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# What is the formula for calculating mean? The mean, or average, is calculated by adding up the scores and dividing the total by the number of scores. Consider the following number set: 3, 4, 6, 6, 8, 9, 11. The mean is calculated in the following manner: 3 + 4 + 6 + 6 + 8 + 9 + 11 = 47. ## How do you calculate the mean step by step? Step-by-Step Process to Find the Mean Step 1: Add up all the numbers. The result is called the sum. Step 2: Count how many numbers there are. This number is called the sample size. ## What is the formula for mean median and mode? Mean Median Mode Formula This is found by adding the numbers in a data set and dividing by the number of observations in the data set. … The mode is the value that occurs the most often in a data set and the range is the difference between the highest and lowest values in a data set. ## How do you work out mean? The mean is the total of the numbers divided by how many numbers there are. 1. To find the mean, add all the numbers together then divide by the number of numbers. 2. Eg 6 + 3 + 100 + 3 + 13 = 125 ÷ 5 = 25. 3. The mean is 25. Read more Why does my online movie keep buffering? ## How can I calculate average? How to Calculate Average. The average of a set of numbers is simply the sum of the numbers divided by the total number of values in the set. For example, suppose we want the average of 24 , 55 , 17 , 87 and 100 . Simply find the sum of the numbers: 24 + 55 + 17 + 87 + 100 = 283 and divide by 5 to get 56.6 . ## What is L in mode formula? The mode of data is given by the formula: Where, l = lower limit of the modal class. h = size of the class interval. f1 = frequency of the modal class. ## What is the formula of medium? The median is the ((n + 1)/2)th item, where n is the number of values. For example, for the list {1, 2, 5, 7, 8}, we have n = 5, so the median is the ((5 + 1)/2)th item. ## How do we calculate mode? To find the mode, or modal value, it is best to put the numbers in order. Then count how many of each number. A number that appears most often is the mode. ## What is mean and mode? The mean (average) of a data set is found by adding all numbers in the data set and then dividing by the number of values in the set. The median is the middle value when a data set is ordered from least to greatest. The mode is the number that occurs most often in a data set. ## How do you find the overall mean? The mean, or average, is calculated by adding up the scores and dividing the total by the number of scores. Read more What TVS are compatible with Miracast? ## How do you find the mean and mode? Mean: Add up all the numbers of the set. Divide by how many numbers there are. Mode: The number that occurs the most. ## What is average in math? In maths, the average value in a set of numbers is the middle value, calculated by dividing the total of all the values by the number of values. When we need to find the average of a set of data, we add up all the values and then divide this total by the number of values. ## What is the formula to calculate average percentage? Calculate the percentage average To find the average percentage of the two percentages in this example, you need to first divide the sum of the two percentage numbers by the sum of the two sample sizes. So, 95 divided by 350 equals 0.27. You then multiply this decimal by 100 to get the average percentage. ## Do you add to find the area? To find the area of a rectangle, multiply its height by its width. For a square you only need to find the length of one of the sides (as each side is the same length) and then multiply this by itself to find the area. … Solution 1 and 2 require that you make two shapes and add their areas together to find the total area. Рубрики Info
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ # 3.2 Quadratic Functions In this section, we will explore the family of 2nd degree polynomials, the quadratic functions. While they share many characteristics of polynomials in general, the calculations involved in working with quadratics is typically a little simpler, which makes them a good place to start our exploration of short run behavior. In addition, quadratics commonly arise from problems involving area and projectile motion, providing some interesting applications. Example 1 A backyard farmer wants to enclose a rectangular space for a new garden. She has purchased 80 feet of wire fencing to enclose 3 sides, and will put the 4th side against the backyard fence. Find a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence have length L. SOLUTION In a scenario like this involving geometry, it is often helpful to draw a picture. It might also be helpful to introduce a temporary variable, W, to represent the side of fencing parallel to the 4th side or backyard fence. Since we know we only have 80 feet of fence available, we know that , or more simply, . This allows us to represent the width, W, in terms of L Now we are ready to write an equation for the area the fence encloses. We know the area of a rectangle is length multiplied by width, so This formula represents the area of the fence in terms of the variable length L. ### Short run Behavior: Vertex We now explore the interesting features of the graphs of quadratics. In addition to intercepts, quadratics have an interesting feature where they change direction, called the vertex. You probably noticed that all quadratics are related to transformations of the basic quadratic function. Example 2 Write an equation for the quadratic graphed below as a transformation of , then expand the formula and simplify terms to write the equation in standard polynomial form. SOLUTION We can see the graph is the basic quadratic shifted to the left 2 and down 3, giving a formula in the form . By plugging in a point that falls on the grid, such as (0,-1), we can solve for the stretch factor: Written as a transformation, the equation for this formula is . To write this in standard polynomial form, we can expand the formula and simplify terms: Notice that the horizontal and vertical shifts of the basic quadratic determine the location of the vertex of the parabola; the vertex is unaffected by stretches and compressions. Try it Now 1. A coordinate grid has been superimposed over the quadratic path of a basketball[1]. Find an equation for the path of the ball. Does he make the basket? Forms of Quadratic Functions • The standard form of a quadratic function is • The transformation form of a quadratic function is • The vertex of the quadratic function is located at (h, k), where h and k are the numbers in the transformation form of the function. Because the vertex appears in the transformation form, it is often called the vertex form. In the previous example, we saw that it is possible to rewrite a quadratic function given in transformation form and rewrite it in standard form by expanding the formula. It would be useful to reverse this process, since the transformation form reveals the vertex. Expanding out the general transformation form of a quadratic gives: This should be equal to the standard form of the quadratic: The second degree terms are already equal. For the linear terms to be equal, the coefficients must be equal: , so This provides us a method to determine the horizontal shift of the quadratic from the standard form. We could likewise set the constant terms equal to find: , so In practice, though, it is usually easier to remember that k is the output value of the function when the input is h, so . Finding the Vertex of a Quadratic For a quadratic given in standard form, the vertex (h, k) is located at: , Example 3 Find the vertex of the quadratic . Rewrite the quadratic into transformation form (vertex form). SOLUTION The horizontal coordinate of the vertex will be at The vertical coordinate of the vertex will be at Rewriting into transformation form, the stretch factor will be the same as the a in the original quadratic. Using the vertex to determine the shifts, Try it Now 2. Given the equation write the equation in standard form and then in transformation/vertex form. In addition to enabling us to more easily graph a quadratic written in standard form, finding the vertex serves another important purpose – it allows us to determine the maximum or minimum value of the function, depending on which way the graph opens. Example 4 Returning to our backyard farmer from the beginning of the section, what dimensions should she make her garden to maximize the enclosed area? SOLUTION Earlier we determined the area she could enclose with 80 feet of fencing on three sides was given by the equation . Notice that quadratic has been vertically reflected, since the coefficient on the squared term is negative, so the graph will open downwards, and the vertex will be a maximum value for the area. In finding the vertex, we take care since the equation is not written in standard polynomial form with decreasing powers. But we know that a is the coefficient on the squared term, so a = -2, b = 80, and c = 0. Finding the vertex: , The maximum value of the function is an area of 800 square feet, which occurs when L = 20 feet. When the shorter sides are 20 feet, that leaves 40 feet of fencing for the longer side. To maximize the area, she should enclose the garden so the two shorter sides have length 20 feet, and the longer side parallel to the existing fence has length 40 feet. Example 5 A local newspaper currently has 84,000 subscribers, at a quarterly charge of $30. Market research has suggested that if they raised the price to$32, they would lose 5,000 subscribers. Assuming that subscriptions are linearly related to the price, what price should the newspaper charge for a quarterly subscription to maximize their revenue? Revenue is the amount of money a company brings in. In this case, the revenue can be found by multiplying the charge per subscription times the number of subscribers. We can introduce variables, C for charge per subscription and S for the number subscribers, giving us the equation Revenue = CS Since the number of subscribers changes with the price, we need to find a relationship between the variables. We know that currently S = 84,000 and C = 30, and that if they raise the price to $32 they would lose 5,000 subscribers, giving a second pair of values, C = 32 and S = 79,000. From this we can find a linear equation relating the two quantities. Treating C as the input and S as the output, the equation will have form . The slope will be This tells us the paper will lose 2,500 subscribers for each dollar they raise the price. We can then solve for the vertical intercept Plug in the point S = 85,000 and C = 30 Solve for b This gives us the linear equation relating cost and subscribers. We now return to our revenue equation. Substituting the equation for S from above Expanding We now have a quadratic equation for revenue as a function of the subscription charge. To find the price that will maximize revenue for the newspaper, we can find the vertex: The model tells us that the maximum revenue will occur if the newspaper charges$31.80 for a subscription. To find what the maximum revenue is, we can evaluate the revenue equation: Maximum Revenue = $2,528,100 ### Short run Behavior: Intercepts As with any function, we can find the vertical intercepts of a quadratic by evaluating the function at an input of zero, and we can find the horizontal intercepts by solving for when the output will be zero. Notice that depending upon the location of the graph, we might have zero, one, or two horizontal intercepts. zero horizontal intercepts one horizontal intercept two horizontal intercepts Example 6 Find the vertical and horizontal intercepts of the quadratic SOLUTION We can find the vertical intercept by evaluating the function at an input of zero: Vertical intercept at (0,-2) For the horizontal intercepts, we solve for when the output will be zero In this case, the quadratic can be factored easily, providing the simplest method for solution or Horizontal intercepts at and (-2,0) Notice that in the standard form of a quadratic, the constant term c reveals the vertical intercept of the graph. Example 7 Find the horizontal intercepts of the quadratic SOLUTION Again we will solve for when the output will be zero Since the quadratic is not easily factorable in this case, we solve for the intercepts by first rewriting the quadratic into transformation form. Now we can solve for when the output will be zero The graph has horizontal intercepts at and Try it Now 3. In Try it Now problem 2 we found the standard & transformation form for the function . Now find the Vertical & Horizontal intercepts (if any). This process is done commonly enough that sometimes people find it easier to solve the problem once in general and remember the formula for the result, rather than repeating the process each time. Based on our previous work we showed that any quadratic in standard form can be written into transformation form as: Solving for the horizontal intercepts using this general equation gives: start to solve for x by moving the constants to the other side divide both sides by a find a common denominator to combine fractions combine the fractions on the left side of the equation take the square root of both sides subtract b/2a from both sides combining the fractions Notice that this can yield two different answers for x Quadratic Formula For a quadratic function given in standard form , the quadratic formula gives the horizontal intercepts of the graph of this function. Example 8 A ball is thrown upwards from the top of a 40 foot high building at a speed of 80 feet per second. The ball’s height above ground can be modeled by the equation 1. What is the maximum height of the ball? 2. When does the ball hit the ground? SOLUTION To find the maximum height of the ball, we would need to know the vertex of the quadratic. , The ball reaches a maximum height of 140 feet after 2.5 seconds. To find when the ball hits the ground, we need to determine when the height is zero – when H(t) = 0. While we could do this using the transformation form of the quadratic, we can also use the quadratic formula: Since the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions: or The second answer is outside the reasonable domain of our model, so we conclude the ball will hit the ground after about 5.458 seconds. Try it Now 4. For these two equations determine if the vertex will be a maximum value or a minimum value. a. b. ### Important Topics of this Section • Quadratic functions • Standard form • Transformation form/Vertex form • Vertex as a maximum / Vertex as a minimum • Short run behavior • Vertex / Horizontal & Vertical intercepts • Quadratic formula Try it Now Answers 1. The path passes through the origin with vertex at (-4, 7). . To make the shot, h(-7.5) would need to be about 4. ; he doesn’t make it. 2. in Standard form; in Transformation form 3. Vertical intercept at (0, 13), NO horizontal intercepts. 4. a. Vertex is a minimum value b. Vertex is a maximum value [1] From http://blog.mrmeyer.com/?p=4778, © Dan Meyer, CC-BY ### Section 3.2 Exercises Write an equation for the quadratic function graphed. 1. 2. 3. 4. 5. 6. For each of the follow quadratic functions, find a) the vertex, b) the vertical intercept, and c) the horizontal intercepts. 7. 8. 9. 10. 11. 12. Rewrite the quadratic function into vertex form. 13. 14. 15. 16. 17. Find the values of b and c so has vertex 18. Find the values of b and c so has vertex Write an equation for a quadratic with the given features 1. x-intercepts (-3, 0) and (1, 0), and y intercept (0, 2) 2. x-intercepts (2, 0) and (-5, 0), and y intercept (0, 3) 3. x-intercepts (2, 0) and (5, 0), and y intercept (0, 6) 4. x-intercepts (1, 0) and (3, 0), and y intercept (0, 4) 5. Vertex at (4, 0), and y intercept (0, -4) 6. Vertex at (5, 6), and y intercept (0, -1) 7. Vertex at (-3, 2), and passing through (3, -2) 8. Vertex at (1, -3), and passing through (-2, 3) 9. A rocket is launched in the air. Its height, in meters above sea level, as a function of time, in seconds, is given by . 1. From what height was the rocket launched? 2. How high above sea level does the rocket reach its peak? 3. Assuming the rocket will splash down in the ocean, at what time does splashdown occur? 10. A ball is thrown in the air from the top of a building. Its height, in meters above ground, as a function of time, in seconds, is given by 11. From what height was the ball thrown? 12. How high above ground does the ball reach its peak? 13. When does the ball hit the ground? 14. The height of a ball thrown in the air is given by , where x is the horizontal distance in feet from the point at which the ball is thrown. 15. How high is the ball when it was thrown? 16. What is the maximum height of the ball? 17. How far from the thrower does the ball strike the ground? 18. A javelin is thrown in the air. Its height is given by , where x is the horizontal distance in feet from the point at which the javelin is thrown. 19. How high is the javelin when it was thrown? 20. What is the maximum height of the javelin? 21. How far from the thrower does the javelin strike the ground? 1. A box with a square base and no top is to be made from a square piece of cardboard by cutting 6 in. squares out of each corner and folding up the sides. The box needs to hold 1000 in3. How big a piece of cardboard is needed? 2. A box with a square base and no top is to be made from a square piece of cardboard by cutting 4 in. squares out of each corner and folding up the sides. The box needs to hold 2700 in3. How big a piece of cardboard is needed? 3. A farmer wishes to enclose two pens with fencing, as shown. If the farmer has 500 feet of fencing to work with, what dimensions will maximize the area enclosed? 1. A farmer wishes to enclose three pens with fencing, as shown. If the farmer has 700 feet of fencing to work with, what dimensions will maximize the area enclosed? 2. You have a wire that is 56 cm long. You wish to cut it into two pieces. One piece will be bent into the shape of a square. The other piece will be bent into the shape of a circle. Let A represent the total area enclosed by the square and the circle. What is the circumference of the circle when A is a minimum? 3. You have a wire that is 71 cm long. You wish to cut it into two pieces. One piece will be bent into the shape of a right triangle with legs of equal length. The other piece will be bent into the shape of a circle. Let A represent the total area enclosed by the triangle and the circle. What is the circumference of the circle when A is a minimum? 4. A soccer stadium holds 62,000 spectators. With a ticket price of$11, the average attendance has been 26,000. When the price dropped to \$9, the average attendance rose to 31,000. Assuming that attendance is linearly related to ticket price, what ticket price would maximize revenue? 5. A farmer finds that if she plants 75 trees per acre, each tree will yield 20 bushels of fruit. She estimates that for each additional tree planted per acre, the yield of each tree will decrease by 3 bushels. How many trees should she plant per acre to maximize her harvest? 6. A hot air balloon takes off from the edge of a mountain lake. Impose a coordinate system as pictured and assume that the path of the balloon follows the graph of . The land rises at a constant incline from the lake at the rate of 2 vertical feet for each 20 horizontal feet. [UW] 1. What is the maximum height of the balloon above water level? 2. What is the maximum height of the balloon above ground level? 3. Where does the balloon land on the ground? 4. Where is the balloon 50 feet above the ground? 7. A hot air balloon takes off from the edge of a plateau. Impose a coordinate system as pictured below and assume that the path the balloon follows is the graph of the quadratic function . The land drops at a constant incline from the plateau at the rate of 1 vertical foot for each 5 horizontal feet. [UW] 1. What is the maximum height of the balloon above plateau level? 2. What is the maximum height of the balloon above ground level? 3. Where does the balloon land on the ground? 4. Where is the balloon 50 feet above the ground? ### Contributors • David Lippman (Pierce College) • Melonie Rasmussen (Pierce College)
Courses Courses for Kids Free study material Offline Centres More Store # What is the weight of a 70 kg body on the surface of a planet whose mass is ${{\dfrac{1}{7}}^{th}}$ that of the earth and radius is half of the earth?A. 20 kgB. 40 kgC. 70 kgD. 140 kg Last updated date: 19th Sep 2024 Total views: 80.7k Views today: 2.80k Verified 80.7k+ views Hint: Weight of a body on any planet is given by mg, where g is the gravitational acceleration on the planet. So, to find the weight of this body on another planet we need to find the gravitational acceleration g’ on the surface of that planet arising due to the gravitational attraction. Which depends on the mass and the radius of the planet. Gravitational attraction force, which is essentially the weight measured on a planet, exerted by the planet on a body of mass m is given as: $\overset{\to }{\mathop{F}}\,=\dfrac{GMm}{{{R}^{2}}}$ Where, M is the mass of the planet m is the mass of body R is the radius of the planet G is the gravitational constant Since we know that weight is also represented mathematically as mg. Let us take the mass of earth as M and radius as R. Therefore, the acceleration experienced by the object of mass m on earth will be g: $g=\dfrac{GM}{{{R}^{2}}}$ According to the question: Mass of that planet M’ $=\dfrac{M}{7}$ where M is the mass of earth Radius of that planet R’ $=\dfrac{R}{2}$ where R is the Radius of earth Now we can put these values in the above given equation as; \begin{align} & g'=\dfrac{GM'}{{{R}^{'2}}} \\ & \Rightarrow g'=\dfrac{4GM}{7{{R}^{2}}} \\ & \Rightarrow g'=\dfrac{4}{7}g \\ \end{align} As we know that mg = 70kg is the weight on earth. Multiplying m on both sides: \begin{align} & mg'=\dfrac{4}{7}mg \\ & \Rightarrow mg'=\dfrac{4}{7}70kg=40kg \\ \end{align} So, the weight of a 70 kg body on the given planet will be 40kg. Option B. Note: Gravitational force is a conservative force, which means that work done by this force doesn’t depend on the path followed. Gravitational force’s effect decreases with increase in distance, and increases with increase in mass.
# Day 8 - Monday, March 30, 2020 • MATH Overview: Math Packet - Week 3, Thursday - - Adding and Subtracting Decimals Estimated Time: Approximately 30 minutes Explanation: Today you will review Adding and Subtracting Decimals. Then, you can play Decimal Victory to practice decimal comparisons. Things to know: • When adding and subtracting decimals, remember to line up the decimals, which will line up your place values. • Please visit https://www.cbsd.org/Page/1511 to access tutorial videos related to accessing online Math in Focus resources. • There are no pre-made worksheets for today. MATH Packet - Week 3, Thursday • Shuffle your set of number cards. • Flip over 3 cards and place them in front of you in a horizontal line. Flip over 3 more cards and place these under each of the first three. Add a decimal to each number either after the first or second digit. • On a separate piece of paper, add and subtract the decimal numbers you created. • Repeat 3 times. You will have 6 problems on your separate of paper. • Decimal Detective: • Locate 8 ways that decimals are used around your house. Share with some else at least 3 of those ways. As a follow up question, “How do decimals (particularly the adding and subtracting of decimals) impact of daily lives?” • You do not need to write this answer down, please just discuss. • Extra: Only do this activity if you want more practice. • Play “Maze: Four Operations of Decimals” on Ed: Your Friend in Learning • Decimal Victory Game: • Shuffle your set of number cards. • Flip over 4 cards and place them in front of you in a horizontal line. • Add a decimal marker between two of the digits to make a number with a decimal. • Considering the remining digits in the pile, make 1 number that would be larger than the number in front of you. (Add the decimal numbers on a separate sheet) • Considering the remining digits in the pile, make 1 number that would be smaller than the number in front of you. (Add and Subtract the decimal numbers on a separate sheet) How is this assignment turned into Mrs. Maduzia Post a picture of your completed decimal problems to Seesaw! There is no assigned reading activity for today. Please be sure you have read up to Pg. 85 in Night of the Twisters WRITING Overview: Choice Writing Estimated Time: Approximately 20 minutes Explanation: Today you will be choosing a short prompt to respond to. Things to know: • You'll need the April writing prompts, which can be found here.
## Subtracting Fractions With Common Denominators ### Learning Outcomes • Model fraction subtraction • Subtract fractions with a common denominator • Subtract fractions with a common denominator that contain variables ## Model Fraction Subtraction Subtracting two fractions with common denominators is much like adding fractions. Think of a pizza that was cut into $12$ slices. Suppose five pieces are eaten for dinner. This means that, after dinner, there are seven pieces (or ${\Large\frac{7}{12}}$ of the pizza) left in the box. If Leonardo eats $2$ of these remaining pieces (or ${\Large\frac{2}{12}}$ of the pizza), how much is left? There would be $5$ pieces left (or ${\Large\frac{5}{12}}$ of the pizza). ${\Large\frac{7}{12}}-{\Large\frac{2}{12}}={\Large\frac{5}{12}}$ Let’s use fraction circles to model the same example, ${\Large\frac{7}{12}}-{\Large\frac{2}{12}}$. Start with seven ${\Large\frac{1}{12}}$ pieces. Take away two ${\Large\frac{1}{12}}$ pieces. How many twelfths are left? Again, we have five twelfths, ${\Large\frac{5}{12}}$. ### Example Use fraction circles to find the difference: ${\Large\frac{4}{5}}-{\Large\frac{1}{5}}$ Solution: Start with four ${\Large\frac{1}{5}}$ pieces. Take away one ${\Large\frac{1}{5}}$ piece. Count how many fifths are left. There are three ${\Large\frac{1}{5}}$ pieces left, or ${\Large\frac{3}{5}}$. ## Subtract Fractions with a Common Denominator We subtract fractions with a common denominator in much the same way as we add fractions with a common denominator. ### Fraction Subtraction If $a,b,\text{ and }c$ are numbers where $c\ne 0$, then ${\Large\frac{a}{c}}-{\Large\frac{b}{c}}={\Large\frac{a-b}{c}}$ To subtract fractions with a common denominators, we subtract the numerators and place the difference over the common denominator. ### Example Find the difference: ${\Large\frac{23}{24}}-{\Large\frac{14}{24}}$ ### Try It Watch the following video for more examples of subtracting fractions with like denominators. ## Subtract Fractions with Variables ### Example Find the difference: ${\Large\frac{y}{6}}-{\Large\frac{1}{6}}$ ### Example Find the difference: ${\Large-\frac{10}{x}-\frac{4}{x}}$ ### Try It Now lets do an example that involves both addition and subtraction. ### Example Simplify: ${\Large\frac{3}{8}}+\left(-{\Large\frac{5}{8}}\right)-{\Large\frac{1}{8}}$ ### Try It In the next video we show more examples of subtracting fractions with a common denominator. Make note of the second example, it addresses a common mistake made by students when simplifying fractions with variables. ## Contribute! Did you have an idea for improving this content? We’d love your input.
Опубліковано Zero Power Law We know that every number is non-zero divided by itself equal to 1. So I can write this: We start by looking at a common division by zero ERROR. To understand the null exponent purpose, we will also rewrite x5x-5 with the negative exponent rule. Therefore, we can conclude that every number, except zero, that is increased to the zero power is 1. Simplify each of the following expressions by using the zero exponent rule for exponents. Write each expression only with positive exponents. The exhibitors seem pretty simple, right? Increasing a number to the power of 1 means you have one of these numbers, increasing to the power of 2 means you have multiplied two of the numbers, power 3 means three of the multiplied number, and so on. This is where things get complicated. The above method breaks because, of course, division by zero is a no-no. Let`s see why. Taking into account several ways to define an exponential number, we can deduce the zero exponent rule by considering: If we generalize this rule, we have the following, where n is a nonzero real number and x and y are also real numbers. Remember that any non-zero real number that is high zero is one, so there is no value! Each number multiplied by zero is equal to zero, it can never be equal to 2. Therefore, we say that division by zero is not defined. There is no solution. But what about zero power? Why is each non-zero number incremented to 1 to the power of zero? And what happens if we increase from zero to zero? Is it still 1? 0° = not defined. This is like dividing a number by zero. In the following example, if we apply the product rule for exponents, we get an exponent of zero. The exponent is attached to the upper right shoulder of the base. It defines how many times the base is multiplied by itself. For example, 4 3 represents an operation. 4 x 4 x 4 = 64. On the other hand, a broken power represents the root of the base, for example (81) 1/2 is equal to 9. When practicing calculus, if you are dealing with an equation that results in an indeterminate form of zero to zero power, be sure to implement techniques for indeterminates, such as Lâhopital`s rule, to correctly evaluate the limit. This makes it easy to explain why any non-zero number is equal to 1. Let`s look again at a concrete example. Including â1 in the definition, we can conclude that any number (including zero) repeated zero times is equal to 1. Negative exponents and null exponents often appear when formulas are applied or expressions are simplified. Since each number divided by itself is always 1;52 * 5-2 = 52 * 1/52 = 52/52 = 25/25 = 1525-2 = 5(2-2) = 5052 5-2 = 52/52 = 1This implies that 50 = 1. This proves the zero exponent rule. This implies that any number is x0 = 1. This proves the zero exponent rule. We can use the same process as in this example with the generalized rule above to show that any nonzero real number increased to zero power must give 1. The zero exponent indicates that there are no factors of a number. This lesson explains how to find the power of a non-zero number or variable that is elevated to zero power. Therefore, we can write the rule as a° = 1. Alternatively, the zero exponent rule can be proved by considering the following cases. Therefore, it is proved that any number or expression raised to zero is always equal to 1. In other words, if the exponent is zero, then the result is 1. The general form of the zero exponent rule is given by: a 0 = 1 and (a/b) 0 = 1. If we try to use the above method with zero as a basis for determining what zero would be at zero power, we stop immediately and cannot continue because we know that 0÷0 â is 1 but indeterminate. Of course, we can take a shortcut and subtract the number of 2 below from the number of 2 above. Since these quantities are represented by their respective exponents, it is sufficient to write the common basis with the difference of the values of the exponents as power. b) Apply the zero exponent rule to each term, and then simplify it. The zero exponent on the first term applies only to the 3 and not to the negative before the 3. It is always true that any non-zero real number raised to zero is one, and we know that ??? 3xy + A??? is really just a representation of a number. This means that the mathematical community is in favor of defining zero to zero as 1, at least in most cases. Any non-zero real number increased to zero is one, that is, all that ??? seems a^0??? is always the same ??? 1??? if??? One??? is non-zero. These limits cannot be evaluated directly, as they are indeterminate forms. Instead, we must use Lâhopital`s rule by taking the derivative of the numerator and denominator separately to find that the solutions are 2 and 3, respectively. Perhaps a useful definition of exponent for the amateur mathematician is as follows: Now, remember, a negative exponent implies that one is divided by the number at the exponent: Now, let`s generalize the formula by calling any number x: You can find that 33 = (34) / 3, 32 = (33) / 3, 31 = (32) / 33 (n-1) = (3n) / 3So 30= (31) / 3 = 3 / 3 = 1 The concept of indefinite forms is common in calculus. A simple example of why 0/0 is indeterminate can be found by looking at some basic limits. x2/x 2 = x 2 – 2 = x 0 But we already know that x2/x2 = 1; so x 0= 1 In this section, we define the negative exponent rule and the zero exponent rule and look at some examples. Good news, the rule still applies if you have more than one variable or a combination of variables and numbers. Since 2/2 = 1, cancel three sentences of 2/2. There are only 2 or 2 squares left. Let`s start by looking at sharing value with exhibitors. x a * x b = x (a + b)If we change one of the exponents to a negative: x a * x-b = x(a-b)And if the exponents have equal sizes, x a * x-b = x a * x-a = x(a-a) = x0 Example 231 = 3 = 332 = 33 = 933 = 333 = 2734 = 3333 = 81And so on. I create online courses to help you rock your math class. Learn more. Recall exponents represent repeated multiplication. So we can paraphrase the above expression as follows: But working with negative exponents is just the rule of exponents that we must be able to use when working with exponential expressions. This issue is hotly debated. Some think it should be defined as 1, while others think it is 0, and some believe it is not defined. There are good mathematical arguments for everyone, and perhaps it is more correctly considered indefinite. This formula works for each number, but not for the number 0. Here we encounter a completely different situation. The solution for x could be any real number! There is no way to determine what x is. Therefore, 0/0 is considered indeterminate, not indefinite. How about 2Ã·0? Let`s see why we can`t do it. x-a = 1/x aRewrite x a x-a in another way:x a * x-a = x a * 1/x a = x a/x aAnd since a number divided by itself is always 1, so:x a * x-a = x a * 1/x a = x a/x a = 1: division is really just a form of multiplication, so what happens, if I paraphrase the above equation as follows: What value could this equation satisfy for x? In this formula, replace one of the exponents with negative:52 * 5-4 = 5(2-4) = 5-2 = 0.04What happens if the exponents are the same size:52 * 5-2 = 5(2-2) = 50 Next lesson: Common misconceptions about probability Stay up to date with everything Math Hacks is doing! Apply the negative exponent rule to the numerator and denominator.
# Proofs for the derivatives of eˣ and ln(x) – Advanced differentiation Last Updated : 15 Dec, 2020 In this article, we are going to cover the proofs of the derivative of the functions ln(x) and ex. Before proceeding there are two things that we need to revise: ### The first principle of derivative Finding the derivative of a function by computing this limit is known as differentiation from first principles. Derivative by the first principle refers to using algebra to find a general expression for the slope of a curve. It is also known as the delta method. The derivative is a measure of the instantaneous rate of change, which is equal to ### e in terms of limit The number e, known as Euler’s number, is a mathematical constant approximately equal to 2.71828. The discovery of the constant itself is credited to Jacob Bernoulli in 1683 who attempted to find the value of the following expression (which is equal to e). ## Proof for the Derivative of ex ### Example 1: Find the derivative of ? Solution: By the chain rule, ### Example 2: Find the derivative of ? Solution: Use here the quotient rule: ## Proof for the Derivative of ln(x) ### Example 1: Find the derivative of 3ln(x)? Solution: 3ln(x)’ = 3(1/x) = 3/x ### Example 2: Find the derivative of ln(x)/5? Solution: (ln(x)/5)’ = 1/5(ln(x))′ = (1/5) (1/x) = 1/5x Previous Article Next Article Article Tags :
# Combinations Combination: Meaning of combination is selection of objects. ### Selection of objects without repetition: The number of selections (combinations or groups) that can be formed from n different objects taken r (0 ≤ r ≤ n) at a time is $\large n_{C_r} = \frac{n!}{r! (n-r)!}$ Explanation: Let the total number of selections (or groups) = x. Each group contains r objects , which can be arranged in r! ways. Hence the number of arrangements of r objects = x × (r!). But the number of arrangements = nPr ⇒ x × (r!) = nPr ⇒ $\large x = \frac{n_{P_r}}{r!} = n_{C_r}$ Illustration : There are two boys B1 and B2. B1 has n1 different toys and B2 has n2 different toys. Find the number of ways in which B1 and B2 can exchange their toys in such a way that after exchanging they still have same number of toys but not the same set. Solution: Total number of toys = n1 + n2 Now let us keep all toys at one place and ask B1 to pick up any n1 toys out of these n1 + n2 toys. He can do it in n1+ n2 Cn1 ways . Out of these ways there is one way when he picks up those n1 toys which he was initially having. Thus required number of ways = n1+ n2 Cn1 – 1 Selection of objects with repetition: The number of combinations of n distinct objects, taken r at a time when each may occur once, twice , thrice , ….. upto r times in any combination is n + r − 1Cr Explanation: Let the n objects be a1, a2, a3 ….. an. In a particular group of r objects, let a1 occurs x1 times, a2 occurs x2 times a3 occurs x3 times ……………. ………….. an occurs xn times such that x1 + x2 + x3 + …. + xn = r …. (1) 0 ≤ xi ≤ r ∀ i ∈ {1 , 2 , 3 , ….., n} Now the total number of selections of r objects out of n = number of non-negative integral solutions of equation (1) = n + r − 1Cn − 1 = n + r − 1Cr Illustration : Let 15 toys be distributed among 3 children subject to the condition that any child can take any number of toys. Find the required number of ways to do this if (i) toys are distinct. (ii) toys are identical. Solution: (i) Toys are distinct Here we have 3 children and we want the 15 toys to be distributed to the 3 children with repetition. In other words, it is same as selecting and arranging children 15 times out of 3 children with the condition that any child can be selected any no. of time, which can be done in 315 ways (n = 3, r = 15). (ii) Toys are identical Here we only have to select children 15 times out of 3 children with the condition that any child can be selected any number of times which can be done in 3H15 = 3 + 15 − 1C15 = 17C2 ways (n = 3 , r = 5) Restricted Selection / Arrangement (a) The number of ways in which r objects can be selected from n different objects if k particular objects are (i) always included = n-k Cr-k (ii) never included = n-k Cr (b) The number of arrangements of n distinct objects taken r at a time so that k particular objects are (i) always included = n-k Cr-k .r! (ii) never included = n-k Cr .r! (c) The number of combinations of n objects, of which p are identical, taken r at a time is = n-pCr + n-pCr-1 + n-pCr-2+ …… + n-pC0 if r ≤ p and it is: = n-pCr + n-pCr-1 + n-pCr -2+ ……+ n-pCr-p if r > p Illustration : A delegation of four students is to be selected from a total of 12 students. In how many ways can the delegation be selected (a) If all the students are equally willing. (b) If two particular students have to be included in the delegation. (c) If two particular students do not wish to be together in the delegation. (d) If two particular students wish to be included together only. (e) If two particular students refuse to be together and two other particular student wish to be together only in the delegation. Solution: (a) Formation of delegation means selection of 4 out of 12. Hence the number of ways = 12C4 = 495. (b) If two particular students are already selected. Here we need to select only 2 out of the remaining 10. Hence the number of ways = 10C2 = 45. (c) The number of ways in which both are selected = 45. Hence the number of ways in which the two are not included together = 495 – 45 = 450. (d) There are two possible cases (i) Either both are selected. In this case, the number of ways in which the selection can be made = 45. (ii) Or both are not selected. In this case all the four students are selected from the remaining ten students. This can be done in 10C4 = 210 ways. Hence the total number of ways of selection = 45 + 210 = 255. (e) We assume that students A and B wish to be selected together and students C and D do not wish to be together. Now there are following 6 cases. (i) (A, B, C) selected, (D) not selected (ii) (A, B, D) selected (C) not selected (iii) (A, B) selected (C, D) not selected (iv) (C) selected (A, B, D) not selected (v) (D) selected (A, B, C) not selected (vi) A, B, C, D not selected For (i) the number of ways of selection = 8C1 = 8 For (ii) the number of ways of selection = 8C1 = 8 For (iii) the number of ways of selection = 8C2 = 28 For (iv) the number of ways of selection = 8C3 = 56 For (v) the number of ways of selection = 8C3 = 56 For (vi) the number of ways of selection = 8C4 = 70 Hence total number of ways = 8 + 8 + 28 + 56 + 56 + 70 = 226 Some results related to nCr Some results related to nCr (i) nCr = nCn-r (ii) If nCr = nCk , then r = k or n-r = k (iii) nCr + nCr-1 = n+1Cr (iv) nCr = (n/r) n-1Cr-1 (v) nCr / nCr-1 = n-r+1/r (vi) (a) If n is even , nCr is greatest for r = n/2 (b) If n is odd, nCr is greatest for r = n-1/2 , n+1/2 Illustration : How many triangles can be formed by joining the vertices of an n- sided polygon. How many of these triangles have (i) exactly one side common with that of the polygon (ii) exactly two sides common with that of the polygon (iii) no sides common with that of the polygon Solution: Number of triangles formed by joining the vertices of the polygon = number of selections of 3 points from n points = nC3 Let the vertices of the polygon be marked as A1, A2,A3,…. An. (i) Select two consecutive vertices A1, A2 of the polygon. For the required triangle, we can select the third vertex from the points A4,A5,….. An-1. This can be done in n-4C1 ways. Also two consecutive points (end points of a side of polygon) can be selected in n ways . Hence the total number of required triangles = n.n-4C1 = n(n – 4). (ii) For the required triangle, we have to select three consecutive vertices of the polygon. i.e. (A1 A2 A3), (A2 A3 A4), (A3 A4 A5), …. ,(An A1 A2). This can be done in n ways. (iii) Triangles having no side common + triangles having exactly one side common + triangles having exactly two sides common (with those of the polygon) = Total number of triangles formed ⇒ Triangles having no side common with those of the polygon = nC3 – n(n-4) – n = n(n−4)(n−5)/6
# Which Description Is Correct for the Polynomial 5X³? [ad_1] Which Description Is Correct for the Polynomial 5x³? Polynomials are an integral part of algebra and are often used to represent various mathematical relationships. They consist of terms, each with a coefficient and an exponent. One such polynomial is 5x³, which can be described in different ways depending on the context. In this article, we will explore the different descriptions and answer some frequently asked questions regarding this polynomial. Description 1: A Cubic Polynomial The polynomial 5x³ is a cubic polynomial because it has the highest degree of 3. The term “cubic” refers to the exponent of the highest power of x in the polynomial. Cubic polynomials are known for their characteristic S-shaped curve when graphed. They often represent real-world situations such as the motion of objects, growth rates, or population dynamics. Description 2: A Monomial The polynomial 5x³ can also be described as a monomial. A monomial is a polynomial with only one term. In this case, the term is 5x³, and it does not have any other terms added or subtracted from it. Monomials are commonly used to represent simple mathematical relationships or to solve equations. Description 3: A Polynomial with a Leading Coefficient Another way to describe the polynomial 5x³ is as a polynomial with a leading coefficient. The leading coefficient refers to the coefficient of the term with the highest degree. In this case, the leading coefficient is 5. This type of polynomial is often used in polynomial division or to find the end behavior of a function. FAQs: Q1: How do I graph the polynomial 5x³? A1: To graph the polynomial 5x³, you can plot a few points and then connect them to form a curve. Choose different values of x, plug them into the polynomial expression, and calculate the corresponding y-values. Plot these points on a graph and connect them to get the shape of the graph. Remember that cubic polynomials tend to have an S-shaped curve. See also What Is a Claim of Policy Apex Q2: Can the polynomial 5x³ have negative values? A2: Yes, the polynomial 5x³ can have negative values. The sign of the polynomial depends on the value of x. For example, if x = -1, then 5(-1)³ = -5. Similarly, if x = -2, then 5(-2)³ = -40. The polynomial can take both positive and negative values depending on the input. Q3: What is the degree of the polynomial 5x³? A3: The degree of a polynomial is the highest exponent of x in any of its terms. In this case, the highest exponent is 3, so the degree of the polynomial 5x³ is 3. Q4: How can I simplify the expression 5x³ + 2x³? A4: To simplify the expression 5x³ + 2x³, you can combine like terms. Since both terms have the same exponent of x (³), you can add their coefficients. In this case, 5x³ + 2x³ simplifies to 7x³. In conclusion, the polynomial 5x³ can be described as a cubic polynomial, a monomial, or a polynomial with a leading coefficient. Each description highlights different aspects of the polynomial’s characteristics and usage. By understanding these descriptions, one can better comprehend the nature and applications of this polynomial in various mathematical contexts. [ad_2]
What Is an Example of Subtracting Mixed Numbers With Regrouping? # What Is an Example of Subtracting Mixed Numbers With Regrouping? An example of a math problem that involves subtracting mixed numbers and regrouping is 4 1/4 - 2 1/3. To solve this type of problem, a student must arrive at a common denominator, regroup, subtract, and simplify the answer, if necessary. The first step in solving the example problem is to obtain a common denominator. The two denominators in the example are 4 and 3, and the common denominator should be the lowest common multiple, or the lowest number that is divisible by both numbers. The lowest common multiple of 4 and 3 is 12. Thus, the denominator of the first fraction is multiplied by 3, and the numerator must also be multiplied by 3. The resulting mixed number is 4 3/12. Multiplying the numerator and denominator of the second fraction by 4 results in 2 4/12. Since 4/12 is greater than 3/12, the student must regroup. This is done by subtracting 1 from the whole number in the first mixed number. The 1 is then converted into a fraction with the same denominator as the other fractions, namely 12, and it is added to the fractional part of the first mixed number. Since 12/12 + 3/12 = 15/12, the first mixed number becomes 3 15/12. Finally, solve 3 15/12 - 2 4/12. This gives the answer 1 11/12. In this example, the fractional part of the mixed number is already in its simplest form, so there is no need to simplify. Similar Articles
Discrete Random Variables # Geometric Distribution OpenStaxCollege [latexpage] There are three main characteristics of a geometric experiment. 1. There are one or more Bernoulli trials with all failures except the last one, which is a success. In other words, you keep repeating what you are doing until the first success. Then you stop. For example, you throw a dart at a bullseye until you hit the bullseye. The first time you hit the bullseye is a “success” so you stop throwing the dart. It might take six tries until you hit the bullseye. You can think of the trials as failure, failure, failure, failure, failure, success, STOP. 2. In theory, the number of trials could go on forever. There must be at least one trial. 3. The probability, p, of a success and the probability, q, of a failure is the same for each trial. p + q = 1 and q = 1 − p. For example, the probability of rolling a three when you throw one fair die is $$\frac{1}{6}$$. This is true no matter how many times you roll the die. Suppose you want to know the probability of getting the first three on the fifth roll. On rolls one through four, you do not get a face with a three. The probability for each of the rolls is q = $$\frac{\text{5}}{\text{6}}$$, the probability of a failure. The probability of getting a three on the fifth roll is $$\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)$$ = 0.0804 X = the number of independent trials until the first success. You play a game of chance that you can either win or lose (there are no other possibilities) until you lose. Your probability of losing is p = 0.57. What is the probability that it takes five games until you lose? Let X = the number of games you play until you lose (includes the losing game). Then X takes on the values 1, 2, 3, … (could go on indefinitely). The probability question is P(x = 5). Try It You throw darts at a board until you hit the center area. Your probability of hitting the center area is p = 0.17. You want to find the probability that it takes eight throws until you hit the center. What values does X take on? 1, 2, 3, 4, … n. It can go on indefinitely. A safety engineer feels that 35% of all industrial accidents in her plant are caused by failure of employees to follow instructions. She decides to look at the accident reports (selected randomly and replaced in the pile after reading) until she finds one that shows an accident caused by failure of employees to follow instructions. On average, how many reports would the safety engineer expect to look at until she finds a report showing an accident caused by employee failure to follow instructions? What is the probability that the safety engineer will have to examine at least three reports until she finds a report showing an accident caused by employee failure to follow instructions? Let X = the number of accidents the safety engineer must examine until she finds a report showing an accident caused by employee failure to follow instructions. X takes on the values 1, 2, 3, …. The first question asks you to find the expected value or the mean. The second question asks you to find P(x ≥ 3). (“At least” translates to a “greater than or equal to” symbol). Try It An instructor feels that 15% of students get below a C on their final exam. She decides to look at final exams (selected randomly and replaced in the pile after reading) until she finds one that shows a grade below a C. We want to know the probability that the instructor will have to examine at least ten exams until she finds one with a grade below a C. What is the probability question stated mathematically? P(x ≥ 10) Suppose that you are looking for a student at your college who lives within five miles of you. You know that 55% of the 25,000 students do live within five miles of you. You randomly contact students from the college until one says he or she lives within five miles of you. What is the probability that you need to contact four people? This is a geometric problem because you may have a number of failures before you have the one success you desire. Also, the probability of a success stays the same each time you ask a student if he or she lives within five miles of you. There is no definite number of trials (number of times you ask a student). a. Let X = the number of ____________ you must ask ____________ one says yes. a. Let X = the number of students you must ask until one says yes. b. What values does X take on? b. 1, 2, 3, …, (total number of students) c. What are p and q? c. p = 0.55; q = 0.45 d. The probability question is P(_______). d. P(x = 4) Try It You need to find a store that carries a special printer ink. You know that of the stores that carry printer ink, 10% of them carry the special ink. You randomly call each store until one has the ink you need. What are p and q? p = 0.1 q = 0.9 # Notation for the Geometric: G = Geometric Probability Distribution Function X ~ G(p) Read this as “X is a random variable with a geometric distribution.” The parameter is p; p = the probability of a success for each trial. Assume that the probability of a defective computer component is 0.02. Components are randomly selected. Find the probability that the first defect is caused by the seventh component tested. How many components do you expect to test until one is found to be defective? Let X = the number of computer components tested until the first defect is found. X takes on the values 1, 2, 3, … where p = 0.02. X ~ G(0.02) Find P(x = 7). P(x = 7) = 0.0177. To find the probability that x = 7, • Enter 2nd, DISTR • Scroll down and select geometpdf( • Press ENTER • Enter 0.02, 7); press ENTER to see the result: P(x = 7) = 0.0177 To find the probability that x ≤ 7, follow the same instructions EXCEPT select E:geometcdf(as the distribution function. The probability that the seventh component is the first defect is 0.0177. The graph of X ~ G(0.02) is: The y-axis contains the probability of x, where X = the number of computer components tested. The number of components that you would expect to test until you find the first defective one is the mean, $$\mu \text{ = 50}$$. The formula for the mean is μ = $$\frac{1}{p}$$ = $$\frac{1}{0.02}$$ = 50 The formula for the variance is σ2 = $$\left(\frac{1}{p}\right)\left(\frac{1}{p}-1\right)$$ = $$\left(\frac{1}{0.02}\right)\left(\frac{1}{0.02}-1\right)$$ = 2,450 The standard deviation is σ = $$\sqrt{\left(\frac{1}{p}\right)\left(\frac{1}{p}-1\right)}$$ = $$\sqrt{\left(\frac{1}{0.\text{02}}\right)\left(\frac{1}{0.\text{02}}-1\right)}$$ = 49.5 Try It The probability of a defective steel rod is 0.01. Steel rods are selected at random. Find the probability that the first defect occurs on the ninth steel rod. Use the TI-83+ or TI-84 calculator to find the answer. P(x = 9) = 0.0092 The lifetime risk of developing pancreatic cancer is about one in 78 (1.28%). Let X = the number of people you ask until one says he or she has pancreatic cancer. Then X is a discrete random variable with a geometric distribution: X ~ G$$\left(\frac{1}{78}\right)$$ or X ~ G(0.0128). 1. What is the probability of that you ask ten people before one says he or she has pancreatic cancer? 2. What is the probability that you must ask 20 people? 3. Find the (i) mean and (ii) standard deviation of X. 1. P(x = 10) = geometpdf(0.0128, 10) = 0.0114 2. P(x = 20) = geometpdf(0.0128, 20) = 0.01 1. Mean = μ = $$\frac{1}{p}$$ = $$\frac{1}{0.0128}$$ = 78 2. Standard Deviation = σ = $$\sqrt{\frac{1-p}{{p}^{2}}}$$ = $$\sqrt{\frac{1-0.0128}{{0.0128}^{2}}}$$ ≈ 77.6234 Try It The literacy rate for a nation measures the proportion of people age 15 and over who can read and write. The literacy rate for women in Afghanistan is 12%. Let X = the number of Afghani women you ask until one says that she is literate. 1. What is the probability distribution of X? 2. What is the probability that you ask five women before one says she is literate? 3. What is the probability that you must ask ten women? 4. Find the (i) mean and (ii) standard deviation of X. 1. X ~ G(0.12) 2. P(x = 5) = geometpdf(0.12, 5) = 0.0720 3. P(x = 10) = geometpdf(0.12, 10) = 0.0380 1. Mean = μ = $$\frac{1}{p}$$ = $$\frac{1}{0.12}$$ ≈ 3333 2. Standard Deviation = σ = $$\sqrt{\frac{\text{1}-p}{{p}^{2}}}$$ = $$\sqrt{\frac{1-0.12}{{0.12}^{2}}}$$ ≈ 7.8174 # References “Millennials: A Portrait of Generation Next,” PewResearchCenter. Available online at http://www.pewsocialtrends.org/files/2010/10/millennials-confident-connected-open-to-change.pdf (accessed May 15, 2013). “Millennials: Confident. Connected. Open to Change.” Executive Summary by PewResearch Social & Demographic Trends, 2013. Available online at http://www.pewsocialtrends.org/2010/02/24/millennials-confident-connected-open-to-change/ (accessed May 15, 2013). “Prevalence of HIV, total (% of populations ages 15-49),” The World Bank, 2013. Available online at http://data.worldbank.org/indicator/SH.DYN.AIDS.ZS?order=wbapi_data_value_2011+wbapi_data_value+wbapi_data_value-last&sort=desc (accessed May 15, 2013). Pryor, John H., Linda DeAngelo, Laura Palucki Blake, Sylvia Hurtado, Serge Tran. The American Freshman: National Norms Fall 2011. Los Angeles: Cooperative Institutional Research Program at the Higher Education Research Institute at UCLA, 2011. Also available online at http://heri.ucla.edu/PDFs/pubs/TFS/Norms/Monographs/TheAmericanFreshman2011.pdf (accessed May 15, 2013). “Summary of the National Risk and Vulnerability Assessment 2007/8: A profile of Afghanistan,” The European Union and ICON-Institute. Available online at http://ec.europa.eu/europeaid/where/asia/documents/afgh_brochure_summary_en.pdf (accessed May 15, 2013). “The World FactBook,” Central Intelligence Agency. Available online at https://www.cia.gov/library/publications/the-world-factbook/geos/af.html (accessed May 15, 2013). “UNICEF reports on Female Literacy Centers in Afghanistan established to teach women and girls basic resading [sic] and writing skills,” UNICEF Television. Video available online at http://www.unicefusa.org/assets/video/afghan-female-literacy-centers.html (accessed May 15, 2013). # Chapter Review There are three characteristics of a geometric experiment: 1. There are one or more Bernoulli trials with all failures except the last one, which is a success. 2. In theory, the number of trials could go on forever. There must be at least one trial. 3. The probability, p, of a success and the probability, q, of a failure are the same for each trial. In a geometric experiment, define the discrete random variable X as the number of independent trials until the first success. We say that X has a geometric distribution and write X ~ G(p) where p is the probability of success in a single trial. The mean of the geometric distribution X ~ G(p) is μ = $$\sqrt{\frac{\text{1}-p}{{p}^{2}}}$$ = $$\sqrt{\frac{1}{p}\left(\frac{1}{p}-1\right)}$$. # Formula Review X ~ G(p) means that the discrete random variable X has a geometric probability distribution with probability of success in a single trial p. X = the number of independent trials until the first success X takes on the values x = 1, 2, 3, … p = the probability of a success for any trial q = the probability of a failure for any trial p + q = 1 q = 1 – p The mean is μ = $$\frac{1}{p}$$. The standard deviation is σ = $$\sqrt{\frac{1\text{ }–\text{ }p}{{p}^{2}}}$$ = $$\sqrt{\frac{1}{p}\left(\frac{1}{p}-1\right)}$$ . Use the following information to answer the next six exercises: The Higher Education Research Institute at UCLA collected data from 203,967 incoming first-time, full-time freshmen from 270 four-year colleges and universities in the U.S. 71.3% of those students replied that, yes, they believe that same-sex couples should have the right to legal marital status. Suppose that you randomly select freshman from the study until you find one who replies “yes.” You are interested in the number of freshmen you must ask. In words, define the random variable X. X = the number of freshmen selected from the study until one replied “yes” that same-sex couples should have the right to legal marital status. X ~ _____(_____,_____) <!– <solution id=”id26594865″> G(0.713) –> What values does the random variable X take on? 1,2,… Construct the probability distribution function (PDF). Stop at x = 6. x P(x) 1 2 3 4 5 6 <!– <solution id=”fs-idp46519888″> –> On average (μ), how many freshmen would you expect to have to ask until you found one who replies “yes?” 1.4 What is the probability that you will need to ask fewer than three freshmen? <!– <solution id=”id26615988″> 0.9176 –> # HOMEWORK A consumer looking to buy a used red Miata car will call dealerships until she finds a dealership that carries the car. She estimates the probability that any independent dealership will have the car will be 28%. We are interested in the number of dealerships she must call. 1. In words, define the random variable X. 2. List the values that X may take on. 3. Give the distribution of X. X ~ _____(_____,_____) 4. On average, how many dealerships would we expect her to have to call until she finds one that has the car? 5. Find the probability that she must call at most four dealerships. 6. Find the probability that she must call three or four dealerships. <!– <solution id=”id18672601″> X = the number of dealers she calls until she finds one with a used red Miata. 1, 2, 3,… X ~ G(0.28) 3.57 0.7313 0.2497 –> Suppose that the probability that an adult in America will watch the Super Bowl is 40%. Each person is considered independent. We are interested in the number of adults in America we must survey until we find one who will watch the Super Bowl. 1. In words, define the random variable X. 2. List the values that X may take on. 3. Give the distribution of X. X ~ _____(_____,_____) 4. How many adults in America do you expect to survey until you find one who will watch the Super Bowl? 5. Find the probability that you must ask seven people. 6. Find the probability that you must ask three or four people. 1. X = the number of adults in America who are surveyed until one says he or she will watch the Super Bowl. 2. X ~ G(0.40) 3. 2.5 4. 0.0187 5. 0.2304 It has been estimated that only about 30% of California residents have adequate earthquake supplies. Suppose we are interested in the number of California residents we must survey until we find a resident who does not have adequate earthquake supplies. 1. In words, define the random variable X. 2. List the values that X may take on. 3. Give the distribution of X. X ~ _____(_____,_____) 4. What is the probability that we must survey just one or two residents until we find a California resident who does not have adequate earthquake supplies? 5. What is the probability that we must survey at least three California residents until we find a California resident who does not have adequate earthquake supplies? 6. How many California residents do you expect to need to survey until you find a California resident who does not have adequate earthquake supplies? 7. How many California residents do you expect to need to survey until you find a California resident who does have adequate earthquake supplies? In one of its Spring catalogs, L.L. Bean® advertised footwear on 29 of its 192 catalog pages. Suppose we randomly survey 20 pages. We are interested in the number of pages that advertise footwear. Each page may be picked more than once. 1. In words, define the random variable X. 2. List the values that X may take on. 3. Give the distribution of X. X ~ _____(_____,_____) 4. How many pages do you expect to advertise footwear on them? 5. Is it probable that all twenty will advertise footwear on them? Why or why not? 6. What is the probability that fewer than ten will advertise footwear on them? 7. Reminder: A page may be picked more than once. We are interested in the number of pages that we must randomly survey until we find one that has footwear advertised on it. Define the random variable X and give its distribution. 8. What is the probability that you only need to survey at most three pages in order to find one that advertises footwear on it? 9. How many pages do you expect to need to survey in order to find one that advertises footwear? X = the number of pages that advertise footwear X takes on the values 0, 1, 2, …, 20 X ~ B(20, $$\frac{29}{192}$$) 3.02 No 0.9997 X = the number of pages we must survey until we find one that advertises footwear. X ~ G($$\frac{29}{192}$$) 0.3881 6.6207 pages Suppose that you are performing the probability experiment of rolling one fair six-sided die. Let F be the event of rolling a four or a five. You are interested in how many times you need to roll the die in order to obtain the first four or five as the outcome. • p = probability of success (event F occurs) • q = probability of failure (event F does not occur) 1. Write the description of the random variable X. 2. What are the values that X can take on? 3. Find the values of p and q. 4. Find the probability that the first occurrence of event F (rolling a four or five) is on the second trial. <!– <solution id=”eip-id1169696227831″> X = the number of times you need to roll the die in order to get the face four or five. X can take on the values 1, 2, 3, and so on. p = 2 6 and q = 4 6 0.2222 –> Ellen has music practice three days a week. She practices for all of the three days 85% of the time, two days 8% of the time, one day 4% of the time, and no days 3% of the time. One week is selected at random. What values does X take on? 0, 1, 2, and 3 The World Bank records the prevalence of HIV in countries around the world. According to their data, “Prevalence of HIV refers to the percentage of people ages 15 to 49 who are infected with HIV.”1 In South Africa, the prevalence of HIV is 17.3%. Let X = the number of people you test until you find a person infected with HIV. 1. Sketch a graph of the distribution of the discrete random variable X. 2. What is the probability that you must test 30 people to find one with HIV? 3. What is the probability that you must ask ten people? 4. Find the (i) mean and (ii) standard deviation of the distribution of X. <!– <solution id=”fs-idm80914272″> X ~ G(0.173) P(x = 30) = geometpdf(0.173, 30) = 0.0007 P(x = 10) = geometpdf(0.173, 10) = 0.0313 Mean = μ = 1 p = 1 0.173 ≈ 5.7804 Standard Deviation = σ = 1−p p 2 = 1−0.173 0.173 2 = 5.2566 –> According to a recent Pew Research poll, 75% of millenials (people born between 1981 and 1995) have a profile on a social networking site. Let X = the number of millenials you ask until you find a person without a profile on a social networking site. 1. Describe the distribution of X. 2. Find the (i) mean and (ii) standard deviation of X. 3. What is the probability that you must ask ten people to find one person without a social networking site? 4. What is the probability that you must ask 20 people to find one person without a social networking site? 5. What is the probability that you must ask at most five people? 1. X ~ G(0.25) 1. Mean = μ = $$\frac{1}{p}$$ = $$\frac{1}{0.25}$$ = 4 2. Standard Deviation = σ = $$\sqrt{\frac{1-p}{{p}^{2}}}$$ = $$\sqrt{\frac{1-\text{0}\text{.25}}{{0.25}^{2}}}$$ ≈ 3.4641 2. P(x = 10) = geometpdf(0.25, 10) = 0.0188 3. P(x = 20) = geometpdf(0.25, 20) = 0.0011 4. P(x ≤ 5) = geometcdf(0.25, 5) = 0.7627 ## Footnotes 1. 1 ”Prevalence of HIV, total (% of populations ages 15-49),” The World Bank, 2013. Available online at http://data.worldbank.org/indicator/SH.DYN.AIDS.ZS?order=wbapi_data_value_2011+wbapi_data_value+wbapi_data_value-last&sort=desc (accessed May 15, 2013). ## Glossary Geometric Distribution a discrete random variable (RV) that arises from the Bernoulli trials; the trials are repeated until the first success. The geometric variable X is defined as the number of trials until the first success. Notation: X ~ G(p). The mean is μ = $$\frac{1}{p}$$ and the standard deviation is σ = $$\sqrt{\frac{1}{p}\left(\frac{1}{p}-1\right)}$$. The probability of exactly x failures before the first success is given by the formula: P(X = x) = p(1 – p)x – 1. Geometric Experiment a statistical experiment with the following properties: 1. There are one or more Bernoulli trials with all failures except the last one, which is a success. 2. In theory, the number of trials could go on forever. There must be at least one trial. 3. The probability, p, of a success and the probability, q, of a failure do not change from trial to trial.
