categories based on how they're formed. Igneous rock (granite, basalt, obsidian) is formed from the hardening of molten rock (lava), sedimentary rock (shale, sandstone, coal) is formed by the gradual despositing and cementing of rock and other debris, and metamorphic rock (marble, slate, quartzite) which is formed when existing rock is altered though pressure, temperature, or chemical processes. ###### Geologic Time Scale The Earth is approximately 4.6 billion years old and its history is divided into time periods based on the events that took place and the forms of life that were dominant during those periods. The largest graduation of time is the eon and each eon is subdivided into eras, eras into periods, periods into epochs, and epochs into ages. ###### Cambrian Period The Cambrian period is one of the most significant geological time periods. Lasting about 53 million years, it marked a dramatic burst of changes in life on Earth known as the Cambrian Explosion. It is from this period that the majority of the history of life on Earth, as documented by fossils, is found. Called the fossil record, the layering of these mineralized imprints of organisms preserved in sedementary rock have allowed geologists to build a historical record of plant and animal life on Earth. ##### Biochemical Cycles ###### Water Cycle The water (hydrologic) cycle describes the movement of water from Earth through the atmosphere and back to Earth. The cycle starts when water evaporates into a gas from bodies of water like rivers, lakes and oceans or transpirates from the leaves of plants. ###### Precipitation Rising into the atmosphere, the water condenses into clouds. When the clouds become too saturated with water, the water is released as snow or ice precipitation which may warm as it falls to reach Earth as rain. ###### Infiltration The water then accumulates as runoff and eventually returns to bodies of water or is absorbed into the Earth (infiltration) and becomes part of the water table, an underground resevoir of fresh water. ###### Carbon Cycle The carbon cycle represents the ciruit of carbon through Earth's ecosystem. Carbon dioxide (CO2) in the atmosphere is absorbed by plants through photosynthesis. Plants then die and release carbon back into the atmosphere during decomposition or are eaten by animals who breathe (respiration) the carbon into the atmosphere they exhale and produce waste which also releases carbon as it decays. ##### Atmosphere ###### Troposphere The Earth's atmosphere has several layers starting with the troposphere which is closest in proximity to the surface. Containing most of the Earth's breathable air (oxygen and nitrogen), it's a region with warmer temperatures closer to the surface and cooler temperatures farther away which results in the rising and falling air that generates weather. ###### Stratosphere The stratosphere is just above the troposphere and is stratified in temperature with warmer layers higher and cooler layers closer to Earth. This increase in temperature is a result of absorption of the Sun's radiation by the ozone layer. ###### Mesosphere In the mesosphere, temperature again drops as altitude increases until the coldest point in the Earth's atmosphere, the mesopause, is reached where temperatures fall to −225 °F (−143 °C). ###### Thermosphere Temperatures again increase with altitude in the thermosphere which is the hottest (4,530 °F / 2,500 °C) atmospheric layer due to direct exposure to the Sun's radiation. However, the gas in this layer is highly diluted so even though the atoms of gas may be very high in temperature, there are too few of them to effectively transfer much heat. ##### Weather ###### Fronts An air mass is a large body of air that has similar moisture (density) and temperature characteristics. A front is a transition zone between two air masses. ###### Cold Front A cold front is a warm-cold air boundary with the colder air replacing the warmer. As a cold front moves into an area, the heavier cool air pushes under the lighter warm air that it is replacing. The warm air becomes cooler as it rises and, if the rising air is humid enough, the water vapor it contains will condense into clouds and precipitation may fall. ###### Warm Front A warm front is the boundary between warm and cool (or cold) air when the warm air is replacing the cold air. Warm air at the surface pushes above the cool air mass creating clouds and storms. ###### Stationary Front When two air masses meet and neither is displaced, a stationary front is created. Stationary fronts often cause persistent cloudy wet weather. ###### Stratus Clouds Clouds are categorized based on their shape, size, and altitude. Stratus clouds are low-altitude clouds characterized by horizontal layering with a broad flat base. When stratus clouds occur on the ground the result is fog. ###### Cumulus Clouds Cumulus clouds are large, puffy, mid-altitude clouds with a flat base and a rounded top. These clouds grow upward and can develop into a cumulonimbus or thunderstorm cloud. ###### Cirrus Clouds Cirrus clouds are thin, wispy high-altitude clouds composed of ice crystals that originate from the freezing of supercooled water droplets. Cirrus clouds generally occur in fair weather and point in the direction of air movement at their elevation. ##### Solar System ###### The Sun The Sun is a G-type main-sequence star (G2V) but is informally known as a yellow dwarf star. Composed of 73% hydrogen and 25% helium, the hot plasma that makes up the Sun reaches 9,900°F (5,505°C) at the surface. It formed approximately 4.6 billion years ago and makes up 99.86% of the mass in the solar system. ###### Terrestrial Planets The four planets closest to the Sun (Mercury, Venus, Earth, and Mars) are called terrestrial (Earth-like) planets because, like the Earth, they're solid with inner metal cores covered by rocky surfaces. ###### Outer Planets In contrast to the solid terrestrial planets, the outer planets (Jupiter, Saturn, Uranus, and Neptune) consist of hydrogen and helium gas and water. ###### Asteroids The solar system also contains over a million rocky fragments of at least 1km in diameter called asteroids as well as millions more with smaller diameters. Many of these asteroids are an asteroid belt between the orbits of Mars and Jupiter. ###### Kuiper Belt The Kuiper Belt is similar to the asteroid belt but much larger. Extending beyond the orbit of Neptune, it contains objects composed mostly of frozen methane, ammonia, and water. Most notably, the Kuiper Belt is home to Pluto, a dwarf planet that, until a 2006 reclassification, was considered the ninth planet of the solar system. ###### Comets A comet is a loose collection of ice, dust, and small rocky particles that, in contrast to an asteroid, has an extended atmosphere surrounding the center. When passing close to the Sun, this atmosphere warms and begins to release gases forming a visible coma or tail. ###### Meteoroids Smaller rocks shed by asteroids and comets are called meteoroids. When these rocks reach Earth's atmosphere, they burn up in the mesosphere and become meteors. If a meteor manages to reach the Earth, it is called a meteorite. ###### Moon Tides are caused by the gravitational interaction of Earth and the Moon. ### Physical Science ##### Metric System ###### Number System The metric system is a number system that designates one base unit for each type of measurement. For example, the base unit for length is the meter and the base unit for mass is the gram. ###### Prefixes A prefix is added to the base units of the metric system to indicate variations in size. Each prefix specifies a value relative to the base unit in a multiple of 10. Common prefixes are: Prefix Symbol Relative Value Example mega M 106 (1,000,000) Mm kilo k 103 (1,000) km base unit N/A 1 m centi c 10-2 (1/100) cm milli m 10-3 (1/1,000) mm ###### Base Units Measurement Base Unit Example length / distance meter (m) km mass gram (g) kg volume liter (L) mL volume (medical) cubic centimeter (cc) cc time second (s), minute (min), hour (h) ms, min, h ###### Fahrenheit Scale More familiar in the United States is the Fahrenheit scale in which the freezing point of water is 32°F (0°C) and the boiling point is 212°F (100°C). To convert from C° to F° use the formula: $$F° = {9 \over 5}C° + 32$$ and to convert from F° to C° use: $$C° = {5 \over 9} (F° - 32)$$ ###### Kelvin Scale In contrast to the Celsius scale (measured in degrees centigrade) that fixes 0° at the freezing point of water and the Fahrenheit scale that uses 32°, the Kelvin scale fixes 0° at absolute zero (-273°C) which is the lowest temperature possible in the universe. ##### Motion ###### Velocity Velocity is the rate at which an object changes position. Rate is measured in time and position is measured in displacement so the formula for velocity becomes $$\vec{v} = { \vec{d} \over t }$$ ###### Vectors Velocity and displacement are vector quantities which means each is fully described by both a magnitude and a direction. In contrast, scalar quantities are quantities that are fully described by a magnitude only. A variable indicating a vector quantity will often be shown with an arrow symbol: $$\vec{v}$$ ###### Momentum Momentum is a measure of how difficult it is for a moving object to stop and is calculated by multiplying the object's mass by its velocity: $$\vec{p} = m\vec{v}$$. Like velocity, momentum is a vector quantity as it expresses force applied in a specific direction. ###### Acceleration Acceleration is the rate of change of velocity per unit of time. In physics, the delta symbol ($$\Delta$$) represents change so the formula for acceleration becomes $$\vec{a} = { \Delta \vec{v} \over t }$$ ##### Forces and Energy ###### Force Force is applied to change an object's speed or direction of motion. ###### Weight Weight is a force that describes the attraction of gravity on an object. Force is measured in newtons (N) with 1 N being the force required to impart an acceleration of 1 m/s2 to a mass of 1 kg. ###### Mass vs. Weight Mass is the amount of matter something has while weight is the force exerted on an object's mass by gravity. So, although a person's mass doesn't change when going from the Earth to the Moon, their weight will decrease because the force of the Moon's gravity is much less than that of Earth. ###### Work Work is performed on an object when an applied force causes displacement along the same vector. Measured in joules (J) or newton-meters (Nm), work is calculated by multiplying force times displacement: $$W = \vec{F}\vec{d}$$ ###### Power Power is the rate at which work is performed or work per unit time: $$P = {w \over t}$$ and is measured in watts (W). ###### Kinetic Energy Kinetic energy is the energy posessed by a moving object. Potential energy is stored energy in a stationary object based on its location, position, shape, or state. ###### Electromagnetic Spectrum The electromagnetic spectrum covers all possible wavelengths and frequencies of radiation. From lowest frequency (longest wavelength) to highest frequency (shortest wavelength) radiation: radio waves → microwaves → infrared waves → visible light → ultraviolet light → X-rays → gamma rays. ###### Magnetism Simple magnets have two poles, north and south, and opposite poles attract each other (N attracts S, S attracts N). Likewise, the same pole of two magnets repel (N repels N, S repels S). The Earth has a magnetic field and North and South Poles which enables the use of a magnetic compass to determine direction. ##### Newton's Laws ###### First Law of Motion Also known as the law of inertia, Newton's first law of motion states that An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force. ###### Second Law of Motion Newton's second law of motion states that The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object. This law basically means that the greater the mass of an object, the more force is needed to overcome its inertia. ###### Acceleration Newton's second law of motion leads to the formula for acceleration which is a measure of the rate of change of velocity per unit time and, if you solve for positive acceleration, reveals how much net force is needed to overcome an object's mass. The formula for acceleration is $$\vec{a} = { \vec{F} \over m }$$ or, solving for force, $$\vec{F} = m\vec{a}$$. ###### Third Law of Motion Newton's third law of motion states that For every action, there is an equal and opposite reaction. When an object exerts a force on another object, the second object exerts a force of equal magnitude in the opposite direction on the first object. ###### Law of Universal Gravitation Newton's law of universal gravitation defines gravity: All objects in the universe attract each other with an equal force that varies directly as a product of their masses, and inversely as a square of their distance from each other. Expressed as a formula: $$\vec{F_{g}} = { Gm_{1}m_{2} \over r^2}$$ where r is the distance between the two objects and G is the gravitational constant with a value of 6.67 x 10-11. ##### Sound Waves ###### Vibration A vibrating object produces a sound wave that travels outwardly from the object through a medium (any liquid or solid matter). The vibration disturbs the particles in the surrounding medium, those particles disturb the particules next to them, and so on, as the sound propagates away from the vibration. ###### Speed The speed of a sound wave will vary with the medium. Sound travels fastest through media that has particles that are very close together, like metal. Thus, it travels faster through water than through air and doesn't travel at all through a vacuum (there are no particles in empty space to vibrate). ###### Frequency The rate of vibration of sound is called frequency and is measured in hertz (Hz). One hertz is one repetition per second and sounds with high frequency have a higher pitch than sounds with lower frequency. Humans can hear sounds in the range of 20 Hz to 20 kHz. ###### Doppler Effect The Doppler effect occurs when the source or listener (or both) of sound waves is moving. If they're moving closer together, the listener perceives the sound with a higher pitch and, when they're moving apart, the listener perceives the sound with a lower pitch. ##### Optics ###### Light Waves Unlike mechanical sound waves that require a physical medium for propagation, light waves are electromagnetic and can travel through empty space. Light waves are also much faster, travelling at 186,000 m/s vs. 343 m/s for sound waves. ###### Refractive Index The speed of light varies based on the material that the waves are passing through. The refractive index of a material indicates how easily light travels through it compared to how easily light travels through a vacuum. For example, the refractive index of water is 1.33, meaning that light travels 1.33 times faster in a vacuum than it does in water. ###### Refraction Because different materials have different refractive indices, light changes speed when passing from one material to another. This causes the light to bend (refraction) at an angle that depends on the change in refractive index between the materials. The greater the difference, the higher the angle of refraction. ###### Reflection The law of reflection specifies how waves, including light waves, bounce off of surfaces. Specifically, the angle of incidence of the approaching wave is equal to the angle of reflection of the reflected wave as measured from a line perpendicular (90°) to the surface. ###### Curved Mirrors A concave (or converging) mirror bulges inward and focuses reflected light on the mirror's focal point where the mirror's angles of incidence converge. In contrast, a convex (or diverging) mirror bulges outward and diffuses the light waves that strike it. A common use of a concave mirror is in a reflecting telescope, a common use of a convex mirror is in the side view mirror of a car. ###### Curved Lenses Unlike curved mirrors that operate on the principle of reflection, lenses utilize refraction. A convex lens is thicker in the middle than on the edges and converges light while a concave lens is thicker on the edges than in the middle and diffuses light. A common use for curved lenses is in eye glasses where a convex lens is used to correct farsightedness and a concave lens is used to correct nearsightedness. ##### Heat ###### Conduction Heat is always transferred from warmer to cooler environments and conduction is the simplest way this transfer can occur. It is accomplished through direct contact between materials and materials like metals that transfer heat efficiently are called conductors while those that conduct heat poorly, such as plastic, are called insulators. ###### Convection Convection is the transfer of heat by the circulation or movement of the heated parts of a liquid or gas. Examples of heat transfer by convection include water coming to a boil on a stove, ice melting, and steam from a cup of coffee. Radiation occurs when electromagnetic waves transmit heat. An example is the heat from the Sun as it travels to Earth. ##### Matter ###### Element An element is matter than cannot be separated into different types of matter by ordinary chemical methods. ###### Atom An atom is the smallest component of an element that still retains the properties of the element. ###### Proton A proton is a subatomic particle found in the nucleus of an atom. It carries a positive electric charge. ###### Neutron A neutron is a subatomic particle found in the nucleus of an atom. It is neutral as it carries no electric charge. ###### Electron An electron is a subatomic particle that orbits the nucleus of an atom. It carries a negative electric charge. Generally, an atom has the same number of negative electrons orbiting the nucleus as it does positive protons inside. ###### Compound A compound is a substance containing two or more different chemical elements bound together by a chemical bond. In covalent compounds, electrons are shared between atoms. In ionic compounds, one atom borrows an electron from another atom resulting in two ions (electrically charged atoms) of opposite polarities that then become bonded electrostatically. ###### Molecule A molecule is the smallest multi-atom particle of an element or compound that can exist and still retain the characteristics of the element or compound. The molecules of elements consist of two or more similar atoms, the molecules of compounds consist of two or more different atoms. ###### Acidity An acid is a substance that gives up positively charged hydrogen ions (H+) when dissolved in water. A base (alkaline) gives up negatively charged hydroxide ions (OH-) when dissolved in water. pH is a scale that measures of how basic or acidic a solution is. Numbered from 0 to 14, solutions with a pH of 7 are neutral, less than 7 are acidic, more than 7 are alkaline. ###### Chemical Change During a chemical reaction molecules and atoms (reactants) are rearranged into new combinations that result in new kinds of atoms or molecules (products). ##### Periodic Table of the Elements ###### Atomic Number The Periodic Table of the Elements categorizes elements primarily by the number of protons in their nucleus (atomic number) and secondarily by the characteristics they exhibit. ###### Periods The rows of the Periodic Table are called periods and contain elements that have the same number of electron shells ordered from lower to higher atomic number. ###### Groups The columns of the Periodic Table are called groups and all elements in a group have the same number of electrons in their outer electron shell. The group that an element occupies generally determines its chemical properties as the number of outer shell electrons establishes the way it reacts with other elements to form molecules. So, because each element has the same number of electrons in its outer shell, each has similar reactivity. ###### Atomic Mass The atomic mass of an element listed in the Periodic Table represents the average mass of a single atom of that element and is measured in atomic mass units (amu). This number is an average as some elements have isotopes with atoms that vary in their number of neturons and, therefore, differ in weight. ##### Physical State ###### Solid An element in a solid state has atoms or molecules that are constricted and do not move freely. Solids maintain a constant volume and shape and exist at a lower temperature than liquids or gases. ###### Liquid In the liquid state, molecules flow freely around each other and exist at a higher temperature range than the same substance in a solid state. Liquids maintain a constant volume but their shape depends upon the shape of their container. ###### Gas The gaseous state occurs at a higher temperature range than the solid and liquid states of the same substance. In this state, molecules flow very freely around each other and will spread out as far as they're able. Gases maintain neither a constant volume nor a constant shape. ###### Phase Transition A substance undergoes a phase transition when it moves from one state of matter to another, for example, when water freezes or boils. # Math Knowledge • 28 Questions • 38 Problems • 55 Flash Cards ### Algebra ##### Expressions ###### Classifications A monomial contains one term, a binomial contains two terms, and a polynomial contains more than two terms. Linear expressions have no exponents. A quadratic expression contains variables that are squared (raised to the exponent of 2). ###### Operations Involving Monomials You can only add or subtract monomials that have the same variable and the same exponent. However, you can multiply and divide monomials with unlike terms. ###### Multiplying Binomials To multiply binomials, use the FOIL method. FOIL stands for First, Outside, Inside, Last and refers to the position of each term in the parentheses. To factor a quadratic expression, apply the FOIL (First, Outside, Inside, Last) method in reverse. ##### Solving Equations ###### One Variable An equation is two expressions separated by an equal sign. The key to solving equations is to repeatedly do the same thing to both sides of the equation until the variable is isolated on one side of the equal sign and the answer on the other. ###### Two Variables When solving an equation with two variables, replace the variables with the values given and then solve the now variable-free equation. (Remember order of operations, PEMDAS, Parentheses, Exponents, Multiplication/Division, Addition/Subtraction.) ###### Two Equations When presented with two equations with two variables, evaluate the first equation in terms of the variable you're not solving for then insert that value into the second equation. For example, if you have x and y as variables and you're solving for x, evaluate one equation in terms of y and insert that value into the second equation then solve it for x. When solving quadratic equations, if the equation is not set equal to zero, first manipulate the equation so that it is set equal to zero: ax2 + bx + c = 0. Then, factor the quadratic and, because it's set to zero, you know that one of the factors must equal zero for the equation to equal zero. Finding the value that will make each factor, i.e. (x + ?), equal to zero will give you the possible value(s) of x. ###### Inequalities Solving equations with an inequality (<, >) uses the same process as solving equations with an equal sign. Isolate the variable that you're solving for on one wide of the equation and put everything else on the other side. The only difference is that your answer will be expressed as an inequality (x > 5) and not as an equality (x = 5). ### Geometry ##### Lines & Angles ###### Line Segment A line segment is a portion of a line with a measurable length. The midpoint of a line segment is the point exactly halfway between the endpoints. The midpoint bisects (cuts in half) the line segment. ###### Right Angle A right angle measures 90 degrees and is the intersection of two perpendicular lines. In diagrams, a right angle is indicated by a small box completing a square with the perpendicular lines. ###### Acute & Obtuse Angles An acute angle measures less than 90°. An obtuse angle measures more than 90°. ###### Angles Around Lines & Points Angles around a line add up to 180°. Angles around a point add up to 360°. When two lines intersect, adjacent angles are supplementary (they add up to 180°) and angles across from either other are vertical (they're equal). ###### Parallel Lines Parallel lines are lines that share the same slope (steepness) and therefore never intersect. A transversal occurs when a set of parallel lines are crossed by another line. All of the angles formed by a transversal are called interior angles and angles in the same position on different parallel lines equal each other (a° = w°, b° = x°, c° = z°, d° = y°) and are called corresponding angles. Alternate interior angles are equal (a° = z°, b° = y°, c° = w°, d° = x°) and all acute angles (a° = c° = w° = z°) and all obtuse angles (b° = d° = x° = y°) equal each other. Same-side interior angles are supplementary and add up to 180° (e.g. a° + d° = 180°, d° + c° = 180°). ##### Polygons ###### Triangle Geometry A triangle is a three-sided polygon. It has three interior angles that add up to 180° (a + b + c = 180°). An exterior angle of a triangle is equal to the sum of the two interior angles that are opposite (d = b + c). The perimeter of a triangle is equal to the sum of the lengths of its three sides, the height of a triangle is equal to the length from the base to the opposite vertex (angle) and the area equals one-half triangle base x height: a = ½ base x height. ###### Triangle Classification An isosceles triangle has two sides of equal length. An equilateral triangle has three sides of equal length. In a right triangle, two sides meet at a right angle. ###### Pythagorean Theorem The Pythagorean theorem defines the relationship between the side lengths of a right triangle. The length of the hypotenuse squared (c2) is equal to the sum of the two perpendicular sides squared (a2 + b2): c2 = a2 + b2 or, solved for c, $$c = \sqrt{a + b}$$ A quadrilateral is a shape with four sides. The perimeter of a quadrilateral is the sum of the lengths of its four sides (a + b + c + d). ###### Rectangle & Square A rectangle is a parallelogram containing four right angles. Opposite sides (a = c, b = d) are equal and the perimeter is the sum of the lengths of all sides (a + b + c + d) or, comonly, 2 x length x width. The area of a rectangle is length x width. A square is a rectangle with four equal length sides. The perimeter of a square is 4 x length of one side (4s) and the area is the length of one side squared (s2). ###### Parallelogram A parallelogram is a quadrilateral with two sets of parallel sides. Opposite sides (a = c, b = d) and angles (red = red, blue = blue) are equal. The area of a parallelogram is base x height and the perimeter is the sum of the lengths of all sides (a + b + c + d). ###### Rhombus A rhombus has four equal-length sides with opposite sides parallel to each other. The perimiter is the sum of the lengths of all sides (a + b + c + d) or, because all sides are the same length, 4 x length of one side (4s). ###### Trapezoid A trapezoid is a quadrilateral with one set of parallel sides. The area of a trapezoid is one-half the sum of the lengths of the parallel sides multiplied by the height. In this diagram, that becomes ½(b + d)(h). ##### Circles ###### Dimensions A circle is a figure in which each point around its perimeter is an equal distance from the center. The radius of a circle is the distance between the center and any point along its perimeter (AC, CB, CD). A chord is a line segment that connects any two points along its perimeter (AB, AD, BD). The diameter of a circle is the length of a chord that passes through the center of the circle (AB) and equals twice the circle's radius (2r). ###### Calculations The circumference of a circle is the distance around its perimeter and equals π (approx. 3.14159) x diameter: c = π d. The area of a circle is π x (radius)2 : a = π r2. ##### Solids ###### Cubes A cube is a rectangular solid box with a height (h), length (l), and width (w). The volume is h x l x w and the surface area is 2lw x 2wh + 2lh. ###### Cylinders A cylinder is a solid figure with straight parallel sides and a circular or oval cross section with a radius (r) and a height (h). The volume of a cylinder is π r2h and the surface area is 2(π r2) + 2π rh. ##### Coordinate Geometry ###### Coordinate Grid The coordinate grid is composed of a horizontal x-axis and a vertical y-axis. The center of the grid, where the x-axis and y-axis meet, is called the origin. ###### Slope-Intercept Equation A line on the coordinate grid can be defined by a slope-intercept equation: y = mx + b. For a given value of x, the value of y can be determined given the slope (m) and y-intercept (b) of the line. The slope of a line is change in y over change in x, $${\Delta y \over \Delta x}$$, and the y-intercept is the y-coordinate where the line crosses the vertical y-axis. # Mechanical Comprehension • 94 Questions • 34 Problems • 100 Flash Cards ### Fundamentals of Mechanics ##### Principles ###### Mechanics Mechanics deals with motion and the forces that produce motion. ###### Mass Mass is a measure of the amount of matter in an object. In general, larger objects have larger mass than smaller objects but mass ultimately depends on how compact (dense) a substance is. ###### Inertia The more mass a substance has the more force is required to move it or to change its direction. This resistance to changes in direction is known as inertia. ###### Net Force In mechanics, multiple forces are often acting on a particular object and, taken together, produce the net force acting on that object. Like force, net force is a vector quantity in that it has magnitude and direction. ###### Newton's Second Law of Motion Newton's Second Law of Motion states that "The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object." This Law describes the linear relationship between mass and acceleration when it comes to force and leads to the formula F = ma or force equals mass multiplied by rate of acceleration. ###### Universal Gravitation Newton's Law of Univeral Gravitation defines the general formula for the attraction of gravity between two objects: $$\vec{F_{g}} = { Gm_{1}m_{2} \over r^2}$$ . In the specific case of an object falling toward Earth, the acceleration due to gravity (g) is approximately 9.8 m/s2. ###### Weight Mass is an intrinsic property of matter and does not vary. Weight is the force exerted on the mass of an object due to gravity and a specific case of Newton's Second Law of Motion. Replace force with weight and acceleration with acceleration due to gravity on Earth (g) and the result is the formula for weight: W = mg or, substituting for g, weight equals mass multiplied by 9.8 m/s2. ###### Boyle's Law Boyle's law states that "for a fixed amount of an ideal gas kept at a fixed temperature, pressure and volume are inversely proportional". Expressed as a formula, that's $$\frac{P_1}{P_2} = \frac{V_2}{V_1}$$ ##### Friction ###### Kinetic Friction Friction resists movement. Kinetic (also called sliding or dynamic) friction resists movement in a direction opposite to the movement. Because it opposes movement, kinetic friction will eventually bring an object to a stop. An example is a rock that's sliding across ice. ###### Static Friction Static friction is friction between two or more solid objects that are not moving relative to each other. An example is the friction that prevents a box on a sloped surface from sliding farther down the surface. ###### Coefficient of Friction Coefficient of friction (μ) represents how much two materials resist sliding across each other. Smooth surfaces like ice have low coefficients of friction while rough surfaces like concrete have high μ. ###### Normal Force Normal force (FN) represents the force a surface exerts when an object presses against it. ###### Force of Friction The formula for force of friction (Ff) is the same whether kinetic or static friction applies: Ff = μFN. To distinguish between kinetic and static friction, μk and μs are often used in place of μ. ###### Kinetic vs. Static Friction For any given surface, the coefficient of static friction is higher than the coefficient of kinetic friction. More force is required to initally get an object moving than is required to keep it moving. Additionally, static friction only arises in response to an attempt to move an object (overcome the normal force between it and the surface). ###### Normal Force vs. Weight Normal force arises on a flat horizontal surface in response to an object's weight pressing it down. Consequently, normal force is generally equal to the object's weight. ##### Other Forces ###### Drag Drag is friction that opposes movement through a fluid like liquid or air. The amount of drag depends on the shape and speed of the object with slower objects experiencing less drag than faster objects and more aerodynamic objects experiencing less drag than those with a large leading surface area. ###### Tension Tension is a force that stretches or elongates something. When a cable or rope is used to pull an object, for example, it stretches internally as it accepts the weight that it's moving. Although tension is often treated as applying equally to all parts of a material, it's greater at the places where the material is under the most stress. ###### Hydraulic Pressure Hydraulics is the transmission of force through the use of liquids. Liquids are especially suited for transferring force in complex machines because they compress very little and can occupy very small spaces. Hydraulic pressure is calculated by dividing force by the area over which it is applied: P = F/A where F is force in pounds, A is area in square inches, and the resulting pressure is in pounds per square inch (psi). ###### Pascal's Law Pascal's law states that a pressure change occurring anywhere in a confined incompressible fluid is transmitted throughout the fluid such that the same change occurs everywhere. For a hydraulic system, this means that a pressure applied to the input of the system will increase the pressure everywhere in the system. ###### Torque Torque measures force applied during rotation: τ = rF. Torque (τ, the Greek letter tau) = the radius of the lever arm (r) multiplied by the force (F) applied. Radius is measured from the center of rotation or fulcrum to the point at which the perpendicular force is being applied. The resulting unit for torque is newton-meter (N-m) or foot-pound (ft-lb). ###### Principle of Moments When a system is stable or balanced (equilibrium) all forces acting on the system cancel each other out. In the case of torque, equilibrium means that the sum of the anticlockwise moments about a center of rotation equal the sum of the clockwise moments. ##### Energy, Work, & Power ###### Joules The Joule (J) is the standard unit of energy and has the unit $${kg \times m^2} \over s^2$$. ###### Kinetic Energy Kinetic energy is the energy of movement and is a function of the mass of an object and its speed: $$KE = {1 \over 2}mv^2$$ where m is mass in kilograms, v is speed in meters per second, and KE is in joules. The most impactful quantity to kinetic energy is velocity as an increase in mass increases KE linearly while an increase in speed increases KE exponentially. ###### Potential Energy Potential energy is the energy of an object by virtue of its position relative to other objects. It is energy that has the potential to be converted into kinetic energy. ###### Gravitational Potential Energy Gravitational potential energy is energy by virtue of gravity. The higher an object is raised above a surface the greater the distance it must fall to reach that surface and the more velocity it will build as it falls. For gravitational potential energy, PE = mgh where m is mass (kilograms), h is height (meters), and g is acceleration due to gravity which is a constant (9.8 m/s2). ###### Conservation of Mechanical Energy As an object falls, its potential energy is converted into kinetic energy. The principle of conservation of mechanical energy states that, as long as no other forces are applied, total mechanical energy (PE + KE) of the object will remain constant at all points in its descent. ###### Work Work is accomplished when force is applied to an object: W = Fd where F is force in newtons (N) and d is distance in meters (m). Thus, the more force that must be applied to move an object, the more work is done and the farther an object is moved by exerting force, the more work is done. ###### Work-Energy Theorem The work-energy theorem states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle. Simply put, work imparts kinetic energy to the matter upon which the work is being done. ###### Power Power is the rate at which work is done, P = w/t, or work per unit time. The watt (W) is the unit for power and is equal to 1 joule (or newton-meter) per second. Horsepower (hp) is another familiar unit of power used primarily for rating internal combustion engines. A 1 hp machine does 550 ft⋅lb of work in 1 second and 1 hp equals 746 watts. ### Machines ##### Simple Machines ###### Types of Simple Machines The six types of simple machines are the lever, wheel and axle, pulley, inclined plane, wedge, and screw. Mechanical advantage is a measure of the force amplification achieved by using a tool, mechanical device or machine system. Such a device utilizes input force and trades off forces against movement to amplify and/or change its direction. ###### Efficiency The efficiency of a machine describes how much of the power put into the machine is turned into movement or force. A 100% efficient machine would turn all of the input power into output movement or force. However, no machine is 100% efficient due to friction, heat, wear and other imperfections that consume input power without delivering any output. ###### Inclined Plane An inclined plane is a simple machine that reduces the force needed to raise an object to a certain height. Work equals force x distance and, by increasing the distance that the object travels, an inclined plane reduces the force necessary to raise it to a particular height. In this case, the mechanical advantage is to make the task easier. An example of an inclined plane is a ramp. ###### Wedge The wedge is a moving inclined plane that is used to lift, hold, or break apart an object. A wedge converts force applied to its blunt end into force perpendicular to its inclined surface. In contrast to a stationary plane where force is applied to the object being moved, with a wedge the object is stationary and the force is being applied to the plane. Examples of a wedge include knives and chisels. ###### First-Class Lever A first-class lever is used to increase force or distance while changing the direction of the force. The lever pivots on a fulcrum and, when a force is applied to the lever at one side of the fulcrum, the other end moves in the opposite direction. The position of the fulcrum also defines the mechanical advantage of the lever. If the fulcrum is closer to the force being applied, the load can be moved a greater distance at the expense of requiring a greater input force. If the fulcrum is closer to the load, less force is required but the force must be applied over a longer distance. An example of a first-class lever is a seesaw / teeter-totter. ###### Second-Class Lever A second-class lever is used to increase force on an object in the same direction as the force is applied. This lever requires a smaller force to lift a larger load but the force must be applied over a greater distance. The fulcrum is placed at one end of the lever and mechanical advantage increases as the object being lifted is moved closer to the fulcrum or the length of the lever is increased. An example of a second-class lever is a wheelbarrow. ###### Third-Class Lever A third-class lever is used to increase distance traveled by an object in the same direction as the force applied. The fulcrum is at one end of the lever, the object at the other, and the force is applied between them. This lever does not impart a mechanical advantage as the effort force must be greater than the load but does impart extra speed to the load. Examples of third-class levers are shovels and tweezers. ###### Fixed Pulley A fixed pulley is used to change the direction of a force and does not multiply the force applied. As such, it has a mechanical advantage of one. The benefit of a fixed pulley is that it can allow the force to be applied at a more convenient angle, for example, pulling downward or horizontally to lift an object instead of upward. ###### Block and Tackle Two or more pulleys used together constitute a block and tackle which, unlike a fixed pulley, does impart mechanical advantage as a function of the number of pulleys that make up the arrangement. So, for example, a block and tackle with three pulleys would have a mechanical advantage of three. ###### Wheel and Axle A wheel and axle uses two different diameter wheels mounted to a connecting axle. Force is applied to the larger wheel and large movements of this wheel result in small movements in the smaller wheel. Because a larger movement distance is being translated to a smaller distance, force is increased with a mechanical advantage equal to the ratio of the diameters of the wheels. An example of a wheel and axle is the steering wheel of a car. ###### Screw A screw is an inclined plane wrapped in ridges (threads) around a cylinder. The distance between these ridges defines the pitch of the screw and this distance is how far the screw advances when it is turned once. The mechanical advantage of a screw is its circumference divided by the pitch. ##### Gears ###### Gear Trains Connected gears of different numbers of teeth are used together to change the rotational speed and torque of the input force. If the smaller gear drives the larger gear, the speed of rotation will be reduced and the torque will increase. If the larger gear drives the smaller gear, the speed of rotation will increase and the torque will be reduced. ###### Gear Ratio The mechanical advantage (amount of change in speed or torque) of connected gears is proportional to the number of teeth each gear has. Called gear ratio, it's the ratio of the number of teeth on the larger gear to the number of teeth on the smaller gear. For example, a gear with 12 teeth connected to a gear with 9 teeth would have a gear ratio of 4:3. ### Applications ##### General ###### Force Lines of Action Collinear forces act along the same line of action, concurrent forces pass through a common point and coplanar forces act in a common plane. ###### Bridge Forms The six basic bridge forms are beam, truss, arch, cantilever, cable, and suspension. ###### Modulus of Elasticity The modulus of elasticity measures how much a material or structure will deflect under stress. Stretch modulus is longitudinal stretch (like stretching raw bread dough), shear modulus is longitudinal deflection (like the horizontal displacement of a stack of magzines when a heavy object is placed upon them), and bulk modulus is compression of volume (like the compression of a loaf of bread under a heavy can at the bottom of a grocery bag). ###### Ceramics Ceramics are mixtures of metallic and nonmetallic elements that withstand exteme thermal, chemical, and pressure environments. They have a high melting point, low corrosive action, and are chemically stable. Examples include rock, sand, clay, glass, brick, and porcelain. ###### Specific Gravity Specific gravity is the ratio of the density of equal volumes of a substance and water and is measured by a hyrdometer.
### Statistics Knowledge Portal A free online introduction to statistics # Correlation ## What is correlation? Correlation is a statistical measure that expresses the extent to which two variables are linearly related (meaning they change together at a constant rate). It’s a common tool for describing simple relationships without making a statement about cause and effect. ## How is correlation measured? The sample correlation coefficient, r, quantifies the strength of the relationship. Correlations are also tested for statistical significance. ## What are some limitations of correlation analysis? Correlation can’t look at the presence or effect of other variables outside of the two being explored. Importantly, correlation doesn’t tell us about cause and effect. Correlation also cannot accurately describe curvilinear relationships. ## Correlations describe data moving together Correlations are useful for describing simple relationships among data. For example, imagine that you are looking at a dataset of campsites in a mountain park. You want to know whether there is a relationship between the elevation of the campsite (how high up the mountain it is), and the average high temperature in the summer. For each individual campsite, you have two measures: elevation and temperature. When you compare these two variables across your sample with a correlation, you can find a linear relationship: as elevation increases, the temperature drops. They are negatively correlated. ## What do correlation numbers mean? We describe correlations with a unit-free measure called the correlation coefficient which ranges from -1 to +1 and is denoted by r. Statistical significance is indicated with a p-value. Therefore, correlations are typically written with two key numbers: r = and p = . • The closer r is to zero, the weaker the linear relationship. • Positive r values indicate a positive correlation, where the values of both variables tend to increase together. • Negative r values indicate a negative correlation, where the values of one variable tend to increase when the values of the other variable decrease. • The p-value gives us evidence that we can meaningfully conclude that the population correlation coefficient is likely different from zero, based on what we observe from the sample. • "Unit-free measure" means that correlations exist on their own scale: in our example, the number given for r is not on the same scale as either elevation or temperature. This is different from other summary statistics. For instance, the mean of the elevation measurements is on the same scale as its variable. ### What is a p-value? A p-value is a measure of probability used for hypothesis testing. It indicates the likelihood of obtaining the data that we are seeing if there is no effect present — in other words, in the case of the null hypothesis. For our campsite data, this would be the hypothesis that there is no linear relationship between elevation and temperature. When a p-value is used to describe a result as statistically significant, this means that it falls below a pre-defined cutoff (e.g., p <.05 or p <.01) at which point we reject the null hypothesis in favor of an alternative hypothesis (for our campsite data, that there is a relationship between elevation and temperature). Once we’ve obtained a significant correlation, we can also look at its strength. A perfect positive correlation has a value of 1, and a perfect negative correlation has a value of -1. But in the real world, we would never expect to see a perfect correlation unless one variable is actually a proxy measure for the other. In fact, seeing a perfect correlation number can alert you to an error in your data! For example, if you accidentally recorded distance from sea level for each campsite instead of temperature, this would correlate perfectly with elevation. Another useful piece of information is the N, or number of observations. As with most statistical tests, knowing the size of the sample helps us judge the strength of our sample and how well it represents the population. For example, if we only measured elevation and temperature for five campsites, but the park has two thousand campsites, we’d want to add more campsites to our sample. ## Visualizing correlations with scatterplots Back to our example from above: as campsite elevation increases, temperature drops. We can look at this directly with a scatterplot. Imagine that we’ve plotted our campsite data: • Each point in the plot represents one campsite, which we can place on an x- and y-axis by its elevation and summertime high temperature. • The correlation coefficient (r) also illustrates our scatterplot. It tells us, in numerical terms, how close the points mapped in the scatterplot come to a linear relationship. Stronger relationships, or bigger r values, mean relationships where the points are very close to the line which we’ve fit to the data. ## What about more complex relationships? Scatterplots are also useful for determining whether there is anything in our data that might disrupt an accurate correlation, such as unusual patterns like a curvilinear relationship or an extreme outlier. Correlations can’t accurately capture curvilinear relationships. In a curvilinear relationship, variables are correlated in a given direction until a certain point, where the relationship changes. For example, imagine that we looked at our campsite elevations and how highly campers rate each campsite, on average. Perhaps at first, elevation and campsite ranking are positively correlated, because higher campsites get better views of the park. But at a certain point, higher elevations become negatively correlated with campsite rankings, because campers feel cold at night! We can get even more insight by adding shaded density ellipses to our scatterplot. A density ellipse illustrates the densest region of the points in a scatterplot, which in turn helps us see the strength and direction of the correlation. Density ellipses can be various sizes. One common choice for examining correlation is a 95% density ellipse, which captures approximately the densest 95% of the observations. If two variables are moving together, like our campsites’ elevation and temperature, we would expect to see this density ellipse mirror the shape of the line. And we can see that in a curvilinear relationship, the density ellipse looks round: a correlation won’t give us a meaningful description of this relationship.
# Peggy Sue decided to enter her famous chili at her local chili cooking contest. Normally, she needs five tomatoes to make enough chili for her family of seven famous recipe. How many tomatoes should she expect to use to make her for 100 people? Oct 22, 2016 72 tomatoes. #### Explanation: You can do it two ways: Method 1: Find the number of tomatoes needed for one person. If five tomatoes are needed to make enough chilli for seven people, then $\frac{5}{7}$ tomatoes are needed for one person. To find the number of tomatoes for one hundred people, simply multiple $\frac{5}{7}$ by 100. $\frac{5}{7} \cdot 100 = 71.4285 \ldots$ Since you can't have 71.4 tomatoes, simply round up to the nearest whole number to get 72 tomatoes. Method 2: Find out how many sevens go into one hundred and multiply that number by five to get the number of tomatoes needed. $\frac{100}{7} = 14.285714286$ $14.285714286 \cdot 5 = 71.4285 \ldots$ Again, round up to the nearest whole number because you can't have 71.4 tomatoes. Oct 22, 2016 It is astonishing just how much ratio is the underlying principle behind many calculations. 72 tomatoes #### Explanation: Using ratio in fraction format giving:" "("tomato count")/("people count") Let the count of tomatoes for 100 people be $x$ $\left(\text{tomato count")/("people count}\right) \to \frac{5}{7} \equiv \frac{x}{100}$ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ $\textcolor{b l u e}{\text{Method 1}}$ Multiply both sides by 100 $\frac{5}{7} \times 100 = \frac{x}{100} \times 100$ $\frac{5 \times 100}{7} = \frac{x \times 100}{100}$ $\frac{500}{7} = x \times \frac{100}{100} \text{ "color(red)(->5xx100/7=x" "larr"see method two}$ $x = 71 \frac{3}{7} \text{ tomatoes exactly}$ As you would not have part of a tomato we have 72 '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ color(blue)("Method 2") color(red)(" "larr" Shows why method 1 works" $\frac{5}{7} = \frac{x}{100}$ We need to change the 7 into 100. This is achieved by multiplying by $\frac{100}{7}$. To maintain the ratio you would need to multiply both top and bottom (numerator and denominator) by the same value. So we have: $\frac{5}{7} \text{ "=" "x/100" " ->" } \frac{\textcolor{red}{5 \times \frac{100}{7}}}{7 \times \frac{100}{7}} = \frac{\textcolor{red}{71 \frac{3}{7}}}{100}$
# Free math solving apps In this blog post, we will show you how to work with Free math solving apps. Let's try the best math solver. ## The Best Free math solving apps One tool that can be used is Free math solving apps. Solving for an exponent variable is similar to solving for a variable that has a coefficient. You can use the same process. You will want to isolate the variable, then simplify the expression. When you isolate the variable, you need to make sure that it can only be one of two values. If it can be more than two values, then you will have to solve for all of those values. You will also want to make sure that you are working with base 10. When you are dealing with exponents in base 10, they will always be between 0 and 9. Once you have isolated your variable, you can simplify the expression by removing all coefficients that are not needed. This will result in a reduced expression that can be simplified further. If there are any variables that are not in the denominator, then they must be set equal to 1. Once they are set equal to 1, then you can simplify your expression again by removing any coefficients that are not needed. Sometimes this process may result in a fraction being placed in front of the expression that was created. You will want to simplify this fraction as well by removing any coefficients that are not needed. Assuming you want a word problem to be solved: A plumber charges \$25 for a service call plus \$50 per hour of service. What will be the total cost for a 4 hour job? The total cost for the 4 hour job would be \$200. Solving proportions is one of the most common skills you'll need to learn as a student. Your teacher may tell you that you need to be able to do this, or they may give you a worksheet with examples of how to solve proportions. With this skill, you'll be able to determine how many parts make up a whole by comparing different amounts of each part. Many math problems involve solving proportions, such as one part of a whole is larger or smaller than another part. For example, if you have one pen and two pencils, the pen is three times as long as the pencils. If there are 6 cookies in a jar, the jar contains 3 parts: 3 cookies, 2 cookies, and 1 cookie. These are just a few examples of how proportions can come up in everyday life! When you first try solving proportions, it can be hard to know what numbers to use for each part. Try using the same number for all your parts (like 1 cookie), but be careful not to forget about any parts (like the lid). Online math tutoring is a great way to improve your understanding of math while also having the chance to ask questions and receive personalized feedback. If you're looking for a reliable and affordable math tutor, you should definitely consider using an online math tutor chat (or video). And if you're looking for a great place to start, we recommend Tutor.com. With this service, you can connect with a local or online math tutor using video chat, text chat, or both. The best part about online math tutoring chat is that it's free! So there's no reason not to give it a try! You can use the following search terms to find a tutoring service that works for you: * Online Math Tutor * Online Math Tutors * Online Math Tutoring Service * Online Math Tutors for Free * Free Online Math Tutors Chat A partial derivative solver is a program that can find the derivative of a function with respect to a variable. This can be useful for solving problems in calculus and physics. There are many different ways to solve partial derivatives, and the best method may vary depending on the problem. ## Math checker you can trust Really like it so far. Have had it for a couple of weeks and it's been really helpful in simplifying equations so I can break them down and understand them better. Would recommend for people who struggle with understanding the processes behind mathematic work. Unity Lee This is a really good app, I have been struggling in math, and whenever I have late work, this app helps me! Plus, there is barely any ads! But you should really add a Eureka Math book thing for 1st, 2nd, 3rd, 4th, 5th, 6th grade, and more! Ursuline Hayes
# What is the slope and y-intercept of the line 4x-5y=20? Mar 1, 2017 The slope is $m = \frac{4}{5}$ and the y-intercept is $b = - 4$. #### Explanation: The slope and y-intercepts are most easily found when you put the equation in the form $y = m x + b$, where $m$ is the slope an $b$ is the y-intercept. To do this, rearrange the equation $4 x - 5 y = 20$ $4 x = 20 + 5 y$ $4 x - 20 = 5 y$ $\frac{1}{5} \left(4 x - 20\right) = y$ $y = \frac{4}{5} x - 4$ Now we can simply read off from our new equation the values $m = \frac{4}{5}$ and $b = - 4$. ($b$ is sometimes called $c$, but I prefer calling it $b$ because it sounds nicer.)
# Construction of positive integers by given rules For a positive integer n there are two operations defined: 1. append one of the digits 0, 4 or 8 at the right end of n 2. n can be divided by 2 if n is even Start number is 4. Is it possible to construct any positive integer with those two rules? Example to construct 55: 4→44→22→220→110→55 yes. Easy proof: Do it back to front: Start with the desired number, multiply by 2 until the last digit is 0,4 or 8 and strike out the last digit and start over if necessary. As we can easily check this is always possible with at most 3 doublings each cycle decreases the number so we must eventually arrive at a single digit. Again, it is easily verified that from there we can get to 4. It is possible to construct every positive integer this way. Proof: Suppose not, and let $$n$$ be the smallest positive integer you can't construct. First of all, $$n$$ is more than one digit long because all single-digit positive integers are constructible. (4 -> 2 -> 1 -> 10 -> 5; 2 -> 24 -> 12 -> 6 -> 3; 1 -> 14 -> 7; 1 -> 18 -> 9; 6 -> 64 -> 32 -> 16 -> 8.) So $$n=10m+d$$ where $$d$$ is its last digit and $$m$$ is also a positive integer. In particular $$m$$ is constructible since $$m and hence so are $$10m+0,4,8$$ so we know that $$d$$ isn't any of those. Likewise, $$2m$$ is constructible since $$2m and hence so is $$(10\cdot2m+4)/2=10m+2$$; $$4m$$ is constructible since $$4m and hence so is $$(10\cdot4m+4)/4=10m+1$$. So $$d$$ isn't any of $$0,1,2,4,8$$. $$4m+2$$ is constructible since $$4m+2 and hence so is $$(10\cdot(4m+2)+4)/4=10m+6$$; $$8m+2$$ is constructible since $$8m+2 and hence so is $$(10\cdot(8m+2)+4)/4=10m+3$$. So $$d$$ isn't any of $$0,1,2,3,4,6,8$$. $$2m+1$$ is constructible, hence so are $$(20m+10,14,18)/2=10m+5,7,9$$ so $$d$$ isn't any of $$5,7,9$$. And that's all the possible last digits, so we're done. • Did you just beat me to it by 11 secs? Apr 1, 2021 at 23:53 • It sure looks like it :-). Your proof is cleaner than mine, though. (Well, of course they're basically the same proof. But I went the route of showing all the cases separately, which I think was suboptimal given that one can summarize the process more neatly and uniformly as you did.) Apr 2, 2021 at 0:17 • Thanks! As pretexts (for not deleting my answer) go this is good enough for me ;-) Apr 2, 2021 at 0:23
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> You are viewing an older version of this Concept. Go to the latest version. # Division of Whole Numbers by Fractions ## Understand how to divide a whole number by a fraction. Estimated7 minsto complete % Progress Practice Division of Whole Numbers by Fractions Progress Estimated7 minsto complete % Division of Whole Numbers by Fractions Remember Julie and her game in the Division of Fractions by Whole Numbers Concept? Julie has 40 inches of paper and she wants to divide this piece of paper in one-half inch strips. How can she do it? Previously we worked on dividing a fraction by a whole number, but in this problem, you are going to work the other way around. To help Julie figure out how to divide this piece of paper into one -half inch strips, you will need to divide a whole number by a fraction. Pay close attention and you will learn all that you to know in this Concept. ### Guidance We can also divide a whole number by a fraction. When we divide a whole number by a fraction we are taking a whole and dividing it into new wholes. Now at first glance, you would think that this answer would be one-half, but it isn’t. We aren’t asking for of one we are asking for 1 divided by one-half. Let’s look at a picture. Now we are going to divide one whole by one-half. Now we have two one-half sections. Our answer is two. We can test this out by using the rule that we learned in the last Concept. Our answer is the same as when we used the pictures. It’s time for you to try a few of these on your own. Find each quotient. Solution: Solution: #### Example C Solution: Now back to Julie and the ribbon. Here is the original problem once again. Remember Julie and her game? Julie has 40 inches of paper and she wants to divide this piece of paper in one - half inch strips. How can she do it? In the last Concept, you divide a fraction by a whole number, but in this problem, you are going to work the other way around. To help Julie figure out how to divide this piece of paper into one -half inch strips, you will need to divide a whole number by a fraction. To figure this out, we first can write an equation. Julie wants to divide 40" of paper into one - half inch strips. Next, we can change this into a multiplication problem. strips of paper. This is our answer. ### Vocabulary Inverse Operation opposite operation. Multiplication is the inverse operation of division. Addition is the inverse operation of subtraction. Reciprocal the inverse of a fraction. We flip a fraction’s numerator and denominator to write a reciprocal. The product of a fraction and its reciprocal is one. ### Guided Practice Here is one for you to try on your own. Answer First, we have to convert this problem to a multiplication problem. Next, we convert this improper fraction to a mixed number. ### Practice Directions: Divide the following whole numbers and fractions. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. ### Vocabulary Language: English Inverse Operation Inverse Operation Inverse operations are operations that "undo" each other. Multiplication is the inverse operation of division. Addition is the inverse operation of subtraction. reciprocal reciprocal The reciprocal of a number is the number you can multiply it by to get one. The reciprocal of 2 is 1/2. It is also called the multiplicative inverse, or just inverse. ### Explore More Sign in to explore more, including practice questions and solutions for Division of Whole Numbers by Fractions. Please wait... Please wait...
# Top 10 PSAT/NMSQT Math Practice Questions Preparing students for the PSAT/NMSQT Math test? The best way to prepare for the PSAT Math test is to work through as many PSAT Math practice questions as possible. Here are the top 10 PSAT Math practice questions to help your students review the most important PSAT Math concepts. These PSAT Math practice questions are designed to cover mathematics concepts and topics that are found on the actual test. The questions have been fully updated to reflect the latest 2021 PSAT guidelines. Answers and full explanations are provided at the end of the post. Help your students start their PSAT Math test prep journey right now with these sample PSAT Math questions. ## PSAT Math Practice Questions 1- A taxi driver earns $9 per 1-hour work. If he works 10 hours a day and in 1 hour he uses 2-liters petrol with price$1 for 1-liter. How much money does he earn in one day? ☐A. $90 ☐B.$88 ☐C. $70 ☐D.$60 2- Five years ago, Amy was three times as old as Mike was. If Mike is 10 years old now, how old is Amy? ☐A. 4 ☐B. 8 ☐C. 12 ☐D. 20 3- What is the solution of the following system of equations? $$\begin{cases}\frac{-x}{2}+ \frac{y}{4} = 1 \\ \frac{-5y}{6}+2x = 4 & \end{cases}$$ ☐A. $$x=48,y=22$$ ☐B. $$x=50,y=20$$ ☐C. $$x=20,y=50$$ ☐D. $$x=22,y=48$$ 4- What is the length of AB in the following figure if AE=4, CD=6 and AC=12? ☐A. 3.8 ☐B. 4.8 ☐C. 7.2 ☐D. 24 5- If a and b are solutions of the following equation, which of the following is the ratio $$\frac{a}{b}$$? $$(a > b)$$ $$2x^2-11x+8=-3x+18$$ ☐A. $$\frac{1}{5}$$ ☐B. 5 ☐C. $$-\frac{1}{5}$$ ☐D. $$-5$$ 6- How many tiles of 8 cm$$^2$$ is needed to cover a floor of dimension 6 cm by 24 cm? ☐A. 6 ☐B. 12 ☐C. 18 ☐D. 24 7- Which of the following is the solution of the following inequality? $$2x+4>11x-12.5-3.5x$$ ☐A. $$x<3$$ ☐B. $$x>3$$ ☐C. $$x≤4$$ ☐D. $$x≥4$$ 8- If a, b and c are positive integers and $$3a = 4b = 5c$$, then the value of $$a + 2b + 15c$$ is how many times the value of a? ☐A. 11.5 ☐B. 12 ☐C. 12.5 ☐D. 15 9- A company pays its employer \$7000 plus $$2\%$$ of all sales profit. If $$x$$ is the number of all sales profit, which of the following represents the employer’s revenue? ☐A. $$0.02x$$ ☐B. $$0.98x-7000$$ ☐C. $$0.02x+7000$$ ☐D. $$0.98x+7000$$ 10- $$\frac{5x^2+75x-80}{x^2-1} =$$? ☐A. $$\frac{5x+75}{ x-1}$$ ☐B. $$\frac{x+16}{ x+1}$$ ☐C. $$\frac{5x+80}{ x+1}$$ ☐D. $$\frac{x+15}{ x-1}$$ ## Best PSAT Math Prep Resource for 2021 1- C $$9×10=90$$ Petrol use: $$10×2=20$$ liters Petrol cost: $$20×1=20$$ Money earned: $$90-20=70$$ 2- D Five years ago, Amy was three times as old as Mike. Mike is 10 years now. Therefore, 5 years ago Mike was 5 years. Five years ago, Amy was: $$A=3×5=15$$ Now Amy is 20 years old: $$15 + 5 = 20$$ 3- D $$\begin{cases}\frac{-x}{2}+ \frac{y}{4} = 1 \\ \frac{-5y}{6}+2x = 4 & \end{cases}$$ Multiply the top equation by 4. then: $$-2x+y=4$$ $$\frac{-5y}{6} + 2x = 4$$ $$\frac{1}{6}y=8→y=48$$ plug in the value of y into the first equation $$→(x=22)$$ 4- B Two triangles ∆BAE and ∆BCD are similar. Then: $$\frac{AE}{CD} = \frac{AB}{BC}$$ $$\frac{4}{6} = \frac{x}{12}$$ then: $$48-4x=6x→10x=48→x=4.8$$ 5- D $$2x^2-11x+8=-3x+18→2x^2-11x+3x+8-18=0→2x^2-8x-10=0$$ $$→2(x^2-4x-5)=0→$$ Divide both sides by 2. Then: $$x^2-4x-5=0$$, Find the factors of the quadratic equation. $$→(x-5)(x+1)=0→x=5$$ or $$x=-1$$ $$a>b$$, then: $$a=5$$ and $$b=-1$$ $$\frac{a}{b} = \frac{5}{-1}= -5$$ 6- C The area of the floor is: 6 cm $$×$$ 24 cm = 144 cm The number is tiles needed $$= 144 ÷ 8 = 18$$ 7- A $$2x+4>11x-12.5-3.5x→$$ Combine like terms: $$2x+4>7.5x-12.5→$$ Subtract $$2x$$ from both sides: $$4>5.5x-12.5$$ Add 12.5 both sides of the inequality. $$16.5>5.5x$$ Divide both sides by 5.5. $$\frac{16.5}{5.5}>x→x<3$$ 8- A $$3a=4b→b=\frac{3a}{4}$$ and $$3a=5c→c= \frac{3a}{5}$$ $$a+2b+15c=a+ (2× \frac{3a}{4})+(15× \frac{3a}{5})=a+1.5a+9a=11.5a$$ 9- C $$x$$ is the number of all sales profit and $$2\%$$ of it is: $$2\%×x=0.02x$$ Employer’s revenue: $$0.2x+7000$$ 10- C First, find the factors of numerator and denominator of the expression. Then simplify. $$\frac{5x^2+75x-80}{x^2-1}=\frac{5(x^2+15x-16)}{(x-1)(x+1)}$$ $$\frac{5(x+16)(x-1)}{ (x-1)(x+1)}=\frac{5(x+16)}{ (x+1)}$$ $$=\frac{(5x+80)}{ (x+1)}$$ ## College Entrance Tests Looking for the best resource to help you succeed on the PSAT Math test? 52% OFF X ## How Does It Work? ### 1. Find eBooks Locate the eBook you wish to purchase by searching for the test or title. ### 3. Checkout Complete the quick and easy checkout process. ## Why Buy eBook From Effortlessmath? Save up to 70% compared to print Help save the environment
# This Division Golf Game Makes it FUN to Practice Division! We know how important it is that students are able to show flexibility with numbers and manipulate them mentally, rather than relying on the standard algorithm. Just like a chef uses what she knows about flavors and pairings to create a meal without a scripted recipe, we want students to be able to naturally choose strategies that work for the situation. But how do we get here? How do we get students to a place where they are able to use numbers and operations flexibly and naturally? Besides modelling for students, we can give them purposeful opportunities to use numbers in flexible ways. For example, instead of giving students a problem to work through. give them an answer and allow them to brainstorm the problem for that answer. Another way we can help students build this flexibility is through strategic games Here is a division game that students love as they learn more complex division skills such as remainders. The game involves strategy and number sense, keeping students engaged and eager to play. ## Division Golf: A Strategic Division Game To play Division golf, your students will work in partners. Each pair of students will need a deck of cards with the J, Q, K, and Aces removed (keep only the numbers 2-9). To keep score, students will need a piece of paper and a pencil or a mini whiteboard. Like real golf, the goal of Division Golf is to be the player with the lowest score (this score will be made up of the remainders you have when you divide). First, shuffle the cards and place them face down. Player #1 draws three cards and uses them to create two numbers (one 1-digit number and one 2-digit number). The goal is the have the fewest remainders when you divide. For example, suppose you draw a 9, 4, and 8. Here are a few of the different division expressions you could build: 94÷8, 98÷4, 48÷9, 49÷8, etc. 94÷8 would leave you with a remainder of 6. 98÷4 would leave you with a remainder of 2. 48÷9 would leave you with a remainder of 3. 49÷8 would leave you with a remainder of 1. Of the four expressions listed here, 49÷8 would be the best one to choose because it has the lowest remainder. This becomes Player #1’s score. Remember, this is a golf game so the goal is to get the lowest score. Players alternate turns. This game can be played for a specific amount of time, or it can be played like traditional golf where each player gets nine turns, and the player with the lowest score at the end is the winner. When playing any game in the classroom, it is essential that your students feel successful. If the game is too difficult, your students will “check out” and resist even trying. This is when classroom management issues arise, and learning is not optimal. To get the most out of this game, ensure that it is a “just right” challenge for your students – not too easy, but also not too difficult. I recommend always starting a bit too easy, and then moving up from there until you find the sweet spot. Here are some ways that you can adapt this game to ensure that it is perfect for your unique classroom. To make this division game easier: • Decrease the number of rounds to 5. • Give your students counting pieces that they can use to model the division problem. You might use counters, beans, base ten blocks, or even simply a personal whiteboard to draw a model. Manipulatives will make the remainders easier to see. • Allow the students to use only 2 of the cards if they wish. For example, if a 9, 4, and 8 are drawn, students could choose to use only the 4 and 8 to make the expression 8÷4. This would give them no remainders, so their score for that round is 0! • If having to choose between the three cards is too challenging right now, have them draw only 2 cards. • If you have not yet introduced the word, “remainder,” you might call it “leftovers” for now. For example, 7 divided by 3 is 2 with 1 leftover. This can be shown with counters so students can see it in real life. To make this division game more challenging: • Add an extra card! Students can then make a 3-digit by 1-digit expression. • Play “18 holes” (or rounds) rather than 9. • Change it up! Make the goal to be the highest score (or most remainders) instead of the lowest! As students play this division golf game, they will likely develop strategies that help them keep their scores low. Have them put their mathematical reasoning into words and share these with the class. The more opportunities students are given to explore and work with numbers, the more flexible their thinking will become. This game is just another way to help make that happen. Looking for more fun math games to add to your collection? Try Salute (always a class favorite) to reinforce addition/subtraction or multiplication/division relationships. Find the instructions here. This site uses Akismet to reduce spam. Learn how your comment data is processed.
# Difference between revisions of "2011 USAMO Problems/Problem 1" ## Problem Let $a$, $b$, $c$ be positive real numbers such that $a^2 + b^2 + c^2 + (a + b + c)^2 \le 4$. Prove that $$\frac{ab + 1}{(a + b)^2} + \frac{bc + 1}{(b + c)^2} + \frac{ca + 1}{(c + a)^2} \ge 3.$$ ## Solutions ### Solution 1 Since \begin{align} (a+b)^2 + (b+c)^2 + (c+a)^2 &= 2(a^2 + b^2 + c^2 + ab + bc + ca) \\ &= a^2 + b^2 + c^2 + (a + b + c)^2, \end{align} it is natural to consider a change of variables: \begin{align} \alpha &= b + c \\ \beta &= c + a \\ \gamma &= a + b \end{align} with the inverse mapping given by: \begin{align} a &= \frac{\beta + \gamma - \alpha}2 \\ b &= \frac{\alpha + \gamma - \beta}2 \\ c &= \frac{\alpha + \beta - \gamma}2 \end{align} With this change of variables, the constraint becomes $$\alpha^2 + \beta^2 + \gamma^2 \le 4,$$ while the left side of the inequality we need to prove is now \begin{align} & \frac{\gamma^2 - (\alpha - \beta)^2 + 4}{4\gamma^2} + \frac{\alpha^2 - (\beta - \gamma)^2 + 4}{4\alpha^2} + \frac{\beta^2 - (\gamma - \alpha)^2 + 4}{4\beta^2} \ge \\ & \frac{\gamma^2 - (\alpha - \beta)^2 + \alpha^2 + \beta^2 + \gamma^2}{4\gamma^2} + \frac{\alpha^2 - (\beta - \gamma)^2 + \alpha^2 + \beta^2 + \gamma^2}{4\alpha^2} + \frac{\beta^2 - (\gamma - \alpha)^2 + \alpha^2 + \beta^2 + \gamma^2}{4\beta^2} = \\ & \frac{2\gamma^2 + 2\alpha\beta}{4\gamma^2} + \frac{2\alpha^2 + 2\beta\gamma}{4\alpha^2} + \frac{2\beta^2 + 2\gamma\alpha}{4\beta^2} = \\ & \frac32 + \frac{\alpha\beta}{2\gamma^2} + \frac{\beta\gamma}{2\alpha^2} + \frac{\gamma\alpha}{2\beta^2}. \end{align} Therefore it remains to prove that $$\frac{\alpha\beta}{2\gamma^2} + \frac{\beta\gamma}{2\alpha^2} + \frac{\gamma\alpha}{2\beta^2} \ge \frac32.$$ We note that the product of the three (positive) terms is 1/8, therefore by AM-GM their mean is at least 1/2, and thus their sum is at least 3/2 and we are done. ### Solution 2 Rearranging the condition yields that $$a^2 + b^2 + c^2 +ab+bc+ac \le 2$$ Now note that $$\frac{2ab+2}{(a+b)^2} \ge \frac{2ab+a^2 + b^2 + c^2 +ab+bc+ac}{(a+b)^2}=\frac{(a+b)^2 + (c+a)(c+b)}{(a+b)^2}$$ Summing this for all pairs of $\{ a,b,c \}$ gives that $$\sum_{cyc} \frac{2ab+2}{(a+b)^2} \ge 3+ \sum_{cyc}\frac{(c+a)(c+b)}{(a+b)^2} \ge 6$$ By AM-GM. Dividing by $2$ gives the desired inequality. 2011 USAMO (Problems • Resources) First Problem Followed byProblem 2 1 • 2 • 3 • 4 • 5 • 6 All USAMO Problems and Solutions 2011 USAJMO (Problems • Resources) Preceded byProblem 1 Followed byProblem 3 1 • 2 • 3 • 4 • 5 • 6 All USAJMO Problems and Solutions
Suggested languages for you: Americas Europe Q11-15E Expert-verified Found in: Page 653 ### Statistics For Business And Economics Book edition 13th Author(s) James T. McClave, P. George Benson, Terry Sincich Pages 888 pages ISBN 9780134506593 # Refer to Exercise 11.14. After the least-squares line has been obtained, the table below (which is similar to Table 11.2) can be used for (1) comparing the observed and the predicted values of y and (2) computing SSE. a. Complete the table. b. Plot the least-squares line on a scatterplot of the data. Plot the following line on the same graph: $\stackrel{\mathbf{^}}{\mathbf{\text{y}}}\mathbf{\text{= 14 - 2.5x.}}$c. Show that SSE is larger for the line in part b than for the least-squares line. 1. Fig. 1 Table. 2. Fig. 2 Table, Fig. 3 Scatterplot diagram. 3. SSE is smaller than the least square. See the step by step solution ## Step-by-Step Solution Step 1: Introduction The line that minimizes the squared sum of residuals is known as the Least Squares Regression Line. By subtracting $\stackrel{^}{\text{y}}$ from y, the residual is the vertical distance between the observed and anticipated points. ## Step 2: Complete the table From exercise 11.14, we have $\stackrel{\mathbf{^}}{\mathbf{\text{y}}}{\mathbf{\text{=1.78 +}}}\mathbf{\left(}\mathbf{-}\mathbf{\text{0.77}}\mathbf{\right)}{\mathbf{\text{x}}}$ ## Step 3: Draw a scatterplot of the data and plot the least-squares line. On the same graph, draw the following line $\stackrel{\mathbf{^}}{\mathbf{\text{y}}}{\mathbf{\text{= 14}}}{\mathbf{-}}{\mathbf{\text{2.5x.}}}$ By putting the x value in the above equation, we get $\stackrel{^}{\text{y}}:$ ## Step 4: Show that SSE is larger for the line in part b than for the least-squares line SSE of $\stackrel{\mathbf{^}}{\mathbf{\text{y}}}{\mathbf{\text{= 14}}}{\mathbf{-}}{\mathbf{\text{2.5x}}}$ is smaller than the least square, i.e. 108 < 153.6.39 Therefore, SSE is smaller than the least square. ## Recommended explanations on Math Textbooks 94% of StudySmarter users get better grades.
# Derivative of 1/x Learn what is the derivative of 1/x with formula. Also understand how to verify the derivative of an algebraic function 1/x by using first principle. Alan Walker- Published on 2023-06-20 ## Introduction to the derivative of 1/x Derivatives have a wide range of applications in almost every field of engineering and science. The 1/x derivative can be calculated by following the differentiation rules. Or, we can directly find the derivative formula of 1/x by applying the first principle of differentiation. In this article, you will learn what the x derivative formula is and how to calculate the derivatives of x^-1 by using different approaches. ## What is the derivative of 1/x? The derivative of 1 by x with respect to the variable x is equal to -1/x^2. It measures the rate of change of the algebraic function 1/x. It is denoted by d/dx(1/x) which is a fundamental concept in calculus. Knowing the formula for derivatives and understanding how to use it can be used in solving problems related to velocity, acceleration, and optimization. ## Differentiation of 1 by x formula The formula for derivative of f(x)=1/x is equal to the 1/x^2, that is; $f'(x) = \frac{d}{dx} (\frac{1}{x}) = -\frac{1}{x^2}$ It is calculated by using the power rule, which is defined as: $\frac{d}{dx} (x^n)=nx^{n-1}$ ## How do you differentiate 1/x? There are multiple ways to prove the differentiation of 1/x. These are; 1. First Principle 2. Product Rule 3. Quotient Rule Each method provides a different way to compute the 1/x differentiation. By using these methods, we can mathematically prove the formula for finding the differential of 1/x. ## Differentiation of 1/x by first principle According to the first principle of derivative, the 1/x derivative is equal to -1/x^2. The derivative of a function by first principle refers to finding a general expression for the slope of a curve by using algebra. It is also known as the delta method. The derivative is a measure of the instantaneous rate of change, which is equal to, $f’(x) = \lim_{h\to 0} \frac{f(x + h) - f(x)}{h}$ This formula allows us to determine the rate of change of a function at a specific point by using the limit definition of derivative calculator. ### Derivative of 1 x proof by first principle To differentiate 1/x by using first principle, replace f(x) by 1/x or you can replace it by 1/x to find the derivative. $f’(x) = \lim_{h\to 0} \frac{\frac{1}{x + h} - \frac{1}{x}}{h}$ Moreover, $f’(x) = \lim_{h\to 0} \frac{\frac{x - x-h}{x(x+h)}}{h}{2}nbsp;$f’(x) = \lim_{h\to 0} \frac{-h}{xh(x+h)}= \lim_{h\to 0} \frac{-1}{x^2+xh}$When h approaches to zero,$f’(x) = \frac{-1}{x^2}$Hence the differentiation of 1/x is equal to -1/x^2. Also calculate the derivative of 1/x graph by using our derivative graph calculator. ## Derivative of x^-1 by product rule The formula for derivative 1/x can be calculated by using the product rule because an algebraic function can be written as the combination of two functions. The product rule derivative is defined as;$[uv]’ = u’.v + u.v’$### Proof of differentiating of 1/x by product rule To differentiate of x^2 by using product rule, we start by assuming that,$f(x) = 1. (\frac{1}{x})$By using product rule of differentiation,$f’(x) = (1)’. \frac{1}{x} + (1)\frac{-1}{x^2}$We get,$f’(x) = 0-\frac{1}{x^2}$Hence,$f’(x) =-\frac{1}{x^2}$Also use our product rule calculator online, as it provides you a step-by-step solution of differentiation of a function. ## Derivative of 1 x formula using quotient rule Another method for finding the differential of x^2 is using the quotient rule, which is a formula for finding the derivative of a quotient of two functions. Since the secant function is the reciprocal of cosine, the derivative of cosecant can also be calculated using the quotient rule. The derivative quotient rule is defined as:$\frac{d}{dx}\left(\frac{f}{g}\right) = \frac{f’(x). g(x) - g’(x).f(x)}{(g(x))^2}$### Proof of differentiating x by quotient rule To differentiate 1/x, we can start by writing it,$f(x) = \frac{1}{x}= \frac{u}{v}$Supposing that u = 1 and v = x. Now by quotient rule,$f’(x) =\frac{vu’ - uv’}{v^2}f’(x) = \frac{x.(1)’ - (x)’}{x^2}f’(x)= \frac{0-1}{x^2}f’(x)= \frac{-1}{x^2}\$ Hence, we have derived the derivative of x^-1 using the quotient rule of differentiation. Use our quotient rule calculator to find the derivative of any quotient function. ## How to find the derivatives of 1/x with a calculator? The easiest way of differentiating 1/x is by using a dy/dx calculator. You can use our derivative calculator for this. Here, we provide you a step-by-step way to calculate derivatives by using this tool. 1. Write the function as 1/x in the enter function box. In this step, you need to provide input value as a function that you want to differentiate. 2. Now, select the variable by which you want to differentiate 1/x. Here you have to choose x. 3. Select how many times you want to calculate the derivatives of 1/x. In this step, you can choose 2 for second, 3 for triple differentiation and so on. 4. Click on the calculate button. After this step, you will get the derivative of 1 by x within a few seconds. 5. After completing these steps, you will receive the differential of 1/x within seconds. Using online tools can make it much easier and faster to calculate derivatives, especially for complex functions. ## Conclusion: In conclusion, the derivative of 1/x. The derivative measures the rate of change of a function with respect to its independent variable, and in the case of 1/x, the derivative can be calculated using the power rule, product rule and quotient rule of differentiation.
# FAQ: What Is The Meaning Of Sum In Math? ## What is the sum in math? A sum is the result of an addition. For example, adding 1, 2, 3, and 4 gives the sum 10, written. (1) The numbers being summed are called addends, or sometimes summands. ## How do you find the sum in math? The result of adding two or more numbers. (because 2 + 4 + 3 = 9). ## What’s the meaning of sum? 1: the result obtained by adding numbers The sum of 4 and 5 is 9. 2: a problem in arithmetic. 3: a quantity of money We donated a small sum. 4: the whole amount Two trips is the sum of my travel experience. ## What is the sum of 9? Adding 8 to 7 results in a sum of digits of 6, and so on down to adding 8 to 1 which gives 9. Number Repeating Cycle of Sum of Digits of Multiples 8 {8,7,6,5,4,3,2,1, 9 } 9 { 9, 9, 9, 9, 9, 9, 9, 9, 9 } 10 {1,2,3,4,5,6,7,8, 9 } 11 {2,4,6,8,1,3,5,7, 9 } ## What is the sum of any two number? (a) The sum of any two odd numbers is an even number. You might be interested: Readers ask: What Is Real Number System In Math? ## What is a sum number? In mathematics, sum can be defined as the result or answer we get on adding two or more numbers or terms. Here, for example, addends 8 and 5 add up to make the sum 13. ## What is the formula of sum of n terms? An example of AP is natural numbers, where the common difference is 1. Therefore, to find the sum of natural numbers, we need to know the formula to find it. Sum of N Terms of AP And Arithmetic Progression. Sum of n terms in AP n /2[2a + ( n – 1)d] Sum of natural numbers n ( n +1)/2 ## What is the sum of the first 100 whole numbers? Clearly, it is an Arithmetic Progression whose first term = 1, last term = 100 and number of terms = 100. Therefore, the sum of first 100 natural numbers is 5050. ## What is another word for sum? Synonym Study In this page you can discover 95 synonyms, antonyms, idiomatic expressions, and related words for sum, like: addition, whole, aggregate, tally, add, gross, entirety, problem, calculate, add together and sum total. ## Does sum multiply? SUM – The sum is the result of adding two or more numbers. PRODUCT – The product of two or more numbers is the result of multiplying these numbers. QUOTIENT – The quotient of two numbers is the result of the division of these numbers. ## What’s the definition of of? English Language Learners Definition of of: belonging to, relating to, or connected with (someone or something) —used to indicate that someone or something belongs to a group of people or things. You might be interested: FAQ: What Is An Experiment In Math? ## What is the sum of the number from 1 to 10? 55 is a sum of number series from 1 to 10 by applying the values of input parameters in the formula. ## What is the sum of 1 20? 210 is a sum of number series from 1 to 20 by applying the values of input parameters in the formula. ## What is the sum of 9 and the difference of 1? The two numbers with the sum of 9 and difference of 1 is 5 and 4. Explanation: Let the two numbers be x and y.