# How Does Blockudoku Differ from Classic Sudoku? views You’ve probably heard of Sudoku, a game loved by many for its challenging puzzles. It’s a classic brain game that has been around for a while, where you fill in numbers on a grid. But have you heard of Blockudoku? It’s a newer game, a twist on the original Sudoku, combining block puzzles with Sudoku-style gameplay. Our goal is to help you understand both games better. Maybe you’ll find a new favorite or get tips to improve your gameplay. So, let’s get started on this exciting journey to compare Blockudoku with traditional Sudoku! ## Understanding the Basics Let’s start off by understanding the fundamental basics of both games, including their history as well as their basic rules and gameplay. ### What is Traditional Sudoku? Sudoku is a puzzle game where you fill a 9×9 grid with numbers 1 to 9. Each number must appear only once in each row, column, and 3×3 square. It’s a test of logic, with part of the grid already filled in as your starting point. #### History and Origin Sudoku is a popular puzzle game that has captured the hearts of many around the world. But where did it come from? The game started in the late 19th century in Europe, originally called “Number Place”. It was a grid-based number game found in newspapers. The version we know today as Sudoku was popularized in Japan in the 1980s. The name “Sudoku” is actually short for a Japanese phrase meaning “numbers must be single” or “numbers must occur only once”. #### Basic Rules and Gameplay The rules of traditional Sudoku are simple but the game can be quite challenging. It’s played on a 9×9 grid, divided into nine 3×3 smaller squares. The goal is to fill every row, column, and 3×3 square with numbers from 1 to 9. The tricky part is that each number can appear only once in each row, column, and square. The game starts with some numbers already placed in the grid; your job is to figure out where the rest of the numbers go. As you play, you use logic and deduction to fill in the blank spaces. ### What is Blockudoku? Blockudoku, a great alternative to Sudoku, combines block puzzles and Sudoku. In a 9×9 grid, you place different shaped blocks to fill rows, columns, or squares, which then clear to score points. The game ends when you can’t fit any more blocks. It’s about strategy and spatial thinking. #### Origin and Development Blockudoku is a modern twist on the classic Sudoku game. It was developed in the 21st century and has gained popularity through mobile game apps. Blockudoku combines elements of traditional Sudoku with block puzzle games (like Tetris). The idea was to create a game that feels familiar but offers a new kind of challenge. #### Basic Rules and Gameplay Blockudoku’s gameplay is different from traditional Sudoku. Instead of a 9×9 grid, you have a 9×9 board where you place blocks of different shapes. The goal is to fill rows, columns, or squares to clear them off the board and earn points. Just like in Sudoku, each move requires careful thought and strategy. The game ends when you run out of space to place new blocks. Blockudoku challenges players to think about space and placement, adding a new layer of strategy to the familiar Sudoku formula. ## Puzzle Layout and Design Okay now that you’ve understood the basics, it’s time to learn more about the differences in the two games, especially in layout and design. ### Traditional Sudoku Grid Structure The layout of a traditional Sudoku puzzle is a 9×9 grid. This grid is further divided into nine smaller squares, each containing 3×3 cells. The design is neat and uniform, with each cell designed to hold a single number from 1 to 9. The challenge arises from how these numbers interact across the rows, columns, and squares. Visually, Sudoku grids are clean and simple, focusing the player’s attention on the numbers and their placement. ### Blockudoku Block Arrangements Blockudoku, on the other hand, presents a different visual approach. While it also uses a 9×9 grid, the gameplay revolves around placing blocks of various shapes onto the grid. These blocks can resemble shapes from classic Tetris-like straight lines, L-shapes, or squares. Unlike Sudoku, where the focus is on numbers, Blockudoku emphasizes the spatial arrangement of these block shapes. The visual challenge here is more about fitting the blocks neatly into the grid to clear space and score points. ## Game Mechanics This section will explore the key differences in the game mechanics, and key strategies that differentiate the two games. ### Strategies in Traditional Sudoku In traditional Sudoku, the strategy is all about deduction and logic. Players must analyze the given numbers and deduce where the missing numbers should go based on the rule that each number appears only once in each row, column, and square. Advanced strategies involve spotting patterns, using process of elimination, and sometimes making educated guesses. The mental challenge is to keep track of multiple possibilities while adhering to the game’s strict rules. ### Blockudoku’s Unique Mechanics Blockudoku introduces a different set of mechanics. The strategy here involves not only the placement of blocks but also planning ahead for future moves. Players must think spatially, considering how placing a block will affect the available space for future moves. The game requires a balance between clearing lines and squares efficiently while ensuring that the board doesn’t get overcrowded. Unlike Sudoku, where each move is more isolated, Blockudoku demands a continuous strategy and foresight, making it dynamic and fast-paced. Section 3: Skills and Challenges ## Cognitive Skills Required While both of the games are somewhat similar, what they teach you and what they require you in the form of skills is quite unique to one another. ### Analysis of the skills needed for Sudoku Sudoku demands several strategies as well as cognitive skills. First, it requires keen logical thinking as players must analyze the existing numbers and deduce the correct placements for the missing ones. Patience is essential because solving some puzzles can take time. Memory plays a role as you need to remember the numbers you’ve already placed and which ones are ruled out for each cell. Concentration is vital to avoid errors. Overall, Sudoku hones your problem-solving skills and enhances your attention to detail. ### Skills developed playing Blockudoku Playing Blockudoku develops skills in spatial reasoning and visual analysis. You must assess the shapes of the blocks and how they fit into the grid, which sharpens your ability to visualize and manipulate objects in your mind. Blockudoku also enhances your planning skills as you strategize where to place blocks for optimal clearance. It encourages quick decision-making and adaptability as the game pace increases. These skills are valuable for improving spatial intelligence and strategic thinking. ## Comparing the Difficulty Levels Traditional Sudoku puzzles come in varying levels of difficulty, from easy to extremely challenging. The difficulty depends on the number and placement of given clues at the beginning. Some puzzles require simple logic, while others demand advanced techniques and tactics. Blockudoku also offers different levels, but its difficulty is more dynamic and based on your ability to manage the grid’s space. It can start easy but become progressively challenging as you advance. ## Type of Challenge Each Game Presents Sudoku challenges your analytical thinking and ability to solve complex logical puzzles. It’s about finding the missing numbers within a fixed framework of rules. In contrast, Blockudoku challenges your adaptability and ability to think on your feet. The game’s speed and the constant need to fit blocks into the grid make it a fast-paced spatial puzzle. While Sudoku’s challenge lies in deduction, Blockudoku offers a unique challenge in spatial organization and time management. Both games provide mental stimulation but in different ways, catering to different preferences. ## Online Presence and Resources Sudoku enthusiasts have a wealth of online resources. Various websites offer Sudoku puzzles of different difficulty levels. Online forums and communities allow players to discuss strategies, share solving techniques, and even create and exchange custom puzzles. Tutorials are available for beginners to learn the basics and for advanced players to master advanced solving techniques. Blockudoku has a strong online presence, primarily through mobile gaming platforms and app stores. Players can easily access the game and receive regular updates and improvements. Online communities, social media groups, and forums dedicated to Blockudoku have emerged. These platforms enable players to connect, share their high scores, exchange tips, and discuss strategies. Additionally, developer support ensures bug fixes and enhancements, enhancing the overall player experience. Blockudoku’s digital nature has facilitated its online presence and engagement with players. ## Summary: Here is a summary of this comparison: Feature Traditional Sudoku Blockudoku Basic Concept Fill a 9×9 grid with numbers 1-9 Place blocks in a 9×9 grid to fill rows, columns, or squares Origin Originated in the late 19th century in Europe Developed in the 21st century, popularized through mobile apps Gameplay Numbers must appear once in each row, column, and 3×3 square Blocks of different shapes are placed to clear lines and score points Grid Structure 9×9 grid divided into nine 3×3 squares 9×9 grid without subdivisions Visual Focus Numbers and their placement Spatial arrangement of block shapes Strategies Deduction, pattern recognition, logical thinking Spatial reasoning, planning ahead, quick decision-making Cognitive Skills Logical thinking, concentration, memory Spatial intelligence, strategic planning, adaptability Difficulty Levels Varies from easy to extremely challenging Dynamic, based on managing grid space Type of Challenge Analytical thinking, solving logical puzzles Spatial organization, time management Online Resources Websites, forums, tutorials Mobile gaming platforms, social media groups, developer support ## Final Words Blockudoku and traditional Sudoku offer unique and engaging puzzle experiences. Sudoku, with its history rooted in logical deduction, challenges players to solve number puzzles using strict rules. On the other hand, Blockudoku introduces a dynamic twist with block placement, demanding spatial thinking and quick decision-making. Whether you prefer the classic logic of Sudoku or the spatial strategy of Blockudoku, both provide mental stimulation and entertainment for puzzle lovers. Try them both to discover your favorite! Tags #### Decoding Slot Symbols: Understanding Wilds, Scatters, and Multipliers Slot machines are not only about spinning reels and matching symbols; they also feature special symbols that can significantly impact gameplay and increase your... #### The Mystery of Scatter Symbols: Your Gateway to Free Spins In the world of online slots, symbols play a pivotal role in determining the outcome of the game. 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• https://me.yahoo.com # Differentiation Differentiation of Simple Algebraic Functions from first principles ## Differentiation From First Principles It is sometimes required that Differentiation be carried out from first principles. Consider the following equation $y=Ax^2+Bx+c$ Let there be small increase in x of $\inline&space;\delta&space;x$ and let the corresponding increase in y be $\inline&space;\delta&space;y$ Rewriting the original equation $y+\delta&space;y=A(x+\delta&space;x)^2+B(x+\delta&space;x)+C$ Multiplying out $y+\delta&space;y=A(x^2+2x\delta&space;x+\delta&space;x^2)+B(x+\delta&space;x)+C$ $y+\delta&space;y=Ax^2+2Ax\delta&space;x+A\delta&space;x^2+Bx+B\delta&space;x+C$ But we know that $y=Ax^2+Bx+C$ Subtracting equation (1) from Equation (2) $\delta&space;y=2Ax\delta&space;x+\delta&space;x^2+B\delta&space;x$ As the value $\inline&space;\delta&space;x$ is reduced and tends towards dx i.e. until it is infinitesimally small, the value of $\inline&space;\delta&space;x&space;^2$ tends towards zero and can be neglected. $\inline&space;\delta&space;y$ is now written as $\inline&space;dy$. It can be seen from the diagram that the value of the tangent at x,y is $\inline&space;\displaystyle&space;\frac{\delta&space;y}{\delta&space;x}$ and at the limit this is written as $\inline&space;\displaystyle&space;\frac{dy}{dx}$ Referring to Equation (1) the value of the tangent is given by: $\frac{dy}{dx}=&space;2Ax+B$ ## The Differential Coefficient (gradient Function) $\inline&space;\mathbf{\frac{dy}{dx}}$ is known as the Gradient function and represents the Derivative of y with respect of x. It is also known as the b{Differential Coefficient}. In the simplest case, if $y=Kx^n+\text{Constant}$ then $\frac{dy}{dx}=Knx^{n-1}$ Example: ##### Example - Differentiation Problem Find the differentiation of $y=3x^5+2x^2-5x+6$ Workings Bringing the power of each x variable down, and subtracting 1 from each power of x yields: $\frac{dy}{dx}=5\times&space;3x^{(5-1)}+2\times&space;2x^{(2-1)}-5x^{(1-1)}$ which is simplified further to $\frac{dy}{dx}=15x^4+4x-5$ Solution $\frac{dy}{dx}=15x^4+4x-5$ ## The Differentiation 0f A Product Of Two Functions Of X It is obvious, that by taking two simple factors such as 5 X 8 that the total increase in the product is Not obtained by multiplying together the increases of the separate factors and therefore the Differential Coefficient is not equal to the product of the d.c's of its factors. If $\mathbf{y=uv}$ then $\mathbf{\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}}$ To Prove the Product Rule let $\inline&space;y=uv$ where u and v are both functions of x. Thus when x increases to $\inline&space;x+\delta&space;x$ u and v will also change to $\inline&space;u+\delta&space;u$ and $\inline&space;v+\delta&space;v$. Their product y will therefore become $(u+\delta&space;u)(v+\delta&space;v)=uv+u\cdot\delta&space;v+v\cdot\delta&space;u+\delta&space;u\;\delta&space;v$ Therefore $\inline&space;\delta&space;y$, the increase in $\inline&space;y=u&space;\cdot&space;\delta&space;v&space;+&space;v&space;\cdot&space;\delta&space;u&space;+&space;\delta&space;u&space;\cdot&space;\delta&space;v$ Thus $\frac{\delta&space;y}{\delta&space;x}=u\cdot\frac{\delta&space;v}{\delta&space;x}&space;+&space;v\cdot\frac{\delta&space;u}{\delta&space;x}+\frac{\delta&space;u}{\delta&space;x}\cdot\delta&space;v$ In the limit as $\inline&space;\delta&space;u$, $\inline&space;\delta&space;v$ and $\inline&space;\delta&space;y$ tend to zero, so the above equation becomes: $\frac{dy}{dx}=u\;\frac{dv}{dx}+v\;\frac{du}{dx}$ Example: ##### Example - Differentiation - Product Rule Problem Differentiate $y=(x^3+1)(x^2+2)$ using the Product Rule Workings if $y=uv$ then the differential is $\frac{dy}{dx}=u\frac{dv}{dx}&space;+&space;v\frac{du}{dx}$ So for $y=(x^3+1)(x^2+2)$ $u=x^3+1$ $v=x^2+2$ Thus $\frac{dy}{dx}=(x^3+1)\cdot&space;2x+(x^2+2)\cdot&space;3x^2$ Solution $\frac{dy}{dx}=5x^4+6x^2+2x$ ## The Differentiation Of A Product Of Any Number Of Functions Of X The rule for finding the differential coefficient of a product of two functions of x can be extended to apply to the product of any finite numbers of functions of x If $y=u\;v\;w$ Where u, v, w are all functions of x, then regarding this as the product of the two factors u and w: $\frac{dy}{dx}=uv\;\frac{dw}{dx}+w\;\frac{d(uv)}{dx}$ $=uv\:\frac{dw}{dx}+\left&space;(&space;u\;\frac{dv}{dx}+v\;\frac{du}{dx}&space;\right&space;)$ $=uv\:\frac{dw}{dx}+uw\;\frac{dv}{dx}+vw\;\frac{du}{dx}$ And similarly for any finite number of factors. Note An important result follows from the above rules. The differential coefficient of $\inline&space;y^2$ with respect to x can be considered to be the product of two factors each of x and hence is given by: $y\;\frac{dy}{dx}+y\;\frac{dy}{dx}=2y\frac{dy}{dx}$ Similarly, if n is any interger, by taking the product of n factors each of y $y\;\frac{dy}{dx}+y\;\frac{dy}{dx}=2y\frac{dy}{dx}$ The differential coefficient of $\inline&space;y^n$ With respect to $\inline&space;x=ny^{n-1}\frac{dy}{dx}$ ## The Differentiation Of A Quotient Of Two Functions Of X Let $y=\frac{u}{v}}$ Then $\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}}$ The proof from first principles of the Quotient Rule. As with previous proofs from first principles x becomes $\inline&space;x+\delta&space;x$, u becomes $\inline&space;u+\delta$ and v becomes $\inline&space;v+\delta&space;v$ Therefore y becomes $\inline&space;\frac{u+\delta&space;u}{y+\delta&space;y}$ Thus $\delta&space;y=\frac{u+\delta&space;u}{v+\delta&space;v}-\frac{u}{v}=\frac{v\delta&space;u-u\delta&space;v}{v(v+\delta&space;v)}$ Therefore $\frac{\delta&space;y}{\delta&space;x}=\frac{v\frac{\delta&space;u}{\delta&space;x}-u\frac{\delta&space;v}{\delta&space;x}}{v(v+\delta&space;v)}$ In the limit when $\inline&space;\delta$ tends to zero the so will $\inline&space;\delta&space;d\;;\;\delta&space;v$ and $\inline&space;\delta&space;y$ Then $\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}}$ Example: ##### Example - Simple example Problem Differentiate using the quotient rule $y=\frac{3x+4}{5x-3}$ Workings Then $\frac{dy}{dx}=\frac{(5x-3)\;3-(3x+4)\;5}{(5x-3)^2}$ Solution $\frac{dy}{dx}=\frac{(-29)}{(5x-3)^2}$
# The length of a rectangle is 4 inches more than its width, and its perimeter is 34 inches. What is the length and width of the rectangle? Mar 2, 2018 Length l = 10.5”, Width w = 6.5” #### Explanation: Perimeter $P = 2 l + 2 w$ Given l =( w + 4)”, P = 34” $\therefore 34 = 2 \left(w + 4\right) + 2 w$ $4 w + 8 = 34$ w = 26/4 = 6.5” l = w + 4 = 6.5 + 4 = 10.5” Mar 2, 2018 length is $10.5$ inches width is $6.5$ inches #### Explanation: Let length be $l$ Let width be $w$ Let perimeter be $P$ First, we must construct an equation for these variables: $l = w + 4$ $P = 34$ But, Perimeter of a rectangle $= l + w + l + w$ $= 2 l + 2 w$ So: $34 = 2 l + 2 w$ But, since $l = w + 4$, we can substitute for $l$, having only the $w$ variable: $34 = 2 \left(w + 4\right) + 2 w$ $34 = 2 w + 8 + 2 w$ $34 = 4 w + 8$ Solve for $w$: $4 w = 34 - 8$ $4 w = 26$ $w = \frac{26}{4}$ $w = 6.5$ inches Now, we can substitute $6.5$ for $w$ in the Perimeter Equation: $34 = 2 l + 2 w$ becomes: $34 = 2 l + 2 \cdot 6.5$ $34 = 2 l + 13$ Solve for $l$: $2 l = 34 - 13$ $2 l = 21$ $l = \frac{21}{2}$ $l = 10.5$ inches Thus, length is $10.5$ inches Thus, width is $6.5$ inches
Divergence Theorem Delve into the world of Engineering Mathematics with an in-depth exploration of the Divergence Theorem. This fundamental theorem is examined in its entirety, from basic meaning to complex applications, with a focus on its key role in engineering studies. Learn to perfect your calculations, understand its connection with Gauss's and Stokes Theorems, and sidestep common mistakes. Equipped with this knowledge, you will be able to apply the Divergence Theorem effectively in various fields and broaden your mathematical proficiency. Discover how this essential theorem underpins many processes in the vast field of engineering. Create learning materials about Divergence Theorem with our free learning app! • Flashcards, notes, mock-exams and more • Everything you need to ace your exams Understanding the Divergence Theorem Embarking on the journey to discover the Divergence Theorem, it's crucial to understand its significance in vector calculus and its real-world applications such as fluid dynamics and electromagnetism. It pertains to the integral of a vector field's divergence, paving the way for easier computation of flow through closed surfaces. Divergence Theorem Meaning: The Basics The Divergence Theorem, also referred to as Gauss's Theorem, is a straightforward but core principle in vector calculus. Essentially, it provides a link between a flux integral over a closed surface and a triple integral over the volume enclosed by the surface. More specifically, it states that the divergence of a vector field integrated over a volume V enclosed by a closed surface S is equal to the flux of the field across S. Flux: In the context of vector calculus, flux refers to the total amount of a field passing through a certain area. In mathematical form, the Divergence Theorem is represented as: $\int\int\int_V (\nabla \cdot \mathbf{F})\, dV = \int\int_S \mathbf{F} \cdot \mathbf{N}\, dS$ Where: • $$\mathbf{F}$$ is the vector field • $$dS$$ is the elementary area • $$\mathbf{N}$$ is the outward normal • And $$\nabla \cdot \mathbf{F}$$ represents the divergence of the vector field $$\mathbf{F}$$ The usage of the Divergence Theorem is vast, including in digital image processing, computational fluid dynamics, and control engineering. Unpacking the Divergence Theorem The crux of the Divergence Theorem lies in the interconnectedness of the flow in and out of a volume. It illustrates that the divergence inside a volume is a quantification of the density of the sources of the field. The theorem fundamentally bridges calculations linked to a volume to calculations dealing with a surface, making intricate mathematical problems more manageable. Let's try to understand the Divergence Theorem using an example. Imagine you have a fluid flowing within a three-dimensional region (like water within a pipe), and you are keen on determining the net flow out of the region. This can be thought of as calculating the fluid's net divergence in the 3D region. Using the Divergence Theorem, we convert the volume integral that represents the net outflow into a surface integral encompassing the volume, simplifying our computations. Divergence Theorem really comes into its own when the vector field from the function is continuously differentiable. Continuously Differentiable: A function is continuously differentiable if it possesses continuous partial derivatives. The Divergence Theorem covers: Brief Description: Flux integral over a closed surface Provides total scalar 'outflow' of a field Divergence of a vector field Gives a scalar field representing the infinitesimal expansion or contraction at a point Volume integral over the divergence Measures total divergence of a field within the volume Remember, the essence of the Divergence Theorem is not just about mastering mathematical equations but truly understanding the significance behind them: the way flow relates to divergences in a 3D plane and how this makes many mathematical tasks much more manageable. Unravelling Gauss's Divergence Theorem When approaching the fascinating world of vector calculus, one is inevitably introduced to Gauss's Divergence Theorem. This vital theorem provides an essential link between a three-dimensional volume integral and a two-dimensional surface integral, thus opening a gateway to solving complex mathematical problems in a simplified and more efficient way. Key Aspects of Gauss's Divergence Theorem The core concept behind Gauss's Divergence Theorem, often known as the Gauss-Ostrogradsky theorem, lies in its delivery of a definitive method to calculate the flux of a vector field across a closed surface. This theorem essentially bridges the gap between divergences within volumetric entities and the net outward flow across their boundaries. The meaning of Gauss's Divergence Theorem can be delineated by understanding its primary components: Vector Field: A depiction of vector values distributed across a region in the space. Divergence: A scalar value indicating the rate at which the vector field diverges at a particular point. Flux: The rate of flow across a given area within the vector field. In the simplest of terms, Gauss's Divergence Theorem maintains that the total divergence of a vector field over some volume is equivalent to the outward flux of the vector field across the boundary of that volume. Total divergence: A measure of the amount by which the vector field diverges at each point in a given region. The mathematical representation of the Divergence Theorem is: $\int\int\int_V (\nabla \cdot \mathbf{F})\, dV = \int\int_S \mathbf{F} \cdot \mathbf{N}\, dS$ Here: • $$\mathbf{F}$$ is the vector field. • $$dV$$ is the volume element. • $$\mathbf{N}$$ is the normal vector to the surface $$S$$. • $$dS$$ is the area element on the surface $$S$$. • And $$\nabla \cdot \mathbf{F}$$ represents the divergence of $$\mathbf{F}$$. The Connection between Divergence Theorem and Gauss's Theorem Gauss's theorem and the divergence theorem are essentially the same concept packaged under different nomenclature. These names represent the same ideas from different viewpoints in electromagnetism or vector calculus. In electromagnetism, Gauss's law (Gauss's theorem) provides a premise stating the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of space inside. Electric flux: A measure of the number of electric lines of force passing through a given area. Translating the aforementioned into mathematical terms, Gauss's law is often expressed as: $\oint_S \mathbf{E} \cdot d\mathbf{A} = \frac{Q}{\varepsilon_0}$ Where: • $$\mathbf{E}$$ represents electric field, • $$d\mathbf{A}$$ is the area vector, • $$Q$$ stands for the total charge enclosed, and • $$\varepsilon_0$$ is the permittivity of free space. When applied in the context of vector calculus and the flow of an arbitrary vector field, Gauss's law transmutes into the divergence theorem. It becomes an expression of flux of a vector field through a closed surface, corresponding to the divergence in the volume it encloses. Therefore, put simply, the divergence theorem is a generalized version of Gauss's law encompassing all types of vector fields, not just electric. A deep grasp of these concepts is pivotal in the realm of physical sciences and engineering where one often deals with quantities distributed across three-dimensional space, ensuing the need to switch between volume and surface integrals effortlessly. Divergence Theorem Applications in Engineering Mathematics The Divergence Theorem is indubitably an integral part of Engineering Mathematics due to its vast utility in various calculations and problems related to fluid dynamics, electromagnetism, and heat transfer. Through simplification of calculations and enabling a flexible switch between dealing with a surface integral or a volume integral, the Divergence Theorem significantly enhances the proficiency of engineering problem-solving. How Divergence Theorem is applied in Different Fields The intrigue of the Divergence Theorem lies in its ability to extend its application to an array of fields, most notably in the sphere of Engineering studies. The pivotal role it plays manifests through the numerous mathematical challenges it helps navigate in the realms of fluid dynamics, electromagnetism, thermodynamics, and control systems to name a few. In Fluid Dynamics, the Divergence Theorem provides a proficient method to analyse and predict the behaviour of fluid systems. By measuring the net outflow, or divergence, of a fluid across a specified volume, one can determine how the fluid interacts with solid boundaries, predict the pressure distribution inside these volumes and also calculate the drag forces exerted on submerged bodies. In the field of Electromagnetism, the Divergence Theorem translates into Gauss's Law which states that the net electric flux passing outwardly through a closed surface is equal to the charge enclosed by that surface. This understanding is instrumental for the calculation of electric field intensity in a variety of scenarios such as the electric field generated by different charge distributions like point charge, uniformly charged sphere, and infinitely long line charge. In Thermodynamics and Heat Transfer, the Divergence Theorem provides an efficient methodology to determine temperature distributions within a system. By evaluating the divergence of the heat flux, one can calculate the net heat transfer out of or into a specified volume, and predict how heat transfer will affect the system’s performance. Similarly, in the realms of Control Systems and Signal Processing, the Divergence Theorem performs a sentinel role in simplifying calculations related to system stability and signal propagation. It allows engineers to isolate radical factors influencing a system and to develop regulation mechanisms accordingly. The Impact of Divergence Theorem in Engineering Studies The adoption of the Divergence Theorem in engineering studies has revolutionised the traditional approach to mathematical problem solving by promoting efficiency and precision. It has considerably streamlined the handling of complex calculations, simplifying them into a more tangible and manageable form. In courses like Fluid Mechanics, the usage of Divergence Theorem becomes indispensable for understanding movement, energy transformation, and fluid behaviour under various conditions. Examining fluid flow using the theorem allows us to predict how fluids interact with their surroundings and to design systems that can effectively control these interactions for optimum utility. In the study of Electromagnetic Theory, the Divergence Theorem simplifies complex calculations related to electric fields. Mapping these fields becomes significantly simpler, steering towards quicker and accurate results in a subject critical for electrical engineering, communications, and photonics. Engineering Subject Importance of Divergence Theorem Fluid Mechanics Empowers understanding of fluid behaviour and interaction with surroundings. Thermodynamics Enables prediction of heat transfer and system performance. Electromagnetic Theory Simplifies electric field mapping and calculations related to electric flux. Control Systems Assists in determining the stability of systems and signal propagation. Moreover, the Divergence Theorem fuels an axiomatic understanding of convolutions in Signal Processing. This theorem plays an instrumental role in simplifying Fourier transforms, initiating the manipulations of signals based on mathematical models for real-life implication in image and sound processing. Pertaining to Thermodynamics, the Harnessing of the Divergence Theorem enables engineers to predict temperature distribution and heat transfer across various media. This knowledge turns out to be pivotal for the performance analysis and design of systems ranging from heat exchangers to climate control systems, supporting an optimal usage of energy. The Divergence Theorem also extends its role to advanced fields like Quantum Mechanics, Reinforcing concepts like Quantum Divergence and the interpretation of vector fields in Quantum Physics. Undeniably, the Divergence Theorem's infiltration across the spectrum of Engineering studies enhances the capability to visualise and understand complex scientific phenomena, thereby making it an uncontested staple in the pursuit of Engineering Mathematics. Getting to Grips with Divergence Theorem Calculations To adeptly manoeuvre through the expansive landscape of engineering and physical sciences, one must get a solid grasp on working with the Divergence Theorem and executing its calculations efficiently. Handling this theorem can seem daunting initially, but with a keen focus on the laid-out steps and careful avoidance of common missteps, one can master these calculations skillfully. Mastering Divergence Theorem Calculations: Step-by-Step Guide To calculate the divergence of a vector field using the Divergence Theorem, follow these steps: 1. Identify the vector field and the volume in question. Ensure that the volume has a piecewise smooth boundary which is oriented outward. 2. Calculate the divergence of the vector field. This is done by taking the dot product of the del operator, $$\nabla = \hat{i} \frac{\partial}{\partial x} + \hat{j} \frac{\partial}{\partial y} + \hat{k} \frac{\partial}{\partial z}$$, with the vector field $$\mathbf{F} = M\hat{i} + N\hat{j} + P\hat{k}$$. This results in: $\nabla \cdot \mathbf{F} = \frac{\partial M}{\partial x} + \frac{\partial N}{\partial y} + \frac{\partial P}{\partial z}$ 1. Integrate the result over the volume. Remember that the result of the integral will be a scalar quantity. 2. Next, evaluate the surface integral of the vector field across the boundary of the volume. This involves taking the dot product of the vector field with the outward normal vector to the surface, and integrating over the surface. 3. Compare the results from steps 3 and 4. They should be equal, as per the Divergence Theorem. As an example, consider evaluating the flux of the vector field $$\mathbf{F} = x^2 \hat{i} + y^2 \hat{j} + z^2 \hat{k}$$ out of the unit sphere. Here, the divergence of the vector field, $$\nabla \cdot \mathbf{F} = 2x + 2y + 2z$$, and the outward normal to the sphere at $$(x, y, z)$$ is $$\mathbf{N} = \hat{i}x + \hat{j}y + \hat{k}z$$. By treating these quantities in the Divergence theorem, one can calculate the net flux out of the sphere painlessly. Common Mistakes in Divergence Theorem Calculations and How to Avoid Them In the quest to master Divergence Theorem calculations and utilise their full potential, it's vital to be cognisant of some of the most common errors that can occur. Here are a few such erroneous steps often observed and the strategies to bypass them: • Incorrect Divergence Calculation: It's imperative to calculate the divergence $$\nabla \cdot \mathbf{F}$$ accurately, as it directly impacts both sides of the Divergence Theorem. Remember that the divergence is the scalar sum of the rate of change of each component of the vector field in its respective direction. • Misidentification of Volume: An incorrect assessment of the volume over which to calculate divergence can lead to erroneous results. Be precise about determining the volume and its boundaries, and remember the requirement of an outward orientation for the boundary. • Handling Non-Smooth Surfaces: The Divergence Theorem applies to volumes with piecewise smooth boundaries. If the boundary is not smooth, it has to be broken down into piecewise smooth parts. • Calculation of Surface Integral: While evaluating the surface integral, the dot product of the vector field and the outward normal to the surface is taken. The 'outward' facing is crucial. Make sure to identify the correct outward direction at each point on the boundary of the volume. Common mistake Solution Incorrect Divergence Calculation Ensure proper application of the calculation of divergence. It's the scalar sum of the rate of change of the vector field's components. Misidentification of Volume Be precise about the volume under consideration and its boundaries, maintaining the outward orientation of the boundary. With the confidence backed by proper understanding, cautious calculation, and an alert mind set to avoid common pitfalls, you can become proficient at Gauss's Divergence Theorem calculations. It's always about the journey towards perfection, therefore make sure to practice extensively, validate your results, and learn from any mistakes along the way! The Link Between Divergence Theorem and Stokes Theorem As you delve deeper into the sphere of calculus in engineering mathematics, two vital theorems you are bound to encounter are the Divergence Theorem and Stokes' Theorem. Beyond their standalone applications, it is their interconnectedness that underpins a plethora of complex mathematical calculations. Understanding the link between these two theorems can elevate your interpretation and manipulation of vector fields, thus optimising your problem-solving proficiency. Understanding the Connection: Divergence Theorem and Stokes' Theorem Serving as the fundamental pillars of vector calculus, both Stokes' Theorem and the Divergence Theorem act as high-dimensional analogues to the Fundamental Theorem of Calculus. Their mutual connection resides in their ability to demonstrate relationships between local properties of a field and global (or integrated) properties of the field, and thus encapsulating the principle of Gauss' Law and Ampere's Law. Stokes' Theorem, in simple terms, relates the surface integral of the curl of a vector field over a surface \< S \> to the line integral of the vector field over its boundary curve (\< C \>). Expressed mathematically: $\oint \mathbf{F} \cdot d\mathbf{R} = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}$ The Divergence Theorem, conversely, communicates the relationship between the flow (or divergence) of a vector field through a surface to the behaviour of the vector field inside the volume bounded by the surface. To represent this as an equation: $\iiint_V \nabla \cdot \mathbf{F} \,dV = \iint_S \mathbf{F} \cdot d\mathbf{S}$ The intriguing linkage between Stokes' Theorem and the Divergence Theorem becomes explicit when you realise that Stokes' Theorem is, in essence, a special case of the Divergence Theorem. Precisely, Stokes' Theorem can be derived from the Divergence Theorem by considering the curl of a vector field, a concept inherently encoding rotation or circulation, as a pseudoscalar field. This elegant connection reinforces the integral consistency of vector calculus and lays the foundation for more advanced mathematical exploration. Practical Use of Divergence Theorem and Stokes' Theorem in Engineering Mathematics Both the Divergence Theorem and Stokes' Theorem serve as powerful mathematical machinery with extensive practical implications. In the realm of engineering mathematics, these theorems are instrumental in understanding fields like electromagnetics, fluid dynamics, heat conduction, and even interpreting the machinations of advanced computer algorithms. • Divergence Theorem: By providing a means to evaluate flux over a volume rather than a surface, the Divergence Theorem can simplify complex electromechanical analyses. In fluid dynamics or heat transfer, engineers use this theorem to evaluate the net flow or diffusion within a specific volume. • Stokes' Theorem: In engineering fields, Stokes' Theorem can be utilised for determining circulation or flux across a boundary without the need for extensive computation over a field. It's crucial in assessing situations which involve curl or rotation, such as in electromagnetism, rigid body dynamics, and certain aspects of fluid dynamics. Theorem Engineering Utility Divergence Theorem Simplification of electromechanical analyses, evaluation of net flow or diffusion in fluid dynamics and heat transfer Stokes' Theorem Calculation of circulation or flux without extensive field computation, evaluation of electromagnetism and rotational dynamics It is crucial to recognise the immense strides you can make in manipulating mathematical problems by understanding the link between Stokes' Theorem and the Divergence Theorem. Through proper comprehension and practice, you can proficiently utilise these powerful tools to optimise your approaches towards multidisciplinary problems in engineering mathematics. The key to implementing these theorems effectively lies in your creativity in interpreting the physical context and system at hand! Divergence Theorem - Key takeaways • Gauss's Divergence Theorem, also known as the Gauss-Ostrogradsky theorem, provides a method to calculate the flux of a vector field across a closed surface and bridges the gap between divergences within volumetric entities and the net outward flow across their boundaries. • Vector Field - Depiction of vector values distributed across a region in space; Divergence - Scalar value indicating the rate at which the vector field diverges at a point; Flux - The rate of flow across a given area within the vector field. • Gauss's theorem and the divergence theorem refer to the same concept from different viewpoints in electromagnetism or vector calculus. In electromagnetism, it is known as Gauss's law and states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of space inside. • Divergence Theorem has vast applications in Engineering Mathematics, notably in fields like fluid dynamics, electromagnetism, and heat transfer. It simplifies various calculations and enables switching between dealing with a surface integral or a volume integral, improving proficiency in problem-solving. • Understanding the Divergence Theorem calculations involves identifying the vector field and volume, calculating divergence, integrating the result over the volume, evaluating the surface integral, and finally comparing the two results which, according to the theorem, should be equal. What is the Divergence Theorem? Please write in UK English. The Divergence Theorem, also known as Gauss's Theorem, is a fundamental principle in vector calculus. It states that the outward flux of a vector field through a closed surface is equal to the volume integral of the divergence of the field over the region inside the surface. When should the Divergence Theorem be used? Divergence theorem is used when you need to calculate the flux of a vector field across a closed surface. It effectively transforms a complicated triple integral into a simpler double integral, making the calculation easier when dealing with complex three-dimensional vector fields. Is there a difference between Gauss's theorem and divergence? Yes, there is a difference. Gauss's Theorem is a special case of the Divergence Theorem. Gauss's theorem applies to the flow across a closed surface in a vector field, while the Divergence theorem is used for calculating the flow from a volume in a vector field. How can I find flux using the divergence theorem? To find flux using the divergence theorem, firstly compute the divergence of the given vector field. Then, integrate this divergence over the volume of the given shape. Finally, multiply the outcome of the integration by the infinitesimal volume element to get the total flux. What does the Divergence Theorem mean? Write in UK English. The Divergence Theorem, in engineering, states that the total outward flux of a vector field across a closed surface is equivalent to the volume integral of the divergence over the region inside the surface. Essentially, it relates surface integrals and volume integrals. Test your knowledge with multiple choice flashcards What is Gauss's Divergence Theorem? What practical example can help in understanding the use of the Divergence Theorem? What is the significance of the Divergence Theorem in vector calculus? StudySmarter is a globally recognized educational technology company, offering a holistic learning platform designed for students of all ages and educational levels. Our platform provides learning support for a wide range of subjects, including STEM, Social Sciences, and Languages and also helps students to successfully master various tests and exams worldwide, such as GCSE, A Level, SAT, ACT, Abitur, and more. We offer an extensive library of learning materials, including interactive flashcards, comprehensive textbook solutions, and detailed explanations. The cutting-edge technology and tools we provide help students create their own learning materials. StudySmarter’s content is not only expert-verified but also regularly updated to ensure accuracy and relevance. StudySmarter Editorial Team Team Engineering Teachers • Checked by StudySmarter Editorial Team
# What Is Trigonometry Used For In Maths? ## What are the 3 trigonometric ratios? There are three basic trigonometric ratios: sine , cosine , and tangent . Given a right triangle, you can find the sine (or cosine, or tangent) of either of the non- 90° angles. Example: Write expressions for the sine, cosine, and tangent of ∠A .. ## What is the concept of trigonometry? Trigonometry (from Greek trigōnon, “triangle” and metron, “measure”) is a branch of mathematics that studies relationships between side lengths and angles of triangles. The field emerged in the Hellenistic world during the 3rd century BC from applications of geometry to astronomical studies. ## What are the basics of trigonometry? There are six trigonometric ratios, sine, cosine, tangent, cosecant, secant and cotangent. These six trigonometric ratios are abbreviated as sin, cos, tan, csc, sec, cot. These are referred to as ratios since they can be expressed in terms of the sides of a right-angled triangle for a specific angle θ. ## Where is trigonometry used? Trigonometry can be used to roof a house, to make the roof inclined ( in the case of single individual bungalows) and the height of the roof in buildings etc. It is used naval and aviation industries. It is used in cartography (creation of maps). Also trigonometry has its applications in satellite systems. ## What is sin equal to? Always, always, the sine of an angle is equal to the opposite side divided by the hypotenuse (opp/hyp in the diagram). The cosine is equal to the adjacent side divided by the hypotenuse (adj/hyp). (1) Memorize: sine = (opposite side) / hypotenuse. cosine = (adjacent side) / hypotenuse. ## Why is trigonometry so difficult? In fact it’s rather basic mathematics. However if all you know is arithmetic and basic linear algebra then yes trigonometry seems more difficult. The reason is that unlike linear algebra (solving linear equations) trigonometry is non-linear which makes the identities between multiples of angles non-trivial. ## How do pilots use trigonometry? What Trigonometry do Pilots use? They must be able to use formulas to find at what angle to lift off and how to get around problems such as mountains and drop of altitude. They have to use trigonometry to find their altitude and to maintain their altitude. ## How do I do trigonometry step by step? There are three steps:Choose which trig ratio to use. – Choose either sin, cos, or tan by determining which side you know and which side you are looking for.Substitute. … Solve. … Step 1: Choose which trig ratio to use. … Step 2: Substitute. … Step 3: Solve. … Step 1: Choose the trig ratio to use. … Step 2: Substitute.More items… ## What is trigonometry and how is it used? Trigonometry is a branch of mathematics that helps us to find the angles and distances of objects. Specifically, it focuses on right-angled triangles – where one angle of the triangle is at 90 degrees. A right-angled triangle means that all sides cannot be the same length. ## What is trigonometry good for? Primarily, trigonometry is a scientific study used to measure bearings and precise distances. Trigonometry uses the three basic functions of Sine, Cosine and Tangent to make its calculations. If two angles or two sides are given, they can be worked out to measure the rest of the details. ## What are the three basic trigonometric functions? The most widely used trigonometric functions are the sine, the cosine, and the tangent. Their reciprocals are respectively the cosecant, the secant, and the cotangent, which are less used in modern mathematics. ## Is trigonometry harder than calculus? If you are talking about the “computational” calculus then that is a lot easier though. On the other hand, computational trig as it’s generally taught in high school is a lot easier than calculus. You usually need to be able to do that sort of trig to be able to do computational calculus. ## What does SOH CAH TOA mean? sine equals opposite over hypotenuse”SOHCAHTOA” is a helpful mnemonic for remembering the definitions of the trigonometric functions sine, cosine, and tangent i.e., sine equals opposite over hypotenuse, cosine equals adjacent over hypotenuse, and tangent equals opposite over adjacent, (1) ## What are the rules of trigonometry? This means that: Sin θ = Cos θ × Tan θ and. Cos θ = Sin θ / Tan θ….Introducing Sine, Cosine and Tangent.NameAbbreviationRelationship to sides of the triangleSineSinSin (θ) = Opposite/hypotenuseCosineCosCos (θ) = Adjacent/hypotenuseTangentTanTan (θ) = Opposite/adjacent ## How do you introduce trigonometry? Introducing TrigonometryMeasure the lengths of the sides of sets of similar right angled triangles and find the ratio of sides.Investigate the relationship between these ratios and the angle size.Use calculators or tables to find the sine, cosine and tangent of angles.More items…
## Worksheet on Ratio and Proportion \| Ratio and Proportion Worksheet with Answers Practicing from Worksheet on Ratio and Proportion helps students to think more of the concept. Solve Ratio and Proportion Question and Answers available to score better grades in the exam. The Questions in this Worksheet are based on Expressing Ratios in their Simplest Form, Simplification of Ratios, Comparison of Ratios, Arranging Ratios in Ascending and Descending Order, Mean Proportional Between Numbers, etc. Solve as many times as possible in order to be familiar with the types of Ratio and Proportion Questions. Answering the Problems over here helps you get a good grip on the entire concept. In addition, you will learn the tips and tricks on how to solve ratio and proportion problems using different methods. For better understanding, we even listed solutions for each and every problem making it easier for you to cross-check whether your answers are correct or not. 1. Express each of the following ratios in the simplest form (a) 5.6 m to 28 cm (b) 6 hours to a day (c) 20 liters to 15 liters (d) 170 : 240 Solution: (a) 5.6 m to 28 cm 1 m = 100cm 5.6 m = 5.6100 = 560 cm Ratio of 5.6m to 20 cm = 560 cm: 20 cm = 140:5 = 28:1 (b) 6 hours to a day In one day there are 24 hrs 6 hrs to a day = 6 hrs: 24 hrs = 1 :4 (c) 20 liters to 15 liters = 20 liters :15 liters = 4:3 (d) 170 : 240 = 170/240 = 17/24 Therefore, ratio of 170:240 in its simplified form is 17/24 2. Simplify the following ratios (a) 1/4 : 1/3 : 1/6 (b) 3.6 : 5.4 (c) 3²/₃ : 4¹/₂ Solution: (a) 1/4 : 1/3 : 1/6 LCM of 4, 3, 6 is 12 Expressing them in terms of a least common factor we have 1/4 = 33/43 = 9/12 1/3 = 14/34 = 4/12 1/6 = 12/62 = 2/12 Therefore 1/4:1/3:1/6 in simplified form is 9:4:2 (b) 3.6 : 5.4 Simplifying it we get the ratio as under Dividing with GCD(3.6, 5.4) i.e. 0.9 we get the simplified form = (3.6/0.9):(5.4/0.9) = 4:6 (c) 3²/₃ : 4¹/₂ = 11/3:9/2 LCM of (3, 2) is 6 Expressing the given ratio in terms of LCM we get the equation as follows 11/3 = 112/32 = 22/6 9/2 = 93/23 = 27/6 Therefore, ratio 3²/₃ : 4¹/₂ in simplified form is 22:27 3. Compare the following ratios (a) 5 : 2 and 4 : 3 (b) 1/3 : 1/5 and 1/5 ∶ 1/6 Solution: (a) 5 : 2 and 4 : 3 Express the given ratio as fraction we get 5:2 = 5/2 4:3 = 4/3 Find the LCM(2, 3) i.e. 6 Making the denominator equal to 6 we get 5/2 = 53/23 = 15/6 4/3 = 42/32 = 8/6 5:2 > 4:3 (b) 1/3: 1/5 and 1/5 ∶ 1/6 1/3:1/5 Finding LCM of 3, 5 we get the LCM as 15 Expressing the ratios given in terms of the LCM as a common denominator 1/3 = 15/35 = 5/15 1/5 = 13/53 = 3/15 thus it becomes 5:3 1/5 ∶ 1/6 Finding LCM of 5, 6 we get the LCM as 30 Expressing the ratios given in terms of the LCM as Common Denominator 1/5 = 16/56 = 6/30 1/6 = 15/65 = 5/30 Therefore, it becomes 6:5 Therefore, 1/3: 1/5 < 1/5 ∶ 1/6 4. In the ratio 3 : 5, the consequent is 20. Find the antecedent? Solution: Let the Antecedent and Consequent be 3x and 5x We know Consequent = 20 5x =20 x= 20/5 = 4 Antecedent = 3x = 34 = 12 Therefore, Antecedent is 12. 5. Divide 2000 among A, B, C in the ratio 2 : 3 : 5? Solution: Let the numbers be 2x, 3x, 5x Sum = 2000 2x+3x+5x = 2000 10x = 2000 x = 2000/10 x = 200 Since the sum is to be split among A, B, C in the ratio of 2:3:5 we get A’s Share = 2x B’s Share = 3x C’s Share = 5x A’s Share = 2200 = 400 B’s Share = 3200 = 600 C’s Share = 5200 = 1000 Therefore, A, B, C’s Share in the amount of 2000 are 400, 600, 1000 respectively. 6. Determine whether the ratios form a Proportion or not (a) 50 cm : 1 m = \$80 : \$160 (b) 200 ml : 2.5 l = \$4 : \$20 Solution: (a) 50 cm : 1 m = \$80 : \$160 50 cm: 1 m We know 1m = 100 cm = 50 cm: 100 cm = 1:2 \$80 : \$160 = 1:2 Since both the ratios are equal they are said to be in Proportion (b)200 ml : 2.5 l = \$4 : \$20 1 liter = 1000 ml 2.5 l = 2.51000 = 2500 200 ml: 2500 ml = 2:25 \$4:\$20 = 1:5 Since both the ratios aren’t equal given values doesn’t form a Proportion. 7. Find the value of x in each of the following (a) 4, 5, x, 48 (b) 7, 21, 30, x (c) x, 28, 24, 4 Solution: (a) 4, 5, x, 48 We know Product of Means = Product of Extremes 448 = 5x 5x = 192 x = 192/5 (b) 7, 21, 30, x We know Product of Means = Product of Extremes 7x = 2130 x = (2130)/7 = 90 (c) x, 28, 24, 4 We know Product of Means = Product of Extremes x4 = 2824 x = (2824)/4 = 168 8. Find the fourth proportional to 54, 27, 18, x Solution: Product of Extremes = Product of Means 54x = 2718 x = (2718)/54 = 9 9. Find the third proportional to (a) 9, 6, x (b) 6, 12, x Solution: To find third proportional we write the expression as 9:6 = 6:x 9x = 66 x = 36/9 = 4 (b) 6, 12, x 6:12 = 12:x 6x = 1212 x = 144/6 x= 24 10. Find the mean proportional between (a) 5 and 20 (b) 1.6 and 0.4 Solution: Mean Proportional between two numbers is defined as the square root of the product of two numbers (a) 5 and 20 = √(520) =√100 = 10 Mean Proportional of 5 and 20 is 10 (b) 1.6 and 0.4 Mean Proportional between two numbers is defined as the square root of the product of two numbers = √(1.60.4) = v6.4 = 0.8 Mean Proportional of 1.6 and 0.4 is 0.8 ## Free Printable Percentage Worksheets \| Percentages Word Problems Worksheets for Practice Percentage is a concept that children often difficulty with while solving related problems. There are several areas in percentages that you need to master in order to get grip on the concept. Our Percentage Worksheets include finding Percentage of a Number, Calculating Percentage Increase, Decrease, changing decimals to and from percents, etc. You will learn percentage is nothing but a fraction over 100. Learn useful tricks, converting between fractions to percentages, percentages, and parts of a whole expressed as a percent value. Solve the Problems in the Percentage Worksheets with Solutions and cross-check where you went wrong. You will no longer feel the concept of percentage difficult once you start practicing these plethora of percent worksheets regularly. In fact, you can find different methods of solving the percentage related problems in no time along with a clear and straight forward description. ### Quick Links for Percentage Concepts Below is the list of Percentage Worksheets available for several underlying concepts. In order to access them, you just need to click on the quick links available and solve the related problems easily. Worksheets on Percentages will make students of different grades familiar with various concepts of Percentages such as Percentage Increase, Percentage Decrease, Conversion from Percentage to Decimal, Fraction, Ratio, and Vice Versa. Percentage Worksheets are free to download, easy to use, and flexible. ### Final Words We wish the information shared regarding Percentage Worksheets has shed some light on you. Make the most out of them to enhance your conceptual knowledge and stand out from the crowd. Bookmark our site to avail more worksheets on topics like Time and Work, Pipes and Cisterns, Exponents, etc. ## Worksheet on Probability \| Free Printable Probability Worksheets to Practice Are you looking for help on the concepts of Probability? Don’t Panic as we have curated the Probability Worksheets with Solutions explaining in detail. Utilize the Worksheet on Probability during your practice sessions and test your knowledge of the concept. Firstly, attempt the Probability Questions in our Worksheets on your own and then cross-check your Answers with the Solutions provided. With consistent practice, you can score better grades in your exams. Worksheets for Probability can be great for learning and practicing the concept. Students of different Grades can utilize these Practice Sheet for Probability and get acquainted with various model questions. Probability Worksheets include concepts like Probability Theory, Applications of Probability, Probability Statistics, etc. Solve more Problems from this Worksheet and attempt the exam with confidence. 1. A coin is tossed 110 times and the tail is obtained 60 times. Now, if a coin is tossed at random, what is the probability of getting a tail? Solution: Probability of an Event to happen = No. of Favourable Outcomes/ Total Number of Outcomes Probability of getting tail when tossed a coin = 60/110 = 6/11 Therefore, the probability of getting tails is 6/11 when tossed a coin. 2. A coin is tossed 200 times and heads are obtained 120 times. Now, if a coin is tossed at random, what is the probability of getting a head? Solution: Probability of an Event to happen = No. of Favourable Outcomes/ Total Number of Outcomes Probability of getting a head = 120/200 = 12/20 = 3/5 Therefore, the probability of getting a head is 3/5. 3. In 120 throws of a dice, 4 is obtained 42 times. In a random throw of a dice, what is the probability of getting 4? Solution: Probability of an Event to happen = No. of Favourable Outcomes/ Total Number of Outcomes Probability of getting 4 = 42/120 = 21/60 = 7/20 The probability of getting 4 is 7/20. 4. What is the Probability of showing neither head nor tail when a coin is tossed? Solution: When a coin is tossed the only possible outcomes are head and tail i.e. 2 Probability of neither head nor tail = 0/2 = 0 Therefore, the Probability of showing neither head nor tail is 0. 5. In 2005, there was a survey of 100 people, it was found that 68 like orange juice while 32 dislike it. From these people, one is chosen at random. What is the probability that the chosen people dislike orange juice? Solution: Probability of people disliking orange juice = number of people disliking orange juice/total number of people = 32/100 = 8/25 Therefore, the Probability of chosen people disliking orange juice is 8/25. 6. In a box there are 20 non-defective and some defective bulbs. If the probability that a bulb selected at random from the box to be defective is 3/4 then find the number of defective bulbs? Solution: Let the number of defective bulbs be x Therefore the total number of bulbs = 20+x Given Probability of Defective Bulbs = 3/4 x/(20+x) = 3/4 4x = 3(20+x) 4x = 60+3x 4x-3x = 60 x = 60 Therefore, the number of defective bulbs is 60. 7. A bag contains 7 white balls and some black balls. If the probability of drawing a black ball from the bag is thrice the probability of drawing a white ball then find the number of black balls? Solution: Let the number of black balls be n Given Number of White Balls = 7 Total Number of Balls = 7+n Probability of Drawing Black Balls = n/7+n Probability of Drawing White Balls = 7/7+n Given Condition is Probability of Drawing Black Ball = 3(Probability of Drawing White Ball) n/(7+n) = 3(7/(7+n)) n/7+n = 21/7+n n = 21 Therefore, the number of black balls is 21. 8. A bag contains 7 red balls, 5 green balls, and some white balls. If the probability of not drawing a white ball in one draw be 2/3 then find the number of white balls? Solution: Let the number of white balls be n Total number of balls in the bag = 7+5+n = 12+n Probability of drawing red ball = 7/12+n Probability of drawing green ball = 5/12+n Probability of drawing white ball = n/12+n Given Probability of not drawing a white ball = 2/3 Thus, the probability of drawing a white ball = 1- 2/3 = 1/3 Therefore, n/12+n =1/3 solving this equation we get the value of n 12+n = 3n 12 = 2n n = 6 Therefore, the number of white balls is 6. 9. One card is drawn at random from a well-shuffled deck of 52 cards. Find the probability that the card is drawn is either a red card or king? Solution: Total Number of Cards = 52 Number of Cards = 26 No. of Kings in Red Cards = 2 Favorable Outcomes for either red card or king = 26+2 Probability of Red Card or King = Favorable Outcomes/Total number of Cards = 28/52 = 7/13 The probability that the card drawn is red or king is 7/13. 10. A die is thrown once, find the probability of getting an odd number and a multiple of 3? Solution: Given a die is thrown once Probability of getting an odd number = {1, 3, 5} Probability of getting an odd number or multiple of 3 is = {1, 3, 5} = 3 Required Probability = 3/6 = 1/2 ## Worksheet on Inverse Variation \| Inverse Variation Worksheet with Answers Worksheet on Inverse Variation includes various questions to practice. Learn how to Solve Inverse Variation Problems by checking out the Sample Problems covering different models. Practice using the Inverse Variation Worksheet as much as possible and get a good grip on the concept. Test your preparation standard using the Worksheet for Inverse Variation and plan your preparation accordingly. Improve your scores in the exam by consistently practicing from the Word Problems on Inverse Variation. 1. If 30 men can reap a field in 12 days, in how many days can 8 men reap the same field? Solution: 30 men – 12 days 8 men – ? Since it is an inverse variation we need to apply the straight multiplication 3012 = 8m m = (3012)/8 = 45 days Therefore, 8 men can reap the same field in 45 days. 2. 10 men can dig a pond in 6 days. How many men can dig it in 5 days? Solution: 10 men – 6 days ? – 5 days Since it is inverse variation apply the straight multiplication 106 = m5 60/5 = m m = 12 Therefore, 12 men can dig the pond in 5 days. 3. A truck covers a particular distance in 2 hours with a speed of 40 miles per hour. If the speed is increased by 10 miles per hour, find the time taken by the truck to cover the same distance? Solution: This is the case of Inverse Variation Because More Speed Less Time Given Speed is 40 miles if it is increased by 10 miles then Speed is 50 miles No. of Hours Speed 2 40 m 50 240 = m50 80 =50m 80/50 = m m = 1.6 hours The truck takes 1.6 hours to cover the same distance. 4. If y varies inversely as x, and y = 16 when x = 3, find x when y = 12? Solution: let y = k/x 16 = k/3 Thus, k = 48 y = k/x 12 = 48/x x = 48/12 x = 4 Therefore, x = 4. 5. The frequency of a vibrating guitar string varies inversely with its length. Suppose a guitar string 0.80 meters long vibrates 4 times per second. What frequency would a string 0.5 meters long have? Solution: We know y = k/x from the given data we can rearrange the equation as f = k/l 4 = k/0.80 k = 0.804 =3.2 f = 3.2/l = 3.2/0.5 = 6.4 times per second. 6. Amar takes 15 days to reduce 20 kilograms of his weight by doing 20 minutes of exercise per day. If he does exercise for 1 hour per day, how many days will he take to reduce the same weight? Solution: More minutes Per Day = Less Days to Reduce Weight Let m be the number of days to reduce weight No. of Days No. of Minutes 15 20 m 60 Since it is Inverse Variation go with straightforward multiplication 1520 = m60 m = (1520)/60 = 5 days Therefore, Amar takes 5 days to reduce weight if he does 1 hour of exercise per day. 7. 12 taps having the same rate of flow, fill a tank in 24 minutes. If three taps go out of order, how long will the remaining taps take to fill the tank? Solution: 12 taps – 24 minutes since three taps went out of order number of taps = 12 -3 =9 12 taps – 24 minutes 9 taps – ? Therefore, applying the inverse variation shortcut we have the equation 1224 = 9m (1224)/9 = m m = 32 Therefore, 9 taps take 32 minutes to fill the tank. 8. 60 patients in a hospital consume 1200 liters of milk in 30 days. At the same rate, how many patients will consume 1440 liters in 28 days? Solution: Given, 60 patients consume 1200 lt of milk in 30 days and say x patients consume 1440 lt of milk in 28 days. 6030/1200 = x28/1440 1800/1200 = 28x/1440 18/12 = 28x/1440 3/2 = 28x/1440 31440/228 = x x = 77 patients(Approx) Therefore, 77 Patients can consume 1440 liters of milk in 28 days. ## Worksheet on Direct Variation \| Word Problems on Direct Variation with Solutions Worksheet on Direct Variation includes different questions to practice. Learn how to solve Word Problems on Direct Variation by referring to the Solved Examples available. We have provided Step by Step Solutions for all the Problems explained in the Direct Variation Worksheet. Practice using them and learn different methods used to approach. Assess your strengths and weaknesses using the Worksheet for Direct Variation and plan your preparation accordingly. 1. If 10 oranges cost \$ 14, how many oranges can be bought for \$ 32? Solution: Given 10 oranges = \$14 ? = \$32 Let the number of oranges that can be bought using the \$32 is x rearranging we get the value of x as x14 = 1032 x = 320/14 x = 22 Oranges(Approx) Therefore, 22 Oranges can be bought for \$32. Solution: From the given data Let the cost of 20 basketballs be m Cross multiplying we get m = (201140)/70 = \$325.71 3. 7 men can complete a work in 42 days. In how many days will 5 men finish the same work? Solution: From given data Men Days 7 42 5 ? Cross multiply to get the no. of days 5 men finish the work no. of days = (542)/7 = 30 Therefore, 5 men take 30 days to finish the work. 4. If a car covers 90 km in 6 liters of petrol, how much distance will it cover in 24 liters of petrol? Solution: 90 Km – 6 Liters ? – 24 Liters distance be d cross multiply to obtain the value of d we have d = (9024)/6 = 360 Km Car Covers 360 Km in 24 Liters of Petrol. 5. If 3 Persons weave 210 Shawls? How many shawls will be Weaven by 7 Persons? Solution: 3 Persons – 210 Shawls 7 Persons – ? Cross multiplying we get No. of Shawls = (2107)/3 = 490 7 Persons can weave 490 Shawls. 6. In 24 Weeks, Michael raised \$240,000 for cancer research. How much money will he raise 36 weeks? Solution: 24 Weeks – \$2,40,000 36 Weeks – ? Money raised by Michael = (2,40,00036)/24 = \$3,60,000 Michael raises \$3,60,000 in 36 Weeks. 7. If 5 men can paint a house in 12 hours, how many men will be able to paint it in 36 hours? Solution: 5 men – 12 hours ? – 36 hours No. of Men = (536)/12 = 15 15 Men can Paint a House in 36 Hours. 8. If the cost of transporting 60 kg of goods for 100 km is Rs 120, what will be the cost of transporting 160 kg of goods for 200 km? Solution: The cost of transporting 60kg of goods for 100 km is 120 cost of 1 kg for 1 km = 120/60100 = Rs. 1/50 cost of 160kg for 200 km = 1/50160200 = Rs. 640 Cost of transporting 160kg of goods for 200km is Rs. 640 ## Ratio and Proportion Worksheets with Answers \| Worksheet on Ratio and Proportion Ratio and Proportion Worksheet available here makes it easy for you to grasp the related concepts. Questions covered in the Worksheet on Ratio and Proportion includes Simplifying Ratios, Arranging Ratios in Ascending or Descending Order, Expressing Ratio in Simplest Form, etc. Practice as many times as possible using the Ratio and Proportion Worksheets and test your preparation standard. To make it convenient for you we have provided detailed solutions for all the Problems provided. 1. Express each of the following Ratios in Simplest Form? a) 2.4 m to 20 cm b) 4 hours to half day c) 10 liters to 7 liters d)144 to 108 Solution: a) 2.4 m to 20 cm We know 1m = 100 cm 2.4 m = 240 cm express it in the form of Ratio = 240: 20 Divide the ratios with their GCF to obtain the simplest form GCF(240, 20) = 20 Dividing with GCF we obtain the simplified ratio as 12:1 Therefore, 2.4 m to 20 cm expressed in simplest form is 12:1 b) 4 hours to half day We know half-day has 12 hours Expressing 4 hours to half-day we have the ratio as 4:12 Divide the ratio with their GCF(4, 12) i.e. 4 = 1:3 4 hours to half-day expressed in simplest form is 1:3 c) 10 liters to 7 liters = 10:7 Given Ratio is in its simplest form since their GCF is 1 and it can’t be simplified further. d) 144 to 108 = 144:108 To obtain the simplified form of the given ratio Divide it with their GCF GCF(144, 108) = 36 = 4:3 2. Simplify the following ratios (a) 1/3: 1/5: 1/8 (b) 5.4: 6.3 Solution: (a) 1/3: 1/5: 1/8 Find the LCM of the Denominators LCM(3, 5, 8) = 120 Express the given ratios with a common denominator = 40/120: 24/120: 15/120 = 40: 24: 15 Therefore, 1/3: 1/5: 1/8 in simplest form is 40:24:15 (b) 5.4: 6.3 Convert the decimals to integers firstly. Multiply with 10 i.e. 5.410 = 54 6.310 = 63 Expressing the integers in terms of ratio we have 54:63 GCF(54, 63) =9 Divide the ratio with GCF to get the simplified form = 6: 7 Therefore, 5.4:6.3 in simplified form is 6:7 3. Determine whether the following Ratios form a Proportion or not a) 4:5 and 8:10 b) 6:7 and 5:3 Solution: a) 4:5 and 8:10 4:5 = 4/5 8:10 = 8/10 = 4/5 Since both the ratios are in the same proportion they are said to be in proportion Therefore, 4:5 and 8:10 are equal. b)6:7 and 5:3 6:7 = 6/7 5:3 = 5/3 Since both proportions aren’t equal they are not in proportion. 4. Compare Ratio 5: 10 and 3: 6 Solution: 5:10 = 5/10 3:6 = 3/6 LCM of 10, 6 is 30 Expressing given ratios as Equivalent Ratios we have 5/10 = 53/103 = 15/30 3/6 = 35/65 = 15/30 Since both the ratio are having same fractions they are equal. 5. Find the Mean Proportional between 9 and 16? Solution: Let the Mean Proportional be x x2= 916 x2 = 144 x = 12 6. If 10, x, x, 10 are in proportion, then find the value of x? Solution: Product of Means= Product of Extremes xx = 1010 x2 = 100 x= 10 7. From the total strength of the class, if the no of boys in the class is 10 and no of girls in the class is 20, then find the ratio between girls and boys? Solution: Given Strength of Boys = 10 Strength of Girls = 20 The ratio of Girls and Boys = 20:10 Simplifying the ratio we get 2:1 8. Suppose 2 numbers are in the ratio 4:3. If the sum of two numbers is 70. Find the numbers? Solution: Given Ratio of Two Numbers is 4:3 As per the given data 4x+3x = 70 7x = 70 x = 10 Numbers are 4x, 3x 4x = 410 = 40 3x = 310 = 30 ## List of Essay Topics for K-12 Students \| English Essay Writing Topics for Thoughtful Learning English Essays can be way more formulaic than you think. Essays are quite common in your elementary school or mid-school as a part of your curriculum. An Essay is a Short Piece of Information to express your views on a particular topic. To help you out we have listed the Most Common Essay Topics for Students of Different Grades all in One Place. You name it and we have it! Make the most out of the Essay Writing Topics and Ideas provided and become a champ in your competitions or speeches. ## Essay Writing Topics & Ideas for Grades Kindergarten, 1, 2, 3, 4,5, 6, 7,8,9, 10, 11, 12 & Competitive Exams Keeping in mind the Students Point of View we have jotted the frequently asked Essay Topics from Different Categories. We have a large collection of Essay Topics right from Kids to College Students. You can use these Sample Essays as a part of the preparation for Competitive Tests or School Speeches. Read and Practice the Long & Short Essays of different word length as per your requirement and bring out the imaginative side in you. ### How to Write an Essay? It’s not as simple as you sit and write down an Essay. A Lot more planning goes into writing an essay successfully. If you are struggling with and want to improve your skills refer to the tips for essay writing. They are as under • Decide what kind of essay you want to write • Brainstorm the Topic and Research on it. • Choose a Writing Style and think of an effective introduction. • Develop a Thesis Statement and Outline your Essay. • Write down the Essay and Proof Read for any Grammar Mistakes. If you follow the aforementioned steps you can always write a Clear and Simple Essay. Use the Tips for Writing Essays and get yourself right on the path of a Well Written Essay. ### Essay Structure \| How to Structure an Essay? An essay needs to be written in the following manner and each sentence needs to be logical and there has to be connectivity between the previous statement. 1. Introduction 2. Main Body 3. Conclusion Introduction: Introduction plays an important role to grab the user’s attention as well to inform them of what will be covered. Provide background information on the topic supporting your argument. After that, Formulate a Thesis Statement. Main Body: In the Body, you can make arguments supporting your thesis statement and develop ideas accordingly. Interpret and analyze the information you have gathered from various sources and frame sentences on your own. The Length of the Body Text depends on the Type of Essay you are writing. Give your essay a clear structure and try to use paragraphs. Conclusion: Conclusion will be the Final Paragraph in the Essay. It shouldn’t take too much text in your essay. A Great Conclusion is necessary to create a strong impact on the user. ### FAQs on Essay Writing Topics 1. Where do I get Simple & Creative Essay Writing Topics for K-12 Grades Students? You can get Simple & Creative Essay Writing Topics for K-12 Grades Students on our page. 2. What are some of the General Essay Writing Tips? Pick up a topic and prepare an overview of your ideas. Write an effective Introduction followed by Thesis Statement. After that write the Body and Conclusion supporting your views. 3. Is there any Website that Provides Essay Topic Suggestions for Students of Grades K to 12? worksheetsbuddy.com is a true and reliable website that provides Essay Topic Suggestions for Students of Grades K to 12. ### Conclusion We wish the information shared regarding the Essay Topics for k-12 Students or Kids has enlightened you and inspired you to write an essay of your own. You can let us know your interests via the comment section and we will add them at the earliest. Stay in touch with our site worksheetsbuddy.com to avail Essay of any Kind. ## Worksheet on Functions or Mapping \| Functions Worksheet with Answers Those seeking help on Functions or Mapping can make the most out of the Worksheet on Functions or Mapping. Most of the questions prevailing here are based on domain, codomain, and range of a function, identifying functions from Mapping Diagrams, determining whether a set of ordered pairs form a function or not, etc. Attempt all the questions in the Identifying Functions from Mappin Diagrams Worksheets and get a better idea of the concepts within. For the sake of your convenience, we even listed the step by step solutions to all the sample provided. Solve the Mapping and Functions Problems on your own and verify with the Answers Provided and test your preparation level. 1. State whether each set of ordered pair represent a function or not? {(1, 9), (-4, –6), (-6, 5), (3, 8), (7, –16), (–10, 7)} Solution: Given Set of Ordered Pairs {(1, 9), (-4, –6), (-6, 5), (3, 8), (7, –16), (–10, 7)} It is a function since any of the first components aren’t repeated and have unique images in the other set. 2. State whether each set of ordered pair represent a function or not? {(-4, 3), (5, -7), (10, 5), (9, 6), (6, -3), (5, 8), (1, 3)} Solution: Given Set of Ordered Pairs {(-4, 3), (5, -7), (10, 5), (9, 6), (6, -3), (5, 8), (1, 3)} It is not function since the first components are repeated. 3. State whether the following relation represents a function or not? Solution: The above arrow diagram represents a function since each element in the set is associated with one element in the other set. Thus it represents a function. 4. A function is given by f(x) = 3x+3. Find the values of (a) f(1) (b) f(0) (c) f(-2) Solution: Given f(x) = 3x+3 To find f(1) substitute 1 in place of x in the given function f(1) = 31+3 = 3+3 = 6 To find f(0) substitute 0 in place of x in the given function f(0) = 30+3 = 0+3 = 3 To find f(-2) substitute -2 in place of x in the given function f(-2) = 3(-2)+3 = -6+3 = -3 5. Determine whether each set of ordered pairs on the graph represents a function or not? Solution: Ordered Pairs = {(-2, 6) (2, 3) (4, -12) (8, 6) (10, 15) (12, -6) (16, 9)} The above graph represents a function since it has first components unique. 6. Determine whether the following Mapping Diagram represents a function or not? Solution: The above mapping diagram represents a function since each element in the set is mapped with only one element in the other set. 7. Which of the following represents mapping? a) {(3, 2); (7, 3); (6, 5); (8, 7)} b) {(4, 8); (5, 12); (4, 14)} Solution: a) {(3, 2); (7, 3); (6, 5); (8, 7)} The Set of Ordered Pairs represent Mapping since all of them have unique images. b) {(4, 8); (5, 12); (4, 14)} Set of Ordered Pairs doesn’t represent Mapping Since doesn’t have unique images and the first components are repeated. 8. Determine whether the relationship given in the Mapping Diagram is a function or not? Solution: Since 4 is mapped with more than one element as output i.e. 0, 9 the relationship given in the above mapping diagram is not a function. 9. Draw Mapping Diagram for the Relation R = {(1, 7) (2, 8) (8, 3)}? Solution: Place all the domain values in the left column and write the range value in the right column and draw arrows between them indicating the relationship. ## Worksheet on Math Relation \| Math Relation Questions with Answers Students can practice various questions from Relation to get grip on the concepts. This Worksheet on Math Relation covers various topics like Mapping for Relations, Representing Relations using a set of ordered pairs, arrow diagrams, etc. Solve the Ample Problems provided in the Math Relation Worksheet and get a better idea of the concepts within it. Practice the problems on your own and cross-check your solutions with the step by step solutions provided to understand where you went wrong. 1. If A= {3, 4, 5, 6, 7, 8}, B = {9, 10, 11, 12}. What is the number of elements in AxB? Solution: Given n(A) = 6 n(B) = 4 n(AxB) = 24 2. Determine the domain and range of the given function {11, -5), (8, -3), (5, 2), (7, 6), (6, -10)} Range =_____ Domain = _____ Solution: Given Function is {11, -5), (8, -3), (5, 2), (7, 6), (6, -10)} Domain is the first component of the ordered pairs and second component in the ordered pairs is the Range. Domain = {11, 8, 5, 7, 6} Range = {-5, -3, 2, 6, -10} 3. Represent Relation {(-3, 3) (0, -5) (2, 0) (6, 0)} using an Arrow Diagram? Solution: Take the Domain Values on the left column and Range Values in the Right Column and mark the relation between them using arrows. 4. Let A = {10, 15, 18, 21, 24} B = { 3, 5, 6, 7} be two sets and let R be a relation from A to B ‘is multiple of’. Represent in the Set of Ordered Pairs? Solution: R = {(10, 5) (15, 3) (18, 3) (18, 6) (21, 3) (21, 7) (24, 6)} 5. Given the relation R = {(3,4), (7,-1), (x,7), (-3,-4)}. Which of the following values for x will make relation R a function? (a) 8 (b) 7 (c) -3 (d) 3 Solution: (a) 8 To make relation R a Function we need to have a value that doesn’t repeat with the first components of the ordered pairs. Therefore, among all the options 8 is the value that is unique. 6. State whether the following statements are true or false (i) All Functions are Relations (ii) All Relations are Functions (iii) A relation is a set of input and output values that are related in some way Solution: (i) True (ii) False (iii) True 7. Express the Relation as a Set of Ordered Pairs? Solution: Expressing the above Relation as a Set of Ordered Pairs we get R = {(-4, 0) (-4, 9) (-4, 11)} 8. If A = {u, v, w} and B = {x, y}, find A × B and B × A. Check whether the two products equal or not? Solution: Given A = {u, v, w} and B = {x, y} AxB = {(u,x) (u, y) (v,x) (v,y) (w, x) (w, y)} BxA = {(x,u) (y, u) (x, v) (y, v) (x, w) (y, w)} AxB and BxA don’t have the same ordered pairs. Therefore, AxB ≠ BxA. 9. If (u/2 + 1, v+2) = (1, 2/5), find the values of u and v? Solution: Given (u/2 + 3, v+2) = (1, 2/5) As per the Equality of Ordered Pairs we have u/2+3 = 1 u/2 = 1-3 u/2 = -2 u = -2*2 u = -4 v+2 = 2/5 v = 2/5-2 v =(2-4) /5 v = -2/5 Therefore, values of u and v are -4, -2/5. 10. Range is the Set of _____ when it comes to Relations in Math? Solution: The range is the set of y-values when it comes to Relations in Math. ## Relations and Mapping Worksheets \| Worksheet on Relations and Functions with Solutions If you need help on Relations and Mapping solve different questions from Relations and Mapping Worksheets. Try to solve various questions on Relations and Mapping and get the concepts underlying easily. The following sections include questions on Ordered Pairs, Cartesian Product of Two Sets, Identifying whether a Mapping Diagram is function or not, Representation of Math Relation, Domain, and Range, etc. The Worksheets on Relations and Mapping include both complex and sample problems. Students can practice them and get step by step solutions to several example problems. The worksheet over here explains how to interpret and present relations as graphs, ordered pairs, arrow diagrams, etc. 1. Express the Relation as a Set of Ordered Pairs? Solution: Ordered Pairs for the given relation {(-2,2), (-2, 6), (-1,1) (0, 2) (4, 6)} 2. Function g(x) = 6x2+2x-2 find the value of g(-1)? Solution: g(x) = 6x2+2x-2 g(-1) = 6.(-1)2+2(-1)-2 = 6.1-2-2 = 6-4 = 2 3. Function f(x) is given by x3. Find the value of f(2)-f(1)/2-1? Solution: f(x) = x3 f(2) = 23 = 8 f(1) = 13 = 1 (f(2)-f(1))/2-1 = (8-1)/2-1 = 7/1 = 7 4. If If (4x + 2, 2y – 3) = (4, 3) find the values of x, y? Solution: Given (4x + 2, 2y – 3) = (4, 3) As per equality of ordered pairs the first components and second components needs to be equal. 4x+2 = 4, 2y-3 = 3 4x= 4-2, 2y = 3+3 4x = 2, 2y = 6 x = 2/4, y = 6/2 x = 1/2, y = 3 Therefore, values of x, y are 1/2 and 3. 5. From the set of Ordered Pairs {(2, 8); (3, 9); (3, 5); (1, 7)} find the Domain and Range? Solution: From the set of ordered pairs given duplicates are not allowed for domain and range. Domain = {2, 3, 1} Range = {8, 9, 5, 7} 6. Following Figure Shows a Relationship from Set A to B. Write the Relation in Roster Form and also provide the Domain and Range? Solution: From the above arrow diagram, we can write the relation from Set A to Set B in Roster Form as R = {(-2, 4) (2, 4) (4, 16) (5, 25) (6, 36)} Domain = { -2, 2, 4, 5, 6} Range = {4, 16, 25, 36} Repetitions are not allowed in the domain and range. 7. For the Relation given in Tabular Form draw the mapping diagram? Solution: The above tabular form relation can be expressed in Mapping as such 8. Let A = {3, 4, 5, 6} B = {x, y, z} find the Cartesian Product of AxB? Solution: Given A = {3, 4, 5, 6} B = {x, y, z} AxB = {(3,x) (3, y) (3, z) (4, x) (4, y) (4, z) (5, x) (5, y) (5, z) (6, x) (6, y) (6, z)} 9. Write the Domain and Range for the following Relations? (a) R₁ = {(6, 2); (6, 5); (3, 5); (0, 8); (7, 3)} (b) R₂ = {(x, 2); (y, 3); (z, 2); (u, 6)} Solution: The domain is the first component of the ordered pairs and range is the second component of the ordered pairs. Repetitions are not allowed in both Domain and Range. (a) R₁ = {(6, 2); (6, 5); (3, 5); (0, 8); (7, 3)} Domain = { 6, 3, 0, 7} Range = { 2, 5, 8, 3} (b) R₂ = {(x, 2); (y, 3); (z, 2); (u, 6)} Domain = { x, y, z, u} Range = { 2, 3, 6} 10. If AxB = {(x, 3); (x, 4); (x, 5); (y, 3); (y, 4); (y, 5)}. Find BxA? Solution: AxB = {(x, 3); (x, 4); (x, 5); (y, 3); (y, 4); (y, 5)} BxA = {(3, x) (4, x) (5, x) (3, y) (4, y) (5, y)}
# Multiplication Chart 35 Learning multiplication after counting, addition, and subtraction is good. Kids learn arithmetic through a organic progression. This advancement of understanding arithmetic is often the following: counting, addition, subtraction, multiplication, and lastly section. This assertion contributes to the question why understand arithmetic in this series? Most importantly, why learn multiplication soon after counting, addition, and subtraction before department? ## These specifics respond to these concerns: 1. Youngsters find out counting initial by associating graphic physical objects because of their fingers. A concrete case in point: Just how many apples are there any in the basket? More abstract case in point is the way outdated are you currently? 2. From counting amounts, another logical phase is addition combined with subtraction. Addition and subtraction tables can be quite helpful instructing assists for children because they are graphic instruments producing the move from counting simpler. 3. Which ought to be discovered after that, multiplication or section? Multiplication is shorthand for addition. At this point, young children have got a company knowledge of addition. For that reason, multiplication may be the up coming reasonable type of arithmetic to understand. ## Evaluate essentials of multiplication. Also, look at the basics utilizing a multiplication table. Allow us to overview a multiplication case in point. Utilizing a Multiplication Table, grow four times three and obtain a solution 12: 4 by 3 = 12. The intersection of row about three and line four of your Multiplication Table is twelve; a dozen is definitely the response. For kids beginning to find out multiplication, this really is effortless. They can use addition to solve the issue therefore affirming that multiplication is shorthand for addition. Illustration: 4 x 3 = 4 4 4 = 12. It is an exceptional introduction to the Multiplication Table. A further gain, the Multiplication Table is graphic and displays back to learning addition. ## Where by can we start understanding multiplication utilizing the Multiplication Table? 1. Initially, get knowledgeable about the table. 2. Get started with multiplying by a single. Commence at row number one. Go on to line number one. The intersection of row one and column the initial one is the answer: one. 3. Perform repeatedly these methods for multiplying by a single. Grow row one particular by posts one particular through 12. The solutions are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 respectively. 4. Recurring these techniques for multiplying by two. Multiply row two by columns one by means of five. The responses are 2, 4, 6, 8, and 10 correspondingly. 5. Let us jump ahead of time. Recurring these methods for multiplying by five. Grow row several by posts one particular via 12. The solutions are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60 respectively. 6. Now let us boost the amount of difficulty. Recurring these actions for multiplying by a few. Grow row about three by posts 1 by means of twelve. The solutions are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, and 36 respectively. 7. When you are comfortable with multiplication up to now, try a test. Remedy the subsequent multiplication issues in your mind after which compare your responses towards the Multiplication Table: grow 6 and two, flourish nine and three, multiply 1 and 11, grow several and a number of, and multiply 7 as well as 2. The trouble responses are 12, 27, 11, 16, and 14 correspondingly. When you got a number of from 5 various troubles correct, create your own multiplication tests. Compute the answers in your mind, and look them using the Multiplication Table.