NSW Syllabuses # Mathematics K–10 - Stage 5.2 - Number and Algebra Ratios and Rates ## Outcomes #### A student: • MA5.2-1WM selects appropriate notations and conventions to communicate mathematical ideas and solutions • MA5.2-2WM interprets mathematical or real-life situations, systematically applying appropriate strategies to solve problems • MA5.2-5NA recognises direct and indirect proportion, and solves problems involving direct proportion ## Content • Students: • Solve problems involving direct proportion; explore the relationship between graphs and equations corresponding to simple rate problems (ACMNA208) • convert between units for rates, eg kilometres per hour to metres per second • identify and describe everyday examples of direct proportion, eg as the number of hours worked increases, earnings also increase • identify and describe everyday examples of inverse (indirect) proportion, eg as speed increases, the time taken to travel a particular distance decreases • recognise direct and inverse proportion from graphs • distinguish between positive and negative gradients when using a graph (Reasoning) • interpret and use conversion graphs to convert from one unit to another, eg conversions between different currencies or metric and imperial measures • use the equation $$\,y = kx\,$$ to model direct linear proportion where $$k$$ is the constant of proportionality • given the constant of proportionality, establish an equation and use it to find an unknown quantity (Communicating, Problem Solving) • calculate the constant of proportionality, given appropriate information, and use this to find unknown quantities (Problem Solving) • use graphing software or a table of values to graph equations representing linear direct proportion ### Language When describing everyday examples involving proportion, teachers should model common words and language structures before independent work is required, eg 'As the speed increases, the time taken to travel a particular distance decreases', 'The greater the speed, the less time is taken to travel a particular distance', 'The time taken to travel a particular distance is reduced when the speed is increased'. ### National Numeracy Learning Progression links to this Mathematics outcome When working towards the outcome MA5.2-5NA the sub-elements (and levels) of Number patterns and algebraic thinking (NPA7) and Comparing units (CoU3) describe observable behaviours that can aid teachers in making evidence-based decisions about student development and future learning. The progression sub-elements and indicators can be viewed by accessing the National Numeracy Learning Progression.
0% Complete 0/73 Steps # Test of Mathematics Solution Subjective 79 -Trigonometric Inequality This is a Test of Mathematics Solution Subjective 79 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance. Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta ## Problem Let ${{\theta}_1}$, ${{\theta}_2}$, … , ${{\theta}_{10}}$ be any values in the closed interval ${[0,\pi]}$. Show that ${F}$ = $${(1 + {\sin}^2 \theta_1)(1 + {\cos}^2 \theta_1)(1 + {\sin}^2 \theta_2)(1 + {\cos}^2 \theta_2)………(1 + {\sin}^2 \theta_{10})(1 + {\cos}^2 \theta_{10})}$$ ${\displaystyle{\le({\frac{9}{4}})^{10}}}$. What is the maximum value attainable by ${F}$ and at what values of ${{\theta}_1}$, ${{\theta}_2}$, … , ${{\theta}_{10}}$, is the maximum value attained? ## Solution ${F}$ = ${\displaystyle{(1 + {\sin}^2 \theta_1)(1 + {\cos}^2 \theta_1)(1 + {\sin}^2 \theta_2)(1 + {\cos}^2 \theta_2).........(1 + {\sin}^2 \theta_{10})(1 + {\cos}^2 \theta_{10})}}$ Now we will show that for any ${\theta \in}$ ${[0,\pi]}$ ${\displaystyle{(1 + {\sin}^2 \theta)(1 + {cos}^2 \theta) < {\frac{9}{4}}}}$ ${\Leftrightarrow}$ ${\displaystyle{2 + {{\sin}^2{\theta}}{{\cos}^2{\theta}}}}$ ${\displaystyle{< {\frac{9}{4}}}}$ ${\Leftrightarrow}$ ${\displaystyle{{{\sin}^2{\theta}}{{\cos}^2{\theta}}}}$ ${\displaystyle{< {\frac{1}{4}}}}$ Now ${\displaystyle{{{\sin}^2{\theta}} + {{\cos}^2{\theta}} = 1}}$ ${\Rightarrow}$ ${\displaystyle{({{\sin}^2{\theta}} + {{\cos}^2{\theta}})^2 = 1}}$ ${\Rightarrow}$ ${\displaystyle{{\sin}^4{\theta} + {\cos}^4{\theta} + 2 {{\sin}^2{\theta}{\cos}^2{\theta}} = 1}}$ ${\Rightarrow}$ ${\displaystyle{4 {{\sin}^2{\theta}}{{\cos}^2{\theta}}}}$ ${\displaystyle{< 1}}$ [ as ${{a^2 + b^2} > {2ab}}$ ] ${\Rightarrow}$ ${\displaystyle{{{\sin}^2{\theta}}{{\cos}^2{\theta}}}}$ ${\displaystyle{< {\frac{1}{4}}}}$ So for any ${\theta \in}$ ${[0,\pi]}$ ${\displaystyle{(1 + {\sin}^2 \theta)(1 + {\cos}^2 \theta) {\le} {\frac{9}{4}}}}$ So ${F {\le} ({\frac{9}{4}})^{10}}$ (proved) Maximum value attained by ${F}$ is ${({\frac{9}{4}})^{10}}$ and will be attained for ${{\theta_i} = {\frac{\pi}{2}} \pm {\frac{\pi}{4}}}$ for ${i = 1, 2, ...., 10}$
Student Question # int sqrt(x)arctan(x^(3/2)) dx Use integration tables to find the indefinite integral. For the given integral problem: int sqrt(x)arctan(x^(3/2))dx , we can evaluate this applying indefinite integral formula: int f(x) dx = F(x) +C . where: f(x) as the integrand function F(x) as the antiderivative of f(x) C as the constant of integration. From the basic indefinite integration table, the problem resembles one of the formula for integral of inverse trigonometric function: int arctan(u) du = u * arctan(u)- ln(u^2+1)/2+C For easier comparison, we may apply u-substitution by letting: u =x^(3/2 ) To determine the derivative of u, we apply the Power rule for derivative:d/(dx) x^n = nx^(n-1) dx. du =d/(dx) x^(3/2) = (3/2) x^(3/2-1) * 1 dx = 3/2x^(1/2) dx =3/2sqrt(x) dx Rearrange du =3/2sqrt(x) dx into (2du)/3 = sqrt(x) dx . Plug-in the values u = x^3/2 and (2du)/3 = sqrt(x) dx , we get: int sqrt(x)arctan(x^(3/2))dx =int arctan(x^(3/2))sqrt(x)dx = int arctan(u) (2du)/3 Apply the basic integration property: int cf(x) dx = c int f(x) dx . int arctan(u) (2du)/3 =2/3int arctan(u)du. Applying the aforementioned formula for inverse trigonometric function, we get: 2/3int arctan(u)du=(2/3) [u arctan(u)- ln(u^2+1)/2]+C =(2u * arctan(u))/3- (2ln(u^2+1))/6+C =(2u * arctan(u))/3- ln(u^2+1)/3+C Plug-in u =x^(3/2) on (2u * arctan(u))/3- ln(u^2+1)/3+C , we get the indefinite integral as: int sqrt(x)arctan(x^(3/2))dx =(2x^(3/2) * arctan(x^(3/2)))/3- ln((x^(3/2))^2+1)/3+C =(2x^(3/2) * arctan(x^(3/2)))/3- ln(x^3+1)/3+C or (2xsqrt(x) arctan(xsqrt(x)))/3- ln(x^3+1)/3+C Note: x^(3/2) = xsqrt(x)
# Theorem, Proof of Equal Sides of an Isosceles Triangle are Produced, the Exterior Angles are equal An isosceles triangle is a special kind of triangle that has two sides of equal length, no matter in what direction the peak of the triangle points. The important properties of an isosceles triangle are it has two equal sides, two equal angles (base angles). According to the isosceles triangle theorem if equal sides of an isosceles triangle are produced, then the exterior angles are equal. The detailed proof for the statement is given here. ## Equal Sides of an Isosceles Triangle are Produced, Exterior Angles are Equal Here is the proof of the statement of equal sides of an isosceles triangle are produced, the exterior angles are equal and obtuse. Proof: Let us take an isosceles triangle ABC. The equal sides AB, AC are produced to D and E. To prove: ∠CBD = ∠BCE. Statement Reason ∠ABC = ∠ACB Angles opposite to equal sides are equal ∠CBD = 180 – ∠ABC Linear pair ∠BCE = 180 – ∠ACB Linear Pair ∠BCE = 180 – ∠ABC Angles opposite to equal sides are equal Therefore, ∠CBD = ∠BCE From 2 & 4 statements Hence, proved. More Related Articles: ### Problems on Isosceles Triangle Problem 1: If the two equal sides of an isosceles triangle are 3x – 1, 2x + 2. The third side is 2x. Find the value of x and the perimeter of the triangle. Solution: Given that, Equal sides of an isosceles triangle are 3x – 1, 2x + 2. So, equate them. 3x – 1 = 2x + 2 3x – 2x = 2 + 1 x = 3 Third side of an isosceles triangle is 2x Third side = 2 x 3 = 6 First two sides are 3x – 1, 2x + 2 = 3(3) – 1, 2(3) + 2 = 9 – 1, 6 + 2 = 8, 8 Perimeter of the triangle = 8 + 8 + 3 = 19. Therefore, the value of x is 3 units, perimeter of the triangle is 19 units. Problem 2: The base BC of triangle ABC has produced both ways and the measure of exterior angles formed are 94°, 126°. Find ∠BAC. Solution: Given that, Base = BC ∠ACD = 126°, ∠ABE = 94° We know that, ∠ABE + ∠ABC = 180° ∠ABC = 180° – ∠ABE = 180° – 94° = 86° Again, ∠ACB + ∠ACD = 180° ∠ACB = 180° – ∠ACD = 180° – 126° = 54° In ΔABC ∠ABC + ∠ACB + ∠BAC = 180° 86° + 54° + ∠BAC = 180° ∠BAC = 180° – 140° = 40° ### Freuently Asked Question’s 1. What are the rules of an isosceles triangle? In an isosceles triangle, two sides are congruent, the third side is called the base. Two angles opposite to equal sides are congruent. 2. What two angles are equal in an isosceles triangle? Two sides are equal in the measure in an isosceles triangle called legs. The angles opposite to legs are also equal. 3. What happens when equal sides of an isosceles triangle are produced? If the equal sides of an isosceles triangle are produced, then the exterior angles are equal and obtuse. 4. What are the 3 properties of an isosceles triangle? • It has two sides of equal length. • The angles opposite to equal sides are equal in measure. • The altitude from one vertex to the base is the perpendicular bisector of the base.
# Types of Algorithm Laws Edited by TheGuyLoveNY, Jen Moreau ## Algorithm Laws ### Rabin Karp Algorithm: Rabin-Karp Algorithm is a searching algorithm used specifically to search strings. Rabin-Karp is also known as Karp-Rabin algorithm. This algorithm is a string searching algorithm which was created by Richard M. Karp and Michael O. Rabin, Since it was created by these two scientists, It was named after the creator. Hence, Rabin-Karp algorithm. This Algorithm uses Hash data structure to find the string pattern. Performance: • Average case : O(n + m) • Best Case : O(n + m) • Worst case : O(m) Where, The length of the text is n. The pattern of the test is p. Combined length of these texts ism. For example: Text = abcdef Pattern = abc h(abc) = x, Formula. Where, a = 1, b = 2, c = 3, d = 4, e = 5, f = 6 • h(abc) = 1+2 x 31 + 3 x 32 = 1 + 6 + 27 = 34 Therefore , h(abc) = x = 34 • h(bcd) = 2 + 3 x 31+ 4 x 32 = 2 + 9 + 36 = 47 Therefore, h(bcd) = 47 • h(cde) = 3 + 4 x 31 + 5 x 32 = 3 + 12 + 45 x = 60 • h(def) = 4 + 5 x 31+ 6 x 32 = 4 + 15 + 54 x = 73 Example 2: Text = abdadaab Patter = ab h(a,b) = x = 7 Where, a = 1, b = 2, d = 3, a = 4, d = 5, a = 6, a = 7, b = 8 h(a,b) = x = 7. • h(ab) = 1 + 2 x 31 = 1+6 = 7 • h(bd) = 2 + 4 x 31 = 2+12 = 14 • h(da) = 4 + 31 = 7. • h(ad) = 1 + 4 x 31 = 7 • h(aa) = 1+3 = 4 • h(ab) = 1 + 6 = 7 (same). Since, h(ab) = 7, The pattern ab is found in the text. ## Knuth-Morris-Pratt Algorithm: The Knuth-Morris-Pratt Algorithm also called KMP algorithm, is a string search algorithm. KMP algorithm searches a substring in a given string. Similar to a Rabin-Karp algorithm, This algorithm searches for a pattern in a given string. As the name implies, This algorithm was invented by three computer scientists namely Donald Knuth, Vaughan Pratt, James H. Morris in the year 1970 but was officially published in 1977. Example: Search the pattern in the given string using KMP Algorithm. Text : "abcxdabxabcdabcdabcy" Search pattern : "abcdabcy" Here, We cannot directly compare the pattern with the text. As KMP algorithm says we compare half of the pattern with prefix and suffix. Now, We do the same for other halves. Hence the pattern is finally matched : APA (American Psychological Association) Types of Algorithm Laws. (2017). In ScienceAid. Retrieved Sep 30, 2023, from https://scienceaid.net/Algorithm_Laws MLA (Modern Language Association) "Types of Algorithm Laws." ScienceAid, scienceaid.net/Algorithm_Laws Accessed 30 Sep 2023. Chicago / Turabian ScienceAid.net. "Types of Algorithm Laws." Accessed Sep 30, 2023. https://scienceaid.net/Algorithm_Laws. ScienceAid welcomes all comments. If you do not want to be anonymous, register or log in. It is free. ## Article Info Categories : Design and Algorithm Analysis Recent edits by: TheGuyLoveNY
Securing Higher Grades Costing Your Pocket? Book Your Assignment at The Lowest Price Now! Statistics probability assignment solution 1. You’re going to catch a fish in the river. You know that its size is normally distributed with mean 56 inches and standard deviation 7 inches. What is the probability that you catch a fish over 60 inches? What is the probability that you catch a fish less than 50 inches? SOLUTION We are given x is normally distributed with mean (µ) = 56 and standard deviation (∂) = 7. 1. P(x> 60) To calculate this probability we shall calculate the z-value for x= (60) then we calculate the area to the right of this value from the z-test tables. Z-value is given by, z = (x - µ)/(∂) where x is sample statistic,µ is the population mean and ∂ is the standard deviation. Z for x = 60, z(60) = (60-56)/7 = 0.5714 to2decimal place is 0.57 For p(x>60) we calculate from z tables the area to the right of 0.57 from z tables, thus p(x>60) = 1-0.7157= 0.2843 Therefore the probability that one catches a fish over 60inches is 0.2843 1. We are given x is normally distributed with mean (µ) = 56 and standard deviation (∂) = 7. P(x<50) to calculate this probability we shall calculate the z-value for x= (57) then we calculate the area to the left of this value from the z-test tables. Z-value is given by, z = (x - µ) / (∂) where x is sample statistic,µ is the population mean and ∂ is the standard deviation. Z for x = 50, z (50) = (50-56)/7 = -6/7 = -0.8571 to 2decimal place = -0.86 For p(x<50) we calculate from z tables the area to the left of -0.86 from z tables, thus p(x<50) = 0.1949 Therefore the probability that one catches a fish over 60inches is 0.1949 1. Construct a 95% confidence interval if the sample size is 30, mean is 60 and the standard deviation is 8. SOLUTION Confidence interval ( CI) = Sample Mean (x) ±( (zα * ∂)/√n) where x is sample mean,zα is the value of z for significance level given, ∂ is the standard deviation and √n is the square root of n Thus, CI = x ± ((zα * ∂)/√n), CI = 60 ± ((Z0.05 * 8) / √30), = 60 ± ((1.96 * 8) / √30 CI = (60 - 2.8628, 60 - 2.8628) CI = (57.1372, 62.8628) Therefore the 95%confidence interval is (57.1372, 62.8628) Assignment Help Features Assignment Help Services • Assignment Help • Homework Help • Writing Help • Academic Writing Assistance • Editing Services • Plagiarism Checker Online
# area of surface of revolution A surface of revolution is a 3D surface, generated when an arc is rotated fully around a straight line. The general surface of revolution is obtained when the arc is rotated about an arbitrary axis. If one chooses Cartesian coordinates, and specializes to the case of a surface of revolution generated by rotating about the $x$-axis a curve described by $y$ in the interval $[a,b]$, its area can be calculated by the formula $A=2\pi\int_{a}^{b}y\,\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}\,dx$ Similarly, if the curve is rotated about the $y$-axis rather than the $x$-axis, one has the following formula: $A=2\pi\int_{a}^{b}x\,\sqrt{1+\left(\frac{dx}{dy}\right)^{2}}\,dy$ The general formula is most often seen with parametric coordinates. If $x(t)$ and $y(t)$ describe the curve, and $x(t)$ is always positive or zero, then the area of the general surface of revolution $A$ in the interval $[a,b]$ can be calulated by the formula $A=2\pi\int_{a}^{b}y\,\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}% \right)^{2}}\,dt$ To obtain a specific surface of revolution, translation or rotation can be used to move an arc before revolving it around an axis. For example, the specific surface of revolution around the line $y=s$ can be found by replacing $y$ with $y\!-\!s$, moving the arc towards the $x$-axis so $y=s$ lies on it. Now, the surface of revolution can be found using one of the formulae above. In this specific case, replacing $y$ with $y=s$, the area of a surface of revolution is found using the formula $A=2\pi\int_{a}^{b}(y-s)\sqrt{\left(\frac{dy}{dx}\right)^{2}}\,dy$ Title area of surface of revolution AreaOfSurfaceOfRevolution 2014-07-24 18:36:37 2014-07-24 18:36:37 rspuzio (6075) pahio (2872) 13 rspuzio (2872) Topic msc 53A05 msc 26B15 area of revolution surface area of revolution SurfaceOfRevolution2 VolumeOfSolidOfRevolution
??? Well done! Reveal All Steps Show Tutorial Clear Progress Data ## We're still working on this.Please check back soon! Solve 6 out of 8 questions correctly + × ÷ ² () # Divisibility and Primes Prime numbers represent both the most basic properties and the most complex unsolved problems. Here you can learn about the building blocks of mathematics. ## Factors and Multiples By now you should be comfortable with basic integer arithmetic and addition, subtraction and multiplication. Division is slightly different, because you can’t always divide any integer by any other. For example 17 divided by 3 is not a whole number – it is somewhere in between 5 and 6. You either have to give a remainder (2), or express the answer as a decimal number (5.66). 12 is divisible by 3 10 is not divisible by 4 If you can divide a number A by a number B, without remainder, we say that B is a factor (or divisor) of A, and that A is a multiple of B. We often write B\|A, where the vertical bar simply means “divides”. For example, 7 × 3 = 21, so 7 is a of 21, 21 is a of 7, and 7\|21. In this short game you have to determine which numbers are factors or multiples, as fast as possible. Click the play button to start. ### Factors and Multiples Quiz \${x} is a of \${y} In the next step we will learn various techniques to easily check if a number is divisible by another. ## Divisibility Rules ### Divisibility by 2 and 5 Every number is divisible by 1. To determine if a number is divisible by 2, we simply have to check if it’s even: any number that ends in 0, 2, 4, 6, or 8 is divisible by 2. 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 26 27 28 29 30 To see if a number is divisibility by 5 we similarly just have to check that its last digit is 0 or 5: 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 26 27 28 29 30 The reason why these rules for 2 and 5 are so simple has to do with our number system. The base of our number system is 10, which means that every digit in a number is worth 10 times as much as the next one to the right. If we take the number 6382 as an example, 6 3 8 2 =6000 =300 =80 =2 Now we can separate the last digit of a number from all its other digits: abcd = abc × 10 + d 6382 = 638 × 10 + 2 Both 2 and 5 are factors of 10, so they will abc × 10, no matter what the values of a, b and c are. Therefore we only have to check the last digit: if d is divisible by 2 then is also divisible by 2. If d is divisible by 5 then the whole number is divisible by 5. ### Divisibility by 4 and 8 Unfortunately 4 doesn’t divide 10, so we can‘t just look at the last number – but 4 does divide 100, so we just have to slightly modify our rule from above. Now we write abcd = ab × 100 + cd. We know that 4 will always divide ab × 100, so we have to look at the last digits to check if a number if divisible by 4. For example, 24 is divisible by 4 so 273524 divisible by 4, and 18 is not divisible by 4 so 194718 divisible by 4. The divisibility rules for 8 get even more difficult, because 100 is not divisible by 8. Instead we have to go up to and look at the last digits of a number. For example, 120 is divisible by 8 so 271120 is also divisible by 8. The easiest is the divisibility rule for 10: we just need to check if the . Practice exercises coming soon … ### Divisibility by 3, 6and 9 The divisibility rule for 3 is rather more difficult. 3 doesn’t divide 10, and it also doesn’t divide 100, or 1000, or any larger power of 10. Simply looking at the last few digits of a number isn’t going to work. Instead we need to use the digit sum of a number, which is simply the sum of all its individual digits. For example, the digit sum of \${13×n+123} is \${digitSumString(123+13×n)} = \${digitSum(123+13×n)} and the digit sum of 3524 is . 1 2 3 3 4 5 6 6 7 8 9 9 10 11 12 3 13 14 15 6 16 17 18 9 19 20 21 3 22 23 24 6 25 26 27 9 28 29 30 3 31 32 33 6 34 35 36 9 37 38 39 12 40 Here we’ve highlighted all numbers which are multiples of three. You can see that their digit sums are always . So to determine if any number is divisible by 3, you just have to calculate its digit sum, and check if the result is also divisible by 3. Next, let’s look at multiples of 9: 9 9 18 9 27 9 36 9 45 9 54 9 63 9 72 9 81 9 90 9 99 18 108 9 117 9 126 9 135 9 144 9 153 9 162 9 171 9 180 9 It seems that all the numbers divisible by 9 have a digit sum which is divisible by 9. For example, the digit sum of 4752 is , so 4752 divisible by 9. Of course, these curious patterns for numbers divisible by 3 and 9 must have some reason – and like before it has to do with our base 10 numbers system. As we saw, writing the number 6384 really means 6 × 1000 + 3 × 100 + 8 × 10 + 4. We can split up each of these products into two parts: 6 × 999 + 6 + 3 × 99 + 3 + 8 × 9 + 8 + 4. Of course, 9, 99, 999, and so on are always divisible by 3 (or by 9). All that remains is to check that what’s left over is also divisible by 3 (or 9): 6 + 3 + 8 + 4 This just happens to be the digit sum! So if the digit sum is a multiple of 3, and we know that everything else is a multiple of 3, then the result must also be a multiple of 3. We’ve still skipped number 6 – but we’ve already done all the hard work. Remember that 6 = 2 × 3. 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 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 To check if a number is divisible by 6 we just have to check that it is divisible by 2 divisible by 3. Note that this happens to work for 6, but certainly not for any number that is the product of two others. More on that later… Unfortunately there is no simple divisibility rule for 7. However we now know the rules for all other numbers from 1 to 10, so let’s try a more advanced version of the divisibility game! Practice exercises coming soon … ## Finding Factors In many cases, it is not enough to find if a number is divisible by another – we need a list of all its divisors. For example, the divisors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60. But we don’t want to have to try 60 different numbers and check which one is a divisor. Instead we can use a simple technique, which relies on the fact that divisors always appear in pairs. In the case of 60 we have 60 = 1 × 60 = 2 × 30 = 3 × 20 = 4 × 15 = 5 × 12 = 6 × 10. Or, in a different notation, 60 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 To find all divisors of a number we simply start at both ends of this list, until we meet in the middle. 42 1, 2, 3, 6, 7, 14, 21, 42 For example, the first divisor pair of 42 is simply 1 and 42, and we write them down with much space in between. After 1 at the beginning, we check if 2 divides 42. It does, and the corresponding pair is 42 ÷ 2 = 21. Next, we check if 3 divides 42. It does, and the corresponding pair if 42 ÷ 3 = 14. Now we check if 4 divides 42. It does not, however, so we move on. 5 also doesn’t divide 42 so we move on. 6 does divide 42 again. Its pair is 42 ÷ 6 = 7. Notice how we’ve met in the middle after only a few attempts, without having to test all numbers from 7 to 42. The only special case with this method is for square number: in that case, you will meet at just a single number in the middle, like 64 = 8 × 8. ## Prime Numbers When calculating these divisor pairs, it can happen that a number doesn’t have any divisors except for the first pair. One example is 13 – its only divisors are 1 and 13 itself. These special numbers are called Prime numbers. They can’t be broken up into products of smaller numbers, which, in a way, makes them the “atoms of numbers”. Note that 1 itself is not a prime number, so the first few prime numbers are 2, 3, 5, 7, 11, 13, … Any number which is not prime can be written as the product of prime numbers: we simply keep dividing it into more parts until all factors are prime. For example, 84 2 × 42 2 × 21 3 × 7 84 = 2 × 2 × 3 × 7 Now 2, 3 and 7 are prime numbers and can’t be divided further. The product 2 × 2 × 3 × 7 is called the prime factorisation of 84, and 2, 3 and 7 are its prime factors. Note that some primes, like 2 in this case, can appear multiple times in a prime factorisation. Every integer has a prime factorisation and no two integers have the same prime factorisation. Furthermore, there is only a single way to write any number as a product of primes – unless we count different orderings of the primes. This is called the Fundamental Theorem of Arithmetic (FTA). Using the FTA can make many problems in mathematics much easier: we divide numbers into their prime factors, we solve the problem for the individual primes, which can often be much easier, and then we combine these results to solve the initial problem. ## The Sieve of Eratosthenes It turned out to be quite difficult to determine if a number is prime: you always had to find all its prime factors, which gets more and more challenging as the numbers get bigger. Instead, the Greek mathematician Eratosthenes of Cyrene came up with a simple algorithm to find all prime numbers up to 100: the Sieve of Eratosthenes. 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 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 First we need to write down all numbers up to 100. We know that 1 is not prime, so we delete it. The smallest prime number is 2. Any multiple of 2 can’t be prime, since it has 2 as a factor. Therefore we can cross out all multiples of 2. The next number in our list is 3 – again a prime number. All multiples of 3 can’t be prime, since they have 3 as a factor. Therefore we can cross these out as well. The next number, 4, is already crossed out so we move on to 5: it is a prime number and again we cross out all multiples of 5. The next prime number must be , since 6 is crossed out. Once more, we cross out all of its multiples. The next prime number is . Notice, however, that all of its multiples are . The same is actually true for all other remaining numbers. Therefore all these remaining numbers must be prime. Now we can count that, in total, there are prime numbers less than 100. ## How many Prime Numbers are there? Of course we can also use the Sieve of Eratosthenes to find larger prime numbers. There are 21 primes between 100 and 200, 16 primes between 200 and 300, 17 primes between 400 and 500 and only 11 between 10,000 and 10,100. The primes seem to keep getting more and more spread out, but do they ever stop? Is there a biggest or a last prime number? The ancient Greek mathematician Euclid of Alexandria first proved that there are infinitely many prime numbers, using the following argument: 1. Suppose there were only finitely many prime numbers. P, P, P, P, P 2. Let us multiply all of them together, to get a very large number which we call N. N = P × P × P × P × P 3. Now let’s think about N + 1. Any prime number that divides N can’t also divide N + 1. And since all prime numbers we have found so far divide N, none of these can also divide N + 1. P, P, P, P, P N P, P, P, P, P N + 1 4. We know from the Fundamental Theorem of Arithmetic that N + 1, must have a prime factor. Either N + 1 is itself prime, or there is some other new prime P’ that divides N + 1. P’ N + 1 5. In both cases we’ve found a new prime not in our original list – but we assumed that all primes were in this list. 6. Clearly something went wrong! But since steps 24 were definitely valid, the only possibility is that our initial assumption in 1 was wrong. This means there must actually be infinitely many primes. Euclid’s explanation is one of the first examples in history of a formal mathematical proof – a logical argument that shows a statement must definitely be true. This example is often called proof by contradiction: we start with an assumption, deduce something impossible, and thus know that our assumption must be incorrect. ## Large Primes It is quite easy for a computer to check if a number is prime – simply by trying to divide it by all smaller numbers. But of course there are much cleverer and faster algorithms. Here you can try it yourself: ### Prime Checker \${result} Throughout history, people have tried to calculate larger and larger primes. In 1460, the largest known prime number was 131,071. In 1772, Leonard Euler showed that 2,147,483,647 is prime. With the arrival of computers in the 20th century, calculating large primes became much faster and easier. Currently, the largest known prime number has 22,338,618 digits – you would need 5000 sheets of paper to print it out! It was discovered in September 2015 using a network of many thousands of computers around the world, run by volunteers, students and universities. The GIMPS Project (Great Internet Mersenne Prime Search) uses a network of computers around the world to find large prime numbers. You might think that calculating these primes is just a waste of time – but, as we will see below, it turns out to be incredibly useful for computers to be able to quickly find large prime numbers. Rather than checking if a large number is definitely prime, there are also much faster algorithms that tell you that a number is almost certainly prime. Here you can generate your own, large prime numbers: ### Prime Generator Number of digits: \${d} \${result} The Polish mathematician Stanisław Ulam came up with a cool way to show the distribution of large prime numbers, while doodling during a “long and very boring” meeting in 1963. 37 36 35 34 33 32 31 38 17 16 15 14 13 30 39 18 5 4 3 12 29 40 19 6 1 2 11 28 41 20 7 8 9 10 27 42 21 22 23 24 25 26 43 44 45 46 47 48 49 We write down all integers in a rectangular grid, starting with 1 in the middle and then spiralling outwards. Then we highlight all numbers which are prime. So far, the Ulam spiral doesn’t look particularly exciting. But if we zoom out, interesting patterns emerge. Here are the primes up to 160,000: Rather than appearing randomly, as one might expect, it seems that certain diagonals are much more popular with primes than others. This creates a curious “plaid” pattern. It turns out that these diagonals all correspond to certain quadratic equations which seem to generate prime numbers more often than average. However it is unknown why that would be the case… Cover of the March 1964 issue of Scientific American ## The Least Common Multiple Two runners are training on a circular racing track. The first runner takes 60 seconds for one lap. The second runner only takes 40 seconds for one lap. If both leave at the same time from the start line, when will they meet again at the start? This question really isn’t about the geometry of the race track, or about velocity and speed – it is about multiples and divisibility. The first runner crosses the start line after 60s, 120s, 180s, 240s, and so on. These are simply the of 60. The second runner crosses the start line after 40s, 80s, 120s, 160s, and so on. The first time both runners are back at the start line is after seconds. What we’ve just found is the smallest number which is both a multiple of 40 and a multiple of 60. This is called the least common multiple or lcm. To find the lcm of any two numbers, it is important to realise that if a divides b, then b needs to have all the prime factors of a (plus some more): 12 60 2 × 2 × 3 2 × 2 × 3 × 5 This is easy to verify: if a prime factor divides a, and a divides b, then that prime factor must also divide b. To find the lcm of 40 and 60, we first need to find the prime factorisation of both: 40 = 2 × 2 × 2 × 5 60 = 2 × 2 × 3 × 5 Suppose that X is the lcm of 40 and 60. Then 40 divides X, so 2, 2, 2 and 5 must be prime factors of X. Also, 60 divides X, so 2, 2, 3 and 5 must be prime factors of X. To find X, we simply combine all the prime factors of 40_ and {.m-blue}60__, but any duplicates we only need once: X = 2 × 2 × 2 × 3 × 5 This gives us that X = 120, just like we saw above. Notice that if the same prime factor appears multiple times, like 2 above, we need to keep the maximum occurrences in one of the two numbers (3 times in 40 is more than 2 times in 60). Now we have a simple method for finding the lcm of two numbers: ol.proof li Find the prime factorisation of each number. li Combine all prime factors, but only count duplicates once. We can use the same method to find the lcm of three or more numbers at once, like 12, 30 and 45: 12 = 2 × 2 × 3 30 = 2 × 3 × 5 45 = 3 × 3 × 5 Therefore the lcm of 12, 30 and 45 is 2 × × 3 × 3 × = 180. A special case are prime numbers: the lcm of two different primes is simply heir , because they don’t have any common prime factors which would get “canceled”. ## The Greatest Common Factor An architect is planning the floor for a large courtyard that measures 18m by 30m. She wants it to be covered in quadratic tiles, without any gaps or overlaps along the sides. What is the largest size of squares she can use? Just like before, this question is not about geometry - it is about divisibility. The length of the sides of the tiles has to divide both 18 and 30, and the largest possible number with that property is . This is called the Greatest Common Divisor or gcd of 18 and 30. Once again, we can use the prime factorisation to calculate the gcd of any two numbers. Remember that any divisor of a number must have some of the prime factors of that number. 18 = 2 × 3 × 3 30 = 2 × 3 × 5 Suppose that X is the gcd of 18 and 30. Then X divides 18 so the prime factors of X must be among 2, 2 and 3. Also, X divides 30 so the prime factors of X must be among 2, 3 and 5. To find X, we simply need to multiply all numbers which are prime factors of 18 and 30: X = 2 × 3 = 6. Now we have a simple method for finding the gcd of two numbers: ol.proof li Find the prime factorisation of each number. li Multiple the prime factors which are in both numbers. Once again prime numbers are special: the gcd of two different primes is always , because they don’t share any prime factors. ## LCM and GCD Applications North America is home to various broods of cicadas. These have the curious property that they only emerge every few years during the summer to breed – the remaining time they spend underground. For example, the cicadas in Florida and Mississippi appear every 13 years. The cicadas in Illinois and Iowa only appear every 17 years. But there are no cicadas with 12, 14, 15 or 16 year cycles. Both 13 and 17 are prime numbers – and that has a very good reason. Imagine that there are predators in the forest which kill cicadas. These predators also appear in regular intervals, say every 6 years. Now imagine that a brood of cicadas appears every \${n} years (\${isPrime(n) ? 'prime' : 'not prime'}). The two animals would meet every \${lcm(n, 6)} years, which is the of 6 and \${n}. This number seems to be much larger if the cicada cycle is a prime number like 13 and 17. That‘s is because prime numbers don‘t share any factors with 6, so when calculating the lcm we don‘t cancel any duplicate factors. Of course, cicadas have no idea what prime numbers are – but over millions of years, evolution has worked out that prime cycles are the safest. The predator animal seems to have gone extinct over time, but the prime number cycles remain. ## Primes in Cryptography One of the most important modern applications of prime numbers is in a field of mathematics called Cryptography. For thousands of years, people have tried to conceal messages so that only the intended recipient could read them – this is called encryption. It is used by everyone from generals exchanging secret orders during wars to personal emails or online banking details. People always tried to come up with better, more secure encryption methods, but after some time, they were all broken using yet more advanced algorithms. In the Second World War, the German army used the Enigma: a complex machine consisting of a keyboard, rotating wheels and plugs. It encrypted messages using one of 158 million million million possibilities (that’s a 158 followed by 18 zeros!). The code was widely belied to be unbreakable, but the British Secret Service, led by mathematician Alan Turing, built some of the first computers that managed to decode it. German four-rotor Enigma machine Today’s computers are much more advanced, capable of trying millions of possibilities every second. To develop better encryption algorithms you have to find a mathematical operations that is difficult for a powerful computers. Computers are incredibly fast at addition, subtraction, multiplication and division. However, as it turns out, computers are very slow at factorising large integers into primes… COMING SOON – RSA example with Alice and Bob This encryption algorithm is called RSA Cryptography, after its three inventors, Ron Rivest, Adi Shamir and Leonard Adleman who published it in 1977. It turns out that a very similar method was known to the British Secret Service since 1973, but remained classified until much later. Today, prime numbers are used by computers all over the world to exchange data. Whenever you send an email or visit a secure website, your phone or laptop quietly generates large prime numbers and exchanges public keys with other computers. ## Mysteries and Unsolved Problems ### The Goldbach Conjecture In 1742, the German mathematician Christian Goldbach made a curious discovery: he noticed that all even integers (except 2) can be written as the sum of two prime numbers. For example, 8 = 5 + 3 and 24 = 13 + 11. This is quite surprising, because primes are defined using multiplication and factors – and shouldn’t have much to do with addition. ### Goldbach Calculator Pick any even number, to calculate how it can be written as the sum of two primes. \${result} Goldbach wrote about his observation in a letter to the famous mathematician Leonhard Euler, but neither of them was able to prove it. It became known as the Goldbach Conjecture. Computers have checked that the Goldbach Conjecture works for every even number up to 4 × 1018 (that’s a 4 with 18 zeros), but mathematicians have still not found a proof that it works for all even integers. And that is a big difference, because there are infinitely many integers, so we couldn’t possibly check all of them. Its apparent simplicity made the Goldbach conjecture one of the most famous unsolved problems in mathematics. ### Twin Primes We have already seen that prime numbers get more spread out as they get bigger. But they always seem to appear completely random, and occasionally we find two primes right next to each other, just one number apart: these are called Twin Primes. 35,1113,4143,101103,20272029,108,377108,379,1,523,6511,523,653 The largest known pair of twin primes has an incredible 58,711 digits! But are there infinitely many twin primes, just like there are infinitely many primes? Nobody knows – the Twin Prime conjecture is another one of the many unsolved problems surrounding the primes. ### The Riemann Hypothesis Mathematicians have spent many centuries exploring the pattern and distribution of prime numbers. They seem to appear completely randomly – sometimes there are huge gaps in between consecutive primes, and sometimes we find twin primes right next to each other. When only 15 years old, the German mathematician Carl Friedrich Gauss had a groundbreaking new idea: he counted the number of primes up to a certain point, and showed the results in a chart: Along the x-axis you can see all integers. Whenever there is a prime, the Prime Counting Function increases by one. As we zoom out, the blue line becomes very smooth. Gauss noticed that the shape of this function looks very similar to the function xlog(x). He predicted that the two functions are always “approximately similar”, and this was proven in 1896. However, as you can see above, there is still a significant error between the actual number of primes, and Gauss’s approximation. In 1859, the mathematician Bernhard Riemann discovered an approximation that looked much better, but he wasn’t able to prove that that would always work. His idea became known as the Riemann Hypothesis. Hundreds of mathematicians have tried to prove Riemann’s hypothesis, but all without success. It is often considered one of the most difficult and most important unsolved problems in mathematics. In 2000, the Clay Mathematics Institute named it one of six Millennium Prize Problems and promised \$1,000,000 to any mathematician who solves it.