# RS Aggarwal Solutions Chapter 1 Real Numbers Exercise - 1D Class 10 Maths Chapter Name RS Aggarwal Chapter 1 Real Numbers Solutions Book Name RS Aggarwal Mathematics for Class 10 Other Exercises Exercise 1AExercise 1BExercise 1CExercise 1E Related Study NCERT Solutions for Class 10 Maths ### Exercise 1D Solutions 1. Define (i) rational numbers (ii) irrational numbers (iii) real numbers Solution (i) Rational numbers: The numbers of the form p/q where p, q are integers and q ≠ 0 are called rational numbers. Example: 2/3 (ii) Irrational numbers: The numbers which when expressed in decimal form are expressible as non-terminating and non-repeating decimals are called irrational numbers. Example: √2 (iii) Real numbers: The numbers which are positive or negative, whole numbers or decimal numbers and rational numbers or irrational number are called real numbers. Example: 2, 1/3, √2 , -3 etc. 2. Classify the following numbers as rational or irrational: (i) 22/7 (ii) 3.1416 (iii) Ï€ (iv) (v) 5.636363 (vi) 2.040040004 (vii) 1.535335333 (viii) 3.121221222 3 (ix) √21 (x) √3 Solution (i) 22/7 is a rational number because it is of the form of p/q, q ≠ 0. (ii) 3.1416 is a rational number because it is a terminating decimal. (iii) Ï€ is an irrational number because it is a non-repeating and non-terminating decimal. (iv)is a rational number because it is a repeating decimal. (v) 5.636363… is a rational number because it is a non-terminating and non-repeating decimal. (vi) 2.040040004… is an irrational number because it is a non-terminating and non-repeating decimal. (vii) 1.535335333… is an irrational number because it is a non-terminating and non repeating decimal. (viii) 3.121221222… is an irrational number because it is a non-terminating and non repeating decimal. (ix) √21 = √3 × √7 is an irrational number because √3 and √7 are irrational and prime numbers. (x)an irrational number because 3 is a prime number. So, √3 is an irrational number. 3. Prove that each of the following numbers is irrational: (i) √6 (ii) 2-√3 (iii) 3+√2 (iv) 2+√5 (v) 5+ 3√2 (vi) 3√7 (vii) 3/√5 (viii) 2- 3√5 (ix) √3+ √5 Solution (i) Let √6 = √2×√3 be rational. Hence, √2, √3 are both rational. This contradicts the fact that√2, √3 are irrational. The contradiction arises by assuming √6 is rational. Hence, √6 is irrational. (ii) Let 2 - √3 be rational. Hence, 2 and 2 - √3 are rational. ∴ (2 - 2 + √3) = √3 = rational [∵ Difference of two rational is rational] This contradicts the fact that √3 is irrational. The contradiction arises by assuming 2 - √3 is rational. Hence, 2 - √3 is irrational. (iii) Let 3 + √2 be rational. Hence, 3 and 3 + √2 are rational. ∴ 3 + √2 – 3 = √2 = rational [∵ Difference of two rational is rational] This contradicts the fact that √2 is irrational. The contradiction arises by assuming 3 + √2 is rational. Hence, 3 + √2 is irrational. (iv) Let 2 + √5 be rational. Hence, 2 + √5 and √5 are rational. ∴ (2 + √5 ) – 2 = 2 + √5 – 2 = √5 = rational [∵ Difference of two rational is rational] This contradicts the fact that √5 is irrational. The contradiction arises by assuming 2 - √5 is rational. Hence, 2 - √5 is irrational. (v) Let, 5 + 3√2 be rational. Hence, 5 and 5 + 3√2 are rational. ∴ (5 + 3√2 – 5) = 3√2 = rational [∵ Difference of two rational is rational] 1/3 ×3√2 = √2 = rational [∵ Product of two rational is rational] This contradicts the fact that √2 is irrational. The contradiction arises by assuming 5 + 3√2 is rational. Hence, 5 + 3√2 is irrational. (vi) Let 3√7 be rational. 1/3 × 3√7 = √7 = rational [∵ Product of two rational is rational] This contradicts the fact that √7 is irrational. The contradiction arises by assuming 3√7 is rational. Hence, 3√7 is irrational. (vii) Let 3/√5 be rational. ∴ 1/3 × 3/√5 = 1/√5 [∵ Product of two rational is rational] This contradicts the fact that is 1/√5 irrational. ∴ (1×√5)/(√5×√5) = 1/5.√5 So, if 1/√5 is irrational, then 1/5.√5 is rational ∴ 5 × 1/5.√5 = √5 rational [∵ Product of two rational is rational] Hence, 1/√5 is irrational The contradiction arises by assuming 3/√5 is rational. Hence, 3/√5 is irrational. (viii) Let 2 - 3√5 be rational. Hence 2 and 2 - 3√5 are rational. ∴ 2 – (2 - 3√5) = 2 – 2 + 3√5 = 3√5 = rational [∵ Difference of two rational is rational] ∴ 1/3 × 3√5 = √5 = rational [∵ Product of two rational is rational] This contradicts the fact that √5 is irrational. The contradiction arises by assuming 2 - 3√5 is rational. Hence, 2 - 3√5 is irrational. (ix) Let √3 + √5 be rational. ∴ √3 + √5 = a, where a is rational. ∴ √3 = a - √5 ...(1) On squaring both sides of equation (1), we get 3 = (a - √5) ⇒ 3 = a2 + 5 - 2√5a ⇒ √5 = (a2 + 2)/2a This is impossible because right-hand side is rational, whereas the left-hand side is irrational. Hence, √3 + √5 is irrational. 4. Prove that 1/√3 is irrational. Solution Let 1/√3 be rational. ∴ 1/√3 = a/b, where a, b are positive integers having no common factor other than 1 ∴ √3 = b/a ...(1) Since a, b are non-zero integers, b/a is rational. Thus, equation (1) shows that √3 is rational. This contradicts the fact that √3 is rational. The contradiction arises by assuming √3 is rational. Hence, 1/√3 is irrational. 5. (i) Give an example of two irrationals whose sum is rational. (ii) Give an example of two irrationals whose product is rational. Solution (i) Let (2 + √3), (2 - √3) be two irrationals. ∴ (2 + √3) + (2 - √3) = 4 = rational number (ii) Let 2√3, 3√3 be two irrationals. ∴ 2√3 × 3√3 = 18 = rational number. 6. State whether the given statement is true or false: (i) The sum of two rationals is always rational (ii) The product of two rationals is always rational (iii) The sum of two irrationals is an irrational (iv) The product of two irrationals is an irrational (v) The sum of a rational and an irrational is irrational (vi) The product of a rational and an irrational is irrational Solution (i) The sum of two rationals is always rational - True (ii) The product of two rationals is always rational - True (iii) The sum of two irrationals is an irrational - False Counter example: 2 + √3 and 2 - √3 are two irrational numbers. But their sum is 4, which is a rational number. (iv) The product of two irrationals is an irrational – False Counter example: 2 √3 and 4 √3 are two irrational numbers. But their product is 24, which is a rational number. (v) The sum of a rational and an irrational is irrational - True (vi) The product of a rational and an irrational is irrational - True 7. Prove that (2 √3 – 1) is irrational. Solution Let x = 2 √3 – 1 be a rational number. x = 2√3 – 1 ⇒ x2 = (2√3 – 1)2 ⇒ x2 = (2√3 )2 + (1)2 – 2(2√3)(1) ⇒ x2 = 12 + 1 - 4√3 ⇒ x– 13 = - 4√3 (13 – x2)/4 = √3 Since x is rational number, x2 is also a rational number. ⇒ 13 - x2 is a rational number (13 – x2)4 is a rational number ⇒ √3 is a rational number But √3 is an irrational number, which is a contradiction. Hence, our assumption is wrong. Thus, (2 √3 – 1) is an irrational number. 8. Prove that (4 - 5√2) is irrational. Solution Let x = 4 - 5√2 be a rational number. x = 4 - 5√2 ⇒ x2 = (4 - 5√2)2 ⇒ x2 = 42 + (5√2)2 – 2(4) (5√2) ⇒ x2 = 16 + 50 – 40√2 ⇒ x2 – 66 = – 40√2 ⇒ (66 – x2)/40 = √2 Since x is a rational number, x2 is also a rational number. ⇒ 66 - x2 is a rational number (66 - x2)/40 is a rational number ⇒ √2 is a rational number But √2 is an irrational number, which is a contradiction. Hence, our assumption is wrong. Thus, (4 - 5√2) is an irrational number. 9. Show that (5 - 2√3) is irrational. Solution Let x = 5 - 2√3 be a rational number. x = 5 - 2√3 ⇒ x2 = (5 - 2√3)2 ⇒ x2 = 52 + (2√3)2 – 2(5) (2√3) ⇒ x2 = 25 + 12 – 20√3 ⇒ x2 – 37 = – 20√3 ⇒ (37 – x2)/20 = √3 Since x is a rational number, x2 is also a rational number. ⇒ 37 - x2 is a rational number ⇒ (37 – x2)/20 is a rational number ⇒ √3 is a rational number But √3 is an irrational number, which is a contradiction. Hence, our assumption is wrong. Thus, (5 - 2√3) is an irrational number. 10. Prove that 5√2 is irrational. Solution Let 5√2 is a rational number. ∴ 5√2 = p/q, where p and q are some integers and HCF (p, q) = 1 …(1) ⇒ 5√2q = p ⇒ (5√2q)2 = p2 ⇒ 2(25q2) = p2 ⇒ p2 is divisible by 2 ⇒ p is divisible by 2 ….(2) Let p = 2m, where m is some integer. ∴ 5√2q = 2m ⇒ (5√2q)2 = (2m)2 ⇒2(25q2) = 4m2 ⇒25q2 = 2m2 ⇒ q2 is divisible by 2 ⇒ q is divisible by 2 ….(3) From (2) and (3) is a common factor of both p and q, which contradicts (1). Hence, our assumption is wrong. Thus, 5√2 is irrational. 11. Show that 2/√7 is irrational. Solution 2/√7 = 2√7 × √7/√7 = 2/7√7 Let 2/7.√7 is a rational number. ∴ 2/7.√7 = p/q, where p and q are some integers and HCF (p, q) = 1 ...(1) ⇒ 2√7q = 7p ⇒ (2√7q) 2 = (7p)2 ⇒ 7(4q2) = 49p2 ⇒ 4q2 = 7p2 ⇒ q2 is divisible by 7 ⇒ q is divisible by 7 ...(2) Let q = 7m, where m is some integer. ∴ 2√7q = 7p ⇒ [2√7(7m)]2 = (7p)2 ⇒ 343(4m2) = 49p2 ⇒ 7(4m2) = p2 ⇒ p2 is divisible by 7 ⇒ p is divisible by 7 ….(3) From (2) and (3), 7 is a common factor of both p and q, which contradicts (1). Hence, our assumption is wrong. Thus, 2/√7 is irrational.
# Wendy is making a blanket with a length of 5.4 ft., a width of 6.3 ft., and an area of 34.02 sq. ft. There is a 2 ft. wide border around the blanket. What is the area of the border? ## I tried doing this, again and again, multiple times, but I kept getting it wrong. Please help me! 27.4 sq. ft. #### Explanation: Let's first talk about the blanket. It's dimensions of 5.4 and 6.3 feet do indeed create an area of 34.02 sq. ft. So far so good. Now to the border. We're told there is a 2ft border around the blanket (I'm going to assume it's on all 4 sides). We're looking for the area of the border. There are a couple of ways to approach this - one is to find the border on each side and then sum up all 4 sides. It's a lot of work. The other way is to find the area of the border and blanket together, subtract out the blanket, and we'll have the border. Not much work - I like this one :) Our borders then become $5.4 + 2 = 7.4$ and $6.3 + 2 = 8.3$ feet, giving a total area of: $7.4 \times 8.3 = 61.42$ sq. ft. We subtract out the blanket area to find the area of the border: $61.42 - 34.02 = 27.4$ sq. ft.
# Math Circles at Home: Two Tricks to Teach Your Parents April 5, 2020Uncategorized I used to be hesitant in building a math circle session around tricks, but I’ve come to appreciate their place. On MC2 Parent Surveys, we often get responses like this: “I can see my child’s curiosity about math is sparked–she tells me about what she did afterwards and has to tried to have me play along with games she learned.” There’s a place for puzzles, tricks, and games that students can share with a parent or sibling. For this ‘At Home Math Circle’, I encourage the kid to read the directions and then try it out with a parent! Note: You’ll need a standard deck of cards for each trick, and a standard die for the 2nd. 1. Piles to King Note: It’s essential that your deck of cards be a complete deck with 52 cards–no cards missing, no jokers! Step 1: Deal a card face up. Say it’s a 6. Continue to deal the cards face up on top of that 6, and in your head count up to king. Count in your head–so you’d deal the 6, then count 7,8, 9, 10, jack, queen, king to make a pile (to be clear, the value of the cards will likely not be be 7, 8, 9, etc.; it’s just a way to keep track of how many cards should be in your pile). Note that if the pile starts with 6, it will have 8 cards. If you start with a 3, you’ll get a stack with 11 cards; a stack that starts with a king will only have that one card! When your count gets to king, that pile is done. Keep making other piles in the same manner. Make 4 or 5 piles. Step 2: Turn all of the piles upside down. Ask the subject/parent/victim of your trick to hand you all but three piles. Take the cards that you are handed and put them with the unused remainder of the deck. Step 3: Have the subject turn over the top card of two of the remaining piles. Take the deck and count out the number of cards corresponding to the cards that are revealed. If you see a 3 and a queen, count out 3 cards, then count out 12 cards (a queen is a 12, a king is 13, and a jack is 11; aces are 1’s). Important: After that, deal out 10 more cards. Step 4: Count the remaining cards in your hand. Suppose there are six; announce that the remaining card on top of the third pile is a six (if there are 11, you’d say jack). Then turn the top card over on the final pile, to the amazement of all (assuming you’ve done your arithmetic right!) Tips: • Do all of the counting in your head. This will make the trick more mysterious. • You can make as many piles as you want, just don’t run out of cards. For example, if you are close to running out, and you deal out an ace, that pile would have 13 cards–if you run out, the trick won’t work if you make that pile! 2. The 3 and a Half of Clubs Note: You can buy a three and a half of clubs card online, but, at the moment that may be hard to do. Alternatively (ask your parents if this is ok), draw on the three of clubs and make it a three and a half. 🙂 Step 1: Put the three and a half of clubs 9th from the top of the deck. Shuffle the deck, but don’t disturb the 9 cards at the top! Step 2: Invite the subject to deal out 20 cards, face down, one on top of the other. Step 3: Have the subject then remove from 1 to 9 cards–their choice–from the top of this new stack. (They shouldn’t tell you the number they chose). Step 4: Have the subject figure out how many cards remain in the pile. This is a two digit number. Have them add those digits together, and then remove that many more cards. (If there were 14 cards left in the pile, they would remove 1+4=5 cards). Step 5: Have the subject roll the die but not show you the result. Tell them that the average of the top and bottom of the die will be the same number as the card at the top of the stack. If they laugh at you because the average is 3 and a half, great, because you know what’s going to happen! Extensions: • Try to think about why each trick works. • For Piles to King, how would you have to adjust the trick if there were four piles and the top card were revealed for three of them?
Zivote Cvetkovica ### Pythagoras Theorem and Irrational Numbers It is interesting to see how Pythagoras Theorem helps in identifying the location of an irrational number on the Number Line. Consider a number x ### Teaching High School Mathematics in One Hour Time Slots In the mid-1990s, the administration of the school in which I taught decided to change from using 40 minute teaching periods to 70 minute periods. Lorem ipsum dolor sit amet consectetur adipiscing elit dolor Lorem ipsum dolor sit amet consectetur adipiscing elit dolor Lorem ipsum dolor sit amet consectetur adipiscing elit dolor Lorem ipsum dolor sit amet consectetur adipiscing elit dolor console.log( 'Code is Poetry' ); It is interesting to see how Pythagoras Theorem helps in identifying the location of an irrational number on the Number Line. Consider a number x which is a rational number but not a perfect square. It follows that the square root of x must be irrational, that is, a non-terminating and non-recurring decimal number. Now, our interest is to determine where this lies on the Number Line. To do this, let us consider a right triangle whose base equals (x-1)/2 the hypotenuse equals (x+1)/2. What would be the height of this right triangle, we mean, the other arm of the right triangle? Let us suppose it is y. Pythagoras Theorem tells us that the sum of squares of the arms of a right triangle equals the square of the hypotenuse. So, in the triangle we have just considered, we can write: [(x – 1)/2]^2 + y^2 = [(x + 1)/2]^2 Or y^2 = [(x + 1)/2]^2 – [(x – 1)/2]^2 = [(x^2 + 2x + 1) / 4] – [(x^2 – 2x + 1) / 4] = [(x^2 + 2x + 1) – (x^2 – 2x + 1)] / 4 = [x^2 + 2x + 1 – x^2 + 2x – 1)] / 4 = 4x / 4 = x i.e. y^2 = x â¹ y = √x This is exactly what we were looking for, the square root of x which is irrational. Now, the length of the third arm of the triangle we constructed can be marked on the Number Line using a compass. So, if you are looking for the size of √x, this is how we go about it. Mark a point A. Mark B such that AB = x units. Mark C such that BC = 1 unit. That is, AC = x+1. Bisect AC. If D is the point of bisection of AC, AD = DC = (x+1)/2. Now, what would be the length of DB? Since DC = (x+1)/2 and BC = 1, DB = DC – BC = [(x+1)/2] – 1. That is (x+1-2)/2 or (x-1)/2. Let’s construct the triangle now. Draw a line perpendicular to AC at B. From D, intersect the vertical line at E such that DE = AD. Now we have a right triangle in which the base is (x-1)/2 and the hypotenuse is (x+1)/2. Can you see what the measure of BE is going to be? Of course, as we have shown above, it’s going to be √x. You may transfer this length of BE to the number line now, using the compass. Try representing √5, √7, √11, √6.8 and √9.5 on the Number Line. Each of them must hardly take a couple of minutes or less. Math can be fun. As you explore, it is exciting to see how arithmetic, algebra and geometry converge eventually.
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ # 3.8: Linear Inequalities (Two Variables) $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ Learning Objectives • Identify and check solutions to linear inequalities with two variables. • Graph solution sets of linear inequalities with two variables. ## Solutions to Linear Inequalities We know that a linear equation with two variables has infinitely many ordered pair solutions that form a line when graphed. A linear inequality with two variables, on the other hand, has a solution set consisting of a region that defines half of the plane. Linear Equation Linear Inequality $$y=\frac{3}{2}x+3$$ $$y\leq \frac{3}{2}x+3$$ Figure 3.8.1 Figure 3.8.2 Table 3.8.1 For the inequality, the line defines one boundary of the region that is shaded. This indicates that any ordered pair that is in the shaded region, including the boundary line, will satisfy the inequality. To see that this is the case, choose a few test points and substitute them into the inequality. $$\color{Cerulean}{Test\:point}$$ $$y\leq\frac{3}{2}x+3$$ $$(0,0)$$ \begin{aligned} 0&\leq \frac{3}{2}(0)+3 \\ 0&\leq 3 \quad\color{Cerulean}{\checkmark} \end{aligned} $$(2,1)$$ \begin{aligned} 1&\leq \frac{3}{2}(2)+3 \\ 1&\leq 3+3 \\ 1&\leq 6\quad\color{Cerulean}{\checkmark}\end{aligned} $$(-2,-1)$$ \begin{aligned} -1&\leq\frac{3}{2}(-2)+3 \\ -1&\leq -3+3 \\ -1&\leq 0\quad\color{Cerulean}{\checkmark} \end{aligned} Table 3.8.2 Also, we can see that ordered pairs outside the shaded region do not solve the linear inequality. $$\begin{array} {c\|c} {\underline{\color{Cerulean}{Test\:point}}}&{\underline{y\leq\frac{3}{2}x+3}}\\{(-2,3)} &{3\leq\frac{3}{2}(-2)+3}\\{}&{3\leq -3+3}\\{}&{3\leq 0\quad\color{red}{x}} \end{array}$$ The graph of the solution set to a linear inequality is always a region. However, the boundary may not always be included in that set. In the previous example, the line was part of the solution set because of the “or equal to” part of the inclusive inequality $$≤$$. If we have a strict inequality $$<$$, we would then use a dashed line to indicate that those points are not included in the solution set. Non-Inclusive Boundary Inclusive Boundary $$y<\frac{3}{2}x+3$$ $$y\leq \frac{3}{2}x+3$$ Figure 8.3.3 Figure 8.3.4 Table 3.8.3 Consider the point $$(0, 3)$$ on the boundary; this ordered pair satisfies the linear equation. It is the “or equal to” part of the inclusive inequality that makes it part of the solution set. $$\begin{array}{c\|c}{\underline{y<\frac{3}{2}x+3}}&{\underline{y\leq \frac{3}{2}x+3}}\\{3<\frac{3}{2}(0)+3}&{3\leq\frac{3}{2}(0)+3}\\{3<0+3}&{3\leq 0+3}\\{3<3\quad\color{red}{x}}&{3\leq 3\quad\color{Cerulean}{\checkmark}} \end{array}$$ So far, we have seen examples of inequalities that were “less than.” Now consider the following graphs with the same boundary: Greater Than (Above) Less Than (Below) $$y\geq \frac{3}{2}x+3$$ $$y\leq \frac{3}{2}x+3$$ Figure 8.3.5 Figure 3.8.6 Table 3.8.4 Given the graphs above, what might we expect if we use the origin $$(0, 0)$$ as a test point? $$\begin{array}{c\|c}{\underline{y\geq\frac{3}{2}x+3}}&{\underline{y\leq\frac{3}{2}x+3}}\\{0\geq\frac{3}{2}(0)+3}&{0\leq\frac{3}{2}(0)+3}\\{0\geq 0+3}&{0\leq0+3}\\{0\geq 3\quad\color{red}{x}}&{0\leq 3\quad\color{Cerulean}{\checkmark}} \end{array}$$ Exercise $$\PageIndex{1}$$ Which of the ordered pairs $$(−2, −1), (0, 0), (−2, 8), (2, 1),$$ and $$(4, 2)$$ solve the inequality $$y>−\frac{1}{2}x+2$$? $$(−2, 8)$$ and $$(4, 2)$$ ## Graphing Solutions to Linear Inequalities Solutions to linear inequalities are a shaded half-plane, bounded by a solid line or a dashed line. This boundary is either included in the solution or not, depending on the given inequality. If we are given a strict inequality, we use a dashed line to indicate that the boundary is not included. If we are given an inclusive inequality, we use a solid line to indicate that it is included. The steps for graphing the solution set for an inequality with two variables are outlined in the following example. Example $$\PageIndex{1}$$ Graph the solution set: $$y>−3x+1$$. Solution: Step 1: Graph the boundary line. In this case, graph a dashed line $$y=−3x+1$$ because of the strict inequality. By inspection, we see that the slope is $$m=−3=−\frac{3}{1}=\frac{rise}{run}$$ and the $$y$$-intercept is $$(0, 1)$$. Figure 3.8.7 Step 2: Test a point not on the boundary. A common test point is the origin $$(0, 0)$$. The test point helps us determine which half of the plane to shade. $$\begin{array}{c\|c} {\underline{\color{Cerulean}{Test\:point}}}&{\underline{y>-3x+1}}\\{(0,0)}&{0>-3(0)+1}\\{}&{0>1\quad\color{red}{x}} \end{array}$$ Step 3: Shade the region containing the solutions. Since the test point $$(0, 0)$$ was not a solution, it does not lie in the region containing all the ordered pair solutions. Therefore, shade the half of the plane that does not contain this test point. In this case, shade above the boundary line. Figure 3.8.8 Consider the problem of shading above or below the boundary line when the inequality is in slope-intercept form. If $$y>mx+b$$, then shade above the line. If $$y<mx+b$$, then shade below the line. Use this with caution; sometimes the boundary is given in standard form, in which case these rules do not apply. Example $$\PageIndex{2}$$ Graph the solution set: $$2x−5y≥−10$$. Solution: Here the boundary is defined by the line $$2x−5y=−10$$. Since the inequality is inclusive, we graph the boundary using a solid line. In this case, graph the boundary line using intercepts. \begin{array}{c\|c} {\underline{\color{Cerulean}{To\:find\:the\:x-intercept,\:set\:y=0.}}}&{\underline{\color{Cerulean}{To\:find\:the\:y-intercept,\:set\:x=0.}}} \\ {\begin{aligned} 2x-5y&=-10 \\ 2x-5(\color{OliveGreen}{0}\color{black}{)}&=-10 \\ 2x&=-10 \\ x&=-5 \end{aligned}}&{\begin{aligned} 2x-5y&=-10 \\ 2(\color{OliveGreen}{0}\color{black}{)-5y}&=-10 \\ -5y&=-10 \\ y&=2 \end{aligned}} \\ {x-intercept:\:(-5,0)}&{y-intercept:\:(0,2)} \end{array} Figure 3.8.9 Next, test a point; this helps decide which region to shade. $$\begin{array} {c\|c} {\underline{\color{Cerulean}{Test\:point}}}&{\underline{2x-5y\geq -10}}\\{(0,0)}&{2(0)-5(0)\geq -10} \\ {}&{0\geq -10\quad\color{Cerulean}{\checkmark}} \end{array}$$ Since the test point is in the solution set, shade the half of the plane that contains it. Figure 3.8.10 In this example, notice that the solution set consists of all the ordered pairs below the boundary line. This may be counterintuitive because of the original $$≥$$ in the inequality. This illustrates that it is a best practice to actually test a point. Solve for $$y$$ and you see that the shading is correct. \begin{aligned} 2x-5y&\geq -10 \\ 2x-5y\color{Cerulean}{-2x}&\geq -10\color{Cerulean}{-2x} \\ -5y&\geq -2x-10 \\ \frac{-5y}{\color{Cerulean}{-5}}&\color{OliveGreen}{\leq}\color{black}{\frac{-2x-10}{\color{Cerulean}{-5}}}\quad\color{Cerulean}{Reverse\:the\:inequality.} \\ y&\leq \frac{2}{5}x+2 \end{aligned} In slope-intercept form, you can see that the region below the boundary line should be shaded. An alternate approach is to first express the boundary in slope-intercept form, graph it, and then shade the appropriate region. Example $$\PageIndex{3}$$ Graph the solution set: $$y<2$$. Solution: First, graph the boundary line $$y=2$$ with a dashed line because of the strict inequality. Figure 3.8.11 Now, test a point. $$\begin{array}{c\|c}{\underline{\color{Cerulean}{Test\:point}}}&{\underline{y<2}}\\{(0,0)}&{0<2\quad\color{Cerulean}{\checkmark}} \end{array}$$ In this case, shade the region that contains the test point. Figure 3.8.12 Exercise $$\PageIndex{2}$$ Graph the solution set: $$5x−y≤10$$. Figure 3.8.13 ## Key Takeaways • Linear inequalities with two variables have infinitely many ordered pair solutions, which can be graphed by shading in the appropriate half of a rectangular coordinate plane. • To graph the solution set of a linear inequality with two variables, first graph the boundary with a dashed or solid line depending on the inequality. If given a strict inequality, use a dashed line for the boundary. If given an inclusive inequality, use a solid line. Next, choose a test point not on the boundary. If the test point solves the inequality, then shade the region that contains it; otherwise, shade the opposite side. • When graphing the solution sets of linear inequalities, it is a good practice to test values in and out of the solution set as a check. Exercise $$\PageIndex{3}$$ Solutions to Linear Inequalities (Two Variables) Is the ordered pair a solution to the given inequality? 1. $$y<5x+1$$; $$(0,0)$$ 2. $$y>−\frac{1}{2}x−4$$; $$(0, −2)$$ 3. $$y≤\frac{2}{3}x+1$$; $$(6, 5)$$ 4. $$y≥−\frac{1}{3}x−5$$; $$(−3, −8)$$ 5. $$y<\frac{1}{5}x-\frac{1}{3}$$; $$(-\frac{1}{2},-1)$$ 6. $$4x-3y\leq 2$$; $$(-2,-1)$$ 7. $$-x+4y>7$$; $$(0, 0)$$ 8. $$7x−3y<21$$; $$(5,-3)$$ 9. $$y>−5$$; $$(−3, −1)$$ 10. $$x≤0$$; $$(0, 7)$$ 1. Yes 3. Yes 5. Yes 7. No 9. Yes Exercise $$\PageIndex{4}$$ Graphing Solutions to Linear Inequalities Graph the solution set. 1. $$y<−3x+3$$ 2. $$y<−\frac{2}{3}x+4$$ 3. $$y≥−\frac{1}{2}x$$ 4. $$y≥\frac{4}{5}x−8$$ 5. $$y≤8x−7$$ 6. $$y>−5x+3$$ 7. $$y>−x+4$$ 8. $$y>x−2$$ 9. $$y≥−1$$ 10. $$y<−3$$ 11. $$x<2$$ 12. $$x≥2$$ 13. $$y≤\frac{3}{4}x−\frac{1}{2}$$ 14. $$y>−\frac{3}{2}x+\frac{5}{2}$$ 15. $$−2x+3y>6$$ 16. $$7x−2y>14$$ 17. $$5x−y<10$$ 18. $$x-y<0$$ 19. $$3x−2y≥0$$ 20. $$x−5y≤0$$ 21. $$−x+2y≤−4$$ 22. $$−x+2y≤3$$ 23. $$2x−3y≥−1$$ 24. $$5x−4y<−3$$ 25. $$\frac{1}{2}x-\frac{1}{3}y<1$$ 26. $$\frac{1}{2}x-\frac{1}{10}y\geq\frac{1}{2}$$ 27. $$x≥−2y$$ 28. $$x<2y+3$$ 29. $$3x−y+2>0$$ 30. $$3−y−2x<0$$ 31. $$−4x≤12−3y$$ 32. $$5x≤−4y−12$$ 33. Write an inequality that describes all points in the upper half-plane above the $$x$$-axis. 34. Write an inequality that describes all points in the lower half-plane below the $$x$$-axis. 35. Write an inequality that describes all points in the half-plane left of the $$y$$-axis. 36. Write an inequality that describes all points in the half-plane right of the $$y$$-axis. 37. Write an inequality that describes all ordered pairs whose $$y$$-coordinates are at least $$2$$. 38. Write an inequality that describes all ordered pairs whose $$x$$-coordinate is at most $$5$$. 1. Figure 3.8.14 3. Figure 3.8.15 5. Figure 3.8.16 7. Figure 3.8.17 9. Figure 3.8.18 11. Figure 3.8.19 13. Figure 3.8.20 15. Figure 3.8.21 17. Figure 3.8.22 19. Figure 3.8.23 21. Figure 3.8.24 23. Figure 3.8.25 25. Figure 3.8.26 27. Figure 3.8.27 29. Figure 3.8.28 31. Figure 3.8.29 33. $$y>0$$ 35. $$x<0$$ 37. $$y≥2$$
# How do you solve abs(3x-4)<=x? Jan 18, 2017 $\| 3 x - 4 \| \le x \implies x \in \left[1 , 2\right]$ or $1 \le x \le 2$ #### Explanation: 2 cases $3 x - 4$ is positive $\| 3 x - 4 \| \ge 0 \implies 3 x - 4 = 3 x - 4 \le x$ or $3 x - 4$ is negative $\| 3 x - 4 \| < 0 \implies \| 3 x - 4 \| = 4 - 3 x \le x$ If $3 x - 4$ is positive then, $3 x - 4 \le x$ $\iff$ $3 x \le x + 4$ Add 4 to both sides $\iff$ $3 x - x = 2 x \le 4$ Subtract x from both sides $\iff$ $x \le 2$ Divide both sides by 2 Or $x \in \left(- \infty , 2\right]$ If $3 x - 4$ is negative then, $4 - 3 x \le x$ $\iff$ $4 \le 4 x$ Add $3 x$ to both sides $\iff$ $1 \le x$ Divide both sides by 4 Or $x \in \left[1 , \infty\right)$ Notice that $\left(- \infty , 2\right] \cap \left[1 , \infty\right) = \left[1 , 2\right]$ Then $x \in \left[1 , 2\right]$ Jan 18, 2017 $x \in \left[1 , 2\right]$ #### Explanation: Here's one method... Given: $\left\mid 3 x - 4 \right\mid \le x$ Note in passing that $\left\mid \ldots \right\mid \ge 0$ and hence $x \ge 0$. Given that $x \ge 0$, we can square both sides of the inequality to get: $9 {x}^{2} - 24 x + 16 \le {x}^{2}$ Subtract ${x}^{2}$ from both sides to get: $8 {x}^{2} - 24 x + 16 \le 0$ Divide both sides by $8$ to get: ${x}^{2} - 3 x + 2 \le 0$ $\left(x - 1\right) \left(x - 2\right) \le 0$ So this is a parabola with positive ${x}^{2}$ coefficient, intersecting the $x$ axis at $\left(1 , 0\right)$ and $\left(2 , 0\right)$ Hence the solution of our inequality is: $1 \le x \le 2$ In interval notation: $x \in \left[1 , 2\right]$ Jan 18, 2017 $x \in \left[1 , 2\right]$. See the segment of of the x-axisl in the Socratic graph. for this solution. #### Explanation: graph{(\|3x-4\|-x-y)<=0 [-5, 5, -2.5, 2.5]} $\| 3 x - 4 \| \le x$ is the combined inequality for $3 x - 4 \le x$, giving $x \le 2$, when $x \ge \frac{4}{3}$ and $- \left(3 x - 4\right) \le x$, giving $x \ge 1$, when $x \le \frac{4}{3}$.
1. Chapter 2 Class 11 Relations and Functions 2. Serial order wise 3. Examples Transcript Example 19, Let R be a relation from Q to Q defined by R = {(a, b): a, b Q and a b Z}. Show that (i) (a, a) R for all a Q Given R = {(a, b): a, b Q and a b Z} Hence we can say that (a, b) is in relation R if a, b Q i.e. both a & b are in set Q a b Z i.e. difference of a & b is an integer We need to prove both these conditions for (a,a) a, a Q , i.e. a is in set Q Also, a a = 0 & 0 is an integer, a a Z Since both conditions are satisfied, (a, a) R Example 19, Let R be a relation from Q to Q defined by R = {(a,b): a,b Q and a b Z}. Show that (ii) (a,b) R implies that (b, a) R Given R = {(a, b): a, b Q and a b Z} Given (a, b) R , i.e. (a, b) is in relation R . So, the following conditions are true a, b Q i.e. both a & b are in set Q a b Z i.e. difference of a & b is an integer We need to prove both these conditions for (b,a) 1. a, b Q,hence b,a Q 2. Given (a b) is an integer, So, negative of integer is also an integer, i.e. (a b) is also integer a + b is integer, b a is integer b a Z Since both conditions are satisfied, (b, a) R Example 19, Let R be a relation from Q to Q defined by R = {(a,b): a,b Q and a b Z}. Show that (iii) (a, b) R and (b, c) R implies that (a, c) R Given R = {(a, b): a, b Q and a b Z} We need to prove both these conditions for (a, c) 1. Given a, b & b, c Q, hence a, c Q 2. (a b) is an integer, So, negative of integer is also an integer, i.e. (a-b) is also integer -a+b is integer, b-a is integer b a Z Since both conditions are satisfied, (b, a) R 2. Given (a b) & (b c) is an integer, Sum of integers is also an integer So, (a b) + (b c) is also integer a b + b c is integer, a c is integer a c Z Since both conditions are satisfied, (a, c) R Examples
USING OUR SERVICES YOU AGREE TO OUR USE OF COOKIES # What Are The Prime Factors Of 96? • The prime factors of number 96 are: 2, 3 • Determined equcation for number 96 factorization is: 2 * 2 * 2 * 2 * 2 * 3 ## Is 96 A Prime Number? • No the number 96 is not a prime number. • Ninety-six is a composite number. Because 96 has more divisors than 1 and itself. ## How To Calculate Prime Number Factors • How do you calculate natural number factors? To get the number that you are factoring just multiply whatever number in the set of whole numbers with another in the same set. For example 7 has two factors 1 and 7. Number 6 has four factors 1, 2, 3 and 6 itself. • It is simple to factor numbers in a natural numbers set. Because all numbers have a minimum of two factors(one and itself). For finding other factors you will start to divide the number starting from 2 and keep on going with dividers increasing until reaching the number that was divided by 2 in the beginning. All numbers without remainders are factors including the divider itself. • Let's create an example for factorization with the number nine. It's not dividable by 2 evenly that's why we skip it(Remembe 4,5 so you know when to stop later). Nine can be divided by 3, now add 3 to your factors. Work your way up until you arrive to 5 (9 divided by 2, rounded up). In the end you have 1, 3 and 9 as a list of factors. ## What is a prime number? Prime numbers or primes are natural numbers greater than 1 that are only divisible by 1 and with itself. The number of primes is infinite. Natural numbers bigger than 1 that are not prime numbers are called composite numbers. ## What Are Prime Factors? • In number theory, the prime factors of a positive integer are the prime numbers that divide that integer exactly. The prime factorization of a positive integer is a list of the integer's prime factors, together with their multiplicities. The process of determining these factors is called integer factorization. The fundamental theorem of arithmetic says that every positive integer has a single unique prime factorization.
# Differential Equations This article explores differential EquationsWe will first look at the definition of a differential equation and how to verify a solution. We then look at how to solve a separable first order ordinary differential equation, sketch families of solutions, and model with a differential equation. #### Create learning materials about Differential Equations with our free learning app! • Flashcards, notes, mock-exams and more • Everything you need to ace your exams ## What is a differential equation? When we have an equation that involves a series of derivatives, we call this a differential equation. When the derivatives are of a function of one variable, we call this an ordinary differential equation (ODE). When talking about a differential equation, we often talk about its order. Order means the highest derivative that is present in the equation. For example, the equation is of order two, as the highest order derivative in the equation is second order. When solving a differential equation, the aim is to find a function that satisfies the equation. This solution will not be unique, as with a derivative, a constant can be added to change the function but still satisfy the equation. The only way that the value of this constant can be found is by the addition of a boundary condition. In a first order ordinary differential equation, we only need one boundary condition to satisfy the unknown. In general, for an ${n}^{th}$ order ordinary differential equation, we need n boundary conditions. A boundary condition specifies the value of the function at a certain point. This allows you to then work out the value of any unknown coefficients. ## Verifying solutions of differential equations When given a differential equation, if we get given a potential solution, we can verify whether it is valid or not. This involves working out all the derivatives used and then filling these in to see if the potential solution is suitable to satisfy the equation. Verify that is a solution to . (Note here that we use ${y}^{\iota }$to represent , and ${y}^{\iota \iota }$ to represent ). Using the Product Rule, let us find the first and second derivative of y with respect to x. Then $y\text{'}\text{'}=\frac{d}{dx}\left(y\text{'}\right)=\frac{d}{dx}\left(3{e}^{2x}+2x{e}^{2x}\right)=6{e}^{2x}+2{e}^{2x}+4x{e}^{2x}=8{e}^{2x}+4x{e}^{2x}$ We can now fill in the values to get Hence the solution is verified. ## Solving differential equations At A level, we only need to know how to solve first order separable ordinary differential Equations. Separable refers to the fact that the two variables (normally x and y) can be separated and then split up to solve. The form of a separable differential equation (for variables y and x, where y is a function of x) is . We can then rearrange this to, and then integrate to get. Once integrated, this is our general solution for the differential equation. We can then apply any boundary conditions to find a specific solution if necessary. It is worth noting that strictly speaking, we cannot manipulate in this way, as it is not a fraction but rather a Notation for the derivative. However, we can treat it like a fraction in this case. Find the general solution to. This is of the form , so that means that the solution is of the form . This means we can fill in the two Functions to get to . Integrating the right-hand side, we get . On the left-hand side, this is a standard integral, given as . This can also be achieved by using the substitution This means that our solution is given as . Note we have combined both the constants into one here. We can simplify this further to give. Find the solution to , with. First, let us separate this equation to get . The left-hand side integrates to , and the right-hand side integrates to. These combine to give. This can simplify further to give. We know that, so we can fill this in to get. This gives C = 1, and our solution is. ## Sketching a family of solution curves for differential equations When we are finding a general solution for a differential equation, then we are left with constants in the general equation. These constants can be any value, and they would still satisfy the differential equation. The family of solution curves is the collection of the Functions with various values for the constants. Find the general solution to , and draw a graph showing this with four different particular solutions. Below is a graph showing when C = -1, 0, 1, 2 A family of solutions, with C = 2 green, C = 1 blue, C = 0 red and C = -1 purple, Tom Maloy - StudySmarter Originals ## Modeling with differential equations The reason we study differential equations is the ability to use them in real-life scenarios. To illustrate this, let's look at an example. Suppose there is a cylindrical water tank of radius 5m. The height of water in the tank at any point is denoted as. Water flows out of the tank at a rate proportional to the square root of the volume in the tank. Find . The volume in the water tank at any one point is given as, meaning. The water flows out at a rate proportional to the square root of the volume, meaning, where a is a constant of proportionality. We can use the earlier expression for volume to give. Then, by the Chain Rule,. ## Differential Equations - Key takeaways • A differential equation is an equation made up of various derivatives. • The order of a differential equation is the highest order of any derivative in the equation. • A separable first-order ordinary differential equation has the form , with solution. • Boundary conditions allow us to put a value on a constant. • A general solution is one that has an unknown constant in it, and a specific solution exists for a specific boundary condition. #### Flashcards in Differential Equations 1 ###### Learn with 1 Differential Equations flashcards in the free StudySmarter app We have 14,000 flashcards about Dynamic Landscapes. What is a differential equation? A differential equation is an equation involving derivatives of a function, and the solution is finding that solution. How do you solve differential equations? At A-Level, we need only solve a first order separable ordinary differential equation of the form dy/dx=f(x)g(y), which has a solution of ∫ 1/g(y )dy=∫ f(x)dx What does a differential equation represent? A differential equation can represent pretty much any system. Anything which changes with time, position, etc. can be represented by differential equations. StudySmarter is a globally recognized educational technology company, offering a holistic learning platform designed for students of all ages and educational levels. Our platform provides learning support for a wide range of subjects, including STEM, Social Sciences, and Languages and also helps students to successfully master various tests and exams worldwide, such as GCSE, A Level, SAT, ACT, Abitur, and more. We offer an extensive library of learning materials, including interactive flashcards, comprehensive textbook solutions, and detailed explanations. 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Sei sulla pagina 1di 5 Unit 1 Introduction Mathematical Background Concepts Much of the work in Unit 1 is a review of key multiplication and division skills and concepts from Grade 3, including the following: • Interpreting products and quotients of whole numbers • Using multiplication and division within 100 to solve story problems in situations involving equal groups, arrays, and measurement quantities • Determining the unknown whole number in a multiplication or division equation • Applying properties of operations as strategies to multiply and divide • Fluently multiplying and dividing within 100 Even though students are required to know from memory all products of two 1-digit numbers by the end of third grade, it’s probably not reasonable to expect that this will be the case for all incoming fourth graders without a few weeks to revisit the strategies and models. Also, basic fact strategies such as doubling to multiply by 2 or using a Double-Doubles strategy to multiply by 4 can be extended to situations in which students are multiplying much larger numbers by single digits. For example, to solve 4 × 125, a student familiar with the Double-Doubles strategy for single-digit multiplication might double 125 to get 250, and double it again to get 500. A student familiar with the Half-Tens strategy might solve 5 × 68 by multiplying 10 × 68 and halving the result: 680 ÷ 2 = 340. While multiplication and division were major topics in Grade 3, the transition from additive to multiplicative thinking is a journey of several years for most learners. To help students make the transition, the authors of the Common Core Standards stipulate that fourth graders learn to interpret a multiplication equation as a comparison. Students who learned to interpret the multiplication equation 4 × 6 = 24 as 4 groups of 6 is equal to 24 in Grade 3 are now expected to interpret that equation to mean 24 is 4 times as many as 6, and 6 times as many as 4. Although this sounds simple—perhaps just a matter of linguistics—understanding what it really means when we say that something is twice as big, three times as tall, or four times as much is not easy. Consider the following task and three responses you might see in a typical fourth grade classroom. Draw a line that is exactly 2 inches long. Then draw a second line that is exactly 3 times as long as the f rst line. Student A I made the line 5 inches long because 2 and then 3 more is 5. 2 inches 5 inches Student B I respectfully disagree with you. I think the line should be 6 inches long because that’s 3 times as many as 2 inches. It’s like 2 and 2 and 2. 2 inches 6 inches Student C I said it was 8 inches long because you have 2 inches, right? And then if you make it 3 more times, like 2, and then 2 and 2 and 2, you get 8. T e f rst and third responses above re f ect two of the most common misconceptions about multiplicative comparisons. Student A has made an additive comparison instead of a multiplica- tive comparison, drawing a line that is 3 inches more than the f rst, rather than 3 times as many. Student C interprets 3 times as many to mean that you add 3 times more to the original length. Both of these students are still employing additive rather than multiplicative reasoning. Central to understanding what it means to say that something is 4 times as much, many, tall, or long, than something else is the idea that 4 of the smaller amount, quantity, height, or length have to ft exactly into the larger amount, quantity, height, or length. T at is the idea shown in this illustration from Module 3: Bridges in Mathematics Grade 4 Teachers Guide ii © The Math Learning Center \| mathlearningcenter.org duction Un t 1 Module 3 Session 3 1 copy for display Jim & the Giant 1 Sam says that the giant is 3 times as tall as Jim. Do you agree with Sam? Why or why not? The giant is actually 4 times as tall as Jim. The giant is the same as 4 Jims standing on top of each other. Jim is one-fourth as tall as the giant. Unit 1 Introduction 12 × 2 4 × 3 = 12 6 6 2 × 6 = 12 × 2 2 ngcenter org 13 gcen er org 13 × 1 13 Twelve is a composite number because it has more factors than just itself and 1. I can see that the factors of 12 are 1, 2, 3, 4, 6, and 12. Thirteen is a prime number. No matter how hard you try, you can only build one rectangle with 13 tiles, so it only has two factors—1 and 13. Strategies T e ability to recall the single-digit multiplication facts is important to free up students’ mental energies for reasoning and problem solving. Students will use these facts as they advance to more complex mathematics, such as solving multi-digit multiplication and division problems and working with fractions. T is automaticity can be accomplished in part by helping students use the facts they know to help them reason about those facts with which they are not yet fuent. In this unit, students consider speci fc strategies for remembering di ferent categories of multiplica- tion facts. Tey use a table of multiplication facts and the array model to review these strategies, which were f rst introduced in third grade Bridges. While some fourth graders may not yet have mastered all of their multiplication facts, many of your returning Bridges students will recall most, if not all, of these strategies. Review the table below so that you are familiar with the strategies as well. Multiplication Strategies Factor Category Example How the strategy works × 0 Zero facts 0 × 3 = 0 The product of any number and 0 is 0. 7 × 0 = 0 × 1 Ones facts 1 × 4 = 4 The product of any number and 1 is that number. 8 × 1 = 8 × 2 Doubles facts 2 × 6 = 12 To multiply any number by 2, double that number. 9 × 2 = 18 × 3 Doubles Plus One facts 3 × 6 = 18 To multiply any number by 3, double the 9 × 3 = 27 number and then add the number. For example, 3 × 6 = (2 × 6) + 6 = 12 + 6 = 18. × 4 Double-Doubles facts 4 × 6 = 24 To multiply any number by 4, double that 9 × 4 = 36 number, and then double the result. For example, 4 × 6 = 2(2 × 6) = 2 × 12 = 24. × 8 Double-Double-Doubles facts 8 × 6 = 48 To multiply any number by 8, double that number, 9 × 8 = 72 double the result, and then double one more time. For example, 8 × 6 = 2(2(2 × 6)) = 2(2 × 12) = 2 × 24 = 48. × 5 Half-Tens facts 5 × 7 = 35 To multiply any number by 5, it may be simples to 8 × 5 = 40 f rst multiply that numbr by 10 and then divide the product by 2. For example, to solve 8 × 5, find 8 × 10 and divide 80 by 2 to get 40. × 6 Half-Tens Plus One Set facts 6 × 7 = 42 To multiply any number by 6, f rst multiply the 8 × 6 = 48 number by 5 and then add the number to the result. For example, 8 × 6 = (8 × 5) + 8 = 48. × 10 Tens facts 10 × 8 = 80 Multiplying by 10 comes naturally for students with a 6 × 10 = 60 solid grasp of skip-counting and place value concepts. × 9 Tens Minus One Set facts 9 × 7 = 63 To multiply any number by 9, think of the related 9 × 9 = 81 Tens fact and then subtract the number. For example, 9 × 7 = (10 × 7) – 7 = 70 – 7 = 63. Bridges in Mathematics Grade 4 Teachers Guide iv © The Math Learning Center \| mathlearningcenter.org Unit 1 Introduction As students shi f away from using additive reasoning (skip-counting) to f nd products, they begin to use multiplicative strategies like doubling (e.g., thinking of 8 × 2 as 8 doubled). T is can lead to use of the doubling and halving strategy (e.g., 12 × 4 = 6 × 8). Students can also use partial prod- ucts with smaller “chunks” of numbers f rst and then use them with bigger chunks (e.g., 32 × 12 = 32 × 10 + 32 × 2). As they reason multiplicatively, they can also use “over strategies” for certain problems (e.g., 99 × 47 = 100 × 47 – 1 × 47). We want to help students build a repertoire of strate- gies based on multiplicative reasoning that they can eventually apply to multi-digit multiplication. 4 × 27 27, 54, 81, 108 Skip-Counting 4 × 27 27, 54, 108 Doubling 4 × 27 = 2 × 54 Doubling/Halving 4 × 27 = (4 × 10) + (4 × 10) + (4 × 7) or or Partial Products 4 × 27 = (4 × 20) + (4 × 7) 4 20 7 80 28 108 Models T e models and strategies that appear in Unit 1 serve to help students review and re-access what they learned in Grade 3. T ese include the open number line, the array or area model, and the ratio table. All three will be extended and greatly expanded in fourth grade, especially the area model and the ratio table. The Open Number Line Because some of your students may be using additive thinking to solve multiplication problems, the open number line is used early in Unit 1 to show repeated addition as a bridge to the array, which encourages multiplicative thinking. T e open number line will resurface later in the year as a way to model addition and subtraction strategies as well. 1 × 3 3 2 × 3 3 × 3 4 × 3 5 × 3 3 3 3 3 0 3 6 9 12 15 The Array or Area Model In the area model for multiplication, the total area of the rectangle represents the product, and the two dimensions represent the factors. factors product 4 × 3 = 12 The dimensions represent the factors. 3 4 1212 The area represents the product. Bridges in Mathematics Grade 4 Teachers Guide v © The Math Learning Center \| mathlearningcenter.org Unit 1 Introduction Because multiplication and division are inverse operations, the same model can be used to illustrate division. divisor dividend quotient 12 ÷ 4 = 3 quotient 3 divisor 4 12 dividend The known The unknown dimension dimension represents represents the divisor. the quotient. 3 4 1212 The area represents the dividend. Bridges helps students use the array model for multiplication by beginning with discrete models in third grade. Students progress over time, using closed arrays, base ten area pieces with linear pieces, and then open arrays. With closed arrays, they can count each square unit by 1s. With base ten area pieces and linear pieces, the area is now modeled in bigger chunks, tens and ones, and the dimensions are def ned with linear pieces, helping students di ferentiate between linear measures and area measures. With open arrays, students can chunk the arrays into pieces that are conve- nient and efcient for the problem. With each model, students can chunk areas into bigger pieces, moving away from counting strategies, to repeated addition, and then to multiplicative thinking. Closed Array Linear Pieces and Base Ten Area Pieces Open Array While students will discover many ways to solve multiplication and division problems, the array model provides a way for them to discuss their strategies with one another, decompose the numbers, apply the distributive property, and identify partial products. The Ratio Table Te ratio table is used in Bridges to simultaneously build multiplicative thinking and proportional reasoning. Te model is introduced in Unit 1 to represent students’ strategies. Students will f ll in tables for situations with a constant ratio such as when one row in a box has 8 crayons and there are 4 rows. Later, the ratio table will become a tool for students to use in problem solving to compute multiplication, division, and fraction problems, as well as make conversions. T is model will also be used for many years to come in higher mathematics to model proportional situations. Rows of Number crayons of crayons 1 8 2 16 × 2 3 24 × 2 6 48 Bridges in Mathematics Grade 4 Teachers Guide vi © The Math Learning Center \| mathlearningcenter.org
NCERT Solutions: Direct & Inverse Proportions - 2 # NCERT Solutions: Direct & Inverse Proportions - 2 - Notes \| Study Mathematics (Maths) Class 8 - Class 8 1 Crore+ students have signed up on EduRev. Have you? Question: Observe the following tables and ⇒nd which pair of variables (here x and y) are in inverse proportion. (i) x so 40 30 20 y 5 6 7 S (ii) x 100 200 300 400 y 60 30 20 15 (iii) x 90 60 45 30 20 5 y 10 15 20 25 30 35 Solution: (i) ∵ x1 = 50 and y1 = 5 ⇒ x1y1 = 50 * 5 = 250 x2 = 40 and y2 = 6 ⇒ x2y2 = 40 * 6 = 240 x3 = 30 and y3 = 7 ⇒ x3y3 = 30 * 7 = 210 x4 = 20 and y4 = 8 ⇒ x4y= 20 * 8 = 160 Also 250 ≠ 240 ≠ 210 ≠ 160 or x1y1 ≠ x2y2 ≠ x3y3 ≠ x4y4 ∴ x and y are not in inverse proportion and x1y1 = x2y2 = x3y3 = x4y4 ∴ x and y are in inverse proportion. and x1y1 = x2y2 = x3y3 ≠ x4y4 ≠ x5y5 ≠ x6y6 ∴ x and y are not in inverse variation. EXERCISE 13.2 Question 1. Which of the following are in inverse proportion? (i) The number of workers on a job and the time to complete the job. (ii) The time taken for a journey and the distance travelled in a uniform speed. (iii) Area of cultivated land and the crop harvested. (iv) The time taken for a fixed journey and the speed of the vehicle. (v) The population of a country and the area of land per person. Solution: REMEMBER If an increase in one quantity brings about a corresponding decrease in the other and vice versa, then the two quantities vary inversely. (i) If number of workers are increased then time to complete the job would decrease. ∴ It is a case of inverse variation. (ii) For longer distance, more time would be required. ∴ It is not a case of inverse variation (iii) For more area of land, more crops would be harvested. ∴ It is not a case of inverse variation. (iv) If speed is more, time to cover a fixed distance would be less. ∴ It is a case of inverse variation. (v) For more population, less area per person would be required. ∴ It is a case of inverse variation. Question 2. In a television game show, the prize money of Rs 1,00,000 is to be divided equally amongst the winners. Complete the following table and find whether the prize money given to an individual winner is directly or inversely proportional to the number of winners? Number of winners 1 2 4 5 8 10 20 Prize for each winner (in Rs) 1,00,000 50,000 ... — ... — Solution: If more the number of winners, less is the prize money. ∴ It is a case of inverse proportion. Thus, the table is completed as under: Number of winners 1 2 4 5 8 10 20 Prize for each winner (in Rs) 1,00,000 50,000 25,000 20,000 12,500 10,000 5,000 Question 3. Rehman is making a wheel using spokes. He wants to fix equal spokes in such a way that the angles between any pair of consecutive spokes are equal. Help him by completing the following table. Number of spokes 4 6 8 10 12 Angle between a pair of consecutive spokes 90o 60o .... .... .... (i) Are the number of spokes and the angles formed between the pairs of consecutive spokes in inverse proportion? (ii) Calculate the angle between a pair of consecutive spokes on a wheel with 15 spokes. (iii) How many spokes would be needed, if the angle between a pair of consecutive spokes is 40°? Solution: (i) Obviously, more the number of spokes, less is the measure of angle between a pair of consecutive spokes. ∴ It is a case of inverse variation. Thus, Thus, the table is completed as under Number of spokes 4 6 8 10 12 Angle between a pair of consecutive spokes 90° 60° 45° 36° 30° (ii) Let the required measure of angle be x°. (iii) Let required number of spokes be n. ∴ The required number of spokes = 9 Question 4. If a box of sweets is divided among 24 children, they will get 5 sweets each. How many would each get, if the number of the children is reduced by 4? Solution: Reduced number of children = 24 – 4 = 20 Since, more the number of children, less is the quantity of sweets. ∴ It is a case of inverse variation. i.e. 24 * 5 = 20 * x or ∴ Each student will get 6 sweets. Question 5. A farmer has enough food to feed 20 animals in his cattle for 6 days. How long would the food last if there were 10 more animals in his cattle? Solution: Number of animals added = 10 ∴ Now, the total number of animals = 10 + 20 = 30 For more number of animals, the food will last less number of days. ∴ It is a case of inverse variation. Thus, we have 30 * x = 20 * 6 or Therefore, the food will now last for 4 days. Question 6. A contractor estimates that 3 persons could rewire jasminder’s house in 4 days. If, he uses 4 persons instead of three, how long should they take to complete the job? Solution: More is the number of persons, less is the time to complete job. ∴ It is a case of inverse variation. Number of persons Number of days to complete the wiring job 34 4x ∴ 3 * 4 = 4 * x or x = 3 * 4 = 3 ∴ The required number of days = 3 Question 7. A batch of bottles were packed in 25 boxes with 12 bottles in each box. If the same batch is packed using 20 bottles in each box, How many boxes would be filled? Solution: Let the number of boxes required = x We have: Number of bottles in a box Number of boxes 1220 25x Since, more the number of bottles in a box, lesser will be the number of boxes required to be filled. ∴ It is a case of inverse variation. i.e. or Thus, the required number of boxes = 15 Question 8. A factory requires 42 machines to produce a given number of articles in 63 days. How many machines would be required to produce the same number of articles in 54 days? Solution: Let the number of machines required be x. We have: Number of machines Number of days 42 63 X 54 or Thus, the required number of machines = 49 Question 9. A car takes 2 hours to reach a destination by travelling at the speed of 60 km/h. How long will it take when the car travels at the speed of 80 km/h? Solution: More the speed, lesser the number of hours to travel a fixed distance. ∴ It is a case of inverse variations. Speed (km/h) Time taken to cover the fixed distance 60 2 80 X or Thus the required number of hours = Question 10. Two persons could fit new window in a house in 3 days. (i) One of the persons fell ill before the work started. How long would the job take now? (ii) How many persons would be needed to fit the windows in one day? Solution: (i) Let the time taken by the remaining persons to complete the job be x. ∵ 2 persons – 1 person = 1 person and lesser the number of person, more will be the number of days to complete the job. ∴ It is a case of inverse proportion. We have: Number of persons Number of (lays 2 3 1 X or ∴ 1 person will complete the job in 6 days. (ii) We have Number of persons Number of days 2 3 X 1 or ∴ 6 persons will be required to complete the job in 1 day. Question 11. A school has 8 periods a day each of 45 minutes duration. How long would each period be, if the school has 9 periods a day, assuming the number of school houres to be the same? Solution: For 9 periods, let the duration per period be x minutes. ∴ We have: Number of periods Duration of a period (in minutes) 89 45X For a fixed duration, more the periods, lesser will be the duration of one period. ∴ It is a case of inverse proportion. i.e. 8 * 45 = 9 * x Thus, the required duration per period = 40 minutes. Question 1. Take a sheet of paper. Fold it as shown in the figure. Count the number of parts and the area of a part in each case. Tabulate your observations and discuss with your friends. Is it a case inverse proportion? Why? Number of parts 1 2 4 8 16 Area of each part area of the paper 1/2 the area of the paper .... .... .... Solution: Number of parts → 1 2 4 8 16 Area of one part → Area of the 1/2 Area of the paper 1/4 Area of the paper 1/8 Area of the paper 1/16 Area of the paper Here, more the number of parts, lesser is the area of each part. ∴ It is a case of “inverse proportion”. Question 2. Take a few containers of different sizes with circular bases. Fill the same amount of water in each container. Note the diameter of each container and the respective height at which the water level stands. Tabulate your observations. Is it a case of inverse proportion? Diameter of container (in cm) Height of water level (in cm) Solution: Diameter of container (in cm) d1 d2 d3 Height of water level (in cm) h1 h2 h3 Here, lesser the diameter, more is the level of water in the container. ∴ It is a case of inverse proportion. The document NCERT Solutions: Direct & Inverse Proportions - 2 - Notes \| Study Mathematics (Maths) Class 8 - Class 8 is a part of the Class 8 Course Mathematics (Maths) Class 8. All you need of Class 8 at this link: Class 8 ## Mathematics (Maths) Class 8 193 videos\|360 docs\|48 tests Use Code STAYHOME200 and get INR 200 additional OFF ## Mathematics (Maths) Class 8 193 videos\|360 docs\|48 tests ### Top Courses for Class 8 Track your progress, build streaks, highlight & save important lessons and more! , , , , , , , , , , , , , , , , , , , , , ;
# Class 10 Maths Chapter 10 Question Answers - Circles Q1. In the adjoining figure, PA and PB are tangents from P to a circle with centre C. If ∠APB = 40° then find ∠ACB. Sol. Since a tangent to a circle is perpendicular to the radius through the point of contact, ∴ ∠1 = 90° and ∠2 = 90° Now, in quadrilateral APBC, we have: ∠1 + ∠ACB + ∠2 + ∠P = 360° ⇒90° + ∠ACB + 90° + 40°= 360° ⇒∠ACB + 220° = 360° ⇒∠ACB = 360° − 220° = 140°. Q2. In the given figure, PA and PB are tangents from P to a circle with centre O. If ∠AOB = 130°, then find ∠APB. Sol. Since a tangent to a circle is perpendicular to the radius through the point of contact, ∴∠1= ∠2 = 90° Now, in quadrilateral AOBP, we have: ∠1 + ∠AOB + ∠2 + ∠APB = 360° ⇒ 90° + 130° + 90° + ∠APB = 360° ⇒ 310° + ∠APB = 360° ⇒∠APB = 360 − 310 = 50° Thus, ∠APB = 50°. Q3. In the given figure, PT is a tangent to a circle whose centre is O. If PT = 12 cm and PO = 13 cm then find the radius of the circle. Sol. Since a tangent to a circle is perpendicular to the radius through the point of contact, ∴∠OTP =90° In rt Δ OTP, using Pythagoras theorem, we get OP2 = OT2 + PT2 ⇒ 132 = OT2 + 122 ⇒ OT2 = 132 − 122 = (13 − 12) (13 + 12) = 1 × 25 = 25 ∴ OT2 = 52 ⇒ OT = 5 Thus, radius (r) = 5 cm. Q4. In the given figure, PT is a tangent to the circle and O is its centre. Find OP. Sol. Since, a tangent to a circle is perpendicular to the radius through the point of contact. ∴∠OTP = 90° In right Δ OTP, using Pythagoras theorem, we get OP2 = OT2 + PT2 = 82 + 152 = 64 + 225 = 289 = 172 Q5. If O is the centre of the circle, then find the length of the tangent AB in the given figure. Sol. ∵ A tangent to a circle is perpendicular to the radius through the point of contact. ∴ ∠OAB = 90° Now, in right Δ OAB, we have OB2 = OA2 + AB2 ⇒ 102 = 62 + AB2 ⇒ AB2 = 102 − 62 = (10 − 6) (10 +6) = 4 × 16 = 64 = 82 Q6. In the figure, PA is a tangent from an external point P to a circle with centre O. If ∠POB = 115° then find ∠APO. Sol. Here, PA is a tangent and OA is radius. Also, a radius through the point of contact is perpendicular to the tangent. ∴ OA = PA ⇒∠PAO = 90° In ∆OAP, ∠POB is an external angle, ∴∠APO + ∠PAO = ∠POB ⇒∠ APO + 90° = 115° ⇒ ∠APO = 115° − 90° = 25° Q7. In the following figure, PA and PB are tangents drawn from a point P to the circle with centre O. If ∠APB = 60°, then what is ∠AOB? Sol. The radius of the circle through the point of contact is perpendicular to the tangent. ∴ OA ⊥ AP and OB ⊥ BP ⇒ ∠PAO = ∠PBO = 90° ∠OAP + ∠APB + ∠PBO + ∠AOB = 360° 90° + 60° + 90° + ∠AOB = 360° ⇒ ∠AOB + 240° = 360° ⇒∠AOB = 360° – 240° = 120° Q8. In the figure, CP and CQ are tangents to a circle with centre O. ARB is another tangent touching the circle at R. If QC = 11 cm, BC = 7 cm then find, the length of BR. Sol. ∵ Tangents drawn from an external point are equal, ∴ BQ = BR ...(1) And CQ = CP Since, BC + BQ = QC ⇒ 7 + BR =11 [∵ BQ = BR] BR = 11 − 7 = 4 cm. Q9. In the figure, ΔABC is circumscribing a circle. Find the length of BC. Sol. Since tangents drawn from an external point to the circle are equal, ∴ AR = AQ = 4 cm ...(1) BR = BP = 3 cm ...(2) PC = QC ...(3) ∴ QC = AC − AQ = 11 − 4 = 7 cm [From (1)] BC = BP + PC [From (3)] = 3 + QC = (3 + 7) cm = 10 cm Q10. In the figure, if ∠ ATO = 40°, find ∠ AOB. Sol. Since the tangent is perpendicular to the radius through the point of contact, ∴ ∠1= ∠4 = 90° Also, OA = OB [Radii of the same circle] OT = OT [Common] ∴∆OAT ≅∆OBT [RHS congruency] ⇒∠3= ∠2 Now, in ΔOAT, ∠3 + ∠4 + ∠5 = 180° ⇒ ∠3 + 90° + 40° = 180° ⇒ ∠3 = 180° − 90° − 40° = 50° ⇒ ∠AOB = 50° + 50° = 100°. Q11. From a point P, the length of the tangent to a circle is 15 cm and distance of P from the centre of the circle is 17 cm, then what is the radius of the circle? Sol. Since radius is perpendicular to the tangent through the point of contact, ∴ OA ⊥ AP ⇒ ∠OAP = 90° In rt ∆OAP, we have: OA2 + AP2 = OP2 ⇒ r2 + (15)2 = (17)2 r2 = 172 − 152 = (17 − 15) (17 + 15) = 2 × 32 = 64 Q12. The two tangents from an external point P to a circle with centre O are PA and PB. If ∠APB = 70°, then what is the value of ∠AOB? Sol. Since tangent is perpendicular to the radius through the point of contact. ∴∠1= ∠2 = 90° ∠AOB + ∠1 + ∠2 + ∠APB = 360° ∠AOB + 90° + 90° + 70° = 360° ⇒∠AOB + 250° = 360° ⇒∠AOB = 360°−250° = 110° The document Class 10 Maths Chapter 10 Question Answers - Circles is a part of the Class 10 Course Mathematics (Maths) Class 10. All you need of Class 10 at this link: Class 10 ## Mathematics (Maths) Class 10 116 videos\|420 docs\|77 tests ## FAQs on Class 10 Maths Chapter 10 Question Answers - Circles 1. What is a circle? Ans. A circle is a closed curve in which all points on the curve are equidistant from a fixed point called the center. 2. What is the formula to find the circumference of a circle? Ans. The formula to find the circumference (C) of a circle is C = 2πr, where r is the radius of the circle and π is a mathematical constant approximately equal to 3.14159. 3. How is the area of a circle calculated? Ans. The area (A) of a circle is calculated using the formula A = πr^2, where r is the radius of the circle. 4. What is the relationship between the diameter and radius of a circle? Ans. The diameter of a circle is twice the length of its radius. In other words, if we have the radius (r), the diameter (d) can be calculated as d = 2r. 5. What is the difference between a circle and a sphere? Ans. A circle is a two-dimensional figure, whereas a sphere is a three-dimensional figure. A circle lies in a plane, while a sphere is a solid object with no edges or vertices. ## Mathematics (Maths) Class 10 116 videos\|420 docs\|77 tests ### Up next Explore Courses for Class 10 exam ### Top Courses for Class 10 Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests. 10M+ students study on EduRev Related Searches , , , , , , , , , , , , , , , , , , , , , ;
# 138 and Divisibility Tricks 4 You 138 is a composite number. Factor pairs: 138 = 1 x 138, 2 x 69, 3 x 46, or 6 x 23. Factors of 138: 1, 2, 3, 6, 23, 46, 69, 138. Prime factorization: 138 = 2 x 3 x 23. 138 is never a clue in the FIND THE FACTORS puzzles. ——————————————————————————— After you learned some basic division facts, you probably realized: • 2 will divide evenly into any EVEN whole number. • 5 will divide evenly into whole numbers ending in 0 or 5. • 10 will divide evenly into whole numbers ending in 0. These three rules are related to each other. All of them are true because we use base ten in our numbering system, and the prime factorization of 10 is 2 x 5. If you needed to find the factors of a 33-digit whole number, you would be able to tell if 2, 5, or 10 divide evenly into it just by looking at the last digit. 33-digits is more than a standard calculator can handle, but no matter how many digits a whole number has, as long as you can see the very last one, you can apply those three simple divisibility rules to know if 2, 5, or 10 are factors. Thus you will be able to do something a calculator can’t. But wait, there are even more divisibility tricks if you can see the last TWO digits of the whole number! • 10 squared, better known as 100, divides evenly into any whole number ending in 00. • 5 x 5 = 25 which divides evenly into any whole number ending in 00, 25, 50, or 75. • 2^2 (AKA 4) divides evenly into a whole number if the final two digits can be divided evenly by 4. How can one tell if the last two digits of a whole number are divisible by 4 (without actually dividing by 4)? I’ll show you how: I’ve put the 25 possible 2-digit multiples of 4 into one of two lists: • 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88 • 12, 16, 32, 36, 52, 56, 72, 76, 92, 96 Notice in the first list ALL the digits are even and the last digit (0, 4, or 8) can be divided evenly by 4. Then look at the second list. The first digit is always odd and the last digit is either 2 or 6 (the only two even digits that are not divisible by 4). Hmm. I think we can rewrite the divisibility rule for 4: • 4 (AKA 2^2) divides evenly into a whole number if the last two digits are even and the final digit is divisible by 4 (the last digit is 0, 4, or 8). • 4 divides evenly into any whole number whose next to the last digit is odd if the final digit is even but not divisible by 4 (the last digit is 2 or 6). The rewritten divisibility rule is longer to read but takes a little less time to implement so you will have to decide which version of the rule works best for you. Either trick takes much less time than dividing some really long whole number by 4 or dividing by 2 twice. Now I’m on to thinking about what the last THREE digits tell us. # 5 Easy as 1-2-3 • 5 is a prime number. • Prime factorization: 5 is prime. • The exponent of prime number 5 is 1. Adding 1 to that exponent we get (1 + 1) = 2. Therefore 5 has exactly 2 factors. • Factors of 5: 1, 5 • Factor pairs: 5 = 1 x 5 • 5 has no square factors that allow its square root to be simplified. √5 ≈ 2.236. How do we know that 5 is a prime number? If 5 were not a prime number, then it would be divisible by at least one prime number less than or equal to √5 ≈ 2.2. Since 5 cannot be divided evenly by 2, we know that 5 is a prime number. 5 is the only number that is the sum of ALL the prime numbers less than itself. 2² + 1² = 5 and 3² + 4² = 5² so 5 is the smallest Pythagorean triple hypotenuse. When 5 is a clue in a FIND THE FACTORS puzzle, use 1 for one of the factors and 5 for the other. Being able to identify factors of a whole number is a very important skill in mathematics. It is a skill that is commonly used in many areas of mathematics ranging from reducing fractions to solving differential equations. The Find the Factors puzzles can help make that skill second nature. Click 10 Factors 2013-11-11 for more puzzles. To solve the puzzles, we are only interested in the limited set of factors that are represented in the following table: What about all the other factors of these numbers? And what about all the other whole numbers not on the chart? How do you find ALL of the factors of a given whole number? For example, suppose you were asked to find all of the factors of 435. Some people might notice right away that it is divisible by 5 because its last digit is 5. While that is true, beginning with 5 is not the best place to start because there is an advantage in considering all possible factors in an organized way. When you are asked to find ALL of the factors of any number, starting at 1 will make finding all of the factors as easy as 1-2-3. So what are the factors of 435? Using a calculator, I notice that the square root of 435 is about 20.85. That means I can find absolutely all of the factors of 435 by considering as divisors just the whole numbers from 1 to 20! Each factor will have a partner that is greater than 20 but will be found at the same time with these few short calculations. To demonstrate my thinking process, I will put each possible factor from 1 to 20 in a chart and write my thoughts as I consider each one. As you may notice, once a possible factor is eliminated, it is not necessary to do any actual division by ANY of the multiples of that number. (4, 6, 8, 10, 12, 14, 16, 18, and 20 are all multiples of 2, which was not a factor, so I didn’t actually divide 435 by any of those multiples.) As I carefully consider each possible factor, I only WRITE DOWN a number if it is an actual factor. Therefore, with only a little bit of effort I would list ALL of the factors of 435 in one tidy list: 1 x 435, 3 x 145, 5 x 87, 15 x 29. See, it was as easy as 1-2-3! Now let’s find all of the factors of 144. Even though 144 is less than 435, it has more factors. One of its factors is paired with itself because the square root of 144 is 12. That fact is also the signal that we can stop looking for more factors, and we can list all the factors of 144 on the following chart: There are 8 multiplication facts that produce 144, but 12 x 12 = 144 is the only fact we consider when solving a Find the Factors 1-12 puzzle with 144 as one of the clues. In every other case one of the pair of numbers in the multiplication fact will be greater than 12 and not eligible to be written in the factor row or factor column. However in solving mathematical problems, any of the factors of a whole number could be the star of the show. Knowing how to find those factors is indeed an important skill and is as easy as 1-2-3.
# How do you solve the system x + 2y -4z = 0, 2x + 3y + z = 1, 4x + 7y + lamda*z = mu? Feb 23, 2017 x = (28-14mu+2 lambda)/(lambda + 7); y = (-16+9mu-lambda)/(lambda + 7) ;z = (mu-1)/(lambda+7) #### Explanation: This system can be solved a number of ways, such as substitution and elimination, matrix row operations, etc. Using Matrix Row Operations we want to get: $\left(\begin{matrix}1 & 0 & 0 & x \\ 0 & 1 & 0 & y \\ 0 & 0 & 1 & z\end{matrix}\right)$ Since we have a 1 in the top left do the following row operations. To row 2: $- 2 {R}_{1} + {R}_{2}$ and to row 3: $- 4 {R}_{1} + {R}_{3}$ ((1,2,-4,0),(2,3,1,1),(4,7,lambda,mu)) => ((1,2,-4,0),(0,-1,9,1),(0,-1,16+lambda,mu)) To get a 1 in the 2nd column of row 2: $- {R}_{2}$ $\left(\begin{matrix}1 & 2 & - 4 & 0 \\ 0 & 1 & - 9 & - 1 \\ 0 & - 1 & 16 + \lambda & \mu\end{matrix}\right)$ To get (0, 1, 0) in the 2nd column: To row 1: $- 2 {R}_{2} + {R}_{1}$ and to row 3: ${R}_{2} + {R}_{3}$ $\left(\begin{matrix}1 & 0 & 14 & 2 \\ 0 & 1 & - 9 & - 1 \\ 0 & 0 & 7 + \lambda & \mu - 1\end{matrix}\right)$ To get a 1 in the 3rd row, 3rd column: ${R}_{3} / \left(\lambda + 7\right)$ $\left(\begin{matrix}1 & 0 & 14 & 2 \\ 0 & 1 & - 9 & - 1 \\ 0 & 0 & 1 & \frac{\mu - 1}{\lambda + 7}\end{matrix}\right)$ To get (0,0,1) in the 3rd column: To row 1: $- 14 {R}_{3} + {R}_{1}$ and to row 2: $9 {R}_{3} + {R}_{2}$ $\left(\begin{matrix}1 & 0 & 0 & - 14 \frac{\mu - 1}{\lambda + 7} + 2 \\ 0 & 1 & 0 & 9 \frac{\mu - 1}{\lambda + 7} - 1 \\ 0 & 0 & 1 & \frac{\mu - 1}{\lambda + 7}\end{matrix}\right) \implies \left(\begin{matrix}1 & 0 & 0 & \frac{28 - 14 \mu + 2 \lambda}{\lambda + 7} \\ 0 & 1 & 0 & \frac{9 \mu - \lambda - 16}{\lambda + 7} \\ 0 & 0 & 1 & \frac{\mu - 1}{\lambda + 7}\end{matrix}\right)$ So x = (28-14mu+2 lambda)/(lambda + 7); y = (-16+9mu-lambda)/(lambda + 7) ;z = (mu-1)/(lambda+7)
# Average Speeds Lesson In the chapter on Travel Graphs, it was suggested that there was a relation between the slope of the graph at some point in the journey and the current speed of the traveller. In this graph showing, we see that the traveller covered $40$40 km in the first $30$30 minutes. At this rate, the traveller would cover $80$80 km in a full hour. So, the speed in the first section was $80$80 km/h. In the next $15$15 minutes, no distance was covered. So, the speed was zero. In the final section, the distance went from $40$40 km to $90$90 km and $30$30 minutes elapsed. The traveller covered $50$50 km in half an hour. This is equivalent to a speed of $100$100 km/h. In reality, a journey does not usually consist of just a few sections each with a constant speed. We would expect a real journey to have fluctuating speeds so that the time-distance graph would look more like the following. If we compare the two graphs above, we see that both start at the point $(0,0)$(0,0) - no distance covered at time zero. Then, both graphs pass through the points $(30,40)$(30,40), $(45,40)$(45,40) and $(75,90)$(75,90) Although in the second graph the small variations in the speed are shown by deviations from the straight line, we can think of the average speed between pairs of points in the journey, as though the graph between these points was indeed a straight line. We ignore the changes of speed along the way and consider only the overall elapsed time and the distance between points on the path. #### Example I left home at $8:45$8:45 a.m. and travelled the $12$12 kilometres to my destination, arriving at $9:05$9:05 a.m. What was my average speed? The elapsed time for the journey was $20$20 minutes, which is $\frac{1}{3}$13 of an hour. In that time I travelled $12$12 kilometres. It is clear that at this average rate I would have travelled $36$36 kilometres in a full hour. So, my average speed was $36$36 km/h. This calculation is equivalent to using the formula $\text{average speed}=\frac{\text{distance travelled}}{\text{elapsed time}}$average speed=distance travelledelapsed time #### Worked Examples ##### Question 1 Maria travels by car for $420$420 km. The trip takes $10$10 hrs. What is the average speed of the trip? ##### Question 2 With respect to the following graph, what was the: 1. Total time taken for the journey? 2. Total distance covered in the journey? 3. Average speed during the journey? Answer to two decimal places. ##### Question 3 Dave states that his trip between Amsterdam and Moscow had an average speed of $80$80 km/h. Use the table, which shows the distances between cities in kilometres, to find how long the trip from Amsterdam to Moscow took Dave. Berlin London Copenhagen Amsterdam Moscow Rome Warsaw Berlin 587 613 354 2133 1301 1086 London 587 362 925 1601 1175 515 Copenhagen 613 362 950 1562 1526 668 Amsterdam 354 925 950 2479 1433 1432 Moscow 2133 1601 1562 2479 2354 1143 Rome 1301 1175 1526 1433 2354 1304 Warsaw 1086 515 668 1432 1143 1304 ### Outcomes #### MS1-12-3 interprets the results of measurements and calculations and makes judgements about their reasonableness
# How do you graph the line 2x+5y=10? Sep 3, 2016 see explanation. #### Explanation: To draw a straight line , we only require 2 points, although 3 is better. When the given line crosses the y-axis, the corresponding x-coordinate will be zero. Substituting x = 0 into the equation allows us to solve for y. $x = 0 \Rightarrow 2 \left(0\right) + 5 y = 10 \Rightarrow 0 + 5 y = 10 \Rightarrow y = 2$ Hence (0 ,2) is a point on the line. Similarly, when the line crosses the x-axis the y-coordinate will be zero. Substitute y = 0 into the equation and solve for x. $y = 0 \Rightarrow 2 x + 0 = 10 \Rightarrow x = 5$ Hence (5 ,0) is another point on the line. Now plot the points (0 ,2) and (5 ,0) and draw a straight line through them. graph{-2/5x+2 [-10, 10, -5, 5]}
Sign up or log in to Magoosh AP Calc Prep. Implicit variation (or implicit differentiation) is a powerful technique for finding derivatives of certain equations. In this review article, we’ll see how to use the method of implicit variation on AP Calculus problems. ## What is Implicit Variation? The usual differentiation rules, such as power rule, chain rule, and the others, apply only to functions of the form y = f(x). In other words, you have to start with a function f that is written only in terms of the variable x. But what if you want to know the slope at a point on a circle whose equation is x2 + y2 = 16, for example? Circle of radius 4. The equation is: x2 + y2 = 16 Here, it would be possible to solve the equation for y and then proceed to take a derivative. However, that’s not really the best way! For one thing, that square root will make finding the derivative more challenging. For another, you really have two separate functions — one using the plus (+), and the other using minus (-) in front of the radical! Which one should you use for finding the derivative? Well, that depends on whether you want the top or bottom semicircle. ### An Easier Way It would be so much simpler to just work with the original equation (x2 + y2 = 16 in this discussion) rather than to solve it out for y. Well we’re in luck! The method of implicit variation does exactly that! ### The Method of Implicit Variation (Differentiation) Given: an equation involving both x and y. Goal: To find an expression for the derivative, dy/dx. Method: 1. Apply the derivative operation to both sides. This means that you should write d/dx before both sides of your equation. This is like an instruction to indicate that you’ll be doing derivatives in the next step. 2. When taking derivatives, treat expressions of x alone as usual. However, if there are any expressions of y, then you must treat y as an unknown function of x. In particular, follow these additional rules: Basically, whenever you take a derivative of a term involving y, then you must tack on dy/dx. 3. Solve (algebraically) for the unknown derivative, dy/dx. At this point, every problem may be different, but there are a few common themes that I’ve found over the years that seem to apply fairly often. 1. Group terms that have a factor of dy/dx on the left side of the equation. Those terms without the derivative should end up on the right side. 2. Factor out by the common dy/dx. 3. Divide by the expression in front of dy/dx 4. Keep in mind, your final answer may involve both x and y. ### Where Do those dy/dx Factors Come From? This is something that had bugged me for a long time after first learning the method myself. Why do we have to tack on an “extra” dy/dx when taking derivatives involving y? The big idea here is that y is actually a function. We just have no idea what that function is! In a typical (or explicit) function, such as y = x3 – 3x + 2, y has already been isolated. In this example, we know that the function is f(x) = x3 – 3x + 2. However, an implicit equation has not been solved for y. In fact, it may be impossible to do so! So we do the next best thing, which is simply to use our rules of calculus, including the Chain Rule, whenever we encounter the unknown function y in our equation. That’s where the extra dy/dx comes from. There is a hidden Chain Rule lurking in the background! ## Example — Free Response Consider the equation x2 – 2xy + 4y2 = 52. (a) Write an expression for the slope of the curve at any point (x, y). (b) Find the equations of the tangent lines to the curve at the point x = 2. (c) Find at (0, √(13)). ### (a) Slope and the Derivative The keyword slope indicates that we must find a derivative. It would be way too difficult to solve the equation explicitly for y. So this is a job for implicit differentiation! First, apply d/dx to both sides. The next few steps are just working out the derivatives. Perhaps the trickiest part is the term involving 2xy. Think of that as the product of 2x with an unknown function y = f(x). That way, it may make more sense why we must use the Product Rule for that term. Finally, solve for the unknown derivative algebraically. Don’t forget to group, factor and divide! We can factor out a common factor of 2 on top and bottom to get a final answer: ### (b) Finding the Tangent Lines There’s a clue in the word lines. You should expect there to be more than one answer. First find the y-coordinate(s) that correspond to x = 2. We do this by plugging x = 2 into the original equation. So there are two solutions: y = 4 and y = -3. This means that there are two different points at which we must find a tangent line. At each point, plug in the (x, y) pair into dy/dx from part (a) to find the slope. Point 1: (2, 4). Slope = . Therefore, using point-slope form for the line, we get y = (1/7)(x – 2) + 4. Point 2: (2, -3). Slope = . Again using point-slope form, we find y = (5/14)(x – 2) – 3. ### (c) Implicit Second Derivatives To find the second derivative of an implicit function, you must take a derivative of the first derivative (of course!). However, all of the same peculiar rules about expressions of y still apply. Note that we are using the Quotient Rule to start things off. Now, the good news is that we don’t have to simplify the expression any further. This is because they are looking for a numerical final answer. So we just have to plug in the given (x, y) coordinates. But what about the two spots where “dy/dx” shows up? Well we already have an expression for dy/dx from part (a). Simply plug in your (x, y) coordinates to find dy/dx …and now you can plug that into the second derivative expression as well. ## Summary On the AP Calculus AB or BC exam, you will need to know the following. • How to find the derivative of an implicitly-defined function using the Method of implicit variation (a.k.a. implicit differentiation). • What the derivative means in terms of slope and how to find tangent lines to a curve defined implicitly. • How to compute second derivatives of implicitly-defined functions. ##### About Shaun Ault Shaun earned his Ph. D. in mathematics from The Ohio State University in 2008 (Go Bucks!!). He received his BA in Mathematics with a minor in computer science from Oberlin College in 2002. In addition, Shaun earned a B. Mus. from the Oberlin Conservatory in the same year, with a major in music composition. Shaun still loves music -- almost as much as math! -- and he (thinks he) can play piano, guitar, and bass. Shaun has taught and tutored students in mathematics for about a decade, and hopes his experience can help you to succeed! Magoosh blog comment policy: To create the best experience for our readers, we will approve and respond to comments that are relevant to the article, general enough to be helpful to other students, concise, and well-written! :) If your comment was not approved, it likely did not adhere to these guidelines. If you are a Premium Magoosh student and would like more personalized service, you can use the Help tab on the Magoosh dashboard. Thanks!
# Difference between Sine and Arcsine There is a significant difference between sine and sine inverse, which is also known as arcsine. They are both basic ratios of trigonometry, and you will encounter them in almost every mathematical coursework you take after high school. Sine helps you calculate the value of an angle if the different sides of a triangle are given. However, in case you want to calculate the sides and you are provided with an angle, you will have to use arcsine. There are two major differences between sine and arcsine: The basic trigonometric function is sine, and arcsine is its inverse. Secondly, the sine function will calculate a number or an angle in radians and between the range of -1 and +1. On the other hand, an arcsine will give the value in terms of a real number and within the range of -1, +1 to -π, +π. ### Instructions • 1 Sine The perfect way to define sine is through a right angled triangle. In the most basic form, sine is a ratio between the perpendicular and the hypotenuse. So basically, sin α = perpendicular/hypotenuse, with α being the angle between these two lines. Sine can also be defined as a function of an angle, but the magnitude is always given in radians. In modern day mathematics, it is known as a solution to certain differential equations. The function, in real numbers, will range between negative infinity to positive infinity. Image Courtesy: conservapedia.com • 2 Arsine (or Sine inverse) The inverse of sine is an arcsine. This function is used to calculate the angle between any two sides which are given in terms of real numbers. The domain and the co-domain in this case is the opposite of a sine function, and shaped backwards. It definitely relates to the sine function as the co-domain of a sine function is the domain of the arcsine function and vice versa. Mostly, the real numbers are mapped between the range of -1, +1 to R. The single problem with this inverse function is in its validity. Mostly, the inverse function is not valid when the whole domain is selected from the original function. Basically, there is a violation in the definition of the function itself. This is the reason why the range of an inverse function of sine is restricted to -π, +π. This way, the different elements in a single domain do not get mixed with the elements included in the co-domain. Hence, when 1 is subtracted from sine, the range is from -1, +1 to -π, +π. Image Courtesy: rapidtables.com
# What is the distance between (13,-14,1) and (12,-21,6)? May 14, 2016 $d = 5 \sqrt{3} \text{ }$ Exactly $d = 8.660 \text{ }$ to 3 decimal places #### Explanation: This is the coordinate system for 3-space. When it is broken down into projections onto the planes you end up with two triangles that are linked by a common side. Consequently we can turn to that good old favourite: Pythagoras. Let the distance between points be $d$ $d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({z}_{2} - {z}_{1}\right)}^{2}}$ $d = \sqrt{{\left(12 - 13\right)}^{2} + {\left(- 21 - \left(- 14\right)\right)}^{2} + {\left(6 - 1\right)}^{2}}$ $d = \sqrt{75} = \sqrt{{5}^{2} \times 3}$ $d = 5 \sqrt{3} \text{ }$ Exactly $d = 8.660 \text{ }$ to 3 decimal places
# 9.4 Perimeter and circumference of geometric figures Page 1 / 1 This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr. This module discusses perimeter and circumference of geometric figures. By the end of the module students should know what a polygon is, know what perimeter is and how to find it, know what the circumference, diameter, and radius of a circle is and how to find each one, know the meaning of the symbol $\pi$ and its approximating value and know what a formula is and four versions of the circumference formula of a circle. ## Section overview • Polygons • Perimeter • The Number $\pi$ • Formulas ## Polygons We can make use of conversion skills with denominate numbers to make measurements of geometric figures such as rectangles, triangles, and circles. To make these measurements we need to be familiar with several definitions. ## Polygon A polygon is a closed plane (flat) figure whose sides are line segments (portions of straight lines). ## Perimeter The perimeter of a polygon is the distance around the polygon. To find the perimeter of a polygon, we simply add up the lengths of all the sides. ## Sample set a Find the perimeter of each polygon. $\begin{array}{ccc}\hfill \text{Perimeter}& =& \text{2 cm}+\text{5 cm}+\text{2 cm}+\text{5 cm}\hfill \\ & =& \text{14 cm}\hfill \end{array}$ Our first observation is that three of the dimensions are missing. However, we can determine the missing measurements using the following process. Let A, B, and C represent the missing measurements. Visualize $A=\text{12m}-\text{2m}=\text{10m}$ $B=\text{9m}+\text{1m}-\text{2m}=\text{8m}$ $C=\text{12m}-\text{1m}=\text{11m}$ ## Practice set a Find the perimeter of each polygon. 20 ft 26.8 m 49.89 mi ## Circumference The circumference of a circle is the distance around the circle. ## Diameter A diameter of a circle is any line segment that passes through the center of the circle and has its endpoints on the circle. A radius of a circle is any line segment having as its endpoints the center of the circle and a point on the circle. The radius is one half the diameter. ## The number $\pi$ The symbol $\pi$ , read "pi," represents the nonterminating, nonrepeating decimal number 3.14159 … . This number has been computed to millions of decimal places without the appearance of a repeating block of digits. For computational purposes, $\pi$ is often approximated as 3.14. We will write $\pi \approx 3\text{.}\text{14}$ to denote that $\pi$ is approximately equal to 3.14. The symbol "≈" means "approximately equal to." ## Formulas To find the circumference of a circle, we need only know its diameter or radius. We then use a formula for computing the circumference of the circle. ## Formula A formula is a rule or method for performing a task. In mathematics, a formula is a rule that directs us in computations. Formulas are usually composed of letters that represent important, but possibly unknown, quantities. If $C$ , $d$ , and $r$ represent, respectively, the circumference, diameter, and radius of a circle, then the following two formulas give us directions for computing the circumference of the circle. ## Circumference formulas 1. $C=\pi d$ or $C\approx \left(3\text{.}\text{14}\right)d$ 2. $C=2\pi r$ or $C\approx 2\left(3\text{.}\text{14}\right)r$ ## Sample set b Find the exact circumference of the circle. Use the formula $C=\pi d$ . $C=\pi \cdot 7\text{in}\text{.}$ By commutativity of multiplication, $C=7\text{in}\text{.}\cdot \pi$ $C=7\pi \text{in}\text{.}$ , exactly This result is exact since $\pi$ has not been approximated. Find the approximate circumference of the circle. Use the formula $C=\pi d$ . $C\approx \left(3\text{.}\text{14}\right)\left(6\text{.}2\right)$ This result is approximate since $\pi$ has been approximated by 3.14. Find the approximate circumference of a circle with radius 18 inches. Since we're given that the radius, $r$ , is 18 in., we'll use the formula $C=2\pi r$ . Find the approximate perimeter of the figure. We notice that we have two semicircles (half circles). The larger radius is 6.2 cm. The smaller radius is $6\text{.}\text{2 cm - 2}\text{.}\text{0 cm}=\text{4}\text{.}\text{2 cm}$ . The width of the bottom part of the rectangle is 2.0 cm. ## Practice set b Find the exact circumference of the circle. 9.1 $\pi$ in. Find the approximate circumference of the circle. 5.652 mm Find the approximate circumference of the circle with radius 20.1 m. 126.228 m Find the approximate outside perimeter of 41.634 mm ## Exercises Find each perimeter or approximate circumference. Use $\pi =3\text{.}\text{14}$ . 21.8 cm 38.14 inches 0.86 m 87.92 m 16.328 cm 0.0771 cm 120.78 m 21.71 inches 43.7 mm 45.68 cm ## Exercises for review ( [link] ) Find the value of $2\frac{8}{\text{13}}\cdot \sqrt{\text{10}\frac{9}{\text{16}}}$ . 8.5 or $\frac{\text{17}}{2}$ or $8\frac{1}{2}$ ( [link] ) Find the value of $\frac{8}{\text{15}}+\frac{7}{\text{10}}+\frac{\text{21}}{\text{60}}$ . ( [link] ) Convert $\frac{7}{8}$ to a decimal. 0.875 ( [link] ) What is the name given to a quantity that is used as a comparison to determine the measure of another quantity? ( [link] ) Add 42 min 26 sec to 53 min 40 sec and simplify the result. 1 hour 36 minutes 6 seconds Is there any normative that regulates the use of silver nanoparticles? what king of growth are you checking .? Renato What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ? why we need to study biomolecules, molecular biology in nanotechnology? ? Kyle yes I'm doing my masters in nanotechnology, we are being studying all these domains as well.. why? what school? Kyle biomolecules are e building blocks of every organics and inorganic materials. Joe anyone know any internet site where one can find nanotechnology papers? research.net kanaga sciencedirect big data base Ernesto Introduction about quantum dots in nanotechnology what does nano mean? nano basically means 10^(-9). nanometer is a unit to measure length. Bharti do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment? absolutely yes Daniel how to know photocatalytic properties of tio2 nanoparticles...what to do now it is a goid question and i want to know the answer as well Maciej Abigail for teaching engĺish at school how nano technology help us Anassong Do somebody tell me a best nano engineering book for beginners? there is no specific books for beginners but there is book called principle of nanotechnology NANO what is fullerene does it is used to make bukky balls are you nano engineer ? s. fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball. Tarell what is the actual application of fullerenes nowadays? Damian That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes. Tarell what is the Synthesis, properties,and applications of carbon nano chemistry Mostly, they use nano carbon for electronics and for materials to be strengthened. Virgil is Bucky paper clear? CYNTHIA carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc NANO so some one know about replacing silicon atom with phosphorous in semiconductors device? Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure. Harper Do you know which machine is used to that process? s. how to fabricate graphene ink ? for screen printed electrodes ? SUYASH What is lattice structure? of graphene you mean? Ebrahim or in general Ebrahim in general s. Graphene has a hexagonal structure tahir On having this app for quite a bit time, Haven't realised there's a chat room in it. Cied what is biological synthesis of nanoparticles how did you get the value of 2000N.What calculations are needed to arrive at it Privacy Information Security Software Version 1.1a Good 7hours 36 min - 4hours 50 min
## Introduction: The Area Problem and the Definite Integral Integration is vital to many scientific areas. Many powerful mathematical tools are based on integration. Differential equations for instance are the direct consequence of the development of integration. So what is integration? Integration stems from two different problems. The more immediate problem is to find the inverse transform of the derivative. This concept is known as finding the antiderivative. The other problem deals with areas and how to find them. The bridge between these two different problems is the Fundamental Theorem of Calculus. What is the "area problem"? We want to find the area of a given region in the plane. It is not hard to see that this problem can be reduced to finding the area of the region bounded above by the graph of a positive function f (x), bounded below by the x-axis, bounded to the left by the vertical line x = a, and to the right by the vertical line x = b. The answer to this problem came through a very nice idea. Indeed, let us split the region into small subregions which we can approximate by rectangles or other simple geometrical figures (whose areas we know how to compute). This is how it goes: split the interval [a, b] into subintervals, preferably with the same width x, x0 = a < x1 < x2 < ... < xn = b with xi + 1 - xi = x = , for i = 0, 1, ... , n - 1. Let be the subregion bounded above by the graph of f (x), bounded below by the x-axis, bounded to the left by x = xi - 1, and to the right by x = xi, for i = 1, ... , n. Clearly we have Area() = Area() + Area() + ... + Area() . So we focus on the subregions , for i = 1, ... , n. Since we want to approximate the regions by rectangles, then we only have to worry about the upper boundary of each region (since on the other sides we already have straight lines). Again: We are looking for good approximations of the regions by rectangles. The easiest way to choose a height for our rectangles is to choose the value of the function at the left (or right) end points of the small intervals [xi - 1, xi]. Let Li be the rectangle defined by the left-end point and Ri be the rectangle defined by the right-end point. Then an approximation to Area() is given by Area(L1) + Area(L2) + ... + Area(Ln) = xf (x0) + xf (x1) + ... + xf (xn - 1) which we will call the left-sum denoted LEFT(n), and Area(R1) + Area(R2) + ... + Area(Rn) = xf (x1) + xf (x2) + ... + xf (xn) which we will call the right-sum denoted RIGHT(n) Example. Consider the function f (x) = x2 for x [0, 1]. Let us split the interval into 4 subintervals. We have x0 = 0 , x1 = , x2 = , x3 = , x4 = 1 . We have x = and LEFT(4) = 02 + + + = and RIGHT(4) = + + + 12 = . Note that Area() equals , a result which we will prove in later pages. Indeed if the function f (x) is not too badly behaved, we will show that when n gets larger, the numbers LEFT(n) and RIGHT(n) get closer to Area(), i.e. Area() = LEFT(n) = RIGHT(n) . This is the main idea described above. The number Area() is called the definite integral (or more simply the integral) of f (x) from a to b and is denoted by f (x) dx . Note that in the expression f (x) dx the variable x may be replaced by any other variable. Example. Let 0. Then we have dx = (b - a) . This is true since the region is simply a rectangle. Example. We have x dx = (b2 - a2) . Indeed, the region is simply the union of two regions: one rectangle and one triangle. The rectangle (depicted in red) is bounded above by x = a and its area is a(b - a). The triangle (in blue) is determined by the points: (a, a), (a, b), and (b, b). Its area is (b - a)2. So we have xdx = a(b - a) + (b - a)2 = (b2 - a2) . A precise definition for the definite integral involves partitions and lower as well as upper sums: Definition. A partition P of the interval [a, b] is a sequence of numbers {xi;i = 0, 1, ... , n} such that x0 = a < x1 < x2 < ... < xn = b For a function f (x) defined on [a, b] and a partition P of [a, b], set mi = inf{f (x); x [xi - 1, xi]} and Mi = sup{f (x); x [xi - 1, xi]} for i = 1, ... , n, provided that f (x) is bounded on [a, b]. The sum Lf(P) = m1(x1 - x0) + m2(x2 - x1) + ... + mn(xn - xn - 1) is called the lower sum for f (x) over the partition P, and Uf(P) = M1(x1 - x0) + M2(x2 - x1) + ... + Mn(xn - xn - 1) is called the upper sum for f (x) over the partition P. Theorem. We have Lf(P) Area() Uf(P) for any partition P of [a, b]. Moreover if f (x) is continuous on [a, b], except maybe at a finite number of points, and I is a number such that Lf(P) I Uf(P) for any partition P of [a, b], then I = Area(). This theorem is fundamental. Let us illustrate this with the following example. Example. Use the above theorem to show x2dx = (b3 - a3) where b a 0. Let P = {x0, x1, ... , xn} be a partition of [a, b]. Since f (x) = x2 is increasing on [a, b], then mi = xi - 12 and Mi = xi2. So we have Uf(P) = x21(x1 - x0) + x22(x2 - x1) + ... + x2n(xn - xn - 1) and Lf(P) = x20(x1 - x0) + x21(x2 - x1) + ... + x2n - 1(xn - xn - 1) . For each i, we have xi - 12 xi - 12 + xi - 1xi + xi2 xi2 since xi - 1 xi. If we multiply by xi - xi - 1, we get xi - 12(xi - xi - 1) xi - 12 + xi - 1xi + xi2(xi - xi - 1) xi2(xi - xi - 1) . But xi - 12 + xi - 1xi + xi2(xi - xi - 1) = xi3 - xi - 13 which implies xi - 12(xi - xi - 1) (xi3 - xi - 13) xi2(xi - xi - 1) . Hence Lf(P) (b3 - a3) Uf(P) since (x13 - x03) + (x23 - x13) + ... + (xn3 - xn - 13) = b3 - a3 . Exercise 1. Use similar ideas as used in the example above to show xn dx = (bn + 1 - an + 1) where b a 0. dx . f (x) = Show that f (x) dx does not exist. 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LCM of 6 and also 7 is the smallest number amongst all usual multiples that 6 and 7. The first couple of multiples the 6 and also 7 room (6, 12, 18, 24, 30, 36, . . . ) and (7, 14, 21, 28, 35, 42, 49, . . . ) respectively. There are 3 typically used methods to find LCM the 6 and also 7 - by department method, by prime factorization, and by listing multiples. You are watching: Common multiples of 7 and 6 1 LCM that 6 and also 7 2 List that Methods 3 Solved Examples 4 FAQs Answer: LCM of 6 and 7 is 42. Explanation: The LCM of two non-zero integers, x(6) and also y(7), is the smallest optimistic integer m(42) the is divisible by both x(6) and y(7) without any remainder. The methods to find the LCM of 6 and 7 are explained below. By division MethodBy Listing MultiplesBy element Factorization Method ### LCM that 6 and 7 by division Method To calculate the LCM of 6 and also 7 by the division method, we will divide the numbers(6, 7) by their prime factors (preferably common). The product of this divisors gives the LCM of 6 and also 7. Step 3: proceed the steps until only 1s are left in the last row. The LCM of 6 and 7 is the product of all prime number on the left, i.e. LCM(6, 7) by division method = 2 × 3 × 7 = 42. ### LCM that 6 and 7 by Listing Multiples To calculate the LCM of 6 and 7 through listing the end the common multiples, we deserve to follow the given below steps: Step 1: list a few multiples that 6 (6, 12, 18, 24, 30, 36, . . . ) and also 7 (7, 14, 21, 28, 35, 42, 49, . . . . )Step 2: The usual multiples native the multiples that 6 and 7 are 42, 84, . . .Step 3: The smallest usual multiple the 6 and 7 is 42. ∴ The least common multiple that 6 and 7 = 42. See more: We Love You, Mom In Spanish With Examples, How Do You Say I Love You Mom In Spanish (Mexico) ### LCM of 6 and also 7 by element Factorization Prime administer of 6 and 7 is (2 × 3) = 21 × 31 and (7) = 71 respectively. LCM the 6 and also 7 have the right to be derived by multiply prime components raised to their respective highest power, i.e. 21 × 31 × 71 = 42.Hence, the LCM of 6 and 7 by element factorization is 42.
Maths How To with Anita # How to multiply fractions – Simple explanation so you can understand fractions now In this article, we will show you how to multiply fractions in a few simple steps. Multiplying fractions is a basic skill that all students should learn. It can be tricky at first, but with a little practice, you’ll be able to do it like a pro! Let’s get started. ## 1. What are fractions and how do they work Fractions are a way of representing parts of a whole. For example, if we have a pizza and we want to share it with our friends, we can cut it into slices and each person can have one slice. We would say that each person has “one-eighth” of the pizza because there are eight slices in total. ## 2. How to multiply two fractions together When we multiply fractions, we simply multiply the numerators (top numbers) and multiply the denominators (bottom numbers). So, if we want to multiply one-half by three-quarters, we would multiply the top numbers together (one times three is three) and the bottom numbers together (two times four is eight): This gives us the answer of three-eighths. And that’s all there is to multiplying fractions! ## 3. Why do you get a smaller fraction when you multiply two fractions together? When you multiply two proper fractions together you end up with a smaller fraction. Why is this? Imagine you had one-quarter of a pizza to share between 2 people. Finding one-half of something is the same as dividing by two. Now visualize cutting that one-quarter pizza into two pieces. How much of the whole pizza is each piece? That’s right. Each piece is now one-eighth. So when you are finding a fraction of another fraction you are breaking that fraction into smaller pieces hence why your answer is a smaller fraction. ## 4. Extra tips and tricks to make multiplying fractions easier • Cancel on the diagonal or vertically before multiplying. This means you will have smaller numbers to simplify after multiplying the numerators and denominators. • A whole number can be written as a fraction with a denominator of one. • Change a mixed number to an improper fraction before multiplying. ## 5. Common misconceptions and errors when multiplying fractions Some students like to multiply a whole number by both the numerator and the denominator. This is incorrect. . Some students cancel horizontally. This is not right. Some students cross multiply. This is incorrect. ## 6. The importance of mastering multiplication skills Multiplication skills are important for many reasons. In everyday life, we use multiplication for things like cooking and shopping. We also use it in more advanced Math concepts like Algebra and Calculus. So it’s important to master this skill now so that you can be successful in the future! We hope this article has helped you understand how to multiply fractions. With a little practice, you’ll be a pro in no time! If you’re still having trouble, there are plenty of online math tutors to help you master this skill. Thanks for reading and good luck!
Common Core Alignment:. 1.NBT.2: Number And Operations In Base Ten Understand Place Value. Understand that the two digits of a two-digit number represent amounts of tens and ones. Understand the following as special cases: a. 10 can be thought of as a bundle of ten ones - called a "ten." b. The numbers from 11 to 19 are composed of a ten and one, two, three, four, five, six, seven, eight, or nine ones. c. The numbers 10, 20, 30, 40, 50, 60, 70, 80, 90 refer to one, two, three, four, five, six, seven, eight, or nine tens (and 0 ones). 1.NBT.2a: Number And Operations In Base Ten Understand Place Value. 10 can be thought of as a bundle of ten ones - called a "ten." b. 1.OA.6: Operations And Algebraic Thinking Add And Subtract Within 20. Add and subtract within 20, demonstrating fluency for addition and subtraction within 10. Use strategies such as counting on; making ten (e.g., 8 + 6 = 8 + 2 + 4 = 10 + 4 = 14); decomposing a number leading to a ten (e.g., 13 - 4 = 13 - 3 - 1 = 10 - 1 = 9); using the relationship between addition and subtraction (e.g., knowing that 8 + 4 = 12, one knows 12 - 8 = 4); and creating equivalent but easier or known sums (e.g., adding 6 + 7 by creating the known equivalent 6 + 6 + 1 = 12 + 1 = 13). K.CC.2: Counting And Cardinality Know Number Names And The Count Sequence. Count forward beginning from a given number within the known sequence (instead of having to begin at 1). K.CC.3: Counting And Cardinality Know Number Names And The Count Sequence. Write numbers from 0 to 20. Represent a number of objects with a written numeral 0-20 (with 0 representing a count of no objects). K.CC.4a: Counting And Cardinality Count To Tell The Number Of Objects. When counting objects, say the number names in the standard order, pairing each object with one and only one number name and each number name with one and only one object. K.NBT.1: Number And Operations In Base Ten Work With Numbers 11-19 To Gain Foundations For Place Value. Compose and decompose numbers from 11 to 19 into ten ones and some further ones, e.g., by using objects or drawings, and record each composition or decomposition by a drawing or equation (e.g., 18 = 10 + 8); understand that these numbers are composed of ten ones and one, two, three, four, five, six, seven, eight, or nine ones.