## Dimensional Analysis Introduction Conversion Factors and Their Reciprocals Use of Conversion Factors: Some Examples Developing the Units Path Summary and Helpful Hints ### Introduction Example 1: Suppose you want to convert \$100 in U.S. dollars to Canadian dollars. You know that the exchange rate is currently \$1.50 Cdn. for every \$1.00 U.S. How much Canadian cash will you get? Solution. You probably solved this in your head: you'll expect to receive \$150 Canadian for your \$100 U.S. But think carefully! This isn't intuitively obvious, you had to learn how to do it. The steps you probably followed, so fast you didn't realize, were these: 1. You recognized that you were starting with \$100 U.S. (the starting quantity) 2. You got a conversion factor from the English sentence that tells you the exchange rate: the conversion factor is 3. You then multiplied the starting quantity by the conversion factor to get your answer (the desired quantity): This gives us a method for solving many problems that aren't so obvious as converting U.S. to Canadian dollars. #### The Dimensional Analysis Method • Identify the starting quantity. Write it down. (In the above example, it was \$100 U.S.) • Decide what units (dimensions) you need for your answer. Develop a path that gets you from your starting units to the units you need. Write down the conversion factor between the units. • Following the "units path," multiply your starting quantity by one or more conversion factors so that you finally obtain the units you need. All the unwanted units will cancel each other when you multiply. To put it in simpler terms, we want to know how to get from what we are GIVEN (the starting quantity) to what we WANT TO KNOW (the desired quantity) Dimensional analysis is sometimes called the factor-label method or the factor-unit method. That is because it involves two quantities: a mathematical proportionality that is thefactor and an expression of the units, or label. ### Conversion Factors and Their Reciprocals Every conversion factor has a reciprocal that can be used for the opposite conversion. To convert \$U.S. to Canadian dollars, we used the factor However, if we wanted to convert Canadian dollars to U.S. dollars, we would use the reciprocal of the above factor, or So, if we had \$500 Cdn., we could determine how much U.S. money we could get in exchange: Below are some other conversion factors you will have seen in your studies of chemistry. Conversion Factor Reciprocal of Conversion Factor 10 mm / cm 1 cm / 10 mm 453.6 g / lb 161 lb / 453.6 g 6.022 x 1023 atoms Cu / mol Cu 1 mol Cu / 6.022 x 1023 atoms Cu 63.546 g Cu / mol Cu 1 mol Cu / 63.546 g Cu 2 mol NaOH / mol H2SO4 1 mol H2SO4 / 2 mol NaOH 36.9 g Fe2O3 / 100 g ore 100 g ore / 36.9 g Fe2O3 ### Use of Conversion Factors: Some Examples Example 1. (a) Write the reciprocal for the following conversion factor: 1 mol NaOH / 40.0 g NaOH. (b) Which of these (the original factor or its reciprocal) do we use if we want to change the mass of NaOH (49.7 g) to moles? Answer: (a) 40.0 g NaOH / mol NaOH (b) the original factor: (49.7 g NaOH) x (1 mol NaOH / 40.0 g NaOH) gives units of moles NaOH Example 2. (a) Write the reciprocal for the following conversion factor: 3 atoms S / molecule P4S3. (b) Which of these (the original factor or its reciprocal) do we use if we want to calculate how many molecules of P4S3 can be made from a 1200 S atoms? Answer: (a) 1 molecule P4S3 / 3 atoms S (b) the reciprocal factor: (1200 S atoms) x (1 molecule P4S3 / 3 atoms S) gives units of molecules of P4S3 Example 3. (a) Write the reciprocal of the following conversion factor: 3 mol AgCl / mol CrCl3. (b) Which of these (the original factor or its reciprocal) do we use if we want to calculate how many moles of AgCl can be prepared from 0.1250 moles of CrCl3? Answer: (a) 1 mol CrCl3 / 3 mol AgCl (b) the original factor: (0.1250 mol CrCl3) x ( 3 mol AgCl / mol CrCl3) has units of moles AgCl Example 4. (a) Write the conversion factor for "36.3% chocolate chips in Chips Ahoy cookies," then write its reciprocal. Which of these (the original or its reciprocal) do we use to calculate how many cookies have to be processed to yield a 500. g of chocolate chips? Answer: (a) conversion factor is 36.3 g chocolate chips / 100 g cookies; reciprocal is 100 g cookies / 36.3 g chocolate chips (b) the reciprocal factor: (500. g chocolate chips) x (100 g cookies / 36.3 g chocolate chips) has units of mass of cookies. ### Developing the Units Path: Extended Examples Example 1. You have 45.5 grams of iron, Fe. What mass (in g) of Fe2O3 can you make if you react the iron with oxygen? The balanced equation is 4 Fe + 3 O2 ---> 2 Fe2O3 Molar masses: Fe 55.85 g / mol; Fe2O3 159.7 g / mol Analysis: We want to know how we can get from what we are given (45.5 g Fe) to what we WANT (mass of Fe2O3). Now you know about chemistry and balanced equations, so you know that the MOLE is the important unit here, and we'll have to work through moles. If we want to get from mass of Fe to mass of Fe2O3, we'll have to have a bunch of conversion factors since no single factor that we've been given can get us there. The balanced equation gives us this conversion factor that relates the product to the reactant: 4 mol Fe / 2 mol Fe2O3. We have two other conversion factors, the molar masses, that allow us to go from masses to moles and vice versa. We also have the reciprocals of these factors. The sequence of operations is this: THIS SEQUENCE IS IMPORTANT! It may take a while to figure out, and you may have to play with conversion factors to realize just how you'll get from one step to another. But by now you'll know that mass/mole conversions are made through the conversion factor molar mass or its reciprocal. And the balanced equation gives you the connection between the moles of reactants and products. So: we begin with 45.5 g Fe, and use the above sequence with its conversion factors chosen so that all the units cancel except "g Fe2O3" which is what we wanted to find: Example 2. You are a chemist in an ore-processing plant that produces iron metal. You are told that the new shipment of ore coming in contains 56.3% Fe2O3. Your company asks you: how many tonnes of iron can be obtained from 5.00 x 102 tonnes of ore? (1 tonne = 1000 kg). Molar masses: Fe 55.85 g / mol; Fe2O3 159.7 g / mol Analysis. You are given the amount of ore that has to be processed. That will be your starting point for this question. Once again, you'll need to work through moles. You'll also have to turn "56.3% Fe2O3" into a conversion factor: 56.3 % Fe2O3 = 56.3 g Fe2O3 / 100 g ore The unit path is then mass of ore ---> mass of Fe2O3 ---> moles Fe2O3 ---> moles Fe ---> mass Fe So we begin with the mass of ore (5.00 x 102 tonnes) and proceed in serval steps: Step 1: mass of ore to mass of Fe2O3 in grams Step 2: mass of Fe2O3 to mass of Fe Step 3: mass of Fe in grams to mass in tonnes Step 1: Step 2: Step 3: ### Summary Learning this method takes practice, but it helps if you remember the basic process: • Write down what you are GIVEN and its units • Write down what you are ASKED TO FIND and its units • Determine the conversion factors that might be needed to develop the units path beteween what you are given and what you are asked to find. • Finally, multiply what you are given by the conversion factors (or their reciprocals if necessary to make the units cancel). If the units of your final answer are correct, and your conversion factors are correct, the answer you obtain almost certainly has to be the correct answer. HELPFUL HINTS: • In the Dimensional Analysis Tutorial that you did in the first lab, the conversion factors were referred to as "fractions" (because they are fractions: one number with its units over another number with its units). • In Section 1 of the Tutorial, "Translation of Statements into Fractions," you are asked to write the fraction in two ways. These two ways correspond to the conversion factor and its reciprocal. Back to Lab Schedule Back to CHEM 1P80 Home Page This page is http://chemiris.labs.brocku.ca/~chemweb/courses/chem180/Dimensional_Analysis.html Created October 10, 2000 by M. F. Richardson © Brock University, 2000
8 Q: # What is 15 of 80 with simple method and examples? A) 12.5% B) 18.75% C) 16.55% D) 17.80% Explanation: Given that to find 15 of 80 => What % of 80 is 15 Let it be 'p' => p% of 80 = 15 => Therefore, 15 is 18.75% of 80 => 18.75% of 80 = 15 Example:: Now, in the similar way we can find 15% of 80 =? => 15x80/100 = p => p = 4 x 3 = 12 Therefore, 15% of 80 = 12 Q: What is 6% of 3,50,000? A) 19500 B) 21000 C) 23600 D) 24000 Explanation: 6% of 350000 Hence, 6% of 3,50,000 = 21,000. 1 65 Q: 20% of 120? A) 20 B) 24 C) 32 D) 36 Explanation: 20% of 120 20 x 120/100 = 4 x 6 = 24 Hence, 20% of 120 = 24. 0 144 Q: Twenty percent of Karthik’s monthly salary is equal to thirty two percent of Nikitha’s monthly salary. Vipin’s monthly salary is four fifth that of Nikitha’s monthly salary. If Vipin’s annual salary is Rs. 3.84 lakhs, what is Karthik’s monthly salary? A) 32,000 B) 64,000 C) 40,000 D) 52,000 Explanation: From the given data, 20% of K = 32% of N V = 4/5 N Given Vipin's salary = 3.84 lpa = 3.84/12 = 32,000 per month Now, N = 5/4 x 32000 = 40000 Now K = 32/20 x 40000 = 64000 Hence, karthik's monthly salary = Rs. 64000. 2 193 Q: 1/4 of 12? A) 4 B) 3 C) 2 D) 1 Explanation: 1/4 of 12 = 1/4 x 12 = 3 Hence, 1/4 of 12 = 3. 0 153 Q: 40% of 75 + 80% of 25 = K% of 250 Find the value of K? A) 10 B) 20 C) 30 D) 35 Explanation: From the given data, 0 336 Q: In a class of 80 students and 5 teachers, each student got sweets that are 15% of the total number of students and each teacher got sweets that are 25% of the total number of students. How many sweets were there? A) 1060 B) 960 C) 860 D) 760 Explanation: From the given data, 80 x 15/100 of 80 + 5 x 80 x 25/100 = 960 + 100 = 1060. 5 363 Q: Sairam spent Rs. 38460 on the renovation of his home and Rs. 24468 on buying TV set and remained with 28% of the total amount. What was the total amount? A) Rs. 87400 B) Rs. 68700 C) Rs. 74000 D) Rs. 94060 Explanation: Let the total amount he had with = Rs. p From the given data, p - 38460 - 24468 = 28p/100 p - 28p/100 = 24468 + 38460 72p/100 = 62928 72p = 62928 x 100 p = 62928 x 100/72 p = 87400 Hence, the total amount he had = p = Rs. 87400. 4 351 Q: A jar can 20 litre milk. From the jar, 4 litres milk was taken out and replaced with an equal quantity of water. If 4 litres of the newly formed mixture is taken out of the can, then what is the final quantity of milk left in the can? A) 14.5 lit B) 12.8 lit C) 11.6 lit D) 10.46 lit
Simplifications \| Laws of Operations \| Grade 3 Mathematics Operations Simplifications for Class 3 Math Here the student will be introduced to the simplifications. Also, they will learn about arithmetic expressions and the order of operations. The student will learn to • Identify arithmetic expressions • Apply order of operations • Use commutative law and associative law • Apply order of operations to solve simplification problems. The learning concept has been explained to the class 3 students with examples, illustrations and a concept map. At the end of the page, two printable simplifications worksheets with solutions are attached for the students. Download the worksheets and solutions to assess our knowledge of the concept. We already know about the operations on numbers. Here we will discuss the simplification of arithmetic expressions involving numbers. Arithmetic Expressions An Arithmetic expression in math is a combination of operations (addition, subtraction, multiplication, division) and terms. There are different types of Arithmetic expressions. Following are examples of some arithmetic expressions. Examples: Orders of Operations To simplify any arithmetic expression, we need to know the order of operations, that should be followed. 1. First, do the operations in brackets. 2. Multiplication and division should be done prior to any operations. 3. Then addition and subtraction should be done after multiplication and division. Examples: Simplify the followings. 1. (62 `Ã·` 2 â€“ 3) Ã— 3 + 6 = 2. 3 + 21 Ã— 6 â€“ (24 â€“ 4) Ã— 2 = 1. (62 `Ã·` 2 â€“ 3) Ã— 3 + 6 2. 3 + 21 Ã— 6 â€“ (24 â€“ 4) Ã— 2 Laws of Operations There are some laws which govern the order in which we perform the operations in arithmetic. • Commutative law • Associative law Commutative Law: The order of the numbers does not affect the result in addition, and in multiplication. Commutative law is not applicable for subtraction and division. Examples: 2. For multiplication, This law is not applicable to subtraction and division. Associative Law This law states that the three numbers are independent of their grouping for addition and multiplication. Clearly, grouping or combination of three numbers doesnâ€™t affect the result. Associative law is not applicable for subtraction and division. Examples: 2. For multiplication, This law is not applicable to subtraction and division. Application: Examples 1: Simplify. Examples 2: 249 â€“ 200 students are not going to the summer camp. 238 â€“ 150 students are not going to the summer camp. 98 â€“ 50 students are not going to summer camp So, the total number of students who donâ€™t want to go to the summer camp is 249 â€“ 200 + 238 â€“ 150 + 98 â€“ 50 = 49 + 88 + 48 = 137 + 48 = 185 Five brothers have given a total of (5 Ã— Rs 60) Two sisters have given a total of (2 Ã— Rs 50) The total amount in the fund (5 Ã— Rs 60) + (2 Ã— Rs 50) Each child will get [(5 Ã— Rs 60) + (2 Ã— Rs 50)] `Ã·` 8 = (Rs 300 + Rs 100) `Ã·` 8 = Rs 400 `Ã·` 8 = Rs 50 • -
# MTH 11203 Algebra PROPERTIES OF THE REAL NUMBER SYSTEM CHAPTER 1 SECTION 10. ## Presentation on theme: "MTH 11203 Algebra PROPERTIES OF THE REAL NUMBER SYSTEM CHAPTER 1 SECTION 10."— Presentation transcript: MTH 11203 Algebra PROPERTIES OF THE REAL NUMBER SYSTEM CHAPTER 1 SECTION 10 If a and b represent any real numbers, then a + b = b + a Commutative property involves a change in order. The order that you add does not matter, same results Exp: 5 + 2 = 2+ 5 7 = 7 Exp:-3 + (-5) = -5 + (-3) -8 = -8 Commutative property will not work for subtraction Commutative Property of Addition If a and b represent any real numbers, then a · b = b · a The order that you multiply does not matter, same results Exp: (5)(6) = (6)(5) 30 = 30 Exp:(r)(g) = (g)(r) rg = rg Commutative property will not work for division Commutative property changes the order Commutative Property of Multiplication If a and b represent any real numbers, then (a + b) + c = a + (b + c) The associative property involves a change in grouping. The order that you add does not matter, same results Exp: (2 + 3) + 4 = 2 + (3 + 4) Exp: 6 + (w + 1) = (6 + w) + 1 5 + 4 = 2 + 7 7 + w = 7 + w 9 = 9 Exp:(3 + 4) + x = 3 + (4 + x) 7 + x = 7 + x Associative property will not work for subtraction Associative Property of Addition If a and b represent any real numbers, then (a · b) · c = a · (b · c) The order that you multiply does not matter, same results Exp: (2 · 6) · 4 = 2 · (6 · 4) 12 · 4 = 2 · 24 48 = 48 Exp:(3 · 7) · x = 3 · (7 · x) 21 · x = 21 · x 21x = 21x Associative property will not work for division The associative property changes grouping Associative Property of Multiplication If a and b represent any real numbers, then a(b + c) = ab + ac The order that you multiply does not matter, same results Exp: 3(2 + 4) = (3)(2) + (3)(4) (3)(6) = 6 + 12 18 = 18 Exp:6(x – 9) = (6)(x) – (6)(9) 6x – (6)(9) = 6x – 54 6x – 54 = 6x – 54 Expand: a (b + c + d + e + f + … + n) = ab + ac + ad + ae + af + … + an Distributive Property Identity Property of Addition If a and b represent any real numbers, then a + 0 = a and 0 + a = a Zero is the identity element of addition Exps: 3 + 0 = 3 6 + 0 = 6 50 + 0 = 50 Identity Property of Multiplication If a and b represent any real numbers, then a · 1 = a and 1 · a = a One is the identity element of Multiplication Exp:3 · 1 = 3 6 · 1 = 6 50 · 1 = 50 Identity Property Inverse Property of Addition If a and b represent any real numbers, then a + (-a) = 0 and -a + a = 0 Additive inverses are any two numbers whose sum is 0 or opposites. Exp: 3 + (-3) = 0 6 + (-6) = 0 50 + (-50) = 0 Inverse Property of Multiplication If a and b represent any real numbers, then a · = 1 and · a = 1 a ≠ 0 Multiplicative inverses are any two numbers whose product is 1 or reciprocal. Exp:3 · = 1 6 · = 1 50 · = 1 Inverse Property #29 pg 84) p · (q · r) = (p · q) · r Associative Property of Multiplication #24 pg 84)3 + y = y + 3 Commutative Property of Addition #31 pg 84)4(d + 3) = 4d + 12 Distributive Property #28 pg 84)-4x + 4x = 0 Inverse Property of Addition #33 pg 85) 3z · 1 = 3z Identity Property of Multiplication Name the property Exp:4(x – 7) = 4x + (4)(-7) 4(x – 7) = 4x – 28 Distributive Property Exp:5y + (-5y) = 0 Inverse Property of Addition Name the property Exp: (3 · 6) · 9 = (3 · 6) · 9 = 3 · (6 · 9) Associative Property of Multiplication Exp:7(x – 4) = 7(x – 4) = 7x – 28 Distributive Property #33 pg 85) 6b + 0 = 6b + 0 = 6b Identity Property of Addition Complete the equation and name the property HOMEWORK 1.10 Page 84 – 85 #23, 25, 27, 37, 39, 41, 43, 53 Download ppt "MTH 11203 Algebra PROPERTIES OF THE REAL NUMBER SYSTEM CHAPTER 1 SECTION 10." Similar presentations
# Geometric Progression • Last Updated : 01 Sep, 2022 A sequence of numbers is called a Geometric progression if the ratio of any two consecutive terms is always the same. In simple terms, A geometric series is a list of numbers where each number, or term, is found by multiplying the previous term by a common ratio r. The general form of Geometric Progression is: GP-series Where, a = First term r = common ratio arn-1 = nth term Example: The sequence 2, 4, 8, 16 is a GP because ratio of any two consecutive terms in the series (common difference) is same (4 / 2 = 8 / 4 = 16 / 8 = 2). ### The geometric progression is of two types: 1. Finite geometric progression 2. Infinite geometric progression. ### 1. Finite geometric progression In finite geometric progression contains a finite number of terms. The last term is always defined in this type of progression. Example: The sequence 1/2,1/4,1/8,1/16,…,1/32768 is a finite geometric series where the first term is 1/2 and the last term is 1/32768. ### 2. Infinite geometric progression Infinite geometric progression contains an infinite number of terms. The last term is not defined in this type of progression. ### Example: Sequence 3, 9, 27, 81, … is an infinite series where the first term is 3 but the last term is not defined. 1. Initial term: In a geometric progression, the first number is called the initial term. 2. Common ratio: The ratio between a term in the sequence and the term before it is called the “common ratio.” 3. The behavior of a geometric sequence depends on the value of the common ratio. If the common ratio is: • Positive, the terms will all be the same sign as the initial term. • Negative, the terms will alternate between positive and negative. • Greater than 1, there will be exponential growth towards positive or negative infinity (depending on the sign of the initial term). • 1, the progression is a constant sequence. • Between -1 and 1 but not zero, there will be exponential decay towards zero. • -1, the progression is an alternating sequence. • Less than -1, for the absolute values there is exponential growth towards (unsigned) infinity, due to the alternating sign. ### The formula for the nth term of a Geometric Progression: If â€˜a1′ is the first term and â€˜râ€™ is the common ratio. Thus, the explicit formula for nth term of finite GP series: Nth term of a Geometric Progression The formula for the sum of the nth term of Geometric Progression: Sum of the Nth term of Geometric Progression How do we check whether a series is a Geometric progression or not? The property of the GP series is that the ratio of the consecutive terms is same. Approach: 1. First calculate the common ratio r by arr[1] / arr[0] 2. Iterate over an array and calculate the ratio of the consecutive terms. 3. Check if the calculated ratio is not equal to the common ratio r • Return false 4. After traversal, if the calculated ratio is equal to the common ratio r every time • Return true Below is the implementation of the above approach: ## C++ `// C++ program to check if a given array``// can form geometric progression` `#include ` `using` `namespace` `std;` `bool` `is_geometric(``int` `arr[], ``int` `n)` `{`` ``if` `(n == 1)`` ``return` `true``;` ` ``// Calculate ratio`` ``int` `ratio = arr[1] / (arr[0]);` ` ``// Check the ratio of the remaining`` ``for` `(``int` `i = 1; i < n; i++) {`` ``if` `((arr[i] / (arr[i - 1])) != ratio) {`` ``return` `false``;`` ``}`` ``}`` ``return` `true``;``}` `// Driven Program``int` `main()``{`` ``int` `arr[] = { 2, 6, 18, 54 };`` ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` ` ``(is_geometric(arr, n)) ? (cout << ``"True"` `<< endl)`` ``: (cout << ``"False"` `<< endl);` ` ``return` `0;``}` ## Java `// Java program to check if a given array``// can form geometric progression``import` `java.util.Arrays;` `class` `GFG {` ` ``// function to check series is`` ``// geometric progression or not`` ``static` `boolean` `is_geometric(``int` `arr[], ``int` `n)`` ``{`` ``if` `(n == ``1``)`` ``return` `true``;` ` ``// Calculate ratio`` ``int` `ratio = arr[``1``] / (arr[``0``]);` ` ``// Check the ratio of the remaining`` ``for` `(``int` `i = ``1``; i < n; i++) {`` ``if` `((arr[i] / (arr[i - ``1``])) != ratio) {`` ``return` `false``;`` ``}`` ``}`` ``return` `true``;`` ``}` ` ``// driver code`` ``public` `static` `void` `main(String[] args)`` ``{`` ``int` `arr[] = { ``2``, ``6``, ``18``, ``54` `};`` ``int` `n = arr.length;` ` ``if` `(is_geometric(arr, n))`` ``System.out.println(``"True"``);`` ``else`` ``System.out.println(``"False"``);`` ``}``}` ## Python3 `def` `is_geometric(li):`` ``if` `len``(li) <``=` `1``:`` ``return` `True`` ` ` ``# Calculate ratio`` ``ratio ``=` `li[``1``]``/``float``(li[``0``])`` ` ` ``# Check the ratio of the remaining`` ``for` `i ``in` `range``(``1``, ``len``(li)):`` ``if` `li[i]``/``float``(li[i``-``1``]) !``=` `ratio:`` ``return` `False`` ``return` `True` `print``(is_geometric([``2``, ``6``, ``18``, ``54``]))` ## C# `// C# program to check if a given array``// can form geometric progression``using` `System;` `class` `Geeks {` ` ``static` `bool` `is_geometric(``int``[] arr, ``int` `n)`` ``{`` ``if` `(n == 1)`` ``return` `true``;` ` ``// Calculate ratio`` ``int` `ratio = arr[1] / (arr[0]);` ` ``// Check the ratio of the remaining`` ``for` `(``int` `i = 1; i < n; i++) {`` ``if` `((arr[i] / (arr[i - 1])) != ratio) {`` ``return` `false``;`` ``}`` ``}`` ``return` `true``;`` ``}` ` ``// Driven Program`` ``public` `static` `void` `Main(String[] args)`` ``{`` ``int``[] arr = ``new` `int``[] { 2, 6, 18, 54 };`` ``int` `n = arr.Length;` ` ``if` `(is_geometric(arr, n))`` ``Console.WriteLine(``"True"``);`` ``else`` ``Console.WriteLine(``"False"``);`` ``}``}` ## PHP `` ## Javascript `` Output `True` Time Complexity: O(n), Where n is the length of the given array. Auxiliary Space: O(1) Basic Program related to Geometric Progression More problems related to Geometric Progression Recent Articles on Geometric Progression! My Personal Notes arrow_drop_up
# Law of cosines (Difference between revisions) Revision as of 08:31, 31 May 2011 (edit)← Previous diff Revision as of 08:34, 31 May 2011 (edit) (undo)Next diff → Line 94: Line 94: $0 = \cos B$ $0 = \cos B$ - Using [[inverse trig functions]], we know that + Using [[Basic Trigonometric Functions#Inverse Trig Functions\| inverse trigonometry]], we know that $B = 90^\circ$ $B = 90^\circ$ ## Revision as of 08:34, 31 May 2011 The law of cosines is a formula that helps in triangulation when two or three side lengths of a triangle are known. The formula relates all three side lengths of a triangle to the cosine of a particular angle. $c^{2} = a^{2} + b^{2} - 2ab \cos C$ When to use it: SAS, SSS. ## Proof Let $\vartriangle ABC$ be oriented so that $C$ is at the origin, and $B$ is at the point$(a,0)$. ### Distance Formula $distance = \sqrt {(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}$ $c$ is the distance from $A$ to $B$. Substituting the appropriate points into the distance formula gives us $c = \sqrt {(a-b \cos C)^{2} + (0-b \sin C)^{2}}$ Squaring the inner terms, we have $c = \sqrt {(a^{2}-2ab \cos C+b^{2} \cos^{2} C) + (b^{2} \sin^{2} C)}$ Since $\cos^{2} C + \sin^{2} C = 1$, $c = \sqrt {(a^{2}+b^{2}-2ab \cos C+b^{2}}$ Square both sides for $c^{2} = (a^{2}+b^{2}-2ab \cos C+b^{2}$ ## Example Triangulation Complete the triangle using the law of cosines. $c^{2} = a^{2} + b^{2} - 2ab \cos C$ ### Solution To find the side length $c$, $c^{2} = 6^{2} + (6 \sqrt{2})^{2} -2 (6) (6 \sqrt{2}) \cos 45^\circ$ Simplify for $c^{2} =36 + 36 (2) - 72 \sqrt{2}) \cos 45^\circ$ Since $\cos 45^\circ = \frac{1}{\sqrt{2}}$, substitution gives us $c^{2} =36 + 36 (2) - 72 \sqrt{2} (\frac{1}{\sqrt{2}})$ Simplify for $c^{2} =36 + 72 - 72$ $c^{2} =36$ Taking the square root of both sides gives us $c =6$ Now we can orient the triangle differently to get get a new version of the law of cosines so we can find angle measure $B$, $b^{2} = a^{2} + c^{2} - 2ab \cos B$ Substituting in the appropriate side lengths gives us $(6 \sqrt{2})^{2} = 6^{2} + 6^{2} - 2(6)(6) \cos B$ Simplify for $36 (2) = 36 + 36 - 72 \cos B$ $72 = 72 - 72 \cos B$ Subtracting $72$ from both sides gives us $0 = - 72 \cos B$ Dividing both sides by $-72$ gives us $0 = \cos B$ Using inverse trigonometry, we know that $B = 90^\circ$ And we can find the last angle measure $A$ by subtracting the other two measures from $180^\circ$ $180^\circ - 90^\circ - 45^\circ = 45^\circ$ $A=45^\circ$
# RATIONAL FUNCTIONS A rational function is a function of the form: ## Presentation on theme: "RATIONAL FUNCTIONS A rational function is a function of the form:"— Presentation transcript: RATIONAL FUNCTIONS A rational function is a function of the form: where p and q are polynomials What would the domain of a rational function be? We’d need to make sure the denominator  0 Find the domain. If you can’t see it in your head, set the denominator = 0 and factor to find “illegal” values. The graph of looks like this: If you choose x values close to 0, the graph gets close to the asymptote, but never touches it. Since x  0, the graph approaches 0 but never crosses or touches 0. A vertical line drawn at x = 0 is called a vertical asymptote. It is a sketching aid to figure out the graph of a rational function. There will be a vertical asymptote at x values that make the denominator = 0 Let’s consider the graph We recognize this function as the reciprocal function from our “library” of functions. Can you see the vertical asymptote? Let’s see why the graph looks like it does near 0 by putting in some numbers close to 0. The closer to 0 you get for x (from positive direction), the larger the function value will be Try some negatives Does the function have an x intercept? There is NOT a value that you can plug in for x that would make the function = 0. The graph approaches but never crosses the horizontal line y = 0. This is called a horizontal asymptote. A graph will NEVER cross a vertical asymptote because the x value is “illegal” (would make the denominator 0) A graph may cross a horizontal asymptote near the middle of the graph but will approach it when you move to the far right or left vertical translation, moved up 3 Graph This is just the reciprocal function transformed. We can trade the terms places to make it easier to see this. The vertical asymptote remains the same because in either function, x ≠ 0 The horizontal asymptote will move up 3 like the graph does. Finding Asymptotes VERTICAL ASYMPTOTES There will be a vertical asymptote at any “illegal” x value, so anywhere that would make the denominator = 0 So there are vertical asymptotes at x = 4 and x = -1. VERTICAL ASYMPTOTES Let’s set the bottom = 0 and factor and solve to find where the vertical asymptote(s) should be. HORIZONTAL ASYMPTOTES We compare the degrees of the polynomial in the numerator and the polynomial in the denominator to tell us about horizontal asymptotes. 1 < 2 degree of top = 1 If the degree of the numerator is less than the degree of the denominator, (remember degree is the highest power on any x term) the x axis is a horizontal asymptote. If the degree of the numerator is less than the degree of the denominator, the x axis is a horizontal asymptote. This is along the line y = 0. 1 degree of bottom = 2 HORIZONTAL ASYMPTOTES The leading coefficient is the number in front of the highest powered x term. If the degree of the numerator is equal to the degree of the denominator, then there is a horizontal asymptote at: y = leading coefficient of top leading coefficient of bottom degree of top = 2 1 degree of bottom = 2 horizontal asymptote at: OBLIQUE ASYMPTOTES If the degree of the numerator is greater than the degree of the denominator, then there is not a horizontal asymptote, but an oblique one. The equation is found by doing long division and the quotient is the equation of the oblique asymptote ignoring the remainder. degree of top = 3 degree of bottom = 2 Oblique asymptote at y = x + 5 SUMMARY OF HOW TO FIND ASYMPTOTES Vertical Asymptotes are the values that are NOT in the domain. To find them, set the denominator = 0 and solve. To determine horizontal or oblique asymptotes, compare the degrees of the numerator and denominator. If the degree of the top < the bottom, horizontal asymptote along the x axis (y = 0) If the degree of the top = bottom, horizontal asymptote at y = leading coefficient of top over leading coefficient of bottom If the degree of the top > the bottom, oblique asymptote found by long division. Acknowledgement I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint. Shawna has kindly given permission for this resource to be downloaded from and for it to be modified to suit the Western Australian Mathematics Curriculum. Stephen Corcoran Head of Mathematics St Stephen’s School – Carramar Download ppt "RATIONAL FUNCTIONS A rational function is a function of the form:" Similar presentations
# Four dogs ## The Puzzle Four dogs in the corners of a square room ## The Analysis So all dogs start off at the same time, with A chasing B, B chasing C, C chasing D and D chasing A. One possible first thought is that they don't ever meet up, they just chase each other round in circles. However, a quick analysis shows that that's not right. They would only go round in circles if they started off running at a tangent to the centre, but they start off towards the adjacent corner of the square. If you think about components of velocity, there is a part of each dog's motion at a tangent, but also a radial component towards the centre. Therefore the dogs get closer together, you can quickly see this with a practical demonstration. Let's look at dog A in the picture. Starting off in the south-east corner of our square, it starts travelling due North towards dog B. At the same time, B starts moving due West, and because A is chasing B, the path of A must bend from due North round to the left as it goes. B's path, of course, is also bending from West as it follows C, and also bends round to the left. From the symmetry of the problem, you can see that the paths of each of the dogs must be identical, the curves are just rotated. So all the dogs spiral inward towards the centre of the room. Note that we still don't know whether they ever meet or not, but we know that the general form is a spiral, not a circle. We can also see by the symmetry of the problem, that if the dogs do meet, then they all meet together (all the dogs are equivalent, so we couldn't have two dogs meeting in one place and the other two meeting somewhere else, for example). And the meeting point must be equivalent for all four dogs, so it couldn't be in one corner of the room or out towards one of the walls - in fact the only point that's the same for all four dogs is the centre of the room, so if they do meet, they must meet exactly in the centre. ## A simple way of expressing it Instead of calculating the distance between two dogs, let's just concentrate on one dog and look at its distance from the centre of the room. After all, we have seen that if the dogs meet, they must meet in the centre because of the exact four-way symmetry. In this diagram you can see that when A runs directly towards B (which it always does), it is running at 45 degrees to the centre of the room O. So we can effectively ignore B and concentrate on the distance from A to O and how that changes. Just considering the components of the dog's velocity, we can see that it has a component cos(45°) towards the centre O and a component sin(45°) in a tangential direction. This component towards O is what is reducing the distance OA. This means that while the dog is running at an angle of 45 degrees, if it runs a distance d, it will get closer to O by a distance d.cos(45°). Or looking at it another way, in order to get 1 metre closer to O, it actually needs to run for 1/cos(45°) metres. (Numerically, cos(45°) is 1/squareroot(2), so the dog needs to run about 1.41m to get 1m closer to O). All the dogs start at the corners of the square of side L, so each dog is initially L.cos(45°) from O. So in order to reach the centre of the room, each dog must run for 1/cos(45°) times this distance, which is L, the length of the side of the room. Interestingly, this is the initial distance between each of the dogs, so it's the same distance A would have to run if B never moved and A just ran straight towards B! ## Breaking it into steps Let's look at the problem again, breaking it up into several linear steps (rather than the curves we know it really produces). Then we can make more and more linear steps until we approximate the real curve. We'll start by letting each dog run halfway towards where the next dog is, with their eyes closed so they don't know that the next dog is also moving during this time. Then they'll each open their eyes, see where the next dog is now, and run halfway towards that point, again with their eyes closed. It should be easy to see that each time the dogs open their eyes, all four dogs are again arranged symmetrically in a square, but in a square which is rotated from the one before (in this example each one is rotated by 45 degrees), and each square is also smaller than the one before. So each time they run with their eyes closed, they repeat the same pattern as last time, only rotated and smaller. So how many steps do we need in this case before the dog reaches the centre of the room? Well, because each dog is only running halfway towards the next dog each time, we've got the famous Zeno's paradox, they will never meet in a finite number of steps. So does that mean the dogs will never meet? Not necessarily. We can calculate the total distance run by each dog in this example (remembering that it will be a different answer to the real problem because they've got their eyes closed). The first stage is simply L/2, half the length of the wall, and the second step is the same but shrunk by a factor 1/squareroot(2). The third stage is shrunk by a further factor 1/squareroot(2) and so on, so the sum is a geometrical progression with a final total sum of . So, although we know this isn't the exact solution, because they're all running with their eyes closed for most of the time, we can see that the dogs spiral inwards, and do travel a finite distance before meeting at the centre (although they do run an infinite number of steps). Now let's make it a bit closer to the real curve, by only letting the dogs run one third of the way to the next dog before opening their eyes. We can see that the curve is still of the same form, still with an infinite number of steps, and we can calculate the total distance travelled in the same way: . So now the distance is only 1.3 times L instead of 1.7 times L for the halfway case. Now we can generalise it using shorter and shorter steps. Instead of splitting the distances into 2, or 3, we'll split it into n sections, so each dog will run one nth of the distance towards the next dog with its eyes closed. Once we've calculated that, we just increase n until it's so large that we'll have the curve we want. Using the same calculations as before, the first section run by our dog is L/n, and the following section is smaller by a factor given by the triangle of 1/n and (n-1)/n). So the total distance is given by the same geometric progression, as , which we can verify gives the same answers as before for n=2 and 3. Now we let n grow to some very large value, and see what the distance tends towards. As n gets larger, any term in n2 will become overwhelmingly large, so (n-1)2+1 can be very well approximated by (n-1)2, and the denominator becomes (n - (n-1)). So as n tends towards an infinitely large number, and the dogs follow their curved paths, the total distance run by each dog becomes L. ## The paths For a visual representation, we can approximate the paths by taking small steps and calculating the positions on a graph, as shown. As we established originally, the forms are symmetric spirals, tightening up as they get closer to the centre. When the dogs are far apart, the angle subtended by the chased dog's movement is small, so the radius of curvature of the paths is large (the paths only bend a bit). As the dogs get nearer, the movement of the dog being chased has a much greater effect on the direction the chasing dog must run, so the radius of curvature gets smaller, and the paths wind in tighter. ## The form of the spirals In the iterations we did, each square was rotated by a fixed angle and shrunk by a fixed factor at each step. That means that the length of each step decreases exponentially as the angle rotates. Because the radius (the distance from the dog to the centre) is proportional to the step length, that means that as the angle is swept around, the radius decreases exponentially also. This is why these spirals are called "logarithmic spirals" (see the links below). However, this doesn't mean that our dogs approach the centre logarithmically! If our dogs run at a steady speed, the angle is not swept around at a constant rate, it accelerates, keeping the radius decreasing at a linear rate as the dogs run. That's how our dogs can reach the centre and not just asymptotically approach it. Once we know that the radius decreases exponentially with angle, we can use a general formula to describe the curve using polar coordinates. We can say quite generally that r = ae-bθ where r is the radius (the distance from the dog to the centre), θ is the angle turned through (measured in radians), and a and b are unknown (positive) constants. The value a determines the size of the spiral, that is if a is twice as big, the spiral looks the same but twice as big. We can fix the value of a by measuring θ from the initial position, so when θ is zero, the radius r is the initial radius, which we can call r0 - we know that the dog is initially in the corner of the room, so r0 is actually L/squareroot(2). The value b determines how tightly-wound the spiral is, that is whether it loops leisurely round the centre without getting much closer to the centre (for small b), or whether it dives more directly towards the centre (for large b). We can calculate our value of b by looking at the angle between the curve and a radial line from the centre. For our four dogs, they all run at an angle of 45 degrees from the centre, as we saw in the simple expression earlier. This means that dr/dθ is the tan of this angle, which is 1. From our expression of the curve, we can see that dr/dθ is just -br, which tells us that in our case b = 1. Now we can write the form of the curve as r = r0e and can calculate the total path length again using this formula. To do this we note that the path length of a small section is given by the change in radius at that point (dr/dθ) multiplied by squareroot(2). And since dr/dθ is now just -r, the total path length is r0squareroot(2) times the total integral of e from θ=0 to infinity. This integral is of course just 1, so the total path length is r0squareroot(2), which is L. ## Extending the problem Once you've solved the puzzle, it should be straightforward to generalize the answer. Instead of a square room with four dogs, suppose we have a regular polygon with n sides (each still of length L), with n dogs at the corners. How far does each dog travel now? Or ask another question - we know that each dog runs a finite distance before it reaches the centre. But how many revolutions does each dog make, that is how many times does each dog go around the centre? After one revolution, how far is the dog from the centre compared to the distance at the start?