# 12.1 The ellipse (Page 6/16) Page 6 / 16 Graph the ellipse given by the equation $\text{\hspace{0.17em}}49{x}^{2}+16{y}^{2}=784.\text{\hspace{0.17em}}$ Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci. Standard form: $\text{\hspace{0.17em}}\frac{{x}^{2}}{16}+\frac{{y}^{2}}{49}=1;\text{\hspace{0.17em}}$ center: $\text{\hspace{0.17em}}\left(0,0\right);\text{\hspace{0.17em}}$ vertices: $\text{\hspace{0.17em}}\left(0,±7\right);\text{\hspace{0.17em}}$ co-vertices: $\text{\hspace{0.17em}}\left(±4,0\right);\text{\hspace{0.17em}}$ foci: $\text{\hspace{0.17em}}\left(0,±\sqrt{33}\right)$ ## Graphing ellipses not centered at the origin When an ellipse is not centered at the origin, we can still use the standard forms to find the key features of the graph. When the ellipse is centered at some point, $\text{\hspace{0.17em}}\left(h,k\right),$ we use the standard forms for horizontal ellipses and for vertical ellipses. From these standard equations, we can easily determine the center, vertices, co-vertices, foci, and positions of the major and minor axes. Given the standard form of an equation for an ellipse centered at $\text{\hspace{0.17em}}\left(h,k\right),$ sketch the graph. 1. Use the standard forms of the equations of an ellipse to determine the center, position of the major axis, vertices, co-vertices, and foci. 1. If the equation is in the form $\text{\hspace{0.17em}}\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1,\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}a>b,\text{\hspace{0.17em}}$ then • the center is $\text{\hspace{0.17em}}\left(h,k\right)$ • the major axis is parallel to the x -axis • the coordinates of the vertices are $\text{\hspace{0.17em}}\left(h±a,k\right)$ • the coordinates of the co-vertices are $\text{\hspace{0.17em}}\left(h,k±b\right)$ • the coordinates of the foci are $\text{\hspace{0.17em}}\left(h±c,k\right)$ 2. If the equation is in the form $\text{\hspace{0.17em}}\frac{{\left(x-h\right)}^{2}}{{b}^{2}}+\frac{{\left(y-k\right)}^{2}}{{a}^{2}}=1,\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}a>b,\text{\hspace{0.17em}}$ then • the center is $\text{\hspace{0.17em}}\left(h,k\right)$ • the major axis is parallel to the y -axis • the coordinates of the vertices are $\text{\hspace{0.17em}}\left(h,k±a\right)$ • the coordinates of the co-vertices are $\text{\hspace{0.17em}}\left(h±b,k\right)$ • the coordinates of the foci are $\text{\hspace{0.17em}}\left(h,k±c\right)$ 2. Solve for $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ using the equation $\text{\hspace{0.17em}}{c}^{2}={a}^{2}-{b}^{2}.$ 3. Plot the center, vertices, co-vertices, and foci in the coordinate plane, and draw a smooth curve to form the ellipse. ## Graphing an ellipse centered at ( h , k ) Graph the ellipse given by the equation, $\text{\hspace{0.17em}}\frac{{\left(x+2\right)}^{2}}{4}+\frac{{\left(y-5\right)}^{2}}{9}=1.\text{\hspace{0.17em}}$ Identify and label the center, vertices, co-vertices, and foci. First, we determine the position of the major axis. Because $\text{\hspace{0.17em}}9>4,$ the major axis is parallel to the y -axis. Therefore, the equation is in the form $\text{\hspace{0.17em}}\frac{{\left(x-h\right)}^{2}}{{b}^{2}}+\frac{{\left(y-k\right)}^{2}}{{a}^{2}}=1,\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}{b}^{2}=4\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{a}^{2}=9.\text{\hspace{0.17em}}$ It follows that: • the center of the ellipse is $\text{\hspace{0.17em}}\left(h,k\right)=\left(-2,\text{5}\right)$ • the coordinates of the vertices are $\text{\hspace{0.17em}}\left(h,k±a\right)=\left(-2,5±\sqrt{9}\right)=\left(-2,5±3\right),$ or $\text{\hspace{0.17em}}\left(-2,\text{2}\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(-2,\text{8}\right)$ • the coordinates of the co-vertices are $\text{\hspace{0.17em}}\left(h±b,k\right)=\left(-2±\sqrt{4},5\right)=\left(-2±2,5\right),$ or $\text{\hspace{0.17em}}\left(-4,5\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(0,\text{5}\right)$ • the coordinates of the foci are $\text{\hspace{0.17em}}\left(h,k±c\right),\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}{c}^{2}={a}^{2}-{b}^{2}.\text{\hspace{0.17em}}$ Solving for $\text{\hspace{0.17em}}c,$ we have: $\begin{array}{l}\begin{array}{l}\\ c=±\sqrt{{a}^{2}-{b}^{2}}\end{array}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=±\sqrt{9-4}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=±\sqrt{5}\hfill \end{array}$ Therefore, the coordinates of the foci are $\text{\hspace{0.17em}}\left(-2,\text{5}-\sqrt{5}\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(-2,\text{5+}\sqrt{5}\right).$ Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse. Graph the ellipse given by the equation $\text{\hspace{0.17em}}\frac{{\left(x-4\right)}^{2}}{36}+\frac{{\left(y-2\right)}^{2}}{20}=1.\text{\hspace{0.17em}}$ Identify and label the center, vertices, co-vertices, and foci. Center: $\text{\hspace{0.17em}}\left(4,2\right);\text{\hspace{0.17em}}$ vertices: $\text{\hspace{0.17em}}\left(-2,2\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(10,2\right);\text{\hspace{0.17em}}$ co-vertices: $\text{\hspace{0.17em}}\left(4,2-2\sqrt{5}\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(4,2+2\sqrt{5}\right);\text{\hspace{0.17em}}$ foci: $\text{\hspace{0.17em}}\left(0,2\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(8,2\right)$ Given the general form of an equation for an ellipse centered at ( h , k ), express the equation in standard form. 1. Recognize that an ellipse described by an equation in the form $\text{\hspace{0.17em}}a{x}^{2}+b{y}^{2}+cx+dy+e=0\text{\hspace{0.17em}}$ is in general form. 2. Rearrange the equation by grouping terms that contain the same variable. Move the constant term to the opposite side of the equation. 3. Factor out the coefficients of the $\text{\hspace{0.17em}}{x}^{2}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{y}^{2}\text{\hspace{0.17em}}$ terms in preparation for completing the square. 4. Complete the square for each variable to rewrite the equation in the form of the sum of multiples of two binomials squared set equal to a constant, $\text{\hspace{0.17em}}{m}_{1}{\left(x-h\right)}^{2}+{m}_{2}{\left(y-k\right)}^{2}={m}_{3},$ where $\text{\hspace{0.17em}}{m}_{1},{m}_{2},$ and $\text{\hspace{0.17em}}{m}_{3}\text{\hspace{0.17em}}$ are constants. 5. Divide both sides of the equation by the constant term to express the equation in standard form. what are you up to? nothing up todat yet Miranda hi jai hello jai Miranda Drice jai aap konsi country se ho jai which language is that Miranda I am living in india jai good Miranda what is the formula for calculating algebraic I think the formula for calculating algebraic is the statement of the equality of two expression stimulate by a set of addition, multiplication, soustraction, division, raising to a power and extraction of Root. U believe by having those in the equation you will be in measure to calculate it Miranda state and prove Cayley hamilton therom hello Propessor hi Miranda the Cayley hamilton Theorem state if A is a square matrix and if f(x) is its characterics polynomial then f(x)=0 in another ways evey square matrix is a root of its chatacteristics polynomial. Miranda hi jai hi Miranda jai thanks Propessor welcome jai What is algebra algebra is a branch of the mathematics to calculate expressions follow. Miranda Miranda Drice would you mind teaching me mathematics? I think you are really good at math. I'm not good at it. In fact I hate it. 😅😅😅 Jeffrey lolll who told you I'm good at it Miranda something seems to wispher me to my ear that u are good at it. lol Jeffrey lolllll if you say so Miranda but seriously, Im really bad at math. And I hate it. But you see, I downloaded this app two months ago hoping to master it. Jeffrey which grade are you in though Miranda oh woww I understand Miranda Jeffrey Jeffrey Miranda how come you finished in college and you don't like math though Miranda gotta practice, holmie Steve if you never use it you won't be able to appreciate it Steve I don't know why. But Im trying to like it. Jeffrey yes steve. you're right Jeffrey so you better Miranda what is the solution of the given equation? which equation Miranda I dont know. lol Jeffrey Miranda Jeffrey answer and questions in exercise 11.2 sums how do u calculate inequality of irrational number? Alaba give me an example Chris and I will walk you through it Chris cos (-z)= cos z . cos(- z)=cos z Mustafa what is a algebra (x+x)3=? 6x Obed what is the identity of 1-cos²5x equal to? __john __05 Kishu Hi Abdel hi Ye hi Nokwanda C'est comment Abdel Hi Amanda hello SORIE Hiiii Chinni hello Ranjay hi ANSHU hiiii Chinni h r u friends Chinni yes Hassan so is their any Genius in mathematics here let chat guys and get to know each other's SORIE I speak French Abdel okay no problem since we gather here and get to know each other SORIE hi im stupid at math and just wanna join here Yaona lol nahhh none of us here are stupid it's just that we have Fast, Medium, and slow learner bro but we all going to work things out together SORIE it's 12 what is the function of sine with respect of cosine , graphically tangent bruh Steve cosx.cos2x.cos4x.cos8x sinx sin2x is linearly dependent what is a reciprocal The reciprocal of a number is 1 divided by a number. eg the reciprocal of 10 is 1/10 which is 0.1 Shemmy Reciprocal is a pair of numbers that, when multiplied together, equal to 1. Example; the reciprocal of 3 is ⅓, because 3 multiplied by ⅓ is equal to 1 Jeza each term in a sequence below is five times the previous term what is the eighth term in the sequence I don't understand how radicals works pls How look for the general solution of a trig function
$$\require{cancel}$$ # 1.2: Combining Probabilities Consider two distinct possible outcomes, $$X$$ and $$Y$$, of an observation made on the system $$S$$, with probabilities of occurrence $$P(X)$$ and $$P(Y)$$, respectively. Let us determine the probability of obtaining either the outcome $$X$$ or the outcome $$Y$$, which we shall denote $$P(X\mid Y)$$. From the basic definition of probability, $P(X\mid Y) =\lim_{{\mit\Omega}({\mit\Sigma})\rightarrow\infty} \frac{ {\mit\Omega}(X \mid Y)}{{\mit\Omega}({\mit\Sigma})},$ where $${\mit\Omega}(X \mid Y)$$ is the number of systems in the ensemble that exhibit either the outcome $$X$$ or the outcome $$Y$$. Now, ${\mit\Omega}(X\mid Y) = {\mit\Omega}(X) + {\mit\Omega}(Y)$ if the outcomes $$X$$ and $$Y$$ are mutually exclusive (which must be the case if they are two distinct outcomes). Thus, $P(X\mid Y) = P(X) + P(Y), \label{x2.4}$ which means that the probability of obtaining either the outcome $$X$$ or the outcome $$Y$$ is the sum of the individual probabilities of $$X$$ and $$Y$$. For instance, with a six-sided die the probability of throwing any particular number (one to six) is $$1/6$$, because all of the possible outcomes are considered to be equally likely. It follows, from what has just been said, that the probability of throwing either a one or a two is simply $$1/6+1/6$$, which equals $$1/3$$. Let us denote all of the $$M$$, say, possible outcomes of an observation made on the system $$S$$ by $$X_i$$, where $$i$$ runs from $$1$$ to $$M$$. Let us determine the probability of obtaining any one of these outcomes. This quantity is unity, from the basic definition of probability, because each of the systems in the ensemble must exhibit one of the possible outcomes. But, this quantity is also equal to the sum of the probabilities of all the individual outcomes, by Equation ([x2.4]), so we conclude that this sum is equal to unity: that is, $\sum_{i=1,M} P(X_i) =1.\label{x2.5}$ The previous expression is called the normalization condition, and must be satisfied by any complete set of probabilities. There is another way in which we can combine probabilities. Suppose that we make an observation on a system picked at random from the ensemble, and then pick a second similar system, completely independently, and make another observation. We are assuming that the first observation does not influence the second observation in any way. In other words, the two observations are statistically independent of one another. Let us determine the probability of obtaining the outcome $$X$$ in the first system and obtaining the outcome $$Y$$ in the second system, which we shall denote $$P(X\otimes Y)$$. In order to determine this probability, we have to form an ensemble of all the possible pairs of systems that we could choose from the ensemble $$ParseError: invalid DekiScript (click for details) Callstack: at (Bookshelves/Quantum_Mechanics/Book:_Introductory_Quantum_Mechanics_(Fitzpatrick)/1:_Probability_Theory/1.2:_Combining_Probabilities), /content/body/p[4]/span[2], line 1, column 1 $$. Let us denote this ensemble $$ParseError: invalid DekiScript (click for details) Callstack: at (Bookshelves/Quantum_Mechanics/Book:_Introductory_Quantum_Mechanics_(Fitzpatrick)/1:_Probability_Theory/1.2:_Combining_Probabilities), /content/body/p[4]/span[3], line 1, column 1 \otimes ParseError: invalid DekiScript (click for details) Callstack: at (Bookshelves/Quantum_Mechanics/Book:_Introductory_Quantum_Mechanics_(Fitzpatrick)/1:_Probability_Theory/1.2:_Combining_Probabilities), /content/body/p[4]/span[4], line 1, column 1 $$. The number of pairs of systems in this new ensemble is just the square of the number of systems in the original ensemble, so ${\mit\Omega}( ParseError: invalid DekiScript (click for details) Callstack: at (Bookshelves/Quantum_Mechanics/Book:_Introductory_Quantum_Mechanics_(Fitzpatrick)/1:_Probability_Theory/1.2:_Combining_Probabilities), /content/body/p[4]/span[5]/span[1], line 1, column 1 \otimes ParseError: invalid DekiScript (click for details) Callstack: at (Bookshelves/Quantum_Mechanics/Book:_Introductory_Quantum_Mechanics_(Fitzpatrick)/1:_Probability_Theory/1.2:_Combining_Probabilities), /content/body/p[4]/span[5]/span[2], line 1, column 1 ) = {\mit\Omega}( ParseError: invalid DekiScript (click for details) Callstack: at (Bookshelves/Quantum_Mechanics/Book:_Introductory_Quantum_Mechanics_(Fitzpatrick)/1:_Probability_Theory/1.2:_Combining_Probabilities), /content/body/p[4]/span[5]/span[3], line 1, column 1 )\, {\mit\Omega}( ParseError: invalid DekiScript (click for details) Callstack: at (Bookshelves/Quantum_Mechanics/Book:_Introductory_Quantum_Mechanics_(Fitzpatrick)/1:_Probability_Theory/1.2:_Combining_Probabilities), /content/body/p[4]/span[5]/span[4], line 1, column 1 ).$ Furthermore, the number of pairs of systems in the ensemble $$ParseError: invalid DekiScript (click for details) Callstack: at (Bookshelves/Quantum_Mechanics/Book:_Introductory_Quantum_Mechanics_(Fitzpatrick)/1:_Probability_Theory/1.2:_Combining_Probabilities), /content/body/p[4]/span[6], line 1, column 1 \otimes ParseError: invalid DekiScript (click for details) Callstack: at (Bookshelves/Quantum_Mechanics/Book:_Introductory_Quantum_Mechanics_(Fitzpatrick)/1:_Probability_Theory/1.2:_Combining_Probabilities), /content/body/p[4]/span[7], line 1, column 1 $$ that exhibit the outcome $$X$$ in the first system and the outcome $$Y$$ in the second system is simply the product of the number of systems that exhibit the outcome $$X$$ and the number of systems that exhibit the outcome $$Y$$ in the original ensemble, so that ${\mit\Omega}(X\otimes Y) = {\mit\Omega}(X) \,{\mit\Omega}(Y).$ It follows from the basic definition of probability that $P(X\otimes Y) =\lim_{{\mit\Omega}({\mit\Sigma})\rightarrow\infty} \frac{{\mit\Omega}(X\otimes Y)}{{\mit\Omega}({{\mit\Sigma}}\otimes {{\mit\Sigma}})}= P(X) \,P(Y).$ Thus, the probability of obtaining the outcomes $$X$$ and $$Y$$ in two statistically independent observations is the product of the individual probabilities of $$X$$ and $$Y$$. For instance, the probability of throwing a one and then a two on a six-sided die is $$1/6 \times 1/6$$, which equals $$1/36$$. # Contributors • Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin) $$\newcommand {\ltapp} {\stackrel {_{\normalsize<}}{_{\normalsize \sim}}}$$ $$\newcommand {\gtapp} {\stackrel {_{\normalsize>}}{_{\normalsize \sim}}}$$ $$\newcommand {\btau}{\mbox{\boldmath\tau}}$$ $$\newcommand {\bmu}{\mbox{\boldmath\mu}}$$ $$\newcommand {\bsigma}{\mbox{\boldmath\sigma}}$$ $$\newcommand {\bOmega}{\mbox{\boldmath\Omega}}$$ $$\newcommand {\bomega}{\mbox{\boldmath\omega}}$$ $$\newcommand {\bepsilon}{\mbox{\boldmath\epsilon}}$$
# How do you solve for r: r = 3p - 5r? Mar 6, 2016 $r = \frac{p}{2}$ #### Explanation: Given:$\text{ } \textcolor{b r o w n}{r = 3 p - 5 r}$ $\textcolor{b l u e}{\text{Method}}$ Collect and manipulate the terms until you have just one $r$ on the right hand side of the equals and everything on the other side. $\textcolor{b l u e}{\text{Solving your question}}$ $\textcolor{g r e e n}{\text{Collecting terms}}$ Add $\textcolor{b l u e}{5 r}$ to both sides $\textcolor{b r o w n}{r \textcolor{b l u e}{+ 5 r} = 3 p - 5 r \textcolor{b l u e}{+ 5 r}}$ $6 r = 3 p + 0$ $\textcolor{g r e e n}{\text{Manipulating so that there is only one } r}$ Divide both sides by $\textcolor{b l u e}{6}$ $\textcolor{b r o w n}{\frac{6}{\textcolor{b l u e}{6}} r = \frac{3}{\textcolor{b l u e}{6}} p}$ But $\frac{6}{6} = 1 \text{ and } \frac{3}{6} = \frac{1}{2}$ $r = \frac{p}{2}$
## Linear Algebra and Its Applications, 4th Edition $$A^{-1}=BC^{-1}$$ $$A^{-1}=U^{-1}L^{-1}P$$ $$AB=C$$ Our goal is to find a formula for $A^{-1}$. Note that matrices are not mutually commutative, i.e. $a\cdot{b}\neq{b}\cdot{a}$. Thus, when multiplying both sides of the equation by any matrix, we must ensure that the inverse is always placed on the same side. Introduce $A^{-1}$ into the equation as a multiple on both sides. $$A^{-1}AB=A^{-1}C$$ Since $\|A^{-1}\cdot{A}\|=1$, $$B=A^{-1}C$$ We proceed to eliminate C from the RHS by multiplying C by its inverse, leaving $A^{-1}$. $$BC^{-1}=A^{-1}CC^{-1}=A^{-1}$$ We now aim to achieve this same objective with the equation $$PA=LU$$ As with earlier, introduce $A^{-1}$ to the equation as a multiple on both sides $$PAA^{-1}=LUA^{-1}$$ $$P=LUA^{-1}$$ $LU$ refers to the product $L\cdot{U}$. To isolate $A^{-1}$, we multiply both sides of the equation by the inverse of this product, $(LU)^{-1}$. $$(LU)^{-1}P=(LU)^{-1}LUA^{-1}=A^{-1}$$ Since $(LU)^{-1}=U^{-1}L^{-1}$, $$U^{-1}L^{-1}P=A^{-1}$$
# Evaluating $\sum _{n=2}^{\infty } \frac{(-5)^n}{8^{2 n}}$ making use of geometric series Evaluate $$\sum _{n=2}^{\infty}\frac{(-5)^n}{8^{2n}}$$ using geometric series. I thought it would be possible to split this series such that we have $$\sum _{n=2}^{\infty } (-5)^n \cdot \sum _{n=2}^{\infty } \left(\frac{1}{8}\right)^{2 n}$$ However, I am not sure that this is actually possible and I also see that the first sum does not converge, so even if it was possible I am not able to solve it. Could someone walk me through the steps? • No, it's not. But note $(-5)^n/ 8^{2n}= (-5/8^2)^n$. – David Mitra Jan 16 '15 at 20:24 • it is a good idea to write out the first few terms. – abel Jan 16 '15 at 20:25 First note that $$\sum\limits_{n=2}^{\infty } \frac{(-5)^n}{8^{2 n}}= \sum\limits_{n=2}^{\infty } \left(-\frac{5}{64}\right)^n$$ Now let's look at the first two terms of the sum $$\left(-\frac{5}{64}\right)^2+ \left(-\frac{5}{64}\right)^3+\dots$$ $$=\left(\frac{5}{64}\right)^2- \left(\frac{5}{64}\right)^3+\dots$$ So now we know that $$a= \left(\frac{5}{64}\right)^2=\frac{25}{4096}$$ And $$r= -\frac{5}{64}$$ Therefore $$\sum\limits_{n=2}^{\infty } \left(-\frac{5}{64}\right)^n=\frac{\frac{25}{4096}}{1-\left(-\frac{5}{64}\right)}=\frac{25}{4416}$$ HINT: It is $\sum_{n=2}^{\infty}\left(\frac{-5}{64}\right)^n$. Try $$\displaystyle \sum _{n=2}^{\infty } \left(-\frac{5}{64}\right)^n = \frac{25}{4096}\sum _{n=0}^{\infty } \left(-\frac{5}{64}\right)^n$$ as a typical geometric series That's not possible, no. But since $8^{2n} = 64^n$, you can rewrite your series as $$\sum \left(\frac{-5}{64}\right)^n,$$ which should get you on your way. • Could you not evaluate it first using the fact that $$\sum\limits_{n=0}^\infty\left(\frac{-5}{64}\right)^n=\frac{1}{1-\left(\frac{-5}{64}\right)}$$ and the subtracting the first two terms from that fraction, since $n=0$ and $n=1$ are not included in the bounds, then simplify? – bjd2385 Jan 16 '15 at 20:29 • Yes, that's exactly right. Alternatively, you could let $a = -5/64$, and call your sum $S$, and note that $\frac{S}{a^2}$ is the sum from $0$ to infinity, so that $S$ must be $a^2$ times that sum. In general, $\sum_0 c r^n = \frac{c}{1 - r}$ lets you handle sums starting at nonzero values (like the "2" in your problem) by factoring out some power of $r$ and including that in $c$. – John Hughes Jan 16 '15 at 20:34 Hint: $$\sum_{n=2}^{\infty} \left(\frac{-5}{64}\right)^n$$
# Partial Differential Equations PDEs are use to model many kinds of problems. Their solutions give evolution of a function as a function of time and space. Boundary conditions involving time and space are used as initial conditions. A method of separation of variables is used for solving them, where it is assumed that . Two other auxiliary ODE results are also needed: Another auxillary ODE are needed for some situations The general process for solving PDEs: • Apply separation of variables • Make an appropriate choice of constant • Nearly always • Solve resulting ODEs • Combine ODE solutions to form general PDE solution • Apply boundary conditions to obtain particular PDE solution • Work out values for the arbitrary constants ## Laplace's Equation Laplace's equation described many problems involving flow in a plane: Find the solution with the following boundary conditions: • and • as Starting with separation of variables: Substituting back into the original PDE: We have transformed the PDE into an ODE, where each side is a function of / only. The only circumstances under which the two sides can be equal for all values of and is if both sides independent and equal to a constant. Since the constant is arbitrary, let it be . Now we have two ODEs and their solutions from the auxiliary results earlier: Substituting the solutions back into , we have a general solution to our PDE in terms of 4 arbitrary constants: We can now apply boundary conditions: • Substituting in gives • Substituting in gives • Using the two together gives , so either: • If , then , so • This is the trivial solution and is of no interest • If , then • This also implies that , so is useless too The issue is that we selected our arbitrary constant badly. If we use instead, then our solutions are the other way round: Checking the boundary conditions again: • First condition, • Gives • Second condition • Gives • Either (not interested) • is an integer, We now have: Where is any integer. Using the other boundary conditions: • as • If is positive, then (otherwise ) • If is negative, then (otherwise ) Taking as positive, the form of the solutions is: The most general form is the sum of these: Applying the final boundary condition: • for all other The complete solution is therefore: ## The Heat Equation The heat equation describe diffusion of energy or matter. With a diffusion coefficient : Solving with the following boundary conditions: Separating variables, , and substituting, exactly the same as Laplace's equation, we have: Setting both sides again equal to a constant , we have two ODEs (one 2nd order, one 1st): The general solution is therefore: Tidying up a bit, let , , : Applying the first boundary condition: • Gives • Since for all , We now have . The second boundary condition: • , so • For the non trivial solution ,and since , • Therefore, for Substituting this in gives: The above equation is valid for any , so summing these gives the most general solution: The last boundary condition is : This is in the form of the a Fourier series: We have: Substituting this into , and letting : ## The Wave Equation The wave equation is used to describe vibrational problems: Solving the equation with the boundary conditions: Doing the usual separation of variables and substitution, and choosing a constant : Solving both ODEs: This is the general solution. Start applying boundary conditions: • implies that • As this is true for all , • implies that • This is also true for all , so • Required that , so • for We now have: Applying the third boundary condition, : As this is for all , , so . We now have: The general solution is then: Applying the final boundary condition of , gives , else . The particular solution is therefore:
US UKIndia Every Question Helps You Learn How many times does the number 144 appear in this quiz? Verbal Reasoning - Related Numbers 4 This Math quiz is called 'Verbal Reasoning - Related Numbers 4' and it has been written by teachers to help you if you are studying the subject at elementary school. Playing educational quizzes is an enjoyable way to learn if you are in the 3rd, 4th or 5th grade - aged 8 to 11. It costs only \$12.50 per month to play this quiz and over 3,500 others that help you with your school work. You can subscribe on the page at Join Us In this series of quizzes, you need to find the relation between the three numbers in brackets. The first and third numbers will have been either added, subtracted, multiplied or divided to get the middle number. Find out which one it is and do the same to the last bracketed numbers. An example has been done for you. Example: (12 [4] 48), (19 [6] 114), (12 [ ? ] 36) 3 4 5 6 This is because 12 and 48 are related by dividing 48 by 12 to get the number in the middle. 19 is related to 114 in the SAME way (dividing 114 by 19 results in 6). The relationship is DIVIDE. 1. Look at the relationship between the first and last number in each set of brackets, and use that relationship to find the missing number. Choose the correct answer from the four choices available. (110 [45] 65), (540 [266] 274), (333 [?] 188) 145 154 122 111 110 and 65 are related by subtracting 65 from 110 to get the number in the middle. 540 is related to 274 in the SAME way (subtracting 274 from 540 results in 266). 333 - 188 = 145. The relationship is SUBTRACT 2. Look at the relationship between the first and last number in each set of brackets, and use that relationship to find the missing number. Choose the correct answer from the four choices available. (16 [192] 12), (20 [340] 17), (11 [?] 27) 270 297 324 351 16 and 12 are related by multiplying 16 by 12 to get the number in the middle. 20 is related to 17 in the SAME way (multiplying 20 by 17 results in 340). 11 x 27 = 297. The relationship is MULTIPLY 3. Look at the relationship between the first and last number in each set of brackets, and use that relationship to find the missing number. Choose the correct answer from the four choices available. (37 [44] 7), (44 [51] 7), (51 [?] 7) 44 58 37 57 37 and 7 are related by adding 7 to 37 to get the number in the middle. 44 is related to 7 in the SAME way (adding 7 to 44 results in 51). 51 + 7 = 58. The relationship is ADD 4. Look at the relationship between the first and last number in each set of brackets, and use that relationship to find the missing number. Choose the correct answer from the four choices available. (49 [51] 2), (78 [85] 7), (64 [?] 9) 69 71 73 75 49 and 2 are related by adding 49 to 2 to get the number in the middle. 78 is related to 7 in the SAME way (adding 7 to 78 results in 85). 64 + 9 = 73. The relationship is ADD 5. Look at the relationship between the first and last number in each set of brackets, and use that relationship to find the missing number. Choose the correct answer from the four choices available. (123 [100] 23), (355 [100] 255), (471 [?] 71) 100 200 300 400 123 and 23 are related by subtracting 23 from 123 to get the number in the middle. 355 is related to 255 in the SAME way (subtracting 255 from 355 results in 100). 471 - 71 = 400. The relationship is SUBTRACT 6. Look at the relationship between the first and last number in each set of brackets, and use that relationship to find the missing number. Choose the correct answer from the four choices available. (44 [30] 14), (61 [38] 23), (79 [?] 26) 51 52 53 54 44 and 14 are related by subtracting 14 from 44 to get the number in the middle. 61 is related to 23 in the SAME way (subtracting 23 from 61 results in 38). 79 - 26 = 53. The relationship is SUBTRACT 7. Look at the relationship between the first and last number in each set of brackets, and use that relationship to find the missing number. Choose the correct answer from the four choices available. (68 [17] 4), (144 [24] 6), (168 [?] 21) 6 8 21 14 This is because 68 and 4 are related by dividing 68 by 4 to get the number in the middle. 144 is related to 6 in the SAME way (dividing 114 by 6 results in 24). 168 ÷ 21 = 8. The relationship is DIVIDE 8. Look at the relationship between the first and last number in each set of brackets, and use that relationship to find the missing number. Choose the correct answer from the four choices available. (5 [380] 76), (7 [147] 21), (9 [?] 18) 162 166 9 2 5 and 76 are related by multiplying 5 by 76 to get the number in the middle. 7 is related to 21 in the SAME way (multiplying 7 by 21 results in 147). 9 x 18 = 162. The relationship is MULTIPLY 9. Look at the relationship between the first and last number in each set of brackets, and use that relationship to find the missing number. Choose the correct answer from the four choices available. (360 [24] 15), (629 [37] 17), (779 [?] 19) 41 14 760 798 This is because 360 and 15 are related by dividing 360 by 15 to get the number in the middle. 629 is related to 17 in the SAME way (dividing 629 by 17 results in 37). 779 ÷ 19 = 41. The relationship is DIVIDE 10. Look at the relationship between the first and last number in each set of brackets, and use that relationship to find the missing number. Choose the correct answer from the four choices available. (11 [154] 14), (12 [144] 12), (13 [?] 13) 0 1 26 169 11 and 14 are related by multiplying 11 by 14 to get the number in the middle. 12 is related to 12 in the SAME way (multiplying 12 by 12 results in 144). 13 x 13 = 169. The relationship is MULTIPLY Author: Stephen O'Hara
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> You are reading an older version of this FlexBook® textbook: CK-12 Math Analysis Go to the latest version. Difficulty Level: At Grade Created by: CK-12 Learning objectives • Learn how to solve a quadratic equation by the three methods: factoring, completing the square, and the quadratic formula. • Understand the properties of quadratic functions and how to use them to make a graph. We know from physics, if you throw a ball up in the air from, say, 10 meters above ground with an initial speed of 12 meters per second, then its height $h$ above the ground $t$ seconds later is given by the function $h(t)=-4.9t^{2}+12t+10$. This function is an example of a quadratic function. A function $f$ defined by $f(x)=ax^{2}+bx+c$, where $a, b,$ and $c$ are real numbers and $a \ne 0$, is called a quadratic function. The defining characteristic of a quadratic function are that it is a polynomial whose highest exponent is 2. The form of the quadratic function above, $f(x)=ax^{2}+bx+c$, is called the standard form of a quadratic function. There are several other ways to write quadratic functions such as in vertex form, $f(x)=a(x-h)^{2}+k$, and in factored form, $f(x)=a(x-r_{1})(x-r_{2})$. You can move between forms of quadratic functions using algebra and you will see in this chapter that we can use each of these forms to graph a quadratic functions. The $y-$intercept of a quadratic function in standard form is $(0, c)$ and it is found by substituting 0 for $x$ in $f(x)=ax^{2}+bx+c$. The shape of the graph of any quadratic function is called a parabola, and it is the same as the shape of $f(x)=x^{2}$ (see below). However, the basic graph may be moved, reflected, or stretched depending on the values of the coefficients $a, b,$ and $c$. Examples of graphs of quadratic functions are shown in below. Notice that they all have the same shape, but some parabolas open upward and have a minimum point (called the vertex point) and some open downward and have a maximum vertex point. In calculus, those minimum and maximum points are usually called the extreme points of a function. As the name implies, at its extreme point(s), a function has a maximum or a minimum height. Graph Equation $f(x)=x^{2}$ $f(x)=x^{2}+2$ $f(x)=-x^{2}+2x$ $f(x)=x^{2}-6x+4$ $f(x)=-x^{2}-x+4$ One way to graph a quadratic function by hand is to translate the basic graph of $f(x)=x^{2}$ using transformations. You can apply rules about vertical shift, horizontal shift, and “stretching” of the basic parabolic shape. This is simplest to do if the quadratic function is written in vertex form by completing the square ((insert cross reference?)). Summary of Vertex Form: Given a quadratic function in the form $f(x)=a(x-h)^{2}+k$: • The vertex is at $(h, k)$ • The parabola opens up if $a>0$ • The parabola opens down if $a<0$ • The parabola is narrower than $y=x^{2}$ if $\|a\|>1$ • The parabola opens wider than $y=x^{2}$ if $\|a\|<1$ Example 1 Graph $g(x)=x^{2}+6x+7$ using transformations Solution: First we need to complete the square to write this function in vertex form. Add and subtract $\left ( \frac{b}{2} \right )^{2}$ to the right hand side of the equation: $g(x) & = x^2 + 6x +7\\& = x^2 + 6x + 9 + 7 - 9$ Now factor the right hand side: $g(x) & = (x+3)^2 -2$ Thus, $a=1$ and the vertex of this parabola is (-3, -2). We know that the parabola opens up with the same width as $y=x^{2}$ and it has a minimum value at the vertex. the graph of the parabola is below. Graph Quadratic Functions Using Vertex, Axis, Intercepts Vertex of a parabola Recall that the graph of a quadratic function is a parabola. In the the standard form of a quadratic function the $x-$coordinate of the vertex of the parabola is given by the equation $x=-\frac{b}{2a}$ The $y-$coordinate of the vertex is found with $y=f \left ( -\frac{b}{2a} \right )$ Axis of Symmetry of a Parabola A parabola has reflective symmetry about a vertical line through the vertex. The vertical line $x=-\frac{b}{2a}$ is also the parabola's axis of symmetry. Example 2 Find the vertex and graph the quadratic function $g(x)=x^{2}-8x+12$ Solution: The $x-$coordinate of the vertex is $x=-\frac{-8}{2}=4$. The $y-$coordinate of the vertex is $g(4)=(4)^{2}-8(4)+12=16-32+12=-4$. Thus the vertex is at (4, -4). To graph the parabola, we will make a table of points staring with the $x-$coordinate of 4: $x$ $y=g(x)$ 4 -4 5 -3 6 0 7 5 Now we can use the symmetry of $g(x)$ to fill in this table for $g(3)$. Note that $g(3)=g(5)=-3$. Likewise, $g(2)=g(6)=0$. The final graph is below. Just like linear functions, the $y-$ and $x-$intercepts of a quadratic function can be calculated by setting $x=0$ for the $y-$intercept and setting $y=0$ for the $x-$intercept. For example, to locate the $y-$intercept, substitute $x=0$ in $f(x)=ax^{2}+bx+c$ to obtain $y=f(0)=c$. The $x-$intercepts are located by setting $y=0$ and then solving the quadratic equation $ax^{2}+bx+c=0$, which can be solved by several methods that you have already studied from your previous mathematics courses: the factoring, completing the square, or using the quadratic formula. As a reminder, the quadratic formula is Given a quadratic function $f(x)=ax^{2}+bx+c$, the $x-$intercepts of the function are: $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ All three methods–factoring, completing the square, or using the quadratic formula–give the $x-$intercepts or the roots or the solution set or the zeros of the quadratic equation (you can tell this is an important concept since there are so many words for it!). The following example gives you a quick review of how to use those three methods. • The quadratic function is defined as $y=f(x)=ax^{2}+bx+c$ and has an extreme point (the vertex) at the $(x, y)$ coordinates $\left( \frac{-b}{2a}, f\left ( \frac{-b}{2a}\right ) \right )$ If $a>0$ (positive), then the extreme point is a minimum and the parabola opens upwards. If $a<0$ (negative), then the extreme point is a maximum and the parabola opens downwards. • Find the $x-$ and $y-$intercepts by setting $f(x)=0$, and $f(0)=y$, respectively. Example 4 Sketch the graph of the function $y=f(x)=x^{2}+2x-3$. Solution Let's first find the intercepts. For the $y-$intercept, if $x=0$, then $f(0)=-3$, or $y=-3$, so the $y-$intercept point is (0, -3). Now, for the $x-$intercepts, if $y=f(x)=0$, then $x^{2}+2x-3=0$, or $x^{2}+2x-3=(x+3)(x-1)=0$ so that $x=-3$ and $x=1$ are the $x-$intercepts, that is (-3, 0) and (1, 0). The vertex (extreme point) is at $x=\frac{-b}{2a}=\frac{-2}{2(1)}=-1$ Since $f(-1) & = (-1)^2 + 2(-1)-3\\& = -4$ The vertex is (-1, -4). Since the coefficient of $x^{2}$ is positive, $a>0$, the extreme point is a minimum and the parabola opens up. From this information, we can make a rough sketch of the parabola containing the points determined above. Notice that the range of the function is $y\ge-4$. Example 5 Sketch the graph of the quadratic function $f(x)=-x^{2}+4x$. Solution To find the $y-$intercept, set $x=0$, and $f(0)=-(0)^{2}+4(0)=0$. Thus the parabola intercepts the $y-$axis at the origin. The $x-$intercept is obtained by setting $y=0$, thus $-x^{2}+4x=0$. Factoring, $-x^{2}+4x = -x(x-4)=0$ so that $x=0$ and $x=4$ are the $x-$intercepts. We have $a=-1$ and $b=4$, so that the extreme point occurs when $x=\frac{-b}{2a}=\frac{-4}{2(-1)}=2$ Since $f(2)=-(2)^{2}+4(2)=-4+8=4$, then (2, 4) is the extreme point. It is a maximum point since $a=-1<0$ and the parabola opens down. Finally, the graph can be obtained by sketching a parabola through the points determined above. From graph, the range of the function is $y \le 4$. Example 6 A projectile is shot from the ground level. If $h(t)=121t-4.9t^{2}$, where $t$ is the time in seconds and $h(t)$ is the height of the projectile (in meters) above the ground at time $t$. Find (a) $h$ when $t=1$ second, (b) the maximum height reached by the projectile, and (c) the graph of the trajectory of the projectile. Solution (a) At $t=1$ second, we have $h(1)=121(1)-4.9(1)^{2}=121-4.9=116.1 \ meters$. (b) To find the maximum height reached by the projectile, we use the formula $x=\frac{-b}{2a}$ to find the time at which the maximum height occurs (but here we use $t$ instead of $x$ for the independent variable): $t=\frac{-(121)}{2(-4.9)}=12.3 \ seconds$ In other words, the maximum height is reached at time 12.3 seconds. That tells us when the projectile reached its maximum height. To find the actual maximum height, we simply substitute this value into the height equation $h(12.3)=121(12.3)-4.9(12.3)^{2}=747 \ meters$ (c) In order to graph it accurately, we need to find the $t-$intercepts. They can be found by setting $h=0$: $121t-4.9t^{2} & = 0\\(121-4.9t)t & = 0$ So the intercepts are at $t=0$ seconds and $t=24.7$ seconds. The $h-$intercept occurs when $t=0$, so $h(0)=121(0)-4.9(0)^{2}=0$. From this information, we can construct the graph, as shown in the Figure 5. Notice that the graph does not extend beyond the interval on the $t-$axis. (why?) Example 7 Find the roots (or $x-$intercepts) of the quadratic equation $x^{2}-5x+6=0$ by using the three methods mentioned above. Solution 1. The Factor Method The factor method is based on writing the quadratic equation in factored form. That is, as a product of two linear expressions. So our equation maybe solved by the following way: $x^{2}-5x+6 & = 0\\(x-3)(x-2) & = 0$ Recall, that $a \cdot b=0 \ \text{if and only if} \ a=0 \ \text{and} \ b=0$ This tells us that $x-3=0$ or $x-2=0$ which give the roots (or zeros) $x=3$ and $x=2$ In other words, the solution set is {2, 3}. 2. Completing the Square Method To solve the above equation by completing the square, first move the “$c$” term to the other side of the equation, $x^{2}-5x=-6$ Next, make the left-hand side a “perfect square” by adding the appropriate number. To do so, take one-half of the coefficient of $x$ (the $b$ coefficient) and square it and then add the result to both sides of the equation: $b & = -5 \\\frac{b}{2} & = \frac{-5}{2}\\\left ( \frac{b}{2} \right ) ^2 & = \frac{25}{4}$ Adding to both sides of the equation, $x^{2}-5x+\frac{25}{4} & = -6 + \frac{25}{4}\\x^2 -5x + \frac{25}{4} & = \frac{1}{4}\\\left ( x-\frac{5}{2} \right )^2 & = \frac{1}{4}$ this last equation can be easily solved by taking the square root of both sides, $\left(x-\frac{5}{2}\right)=\sqrt{\frac{1}{4}}=\pm\frac{1}{2}$ Hence $x=\pm\frac{1}{2}+\frac{5}{2}$ and the solutions are $x=3$ and $x=2$ which are identical to answers of the factor method. The quadratic formula is the most useful method in applications for solving quadratic functions because it works for all quadratic equations. Therefore, it is highly encouraged that you memorize the formula. Recall, if $ax^{2}+bx+c=0$ where $a, b,$ and $c$ are real numbers and $a\ne0$, then the roots of the equation can be determined by the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}.$ Here the coefficients are $a=1, b=-5,$ and $c=6$. Substituting into the quadratic formula, we get $x & = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(6)}}{2(1)}\\& = \frac{5 \pm \sqrt{25-24}}{2}\\& = \frac{5 \pm 1}{2}\\& = 3 \ \text{or} \ 2$ which is, again, identical to our two solutions above. Applications, Technological Tools You can graph quadratic functions using your computer's graphing program or with a TI-83/84 Calculator. Below we give basic directions for graphing a quadratic function and finding the vertex and roots ($x-$intercepts) using the functions on a TI-83 or TI-84 Graphing calculator. Graphing 1. As with graphing any function on the calculator, you enter the function's equation by using the $Y=$ button. 2. Set an appropriate window using the WINDOW menu. You can use the vertex and $y-$intercept to help you find the best values for XMAX, XMIN, YMAX, YMIN, and the $X-$ and $Y-$scales for viewing the graph of the parabola. Alternatively, you can set a standard window and use zooming to view the graph. 3. Press GRAPH. 4. Adjust and refine the view using ZOOM or by changing the WINDOW settings. ((Note: Insert 3 screen shots: 1) the $Y=$ menu with a quadratic function in $Y_{1}$, 2) The WINDOW menu, and 3) A view of the graph.)) Finding the Vertex You can use functions built into the calculator to find the vertex of any parabola you graph. 2. From the Graph screen, press 2ND TRACE (This is the CALC Menu). 3. Scroll down and choose MINIMUM or MAXIMUM depending on whether the vertex is a minimum or maximum. 4. The calculator takes you to the graph and prompts LEFT BOUND? Use the arrow keys (< or >) to place the cursor to the left of the vertex (or enter an $x-$coordinate by typing one in), and press ENTER 5. The calculator then prompts RIGHT BOUND? Use the arrows to place the cursor to the right of the vertex and press ENTER. 6. Finally, the calculator will prompt you for a guess with GUESS? You can either enter an $x-$value close to the vertex, or press ENTER again if your bounds were relatively good. 7. The calculator will display the $x-$ and $y-$coordinates of the vertex. ((Note: Add 2 screen shots: 1) the CALC menu in step #2, 2) A graph with LEFTBOUND? prompt (step #4), 3) A graph showing the vertex as MAX/MIN (i.e. after doing step 7))) Finding the Roots 1. From the Graph screen, press 2ND TRACE (This is the CALC Menu). 2. Choose ZERO. 3. The calculator takes you to the graph and prompts LEFT BOUND? Use the arrow keys (< or >) to place the cursor to the left of the zero (or enter an $x-$coordinate by typing one in), and press ENTER. 4. The calculator then prompts RIGHT BOUND? Use the arrows to place the cursor to the right of the zero and press ENTER. 5. Finally, the calculator will prompt you for a guess with GUESS? You can either enter an $x-$value close to the zero, or press ENTER again if your bounds were relatively good. 6. The calculator will display the $x-$coordinate of the zero (root). Note: Repeat steps 1-6 to find the other root of the quadratic. ((Note: Add 3 screen shots showing the calculator after Step 2, Step 4, and Step 6.)) Exercises 1. Solve each equation by factoring. 1. $x^{2}=64$ 2. $4x^{2}+7x=2$ 3. $4x^{2}-17x=-4$ 4. $x^{2}+1=-\frac{13}{6}x$ 2. Solve each equation by completing the square 1. $x^{2}-4x+1=0$ 2. $\frac{2}{3}x^{2}-x+\frac{1}{3}=0$ 3. $x^{2}+2.8=4.7x$ 4. $2x^{2}-3x-\frac{23}{8}=0$ 3. Solve each equation by using the quadratic formula 1. $0.4x^{2}+x-0.3=0$ 2. $25x^{2}+80x+61=0$ 3. $(z+6)^{2}+2z=0$ 4. $\left(\frac{5}{7}w-14\right)^{2}=8w$ 4. Sketch the graph of the quadratic function $f(x)=-x^{2}-4x-4$ by locating the $x-$ and $y-$intercepts and the vertex point. Use the graph to determine the range of the parabola. 5. Sketch the graph of the quadratic function $f(x)=2x^{2}+4x-3$ by locating the $x-$ and $y-$intercepts and the vertex point. Use the graph to determine the range of the parabola. 6. Sketch the graph of the quadratic function $f(x)=x^{2}+x+5$ by locating the $x-$ and $y-$intercepts and the vertex point. Use the graph to determine the range of the parabola. 7. A projectile is fired from a cliff. If the height, $h$ of the projectile after $t$ seconds is given by $h(t)=-16t^{2}+96t+256$, find 1. $h$ when $t=0$. 2. The maximum height reached by the projectile. 3. Graph $h(t)$ and show its range and domain. 8. A surface-to-surface missile is fired and follows a parabolic path. Its path as a function of time is described by the function $p(t)=-t^{2}+t+1$. At what time the missile reaches the highest point and when does it hit the ground? 9. Use a graphing calculator to find the vertex and roots of $q(x)=-0.035x^{2}-7.25x-12.3$. Round your answer to three decimal places. 1. $\pm8$ 2. $x=\frac{1}{4}, -2$ 3. $x=\frac{1}{4}, 4$ 4. $x=\frac{-2}{3}, \frac{-3}{2}$ 1. $2\pm\sqrt{3}$ 2. $x=\frac{1}{2}, 1$ 3. $x=\frac{7}{10}, 4$ 4. $\frac{3}{4}\pm\sqrt{2}$ 1. $x=0.275, -2.78$ 2. $-\frac{8}{5}\pm\frac{\sqrt{3}}{5}$ 3. $-7\pm\sqrt{13}$ 4. $\frac{686\pm196\sqrt{6}}{25}$ 1. The vertex and the $x-$intercept is (-2, 0). The $y-$intercept is (0, -4). 2. Vertex: (-1, -5). $x-$intercepts: $\left(\frac{-2+\sqrt{10}}{2}, 0\right)$ and $\left(\frac{-2-\sqrt{10}}{2}, 0\right)$. $y-$intercept: (0, -3). 3. Vertex: $\left(\frac{-1}{2}, \frac{19}{4}\right)$. No $x-$intercepts. $y-$intercept: (0, 5) 1. 256 ft 2. 400 ft 4. At $\frac{1}{2}$ second; after 1 second 5. Vertex: (-103.571, 363.146); roots at (-205.432, 0) and (-1.711, 0). Feb 23, 2012 Jun 08, 2015
# How do you solve "-8-x=x-4x"? Feb 23, 2018 #### Explanation: $- 8 - x = x - 4 x$ You can subtract $4 x$ from $x$ and get: $- 8 - x = - 3 x$ Then you add the $x$ to both sides, which cancels out the $x$ on the left side and get: $- 8 = - 2 x$ Then divide both sides by $- 2$ and get: $4 = x$ Hope that helps. Feb 24, 2018 $x = 4$ #### Explanation: $- 8 - x = x - 4 x$ $- 8 - x = - 3 x$ Add $3 x$ to both sides# $- 8 - x + 3 x = - 3 x + 3 x$ $- 8 - x + 3 x = 0$ $- 8 + 2 x = 0$ Then add $- 8$ to both sides $- 8 + 2 x + 8 = 0 + 8$ $2 x = 8$ Finally, divide both sides by $2$ $\frac{2 x}{2} = \frac{8}{2}$ $x = 4$
## NCERT Solutions for Class 11 Maths Chapter 9 Sequence and Series Exercise 9.1 Chapter 9 Sequence and Series Exercise 9.1 NCERT Solutions for Class 11 Maths is very important topic which you need to improve your marks in the examinations. Here you will find Class 11 Maths NCERT Solutions that will help you a lot in understanding the basics things of the chapter and implementing them. It will save your precious time if you want to complete your homework on time. 1. Write the first five terms of each of the sequence defined by the following: an = n (n + 2) an = n (n + 2) For n = 1, a1 = 1(1 + 2) = 3 For n = 2, a2 = 2(2 + 2) = 8 For n = 3, a3 = 3(3 + 2) = 15 For n = 4, a4 = 4(4 + 2) = 24 For n = 5, a5 = 5(5 + 2) = 35 Thus first five terms are 3, 8, 15, 24, 35. 2. Write the first 5 terms of the sequence defined by the following: an = n/(n + 1) an = n/(n + 1) for n = 1, a1 = 1/(1 + 1) = 1/2 for n = 2, a2 = 2/(2 +1 ) = 2/3 for n = 3, a3 = 3/(3 + 1) = 3/4 for n = 4, a4 = 4/(4 + 1) = 4/5 for n = 5, a5 = 5/(5 + 1) = 5/6 Hence, first 5 terms of the given sequence are 1/2, 2/3, 3/4, 4/5 and 5/6. 3. Write the first five terms of the sequence whose nth term is: an = 2n an = 2n, Putting n = 1, 2, 3, 4, 5 a1 = 21 = 2, a2 = 22 = 4, a3 = 23 = 8 a 4 = 24 = 16, a5 = 25 = 32 Required first five term of sequences are 2, 4, 8, 16, 32. 4. Write the first five terms of the sequence whose n th term is: an = (2n – 3)/6. Here an = (2n – 3)/6 Putting n = 1, 2, 3, 4, 5, we get a1 = (2 × 1 – 3)/6 = (2 – 3)/6 = -(1/6); a2 = (2 × 2 – 3)/6 = (4 – 3)/6 = 1/6; a3 = (2 × 3 – 3)/6 = (6 – 3)/6 = 3/6 = 1/2; a4 = (2 × 4 – 3)/6 = (8 – 3)/6 = 5/6 and a5 = (2 × 5 – 3)/6 = (10 – 3)/6 = 7/6 ∴ the first five terms are –(1/6), 1/6, 1/2, 5/6 and 7/6. 5. Write the first five terms of the sequence whose n th term is an = (–1)n – 1 5 n + 1 Putting n = 1, 2, 3, 4, 5 a1 = (–1)0. 5 1 + 1 = 52 = 25 a2 = (–1)1. 5 2 + 1 = – 53 = –125 a3 = (–1)2. 5 3 + 1 = 54 = 625 a4 = (–1)3. 5 4 + 1 = – 55 = – 3125 a5 = (–1)4. 5 5 + 1 = 56 = 15625 6. Write the first 5 terms of the sequence defined by the following: an = (n(n2 + 5))/4 for n = 1, a1 = (1(12 + 5))/4 = 6/4 = 3/2 for n =2 , a2 = (2(22 + 5))/4 = (2 × 9)/4 = 9/2 for n = 3, a3 = (3(32 + 5))/4 = (3 × 14)/4 = 21/2 for n = 4, a4 = (4(42 + 5))/4 = (4 × 21)/4 = 21 for n = 5, a5 = (5(52 + 5))/4 = (5 × 30)/4 = 75/2 hence the first 5 terms are 3/2, 9/2, 21/2, 21 and 75/2 7. Find the indicated terms in the following sequence whose n th term is: an = 4n – 3, a17, a24 an = 4n – 3 For n = 17, a17 = 4 × 17 – 3 = 68 – 3 = 65 For n = 24, a24 = 4 × 24 – 3 = 96 – 3 = 93 Hence a17 = 65 and a24 = 93 8. Find the indicated terms in the following sequence whose n th term is: an = n2/2n ; a7 an = n2/2n putting n = 7 a7 = 72/27 = 49/128 9. Find the indicated term in the following sequence whose n th term is: an = (–1)n – 1 n 3, a9 an = (–1)n – 1 n3, Putting n = 9 a9 = (–1)9 – 1 93 = 729 10. Find the indicated terms in the sequence whose n th term is an = (n(n -2))/(n + 3) ; a20 an = (n(n -2))/(n + 3), Putting n = 20, a20 = (20(20 – 2))/(20 + 3) = (20 × 18)/23 = 360/23 11. Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series. a1 = 3, an = 3, an – 1 + 2 for all n > 1. a1 = 3, an = 3 an –1 + 2. Putting n = 2, 3, 4, 5 a2 = 3.a1 + 2 = 3. 3 + 2 = 9 + 2 = 11 a3 = 3.a2 + 2 = 3. 11 + 2 = 33 + 2 = 35 a4 = 3.a3 + 2 = 3. 35 + 2 = 105 + 2 = 107 a5 = 3.a4 + 2 = 3. 107 + 2 = 321 + 2 = 323 The first five terms of the sequences are 3, 11, 35, 107, 323 ∴ The corresponding series is 3 + 11 + 35 + 107 + 323 +.... 12. a1 = -1, an = (an -1)/n , n ≥ 2 Here a1 = -1, an = (an – 1)/n , n ≥ 2 Putting n = 2, 3, 4, 5 and 6 a2 = (a2 – 1)/2 = a1/2 = -1/2 a3 = (a3 – 1)/3 = a2/2 = -(1/2)/3 = -1/6; a4 = (a4 – 1)/4 = a3/4 = -(1/6)/4 = -1/24; a5 = a5-1/5 = a4/5 = -(1/24)/5 = -1/120 The first five terms of the sequences are -1, -(1/2), -(1/6), -(1/24), -1/120 The corresponding series is -1, -(1/2), -(1/6), -(1/24), -1/120 ….. 13. a1 = a2 = 2, an = an– 1 – 1, n > 2. Given: a1 = a2 = 2 and an = an– 1 – 1, n > 2. For n = 3, a3 = a2 – 1 = 2 – 1 = 1 For n = 4, a4 = a3 – 1 = 1 – 1 = 0 For n = 5, a5 = a4 – 1 = 0 – 1 = –1 Hence first five terms are 2, 2, 1, 0, –1 and the corresponding series = 2 + 2 + 1 + 0 + (–1) +.... 14. The Fibonacci sequence is defined by 1 = a1 = a2 and an = an – 1 + an – 2, n > 2 Find (an+ 1)/an , for n = 1, 2, 3, 4, 5
If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. ## MAP Recommended Practice ### Course: MAP Recommended Practice>Unit 34 Lesson 15: Adding and subtracting fractions with unlike denominators # Adding fractions with unlike denominators Learn how to add two fractions with different denominators. It can be challenging to combine fractions when the denominators don't match. It is important to find a common denominator. Finally, the resource shows how to find a common multiple of the two denominators in order to convert the fractions so they can be added together. ## Want to join the conversation? • How do I find the common denominator for 3/4+3/8?? • First, you'll want to figure out whether or not the larger denominator (3/8) is divisible by the smaller one (3/4). In this case, the 8 is divisible by 4, (8/4 = 2), so you're going to multiply the smaller one (4) by 2, bringing it to 8. Now, you have to remember, whenever you scale fractions, you have to multiply both, so now it's 6/8+3/8 = 9/8 = 1 1/8. In other cases where the larger denominator isn't divisible by the smaller one, find the LCM (least/smallest common multiple), and scale both fractions so the denominators are equal. Hope this helped! • khan academy needs to add dark mode for people who spend along time on their website because your eyes are gonna start to hurt. UPVOTE IF YOU AGREE LETS START A PETITION TOGETHER! • the white is the theme for the website because it most likely is used to catch teachers or schools eyes and they will see that its used for learning and the website will be used more [my theory for this] • Can the Rule of Four be used for what he teaches at ? Or is this something different? • yes it can • At to . Since Sal says that 9/10 and 27/10 are the same number, could you have written only 9/10 or 27/30. • Good Question! You can do it either ways! But sal is not switching the numbers in this video! You could do both but only can do 27/30 when switching numbers! • Play it at speed 0.25. Upvote if you tried :D • No don't comment it's meant for learning. • I do not understand what if it is a whole number times a fraction • If there is a whole number, you would make the whole number look like this: (this is just an example) 3/1. If the whole number is 3, then you just make the whole number over one. 3/1. I hope this helped you!! P.S. Sorry for the three-year-late response!! • I don't understand what he did in , someone please explain • the fraction was a improper fraction so he made it a proper fraction by saying 15 goes into 16 1 time and there's a re remainder of 1 • How do you make fractions into decimals. • To convert fractions into decimals, divide the denominator into 100, then multiply the answer by the numerator, and finally add the decimal point, i guess. For example, 3/5 is equal to 0.6 because 5 going into 100 is 20 which would make the decimal 0.2. The numerator is 3 and 0.2 x 3 = 0.6. I hope this helped!
On this post you will find the explanation of how to do the addition of polynomials. Also, you will see examples of sums of polynomials and even exercises solved step by step. Finally, we also explain what are the properties of this type of operation with polynomials. To add two or more polynomials, add the terms of the polynomials that are like terms. That is, the addition of polynomials consists of adding the terms that have the same variables and the same exponents. Thus, an addition of polynomials can be done with two different methods: the horizontal method or the vertical method. Here is an explanation of both procedures. Let’s see how to do an addition of polynomials with the horizontal method solving an example: • Find the sum of the following two polynomials: First of all, we have to place the two polynomials in the same operation, in other words, one polynomial after another: And now we add the terms that have the same variables (letters) with the same exponents. Terms that are not like terms cannot be added. So the result of the addition of polynomials is: We have just seen how to add polynomials horizontally, but now we are going to see the other method that there is to do an addition of polynomials: adding polynomials vertically. And so that you can see the differences between the two methods, we will add the same polynomials as in the previous example: • Add the following two polynomials: The first thing we must do is place one polynomial below another, so that the like terms of the two polynomials are aligned by columns: Note: if a polynomial does not have a term of a certain degree, we must leave a blank space. For example, the polynomial does not have a monomial of second degree, so there is a blank space in its place. Once we have arranged all the terms from highest to lowest degree, we add the coefficients of each column, keeping the variables and exponents the same: Therefore, the result obtained from the addition of the 2 polynomials is: As you can see, we have obtained the same result with both methods, so when you do an addition of polynomials you can use the method you prefer. ## Practice problems on adding polynomials We leave you with several practice problems of additions of polynomials. If you have any questions, you can ask them below in the comments. ### Problem 1 In this case, we will add the two polynomials vertically. To do this, we order the polynomials by degree and add the terms in the same column: ### Problem 2 Solve the addition of the following two polynomials: We will do the sum of the two polynomials using the vertical method. So we order the polynomials by degree and add the terms in the same column: Note that in this particular case we must leave an empty space in the second degree column of the second polynomial, because it does not have a second degree term. ### Problem 3 Simplify the following addition of two polynomials: We will add the two polynomials using the vertical method. Therefore: ### Problem 4 We will calculate the addition of the 3 polynomials using the vertical method. So we put the polynomials ordered by degree and add the terms that are in the same column: ## Properties of the addition of polynomials The addition of polynomials has the following characteristics: • Associative property: when 3 or more polynomials are added, it does not matter how the polynomials are grouped, since the result is always the same. That is, the following equation is true: • Commutative property: in the sum of polynomials the order of the addends does not alter the result of the sum. • Neutral element: logically, adding a polynomial plus the zero polynomial is equivalent to the first polynomial. • Opposite element: the result of adding any polynomial plus its opposite polynomial is always null.
# Summing Consecutive Numbers ## Consecutive numbers, consecutive odd numbers, consecutive even numbers. The finite sum of the consecutive numbers. The formulas of the finite sums of the consecutive numbers. The subject expression and solved questions. Consecutive Numbers Consecutive numbers are numbers that increase according to a certain rule. This amount of increase may be in the form 1, 2, 3, 9, …, etc. If consecutive numbers increase one by one and cannot be divided exactly by 2, this numbers are said "consecutive odd numbers". If consecutive numbers increase by twos and can be divided exactly by 2, this numbers are said "consecutive even numbers". Consecutive Numbers That Increasing One By One n is an integer, 1, 2, 3, 4, 5, 6, … , n Numbers are consecutive counting numbers. Also this numbers are consecutive positive integers. The formula of the sum of numbers from 1 to n. T = n.(n + 1) 2 Example: Find the sum of numbers from 1 to 35. Solution: The formula of the sum of numbers from 1 to n. T = n•(n+1) 2 For n = 35, T = 35•36 2 T = 630 Example: Find the sum of consecutive numbers from 30 to 60 (inclusive of 30 and 60) Solution: Firstly, let we find the sum of numbers from 1 to 29. Later, let we find the sum of numbers from 1 to 60. The difference between these two numbers will give us the sum of numbers from 30 to 60. T1 = 29•30 2 T1 = 435 T2 = 60•61 2 T2 = 1830 T = 1830 – 435 T = 1395 Consecutive Odd Numbers n is an integer, 1,3, 5, 7, 9, 11, 13,15, 17, ... (2n-1) Numbers are called "consecutive odd numbers". The formula for the sum of consecutive odd numbers. T = n•n T = n2 Finding n, n = Last term - first term + 1 2 Example: Find the sum of consecutive odd numbers from 1 to 15. Solution: The formula is T = n2 n = Last term - first term + 1 2 n = 15 - 1 + 1 2 n = 8 Sum = 82 Sum = 64 Example: Find the sum of consecutive odd numbers from 19 to 75. Solution: Firstly, we find the sum of consecutive odd numbers drom 1 to 17. Later, we find the sum of consecutive odd numbers 1 to 75. he difference of these two sum gives us the sum of consecutive odd numbers from 19 to 75. n = 17 - 1 + 1 2 n = 9 T1 = 92 T1 = 81 m = 75 - 1 + 1 2 m = 38 T2 = 382 T2 = 1444 T = T2 – T1 T = 1444 – 81 T = 1363 Consecutive Even Numbers If consecutive numbers increase by twos and can be divided exactly by 2, this numbers are said "consecutive even numbers". Consecutive even numbers are in the form of 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, … 2n The sum of consecutive even numbers is founds by equality below. Total = n•(n + 1) Finding n, n = Last term - first term + 1 2 Example: Find the sum of consecutive even numbers from 2 to 90. Solution: Finding the n, 2n = 90 n = 45 Or n = Last term - First term + 1 2 Last term = 90 First term = 2 n = 90 - 2 + 1 2 n = 45 Total = 45 • 46 = 2070 The General Formula Of the Sum Of The Consecutive Numbers The consecutive numbers may not always be as regular as above explanations. Now, we will give a general formula that can used for the sum of the all consecutive numbers. T = The sum of the consecutive numbers The Number Of Terms r + (x•1 + r) + (x•2 + r) + … + (x•n + r) in the number array, m = Last term m = x•n + r Nt = Number of terms, Example: What is the sum of the numbers that starting from 1 and continuing up to 64 and increasing three by three? Solution: The first term = 1 The last term = 64 The amount increase = 3 Nt = Number of terms, Nt = 64 - 1 + 1 3 Nt = 22 T = The sum of the consecutive numbers T = 22 •65 2 T = 11•65 T = 715 Example: What is the sum of the numbers that starting from 12 and continuing up to 117 and increasing seven by seven. Solution: The first term = 12 The last term = 117 The amount of increase = 7 Nt = The number of terms Nt = 117 – 12 + 1 7 Nt = 15 + 1 Nt = 16 Sum = 16 • (117 + 12) 2 Total = 1032 Example: The sum of three consecutive integers is 72. The amount of increase of these numbers is 4. What is the smallest these numbers. Solution: Suppose that smallest of these numbers is x. In this case, Second number is (x + 4) Third number is (x + 8) Sum = 72 72 = x + (x + 4) + (x + 8) 72 = 3x + 12 60 = 3x x = 20 RISE KNOWLEDGE 17/07/2018 • WRITE COMMENT • NAME SURNAME(or nick) • COMMENT
# Sample Report On Numerical Solutions To Odes Type of paper: Report Pages: 3 Words: 825 Published: 2020/10/15 ## The task is to solve ordinary differential equation using numerical methods. The equation is:dydx=2xy+x3 (1) ## Equation (1) is linear non-ordinary differential equation, let`s solve it. First we need to solve the linear ordinary part, let`s rewrite it: dydx=2xy dyy=2dxx lny=2lnx+ lnC ## Solve for y by exponentiating, and we get the general solution y=x2C As we found the general solution of linear ordinary part let`s find solution for non-oridnary part, to do this we should rewrite the general solution in next way y=x2φx (2) Where φx is an arbitrary function. Now we have to substitute (2) into (1), to find the unknown function φx. 2x φx+x2 φ'x=2xx2 φx+x3 (3) φx=x22+A ## So the general solution of (1) is: y=x2A+x42 (4) 2Ax+2x3=2xAx2+x42+x3 ## After simple operations we get that left part is equal to right part, so the general solution (4) is correct. Using the initial condition y0.1=1 we see that A = 99.995, so the actual solution is: y=12 (199.99x2+x4) ## Let`s solve equation (1) using numerical methods. The following code is written in Python: import math def euler(f, y0, a, b, h): t, y = a, y0 result = [] while t < b + h: result.append(y) y += h * f(t, y) t += h return result def midpoint(f, y0, a, b, h): t, y = a, y0 result = [] while t < b + h: result.append(y) y += h * f(t + h / 2, y + h / 2 * f(t, y) ) t += h return result def rk4(f, y0, a, b, h): t, y = a, y0 result = [] while t < b + h: result.append(y) k1 = h * f(t, y) k2 = h * f(t + h / 2, y + k1 / 2) k3 = h * f(t + h / 2, y + k2 / 2) k4 = h * f(t + h, y + k3) y += (k1 + 2 * k2 + 2 * k3 + k4) / 6 t += h return result def actual(f, y0, a, b, h): t, y = a, y0 result = [] while t < b + h: result.append(y) t += h y = (t * t * 199.99 + t ** 4) / 2 return result def function(time, temp): return (2 / time * temp + time ** 3) h = 0.1 a = 0 + h b = 2 y0 = 1 results = [ [], [], [], [] ] results[0] = euler(function, y0, a, b, h) results[1] = midpoint(function, y0, a, b, h) results[2] = rk4(function, y0, a, b, h) results[3] = actual(function, y0, a, b, h) f = open('output.txt', 'w+') t, y = a, y0 print "x\tEuler\t\tMidpoint\tRunge-Kutta-4\tActual" for i in xrange(0, len(results[0])): temp = "%2.1f\t%2.8f\t%2.8f\t%2.8f\t%2.8f" % (t, results[0][i], results[1][i], results[2][i], results[3][i]) print temp f.write(temp + '\n') t += h print print "Euler\t\tMidpoint\tRunge-Kutta-4" for i in xrange(0, len(results[0])): temp = "%2.8f\t%2.8f\t%2.8f" % (abs(results[0][i] - results[3][i]), abs(results[1][i] - results[3][i]), abs(results[2][i] - results[3][i])) print temp f.write(temp + '\n') t += h r = raw_input() ## The results we can view at Figure 1, and the errors in Figure 2 Figure 1 Figure 2 As we can see the Euler method gives the worst performance, and the RK-4 method gives the best performance, because it`s error is the smallest. All results were checked using wolfram alpha online tool. Choose cite format: • APA • MLA • Harvard • Vancouver • Chicago • ASA • IEEE • AMA WePapers. (2020, October, 15) Sample Report On Numerical Solutions To Odes. Retrieved June 20, 2024, from https://www.wepapers.com/samples/sample-report-on-numerical-solutions-to-odes/ "Sample Report On Numerical Solutions To Odes." WePapers, 15 Oct. 2020, https://www.wepapers.com/samples/sample-report-on-numerical-solutions-to-odes/. Accessed 20 June 2024. WePapers. 2020. Sample Report On Numerical Solutions To Odes., viewed June 20 2024, <https://www.wepapers.com/samples/sample-report-on-numerical-solutions-to-odes/> WePapers. Sample Report On Numerical Solutions To Odes. [Internet]. October 2020. [Accessed June 20, 2024]. Available from: https://www.wepapers.com/samples/sample-report-on-numerical-solutions-to-odes/ "Sample Report On Numerical Solutions To Odes." WePapers, Oct 15, 2020. Accessed June 20, 2024. https://www.wepapers.com/samples/sample-report-on-numerical-solutions-to-odes/ WePapers. 2020. "Sample Report On Numerical Solutions To Odes." Free Essay Examples - WePapers.com. Retrieved June 20, 2024. (https://www.wepapers.com/samples/sample-report-on-numerical-solutions-to-odes/). "Sample Report On Numerical Solutions To Odes," Free Essay Examples - WePapers.com, 15-Oct-2020. [Online]. Available: https://www.wepapers.com/samples/sample-report-on-numerical-solutions-to-odes/. [Accessed: 20-Jun-2024]. Sample Report On Numerical Solutions To Odes. Free Essay Examples - WePapers.com. https://www.wepapers.com/samples/sample-report-on-numerical-solutions-to-odes/. Published Oct 15, 2020. Accessed June 20, 2024. Copy Share with friends using:
Edit Article # wikiHow to Divide Mixed Fractions A mixed number, or mixed fraction, is a number that combines a whole number and a fraction.[1] It is possible to divide mixed numbers; however, doing so requires converting them to improper fractions first. Once the mixed numbers are converted, you can divide as you would divide any other fractions. ### Part 1 Converting Mixed Numbers to Improper Fractions 1. click to unmute.. to Divide Mixed Fractions 1 Multiply the whole number by the denominator of its combined fraction.[2] Do this for both mixed numbers. Set these products aside. They are only part of your new numerator. • For example, if you want to calculate ${\displaystyle 6{\frac {1}{2}}\div 2{\frac {1}{4}}}$, you would multiply ${\displaystyle 6\times 2=12}$ and ${\displaystyle 2\times 4=8}$. 2. click to unmute.. to Divide Mixed Fractions 2 Add the numerator to the product.[3]Do this for both mixed numbers. This sum will be the numerator of your improper fraction. for • For example, ${\displaystyle 12+1=13}$ and ${\displaystyle 8+1=9}$. 3. click to unmute.. to Divide Mixed Fractions 3 Place the sum over the original denominator.[4] Complete this step for both fractions, making sure you use the correct denominators. These are your improper fractions that you will use to complete the division. • For example, ${\displaystyle 6{\frac {1}{2}}}$ becomes ${\displaystyle {\frac {13}{2}}}$ and ${\displaystyle 2{\frac {1}{4}}}$ becomes ${\displaystyle {\frac {9}{4}}}$. 4. click to unmute.. to Divide Mixed Fractions 4 Convert whole numbers to fractions. If you are working with any whole numbers, you need to convert them to fractions. To do this, turn the number into the numerator of a fraction. The denominator will be 1.[5] • For example, ${\displaystyle 3={\frac {3}{1}}}$. ### Part 2 Dividing Improper Fractions 1. click to unmute.. to Divide Mixed Fractions 1 Write the new division problem. Use the improper fractions you found by completing the calculations in Part 1. • For example, ${\displaystyle {\frac {13}{2}}\div {\frac {9}{4}}}$. 2. click to unmute.. to Divide Mixed Fractions 2 Take the reciprocal of the second fraction.[6] To find a reciprocal of a fraction, you need to “flip” it, so that the numerator becomes the denominator, and the denominator becomes the numerator.[7] Then, change the problem to a multiplication problem.[8] • For example, if you take the reciprocal of ${\displaystyle {\frac {9}{4}}}$, it becomes ${\displaystyle {\frac {4}{9}}}$. So ${\displaystyle {\frac {13}{2}}\div {\frac {9}{4}}}$ becomes ${\displaystyle {\frac {13}{2}}\times {\frac {4}{9}}}$ 3. click to unmute.. to Divide Mixed Fractions 3 Multiply the numerators. To do this, multiply them as if they were whole numbers. This product will be the numerator of your answer.[9] • For example, if calculating ${\displaystyle {\frac {13}{2}}\times {\frac {4}{9}}}$, you would multiply the numerators: ${\displaystyle 13\times 4=52}$. 4. click to unmute.. to Divide Mixed Fractions 4 Multiply the denominators. To do this, multiply them as if they were whole numbers. This product will be the denominator of your answer.[10] • For example, if calculating ${\displaystyle {\frac {13}{2}}\times {\frac {4}{9}}}$, you would multiply the denominators: ${\displaystyle 2\times 9=18}$. Putting together your numerator and denominator, your answer becomes ${\displaystyle {\frac {52}{18}}}$. 5. click to unmute.. to Divide Mixed Fractions 5 Simplify your answer, if possible. To simplify, or reduce, a fraction, you need to find the greatest factor (besides 1) that is common to the numerator and the denominator. Then, divide the numerator and denominator by that factor. For more information on this process, read Reduce Fractions. • For example, ${\displaystyle 52}$ and ${\displaystyle 18}$ are both divisible by ${\displaystyle 2}$. ${\displaystyle 52\div 2=26}$ ${\displaystyle 18\div 2=9}$ So, ${\displaystyle {\frac {52}{18}}={\frac {26}{9}}}$ ### Part 3 Converting Improper Fractions Back Into Mixed Numbers 1. click to unmute.. to Divide Mixed Fractions 1 Divide the numerator by the denominator. If there is no remainder, then your answer is a whole number rather than a mixed number, and you need not do anything further. Likely, though, you will have a remainder. Set this aside for now. The quotient you found when Dividing the numerator by the denominator will be the whole number of your mixed number.[11] • For example, ${\displaystyle 26\div 9=2}$ with a remainder of ${\displaystyle 8}$. Thus, the whole number of your mixed number will be 2. 2. click to unmute.. to Divide Mixed Fractions 2 Turn the remainder into the numerator of your fraction. Place this numerator over the original denominator. This will give you the fraction of your mixed number.[12] • For example, if your original denominator is ${\displaystyle 9}$ and your remainder is ${\displaystyle 8}$, the fraction of your mixed number is ${\displaystyle {\frac {8}{9}}}$. 3. click to unmute.. to Divide Mixed Fractions 3 Combine the whole number and the fraction. This gives you the final answer to your original division problem. ## Community Q&A Search • To find the reciprocal, I flip the second fraction. What if I have fraction divided by fraction divided by fraction? Would I flip the last fraction, or the second one? Flip both the second and third fractions. • What is 2 1/4 divided by 3? Before we can divide, we need to turn these numbers into improper fractions. So 2 1/4 = (2·4)+1 [9] over 4, the original denominator. Then we simply place 3 over 1 because it is a whole number. Now we can get to the equation. 9/4 divided by 3/1 really means turn the division sign into a multiplication sign and invert one of the fractions. So 9/4·1/3 [(9·1) over (4·3)]. This gives us 9/12, which we can simplify by finding the highest common factor and then dividing both numerator and denominator by it. The hcf is 3 so 9 divided by 3 [3] over 12 divided by 3 [4]. Your final answer is 3/4. • How do I multiply fractions? wikiHow Contributor Just multiply across the top and the bottom. • How can I divide mixed numbers easily? wikiHow Contributor First, change the mixed number to an improper fraction. Next, multiply the divisor and simplify if needed. Then, put your answer in the lowest term. Make sure to check your work. • Do I always divide the mixed fraction by 2? wikiHow Contributor Not always. It depends on the greatest common factor of the numerator and denominator. If the fraction was 3/9, you would divide by 3 because that is the greatest common factor. • Why do you flip the fractions and then multiply in Part 2, step 2? wikiHow Contributor Any fraction multiplied by its reciprocal is equal to 1. For example, 5/10 x 10/5 = 50/50 = 1. When you divide by a fraction, you want to cancel the second fraction by changing it to a 1. So, to do this, you have to multiply it by its reciprocal (or flip it). But, if you're going to multiply the second fraction by its reciprocal, you also have to multiply the other number by its reciprocal. So when you "flip and multiply" you are really cancelling out the second fraction, which leaves you with just multiplying the first fraction by the second fraction's reciprocal. • 12 people calls for 3 3/4 pound burger. How many pounds of burger are needed to serve 18 people? Set up a proportion: 12 is to 3 3/4 as 18 is to x, where x is the number of pounds needed. Change the 3 3/4 to 15/4, then multiply both the 12 and the 15/4 by four to clear the fraction. The proportion is now 48 is to 15 as 18 is to x. Cross-multiply to solve for x. • How do I answer this question?: Billie needed three two-thirds feet of thread to finish a pillow. If she has two times as much thread as she needs, what is the length she has? Is the answer 7 and one third? wikiHow Contributor Three plus three is six and two thirds plus two thirds equals one and a third feet of thread. six plus one and a third equals seven and one third, so yes, that is right. • What is 34 divided by 5 1/2, multiplied by 21/8? wikiHow Contributor You would first follow the steps above to divide 34 by 5 1/2: 5 x 2 = 10 10 + 1 =11 So, 5 1/2 as an improper fraction is 11/2. 32 is a whole number, so converting it to a fraction, you get 32/1. So, your problem is now 32/1 divided by 11/2. Take the reciprocal of 11/2, then multiply the fractions. 32/1 x 11/2 = 352/2 Now you need to multiply 352/2 x 21/8 = 7392/16 Next you need to simplify. 7392 and 16 are both divisible by 16, so 7392/16 = 462/1, which simplifies to just 462. • When you convert the mixed fractions with a whole number your supposed to use multiplication instead of division? • How do I divide three or more mixed numbers in an equation? 200 characters left ## Article Info Categories: Multiplication and Division \| Fractions In other languages: Italiano: Dividere le Frazioni Miste, Español: dividir fracciones mixtas, Português: Dividir Frações Mistas, Français: diviser des nombres mixtes, Deutsch: Dividieren von gemischten Brüchen, Nederlands: Gemengde breuken delen, 中文: 求带分数的除法, Русский: делить смешанные дроби, Bahasa Indonesia: Membagi Pecahan Campuran Thanks to all authors for creating a page that has been read 175,954 times.
Is 3/8 bigger than 1/2? Many times we have to compare which is greater among 2 or more fractions. One such common question is 3/8 Bigger than 1/2? When two values are present in the form of a fraction then you have to learn to compare the values and determine which fractional value is small and which fraction is greater in quantity. Here are two fractions, 1/2 and 3/8 and we will teach you how to understand which one is the greater fraction. Basic concept 1) A fraction consists of two parts. The digits above the line of division are called numerators and the ones below the line of division are called the denominator. 2) When two denominators are the same then you can compare the values of the numerator as single whole numbers and say which fraction is great. In this case, the greater the numerator greater is the fraction. 3)However, the problem arises when the denominators and different, and some basic calculating steps need to be followed to find the value of the sum. Is 3/8 bigger than 1/2? No, 1/2 is greater than 3/8 Calculation Here we can see that the denominators are different and thus you have to initiate the following calculation to yield the results. Method #1: Converting Denominators Step 1: First, we need to find the smallest number that can be divided by both 2 and 8. By using the prime factorization method, it is seen that 8 is the smallest number that can be divided by both 2 and 8. Now you need to balance both the given fractions. (1×4)/(2×4) = 4/8 [latexpage] $\frac{(1\times4)}{(2\times4)}=\frac{4}{8}$ (3×1)/(8×1) = 3/8 [latexpage] $\frac{(3\times1)}{(8\times1)}=\frac{3}{8}$ This is done by checking how many times the two denominators will go in number 8. In the first case, number 2 goes 4 times to result in 8, and in the case of the second denominator 8, eight only goes one time in 8. Step 2: Now both the numerators are multiplied accordingly to 4 and 1 respectively to yield the two final fractions which have the same denominator. Here, 4/8 and 3/8 [latexpage] $\frac{4}{8}>\frac{3}{8}$ Now since both denominators are the same you can simply compare the two numerators as whole numbers and find out which fraction is greater. which means that [latexpage] $\frac{1}{2}>\frac{3}{8}$ Also read: What is 13/20 as a Percentage? Method #2: Decimal Conversion Method This is another simple method that you can acquire to instantly compare the two fractional values. When 1/2 is converted to decimal it results as 0.5 and when 3/8 is converted to fractions it results as 0.375 3 ÷ 8 = 0.375 Now since 0.5 is greater than 0.375, the fraction 1/2 is greater than 3/8 Result: 1/2>3/8 Now here you know both the methods of comparing two fractions which can be either done by equating the denominators or by changing the fractions to decimal values.
# Understanding Probability Space: A Comprehensive Example In the field of probability theory, a probability space is a fundamental concept that serves as the foundation for understanding and analyzing random phenomena. By providing a formal framework to represent and analyze the outcomes of random experiments, probability spaces are essential for assessing the likelihood of different events and making informed decisions in a wide range of fields, including statistics, finance, and engineering. In this article, we will delve deep into the concept of probability space, exploring its key components and properties through a detailed example. By the end of this guide, you will have a solid understanding of how probability spaces are constructed and how they can be used to model uncertain events. ## The Components of a Probability Space Before diving into an example, let's briefly review the essential components of a probability space. A probability space consists of three key elements: ### 1. Sample Space The sample space, denoted by S, is the set of all possible outcomes of a random experiment. It is the complete set of all elementary events that can occur. For example, when rolling a fair six-sided die, the sample space S = {1, 2, 3, 4, 5, 6}. ### 2. Events Events are subsets of the sample space representing specific outcomes or combinations of outcomes. An event can consist of a single outcome, multiple outcomes, or even the entire sample space. For instance, the event of rolling an even number on the six-sided die can be represented as E = {2, 4, 6}. ### 3. Probability Measure The probability measure assigns a probability to each event in the sample space, indicating the likelihood of the event occurring. It is a function that takes an event as input and returns a real number between 0 and 1, inclusive. The sum of the probabilities of all events in the sample space is 1. ## An Example of Probability Space To illustrate the concept of probability space, let's consider a simple example involving the tossing of a fair coin. In this example, we will construct the probability space for the coin-tossing experiment and explore its properties. ### Sample Space The sample space for the coin-tossing experiment consists of all possible outcomes of a single coin toss. Since a coin can land either heads (H) or tails (T), the sample space is given by S = {H, T}. ### Events Now, let's define some events based on the outcomes of the coin toss: • The event of getting heads: EH = {H} • The event of getting tails: ET = {T} • The event of getting either heads or tails: EHT = {H, T} ### Probability Measure Since the coin is fair, the probability of getting heads (H) is equal to the probability of getting tails (T), and both are 0.5. Mathematically, this can be expressed as: P(EH) = P(ET) = 0.5 Furthermore, the probability of getting either heads or tails is the sum of their individual probabilities, which equals 1: P(EHT) = P(EH) + P(ET) = 0.5 + 0.5 = 1 Thus, the probability space for the coin-tossing experiment is fully defined by the sample space, the events, and the probability measure. ### What is the importance of probability spaces? Probability spaces provide a rigorous framework for quantifying uncertainty and randomness, enabling the mathematical treatment of uncertain events in various fields, including science, engineering, and finance. ### Can the sample space contain an infinite number of outcomes? Yes, the sample space can be finite, countably infinite, or uncountably infinite, depending on the nature of the random experiment. For example, the sample space of all possible real numbers in a given interval is uncountably infinite. ### How are probabilities assigned to events in a continuous sample space? In the case of continuous sample spaces, such as the real number line, probabilities are assigned using techniques from calculus, and the concept of probability density functions is employed to quantify the likelihood of events. ## Conclusion In conclusion, understanding probability spaces is crucial for modeling and analyzing uncertain events in a rigorous and systematic manner. By carefully defining the sample space, events, and associated probabilities, we can gain valuable insights into the likelihood of different outcomes and make sound decisions based on probabilistic reasoning. Whether it's predicting stock market trends or estimating the likelihood of a medical diagnosis, probability spaces are indispensable tools for dealing with uncertainty in the real world. 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## Euler Path: 8 Lines Solution The problem of Euler path marked a very fundamental moment in algorithm studies. When Euler posed the 7-bridge problem, there was no mathematical tool to solve it, hence he created the tool – graph theory. It sounds like how Newton invented calculus to solve his gravity problems and how Bernoulli invented calculus of variations to solve his brachistochrone problem. As I quote Sutherland, “Well, I didn’t know it was hard.” ### Problem statement Say you have arrived in a new country with a bunch of islands and one way bridges between some pairs of islands. As a tourist you would like to visit all the bridges once and only once. Abstractly, an Euler path is a path that traverses all the edges once and only once. Is it always possible? If possible, how do we find such a path? ### Existence of path Unlike many other problems, we can determine the existence of the path without actually finding the path. There are 2 conditions for it. The first has to do with in and out degrees, and the second is about the connectivity of the graph. For a directed graph, the in degree is the number of incoming edges to a vertex, and the out degree is the number of outgoing edges to a vertex. Easily, if we have an Euler path, then there will be a start and end vertex. The start vertex will have an out degree – in degree of 1, the end vertex will have in degree – out degree of 1. Every other vertex will have an equal number of out degree and in degree, because if you have a different number, you obviously cannot go through all of those edges in one path. The only exception is when the start and end vertices are the same, in which case all the vertices have the same number of in and out degrees. The second condition is that all vertices have to be weakly connected, meaning that by treating all edges as undirected edges, there exists a path between every pair of vertices. The only exception is for the vertices that have no edges – those can just be removed from the graph, and the Euler path trivially does not include them. Given the above two properties, you can prove there is an Euler path by the following steps. First it is easy to see that if you start walking from the start vertex (out – in = 1) and removing edges as you walk through them, you can only end up at the end vertex (in – out = 1). Then, the remaining graph is full of cycles that can be visited through some vertex in the path we already have, and you just have to merge the cycles and the path to get the Euler path. It is easier to see the other direction of the proof: if there is an Euler path, both conditions have to be met. ### Implementing existence of path As mentioned above, we code up the two conditions and check whether we can find a valid starting position, otherwise return -1. Vertices are numbered 0 to n-1. The graph is stored as an adjacency list, meaning that adj[i] has all the neighbors (edges go from i to those). So the out degree of i is adj[i].size(). ```int start(vector<vector<int> >& adj) { // condition 1: in and out degrees for (int i = 0; i < adj.size(); i++) { deg[x]--; } int ans = -1; for (int i = 0; i < deg.size(); i++) if ((ans != -1 && deg[i] != 0 && deg[i] != -1) \|\| deg[i] > 1) return -1; else if (deg[i] == 1) ans = i; if (ans == -1) // start and end vertices are the same for (int i = 0; i < adj.size() && ans == -1; i++) ans = i; if (ans == -1) // there is no edge at all return -1; // condition 2: connectivity vector<int> bfs{ans}; vis[ans] = true; for (int i = 0; i < bfs.size(); i++) if (!vis[x]) { bfs.push_back(x); vis[x] = true; } for (int i = 0; i < adj.size(); i++) return -1; return ans; } ``` That is slightly more clumsy that I would like it to be, but it should be clear. Basically just counting in and out degrees and running a BFS on the starting vertex. ### Actually finding the path As mentioned above in the sketch of proof, finding a path consists of the 3 steps: 1. Walk from the start vertex, removing edges as we use them, until there is nowhere to go. Then we have a path from start to end. 2. For the remaining edges, start at a vertex on the path and randomly walk until we go back to the same vertex. Then we have a cycle. Merge the cycle on the path. For example, if we had a path s->a->b->c->…->t and a cycle c->alpha->beta->c, we merge them to become s->a->b->c->alpha->beta->c->…->t. 3. Repeat step 2 until there is no remaining edge. Well, that does not sound very easy to write nor very efficient to run, if we literally implemented the above. How many lines of code would that be? ### The 8 lines solution The answer is 8. Here’s the code: ```void euler(vector<vector<int> >& adj, vector<int>& ans, int pos) { } ans.push_back(pos); } ``` This is a very beautiful solution using recursion, in my opinion. I was quite surprised when I first saw this. Where is everything? Where is getting the main path? Where is getting the cycles? Where are we merging them? The gist of this recursion is a post-order DFS, meaning that we visit the end of the graph first, and then backtrack. This comes from a very crucial observation: we can always be sure what could be at the end of the path, but not the front. It is important to know that the answer array is in reverse order of visit, i.e. we need to reverse it to get the Euler path. Let’s go through the stages of how the program works. 1. Path from start to end vertex. The first time we push back is when we run out of outgoing edges, which can only be the case of the end vertex, with in-out = 1. In all other vertices, the numbers are the same so if you can go in, you can definitely go out of that vertex. Hence the first push back occurs with the end vertex, and at that point the program execution stack has the entire path. 2. As we return from a recursive call of the function, we are essentially going back from the end to the start. If a vertex on the main path does not have any outgoing edge, we know we will visit it next, so we push it to the ans vector and return from the function. 3. But if a vertex on the path does have an outgoing edge, that means there is at least one cycle including this vertex. Then in the next iteration of the while loop, we will visit one of the outgoing edges and start another round of recursion. Again, this recursion must end on the same vertex because it is the only one with in-out = 1. This recursion gets you a cycle and by the time it returns, the cycle would have been pushed to the ans array already, finishing the merge operation. By studying this code, there is one interesting point to note. That is, the while loop will only be executed 0 to 2 times in any recursive call. It will only be 0 at the end vertex, and on the main path with no branches, it will be 1. On the main path with branches, it will only be 2 but not more regardless of the out degree, because surely the recursive call for cycle needs to end on that vertex but it will not end until all outgoing edges are used up. Therefore to the program, multiple cycles on one vertex is just one big cycle. On the other vertices on cycles, it works the same way whether they have branches or not. In case you want to see how to run it, here is the main function I wrote to test it: ```int main() { int n, q; cin >> n >> q; for (int i = 0; i < q; i++) { int u, v; cin >> u >> v; } if (s == -1) { cout << "no path" << endl; return 0; } vector<int> ans; reverse(ans.begin(), ans.end()); for (int x : ans) cout << x << endl; return 0; } ``` That’s it – a simple problem with a simple solution. Leetcode has one Euler path problem, and the algorithm in this blog post comes from the discussion of that problem. Everything is pretty much the same for undirected graphs – you just have to use different data structures to store the edges. The proof is mostly the same with the first condition now about odd/even number of edges at each vertex, as there is no distinction between in and out degree. ## TIW: Binary Indexed Tree Binary indexed tree, also called Fenwick tree, is a pretty advanced data structure for a specific use. Recall the range sum post: binary indexed tree is used to compute the prefix sum. In the prefix sum problem, we have an input array v, and we need to calculate the sum from the first item to index k. There are two operations. Update: change the number at one index by adding a value (not resetting the value), and query: getting the sum from begin to a certain index. How do we do it? There are two trivial ways: 1. Every time someone queries the sum, just loop through it and return the sum. O(1) update, O(n) query. 2. Precompute the prefix sum array, and return the precomputed value from the table. O(n) update, O(1) query. To illustrate the differences and better explain what we’re trying to achieve, I will write the code for both approaches. They are not the theme of this post though. ```class Method1 { private: vector<int> x; public: Method1(int size) { x = vector<int>(size); } void update(int v, int k) { x[k] += v; } int query(int k) { int ans = 0; for (int i = 0; i <= k; i++) ans += x[i]; return ans; } }; class Method2 { private: vetor<int> s; public: Method2(int size) { s = vector<int>(size); } void update(int v, int k) { for (; k < s.size(); k++) x[k] += v; } int query(int k) { return s[k]; } }; ``` Read through this and make sure you can write this code with ease. One note before we move on: we’re computing the sum from the first item to index k, but in general we want the range sum from index i to index j. To obtain range sum, you can simply subtract the prefix sums: query(j)-query(i-1). OK, that looks good. If we make a lot of updates, we use method 1; if we make a lot of queries, we use method 2. What if we make the same amount of updates and queries? Say we make n each operations, then no matter which method we use, we end up getting O(n^2) time complexity (verify!). We either spend too much time pre-computing or too much time calculating the sum over and over again. Is there any way to do better? Yes, of course! Instead of showing the code and convincing you that it works, I will derive it from scratch. The quest of log(n) The problem: say we have same amount of updates and queries, and we do not want to bias the computation on one of them. So we do a bit of pre-computation, and a bit of summation. That’s the goal. Say we have an array of 8 numbers, {1, 2, 3, 4, 5, 6, 7, 8}. To calculate the sum of first 7 numbers, we would like to sum up a bunch of numbers (since there has to be a bit of summation). But the amount of numbers to be summed has to be sub-linear. Let’s say we want it to be log(n). log2(7) is almost 3, then maybe we can sum 3 numbers. In this case, we choose to sum the 3 numbers: sum{1, 2, 3, 4}, sum{5, 6} and sum{7}. Assume that we have these sums already pre-computed, we have log(n) numbers to sum, hence querying will be log(n). For clarity, let me put everything in a table: Table 1a sum{1} = sum{1} sum{1, 2} = sum{1, 2} sum{1, 2, 3} = sum{1, 2} + sum{3} sum{1, 2, 3, 4} = sum{1, 2, 3, 4} sum{1, 2, 3, 4, 5} = sum{1, 2, 3, 4} + sum{5} sum{1, 2, 3, 4, 5, 6} = sum{1, 2, 3, 4} + sum{5, 6} sum{1, 2, 3, 4, 5, 6, 7} = sum{1, 2, 3, 4} + sum{5, 6} + sum{7} sum{1, 2, 3, 4, 5, 6, 7, 8} = sum{1, 2, 3, 4, 5, 6, 7, 8} The left hand side of the table is the query, and all the terms on the right hand side are pre-computed. If you look closely enough you will see the pattern: for summing k numbers, first take the largest power of 2, 2^m, that is ≤ k, and pre-compute it. Then for the rest of the numbers, k-2^m, take the largest power of 2, 2^m’ such that 2^m’ ≤ k-2^m, and pre-compute it, and so on. There are two steps to do: show that querying (adding terms on the right hand side) is log(n) and show that pre-computing the terms on the right hand side is log(n). Querying is log(n) is easily seen, because by taking out the largest power of 2 each time, we will at least take out half of the numbers (Use proof by contradiction). Taking out no less than one half each time, after O(log(n)) time we would have taken out all of it. Now we are one step from finishing on the theoretical side: how do we pre-compute those terms? Let’s say we want to change the number 1 into 2, essentially carrying out update(1, 0). Look at the terms above: we need to change sum{1}, sum{1, 2}, sum{1, 2, 3, 4} and sum{1, 2, 3, 4, 5, 6, 7, 8}. Each time we update one more pre-computed term, we cover double the number of elements in the array. Therefore we also only need to update log(n) terms. Let’s see it in a table: Table 1b update 1: sum{1}, sum{1, 2}, sum{1, 2, 3, 4}, sum{1, 2, 3, 4, 5, 6, 7, 8} update 2: sum{1, 2}, sum{1, 2, 3, 4}, sum{1, 2, 3, 4, 5, 6, 7, 8} update 3: sum{3}, sum{1, 2, 3, 4}, sum {1, 2, 3, 4, 5, 6, 7, 8} update 4: sum{1, 2, 3, 4}, sum{1, 2, 3, 4, 5, 6, 7, 8} update 5: sum{5}, sum{5, 6}, sum{1, 2, 3, 4, 5, 6, 7, 8} update 6: sum{5, 6}, sum{1, 2, 3, 4, 5, 6, 7, 8} update 7: sum{7}, sum{1, 2, 3, 4, 5, 6, 7, 8} update 8: sum{1, 2, 3, 4, 5, 6, 7, 8} Cool, now we have a vague idea about what to pre-compute for update and what to add for query. Now we should figure out the details of the code. How is the code written? First, we need to determine the representation of the pre-computed terms. Here is a list of all pre-computed terms: {1}, {1, 2}, {3}, {1, 2, 3, 4}, {5}, {5, 6}, {7}, {1, 2, 3, 4, 5, 6, 7, 8} The last number of each term is unique and covers the range 1-8. That’s great news! We can use a vector to store these terms easily, and let the index of the array be the last number of the term. For example, the sum of {5, 6} will be stored at bit[6]. First, the query operation. Let’s revisit the table with binary representation of numbers: Table 2a: revised version of table 1a, with sums written as bit elements, indices in binary query 0001: bit[0001] query 0010: bit[0010] query 0011: bit[0011]+bit[0010] query 0100: bit[0100] query 0101: bit[0101]+bit[0100] query 0110: bit[0110]+bit[0100] query 0111: bit[0111]+bit[0110]+bit[0100] query 1000: bit[1000] Do you see the pattern yet? Hint: for queries that have k ones, we have k terms on the right. The pattern is that while the index has at least 2 ones, we remove the lowest bit that is one, then move on to the next term. 0111->0110->0100. Finally, here’s the code: ```int query(vector<int>& bit, int k) { int ans = 0; for (k++; k; k -= k & (-k)) ans += bit[k]; return ans; } ``` After all the work we’ve been through, the code is extremely concise! Two things to notice: the k++ is to change the indexing from 0-based to 1-based, as we can see from the above derivation we go from 1 to 8, instead of 0 to 7. The second thing is the use of k & (-k) to calculate the lowest bit. You can refer to the previous blog post on bitwise operations. OK, we’re almost done. What about update? Another table: Table 2b: revised version of table 1b update 0001: bit[0001], bit[0010], bit[0100], bit[1000] update 0010: bit[0010], bit[0100], bit[1000] update 0011: bit[0011], bit[0100], bit[1000] update 0100: bit[0100], bit[1000] update 0101: bit[0101], bit[0110], bit[1000] update 0110: bit[0110], bit[1000] update 0111: bit[0111], bit[1000] update 1000: bit[1000] What’s the pattern this time? Hint: again, look for the lowest bit! Yes, this time instead of removing the lowest bit, we add the lowest bit of the index to itself. This is less intuitive than the last part. For example, lowest bit of 0101 is 1, so the next index is 0101+1 = 0110; lowest bit is 0010, next index is 0110+0010 = 1000. So here’s the code, note also the k++: ```void update(vector<int>& bit, int v, int k) { for (k++; k < bit.size(); k += k & (-k)) bit[k] += v; } ``` This is deceivingly easy! That can’t be right. It can’t be that easy… Actually it can, if you look at the category of this post; nothing I write about is hard. Actually it was easy because we were just matching patterns and assuming it would generalize. Let’s study the for loops a little more to understand why and how low bits are involved in this. This is rather complicated, so for practical purposes you might as well skip them. First, observe that the low bit of each index indicates the number of integers that index sums over. Say, 0110 has low bit 0010 which is 2, so bit[6] is a sum of two numbers: 5 and 6. This is by design, since this is exactly how we picked the pre-computed terms, so there is no way of explanation. Second, bit[k] is the sum from indices k-lowbit(k)+1 to k. This is a direct consequence from (1) bit[k] is a summation that ends at the kth number and (2) bit[k] sums over lowbit(k) numbers. In light of this fact, the code for querying becomes clear: for an index k, we first get the sum from k-lowbit(k)+1 to k from bit[k], then we need to find the sum from 1 to k-lowbit(k). The latter becomes a sub-problem, which is solved by setting k-lowbit(k) as the new k value and going into the next iteration. For updating, it is much trickier. From the above, we have l-lowbit(l) < k ≤ l, iff bit[l] includes k. Below is a sketch of proof, the actual proof will include more details and be more tedious and boring to go through. For the kth number, bit[k+lowbit(k)] must include it. This is because the lowbit of k+lowbit(k) must be at least 2 times lowbit(k), so k+lowbit(k)-lowbit(k+lowbit(k)) ≤ k-lowbit(k) < k ≤ k+lowbit(k), satisfying the inequality. Also, k can be updated to k+lowbit(k) in the next iteration, because given lowbit(k) < lowbit(m) < lowbit(n) and that bit[m] includes k and bit[n] includes m, bit[n] must include k as well. Till now, we have shown that the bit[k] values we have modified in the for loop must include k. Then, we also need to show that all bit[l] values that include k are modified in our loop. We can actually count all the bit[l] values that include k: it is equal to one plus the number of zeros before lowbit(k). It is not difficult to see how the for loop reduces the number of zeros before lowbit(k) each time the loop moves on to the next iteration. The only question remaining is why that number? Let’s look at the table 2b again. The numbers of terms for the first four entries, i.e. {4, 3, 3, 2}, are one more than the number of terms for the second four entries, i.e. {3, 2, 2, 1}. This is by design, because bit[4] covers the first four but not the second four, and everything else is pretty symmetric. Again, the first two entries have one more coverage than the second two entries, because bit[2] records the first two but not the second two. Hence, each time we “go down the tree” on a “zero edge” (appending a 0 to the index prefix), the numbers will be covered once more than if we “go down the tree” on a “one edge” (appending a 1 to the index prefix). After we hit the low bit, no more terms of smaller low bits will cover this index, and of course the index itself includes itself, thus the plus one. This is a basic and very rough explanation on how the numbers of zeros relate to the number of terms including a certain index. Here we have argued semi-convincingly the update loop is valid and complete. OK, anyway, time for practice: Range Sum Query – Mutable It’s literally implementing a binary indexed tree, nothing more. ```class NumArray { private: vector<int> bit; void update_helper(int v, int k) { for (k++; k < bit.size(); k += k & (-k)) bit[k] += v; } int query_helper(int k) { int ans = 0; for (k++; k; k -= k & (-k)) ans += bit[k]; return ans; } public: NumArray(vector<int> &nums) { bit.resize(nums.size()+1); for (int i = 0; i < nums.size(); i++) update_helper(nums[i], i); } void update(int i, int val) { update_helper(val-query_helper(i)+query_helper(i-1), i); } int sumRange(int i, int j) { return query_helper(j)-query_helper(i-1); } }; ``` It got a little complicated because I didn’t store the original values, so we need some work on line 21 to calculate the change at a certain index given the new value and the old range sums. But that’s nothing important. That’s it for the basic introduction of binary indexed trees. There are some variants to it, such as replacing the + sign in update function to a min or max function to take prefix min or max, or extending the algorithm to a 2D matrix, aka 2D binary indexed tree. We can even use it for some dynamic programming questions. There are in fact a few more questions on Leetcode that uses this data structure. But that’s for later. I learned binary indexed tree through the TopCoder tutorial. If you think I did a really bad job and you do not understand at all, you can refer to it as well.
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # 4.3: Circles and Arrows 2 Difficulty Level: At Grade Created by: CK-12 Can you figure out the value of each letter in the diagram below? In this concept, we will use problem solving steps to help us find the value of letters in circles and arrows diagrams. ### Guidance In order to solve the problem above, use the problem solving steps. • Start by describing what you see in the diagram. • Next, figure out what your job is in this problem. In all of these problems your job will be to figure out the value of the two letters in the diagram. • Then, make a plan for how you will solve. There is usually more than one way to solve the problem. You will want to figure out one letter first and then the next letter. • Next, solve the problem. • Finally, check to make sure that the values you found work for all 5 of the arrows. #### Example A Figure out the value of each letter. Solution: We can use the problem solving steps to help us. \begin{align}& \mathbf{Describe:} && \text{There are two rows and three columns of numbers and letters.}\\ &&& \text{Arrows point to sums.}\\ & \mathbf{My \ Job:} && \text{Figure out the value of the letters} \ C \ \text{and} \ D.\\ & \mathbf{Plan:} && \text{Start with column 2. Solve for}\ C.\\ &&& \text{Replace}\ C \ \text{with its value in row 1. Solve for}\ D.\\ & \mathbf{Solve:} && \text{Column} \ 2: C + C = 10. \ \text{Therefore}, \ C = 5.\\ &&& \text{Row} \ 1: 6 + C + D = 15, \ \text{so}, \ 6 + 5 + D = 15. \ \text{So}, \ D = 15 - 11, \ \text{or}\ 4.\\ & \mathbf{Check:} && \text{Replace all}\ C\text{s with 5. Replace all}\ D\text{s with 4.}\\ &&& \text{Add rows and columns. Check the sums.}\\ &&& \text{Row} \ 1: 6 + 5 + 4 = 15 \quad \text{Column} \ 1: 6 + 4 = 10\\ &&& \text{Row} \ 2: 4 + 5 + 3 = 12 \quad \text{Column} \ 2: 5 + 5 = 10\\ &&& \qquad \qquad \qquad \qquad \qquad \quad \text{Column} \ 3: 4 + 3 = 7\end{align} #### Example B Figure out the value of each letter. Solution: We can use the problem solving steps to help us. \begin{align}& \mathbf{Describe:} && \text{There are two rows and three columns of numbers and letters.}\\ &&& \text{Arrows point to sums.}\\ & \mathbf{My \ Job:} && \text{Figure out the value of the letters} \ E \ \text{and} \ F.\\ & \mathbf{Plan:} && \text{Start with column 1. Solve for}\ E.\\ &&& \text{Replace}\ E \ \text{with its value in row 1. Solve for}\ F.\\ & \mathbf{Solve:} && \text{Column} \ 1: E + E = 6. \ \text{So}, \ E = 3.\\ &&& \text{Row} \ 1: E + F + 2 = 13, \ \text{so}, \ 3 + F + 2 = 13. \ \text{Therefore}, \ F = 13 - 5, \ \text{or}\ 8.\\ & \mathbf{Check:} && \text{Replace all}\ E\text{s with 3. Replace all}\ F\text{s with 8.}\\ &&& \text{Add rows and columns. Check the sums.}\\ &&& \text{Row} \ 1: 3 + 8 + 2 = 13 \quad \text{Column} \ 1: 3 + 3 = 6\\ &&& \text{Row} \ 2: 3 + 7 + 8 = 18 \quad \text{Column} \ 2: 8 + 7 = 15\\ &&& \qquad \qquad \qquad \qquad \qquad \quad \text{Column} \ 3: 2 + 8 = 10\end{align} #### Example C Figure out the value of each letter. Solution: We can use the problem solving steps to help us. \begin{align}& \mathbf{Describe:} && \text{There are two rows and three columns of numbers and letters.}\\ &&& \text{Arrows point to sums.}\\ & \mathbf{My \ Job:} && \text{Figure out the value of the letters} \ G \ \text{and} \ H.\\ & \mathbf{Plan:} && \text{Start with row 2. Solve for}\ H.\\ &&& \text{Replace}\ H \ \text{with its value in Column 3. Solve for}\ H.\\ & \mathbf{Solve:} && \text{Row} \ 2: 3 + H + H = 13, \ \text{so} \ H + H = 10. \ \text{Therefore}, \ H = 5.\\ &&& \text{Column} \ 3: G + H = 7, \ \text{so} \ G + 5 = 7. \ \text{Therefore}, \ G = 7 - 5, \ \text{or}\ 2.\\ & \mathbf{Check:} && \text{Replace all}\ H\text{s with 5. Replace all}\ G\text{s with 2.}\\ &&& \text{Add rows and columns. Check the sums.}\\ &&& \text{Row} \ 1: 2 + 9 + 2 = 13 \quad \text{Column} \ 1: 2 + 3 = 5\\ &&& \text{Row} \ 2: 3 + 5 + 5 = 13 \quad \text{Column} \ 2: 9 + 5 = 14\\ &&& \qquad \qquad \qquad \qquad \qquad \quad \text{Column} \ 3: 5 + 2 = 7\end{align} #### Concept Problem Revisited Solution: We can use the problem solving steps to help us. \begin{align}& \mathbf{Describe:} && \text{There are two rows and three columns of numbers and letters.}\\ &&& \text{Arrows point to sums.}\\ & \mathbf{My \ Job:} && \text{Figure out the value of the letters} \ A \ \text{and} \ B.\\ & \mathbf{Plan:} && \text{Start with column 3. Solve for}\ B.\\ &&& \text{Replace}\ B \ \text{with its value in Column 2. Solve for}\ A.\\ & \mathbf{Solve:} && \text{Column} \ 3: B + B = 12. \ \text{So}, \ B = 6.\\ &&& \text{Column} \ 2: A + 6 = 9. \ \text{So}, \ A = 9 - 6, \ \text{or}\ 3.\\ & \mathbf{Check:} && \text{Replace all}\ A\text{s with 3. Replace all}\ B\text{s with 6.}\\ &&& \text{Add rows and columns. Check the sums.}\\ &&& \text{Row} \ 1: 3 + 3 + 6 = 12 \quad \text{Column} \ 1: 3 + 5 = 8\\ &&& \text{Row} \ 2: 5 + 6 + 6 = 17 \quad \text{Column} \ 2: 3 + 6 = 9\\ &&& \qquad \qquad \qquad \qquad \qquad \quad \text{Column} \ 3: 6 + 6 = 12\end{align} ### Vocabulary In math, an unknown is a letter that stands for a number that we do not yet know the value of. In this concept, when you figured out the value of the letters in the circles and arrows diagrams you were solving for unknowns. ### Guided Practice For each problem, figure out the value of each letter. 1. 2. 3. 4. 1. \begin{align}J = 6, K = 4\end{align} 2. \begin{align}L = 8, M = 4\end{align} 3. \begin{align}N = 3, P = 7\end{align} 4. \begin{align}Q = 4, R = 2\end{align} ### Practice For each problem, figure out the value of each letter. ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Please to create your own Highlights / Notes Show Hide Details Description Difficulty Level: Tags: Date Created: Jan 18, 2013
# Solve the Following Systems of Equations: 2/X + 3/Y = 9/(Xy) 4/X + 9/Y = 21/(Xy), X != 0, Y != 0 - Mathematics Solve the following systems of equations: 2/x + 3/y = 9/(xy) 4/x + 9/y = 21/(xy), where x != 0, y != 0 #### Solution The system of given equation is 2/x + 3/y = 9/(xy) ....(i) 4/x + 9/y = 21/(xy), x != 0, y != 0 ......(ii) Multiplying equation (i) adding equation (ii) by ,xy we get 2y + 3x = 9 ....(iii) 4y + 9x = 21 .....(iv) From (iii), we get 3x = 9 - 2y => x = (9 - 2y)/3 Substituting x = (9 - 2y)/3 in equation (iv) weget 4x + 9((9 - 2y)/3) = 21 => 4y + 3(9 - 2y) = 21 => 4y + 27 - 6y = 21 => -2y = 21 - 27 => -2y = -6 => y = (-6)/(-2) = 3 Putting y = 3 in x = (9 - 2y)/3 we get x = (9 - 2xx3)/3 = (9-6)/3 = 3/3 = 1 Hence, solution of the system of equation is x = 1, y = 3 Concept: Algebraic Methods of Solving a Pair of Linear Equations - Substitution Method Is there an error in this question or solution? #### APPEARS IN RD Sharma Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Q 22
# Video: GCSE Mathematics Foundation Tier Pack 5 • Paper 2 • Question 3 GCSE Mathematics Foundation Tier Pack 5 • Paper 2 • Question 3 04:05 ### Video Transcript Five digits are shown in the grid below: seven, two, nine, five, three. Part a) Write down the largest three-digit number that can be made by arranging three of these digits without repeating any digit. Part b) Write down the closest number to 3000 that can be made by arranging four of these digits without repeating any digit. So the first thing I’m gonna do before I answer any of the parts of this question, so a) or b), is rearrange our numbers or our digits from smallest to largest. So the smallest digit is two. Then we have three, then five, then seven, and finally the digit nine. Okay, so now let’s look at part a). What we want to do is write down the largest three-digit number that can be made by arranging three of these digits without repeating any digit. So to enable us to do this, what I want to do is use the three largest digits. So we’ve got five, seven, and nine. And then what I want to do is rearrange these three digits to have the largest digit first and the smallest digit of of these three last. And when we do that, we’re gonna have nine in the hundreds place value, seven in the tens place value, and five in the ones place value or units place value. So therefore, the largest three-digit number that we can make by arranging three of these digits without repeating any is 975. So now what we want to do is answer part b). And for part b), we need to write down the closest number to 3000 that could be made by arranging four of the digits without repeating any digit. So what we want to do, because we want to get the number that’s closest to 3000, is write down the greatest number we can write down that is in the 2000s and the smallest that is in the 3000s. And then we want to compare them to see which is actually closer to 3000 itself. So I’m gonna start with the largest number in the 2000s. So we’re gonna start with the number two. And then we’re gonna select the three largest digits. And then again, as we did in part a), we’re gonna arrange them from the largest digit to the smallest digit. And when we do that, we’re gonna get 2975. And then what we can see is how close this is to 3000. Well, we can see that we’d have to add 25 to 2975 to get to 3000. And we can work that out by counting up, because to go from 2975, if you add five, you get to 2980, then add 20 we get to 3000. So now what we want to do is list the smallest number in the 3000s. So first of all, we’re gonna start with three. And then what we want are the next three smallest digits, because we want it to be the lowest number in the 3000s. And those three digits are two, five, and seven. And this time, we’re gonna put the digits in order from smallest to largest, because we want to make the smallest number that’s in the 3000s. So when we do that, we get 3275 [3257]. So again, what we do is we compare it to the 3000. And we can see that we’d have to add 257 to 3000 to get to 3257. Well, if we now compare the differences, we can see that 25 is less than 257. So therefore, the closest number to 3000 that could be made by arranging four of the digits without repeating any digit is 2975.
Search IntMath Close How to Construct an Angle Bisector in Geometry An angle bisector is a line or ray which divides an angle into two equal parts. The process of constructing an angle bisector is an important skill for any student of geometry, so let’s explore the steps necessary to construct an angle bisector. Step 1: Draw the Angle and Its Vertex Point The first step in constructing an angle bisector is to draw the given angle and its vertex point. To do this, you will need a ruler and a protractor. With these tools, you can draw the given angle accurately, as well as measure it if necessary. Step 2: Mark the Center Point of the Angle Once you have drawn the given angle, you must find and mark its center point (also known as its “vertex”). The vertex can be found by drawing two rays from opposite sides of the angle until they meet at a single point. This will be your center point; mark it with a dot or circle to denote that it is special. Step 3: Construct Two Line Segments Through the Center Point The third step in constructing your angle bisector is to draw two line segments through your center point that are at equal angles to each side of your given angle. To make sure these line segments are equal angles, use your protractor to measure them; they should both measure exactly half of your original given angle. Once both line segments have been constructed properly, they will form two smaller angles which together will form your original given angle. These two new lines are now considered part of your original given ray and constitute your newly constructed “angle bisector”! Conclusion: Constructing an angle bisector is an important skill for students studying geometry, as it helps them understand how angles work and can be used in many other equations and proofs involving angles. The process consists of three main steps: first drawing the given angle with its vertex point, then finding and marking its center point (its vertex), and finally drawing two line segments through that center point at equal angles from each side of the original given ray. With practice, students can become adept at constructing their own accurate angles bisectors! FAQ How do you construct an angle bisector in geometry? The process of constructing an angle bisector in geometry consists of three main steps: first drawing the given angle with its vertex point, then finding and marking its center point (its vertex), and finally drawing two line segments through that center point at equal angles from each side of the original given ray. With practice, students can become adept at constructing their own accurate angles bisectors! What tools do I need to construct an angle bisector? To construct an angle bisector, you will need a ruler and a protractor. With these tools, you can draw the given angle accurately, as well as measure it if necessary. Additionally, you may want to use a pencil, eraser, and compass to help you draw the angles more precisely. Why is it important to know how to construct an angle bisector? Knowing how to construct an angle bisector is an important skill for students studying geometry because it helps them understand how angles work and can be used in many other equations and proofs involving angles. Additionally, it can help them with other types of math problems that involve angles. By understanding how to construct an angle bisector accurately, students will be better prepared for more advanced mathematical tasks.
# Verbal Reasoning - Blood Relation Test @ : Home > Verbal Reasoning > Blood Relation Test > Introduction • Introduction ### Exercise "Everything should be made as simple as possible, but not simpler." - Albert Einstein Introduction: The questions which are asked in this section depend upon Relation. You should have a sound knowledge of the blood relation in order to solve the questions. To remember easily the relations may be divided into two sides as given below: 1. Relations of Paternal side: Father's father → Grandfather Father's mother → Grandmother Father's brother → Uncle Father's sister → Aunt Children of uncle → Cousin Wife of uncle → Aunt Children of aunt → Cousin Husband of aunt → Uncle 2. Relations of Maternal side: Mother's father → Maternal grandfather Mother's mother → Maternal grandmother Mother's brother Maternal uncle Mother's sister → Aunt Children of maternal uncle → Cousin Wife of maternal uncle → Maternal aunt Relations from one generation to next: Differenct types of questions with explanation: Type 1: If A + B means A is the mother of B; A x B means A is the father of B; A \$ B means A is the brother of B and A @ B means A is the sister of B then which of the following means P is the son of Q? (A) Q + R @ P @ N (B) Q + R * P @ N (C) Q x R \$ P @ N (D) Q x R \$ P \$ N Solution: (D) Q x R = Q is the mother of R [-Q, ±R] R \$ P = R is the brother of P [+ R, ±P] P \$ N = P is the brother of N [+ P, ±N] Therefore P is the son of Q. Type 2: A has 3 children. B is the brother of C and C is the sister of D, E who is the wife of A is the mother of D. There is only one daughter of the husband of E. what is the relation between D and B? Solution: With the chart Therefore, D is a boy because there is only one daughter of E. Hence, B is the brother of D. Type 3: Pointing to a photograph, Rekha says to Lalli, "The girl in the photo is the second daughter of the wife of only son of the grandmother of my younger sister." How this girl of photograph is related to Rekha? Solution: First Method - By Generating Charts: Second method: Grandmother of younger sister of Rekha → Grandmother of Rekha Wife of only son of grandmother → Mother of Rekha Younger daughter of the mother → Younger sister. Note: While solving the question (+) can be used for male and (-) can be used for female.
# Quadratic equation to vertex form solver ## The Best Quadratic equation to vertex form solver This Quadratic equation to vertex form solver helps to quickly and easily solve any math problems. Then, take the square root of this number to find the length of the hypotenuse. For example, if you know that one side is 3 feet long and another side is 4 feet long, you would first square these numbers to get 9 and 16. Then, you would add these numbers together to get 25. Taking the square root of 25 gives you 5, so you know that the hypotenuse is 5 feet long. Solving for x in a right triangle is a simple matter of using the Pythagorean theorem. With a little practice, you'll be able to do it in your sleep! There are a number of ways to solve quadratic equations, but one of the most reliable methods is to factor the equation. This involves breaking down the equation into its component parts, which can then be solved individually. For example, if the equation is x2+5x+6=0, it can be rewritten as (x+3)(x+2)=0. From here, it is a simple matter of solving each individual term and finding the value of x that makes both terms equal to zero. While it may take a bit of practice to become proficient at factoring equations, it is a valuable skill to have in your mathematical toolkit. Absolute value is a concept in mathematics that refers to the distance of a number from zero on a number line. The absolute value of a number can be thought of as its magnitude, or how far it is from zero. For example, the absolute value of 5 is 5, because it is five units away from zero on the number line. The absolute value of -5 is also 5, because it is also five units away from zero, but in the opposite direction. Absolute value can be represented using the symbol "\| \|", as in "\|5\| = 5". There are a number of ways to solve problems involving absolute value. One common method is to split the problem into two cases, one for when the number is positive and one for when the number is negative. For example, consider the problem "find the absolute value of -3". This can be split into two cases: when -3 is positive, and when -3 is negative. In the first case, we have "\|-3\| = 3" (because 3 is three units away from zero on the number line). In the second case, we have "\|-3\| = -3" (because -3 is three units away from zero in the opposite direction). Thus, the solution to this problem is "\|-3\| = 3 or \|-3\| = -3". Another way to solve problems involving absolute value is to use what is known as the "distance formula". This formula allows us to calculate the distance between any two points on a number line. For our purposes, we can think of the two points as being 0 and the number whose absolute value we are trying to find. Using this formula, we can say that "the absolute value of a number x is equal to the distance between 0 and x on a number line". For example, if we want to find the absolute value of 4, we would take 4 units away from 0 on a number line (4 - 0 = 4), which tells us that "the absolute value of 4 is equal to 4". Similarly, if we want to find the absolute value of -5, we would take 5 units away from 0 in the opposite direction (-5 - 0 = -5), which tells us that "the absolute value of -5 is equal to 5". Thus, using the distance formula provides another way to solve problems involving absolute value. Once this has been accomplished, the resulting equation can be solved for the remaining variable. In some cases, it may not be possible to use elimination to solve a system of linear equations. However, by understanding how to use this method, it is usually possible to simplify a system of equations so that it can be solved using other methods. Solving for an exponent can be a tricky business, but there are a few tips and tricks that can make the process a little bit easier. First of all, it's important to remember that an exponent is simply a number that tells us how many times a given number is multiplied by itself. For instance, if we have the number 2 raised to the 3rd power, that means that 2 is being multiplied by itself 3 times. In other words, 2^3 = 2 x 2 x 2. Solving for an exponent simply means finding out what number we would need to raise another number to in order to get our original number. For instance, if we wanted to solve for the exponent in the equation 8 = 2^x, we would simply need to figure out what number we would need to raise 2 to in order to get 8. In this case, the answer would be 3, since 2^3 = 8. Of course, not all exponent problems will be quite so simple. However, with a little practice and perseverance, solving for an exponent can be a breeze! ## We cover all types of math issues I am very impressed with this app. It’s smooth and the computer vision is extremely quick. I'm also impressed how quickly it compiles the steps, even if the types of problems it can do seem limited to a single variable. 10/10 keep up the great app development. Harriet Baker this is an amazing app. like, I really like it. good stuff. Except, one excruciatingly major thing. where? where is the dark mode? my eyes. my eyes are constantly blinded by the bright white theme of this calculator in my morning class. they cannot tolerate this suffering any longer. this is only a suggestion, but with implementing a dark mode, you could save lives. thank you. Helen Howard Quotient of a binomial and polynomial solver Websites that help you with math Homework cheats math Maths puzzle solver Math tutors san francisco
\| Using the annihilation model \| Movies of addition of directed numbers \| \| Quick quiz \| Using the annihilation model We will use the annihilation model to give meaning to the addition of directed numbers. To refresh how we use the model, let's look at how some integers are represented. This model is introduced in Meaning and Models. Recall that any number can be represented in many ways. Collection Model Symbol Thinking +1, +1, +1, +1 +4 Four positive chips represent +4 -1, -1, -1 -3 Three negative chips represent -3. -1, +1, -1, -1, +1 -1 Two +1s and three -1s represent -1, since one negative chip and one positive chip annihilate one another, leaving one negative. The operation of addition is modelled by combining the collections of chips. Example 1: 5 + 3 5 + 3 can be represented as 5 positive chips plus 3 positive chips, which when combined, gives 8 positive chips. + = +5 + (+3) = +8 Example 2: (-5) + 3 (-5) + 3 is represented as a collection of 5 negative chips combined with a collection of 3 positive chips. When we put the collection together we can see that 3 positive chips and 3 negative chips will annihilate each other, leaving 2 negative chips. + = -5 + (+3) = -2 Example 3: 5 + (-3) When we add -3 to 5 we can see that 3 positive chips and 3 negative chips annihilate each other, and we are left with 2 positive chips. + = +5 + (-3) = +2 This example shows how the addition of a negative number is the same as subtraction of a positive number. We have added 3 negative chips which is the same as taking away 3 positive chips. We can show this using the model below: Example 4: 5 - +3 = +5 - (+3) = +2 To add a negative number, you can subtract the positive value of the number Example: 5 + (-3) = 5 - (+3) = 2 If a and b are any integers, then a + (-b) = a - (+b) Example 5: (-5) + (-3) + = -5 + (-3) = -8 We can draw a table of basic addition facts to show the patterns that form when we add positive and negative integers. + -5 -4 -3 -2 -1 0 +1 +2 +3 +4 +5 +5 0 1 2 3 4 5 6 7 8 9 10 +4 -1 0 1 2 3 4 5 6 7 8 9 +3 -2 -1 0 1 2 3 4 5 6 7 8 +2 -3 -2 -1 0 1 2 3 4 5 6 7 +1 -4 -3 -2 -1 0 1 2 3 4 5 6 0 -5 -4 -3 -2 -1 0 1 2 3 4 5 -1 -6 -5 -4 -3 -2 -1 0 1 2 3 4 -2 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 -3 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 -4 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 -5 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 A negative number added to a negative number results in a negative number. (-a) + (-b) = -(a + b) Write down some more patterns that you can see. Movies of addition of directed numbers When we are adding and subtracting many directed numbers we sometimes put brackets around the negative numbers so that the example is easier to read. For Example, -6 + -7 - -9 + -21 - -4 - 3 + -12 is better written as, -6 + (-7) - (-9) + (-21) - (-4) - 3 + (-12) Grouping the number with its negative sign makes it clearer to see what is going on and leaves less chance to make mistakes. Example 6: movie - using the annihilation model, (-10) + 6 Example 7: movie - using the annihilation model, -14 + 11 + (-6) + (-7) + 5 Example 8: movie - using the number line model, 2 + 3 + (-3) = 2 Quick quiz 1. Using the annihilation model, or another model if you prefer, find answers to the following examples: a) 6 + (-3) b) -7 + 4 c) -8 + (-5) d) -12 + (-3) 2. Without using a model find answers to the following examples: a) -14 + 6 b) -8 + 10 c) -22 + (-13) d) -7 + (-14) e) -15 + 5 + (-8) + (-7) + 10 f) 13 + -9 + (-11) - 3 g)-22 + (-19) + 4 + (-6) + 2 h) -10 + (-11) + 1 + 2 - 3 3. Heavy rain in the catchment area of the local water reservoir has filled it during the winter. Since then, after a very hot summer, the water level dropped by 5 metres. Subsequent autumn rain has now raised the water level by 1 metre. What is the current level of the reservoir? 'Talking through' questions The 'talking through' questions and answers below have been provided to enable you to see how an 'expert' might tackle these questions. The annihilation model has been used in the explanations where appropriate. 1. Find the answer to - 4 + 9 This reads, negative four plus positive nine. Using the annihilation method, we have 4 negative chips and 9 positive chips. When 4 negative chips annihilate 4 of the positive chips we are left with 5 positive chips. So, - 4 + 9 = +5 or just 5 2. Find the answer to - 6 + (- 5) This reads, negative six plus negative five. Using the annihilation method, we have 6 negative chips and another 5 negative chips. Because we have no positive chips we cannot annihilate any chips. We just have a total of 11 negative chips. So, - 6 + (- 5) = - 11
Courses Courses for Kids Free study material Offline Centres More Store # Classify the following number as rational and irrational. $\sqrt {1.44}$. Last updated date: 13th Jun 2024 Total views: 383.7k Views today: 4.83k Verified 383.7k+ views Hint: Here, we will find the number whether rational or irrational. First we will find the square root of the given number and then we will use the definition to find whether the number is rational or irrational. A rational number is defined as a number which can be expressed in the form of fractions. An irrational number is defined as a number which cannot be expressed in the form of fractions. We are given a number $\sqrt {1.44}$to find whether it is a rational or irrational number. We will convert the decimal number into fraction. $\sqrt {1.44} = \sqrt {\dfrac{{144}}{{100}}}$ ……………….$\left( 1 \right)$ We will use factorization method, to find the square root of a number. $\begin{array}{l}2\left\| \!{\underline {\, {144} \,}} \right. \\2\left\| \!{\underline {\, {72} \,}} \right. \\2\left\| \!{\underline {\, {36} \,}} \right. \\2\left\| \!{\underline {\, {18} \,}} \right. \\3\left\| \!{\underline {\, 9 \,}} \right. \\3\left\| \!{\underline {\, 3 \,}} \right. \end{array}$ and $\begin{array}{l}5\left\| \!{\underline {\, {100} \,}} \right. \\5\left\| \!{\underline {\, {20} \,}} \right. \\2\left\| \!{\underline {\, 4 \,}} \right. \\2\left\| \!{\underline {\, 2 \,}} \right. \end{array}$ Now we can write 144 and 100 in form of the factors: $144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3$ $100 = 5 \times 5 \times 2 \times 2$ Now , $\sqrt {\dfrac{{144}}{{100}}} = \sqrt {\dfrac{{2 \times 2 \times 2 \times 2 \times 3 \times 3}}{{5 \times 5 \times 2 \times 2}}}$ $\Rightarrow \sqrt {\dfrac{{144}}{{100}}} = \dfrac{{2 \times 2 \times 3}}{{5 \times 2}}$ Multiplying the terms, we get $\Rightarrow \sqrt {\dfrac{{144}}{{100}}} = \dfrac{{12}}{{10}}$ Simplifying the terms, we get $\Rightarrow \sqrt {\dfrac{{144}}{{100}}} = \dfrac{6}{5}$ Since the number can be expressed as $\dfrac{p}{q}$, then the number $\sqrt {1.44}$ is a rational number. Therefore, the number $\sqrt {1.44}$ is a rational number. Note: First, we will count the number of digits after the decimal point. If there $n$ is the number of digits after the decimal point, then we have to multiply and divide ${10^n}$ to remove the decimal and to convert the decimal into fraction. A rational number can also be defined as the ratio of two integers. An irrational number can also be defined which cannot be expressed as the ratio of two integers. We might make a mistake by considering the given number as irrational because it is expressed as a square root. Before coming to a conclusion, we need to simplify the number and then if the number cannot be expressed in fraction then it is irrational.