# What type of proportion shows that as one quantity increases the other quantity increases also? ## What type of proportion shows that as one quantity increases the other quantity increases also? Direct Proportion Direct Proportion In simple words, if one quantity increases, the other quantity also increases and vice-versa. For example, if the speed of a car is increased, it covers more distance in a fixed amount of time. In notation, the direct proportion is written as y ∝ x. When one quantity increases the other quantity also increases at the same rate and vice versa? The opposite of inversely proportional is directly proportional. It means when an increase in one quantity brings an increase in the other and vice versa then they are said to be directly proportional. ### When one quantity increases while the other quantity decreases What is an example of it? But when quantities X and Y are inversely proportional to each other or in the inverse proportion, one quantity decreases when the other quantity increases or when one quantity increases the other quantity decreases. It is also known as inverse variation. The ratio of these values varies inversely. When things increase other increases? In general, direct variation suggests that two variables change in the same direction. As one variable increases, the other also increases, and as one decreases, the other also decreases. In contrast, inverse variation suggests that variables change in opposite directions. ## What do we call when the other quantity increases the other quantities increases as well? When two quantities are proportional, it means that as one quantity increases the other will also increase and the ratio of the quantities is the same for all values. When one quantity is increased the other quantity is also increased? Answer: When two quantities X and Y increasetogether or decrease together, they are said to be directly proportional or they are in direct proportion with each other. It is also known as a direct variation. ### Which relations between two quantities States if one quantity increases the other also increases? Directly proportional: as one amount increases, another amount increases at the same rate. When one quantity is increased the other quantity is also increased this proportion is called * 1 point? ## What is a function which occurs between two quantities where one increases when the other increases and decreases when the other decreases? When is one quantity increases the other quantity decreases? When one quantity increases the other quantity increases too. When one quality decreases the other quantity decreases too. The corresponding ratios always remain constant. It is also called a direct variation. Let’s say: X is directly proportional to Y here. ### When do two quantities increase in direct proportion? When two quantities X and Y increase together or decrease together, they are said to be directly proportional or they are in direct proportion with each other. It is also known as a direct variation. The ratio of these values will remain constant. When are quantities x and Y inversely proportional to each other? But when quantities X and Y are inversely proportional to each other or in the inverse proportion, one quantity decreases when the other quantity increases or when one quantity increases the other quantity decreases. It is also known as inverse variation. The ratio of these values varies inversely. 3. ## What happens when two variables are multiplied by the same constant? So, if two variables vary directly and one variable is multiplied by a constant, then the other variable is also multiplied by the same constant. If one variable doubles, the other doubles; if one triples, the other triples; if one is cut in half, so is the other.
```2.6 – Combinations of Functions    We can add, subtract, multiply and divide functions. We can find the composition of one function with another and use them to model and solve real-life problems. Arithmetic Combinations Using () = 2 − 3 and () = 2 − 1:  Sum of functions: ( + )() = () + () o () + () = (2 − 3) + ( 2 − 1) = 2 + 2 − 4 o Add the two functions together and combine like terms  Difference of functions: ( − )() = () − () o () − () = (2 − 3) − ( 2 − 1) = − 2 + 2 − 2 o Subtract the two functions, REMEMBER TO PUT THE SECOND IN PARENTHESIS AND DISTRIBUTE THE NEGATIVE, then combine like terms  Product: ()() = () ∙ () o ()() = (2 − 3)( 2 − 1) = 2 3 − 3 2 − 2 + 3 o FOIL the two functions together  Division:( ) () = o () () = 2−3 2 −1 () () , ≠ ±1 o Write the ratio of the two functions as a fraction o Remember to inspect the domain of each to find the excluded x-values A. () = + & () = √ − B. () = √ − & () = a.) ( + )() a.) ( + )() b.) ( − )() b.) ( − )() c.) ()() c.) ()() d. ( ) () and find the domain FROM A GRAPH: d. ( ) () and find the domain +  Composition of Functions The composition of the function f with the function g is: ( ∘ )() = (()) Which is read f of g of x. The domain of ( ∘ ) is the set of all x in the domain of g such that g(x) is the domain of f. Given () = + 2 & () = 4 − 2 : a.) ( ∘ )() b.) ( ∘ )() c.) ( ∘ )() d.) ( ∘ )(−2) ```
Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Additional Questions Students can Download Maths Chapter 3 Algebra Additional Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations. Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Additional Questions Exercise 3.1 Question 1. Write any three expressions each having 4 terms: Solution: (i) 2x3 – 3x2 + 3xy + 8 (ii) 7x3 + 9y2 – 2xy2 – 6 (iii) 9x2 – 2x + 3xy – 1 Question 2. Identify the co-efficients of the terms of the following expressions (i) 2x – 2y (ii) x + y +3 Solution: (i) 2x – 2y The co-efficient of x in 2x is 2 The co-efficient of y in – 2y is – 2 (ii) x + y + 3 The co-efficient of x is 1 The co-efficient ofy is 1 The constant term is 3 Question 3. Group the like terms together from the following: 6x, 6, -5x, – 5, 1, x, 6y, y, 7y, 16x, 3 Solution: We have 6x, -5x, x, 16x are like terms 6y, y, 7y, are like terms 6, – 5, 1, 3 are like terms Question 4. Give the algebraic expressions for the following cases: (i) One half of the sum of a and b. (ii) Numbers p and q both squared and added Solution: (i) $$\frac{1}{2}$$ (a + b) (ii) p2 + q2 Exercise 3.2 Question 1. If A = 2a2 – 4b – 1 ; B = 5a2 + 3b – 8 and C = 2a2 – 9b + 3 then find the value of A – B + C. Solution: Given A = 2a2 – 4b – 1 ; B = 5a2 + 3b – 8 ; C = 2a2 – 9b + 3 A – B + C = (2a2 – 4b – 1) – (5a2 + 3b – 8) + (2a2 – 9b + 3) = 2a2 – 4b – 1 + (-5a2 – 3b + 8) + 2a2 – 9b + 3 = 2a2 – 4b – 1 – 5a2 – 3b + 8 + 2a2 – 9b + 3 = 2a2 – 5a2 + 2a2 – 4b – 3b – 9b – 1 + 8 + 3 = (2 – 5 + 2) a2 + (-4 – 3 – 9) 6 + (-1 + 8 + 3) = -a2 – 16b + 10 Question 2. How much 2x3 – 2x2 + 3x + 5 is greater than 2x3 + 7x2 – 2x + 7? Solution: The required expression can be obtained as follows. = 2x3 – 2x2 + 3x + 5 – (2x3 + 7x2 – 2x + 7) = 2x3 – 2x2 + 3x + 5 + (-2x3 – 7x2 + 2x – 7) = 2x3– 2x2 + 3x + 5 – 2x3 – 7x2 + 2x – 7 = (2 – 2) x3 + (-2 – 7) x2 + (3 + 2) x + (5 – 7) = 0x3 + (-9x2) + 5x – 2 = -9x2 + 5x – 2 ∴ 2x3 – 2x2 + 3x + 5 is greater than 2x3 + 7x2 – 2x + 7 by -9x2 + 5x – 2 Question 3. What should be added to 2b2 – a2 to get b2 – 2a2 Solution: The required expression is obtained by subtracting 2b2 – a2 from b2 – 2a2 b2 – 2a2 – (2b2 – a2) = b2 – 2a2 + (-2b2 + a2) = b2 – 2a2 – 2b2 + a2 = (1 – 2) b2 + (-2 + 1) a2 = -b2 – a2 So -b2 – a2 must be added Exercise 3.3 Question 1. Length of one side of an equilateral triangle is 3x – 4 units. Find the perimeter. Solution: Equilateral triangle has three sides equal. Perimeter = Sum of three sides = (3x – 4) + (3x – 4) + (3x – 4) = 3x – 4 + 3x – 4 + 3x – 4 = (3 + 3 + 3)x + [(-4) + (-4) + (-4)] = 9x + (-12) = 9x – 12 ∴ Perimeter = 9x – 12 units. Question 2. Find the perimeter of a square whose side is y – 2 units. Solution: Perimeter = (y – 2) + (y – 2) + (y – 2) + (y – 2) = y – 2 + y – 2 + y – 2 + y – 2 = 4y – 8 Perimeter of the square = 4y – 8 units. Question 3. Simplify 3x – 5 – x + 9 if x = 3 Solution: 3x – 5 – x + 9 = 3(3) – 5 – 3 + 9 = 9 – 5 – 3 + 9 = 18 – 8 = 10
<meta http-equiv="refresh" content="1; url=/nojavascript/"> Evaluate Numerical and Variable Expressions Involving Powers \| CK-12 Foundation You are reading an older version of this FlexBook® textbook: CK-12 Middle School Math Concepts - Grade 8 Go to the latest version. # 1.6: Evaluate Numerical and Variable Expressions Involving Powers Difficulty Level: Basic Created by: CK-12 0 0 0 % Best Score Practice Algebra Expressions with Exponents Best Score % Do you know how to evaluate a numerical expression when it has powers in it? Casey is having a difficult time doing exactly that. When Casey arrived home from school he looked at this homework. He immediately noticed this problem. $5^4 + -2^4 + 12$ Casey isn't sure how to evaluate this expression. Do you know how to do it? This Concept will show you how to evaluate numerical expressions involving powers. Then you will be able to help Casey at the end of the Concept. ### Guidance Did you know that you can evaluate numerical and variable expressions involving powers? First, let's identify a numerical and a variable expression. A numerical expression is a group of numbers and operations that represent a quantity, there isn’t an equal sign. A variable expression is a group of numbers, operations and variables that represents a quantity, there isn’t an equal sign. We can combine the order of operations, numerical expressions and variable expressions together with powers. A power is a number with an exponent and a base . An exponent is a little number that shows the number of times a base is multiplied by itself. The base is the regular sized number that is being worked with. Take a minute to write these definitions in your notebook. Now let's apply this information. $4^2 = 16$ What happened here? We can break down this problem to better understand powers and exponents. In the power $4^2$ or four-squared, four is the base and two is the exponent. $4^2$ means four multiplied two times or $4 \times 4$ . Therefore, $4^2$ is sixteen. Let's go back a step and evaluate an expression with a power in it. Take a look. Evaluate $6^3$ First, we have to think about what this means. It means that we take the base, 6 and multiply it by itself three times. $& 6 \times 6 \times 6\\& 36 \times 6 \\& 216$ Here is another one with a negative number in it. Evaluate $-8^2$ To work on this one, we have to work on remembering integer rules. Think back remember that we multiply a negative times a negative to get a positive. $-8 \cdot -8 = 64$ This is called evaluating a power. Let’s look at evaluating powers within expressions. Simplify the expression $6^4 + 2^5 + 12$ . Step 1: Simplify $6^4$ . $6^4 = 6 \times 6 \times 6 \times 6 = 1,296$ Step 2: Simplify $2^5$ . $2^5 = 2 \times 2 \times 2 \times 2 \times 2= 32$ $1,296 + 32 + 12 = 1,340$ We can also evaluate variable expressions by substituting given values into the expressions. Evaluate the expression $4a^2$ when $a = 3$ . Step 1: Substitute 3 for the variable “ $a$ .” $4(3)^2$ Step 2: Simplify the powers. $& 4(3)^2\\& 4(3 \cdot 3)\\& 4(9)$ Step 3: Multiply to solve. $& 4(9)\\& 36$ #### Example A $6^3 + 5^2 + 25$ Solution: $266$ #### Example B $16(12^3)$ Solution: $27,648$ #### Example C $6^2 + 5^3 + 15 - 11$ Solution: $165$ Now let's go back to the dilemma at the beginning of the Concept. Here is the problem that was puzzling to Casey. $5^4 + -2^4 + 12$ First, Casey will need to evaluate the powers. $5^4 = (5)(5)(5)(5) = 625$ $-2^4 = (-2)(-2)(-2)(-2) = 16$ Now we can substitute these values back into the expression. $625 + 16 + 12 = 653$ This is the answer to Casey's problem. ### Vocabulary Numerical Expression a group of numbers and operations used to represent a quantity without an equals sign. Variable Expression a group of numbers, operations and variables used to represent a quantity without an equals sign. Powers the value of a base and an exponent. Base the regular sized number that the exponent works upon. Exponent the little number that tells you how many times to multiply the base by itself. ### Guided Practice Here is one for you to try on your own. Evaluate the expression $5b^4 + 17$ . Let $b=5$ . Solution Step 1: Substitute 5 for “ $b$ .” $5(5)^4 + 17$ Step 2: Simplify the powers. $& 5(5 \cdot 5 \cdot 5 \cdot 5) + 17\\& 5(625) + 17$ Step 3: Multiply then add to solve. $& 5(625) + 17\\& 3,125 + 17 = 3,142$ ### Practice Directions: Evaluate each power. 1. $3^3$ 2. $4^2$ 3. $-2^4$ 4. $-8^2$ 5. $5^3$ 6. $2^6$ 7. $-9^2$ 8. $-2^6$ Directions: Evaluate each numerical expression. 1. $6^2 + 22$ 2. $-3^3 + 18$ 3. $2^3 + 16 - 4$ 4. $-5^2 - 19$ 5. $-7^2 + 52 - 2$ 6. $18 + 9^2 - 3$ 7. $22 - 3^3 + 7$ Directions: Evaluate each variable expression using the given values. 1. $6a + 4^2 - 2$ , when $a = 3$ 2. $a^3 + 14$ , when $a = 6$ 3. $2a^2 - 16$ , when $a = 4$ 4. $5b^3 + 12$ , when $b = -2$ 5. $2x^2 + 52$ , when $x = 4$ Basic Dec 19, 2012 Sep 16, 2014 # Reviews Image Detail Sizes: Medium \| Original MAT.ALG.132.L.3 ShareThis Copy and Paste
0 energy points # Dividing rational expressions Video transcript Divide and express as a simplified rational. State the domain. We start off with this expression. We actually have one rational expression divided by another rational expression. And like we've seen multiple times before, these rational expressions aren't defined when their denominators are equal to 0. So p plus 5 cannot be equal to 0, or if we subtract both sides of this-- we can't call it an equation, but we could call it a not-equation-- by negative 5, if we subtract negative 5 from both sides, you get p cannot be equal to-- these cancel out-- negative 5. That's what that tells us. Over here, we could do the same exercise 4p plus 20 also cannot be equal to 0. If it was, this expression would be undefined. Subtract 20 from both sides. 4p cannot be equal to negative 20. Divide both sides by 4. p cannot be equal to negative 5. So in both situations, p being equal to negative 5 would make either of these rational expressions undefined. So the domain here is the set of all reals such-- or p is equal to the set of all reals such that p does not equal negative 5, or essentially all numbers except for negative 5, all real numbers. We've stated the domain, so now let's actually simplify this expression. When you divide by a fraction or a rational expression, it's the same thing as multiplying by the inverse. Let me just rewrite this thing over here. 2p plus 6 over p plus 5 divided by 10 over 4p plus 20 is the same thing as multiplying by the reciprocal here, multiplying by 4p plus 20 over 10. I changed the division into a multiplication and I flipped this guy right here. Now, this is going to be equal to 2p plus 6 times 4p plus 20 in the numerator. I won't skip too many steps. Let me just write that. 2p plus 6 times 4p plus 20 in the numerator and then p plus 5 times 10 in the denominator. Now, in order to see if we can simplify this, we need to completely factor all of the terms in the numerator and the denominator. In the numerator, 2p plus 6, we can factor out a 2, so the 2p plus 6 we can rewrite it as 2 times p plus 3. Then the 4p plus 20, we can rewrite that. We can factor out a 4 as-- so 4 times p plus 5. Then we have our p plus 5 down there in the denominator. We have this p plus 5. We can just write it down in the denominator. Even 10, we can factor that further into its prime components or into its prime factorization. We can factor 10 into 2 times 5. That's the same thing as 10. Let's see what we can simplify. Of course, this whole time, we have to add the caveat that p cannot equal negative 5. We have to add this restriction on the domain in order for it to be the same expression as the one we started off with. Now, what can we cancel out? We have a 2 divided by a 2. Those cancel out. We have a p plus 5 divided by a p plus 5. We know that p plus 5 isn't going to be equal to 0 because of this constraint, so we can cancel those out. What are we left with? In the numerator, we have 4 times p plus 3, and in the denominator, all we have is that green 5, and we're done! We could right this as 4/5 times p plus 3, or just the way we did it right there. But we don't want to forget that we have to add the constraint p cannot be equal to negative 5, so that this thing is mathematically equivalent to this thing right here.
# Volumes of composite solid shapes It’s not really that different finding volumes of composite shapes compared with finding surface areas. You’ve just to remember to not mix up formulas between the two! A lot of questions will give you two dimensional diagrams of a 3 dimensional shape, and tell you the information about the third dimension. For instance, you might get an overhead diagram of a pool (what it would look like as if you were hovering in the air directly over it): The question might say something like “the pool linearly increases in depth from the shallow end to the deep end”. Now, for you to solve the question, you really need to have a 3-dimensional image in your head of what the pool looks like. We know that the top of the pool is a rectangle 8 m by 4 metres, and we can assume that the bottom of the pool is also a rectangle, since there’s no information in the question to suggest otherwise. Also, we can work out that the shallow end and deep end sides must be rectangles. The shallow end side will be a rectangle 1 metre by 4 metres, the deep end rectangle 2 metres by 4 metres. The hard bits to work out are the sides of the pool that run along the 8 metre length. But if we draw the rest of the diagram first, they’re not so hard to work out. Here’s the rest of the diagram first: We can add in the sides that run along the length of the pool simply by drawing two more lines in our diagram. Then we can label all the dimensions that we know: Now, with a 3D diagram, the question becomes easy to solve. Looking at the diagram, we can see that the long sides of the pool form the bases of a prism. To work out the volume of a prism, all you need to do is work out the area of the base, and multiply that by the ‘height’ or ‘length’ of the prism (distance between the two bases). The base shape is a trapezoid, we can work out its area this way: Now all we need to do is multiply this base area by the ‘length’ or ‘height’ of the prism – the distance between the two bases. In this case, this distance corresponds to the width of the pool – 4 metres:
# Difference of Two Squares In the difference of two squares when the algebraic expression is to be factorized in the form a2 – b2, then the formula a2 – b2 = (a + b) (a – b) is used. Factor by using the formula of difference of two squares: 1. a4 – (b + c)4 Solution: We can express a4 – (b + c)4 as a2 – b2. = [(a)2]2 - [(b + c)2]2 Now we will apply the formula of a2 – b2 = (a + b) (a – b) we get, = [a2 + (b + c)2] [a2 - (b + c)2] = [a2 + b2 + c2 + 2ac] [(a)2 - (b + c)2] Now again, we can express (a)2 - (b + c)2 using the formula of a2 – b2 = (a + b)(a - b) we get, = [a2 + b2 + c2 + 2ac] [a + (b + c)] [a - (b + c)] = [a2 + b2 + c2 + 2ac] [a + b + c] [a – b – c] 2. 4x2 - y2 + 6y - 9. Solution: 4x2 - y2 + 6y - 9 = 4x2 - (y2 - 6y + 9), Rearrange the terms We can write y2 - 6y + 9 as a2 – 2ab + b2. = (2x)2 - [(y)2 - 2(y)(3) + (3)2] Now using the formula a2 – 2ab + b2 = (a – b)2 we get, = (2x)2 - (y - 3)2 Now we will apply the formula of a2 – b2 = (a + b) (a – b) we get, = (2x + y - 3) {2x - (y - 3)}, simplifying = (2x + y - 3) (2x - y + 3). 3. 25a2 - (4x2 - 12xy + 9y2) Solution: 25a2 - (4x2 - 12xy + 9y2) We can write 4x2- 12xy + 9y2 as a2 – 2ab + b2. = (5a)2 - [(2x)2 - 2(2x)(3y) + (3y)2] Now using the formula a2 – 2ab + b2 = (a – b)2 we get, = (5a)2 - (2x - 3y)2 Now we will apply the formula of a2 – b2 = (a + b) (a – b). = [5a + (2x - 3y)] [5a - (2x - 3y)] = (5a + 2x - 3y)(5a - 2x + 3y) Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
# How do you factor -50x^2 + x^4 + 49? Dec 28, 2015 Factor as a quadratic in ${x}^{2}$, then use the difference of squares identity twice to find: $- 50 {x}^{2} + {x}^{4} + 49 = \left(x - 7\right) \left(x + 7\right) \left(x - 1\right) \left(x + 1\right)$ #### Explanation: $- 50 {x}^{2} + {x}^{4} + 49$ $= {\left({x}^{2}\right)}^{2} - 50 \left({x}^{2}\right) + 49$ $= {\left({x}^{2}\right)}^{2} - \left(49 + 1\right) \left({x}^{2}\right) + \left(49 \times 1\right)$ $= \left({x}^{2} - 49\right) \left({x}^{2} - 1\right)$ $= \left({x}^{2} - {7}^{2}\right) \left({x}^{2} - {1}^{2}\right)$ $= \left(x - 7\right) \left(x + 7\right) \left(x - 1\right) \left(x + 1\right)$ using the difference of squares identity: ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$
Future Value Formula And Calculator \| Last update: 30 October 2022 The future value formula helps you calculate the future value of an investment (FV) for a series of regular deposits at a set interest rate (r) for a number of years (t). Using the formula requires that the regular payments are of the same amount each time, with the resulting value incorporating interest compounded over the term. In this article we'll delve into the formulae available and then go through a couple of examples. At the bottom of this article, you'll find an interactive formula, which will allow you to enter figures of your choosing and see how the calculation is made. Should you wish to read it, we also have an article discussing the compound interest formula. Future value of a series formula Formula 1: A = PMT × (((1 + r/n)^(nt) - 1) ÷ (r/n)) The formula above assumes that deposits are made at the end of each period (month, year, etc). Below is a variation for deposits made at the beginning of each period: Alternative formula: A = PMT × (((1 + r/n)^(nt) - 1) ÷ (r/n)) × (1+r/n) Where: A = the future value of the investment, including interest PMT = the payment amount per period r = the annual interest rate (decimal) n = the number of compounds per period t = the number of periods the money is invested for ^ means 'to the power of' Future value formula example 1 An investment is made with deposits of \$100 per month (made at the end of each month) at an interest rate of 5%, compounded monthly (so, 12 compounds per period). The value of the investment after 10 years can be calculated as follows... PMT = 100. r = 5/100 = 0.05 (decimal). n = 12. t = 10. If we plug those figures into formula 1, we get: Total = [ PMT × (((1 + r/n)^nt - 1) ÷ (r/n)) ] Total = [ 100 × (((1 + 0.00416)^(120) - 1) ÷ (0.00416)) ] Total = [ 100 × (0.647009497690848 ÷ 0.00416) ] Total = [ 15528.23 ] So, the investment figure after 10 years will stand at \$15,528.23. Future value formula example 2 An individual decides to invest \$10,000 per year (deposited at the end of each year) at an interest rate of 6%, compounded annually. The value of the investment after 5 years can be calculated as follows... PMT = 10000. r = 6/100 = 0.06 (decimal). n = 1. t = 5. Total = [ PMT × (((1 + r/n)^(nt) - 1) ÷ (r/n)) ] Total = [ 10000 × (((1 + 0.06)^5 - 1) ÷ 0.06) ] Total = [ 10000 × (0.3382255776 ÷ 0.06) ] Total = [ 10000 × 5.63709296 ] Total = [ 56370.9296 ] Our investment balance after 5 years is therefore \$56,370.93. This would be comprised of \$50,000 in investment and \$6,370.93 in interest. Interactive future value formula Use the calculator below to show the formula and resulting calculation for your chosen figures. Note that this calculator requires JavaScript to be enabled in your browser. Also, note that ^ means 'to the power of' PMT × ((( 1 + r/n) ^ (n × t) - 1) ÷ (r/n)) PMT × ((( ^ ) - 1) ÷ ) = A Should you wish to have a visual breakdown of deposits and interest over time, give our compound interest calculator a try.
# An altitude of an equilateral triangle is 30 inches. How do you find the perimeter of the triangle? Mar 4, 2016 The perimeter $= 60 \sqrt{3}$ inches. #### Explanation: The formula for the altitude of an equilateral triangle is: color(blue)(sqrt3/2 xx (side) The altitude of the equilateral triangle is $30$ inches. Let the side of the triangle be denoted as $a$ $\frac{\sqrt{3}}{2} \times a = 30$ $a = 30 \times \frac{2}{\sqrt{3}}$ $a = \frac{60}{\sqrt{3}}$ Rationalising the denominator. a = (60 xx sqrt3)/ (sqrt3 xx sqrt3 a = (60 sqrt3)/ (3 $a = 20 \sqrt{3}$ The side, $a = 20 \sqrt{3}$ inches. The perimeter of an equailateral triangle $= 3 \times a$ (side) $= 3 \times 20 \sqrt{3} = 60 \sqrt{3}$ inches.
# Selina solutions for Concise Mathematics Class 9 ICSE chapter 18 - Statistics [Latest edition] ## Chapter 18: Statistics Exercise 18 (A)Exercise 18 (B) Exercise 18 (A) [Pages 227 - 228] ### Selina solutions for Concise Mathematics Class 9 ICSE Chapter 18 Statistics Exercise 18 (A) [Pages 227 - 228] Exercise 18 (A) \| Q 1.1 \| Page 227 State, the following variable is continuous or discrete: number of children in your class. • Continuous variable. • Discrete variable. Exercise 18 (A) \| Q 1.2 \| Page 227 State, the following variable is continuous or discrete: Distance travelled by car. • Continuous variable. • Discrete variable. Exercise 18 (A) \| Q 1.3 \| Page 227 State, the following variable is continuous or discrete: Sizes of shoes. • Continuous variable. • Discrete variable. Exercise 18 (A) \| Q 1.4 \| Page 227 State, the following variable is continuous or discrete: Time. • Continuous variable. • Discrete variable. Exercise 18 (A) \| Q 1.5 \| Page 227 State, the following variable is continuous or discrete: Number of patients in a hospital. • Continuous variable. • Discrete variable. Exercise 18 (A) \| Q 2 \| Page 227 Given below are the marks obtained by 30 students in an examination: 08 17 33 41 47 23 20 34 09 18 42 14 30 19 29 11 36 48 40 24 22 02 16 21 15 32 47 44 33 01 Taking class intervals 1 - 10, 11 - 20, ....., 41 - 50; make a frequency table for the above distribution. Exercise 18 (A) \| Q 3 \| Page 227 The marks of 24 candidates in the subject mathematics are given below: 45 48 15 23 30 35 40 11 29 0 3 12 48 50 18 30 15 30 11 42 23 2 3 44 The maximum marks are 50. Make a frequency distribution taking class intervals 0 - 10, 10-20, ....... Exercise 18 (A) \| Q 4.1 \| Page 227 Fill in the blank : A quantity which can very from one individual to another is called a ............. Exercise 18 (A) \| Q 4.2 \| Page 227 Fill in the blank : Sizes of shoes are ........... variables. Exercise 18 (A) \| Q 4.3 \| Page 227 Fill in the blank : Daily temperatures is ........... variable. Exercise 18 (A) \| Q 4.4 \| Page 227 Fill in the blank : The range of data 7, 13, 6, 25, 18, 20, 16 is ............ Exercise 18 (A) \| Q 4.5 \| Page 227 Fill in the blank : In the class interval 35 - 46; the lower limit is .......... and the upper limit is ......... Exercise 18 (A) \| Q 4.6 \| Page 227 Fill in the blank : The class mark of class interval 22 - 29 is ........... Exercise 18 (A) \| Q 5 \| Page 228 Find the actual lower class limits, upper-class limits and the mid-values of the classes: 10 - 19, 20 - 29, 30 - 39 and 40 - 49. Exercise 18 (A) \| Q 6 \| Page 228 Find the actual lower and upper-class limits and also the class marks of the classes: 1.1 - 2.0, 2.1 -3.0 and 3.1 - 4.0. Exercise 18 (A) \| Q 7 \| Page 228 Use the table given below to find: (a) The actual class limits of the fourth class. (b) The class boundaries of the sixth class. (c) The class mark of the third class. (d) The upper and lower limits of the fifth class. (e) The size of the third class. Class Interval Frequency 30 - 34 7 35 - 39 10 40 - 44 12 45 - 49 13 50 - 54 8 55 - 59 4 Exercise 18 (A) \| Q 8.1 \| Page 228 Construct a cumulative frequency distribution table from the frequency table given below: Class Interval Frequency 0 -8 9 8 - 16 13 16 - 24 12 24 - 32 7 32 - 40 15 Exercise 18 (A) \| Q 8.2 \| Page 228 Construct a cumulative frequency distribution table from the frequency table given below: Class Interval Frequency 1 - 10 12 11 - 20 18 21 - 30 23 31 - 40 15 41 - 50 10 Exercise 18 (A) \| Q 9.1 \| Page 228 Construct a frequency distribution table from the following cumulative frequency distribution: Class Interval Cumulative Frequency 10 - 19 8 20 - 29 19 30- 39 23 40- 49 30 Exercise 18 (A) \| Q 9.2 \| Page 228 Construct a frequency distribution table from the following cumulative frequency distribution: C.I C.F 5 - 10 18 10 - 15 30 15 - 20 46 20 - 25 73 25 - 30 90 Exercise 18 (A) \| Q 10 \| Page 228 Construct a frequency table from the following data: Marks No. of students less than 10 6 less than 20 15 less than 30 30 less than 40 39 less than 50 53 less than 60 70 Exercise 18 (A) \| Q 11 \| Page 228 Construct the frequency distribution table from the following cumulative frequency table: Ages No. of students Below 4 0 Below 7 85 Below 10 140 Below 13 243 Below 16 300 (i) State the number of students in the age group 10 - 13. (ii) State the age-group which has the least number of students. Exercise 18 (A) \| Q 12 \| Page 228 Fill in the blank in the following table: Class interval Frequency Cumulative Frequency 25 - 34 ...... 15 35 - 44 ...... 28 45 - 54 21 ...... 55 - 64 16 ...... 65 - 74 ...... 73 75 - 84 12 ...... Exercise 18 (A) \| Q 13 \| Page 228 The value of π up to 50 decimal place is 3.14159265358979323846264338327950288419716939937510 (i) Make a frequency distribution table of digits from 0 to 9 after the decimal place. (ii) Which are the most and least occurring digits? Exercise 18 (B) [Page 233] ### Selina solutions for Concise Mathematics Class 9 ICSE Chapter 18 Statistics Exercise 18 (B) [Page 233] Exercise 18 (B) \| Q 1 \| Page 233 Construct a frequency polygon for the following distribution: Class-intervals 0-4 4 - 8 8 - 12 12 - 16 16 - 20 20 - 24 Frequency 4 7 10 15 11 6 Exercise 18 (B) \| Q 2 \| Page 233 Construct a combined histogram and frequency polygon for the following frequency distribution: Class-Intervals 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 Frequency 3 5 6 4 2 Exercise 18 (B) \| Q 3 \| Page 233 Construct a frequency polygon for the following data: Class-Intervals 10 - 14 15 - 19 20 - 24 25 - 29 30 - 34 Frequency 5 8 12 9 4 Exercise 18 (B) \| Q 4 \| Page 233 The daily wages in a factory are distributed as follows: Daily wages (in Rs.) 125 - 175 175 - 225 225 - 275 275 - 325 325 - 375 Number of workers 4 20 22 10 6 Draw a frequency polygon for this distribution. Exercise 18 (B) \| Q 5.1 \| Page 233 Draw frequency polygons for each of the following frequency distribution: (a) using histogram (b) without using histogram C.I 10 - 30 30 - 50 50 - 70 70 - 90 90 - 110 110 - 130 130 - 150 ƒ 4 7 5 9 5 6 4 Exercise 18 (B) \| Q 5.2 \| Page 233 Draw frequency polygons for each of the following frequency distribution: (a) using histogram (b) without using histogram C.I 5 -15 15 -25 25 -35 35 - 45 45-55 55-65 ƒ 8 16 18 14 8 2 ## Chapter 18: Statistics Exercise 18 (A)Exercise 18 (B) ## Selina solutions for Concise Mathematics Class 9 ICSE chapter 18 - Statistics Selina solutions for Concise Mathematics Class 9 ICSE chapter 18 (Statistics) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CISCE Concise Mathematics Class 9 ICSE solutions in a manner that help students grasp basic concepts better and faster. Further, we at Shaalaa.com provide such solutions so that students can prepare for written exams. Selina textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students. Concepts covered in Concise Mathematics Class 9 ICSE chapter 18 Statistics are Variable, Tabulation of Data, Frequency, Frequency Distribution Table, Class Intervals and Class Limits, Cumulative Frequency Table, Graphical Representation of Data, Concepts of Statistics, Frequency Distribution Table, Graphical Representation of Continuous Frequency Distribution. Using Selina Class 9 solutions Statistics exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in Selina Solutions are important questions that can be asked in the final exam. Maximum students of CISCE Class 9 prefer Selina Textbook Solutions to score more in exam. Get the free view of chapter 18 Statistics Class 9 extra questions for Concise Mathematics Class 9 ICSE and can use Shaalaa.com to keep it handy for your exam preparation
Montessori # Montessori Honeybee Math Activity for Preschoolers Math should be fun, especially for young children, right? Absolutely! Math in early childhood Montessori focuses on preparing the mathematical mind for problem-solving, logical and critical thinking. Young children are exposed to shapes, patterns and spatial relationships that lend to more complex mathematical concepts later in life. This Montessori honeybee Math Activity is an original work I created during my Montessori training. For my Montessori training, we have to prepare and execute a unit in the classroom. I chose honeybees because honeybees are an extraordinary little creature and all children should learn of their purpose. Below is one of my original math materials with a honeybee theme. Included are a few free printables. Enjoy! ## Montessori Honeybee Math Activity Purpose: To practice addition and subtraction skills with real objects Age: 4+ ### Montessori Math Activity Presentation: 1. Tell the child: “Remember when you worked with the short bead stair to do addition and subtraction? Today you are going to do addition and subtraction with honeybees.” 2. Take the tray to a work table 3. Identify the objects on the tray: “These are mini honeybees. These pieces represent honeycomb. Honeycomb has six sides and is called a hexagon. These are dice. Each die has six sides.” 5. “Roll the dice. You will see two numbers.” Demonstrate and say: “We rolled the numbers 2 and 4. Choose the bigger number (“4”) and write it in the hexagon on the left. Then, choose the smaller number (“2”) and write it in the middle hexagon. You will write the total in the hexagon on the right. 6. Can you count 2 mini honeybees place them in the middle beehive? Do the same for 4 but place the mini honeybees in the left or first beehive. 7. “The next step we count the total number of honeybees by first taking the two in the left most hexagon, adding them to the four in the next hexagon and moving the total number of honeybees, in this case, six, to the result place after the equals sign.” 8. “Write ‘6’ on the print out.” 9. Offer the child a turn. Control of Error: 1. Teacher Points of Interest: 1. Honeybees and Honeycomb 2. Rolling Dice 3. Writing Numbers Direct Aims: 2. One to One Correspondence Variations: 1. Prepare equations ahead of time 2. Use dice with more than six sides 3. Use dice with the option to roll either addition or subtraction 4. Partner Work ### Check out additional Montessori posts on Honeybees: Have fun! Marnie Marnie Craycroft Marnie hails from Maine where she spent summers buried in sand and winters buried in snow. She is the daughter of a nearly four decade veteran of the public school systems. Teaching has always been a part of her life. She founded Carrots Are Orange in 2010. Carrots Are Orange is a Montessori learning and living website for parents and teachers. Marnie graduated from Wesleyan University in 1999 with a BA in Economics. She spent nearly a decade working in investment management. In 2006, she earned her MA in business from the University of Washington’s Foster School of Business. Marnie moved to the west coast in 1999 and currently lives in Boulder with her husband and three sons. She is Montessori trained. Her work has been featured on Apartment Therapy, Buzzfeed, PBS Kids, BabyCenter, the Melissa & Doug blog, Huffington Post, and WhattoExpect.com. Besides writing, passions include running (usually after her three sons), photography, and outdoor adventures. Share ## Montessori Back to School Gifts for Parents Welcome to the start of a new school year! I have been busy preparing the… Read More 1 week ago ## How to Empower Young Children Positive discipline is a hot topic these days. The approach sounds ideal but how on… Read More 3 weeks ago ## 7 Powerful Benefits of Gardening for Kids Growing food is a powerful way to improve health, spend time outdoors, and enjoy learning… Read More 2 months ago When I think about being happy, purposeful, and sure of oneself, I believe self-esteem is… Read More 2 months ago ## How to Make a Worm Tower – Outdoor Learning for Kids Kids will absolutely love this gardening activity! Make a Worm Tower! You can do this activity… Read More 2 months ago ## How to Present a Montessori Brown Stair Sensorial Lesson The Montessori Brown Stair is work included within the Sensorial sequence. Specifically, it's located within… Read More 2 months ago
# How do you solve \frac {p - 9} { 3} = 2? Mar 9, 2017 See the entire solution process below: #### Explanation: First, multiply each side of the equation by $\textcolor{red}{3}$ to eliminate the fraction while keeping the equation balanced: $\textcolor{red}{3} \times \frac{p - 9}{3} = \textcolor{red}{3} \times 2$ $\cancel{\textcolor{red}{3}} \times \frac{p - 9}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}} = 6$ $p - 9 = 6$ Now, add $\textcolor{red}{9}$ to each side of the equation to solve for $p$ while keeping the equation balanced: $p - 9 + \textcolor{red}{9} = 6 + \textcolor{red}{9}$ $p - 0 = 15$ $p = 15$
# Solve algebra with steps Are you ready to learn how to Solve algebra with steps? Great! Let's get started! Our website can solve math word problems. ## Solving algebra with steps This can help the student to understand the problem and how to Solve algebra with steps. There are a lot of different types of calculators, but the simplest ones are probably the most effective. There are also programs that help you create your own calc sheets and other functions. The best calculators will be able to handle any type of precalculus problem and there will be an option for graphing as well. Most calculators will have a function that allows you to create your own equations, but this can only be done on certain models. But there are some special cases where it can be more complicated. If you're dealing with a number like x or y that's between 0 and 1, it's usually easiest to use the properties of logarithms to solve for x: Assume that \|x\| 1: Subtract log C from both sides: ⌊log C⌋ - ⌊log A⌋ Solve for x on both sides: x = −C / log A The absolute value on the left makes this an easier task than it would be if you didn't take into account whether or not \|x\| 1. Assume that \|x\| > 1: Subtract log C from both sides: ⌊log C⌋ - ⌊log A⌋ Solve for x on both sides: x = −C / log A + 1 The absolute value on the right makes this an easier task than it would be if To solve this equation, we start by first converting the left-hand side to a ratio: Similarly, since the right-hand side is a fraction, we can convert this to a decimal: We then multiply both sides of the equation by 1/10 , and then divide by 10 : Finally, we convert back to the original form of the equation, and solve for x . There are no exact formulas for how to solve logarithmic equations. However, there are some useful tricks and techniques that can be used to help you solve these types of equations. One good way to solve logarithmic equations is to use a table. One easy way to do this is to look at what other logarithmic equations look like. Since logarithms follow an exponential pattern, it is usually possible to find a similar equation on which the base can be found. Another trick is to try doing all comparisons in your head before you write them down. If you have trouble coming up with a number that works for both sides of the equation then try using numbers from previous Math problems can be extremely frustrating. Sometimes, it can feel like your brain is not working properly. This is especially true when you are trying to do math in your head. Using a math sheet can help you by allowing you to write out the problem and then rearranging the numbers until you get the answer. However, this method may not be the best option for everyone. There are some people who find it difficult to write out math problems on paper. If this is the case for you, you might want to consider using an app that can solve math problems for you. One of the most popular apps for solving math problems is called Sudoku Jack. This app allows you to solve sudoku puzzles by simply touching the numbers on your phone screen. The app will then fill in all of the numbers automatically, giving you one solution after another as you go along. This method could be very helpful if you need to practice forming equations or if you struggle with basic arithmetic. That way you can check your answer without having to recalculate the problem on paper first. Many people like this tool because it saves them time—they can just scan their paper and get their answer instantly! The other benefit is that it helps you stay organized, because you can quickly scan all your papers together in one place. Another great thing about a math problem scanner is that you don’t have to buy any special equipment. So it’s perfect for anyone who wants to try it out for the first time! It shows me the solving steps and sometimes it doesn't but then it will show me the correct answer this my work was an algebra and pre-algebra I got a question is the pre-algebra same thing? Octavia Smith Amazing! This is the best app for math homework! Why sit there for 1 hour doing hard math, when the app gets it done in seconds! This tool helped me get stressful math problems done since I was in 5th grade. I'd give this a 100/10 Charleigh Barnes Maths Need Question App to help solve math problems Solve by square roots Complete the square solver with steps Solving the equation for y Math tutor chat
## Two-Digit Subtraction With Regrouping: When Students Ask • Will I always regroup when I subtract? Explain to children that they will not always regroup. To decide whether or not they should regroup, they should look at the two numbers in the ones column. If the number on top in the ones column is less than the number they are subtracting from it, they should regroup. Remind children that if they have to regroup, they must regroup 1 ten as 10 ones. Explain that if the number on top in the ones column is larger than the number they are subtracting from it, they can subtract without regrouping. Also remind them that they will not take 1 ten from the tens column if they do not need to regroup. • How can I subtract two-digit numbers correctly? Explain to children that they first need to decide whether or not they need regroup so they should check the number of ones in the ones column carefully. Remind children that it is necessary to record each step as they regroup. First, they cross out the number of tens and write the new number of tens above the tens column. Then they cross out the number of ones at the top of the ones column and write the new number of ones. Finally, they subtract the ones column, and then they subtract the tens column. • Is regrouping for subtraction different from regrouping for addition? Yes. In addition, children add the ones and regroup the ones to make a ten if there are more than 10 ones. In regrouping for subtraction, they take away the ones. If there are not enough ones to subtract, they have to regroup a ten to make 10 ones and add those ones to the number of ones at the top in the ones column. Further explain that in regrouping for addition, they regrouped ones to make a ten; for subtraction, they take away a ten and regroup it to make 10 ones. • Why do I need to know this? Explain to children that they will be using subtraction all through their lives. Tell children that they will need to know how to regroup in subtraction when they learn division. Point out that one very important use for subtraction will be keeping track of their money in a checkbook. • Why should I know how to estimate a difference? Tell children that it is not always necessary to have an exact answer. When they go to the store, they will want to know approximately how much money they should get in change from their purchase. If they can estimate quickly, they can tell if their change is close to the amount they should have received, or whether they should check the amount more carefully.