# Jen takes 7 minutes more to complete an illustration than Jon. The total time taken by both of them is 6 hours. How d you form an algebraic expression to express this and identify the variable, constant and coefficient of the expression? Jun 9, 2016 $2 x + 7 = 360$ #### Explanation: Start by defining the time taken by one of the people and writing an expression using the information given.. It is easier to let $x$ be the smaller value. (Jon's time) Let $x$ be the time taken by Jon (in minutes). So, $x + 7$ is Jen's time. (Jen takes MORE time than Jon.) $x$ is the variable and 7 is the constant To form an equation, use the expressions we have written. The total time for both people is 6 hours. However, the unit of the 7 is minutes, so we need to make sure that the same unit is used. throughout. (Either change 6 hours to 360 minutes, or write the 7 as $\frac{7}{60}$ hours - minutes seems easier.) $x + x + 7 = 360$ $2 x + 7 = 360 \text{ 2 is the coefficient of } x$
Center Home -> Content Areas Home -> Math Home -> Project Activities -> Excel Activities -> Relationships Between Area and Perimeter Activity Description Activity Guide • Discuss the following scenario in small groups. Scenario: (from Ma, L. 1999. Knowing and teaching elementary mathematics: Teachers’ understanding of fundamental mathematics in China and the United States. Mahwah, NJ: Erlbaum, p. 84) Imagine that one of your students comes to class very excited. She tells you that she has figured out a theory that you never told the class. She explains that she has discovered that as the perimeter of a closed figure increases, the area also increases. She shows you this picture to prove what she is doing: • How would you respond to this student? • Is this student’s “theory” valid? After initial discussions in small groups, write a short journal entry about your reaction to this student and a brief mathematical discussion either validating or disconfirming this “theory.” Part 2: Exploring perimeter with a fixed area with manipulatives • With 12 square tiles, arrange the tiles (flat on a surface) so as to obtain the maximum perimeter. The tiles should touch completely side-to-side, not by corners. Examples include: • What is the maximum perimeter? What general observations did you make about how to make your figure have a larger perimeter? Is there more than one figure that can have the same maximum perimeter? Why or why not? • Restrict the figure to a solid rectangle. Find all possible rectangles with an area of 12 square tiles. Record the respective dimensions and perimeter for the rectangles. Which rectangle has the maximum perimeter? Why? • Given any fixed area for a rectangle, make a general statement about which dimensions maximize the perimeter. Part 3: Exploring area with a fixed perimeter using manipulatives • Consider the following problem: Farmer Ted wants to build a rectangular pigpen for his pigs. He has 24 meters of fencing left from his last project. What size rectangle should he build so that the pigs have the maximum amount of play area? • Use square tiles to model different-sized solid rectangular pigpens. Build all possible rectangular pigpens that can be enclosed with 24 meters of fencing. Record the corresponding dimensions and areas for those pigpens. Which pigpen gives the pig the most play area? • As the length of the rectangular pigpen changes, how does that affect the width and area of the pigpen? What patterns do you notice? Explore this relationship numerically and express this algebraically. Use this relationship to explain why the maximum area occurs with a square pigpen. • Given any amount of fencing, make a general statement about the dimensions of the rectangular pigpen that give the maximum area. • Open the Excel file called maxarea.xls. (You will need to ENABLE MACROS when prompted.) This spreadsheet allows the user to change the perimeter (P) and then vary the length of side of X using the scroll bars. As these values change, the yellow box will display the current values of P, X, Y, and the Area (A) of the rectangle. The rectangle and the data point (X, A) on the graph of the Length of side X vs. Area will also change accordingly. • Fix the perimeter (P) to 24 as in the pigpen problem. Then incrementally increase the length of side X from 1 to 12. Describe the changes that occur in the numerical, graphical, and geometrical representations. • Click on the Show All Points tab to display the next worksheet. What is the shape of the graph of all the data points (X, A)? Why does this make sense? Which Length of X gives the maximum value for A? Why? • How does this experience differ from the manipulative experience of physically building each rectangle? Discuss the benefits and drawbacks of the physical and spreadsheet approach to the pigpen problem. • Do the same exploration with the following perimeters and record the results in a table. Make a prediction of the dimensions and maximum area before using the spreadsheet to simulate the changes in the rectangle for each perimeter. • If Farmer Ted had 158 meters of fencing, predict what the dimensions would be of the rectangle with the maximum area. Explain how you made this prediction. Create a rule as a function of P to obtain the length of side X which gives the maximum area. Confirm this rule with several other examples. • Create a general equation for determining the Area of a rectangle given both a fixed P and length of side X. How does this equation relate to the shape of the graph in the spreadsheet? With this Area equation, use the first derivative to verify the value of X which gives the maximum value of A. How does this compare to the rule you created above (a function of P to obtain the length of side of X which gives the maximum area)? • Discuss the non-calculus and calculus based approaches to making this generalization, including students’ accessibility, advantages, and disadvantages. • Why is the maximum value for Area achieved when the first derivative is equal to zero? For a graphical exploration of this, use the following java applet: http://www.ies.co.jp/math/java/calc/doukan/doukan.html Part 6: Revisiting the Opening Scenario • Revisit the scenario and subsequent group discussions from Part 1. Now that you have explored the relationship between perimeter and area at several different levels and with various tools, how would you use the student’s “theory” to provide relevant classroom experiences if you were teaching: 1. A sixth grade class 2. An Algebra I course 3. A Calculus course Either discuss this in small groups or write an individual journal entry that addresses how you would approach this topic at each of the three levels, including justifications as to why you believe your approach is pedagogically and mathematically appropriate. Extensions: • Consider the generalization you made in Part 2 when a rectangle has a fixed area and you want to maximize perimeter. Create an interactive spreadsheet template that shows the numerical and graphical representations of this problem. Use scroll bars to fix the value of A and change the values of X, and create a graph of X vs. P. • View the following java applet: http://www.ma.utexas.edu/users/kawasaki/mathPages.dir/calculus.dir/max-min.dir/ problem1.html Discuss the benefits and drawbacks of using the applet or the spreadsheet template to explore the relationship between perimeter, area, and the length of side X. Back to Project Activities \| Back to Math Homepage Send questions or comments here.
# For what values of x is f(x)= 2x^3-9x concave or convex? Jan 6, 2017 $x = \pm \sqrt{\frac{3}{2}}$. Read on to see which is which. I try not to confuse myself over "concave" vs. "convex". Instead I think about it as concave up or concave down. The first derivative equals $0$ at an extremum, which may be either an inflection point, concave up, or concave down. It is the second derivative at each of these points that tells you which of these three they are. Positive, if concave up, and negative, if concave down. $f ' \left(x\right) = 6 {x}^{2} - 9$ (power rule; $\frac{d}{\mathrm{dx}} \left[{x}^{n} = n {x}^{n - 1}\right]$) $f ' ' \left(x\right) = 12 x$ For us to find where the extrema are: $0 = 6 {x}^{2} - 9$ $\implies {x}^{2} = \frac{9}{6} = \frac{3}{2}$ $\implies x = \pm \sqrt{\frac{3}{2}}$ And to find which one is concave up/down or an inflection point, we take $f ' ' \left(\pm \sqrt{\frac{3}{2}}\right)$. $f ' ' \left(\sqrt{\frac{3}{2}}\right) = 12 \left(\sqrt{\frac{3}{2}}\right) > 0$ $f ' ' \left(- \sqrt{\frac{3}{2}}\right) = 12 \left(- \sqrt{\frac{3}{2}}\right) < 0$ Thus, $f \left(x\right)$ is concave up (a minimum) at $x = \sqrt{\frac{3}{2}}$ and concave down (a maximum) at $x = - \sqrt{\frac{3}{2}}$. Indeed, $f \left(x\right) = 2 {x}^{3} - 9 x$ has features of ${x}^{3}$ and $- x$. We can expect it to look like a cubic function that has a $y = - x$ close to the origin, like so: graph{2x^3 - 9x [-10,10, -10.14, 10.13]}
##### ASVAB AFQT A great way to flex your algebra muscles is by tackling number sequence problems. On the Arithmetic Reasoning subtest on the ASVAB, these problems will usually require you to find consecutive numbers (odd or even) based on, for example, their sum. ## Practice questions 1. Two consecutive numbers have a sum of 109. What are the numbers? A. 64 and 65 B. 54 and 55 C. 53 and 54 D. 52 and 56 2. Three consecutive odd numbers have a sum of 21. What are the three numbers? A. 1, 3, and 5 B. 7, 9, and 11 C. 3, 5, and 7 D. 5, 7, and 9 1. The correct answer is Choice (B). Let x represent the first number, and let x + 1 represent the second number. Then create your equation: The first number is 54, and because the problem tells you that you need consecutive numbers (and because the second number is x + 1), you know the second number is 55. Sometimes on the ASVAB, running through the answer choices is a good idea. You can immediately rule out Choice (D) because the numbers aren't consecutive, and you can rule out Choices (A) and (C) because they don't add up to 109. 2. The correct answer is Choice (D). Let n represent the first odd number. Because the numbers are consecutive odd numbers, let n = 2 represent the second consecutive odd number, and then let n = 4 represent the third consecutive odd number. Create an equation and solve: The first number is 5, which tells you that the next odd numbers in the sequence are 7 and 9. On problems like this one, plugging in the likely answer choices may be faster. If they add up to the correct amount, you can choose quickly and move to the next problem. You have only 39 minutes to answer 16 questions on the Arithmetic Reasoning subtest of the CAT-ASVAB and 36 minutes to answer 30 questions on the paper-and-pencil version, so save time where you can.
## Linear Algebra and Its Applications, 4th Edition Multiply by l = $\frac{10}{2}$ = 5, and subtract to find 2x + 3y = 1 and -6y = 6. Pivots 2, -6. Our goal should be to form a triangular system of equations, such that in equation 1 all of our coefficients are non-zero, and in equation two our first coefficient is zero and second coefficient is non-zero. Start with your given system of equations: 2x+3y=1 10x+9y=11 Rewrite the equations as an augmented matrix: {2 3 \| 1} {10 9 \| 11} Multiply the first row vector of the matrix by 5 (the value of l), and subtract the first row vector from the second row vector to create a new value for the second row vector of your matrix. ($R_{2}$-$R_{1}$-->$R_{2}$) {10 9 \| 11} - {10 15 \| 5} = {0 -6 \| 6} Our new matrix is now: {2 3 \| 1} {0 -6 \| 6} Which means our system of linear equations is: 2x+3y=1 (0x)-6y=6 This system of linear equations is now triangular and has pivot points. As a reminder, a pivot point is the first non-zero entry in a row when dealing with elimination. This would make our pivot coefficients 2 and -6.
# Lesson 3: Changing Elevation ## 3.1: That's the Opposite 1. Draw arrows on a number line to represents these situations: 1. The temperature was -5 degrees. Then the temperature rose 5 degrees. 2. A climber was 30 feet above sea level. Then she descended 30 feet. 2. What’s the opposite? 1. Running 150 feet east. 2. Jumping down 10 steps. 3. Pouring 8 gallons into a fish tank. ## 3.2: Cliffs and Caves Explore the applet and then answer the questions. GeoGebra Applet Vxv48Gtz 1. A mountaineer is climbing on a cliff. She is 200 feet above the ground. If she climbs up, this will be a positive change. If she climbs down, this will be a negative change. 1. Complete the table. starting elevation (feet) change (feet) final elevation (feet) row 1 +200 +70 row 2 +200 -120 row 3 +200 +300 row 4 +200 +110 row 5 +200 +35.2 2. Select three rows from the table and draw a number line diagram for each. Include the starting elevation, change, and final elevation on each diagram. 2. A spelunker is down in a cave next to the cliff. If she climbs down deeper into the cave, this will be a negative change. If she climbs up, whether inside the cave or out of the cave and up the cliff, this will be a positive change. 1. Complete the table. starting elevation (meters) change (meters) final elevation (meters) row 1 -20 -15 row 2 -20 +130 row 3 -20 +61 row 4 -20 -5 row 5 -20 -50 row 6 -20 +27.5 row 7 -20 +73.5 2. Select three rows from the table and draw a number line diagram for each. Include the starting elevation, change, and final elevation on each diagram. 3. Complete the table and draw a number line diagram to represent each row in this table. starting elevation (meters) change (meters) final elevation (meters) row 1 +200 -200 row 2 -20 0 Find the sums. 1. $\text- 35 + (30+ 5)$ 2. $\text- 0.15 + (\text- 0.85) + 12.5$ 3. $\frac{1}{2} + (\text- \frac{3}{4})$ ## 3.4: School Supply Number Line Your teacher will give you a long strip of paper. Follow these instructions to create a number line. 1. Fold the paper in half along its length and along its width. 2. Unfold the paper and draw a line along each crease. 3. Label the line in the middle of the paper 0. Label the right end of the paper $+$ and the left end of the paper $-$. 4. Select two objects of different lengths, for example a pen and a gluestick. The length of the longer object is $a$ and the length of the shorter object is $b$. 5. Use the objects to measure and label each of the following points on your number line. $a$ $b$ $2a$ $2b$ $a + b$ $\text-a$ $\text-b$ $a + \text-b$ $b + \text-a$ 6. Complete each statement using <, >, or =. Use your number line to explain your reasoning. 1. $a$ _____ $b$ 2. $\text-a$ _____ $\text-b$ 3. $a + \text-a$ _____ $b + \text-b$ 4. $a + \text-b$ _____ $b + \text-a$ 5. $a + \text-b$ _____ $\text-a + b$ ## Summary The opposite of a number is the same distance from 0 but on the other side of 0. The opposite of -9 is 9. When we add opposites, we always get 0. This diagram shows that $9 + \text-9 = 0$. When we add two numbers with the same sign, the arrows that represent them point in the same direction. When we put the arrows tip to tail, we see the sum has the same sign. To find the sum, we add the magnitudes and give it the correct sign. For example, $(\text-5) + (\text-4) =\text - (5 + 4)$. On the other hand, when we add two numbers with different signs, we subtract their magnitudes (because the arrows point in the opposite direction) and give it the sign of the number with the larger magnitude. For example, $(\text-5) + 12 = +(12 - 5)$.
## 8.2 A Study in Unraveling ### A. The Number of Elements in each Unraveling #### Prior Within the simple Seed Equation are the instructions for unfolding our structure. Just as a seed unfolds into a tree or plant, this equation unfolds into a bigger equation, a Raveled and Furled Equation. Unraveling the Equation is no big thing. (Maybe Raveling is an unnecessary middle step. It could have easily been bundled up with Furling, but we wanted to look at the Furling/Unfurling operation all by itself.) But it is a big deal to Unfurl our Unfolded, Unraveled Equation. This section looks at a few reasons why. #### Background In the Root & Seed Notebook, we derived a seed equation, which had to first be unfolded. After the unfolding, we were left with terms that need to be unraveled and unfurled, in order to determine any of the directionals. Even for the 2nd Directional, shown below, a double, triple, and quadruple unraveling are necessary to determine specific values. The current study is to determine how many terms there are in each unraveling. #### Elements in Unravelings, 1 through 3 The number of terms in a single unraveling is, of course, 1, because it is the data itself. It is already unraveled. For a double unraveling the number of terms is N, because each piece of Data is necessary in determining the Decaying Average, which is a double unraveling or the data unfurled once divided by D. For a triple unraveling, the number of terms increases significantly. It is the sum of all the integers from 1 to N. #### Individual Coefficients increase in complexity with multiple Unravelings Let us mention at this point that there are still only N data points needed. If all the terms are combined then there are only N terms. However, the coefficient of each term increases in complexity with each additional unraveling. (See the terms below for an elementary directional.) #### The number of terms in a quadruple unraveling For a quadruple unraveling, which is necessary for the 2nd Directional, the number of terms is equal to the sum of all the integers from 1 to N, plus the sum of all the integers from 1 to N-1, plus and so forth down to one. The number of terms is incredible, with a subsequent increase in the complexity of the coefficients of the individual terms. #### A general expression for the number of terms in a multiple Unraveling The pattern is easy to see. Below is the general expression. It is easy to see that when N is even moderately large that the number of terms is out of control for even the simplest directional. #### A Fatal flaw for the content-based approach to individual derivatives Before proceeding any further, we would like to hammer on the content-based approach for individual derivatives in Living Data Streams. While the Seed Equation is simplicity itself, to make practical use of it requires so many unfoldings, unravelings and unfurlings, that the number of terms to be dealt is prohibitive. With the contextual based approach, only a handful of terms need be stored to find even the current 4th directional after 100 data points. With the content-based approach, 100 terms need to be stored to even calculate the Decaying Average, the zeroth directional. #### Context-Individual; Content-General We have beat upon the content-based approach to individual derivatives enough. We are going to let it go now. We are now going to look at some marvelous implications of the content-based approach on general levels. While the contextual approach is superior for individual derivatives, the content-based approach is superior to discover general characteristics of the foundations of the derivative approach to Live Data Streams. The content-based approach helps us understand the Nature of Derivatives, while the contextual approach helps us calculate the Derivatives, and give the individual derivatives meaning in the context of their occurrence. Meaning vs. Nature ### B. Dimensionality & Raveling #### Diagrams of Dimensionality: Q=1&2 Below are some diagrams of the number of elements in each Unraveling. When Q = 1, from Equation #5 above, we have a point, zero dimensions. When Q = 2 and N = 5, we have a series of 5 points, which we will connect in a line, 1 dimension. #### Q=3 In the third level of Unraveling, when Q = 3 and N = 5, we have a series of 5, 4, 3, 2, & 1 points. First we connect them into lines. Then we connect these lines into a 2 dimensional area. This area is a right triangle. #### Q=4 In the fourth level of Unraveling, when Q = 4 and N = 5, we have a series of 5, 4, 3, 2, & 1 points, plus a series of 4, 3, 2, & 1 points, plus a series of 3, 2, & 1 points, plus a series of 2 & 1 points, plus a series of 1 point. First we connect them into lines, then we connect these lines into triangles, finally we connect the triangles into a solid, 3 dimensions. This solid is a right triangular prism. #### Q=5 In the fifth level of Unraveling, when Q = 5 and N = 5, we have a solid, a right triangular prism of altitude & base 5, plus a similar prism with height and base of 4, also one of 3, & 2, & 1. If we could we would nest these solids within each other to suggest a four dimensional object. Instead we'll just show the solids. #### Q=6 In the sixth level of Unraveling, when Q = 6 and N = 5, we have a series of 5 four-dimensional objects, which we will show broken into their nested solids, right triangular prisms. If we could we would nest these four dimensional object within each other to suggest a five dimensional object. Instead we'll just show the solids. It is easy to get the idea. With each level of unraveling comes another level of dimensionality. #### Beware of Brain Puns Remember, however, that raveling and dimensionality are only connected in the construction of the individual directionals. If there are other connections, they are indirect. Here we are only discovering the number of elements in each unraveling and have found out that a dimensional analogy helps to visualize the rapid growth of elements in each Raveling ### C. Impact Diagrams #### Potential Impact Decreases Proportionally with Time We must also remember that although the number of points is 5, that their potential impact on the derivative is not equal, but varies according their proximity to the derivative. The below Diagram gives a sketch of their potential impact upon the zeroth derivative, the Decaying Average. It spirals inward, diminishing in impact with each successive layer. #### D = 3, N = 20: Impact Diagram With 20 Data points and a D of 3, we get the diagram below. Notice how the most recent data point, 20, represents a third of the potential impact while the other 19 points represent the remaining two thirds. #### D=10, N = 50, Potential Impact Diagram When D = 10, the most recent Data represents only 10% of the impact, while the rest of the Data added up yields 90% of the impact. The density of the Data Derivatives increases as D increases. D can only increase when N is sufficiently large. {See the Notebook, Building a Derivative System from a Discontinuous Data Stream.} So the greater that N, the number of Data points, is, the greater D can be and resultantly the greater the Data Density of the Derivatives can be. ### D. Fractional Dimensions #### Quantized Points Pretending to be a Line Let it be stated that the points are only pretending to be a line. They are quantized, into separate events. {See the Spiral Time Notebook for a fuzzy discussion of Events.} They are separate, individual. When merged into a point, which is the Decaying Average, which is a scalar magnitude, their influence is diminished in the moment, while spreading its influence over time. See Diagram. The line represents the event, while the curve represents its manifestation over time. The scalar of the line equals the scalar of the area under the curve. Immediate impact has been sacrificed for spatial continuity. #### Derivative Density It is a bunchy line when there are only 5 points. As N, the number of Data points, increases, the density of the Derivative increases. See Diagram below. So eventually, a bunch of points becomes a line? No actually, if enough data points are bunched closely enough together they appear as a line. So even though each data point exists separately, independent from the rest, broken up by the days and nights, they appear as a continuous whole, when viewed from a grand enough perspective. They appear to take on a continuous life of their own. It is this continuity that we refer to in the Notebook, Spiral Time. ### E. Dimensional Equations #### The Combination of Dimensional Numbers But as we've seen above, each of the levels is only pretending to be dimensional. Just as the initial points were only pretending to be a line by clustering together, so on each higher level do they pretend to be real by clustering together in vast amounts. What we're pointing out is that each individual derivative is only one number. But they are constructed in the manner shown above. The unfurled Data is a single point. #### A Linear Response The Decaying Average is a linear number based upon a line of points. Remember the more points the denser the line. Unfurling a piece of data gives it dimensionality. Unfurling diminishes the individual data, but combines it with the rest of the Data to give it linearity, and its first dimensionality. But remember it is only a logarithmic fractal response, because it is based upon Decay, hence it only achieves partial dimensionality. #### The Difference between Linear & Planar With each unfurling the data achieves another dimension, or at least partial dimensionality. The First Derivative, Directional or Deviational, is the difference between a linear number and a planar number. With each Unraveling comes another Unfurling and another Dimension. All of our Derivatives are based upon differences. The First Derivative is based upon the difference between the present linear fractal response and the most recent planar fractal response. #### The Difference between a Linear/Planar Difference and a Planar/Spatial Difference The 2nd Derivative, in Seed Equation terminology, is the difference between two differences: the difference between a linear and planar number and the difference between a planar and a spatial number. Combining terms, we get a more concise expression. In different terms, the 2nd Directional is equal to the difference between the sum of the linear and spatial dimensions of the data and twice the planar dimension. What does this mean? Who knows? We're just having fun. #### Introducing a 4th Dimensional Response In some ways, each Unraveling or Unfurling gives the Data another dimension. With the 3rd Directional a fourth dimensional response is introduced. First we show the 3rd Directional unfolded in dimensional terminology. We combine terms to compress the equation and reveal Pascal's Triangle hiding underneath. ### F. Collapsible Dimensional Numbers #### Interchangeable Dimensions or more relativity We have spoken about the Data as a point, the Decaying Average as a line, the 1st Directional as a plane, and the 2nd Directional as a solid. We could just as easily, maybe more so, have talked about the Data as a scalar line. Then the Decaying Average would be a plane, the 1st Directional a solid, and the 2nd Directional as a 4th dimensional solid. In some ways this would have been easier because our dimensions would have coincided with our Q numbers for Unraveling. When Q = 2, our measure would be constructed in the 2 dimensional plane and then collapsed into a 1 dimensional scalar line for comparison. This coincides with our view of the Decaying Average as constructed by a Spiral Square. Thus this expanded view of the Data as a scalar line rather than a point would have coincided with our geometric understanding of the construction of the measures. #### A Geometrical Understanding of the 3rd Unraveling, Q=3 The cube below is the geometrical representation of the 3rd Unraveling, Q=3. When viewed straight on as a square, each rectangle represents the potential impact of a Decaying Average upon the 3rd Unraveling. The largest represents the most recent Decaying Average. As the rectangles spiral in they represent the potential impact of previous Decaying Averages, more and more separated from the most recent. The potential impact of distant Decaying Averages shrinks to almost nothing. From discussions above we also know that each Decaying average can be represented as planar spiral of Data. So each Decaying Average rectangle can be constructed or visualized as dimensional extension into another plane. Remember that the most recent Decaying Average will have the same impact upon the 3rd Unraveling that the most recent Data will have upon the Decaying Average. But while the Data is made up of only one measure, the Decaying Average is made up of all the Data. So the cube below represents the construction of the 3rd unraveling, with each rectangular prism representing the influence of each Data Point. Similarly a tesseract, a 4th Dimensional cube, could be drawn to represent the construction of the 4th Unraveling. #### Collapsible Dimensions There are no real dimensions associated with our geometrical representations. They only represent the whole Impact. The square represents the whole impact. The cube represents the whole impact. The tesseract represents the whole impact. Each dimension can be collapsed to the dimension before along the lines of Unraveling. A second dimensional 3rd unraveling is equivalent to a 1 dimensional 2nd Unraveling, or a 3rd dimensional 4th unraveling. The above square can be viewed as a 6th Unraveling square with 5th level unraveling rectangles. The cube can be viewed as a 6th level unraveling with 4th level unraveling prisms. The other levels are locked up inside the prisms. We can collapse or expand our structures up and down to understand the basic construction of each Raveling measure. So whether we view the Data as points or scalar lines makes no difference. The Data as a point made more sense when exploring the number of terms making up a Raveling, because each piece of Data represented only one term. Viewing the Data as a line makes more sense when speaking about its impact as a scalar measure. It is irrelevant which view is taken because of the collapsibility of the dimensions in this scheme.
# Infinite sums of series If instead of adding only the $$n$$ first terms of a succession, we want to add them all, we will write: $$S=\sum_{n \geq 1} a_n$$$To indicate that we are adding all the terms from the first one. This sum $$S$$ is called series. If the succession which series we are calculating is a geometric progression, we can extend the formula: $$S_n=\dfrac{a_1\cdot (1-r^n)}{1-r}$$$ tending $$n$$ to infinity, two situations are possible, $$\left\{ \begin{array}{l} \mbox{si} \ r\leq 1 \Rightarrow r^n\rightarrow \infty \\\\ \mbox{si} \ r < 1 \Rightarrow r^n\rightarrow 0 \end{array} \right.$$$Then we have two choices: • In a geometric progression of ratio $$r \geq 1$$, the sums $$S_n$$ grow arbitrarily as the value of $$n$$ increases, and it is said that they tend to infinity, or that the series is divergent. • On the contrary, a geometric progression of ratio $$r < 1$$ the sums $$S_n$$ stabilize and they increasingly approach the quantity: $$S=\dfrac{a_1}{1-r}$$$ which is what we will call sum of the series. In this case we will say that the series is convergent. Continuing with the previous examples, $$\sum_{n \geq 1}a_n = \sum_{n \geq 1}3\cdot 2^{n-1}$$ it is divergent because it is a series of a geometric progression of ratio $$r=2 \geq 1$$, while the series $$\sum_{n \geq 1}b_n = \sum_{n \geq 1}\dfrac{7}{3^{n-1}}=\dfrac{b_1}{1-r}=\dfrac{7}{1-\dfrac{1}{3}}=\dfrac{21}{2}$$ is convergent since it is the series of a geometric progression of ratio $$r=\dfrac{1}{3} < 1.$$
If three cards are drawn without replacement, what is the probability of each subsequent card being larger than the previous card? Suppose we have a deck of $$500$$ cards numbered from 1 to 500. If the cards are shuffled randomly and you are asked to pick three cards (without replacement), one at a time, what's the probability of each subsequent card being larger than the previous drawn card? My solution: Let $$i$$ be the second card that is picked, then $$i-1$$ cards will be less that $$i$$ and $$500 - i$$ cards will be greater than $$i$$. Thus: P(subsequent card being larger than the previous card) $${\displaystyle = \sum_{i=1}^{500} \frac {(i -1)(500 - i)}{500 \cdot 499 \cdot 498}}$$ I'm not sure if my answer is correct. • Please read this tutorial on how to typeset mathematics on this site. Sep 26, 2019 at 14:32 Suppose you draw three cards. We don't care what they actually are... Now... without loss of generality, suppose the cards were $$1,2,3$$. Recognize that each of the six possible orders that you could have drawn them in were equally likely to occur: $$123,132,213,231,312,321$$. Exactly one of those six equally likely scenarios will result in the cards occurring in increasing order. The probability then: $$\frac{1}{6}$$ As an aside, your answer is correct., however is rather tedious to calculate directly without a calculator. We can use permutations and combinations to solve this problem. Number of ways of picking cards in increasing order will be 500 choose 3 (unordered) Total number of ways to pick 3 cards will be 500 permute 3 (ordered) $$P = {500C3\over 500P3}$$ $$P = {500C3\over 500C3}{\times}{1\over 3!}$$ $$P = {1\over 6}$$ • It seems the result is correct but I don't get why this works. Could you elaborate on why this method can solve the problem? Dec 15, 2019 at 20:06 • The total ways in which you could pick $3$ cards is $500P3$. Now when you choose any $3$ non-ordered cards, you can arrange them such that they are in increasing order. Also this arrangement is unique with respect to the $3$ cards chosen. Hence $500C3$ is the number of valid cases. – Sam Dec 17, 2019 at 14:50 An easier way to think about the problem using combination and permutation is the following: For any given three random distinct cards drawn from the pile, there is exactly one ordered combination such that it is in an increasing order while there is six possible ordered combinations of three cards. Hence, the probability of that three cards drawn are in an increasing order is 1/6. We can generalize this problem to any number of cards drawn -- If N cards drawn then the probability of interest is 1/P(N,N) = 1/(N!).
# Solve Two-Stage Equations (1) In this worksheet, students solve simple two-stage equations. Key stage: KS 3 Curriculum topic: Algebra Curriculum subtopic: Solve Linear Equations (One Variable) Difficulty level: ### QUESTION 1 of 10 When we solve algebraic equations, our aim is to end up with one letter on one side of the equals sign and one number on the other. This is the solution. We do this by using inverse operations to undo things that get in the way, but remember that we must do the same thing to both sides. Example 5b + 3 = 23 Subtract 3 from both sides. 5b + 3 - 3 = 23 - 3 Simplify 5b = 20 Divide both sides by 5 5b ÷ 5 = 20 ÷ 5 Simplify b = 4 Example b/5 - 3 = 17 b/5 - 3 + 3 = 17 + 3 Simplify b/5 = 20 Multiply both sides by 5 b ÷ 5 x 5 = 20 x 5 Simplify b = 100 Example Solve for a 5 - a = 10 5 - a + a = 10 + a Simplify 5 = 10 + a Subtract 10 from both sides. 5 -10 = 10 - 10 + a Simplify -5 = a a = -5 Solve for a: 3a + 4 = 19 a = 5 a = 23 a = 23/3 Solve for a: 3a - 4 = 5 a = 27 a = ⅓ a = 3 Solve for a: 7a + 2 = 30 a = 28 a = 4 a = 23 Solve for a: 4a - 2 = 22 a = 6 a = 5 a = 24 Solve for a: a/2 + 3 = 7 a = 20 a = 2 a = 8 Solve for a: a/3 - 7 = 2 a = 15 a = 3 a = 27 Solve for a: 5a - 8 = 92 a = 20 a = 95 a = 84 Solve for a: 3a - 2 = 190 a = 64 a = 189 a = 54 Solve for a: 15 + 5a = 20 a = -5 a = -1 a = 1 Solve for a: 15 - 5a = 20 a = -1 a = ½ a = 1 • Question 1 Solve for a: 3a + 4 = 19 a = 5 • Question 2 Solve for a: 3a - 4 = 5 a = 3 • Question 3 Solve for a: 7a + 2 = 30 a = 4 • Question 4 Solve for a: 4a - 2 = 22 a = 6 • Question 5 Solve for a: a/2 + 3 = 7 a = 8 • Question 6 Solve for a: a/3 - 7 = 2 a = 27 • Question 7 Solve for a: 5a - 8 = 92 a = 20 • Question 8 Solve for a: 3a - 2 = 190 a = 64 • Question 9 Solve for a: 15 + 5a = 20 a = 1 • Question 10 Solve for a: 15 - 5a = 20 a = -1 ---- OR ---- Sign up for a £1 trial so you can track and measure your child's progress on this activity. ### What is EdPlace? We're your National Curriculum aligned online education content provider helping each child succeed in English, maths and science from year 1 to GCSE. With an EdPlace account you’ll be able to track and measure progress, helping each child achieve their best. We build confidence and attainment by personalising each child’s learning at a level that suits them. Get started
# CK-12 Geometry: Perpendicular Bisectors in Triangles Save this PDF as: Size: px Start display at page: ## Transcription 1 CK-12 Geometry: Perpendicular Bisectors in Triangles Learning Objectives Understand points of concurrency. Apply the Perpendicular Bisector Theorem and its converse to triangles. Understand concurrency for perpendicular bisectors. Review Queue a. Construct the perpendicular bisector of a 3 inch line. Use Investigation 1-3 from Chapter 1 to help you. b. Find the value of. a. b. c. Find the value of and. Is the perpendicular bisector of? How do you know? Know What? An archeologist has found three bones in Cairo, Egypt. The bones are 4 meters apart, 7 meters apart and 9 meters apart (to form a triangle). The likelihood that more bones are in this area is very high. The archeologist wants to dig in an appropriate circle around these bones. If these bones are on the edge of the digging circle, where is the center of the circle? Can you determine how far apart each bone is from the center of the circle? What is this length? Perpendicular Bisectors 2 In Chapter 1, you learned that a perpendicular bisector intersects a line segment at its midpoint and is perpendicular. In #1 in the Review Queue above, you constructed a perpendicular bisector of a 3 inch segment. Let s analyze this figure. is the perpendicular bisector of. If we were to draw in and, we would find that they are equal. Therefore, any point on the perpendicular bisector of a segment is the same distance from each endpoint. Perpendicular Bisector Theorem: If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment. The proof of the Perpendicular Bisector Theorem is in the exercises for this section. In addition to the Perpendicular Bisector Theorem, we also know that its converse is true. Perpendicular Bisector Theorem Converse: If a point is equidistant from the endpoints of a segment, then the point is on the perpendicular bisector of the segment. Proof of the Perpendicular Bisector Theorem Converse Given: Prove: Statement is the perpendicular bisector of Reason 1. Given 2. is an isosceles triangle Definition of an isosceles triangle 3. Isosceles Triangle Theorem 4. Draw point, such that is the midpoint of. Every line segment has exactly one midpoint 5. Definition of a midpoint 6. SAS 7. CPCTC 8. Congruent Supplements Theorem 3 Statement is the perpendicular bisector of Reason Definition of perpendicular lines Definition of perpendicular bisector Let s use the Perpendicular Bisector Theorem and its converse in a few examples. Example 1: Algebra Connection Find and the length of each segment. Solution: From the markings, we know that is the perpendicular bisector of. Therefore, we can use the Perpendicular Bisector Theorem to conclude that. Write an equation. To find the length of and, substitute 8 into either expression,. Example 2: is the perpendicular bisector of. a) Which segments are equal? b) Find. c) Is on? How do you know? Solution: a) because they are both 15. because is the midpoint of 4 because is on the perpendicular bisector of. b) c) Yes, is on because (Perpendicular Bisector Theorem Converse). Perpendicular Bisectors and Triangles Two lines intersect at a point. If more than two lines intersect at the same point, it is called a point of concurrency. Point of Concurrency: When three or more lines intersect at the same point. Investigation 5-1: Constructing the Perpendicular Bisectors of the Sides of a Triangle Tools Needed: paper, pencil, compass, ruler 1. Draw a scalene triangle. 2. Construct the perpendicular bisector (Investigation 1-3) for all three sides. The three perpendicular bisectors all intersect at the same point, called the circumcenter. Circumcenter: The point of concurrency for the perpendicular bisectors of the sides of a triangle. 3. Erase the arc marks to leave only the perpendicular bisectors. Put the pointer of your compass on the circumcenter. Open the compass so that the pencil is on one of the vertices. Draw a circle. What happens? 5 The circumcenter is the center of a circle that passes through all the vertices of the triangle. We say that this circle circumscribes the triangle. This means thatthe circumcenter is equidistant to the vertices. Concurrency of Perpendicular Bisectors Theorem: The perpendicular bisectors of the sides of a triangle intersect in a point that is equidistant from the vertices. If, and are perpendicular bisectors, then. Example 3: For further exploration, try the following: a. Cut out an acute triangle from a sheet of paper. b. Fold the triangle over one side so that the side is folded in half. Crease. c. Repeat for the other two sides. What do you notice? Solution: The folds (blue dashed lines)are the perpendicular bisectors and cross at the circumcenter. Know What? Revisited The center of the circle will be the circumcenter of the triangle formed by the three bones. Construct the perpendicular bisector of at least two sides to find the circumcenter. After locating the circumcenter, the archeologist can measure the distance from each bone to it, which would be the radius of the circle. This length is approximately 4.7 meters. Review Questions Construction Construct the circumcenter for the following triangles by tracing each triangle onto a piece of paper and using Investigation 5-1. 6 Can you use the method in Example 3 to locate the circumcenter for these three triangles? 5. Based on your constructions in 1-3, state a conjecture about the relationship between a triangle and the location of its circumcenter. 6. Construct equilateral triangle (Investigation 4-6). Construct the perpendicular bisectors of the sides of the triangle and label the circumcenter. Connect the circumcenter to each vertex. Your original triangle is now divided into six triangles. What can you conclude about the six triangles? Why? Algebra Connection For questions 7-12, find the value of. bisector of. is the perpendicular 7 is the perpendicular bisector of. a. List all the congruent segments. b. Is on? Why or why not? c. Is on? Why or why not? For Questions 14 and 15, determine if is the perpendicular bisector of. Explain why or why not For Questions 16-20, consider line segment with endpoints and. 16. Find the slope of. 17. Find the midpoint of. 18. Find the equation of the perpendicular bisector of. 8 19. Find. Simplify the radical, if needed. 20. Plot, and the perpendicular bisector. Label it. How could you find a point on, such that would be the third vertex of equilateral triangle? You do not have to find the coordinates, just describe how you would do it. For Questions 21-25, consider with vertices and. Plot this triangle on graph paper. 21. Find the midpoint and slope of and use them to draw the perpendicular bisector of. You do not need to write the equation. 22. Find the midpoint and slope of and use them to draw the perpendicular bisector of. You do not need to write the equation. 23. Find the midpoint and slope of and use them to draw the perpendicular bisector of. You do not need to write the equation. 24. Are the three lines concurrent? What are the coordinates of their point of intersection (what is the circumcenter of the triangle)? 25. Use your compass to draw the circumscribed circle about the triangle with your point found in question 24 as the center of your circle. 26. Repeat questions with where and. 27. Repeat questions with where and. 28. Can you explain why the perpendicular bisectors of the sides of a triangle would all pass through the center of the circle containing the vertices of the triangle? Think about the definition of a circle: The set of all point equidistant from a given center. 29. Fill in the blanks: There is exactly circle which contains any points. 30. Fill in the blanks of the proof of the Perpendicular Bisector Theorem. Given: is the perpendicular bisector of Prove: Statement Reason is the midpoint of 3. Definition of a midpoint 4. and are right angles 9 Statement 5. Reason 6. Reflexive PoC Write a two column proof. Given: is a right isosceles triangle and is the bisector of Prove: and are congruent. 32. Write a paragraph explaining why the two smaller triangles in question 31 are also isosceles right triangles. Review Queue Answers a. Reference Investigation 1-3. a. b. b. Yes, is the perpendicular bisector of because it is perpendicular to and passes through the midpoint. ### CAIU Geometry - Relationships with Triangles Cifarelli Jordan Shatto CK-12 FOUNDATION CAIU Geometry - Relationships with Triangles Cifarelli Jordan Shatto CK-12 Foundation is a non-profit organization with a mission to reduce the cost of textbook materials for the K-12 ### Duplicating Segments and Angles CONDENSED LESSON 3.1 Duplicating Segments and ngles In this lesson, you Learn what it means to create a geometric construction Duplicate a segment by using a straightedge and a compass and by using patty ### 1. A student followed the given steps below to complete a construction. Which type of construction is best represented by the steps given above? 1. A student followed the given steps below to complete a construction. Step 1: Place the compass on one endpoint of the line segment. Step 2: Extend the compass from the chosen endpoint so that the width ### Geometry Chapter 5 Relationships Within Triangles Objectives: Section 5.1 Section 5.2 Section 5.3 Section 5.4 Section 5.5 To use properties of midsegments to solve problems. To use properties of perpendicular bisectors and angle bisectors. To identify ### 2. Construct the 3 medians, 3 altitudes, 3 perpendicular bisectors, and 3 angle bisector for each type of triangle Using a compass and straight edge (ruler) construct the angle bisectors, perpendicular bisectors, altitudes, and medians for 4 different triangles; a, Isosceles Triangle, Scalene Triangle, and an. The ### Lesson 2: Circles, Chords, Diameters, and Their Relationships Circles, Chords, Diameters, and Their Relationships Student Outcomes Identify the relationships between the diameters of a circle and other chords of the circle. Lesson Notes Students are asked to construct ### CK-12 Geometry: Parts of Circles and Tangent Lines CK-12 Geometry: Parts of Circles and Tangent Lines Learning Objectives Define circle, center, radius, diameter, chord, tangent, and secant of a circle. Explore the properties of tangent lines and circles. ### Lesson 5-3: Concurrent Lines, Medians and Altitudes Playing with bisectors Yesterday we learned some properties of perpendicular bisectors of the sides of triangles, and of triangle angle bisectors. Today we are going to use those skills to construct special ### Congruence. Set 5: Bisectors, Medians, and Altitudes Instruction. Student Activities Overview and Answer Key Instruction Goal: To provide opportunities for students to develop concepts and skills related to identifying and constructing angle bisectors, perpendicular bisectors, medians, altitudes, incenters, circumcenters, ### Student Name: Teacher: Date: District: Miami-Dade County Public Schools. Assessment: 9_12 Mathematics Geometry Exam 1 Student Name: Teacher: Date: District: Miami-Dade County Public Schools Assessment: 9_12 Mathematics Geometry Exam 1 Description: GEO Topic 1 Test: Tools of Geometry Form: 201 1. A student followed the ### Contents. 2 Lines and Circles 3 2.1 Cartesian Coordinates... 3 2.2 Distance and Midpoint Formulas... 3 2.3 Lines... 3 2.4 Circles... Contents Lines and Circles 3.1 Cartesian Coordinates.......................... 3. Distance and Midpoint Formulas.................... 3.3 Lines.................................. 3.4 Circles.................................. ### A geometric construction is a drawing of geometric shapes using a compass and a straightedge. Geometric Construction Notes A geometric construction is a drawing of geometric shapes using a compass and a straightedge. When performing a geometric construction, only a compass (with a pencil) and a ### Unit 2 - Triangles. Equilateral Triangles Equilateral Triangles Unit 2 - Triangles Equilateral Triangles Overview: Objective: In this activity participants discover properties of equilateral triangles using properties of symmetry. TExES Mathematics ### 39 Symmetry of Plane Figures 39 Symmetry of Plane Figures In this section, we are interested in the symmetric properties of plane figures. By a symmetry of a plane figure we mean a motion of the plane that moves the figure so that ### Isosceles triangles. Key Words: Isosceles triangle, midpoint, median, angle bisectors, perpendicular bisectors Isosceles triangles Lesson Summary: Students will investigate the properties of isosceles triangles. Angle bisectors, perpendicular bisectors, midpoints, and medians are also examined in this lesson. A ### Chapter 1: Essentials of Geometry Section Section Title 1.1 Identify Points, Lines, and Planes 1.2 Use Segments and Congruence 1.3 Use Midpoint and Distance Formulas Chapter 1: Essentials of Geometry Learning Targets I Can 1. Identify, ### GEOMETRY. Constructions OBJECTIVE #: G.CO.12 GEOMETRY Constructions OBJECTIVE #: G.CO.12 OBJECTIVE Make formal geometric constructions with a variety of tools and methods (compass and straightedge, string, reflective devices, paper folding, dynamic ### The Four Centers of a Triangle. Points of Concurrency. Concurrency of the Medians. Let's Take a Look at the Diagram... October 25, 2010. Points of Concurrency Concurrent lines are three or more lines that intersect at the same point. The mutual point of intersection is called the point of concurrency. Example: x M w y M is the point of ### Lesson 3.1 Duplicating Segments and Angles Lesson 3.1 Duplicating Segments and ngles In Exercises 1 3, use the segments and angles below. Q R S 1. Using only a compass and straightedge, duplicate each segment and angle. There is an arc in each ### Sec 1.1 CC Geometry - Constructions Name: 1. [COPY SEGMENT] Construct a segment with an endpoint of C and congruent to the segment AB. Sec 1.1 CC Geometry - Constructions Name: 1. [COPY SEGMENT] Construct a segment with an endpoint of C and congruent to the segment AB. A B C Using a ruler measure the two lengths to make sure they have ### Geometry Course Summary Department: Math. Semester 1 Geometry Course Summary Department: Math Semester 1 Learning Objective #1 Geometry Basics Targets to Meet Learning Objective #1 Use inductive reasoning to make conclusions about mathematical patterns Give ### Geometry Enduring Understandings Students will understand 1. that all circles are similar. High School - Circles Essential Questions: 1. Why are geometry and geometric figures relevant and important? 2. 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Any line segment ### New York State Student Learning Objective: Regents Geometry New York State Student Learning Objective: Regents Geometry All SLOs MUST include the following basic components: Population These are the students assigned to the course section(s) in this SLO all students ### Chapters 6 and 7 Notes: Circles, Locus and Concurrence Chapters 6 and 7 Notes: Circles, Locus and Concurrence IMPORTANT TERMS AND DEFINITIONS A circle is the set of all points in a plane that are at a fixed distance from a given point known as the center of ### Mathematics Geometry Unit 1 (SAMPLE) Review the Geometry sample year-long scope and sequence associated with this unit plan. Mathematics Possible time frame: Unit 1: Introduction to Geometric Concepts, Construction, and Proof 14 days This ### GEOMETRY COMMON CORE STANDARDS 1st Nine Weeks Experiment with transformations in the plane G-CO.1 Know precise definitions of angle, circle, perpendicular line, parallel line, and line segment, based on the undefined notions of point, ### The Geometry of Piles of Salt Thinking Deeply About Simple Things The Geometry of Piles of Salt Thinking Deeply About Simple Things PCMI SSTP Tuesday, July 15 th, 2008 By Troy Jones Willowcreek Middle School Important Terms (the word line may be replaced by the word ### Topics Covered on Geometry Placement Exam Topics Covered on Geometry Placement Exam - Use segments and congruence - Use midpoint and distance formulas - Measure and classify angles - Describe angle pair relationships - Use parallel lines and transversals ### 55 questions (multiple choice, check all that apply, and fill in the blank) The exam is worth 220 points. 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Therefore, If a parallelogram ### Tangents and Chords Off On a Tangent SUGGESTED LERNING STRTEGIES: Group Presentation, Think/Pair/Share, Quickwrite, Interactive Word Wall, Vocabulary Organizer, Create Representations, Quickwrite circle is the set of all points in a plane ### The Use of Dynamic Geometry Software in the Teaching and Learning of Geometry through Transformations The Use of Dynamic Geometry Software in the Teaching and Learning of Geometry through Transformations Dynamic geometry technology should be used to maximize student learning in geometry. Such technology ### The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Tuesday, August 13, 2013 8:30 to 11:30 a.m., only. GEOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY Tuesday, August 13, 2013 8:30 to 11:30 a.m., only Student Name: School Name: The possession or use of any communications ### Objectives. 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This lesson plan ### IMO Training 2008 Circles Yufei Zhao. Circles. Yufei Zhao. ircles Yufei Zhao yufeiz@mit.edu 1 Warm up problems 1. Let and be two segments, and let lines and meet at X. Let the circumcircles of X and X meet again at O. Prove that triangles O and O are similar. ### DEFINITIONS. Perpendicular Two lines are called perpendicular if they form a right angle. DEFINITIONS Degree A degree is the 1 th part of a straight angle. 180 Right Angle A 90 angle is called a right angle. Perpendicular Two lines are called perpendicular if they form a right angle. Congruent ### The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Thursday, August 13, 2015 8:30 to 11:30 a.m., only. 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The (degree) measure of ù DE is the measure of DOE. efinition: circle is the set of all points in a plane that are equidistant from a given point called the center of the circle. We use the symbol to represent a circle. The a line segment from the center ### Algebra Geometry Glossary. 90 angle lgebra Geometry Glossary 1) acute angle an angle less than 90 acute angle 90 angle 2) acute triangle a triangle where all angles are less than 90 3) adjacent angles angles that share a common leg Example: ### Geometry 1. Unit 3: Perpendicular and Parallel Lines Geometry 1 Unit 3: Perpendicular and Parallel Lines Geometry 1 Unit 3 3.1 Lines and Angles Lines and Angles Parallel Lines Parallel lines are lines that are coplanar and do not intersect. Some examples ### Geometry of 2D Shapes Name: Geometry of 2D Shapes Answer these questions in your class workbook: 1. Give the definitions of each of the following shapes and draw an example of each one: a) equilateral triangle b) isosceles ### The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Student Name: GEOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY Thursday, June 17, 2010 1:15 to 4:15 p.m., only Student Name: School Name: Print your name and the name of your ### Special Segments in Triangles HPTER 10 Special Segments in Triangles c GOL Identify the altitudes, medians, and angle bisectors in a triangle. You will need a protractor a ruler Learn about the Math Every triangle has three bases and ### Triangles. (SAS), or all three sides (SSS), the following area formulas are useful. Triangles Some of the following information is well known, but other bits are less known but useful, either in and of themselves (as theorems or formulas you might want to remember) or for the useful techniques ### Semester Exam Review. Multiple Choice Identify the choice that best completes the statement or answers the question. Semester Exam Review Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Are O, N, and P collinear? If so, name the line on which they lie. O N M P a. No, ### Geometry - Semester 2. Mrs. Day-Blattner 1/20/2016 Geometry - Semester 2 Mrs. Day-Blattner 1/20/2016 Agenda 1/20/2016 1) 20 Question Quiz - 20 minutes 2) Jan 15 homework - self-corrections 3) Spot check sheet Thales Theorem - add to your response 4) Finding ### The Inscribed Angle Alternate A Tangent Angle Student Outcomes Students use the inscribed angle theorem to prove other theorems in its family (different angle and arc configurations and an arc intercepted by an angle at least one of whose rays is ### 56 questions (multiple choice, check all that apply, and fill in the blank) The exam is worth 224 points. 6.1.1 Review: Semester Review Study Sheet Geometry Core Sem 2 (S2495808) Semester Exam Preparation Look back at the unit quizzes and diagnostics. Use the unit quizzes and diagnostics to determine which
Solve for r Solve for s ## Share 3r+6s=2t-4 Use the distributive property to multiply 3 by r+2s. 3r=2t-4-6s Subtract 6s from both sides. 3r=-6s+2t-4 The equation is in standard form. \frac{3r}{3}=\frac{-6s+2t-4}{3} Divide both sides by 3. r=\frac{-6s+2t-4}{3} Dividing by 3 undoes the multiplication by 3. r=\frac{2t}{3}-2s-\frac{4}{3} Divide -4+2t-6s by 3. 3r+6s=2t-4 Use the distributive property to multiply 3 by r+2s. 6s=2t-4-3r Subtract 3r from both sides. 6s=-3r+2t-4 The equation is in standard form. \frac{6s}{6}=\frac{-3r+2t-4}{6} Divide both sides by 6. s=\frac{-3r+2t-4}{6} Dividing by 6 undoes the multiplication by 6. s=\frac{t}{3}-\frac{r}{2}-\frac{2}{3} Divide 2t-4-3r by 6.
# Differentiation Sub Topics An important tools that is used in differential calculus is the derivative of a function. Let us assume the given function is denoted as y = f(x). If the independent variable changes from the value ‘x’ to some other value x + $\partial$ x then $\partial$x is called an increment of ‘x’. Similarly $\partial$y denotes an increment in ‘y’. If ‘x’ changes x + x then the function y = f(x) will change from ‘y’ to y + $\partial$y so that y +$\partial$ y = f( x +$\partial$ x ) $\partial$ y = f( x + $\partial$x ) – y $\partial$y = f( $\partial$x + x) – f(x) Divide both sides with Δx $\frac{\partial y}{\partial x}$ = $\frac{f(x+\partial k)-f(x)}{\partial k}$ Applying limit $\partial$ x tends to ‘0’ on both sides and If we denote $\lim_{\partial x \rightarrow 0}$ $\frac{\partial y}{\partial x}$=0 $\frac{dy}{dx}$=$\lim_{\partial x \rightarrow 0}$$\frac{f(x+\partial k)-f(x)}{\partial k}$ $\frac{dy}{dx}$ is called derivative of ‘y’ with respect to ‘x’. The derivative of f(x) is denoted as f’(x), D’. Differentiation is the process of finding the derivative of a function. ## Numerical Differentiation The method that is used to estimate the derivative of a function is known as Numerical Differentiation. The simplest way to estimate the derivative of a function is given by the formula If f ( x ) is a mathematical function, then it’s derivative is given by f(x)=$\lim_{h\rightarrow 0}$ $\frac{f(x+h)-f(x)}{h}$ ## Differentiation Formulas 1. Derivative of ( xn ) = n xn – 1 2. Derivative of ( ex ) = ex 3. Derivative of log( x ) = $\frac{1}{x}$ 4. Derivative of log( x ) to the base ‘a’ = $\frac{log a}{x}$ 5. Derivative of ( ax ) = ax * ln( a ) List of derivatives of trigonometric functions (i) Derivative of sin ($\theta$) is equal to cos ($\theta$) (ii) Derivative of cos ($\theta$) is equal to - sin ($\theta$) (iii) Derivative of tan ($\theta$) is equal to sec2 ($\theta$) (iv) Derivative of cot ($\theta$) is equal to – csc2 ($\theta$) (v) Derivative of sec ($\theta$) is equal to sec ($\theta$) . tan ($\theta$ ) (vi) Derivative of csc ($\theta$) is equal to - csc ($\theta$) . cot ($\theta$) ## Derivatives of Inverse Trig Functions (i) Derivative of sin-1 ( u ) is equal to $\frac{1}{\sqrt{(1 – u^{2})}}$ or $\frac{1}{(1 – u^{2} )^{\frac{1}{2}}}$ (ii) Derivative of cos-1 (u ) is equal to $\frac{- 1}{\sqrt{(1 – u^{2})}}$ or $\frac{-1}{(1 – u^{2} )^{\frac{1}{2}}}$ (iii) Derivative of tan-1 ( u ) is equal to $\frac{1}{(1 + u^{2})}$ (iv) Derivative of cot-1 ( u ) is equal to $\frac{-1}{(1 + u^{2})}$ (v) Derivative of sec-1 (u ) is equal to $\frac{1}{[\|u\| * \sqrt{(u^{2} - 1)}]}$ or $\frac{1}{[\|u\| (u^{2} - 1)^{\frac{1}{2}}]}$ (vi) Derivative of csc-1 (u) is equal to $\frac{-1}{[\|u\| * \sqrt{(u^{2} - 1)}]}$ or $\frac{-1}{[\|u\| (u^{2} - 1)^{\frac{1}{2}}]}$ ## Differentiation by Parts If ‘u(x)’ and ‘v(x)’ are differentiable functions of ‘x’ then the product ‘u * v’ is also differentiable such that $\frac{dy}{dx}$(uv) = u $\frac{dv}{dx}$ (v) + v $\frac{dv}{dx}$ (u) Let us see a worked out example that will clearly explain the concept of the product rule for derivatives Differentiation by parts Example ## Derivative of Implicit Functions If the variable ‘x’ and ‘y’ are connected by a relation of the form f(x, y) = 0 and it is not possible or convenient to express ‘y’ as a function of ‘x’ in the form y $\Psi$= (x), then ‘y’ is said to be an implicit function of ‘x’. To find the derivative of ‘y’ in such a case, we differentiate both sides of the given relation with respect to ‘x’, keeping in mind that the derivative of (y) with respect to ‘x’ is $\frac{d\Psi}{dx}$* $(\frac{dy}{dx})$. Example of Implicit Differentiation: ## Logarithmic Differentiation We have seen derivatives of the functions of the form [f (x) ]n, nf(x) where f(x) is a function of ‘ x ‘ and ‘ n ‘is a constant. Here we shall mainly discuss derivatives of the functions that are in the form of [f(x)]g(x) where f ( x ) and g ( x ) are functions of ‘x’. To find the derivative of this type of functions we proceed by applying logarithm on both sides and then differentiate. Let us understand the concept of logarithmic differentiation by the following examples Example of Logarithmic Differentiation: ## Differential Equation An equation involving one dependent variable and its derivatives with respect to one or more independent variables is called a differential equation. If a differential equation contains only one independent variable, then it is called an Ordinary differential equation and if it contains more than one independent variable then it is called a Partial differential equation. The exponent of x and y need not be an integer. Order of a Differential Equation: The Order of a differential equation is the order of the highest derivative occurring in it. First Order Differential Equation A first degree first order differential equation contains terms like dy/dx and some terms involving ‘x’ and ‘y’, a general first degree first order differential equation is of the form $\frac{dy}{dx}$= F ( x, y ), where F is a function of ‘x’ and ‘y’ ## Solving First Order Differential Equation If the given first order differential equation can be put in the form f (x ) dx + g ( y ) dy = 0 Then its solution can be obtained by integrating each term. This method of solving the first order differential equation is called variable separable method. First Order Differential Equation Example: Solution: We are given that, x dy – y dx = 0. This can be written as x dy = y dx $\left(\frac{dx}{x} \right )$ = $\left (\frac{dy}{y} \right )$ On integrating both sides with respect to ‘ x ‘, we get log (x) = log (y) + log ( c ) log (x) = log ( yc ), where ‘ c ‘ being an arbitrary positive constant. That is, x = yc is the required solution of the given first order differential equation. Second Order Differential Equation A second order differential equation will be of the form f(x)$\frac{d^y}{dx^2}$ + G(x) $\frac{dy}{dx}$ + H(x) y = P(x) Where F ( x ), G ( x ), H ( x ) and P ( x ) are continuous functions in ‘x’ ## Homogeneous Differential Equation Definition: A function f ( x, y ) is called a homogeneous function of degree ‘ n ‘, if f ( cx, cy ) = cn f(x, y) Definition of a homogeneous differential equation A first order first degree differential equation is expressible in the form $\frac{dy}{dx}$ = $\frac{f(x,y)}{g(x,y)}$ Where f ( x, y ) and g ( x, y ) are homogeneous functions of the same degree, then it is called a Homogeneous Differential Equation Let us write an algorithm to solve a Homogeneous Differential equation Where f ( x, y ) and g ( x, y ) are homogeneous functions of the same degree, then it is called a Homogeneous Differential Equation Let us write an algorithm to solve a Homogeneous Differential equation Algorithm: Step 1: Put the given differential equation in the form of $\frac{dy}{dx}$ = $\frac{f(x,y)}{g(x,y)}$ Step 2: Put y = vx and $\frac{dy}{dx}$ = v + x$\frac{dy}{dx}$ in the equation in step 1 and cancel out ‘x’ from the right hand side. Then the equation reduces to the form v + x$\frac{dy}{dx}$ = F ( v ) Step 3: Shift ‘v’ on right hand side and separate the variables ‘v’ and ‘x’ Step 4: Integrate both sides to obtain the solution in terms of ‘v’ and ‘x’ Step 5: Replace ‘v’ by ($\frac{y}{x}$) in the solution obtained in step 4 to obtain the solution in terms of ‘x’ and ‘y’.
# Chapter 2 : Polynomials Ncert maths solutions for class 10 chapter 2 Polynomials , serve as an invaluable companion for students embarking on their mathematical journey. This chapter introduces students to the world of polynomials. NCERT solutions for Chapter 2 are designed to simplify the learning process and foster a genuine appreciation for the elegance of polynomials in mathematics. ## Exercise 2.1 Polynomials 1. The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case. ### Solution: (i) The graph is parallel to x-axis and not cut it at any point hence the number of zeroes of p(x) is 0 . (ii) The graph intersects the x-axis at only one point hence the number of zeroes of p(x) is 1 . (iii) The graph intersects the x-axis at three points hence the number of zeroes of p(x) is 3 . (iv) The graph intersects the x-axis at two points hence the number of zeroes of p(x) is 2 . (v) The graph intersects the x-axis at four points hence the number of zeroes of p(x) is 4 . (vi) The graph intersects the x-axis at three points hence the number of zeroes of p(x) is 3 . ## Exercise 2.2 Polynomials 1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. Solutions: (i) x2–2x –8 ⇒ x2– 4x+2x–8 = x(x–4)+2(x–4) = (x-4)(x+2) x – 4 = 0 or x + 2 = 0 ⇒ x = 4 or x = -2 Therefore, zeroes of polynomial = 4, -2 Sum of zeroes = 4–2 = 2 = Again sum of the zeroes using formula = -(Coefficient of x)/(Coefficient of x2) = -(-2)/1 = 2 Product of zeroes = 4×(-2) = -8 Again product of zeroes using formula = (Constant term)/(Coefficient of x2) =-(8)/1 = -8 verify (ii) 4s2–4s+1 ⇒4s2–2s–2s+1 = 2s(2s–1)–1(2s-1) = (2s–1)(2s–1) Therefore, zeroes of polynomial = 1/2, 1/2 Sum of zeroes = (½)+(1/2) = 1 Again sum of the zeroes using formula = -(Coefficient of s)/(Coefficient of s2) = -(-4)/4 = 1 Product of zeros = (1/2)×(1/2) = 1/4 Again product of the zeroes using formula = (Constant term)/(Coefficient of s) = 1/4 verify (iii) 6x2–3–7x ⇒ 6x2–7x–3 = 6x– 9x + 2x – 3 = 3x(2x – 3) +1(2x – 3) = (3x+1)(2x-3) Therefore, zeroes of polynomial = -1/3, 3/2 Sum of zeroes = -(1/3)+(3/2) = (7/6) Again sum of the zeroes using formula = -(Coefficient of x)/(Coefficient of x2) = -(-7)/6 = 7/6 Product of zeroes = -(1/3)×(3/2) = -3/6 Again product of the zeroes using formula = (Constant term) /(Coefficient of x) = -3/6 (iv) 4u2+8u ⇒ 4u(u+2) Therefore, zeroes of polynomial = 0, -2 Sum of zeroes = 0+(-2) = -2 Again sum of the zeroes using formula = -(Coefficient of u)/(Coefficient of u2) = -(8/4) = -2 Product of zeroes = 0×-2 = 0 Again product of the zeroes using formula = (Constant term)/(Coefficient of u) = 0/4 verify (v) t2–15 ⇒ t2 = 15 or t = ±√15 Therefore, zeroes of polynomial equation t2 –15 are (√15, -√15) Sum of zeroes =√15+(-√15) = 0= -(0/1)= -(Coefficient of t) / (Coefficient of t2) Product of zeroes = √15×(-√15) = -15 = -15/1 = (Constant term) / (Coefficient of t) verify (vi) 3x2–x–4 ⇒ 3x2–4x+3x–4 = x(3x-4)+1(3x-4) = (3x – 4)(x + 1) Therefore, zeroes of polynomial = (4/3, -1) Sum of zeroes = (4/3)+(-1) = (1/3) Again sum of the zeroes using formula = -(Coefficient of x) / (Coefficient of x2) = -(-1/3) = 1/3 Product of zeroes=(4/3)×(-1) = (-4/3) Again product of the zeroes using formula = (Constant term) /(Coefficient of x) = (-4/3) verify 2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes, respectively. (i) 1/4 , -1 Solution: Let zeroes of the quadratic polynomial Sum of zeroes = α+β = 1/4 Product of zeroes = α β = -1 The quadratic polynomial equation can be written as:- x2–(α+β)x +αβ = 0 ⇒ x2–(1/4)x +(-1) = 0 ⇒ 4x2–x-4 = 0 (ii)√2, 1/3 Solution: Sum of zeroes = α + β =√2 Product of zeroes = α β = 1/3 The quadratic polynomial equation can be written as:- x2–(α+β)x +αβ = 0 ⇒ x2 –(√2)x + (1/3) = 0 ⇒ 3x2-3√2x+1 = 0 Thus, 3x2-3√2x+1 is the quadratic polynomial. (iii) 0, √5 Solution: Given, Sum of zeroes = α+β = 0 Product of zeroes = α β = √5 The quadratic polynomial equation can be written as:- x2–(α+β)x +αβ = 0 ⇒ x2–(0)x +√5= 0 Thus, x2+√5 is the quadratic polynomial. (iv) 1, 1 Solution: Given, Sum of zeroes = α+β = 1 Product of zeroes = α β = 1 The quadratic polynomial equation can be written as:- x2–(α+β)x +αβ = 0 ⇒ x2–x+1 = 0 Thus, x2–x+1 is the quadratic polynomial. (v) -1/4, 1/4 Solution: Given, Sum of zeroes = α+β = -1/4 Product of zeroes = α β = 1/4 Then the quadratic polynomial equation can be written directly as:- x2–(α+β)x +αβ = 0 ⇒ x2–(-1/4)x +(1/4) = 0 ⇒ 4x2+x+1 = 0 Thus, 4x2+x+1 is the quadratic polynomial. (vi) 4, 1 Solution: Given, Sum of zeroes = α+β =4 Product of zeroes = αβ = 1 Then the quadratic polynomial equation can be written directly as:- x2–(α+β)x+αβ = 0 ⇒ x2–4x+1 = 0 Thus, x2–4x+1 is the quadratic polynomial.
# Simplify (2x)2 • Last Updated : 06 Oct, 2021 Mathematics is not only about numbers but it is about dealing with different calculations involving numbers and variables. This is what basically is known as Algebra. Algebra is defined as the representation of calculations involving mathematical expressions that consist of numbers, operators, and variables. Numbers can be from 0 to 9, operators are the mathematical operators like +, -, ×, ÷, exponents, etc, variables like x, y, z, etc. ### Exponents and Powers Exponents and powers are the basic operators used in mathematical calculations, exponents are used to simplify the complex calculations involving multiple self multiplications, self multiplications are basically numbers multiplied by themselves. For example, 7 × 7 × 7 × 7 × 7, can be simply written as 75. Here, 7 is the base value and 5 is the exponent and the value is 16807. 11 × 11 × 11, can be written as 113, here, 11 is the base value and 3 is the exponent or power of 11. The value of 113 is 1331. Exponent is defined as the power given to a number, the number of times it is multiplied by itself. If an expression is written as cxy where c is a constant, c will be the coefficient, x is the base and y is the exponent. If a number say p, is multiplied n times, n will be the exponent of p. It will be written as p × p × p × p … n times = pn ### Basic rules of Exponents There are certain basic rules defined for exponents in order to solve the exponential expressions along with the other mathematical operations, for example, if there are the product of two exponents, it can be simplified to make the calculation easier and is known as product rule, let’s look at some of the basic rules of exponents, • Product Rule ⇢ an + am = an + m • Quotient Rule ⇢ an / am = an – m • Power Rule ⇢ (an)m = an × m or m√an = an/m • Negative Exponent Rule ⇢ a-m = 1/am • Zero Rule ⇢ a0 = 1 • One Rule ⇢ a1 = a ### Simplify (2x)2. Solution: As it is clearly seen, the entire problem statement is asking for a simplification using exponent rules, looking at the expression (2x)2, it is observed that the exponent 2 is the exponent for both 2 and x, therefore, simply apply the power for both 2 and x, (2x)2 = 22 × x2 = 4x2 Therefore, 4x2 is the value obtained. ### Similar Problems Question 1: Simplify 7(y1)5 Solution: It is observed that 1 is the exponent of y and 5 is the exponent of y1, and 7 is constant, using the power rule of exponents, it can be written as, Power Rule ⇢ (an)m = an × m 7(y1)5 = 7y(1 × 5) = 7y5 Question 2: Simplify 5(ex)2 Solution: As it is clearly seen, the entire problem statement is asking for a simplification using exponent rules, looking at the expression 5(ex)2, it is observed that x is the exponent of e and 2 is the exponent of ex, and 5 is constant, using the power rule of exponents, it can be written as, Power Rule ⇢ (an)m = an × m 5(ex)2 = 5(ex × 2) = 5(e2x) Question 3: Simplify 20(z6)0 Solution: It is observed that 6 is the exponent of z and 0 is the exponent of z6, and 20 is constant, using the power rule of exponents, it can be written as, Power Rule ⇢ (an)m = an × m 20(z6)0 = 20(z6 × 0) Applying Zero Rule ⇢ a0 = 1 = 20(1) = 20 My Personal Notes arrow_drop_up
Set Theory Set theory is a fundamental concept throughout all of mathematics. This branch of mathematics forms a foundation for other topics. Intuitively a set is a collection of objects, which are called elements. Although this seems like a simple idea, it has some far-reaching consequences. Elements The elements of a set can really be anything - numbers, states, cars, people or even other sets are all possibilities for elements. Just about anything that can be collected together may be used to form a set, though there are some things we need to be careful about. Equal Sets Elements of a set are either in a set or not in a set. We may describe a set by a defining property, or we may list the elements in the set. The order that they are listed is not important. So the sets {1, 2, 3} and {1, 3, 2} are equal sets, because they both contain the same elements. Two Special Sets Two sets deserve special mention. The first is the universal set, typically denoted U. This set is all of the elements that we may choose from. This set may be different from one setting to the next. For example, one universal set may be the set of real numbers whereas for another problem the universal set may be the whole numbers {0, 1, 2,… }. The other set that requires some attention is called the empty set. The empty set is the unique set is the set with no elements. We can write this as { } and denote this set by the symbol ∅. Subsets and the Power Set A collection of some of the elements of a set A is called a subset of A. We say that A is a subset of B if and only if every element of A is also an element of B. If there are a finite number n of elements in a set, then there are a total of 2n subsets of A. This collection of all of the subsets of A is a set that is called the power set of A. Set Operations Just as we can perform operations such as addition - on two numbers to obtain a new number, set theory operations are used to form a set from two other sets. There are a number of operations, but nearly all are composed from the following three operations: • Union - A union signifies a bringing together. The union of the sets A and B consists of the elements that are in either A or B. • Intersection - An intersection is where two things meet. The intersection of the sets A and B consists of the elements that in both A and B. • Complement - The complement of the set A consists of all of the elements in the universal set that are not elements of A. Venn Diagrams One tool that is helpful in depicting the relationship between different sets is called a Venn diagram. A rectangle represents the universal set for our problem. Each set is represented with a circle. If the circles overlap with one another, then this illustrates the intersection of our two sets. Applications of Set Theory Set theory is used throughout mathematics. It is used as a foundation for many subfields of mathematics. In the areas pertaining to statistics, it is particularly used in probability. Much of the concepts in probability are derived from the consequences of set theory. Indeed, one way to state the axioms of probability involves set theory.
# Lesson 9Solutions of Inequalities Let’s think about the solutions to inequalities. ### Learning Targets: • I can determine if a particular number is a solution to an inequality. • I can explain what it means for a number to be a solution to an inequality. • I can graph the solutions to an inequality on a number line. ## 9.1Unknowns on a Number Line The number line shows several points, each labeled with a letter. 1. Fill in each blank with a letter so that the inequality statements are true. a. _______ > _______ b. _______ < _______ 2. Jada says that she found three different ways to complete the first question correctly. Do you think this is possible? Explain your reasoning. 3. List a possible value for each letter on the number line based on its location. ## 9.2Amusement Park Rides Priya finds these height requirements for some of the rides at an amusement park. to ride the . . . you must be . . . High Bounce between 55 and 72 inches tall Climb-A-Thon under 60 inches tall Twirl-O-Coaster 58 inches minimum 1. Write an inequality for each of the three height requirements. Use for the unknown height. Then, represent each height requirement on a number line. High Bounce Climb-A-Thon Twirl-O-Coaster 2. Han’s cousin is 55 inches tall. Han doesn’t think she is tall enough to ride the High Bounce, but Kiran believes that she is tall enough. Do you agree with Han or Kiran? Be prepared to explain your reasoning. 3. Priya can ride the Climb-A-Thon, but she cannot ride the High Bounce or the Twirl-O-Coaster. Which, if any, of the following could be Priya’s height? Be prepared to explain your reasoning. 59 inches 53 inches 56 inches 4. Jada is 56 inches tall. Which rides can she go on? 5. Kiran is 60 inches tall. Which rides can he go on? 6. The inequalities and represent the height restrictions, in inches, of another ride. Write three values that are solutions to both of these inequalities. ### Are you ready for more? 1. Represent the height restrictions for all three rides on a single number line, using a different color for each ride. 2. Which part of the number line is shaded with all 3 colors? 3. Name one possible height a person could be in order to go on all three rides. ## 9.3What Number Am I? Your teacher will give your group two sets of cards—one set shows inequalities and the other shows numbers. Arrange the inequality cards face up where everyone can see them. Stack the number cards face down and shuffle them. To play: • Nominate one member of your group to be the detective. The other three players are clue givers. • One clue giver picks a number from the stack and shows it only to the other clue givers. Each clue giver then chooses an inequality that will help the detective identify the unknown number. • The detective studies the inequalities and makes three guesses. • If the detective cannot guess the number correctly, the clue givers must choose an additional inequality to help. Add as many inequalities as needed to help the detective identify the correct number. • When the detective succeeds, a different group member becomes the detective and everyone else is a clue giver. • Repeat the game until everyone has had a turn playing the detective.
# A race boat covers a distance of 60 km downstream in one and a half hour.It covers this distance upstream in 2 hrs. The speed of race boat in still water is 35 km/hr.Find the speed of the stream ? Nov 28, 2016 $5$ km/hr #### Explanation: As the speed of the boat in still water is known to us we can calculate the speed of the stream either any stream i.e. down stream or up stream. Say, the speed of the stream is $x$ km/hr. Going downstream the speed of the boat and the speed of the stream will be added together i. e. $\left(35 + x\right)$ km/hr Hence the boat goes $\left(35 + x\right)$ km in 1 hr. For a distance of 60 km, the Time $= \frac{60}{35 + x}$hr so, $\frac{60}{35 + x} = 1 \frac{1}{2} = \frac{3}{2}$ or, $3 \left(35 + x\right) = 2 \cdot 60$ [cross multiplication] or, $105 + 3 x = 120$ or $3 x = 120 - 105$ or, $x = \frac{15}{3} = 5$ Speed of the stream is 5 km/hr. Another way Going upstream, the speed of the stream is deducted from speed of the boat i.e $\left(35 - x\right)$ km/hr Hence goes $\left(35 - x\right)$ km in 1 hr. Therefore for a distance of $60$ km Time = $\frac{60}{35 - x}$ hrs. As per question, Time is 2 hours, so $\frac{60}{35 - x} = 2$ or $60 = 2 \left(35 - x\right)$ [ cross multiplication] or, $60 = 70 - 2 x$ or $2 x = 70 - 60$ or, $x = \frac{10}{2} = 5$ So, speed of the stream is $5$ km/hr. Dec 3, 2016 5 km/h #### Explanation: To solve this you only need one of the directions. That is; upstream or downstream. This is because you are given the speed of the boat in still water. Of you did not have the still water speed you would have to use both upstream and downstream. Using ratio but in fraction FORMAT For downstream we are given that: $\left(\text{distance")/("time}\right) \to \frac{60 k m}{1.5 h}$ But we need the distance in 1 hour (speed or velocity) $\left(\text{distance")/("time}\right) \to \frac{60 k m \div 1.5}{1.5 h \div 1.5} = \frac{40 k m}{1 h} \to 40 \frac{k m}{h}$ We are told that the speed of the boat in still water is $35 \frac{k m}{h}$ So the speed of the river has to be the difference between the two giving: $\left(40 - 35\right) \frac{k m}{h} = 5 \frac{k m}{h}$ You normally see this written as 5 km/h
1. ## Step by step Walking slowly down the down-moving escalator, it requires 50 steps to reach the bottom. Running up the escalator one step at a time, it takes 125 steps to reach the top. Assuming that running pace is five times as fast as walking pace (that is, five steps are taken at running pace in the same time interval as one step at walking pace), and that each trip was made at uniform speed, how many steps would be visible if the escalator stopped running? Alan 2. ## Re: Step by step I calculate that there would be <span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>75 </font color=yellow></span hi> steps visible. Solution: <span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>Let the velocity of of the excalator be Ve Let the Time to walk the escalator = Tw Let the Time to run the escalator = Tr Let the walking velocity = Vw Let the running velocity = Vr We know that: Vw = 50 /Tw Vr = 125/ Tr The length of the escalator is then: 50 + VeTw = 125 - VeTr And we are given that 5Vw = Vr 5 50 /Tw = 125 /Tr Rearranging: 2 * Tw = Tr Replacing in the other equation: 50 + VeTw = 125 - 2VeTw Rearranging: VeTw = 25 So length = 50 + 25 = 125 - 225 = 75</font color=yellow></span hi> Steve 3. ## Re: Step by step I calculate <span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>100</font color=yellow></span hi> steps. My solution: <span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>L = escalator length in steps Ve = escalator speed Vd = speed pacing down escalator, td = time taken for trip down Vu = speed running up escalator, tu = time taken for trip up Vu = 5 x Vd Trip down: L = (Ve + Vd)td, td = 50/Vd => L = 50(Ve/Vd + 1) => Ve = Vd(L/50 - 1) ...(A) Trip up: L = (Vu - Ve)tu = (5Vd - Ve)tu , tu = 125/Vu = 25/Vd => L = 125(1 - Ve/(5Vd)) ... substitute (A) -> => L = 125(1 - (L/50 - 1)/5) = 25(5 - L/50 + 1) = 25(6 - L/50) => 50L = 25(300 - L) => 2L = 300 - L => L = 100</font color=yellow></span hi> Alan 4. ## Re: Step by step <P ID="edit" class=small>(Edited by KTYorke on 08-Sep-03 12:37. to fix the hidden answer... NOTE: you can now use the "hide" tag to hide answers instead of a combination of "highlight" and "yellow" :-))</P>OOPS. You are correct. I made a math error: I went from <span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow> 5 50 /Tw = 125 /Tr and Rearranging: 2 * Tw = Tr It SHOULD have been: 2 * Tr = Tw !! Giving by replacing in the other equation: 50 + 2* VeTr = 125 - VeTr Rearrangement Rearranging: Ve*Tr = 25 So steps = 100 as you stated</font color=yellow></span hi> God I hate to make math errors! Steve #### Posting Permissions • You may not post new threads • You may not post replies • You may not post attachments • You may not edit your posts •
How do you differentiate y=x*sqrt(16-x^2)? Sep 26, 2015 $\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{2} / \sqrt{16 - {x}^{2}} + \sqrt{16 - {x}^{2}}$ Explanation: This is a product of 2 functions so we use the product rule which states that the derivative of the product of 2 functions is the first function times the derivative of the second, plus the second function times the derivative of the first. Inside this product rule we will also require the power rule which states that the derivative of a function to a power is the power times the function raised to 1 less than the given power, multiplied by the derivative of the function. Putting this all together, we get : $\frac{\mathrm{dy}}{\mathrm{dx}} = x \cdot \frac{1}{2} {\left(16 - {x}^{2}\right)}^{- \frac{1}{2}} \cdot 2 x + {\left(16 - {x}^{2}\right)}^{\frac{1}{2}} \cdot \left(1\right)$ $= {x}^{2} / \sqrt{16 - {x}^{2}} + \sqrt{16 - {x}^{2}}$ $\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{2} / \sqrt{16 - {x}^{2}} + \sqrt{16 - {x}^{2}} = {x}^{2} / \sqrt{16 - {x}^{2}} + \frac{16 - {x}^{2}}{\sqrt{16 - {x}^{2}}}$ $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{16}{\sqrt{16 - {x}^{2}}}$
# Lesson 1 Representaciones de fracciones (parte 1) ### Lesson Purpose The purpose of this lesson is for students to make sense of unit fractions with denominators 2, 3, 4, 5, 6, 8, 10, and 12, using physical and visual representations. ### Lesson Narrative In grade 3, students were introduced to fractions as numbers. They learned to name and represent fractions, to recognize simple equivalent fractions, and to compare fractions with like numerators and denominators (limited to 2, 3, 4, 6, and 8). They used fraction strips, area diagrams, tape diagrams, and number lines to support their reasoning with fractions. This lesson activates students’ prior knowledge of unit fractions and includes fractions with new denominators 5, 6, 10, and 12. Students revisit the meaning of numerator and denominator, name unit fractions, create representations for them, and recall some strategies and tools for reasoning about fractions. The idea of equivalence may naturally come up (and will help to prepare students for upcoming work), but it is not the focus of this lesson. • Engagement ### Learning Goals Teacher Facing • Make sense of the numerator and denominator of unit fractions that have denominators 2, 3, 4, 5, 6, 8, 10, and 12. • Use physical and visual representations to reason about fractions. ### Student Facing • Nombremos algunas fracciones y representémoslas visualmente. ### Required Materials Materials to Gather Materials to Copy • Fraction Strips ### Required Preparation Activity 1: • Each group of 2 needs 4 strips of equal-size paper (cut lengthwise from letter-size or larger paper or use the provided blackline master). Building On Building Towards ### Lesson Timeline Warm-up 10 min Activity 1 20 min Activity 2 15 min Lesson Synthesis 10 min Cool-down 5 min ### Teacher Reflection Questions What did you learn about each student and their foundational understanding of fractions based on their work today? ### Suggested Centers • Get Your Numbers in Order (1–5), Stage 3: Denominators 2, 3, 4, or 6 (Addressing) • Mystery Number (1–4), Stage 3: Fractions with Denominators 2, 3, 4, 6 (Supporting) ### Print Formatted Materials Teachers with a valid work email address can click here to register or sign in for free access to Cool Down, Teacher Guide, and PowerPoint materials. Student Task Statements pdf docx Lesson Cover Page pdf docx Cool Down Log In Teacher Guide Log In Teacher Presentation Materials pdf docx Blackline Masters zip
How to Compare Large Numbers # How to Compare Large Numbers How to Compare Large Numbers ### Rules for Comparing Large Numbers 1. If the number of digits in the numbers to be compared are different, the number having more number of digits is larger. For example, 8,35,27,725 > 92,41,785 [Since 92,41,785 has 7 digits and 8,35,27,725 has 8 digits]. 2. If both the numbers have the same number of digits, then compare the digits starting from the highest place value till the digits are different. For example, a. 5,32,47,819 < 8,21,53,458 because 5 < 8 b. 59,73,153 > 58,32,171 because 5 = 5 but 9 > 8 ## Ordering Large Numbers ### Ascending Order When numbers are arranged in the order of the smallest to the largest, they are in ascending order. Example: Arrange 67,64,645, 27,36,389, 54,53,927 and 1,08,74,739 in ascending order. Solution: Here, three numbers are 7-digit numbers and one number is an 8-digit number. So, the 8-digit number will be the largest number. Now, compare the digits with the highest place value among the other three numbers. Clearly, 2 < 5 < 6. Thus, the ascending order is 27,36,389 < 54,53,927 < 67,64,645 < 1,08,74,739. ### Descending Order When numbers are arranged in the order of the largest to the smallest, they are in descending order. Example: Arrange 78,68,974, 73,43,638, 73,42,903 and 79,55,822 in descending order. Solution: Arrange 98,68,974, 93,43,638, 93,42,903 and 99,55,822 in descending order. All the numbers are 7-digit numbers. For all the numbers, the digit at the highest place value is also same, that is, 7. So, compare the digits at the next place value. One number has 8, two numbers have 3 and one number has 9 as the digit. Clearly, 9 > 8 > 3. Hence, 99,55,822 is the largest number and 98,68,974 is the second largest number. Now, let us compare 93,43,638 and 93,42,903. In this case, the three left most digits are same. Compare the fourth digit from the left. Clearly, 3 > 2. Hence, 93,43,638 > 93,42,903 Thus, the descending order is 99,55,822 > 98,68,974 > 93,43,638 > 93,42,903. Related Topics: How to form the greatest and the smallest numbers using the digits? ---- Click Here!
# Fractions, Decimals, Percent, Ratio, and Proportion Development ## Fraction to decimal to percent connections Develop connections like: • 1/2, .5, .50, 50%; • 1/10, .1, .10, 10%,; • 1/4, .25, 25%; • 1/5, 2/10, .2, .20, 20% ## What are fractions? • Fractions are unsolved division problems. 3/5, three divided by five. • Fractions are parts of a whole. 3/5, three parts of a whole ... • Fractions are multiplication and division. 3/5 are a whole divided into fifths and one-fifth multiplied three times. A fraction is a relationship of parts to a whole expressed as division and multiplication with a numerator and denominator. ## Comparing fractions If fractions are relationships of parts to a whole expressed with multiplication and division, then how do we compare fractions? They must be compared by using the same whole. Here are three ways to compare fractions: 1. If the denominators are the same number, then the denominators are equal and the numerators can be compared to the same whole. For example: 2/3 compared to 1/3. Thirds are the same and two is twice one, therefore 2/3 > 1/3. 2. If the numerators are the same numbers, then the numerators are equal and the denominators can be compared to a whole. • Example 1: 2/3 compared to 2/5. The numerators are the same so compare thirds and fifths to a whole. Thirds are more than fifths, therefore, 2/3 > 2/5. • Example 2: Compare 1/3 to 1/4 to 1/2. The numerators are all the same so when the denominators are compared to a whole, fourths are smaller than thirds, and both are smaller than halfs. Therefore, 1/4 < 1/3 < 1/2. 3. Neither the numerators or denominators are the same. 2/3 & 4/5. They must be compared using the same whole so make the numerators or the denominators the same so they can be compared. • Example 1: Compare 2/3 with 4/5. Make the numerators the same, 4/6 & 4/5 and compare the denominators to the same whole. Fifths are more than sixths. Therefore, 4/5 > 4/6. • Example 2: Compare 2/3 with 4/5. Make the denominators the same, 10/15 & 12/15 and compare numerators. Ten fifteenths are less than twelve fifteenths. Therefore, 12/15 > 10/15. The whole matters because fractions are relations. ## Equivalent fractions Use multiples of one to divide or multiply • 1/2 * 2/2 = 2/4; • 2/4 / 2/2 = 1/2; Chck equivalent fractions by using a/b = c/d, therefore ad/b = c? • 1/2 = 2/4, therefore (1 * 4)/ 2 = 2 • 3/5 = 6/10, therefore (3 * 10)/5 = 6 ## Connecting multiplication and division to fractions. ### Partative 3/4, or three divided among or partitioned into four groups. (3/4 = 3/4). The equation seems like its saying the same thing so let's make it a more concrete. Three sandwiches shared with four people. Each person gets 3/4 of a sandwich. Three dollars shared with four people. Each person gets three quarters or 75 cents. Review with whole numbers. Partative is thought of as division where a set is put into a certain number of groups. For example 24 cookies are to be wrapped and put into lunch sacks for a team of eight. How many cookies can be wrapped for each player? (24 / 8 = 3) or (24 / ? = 8) Or if a set a is put into b groups, it results in c number in each group. c, is the quotient of a and b. (a / b = c ). (Quotient is the answer to a division problem) If divison is introduced as fair sharing it is more likely seen as division and division with multiplication when 3/4 is understood as 3 * 1/4. • A pie is cut into four equal pieces. One piece or 1/4 of the pie is removed. There are three equal pieces of one-four left. Three pieces are three times one piece, 3/4 = 3 * 1/4. ### Quotative and Measurement Quotative 3/4, or three out of four parts of the same one/whole are shaded. Can be thought of as both multiplication and division. One whole divided into four parts with one-fourth shaded three times. Three-fourths put into one group. Review with whole numbers. Quotative is thought of division where a set is put into groups. For example 12 sandwiches are packaged 3 sandwiches to a bag, resulting in 4 bags. Or Set a is put into groups the size of b, it results in a certain number of groups, c of size b being formed. c, is the quotient of a and b. (a / b = c ) Measurement When fractions are experienced as measurement/quotative (shaded parts of a whole) the idea of division and multiplication can be lost. Review with whole numbers. Measurement is thought of division where a value or set is used as a unit of measure to measure another group. If you want teams of five, and you have 20 kids, then how many teams will be needed? Or where set a is measured by the size of b, it results in a certain number of groups, c of size b. c, is the quotient of a and b. (a / b = c ) Sample problems 3 1/3 / 1/2 = 6 2/3 Solution 1: Quotative This procedure is easy to memorize. However, it is much harder to model and explain how it works. 1. Divide 3 1/3 into groups of 1/2. How many groups? 2. Three whole in half to make 6, 3. The 1/3 is less than 1/2. 4. So the answer is between 6 and 7. 5. How do we find how much 1/3 is of 1/2? 6. Draw a unit of one, divide it into thirds and mark 1/3 and 2/3. 7. Then divide the same unit of one in half and mark 1/2. 8. Notice that the 1/2 of the whole divides the middle third in half. Which tells us that piece is 1/3 of the 1/2 of the whole. This gets tricky so make sure you argree with it because you are going to need to use it soon. 9. This next part is the tricky part. We want to know how much of the 1/2 of the whole does the 1/3 of the whole represent? Well we just saw what 1/3 of to 1/2 of the whole was so the 1/3 of the whole is twice it or 2/3. 10. Therefore, 3 1/3 divided into halves will result in 6 2/3 halves. 11. Another way of thinking about it is the thirds of 1/2 are 1/6 of one whole. Therefore, 2/6 of one = 1/3 but the 2/6 of one is 2/3 of the 1/2. This is fairly confusing, but is easier seen with diagrams. Solution 2: Measurement Or how many 1/2 pieces are in 3 1/3? This way takes a 1/2 piece and measures the 3 1/3. Again pretty easy for the 3 (six pieces of 1/2). But when the 1/2 is placed beside the left over 1/3 it becomes harder to see. However, a picture as described above should show that the 1/2 would measure 2/3 of the 1/2. Solve this one both ways: 3 1/3 / 1/3 (=10) Sample problem 2 2/3 / 1/4 1. Ask, how many fourths in two two-thirds? (10). 2. Can measure with 1/4. Four fourths in each one. For a total of 8., but how many fourths in two-thirds? Solve with decimals 1. 2/3 = .66 2/3, and .66 2/3 = .25 + .25 + .16 2/3, and 2/3 = 1/4 +1/4 + 1/6, because .16 2/3 is 2/3 of .25, because .08 1/3 + .08 1/3 + .08 1/3 = .25, 2. But what is it of one? (.16 2/3) / 1 or (16 2/3) / 100 = 1/6, or how many 1/4 in 1/6 is the same as 1/6 / 1/4 = 2/3. Or 1/4 (4) + 1/4 (4) + 1/6 (4) = 1 + 1 + 2/3. The answer (2 2/3) is in relation to 1/4 of an inch (first whole to consider), but there is also the inch (second whole to consider). Another way to think of it is 2/3 / 1/4, find a common multiple /denominator of three and four and get, 8/12 / 3/12, then ask how many 3/12 (1/4) in 8/12 (2/3), 3/12 + 3/12 + 2/12, these three numbers represent the one whole to consider 2/3 put into 2 piles of 1/4 (3/12) and a leftover pile 2/12, To find the value in relationship to the second whole (one) we multiply all three by 4. 4 * 3/12 + 4 * 3/12 + 4* 2/12, and get 12/12 + 12/12 + 8/12, that is 2 2/3. ## Multiplication. As in division of fractions, multiplication or fractions requires that two wholes be consider for each problem: A relation of a relation. Example 1/6 of 1/2, (1/12) 1. Asks what is 1/6 of one-half 2. Need to consider one whole to take one-half of. 3. Need to consider 1/2 as one whole to take 1/6 of. 4. Then finally need to consider 1/12 as part of one whole that was used to start. ## Decimals and percent Decimals and percentage are specific equivalents of fractional relations. Decimals are fractional base-ten equivalents using place value. Percents are relationships based on a one-hundred-part whole. In addition and subtraction of fractions there is one whole that needs to be considered. Distributive and quotative division models each are likely to bring different ideas. Sample problem A group of six people have five candy bars and want to share them equally. How much would each person get? a ratio of candy bars to children. Cut each candy bar into six pieces and pass them out to five people (distributed). Each person gets five pieces or 5 x 1/6 or 5/6. Partitive take five candy bars and cut three in half. If each person takes half there will be two pieces left. Take the two pieces and cut each into thirds. So each person gets 1/2 + 1/3 or 5/6. Sample problem I used 2/3 of a can of paint to cover 1/2 of the porch floor. How much paint will be needed for the whole floor? It is a ratio of 2/3 can to 1/2 floor. It is hard to see that it is distributive since there is only part of a group (the floor). ## Measurement quotative division Sample problem How many bags of cookies can be make with 12 cookies and three bags? Shade three of four parts and label. Starting with a quotative model is not the place to start. It does not emphasize the whole and the relationship to the whole is missed. Sample problem John is baking a cake and only has a 1/2-cup measuring tool. The recipe calls for 2/3 of a cup of flour. How many times should he fill the 1/2-cup? This is quotative, not just because it involves measurement. It asks how many 1/2 cups fit in 2/3rds a cup. This is more difficult for children to understand and is usually done by making common denominators. Eventually students should link the two ideas and see they are the same. Five bars divided by six kids is 5 x 1/6; The 5/6 mark on a strip is also 5 x 1/6. That shows how multiplication and division are related. Fractions are multiplication and division. Sample problem Time on a clock. Mark a clock with fourths, thirds, fifths, sixths, and tenths. Try adding different fractions. 1/3 + 1/4 is how many minutes plus how many minutes? 1/3 (20 minutes) + 1/4(15 minutes) = (35 minutes) 7/12. Or 1 1/4 - 2/3; 1:15 back 40 minutes to 35 or 7/12. Sample problem Chris told Pat that s/he was .8 of the way to their goal of each saving a certain amount of money Pat said that s/he had only .5 of what Chris had. What part of their goal did Pat have? Arrays make a ten by ten grid and shade .4 of .8. Can also show that 1/2 * 4/5 is the same by making a two by ten array. Algorithms vs. number sense 6/16 * 8/18 (I would switch the numbers and reduce (1/2 * 1/3 = 1/6.) Chinese teachers would decompose the numbers to expanded notation and multiply the parts like 6 * 8 * 1/16 * 1/18. American teachers teach the algorithm (multiply top numbers and bottom numbers) without conceptual understanding and treat errors as procedural errors of digit manipulation. Use arrays to multiply. 3/5 * 5/6 Show in arrays three ways. 3x5 and 5x3 showing 3 * 1/5 with 5 * 1/6 shaded to show four different parts of problem and whole. Then switch the directions of the two numbers and repeat. Then show the answer (15) in relation to the whole (30). Write problems as 2 * 3 * 1/3 * 1/5 for 2/3 * 3/5 … ## Use distributive property Problem 3 1/2 * 14; 1. Think of the problem as (3 + 1/2) * 14; then 2. (3 * 14) + (1/2 * 14); 3. 42 + 7; 4. 49 Arrang dots in a triangular pattern with Three dots in the top row ... for a total of 24. Can do different multiplication facts and fractions 1/8, 3/24, 24/8, 1/4, 1/6, 6 * 4, 8 * 3,…3/8, It’s like the three is one and the eight is the whole (8 groups of 3 made from the 24). It's really cool how multiplication and division helps us with fractions. (It's the relationship.) May want to support younger students with grouping of numbers more. E.g. placing four 8-dot dominoes in a rectangular pattern for the same kind of problem (4 * 8) Or three groups of pennies in clusters of six (die pattern). ## Using time and the clock for problems Problems 1. My son suggested that a good way to exercise is to jog for 1/3 of an hour and walk for 1/4 of an hour. How much time is that altogether? 2. Or I could walk for 1/3, jog 1/2, and walk for 1/3. How much time? 3. Or 1/4, 1/3, 1/4… Write as reduced fraction and fraction with denominator of 60, could make ratio table. ## Clock chart or pie chart. Again 1/2 + 1/6 = 2/3, 30/60 + 10/60 = 40/60 = 4/6… Would the clock be useful for sevenths? Make a list of fractions for which the clock would be useful. ## Lines 1/4 + 1/5 Lets say that we are going on an imaginary trip. What would be a good distance to use? 100 Draw a line and put 0 and 100 at the ends. Why 100? I thought of 1/4 as 25 and 1/5 as 20. Put on double number line (fractions and whole numbers). How can we add it? 25/100 + 20/100 = 45/100. I would make the track 20 miles. Again make double number line and put both sets of fractions on both lines. 1/4, 9/20,… 25/100, 45/100,… What about 10? 2.5 is 1/4 because 5 is 1/2, 2 is 1/5 because 2 * 5 = 10, therefore 2.5/10 + 2/10 = 4.5/10. Put on line with 9/20 and 45/100 2/4 + 1/5 Do with 100, 10, 20, ## Open arrays Are arrays too easily constructed procedurally so that students do not see the conceptual? A person really has to think to see the fractional relationship (an array within an array). 1/3 * 1/4 (array 3 x 4) 2/3 * 1/4 (use same array, is it twice the other? Why?) 2/3 * 3/4 (How does this compare with the two preceding?) 4/3 * 3/2 Make a 3 x 4 array 2 x 3 would be the whole… Might try 1/2 of 4/3 first, then 2/2 * 4/3, then 3/2 * 4/3 Then do these and transition to swapping numerators and denominators 1/5 * 1/7 3/5 * 4/7 Swapping numerators and denominators 4/5 * 3/7 Or 4 * 3 * 1/5 * 1/7 If only an array is used students will notice that the inside array and the outside array are the same, only rotated 90 degrees, and see that multiplication is commutative (doesn't matter if multiply 1/2 * 2/3 or 2/2 * 2/3) but not understand multiplication of fractions conceptually. 3/8 * 4/9 5/6 * 3/5 4/5 * 5/8 Use to find when swapping is a useful strategy. ## Getting rid of a fraction or decimal or percent Multiplication of fractions • 3 1/2 * 18 (double to get rid of the fraction and halve to maintain equality); 7 * 9 = 63 • 3 1/4 * 28 (multiply by four to get rid of the fraction and divide by four to maintain equality); 13 * 7 = 70 + 21 = 91 • 3 1/2 * 14; (multiply by 2 and divide by 2) (3 1/2 * 2) * (14 / 2); 7 * 7 = 49. • 2 1/4 * 16 (multiply and divide by 4) 9 * 4 = 36 • 3 1/5 * 45 (multiply and divide by 5) = 16 * 9 ; 8 * 18; 4 * 36; 2 * 72; 144 • 3 1/5 * 50 Division Remember this earlier problem? See quotative and measurement above. 1. 3 1/3 / 1/2 (multiply both factors by two or double each factor) 2. 6 2/3 / 2/2 3. 6 2/3 / 1 4. 6 2/3 Percentage Chris wanted to buy a leather coat for \$350. Mom said she would help but Chris would have to pay 80%. How much would Chris have to pay? • .8 * 350 (multiply and divide by 10) 8 * 35 then (multiply by 2 and divide by 2) 4 * 70, and again 2 * 140, one last time 280 • Or 4/5 * 350, (multiply and divide by 5) 4 * 70 = 280 • Or 4/5 * 350 = 4 * (1/5 * 350) = 4 * 70 = 280 ## Developing strategies for computation with decimals All strategies for whole numbers can work with decimals. ## Friendly numbers 71.87 + 28.2 = 72.07 + 28 (compensation subtract .2 and add .2) 71.87 + 28.2 = 71.07 + 29 (compensation subtract .8 and add .8) 71.87 + 28.2 = 72 + 28 + .07 ## Using money .20 * 9 = (5 * .20 = 1.00 and 4 * .20 = .80) 1.80 .20 * 9 = (.20 * 10 - .20) 1.80 .25 * 9 16 * .25 = 4 How many students would know that a 4x4 array of quarters is 4 dollars? ## Using fractions and decimals interchangeably 75 * 80 = 3/4 * 80 * 100 ( because 75 was treated as 75/100, divide 100 need to multiply 100) 1/4 * 80 .25 * 80 25 * 80 1/2 * 60 .5 * 60 .50 * 60 .50 * .60 ## Sample problems Try to use different strategies to solve these: 1/3 * 1/4 2/3 * 1/4 2/3 * 3/4 1/5 * 1/7 3/5 * 4/7 4/5 * 3/7 3/8 * 4/9 5/6 * 3/5 4/5 * 5/8 6 * 10 12 * 5 24 * 2 1/2 8 * 30 16 * 15 32 * 7 1/2 64 * 3 3/4 18 * 5 1/2 9 * 11 4 1/2 * 22 14 * 3 1/2 4 / 1/2 8 / 1 16 / 1/4 32 / 1/2 64 / 1 5 1/2 / 1/3 16 1/2 / 1 2 1/2 / 1/5
# Please explain $4z + 2 = 2z + 1$ I'm doing this course online and unfortunately it did not explain the steps. Find z if: $$x = 2z + 1, \quad x = y, \quad y = 4z + 2.$$ The solution: $4z + 2 = 2z + 1$ $4z − 2z + 2 = 2z − 2z + 1$ //why did the teacher put 4z on one side and 2z on the other? $2z + 2 = 1$ $2z + 2 − 2 = 1 − 2$ $2z = − 1$ $2z/2 = −1/2$ $z = −1/2$ Can someone try to explain how this works? Sorry if im stupid but I have not done math in forever. thanks • Since $y=x$ you basically have $z$ in terms of $y$ in two different ways. So you can set them equal to each other and then you have an equation just in $z$ that you can then solve for $z$ by getting all $z$ terms to one side and everything else to the other side. – Gregory Grant Jan 5 '16 at 7:23 • Exactly where is your problem? Any equality, such as $4z+2=2z+1$, says that two things or expressions are equal. When you do the same operation on both sides of the equality sign, the results will be equal to each other. By choosing the operations wisely, you end up with an equality saying that $z$ is equal to some number. – Per Manne Jan 5 '16 at 7:44 You have: \begin{align} 4z+2&=2z+1\\ 4z−2z+2&=2z−2z+1& \text{why did the teacher put 4z on one side and 2z on the other?} \end{align} It's like this, assuming you can see colors. \begin{align} \color{blue}{4z+2}&=\color{blue}{2z+1}\\ \color{blue}{4z}\color{red}{−2z}\color{blue}{+2}&=\color{blue}{2z}\color{red}{−2z}\color{blue}{+1}& \text{the red terms were added to each side} \end{align} Answering your question in the side comment, the $4z$ and $2z$ in the second line are just coming from the first line. It's the instances of ${}-2z$ that are new in the second line. We are given that $x=2z+1$, $y=4z+2$ and $x=y$ so lets rearrange and just write $x=2z+1$ and $x=4z+2$ ( since $x=y$) and so $x=2z+1=4z+2=y$ $4z+2=2z+1$ $2z+2=1$ $2z=-1$ $z=\frac{-1}{2}$ $x=y=4z+2=4(\frac{-1}{2})+2=2$ We basically just use that we are given that x=y, and from that we can simplify the equations
# What is 122/99 as a decimal? ## Solution and how to convert 122 / 99 into a decimal 122 / 99 = 1.232 Convert 122/99 to 1.232 decimal form by understanding when to use each form of the number. Decimals and Fractions represent parts of numbers, giving us the ability to represent smaller numbers than the whole. But in some cases, fractions make more sense, i.e., cooking or baking and in other situations decimals make more sense as in leaving a tip or purchasing an item on sale. After deciding on which representation is best, let's dive into how we can convert fractions to decimals. ## 122/99 is 122 divided by 99 Teaching students how to convert fractions uses long division. The great thing about fractions is that the equation is already set for us! Fractions have two parts: Numerators on the top and Denominators on the bottom with a division symbol between or 122 divided by 99. We use this as our equation: numerator(122) / denominator (99) to determine how many whole numbers we have. Then we will continue this process until the number is fully represented as a decimal. Here's 122/99 as our equation: ### Numerator: 122 • Numerators sit at the top of the fraction, representing the parts of the whole. Any value greater than fifty will be more difficult to covert to a decimal. The good news is that 122 is an even number which can simplify equations (sometimes). Values closer to one-hundred make converting to fractions more complex. Now let's explore the denominator of the fraction. ### Denominator: 99 • Unlike the numerator, denominators represent the total sum of parts, located at the bottom of the fraction. Larger values over fifty like 99 makes conversion to decimals tougher. But 99 is an odd number. Having an odd denominator like 99 could sometimes be more difficult. Overall, two-digit denominators are no problem with long division. Next, let's go over how to convert a 122/99 to 1.232. ## Converting 122/99 to 1.232 ### Step 1: Set your long division bracket: denominator / numerator $$\require{enclose} 99 \enclose{longdiv}{ 122 }$$ To solve, we will use left-to-right long division. This method allows us to solve for pieces of the equation rather than trying to do it all at once. ### Step 2: Solve for how many whole groups you can divide 99 into 122 $$\require{enclose} 00.1 \\ 99 \enclose{longdiv}{ 122.0 }$$ We can now pull 99 whole groups from the equation. Multiple this number by our furthest left number, 99, (remember, left-to-right long division) to get our first number to our conversion. ### Step 3: Subtract the remainder $$\require{enclose} 00.1 \\ 99 \enclose{longdiv}{ 122.0 } \\ \underline{ 99 \phantom{00} } \\ 1121 \phantom{0}$$ If your remainder is zero, that's it! If you have a remainder over 99, go back. Your solution will need a bit of adjustment. If you have a number less than 99, continue! ### Step 4: Repeat step 3 until you have no remainder Sometimes you won't reach a remainder of zero. Rounding to the nearest digit is perfectly acceptable. ### Why should you convert between fractions, decimals, and percentages? Converting between fractions and decimals is a necessity. They each bring clarity to numbers and values of every day life. And the same is true for percentages. So we sometimes overlook fractions and decimals because they seem tedious or something we only use in math class. But they all represent how numbers show us value in the real world. Here are examples of when we should use each. ### When you should convert 122/99 into a decimal Speed - Let's say you're playing baseball and a Major League scout picks up a radar gun to see how fast you throw. Your MPH will not be 90 and 122/99 MPH. The radar will read: 90.123 MPH. This simplifies the value. ### When to convert 1.232 to 122/99 as a fraction Progress - If we were writing an essay and the teacher asked how close we are to done. We wouldn't say .5 of the way there. We'd say we're half-way there. A fraction here would be more clear and direct. ### Practice Decimal Conversion with your Classroom • If 122/99 = 1.232 what would it be as a percentage? • What is 1 + 122/99 in decimal form? • What is 1 - 122/99 in decimal form? • If we switched the numerator and denominator, what would be our new fraction? • What is 1.232 + 1/2?
# How do you factor x^2+11=300? Sep 16, 2015 $\left(x + 17\right) \left(x - 17\right) = 0$ #### Explanation: Subtract $300$ from both sides ${x}^{2} + 11 - 300 = 300 - 300$ ${x}^{2} - 289 = 0$ The left hand side is the difference of two squares. We factor as follows $\left(x + 17\right) \left(x - 17\right) = 0$ Or we could say ${x}^{2} = 289$ by subtracting $11$ from both sides Taking the square root we have $x = \pm \sqrt{289} = \pm 17$ but you want the factorization so we go with what i did first Sep 16, 2015 (x-17)(x+17) #### Explanation: To factorise this, recollect that ${17}^{2} = 289$ The give expression can be simplified to ${x}^{2} - 300 + 11 = 0$ ${x}^{2} - 289 = 0$ $\left({x}^{2} - {17}^{2}\right) = 0$ (x-17)(x+17)=0
## Balance Equations: Momentum Balance #### Learning Outcomes • Recall Newton’s equation of equilibrium applied to a particle. • Describe how Euler extended the equations of equilibrium to be applied to a continuum. • Identify that in static equilibrium, the stress distribution inside a continuum at equilibrium can be obtained by solving the three equations of equilibrium developed by Euler. • Compute the solution of static equilibrium in simple cases. ### Newton and Euler Laws of Motion Newton’s second law of motions states that the rate of change with respect to the time of the linear momentum of an object is equal to the net force acting on that object. If is the force vector acting on an object with mass moving with a velocity vector , then, by denoting the momentum by , we have: If the mass is constant then: where is the acceleration vector of the object. Euler extended Newton’s second law of motion to a continuum object and introduced two laws. The first law is the balance of linear momentum while the second is the balance of angular momentum of a continuum object. These are as follows: #### Euler’s Law of the Balance of Linear Momentum The first equation is a direct extension of Newton’s first law to an arbitrary volume of a continuum body. Given an arbitrary volume of a continuum body represented by the set , then the net force acting on the arbitrary volume is equal to the rate of change of the linear momentum of that volume. The net force is the resultant of the traction vectors acting on the boundary and the body forces acting on the continuum points. Thus, according to Euler’s Law: where, is time, is the mass density distribution, is the velocity vector distribution, is the body force distribution, is the traction vector distribution acting on the surface of , and and are the differential volume and surface elements of the arbitrary volume . #### Euler’s Law of the Balance of Angular Momentum Euler’s law of angular momentum preservation states that the rate of change of the angular momentum of an arbitrary volume of a continuum body around a fixed origin is equal to the torque (measured around the same fixed point) applied by the traction vectors on the boundary and the body forces acting on the continuum points. Thus, according to Euler’s Law: where, is the position vector relative to the fixed origin. It should be noted that the fixed origin can be arbitrary. If is the position vector relative to another fixed point such that , then, by replacing with and using the fact that is a fixed vector: By using Euler’s law of the balance of linear momentum we reach to the same statement of the balance of angular momentum with replacing : ### The Differential Equation of Equation of Equilibrium: Derivation of Using a Differential Volume #### Balance of Linear Momentum Newton’s laws of motion were originally formulated for systems of rigid particles mechanically interacting with each other. Newton’s laws can be extended to a continuum by considering a differential volume and the stresses acting on it. Let be a set representing a deformed configuration. Consider a differential cube of mass inside the body with dimensions ,, and whose mass density is . The differential volume of the cube is given by . Let be the resultant force acting on the cube which is due to a body forces vector per unit mass acting on the cube and the stresses on the different faces of the cube. To write the differential form of the equilibrium equations, the stresses are assumed to be continuous functions. Such assumption allows the use of the first component in the Taylor expansion for the stresses. Thus, the resultant force component acting on the cube in the direction of the first basis vector can be evaluated as follows: Similarly, If is the acceleration vector of the differential cube, then the statement of Newton’s second law dictates that the resultant force is related to the acceleration via: Thus, in component form, the equilibrium equations can be written as: (1) The above equations are usually written in the following compact form with : If the divergencer operator is used, the equations of equilibrium can be written in the following vector form: #### Balance of Angular Momentum The three angular momentum balance equations can be used to derive the symmetry of the stress tensor. Consider a differential rectangular volume oriented with the coordinate system with volume . Let the stresses on the sides with the negative normals have stresses and the stresses on the sides with positive normals have stresses . The first two equations for the balance of linear momentum dictate the following: The moment of the external forces (stresses on the boundary and body forces) acting on the rectangular volume around the axis have the following form: The rate of change of angular momentum around the axis is denoted and has the form: By equating , using the equilibrium equations, and neglecting the smaller terms we get: Therefore, . By repeating the procedure for the moment around other axes we reach: ### The Differential Equation of Equilibrium: Derivation Using Integrals Over Arbitrary Volumes The same differential equations of motion can be obtained using an equivalent but simpler derivation technique that relies on the integration over an arbitrary volume . #### Balance of Linear Momentum Using Euler’s law of the balance of linear momentum we have: The differential operator on the left hand side cannot be interchanged with the integration operator unless the volume of integration is independent of time. Therefore, we can replace the differential volume with where is the determinant of the deformation gradient and is the corresponding differential volume in the reference configuration, which is independent of time. We will also utilize the relationship between the traction vector and the stress tensor: , and the divergence theorem which lead to: The mass balance equations can be used to simplify the left hand side: Since the integration domain is arbitrary, then the integrand is equal to zero. Therefore, the linear momentum balance equation is: #### Balance of Angular Momentum Using Euler’s law of the balance of angular momentum we have: The differential operator on the left hand side cannot be interchanged with the integration operator unless the volume of integration is independent of time. Therefore, we can replace the differential volume with where is the determinant of the deformation gradient and is the corresponding differential volume in the reference configuration, which is independent of time. We also utilize the mass balance equations: Therefore, we have: Using the balance of linear momentum equation and replacing with we get: The last equation can be written in component form to show the symmetry of the stress matrix. The component of the above vector equation can be written as: Using the divergence theorem , the right hand side can be converted into a volume integral as follows: The right hand side can be further simplified as follows: Therefore, the integrand is equal to zero. Setting , and , the three components of the above vector equation have the form: Therefore: It should be noted that the book by P. Chadwick has another proof showing the same result without relying on the component form of the vector equation. ### Solution of the Equilibrium Equations in Static Problems A continuum problem is solved if a stress field (the stress at every point inside the body) and an acceleration field are found such that the three equations of force equilibrium (conservation of linear momentum equations) are satisfied at every point. On the other hand, the restriction that ensures the conservation of angular momentum and no additional information can be obtained from considering the three equations of moment equilibrium. The unknown stress and velocity fields are required to also satisfy certain boundary conditions, which are traditionally given as either displacements or external traction forces on the exterior of the body and initial conditions for velocities. In static equilibrium, which is the purpose of many engineering applications, the acceleration vector is assumed to be zero and, thus, the equilibrium equation is simplified and the stress field becomes the only unknown. In that case, the equations become: (2) and . The problem is mathematically formulated as follows: Let be a set representing a reference configuration of a continuum body. Given a constant or variable body forces vector , find the distribution of the stresses inside the continuum body that would satisfy the equations of equilibrium (Eq. 2) above. The boundary conditions for the equations of equilibrium are given on two parts of the boundary of . On the first part, , the external traction vectors are known so we have the boundary conditions for since . On the second part, , the displacement of the continuum is given. The boundary of is . #### Difficulties Associated with Obtaining a Solution to the Equilibrium Equations There are two difficulties associated with solving Eq. 2. First, these are three equations of static equilibrium. However, there are six unknowns (six stress variables). Thus, in their current form, many solutions could possibly satisfy the equilibrium equations! The second issue is that some of the boundary conditions contain expressions of displacements, while the equations themselves in this form do not have displacements as variables! These two major issues impel replacing the six unknown stress variables in the above equations with three unknown variables (usually the three displacements , , and . This can be performed by using a “constitutive equation” that describes the relationship between the stress and the strains inside the material. By replacing the stresses with the strains, and by using the relationship between the strains and the displacements described in the strain measures section, the problem becomes well posed, i.e., a solution can be obtained. The following few problems illustrate the applications and the solutions of the equilibrium equations given simplified assumptions allowing such solutions to exist. ### The Differential Equation of Equilibrium: Lagrangian Formulation In sections 6.3.2 and 6.3.3, the Eulerian form of the differential equation of equilibrium was derived. In the Eulerian form, the independent variables are the spatial position coordinates in the deformed configuration . One difficulty arises from the fact that the set representing the embedding of the a deforming object is in itself unknown. It is, therefore, desirable to have the differential equations written written in terms of the referential position coordinates represented by the set . In the following two subsections, I will present the derivation of these equations using two approaches, a component form, and a vector form approaches. #### Component Form Approach In the interest of brevity of the derivation, I will be utilizing Einstein Summation Convention. The Eulerian form of the differential equation is given above as: My intention now is to replace the variables with the variables . For the acceleration and the body forces vectors, that only means a simply change of the argument with the capital letters used for the mappings of the referential coordinates: I will recall the density equation that relates the density in the reference configuration with the spatial density: I will also recall the equation that relates the First Piola-Kirchhoff stress tensor with the Cauchy stress tensor: Substituting the above equations in the Eulerian form of the differential equation yields: I will decompose the first term, replace the derivation with respect to the components of with the derivation with respect to the components of , and will utilize the fourth statement in the matrix calculus section: Therefore, the Lagrangian form of the equilibrium equation can be written in terms of the First Piola-Kirchhoff stress tensor as follows: The balance of angular moment imposes the following restriction on the components of : #### Vector Form Approach Euler’s law of the balance of linear moment that I showed above is repeated here: By a simple change of variables, can be replaced by and the derivative operator is used instead of to indicate the Lagrangian formulation (material time derivatives with respect to the reference position components): I will use the mass balance equations to simplify the first expression and will use the First Piola-Kirchhoff stress tensor and the reference area vector for the traction vector: Considering that can be chosen arbitrarily, the vector form of the differential equation has the form: The balance of angular moment imposes the following restriction on :
# How to write inequalities with interval notation Writing the set for this figure in interval notation can be confusing. x can belong to two different intervals, but because the intervals dont overlap, you have to write them separately: The first interval is x 4. This interval includes all numbers between negative infinity and 4. Interval Notation. Back Miscellaneous Mathematics Mathematics Contents Index Home. Interval notation is a method of writing down a set of numbers. Usually, this is used to describe a certain span or group of spans of numbers along a axis, such as an xaxis. The simpler form of this notation would be something like" x x 3"which is pronounced as" all x such that x is less than minus three". " Interval notation" writes the solution as an interval (that is, as a section or length along the number line). Now we simply graph and write the answer in interval notation. 1 0 1 (So the solution is ()1. b. Again, we solve as we did with equations and flip the inequality symbol if needed. We get 2 5 10 5 4 6 7 4 2 6 x x x x x Subtract 2x on both sides Subtract 4 on both sides Divide by 5 on both sides So we graph and write as an interval. Exercise Set 1. 7: Interval Notation and Linear Inequalities 94 University of Houston Department of Mathematics For each of the following inequalities: (a) Write the inequality algebraically. (b) Graph the inequality on the real number line. (c) Write the inequality in interval notation. 1. x is greater than 5. 2. x is less than 4. Interval notation is a simplified form of writing the solution to an inequality or system of inequalities, using the bracket and parenthesis symbols in lieu of the inequality symbols. Intervals with parentheses are called open intervals, meaning the variable cannot have the value of the endpoints. An Interval is all the numbers between two given numbers. Showing if the beginning and end number are included is important; There are three main ways to show intervals: Inequalities, The Number Line and Interval Notation. Writing Compound Inequalities Using Interval Notation Ex 3 Using Interval Notation to Express Inequalities Ex 1 Using Interval Notation to Express Inequalities Ex 2 Use interval notation to express the range of numbers making your inequality a true statement. The solution set describing all numbers between 2 and 3 is expressed as: (2, 3). For the inequality x 2 4, the solution set includes all numbers less than 2. Algebra Solving Inequalities Interval Notation. Page 1 of 4. Interval Notation. This notation is my favorite for intervals. It's just a lot simpler! Let's look at the intervals we did with the setbuilder notation: Let's start with the first one: This is what it means; So, we write it like this: Use Phone: (255) 645-7967 x 7908 Email: [email protected]
Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too! # The mean and standard deviations of a group of 100 observations were found to be 20 and 3 respectively. Question: The mean and standard deviations of a group of 100 observations were found to be 20 and 3 respectively. Later on it was found that three observations 21, 12 and 18 were incorrect. Find the mean and standard deviation if the incorrect observations were omitted. Solution: Given that number of observations (n) = 100 Incorrect Mean $(\bar{x})=20$ and Incorrect Standard deviation, $(\sigma)=3$ We know that, $\overline{\mathrm{x}}=\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{x}_{\mathrm{i}}$ $\Rightarrow 20=\frac{1}{100} \sum x_{i}$ $\Rightarrow 20 \times 100=\sum x_{i}$ $\Rightarrow 2000=\sum x_{i}$ $\Rightarrow \sum x_{i}=2000$ ...........(i) $\therefore$ Incorrect sum of observations $=2000$ Finding correct sum of observations, incorrect observations 21,12 and 18 are removed So, Correct sum of observations = Incorrect Sum $-21-12-18$ $=2000-51$ $=1949$ Hence, Correct Mean $=\frac{\text { Correct Sum of Observations }}{\text { Total number of observations }}$ $=\frac{1949}{100-3}$ $=\frac{1949}{97}$ $=20.09$ Now, Incorrect Standard Deviation ( $\sigma$ ) $=\frac{1}{\mathrm{~N}} \sqrt{\mathrm{N} \times\left(\text { Incorrect } \sum \mathrm{x}_{\mathrm{i}}{ }^{2}\right)-\left(\text { Incorrect } \sum \mathrm{x}_{\mathrm{i}}\right)^{2}}$ $3=\frac{1}{100} \sqrt{100 \times\left(\text { Incorrect } \sum \mathrm{x}_{\mathrm{i}}{ }^{2}\right)-(2000)^{2}}$ $3 \times 100=\sqrt{100 \times\left(\text { Incorrect } \sum x_{i}^{2}\right)-4000000}$ $300=\sqrt{100 \times\left(\text { Incorrect } \sum x_{i}^{2}\right)-4000000}$ Squaring both the sides, we get $(300)^{2}=100 \times\left(\right.$ Incorrect $\left.\sum x_{i}^{2}\right)-4000000$ $\Rightarrow 90000=100 \times\left(\right.$ Incorrect $\left.\sum x_{i}^{2}\right)-4000000$ $\Rightarrow 90000+4000000=100 \times\left(\right.$ Incorrect $\left.\sum x_{i}^{2}\right)$ $\Rightarrow 4090000=100 \times\left(\right.$ Incorrect $\left.\sum x_{i}^{2}\right)$ $\Rightarrow \frac{4090000}{100}=\left(\right.$ Incorrect $\left.\sum x_{i}^{2}\right)$ $\Rightarrow 40900=\left(\right.$ Incorrect $\left.\sum x_{i}{ }^{2}\right)$ Since, 21, 12 and 18 are removed So, Correct $\sum x_{i}^{2}=40900-(21)^{2}-(12)^{2}-(18)^{2}$ $=40900-441-144-324$ $=40900-909$ $=39991$ Now, Correct Standard Deviation $=\sqrt{\frac{\left(\text { Correct } \sum x_{i}^{2}\right)}{N}-\left(\frac{\text { Correct } \sum x_{i}}{N}\right)^{2}}$ $=\sqrt{\frac{39991}{97}-(20.09)^{2}}$ $\left[\because \overline{\mathrm{x}}=\frac{\text { Correct } \sum \mathrm{x}_{\mathrm{i}}}{\mathrm{N}}=20.09\right]$ $=\sqrt{412.27-403.60}$ $=\sqrt{8.67}$ $=2.94$ Hence, Correct Mean $=20.09$ and Correct Standard Deviation $=2.94$
Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis : # Table of 424 The Multiplication Table for 424 is nothing more than the repetitive addition of the number 424.Â Repetitive addition is adding the same number multiple times. For example, adding 4 times 424, which is 424 + 424 + 424 + 424 is the same as 424 x 4 and that equals 1696. But adding the same number multiple times, say 8 times, can result in mistakes. Instead, the below-given chart on Table 424 can be referred to. Any multiplication table can be computed using the Multiplication Calculator. ## What is the 424 Times Table? In the table of 424, we are repeatedly adding 424 multiple times. And the table below gives a repetitive addition 10 times for the number 424. 424 Ã— 1 = 424 424 424 Ã— 2 = 848 424 + 424 = 848 424 Ã— 3 = 1272 424 + 424 + 424 = 1272 424 Ã— 4 = 1696 424 + 424 + 424 + 424 = 1696 424 Ã— 5 = 2120 424 + 424 + 424 + 424 + 424 = 2120 424 Ã— 6 = 2544 424 + 424 + 424 + 424 + 424 + 424 = 2544 424 Ã— 7 = 2968 424 + 424 + 424 + 424 + 424 + 424 + 424 = 2968 424 Ã— 8 = 3392 424 + 424 + 424 + 424 + 424 + 424 + 424 + 424 = 3392 424 Ã— 9 = 3816 424 + 424 + 424 + 424 + 424 + 424 + 424 + 424 + 424 = 3816 424 Ã— 10 = 4240 424 + 424 + 424 + 424 + 424 + 424 + 424 + 424 + 424 + 424 = 4240 ## Multiplication Table of 424 Given below is the multiplication table of 424 being done 20 times. The below table helps you with multiplication problems that require the tables of 424. 424 Ã— 1 = 424 424 Ã— 2 = 848 424 Ã— 3 = 1272 424 Ã— 4 = 1696 424 Ã— 5 = 2120 424 Ã— 6 = 2544 424 Ã— 7 = 2968 424 Ã— 8 = 3392 424 Ã— 9 = 3816 424 Ã— 10 = 4240 424 Ã— 11 = 4664 424 Ã— 12 = 5088 424 Ã— 13 = 5512 424 Ã— 14 = 5936 424 Ã— 15 = 6360 424 Ã— 16 = 6784 424 Ã— 17 = 7208 424 Ã— 18 = 7632 424 Ã— 19 = 8056 424 Ã— 20 = 8480 ## Solved Example on Table of 424 Example: Solve for y: 424 y – 300 = 6060 Solution: y = 15 424 y – 300 = 6060 424 y = 6060 + 300 424 y = 6360 y = 6360 / 424 y = 15. ## Frequently Asked Questions on Table of 424 ### Question 1: What is 16 times 424? Answer: 16 times 424 is 6784. ### Question 2: What are the factors of 424? Answer: The factors of 424 are 1, 2, 4, 8, 53, 106, 212 and 424. ### Question 3: What are the prime factors of 424? Answer: 2 and 53 are the prime factors of 424.
# FRACTIONS Essay by lilman_17_08Junior High, 8th grade April 2004 Fractions are ways to represent parts of a whole. Common fractions are ÃÂ½ and ÃÂ¾. These are proper or regular fractions. Some fractions are called mixed numbers. These are represented by a whole number with a fraction (proper fraction). 1 ÃÂ½ and 2 ÃÂ¾ are good examples. An improper fraction has a larger number on the top than on the bottom, such as 9/8. I will explain how to convert these fractions to decimals. I will show you how to change an improper fraction to a mixed number. Operations (addition, subtraction, multiplication, and division) will be explained as well. CONVERSIONS This section will explain how to convert a fraction into a decimal. First, let's get an example fraction. How about 3/8? To find a decimal, divide the numerator (top number) by the denominator (bottom number). So we would divide 3 by 8. 3)8=0.375 or .38. Lets do another. Try 1/3. 1)3=.33333... So 1/3 is equal to about.33. Next comes the mixed numbers to the improper fractions . All you do is multiply the denominator by the whole number and add the numerator. This is the numerator of the improper fraction. You keep the same denominator. Let's try 2ÃÂ¾. Four times two is eight. Eight plus three is eleven. Keep the denominator and you have 11/4. Now to convert the improper fraction into a mixed number. You need to remember, keep the same denominator. 9/2=4ÃÂ½. 2 goes into 9 4 times. There is 1 left over. So, it equals 4ÃÂ½. OPERATIONS First, I will explain how to add like fractions, then, unlike fractions. Let's say you have ÃÂ¾ of a pizza and ÃÂ¼ of a pizza. How much pizza do you have? When adding fractions, you NEVER add the denominators. Just the...
# How to Solve Word Problems on Dividing Whole Numbers by Unit Fractions In real-life scenarios, we often encounter situations where we need to divide whole numbers by unit fractions. These word problems help us understand the practical applications of such divisions. Let's explore some word problems that involve dividing whole numbers by unit fractions. ## Dividing Whole Numbers and Unit Fractions: Word Problems ### Example 1: A chocolate bar is divided into sections, each representing $$\frac{1}{4}$$ of the bar. If there are 8 sections in total, how long is the entire chocolate bar? Solution Process: To find the length of the whole chocolate bar, divide the number of sections (8) by the size of each section ($$\frac{1}{4}$$). The chocolate bar is 2 times as long. The Absolute Best Book for 5th Grade Students ### Example 2: A rope is cut into pieces, each measuring $$\frac{1}{3}$$ of a meter. If there are 9 pieces, how long was the rope originally? Solution Process: To find the original length of the rope, divide the number of pieces (9) by the length of each piece ($$\frac{1}{3}$$). The rope was originally 3 meters long. Word problems involving the division of whole numbers by unit fractions help bridge the gap between abstract math concepts and real-world applications. By understanding the context and using systematic division, you can solve these problems with ease. Whether you’re planning a party, dividing resources, or just curious about everyday scenarios, these skills will come in handy. So, the next time you’re faced with a word problem involving division by unit fractions, approach it methodically and confidently, knowing you have the tools to find the solution! ### Practice Questions: 1. A book has chapters, each $$\frac{1}{5}$$ of the total pages. If there are 10 chapters, how many pages are in the book? 2. A cake is sliced into pieces, each representing $$\frac{1}{6}$$ of the cake. If there were 12 slices, how big was the cake? 3. A tank is divided into sections, each holding $$\frac{1}{8}$$ of the total water. If there are 16 sections, what’s the tank’s total capacity? 4. A playlist has songs, each $$\frac{1}{7}$$ of the total duration. If there are 14 songs, how long is the playlist? A Perfect Book for Grade 5 Math Word Problems! 1. 2 pages 2. 2 cakes 3. 2 tanks 4. 2 hours (or whatever unit of time is being used) The Best Math Books for Elementary Students ### What people say about "How to Solve Word Problems on Dividing Whole Numbers by Unit Fractions - Effortless Math: We Help Students Learn to LOVE Mathematics"? No one replied yet. X 45% OFF Limited time only! Save Over 45% SAVE $40 It was$89.99 now it is \$49.99

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