# The substitution rule ## A first example • Let's calculate the indefinite integral $$\int x\sqrt{1+x^2}\,\mathrm dx$$. • So far, we could integrate noninteger powers of $$x$$ only using the power rule. • We can vastly increase the number of functions using the substitution rule, which for antiderivatives what the chain rule is for derivatives. • We want to find the antiderivative of $$f(x)=x\sqrt{1+x^2}$$, that is a function $$F(x)$$ so that $$F'(x)=f(x)$$. • When using the substitution rule, we want to change variables from $$x$$ to $$u(x)$$. • That is, we are searching for functions $$F(u)$$ and $$u(x)$$ such that $\frac{\mathrm dF(u(x))}{\mathrm dx}=f(x).$ • Using the chain rule, we see that we need $F'(u(x))u'(x)=f(x)=x\sqrt{1+x^2}$ • Upon inspection, we see that we should have $$u(x)=1+x^2$$, and $$F(u)=ku^{3/2}$$, where $$k$$ is a constant. • To find $$k$$: we need $F'(u(x))u'(x)=\frac32k\sqrt{1+x^2}2x=x\sqrt{1+x^2},$ that is $$k=\frac13$$. • This means that we get $$\int x\sqrt{1+x^2}\,\mathrm dx=F(u(x))=\frac13(1+x^2)^{3/2}$$. ## General formulation and a more streamlined approach • In the previous example, we have searched for the antiderivative at once. • It is possible, to change variables in an indefinite integral, without immediately evaluating it. • For this, consider the following equalities. $\int F'(u(x))u'(x)\,\mathrm dx=F(u(x))+C=F(u)+C=\int F'(u)\,\mathrm du.$ • Substitution rule. Suppose that $$I$$ is an interval, $$u$$ is a function differentiable on $$I$$, and $$f$$ is a function continuous on $$I$$. Then we have the following formula. $\int f(u(x))u'(x)\,\mathrm dx=\int f(u)\,\mathrm du$ • The following is an alternative way of expressing the formula. $\int f(u(x))\frac{\mathrm du}{\mathrm dx}\mathrm dx=\int f(u)\mathrm du.$ • Just as in the case of the chain rule, it looks like as if we were dealing with fractionals with differentials in them, and we could simplify. This is not true, but the similarity can help to remember the formula. • Exercises. 5.5: 2, 6, 10, 12, 18, 20, 24, 26, 30, 34, 36, 42, 44, 46, 48 ## Substitution rule for definite integrals • Consider now a definite integral $$\int_a^bF'(u(x))u'(x)\,\mathrm dx$$. • Since we have $$\int F'(u(x))u'(x)\,\mathrm dx=F(u(x))+C$$, we get $\int_a^bF'(u(x))\,\mathrm dx=F(u(x=b))-F(u(x=a)).$ • But note that as $$\int F'(u)\,\mathrm du=F(u)$$, we also have $\int_u(x=a)^u(x=b)F'(u)\,\mathrm du=F(u=u(x=b))-F(u=u(x=a)).$ • Substitution rule for definite integrals. Suppose that $$u$$ is a differentiable function on an open interval containing $$[a,b]$$ and $$f$$ is a continuous function on $$[a,b]$$. Then we have $\int_a^bf(u(x))u'(x)\,\mathrm dx=\int_{u(x=a)}^{u(x=b)}f(u)\,\mathrm du.$ • That is, we can substitute in the boundaries before evaluating the integrals. • Warning. When you use the substitution rule for definite integrals, never forget to change the boundaries. • Exercises. 5.5: 54, 56, 62, 78, 84
Putting It Together: Equations and Inequalities At the beginning of this module, we explored a few problems associated with the construction of a school. First we had to set up a mathematical model that would accurately reflect the following requirements for the foundation of the building. • The shape must be rectangular. • The length must be 100 ft. longer than the width. • There must be a total of 120,000 sq. ft. in the foundation. Letting $x$ stand for the width of the rectangle, the length must then be $x+100$. The area of any rectangle is equal to the product of its length and width, and since the area must be 120,000 sq. ft., this leads to the equation, $\left(x+100\right)\left(x\right)=120,000$ $x^2+100x=120,000$ $x^2+100x-120,000=0$ The equation is quadratic, that is, degree 2. In this case, the trinomial can be factored. $\left(x-300\right)\left(x+400\right)=0$ This implies either $x=300$, or $x=-400$. However, since negative width does not make much sense (have you ever tried to order a negative 8 foot fence post?), we only use the positive answer, $x=300$. This means the width is 300 ft., and the length is 300 + 100 = 400 ft. A quick check verifies that the dimensions yield the correct square footage: $300\times400=120,000$ Ok, now that we have our foundation planned, let’s move on to the walls. Recall, this project requires at least 1800 concrete blocks. The budget for this material is $2,400, and each block costs$1.20. Furthermore, there is a fee of \$100 for hauling the blocks to the construction site, regardless of how many blocks are ordered. Let $n$ stand for the number of blocks that would be purchased. Then two inequalities can be set up based on the constraints in the problem. $n\ge1800$ $(1.20)n+100\le2400$ What is the maximum number of blocks that you can purchase? Let’s solve the second inequality to find out. $(1.20)n\le2400-100=2300$ $n\le{\Large\frac{2300}{1.20}}\approx1916.7$ There is no use for a fraction of a block, so we might as well say that $n\le1916$. A maximum of 1916 whole blocks can be purchased. Notice that $1916\ge1800$, so the first inequality is also satisfied, meaning that our budget allows for more than enough blocks to complete this phase of the project. With the help of equations and inequalities, as well as specialized techniques for solving them, the plans for the new school are coming together nicely!
# Intersections, Unions, and Compound Inequalities - PowerPoint PPT Presentation Intersections, Unions, and Compound Inequalities 1 / 26 Intersections, Unions, and Compound Inequalities ## Intersections, Unions, and Compound Inequalities - - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - - ##### Presentation Transcript 1. Intersections, Unions, and Compound Inequalities • Two inequalities joined by the word “and” or the word “or” are called compound inequalities. 2. Intersections of Sets and Conjunctions of Sentences A B A The intersection of two set A and B is the set of all elements that are common in both A and B. 3. Example 1 Find the intersection. {1, 2, 3, 4, 5} Solution: The numbers 1, 2, 3, are common to both set, so the intersection is {1, 2, 3} 4. Conjunction of the intersection When two or more sentences are joined by the word and to make a compound sentence, the new sentence is called a conjunction of the intersection. The following is a conjunction of inequalities. A number is a solution of a conjunction if it is a solution of both of the separate parts. The solution set of a conjunction is the intersection of the solution sets of the individual sentences. 5. -7 -7 -7 -2 -2 -2 -1 -1 -1 1 1 1 3 3 3 5 5 5 7 7 7 -6 -6 -6 -5 -5 -5 -4 -4 -4 -3 -3 -3 0 0 0 4 4 4 6 6 6 8 8 8 2 2 2 Example 2 Graph and write interval notation for the conjunction ) ) ) ) 6. Mathematical Use of the Word “and” The word “and” corresponds to “intersection” and to the symbol ““. Any solution of a conjunction must make each part of the conjunction true. 7. Example 3 Graph and write interval notation for the conjunction SOLUTION: This inequality is an abbreviation for the conjunction true Subtracting 5 from both sides of each inequality Dividing both sides of each inequality by 2 8. -7 -7 -7 -2 -2 -2 -1 -1 -1 1 1 1 3 3 3 5 5 5 7 7 7 -6 -6 -6 -5 -5 -5 -4 -4 -4 -3 -3 -3 0 0 0 4 4 4 6 6 6 8 8 8 2 2 2 Example 3 Graph and write interval notation for the conjunction [ ) ) [ 9. The steps in example 3 are often combined as follows Subtracting 5 from all three regions Dividing by 2 in all three regions Caution: The abbreviated form of a conjunction, like -3 can be written only if both inequality symbols point in the same direction. It is not acceptable to write a sentence like -1 > x < 5 since doing so does not indicate if both -1 > x and x < 5 must be true or if it is enough for one of the separate inequalities to be true 10. Example 4 Graph and write interval notation for the conjunction SOLUTION: We first solve each inequality retaining the word and Add 5 to both sides Subtract 2 from both sides Divide both sides by 2 Divide both sides by 5 11. -7 -7 -7 -2 -2 -2 -1 -1 -1 1 1 1 3 3 3 5 5 5 7 7 7 -6 -6 -6 -5 -5 -5 -4 -4 -4 -3 -3 -3 0 0 0 4 4 4 6 6 6 8 8 8 2 2 2 Example 4 Graph and write interval notation for the conjunction [ [ [ 12. Example 5 Sometimes there is no way to solve both parts of a conjunction at once When A A and B are said to be disjoint. B A A 13. Example 5 Graph and write interval notation for the conjunction SOLUTION: We first solve each inequality separately Add 3 to both sides of this inequality Add 1 to both sides of this inequality Divide by 2 Divide by 3 14. -7 -7 -7 -2 -2 -2 -1 -1 -1 1 1 1 3 3 3 5 5 5 7 7 7 -6 -6 -6 -5 -5 -5 -4 -4 -4 -3 -3 -3 0 0 0 4 4 4 6 6 6 8 8 8 2 2 2 Example 5 Graph and write interval notation for the disjunction ) ) ) ) 15. Unions of sets and disjunctions of sentences A B A The unionof two set A and B is the collection of elements that belong to A and / or B. 16. Example 6 Find the union. {2, 3, 4 Solution: The numbers in either or both sets are 2, 3, 4, 5, and 7, so the union is {2, 3, 4, 5, 7} 17. disjunctionsof sentences When two or more sentences are joined by the word orto make a compound sentence, the new sentence is called a disjunctionof the sentences. Here is an example. A number is a solution of a disjunction if it is a solution of at least one of the separate parts. The solution set of a disjunction is the union of the solution sets of the individual sentences. 18. -7 -7 -7 -2 -2 -2 -1 -1 -1 1 1 1 3 3 3 5 5 5 7 7 7 -6 -6 -6 -5 -5 -5 -4 -4 -4 -3 -3 -3 0 0 0 4 4 4 6 6 6 8 8 8 2 2 2 Example 7 Graph and write interval notation for the conjunction ) ) ) ) 19. Mathematical Use of the Word “or” The word “or” corresponds to “union” and to the symbol ““. For a number to be a solution of a disjunction, it must be in at least one of the solution sets of the individual sentences. 20. Example 8 Graph and write interval notation for the conjunction SOLUTION: We first solve each inequality separately Subtract 7 from both sides of inequality Subtract 13 from both sides of inequality Divide both sides by 2 Divide both sides by -5 21. -7 -7 -7 -2 -2 -2 -1 -1 -1 1 1 1 3 3 3 5 5 5 7 7 7 -6 -6 -6 -5 -5 -5 -4 -4 -4 -3 -3 -3 0 0 0 4 4 4 6 6 6 8 8 8 2 2 2 Example 8 Graph and write interval notation for the conjunction ) [ ) [ 22. Caution: A compound inequality like: As in Example 8, cannot be expressed as because to do so would be to day that x is simultaneously less than -4 and greater than or equal to 2. No number is both less than -4 and greater than 2, but many are less than -4 or greater than 2. 23. Example 9 Graph and write interval notation for the conjunction SOLUTION: We first solve each inequality separately Add 5 to both sides of this inequality Add 3 to both sides of this inequality Divide both sides by -2 24. -7 -7 -7 -2 -2 -2 -1 -1 -1 1 1 1 3 3 3 5 5 5 7 7 7 -6 -6 -6 -5 -5 -5 -4 -4 -4 -3 -3 -3 0 0 0 4 4 4 6 6 6 8 8 8 2 2 2 Example 9 Graph and write interval notation for the conjunction ) ) ) ) 25. Example 10 Graph and write interval notation for the conjunction SOLUTION: We first solve each inequality separately Add 11 to both sides of this inequality Subtract 9 from both sides of this inequality Divide both sides by 4 Divide both sides by 3 26. -7 -7 -7 -2 -2 -2 -1 -1 -1 1 1 1 3 3 3 5 5 5 7 7 7 -6 -6 -6 -5 -5 -5 -4 -4 -4 -3 -3 -3 0 0 0 4 4 4 6 6 6 8 8 8 2 2 2 Example 10 Graph and write interval notation for the conjunction ) [
# Fraction Operations Students start to operate on fractions, learning how to add fractions with like denominators and multiply a whole number by any fraction. Math Unit 5 ## Unit Summary In this unit, 4th grade students begin their work with operating with fractions by understanding them as a sum of unit fractions or a product of a whole number and a unit fraction. Students will then add and subtract fractions with like denominators and multiply a whole number by a fraction, including mixed numbers. Students will apply this knowledge to word problems and line plots. In 3rd Grade Math, students developed their understanding of the meaning of fractions, especially using the number line to make sense of fractions as numbers themselves. They also did some rudimentary work with equivalent fractions and comparison of fractions. In Unit 4, 4th grade students deepened this understanding of equivalence and comparison, learning the fundamental property that “multiplying the numerator and denominator of a fraction by the same non-zero whole number results in a fraction that represents the same number as the original fraction” (NF Progression, p. 6). Thus, in this unit, armed with a deep understanding of fractions and their value, students start to operate on them for the first time. The unit is structured so that students build their understanding of fraction operations gradually, first working with the simplest case where the total is a fraction less than 1, then the case where the total is a fraction between 1 and 2 (to understand regrouping when operating in simple cases), and finally the case where the total is a fraction greater than 2. With each of these numerical cases, they first develop an understanding of non-unit fractions as sums and multiples of unit fractions. Next, they learn to add and subtract fractions. And finally, they apply these understandings to complex cases, such as word problems or fraction addition involving fractions where one denominator is a multiple of the other, which helps prepare students for similar work with decimal fractions in Unit 6. After working with all three numerical cases in the context of fraction addition and subtraction, they work with fraction multiplication, learning methods for multiplying a whole number by a fraction and a mixed number and using those skills in the context of word problems. Finally, students apply this unit’s work to the context of line plots. Students will solve problems by using information presented in line plots, requiring them to use their recently acquired skills of fraction addition, subtraction, and even multiplication, creating a contextual way for this supporting cluster content to support the major work of the grade. The unit provides lots of opportunity for students to reason abstractly and quantitatively (MP.2) and construct viable arguments and critique the reasoning of others (MP.3). Students’ understanding of fractions is developed further in Unit 6, in which students explore decimal numbers via their relationship to decimal fractions, expressing a given quantity in both fraction and decimal forms (4.NF.5—7). Then, in 5th grade, students extend their understanding and ability with operations with fractions (5.NF.1—7), working with all cases of fraction addition, subtraction, and multiplication and the simple cases of division of a unit fraction by a whole number or vice versa. Students then develop a comprehensive understanding of and ability to compute fraction division problems in all cases in 6th grade (6.NS.1). Beyond these next few units and years, it is easy to find the application of this learning in nearly any mathematical subject in middle school and high school, from ratios and proportions in the middle grades to functional understanding in algebra. Pacing: 25 instructional days (21 lessons, 2 flex days, 1 assessment day) Fishtank Plus for Math Unlock features to optimize your prep time, plan engaging lessons, and monitor student progress. ## Assessment The following assessments accompany Unit 5. ### Pre-Unit Have students complete the Pre-Unit Assessment and Pre-Unit Student Self-Assessment before starting the unit. Use the Pre-Unit Assessment Analysis Guide to identify gaps in foundational understanding and map out a plan for learning acceleration throughout the unit. ### Mid-Unit Have students complete the Mid-Unit Assessment after lesson 11. ### Post-Unit Use the resources below to assess student understanding of the unit content and action plan for future units. Expanded Assessment Package Use student data to drive instruction with an expanded suite of assessments. Unlock Pre-Unit and Mid-Unit Assessments, and detailed Assessment Analysis Guides to help assess foundational skills, progress with unit content, and help inform your planning. ## Unit Prep ### Intellectual Prep Unit Launch Before you teach this unit, unpack the standards, big ideas, and connections to prior and future content through our guided intellectual preparation process. Each Unit Launch includes a series of short videos, targeted readings, and opportunities for action planning to ensure you're prepared to support every student. #### Intellectual Prep for All Units • Read and annotate “Unit Summary” and “Essential Understandings” portion of the unit plan. • Do all the Target Tasks and annotate them with the “Unit Summary” and “Essential Understandings” in mind. • Take the Post-Unit Assessment. #### Unit-Specific Intellectual Prep number line Example: Use a number line to solve $$1\frac{1}{3}-\frac{2}{3}$$ tape diagram Example: Use a tape diagram to solve $$2\times \frac{3}{10}$$ line plot ### Essential Understandings • “The meaning of addition is the same for both fractions and whole numbers, even though algorithms for calculating their sums can be different. Just as the sum of $$4$$ and $$7$$ can be seen as the length of the segment obtained by joining together two segments of lengths 4 and 7, so the sum of $${{{{2\over3}}}}$$ and $${{{{8\over5}}}}$$ can be seen as the length of the segment obtained joining together two segments of length $${{{{2\over3}}}}$$ and $${{{{8\over5}}}}$$” (Progressions for the Common Core State Standards in Mathematics, 3-5 Numbers and Operations - Fractions, p. 7). • Quantities cannot be added or subtracted if they do not have like units. Just like one cannot add 4 pencils and 3 bananas to have 7 of anything of meaning (unless one changes the unit of both to “objects”), the same applies for the units of fractions (their denominators). This explains why one must find a common denominator to be able to add fractions with unlike denominators when adding and subtracting fractions. Further, when you add or subtract quantities with like units, their units do not change. Just like one adds 5 bananas and 2 bananas and gets 7 bananas, one adds 5 eighths and 2 eighths and gets 7 eighths. • “Converting a mixed number to a fraction should not be viewed as a separate technique to be learned by rote, but simply as a case of fraction addition. Similarly, converting an improper fraction to a mixed number is a matter of decomposing the fraction into a sum of a whole number and a number less than 1” (Progressions for the Common Core State Standards in Mathematics, 3-5 Numbers and Operations - Fractions, p. 8). • “It is possible to over-emphasize the importance of simplifying fractions. There is no mathematical reason why fractions must be written in simplified form, although it may be convenient to do so in some cases” (Progressions for the Common Core State Standards in Mathematics, 3-5 Numbers and Operations - Fractions, p. 6). Thus, students should not be expected to simplify fractions in all cases where it’s possible to do so. ### Vocabulary fraction greater than one mixed number To see all the vocabulary for Unit 5, view our 4th Grade Vocabulary Glossary. ### Materials • Buttons (About 20 per small group) — These should be of various diameters. There are other options to use in place of this material. See Lesson 20 Anchor Task 3 for more information. • Rulers (1 per student) — These should measure to the nearest eighth inch and ideally the 0 inch mark is not flush with the end of the ruler. ## Unit Practice Word Problems and Fluency Activities Help students strengthen their application and fluency skills with daily word problem practice and content-aligned fluency activities. ## Lesson Map Topic A: Building, Adding, and Subtracting Fractions Less Than or Equal to 1 Topic B: Building, Adding, and Subtracting Fractions Less Than 2 Topic C: Building, Adding, and Subtracting Fractions Greater Than or Equal to 2 Topic D: Multiplication of Fractions Topic E: Line Plots ## Common Core Standards Key Major Cluster Supporting Cluster ### Core Standards #### Measurement and Data • 4.MD.B.4 — Make a line plot to display a data set of measurements in fractions of a unit (1/2, 1/4, 1/8). Solve problems involving addition and subtraction of fractions by using information presented in line plots. For example, from a line plot find and interpret the difference in length between the longest and shortest specimens in an insect collection. #### Number and Operationsâ€”Fractions • 4.NF.B.3 — Understand a fraction a/b with a > 1 as a sum of fractions 1/b. • 4.NF.B.3.A — Understand addition and subtraction of fractions as joining and separating parts referring to the same whole. • 4.NF.B.3.B — Decompose a fraction into a sum of fractions with the same denominator in more than one way, recording each decomposition by an equation. Justify decompositions, e.g., by using a visual fraction model. Examples: 3/8 = 1/8 + 1/8 + 1/8 ; 3/8 = 1/8 + 2/8 ; 2 1/8 = 1 + 1 + 1/8 = 8/8 + 8/8 + 1/8. • 4.NF.B.3.C — Add and subtract mixed numbers with like denominators, e.g., by replacing each mixed number with an equivalent fraction, and/or by using properties of operations and the relationship between addition and subtraction. • 4.NF.B.3.D — Solve word problems involving addition and subtraction of fractions referring to the same whole and having like denominators, e.g., by using visual fraction models and equations to represent the problem. • 4.NF.B.4 — Apply and extend previous understandings of multiplication to multiply a fraction by a whole number. • 4.NF.B.4.A — Understand a fraction a/b as a multiple of 1/b. For example, use a visual fraction model to represent 5/4 as the product 5 Ă— (1/4), recording the conclusion by the equation 5/4 = 5 Ă— (1/4). • 4.NF.B.4.B — Understand a multiple of a/b as a multiple of 1/b, and use this understanding to multiply a fraction by a whole number. For example, use a visual fraction model to express 3 Ă— (2/5) as 6 Ă— (1/5), recognizing this product as 6/5. (In general, n Ă— (a/b) = (n Ă— a)/b.) • 4.NF.B.4.C — Solve word problems involving multiplication of a fraction by a whole number, e.g., by using visual fraction models and equations to represent the problem. For example, if each person at a party will eat 3/8 of a pound of roast beef, and there will be 5 people at the party, how many pounds of roast beef will be needed? Between what two whole numbers does your answer lie? • 3.MD.B.4 • 3.NF.A.1 • 3.NF.A.2 • 4.NF.A.1 • 4.NF.A.2 • 4.MD.A.2 • 5.MD.B.2 • 4.NF.C.5 • 5.NF.A.1 • 5.NF.B.4 • 5.NF.B.7 ### Standards for Mathematical Practice • CCSS.MATH.PRACTICE.MP1 — Make sense of problems and persevere in solving them. • CCSS.MATH.PRACTICE.MP2 — Reason abstractly and quantitatively. • CCSS.MATH.PRACTICE.MP3 — Construct viable arguments and critique the reasoning of others. • CCSS.MATH.PRACTICE.MP4 — Model with mathematics. • CCSS.MATH.PRACTICE.MP5 — Use appropriate tools strategically. • CCSS.MATH.PRACTICE.MP6 — Attend to precision. • CCSS.MATH.PRACTICE.MP7 — Look for and make use of structure. • CCSS.MATH.PRACTICE.MP8 — Look for and express regularity in repeated reasoning. Unit 4 Fraction Equivalence and Ordering Unit 6 Decimal Fractions ## Request a Demo See all of the features of Fishtank in action and begin the conversation about adoption. Yes No
<meta http-equiv="refresh" content="1; url=/nojavascript/"> % Progress Progress % What if you had a quadratic equation like $x^2 + 5x + 2$ that you could not easily factor? How could you use its coefficient values to solve it? After completing this Concept, you'll be able to use the quadratic formula to solve equations like this one. ### Watch This For more examples of solving quadratic equations using the quadratic formula, see the Khan Academy video at ### Guidance The Quadratic Formula is probably the most used method for solving quadratic equations. For a quadratic equation in standard form, $ax^2 + bx + c = 0$ , the quadratic formula looks like this: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ This formula is derived by solving a general quadratic equation using the method of completing the square that you learned in the previous section. We start with a general quadratic equation: $ax^2 + bx + c = 0$ Subtract the constant term from both sides: $ax^2 + bx = -c$ Divide by the coefficient of the $x^2$ term: $x^2 + \frac{b}{a} x = - \frac{c}{a}$ Rewrite: $x^2 + 2 \left (\frac{b}{2a} \right ) x = - \frac{c}{a}$ Add the constant $\ \left (\frac{b}{2a} \right )^2$ to both sides: $x^2 + 2 \left (\frac{b}{2a} \right )x + \left (\frac{b}{2a} \right )^2 = - \frac{c}{a} + \frac{b^2}{4a^2}$ Factor the perfect square trinomial: $\left (x + \frac{b}{2a} \right )^2 = - \frac{4ac}{4a^2} + \frac{b^2}{4a^2}$ Simplify: $\left (x + \frac{b}{2a} \right )^2 = \frac{b^2 - 4ac}{4a^2}$ Take the square root of both sides: $x + \frac{b}{2a} = \sqrt{\frac{b^2 - 4ac}{4a^2}} \ \text{and} \ x + \frac{b}{2a} = - \sqrt{\frac{b^2 - 4ac}{4a^2}}$ Simplify: $x + \frac{b}{2a} = \frac{\sqrt{b^2 - 4ac}}{2a} \ \text{and} \ x + \frac{b}{2a} = - \frac{\sqrt{b^2 - 4ac}}{2a}$ $x = - \frac{b}{2a} + \frac{\sqrt{b^2 - 4ac}}{2a} \ \text{and} \ x = - \frac{b}{2a} - \frac{\sqrt{b^2 - 4ac}}{2a}$ $x = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \ \text{and} \ x = \frac{-b - \sqrt{b^2 - 4ac}}{2a}$ This can be written more compactly as $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ . You can see that the familiar formula comes directly from applying the method of completing the square. Applying the method of completing the square to solve quadratic equations can be tedious, so the quadratic formula is a more straightforward way of finding the solutions. Solve Quadratic Equations Using the Quadratic Formula To use the quadratic formula, just plug in the values of $a, b,$ and $c$ . #### Example A Solve the following quadratic equations using the quadratic formula. a) $2x^2 + 3x + 1 = 0$ b) $x^2 - 6x + 5 = 0$ c) $-4x^2 + x + 1 = 0$ Solution Start with the quadratic formula and plug in the values of $a, b$ and $c$ . a) $\text{Quadratic formula:} && x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\\text{Plug in the values} \ a = 2, \ b = 3, \ c = 1 && x &= \frac{-3 \pm \sqrt{(3)^2 - 4(2)(1)}}{2(2)}\\\text{Simplify:} && x &= \frac{-3 \pm \sqrt{9-8}}{4} = \frac{-3 \pm \sqrt{1}}{4}\\\text{Separate the two options:} && x &= \frac{-3 + 1}{4} \ \ \text{and} \ \ x = \frac{-3 - 1}{4}\\\text{Solve:} && x &= \frac{-2}{4} = - \frac{1}{2} \ \text{and} \ x = \frac{-4}{4} = -1$ Answer: $x = - \frac{1}{2}$ and $x = -1$ b) $\text{Quadratic formula:} && x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\\text{Plug in the values} \ a = 1, \ b = -6, \ c = 5 && x &= \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(5)}}{2(1)}\\ \text{Simplify:} && x &=\frac{6 \pm \sqrt{36 - 20}}{2} = \frac{6 \pm \sqrt{16}}{2}\\\text{Separate the two options:} && x &= \frac{6 + 4}{2} \ \text{and} \ x = \frac{6 -4}{2}\\\text{Solve:} && x &= \frac{10}{2} = 5 \ \text{and} \ x = \frac{2}{2} = 1$ Answer: $x = 5$ and $x = 1$ c) $\text{Quadratic formula:} && x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\\text{Plug in the values} \ a = -4, \ b = 1, \ c = 1 && x &= \frac{-1 \pm \sqrt{(1)^2 - 4(-4)(1)}}{2(-4)}\\\text{Simplify:} && x &= \frac{-1 \pm \sqrt{1 + 16}}{-8} = \frac{-1 \pm \sqrt{17}}{-8}\\\text{Separate the two options:} && x &= \frac{-1 + \sqrt{17}}{-8} \ \text{and} \ x = \frac{-1 - \sqrt{17}}{-8}\\\text{Solve:} && x &= -.39 \ \text{and} \ x = .64$ Answer: $x = -.39$ and $x = .64$ Often when we plug the values of the coefficients into the quadratic formula, we end up with a negative number inside the square root. Since the square root of a negative number does not give real answers, we say that the equation has no real solutions. In more advanced math classes, you’ll learn how to work with “complex” (or “imaginary”) solutions to quadratic equations. #### Example B Use the quadratic formula to solve the equation $x^2 + 2x + 7 = 0$ . Solution $\text{Quadratic formula:} && x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\\text{Plug in the values} \ a = 1, \ b = 2, \ c = 7 && x &= \frac{-2 \pm \sqrt{(2)^2 - 4(1)(7)}}{2(1)}\\\text{Simplify:} && x &= \frac{-2 \pm \sqrt{4 - 28}}{2} = \frac{-2 \pm \sqrt{-24}}{2}$ Answer: There are no real solutions. To apply the quadratic formula, we must make sure that the equation is written in standard form. For some problems, that means we have to start by rewriting the equation. Finding the Vertex of a Parabola with the Quadratic Formula Sometimes a formula gives you even more information than you were looking for. For example, the quadratic formula also gives us an easy way to locate the vertex of a parabola. Remember that the quadratic formula tells us the roots or solutions of the equation $ax^2 + bx + c = 0$ . Those roots are $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$ , and we can rewrite that as $x = \frac{-b}{2a} \pm \frac{\sqrt{b^2 - 4ac}} {2a}.$ Recall that the roots are symmetric about the vertex. In the form above, we can see that the roots of a quadratic equation are symmetric around the $x-$ coordinate $\frac{-b}{2a}$ , because they are $\frac{\sqrt{b^2 - 4ac}} {2a}$ units to the left and right (recall the $\pm$ sign) from the vertical line $x = \frac{-b}{2a}$ . #### Example C In the equation $x^2 - 2x - 3 = 0$ , the roots -1 and 3 are both 2 units from the vertical line $x = 1$ , as you can see in the graph below: Watch this video for help with the Examples above. ### Vocabulary • For a quadratic equation in standard form, $ax^2 + bx + c = 0$ , the quadratic formula looks like this: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ • The quadratic formula tells us the roots or solutions of the equation $ax^2 + bx + c = 0$ . Those roots are $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , and we can rewrite that as $x = \frac{-b}{2a} \pm \frac{\sqrt{b^2 - 4ac}} {2a}.$ • The roots are symmetric about the vertex . In the form above, we can see that the roots of a quadratic equation are symmetric around the $x-$ coordinate $\frac{-b}{2a}$ , because they are $\frac{\sqrt{b^2 - 4ac}} {2a}$ units to the left and right (recall the $\pm$ sign) from the vertical line $x = \frac{-b}{2a}$ . ### Guided Practice Solve the following equations using the quadratic formula. a) $x^2 - 6x = 10$ b) $-8x^2 = 5x + 6$ Solution a) $\text{Re-write the equation in standard form:} && x^2 - 6x - 10 &= 0\\\text{Quadratic formula:} && x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\\text{Plug in the values} \ a = 1, \ b = -6, \ c = -10 && x &= \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-10)}}{2(1)}\\\text{Simplify:} && x &= \frac{6 \pm \sqrt{36 + 40}}{2} = \frac{6 \pm \sqrt{76}}{2}\\\text{Separate the two options:} && x &= \frac{6 + \sqrt{76}}{2} \ \text{and} \ x = \frac{6 - \sqrt{76}}{2}\\\text{Solve:} && x &= 7.36 \ \text{and} \ x = -1.36$ Answer: $x = 7.36$ and $x = -1.36$ b) $\text{Re-write the equation in standard form:} && 8x^2+5x+6 &= 0\\\text{Quadratic formula:} && x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\\text{Plug in the values} \ a = 8, \ b = 5, \ c = 6 && x &= \frac{-5 \pm \sqrt{(5)^2 - 4(8)(6)}}{2(8)}\\\text{Simplify:} && x &= \frac{-5 \pm \sqrt{25 - 192}}{16} = \frac{-5 \pm \sqrt{-167}}{16}$ Answer: no real solutions ### Explore More Solve the following quadratic equations using the quadratic formula. 1. $x^2 + 4x - 21 = 0$ 2. $x^2 - 6x = 12$ 3. $3x^2 - \frac{1}{2}x = \frac{3}{8}$ 4. $2x^2 + x - 3 = 0$ 5. $-x^2 - 7x + 12 = 0$ 6. $-3x^2 + 5x = 2$ 7. $4x^2 = x$ 8. $x^2 + 2x + 6 = 0$ 9. $5x^2 -2x + 100 = 0$ 10. $100x^2 +10x + 70 = 0$ ### Vocabulary Language: English Binomial Binomial A binomial is an expression with two terms. The prefix 'bi' means 'two'. Completing the Square Completing the Square Completing the square is a common method for rewriting quadratics. It refers to making a perfect square trinomial by adding the square of 1/2 of the coefficient of the $x$ term. The quadratic formula states that for any quadratic equation in the form $ax^2+bx+c=0$, $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$. Roots Roots The roots of a function are the values of x that make y equal to zero. Square Root Square Root The square root of a term is a value that must be multiplied by itself to equal the specified term. The square root of 9 is 3, since 3 * 3 = 9. Vertex Vertex The vertex of a parabola is the highest or lowest point on the graph of a parabola. The vertex is the maximum point of a parabola that opens downward and the minimum point of a parabola that opens upward.
# Volume of a Sphere: Definition, Formula and Examples A sphere is a geometrical figure whose set of all points in three-dimensional space are at the same distance (the radius) from a given point (the centre), or the result of rotating a circle about one of its diameters. In order to determine the volume of a sphere, one must apply the formula required to calculate it. The sphere is also the one of the most important topics of geometry that is asked in the government recruitment exams. Generally, there are questions asked are related to basic concepts and formulas. To let you make the most of the Mathematics section Here we are providing the definition, properties, formula with examples of a sphere to explain the students on the applications of the same. ### The volume of a sphere: Definition sphere is a perfectly round geometrical object in 3-dimensional space. It can be characterized as the set of all points located equidistance (radius) away from a given point (center). It is perfectly symmetrical and has no edges or vertices. ### Volume of a sphere: Formula Greek philosopher Archimedes discovered The formula for the Volume of a sphere over two thousand years ago. He was the first person to state and prove that the volume of a sphere is exactly two thirds the volume of its circumscribed cylinder, which is the smallest cylinder that can contain the sphere. The spherical object is placed inside a solid container where the radius of the spherical object is equal to the radius of the circular bases of the cylinder. The Diameter of a sphere is equal to the height of the cylinder. Volume of a sphere= 2/3 volume of a Cylinder Volume of a Sphere = 2/3 (πr2h); where r is the radius and h is the height of the cylinder Now we know that height of cylinder= diameter of the sphere Therefore, Volume of a Sphere= 2/3 (πr2.2r); Diameter= 2x Radius Volume of a Sphere= 4/3(πr3) ### Volume of a sphere: Example Q. Radius of a sphere is 11cm. Calculate its volume Solution: We are given the radius of the sphere= 11cm We know that Volume of Sphere= 4/3(πr3) Volume of sphere= 4/3 (3.14×113) Volume of sphere= 5572.45 cm3 ### Click here for more Maths Study Notes You may also like to read: ## Download Upcoming Government Exam Calendar 2021 × Thanks for downloading the guide. For similar guides, free study material, quizzes, videos and job alerts you can download the Adda247 app from play store. Thank You, Your details have been submitted we will get back to you. Join India largest learning distination What You Will get ? • Daily Quizes • Subject-Wise Quizes • Current Affairs • previous year question papers • Doubt Solving session OR Join India largest learning distination What You Will get ? • Daily Quizes • Subject-Wise Quizes • Current Affairs • previous year question papers • Doubt Solving session OR Join India largest learning distination What You Will get ? • Daily Quizes • Subject-Wise Quizes • Current Affairs • previous year question papers • Doubt Solving session Enter the email address associated with your account, and we'll email you an OTP to verify it's you. Join India largest learning distination What You Will get ? • Daily Quizes • Subject-Wise Quizes • Current Affairs • previous year question papers • Doubt Solving session ## Enter OTP Please enter the OTP sent to /6 Did not recive OTP? Resend in 60s Join India largest learning distination What You Will get ? • Daily Quizes • Subject-Wise Quizes • Current Affairs • previous year question papers • Doubt Solving session Join India largest learning distination What You Will get ? • Daily Quizes • Subject-Wise Quizes • Current Affairs • previous year question papers • Doubt Solving session ## Almost there Please enter your phone no. to proceed +91 Join India largest learning distination What You Will get ? • Daily Quizes • Subject-Wise Quizes • Current Affairs • previous year question papers • Doubt Solving session ## Enter OTP Please enter the OTP sent to Edit Number Did not recive OTP? Resend 60 ## By skipping this step you will not recieve any free content avalaible on adda247, also you will miss onto notification and job alerts Are you sure you want to skip this step? ## By skipping this step you will not recieve any free content avalaible on adda247, also you will miss onto notification and job alerts Are you sure you want to skip this step?
# RS Aggarwal Solutions for Class 8 Maths Chapter 15 - Quadrilaterals RS Aggarwal Solutions For Class 8 Maths Chapter 15 Quadrilaterals are provided here. You can download the pdf of RS Aggarwal Solutions for Class 8 Maths Chapter 15 Quadrilaterals from the given links. Class 8 is an important phase of a studentâ€™s life. It is critical to thoroughly understand the concepts taught in Class 8 as they are continued in Class 9 and 10, and also create a foundation. Here we will learn about the quadrilaterals, types of quadrilaterals and properties of quadrilaterals. RS Aggarwal Solutions helps students to get a good score in the examinations, while also providing extensive knowledge about the subject, as Class 8 is a critical stage in their academic career. Therefore, we at BYJUâ€™S provide answers to all questions uniquely and briefly. ## Exercise 15 Page 1. Fill in the blanks: 1. A quadrilateral has â€¦â€¦. Sides 2. A quadrilateral has â€¦â€¦ angles. 3. A quadrilateral has â€¦â€¦. Vertices, no three of which areâ€¦â€¦ 4. A quadrilateral has â€¦â€¦.. diagonals. 5. A diagonal of a quadrilateral is a line segment that joins two â€¦â€¦. Vertices of the quadrilateral. 6. The sum of the angles of a quadrilateral is â€¦â€¦â€¦ Solution: 1. Four Explanation: AB, BC, CD, and DA are the four sides of quadrilateral. 1. Solution: Four Explanation: âˆ A, âˆ B, âˆ C and âˆ D are the four angles of quadrilateral. 1. Solution: Four, collinear Explanation: A, B, C and D are four vertices of quadrilateral. Which has no three of which are collinear, if it is collinear it will becomes line segment 1. Solution: Two Explanation: AC and BD are the two diagonals of quadrilateral. 1. Solution: Opposite Explanation: Diagonals connect opposite vertices like A to c and B to D. 1. Solution: 360o Explanation: If we draw one diagonal to quadrilateral we will get two triangles i.e. â–³ABC and â–³ACD But we know that sum of angles of triangle is 180o Sum of angles of quadrilateral ABCD= â–³ABC + â–³ACD = 180o+180o = 360o 1. How many pairs of adjacent sides are there? Name them. 2. How many pairs of opposite sides are there? Name them. 3. How many pairs of adjacent angles are there? Name them. 4. How many pairs of opposite angles are there? Name them. 5. How many diagonals are there? Name them. Solution: 1. Adjacent sides is nothing but two sides of a quadrilateral having same end i.e. there are four pairs of adjacent sides. They are (AB, BC), (BC, CD), (CD, DA) and (DA, AB) 1. Two sides of a quadrilateral have same end point are called opposite sides. There are two opposite sides. They are There are four adjacent angles. They are (âˆ A, âˆ B), (âˆ B, âˆ C), (âˆ C, âˆ D) and (âˆ D, âˆ A) 1. When angles of a quadrilateral are not adjacent ten it is called as opposite angles. There are two pairs of opposite angles. They are, (âˆ A, âˆ C), (âˆ B, âˆ D) 1. A diagonal is a line segment which joins two opposite vertices. Here there are 2 diagonals are there. They are (AC, BD) 3. Prove that the sum of the angles of a quadrilateral is 360o Solution: If we draw one diagonal to quadrilateral we will get two triangles i.e. â–³ABC and â–³ACD But we know that sum of angles of triangle is 180o Sum of angles of quadrilateral ABCD= sum of angles â–³ABC +sum of angles â–³ACD = 180o+180o = 360o 4. The three angles of a quadrilateral are 76o, 54o and 108o. Find the measure of the fourth angle. Solution: We know that there are four angles in quadrilateral. Also know that sum of the angles of a quadrilateral is 360o Let unknown angle be x Sum of angles of quadrilateral=76o+ 54o +108o+x 360o=238o+x X=122o Therefore the fourth angle is 122o ## RS Aggarwal Solutions for Class 8 Maths Chapter 15- Quadrilaterals Chapter 15, Quadrilaterals, contains one Exercise. RS Aggarwal Solutions given here contains the answers to all the questions present in this exercise. Let us have a look at some of the concepts that are being discussed in this chapter.
# Fraction calculator The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression. ## Result: ### 3/11 * 4/9 = 4/33 ≅ 0.1212121 Spelled result in words is four thirty-thirds. ### How do you solve fractions step by step? 1. Multiple: 3/11 * 4/9 = 3 · 4/11 · 9 = 12/99 = 4 · 3/33 · 3 = 4/33 Multiply both numerators and denominators. Result fraction keep to lowest possible denominator GCD(12, 99) = 3. In the next intermediate step, , cancel by a common factor of 3 gives 4/33. In words - three elevenths multiplied by four ninths = four thirty-thirds. #### Rules for expressions with fractions: Fractions - use the slash “/” between the numerator and denominator, i.e., for five-hundredths, enter 5/100. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part. The slash separates the numerator (number above a fraction line) and denominator (number below). Mixed numerals (mixed fractions or mixed numbers) write as non-zero integer separated by one space and fraction i.e., 1 2/3 (having the same sign). An example of a negative mixed fraction: -5 1/2. Because slash is both signs for fraction line and division, we recommended use colon (:) as the operator of division fractions i.e., 1/2 : 3. Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45. The colon : and slash / is the symbol of division. Can be used to divide mixed numbers 1 2/3 : 4 3/8 or can be used for write complex fractions i.e. 1/2 : 1/3. An asterisk * or × is the symbol for multiplication. Plus + is addition, minus sign - is subtraction and ()[] is mathematical parentheses. The exponentiation/power symbol is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2 #### Examples: subtracting fractions: 2/3 - 1/2 multiplying fractions: 7/8 * 3/9 dividing Fractions: 1/2 : 3/4 exponentiation of fraction: 3/5^3 fractional exponents: 16 ^ 1/2 adding fractions and mixed numbers: 8/5 + 6 2/7 dividing integer and fraction: 5 ÷ 1/2 complex fractions: 5/8 : 2 2/3 decimal to fraction: 0.625 Fraction to Decimal: 1/4 Fraction to Percent: 1/8 % comparing fractions: 1/4 2/3 multiplying a fraction by a whole number: 6 * 3/4 square root of a fraction: sqrt(1/16) reducing or simplifying the fraction (simplification) - dividing the numerator and denominator of a fraction by the same non-zero number - equivalent fraction: 4/22 expression with brackets: 1/3 * (1/2 - 3 3/8) compound fraction: 3/4 of 5/7 fractions multiple: 2/3 of 3/5 divide to find the quotient: 3/5 ÷ 2/3 The calculator follows well-known rules for order of operations. The most common mnemonics for remembering this order of operations are: PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction. BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction. GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction. Be careful, always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) has the same priority and then must evaluate from left to right. ## Fractions in word problems: • Denominator 2 Denominator of a fraction is 5 and numerator is 7. Write the fraction . • In the cafeteria There are 18 students in Jacob’s homeroom. Six students bring their lunch to school. The rest eat lunch in the cafeteria. In simplest form, what fraction of students eat lunch in the cafeteria? • In fractions An ant climbs 2/5 of the pole on the first hour and climbs 1/4 of the pole on the next hour. What part of the pole does the ant climb in two hours? • Fraction to decimal Write the fraction 3/22 as a decimal. 7 is added to the sum of 4/5 and 6/7 • Cupcakes 2 Susi has 25 cupcakes. She gives 4/5. How much does she have left? • One fourth One fourth of an apple pie is left for 2 family members to share equally. What fraction of the original pie will each member get? • A quotient What is the quotient of 3/10 divided by 2/4 as a fraction? • One-third 2 One-third of the people in a barangay petitioned the council to allow them to plant in vacant lots, and another 1/5 of the people petitioned to have a regular garbage collection. What FRACTION of the barangay population made the petition? • New bridge Thanks to the new bridge, the road between A and B has been cut to one third and is now 10km long. How much did the road between A and B measure before? • Slices of pizza Maria ate 1/4 of a pizza. If there were 20 slices of pizza, how many slices did Maria eat? • There 12 There are 42 students in the class and 2/3 of them are girls. How may girls are there in the class? • Frac equation v2 3.80 is two fifths of a quarter sum of money. How much money is there?
# Dividing word problem worksheet with answer Q.1 The product of two numbers is 21\frac{3}{4}.If one of the number is \frac{15}{32},find the other Solution: Let the other number be x If the product of two numbers is 21\frac{3}{4} , then \frac{15}{32} × x= 21\frac{3}{4} \frac{15x}{32} = \frac{87}{4} Multiplying both side by \frac{32}{15} ,we get \frac{15}{32} × \frac{32}{15} × x = \frac{87}{4} ⇒ x =\frac{232}{5} or 46\frac{2}{5} Hence , required number is 46\frac{2}{5} Q.2 By what number should 1\frac{3}{4} be divided to get 1\frac{5}{16}? Solution: Let the required number be x ∴ 1\frac{3}{4} ÷ x = 1\frac{5}{16} \frac{7}{4} × \frac{1}{x} = \frac{21}{16} \frac{7}{4x} = \frac{21}{16} ⇒ 7 × 16 = 21 × 4x ⇒ 112 = 84 x x=\frac{112}{84}=\frac{4}{3} Hence , required number is \frac{4}{3} [Best sample paper for class 7 ] Q.3 If the cost of 7\frac{1}{2} kg apples is 600 Rs, find the cost of 1 kg apples Solution: Cost of 7\frac{1}{2} kg apples = 600 Rs or Cost of \frac{15}{2} kg apples = 600 Rs ∴ Cost of 1 kg apple = 600 ÷ \frac{15}{2} ⇒ Cost of 1 kg apple = 600 × \frac{2}{15} ⇒ Cost of 1 kg apple = 40×2=80 Rs Q.4 A rope of the length 8\frac{1}{3} m is cut into 5 pieces of equal length. Find the length of each piece. Solution: Length of the rope = 8\frac{1}{3} m or \frac{25}{3} If rope cut into 5 equal pieces, then Length of each piece is given by = \frac{25}{3} ÷ 8 ⇒ Length of each piece = \frac{25}{3} × \frac{1}{8} ⇒ Length of each piece = \frac{25}{24} Q.5 The quotient of two numbers is 2 . If the denominator is 7\frac{2}{5}, find the numerator. Solution: Quotient of two numbers =2 If the denominator of the fraction = 7\frac{2}{5} or \frac{37}{5}. We know , \frac{Numerator}{Denominator}= Quotient \frac{numerator}{\frac{37}{5}}= 2 ⇒ Numerator × \frac{5}{37} = 2 ⇒ 5 Numerator = 74 Hence , Numerator = \frac{74}{5}. Check these stuff as well Don’t forget to comment your feedback in the comment section
# Find a point on the curve Question: Find a point on the curve $y=x^{3}-3 x$ where the tangent is parallel to the chord joining $(1,-2)$ and $(2,2)$. Solution: Let (x1y1) be the required point. Slope of the chord $=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{2+2}{2-1}=4$ $y=x^{3}-3 x$ $\Rightarrow \frac{d y}{d x}=3 x^{2}-3 \ldots(1)$ Slope of the tangent $=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=3 \mathrm{x}_{1}^{2}-3$ It is given that the tangent and the chord are parallel. $\therefore$ Slope of the tangent $=$ Slope of the chord $\Rightarrow 3 x_{1}^{2}-3=4$ $\Rightarrow 3 x_{1}^{2}=7$ $\Rightarrow x_{1}^{2}=\frac{7}{3}$ $\Rightarrow x_{1}=\pm \sqrt{\frac{7}{3}}=\sqrt{\frac{7}{3}}$ or $-\sqrt{\frac{7}{3}}$ Case 1 When $x_{1}=\sqrt{\frac{7}{3}}$ On substituting this in eq. (1), we get $y_{1}=\left(\sqrt{\frac{7}{3}}\right)^{3}-3\left(\sqrt{\frac{7}{3}}\right)=\frac{7}{3} \sqrt{\frac{7}{3}}-3 \sqrt{\frac{7}{3}}=\frac{-2}{3} \sqrt{\frac{7}{3}}$ $\therefore\left(x_{1}, y_{1}\right)=\left(\sqrt{\frac{7}{3}}, \frac{-2}{3} \sqrt{\frac{7}{3}}\right)$ Case 2 When $x_{1}=-\sqrt{\frac{7}{3}}$ On substituting this in eq. (1), we get $y_{1}=\left(-\sqrt{\frac{7}{3}}\right)^{3}-3\left(-\sqrt{\frac{7}{3}}\right)=\frac{-7}{3} \sqrt{\frac{7}{3}}+3 \sqrt{\frac{7}{3}}=\frac{2}{3} \sqrt{\frac{7}{3}}$ $\therefore\left(x_{1}, y_{1}\right)=\left(-\sqrt{\frac{7}{3}}, \frac{2}{3} \sqrt{\frac{7}{3}}\right)$
# 78 as a Decimal When we talk about decimal numbers, it is important to understand the base-10 system. This system is based off of powers of 10, meaning that each place value is 10 times greater than the place value to its right. For example, the number 78 in this system would be written as 7 × 101 + 8 × 100, where the first number is the number of tens and the second number is the number of ones. This is how 78 is written in the base-10 system, which is also known as the decimal system. When we convert 78 to its decimal form, we use a simple process. First, we look at the number of tens and ones. In this case, there are 7 tens and 8 ones. Then, we multiply the number of tens by 10 and the number of ones by 1. So, 7 × 10 + 8 × 1 = 78. This is how 78 is written in decimal form. When written in decimal form, 78 is a very important number. It is the smallest 3-digit number that is divisible by both 2 and 3. It is also the smallest prime number that is divisible by 4. In addition to this, 78 is the smallest number that is divisible by all of its digits. As such, it is important to understand how to convert numbers to decimal form. Converting 78 to decimal form is also important for other operations. For example, when adding or subtracting two numbers, it is important to convert them to decimal form. This ensures that the numbers are written in the same form, which makes it easier to do the calculations. In addition, when multiplying or dividing two numbers, it is important to convert them to decimal form in order to simplify the operations. This is why understanding how to convert numbers to decimal form is important. In conclusion, 78 is written as 78 in decimal form. This number is important because it is the smallest 3-digit number that is divisible by both 2 and 3. It is also the smallest prime number that is divisible by 4. In addition to this, 78 is the smallest number that is divisible by all of its digits. As such, understanding how to convert numbers to decimal form is important for various operations. ## 78 is an important number that is written as 78 in decimal form. This number is important because it is the smallest 3-digit number that is divisible by both 2 and 3. It is also the smallest prime number that is divisible by 4. In addition to this, 78 is the smallest number that is divisible by all of its digits. As such, understanding how to convert numbers to decimal form is important for various operations.
1. Home 2. \| 3. MBA\| # Learn Number System in Quantitative Aptitude Jul 4, 2014 17:20 IST We are talking about a very fundamental and basic topic here in this article, Number System. Quantitative Aptitude begins with the concept of Number System. Let us understand the most important points in Number System first. What all you should know? You should know the following things to get a hold on Number System: 1. Types of Numbers 2. What are Prime numbers 3. Properties Of Numbers 4. Properties of Integers 5. Decimals and Fractions 6. Exponents 7. Divisibility Rules 8. Factorization 9. LCM and HCF 10. Remainder Theorem Introduction: The number system that we use is the decimal number system that has ten numbers from 0 to 9. It can be represented in a number line as shown below. Types of numbers: There are many kinds of numbers each having its own properties and hence different from others. The following are the different kinds. • Complex Numbers - Real and Imaginary Numbers • Real Numbers - Rational and Irrational numbers (also Decimals) • Rational numbers - Integers and Fractions • Integers - Even and Odd (Negative & Positive) and Whole numbers • Whole Numbers - Zero and Natural Numbers • Natural Numbers - Even & Odd Numbers (and also Prime & Composite Numbers) Fractions: A fraction has two parts namely a numerator and a denominator to denote the parts of a whole number. Points to remember: - 1 is neither a prime nor a composite number but an odd number not even. - 2 is the lowest prime number and the only even prime number - 3 is the lowest odd prime number - When a prime number greater than or equal to 5 (>5) is divided by 6, it gives a remainder of either 1 or 5. But any number greater than or equal to 5 (>5) when divided by 6, giving remainder of either 1 or 5 is not necessarily prime. - All prime numbers can be represented by the form (6x + 1) or (6x - 1). Any number that can be represented in the form (6x + 1) or (6x - 1) is not necessarily a prime number. Remainder Theorem: If the product of numbers (P X Q X R X S) is divided by N, then the remainder will be equal to that of the product of the individual remainders when divided by the same dividend. Few Divisibility Rules: All whole numbers are divisible by 1. All even numbers are divisible by 2. A number is divisible by 5 if it ends in 0 or 5 and is a non-zero number. In order to check the divisibility of a number by a composite number, divide the composite divisor into prime factors and then check for its divisibility with each. For example, to check the divisibility of a number with 12, break down 12 into 3 and 4. Check what have you learnt here “Jagran Josh also helps to get information about Top MBA Institutes based upon your choice of Location and Specialization. Click here to get details & the Best Seller “MBA Pedia” eBook, absolutely FREE. DISCLAIMER: JPL and its affiliates shall have no liability for any views, thoughts and comments expressed on this article. ## Register to get FREE updates All Fields Mandatory • (Ex:9123456789)
# Given positive integers $x , y$ and that $3x^2 + x = 4y^2 +y$ Prove that $x-y$ is a perfect square This is some old Olympiad problem (I vaguely remember its Iranian ) I factorised it as $(x-y)(3x + 3y +1) = y^2$ Then I fruitlessly tried to prove $\gcd(x-y, 3x + 3y +1 ) = \gcd (6x+1,6y+1) = 1$ for proving that $x-y$ is a perfect square. How do I continue my method or are there other ways to approach this intuitively? Defining $z = x-y$, from $y^2 = (x-y)(3x + 3y +1)$, we have $$y^2 = 3z^2 + 6yz+ z$$ which is equivalent to $$(y - 3z)^2 = z(12z+1).$$ Since $\gcd(z, 12z+1)=1$, $z$ and $12z+1$ both must be perfect squares. Here is the method to approach this sort of problem. Note that $x>y$. Let $\gcd(x,y)=d$. Now, first note that $$x-y=4y^2-3x^2 \equiv 0 \pmod{d^2} \tag{1}$$ Now, $$4y^2-3x^2=4(y-x)(y+x)+x^2 \equiv 0 \pmod {x-y}$$ And $$4y^2-3x^2=3(y-x)(y+x)+y^2 \equiv 0 \pmod {x-y}$$ This gives us that $x-y$ is a common factor of $x^2$ and $y^2$, so it must divide their greatest common divisor, $d^2$. So we have $$d^2 \equiv 0 \pmod{x-y}$$ So $d^2$ and $x-y$ both divide each other. Since $x>y$, we have that $x-y=d^2$. Our proof is done.
# Find all the factors of 120 by a simple method Learn the process of finding factors of 120. As the given number 120 is a somewhat big number so it is difficult to write or to find all the factors of 120. Before finding the factors of 120 first we will know about the meaning of a factor. Do you know the meaning of a factor? If you don’t know, let’s see what is a factor. A factor is a number which divides the given number exactly. For example, 2 is a factor of 4 because 2 divides 4 exactly. # Method to find the number of factors of 120: Write the given number 120 as the product of prime numbers. 120 = 23 × 31 × 51 Total number of factors of the number 120 = (3+1) × ( 1 + 1) × ( 1 + 1) = 4 × 2 × 2 = 16 So 120 has a total of 16 factors. Now we need to find what are that 16 factors of the number 120. ## Method to find all the factors of 120 without missing any factor: First, write 120 as a product of 1 and 120 because 1 divides any number. After 1 we have number 2. Check whether 120 is divisible or not. As 120 is divisible by 2 so write 20 as a product of 2 and 60. Next take 3, in third table 120 goes 40 times so write 120 as a product of 3 and 40. Now we have 4, in fourth table 120 goes 30 times so now we can write 120 as a product of 4 and 30. After 4 we have 5, in 5th table 120 goes 24 times. So write 120 as a product of 24 and the number 5. Now take 6 and 120 goes 20 times in 6th table. Write 120 as a product of 20 and the number 6. 7 can’t divide 120 exactly so leave it. Take 8 and in 8th table 120 goes 15 times so write 120 as a product of 8 and 15. Next, we have 9 as it can’t divide 120 exactly leave that and take next number. Next 10 as we know 120 goes 12 times so write 120 as a product of 10 and 12. Finally, we can write as shown in below table. Therefore the factors of 120 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120 Please follow and like us: ### 2 Leave a Reply 1 Comment threads 1 Thread replies 0 Followers Most reacted comment Hottest comment thread 2 Comment authors Recent comment authors Subscribe Notify of Guest i need it so much
# 180 Days of Math for Sixth Grade Day 175 Answers Key By accessing our 180 Days of Math for Sixth Grade Answers Key Day 175 regularly, students can get better problem-solving skills. ## 180 Days of Math for Sixth Grade Answers Key Day 175 Directions: Solve each problem. Question 1. Answer: Explanation: Perform subtraction operation on above two given numbers. Subtract 58 from 137 the difference is 79. Question 2. 16 × 50 = ___________ Answer: 16 × 50 = 800 Explanation: Perform multiplication operation on above two given numbers. Multiply 16 with 50 the product is 800. Question 3. Divide 723 by 9. Answer: 723/9 = 80.33 Explanation: Perform division operation on above two given numbers. Divide 723 by 9 the quotient is 80.33. Question 4. What is the last odd number before 300,000? Answer: The last odd number before 300,000 is 299,999. Question 5. $$\frac{5}{6}$$ – $$\frac{1}{3}$$ = ____________ Answer: $$\frac{5}{6}$$ – $$\frac{1}{3}$$ = (5 – 2)/6 = 3/6 or 1/2 Explanation: Perform subtraction operation on above two given numbers. Subtract 1/3 from 5/6 the difference is 3/6 or 1/2. Question 6. Write the next number in the sequence. 160, 240, 320, ___________ Answer: The given sequence is 160, 240, 320, The difference between first two numbers in the sequence is 80. 240 – 160 = 80 The difference between next two numbers in the sequence is 80. 320 – 240 = 80 So add 80 to the next number in the sequence number 320 the sum is 400. 320 + 80 = 400 The next number in the sequence is 160, 240, 320, 400. Question 7. Write the expression for seven more than x squared. Answer: The expression for seven more than x squared is x² + 7. Question 8. Find y. 18y = 72 y = ___________ Answer: 18y = 72 y = 72/18 y = 4 Explanation: To calculate y value we have to perform division operation on given equation. Divide 72 by 18 the quotient is 4. Question 9. 126 inches = ___________ yards Answer: 1 yard = 36 inches 1 inch = 1/36 yards To convert inches to yards we have to multiply 126 with 1/36 yards. 126 x 1/36 yards= 3.5 yards So, 126 inches = 3.5 yards Question 10. How many total degrees are in three right angles? Answer: A right angle has 90 degrees. Here we need to calculate for three right angles. Add 90 degrees as three times the sum is 270 degrees. 90° + 90° + 90° = 270° The total degrees in three right angles are 270°. Question 11. What is the mode of this set of data? 2,216; 2,443; 2,341; 2,443; 2,401 Answer: Mode is defined as the number that occurs the most in a given set of numbers. We can have more than one mode or no mode at all. In the given series 2,216; 2,443; 2,341; 2,443; 2,401 The number 2,216 is 1. The number 2,443 are 2. The number 2,341 is 1. The number 2,401 is 1. The mode for the given set of data is 2,443. Question 12. The deep end of a swimming pool is 7 meters deep. If the shallow end is half as deep, how many centimeters deep is the shallow end? Answer: Given, The deep end of a swimming pool is 7 meters deep. 7 ÷ 2 = 3.5 meters The shallow end is 3.5 meters deep. Scroll to Top
Week 1 – Precalc 11 This week in math we learned about arithmetic sequences, to be specific how to find a specific term in a sequence using an equation. To begin, a sequence is a set of numbers. For a sequence to be arithmetic, it has to go up by the same number each time, or in other words, have a common difference. Say we have a sequence: 2, 4, 6, 8, 10, 12. The common difference is 2. If we want to find the 100th term, we need to use the equation: $t_n$ = $t_1$+(n-1)d. We are trying to find the 100th term, so $t_n$=$t_{100}$. Next, $t_1$ is the first term in the sequence, so it’s $t_1$=2. And we already know what n is, 100. Now, we plug into these numbers into the equation. $t_{100}$=2+(100-1)(2). You can now try to solve this and will find what the 100th term in the sequence is. To check your answer: the 100th term in the sequence is 200, which is found by solving the equation. You can also use the equation to find different things, just plug all the numbers you have into the equation and as long as there is only one unknown variable, you can find the answer. Some important notes to take from this: It is essential to have a common difference in a sequence of numbers for it to be arithmetic, if there is no common difference this equation won’t work. The equation is $t_n$=$t_1$+(n-1)d, this equation is very important! Answer it step by step if needed too, it can help you organize your thoughts if you find this confusing. I’ve included a picture example below of another question this equation can be used to answer.
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # HL Triangle Congruence ## Hypotenuse-leg pairs show right triangle congruence. Estimated6 minsto complete % Progress Practice HL Triangle Congruence MEMORY METER This indicates how strong in your memory this concept is Progress Estimated6 minsto complete % HL Triangle Congruence ### Hypotenuse-Leg Congruence Theorem If the hypotenuse and leg in one right triangle are congruent to the hypotenuse and leg in another right triangle, then the two triangles are congruent. This is called the Hypotenuse-Leg (HL) Congruence Theorem. Note that it will only work for right triangles. If ABC\begin{align}\triangle ABC\end{align} and XYZ\begin{align}\triangle XYZ\end{align} are both right triangles and AB¯¯¯¯¯¯¯¯XY¯¯¯¯¯¯¯¯\begin{align}\overline{AB} \cong \overline{XY}\end{align} and BC¯¯¯¯¯¯¯¯YZ¯¯¯¯¯¯¯\begin{align}\overline{BC} \cong \overline{YZ}\end{align} then \begin{align}\triangle ABC \cong \triangle XYZ\end{align}. What if you were given two right triangles and provided with only the measure of their hypotenuses and one of their legs? How could you determine if the two right triangles were congruent? ### Examples #### Example 1 Fill in the blanks in the proof below. Given: \begin{align}\overline{SV} \perp \overline{WU}\end{align} \begin{align}T\end{align} is the midpoint of \begin{align}\overline{SV}\end{align} and \begin{align}\overline{WU}\end{align} Prove: \begin{align}\overline{WS} \cong \overline{UV}\end{align} Statement Reason 1. 1. 2. \begin{align}\angle STW\end{align} and \begin{align}\angle UTV\end{align} are right angles 2. 3. 3. 4. \begin{align}\overline{ST} \cong \overline{TV}, \ \overline{WT} \cong \overline{TU}\end{align} 4. 5. \begin{align}\triangle STW \cong \triangle UTV\end{align} 5. 6. \begin{align}\overline{WS} \cong \overline{UV}\end{align} 6. Statement Reason 1. \begin{align}\overline{SV} \perp \overline{WU}\end{align} 1. Given 2. \begin{align}\angle STW\end{align} and \begin{align}\angle UTV\end{align} are right angles 2. Definition of perpendicular lines. 3. \begin{align}T\end{align} is the midpoint of \begin{align}\overline{SV}\end{align} and \begin{align}\overline{WU}\end{align} 3. Given 4. \begin{align}\overline{ST} \cong \overline{TV}, \ \overline{WT} \cong \overline{TU}\end{align} 4. Definition of midpoint 5. \begin{align}\triangle STW \cong \triangle UTV\end{align} 5. SAS 6. \begin{align}\overline{WS} \cong \overline{UV}\end{align} 6. CPCTC Note that even though these were right triangles, we did not use the HL congruence shortcut because we were not originally given that the two hypotenuses were congruent. The SAS congruence shortcut was quicker in this case. #### Example 2 Explain why the HL Congruence shortcut works. The Pythagorean Theorem, which says, for any right triangle, this equation is true: \begin{align}(leg)^2 + (leg)^2 = (hypotenuse)^2\end{align} What this means is that if you are given two sides of a right triangle, you can always find the third. Therefore, if you know two sides of a right triangle are congruent to two sides of another right triangle, then you can conclude that the third sides are also congruent. If three pairs of sides are congruent, then we know the triangles are congruent by SSS. #### Example 3 What additional information would you need to prove that these two triangles were congruent using the HL Theorem? For HL, you need the hypotenuses to be congruent. \begin{align}\overline{AC} \cong \overline{MN}\end{align}. #### Example 4 Determine if the triangles are congruent. If they are, write the congruence statement and which congruence postulate or theorem you used. We know the two triangles are right triangles. The have one pair of legs that is congruent and their hypotenuses are congruent. This means that \begin{align}\triangle ABC \cong \triangle RQP\end{align} by HL. #### Example 5 Determine the additional piece of information needed to show the two triangles are congruent by HL. We already know one pair of legs is congruent and that they are right triangles. The additional piece of information we need is that the two hypotenuses are congruent, \begin{align}\overline{UT} \cong \overline{FG} \end{align}. ### Review Using the HL Theorem, what additional information do you need to prove the two triangles are congruent? The triangles are formed by two parallel lines cut by a perpendicular transversal. \begin{align}C\end{align} is the midpoint of \begin{align}\overline{AD}\end{align}. Complete the proof to show the two triangles are congruent. Questions 4-7 are within the proof. Statement Reason 1. \begin{align}\angle ACB\end{align} and \begin{align}\angle DCE\end{align} are right angles. (4.) 2. (5.) Definition of midpoint 3. (6.) Given 4. \begin{align}\Delta ACD \cong \Delta DCE\end{align} (7.) Based on the following details, are the two right triangles definitively congruent? The statements do not build off of each other. 1. The hypotenuses of two right triangles are congruent. 2. Both sets of legs in the two right triangles are congruent. 3. One set of legs are congruent in the two right triangles. 4. The hypotenuses and one pair of legs are congruent in the two right triangles. 5. One of the non right angles of the two right triangles is congruent. 6. All of the angles of the two right triangles are congruent. 7. All of the sides of the two right triangles are congruent. 8. Both triangles have one leg that is twice the length of the other. To see the Review answers, open this PDF file and look for section 4.9. ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English Spanish hypotenuse The side opposite the right angle in a right triangle. legs The two sides adjacent to the right angle. right triangle A triangle with exactly one right ($90^\circ$) angle. The two sides adjacent to the right angle are called legs and the side opposite the right angle is called the hypotenuse. H-L (Hypotenuse-Leg) Congruence Theorem If the hypotenuse and leg in one right triangle are congruent to the hypotenuse and leg in another right triangle, then the two triangles are congruent.
kolutastmr 2022-07-02 Try to visualize process of multiplication fraction addition is obvious, need to split each part to the same size - "reduce to a common denominator" for example $\frac{2}{3}+\frac{2}{4}=\frac{8}{12}+\frac{6}{12}$ Multiplication rule numerator * numerator, denominator * denominator - how to visualize this rule like addition way which I explained above ? Thanks. Kiana Cantu Expert Let us say you have 2 fractions you need to multiply, $\frac{2}{7}$ and $\frac{4}{5}$ for example. Then the multiplication rule you explained above means that: $\frac{2}{7}×\frac{4}{5}=\frac{2×4}{7×5}=\frac{8}{35}$ In general: $\frac{a}{b}×\frac{c}{d}=\frac{ac}{bd}$ Now, to explain why this works, you need to consider what a fraction is or means. $\frac{2}{7}$ actually means $2÷7$. So above, when I did $\frac{2}{7}×\frac{4}{5}$, I was actually doing the same thing as $\left(2÷7\right)×\left(4÷5\right)$ Now: visualise some cakes (or pizzas or pies). Let's say I have 3 cakes. If I multiply the number of cakes I have by 5, then I take 5 sets of 3 cakes to make 15 cakes. If I divide the number of cakes I have by 6, I am trying to split up the cakes into 6 groups of equal size. So I will end up with 6 halves. Now notice what happens if I first multiply my 3 cakes by 5 and then divide by 6. I get the same answer as first dividing by 6 and then multiplying by 5. So order does not matter when it comes to multiplication and division (Note you still have to pay attention to brackets though). Therefore, we could rearrange what we had above: $\frac{2}{7}×\frac{4}{5}=\left(2÷7\right)×\left(4÷5\right)=2÷7×4÷5=2×4÷7÷5=\left(2×4\right)÷\left(7×5\right)=\frac{2×4}{7×5}$ Another helpful way of understanding what I said above is to think about the analogy between addition/subtraction and multiplication/division. What I said above is still true if I replace all of the + and − with × and ÷ respectively (ignoring all of the fractions, as there isn't a "fraction" sign for subtraction). The way in which these 2 pairs are related are very similar. Show me why this rule works: $\left(5-3\right)+\left(4-9\right)=\left(5+4\right)-\left(3+9\right)$ $\left(a-b\right)+\left(c-d\right)=\left(a+c\right)-\left(b+d\right)$ It is because you can change around the order of things without changing the meaning. In mathematics, this means that an operation is "commutative". Yet another way of understanding this is to consider the idea of "inverses". An inverse operation undoes an operation. So for example, if I add 5 to a number, the inverse operation to this would be subtracting 5 from the new number, as this gets me back to my original number. Subtraction is the inverse operation of addition. Division is the inverse operation of multiplication. With inverses, you can change around the order of which you do things. This is actually related to basic Group Theory. Group theory replaces the operations $+$ $-$ $×$ $÷$ with other symbols like $\ast$ to prove generic properties about these operations without it being necessary for you to know what that particular operation is. I hope this helps! Expert I'm unsure if this helps, but: $\frac{a}{b}=a\cdot \frac{1}{b}$ $\frac{a}{b}\cdot \frac{c}{d}=a\cdot \frac{1}{b}\cdot c\cdot \frac{1}{d}$ $\begin{array}{}\text{(1)}& =ac\cdot \frac{1}{b}\cdot \frac{1}{d}\end{array}$ $\begin{array}{}\text{(2)}& =ac\cdot \frac{1}{bd}\end{array}$ $=\frac{ac}{bd}$ Now the problem is really getting from (1) to (2), visually. You can think of it as if you have a slice of pizza, $\frac{1}{b}$ out of the entire pizza. You cut that slice into $d$ pieces and now you have $\frac{1}{bd}$ of the whole slice. Do you have a similar question?
Top # Simplification 1. 5×5/5 = ? $$5 \times 5/5 = 25/5 = 5$$ #### Enter details here 2. The price of 80 apples is equal to that of 120 oranges. The price of 60 apples and 75 oranges together is Rs. 1320. The total price of 25 apples and 40 oranges is: Let price of one apple $$=a$$ price of one orange $$=b$$ Price of 80 apples = Price of 120 oranges $$\Rightarrow 80a=120b\\ \Rightarrow 2a=3b\\ \Rightarrow b = \dfrac{2a}{3}~\cdots(1)$$ Price of 60 apples + Price of 75 oranges =Rs. 1320 $$\Rightarrow 60a+75b=1320\\ \Rightarrow 4a+5b=88\\ \Rightarrow 4a+\dfrac{5(2a)}{3}=88 \quad \text{[ Because substituted value of } b \text{ from }(1)]\\ \Rightarrow 12a+10a=88 \times 3\\ \Rightarrow 6a+5a=44 \times 3\\ \Rightarrow 11a=44 \times 3\\ \Rightarrow a=4 \times 3=12$$ $$b=\dfrac{2a}{3}=\dfrac{2 \times 12}{3}=8$$ Total price of 25 apples and 40 oranges $$=25a+40b\\=25\times12+40\times8\\=300+320\\=620$$ #### Enter details here 3. A man has Rs. 480 in the denominations of one-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has ? Let number of notes of each denomination be $$x$$ Then $$x + 5x + 10x = 480$$ $$\Rightarrow 16x = 480$$ $$\therefore x = 30$$ Hence, total number of notes $$= 3x = 90$$ #### Enter details here 4. There are 6 working days in a regular week and for each day, the working hours are 10. A man earns Rs. 2.10 per hour for regular work and Rs. 4.20 per hour for overtime. If he earns Rs. 525 in 4 weeks, how many hours did he work? Regular working hours in 4 weeks $$=4 \times 6 \times 10=240$$ hours Amount earned by working in these regular working hours $$=240 \times 2.10= \text {Rs. } 504$$ $$=525-504=\text{Rs. }21$$ Hours he worked overtime $$=\dfrac{21}{4.2}=\dfrac{210}{42}=5$$ hours Total hours he worked $$=240+5=245$$ #### Enter details here 5. Simplfy: $$b - [b -(a+b) - {b - (b - a+b)} + 2a]$$ $$b-[b-(a+b)-{b-(b-a+b)}+2a]$$ $$=b-[b-a-b-{b-(2b-a)}+2a]$$ $$=b-[-a-{b-2b+a}+2a]$$ $$=b-[-a-{-b+a}+2a]$$ $$=b-[-a+b-a+2a]$$ $$=b-[-2a+b+2a]$$ $$=b-b$$ $$=0$$
Graphs of Trigonometric Functions A function f is periodic if there exists a nonzero number p such that f(x + p) = f(s) for all x in the domain of f. The smallest such positive value of p (if it exists) is the period of f. The sine, cosine, secant, and cosecant functions each have a period of 2π, and the other two trigonometric functions have a period of π, as shown in the figure below. Note in the previous figure that the maximum value of sin x and cos x is 1 and the minimum value is -1. The graphs of the functions y = a sin bx and y = a cos bx oscillate between -a and a, and hence have an amplitude of \| a \|. Furthermore, because bx = 0 when x = 0 and bx = 2π when x = 2π/b, it follows that the functions y = a sin bx and y = a cos bx each have a period of 2π/\| b \|. The table below summarizes the amplitudes and periods for some types of trigonometric functions. Example 1 Sketching the Graph of a Trigonometric Function Sketch the graph of f(x) = 3 cos 2x. Solution The graph of f(x) = 3 cos 2x has an amplitude of 3 and a period of 2π/2 = π. Using the basic shape of the graph of the cosine function, sketch one period of the function on the interval [0, π], using the following pattern. Maximum: (0, 3) Minimum: Maximum: [π, 3] By continuing this pattern, you can sketch several cycles of the graph, as shown in the following figure. Horizontal shifts, vertical shifts, and reflections can be applied to the graphs of trigonometric functions, as illustrated in Example 2. Example 2 Shifts of Graphs of Trigonometric Functions Sketch the graphs of the following functions. Solution a. To sketch the graph of f(x) = sin(x + π/2), shift the graph of y = sinx to the left π/2 units, as shown in the figure (a) below. b. To sketch the graph of f(x) = 2 + sin x, shift the graph of y = sin x up two units, as shown in figure (b) below. c. To sketch the graph of f(x) = 2 + sin(x – π/4), shift the graph of y = sin x up two units and to the right π/4 units, as shown in the figure (c) below. You are watching: Graphs of Trigonometric Functions. Info created by GBee English Center selection and synthesis along with other related topics.
# Trigonometry Triangles may seem like simple figures, but the mathematics behind them is deep enough to be considered its own subject: trigonometry. As the name suggests, trigonometry is the study of triangles. More specifically, trigonometry deals with the relationships between angles and sides in triangles. Somewhat surprisingly, the trigonometric ratios can also provide a richer understanding of circles. These ratios are often used in calculus as well as many branches of science including physics, engineering, and astronomy. The resources in this guide cover the basics of trigonometry, including a definition of trigonometric ratios and functions. They then go over how to use these functions in problems and how to graph them. Finally, this resource guide concludes with an explanation of the most common trigonometric identities. ## Basic Trigonometry Trigonometry especially deals with the ratios of sides in a right triangle, which can be used to determine the measure of an angle. These ratios are called trigonometric functions, and the most basic ones are sine and cosine. These two functions are used to define the other well-known trigonometric functions: tangent, secant, cosecant, and cotangent. This section begins by reviewing right triangles and explaining the basic trigonometric functions. It also explains their reciprocals. The topic also covers how to evaluate trigonometric angles, especially the special angles of 30-, 45-, and 60-degrees. Finally, the guide to this topic covers how to deal with the inverses of trigonometric functions and the two most common ways to measure angles. • Identify the Sides of Right Triangles • Trigonometric Functions or Trig. Ratios • Sine • Cosine • Tangent • Review of Sine, Cosine, and Tangent • Secant, Cosecant, Cotangent • Sin, Cos, Tan, Sec, Csc, Cot • Co-Functions • Evaluate Trigonometric Angles • Special Angles: 30-Degrees, 45-Degrees, 60-Degrees • Using a Calculator • Inverse Trigonometry ## Applications of Trigonometry There are actually a wide variety of theoretical and practical applications for trigonometric functions. They can be used to find missing sides or angles in a triangle, but they can also be used to find the length of support beams for a bridge or the height of a tall object based on a shadow. This topic covers different types of trigonometry problems and how the basic trigonometric functions can be used to find unknown side lengths. It also covers how they can be used to find angles and even the area of a triangle. Finally, this section concludes with subtopics on the Laws of Sines and the Law of Cosines. • Trigonometry Problems • Sine Problems • Cosine Problems • Tangent Problems • Find Unknown Sides of Right Angles • Find Height of Object Using Trigonometry • Trigonometry Applications • Angle of Elevation and Depression • Area of Triangle Using the Sine Function • Law of Sines or Sine Rule • Law of Cosines or Cosine Rule ## Trigonometry in the Cartesian Plane Trigonometry in the Cartesian Plane is centered around the unit circle. That is, the circle centered at the point (0, 0) with a radius of 1. Any line connecting the origin with a point on the circle can be constructed as a right triangle with a hypotenuse of length 1. The lengths of the legs of the triangle provide insight into the trigonometric functions. The cyclic nature of the unit circle also reveals patterns in the functions that are useful for graphing. This topic begins with a description of angles at the standard position and coterminal angles before explaining the unit circle and reference angles. It then covers how the values of the trigonometric functions change based on the quadrant of the Cartesian Plane. Finally, this section ends by explaining how the unit circle and the xy-plane can be used to solve trigonometry problems. • Angles at Standard Position and Coterminal Angles • Unit Circle • Reference Angle • Trigonometric Ratios in the Four Quadrants • Finding the Quadrant in Which an Angle Lies • Coterminal Angles • Trigonometric Functions in the Cartesian Plane • Evaluating Trigonometric Functions for an Angles, Given a Point on the Angle • Evaluating Trigonometric Functions Using the Reference Angle • Finding Trigonometric Values Given One Trigonometric Value/Other Info • Evaluating Trigonometric Functions at Important Angles ## Graphs of Trigonometric Functions Although the unit circle in the Cartesian plane provides into trigonometric functions, each of these functions also has its own graph. These graphs are cyclic in nature. Typically, graphs of trig functions make the most sense when the x-axis is divided into intervals of pi radians while the y-axis is still divided into intervals of whole numbers. This topic covers the basic graphs of sine, cosine, and tangent. It then discusses transformations of those graphs and their properties. Finally, the topic concludes with a subtopic about the graphs of the reciprocals of the basic trig functions. • Trigonometry Graphs • Sine Graph • Cosine Graph • Tangent Graph • Transformations of Trigonometric Graphs • Graphing Sine and Cosine with Different Coefficients • Maximum and Minimum Values of Sine and Cosine Functions • Graphing Trig Functions: Amplitude, Period, Vertical, and Horizontal Shifts • Tangent, Cotangent, Secant, Cosecant Graphs ## Trigonometric Identities This is the point where trigonometric functions take on a life of their own apart from their basis in triangle side ratios. The functions contain numerous identities that illuminate the relationship between different types of trig functions. These identities can be used to find the values of angles outside the common reference angles. In fact, they were the main tool available for doing that before calculators. This topic explains trigonometric identities and how to find and remember them. It also explains how to use the identities to simplify expressions, which involves a fair amount of algebraic manipulation. The guide goes on to explain how to find the values of different angles based on reference angles with the sum and difference identities and the double-angle and half-angle formulas. The topic continues and concludes with more ways to simplify, factor, and solve trigonometric equations. • Trigonometric Identities • Trigonometric Identities: How to Derive/ Remember Them • Using Trigonometric Identities to Simplify Expressions • Sum and Difference Identities • Double-Angle and Half-Angle Formulas • Trigonometric Equations • Simplifying Trigonometric Expressions Using Trig Identities • Simplifying Trigonometric Expressions Involving Fractions • Simplifying Products of Binomials Involving Trigonometric Functions • Factoring and Simplifying Trigonometric Expressions • Solving Trigonometric Equations • Solving Trigonometric Equations Using Factoring • Examples with Trigonometric Functions: Even, Odd, or Neither • Proving a Trigonometric Identity
Request a call back # Binomial Theorem and its Simple Applications Binomial Theorem and its Simple Applications PDF Notes, Important Questions And Synopsis ## Bionomial Theorem and its Simple Applications PDF Notes, Important Questions and Synopsis SYNOPSIS 1. A binomial is a polynomial having only two terms. For e.g 2y2- 1 2. (x + y)n can be expanded using the Binomial theorem without actually multiplying it n times. 3. Properties of Binomial Expansion (x + y)n i.Total number of terms in this expansion is n + 1. ii.The exponent of x decreases by 1, while the exponent of y increases by 1 in subsequent terms. iii.The first term is and the final/last term is iv.The general term in this expansion is given by 4. Binomial Coefficients: Binomial coefficients in the expansion of (x + y)n are simply the number of ways of choosing x from the brackets and y from the rest. 5. Pascal’s Triangle: Binomial coefficients can be found using Pascal’s triangle given below. We can also say that . Also, the binomial coefficient is given by 6. Terms in the binomial expansion of (x + y)n i.General Term (rth term): General term in the expansion of (x + y)n is given by ii.Middle Term(s): 1. When n is even, 2. When n is odd, and iii.Greatest Term: In any binomial expansion, the values of the terms increase, reach a maximum and then decrease. So, to find the greatest term, find the value of r till So, the greatest term occurs when r = iv.Term independent of x: The term independent of x is the term not containing x. So, find the value of r such that the exponent of x is zero. 7. Applications of Binomial Expansion: i. We have a very important result , n ≥ 1, n ∊ N ii. Finding the remainder using Binomial Theorem: To find the remainder when pn is divided by q, adjust the power of p to pm which is very close to b, say with difference 1 and then divide by taking the remainder always positive. JEE Main - Maths Asked by universitymails6 \| 22 Jun, 2022, 07:15: PM JEE Main - Maths if x + 1/x = 2 , then what is the value of x⁹⁵ + 1/x⁹⁵. Asked by ria102680 \| 19 May, 2022, 07:41: AM JEE Main - Maths Plz solve with proper steps Asked by rinamahuya \| 25 Jan, 2020, 01:42: PM
# Transversal (geometry) In geometry, a transversal is a line that passes through two lines in the same plane at two distinct points. Transversals play a role in establishing whether two other lines in the Euclidean plane are parallel. The intersections of a transversal with two lines create various types of pairs of angles: consecutive interior angles, corresponding angles, and alternate angles. As a consequence of Euclid's parallel postulate, if the two lines are parallel, consecutive interior angles are supplementary, corresponding angles are equal, and alternate angles are equal. Eight angles of a transversal.(Vertical angles such as ${\displaystyle \alpha }$ and ${\displaystyle \gamma }$ are always congruent.) Transversal between non-parallel lines.Consecutive angles are not supplementary. Transversal between parallel lines.Consecutive angles are supplementary. ## Angles of a transversal A transversal produces 8 angles, as shown in the graph at the above left: • 4 with each of the two lines, namely α, β, γ and δ and then α1, β1, γ1 and δ1; and • 4 of which are interior (between the two lines), namely α, β, γ1 and δ1 and 4 of which are exterior, namely α1, β1, γ and δ. A transversal that cuts two parallel lines at right angles is called a perpendicular transversal. In this case, all 8 angles are right angles [1] When the lines are parallel, a case that is often considered, a transversal produces several congruent and several supplementary angles. Some of these angle pairs have specific names and are discussed below:[2][3]corresponding angles, alternate angles, and consecutive angles. ### Alternate angles One pair of alternate angles. With parallel lines, they are congruent. Alternate angles are the four pairs of angles that: • have distinct vertex points, • lie on opposite sides of the transversal and • both angles are interior or both angles are exterior. If the two angles of one pair are congruent (equal in measure), then the angles of each of the other pairs are also congruent. Proposition 1.27 of Euclid's Elements, a theorem of absolute geometry (hence valid in both hyperbolic and Euclidean Geometry), proves that if the angles of a pair of alternate angles of a transversal are congruent then the two lines are parallel (non-intersecting). It follows from Euclid's parallel postulate that if the two lines are parallel, then the angles of a pair of alternate angles of a transversal are congruent (Proposition 1.29 of Euclid's Elements). ### Corresponding angles One pair of corresponding angles. With parallel lines, they are congruent. Corresponding angles are the four pairs of angles that: • have distinct vertex points, • lie on the same side of the transversal and • one angle is interior and the other is exterior. Two lines are parallel if and only if the two angles of any pair of corresponding angles of any transversal are congruent (equal in measure). Proposition 1.28 of Euclid's Elements, a theorem of absolute geometry (hence valid in both hyperbolic and Euclidean Geometry), proves that if the angles of a pair of corresponding angles of a transversal are congruent then the two lines are parallel (non-intersecting). It follows from Euclid's parallel postulate that if the two lines are parallel, then the angles of a pair of corresponding angles of a transversal are congruent (Proposition 1.29 of Euclid's Elements). If the angles of one pair of corresponding angles are congruent, then the angles of each of the other pairs are also congruent. In the various images with parallel lines on this page, corresponding angle pairs are: α=α1, β=β1, γ=γ1 and δ=δ1. ### Consecutive interior angles One pair of consecutive angles. With parallel lines, they add up to two right angles Consecutive interior angles are the two pairs of angles that:[4][2] • have distinct vertex points, • lie on the same side of the transversal and • are both interior. Two lines are parallel if and only if the two angles of any pair of consecutive interior angles of any transversal are supplementary (sum to 180°). Proposition 1.28 of Euclid's Elements, a theorem of absolute geometry (hence valid in both hyperbolic and Euclidean Geometry), proves that if the angles of a pair of consecutive interior angles are supplementary then the two lines are parallel (non-intersecting). It follows from Euclid's parallel postulate that if the two lines are parallel, then the angles of a pair of consecutive interior angles of a transversal are supplementary (Proposition 1.29 of Euclid's Elements). If one pair of consecutive interior angles is supplementary, the other pair is also supplementary. ## Other characteristics of transversals If three lines in general position form a triangle are then cut by a transversal, the lengths of the six resulting segments satisfy Menelaus' theorem. ## Related theorems Euclid's formulation of the parallel postulate may be stated in terms of a transversal. Specifically, if the interior angles on the same side of the transversal are less than two right angles then lines must intersect. In fact, Euclid uses the same phrase in Greek that is usually translated as "transversal".[5] Euclid's Proposition 27 states that if a transversal intersects two lines so that alternate interior angles are congruent, then the lines are parallel. Euclid proves this by contradiction: If the lines are not parallel then they must intersect and a triangle is formed. Then one of the alternate angles is an exterior angle equal to the other angle which is an opposite interior angle in the triangle. This contradicts Proposition 16 which states that an exterior angle of a triangle is always greater than the opposite interior angles.[6][7] Euclid's Proposition 28 extends this result in two ways. First, if a transversal intersects two lines so that corresponding angles are congruent, then the lines are parallel. Second, if a transversal intersects two lines so that interior angles on the same side of the transversal are supplementary, then the lines are parallel. These follow from the previous proposition by applying the fact that opposite angles of intersecting lines are equal (Prop. 15) and that adjacent angles on a line are supplementary (Prop. 13). As noted by Proclus, Euclid gives only three of a possible six such criteria for parallel lines.[8][9] Euclid's Proposition 29 is a converse to the previous two. First, if a transversal intersects two parallel lines, then the alternate interior angles are congruent. If not, then one is greater than the other, which implies its supplement is less than the supplement of the other angle. This implies that there are interior angles on the same side of the transversal which are less than two right angles, contradicting the fifth postulate. The proposition continues by stating that on a transversal of two parallel lines, corresponding angles are congruent and the interior angles on the same side are equal to two right angles. These statements follow in the same way that Prop. 28 follows from Prop. 27.[10][11] Euclid's proof makes essential use of the fifth postulate, however, modern treatments of geometry use Playfair's axiom instead. To prove proposition 29 assuming Playfair's axiom, let a transversal cross two parallel lines and suppose that the alternate interior angles are not equal. Draw a third line through the point where the transversal crosses the first line, but with an angle equal to the angle the transversal makes with the second line. This produces two different lines through a point, both parallel to another line, contradicting the axiom.[12][13] ## In higher dimensions In higher dimensional spaces, a line that intersects each of a set of lines in distinct points is a transversal of that set of lines. Unlike the two-dimensional (plane) case, transversals are not guaranteed to exist for sets of more than two lines. In Euclidean 3-space, a regulus is a set of skew lines, R, such that through each point on each line of R, there passes a transversal of R and through each point of a transversal of R there passes a line of R. The set of transversals of a regulus R is also a regulus, called the opposite regulus, Ro. In this space, three mutually skew lines can always be extended to a regulus. ## References 1. ^ "Transversal". Math Open Reference. 2009. (interactive) 2. ^ a b Rod Pierce (2011). "Parallel Lines". MathisFun. (interactive) 3. ^ Holgate Art. 87 4. ^ C.Clapham, J.Nicholson (2009). "Oxford Concise Dictionary of Mathematics" (PDF). Addison-Wesley. p. 582. 5. ^ Heath p. 308 note 1 6. ^ Heath p. 307 8. ^ Heath p. 309-310 10. ^ Heath p. 311-312 12. ^ Heath p. 313 13. ^ A similar proof is given in Holgate Art. 93 • Holgate, Thomas Franklin (1901). Elementary Geometry. Macmillan. • Thomas Little Heath, T.L. (1908). The thirteen books of Euclid's Elements. 1. The University Press. pp. 307 ff.
1. Home 2. Knowledge Base 3. Sum of Squares – Definition # Sum of Squares – Definition ### Sum of Squares Definition In regression analysis, sum of squares refers to a statistical method of analyzing how data series are generated and how they disperse. The goal of the analysis is to uncover how the data points or series are alloted a fitting function. Variation is another term that describes the sum of squares. To calculate sum of squares, the formula below will be used; Sum of squares=i=0∑n (XiX)2 In the above formula, Xi =The ith item in the set X=The mean of all items in the set (XiX)=The deviation of each item from the mean (The above formula is applicable for a set X of n items.) ### A Little More on What is the Sum of Squares Statistically, sum of squares evaluate how data series are dispersed and how they deviate from the mean. This analysis is important in the calculation of mean for a set of data points. Also, sum of squares tells us the extent of variation that has occurred to a set of measurement. The variation of data series is helpful in knowing how the values of data points fit into the available regression model. While mean is defined as the average of a set of numbers, variation refers to a spread between individual values and the mean. Sum of squares is an important method of analysis that enable analysts find out the variation between data points as well as how each data point is fitted into a function before the points are summed up. Also, the relationship between two variables be it a linear relationship or otherwise is determined through sum of squares. A variation between two variables that is unexplainable is known as residual sum of squares. Here are some vital things to know about sum of Squares; • In regression analysis, sum of squares is an analysis technique used in measuring the dispersion of data points. • The sum of squares examine the deviation of data series from mean. • When the sum of squares is high, it indicates that there is a huge variability within the data points, while a low sum of squares mean that there isn’t much variation between data points and the mean value. ### Limitations of Using the Sum of Squares Sum of squares might be difficult to analyze or interpret when additional data points are included in the set, this is a limitation because when this happens the sum of squares expands significantly. Oftentimes, sum of squares cannot be used in making decisions on investment. Analysts and investors are required to make detailed findings and various data analysis to know the degree of variability of an asset. Sum of squares is important when using variance and standard deviation as measurements of variation. Also, the linear least squares method and non-linear least squares method of regression analysis make use of sum of squares. ### References for “Sum Of Squares” https://www.investopedia.com/terms/s/sum-of-squares.asp https://en.wikipedia.org/wiki/Sum_of_squares https://sciencing.com › Math › Probability & Statistics › Mean/Median/Mode
# Chapter 7: Recursion Recursion is a deceptively simple concept: Any routine that calls itself is recursive. Despite this apparent simplicity, understanding and applying recursion can be surprisingly complex. One of the major barriers to understanding recursion is that general descriptions tend to become highly theoretical, abstract, and mathematical. Although there is certainly value in that approach, this chapter will instead follow a more pragmatic course, focusing on example, application, and comparison of recursive and iterative (nonrecursive) algorithms. ## Understanding Recursion Recursion is most useful for tasks that can be defined in terms of similar subtasks. For example, sort, search, and traversal problems often have simple recursive solutions. A recursive routine performs a task in part by calling itself to perform the subtasks. At some point, the routine encounters a subtask that it can perform without calling itself. This case, in which the routine does not recurse, is called the base case; the former, in which the routine calls itself to perform a subtask, is referred to as the recursive case. Tip Recursive algorithms have two types of cases: recursive cases and base cases. These concepts can be illustrated with a simple and commonly used example: the factorial operator. n! (pronounced “n factorial”) is essentially the product of all integers between n and 1. For example, 4! = 4 _ 3 _ 2 _ 1 = 24. n! can be more formally defined as follows: ` n! = n (n – 1)! 0! = 1! = 1` This definition leads easily to a recursive implementation of factorial. The task is determining the value of n!, and the subtask is determining the value of (n – 1)! In the recursive case, when n is greater than 1, the routine calls itself to determine the value of (n – 1)! and multiplies that by n. In the base case, when n is 0 or 1, the routine simply returns 1. Rendered in any C-like language, this looks like the following: ` int factorial( int n ){ if (n > 1) { /* Recursive case / return factorial(n-1) n; } else { /* Base case / return 1; } }` Figure 7-1 illustrates the operation of this routine when computing 4!. Notice that n decreases by 1 each time the routine recurses. This ensures that the base case will eventually be reached. If a routine is written incorrectly such that it does not always reach a base case, it will recurse infinitely. In practice, there is usually no such thing as infinite recursion: Eventually a stack overflow occurs and the program crashes - a similarly catastrophic event. (There is a form of recursion, called tail recursion, that can be optimized by the compiler to use the same stack frame for each recursive call. An appropriately optimized tail recursive algorithm could recurse infinitely because it wouldn’t overflow the stack.) Figure 7-1 Tip Every recursive case must eventually lead to a base case. This implementation of factorial represents an extremely simple example of a recursive routine. In many cases, your recursive routines may need additional data structures or an argument that tracks the recursion level. Often the best solution in such cases is to move the data structure or argument initialization code into a separate routine. This wrapper routine, which performs initialization and then calls the purely recursive routine, provides a clean, simple interface to the rest of the program. For example, if you need a factorial routine that returns all of its intermediate results (factorials less than n), as well as the final result (n!), you most naturally return these results as an integer array, which means the routine needs to allocate an array. You also need to know where in the array each result should be written. These tasks are most easily accomplished using a wrapper routine, as follows: ` int[] allFactorials( int n ){ / Wrapper routine / int[] results = new int[ n == 0 ? 1 : n ]; doAllFactorials( n, results, 0 ); return results; } int doAllFactorials( int n, int[] results, int level ){ if( n > 1 ){ / Recursive case / results[level] = n doAllFactorials( n - 1, results, level + 1 ); return results[level]; } else { /* Base case / results[level] = 1; return 1; } }` You can see that using a wrapper routine enables you to hide the array allocation and recursion level tracking to keep the recursive routine very clean. In this case, it’s possible to determine the appropriate array index from n, avoiding the need for the level argument, but in many cases there is no alternative to tracking the recursion level as shown here. Tip It may be useful to write a separate wrapper routine to do initialization for a complex recursive routine. Although recursion is a very powerful technique, it is not always the best approach, and rarely is it the most efficient approach. This is due to the relatively large overhead for routine calls on most platforms. For a simple recursive routine like factorial, many computer architectures spend more time on call overhead than the actual calculation. Iterative routines, which use looping constructs instead of recursive routine calls, do not suffer from this overhead and are frequently more efficient. Tip Iterative solutions are usually more efficient than recursive solutions. Any problem that can be solved recursively can also be solved iteratively. Iterative algorithms are often quite easy to write, even for tasks that might appear to be fundamentally recursive. For example, an iterative implementation of factorial is relatively simple. It may be helpful to expand the definition of factorial, such that you describe n! as the product of every integer between n and 1, inclusive. You can use a for loop to iterate through these values and calculate the product: ` int factorial( int i ){ int n, val = 1; for( n = i; n > 1; n-- ) / n = 0 or 1 falls through / val = n; return val; }` This implementation is significantly more efficient than our previous recursive implementation because no routine calls are involved. Although it represents a different way of thinking about the problem, it’s not really any more difficult to write than the recursive implementation. For some problems, there are no obvious iterative alternatives like the one just shown. It’s always possible, though, to implement a recursive algorithm without using recursive calls. Recursive calls are generally used to preserve the current values of local variables and restore them when the subtask performed by the recursive call is completed. Because local variables are allocated on the program’s stack, each recursive instance of the routine has a separate set of the local variables, so recursive calls implicitly store variable values on the program’s stack. You can therefore eliminate the need for recursive calls by allocating your own stack and manually storing and retrieving local variable values from this stack. Implementing this type of stack-based interactive routine tends to be significantly more complicated than implementing an equivalent routine using recursive calls. In languages like Java or C#, a stack-based iterative implementation of a recursive algorithm won’t be more efficient than the recursive version. Given the large increase in complexity, you should implement recursive algorithms with recursive calls unless instructed otherwise. An example of a recursive algorithm implemented without recursive calls is given in the solution to the “Preorder Traversal, No Recursion” problem in Chapter 5. Tip A recursive algorithm can be implemented without recursive calls by using a stack, but it’s usually more trouble than it’s worth. In an interview, a working solution is of primary importance; an efficient solution is secondary. Unless you’ve been told otherwise, go with whatever type of working solution comes to you first. If it’s a recursive solution, you might want to mention the inefficiencies inherent in recursive solutions to your interviewer, so it’s clear that you know about them. In the rare instance that you see a recursive solution and an iterative solution of roughly equal complexity, you should probably mention them both to the interviewer, indicating that you’re going to work out the iterative solution because it’s likely to be more efficient. Programming Interviews Exposed: Secrets to Landing Your Next Job, 2nd Edition (Programmer to Programmer) ISBN: 047012167X EAN: 2147483647 Year: 2007 Pages: 94 Similar book on Amazon
# RS Aggarwal Class 8 Maths Chapter 7 Ex 7.3 Solutions 2022 \| Download Free PDF RS Aggarwal Class 8 Maths Chapter 7 Ex 7.3 Solutions: In this exercise, the students will learn the factorization of algebraic expressions when a binomial is a common factor. These solutions make students stress free during exam preparation. The answers are formulated with the objective of assisting the students to understand each topic in detail so as to attain excellent ranks. The students must solve the exercise problems regularly that enable them to solve any question in the exams effortlessly. RS Aggarwal Class 8 Maths Chapter 7 Ex 7.3 Solutions assist the students to clear their queries while solving problems & to prepare more efficiently for the Class 8th Maths final exams. The solutions are created by Mathematics experts in order to improve logical thinking among students. In this exercise, important formulas are explained in a structured manner so that the students perform best in the final exams. ## Download RS Aggarwal Class 8 Maths Chapter 7 Ex 7.3 Solutions RS Aggarwal Class 8 Maths Chapter 7 Ex 7.3 Solutions ## Important Definition for RS Aggarwal Class 8 Maths Chapter 7 Ex 7.3 Solutions The students will get well-prepared to answer any questions from this exercise in the exam accurately with the help of practicing RS Aggarwal Class 8 Maths Chapter 7 Ex 7.3 Solutions. The important topics are discussed in this exercise in detail. These solutions are considered as a prepared refresher that offers a quick revision of all the topics. • Factorization of algebraic expressions when a binomial is a common factor In factorization, when binomial is common, an algebraic expression having a binomial as a common factor. We write the expression as the products of the binomial & the quotient acquired on dividing the given expression by the binomial in order to factorize. Follow the following steps in order to factorize: Step 1: Find the common binomial. Step 2: Write the given expression as the product of this binomial & the quotient acquired on dividing the given expression by this binomial. For instance: Factorize the algebraic expressions of 5a(2x – 3y) + 2b(2x – 3y) 5a(2x – 3y) + 2b(2x – 3y) Here, we notice that the binomial (2x – 3y) is common to both the terms. = (2x – 3y)(5a + 2b) ## Benefits of RS Aggarwal Class 8 Maths Chapter 7 Ex 7.3 Solutions • The students can improve their problem-solving skills with the help of these solutions which students can easily access anytime and anywhere. • The solutions assist the students to develop the proper knowledge of topics covered in this exercise which are presented in a simple and interactive way. • RS Aggarwal Class 8 Maths Chapter 7 Ex 7.3 Solutions are created according to the latest CBSE syllabus pattern & guidelines. • The solutions provide detailed & explanative answers for all questions in easily understandable language. • The students can easily download the solutions PDF and these are available online as well as offline mode. Know more at the official website.
Vous êtes sur la page 1sur 5 # LCM/HCF Factors: Factors of a number are the numbers which will divide the given number exactly without leaving any reminders. Eg) Factors of 6 are 1, 2, 3, 6 Multiples: Multiple of a number are generated by multiplying other integers with the given number. Eg) multiples of 6 are 6, 12,18,24 etc. LEAST COMMON MULTIPLE(LCM) The least common multiple of two numbers is the lowest number that is a multiple of both the numbers. Patterns of LCM/HCF 1. LCM: Division Method Step:1 Given numbers are written in a line separated by comma. Step 2 Divide by any one of the prime numbers 2,3,5,7,11, which will divide at least any two numbers exactly. The quotient and the undivided numbers are written in a line below the first . Step 3: ( step 2) is repeated until a line of numbers ( prime to each other) appears. Step 4: Find the product of all divisors and numbers in the last line which is required LCM. HIGHEST COMMON FACTOR: HCF of two or more numbers is the greatest number that divides each one of them exactly. 2.HCF: Continuous Division Method until the reminder comes zero: Step 1: Greater number is divided by the smaller one. Step 2: Divisor of (1) is divided by the reminder. Step 3: Divisor of (2) is divided by its reminder. This is continued until no reminder is left. 3.LCM/HCF of Fractions: LCM of fractions = LCM of numerators / HCF of denominators. HCF of fractions = HCF of numerators / LCM of denominators. 4.Product of two numbers: Product of two numbers = LCM * HCF 5.Applications of LCM & HCF: S.No Patterns Methods 1. Find the greatest number that will exactly divide a,b,c HCF (a,b,c) 2. Find the greatest number that will divide a,b,c leaving reminders of x,y,z. HCF (a-x, b-y , c-z) 3. Find the greatest number which when id divides a,b,c will leave the same reminder in each case If reminder value is not given Then HCF (a-b, b-c, c-a). If reminder value is given HCF( a-r,b-r,c-r) 4. Find the least number which is exactly divisible by a,b,c LCM of(a,b,c) 5 Find the least number which when divided by a,b,c leaves the same reminder r in each case LCM (a,b,c)=r 6 Find the least number which when divided by a,b,c leaves the reminders x,y,z It is a-x=b-y=c-z=K(say), then LCM(a,b,c) - K 6. To find n- digit greatest number which when divided by x,y,z a) Leaves no reminder 1. LCM of x,y,z = L 2. L divideds n digit greatest number , find reminder 3. Req no= n digit greastest number - R b) Leaves reminder K in each case . Req no=(n digit greatest number R)+ K 7. To find the n digit smallest number which when divided by x,y,z a) Leaves no reminder 1. LCM of x,y,z = L 2. L divideds n digit Smallest number , find reminder 3. Req no= n digit Smallest number +(L R) b) Leaves reminder K in each case . Req no=(n digit Smallest number + (L R)+ K) 1. Find the LCM of 12,27,40 a) 1880 b)1080 c) 1750 d) none Solution: 2 12, 27, 40 2 6, 27, 20 3 3, 27, 10 1, 9, 10 LCM= 2x2x3x9x10 = 1080. 2. Find the HCF of 777 and 1147 a) 47 b) 49 c)37 d)59 777)1147(1 777 370)777(2 740 37)370(10 370 0 Therefore HCF= 37 3.Find the LCM of 1/3, 5/6, 5/9, 10/27? a) 3/4 b) 10/3 c) 60/3 d) 7/5 solution: option b: 10/3 In the above LCM of Nrs 1,5,5,10 is 10. HCF of Drs 3,6,9,27 is 3. 4)Find the HCF of 1/2, 3/4,5/6,7/8,9/10? a) 2/127 b) 1/120 c) 3/130 d)none Solution: Option b: 1/120 In the above, HCF of Nrs is 1 and LCM of Drs is 120. 5)LCM of two number is 2310. Its HCF is 30. If one number is 210 find the other number? a) 40 b)60 c)30 d)50 solution: option c:30 Second number= (231030)/210 6) LCM of two numbers is 495 and their HCF is 5. IF the sum of the two numbers is 100, then difference is? A) 20 b) 10 c)30 d)50 Solution: Option b: 10 Since LCMHCF= Product of two nos 4955= x(100-x) x 2 -100x+2475=0 Hence x=45 and 55 Difference is 10 7) Find the least number which when divided by 5,6,7,8 gives the reminder 3 but is divisible by 9? a) 1782 b)1683 c)1825 d)none Solution: Option b: 1683 LCM = 840 , hence rem no= 840x +3 which is exactly divisible by 9. If x=2 req no= 8402 +3=1683 8)The smallest positive number x, which leaves a reminder 1 when divided by 2,3,4 and 5? a) 71 b)11 c)7 d)61 Solution: option d:61 LCM= 60, Hence small no is 60+1=61. 9)Find the greatest number of five digits which when divided by 8,9 and 10 leaves 3 as reminder in each case? A) 99620 b)99825 c)99723 d)99743 Solution: option c: 99723 LCM = 360 Greatest 5 digit number is 99999. Divide 99999/360 . Rem = 99720 hence required no= 99720+3 = 99723 10) Find the greatest number of 4 digits which is exactly divisible by 24, 28, 30, and 35? a) 9720 b)9240 c)9840 d)9540 Solution: option b: 9240 LCM= 840 Rem = 759 Req no = 9999-759=9240 11) Find the greatest number of Four digits which must be added 5231 so that the final number becomes exactly divisible by 12,15,27,32,40? a) 7729 b)8720 c)9782 d)8569 Solution: option a: 7729 LCM of 12,15,27,32,40 is 4320. 9999+5231= 15230; 15230/4320 rem =2270 Hence 9999-759= 7729 PROBLEMS BASED ON BELLS 12) Six bells commence tolling together and tolls at intervals of 2s,4s,6s,8s,10s,12s. In 30 min, how many times do they toll together? a) 16 b)15 c)20 21 Solution: Option a: 16 LCM = 120 Total no of times = (30x60)/120 = 15. Hence 15+1= 16 13)The three traffic lights at three different road crossing change after every 48s,72s and 108s. If they all change simultaneously at 8.20 h, Then they will again change simultaneously at? a) 8 h 27 min b 9 h 25 m 16s c)8. 27h 12 sec d) 5.27hr 12 sec Solution: Option c: 8h 27min 12 sec LCM= 432s, hence 8.20+ 7 min+12 sec = 8.27 h and 12 sec 14) Three rings complete 60,36 and 24 revolutions in a minute.. They start from a certain point in their circumference downwards. By what time , they come together again in same position? a) 10s b)15s c)5s d)25s Solution: option c: 5s Time taken by each ring = 60/60,60/36,60/24 is 1,5/3,5/2 sec LCM (1,1/5,5/2) = 5 sec

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